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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , PHYSICAL WORLD, , PHYSICAL WORLD, SYNOPSIS, PHYSICS, , , Physics deals with the study of the basic laws of nature and their manifestation in different phenomena. The, basic laws of physics are universal and are applied in widely different contexts and conditions., , Physics, Technology and Society, , , Science,Technology and Society have strong relationships among one on other. Science is the mother of, technology and both of them are the reasons for the creation and development of the society., , , , Science and technology issues are actually discussed worldwide today. Progress in this has led to produce, the ability to integrate different types of physical products., , , , Physics is a basic discipline in the category of natural sciences which also includes other disciplines like, Chemistry and Biology. The word physics comes from a Greek word meaning nature., , (1) Some physicists from different countries of the world and their major contributions, Name, Archimedes, Galileo Galilei, Christiaan Huygens, Isaac Newton, Michael Faraday, James Clerk Maxwell, Heinrich Rudolf Hertz, J.C. Bose, W.K.Roentgen, J.J. Thomson, Marie sklodowska Curie, Albert Einstein, Victor Francis Hess, R.A. Millikan, Ernest Rutherford, Niels Bohr, C.V. Raman, Louis Victor de Broglie, M.N. Saha, S.N. Bose, NARAYANA GROUP, , Major contribution /Discovery, Principle of buoyancy; Principle of the lever, Law of inertia, Wave theory of light, Universal law of gravitation ; Laws of motion ;, Reflecting telescope, Laws of electromagnetic induction, Electromagnetic theory; Light - an electromagnetic wave, Generation of electromagnetic waves, Short radio waves, X-rays, Electron, Discovery of radium and polonium;, Studies on natural radio activity, Explanation of photoelectric effect;Theory of relativity, Cosmic radiation, Measurement of electronic charge, Nuclear model of atom, Quantum model of hydrogen atom, Inelastic scattering of light by molecules, Wave nature of matter, Thermal ionisation, Quantum statistics, , Country of, , Origin, Greece, Italy, Holland, U.K., U.K., U.K., Germany, India, Germany, U.K., poland, Germany, Austria, U.S.A., New Zealand, Denmark, India, France, India, India, 1
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PHYSICAL WORLD, Name, Wolfgang Pauli, Enrico Fermi, Werner Heisenberg, Paul Dirac, Edwin Hubble, Ernest Orlando Lawrence, James Chadwick, Hideki Yukawa, Homi Jehangir Bhabha, Lev Davidovich Landau, S.Chandrasekhar, John Bardeen, C.H. Townes, Abdus Salam, , JEE-ADV PHYSICS-VOL - I, Major contribution /Discovery, , Country of, , Exclusion principle, Controlled nuclear fission, Quantum mechanics; Uncertainity principle, Relativistic theory of electron; Quantum statistics, Expanding universe, Cyclotron, Neutron, Theory of nuclear forces, Cascade process of cosmic radiation, Theory of condensed matter; Liquid helium, Chandrasekhar limit, structure and evolution of stars, Transistors ; Theory of super conductivity, Maser; Laser, Unification of weak and electromagnetic interactions, , Origin, Austria, Italy, Germany, U.K., U.S.A., U.S.A., U.K., Japan, India, Russia, India, U.S.A., U.S.A., Pakistan, , 2) Link between technology and physics, Technology, Steam engine, Nuclear reactor, Radio and Television, Computers, Lasers, Production of ultra high, magnetic fields, Rocket propulsion, Electric generator, Hydroelectric power, Aeroplane, Particle accelerators, Sonar, Optical fibres, Non-reflecting coatings, Electron microscope, Photocell, Fusion test reactor (Tokamak), Giant Metrewave Radio, Telescope ( GMRT), Bose-Einstein condensate, , 2, , Scientific prionicple(s), Laws of thermodynamics, Controlled nuclear fission, Generation,propagation and detection of electromagnetic, waves, Digital logic, Light amplification by stimulated emission of radiation, Superconductivity, Newton’s laws of motion, Faraday’s laws of electromagnetic induction, Conversion of gravitational potential energy into electrical, energy, Bernoulli’s principle in fluid dynamics, Motion of charged particles in electromagnetic fields, Reflection of ultrasonic waves, Total internal reflection of light, Thin film optical interference, Wave nature of electrons, Photoelectric effect, Magnetic confinement of plasma, Detection of cosmic radio waves, Trapping and cooling of atoms by laser beams and, magnetic fields, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , Fundamental forces in nature :, , , There are four fundamental forces in nature. They, are the ‘gravitational force’, the ‘electromagnetic, force’, the ‘strong nuclear force’, and the ‘weak, nuclear force’. Unification of different forces/, domains in nature is a basic quest in physics., , Nature of physical laws :, , , PHYSICAL WORLD, 7), , 8), , (i) The physical quantities that remain unchanged, in a process are called conserved quantities. Some, of the general conservation laws in nature include, 9), the laws of conservation of mass, energy, linear, momentum, angular momentum, charge, etc., Some conservation laws are true for one 10), fundamental force but not for the other., (ii) Conservation laws have a deep connection, with symmetries of nature. Symmetries of space 11), and time, and other types of symmetries play a, central role in modern theories of fundamental, forces in nature., , What is the principle involved in the working, of Rocket?, 1) Newton’s laws of motion, 2)Bernoulli’s principle, 3) Photoelectric effect, 4) Faraday’s laws of EMI, Name the branch of science which deals with, the study of stars?, 1) Astronomy, 2) Biology, 3) Geology, 4) Chemistry, Who Discovered the principle of inertia?, 1) Newton, 2) Galileo, 3) Tycho Brahe, 4) Kepler, Who invented wireless telegraphy?, 1) Maxwell, 2) Marconi, 3) Hertz, 4) Faraday, Match the following, A-Force, B-relative strength, 1) Gravitational force a) 1, 2) Weak force, b) 1025, , 3) Electromagnetic, c) 1036, force, 4) Nuclear force, d) 1038, What is the discovery of CV. Raman?, 1) 1-a, 2-b, 3-c, 4-d, 2) 1-b, 2-c, 3-a, 4-d, 1) Inelastic scattering of light by molecules, 3) 1-d, 2-c, 3-b, 4-a, 4) 1-a, 2-c, 3-b, 4-d, 2) Steam engine, 12) Match the technology in column A to its, 3) Propagation of EM Waves, related scientific principle in column B, 4) Reflection of Ultrasonic waves, B-Scientific Principle, A- Technology, What is the contribution of S.Chandra Sekhar, 1), Steam, engine, a), Faraday’s laws, to physics?, 2) Nuclear Reactor, b)Thermodynamic laws, 1) cosmic radiation 2) Nuclear model & atom, 3) Computer, c) Nuclear Fission, 3) LASER, 4) Electric generation d) Digital logic, 4) structure and evolution of stars, 1) 1-b;2-d;3-c;4-a, 2) 1-b;2-c;3-d;4-a, Who discovered electron?, 3), 1-a;2-c;3-d;4-b, 4) 1-d;2-a;3-c;4-b, 1) Albert Einstein, 2) J.C. Bose, 13) Match the scientist’s name against discovery, 3) J.J.Thomson, 4) Bohr, B-Discovery, A-Scientist, Who discovered Neutron?, 1) Faraday, a) Expansion of Universe, 1) James chadwick, 2) Fermi, 2) Newton, b) Law of EMI, 3) S.N.Bose, 4) Millikan, 3), Einstein, c) Law of gravitation, What is the working principle of Steam, 4) Hubble, d) Theory of relativity, engine?, 1) 1-b;2-c;3-d;4-a, 2) 1-b;2-d;3-c;4-a, 1) Digital logic, 2) Super conductivity, 3)1-b;2-c;3-a;4-d, 4) 1-b;2-a;3-c;4-d, 3) Laws of thermodynamics, 4) Nuclear fission, C.U.Q- KEY, Photocell works on the principle of, 1)1, 2) 4 3) 3 4) 1 5) 3, 1) Raman effect, 2) Compton effect, 6) 4, 7) 1 8)1, 9) 2 10) 2, 3) Seebeck effect 4) photoelectric effect, 11) 1, 12)2 13)1, , C.U.Q, , 1), , 2), , 3), , 4), , 5), , 6), , NARAYANA GROUP, , 3
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UNITS AND MEASUREMENTS, , JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, SYNOPSIS, , , , , , , , , Fundamental Quantities and their SI Units, , There are seven fundamental quantities and two, supplementary quantities in S. I. system. These, Physical Quantity:, quantities along with their unit and symbols are given, Any quantity which can be measured directly (or), below:, indirectly (or) interms of which the laws of physics, S.No, Physical Quantity SI unit, Symbol, can be expressed is called physical quantity., There are two types of physical quantities, 1., Length, metre, m, 1) Fundamental quantities2) Derived quantities, 2., Mass, kilogram, kg, Fundamental Quantities: Physical quantities, 3., Time, second, s, which cannot be expressed interms of any other, Thermo dynamic, physical quantities are called fundamental physical 4., temperature, kelvin, K (or) , quantities., 5., Luminous, Ex. length, mass, time, temperature etc.., intensity, candela, Cd, Derived Quantities: Physical Quantities which, Electric current, ampere, A, are derived from fundamental quantities are called 6., 7., Amount, of, derived quantities., substance, Ex. Area, density, force etc..., (or) quantity of, Unit of physical quantity:, matter, mole, mol, A unit of measurement of a physical quantity is the, Supplementary, quantities, standard reference of the same physical quantity, Plane angle, radian, rad, which is used for comparison of the given physical 1., 2., Solid, angle, steradian, sr, quantity., Fundamental unit :The unit used to measure the, fundamental quantity is called fundamental unit., Measurement of length, Ex: metre for length, kilogram for mass etc.., The length of an object can be measured by using, Derived unit : The unit used to measure the, different units. Some practical units of length are, derived quantity is called derived unit., angstrom( Ao )=10 10 m=108 cm, Ex: m2 for area, gm cm-3 for density etc..., nanometre(nm) 109 m 10 A0, The numerical value obtained on measuring a, physical quantity is inversely proportional to the, fermi 1015 m, magnitude of the unit chosen., micron 106 m, 1, n U = constant, X-ray unit 1013 m, U, 1 A.U. = distance between sun & earth, n1U n2U, = 1.496×1011 m, Where n1 and n2 are the numerical values and , One light year is the distance travelled by light in, one year in vacuum . This unit is used in astronomy., U1 and U 2 are the units of same physical quantity, in different systems., Light year 9.46 1015 m, System of units, parsec 3.26 light years 30.84 1015 m, There are four systems of units, Bohr radius 0.5 1010 m, 1) F.P.S, 2) C.G.S, Mile=1.6 km, 3) M.K.S, 4) SI, Measurement of mass:, Based on SI system there are three categories of, The mass of an object can be measured by using, physical quantities., different units.Some practical units of mass are, 1)fundamental quantities, Quintal = 100 kg, 2)supplementary quantities and, Metric ton = 1000 kg, 3)derived quantities, Atomic mass unit (a.m.u) = 1.67 10 27 kg, , n, , 1, , , , , 4, , 2, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , Measurement of time:, One day = 86400 second, Shake 108 second, , Some physical constants and their values:, 1 amu = 1.67 1027 kg 931.5MeV, 1 atm pressure = pressure exerted by 76cm of, Hg column 1.013 105 Pa, Avagadro number (N)= 6.023 1023, Permittivity of free space= 8.854 1012 Fm 1 or, , , , Abbreviations for multiples and sub multiples:, , MACRO Prefixes, Multiplier, 101, 102, 103, 106, 109, 1012, , , , Symbol, da, h, k, M, G, T, 15, P, 10, E, 1018, Z, 1021, 24, Y, 10, MICRO Prefixes, Multiplier, Symbol, 10-1, d, 10-2, c, -3, 10, m, , 10-6, 10-9, n, 10-12, p, f, 10 15, a, 10 18, 21, z, 10, 24, y, 10, , Prefix, Deca, Hecto, Kilo, Mega, Giga, Tera, Peta, Exa, Zetta, Yotta, Prefix, deci, centi, milli, micro, nano, pico, femto, atto, zepto, yocto, , Some important conversions:, , , 5, 1kmph ms 1, 18, 1 newton=105 dyne, 1 joule= 107 erg, 1 calorie=4.18 J, 1eV= 1.6 1019 J, 1gcm 3 1000kgm 3, 1 lit=1000cm3 10 3 m 3, 1KWH 36 105 J, 1 HP=746 W, 1 degree=0.017 rad, 1cal g 1 4180JKg 1, 1kgwt= 9.8 N, 1 telsa= 104 gauss, 1Am 1 4 103 oersted, 1 weber=108 maxwell, , NARAYANA GROUP, , C 2 / Nm2, Permeability of free space, , 0 4 107 Hm1, Joule’s constant (J)= 4.186 Jcal 1, Planck’s constant(h)= 6.62 1034 Js, Rydberg’s constant(R)= 1.0974 107 m 1, Boltzmann’s constant(KB)=1.38 10 23 JK 1, Stefan’s constant 5.67 108Wm 2 K 4, Universal gas constant(R)= 8.314Jmol 1 K 1, = 1.98cal mol 1 K 1, Wien’s constant(b)= 2.93 10 3 metre kelvin, , Accuracy and precision of instruments :, , , The numerical values obtained on measuring, physical quantities depend upon the measuring, instruments, methods of measurement., Accuracy refers to how closely a measured value, agrees with the true value., Precision refers to what limit or resolution the given, physical quantity can be measured., Precision refers to closeness between the different, observed values of the same quantity ., High precision does not mean high accuracy., The difference between accuracy and precision, can be understood and by the following example:, Suppose three students are asked to find the length, of a rod whose length is known to be 2.250cm.The, observations are given in the table ., , , , , , , , Student, , Measurement1, , Measurement2, , Measurement3, , Average, length, , A, , 2.25cm, , 2.27cm, , 2.26cm, , 2.26cm, , B, , 2.252cm, , 2.250cm, , 2.251cm, , 2.251cm, , C, , 2.250cm, , 2.250cm, , 2.250cm, , 2.250cm, , It is clear from the above table , that the, observations taken by a student A are neither, precise nor accurate. The observations of student, B are more precise . The observations of student, C are precise as well as accurate., , Error:, , , The result of every measurement by any measuring, instrument contains some uncertainty. This, uncertainty in measurement is called error., 5
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, , , , , Mathematically, Error = True value - Measured value, Correction =-error, True value means, standard value free of errors., Errors are broadly classified into 3 types :, i) Systematic errors, ii) Random errors, iii) Gross errors, , probable error , Ex: Parallax error, , Random Errors:, , , Systematic Errors, , , , , , , , The errors due to a definite cause and which follow, a particular rule are called systematic errors. They, always occur in one direction (either +ve or -ve ), Systematic errors with a constant magnitude are, called constant errors., The constant arised due to imperfect design, zero, error in the instrument or any other such defects., These are also called instrumental errors., Example for the error due to improper designing, and construction., If a screw gauge has a zero error of -4 head scale, divisions, then every reading will be 0.004cm less, than the true value., The error arised due to external conditions like, changes in environment, changes in temperature,, pressure, humidity etc., Ex: Due to rise in temperature, a scale gets, expanded and this results in error in measurement, of length., , , , , , , , , The error due to experimental arrangement, , procedure followed and experimental technique, , is called imperfection error., Ex: In calorimetric experiments, the loss of heat, due to radiation, the effect on weighing due to, buoyancy of air cannot be avoided., , Personal errors or observational errors:, , , , , These are entirely due to the personal peculiarities , of the experimenter. Individual bias, lack of proper, setting of the apparatus, carelessness in taking, observations (without taking the required, necessary precautions.) etc. are the causes for, these type of errors. A person may be habituated, to hold his eyes (head) always a bit too far to the, right (or left) while taking the reading with a scale., This will give rise to parallax error., If a person keeps his eye-level below the level of , mercury in a barometer all the time, his readings, will have systematic error., These errors can be minimised by obtaining, several readings carefully and then taking their, arithmetic mean.., , 6, , They are due to uncontrolled disturbances which, influence the physical quantity and the instrument., these errors are estimated by statistical methods., 1, Random error , no. of observations, Ex-:The errors due to line voltage changes and, backlash error., Backlash errors are due to screw and nut., , Gross Errors, , Imperfection in Experimental, technique or Procedure:, , , 1, no. of readings, , The cause for gross errors are improper recording,, neglecting the sources of the error, reading the, instrument incorrectly, sheer carelessness, Ex: In a tangent galvanometer experiment, the coil, is to be placed exactly in the magnetic meridian, and care should be taken to see that no any other, magnetic material is present in the vicinity., No correction can be applied to these gross errors., When the errors are minimised, the accuracy, increases., The systematic errors can be estimated and, observations can be corrected., Random errors are compensating type. Aphysical, quantity is measured number of times and these values, lie oneitherside ofmean value. These errors are estimated, by statistical methods and accuracy is achieved., Personal errors like parallax error can be avoided, by taking proper care., The instrumental errors are avoided by calibrating, the instrument with a standard reference and by, applying proper corrections., , Errors in measurement., True Value :, In the measurement of a physical quantity the, arithmetic mean of all readings which is found to, be very close to the most accurate reading is to, be taken as True value of the quantities., If a1, a2 , a3 ..................an are readings then true, value amean , , 1 n, ai, n i 1, , Absolute Error :, The magnitude of the difference between the true, value of the measured physical quantity and the, value of individual measurement is called absolute, error., Absolute error =|True value - measured values|, ai amean ai, The absolute error is always positive., NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , Mean absolute error:, , , The arithmetic mean of all the absolute errors is, considered as the mean absolute error of the, physical quantity concerned., amean , , a1 a2 an, n, , , , 1 n, ai, n i1, , The mean absolute error is always positive., , Relative error:, , , The relative error of a measured physical quantity, is the ratio of the mean absolute error to the mean, value of the quantity measured., a, , Relative error=, , WE-2 : The length and breadth of a rectangle are, (5.7 0.1) cm and (3.4 0.2) cm. Calculate, the area of the rectangle with error limits., Sol. Here l 5.7 0.1 cm, b 3.4 0.2 cm, Area : A l b 5.7 3.4 19.38 cm 2 19 cm 2, (rounding off to two significant figures), , , A, l b , 0.1 0.2 , , , , , , A, b , l, 5.7 3.4 , 1.48, 0.34 1.14 , , , 19.38, 5.7 3.4 , , m ean, , a mean, , It is a pure number having no units., , Percentage error:, a, , a mean 100 %, amean, , , A , , 1.48, 1.48, 19.38 1.48 1.5, A , 19.38, 19.38, , (rounding off to two significant figures), So, Area 19.0 1.5 cm 2, WE-3: The distance covered by a body in time, , 5.0 0.6 s is 40.0 0.4 m. Calculate the, Relative error and percentage error give a measure, speed of the body. Also determine the, of accuracy i.e. if percentage error increases, percentage error in the speed., accuracy decreases., WE- 1:Repetition in the measurements of a certain Sol. Here, s 40.0 0.4 m and t 5.0 0.6 s, quantity in an experiment gave the following, s, s 40.0, 8.0 ms 1 As v , Speed v , values: 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30,, t, 5.0, t, and 1.33. Calculate the mean value, mean, , v, , s, , t, absolute error, relative error and percentage, , , , error., v, s, t, Here s 0.4 m, s=40.0 m, t 0.6 s, t=5.0 s, Sol. Here, mean value, v 0.4 0.6, , , , 0.13, 1.29 1.33 1.34 1.35 1.32 1.36 1.30 1.33, v 40.0 5.0, xm , 8, v 0.13 8.0 1.04, = 1.3275=1.33 (rounded off to two places of decimal), Hence, v 8.0 1.04 ms 1, Absolute errors in measurement are, x1 1.33 1.29 0.04; x2 1.33 1.33 0.00;, x3 1.33 1.34 0.01; x4 1.33 1.35 0.02;, x5 1.33 1.32 0.01; x6 1.33 1.36 0.03;, x7 1.33 1.30 0.03; x8 1.33 1.33 0.00;, mean absolute error, 0.04 0.00 0.01 0.02 0.01 0.03 0.03 0.00, xm , 8, = 0.0175, = 0.02 (rounded off to two places of decimal), xm, 0.02, , 0.01503 0.02, xm, 1.33, (rounded off to two places of decimal), Percentage error = 0.01503100 1.503 1.5%, , Relative error , , v, , 1, , value of 1 main scale division, , , Least count =, , Total divisions on circular scale, , mm, , 100, , Diameter of wire = M.S.R +( C.S.R x L.C), 0 52 , , NARAYANA GROUP, , , , Percentage error v 100 0.13 100 13%, WE- 4 : A screw gauge gives the following reading, when used to measure the diameter of a wire., Main scale reading : 0 mm, Circular scale reading : 52 divisions, Given that 1 mm on main scale corresponds, to 100 divisions of the circular scale., [AIEEE 2011], Sol. Main scale reading = 0 mm, Circular scale reading = 52 divisions, , 1, , 100, , mm 0.52mm 0.052cm, 7
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, WE-5:The current voltage relation of diode is given , , Whether it is addition or subtraction, absolute error, is same., In subtraction the percentage error increases., Error due to Multiplication:, , by I e1000V /T 1 mA,where the applied , , voltage V is in volt and the temperature T is, Z A B, , , If Z = AB then, in kelvin.If a student makes an error, Z, A, B, measuring 0.01V while measuring the, Z, is called fractional error or relative error.., current of 5mA at 300K,what will be the, Z, error in the value of current in mA?, Z, A, B, , 100 , 100 , 100 , Percentage error , (JEE MAIN-2014), A, B, Z, , , , , Here, percentage, error, is, the, sum, of, individual, 1000V / T, 1 mA, Sol. I e, percentage errors., dV= 0.01V, T=300K,I=5mA, A, , Error due to division: if Z , 1000V / T, B, I 1 e, Z A B, 1000V, , , Maximum possible relative error, log I 1 , Z, A, B, T, Max. percentage error in division, dI, 1000, A, B, , dV dI=0.2mA, , , 100, , 100, I 1, T, A, B, WE-6 : In an experiment the angles are required, Z, A, to be measured using an instrument. 29 , n, Error due to Power: If Z= An ;, Z, A, divisions of the main scale exactly coincide, p q, A B, with the 30 divisions of the vernier scale. If, Z, , , In, more, general, form, :, If, the smallest division of the main scale is halfCr, then maximum fractional error in Z is, a-degree(= 0.50 ), then the least count of the, Z, A, B, C, instrument is, (AIEEE-2009), p, q, r, Z, A, B, C, Valueof main scaledivision, As, we, check, for, maximum, error a +ve sign is to, Sol. Least count = No.of divisions of vernier scale, 0, , =, , 0, , 1, 1 1 1, MSD , 1 min, 30, 30 2 60, , Combination of Errors:, , , Error due to addition, If Z A B ;, Z A B (Max. possible error), Z Z A B A B , A B, A B, A B, 100, Percentage error=, A B, Error due to subtraction, If Z=A-B, Z A B (Max. possible error ), Z Z A B A B , , Relative error=, , , , A B, A B, A B, 100, Percentage error =, A B, , Relative error =, , 8, , be taken for the term r, , C, C, , Maximum Percentage error in Z is, Z, A, B, C, 100 p, 100 q, 100 r, 100, Z, A, B, C, WE-7: A physical quantity is represented by x, =Ma LbT-c. The percentage of errors in the, measurements of mass,length and time are, %, %, % respectively then the maximum, percentage error is, x, M, L, T, 100 a., 100 b. 100 c. 100, Sol., x, M, L, T, a b c, WE-8:Resistance of a given wire is obtained by, measuring the current flowing in it and the, voltage difference applied across it. If the, percentage errors in the measurement of the, current and the voltage difference are 3%, each, then error in the value of resistance of, the wire is, [AIEEE 2012], NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, V, log R log V log I , I, R, V I , R 100 V I 100 , , , = 3% + 3% = 6%, , UNITS AND MEASUREMENTS, , , Sol. R , , Rules for determining the number of, significant figures:, , , WE-9: Two resistors of resistances R1 100 3, ohm and R2 200 4 ohm are connected (a), in series, (b) in parallel. Find the equivalent , resistance of the (a) series combination, (b), parallel combination. Use for (a) the relation, R R1 R2 and for (b), 1, 1, 1, R' R1 R2, , , 2 2, R ' R1 R2 and R'2, R1, R2, , , , Sol. (a) The equivalent resistance of series, combination, , R R1 R2 100 3 ohm 200 4 ohm, , , , = 300 7 ohm., , (b) The equivalent resistance of parallel, combination, , R1 R2, 200, ', R , , 66.7 ohm, 3, R1 R2, 1, , 1, , 1, , Then, from R ' R R, 1, 2, , R1, R, R '2 22, 2, R1, R2, , 2, , , , , , Then, R ' 66.7 1.8 ohm, , Significant Figures :, , , A significant figure is defined as the figure, which, is considered reasonably, trust worthy in number., Ex:, = 3.141592654, (upto 10 digits), =3.14 (with 3 figures ), =3.1416 (upto 5 digits ), , NARAYANA GROUP, , The result of computation with approximate, numbers, which contain more than one uncertain, digit,should be rounded off., , Rules for rounding off numbers:, , 2, , 66.7 , 66.7 , , 3 , 4 1.8, 100, , , 200 , , All the non-zero digits in a given number are, significant without any regard to the location of, the decimal point if any., Ex: 18452 or 1845.2 or 184.52 all have the, same number of significant digits,i.e. 5., All zeros occurring between two non zero digits, are significant without any regard to the location, of decimal point if any., Ex: 106008 has six significant digits., 106.008 or 1.06008 has also got six significant, digits., If the number is less than one, all the zeros to the, right of the decimal point but to the left of first, non-zero digit are not significant., Ex: 0.000308, In this example all zeros before 3 are insignificant., a)All zeros to the right of a decimalpoint are significant, if they are not followed by a non-zero digit., Ex: 30.00 has 4 significant digits, b) All zeros to the right of the last non-zero digit, after the decimal point are significant., Ex: 0.05600 has 4 significant digits, c) All zeros to the right of the last non-zero digit in, a number having no decimal point are not, significant., Ex: 2030 has 3 significant digits, , Rounding off numbers:, , , R ' R1 R2, we get, R '2 R 2 R 2, 1, 2, , R ' R '2 , , The significant figures indicate the extent to which, the readings are reliable., , , , The preceding digit is raised by 1 if the immediate, insignificant digit to be dropped is more than 5., Ex: 4728 is rounded off to three significant figures as, 4730., The preceding digit is to be left unchanged if the, immediate insignificant digit to be dropped is less, than 5., Ex: 4723 is rounded off to three significant figures, as 4720, If the immediate insignificant digit to be dropped, is 5 then there will be two different cases, a) If the preceding digit is even then it is to be, unchanged and 5 is dropped., Ex: 4.7253 is to be rounded off to two decimal, places. The digit to be dropped here is 5 (along, with 3) and the preceding digit 2 is even and hence, to be retained as two only 4.7253=4.72, 9
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, b)If the preceding digit is odd, it is to be raised by 1, Ex: 4.7153 is to be rounded off to two decimal , places. As the preceding digit ‘1’ is odd, it is to, be raised by 1., 4.7153=4.72, , Rules for Arithmetic Operations with, significant Figures:, , , In multiplication or division, the final result should, retain only that many significant figures as are there, in the original number with the least number of, significant figures., Ex: 1.2 2.54 3.26 9.93648 .But the result, should be limited to the least number of significant, digits-that is two digits only. So final answer is, 9.9., , In addition or subtraction the final result should, retain only that many decimal places as are there, in the number with the least decimal places., Ex:2.2+4.08+3.12+6.38=15.78.Finally we, should have only one decimal place and hence, 15.78 is to be rounded off as 15.8., WE-10:The respective number of significant, figures for the numbers 23.023,0.0003and, (AIEEE-2010), 21 103 are, Sol.(i)All non -zero numbers are significant figures. Zeros, occurring between zero digits are also significant., (ii) If the number is less than one,zero between the, decimal and first non zero digit are not significant., (iii) Powers of 10 is not a significant figure., 5,1,2, , , , Dimensional variables are those physical, quantities which have dimensions and do not have, fixed value., Ex:velocity, acceleration, force, work, power.etc., , Dimensionless variables:, , , Dimensionless variables are those physical, quantities which do not have dimensions and do, not have fixed value.,, Ex: Specific gravity, refractive index, Coefficient, of friction, Poisson's Ratio etc.,, , Limitations of dimensional analysis method:, , , , , , Dimensions of a physical quantity are the powers, to which the fundamental quantities are to be , raised to represent that quantity., , Dimensional Formula :, , , Dimensionless quantities are those which do not, have dimensions but have a fixed value., (a):Dimensionless quantities without units., Ex:Pure numbers,angle trigonometric functions ,, logarthemic functions etc.,, (b)Dimensionless quantities with units., Ex:Angular displacement - radian, Joule's, constant etc.,, , Dimensional variables:, , Dimensions of physical quantities:, , , Dimensionless Quantities:, , An expression showing the powers to which the , fundamental quantities are to be raised to, represent the derived quantity is called dimensional, formula of that quantity., In general the dimensional formula of a quantity, , Dimensionless quantities cannot be determined by, this method. Constant of proportionality cannot, be determined by this method. They can be found, either by experiment (or) by theory., This method is not applicable to trigonometric,, logarithmic and exponential functions., In the case of physical quantities which are, dependent upon more than three physical, quantities, this method will be difficult., In some cases, the constant of proportionality also, possesses dimensions. In such cases we cannot, use this system., If one side of equation contains addition or, subtraction of physical quantities, we cannot use, this method., , can be written as M x Ly T z . Here x,y,z are, dimensions., , Dimensional Constants:, , , 10, , The physical quantities which have dimensions and, have a fixed value are called dimensional constants., Ex:Gravitational constant (G), Planck's constant, (h), Universal gas constant (R), Velocity of light in, vacuum (c) etc.,, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , The following is the list of some physical quantities, with their formulae and dimensional formulae with units, S.No. Physical Quantity, , 1., , 2., , Distance ,, Displacement,, Wave length,, Radius of gyration,, Circumference,, Perimeter,Light year,, Mass, , 3., , Period of oscillation,, , Explanation or Formulae, , Dimensional, Formulae, , S.I.Unit, , M 0 L1 T 0 , , m, , M 1 L0 T 0 , , kg, , M 0 L0 T 1 , , s, , M 0 L0T 1 , , hertz ( Hz), , total time, no.of oscillations, , Time,, Time constant, , T = Capacity Resistance, , 4., , Frequency, , Reciprocal of time period n , , 5., , Area, , A = length breadth, , M 0 L2T 0 , , m2, , 6., , Volume, , V=length breadth height, , M 0 L3T 0 , , m3, , 7., , Density, , d=, , M 1 L3T 0 , , kgm-3, , 8., , Linear mass density, , M 1 L1T 0 , , kgm-1, , 9., , Speed, Velocity, , M 0 L1T 1 , , ms-1, , M 0 L1T 2 , , ms-2, , M 1 L1T 1 , , kgms-1, , M 1 L1T 2 , , N, , mass, volume, mass, λ=, length, , 11. Linear momentum, , displacement, time, change in velocity, a=, time, P= mass velocity, , 12. Force, , F = Mass acceleration, , 10. Acceleration, , 13. Impulse, 14. Work,Energy,PE, KE,, Strain energy,, Heat energy, , v=, , J= Force time, W = Force displacement, P.E= mgh, KE =, SE=, , 1, 2, , (Mass) (velocity)2, , NARAYANA GROUP, , P=, , 1 1, , 1, , M L T , , Ns, , M 1 L2T 2 , , J(or) N.m, , M 1 L2T 3 , , watt, , M 1 L1T 2 , , pascal or Nm 2, , 1, ×Stress×Strain×volume, 2, , Work, time, Force, 16. Pressure , Stress,, Area, Stress, Modulus of elasticity (Y, , k) Y=, Strain, , 15. Power, , 1, T, , 11
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 17., , Strain, , 18., , Strain energy density, , 19., , Angular displacement, , 20., , Angular velocity, , 21., , Angular acceleration, , 22., , Angular momentum, , 23., , Planck's constant, , 24., , Angular impulse, , 25., 26., , change in dimension, = original dimension, work, E=, volume, length of arc, θ=, radius, angular dispacement, ω=, time, changein angular velocity, α=, time, L=linear momentum, perpendicular distance, energy, h=, frequency, Torque time, , τ=force× distance, , gravity(g), , g=, , no units, , M 1 L1T 2 , , Jm-3, , M 0 L0T 0 , , rad, , M 0 L0T 1 , , rads-1, , M 0 L0T 2 , , rads-2, , M 1 L2T 1 , , Js, , M 1 L2T 1 , , Js, , M 1 L2T 1 , , Js, , 1 2, , 2, , M L T , , Nm, , M 0 LT 2 , , ms-2 or Nkg-1, , M 1 L3T 2 , , Nm2 kg-2, , M 1 L2T 0 , , kgm2, , M 0 L0T 1 , , S 1, , M 1 L0T 2 , , Nm-1 or Jm-2, , 31., , force, elongation, tangential stress, Coefficient of viscosity η= velocity gradient, , M 1 L1T 1 , , Pa s (or) Nm 2 s, , 32., , Gravitational potential Gravitational field distance, , M 0 L2T 2 , , J/Kg, , 33., , Heat energy, , msθ, , M 1 L2T 2 , , joule, , 34., , Temperature, , θ, , M 0 L0T 0 1 , , kelvin( K), , 35., , Specific heat capacity S (or) C= mass×temp., , M 0 L2T 2 1 , , Jkg-1 K-1, , 36., , Thermal capacity, , dQ, =mass×specific heat, dθ, , M 1 L2T 2 1 , , JK-1, , 27., , Torque, Acceleration due to, , M 0 L0T 0 , , Universal gravitational, , weight, mass, , G=, , Force distance , , 2, , Mass1 Mass 2, , Constant, 28., , Moment of inertia, , I=Mass (radius of gyration)2, , 29., , Velocity gradient, , =, , 30., , Surface tension,, , S=, , dv, dx, surface energy force, =, changein area length, , Surface energy, Spring constant, Force constant, , 12, , K=, , heat energy, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , 37. Latent heat (or), Calorific value, 38. Water equivalent, 39. Coefficient of thermal, expansion, 40. Universal gas constant, 41. Gas constant (for 1 gm), 42. Boltzmann’s constant, (for 1 Molecule), , heat energy, mass, W=Mass specific heat, L=, , l, A V, , ; , ; V , l , A, , PV, nT, R, r=, Mol.wt, , R=, , k=, , R, Avagadro number, , M 0 L2T 2 , , Jkg-1, , M 1 L0T 0 , , kg, , M 0 L0T 0 1 , , K -1, , M 1 L2T 2 1mol 1 , , Jmol-1K-1, , M 0 L2T 2 1mol 1 , , Jkg-1K-1, , M 1 L2T 2 1 , , JK-1molecule-1, , 43. Mechanical equivalent, of heat, , J, , W, H, , M 0 L0T 0 , , no SI units, , 44. Coefficient of thermal, , K=, , Qd, A Δθt, , M 1 L1T 3 1 , , Js-1 m-1 K-1 (or) Wm-1 K-1, , M 1 L2T 2 1 , , JK-1, , ΔE, ΔAΔTθ 4, , M 1 L0T 3 4 , , Js-1m-2K-4 (or) Wm-2K-4, , dθ, temp×time, =, dQ, Heat, , , , , dt , , M 1 L2T 3 1 , , KsJ-1, , 48. Temperature gradient, , d, KA, Change in temp dθ, =, length, dl, , M 0 L1T 0 1 , , Km-1, , 49. Pressure gradient, , Change in pressure dp, =, length, dl, , M 1 L2T 2 , , pascal m-1, , 50. Solar constant, , Energy, ΔE, =, area × time AT, , M 1 L0T 3 , , Js-1m-2 (or) Wm-2, , 51. Enthalpy, , heat ( Q ), , M 1 L2T 2 , , joule, , conductivity, 45. Entropy, , dQ heat energy, =, T temperature, , 46. Stefan's constant, , σ=, , R=, , 47. Thermal resistance, , ( or) R=, , 52. Pole strength, , m =IL ( or), , 0, , 0, , M LT A, , Am, , M 0 L2T 0 A , , Am2, , Magnetic Momement, Mag.Length, , 53. Magnetic moment, , NARAYANA GROUP, , M= 2 l ×m, , 13
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 54. Magnetic intensity (or), Magnetising field, 55. Intensity of magnetisation, 56. Magnetic flux, , H=, , m, 4πd 2, , Magnetic moment, Volume, , = B×A, I=, , M 0 L1T 0 A, , Am-1, , M 0 L1T 0 A, , Am-1, , M 1 L2T 2 A1 , , Wb, , =(Magnetic induction Area), 57. Magnetic induction, , Magnetic flux F, B , =, A, Area, il, , 58. Magnetic permeability, , µ=, , 59. Magnetic susceptibility, , χ=, , 60. Electric current, 61. Charge, , 4πFd 2, m1 m 2, , Tesla (or) Wbm-2 (or) NA-1m-1, , M 1 L1T 2 A2 , , Hm-1, , M 0 L0T 0 , , no units, , I, , M 0 L0T 0 A, , A, , Q =Current Time, , M 0 L0TA, , C, , M 0 L1 AT , , Cm, , M 1 LT 3 A1 , , NC -1, , M 1 L3T 3 A1 , , Nm2 C-1, , M 1 L2T 3 A1 , , V, , M 1 L2T 3 A2 , , , , M 1 L2T 3 A2 , , mho (or) Siemen (S), , M 1 L3T 3 A2 , , Ohm-m, , I, H, , 62. Electric dipole moment P=Charge Distance, 63. Electric field strength (or), Force, E=, Electric field intensity, Charge, 64. Electrical flux ( E ), Electrical intensity area, 65. Electric potential (or), , M 1 L0T 2 A1 , , V=, , Work, Charge, , Potential difference, 66. Electrical resistance, 67. Electrical conductance, , Pot.diff, Current, 1, 1, C= =, R Resistance, R=, , 68. Specific resistance (or, Resistivity, , (or) s, , ρ=, , RA, l, , 1, , 69. Electrical conductivity, , = R esistivity, , M 1 L3T 3 A2 , , Ohm-1 m-1 (or) Siemen m-1, , 70. Current density, ( current per unit area, of cross section), , J = Electrical intensity, Conductivity, , M 0 L2T 0 A, , Am-2, , M 1 L2T 4 A2 , , F, , M 1 L2T 2 A2 , , H(or) Wb/amp, , Current , , or Area , , , 71. Capacitance, , C=, L=, , 72. Self (or) Mutual, , Q Charge, =, V Potential, dε Voltage×Time, =, Current, dI , , dt, , , inductance, 14, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , ε=, , 73. Electrical permittivity, , UNITS AND MEASUREMENTS, , q 1q 2, 4πFd 2, , 74. Surface charge density, Charge, Area, Light energy, Time, ΔE, Luminous flux , I=, =, , ΔtΔA , Area, , , 75. Luminous flux, 76. Intensity of illumination, , M 1 L3T 4 A2 , , farad/m, , M 0 L2T 1 A1 , , Cm-2, , M 1 L2T 3 , , lumen, , M 1 L0T 3 , , lumen m-2 (or) lux., , (or) Iluminance, 77. Focal power, , P=, , 1, Focal length, , M 0 L1T 0 , , dioptre, , 78. Wave number, (Propagation constant), , =, , 1, λ, , M 0 L1T 0 , , m-1, , 79., , R=, , Z2e4m, 8ε 20 ch 3, , M 0 L1T 0 , , m-1, , Rydberg’s constant, , Physical Quantities Having Same, WE-11: Let 0 denote the dimensional formula Dimensional Formulae:, of permittivity of vacuum .If M is mass ,L is, length,T is time and A is electric current,then, (JEE-MAIN 2013), 1 q1q2, Sol: From coulomb’s law F 4 R 2, 0, q1q2, 4 FR 2, Substituting the units, , , , , , L T 2 , , , , 0 , , , 2, , AT , c2, 0 , , 2, N m MLT 2 L2 , 1 3, , 4, , 2, , M L T A , , WE-12:The dimensional formula of magnetic field, strength in M, L, T and C (coulomb) is given, as (AIEEE 2008), Sol: From F = Bqv, 2, F MLT , B, , M 1 L0T 1C 1 , qv C LT 1 , , Distance, Displacement, radius,wavelength, radius, of gyration [L], Speed, Velocity, Velocity of light LT 1 , acceleration ,acceleration due to gravity, intensity, of gravitational field, centripetal acceleration, Impulse, Change in momentum M LT 1 -size, changed, Force, Weight, Tension,energy gradient, Thrust, M LT 2 , , , , , , -- size changed, Work, Energy, Moment of force or Torque,, , , , Moment of couple M L2 T 2 -- size changed, Force constant, Surface Tension, Spring constant,, , , , surface energy i.e. Energy per unit area M T 2 size changed, Angular momentum, Angular impulse, Planck's, , , , , , constant M L2 T 1 - size changed, Angular velocity, Frequency, angular, frequency,Velocity gradient,, Decay constant, rate of disintegration [T–1], Stress, Pressure, Modulus of Elasticity, Energy, density M L1 T 2 , , , NARAYANA GROUP, , Latent heat, Gravitational potential L2 T 2 , 15
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, Specific heat, Specific gas constant L2 T 2 1 , Thermal capacity, Entropy, Boltzmann constant,, Molar thermal capacity, M L2 T 2 1 , Wave number, Power of a lens, Rydberg’s constant, L1 , , , L, Time, RC, ,, R, , a, bx should represent pressure, , a , a 1, 2, 1 2, b L ML T b MT , , , Uses of dimensional analysis method:, , To check the correctness of the given equation., (This is based on the principle of homogeneity), Power, Rate of dissipation of energy, ML2T 3 , To convert one system of units into another, Intensity of sound, Intensity of radiation [ MT 3 ], system., Electric potential, potential difference, electromotive To derive the equations showing the relation, force [ ML2T 3 I 1 ], between different physical quantities., Intensity of magnetic field, Intensity of magnetization, 1 2, WE-14:Check whether the relation S ut at, I L1 , 2, is, dimensionally, correct, or, not,, where, symbols, 3 1, Electric field and potential gradient MLT A , have their usual meaning., Rydberg’s constant and propagation constant, 1 2, Sol: We have S ut at . checking the dimensions, M 0 L1T 0 , 2, Strain , Poisson’s ratio, refractive index, dielectric, on both sides, LHS= S M 0 L1T 0 ,, constant, coefficient of friction, relative permeability, magnetic susceptibility, electric susceptibility,, 1 2 , 1, 2, 2, RHS= ut 2 at LT T LT T , angle, solid angle, trigonometric ratios,logarithm func, , tion, exponential constant are all dimensionless., 1 0, 1 0, 1 0, , M 0 LT, M 0 LT, M 0 LT, If L,C and R stands for inductance, capacitance and, L, we find LHS=RHS., resistance respectively then , LC , RC and time, R, Hence, the formula is dimensionally correct., 0 0, M L T , WE-15:Young’s modulus of steel is 19 1010 N / m 2 ., Coefficient of linear expansion, coefficient of superExpress it in dyne / cm 2 . Here dyne is the CGS, ficial expansion and coefficient of cubical, unit of force., expansion,temperature coefficient of resistance, Sol: The SI unit of Young’s modulus is N / m 2 . ., M 0 L0T 0 K 1 , 5, , 0 3, 10 dyne , 10, Solar constant and poynting vector ML T , 10 N 19 10 , Given Y 19 10, 102 cm 2 , m2, Principle of homogeneity:, , , It states that only quantities of same dimensions can, dyne , be added, subtracted and equated., 19 1011 , 2 , cm , a, WE-13: The dimensional formula of in the WE-16 : For a particle to move in a circular orbit, b, uniformly, centripetal force is required,, 2, which depends upon the mass (m), velocity, a ct, equation P , where P = pressure,, (v) of the particle and the radius (r) of the, bx, circle. Express centripetal force in terms of, x = displacement and t = time, these quantities, 2, a ct , Sol: According to the provided information,, Sol : P bx bx , , let F m a vb r c . F km a vb r c, b, By principle of Homogeneity,, 1 2, M a LT 1 Lc , M 1 LT, 1, , LC ,[T ], , , , 16, , , , , , , , , , , , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 1 2, M a Lb cT b , M 1 LT, , using principle of homogeneity we have, a = 1 ,b + c = 1 ,b = 2, on solving we have a = 1, b = 2, c = -1, using these values we get F = km1v 2 r 1, , UNITS AND MEASUREMENTS, , EJ 2, M 5G 2, Substituting D.F. of E, J, M, and G in above formula, , Sol :, , , , D.F. of, , ML2T 2 ML2T 1 , M 5 M 1 L3T 2 , , 2, , 2, , M 0 L0T 0 , , mv 2, F k, 1 , y, r, WE20: In the equation p k T where p is the, , , B, Note: The value of the dimensionless constant k, pressure, y is the distance, k B is Boltzmann, is to be found experimentally., constant and T is the temperature. DimenWE-17: Derive an expression for the time period, sions of are (Med- 2013), of a simple pendulum of mass(m), length (l) at, a place where acceleration due to gravity is (g)., 1, y, , Sol: Let the time period of a simple pendulum depend Sol., p k BT, upon the mass of bob m, length of pendulum l ,, Dimension of, and acceleration due to gravity g, then, Dimensional formula of kB Dimensional formula of T , , Dimensional formula of p Dimensional formula of y, t m a l b g c t km a l b g c, 0 0, , 1, , a b, , 2, , ML2T 3 T , , M 0 L2T 0 , ML1T 2 L , , c, , M L T M L LT , M 0 L0T 1 M a Lb cT 2 c, Dimensions of M,L,T in are 0,2,0, comparing the powers of M, L, and T on, both sides, we get a = 0, b + c = 0, -2c=1, WE21: The vander Waal’s equation for n moles of, a, a = 0, b = 1/2 and c = -1/2. Putting these values,, , a real gas is p V 2 V b nRT where p is pres1, , , 2, l, 0 l, sure,, V, is, volume,, T, is absolute temperature, R, we get T km 12 T k g ,, is molar gas constant a, b and c are vander, g, Waal’s constants. The dimensional formula for, which is the required relation., ab is (Med- 2012), WE18: If C is the velocity of light, h is Planck’s Sol.By principle of homogenity of dimensions P can, constant and G is Gravitational constant are, a, taken as fundamental quantities, then the diadded to P only. It means 2 also gives pressure., mensional formula of mass is.(Eamcet - 2014), V, 1, 2 1, h ML T (2), Sol: C LT (1) ;, Dimension formulae for pressure P M 1L1T 2 , G M 1 L3T 2 (3), , and Volume V M 0 L3T 0 , , Solving (2) and (3), h ML2T 1 , , M 2 L1T 1 , G M 1 L3T 2 , , Substituting (1) in above, 1, 1 1, h M 2 M h 2 G 2 C 2 , , , , G, C, , , WE19: If E, M, J and G respectively denote energy,, mass, angular momentum and universal gravitational constant, the quantity, which has the, same dimensions as the dim ensions of, 2, , EJ, (Eamcet - 2013), M 5G 2, NARAYANA GROUP, , Since, , , a, = pressure, V2, , a, a, M 1L1T 2 0 6 0 M 1L1T 2 , 0, M LT, M L T , 0 3, , a M 1 L5T 2 , similarly, b will have same dimensions as, volume V b volume, b M 0 L3T 0 , , ab M 1 L5T 2 M 0 L3T 0 M 1 L8T 2 , , 17
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, W.E-22:A screw gauge having 100 equal divisions, and a pitch of length 1 mm is used to measure, the diameter of a wire of length 5.6 cm. The, main scale reading is 1 mm and 47th circular, division coincides with the main scale. Find the, curved surface area of the wire in cm 2 to, appropriate significant figures.(Use = 22/7), 1 mm, 0.01 mm, Sol. Least Count =, 100, Diameter = MSR + CSR(LC) = 1 mm+47 (0.01), mm = 1.47 mm, Surface area = Dl , , 22, 1.47 56 mm 2, 7, , = 2.58724 cm 2 = 26cm 2, W.E-23: In Searle’s experiment, the diameter of the, wire as measured by a screw gauge of least, count 0.001 cm is 0.050 cm. The length,, measured by a scale of least count 0.1 cm, is, 110.0 cm. When a weight of 50 N is suspended, from the wire, the extension is measured to be, 0.125 cm by a micrometer of least count 0.001, cm. Find the maximum error in the, measurement of Young’s modulus of the, material of the wire from these data., Sol.Maximum percentage error in Y is given by, Y, , W, L, Y , D x L, , , 2, 2, , D x, L, Y , D x, 4, , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 0.001 0.001 0.1 , 2, , , 0.0489, 0.05 0.125 110 , , W.E24:The side of a cube is measured by vernier, calipers (10 divisions of the vernier scale, coincide with 9 divisions of the main scale,, where 1 division of main scale is 1 mm). The, main scale reads 10 mm and first division of, vernier scale coincides with the main scale., Mass of the cube is 2.736 g. Find the density of, the cube in appropriate significant figures., Sol.Least count of vernier calipers, , , 09., , 10., , 1 division of main scale, 1, , 0.1 mm, Number of divisions in vernier scale 10, , The side of cube = 10 mm + 1 0.1 mm 1.01 cm, Mass, , 2.736 g, , 3, Now, density = Volume 1.013 cm3 2.66 g cm, , C.U.Q, UNITS & MEASUREMENTS, 1., , 18, , The reliability of a measurement depends on, 1) precision, 2) accuracy, 3) systematic error, 4) random error, , 11., , 12., , The error due to resolution of a measuring, instrument is, 1) personal error, 2) random error, 3) systematic error, 4) gross error, The error due to resolution of a measuring, instrument is, 1) random error 2) personal error, 3) gross error, 4) least count error, The random error which exists invariably in, screw gauge is, 1) least count error, 2) Zero error, 3) gross error, 4) backlash error, The errors which are estimated by statistical, methods are, 1) systematic errors, 2) random errors, 3) theoretical errors, 4) gross errors, The measure of accuracy is, 1) absolute error, 2) relative error, 3) percentage error, 4) both 2 and 3, The decrease in percentage error, 1) increases the accuracy, 2) does not effect the accuracy, 3) decreases the accuracy, 4) both 1 and 3, In a measurement, both positive and negative, errors are found to occur with equal, probability. The type of errors is, 1) proportional errors 2) systematic errors, 3) determinate errors, 4) random errors, The errors that always occur in the, measurement with screw gauge is, 1) random errors 2) systematic errors, 3) gross errors 4) negligible errors, A physicist performs an experiment and takes, 200 readings.He repeats the same experiment, and now takes 800 readings. By doing so, 1) the probable error remains same, 2) the probable error is four times, 3) the probable error is halved, 4) the probable error is reduced by a factor ¼, More the number of significant figures shows, more the, 1)accuracy 2)error 3)number of figures 4)value, If a measured quantity has n significant, figures, the reliable digits in it are, 1) n, 2) n-1, 3) n 1 4) n/2, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , 13. If the significant figures are more,, 1)percentage error is more and accuracy is less, 2)percentage error is less and accuracy is more, 3)percentage error is less and accuracy is less, 4)percentage error is more and accuracy is more, 14. The mathematical operation in which the, accuracy is limited to least accurate term is, 1) addition, 2) subtraction, 3) multiplication & division 4) both 1 and 2, 15. The time period of a seconds pendulum is, measured repeatedly for three times by two stop, watches A,B. If the readings are as follows, then, S.NO, A, B, 1., 2.01 sec, 2.56 sec, 2., 2.10 sec, 2.55 sec, 3., 1.98 sec, 2.57 sec, 1) A is more accurate but B is more precise, 2) B is more accurate but A is more precise, 3) A,B are equally precise, 4) A,B are equally accurate, 16. If Y = a + b, the maximum percentage error in, the measurement of Y will be, 1), 3), , a b , , , 100, b , a, a b , a b 100, , , , a, , b , , 2) a b a b 100, b , a, , , 100, a, , b, a, b , , , 4), , 17. If Y = a - b, the maximum percentage error in, the measurement of Y will be, 1), , a b , , , 100, b , a, , a, , a, , , b , , , a, , b , , 2) a b a b 100, , b , , 4) a b a b 100, 3) a b 100, , , , , 18. If Y = a x b, the maximum percentage error in, the measurement of Y will be, a b , a, b, , 1) a 100 / b 100 2) 100, , , , a b, , a b , 100, a b , 19. If Y = a/b, the maximum percentage error in, the measurement of Y will be, 3), , a, b, , 100 , 100 4), , a, b, , , a b , a, b, , 1) a 100 / b 100 2) 100, b , , , , a, , a, , b, , , , 3) 100 100 , a, b, , a b , , 4) , 100, b , a, 20. Of the following the dimensionless error is, 1) Systematic error, 2) Gross error, 3) Random error, 4) Relative error, NARAYANA GROUP, , 21. In determining viscosity by the equation, , pr 4, which of the quantities must be, 8vl, measured more accurately, 1) P, 2) r, 3) v, 4) l, 22. The number of significant figures in 0.007 is, 1) 4, 2) 2 3) 3 4) 1, 23. Round off 20.96 to three significant figures, 1) 20.9, 2) 20 3) 21.0 4) 21, UNITS AND DIMENSIONAL FORMULA, 24. The dimensional formula for strain energy, density is, 2) [ M 1 L2T 3 ], 1) [ M 1 L2T 3 ], 1, , 1, , 2, 3) [ M L T ], 4) [ M 1 L2T 2 ], 25. The dimensional formula for areal velocity is, 2) [ M 0 L2T 1 ], 1) [ M 0 L2T 1 ], 0, 2, , 1, 3) [ M L T ], 4) [ M 0 L2T 1 ], 26. The physical quantity having the same, dimensional formula as that of force is, 1) Torque 2)work 3) pressure 4) thrust, 27. Nm-1 is the SI unit of, 1) velocity gradient, 2) Rydberg’s constant, 3) coefficient of viscosity 4) Spring constant, 28. If P is the X-ray unit and Q is micron then P/Q, is, 1) 105, 2) 105, 3) 107, 4) 107, 29. The dimension of mass is zero in the following, physical quantities., 1)Surface tension 2)coefficient of viscosity, 3)heat, 4) Specific heat capacity, 30. The SI unit of a physical quantity is, [J m-2 ]. The dimensional formula for that, quantity is, 1)[ M 1 L2 ], 2)[ M 1 L0T 2 ], 3)[ M 1 L2T 1 ], 4)[ M 1 L1T 2 ], -2, 31. [Jm ] is the unit of, 1) Surface tension, 2) Viscosity, 3) Strain energy, 4) Intensity of energy, 32. The set of quantities which can form a group, of fundamental quantities in any system of, measurement is, 1) Length,mass and time, 2)Length,mass and velocity, 3)Length,velocity and time, 4)velocity,mass and time, 33. The fundamental unit which is common in, C.G.S. and S.I system is, 1) metre 2) second 3) gram 4) all the above, 34. 1 a.m.u is equal to, 1) 1.66 x 10-24 g, 2) 1.66 x 10-27 g, 24, 3) 1.66 x 10 g, 4) 1.66 x 1027 g, , , 19
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 35. Modulus of Elasticity is dimensionally, equivalent to, (1996 E), 1) Stress, 2) Surface tension, 3) Strain, 4)Coefficient of viscosity, 36. If x times momentum is work, then the, dimensional formula of x is, 1 1, 1) [L1T] 2) [LT 1] 3) [ML1T1] 4) [MLT, ], 37. The following does not give the unit of energy, 1) watt second, 2) kilowatt hour, 3) newton metre, 4) pascal metre, 38. 1 fermi is equal to, 1) 1012 m 2) 109 m 3) 106 A0 4)10-9 micron, 39. "Impulse per unit area " has same dimensions, as that of, 1)coefficient of viscosity 2) surface tension, 3) bulk modulus, 4) gravitational potential, 40. The following pair does not have same, dimensions, 1) Pressure, modulus of elasticity, 2) Angular velocity, velocity gradient, 3) Surface tension and force constant, 4) Impulse and torque, 41. Dimensions of solar constant are, 1 2, , 1) M 0 L0T , 2) M 1 LT, 42., , 43., 44., , 45., , 46., 47., , 48., , 20, , 3) M 1L1T 2 , 4) M 1T 3 , The following is a unitless and dimensionless, quantity, 1) Angle, 2) Solid angle, 3) Mechanical equivalent of heat, 4) Coefficient of friction, The unitless quantity is, 1) Velocity gradient, 2) Pressure gradient, 3) Displacement gradient 4) Force gradient, If the unit of tension is divided by the unit of, surface tension the derived unit will be same, as that of, 1) Mass 2) Length, 3) Area 4) Work, Atto is ___________, 1) An instrument used to measure gradient, 2) An instrument used to measure the altitude, 3) 1018, 4) 10-18, -1, N m s is the unit of, 1) Pressure, 2) Power, 3) Potential, 4) Pressure gradient, Which one of the following represents the, correct dimensions of the coefficient of, viscosity?, (AIEEE 2004), 1) [ ML1T 2 ], 2) [ MLT 1 ], 3) [ ML1T 1 ] 4) [ ML2T 2 ], Stefan's constant has the unit as, 1) J s-1 m-2 K4, 2) Kg s-3 K4, -2, -4, 3) W m K, 4) Nms-2 K-4, , 49. Which one of the following is not measured in, the units of energy, 1) (couple) x (angle turned through), 2) moment of inertia x ( angular velocity)2, 3) force x distance, 4) impulse x time, 50. An example to define length in the form of, time at a place is, 1) Wrist watch 2) Linear expansion of iron rod, 3) Frequency of ripples on the surface of water, 4) Seconds pendulum, 51. The one which is not the unit of length is, 1) Angstrom unit, 2) Micron, 3) Par-sec, 4) Steradian, 52. The physical quantity having the same, dimensional formula as that of entropy is :, 1) Latent heat, 2) Thermal capacity, 3) Heat, 4) Specific heat, 53. Js is the unit of, 1) Energy, 2) Angular Momentum, 3) Momentum, 4) Power, 54. Which of the following cannot be expressed, as dyne cm-2?, 1) Pressure, 2) Longitudinal stress, 3) Longitudinal strain, 4) Young's modulus of elasticity, 55. The unit of atmospheric pressure is :, 1) Metre 2) kgwt, 3) g cm-2, 4) bar, 56. The ratio between pico and giga is, 1) 1021, 2) 10-21, 3) 1014, 4) 108, 57. 1 micron =___ nanometer, 1) 10-6, 2)10-10, 3) 103, 4) 10-3, 58. Which of the following has smallest value?, 1) peta, 2)femto, 3) kilo, 4)hecto, 59. The physical quantity having dimension 2 in, length is, 1) Power, 2) Acceleration, 3) Force constant, 4) Stress, 60. If m is the mass of drop of a liquid of radius 'r', mg, , then r has the same dimensions of :, 1) Surface tension, 2) Tension, 3) Young's Modulus 4) Coefficient of viscosity, 61. The intensity of a wave is defined as the energy, transmitted per unit area per second. Which of, the following represents the dimensional, formula for the intensity of the wave?, 1) ML0T 2 , 2) ML0T 3 , 3) ML0T 1 , 4) [ ML4T ], 62. The fundamental unit which has the same, power in the dimensional formula of surface, tension and coefficient of viscosity is(1989 E), 1) mass 2) length 3) time 4) none, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 63. Electron volt is the unit of, (1988 E), 1) Power, 2) Potential difference, 3) Charge, 4) Energy, 64. One shake is equal to, 2) 10 9 s 3) 10 10 s 4) 109 s, 1) 10 8 s, 65. Torr is the unit of physical quantity, 1) density, 2) pressure, 3) torque, 4) None, 66. The S.I. value of Mechanical equivalent of, heat is:, 1) 4.2, 2) 1, 3) 2.4, 4) 2, 67. The physical quantity that has no dimensions is:, 1) angular velocity, 2) linear momentum, 3) angular momentum 4) strain, 68. The physical quantities not having same dimensions are, 1) torque and work, 2) momentum and Planck’s constant, 3) stress and Young’s modulus, , , 4) speed and 0 o , , , , 77. Of the following quantities which one has the, dimensions different from the remaining, three?, 1) energy density2) force per unit area, 3) product of charge per unit volume and voltage, 4) Angular momentum per unit mass, 78. The dimensional formula of resistivity in terms, of M, L, T and Q, where Q stands for the dimensions of charge is, 1) [ ML3T 1Q 2 ] 2) [ ML3T 2Q 1 ], 79., , 80., , 1/ 2, , 69. A pair of physical quantities having the same, dimensional formula are (1992 M), 1) Force and Work, 2) Work and energy, 3) Force and Torque, 4) Work and Power, 70. The dimensional formula of calorie are, 1) [ ML2T 2 ], 2) [ MLT 2 ], 2, , 1, 3) [ ML T ], 4)[ ML T 1 ], 71. The dimensional formula for coefficient of, kinematic viscosity is :(2002M), 1. [ M 0 L1T 1 ], 2. [ M 0 L2T 1 ], 3. [ ML2T 1 ], 4. [ ML1T 1 ], 72. The product of energy and time is called action., The dimensional formula for action is same, as that for, 1) force velocity, 2) impulse distance, 3) power, 4) angular energy, 73. Specific heat is in joule per kg per 0C rise of, temperature. Its dimensions are:, 2) [ ML2T 2 K 1 ], 1) [ MLT 1 K 1 ], 0, 2, , 2, , 1, 3)[ M LT K ], 4) [ MLT 2 K1 ], 74. The dimensional formula for Magnetic, Moment of a magnet is, 1) [ M 0 L2T 0 A1 ], 2) [ M 0 L2T 0 A1 ], 3)[ M 0 L2T 0 A1 ], 4) [ M 0 L2T 0 A1 ], 75. Dimensions of C x R (Capacity x Resistance), is, (1995 E), 1) frequency, 2) energy, 3) time period, 4) current, 76. Dimensional formula for capacitance is (1997E), 1) [ M 1 L2T 4 I 2 ], 2) [ M 1 L2T 4 I 2 ], 3) [ M 1 L2T 2 ], 4) [MLT–1 ], , NARAYANA GROUP, , UNITS AND MEASUREMENTS, , 81., , 82., , 83., , 3) [ ML2T 1Q 1 ] 4) [ MLT 1Q 1 ], The dimensional formula for Magnetic, induction is, (2000 M), , 1, , 1, 2) [ MT 2 A1 ], 1) [ MT A ], , 1, 3) [ MLA ], 4)[ MT 2 A ], The dimensional formula for magnetic flux is, (2003M), 2, , 2, , 1, 1) [ ML T I ] 2) [ ML2T 2 I 2 ], 3) [ ML2T 2 I 1 ]4) [ ML2T 2 I 2 ], The SI unit of a physical quantity having the, dimensional formula of [ ML0T 2 A1 ], 1) tesla, 2)weber, 3)amp metre, 4)amp m2, , What are the units of 0, 4, , 1, 2, 1) NA m, 2) NA2, 3) Nm 2C 2, 4) unitless, If is the permeability and is the, 1, , permittivity then is equal to, 1. speed of sound, 2. speed of light in vacuum, 3. speed of sound in medium, 4. speed of light in medium, Permeability , , 84. Permittivity will have the dimensional, , , formula of :, 1) [ M 0 L0T 0 A0 ], 2) [ M 2 L2T 4 A2 ], 3) [ M 2 L4T 6 A4 ], 4)[ M 2 L4T 6 A4 ], 85. Siemen is the S.I unit of, (1991 E), 1)Electrical conductance 2) Electrical conductivity, 3)Potential difference, 4)Inductance, 86. Which of the following quantities has the units, Kg m2 s-3 A-2?, 1) Resistance, 2) Inductance, 3) Capacitance, 4) Magnetic flux, 87. The SI unit of magnetic permeability is, 1) Am1 2) Am 2, 3) Hm 2 4) Hm 1, 88. The dimensions of time in Electrical intensity is, 1) -1, 2) -2, 3) -3 4)3, 21
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 89. SI Unit of a physical quantity whose, dimensional formula is M 1 L2T 4 A2 is, 1.ohm 2. volt 3. siemen, 4. farad, 1, 90., Capacitance Inductance have the same, unit as, 1) time, 2) velocity, 3)velocity gradient 4) none of the above, 1, 91. What are the units of K 4 ? (AIEEE 2004), , 2, , 1, , 2, , 2, 1, 2, 1) C N m, 2) C N m, 2, 1, 2, 4) unitless, 3) C N m, 92. [M1L2T-3A-2] is the dimensional formula of :, 1) electric resistance, 2) capacity, 3) electric potential, 4) specific resistance, 93. If L is the inductance, 'i' is current in the, 1, , circuit, Li 2 has the dimensions of, 2, 1. Work 2. Power 3. Pressure 4. Force, 94.The dimension of length in electrical resistance is, 1) 2, 2) 1, 3) -2, 4) -1, 95. If m is the mass, Q is the charge and B is the, magnetic induction, m/BQ has the same, dimensions as :(1999 M), 1)Frequency 2)Time 3)Velocity 4)Acceleration, 96. If L has the dimensions of length, V that of, potential and 0 is the permittivity of free space, then quantity 0 LV, V has the dimensions of, 1) current 2) charge 3) resistance 4) voltage, 97. Dimensional formula of ‘ohm’ is same as, h, h, h2, h2, 1), 2), 3) 2, 4) 2, e, e, e, e, 98. If 'm' is the mass of a body, 'a' is amplitude of, vibration, and ' ' is the angular frequency,,, 1, ma 2 2 has same dimensional formula as, 2, 1) impulse, 2) moment of momentum, 3) moment of inertia 4) moment of force, 99. If C, R, L and I denote capacity, resistance,, inductance and electric current respectively,, the quantities having the same dimensions of, time are, (2006 E), a) CR b) L/R c) LC, d) LI 2, 1) a and b only, 2) a and c only, 3) a and d only, 4) a, b and c only, 100. Which of the following do not have the same, dimensions as the other three? Given that, l = length, m = mass, k= force constant,, I = moment of inertia, B = magnetic, induction, Pm magnetic dipole moment,, R= radius, g = acceleration due to gravity, 22, , 1) l / g 2) I / PmB 3) k / m 4) R / g, 101. Given that I= moment of inertia,, Pm magnetic dipole moment and, B= magnetic induction, then the dimensional, formula for I / Pm B is same as that of, 1) time 2) length 3) time2 4) length 2, 102. Given that m = mass, l = length, t = time and i, = current. The dimensional formula of ml 2 / t 3i, are the same as that of, 1) electric field, 2) electric potential, 3) capacitance, 4) inductance, 103. If F is the force, is the permeability, H is the, intensity of magnetic field and i is the electric, F, , current, then Hi has the dimensions of, 1) mass 2) length, 3) time 4) energy, 104. If e,0 , h and c respectively represent electric, charge, permittivity of free space, Planck’s, e2, constant and speed of light then, has the, 0 hc, dimensions of, a) angle, b) relative density, c) strain, d) current, 1) a & b are correct, 2) d & c are correct, 3) a, b & c are correct 4) a,b,c & d are correct, 105. Two physical quantities are represented by P, and Q. The dimensions of their product is, [ M 2 L4T 4 I 1 ] and the dimensions of their ratio, is [ I 1 ]. Then P and Q respectively are, 1. magnetic flux and Torque acting on a magnet., 2. torque and Magnetic flux., 3. magnetic moment and Pole strength, 4. magnetic moment and Magnetic permeability, , C.U.Q-KEY, 1) 2 2) 3, 7) 1 8) 4, 13) 2 14) 4, 19) 2 20) 4, 25) 3 26) 4, 31) 1 32) 1, 37) 4 38) 4, 43) 3 44) 2, 49) 4 50) 4, 55) 4 56) 2, 61) 2 62) 1, 67) 4 68) 2, 73) 3 74) 1, 79) 2 80) 1, 85) 1 86) 1, 91) 2 92) 1, 97) 3 98) 4, 103) 2 104) 3, , 3) 4, 9) 2, 15) 1, 21) 2, 27) 4, 33) 2, 39) 1, 45) 4, 51) 4, 57) 3, 63) 4, 69) 2, 75) 3, 81) 1, 87) 4, 93) 1, 99) 4, 105) 1, , 4) 4, 10) 4, 16) 2, 22) 4, 28) 4, 34) 1, 40) 4, 46) 2, 52) 2, 58) 2, 64) 1, 70) 1, 76) 1, 82) 2, 88) 3, 94) 1, 100) 3, , 5) 2, 11) 1, 17) 2, 23) 3, 29) 4, 35) 1, 41) 4, 47) 3, 53) 2, 59) 1, 65) 2, 71) 2, 77) 4, 83) 4, 89) 4, 95) 2, 101) 3, , 6) 4, 12) 2, 18) 2, 24) 3, 30) 2, 36) 2, 42) 4, 48) 3, 54) 3, 60) 1, 66) 2, 72) 2, 78) 1, 84) 3, 90) 3, 96) 2, 102) 2, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , LEVEL-I (C.W), ACCURACY, PRECISION, TYPES OF, ERRORS AND COMBINATION OF, ERRORS, 1., , 2., , 3., , The accuracy in the measurement of the, diameter of hydrogen atom as 1.06 x 10-10 m is, 1, 1) 0.01 2)106 x 10-10 3), 4)0.01 x 10-10, 106, The length of a rod is measured as 31.52 cm., Graduations on the scale are up to, 1) 1 mm 2) 0.01 mm 3) 0.1 mm 4) 0.02 cm, If L 20 0.01 m and B 10 0.02 m, then L/B is, 1), , 4., , 2 0.03 m, 2 0.01 m, , 2), , 2 0.015 m, 2 0.005 m, , UNITS AND MEASUREMENTS, SIGNIFICANT FIGURES & ROUNDING OFF, 10. If the value of 103.5 kg is rounded off to three, significant figures, then the value is, 1) 103 2) 103.0 3) 104 4) 10.3, 11. The number of significant figures in, , 12., 13., , 14., 15., , 3), 4), The radius of a sphere is measured as, , 10 0.02% cm ., , 5., , 6., , 7., , The error in the, measurement of its volume is, 1) 25.1cc 2)25.12cc 3)2.51cc 4)251.2cc, If length and breadth of a plate are, 40 0.2 cm and 30 0.1 cm , the absolute, error in measurement of area is, 1) 10 cm 2 2) 8 cm 2 3) 9 cm 2 4) 7 cm 2, If the length of a cylinder is measured to be, 4.28 cm with an error of 0.01 cm, the, percentage error in the measured length is, nearly, 1) 0.4 % 2) 0.5 %, 3) 0.2 % 4) 0.1 %, When 10 observations are taken, the random, error is x. When 100 observations are taken,, the random error becomes, 1) x/10, 2) x 2, 3) 10 x 4) x, , 8., , If L1 2.02 0.01 m and L2 1.02 0.01 m, , 9., , then L1 2 L2 is (in m), 1) 4.06 0.02, 2) 4.06 0.03, 3) 4.06 0.005, 4) 4.06 0.01, A body travels uniformly a distance of, , 20.0 0.2 m, , in time 4.0 0.04 s . The, velocity of the body is, 1) 5.0 0.4 ms 1, , 2) 5.0 0.2 ms 1, , 3) 5.0 0.6 ms 1, , 4) 5.0 0.1 ms 1, , NARAYANA GROUP, , 16., 17., , 18., , 19., 20., , 6.023 10 23 mole 1 is, 1) 4, 2) 3, 3) 2, 4) 23, The side of a cube is 2.5 metre. The volume, of the cube to the significant figures is, 1) 15, 2) 16, 3) 1.5 4) 1.6, When a force is expressed in dyne, the number, of significant figures is four. If it is expressed, in newton, the number of significant figures, will become, ( 105 dyne =1N ), 1) 9, 2) 5, 3)1, 4) 4, 2.0 is, 1) 1.414 2) 1.4, 3)1.0, 4) 1, The mass of a box is 2.3 kg. Two marbles of, masses 2.15 g and 12.48 g are added to it. The, total mass of the box is, 1) 2.3438 kg, 2) 2.3428 kg, 3) 2.34 kg, 4) 2.31 kg, The number of significant figures in 0.010200 is, 1) 6, 2) 5, 3) 3, 4) 2, When the number 0.046508 is reduced to 4, significant figures, then it becomes, 1) 0.0465, 2) 4650.8 x 10-5, -2, 3) 4.651 x 10, 4) 4.650 x 10-2, With due regard to significant figures, the, value of (46.7 – 10.04) is, 1) 36.7, 2) 36.00 3) 36.66 4) 30.6, The value of / 53.2 with due regard to, significant figures is,, 1) 0.0591 2) 0.0590 3) 0.590 4) 0.5906, By rounding off, a) 20.96 and b) 0.0003125, to 3 significant figures, we get, 1) 21.0 ; 312 × 104, 2) 21.0 ; 3.12 × 104, 3) 2.10 ; 3.12 × 104, 4) 210; 3.12 × 104, , UNITS AND DIMENSIONAL, FORMULAE, 21. If the unit of length is doubled and that of mass, and time is halved, the unit of energy will be, 1) doubled 2)4 times 3)8 times 4) same, 22. Given M is the mass suspended from a spring, of force constant. k. The dimensional formula, 1/ 2, , for M / k is same as that for, 1) frequency, 2) time period, 3) velocity, 4) wavelength, 23
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 23. The dimensional formula for the product of two, physical quantities P and Q is [ ML2T 2 ]. The, P, , dimensional formula of Q is [ MT 2 ]. Then P, and Q respectively are(2001 M), 1) Force and velocity, 2) Momentum and displacement, 3) Force and displacement, 4) Work and velocity, 24. The fundamental physical quantities that have, same dimension in the dimensional formula of, Torque and Angular Momentum are(2000 E), 1) mass, time, 2) time, length, 3) mass, length, 4)time, mole, 25. The physical quantity which has the, energy, dimensional formula as that of mass length, is, (2000 M), 1) Force 2) Power 3) Pressure 4) Acceleration, 26. If J and E represent the angular momentum, and rotational kinetic energy of a body,, , J2, 2E, , represents the following physical quantity., 1) Moment of couple, 2) Moment of force, 3) Moment of inertia, 4) Force, 27. If the fundamental units of length, mass and, time are doubled, the unit of force will, 1) doubled, 2)halved, 3) remain same, 4) four times, , PRINCIPLE OF HOMOGENEITY, B C, , is dimensionally correct. The, 2, dimensions of A, B and C respectively are ( ,, , 28. A , , A, B, C are constants) where is wave length, of wave, 1)No dimensions, L, L2 2)L2, No dimensions, L, 3) L, L2, No dimensions 4)L,No dimensions,L2, 29. According to Bernoulli’s theorem, , p v2, gh constant. The dimensional, d 2, formula of the constant is ( P is pressure, d is, density, h is height, v is velocity and g is, acceleration due to gravity) (2005 M), 1) [ M 0 L0T 0 ], 2) [ M 0 LT 0 ], 3) [ M 0 L2T 2 ], 4) [ M 0 L2T 4 ], , CONVERSION OF UNITS, 30. The surface tension of a liquid in CGS system, is 45 dyne cm-1. Its value in SI system is, 1) 4.5 Nm-1, 2) 0.045 Nm-1, -1, 3) 0.0045 Nm, 4) 0.45 Nm-1, 24, , 31. If minute is the unit of time, 10 ms-2 is the unit, of acceleration and 100 kg is the unit of mass,, the new unit of work in joule is, 1) 105 2) 106 3) 6 x 106, 4) 36x 106, 32. The magnitude of force is 100 N. What will, be its value if the units of mass and time are, doubled and that of length is halved?, 1) 25 2)100 3) 200 4) 400, 33. A motor pumps water at the rate of V m3 per, second, against a pressure P Nm-2. The power, of the motor in watt is, 1) PV 2) (P / V), 3) (V/P) 4) V P , 34. If the units of length and force are increased, by four times the unit of energy will be, increased by, 1) 16% 2)1600% 3)1500% 4) 400%, 35. SI unit and CGS unit of a quantity vary by 103, times, it is :, (1994 E), 1) Boltzmann constant 2)Gravitational constant, 3) Planck's constant, 4) Angular Momentum, 36. The value of universal gravitational constant, G in CGS system is 6.67 108 dyne cm2 g-2. Its, value in SI system is, 1)6.67 x 10-11Nm2 kg-2 2)6.67 x 10-5 Nm2 kg-2, 3)6.67 x 10-10Nm2 kg-2 4)6.67 x 10-9 Nm2 kg-2, , TO CHECK THE CORRECTNESS OF, PHYSICAL RELATION AND DERIVING, THE EQUATIONS, 37. The final velocity of a particle falling freely, under gravity is given by V 2 u 2 2 gx where, x is the distance covered. If v = 18 kmph,, g = 1000 cm s-2, x = 120 cm then u = ----ms-1., 1) 2.4 2) 1.2 3) 1 4) 0.1, 38. The equation which is dimensionally correct, among the following is, 1) v u at 2, 2) s ut at 3, 3) s ut at 2, 4) t s av, 39. The dimensions of 'k' in the relation V = k avt, (where V is the volume of a liquid passing, through any point in time t, 'a' is area of cross, section, v is the velocity of the liquid) is, 1) [ M 1 L2T 1 ], 2) [ M 1 L1T 1 ], 3) [ M 0 L0T 1 ], 4) [ M 0 L0T 0 ], 40. If force (F), work (W) and velocity (V) are, taken as fundamental quantities then the, dimensional formula of Time (T) is (2007 M), 1) [ W 1 F 1V 1 ], 2) [ W 1 F 1V 1 ], 3) [ W 1 F 1V 1 ], 4) [ W 1 F 1V 1 ], , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , 41. If Force F, Mass M and time T are chosen as From 13 to 20 follow the rules of significant figures, and rounding off numbers, fundamental quantities the dimensional, 2, 2, formula for length, is, E2 M 2 L2 T2 , -1 2, 2 -2, -1 -2 -2, , 1)[FMT] 2)[FM T ] 3)[FL T ] 4)[F L T ] 21., , E1 M 1 L1 T1 , 42. If force F, Length L and time T are chosen as, fundamental quantities,the dimensional 22. Here [k] = force/ length = ML0T 2, 1/ 2, formula for Mass is, M , 0 0, -1 -1 -2, Hence, 1) [FLT], 2) [F L T ], k M LT, , 3) [F-2L-2T-2], 4) [F1L-1T2], P, 2, LEVEL-I(C.W)-KEY, -----(2), 23. PQ ML2T 2 ----(1); Q MT, 01)3 02)3 03)4 04)3, 05)1 06)3, (1) × (2) = P 2 M 2 L2T 4, 07)1 08)2 09)4 10)3, 11)1 12)2, 13)4 14)2 15)4 16)2, 17)3 18)1, P MLT 2 FORCE, (1) (2) = Q2 = L2, 19)2 20)2 21)3 22)2, 23)3 24)3, 24. By dimensional formula, 25)4 26)3 27)3 28)1, 29)3 30)2, 25. Substitute D.F. of quantities, 31)4 32)1 33)1 34)3, 35)2 36)1, 37)3 38)3 39)4 40)4, 41)2 42)4, 26. J ML2T 1 ; E ML2T 2, 27. n1u1 n2u2 28. Substitute D.F. of quantities, LEVEL-I (C.W) - HINTS, 29. Use principle of homogenity, d 0.0110 10, 1, Dyne 10 5 N, N, , , 1., 2 103, 30., 10, d 1.06 10, 106, cm, 10 m, m, 2., 0.01cm is the least count of varnier caliperse., W2 M 2 a2 2T2 2, , 2 2 ;, 31., W Ma T, x L B, L B , W1, M1a12T12, , , x x , 3., x, L, B, B , L, 32. n1[ M1L1T12 ] n2 [M 2 L2T22 ], 20 0.01 0.02 , 33. Power P aV b 34. Energy = Force x length, , , , , 10 20, 10 , 35. n1u1 n2u2, 2, x x 2 0.005 m, 36. 6.67 10 8 dyne cm 2 gm , 4., , 5., , 6., 7., 8., , 9., 10., 11., 12., , 4, v, r, V r3 , 100 3 100, 3, v, r, r, v 3 v, r, A lb , , A l b, l b , , , A A , A, l, b, b , l, , A bl l b 10cm 2, l, 0.01, 100 , 100 0.2%, l, 4.28, X1, N2, 10, 1, X, X N 100, N, 2, 1, L1 2 L2 2.02 2 1.02 4.06, , 2, , 6.67 108 105 N 102 m 103 kg , 37., 38., 39., 40., 41., , L1 2L2 0.01 2 0.01 0.03, 1., S, V S T, V , , , T, V, S, T, If last digit is 5, if the preceding digit is odd then it, should be increased by adding 1 and last digit 5, 2., has to be ignored., Use limitation of significant figures, V l 3 and rounded off to minimum significant, , NARAYANA GROUP, , 2, , v 2 u 2 2gx and change into S.I, Substitute D.F. of quantities, Substitute D.F. of quantities, 0 0 1, 2 x, 2 2 y, 1 z, T F xW yV z ; M L T [MLT ] [ ML T ] [ LT ], 42. M Fa Lb Tc, L F a M bT c, , LEVEL - I (H.W), ACCURACY, PRECISION, TYPES OF, ERRORS AND COMBINATION OF, ERRORS, The Accuracy of a clock is one part in 1010 ., The maximum difference between two such, clocks operating for 1010 seconds is ______, 1) 1 s, 2) 5 s, 3)10 s, 4) 1010 s, The length of a rod is measured as 35.3 cm, then the graduations on the scale are up to, 1) 1 cm, 2) 1 mm 3)0.01 mm 4)0.1 mm, 25
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 3., , 4., , 13., If L 2.06cm 0.02cm,, B 1.11cm 0.03cm, then L+B equals to, 1) 3.17cm 0.05cm, 2) 2.06cm 0.05cm,, 3) 3.17cm 0.02cm, 4) 3.17cm 0.03cm,, 14., The radius of sphere is measured as, , 5.2 0.2 , , The radius of disc is 1.2 cm, its area, according to idea of significant figures is ___, 1) 4.5216cm 2, 2) 4.521cm 2, 3) 4.52cm 2, 4) 4.5cm 2, When Energy is expressed in erg the no of, significant figure is four. If it is expressed in, joule the no of significant figures will become, 1) 9, 2) 5, 3) 1, 4) 4, 58.97 is, 1) 7.679 2) 7.68, 3)7.6, 4)7.7, A stick has a length of 12.132 cm and another stick has a length of 12.4 cm then the, total length of the stick is ___, 1)24.53 cm 2)24.5 cm 3)2.45 cm 4)2.453 cm, The respective number of significant figures for, the number 23.023, 0.0003 and 21 x 10-3 are, 1)5,1,2, 2)5,1,5, 3)5,5,2, 4)4,4,2, The Number of significant figures in, 5.69 1015 kg is, 1) 1, 2) 2, 3) 3, 4) 4, The value of 124.2 + 52.487 with due regard, to significant places is ___, 1) 176.69 2) 176.7 3)176, 4)177, , 8., , cm then the percentage error in, volume of the ball is _, 15., 1) 11%, 2) 4%, 3) 7%, 4) 9%, If the length and breadth of a plate are, 16., 5.0 0.2 cm and 4.0 0.1 cm then the, absolute error in measurement of area is _, 1) 10cm 2 2) 11cm 2 3) 12cm 2 4) 1.3cm2, 17., If the length of a cylinder is measured to be, 8.28 cm with an error of 0.01 cm then the, percentage error in measured length is nearly 18., 1) 0.4 % 2)0.2 % 3) 0.1 % 4) 0.5%, A student performs experiment with simple, pendulum and measures time for 10, vibrations. If he measures the time for 100 19., vibrations, the error in measurement of time, period will be reduced by a factor of _, 1) 10, 2) 90, 3) 100, 4)1000, 20., If L1 (3.03 0.02)m and L2 (2.01 0.02)m, , 9., , then L1 2 L2 is (in m), 21., 2) 6.05 0.06, 1) 7.05 0.06, 3) 6.05 0.02, 4) 7.05 0.02, A body travels uniformly a distance of, UNITS AND DIMENSIONAL FORMULAE, , 5., , 6., , 7., , 13.8 0.2 m in a time 4.0 0.3 s then the, velocity of the body is ___, 1) 3.45 0.2 ms 1, 10., , 11., 12., , 26, , 2) 3.45 0.3 ms1, , 22., , 9.27, , The value of, with due regard to signifi41, cant figures is ___, 1)0.226, 2)0.23, 3) 0.2, 4)0.2261, When 57.986 is rounded off to 4 significant, figures, then it becomes ___, 1) 58, 2) 57.00 3) 57.90 4) 57.99, If ‘L’ is length of simple pendulum and ‘g’ is, acceleration due to gravity then the dimen1, , l 2, sional formula for is same as that for, g, , 3) 3.45 0.4 ms 1, 4) 3.45 0.5 ms 1, 1)Frequency 2)Velocity3)Time period 4)wavelength, The pressure on a square plate is measured by 23. The dimensional formula for the product of, measuring the force on the plate and the length, two physical quantities P and Q is L2T 2 , of the sides of the plate. If the maximum error, in measurement of force and length are respecthe dimensional formula of P/Q is T 2 the P, tively 4% and 2% then the maximum error in, and Q respectively are ___, Measurement of pressure is _____, 1) distance and velocity, 1) 1%, 2) 2%, 3) 6%, 4) 8%, 2) distance and acceleration, SIGNIFICANT FIGURES &, 3) displacement and velocity, 4) displacement and force, ROUNDING OFF, 24., The fundamental physical quantities that have, 2.34 is obtained by rounding off the number, same dimensions in the dimensional formula, 1) 2.346 2) 2.355 3) 2.335 4) 2.334, of force and Energy are ___The number of significant figures in 0.0006032, 1) mass, time, 2) time, length, is, 3), mass,, length, 4) time, mole, 1) 7, 2) 4, 3) 5, 4) 2, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 25., , 26., , If is rigidity modulus, r is the radius, l is 34., the length and C is the moment of the couple, 2lc, then, has the dimensions of _, r 4, 1) Angle 2) Mass 3) Length 4) Frequency, 35., PRINCIPLE OF HOMOGENEITY, The acceleration of an object varies with time, as a AT 2 BT C taking the unit of time, as 1 sec and acceleration as ms 2 then the, units of A,B,C respectively are __, 2) ms 2 , ms 1 , ms, 1) ms 3 , ms 2 , ms 1, 36., 4) ms 4 , ms 3 , ms 2, 3) ms 1 , ms 2 , ms 3, , A, log( Bx C ) is dimensionally true,, B, then (here is the coefficient of viscosity and, x is the distance), 1) C is dimensionless constant, 37., 2) B has dimensions of -1 in length, 3) The dimensional formula of A is ML2T 1, 4) All are true, 28. If the velocity (v) of a body in time ‘t’ is, given by V AT 3 BT 2 CT D then the, dimensions of C are ____, 38., 1, 2, 3, 4, 1) LT 2) LT 3) LT 4) LT , , 27., , 29., , 30., , 31., , 32., , 33., , If , , pr 4, where the letters, 8 l, have there usual meanings the dimensions of, 39., V are ___, 0, 3, 0, 0, 3, , 1, 1) M L T, 2) M L T, 0, , 3, , 1, 3) M L T, 4) M 1L3T 0, If the acceleration due to gravity is 10 ms 2, and the units of length and time are changed, to kilometre and hour respectively the, numerical value of acceleration is _____, 1) 36000 2) 72000 3) 36000 4) 129600, The magnitude of Energy is 100J. What will, be its value if the units of mass and time are, doubled and that of length is halved?, 1) 100 J 2) 200 J 3) 400 J 4) 800 J, If the units of mass and velocity are increased, by two times then the unit of momentum will, be increased by __, 1) 400% 2) 200% 3) 300% 4) 100%, SI unit and CGS unit of a quantity vary by, 107 times, it is ___, 1., 1) Boltzmann’s constant2) Gravitational constant, 3) Planck’s constant 4) Angular momentum., 3., In the relation V , , NARAYANA GROUP, , UNITS AND MEASUREMENTS, The initial velocity of a particle is given by, u 2 v 2 2 gx where x is the distance, covered. If u = 18 km h 1 , g = 1000 cm / s 2 x, = 150 cm then v = ____ m/s, 1) 45 2) 55, 3) 35, 4) 65, The equation which is dimensionally correct, among the following is, 1, 2) v ut at, 1) v u at, 2, 3) s ut at 3, 4) t s av, p, The dimensions of in the relation v , , (where v is velocity, p is pressure , is, density), 1) Dimensionless, 2) LT 1 , 3) ML1T 2 , 4) ML3 , Taking frequency f, velocity (v) and Density, ( ) to be the fundamental quantities then the, Dimensional formula for momentum will be, 1) v4 f 3 , 2) v3 f 1 , 3) vf 2 , 4) 2 v2 f 2 , If momentum (p), Mass (M), Time (T) are, chosen as fundamental quantities then the, dimensional formula for length is ___, 1) P1T 1 M 1 , 2) P1T 1M 2 , 2 2, 1, 3) P1T 1M 1 , 4) P T M , If pressure (P), velocity (V) and time (T) are, taken as the fundamental quantities, then the, dimensional formula of force is ___, 1) P1V 1T 1 , 2) P1V 2T 1 , , 3) P1V 1T 2 , , 4) P1V 2T 2 , , LEVEL-I (H.W) - KEY, 01)1, 07)1, 13)4, 19)2, 25)1, 31)4, 37)1, , 02) 4, 08) 1, 14)4, 20)2, 26)4, 32)3, 38)3, , 03) 1, 09) 2, 15) 1, 21)4, 27)4, 33)3, 39)4, , 04) 1, 10)4, 16) 2, 22)3, 28)2, 34)2, , 05)4, 11)3, 17)1, 23)2, 29)2, 35)1, , 06)3, 12)2, 18)3, 24)1, 30)4, 36)1, , LEVEL-I (H.W) - HINTS, d, 2) 0.01cm is the L.C of vernier caliperse., d, Let x=L+B=3.17 ; x L B 0.05, 27
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 2., , x x 3.17 0.05, 4., , 4, V, r, V r3 &, 100 3 100, 3, V, r, , 5., , A lb , , A l b, l b , , , A A , A, l, b, b , l, X1 N2, l, , 100, 7., X, N1, 2, l, , 6., , 3., , x L1 2 L2 7.05 ; x L1 2L2, , 8., , V, , 9., , S, V S T, S T , , , , ; V V S T , T, V, S, T, , , , 4., , F F P, F 2L , = 2 ; P 100 = F L 100 , , , A L, 11. If last digit is 5 and if the preceding digit is odd, then it should be increased by adding 1 and last, 5., digit 5 has to be ignored., 2, 12. Use limitation of significant figures 13) A r, From 14 to 21 follow the rules of significant, figures and rounding off numbers, , 10., , P, , 1/2, , l , 22. Hence g , , , 1/2, , M 0 L1T 0 , 0 1 2 , M LT , , M 0 L0T 1, , P, , SIGNIFICANT FIGURES &, ROUNDING OFF, , 2, , T -----(2), 23. PQ L2T 2 ----(1) ;, Q, 24. Use dimensional analysis, 25. Using dimensional formula, 26. Principle of homogenity, 27. Using dimensional formula, 28 & 29. Use principle of homogenity, ML2, , 30., , a LT 2 , , , , 33., , N1 U1 N 2 U 2, , 35., , using dimensional analysis 36., , 37., 38., , 31. E , , T2, , 6., , 7., , 32. P MV, , 34. v 2 u 2 2 gx, , 1 a, , V, 1 b, , p, , , 8., , 3 c, , P f a v b c ; MLT 1 k T LT ML , L P, , a, , b, , M T , , c, , The least count of a stop watch is (1/5) s. The, time of 20 oscillations of a pendulum is, measured to be 25 s. The maximum, percentage error in this measurement is, 1) 8 % 2) 1 %, 3) 0.8 % 4) 16 %, The diameter of a wire as measured by a, screw gauge was found to be 1.002 cm, 1.004, cm and 1.006 cm. The absolute error in the, third reading is, 1) 0.002 cm, 2) 0.004 cm, 3) 1.002 cm, 4) zero, Force and area are measured as 20 N and, 5m2 with errors 0.05 N and 0.0125m2. The, maximum error in pressure is (SI unit), 2) 4 0.05, 1) 4 0.0625, 3) 4 0.125, 4) 4 0.02, The length and breadth of a rectangular, object are 25.2cm and 16.8cm respectively, and have been measured to an accuracy of, 0.1cm. Relative error and percentage error, in the area of the object are, 1) 0.01 & 1%, 2) 0.02 & 2%, 3) 0.03 & 3%, 4) 0.04 & 4%, , The velocity of light in vacuum is 30 crore m/, s. This is expressed in standard form up to 3, significant figures as, 1) 0.003 x 1011 m/s 2)300 x 106 m/s, 3) 3.00 x 108 m/s, 4) 0.030 x 1010 m/s, The length, breadth and thickness of a, rectangular lamina are 1.024 m, 0.56 m, and, 0.0031 m. The volume is …………..m3, 1) 1.8 x 10-3, 2) 1.80 x 10-3, -4, 3) 0.180 x 10, 4) 0.00177, The initial and final temperatures of a liquid, are measured to be, , 76.3 0.3, , 39. F P a V b T c, , 0, , 28, , c and, , c then rise in temperature with, 0, , 1., , 0, , error limit is, 1) 8.6 0.2 C, , LEVEL - II (C.W), ACCURACY, PRECISION,TYPES OFERRORS, AND COMBINATION OFERRORS, , 67.7 0.2 , 0, , 2) 8.6 0.3 C, , 0, , 9., , The error in the measurement of the length, of the simple pendulum is 0.2 % and the error, in time period 4%. The maximum possible, 10., L, error in measurement of 2 is, T, 1) 4.2% 2) 3.8% 3) 7.8% 4) 8.2%, , 0, 3) 8.6 0.5 C, 4) 8.6 0.6 C, Less accurate of the four options given below, 1) 9.27 2)41 3) 1.01 4) 9.00 100, , UNITS AND DIMENSIONAL, FORMULAE, If the ratio of fundamental units in two, systems is 1 :3, then the ratio of momenta in, the two systems is, 1) 1:3, 2) 1:9, 3) 1:27, 4) 3:1, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 11., , UNITS AND MEASUREMENTS, , The velocity of the waves on the surface of 19., water is proportional to g where, =wave length, = density and g =, acceleration due to gravity. Which of the, following relation is correct?, 1) , 2) , 3) , 4) , 20., , PRINCIPLE OF HOMOGENITY, 12., , The work done ‘w’ by a body varies with, displacement 'x' as w Ax , , B, , C x 2, , . The, , 21., , dimensional formula for 'B' is, 1. [ ML2T 2 ] 2. [ ML4T 2 ] 3. [ MLT 2 ] 4. [ ML2T 4 ], , CONVERSION OF UNITS, , 13., , 14., , If the units of mass, time and length are 100 22., g, 20 cm and 1 minute respectively the, equivalent energy for 1000 erg in the new, system will be, 1. 90, 2. 900, 3. 2 x 106, 4. 300, The ratio of SI unit to the CGS unit of planck's, constant is, 23., 1. 107:1 2. 104 :1 3. 106 :1, 4. 1 :1, , TO CHECK THE CORRECTNESS OF, PHYSICAL RELATION & DERIVING, THE EQUATIONS, 15., , The velocity of a body is expressed as, V = G a M b R c where G is gravitational 24., constant. M is mass, R is radius. The values, of exponents a, b and c are :, 1 1, 2 2, , 1) , , , 16., , 17., , 1, 2, , 1 1 1, 2 2 2, , 2) 1, 1, 1 3) , ,, , 4)1,1,, , 1, 2, , 25., , The velocity of a spherical ball through a, viscous liquid is given by v= v0(1-ekt), where, v0 is the initial velocity and t represents time., If k depends on radius of ball (r), coefficient, of viscosity ( ) and mass of the ball (m), then, 1) k = mr/, 2) k = m/r, 1), 3) k = r /m, 4) k = mr, Dimensional analysis of the equation 2), x, , 3, , -3, , Velocity = Pressure difference 2 . density 2, 18., , 3), , gives the value of x as:, (1986 E), 4), 1) 1, 2) 2, 3) 3, 4)-3, For the equation F =Aavbdc where F is force, 26., A is area, v is velocity and d is density, with, the dimensional analysis gives the following, values for the exponents., (1985 E), 1) a=1, b = 2, c =1 2) a =2, b =1, c= 1, 3)a =1, b =1, c= 2, 4) a = 0, b =1 , c = 1, , NARAYANA GROUP, , The length of pendulum is measured as 1.01m, and time for 30 oscillations is measured as, one minute 3 seconds. Error in length is 0.01, m and error in time is 3 secs. The percentage, error in the measurement of acceleration due, to gravity is., (Eng - 2012), 1) 1, 2) 5, 3) 10, 4) 15, 1, , 2, The dimensional formula of 0 H ( 0 -per-2, meability of free space and H-magnetic field, intensity) is: (Eng - 2011), 1) MLT 1 2) ML2T 2 3) ML1T 2 4) ML2T 1, If the force is given by F at bt 2 with t as, time.The dimensions of a and b are (Eng-10), 1) MLT 4 , MLT 2, 2) MLT 3 , MLT 4, 3) ML2T 3 , ML2T 2, 4) ML2T 3 , ML3T 4, When a wave traverses a medium, the displacement of a particle located at ‘x’ at a time ‘t’ is, , given by y a sin bt cx , where a,b and c are, constants of the wave, which of the following is, a quantity with dimensions? (Eng - 2009), 1) y/a, 2) bt, 3) cx, 4) b/c, The Energy (E), angular momentum (L) and, universal gravitational constant (G) are, chosen as fundamental quantities. The, dimensions of universal gravitational constant, in the dimensional formula of Planks constant, (h) is (Eng - 2008), 1) 0, 2) -1, 3) 5/3, 4) 1, If the absolute errors in two physical quantities, A and B are a and b respectively, then the, absolute error in the value of A-B is(Med- 2014), 1) a-b, 2) b-a, 3) a b 4) a+b, If the velocity v (in cm/s) of a particle is, given in terms of time t (in sec) by the, b, equation v at , , then the dimensions, tc, of a, b and c are (Med- 2011), a, b, c, 2, L , LT 2 , T , LT 2 , LT 2 , , LT , L, LT , , L, T , , T 2 , A body weighs 22.42 g and has a measured, volume of 4.7 cc the possible errors in the, measurement of mass and volume are0.01g, and 0.1 cc. Then the maximum percentage, error in the density will be(Med- 2010), 1) 22% 2) 2.2% 3) 0.22% 4) 0.022%, , L, , 29
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 27., , 28., , 1., 2., 3., 4., , If energy E, velocity v and time T are taken, as fundamental quantities, the dimensional, formula for surface tension is (Med-2009), 1) Ev 2T 2 , 2) E 2 vT 2 , , b, , 16., , 1, a, 1 1, c, k r a b m c ; T L ML T M, , 17., , Substitute dimension formulae, , 18., , F = Aa vb d c ; MLT 2 L2 LT 1 ML3 , comparing the powers on both sides, , a, , b, , c, , 3) Ev 2T 1 , 4) E 2 v 2T 1 , If power (p), surface tension (T) and Planck’s, l, T, l g, constant (h) are arranged, so that the dimen100 100 2, 100, T 2, 19., ;, sions of time in their dimensional formulae, l, T, g g, are in ascending order, then which of the fol20. Substitute dimensional formula of µ 0 and H, lowing is correct? (Med- 2008), 1) P. T, h 2) P, h, T 3) T, P, h 4) T, h, P, 21. MLT 2 at ; MLT 2 bt 2, 22. by dimensional formulae, LEVEL-II (C.W) - KEY, 23. h E , L, G, 1) 4 2) 3 3) 1 4) 4, 5) 1 6) 3, a, b, c, ML2T 1 ML2T 2 ML2T 1 M 1 L3T 2 , 7) 1 8) 3 9) 4 10) 1 11)3 12) 2, 13) 1 14) 1 15) 1 16) 3 17) 3 18) 1, Comparing the powers on both sides we get a,b,c, 24., If Z A B ; Z A B (Max possible error), 19) 3 20) 3 21) 2 22) 4 23) 1 24) 4, 25) 3 26) 2 27) 1 28) 1, Z a b, 25. Use principal of homogenity, LEVEL-II (C.W) - HINTS, M, 26. The density of d , ; % Error of density, L, L, T, V, x, 2, Let x 2 ; x =, T, L, T, Δd, ΔM, ΔV, ×100=, ×100+, ×100, 1/ 5, 25, T, d, M, V, T , and T , 100, ; % error , 20, 20, T, a, b, c, 27. S E v T , x3 x3 xmean, a, b, c, MT 2 ML2T 2 LT 1 T , P F A, F, F A , Comparing the powers on both sides we get a,b,c, P P , , P , , , , A, , P, , F, , A, , F, , A , , A l b, 28. Use dimensional analysis, A l b ; A l b, LEVEL - II (H.W), A, l b , 100 , 100, A, b , l, 8., t t2 t1, ACCURACY, PRECISION, TYPES OF, 9., Less no. of significant figures represent less, ERRORS AND COMBINATION OF, accuracy., ERRORS, M 1 L1 T1 1 P1 M1 L1 T1 1, 10. M L T 3 ; P M L T , 1., The error in the measurement of length of a, 2 2 2 2 , 2, 2, 2, simple pendulum is 0.1 % and error in the, 11. v g ; LT 1 L M L3 L T 2 ., time period is 2% . The possible maximum, Comparing the powers on both sides, we get, error in the quantity having dimensional, , and , formula LT 2 is, B, 1) 1.1 % 2) 2.1 % 3) 4.1 % 4) 6.1 %, 12. w = Ax C x 2 ( principle of homogenity), 2., The length of a cylinder is measured as 5cm, using a vernier calipers of least count 0.1mm., 13. n1[ M 1 L12T12 ] n2 [ M 2 L2 2T2 2 ], The percentage error in the measured length, a, , 1, , 1, 3, , 2, b, c, is nearly, 14. h ML2T 1 15. [ LT ] M L T M L, 1) 0.5 % 2) 2 %, 3) 20 % 4) 0.2 %, , 5., , 30, , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 3., , 4., , 5., , 6., , 7., , 8., , 9., , The diameter of a wire as measured by a, screw gauge was found to be 1.002 cm,, 1.000cm, 1.006cm, the absolute error in the, first reading., 1) 0.001cm 2)0.004 cm 3)0.006m 4)0.003cm, The number of particles crossing per unit area, perpendicular to x-axis in unit time is, n n , N D 2 1 Where n and n are, 1, 2, x2 x1 , number of particles per unit volume for the, value of x1 and x2 respectively.The, dimension of diffusion constant D is, 1 3 4), 1) M 0 L1T 2 2) M 0 L2T 4 3) M 0 LT, M 0 L2T 1, The external and internal diameters of a hollow, cylinder are determined with vernier calipers, and the results are recorded as (4.23 0.001)cm, and (3.89 0.01)cm. The thickness of the cylinder, wall within the limits of error is, 1) 0.34 0.01 cm, 2) 0.34 0.02 cm, 3) 0.34 0.04 cm, 4) 0.17 0.01 cm, The density of a cube can be measured by, measuring its mass and the length of its side., If the maximum errors in the measurement, of mass and length are 3% and 2%, respectively, the maximum error in the, measurement of the density of the cube is, 1) 9% 2) 19% 3) 10%, 4) 90%, the diameter of a sphere is 3.34m Calculate, its volume with due regard to significant, figures ( in m3 ) ., 1)19.5169, 2)9.516 3)19.5 4) 19.51, The length, breadth and thickness of a metal, sheet are 4.234 m, 1.005m, and 2.01 cm, respectively then the volume of the sheet is, 1) 0.08 m3, 2) 0.0855 m3, 3)0.085 m3, 4) 0.087 m3, The sides of rectangle are 10.5 0.2 cm, , UNITS AND MEASUREMENTS, 12., , A, , 13., , 14., 15., , 16., , 17., , 18., , and 5.2 0.1 cm then its perimeter with, error limit., 19., 1) 31.4 0.6 cm, 2) 31.4 0.2 cm, 10., , 11., , 3) 31.4 0.1 cm, 4) 31.4 0.9 cm, If the ratio of fundamental units in two, systems are 2:3 the ratio of force in these, two systems is, 20., 1) 1:3, 2) 1:1, 3) 3:1 4) 1:27, If L, R, C, and V, respectively, represent, inductance, resistance, capacitance and, potential difference, then the dimensions of, L/RCV are the same as those of, 1) Charge 2)1/Charge 3)Current 4)1/Current, , NARAYANA GROUP, , Hydrostatic pressure ‘P’ varies with, displacement 'x' as P log Bx 2 C where, B, A, B and C are constants. The dimensional, formula for 'A' is, 2) [ MLT 2 ], 1) [ M 1 L1T 2 ], 3) [ ML2T 2 ], 4) [ ML3T 2 ], The units of force, velocity and energy are, 100 dyne, 10 cm s-1 and 500 erg respectively., The units of mass, length and time are, 1) 5 g, 5 cm, 5 s, 2) 5 g, 5 cm, 0.5 s, 3) 0.5 g, 5 cm, 5 s, 4) 5 g, 0.5cm, 5 s, The ratio of SI unit to CGS unit of, gravitational constant is, 1) 1:103 2) 103 :1 3) 1:1, 4) 1:107, The frequency f of vibrations of a mass m, suspended from a spring of spring constant k, is given by f Cm x K y , where C is a, dimensionless constant. The values of x and, y are, respectively., 1 1, 1 1, 1 1, 1 1, 1) ,, 2) , , 3) , 4) ,, 2 2, 2 2, 2 2, 2 2, If the time period 'T' of a drop under surface, tension 's' is given by T = d a r b s c where, d is the density, r is the radius of the drop., If a =1, c = -1 then the value of b is (1993 E), 1) 1, 2) 2, 3) 3, 4) -1, If the velocity (V), acceleration (A), and force, (F) are taken as fundamental quantities instead, of mass (M), length (L), and time (T), the, dimensions of Young’s modulus (Y) would be., 1) FA2V 4 2) FA2V 5 3) FA2V 3 4) FA2V 2, The time dependence of a physical quantity, , 2, , P is given by P P e t , where is a, 0, constant and t is time. Then constant , 1)is dimensionless 2)has dimensions of T 2, 3)has dimensions of P 4)has dimensions of T 2, The value of x in the formula Y , , 2mgl x, 5bt 3e, , where m is the mass, 'g' is acceleration due, to gravity, l is the length, 'b' is the breadth,, ‘t’ is the thickness and e is the extension, and Y is Young's Modulus, is, 1) 3, 2) 2, 3) 1, 4) 4, The velocity of sound in air (V) pressure (P), and density of air (d) are related as V p x d y ., The values of x and y respectively are, 1) 1,, , 1, 2, , 1, 2, , 1, 2, , 2) , 3), , 1 1, ,, 2 2, , 4), , 1, 1, , , 2, 2, , 31
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 2., , LEVEL-II (H.W) - KEY, 1) 3, 7) 3, 13)2, 19) 1, , 2) 4 3) 1 4) 4, 8) 2 9) 1 10) 2, 14) 1 15) 4 16) 3, 20) 4, , 5) 4 6) 1, 11)4 12)4, 17)1 18)2, 3., , LEVEL-II (H.W) - HINTS, 1., 3., 5., , l, L, T, 100, 2, 2., L, T, l, a a a3, amean 1 2, ; a3 amean a3, 3, , t, , d 2 d1, 2, , ; t t2 t1, d, , M, , 3L, , 7., , , , ; d 100 M L 100, , , 3, 4, V 3R, 8. V lbw, , 9., , p 2 l b p 2 l b , , 10., , M 1 L1 T1 2, , , M 2 L2 T2 3, , 11., , L, L, , RCV, di , t L. , dt , , 6., , d, , M M, 3, V, L, , and, , 4., , F1 M 1 L1T12, , F2 M 2 L2T2 2, , 13., , A, Bx C Constant ; BL2 M 0 L0T 0 ; P B, F MLT 2 100 dyne ;, V LT 1 10 cm / sec ; E ML2T 2, , 14., , n, , 16., , T, , 12., , 17., 18., 19., 20., , 5., , 2, , 1, n1u1 n2 u2, u, 3 a, , ML , , .Lb MT 2 , , 15., , f, , 1, 2, , k, m, , c, , 1 2, 1 a, 2 b, 2 c, Y V a Ab F c ; ML T LT LT MLT , t 2 M 0 L0T 0, Dimensional formula of Y ML1T 2, Dimension of L,b,t,e=L, p, V, ; LT1 V, d, , 6., , LEVEL - III, ACCURACY, PRECISION, TYPES OF, ERRORS AND COMBINATION OF, ERRORS, 7., , 1., , The measured mass and volume of a body, are 53.63 g and 5.8 cm3 respectively, with, possible errors of 0.01 g and 0.1 cm3. The, maximum percentage error in density is about, 1) 0.2%, 2) 2% 3) 5% 4) 10%, , A vernier calipers has 1 mm marks on the, main scale . It has 20 equal divisions on the, vernier scale,which match with 16main scale, divisions. For this vernier calipers the least, count is, 1) 0.02mm 2) 0.05 mm 3) 0.1mm 4) 0.2mm, The resistance of metal is given by V=IR., The voltage in the resistance is V 8 0.5 , V and current in the resistance is, I 2 0.2 A, the value of resistance with, its percentage error is, 1) 4 16.25% , 2) 4 2.5% , 3) 4 0.04% , 4) 4 1% , In an experiment, the values of refractive, indices of glass were found to be 1.54, 1.53,, 1.44, 1.54, 1.56 and 1.45 in successive, measurements i) mean value of refractive, index of glass ii) mean absolute error, iii) relative error and iv) percentage error, are respectively,, 1)1.51,0.04,0.03,3% 2)1.51,0.4,0.03,3 %, 3)15.1,0.04,0.03,3% 4)15.1,0.04,0.3,3 %, A student performs an experiment for, 4 2 L , g, determination of T 2 ,L 1m, and he, , , commits an error of L for T he tajes the, time of n oscillations with the stop watch of, least count T .For which of the following, data the measurement of g will be most, accurate?, 1) L 0.5, T 0.1, n 20, 2) L 0.5, T 0.1, n 50, 3) L 0.5, T 0.01, n 20, 4) L 0.5, T 0.05, n 50, A rectangular metal slab of mass 33.333 has, its length 8.0 cm, breadth 5.0 cm and thickness, 1mm. The mass is measured with accuracy, up to 1 mg with a sensitive balance. The, length and breadth are measured with vernier, calipers having a least count of 0.01 cm. The, thickness is measured with a screw gauge of, least count 0.01 mm. The percentage, accuracy in density calculated from the above, measurements is, 1) 13 % 2)130 % 3)1.6 % 4)16 %, The initial and final temperatures are, recorded as 40.6 0.30 C and 50.7 0.2 0 C ., The rise in temperature is, 0, , 2) 10.1 0.3 C, , 1) 10.10 C, 0, , 3) 10.1 0.5 C, 32, , 0, , 4) 10.1 0.1 C, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 8., , UNITS AND MEASUREMENTS, pendulum is 112 s with an accuracy of 0.01s., The percentage accuracy in g is, 1) 1, 2) 2.8, 3) 1.3, 4) 2.1, , In the measurement of a physical quantity, 2, , X, , 9., , 10., , AB, . The percentage errors introduced, C1/ 3 D 3, , in the measurements of the quantities A,B,C and, D are 2%, 2%, 4% and 5% respectively. Then 16., the minimum amount of percentage of error in, the measurement of X is contributed by, 1) A, 2) B, 3) C, 4) D, There are atomic (Cesium) clocks capable of 17., measuring time with an accuracy of 1 part in, 1011 . If two such clocks are operated to, precision, then after running for 5000 years,, these will record a difference of, 1) 1 day 2) 1 s 3) 1011 s, 4) 1year, If the length of a simple pendulum is, 18., recorded as 90.0 0.02 cm and period as, , 1.9 0.02 s, the percentage of error in the, 11., , 12., , 13., , 14., , 5.0 10 4.5 10 , 8, , 6, , 1) 4.55 10 6, 2) 4.5 10 6, 3) 4.6 10 6, 4) 4 106, The dimensions of a wooden block are, The number of, 1.1m 2.36m 3.1m ., significant figures in its volume should be, 1) 1, 2) 2, 3) 3 4) 4, , measurement of acceleration due to gravity is, PRINCIPLE OF HOMOGENITY, 1) 4.2, 2) 2.1 3) 1.5, 4) 2.8, In the determination of the Young’s modulus, z / K, of a given wire, the force, length, radius and 19. In the relation P e, ; P is, , extension in the wire are measured as, pressure, K is Boltzmann’s constant, Z is, 100 0.01 N , 1.25 0.002 m,, distance and is temperature. The, dimensional formula of will be, 0.001 0.00002 m, and 0.01 0.00002 m,, 0 2, 0, respectively. The percentage error in the, 2) M 1 L2 T 1 , 1) M L T , measurement of Young’s modulus is, 1) 4.37, 2) 2.37 3) 0.77 4) 2.77, 3) ML0 T 1 , 4) M 0 L2 T 1 , The radius ( r ) , length ( / ) and resistance, 20. The Richardson equation is given by, (x) of a thin wire are, I AT 2 e B / kT . The dimensional formula for, 0.2 0.02 cm, 80 0.1 cm, and 30 1 , AB 2 is same as that for A and B are constants, respectively . The percentage error in the, 1) IT 2 2) k T, 3) Ik 2 4) Ik 2 / T, specific resistance is, 21. The heat generated in a circuit is given by, 1) 23.4% 2) 25.4% 3) 26% 4) 27.5 %, Q = i2 Rt joule , where ‘i’ is current, R is, When a current of 2.5 0.5 ampere flows, resistance and t is time. If the percentage, through a wire, it develops a potential, errors in measuring i, R and t are 2%, 1%, and 1% respectively, the maximum error in, difference of 20 1 volt, the resistance of, measuring heat will be, the wire is, 1) 2 % 2) 3 %, 3) 4 % 4) 6 %, 1) 8 2 , 2) 10 3 , , LEVEL - III -KEY, , 3) 18 4 , 4) 20 6 , Two objects A and B are of lengths 5 cm and, 7 cm determined with errors 0.1 cm and 0.2, cm respectively. The error in determining (a), the total length and (b) the difference in their, lengths are, 1) 12 0.3 , 2 0.3 2) 7 0.3 , 2 0.3, , 1)2, 7)3, 13)1, 19)1, , 3) 12 0.3 , 12 0.3 4) 12 0.3 , 2 0.6 , In a simple pendulum experiment, length is 2., measured as 31.4 cm with an accuracy of, 1mm. The time for 100 oscillations of, , NARAYANA GROUP, , 2)4, 8)3, 14)1, 20)3, , 3)1, 9)2, 15)4, 21)4, , 4)1, 10)2, 16)4, , 5)4 6)3, 11)1 12)1, 17)3 18)2, , LEVEL - III - HINTS, 1., , 15., , SIGNIFICANT FIGURES, Three pieces of silver have masses 2.3 kg,, 41.15 g and 30.19 g. The total mass of correct, significant figures is ( in kg), 1)2.37032 2)2.370, 3)2.37 4) 2.4, The sum of the given two numbers with, regard to significant figures is, , M , M V , ; 100 M V 100, , , V, 16 M.S.D = 20 V.S.D 1V .S .D 4 / 5 M .S .D, L.C = 1M.S.D - 1 V.S.D, , Density , , 33
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 3., , 4., , V, R V I , ; 100 R V I 100, , , I, R, , , Resistance = R R 100 , , , R, , mean, , ; , , 6, , mean, , , , , mean, , i , , 6, , LEVEL - IV, Matching Questions, 1., , ;, , , , mean, 100, relative % error in = , mean, , 5., , g l, T, 2, ( l and T are least, and the, g, l, T, , number of readings are maximum), 6., , Percentage error gives percentage accuracyd , relative error,, , 2., , m, lbh, , d m l b h, , , , d, m, l, b, h, , d , 100, d , , and calculate , 7., , t2 t1 50.7 40.6 0.3 0.2 , , 8., , X, A, B, 100 2, 100 , 100, X, A, B, , , , 09., , 10., 11., 12., , 13., 14., , 3., , 1 C, 3D, 100 , 100, 3 C, D, , 1, years rounded off to minimum 4., 1011, significant figures, l, 2T, l g, A), 100, g 4 2 2 ; 100 100 , B), g, l, T, T, C), FL, FL Y, F L 2r e , 100 , , , 100, Y, 2 ;, D), Y, F, L, r, e, Ae r e, , , t 5000 , , r2x, Specific Resistance , L, 2r L x , , 100, Total % error is , L, x , r, V V I , R R , R, I V, I , x a b and x a b, , 5., , x a b and x a b, 6., l, 2 T, l g, , 100, , , 100, , , 100, ;, l, T, T2 g, From 16 to 18 follow the rules of significant figures, and rounding off numbers, z , 19. 1 ; 20)Here A IT 2 and B KT, k , , 2, 15. g 4, , 21., , 34, , Q, , 2 i, , R, , t, , Q i 2 Rt ; Q 100 i 100 R 100 t 100, , Column-I, a) Backlash error, b) Zero error, , Column-II, p) Always subtracted, q)Least count, =1M.S.D-1V.S.D, c) Vernier callipers, r) May be -ve or +ve, d) Error in screw gauge s) Due to loose fittings, There are four vernier scales, whose specification are given in column-I and the least, count is given in column-II ( S=value of main, scale division,n=number of marks on vernier), Column-I, Column-II, a) S=1 mm ,n=10, p) 0.05 mm, b) S=0.5mm,n=10, q) 0.01 mm, c) S=0.5 mm,n=20 r) 0.1 mm, d) S=1 mm , n=100 s) 0.025 mm, Using signification figures, match the following, Column-I, Column-II, a) 0.12345, p) 5, b) 0.1210 cm, q) 4, c) 47.23/2.3, r) 3, d) 3 108, s) 2, t) 1, Match List I with List II and select the correct, answer using the codes given below the Lists., List - I, List - II, Distance between earth and stars I) Micron, Inter atomic distance in a solid II) angstrom, Size of the nucleus, III) Light year, Wave length of infrared laser, IV) fermi, V) kilometre, Some physical constants are given in, List - I and their dimensional formulae are, given in List- 2.Match the following (2007 E), List - I, List - II, 1 2, a) Planck’s constant, e) ML T , b) Gravitational constant, c) Bulk modulus, , f) ML1T 1 , g) ML2T 1 , , d) Coefficient of Viscosity, h) M 1 L3T 2 , Names of units of some physical quantities, are given in List - I and their dimensional, formulae are given in List - II. Match the, correct pair of the lists., (2005 E), List - I, List - II, a) Pa s, e) L2T 2 K 1 , b) NmK-1, f) [ MLT 3 K 1 ], 1, 1, c) J kg K, g) [ ML1T 1 ], d) Wm1K 1, h) ML2T 2 K 1 , NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, , UNITS AND MEASUREMENTS, , 7., , Match List I with List II and select the, correct answer using the codes given below, the lists., 15., List - I, List - II, a) joule, e) henry amp/s, b) watt, f) farad volt, c) volt, g) coulomb volt, 16., d) coulomb, h) oersted cm, i) ampere gauss, j) (ampere)2 ohm, 8., Match List I with List II and select the, correct answer using the codes given below, the lists., List - I, List - II, a) Same negative, I) pressure,, dimensions of mass, Rydberg’s constant, b) same negative, II) Magnetic, dimensions of length, induction field,potential, 17., c) same dimensions, III) Capacity, universal, of time, gravitational constant, d) Same dimension, IV) Energy density,, of current, surface tension, 18., , Assertion & Reasoning Questions, , 9., , 10., , 11., , 12., , 13., , 14., , Options :, 1. A and R are correct and R is correct, explanation of A, 2. A and R are correct and R is not correct, explanation of A, 3. A is true and R is false, 4. Both A and R are false, Assertion(A) : The equation y = x + t cannot be, true where x,y are the distances and t is time, Reason(R) : quantities with different dimensions, can not be added, Assertion(A) : Plane angle is dimensionless, quantity., Reason(R) : All unitless quantities are, dimensionless, Assertion(A) : Dimensions of constant of, proportionality of constants can be derived from, dimensional method, Reason(R) : Numerical value of constant of, proportionality can be found from experiments only, Assertion(A) : Solid angle is dimensionless, quantity and it is a supplementary quantity., Reason(R) : All supplementary quantities are, dimensionless., Assertion(A) : When we change the unit of measurement of a quantity, its numerical value changes., Reason(R) : Smaller the unit of measurement,, smaller is its numerical value., Assertion(A) : If u1 and u2 are units and n1 , n2, are their numerical values in two different systems, then n1 n2 u1 u2 ., , NARAYANA GROUP, , 19., , 20., , 21., 22., , 23., , Reason(R) : The numerical value of physical, quantity is inversely proportional to unit., Assertion(A) : Surface tension and spring constant have the same dimensions., Reason(R) : Both are equivalent to force per, unit length, Assertion(A) : Method of dimensions cannot, be used for deriving formulae containing trigonometrical ratios., Reason(R) : Trigonometrical ratios have no dimensions., , Statement Type Questions, Options :, 1. Statement-1 is true and statement-2 is true, 2. Statement-1 is true and statement-2 is false, 3. Statement-1 is false and statement-2 is true, 4. Statement-1 is false and statement-2 is false, Statement-1: Plane angle is a dimensionless, quantity., Statement-2: All supplementary quantities are, dimensionless., Statement-1 :The size (u) of the unit of physical, quantity and its numerical magnitude (n) are, related to each other by the relation, nu = constant, Statement-2: The choice of mass, length and time, as fundamental quantities is not unique., Statement-1: The MKS system is a coherent, system of units, Statement-2:In SI, joule is the unit for all forms, of energy, Statement-1: Two quantities which are to be, added must have the same dimensions, Statement-2: Two quantities which are to be multiplied may have the same dimensions., Statement-1:Susceptibility is expressed as Am-1., Statement-2:Magnetic flux is expressed as JA-1, Statement-1 :Electromotive force is expressed, in newton., Statement-2:Electric intensity is expressed in NC-1, e2, , Statement-1:The quantity ch is dimensionless, 0, 1, , 24., 25., , 26., , Statement 2: has the dimensions of ve0, 0, locity and is numerically equal of velocity of light., Statement-1 : Electric current is a scalar, Statement-2 : All fundamental physical quantities, are scalars, Statement-1 : Pressure can be subtracted from, pressure gradient, Statement-2: Only like quantities can be added, or subtracted from each other, Statement-1 : Energy cannot be divided by, volume, Statement-2 : Dimensions of energy and volume, are different, 35
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JEE MAINS, - C.W - VOL -- III, JEE-ADV, PHYSICS-VOL, , UNITS AND MEASUREMENTS, 27., 28., , 29., , 30., , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 36, , Statement-1: Light year is a unit of time, Statement-2: Light year is the distance traveled, by light in vacuum in one year., Statement-I: Dimensional analysis can give us the, numerical value of proportionality constants that, may appear in an algebraic expression., Statement-II: Dimensional analysis make use of, the fact that dimensions can be treated as algebraic quantities., Statement-I: The product of the numerical value, and unit of physical quantity remains same in every system of unit., Statement-II: magnitude of a physical quantity, remains same in every system of units., Statement-I: Systematic errors can be removed, completely., Statement-II: the cause of systematic errors can, be known., Statement-I: Random errors can be positive or, negative., Statement-II: Cause of random errors are uncertain., Statement-I:In the measurement of g using simple, pendulum generally we take central position (mean, position) of the oscillation as reference position, for measuring time of oscillation., Statement-II: This reduces the human error in, measurement of time., Statement-I: When a length of 2.0 m is converted, into centimeter, the result is 200cm, Statement-II: The numerical value of a measurement is proportional to reciprocal of the size of, unit used., Statement-I:The length of an object is measured, with two instruments as l =4.01cm and, 1, l =4.009cm.The second instrument, has a better, 2, resolution., Statement-II: More value is the least count of, an instrument , better is the resolution., Statement-I:If a physical quantity has a unit , it, must not be dimensionless., Statement-II: No physical quantity exists which, has dimension but no unit., Statement-I: A formula derived using dimensional, analysis obeys principle of homogenity ., Statement-II: A physically correct relation is, always in accordance with principle of homogenity, Statement-I: Mass, length and time are fundamental quantities., Statement-II:Mass,length and time are independent of on another., Statement-I: The number of significant figures in, 0.001 is 1 while in 0.100 it is 3., Statement-II:Zeros before a non-zero significant, digit are not counted while zeros after a non-zero, significant digit are counted., , 39., , Statement-I: If error in measurement of mass is, 2% and that in measurement of velocity is 5% than, error in measurement of kinetic energy is 6%., Statement-II:Error in kinetic energy is, K m, v , , 2 ., K m, v , , More than One Answer Questions, 40., , A book with many printing errors contains, four different expressions for the displacement ‘y’ of a particle executing simple, harmonic motion. The wrong formula on dimensional basis (v=velocity), i. y A sin 2 t / T ii. y A sin Vt , A, , 41., , 42., 43., , 44., , 45., , 46., , sin t cos t , iii. y A/ T sin t / A iv.. y , 2, 1)ii only 2)ii and iii only 3)iii only 4)iii and iv only, Three of the quantities defined below have, the same dimensional formula. Identify them., i) Energy / mass, ii) pressure / density, iii) Force / linear density, iv) Angular frequency / radius, 1) i,ii,iii, 2) ii,iii,iv 3) iii,iv,i 4) iv,i,ii, Which of the following is not a unit of time?, a) parsec b)light year c) micron d) sec, 1) a and c 2) a and b 3) a,b and c, 4) all, Which of the following is dimensionless?, a)Boltzmann’s constant b)Planck’s constant, c) Poisson’s ratio, d) Relative density, 1) a and b 2) c and b 3) c and d 4) d and a, Which of the following pairs have same, dimensions., a) Torque and work, b) Angular momentum and work, c) Energy and Young’s modulus, d) Light year and wavelength, 1) a and b 2) b and c 3) c and d 4) a and d, The pair of physical quantities that have the, same dimensions are, a) Reynold’s number and coefficient of friction, b) Latent heat and gravitational potential, c)angular velocity and frequency of light wave, d) Planck’s constant and torque, 1) b and c are correct 2) a and b are correct, 3) a,b and c are correct 4) all are correct, Choose the false statement from given, statements., I.Relative permittivity is dimensionless variable, II. Angular displacement has neither units nor, dimensions, III.Refractive index is dimensionless variable, IV. Permeability of vacuum is dimensional, constant, 1)only I and II 2)Only II 3)Only III 4)Only IV, NARAYANA GROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, -I, 47., , UNITS AND MEASUREMENTS, , The SI unit of inductance, henry can be 54., written as, a) weber/ampere b) volt second/ampere, c) joule/(ampere)-2 d) ohm/second, 1) a & c are correct 2) a & d are correct, 3)a, b, & c are correct 4) a & b are correct, , Ascending & Descending Order, 48., , 49., , 50., , 51., , 52., , 53., , Arrange the following lengths in increasing, order, I. 1 angstrom, II. 1 Micron, III. 1 fermi, IV. 1 light year, 1. III, I, II, IV, 2. I, II, III, IV, 3. III, II, I, IV, 4. II, III, I, IV, Arrange the following multiples in, decreasing order, I. milli II. centi III. nano IV. pico, 1. IV, II, I, III, 2. II, I, III,IV, 3. I, III, II, IV, 4. II,I,IV,III, Arrange the following physical quantities in, increasing order of their magnitudes, I. 106 dyne, II. 1 N, III. 3 kg ms 2, IV. 107 gm cm s 2, 1., II, I, III, IV, 2., IV, I, III, II, 3., II, III, I, IV, 4., I, II, III, IV, Arrange the following physical quantities in, the decreasing order of dimension of length, I. Density, II. Pressure, III. Power, IV. Impulse, 1. I, II, III, IV, 2. III, II, I, IV, 3. IV, I,II, III, 4. III, IV, II, I, The correct order in which the dimensions of, length increases in the following physical, quantities is, a) permittivity, b) resistance, c) magnetic permeability d) stress, 1) a, b, c, d, 2) d, c, b, a, 3) a, d, c, b, 4) c, b, d, a, The correct order in which the dimensions of, “length “ decreases in the following, physical quantities is, a) Coefficient of viscosity, b) Thermal capacity c) Escape velocity, d) Density, 1) b,c,a,d 2) a,b, c,d 3) c,d,b,a 4) a,d,c,b, , NARAYANA GROUP, , The correct order in which the dimensions of, “time” increases in the following, physical quantities is, a) Stress b) Period of revolution of satellite, c) Angular displacement, d) Coefficient of thermal conductivity, 1), a, b, c, d, 2), d, c, b, a, 3), a, d, c, b, 4), d, a, c, b, , LEVEL-IV- KEY, Matching Questions, 1) a-s,, 2) a-r,, 3) a-p,, 4) a-III,, 5) a-g, 6) a-g, 7) a-g, 8) a-III, , b-p,r,, b-p,, b-q,, b-II,, b-h, b-h, b-j, b-I, , c-q,, c-s,, c-s,, c-IV, c-e, c-e, c-e, c-IV, , d-r,s, d-q, d-t, d-I, d-f, d-f, d-f, d-II, , Assertion & Reason Type, 9)1, 15)1, , 10)2, 16)1, , 11)2, , 12)1, , 13)3 14)1, , Statement Type, 17)1, 23)1, 29)1, 35)3, , 18)3, 24)1, 30)1, 36)1, , 19)1, 25)3, 31)1, 37)1, , 20)1, 26)3, 32)1, 38)1, , 21)3 22)3, 27)3 28)3, 33)1 34)2, 39)3, , More than one answer type questions, 40)2, 46)2, , 41)1, 47)4, , 42)3, , 43)3, , 44)4 45)1, , Ascending & Descending Order, 48) 1 49)2, 54)4, , 50)3, , 51)4, , 52)3 53)1, , LEVEL - IV - HINTS, from principle of homogenity., arc length, M 0 L0T 0, 10., Plane angle =, radius, 11., dimensional method is not useful for deriving, proportional constants., 12., supplementary quantities have no dimensional, formula., 1, 13 & 14 N1U1 N 2 U 2 and N , U, 15., by dimensional method, 16., Method of dimensions can not be used for, trignometric ,logarthemic and exponential, functions, 9., , 37
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , VECTORS, unit vector is dimensionless physical quantity., , SYNOPSIS, , The unit vector along R = xi$ + y $j + zk$ is given, , Physical Quantities:, Ø The quantities that are measurable are called physical, quantities, Ex: Length, Mass, Time, Velocity, Force, etc.., Physical quantities are mainly classified into three, types. a) Scalars b) Vectors c) Tensors, Ø Vectors are those which have both magnitude and, direction and also obey laws of vector addition., Ex : Velocity, Force, Momentum , Torque etc, Ø Scalars are those which have only magnitude., Ex : Mass, Time, Distance, Flux etc, Note : A physical quantity having magnitude and, direction but not obeying laws of vector addition is, treated as a scalar., Ex : Electric current is a scalar quantity., Electric current is always associated, with direction, but it is not a vector quantity.It does, not obey laws of vector addition for its addition., , i1, θ, i2, , (i1+i2), , ur, ˆ ˆ, R, xiˆ + yj+zk, R̂ = ur =, by, x2 + y2 +z2, R, , Orthogonal Unit Vectors / Base Vectors, $i, $j and k$ are called orthogonal unit vectors., (It is an X, Y and Z axes of Cartesian co-ordinate, system ), , Position Vector, It is a vector that represents the position of a particle, with respect to the origin of a coordinate system., The Position Vector of a point (x, y, z) is, ur, R = x ˆi + y ˆj + z kˆ, , y, P, R, , O, , x, , z, , Null Vector (or) Zero Vector, It is a vector of zero magnitude., Its direction is indeterminate, r, It is represented as 0, Ex: velocity of simple pendulum at extreme position,, acceleration of particle moving with uniform velocity, etc, , The resultant of i1 and i2 is ( i1 + i2 ) by Kirchhoff’s, f’s, current law.The resultant does not depend on angle, between currents i1 and i2 ., Representation of angle between the, Ø Tensors are those quantities having different, two vectors :, magnitudes in different directions. These do not obey, Ø The angle between two vectors is represented by, the laws of vector addition., the smaller of the two angle between the vectors when, Ex: Moment of inertia , Stress etc.., they are placed tail to tail by displacing either of the, Ø Unit Vector :It is a vector whose magnitude is, ur, vectors parallel to it self., unity. A unit vector parallel to a given vector R is, ur, ur, ur, Ex: The angle between A and B is correctly, R, µ = ur, R, represented in the following figures, given by, R, ur, µ gives the sense of direction of the vector R., R, 196, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, Sol. To find the angle between two vectors we connect, the tails of the two vectors. We can shift the vectors, ur, ur ur, parallel to themselves such that tails of A , B and C, are connected as shown in figure., , B, , θ, A, B, , θ, , y, , B, , θ, , (or ) B, , A, A, ur, ur, a) If the angle between A and B is θ , then the, ur, ur, angle between A and K B is also θ . Where `K’ is, a positive constant., , θ, , 0, , 30, 0, 45, 30, , x, , 0, , B, C, , KB, B, , A, , θ, , A, A, ur, ur, b) If the angle between A and B is θ , then the, ur, ur, angle between A and - K B is (180- θ ). `K’ is, positive constant., , ur, ur, Now we observe that angle between A and B, ur, ur, ur, ur, is 600 , B and C 150 ,between A and C is 750, ur ur ur, W.E-2: If A , B , C represents the three sides of an, equilateral triangle taken in the same order, then find the angle between, ur, ur, ur, ur, ur, ur, i) A and B ii) B and C iii) A and C ., , B, , B, B, , θ, θ, , 180 -θ, 0, , A, 0, , 120, C, , B, − KB, 0, A, 120, c) Angle between collinear vectors is always zero or, A, 0, A, 1800, C 120, Q, Sol. From the diagram the angle between vectors, Q, ur, ur, ur, ur, and B is 1200 , between B and C is 1200 and, A, or, ur, 0, ur, 0, θ = 180 P, between A and C is 1200, θ=0, P, Note: Angle between vectors in anticlockwise direction, is taken as positive and clockwise direction as W.E-3 : A man walks towards east with certain, velocity. A car is travelling along a road, negative, ur ur ur, which is 300 west of north.While a bus is, W.E-1:Three vectors A , B , C are shown in the, travelling in another road which is 600 south, figure. Find angle between, ur, ur, ur, ur, ur, ur, of west.Find the angle between velocity vector, (i) A and B (ii) B and C (iii) A and C ., of, a) man and car b) car and bus, c) bus and man., 0, A 30, x, x, x, 0, 0, 45, 30, C, B, , NARAYANAGROUP, , 197
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , North, , VC, , 0, , 30, West, , Vm, , 0, , O, , 60, , East, , Vb, , South, Sol. From the diagram the angle between velocity vector, of man and car is 900 + 300 = 1200, The angle between velocity vector of car and bus is, 600 + 600 = 1200, The angle between velocity vector of bus and man, is 300 + 900 = 1200, ur, W.E -4 : A vector A makes an angle 300 with the, Y-axis in anticlockwise direction.Another, ur, vector B makes an angle 300 with the x-axis, in clockwise direction. Find angle between, ur, ur, vectors A and B ., y, 0, , 30, , ur, The projection of R along x-axis is called, , ( ), , horizontal component (Rx) R x =Rcos?, ur, The projection of R along y-axis is called vertical, , ( ), , R y =Rsin?, component (Ry), Component of a vector is a scalar quantity., r, Magnitude of the resultant R = R x 2 + R y 2, , Direction of the resultant with x-axis is, , R , θ = tan −1 y , Rx , Ø Note :, , y, , A, 30, , 0, , x, , B, , ur, ur, Sol. From the diagram the angle between A and B is, , 300 + 900 + 300 = 1500, , Resolution of a Vector into Components, in two Dimensions, , ur, A vector R can be resolved into two mutually, perpendicular components Rx and Ry in a plane say x - y, , ( ), , B, , -x, , o, C, , A, , x, D, , -y, r, a) If the vector A is in first quadrant then it can be, r, written as A = Ax iˆ + Ay ˆj, r, b) If the vector B is in second quadrant then, r, B = − Bx iˆ + By ˆj, r, c) If the vector C is in third quadrant then, r, C = −C xiˆ − C y ˆj, r, d) If the vector D is in fourth quadrant then, r, D = Dxiˆ − Dy ˆj, , Applications on resolution of vector :, , 198, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, than integral multiple of 2 π (or 360 0 ) its direction, changes, but magnitude does not change. so vector, changes., , 1), , A book is placed on smooth horizontal surface and, ur, pulled by a force F making an angle ` θ ’ with, horizontal., Ø Note : If the frame of reference is rotated the vector, Component of force along horizontal =Fcosθ ., does not change (though its components may, Component of force along vertical =Fsin θ ., change)., 2), vector, S, O, , A block of mass `m’ is placed on an inclined plane, of angle ` θ ’then the component of weight parallel, to the inclined plane is `mg sin θ ’ , the component, of weight perpendicular to the inclined plane is, mgcosθ ., 3), , O, , θ, , WE-5: The components of a vector along the x, and y directions are (n+1) and 1 respectively. If, the coordinate system is rotated by an angle 600, then the components changes to n and 3. Find, the value of n., Sol. Length of the vector does not change on rotation., 7, 2, ( n + 1) + 12 = n2 + 32 ⇒ n = 2 = 3.5, , WE-6: A weight mg is suspended from the middle, of a rope whose ends are rigidly clamped at, l −x, x, F, the same level.The rope is no longer, horizontal.What is the minimum tension, mg, required to completely straighten the rope, A simple pendulum having a bob of mass `m’ is, suspended from a rigid support and it is pulled by a Sol. From the diagram, horizontal force `F’ . The string makes an angle θ, θ, θ, with the vertical as shown in figure., T, T, The horizontal component of tension = T sin θ, θ, θ, The vertical component of tension = T cos θ, when the bob is in equilibrium, mg, T sin θ =F ................(1), mg, 2T sinθ = mg ⇒ T =, T cos θ = mg .................(2), 2sinθ, The rope will be straight when θ = 00, mg, mgl, T=, =, mg, =∞, T=, cos θ, l 2 − x2, 2sin 00, The tension required to completely straighten the, From equation (1) and (2, rope is infinity., F, x, Tanθ =, ⇒ F = mgTanθ = mg, WE-7: The sum of magnitudes of two forces acting, mg, l 2 − x2, at a point is 16 N.If their resultant is normal, 2, 2, to the smaller force and has a magnitude of, T = F + ( mg ), 2, , 2, , T, , l, , Ø Note: If a vector is rotated through an angle other, , NARAYANAGROUP, , 199
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, 8N.Then the forces are, , F2, , θ, , R, R, R, cosa = urx , cosß = ury and cos? = urz, R, R, R, sum of the squares of direction cosines = 1, i.e cos 2 a + cos2 ß + cos 2 ? = 1, , F, , 90, , 0, , F1, , uur, uur, ur, Sol. let F be the resultant of two forces F1 and F2, as shown in figure with F2 > F1, F2 sin θ = F1 ...(i) F2 cos θ = F = 8 ...(ii), Squaring and adding Eqs (i) and (ii), we get, F2 2 = F12 + 64 ...................(iii), Given F1 + F2 = 16 ..............(iv), Solving Eqs. (iii) and (iv), we get, F1 = 6 N and F2 = 10 N ., , If l = cos α , m = cos β and n = cos γ ,, then l 2 + m 2 + n 2 = 1 ., Now, sin 2 a + sin 2 ß+ sin 2 ?=2, W.E-8: A bird moves with velocity 20m/s in a, direction making an angle of 600 with the, eastern line and 600 with the vertical upward., Represent the velocity vector in rectangular, form, Sol. Let eastern line be taken as x-axis, northern as yaxis and vertical upward as z-axis., r, Let the velocity v makes angle α , β and γ with, x,y and z axis respectively, then α = 600 , γ = 600, we have, , Resolution in 3D Space, A point in space can be specified by a position vector, ur, R = Rx $i + R y $j + Rz k$ where Rx , Ry and Rz being, the coordinates of the point in Cartesian coordinate, system., ur, Magnitude of position vector R is, R = Rx2 + Ry2 + Rz2, , cos 2 α + cos 2 β + cos 2 γ = 1, cos 2 600 + cos 2 β + cos 2 600 = 1, , 1, 1, ; cos β =, 2, 2, r, so v = v cos α i$ + v cos β $j + v cos γ kˆ, cos 2 β =, , 1, = 20 $i +, 2, = 10$i + 10, , y, , 1 $ 1 ˆ, j + k, 2 , 2, 2 $j + 10kˆ, , Mathematical operations with vectors, Ry, γ, , z, , β, α, , r, r, Consider two vectors A and B in x-y plane, r, r, A = Ax $i + Ay $j and B = Bx $i + By $j, , R, Rx, , 1) Vector Addition ( Analytical Method ), x, , ur, , (, , RZ, , Direction Cosines, , ur, If the position vector R makes angles a, ß, ? with, x,y and z axes respectively, then, cosa , cosß, cos? are called direction cosines., , ( ), , ur, , ur, , ur, , Let R be their sum. We have R = A + B, r, R = Ax $i + Ay $j + Bx $i + By $j, , ) (, , ), , = ( Ax + Bx ) $i + ( Ay + By ) $j, , 2) Vector Subtraction ( Analytical Method), , ur, ur ur ur, Let R be their difference. We have R = A − B, r, R = Ax $i + Ay $j − Bx $i + By $j, , (, , )(, , ), , = ( Ax − Bx ) $i + ( A y − By ) $j, , 200, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, ur, ur, W.E - 14: If A = 3iˆ + 4 ˆj and B = 7iˆ + 24 ˆj, find a, ur, vector having the same magnitude as B and, ur, parallel and same direction as A ., ur, Sol . The vector parallel to A and having magnitude, ur ur, ur, C = B Aˆ, of B is, , ur ur ur ur, given by A. B = A B cos? = ABcosθ, ur, ur, where θ is angle between A and B, ur, ur, A cos θ is component of A along B and, ur, ur, B cos θ is component of B along A, The dot product of two vectors is a scalar., Properties of Scalar Product, Ø a) Scalar product is commutative, ur ur uurur, i.e. A.B = B. A, b) Scalar product is distributive, , B = 7 2 + 242 = 25, ur, A 3iˆ + 4 ˆj, 1, µ, A= =, = 3iˆ + 4 ˆj, 2, 2, A, 5, 3 +4, ur ur ur, ur ur ur ur, ur, 1 ˆ, A., B, +, C, =, A.B, + A.C, i.e., (, ), C = 25 × 3i + 4 ˆj = 15iˆ + 20 ˆj, 5, c) It does not obey associative law, ur, ur, W.E -15: The resultant of two vectors A and B is Ø Scalar product of two parallel vectors is maximum, ur ur ur ur, ur, perpendicular to A and equal to half of the, A.B = A B cos? = AB (Q ? = 00 ), ur, ur, magnitude of B .Find angle between A and, The scalar product of two opposite vectors is, ur, negative and minimum, B?, ur, ur, The scalar product of two perpendicular vectors is, Sol : Since R is perpendicular to A ., ur ur, ur, zero, fig shows the three vectors A , B and R ., ur ur, ur, ur, i.e. A.B = ABcos? = 0 ( Q ?=900 ), angle between A and B is π − θ, , (, , (, , ), , ), , Scalar product is negative if θ>900 and <1800, , In case of orthogonal unit vectors, $i.i$ = $j. $j = k$ .k$ = 1 ; $i. $j = $j.k$ = k$ .$i = 0, , R, B 1, =, sin θ = =, B 2B 2, , B, , R, , θ, , π−θ, , A, , A, , ⇒ θ = 300, ur, ur, ⇒ angle between A and B is 1500 ., , Multiplication of Vectors, Ø A vector multiplied by another vector may give, a scalar or a vector. Hence there are two types of, products for multiplication of two vectors., a) dot product or scalar product, b) cross product or vector product, , Scalar Product or Dot Product, Ø, , 202, , ur, ur, The scalar product of two vectors A and B is, , Ø In terms of Components of vectors, ur, If A = Ax $i + Ay $j + Az k$, ur, and B = Bx $i + B y $j + Bz k$, ur ur, Then, A.B=A x B x +A y B y +A z B z, uruur 2, A.A=A x +A 2 y +A 2 z = A2, and magnitude of any vector is, r, A = A = A 2 x +A 2y +A 2z, for perpendicular vectors, A x Bx +A y By +A z Bz = 0, , Applications of Dot Product, , ur, i) Angle between two vectors ur, A and B can be found, ur ur, A.B, ur ur, from cos θ = A, B, r, r, ii) Vector component of A along B is, r r, r, A. B ˆ, B = A. Bˆ Bˆ, B, r, r, iii) Vector component of B along A is, , (, , ), , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , r, W.E 19: If ar and b are two unit vectors such that, , Ø Examples of cross product :, →, , →, , r, r, r, r, a + 2 b and 5 a − 4 b are perpendicular to each, r, other, then the angle between ar and b is, , →, , i) Angular momentum L = r × P, →, → →, ii) Linear velocity V = ω × r, →, , →, , →, , iii) Torque τ = r × F, uur, iv) Torque on a magnet of magnetic moment M in, →, →, →, ur, magnetic field induction B is τ = M × B, v) Force on a conductor of length l carrying current i, , ur, , , in magnetic field induction B is F = i l × B , , , →, , →, , r, , , , ur, magnetic field induction B is F = q v× B , , , →, , →, , ur, , vii) Torque on a coil of area A carrying current i in, →, → →, ur, magnetic field induction B is τ = i A× B , , , r, r, viii) A force F acts at P and τ is torque produced, , r, about Q. If position vector of P is r1 and position vector, r r r, ur ur, r, ur, of Q is r2 then τ = r × F = r1 − r2 × F ., , (, , r, , ), , Ø Applications of cross product :, ur ur, (i) The area of triangle formed by A & B as, 1 ur ur, adjacent sides is A × B, 2, ur ur, (ii) The area of parallelogram formed by A& B, ur ur, as adjacent sides is A× B, , r, , r, , r, , ( a + 2b ). (5a − 4b ) = 0, 5a2 − 4ab cos θ+ 10ab cos θ− 8b2 = 0, 5 – 4 cosθ + 10 cosθ – 8 = 0, – 3 + 6 cosθ = 0 ⇒ cosθ = 1/2, θ = 600, , →, , vi) Force on a charge q moving with velocity v in, →, , Sol. a = b = 1, , W.E 20: If, Sol., , r, F = i$ + 2 $j − 3k$, , r r, and rr = 2$i − $j + $, k find r × F, , i j k, r r, r × F = 2 −1 1, 1 2 −3, , = ( 3 − 2 ) $i − ( −6 − 1) $j + ( 4 + 1) k$ = i$ + 7 $j + 5k$, W.E 21: Find the vector components of vector, r, A = 2 î + 3 ĵ along the directions of, , r, B = î + ĵ ., , r r r, r, A.B B , ˆ, Sol. C= ( Acosθ ) B= Br Br , , , r 2+3 ˆi+jˆ 5, C= , , = ˆi+jˆ ., , 1+1 1+1 2, , ( ), , W.E 22: Find the unit vector perpendicular to, ur, ur, A = 3$i + 2 $j − k$ and B = $i − $j + k$ ., , ur ur, (iii) The area of parallelogram formed by, $i − 4 $j − 5k$, A × B i$ − 4 $j − 5k$, ˆ, n, =, =, =, uu, r, uu, r, u, r, u, r, Sol, :, uur uur, 1, 1 + 16 + 25, 42, A× B, d1 & d 2 as its diagonals is d1 × d 2, 2, W.E 17: Find the angle between two vectors W.E 23 : Find the angle between the diagonals of, r, a cube with edges of length “a”., r, $ $, A = 2i$ + $j − k$ and B = i − k, Z, r r, A.B, , 2 + 0 +1, , A B, , 6 2, , Sol : cos θ = r r =, r, , =, , r, , 3, , 2 3, , =, , 3, 2, , ⇒ θ = 300, , W.E 18: If a1 and a2 are two non collinear unit, vectors inclined at 600 to each other then the, r, r, r, r, value of ( a1 − a2 ).(2 a1 + a2 ) is, Sol. a1 = a2 = 1, Sol., r r, r r, a 1 − a 2 . 2a 1 + a 2 = 2a12 − a22 − a1 a2 cos θ, Now, , (, , )(, , 1, 2, , = 2 −1− =, 204, , 1, 2, , ), , F, , A, B, O, , G, E, , Y, D, , C, X, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , From the diagram, uuur uur, uuur uur, OG = d1 = ai$ + a $j + ak$ , CF = d 2 = − ai$ + a $j + ak$, , 1., 2., 3., , 4., , 5., , 6., , 7., , 8. Choose the correct statement., 1) Scalar + vector = scalar/vector, vector, 2), = scalar, \ The angle between the diagonals is, vector, r r, 3) Scalar/vector = scalar (or) vector, d1. d 2, 1, 1, , , 4) vector - vector = vector., cos θ = r r = ⇒ θ = cos −1 , p, 3, r, r, 3, d1 d 2, 9. If angle between a and b is , then angle, 3, r, r, between 2a and -3b is, C.U.Q, p, 2p, p, 5p, 1), 2), 3), 4), 3, 3, 6, 3, SCALARS,VECTORS AND, 10. If component of one vector in the direction of, another vector is zero, then those two vectors, RESOLUTION OF VECTORS, 1) are parallel to each other, Of the following the vector quantity is, 1) Time, 2) Electric Current, 2) are perpendicular to each other, 3) Velocity of light, 4) Gravitational force, 3) are opposite to each other, Of the following the scalar quantity is, 4) are coplanar vectors., 1) Temperature, 2) Moment of force, 11. The component of a vector is, 3) Moment of couple 4) Magnetic moment, 1) always less than its magnitude, Choose the correct statement, 2) always greater than its magnitude, 1) Temperature is a scalar but temperature gradient, 3) always equal to its magnitude, is a vector, 2) Velocity of a body is a vector but velocity of light, 4) less than or equal to its magnitude, is a scalar, 12. The horizontal component of the weight of a, 3) Electric intensity and Electric current density are, body of mass m is, vectors, mg, 1) mg, 2), 3) zero, 4) infinity, 4) All the above, 2, Choose the false statement :, DOT PRODUCT AND CROSS PRODUCT, 1) Electric current is a vector because it has both, 13. Cross product of vectors obeys, magnitude and direction, 1) commutative law, 2) associative law, 2) Time is a vector which has direction always in the, 3), distributive, law, 4) all the above, forward direction, 14., Distributive, law, is, obeyed, by, 3) All quantities having magnitude and direction are, 1), scalar, product, 2), vector, product, vector quantities, 3) both, 4) none, 4) All the above, 15., Choose, the, false, statement, Which of the following units could be associated, 1) Scalar product and vector product obey, with a vector quantity ?, commutative law, 1) newton/metre 2) newton metre / second, 2) Scalar product does not obey distributive law where, 3) kg m2 s-2, 4) newton second, as vector product obeys commutative law, 3) Scalar product and vector product obey, A vector is not changed if, associative law, 1) it is rotated through an arbitrary angle, 4) All the above, 2) it is multiplied by an arbitrary scalar, ur ur, 3) it is cross multiplied by a unit vector, 16. Three vectors satisfy the relation A . B = 0 and, uur ur, ur, 4) it slides parallel to itself, ,, then, A, ., C, =, 0, A is parallel to, Which of the following is meaningful?, ur, ur ur, ur, ur ur, 1) C, 2) B, 3) B ´C 4) B.C, 1) Vector / Vector, 2) Scalar / Vector, , 3) Scalar + Vector, NARAYANAGROUP, , 4) Vector / Scalar, 205
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , r, , 27. The position vector rr and linear momentum P are, 17. Let F be the force acting on a particle having, r, r $, r = i and P = 4 $j the angular momentum vector is, →, →, perpendicular to, position vector r and τ be the torque of this, 1) x-axis 2) y-axis 3) z-axis, 4) xy-plane, foce about the origin. Then (AIEEE-2003), rr r, 28. The vector area of triangle whose sides are a, b, c is, → →, →, → →, 1) →r . F = 0 and →r . →, 1r r r r r r, 1r r r r r r, τ ≠ 0 2) r . τ ≠ 0 and F . τ = 0, 1) b´ c + c ´a + a´b 2) b´ c + c ´a + a´b, 2, 6, → →, →, → →, 3) →r . →τ ≠ 0 and F . →τ ≠ 0 4) r . τ = 0 and F . τ = 0, 1r r r r r r, 1 r r r r r r, 3) 3 b´ c + c ´a + a´ b 4) 2 -b´ c + c ´a + a´b, r r, r r, 18. ( A´ B) + ( B´ A) is equal to, 29. Set the following vectors in the increasing order of, their magnitude., 1) 2 AB 2) A2 B2 3) zero 4) null vector, r r r, r, 19. If C = A´ B , then C is, a ) 3$i + 4 $j b ) 2i$ + 4 $j + 6k$ c )2$i + 2 $j + 2 k$, r, r, 1) a, b, c 2) c, a, b 3) a, c, b 4) b, c, a, 1) parallel to A, 2) parallel to B, 30., Arrange, the vectors additions so that their magnitudes, r, r, are in the increasing order., 3) perpendicular to A and parallel to B, r, r, r, r, a) Two vector A and B are parallel, 4) perpendicular to both A and B, r, r, r, r, b), Two, vectors, and, are antiparallel, B, A, 20. If A and B are two vectors, then which of the, r, r, c) Two vectors A and B making an angle 600, following is wrong ?, r, r r r r, r r r r, r, d) Two vectors A and B making 1200., 1) A.B = B. A, 2) A + B = B + A, r r r r, r r, r r, 1)b, d, c, a 2)b, c, d, a 3)a, c, d, b 4)c, d, a, b, 3) A´ B = B ´ A, 4) A´ B = -B ´ A, 31., Arrange, the vector subtractions so that their magnitudes, r r, r r, r, r, 21. The angle between ( A´ B) and ( B ´ A) is (in, are in decreasing order. If the two vectors A and B, ur ur, radian), are acting at an angle (| A |>| B |) ., 1) π / 2, 2) π, 3) π / 4, 4) zero, a) 600, b) 900, c) 1800, d) 1200, ur ur, ur, 1) d,c,b,a 2) a,b,d,c 3) c,d,b,a 4) c,d,a,b, 22. If none of the vectors A, B and C are zero and 32. Set the angles made by following vectors with, ur ur, ur ur, ur ur, x-axis in the increasing order., if A´B = 0 and B´C = 0 the value of A´C is, 2, a ) 3$i + 4 $j, b ) 4i$ + 3 $j, c )i$ + $j, 1) unity, 2) zero, 3) B, 4) AC cos q, 1) a, b, c 2) c, b, a 3) b, c, a, 4) a, c, b, 23. Choose the false statement, 1) A vector having zero magnitude can have a 33. Arrange the dot products in increasing order, ur, ur, direction, a) A and B are parallel, ur, ur, r, r r →, r, b) A and B are making an angle 600, 2)If A´ B = 0 , then either A or B or both must, ur, ur, c) A and B making, an angle 1800, have zero magnitude, 1) c, b, a 2) a, b, c 3) b, c, a 4) c, a, b, 3)The component of a vector is a vector, 34., Arrange the magnitude of cross products in the, 4)all the above, r, r r, decreasing, order., 24. If A , B and C are coplanar vectors, then, ur, ur, r r r, r r r, a) A and B making angle zero, ur, ur, 1) ( A.B)´C = 0, 2) ( A´ B).C = 0, b), and B making angle 300, A, r r r, ur, ur, 3) ( A.B).C = 0, 4) all the above are true, c) A and B making angle 1200, r, r, 1) a, b, c 2) b, c, a 3) c, a, b, 4) c, b, a, 25. If A along North and B along vertically upward, →, , →, , then the direction of A × B is along, 1) west 2) south 3) east 4) vertically downwards, r, r r, r, 26. The angle between (A + B) & (A ´ B ), 1) 0, 2) π / 4, 3) π / 2, 4) π, , 206, , C.U.Q - KEY, , →, , 01) 4, 07) 4, 13) 3, 19) 4, 25) 3, 31) 3, , 02) 1, 08) 4, 14) 3, 20) 3, 26) 3, 32) 3, , 03) 4, 09) 2, 15) 4, 21) 2, 27) 4, 33) 1, , 04) 4, 10) 2, 16) 3, 22) 2, 28) 1, 34) 4, , 05) 4, 11) 4, 17) 4, 23) 4, 29) 2, , 06) 4, 12) 3, 18) 4, 24) 2, 30) 1, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , B, , LEVEL - I (C.W), 6N, , ADDITION,SUBTRACTION AND, 1., , 2., , 3., , 4., , 5., , 6., , RESOLUTION OF VECTORS, r, r, If A = 3iˆ − 4 ˆj and B = −iˆ − 4 ˆj , calculate the dir r, rection of A + B, 1) tan −1 ( 4 ) with positive X − axis in clock wise, 2) tan −1 ( 4 ) with negative X − axis in clock wise, 3) tan −1 ( 4 ) with positive X − axis in anticlock wise, 4) tan −1 ( 4 ) with negative X − axis in anticlock wise, r, ˆ, ˆ ˆ, Two, r vectors are givenr by ar = r−2i + j − 3k and, b = 5iˆ + 3 ˆj − 2kˆ . If 3a + 2b − c = 0 then third, vector cr is, 1) 4iˆ + 9 ˆj − 13 kˆ, 2) −4iˆ − 9 ˆj + 13kˆ, 3) 4iˆ − 9 ˆj − 13kˆ, 4) 2iˆ − 3 ˆj + 13kˆ, The vector sum of two vectors of magnitudes, 10 units and 15 units can never be, 1) 28 units 2) 22 units 3) 18 units 4) 8 units, The car makes a displacement of 100 m towards, east and then 200 m towards north. Find the, magnitude and direction of the resultant., 1) 223.7m, tan −1 ( 2 ) , N of E, 2) 223.7m, tan −1 ( 2 ) , E of N, 3) 300m, tan −1 ( 2 ) , N of E, 4) 100m, tan −1 ( 2 ) , N of E, If a vector has an x -component of -25.0units, and a y- component of 40.0 units, then the, magnitude and direction of this vector is, −1 5, with -ve x-axis, 1) 5 89 units;sin, 89, −1 5, 2) 5 89 units; cos, with -ve x-axis, 89, −1 −5, 3) 45 units; cos, with x-axis, 9, −1 −5, 4) 45 units;sin, with x-axis, 9, A force of 10N is resolved into the perpendicular, components. If the first component makes 300 with, the force, the magnitudes of the components are, 1) 5N, 5N, 2) 5 2 N, 5 N, , 3) 5 3 N, 5N, 4) 10 N, 10 3 N, 7. If the system is in equilibrium, (cos 530 = 3/5), then the value of 'P' is, , NARAYANAGROUP, , 0, , A, , 53, O, , 10N, , P, 1) 16N, 2) 4N, 3) 208N 4) 232N, 8. Two billiard balls are moving on a table and the, component velocities along the length and, breadth are 5,5 ms-1 for one ball 2 3 , 2ms-1 for, the other ball the angle between the motion of, balls is, 1) 300, 2)600, 3)400, 4) 150, r, 9. If A = 2iˆ − 3 ˆj + 4kˆ , its components in YZ - plane, and ZX- plane are respectively, 1) 13 and 5, 2) 5 and 2 5, 3) 2 5 and 13, 4) 13 and 29, 10. A car weighing 100kg is on a slope that makes an, angle 300 with the horizontal. The component of, car’s weight parallel to the slope is ( g = 10ms −2 ), 1) 500N 2) 1000N 3) 15,000N 4) 20,000N, 11. A room has dimensions 3m × 4m × 5m. A fly, starting at one corner ends up at the, diametrically opposite corner. The magnitude, of the displacement of the fly is, 1) 12m, 2) 60 m 3) 2 5m 4) 5 2m, r, 12. If P = iˆ + 2 ˆj + 6kˆ , its direction cosines are, 1, 2, 6, 1 2, 6, ,, ,, 1), and, 2), and, 41 41, 41, 41 41, 41, 3, 8, 7, ,, 3), and, 4) 1, 2 and 6, 41 41, 41, , MULTIPLICATION OFA VECTOR BYA, SCALAR, 13. The value of 'm', if $i+2 $j-3k$ is parallel to, $, $j-9 k$ is, 3i+m, 1) 12, 2) 9, 3) 6, 4) 3, ˆ, ˆ, ˆ, 14. A force 2i + j − k newton acts on a body which, is initially at rest. If the velocity of the body at, the end of 20seconds is 4iˆ + 2 ˆj − 2kˆ ms −1 , the, mass of the body, 1) 20kg, 2) 15kg 3) 10kg 4) 5kg, 207
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , SCALAR PRODUCT (OR) DOT PRODUCT, , VECTOR PRODUCT (OR )CROSS, , ur, PRODUCT, 15. When a force vector F= $i+2$j+k$ N acts on a, body and produces a displacement of 23. Find the torque of a force Fr = −3i$ + 2 $j + k$, r, , (, , (, , ), , ), , $ $j+7k$ m , then the work done is, S= 4i+, acting at the point rr = 8$i + 2 $j + 3k$ about origin, 1) 9J, 2) 13J, 3) 5J 4) 1J, 2) 4$i + 4 $j + 6k$, 1) 14$i − 38 $j + 3k$, 16. The angle between the two vectors, ur, ur, A = i$ + 2 $j − k$ and B = −$i + $j − 2k$ is, 3) −14$i + 38 $j − 16k$ 4) −4$i − 17 $j + 22k$, 0, 0, 0, 0, 1) 90, 2) 30, 3) 45, 4) 60, r 24. The area of the triangle whose adjacent sides, 17. In a right angled triangle the three vectors ar, b, are represented by the vectors 4$i + 3$j + 4k$ and, r, r r, and c add to zero. Then a. b is, 5$i in sq. units is, 1) 25, 2) 12.5, 3) 50, 4) 45, 25. The magnitude of scalar and vector products of, c, two vectors are 48 3 and 144 respectively. What, 5, 3 b, is the angle between the two vectors ?, 1) 300, 2) 450, 3) 600, 4) 900, 4, 26. The adjacent sides of a parallelogram are, a, r, r, A = 2$i − 3 $j + k$ and B = − 2i$ + 4 $j − k$ What is the, 1) −9, 2) +9 3) 0, 4) −3, area of the parallelogram ?, 18. A vector perpendicular to the vector $i + 2 $j, 1) 4 units 2) 7 units 3) 5 units 4) 8 units, and having magnitude 3 5 units is, 27. What is the condition for the vectors, 1) 3 iˆ + 6 ĵ 2) 6 iˆ –3 ĵ 3) 4 iˆ –2 ĵ 4) iˆ –2 ĵ, 2$i + 3 $j − 4k$ and 3i$ − a $j + bk$ to be parallel ?, ur, r, 19. If A = 2i$ + 3$j and B = 2 $j + 3k$ the component, 1) a = –9/2, b = – 6 2) a = –6, b = –9/2, ur, ur, 3) a = 4, b = 5, 4) a = 8, b = 2, of B along A is, , (, , (, , ), , ), , 6, 1, 6, 1) 6, 2), 3), 4), 13, 6, 13, r, r, ˆ, 20. If the vectors A = aiˆ+ ajˆ + 3k and B = aiˆ −2ˆj −kˆ are, perpendicular to each other then the positive, value of 'a' is, 1) Zero, 2) 1, 3) 2, 4) 3, 21. A force of 2i +3j + 2k N acts on a body for 4 s, 1., and produces a displacement of 3i$ + 4$j + 5k$ m, , LEVEL -I (C.W) - KEY, , 01) 1, 07) 3, 13) 3, 19) 4, 25) 3, , 02) 1, 08) 4, 14 ) 3, 20) 4, 26) 3, , 03) 1, 09) 2, 15) 2, 21) 3, 27) 1, , 04) 1, 10) 1, 16) 4, 22) 1, , 05) 2, 11) 4, 17) 3, 23) 4, , 06) 3, 12) 2, 18) 2, 24) 2, , LEVEL-I (C .W) - HINTS, , ur ur ur ur, R, R = A + B , R = Rxiˆ + Ry ˆj ; Tanα = y, Rx, r r r, 2. c = 3a + 2b, calculate the power ?, ur ur, 1) 5 w, 2) 6 w, 3) 7 w, 4) 9 w, 3. P, Q are two vectors ; P + Q ≥ R ≥ P − Q, r, 22. If θ is the angle between unit vectors A and 4., r r, s1 = 100iˆ , s2 = 200 ˆj, (1 − A.B, ), r, r r, s, B , then (1 + A.B ) is equal to, s = s12 + s2 2 , Tanθ = 2 → N of E, s1, 1) tan2 ( θ / 2), 2) sin2 ( θ / 2), y, 2, 2, , Tanθ =, 3) cot2 ( θ / 2), 4) cos2 ( θ / 2), 5. A = x + y, x, , 208, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 6., , VECTORS, x1 y1 z1, 27. x = y = z, 2, 2, 2, , F2, F, , F1 = F cos 300, , ,, , F2 = F sin 30 0, , LEVEL - I (H.W), , 0, , 30, , F1, , 7., , ADDITION,SUBTRACTION AND, , 6N, 53, , 10 N, , 0, 0, , 37, , p, , ur r, 6 ˆj + 10 cos 53 ˆj + 10 cos 37iˆ + P = 0 solve for P, ur, uur, 8. V 1 = 5iˆ + 5 ˆj , V2 = 2 3iˆ + 2 ˆj, Y, V1, , V2, , Tanθ1 =, , θ1 θ2, , y1, y, ,Tan θ 2 = 2, x1, x2, , X, 2, 2, 9. → InYZ plane x = 0 , ur, A = Ay + Az, , ur, 2, 2, → In XZ plane y = 0 , A = Ax + Az, , 10., , mg sin θ, , θ, , 11. S = x + y 2 + z 2, A, A, A, cos α = urx , cos β = ury , cos γ = urz, 12., A, A, A, 2, , x1 y1 z1, 13. x = y = z, 2, 2, r r2, A.B, 16. cosθ = Ar Br, , r, r, 14. Ft = mV, , r r, 15. W = F .S, , rr, 17. a.b = ab cos 900, r r, r, r r, a .b, 18. a .b = 0 , b = 3 5, 19. b cos θ =, a, r r 2, 20. A . B = a − 2a − 3 = 0 ⇒ ( a − 3 )( a + 1) = 0, r r, r r, r r F .S, 1 − A . B 1 − cos θ, r r=, 21. P = F .V =, 22., 1 + A . B 1 + cos θ, t, r r ur, 23. Torque of the force, τ = r × F, 1 r r, 24. Area of triangle = A × B, 2, 25. ab cos θ = 48 3 , ab sin θ = 144, r r, 26. Area of parallelogram = A × B, NARAYANAGROUP, , RESOLUTION OF VECTORS, ur, ur, 1. If A = 3iˆ − 4 ˆj and B = −iˆ − 4 ˆj , calculate the, ur uur, direction of A − B., 1) along positive x- axis 2) along negative x-axis, 3) along positive y- axis 4) along negative y -axis, 2. The resultant of the forces, uur, uur, F1 = 4iˆ − 3 ˆj and F2 = 6iˆ + 8 ˆj is, 1) 5 5, 2) 10iˆ − 5 ˆj 3) 125, 4) −2iˆ − 3 ˆj, 3. The vector sum of two vectors of magnitudes 10, units and 15 units can never be, 1) 20 units 2) 22 units 3) 18 units 4) 3 units, 4. A car moves 40m due east and turns towards, north and moves 30m then turns 450 east of north, and moves 20 2m. The net displacement of car, is ( east is taken positive x -axis, North as positive, y - axis), 1) 50iˆ + 60 ˆj 2) 60iˆ + 50 ˆj 3) 30iˆ + 40 ˆj 4) 40iˆ + 30 ˆj, 5. A bird moves in such a way that it has a, displacement of 12 m towards east, 5 m towards, north and 9 m vertically upwards. Find the, magnitude of its displacement, 2) 5 10m 3) 5 5m 4) 5m, 1) 5 2m, 6. An aeroplane is heading north east at a speed, of 141.4 ms −1 . The northward component of its, velocity is, 1) 141.4 ms −1 2) 100 ms −1 3) zero 4) 50 ms −1, 7. The unit vector parallel to the resultant of the, ur, ur, vectors A = 4i$ + 3j$ + 6k$ and B = −$i + 3j$ − 8k$ is, 1 ^, , ^, , ^, , 1) 7 3 i + 6 j− 2 k , , , , 1 ^ ^ ^ , 3) 49 3 i + 6 j− 2 k , , , , 1 ^, , ^, , ^, , 2) 7 3 i + 6 j+ 2 k , , , , 1 ^ ^ ^ , 4) 49 3 i − 6 j+ 2 k , , , , 8. The vector parallel to 4iˆ − 3 ˆj + 5kˆ and whose, length is the arithmetic mean of lengths of two, vectors 2iˆ − 4 ˆj + 4kˆ and iˆ + 6 ˆj + 3kˆ is, 1) 4iˆ − 3 ˆj + 5kˆ, 2) (4iˆ − 3 ˆj + 5kˆ) / 3, 3) (4iˆ − 3ˆj + 5kˆ)/ 2, 4) (4iˆ − 3ˆj + 5kˆ) / 5, r, 9. The direction cosines of a vector A are, cos α =, , 4, 5 2, , , cos β =, , r, vector A is, 1) 4iˆ + ˆj + 3kˆ, , 1, 3, and cos γ =, then the, 2, 5 2, , 2) 4iˆ + 5 ˆj + 3kˆ, 209
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, r, 20. If the vectors A = aiˆ + ˆj − 2kˆ and, 4) iˆ + 5 ˆj − kˆ, 3) 4iˆ − 5 ˆj − 3kˆ, r, r, 10. Given two vectors A = ˆi - 2 ˆj - 3kˆ and, B = aiˆ − ajˆ + kˆ are perpendicular to each, r r, r, other then the positive value of 'a' is, B = 4iˆ - 2ˆj + 6kˆ . The angle made by (A + B), VECTORS, , with the X - axis is, (2007 M), 1) Zero, 2) 1, 3) 2, 4) 3, 0, 0, 0, 1) 30, 2) 45, 3) 60, 4) 900, 11.To go from town A to town B a plane must fly 21. When a force 8iˆ + 4 ˆj newton displaces a, about1780 km atan angleof300 West of north., particle through 3iˆ − 3 ˆj metre, the power is, How far West of A is B ?, 0.6W. The time of action of the force is, 1) 1542km 2) 1452 km 3) 1254 km 4) 890 km, 12. A vector iˆ + 3 ˆj rotates about its tail through, 1) 20s, 2) 7.2s, 3) 72s, 4) 2s, an angle 600 in clockwise direction then the new, r, vector is, 22. If ar and b are two unit vectors and the angle, r r, 1) iˆ + 3 ˆj 2) 3$i - 4 $j 3) 2 ˆj 4) 2iˆ, (1 + a.b), MULTIPLICATION OFA VECTOR BYA, r r is, between them is 600 then, , (, , (, , ), , ), , (1- a.b ), , SCALAR, , r, r, 13. If a = 2iˆ + 6 ˆj + mkˆ and b = niˆ + 18 ˆj + 3kˆ are, parallel to each other then values of m,n are, 1) 1,6, 2) 6,1, 3) -1,6 4) -1,-6, , 14. A particle has an initial velocity (6iˆ +8 ˆj) ms–1, , 1) 2, , 2) 3, , 3) 0, , 4) 1/2, , VECTOR PRODUCT (OR) CROSS, PRODUCT, , ur, r, ˆ + 3 ˆj − kˆ and r = iˆ − ˆj + 6kˆ find rr × ur, 23., If, $, $, F, =, 2, i, –2, F, and an acceleration of 0.8i + 0.6 j ms . Its, speed after 10s is, 1) −17$i + 13 $j + 5k$, 2) −17$i − 13 $j − 5k$, 1)20 ms–12) 7 2 ms–1 3) 10 ms–1 4) 14 2 ms-1, 3) 3$i + 4 $j − 5k$, 4) −3$i − 4 $j + 5k$, , (, , ), , SCALAR PRODUCT (OR) DOT PRODUCT, , 15. A motor boat is going in a river with velocity 24. Two sides of a triangle are given by $i + $j + k$, ur, V = 4$i − 2 $j + k$ ms −1 If the resisting force due, and −$i + 2j$ + 3k$ , then area of triangle is, ur, to stream is, F = 5$i − 10 $j + 6k$ N . Then the, 1) 26, 2) 26 / 2 3) 46, 4) 26, power of the motor boat is., 25. The magnitude of scalar and vector products of, 1) 100 w, 2) 50 w 3) 46 w, 4) 23 w, two vectors are 144 and 48 3 respectively. What, $ kˆ, 16. The angle between the two vectors -2i$ + 3jis the angle between the two vectors ?, , (, , ), , and ˆi + 2jˆ + 4kˆ is, 1) 300, 2) 450, 3) 600, 4) 900, 1) 00, 2) 900, 3)1800, 4)450, 26. Area of a parallelogram formed by vectors, ur, ur, ˆ, ˆ, ˆ, ˆ, ˆ, ˆ, 17. If a vector A = 2i + 2 j + 3k, and B = 3i + 6 j + nk ,, 3iˆ - 2jˆ + kˆ m and ˆi + 2jˆ + 3kˆ m as adjacent sides, are perpendicular to each other then the value of, is, `n’ is, 1) 4, 2) 12, 3) 6, 4) -6, 1) 3 8 m2 2) 24 m2 3) 8 3m2 4) 4 3m 2, $, $, 18. A vector parallel to the vector (i + 2 j ) and, 27. Find the values of x and y for which vectors, ur, ur, having magnitude 3 5 units is, may be, A = 6$i + x $j - 2k$ and B = 5$i - 6 $j - yk$, 1) 3 iˆ + 6 ĵ 2) 6 iˆ –3 ĵ 3) 4 iˆ –2 ĵ 4) iˆ –2 ĵ, parallel, r, r, 19. If A = 5iˆ − 2 ˆj + 3kˆ, and B = 2iˆ + ˆj + 2kˆ ,, 2, −36, 5, r, r, ,y =, 1) x = 0, y =, 2) x =, component of B along A is, 3, 5, 3, 28, 14, 14, 28, −15, 23, 36, 15, 1), 2), 3), 4), ,y=, 3) x =, 4) x = , y =, 38, 38, 38, 38, 3, 5, 3, 14, , (, , 210, , ), , (, , ), , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , LEVEL -I (H.W) - KEY, 01) 1, 07) 1, 13)1, 19)4, 25)1, , 02) 1, 08) 3, 14) 4, 20) 3, 26) 3, , 03) 4, 09) 2, 15) 3, 21) 1, 27)2, , 04) 2, 10) 2, 16) 2, 22) 2, , 05) 2, 11)4, 17) 4, 23)1, , 06) 2, 12)4, 18)1, 24) 2, , LEVEL -I (H.W) - HINTS, 1., 2., 3., , ur ur, Ar− Buur= 4iˆ, uu, F +F, ur1 ur 2, P, Q are two vectors ; P + Q ≥ R ≥ P − Q, , 4., , 8., 9., , LEVEL - II (C.W), , 20 2, , ADDITION,SUBTRACTION AND RESO-, , 450, X, , 40iˆ, , 7., , 1 r r, A× B, 2, 25. ab cos θ = 144 , ab sin θ = 48 3, r r, 26. Area of parallelogram = A × B, 6 x −2, 27. 5 = −6 = − y, , 24. Area of triangle =, , Y, , 30 ˆj, , 5., 6., , iˆ ˆj kˆ, r r, r × F = 1 −1 6, 23., 2 3 −1, , 1., , S = x 2 + y2 + z 2, , 141.4r sinr 450, , A+ B, nˆ = r r ;, A+ B, ur, ur, ur, | B|+|C |, K=, and D = K, 2, , r, A = Ax iˆ + Ay ˆj + Az kˆ, , ur, A, ur, A, , 2., , x, , 10. cos θ = x 2 + y2 + z 2, 11. x = d sin θ, , r, 0, 12. w.r.t x-axis initially θ = 60 , A = 2 units, r1, on rotation θ1 = 00 , A = 2 units, 13., , 3., , 2 6 m, =, =, n 18 3, , r, r, 14. V = ur + at, ur ur, 15. P = F .V, r r, , A.B, 16. cos θ =, rAB, r, A, 18. B = K Ar, , r, , r r, 17. A . B = 0, , 1 + cos θ, , θ, , , 2, 22. 1 − ar . br = 1 − cos θ = cot 2 , , , NARAYANAGROUP, , 4., , ), , r, , 2) 2iˆ + 2 3 ˆj, , 3) 2 ˆj + 2 3kˆ, 4) 2 3 ˆj + 2kˆ, Cosines of angles made by a vector with X, Y, axes are 3 / 5 2 , 4 / 5 2 respectively. If the, magnitude of the vector is 10 2 then that, vector is, 1) 8iˆ + 6 ˆj − 10kˆ, , 5., , (, , Three forces F1 = a iˆ − ˆj + kˆ , F2 = 2iˆ − 3 ˆj + 4kˆ, r, and F3 = 8iˆ − 7 ˆj + 6kˆ act simultaneously on a, particle. If the particle is in equilibrium, the, value of a is, 1) 10, 2) -10, 3) 8, 4) 2, If a particle is displaced from (0,0,0) to a point, in XY - plane which is at a distance of 4 units in, a direction making an angle clock wise 600 with, the negative x-axis. What is the final position, vector of the particle ?, 1) −2iˆ + 2 3 ˆj, , r r, a .b, 19. b cos θ =, a, r r, 2, 20. A . B = a − a − 2 = 0 ; ( a − 2 )( a + 1) = 0, r r, F .S, 21. t =, P, r r, 1+ a .b, , LUTION OF VECTORS, A man travels 1 mile due east, then 5 miles, due south, then 2 miles due east and finally 9, miles due north. His displacement is, 1) 3 miles, 2) 5 miles, 3) 4 miles, 4) between 5 and 9 miles, , 2) 6iˆ − 8 ˆj − 10kˆ, , 3) −6iˆ − 8 ˆj + 10kˆ, 4) 6iˆ + 8 ˆj + 10kˆ, ur, If a vector A makes angles 450 and 600 with x, and y axis respectively then the angle made, by it with z - axis is, 1) 300, , 2) 600, , 3) 900, , 4) 1200, 211
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JEE-ADV, PHYSICS-VOL, -I, VECTORS, JEE MAINS, - VOL - VI, r, 6. A vector Q which has a magnitude of 8 is added 13. Velocity and acceleration vectors of charged, r, particle moving perpendicular to the direction, to the vector P which lies along the, of magnetic field at a given instant of time are, r, r, X-axis. The resultant of these two vectors is a, $ + cj$ and a = 3i$ + 4$j respectively. Then, r, v, =, 2i, third vector R which lies along the Y-axis and, r, the value of ‘c’ is, (2007 E), has a magnitude twice that of P . The magnitude, 1) 3, 2) 1.5, 3) – 1.5, 4) –3, r, of P is, 14. Dot product is used in the determination of, a) Work done by a force, 6, 8, 12, 16, 1), 2), 3), 4), b) Power developed by an automobile moving with, 5, 5, 5, 5, uniform velocity., c) The normal flux linked with a coil kept in magnetic, SCALAR PRODUCT (OR) DOT PRODUCT, field., ur, ˆ, ˆ, 7. If V = 3i + 4j then, with what scalar ‘C’ must, d) The force acting on a conductor carrying current, kept in a magnetic field., it be multiplied so that C V = 7 . 5, 1)a,d aretrue, 2)b,d aretrue, 1) 0.5, 2)2.5, 3)1.5, 4)3.5, 3)a,b,c aretrue, 4)c,d aretrue, 8. The angle between the diagonals of a cube with, VECTOR PRODUCT OR CROSS PRODUCT, edges of unit length is, 15. The, unit vectorur perpendicular to, ur, 2) cos−1 (1 3), 1) sin −1 (1 3), ˆ, A = 2i + 3 ˆj + kˆ and B = iˆ − ˆj + kˆ is, 3) tan −1 (1 3), 9., , 4) cot −1 (1 3), , ur, The angle made by the vector A = 2iˆ + 3 ˆj, with Y-axis is, −1, , 3, , −1, , 2, , 1) tan 2 , , , 2) tan 3 , , , −1 2 , 3) sin 3 , , , −1 3 , 4) cos 2 , , , 1), 3), , 4 iˆ − ˆj − 5 kˆ, 42, 4 iˆ + ˆj + 5kˆ, , 2), , 4iˆ + ˆj − 5kˆ, , 4), 42, 42, r r, r r, 16. Find the value of a + b × a − b =, r r, r r, 1) a × b, 2) 2 a × b, rr, r r, 3) −2 a.b, 4) −2 a × b, , (, , 10. If l1,m1,n1 and l2,m2,n2 are the directional cosines, of two vectors and θ is the angle between, them,then their value of cos θ is, , (, , ), , ( ), , 1) 2 2) 2 3) 1, 7) 3 8) 2 9)2, 13) 3 14) 3 15) 1, , 2) l1m1 + m1n1 + n1l1, , 4) 4, 10) 1, 16) 4, , 6)2, 12) 1, , 9m, , 4) m1l2 + l2 m2 + n1m2, , 4m, , ur ur ur, ur, 11. If A + B = C , then magnitude of B is, ur ur, 1) C − A, 2) C - A, , 4), , uurur ur ur, C. A − B.A, , r, r r, 12. If a = mb + c . The scalar m is, ur r r r, ur r r r, a.b - b.c, c.b - a.c, 1), 2), b2, a2, ur r r r, ur r r r, c.a - b.c, a.b - b.c, 3), 4), c2, a2, 212, , 5) 2, 11) 3, , LEVEL - II (C.W) - HINTS, , 3) l2 m2 + m2 n2 + n2 l2, , 3), , ) ( ), ( ), ( ), , LEVEL -II (C.W) - KEY, , 1) l1l2 + m1m2 + n1n2, , uur ur ur ur, C.B − A.B, , 4 iˆ − ˆj + 5 kˆ, 42, , 1., 2., 3., 4., , 5., , 1m, , 5m, , =, 5m, , 3m, , r r r2m, F1 + F2 + F3 = 0, ( x, y, z ) = (−4cos 600 , 4 sin 600 , 0), cos 2 α + cos 2 β + cos 2 γ = 1, Ay, A, A, , cos γ = z, where cos α = x , cos β =, A, A, A, ur, ˆ, ˆ, ˆ, ∴ A = Ax i + Ay j + Az k, cos 2 α + cos 2 β + cos 2 γ = 1, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, r, r r r, 6. Q = 8, P + Q = R ;, , VECTORS, 3., , Ry = 2 Px ; ( Px + Qx ) + Qy = Ry, Qx2 + Qy2 = 64 ⇒ Px2 + 4 Px2 = 64, 7., , r, 7.5 3, =, V = 32 + 42 = 5 ; CV = 7.5 ⇒ C =, 5, 2, , 8., , uur, uur, d1.d 2, d1 = $i + $j + k$ ; d 2 = $i + $j − k$ ; cosθ = d d, 1 2, , 9., , cos θ =, , 4., , uur uur, , ( 2 iˆ + 3 ˆj ) . ( ˆj ), , ur ur, , ur ur, , 1., , 4+9, , ur ur, A.B, 10. cos θ = uAr uBr, , O is a point on the ground chosen as origin. A, body first suffers a displacement of 10 2 m, North - East, next 10 m north and finally 10 2, North-West. How far it is from the origin ?, 1) 30 m north, 2) 30 m south, 3) 30 m west, 4) 30 m east, If the two directional cosines of a vectors are, 1, 1, and, then the value of third directional, 2, 3, cosine is, , 1, 6, , 2., , rr, , ur ur, , 5., , r r r, , 12. mb.b = ( a − c ) .b, , r r, , r, , r r, , r, , r, , r, , 14. a ) W = F .S ; b) P = F .vr c ) φ = B. A; b) F = idl × B, r r, A× B, ˆ, 15. n = Ar × Br, , 6., , 16.properties of vector product, 7., , LEVEL - II (H.W), , 8., , A particle has a displacement of 12 m towards, east then 5 m towards north and then 6 m vertically upwards the resultant displacement is nearly, 1) 10.04 m 2) 12.10 m 3) 14.32 m 4) 13.06 m 9., Four co-planar concurrent forces are acting on, a body as shown in the figure to keep it in, equilibrium. Then the values of P and θ are., 3N, , P, 0, , 90, O, , 4., , 1, 10, , 1) 1200 2) 900, 3) 600, 4) 450, ur, ur, If A = 9i$ − 7$j + 5k$ and B = 3i$ − 2j$ − 6k$, ur ur ur ur, then the value of A + B . A − B is, 1)206, , θ, , 1, 7, , ur, ˆ acting on a particle, A force F = 3iˆ + cjˆ + 2kN, ur, causes a displacement S = −4iˆ + 2 ˆj − 3kˆ m . If, the workdone is 6 joule, the value of c is, 1) 0, 2) 1, 3) 12, 4) 6, r, r, r, r, If a and b are two unit vector such that a + 2b, r r, and 5a − 4b are perpendicular to each other, r, r, then the angle between a and b is., , (, , ADDITION,SUBTRACTION &RESOLUTION OF VECTORS, , 2., , 3., , SCALAR PRODUCT (OR) DOT PRODUCT, , 11. A.B + B.B = C.B, r r, 13. v . a = 0, , 1., , 1, 5, , 1N, , 0, , 60, , 2N, , 1) P = 4N , θ = 00, 2) P= 2N, θ =900, 3) P = 2N , θ = 00, 4) P= 4N, θ =900, , NARAYANAGROUP, , 2)128, , )(, , ), , 3)106, , 4) -17, , The work done by a force 2i$ − $j + 5k$ when it, displaces the body from a point (3,4,6) to a point, (7,2,5) is, 1)5units 2) 7units 3)1units 4)15units, ur, ur, The component of A along B is 3 times that, ur, ur, of the component of B along A .Then A:B is, , 1) 1: 3 2) 3 :1 3) 2: 3 4) 3 : 2, ur, ur, 10. If A = 2i$ + 3j$ and B = $i - $j then, ur, component of A perpendicular to vector, ur, B and in the same plane is, , (, , 1), 3), , ( ), , 5 $ $, i+ j, 2, , ( ), , 5 $ $, i+ j, 2, , ( ), , ), , ( ), ( ), , 5 $ $, i+ j, 2, 5 $ $, i+k, 4), 2, , 2), , 213
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, , r, , r, , r, , r, , r, , 11. If A + B = R and 2A + B is perpendicular to, r, B then, 1) A = R 2) B = 2R 3) B = R, , 4) B = A, , Vector product (or) cross product, , 9., , ur ur, A.B, A cos θ = ur and Bcos θ =, B, , ur ur, A.B, ur ;, A, , A cos θ = 3B cos θ, ur ur ur, µ B, µ, 10. C = A - A.B, , (, , ), , ur, 1, 1, cos θ ˆi +, sin θ ˆj , what will be the 11. ( 2 Ar + Br ) . Br = 0 ⇒ 2 AB cosθ = − B 2, 12. If A =, 2, 2, r, ur, unit vector perpendicular to A, ur ur, B, ˆ, 12. A × B = ABnˆ ; B =, B, 1) cos θiˆ + sin θˆj, 2) − cos θˆi + sin θˆj, 13. properties of vector product, 1 r r, cos θˆi + sin θˆj, 3), 4) sin θiˆ − cos θˆj, 14. area of parallelogram = d1 × d 2, 2, 2, 13., , (iˆ + ˆj ) × (iˆ − ˆj ) =, 1) -2 k̂, , 2) 2 k̂, , NOTE : LEVEL-III MODELS ARE INCLUDED IN MOTION, IN A PLANE CHAPTER, , 3) zero, , 4) 2 iˆ, , LEVEL - IV, , 14. The, diagonals of ra parallelogram are, r, Matching Type, ˆ, A = 2i − 3 ˆj + kˆ and B = − 2iˆ + 4 ˆj − kˆ What is, r, r, 1. Given two vectors a = xi$ − 4 $j and b = 6$i + 2 $j ., the area of the parallelogram ?, For value of x match the following, 5, Column-I, Column-II, 1) 2 units 2) 4 units 3) 5 units 4), units, r, r, 2, a) a is perpendicular to b, p) x = −12, LEVEL - II (H.W) - KEY, r, b) a is either parallel or, 1) 3 2) 2, 3) 1, 4) 1 5) 3 6) 3, r, 7) 3 8) 1, 9) 2, 10)1 11)1 12) 4, antiparallel to b, q) x = 2 6, 13) 1 14) 4, r r, c) a + b is perpendicular, , ( ), r r, to ( a − b ), r r, d) ( a + b ) is parallel to, r r, (a − b), , LEVEL - II (H.W) -HINTS, 1., , S = x 2 + y2 + z 2, , 2., , P cos θ + 3 = 2sin 600, , P sin θ = 1 + 2cos 600, 3. s1 = 10$i + 10 $j , s2 = 10 $j , s3 = −10$i + 10 $j, →, , →, , →, , →, , →, , →, , →, , s = s1 + s 2 + s3, , 4., , cos 2 γ = 1 − cos 2 α − cos2 β ;, , 5., , W = F.S, , → →, , 7., , , , , a + 2b . 5a − 4b = 0, , , , ur ur, ur ur, A + B . A − B = A2 − B 2, , 8., , ur ur ur ur ur, W = F .S = F . r2 − r1, , →, , 6., , 214, , (, , →, , →, , →, , ) (, , ), , (, , ), , 2., , r) x =, , 4, 3, , s) x = −6, r r, r, If a , b and c represent the sides of a, uur, uur, triangle and d1 and d 2 are the diagonals of, a parallelogram generated, by the triangle, r r, r, formed by the vectors a , b and c ., Column-I, Column-II, r r, a) a + b, p) Diagonal, b), , r, , r r, , ( a × b ) .c, , 1 uur uur, d1 × d 2, 2, r r, d) a − b, , c), , q) Area, r) zero, r r, s) a × b, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, ur, 3., The component of A = 5$i + 4 $j + k$ along, Column-I, Column-II, a) x-axis, p) 1 unit, b) y-axis, q) 5 units, c) z-axis, r) 4 units, ur, d) B = 2$i − 2 $j + k$, s) 2 units, 4., Column-I, Column-II, r r, p) 0, a) a, b are unit, , (, , r, , r, , ) (, , r, , r, , ), , VECTORS, 7., , 8., , vectors and a + 2b ⊥ 5a − 4b ,, rr, then 2 a.b is equal to, , 6., , r, , r, , r, , r, , r, , a) Volume of the parallelepiped p), , π, 4, , r r, rr, c) a × b = a.c, r r rr, d) a × b = a.c, Column-I, r, r a, a) a = $, a, rr rr, b) a.b = a.c, r r r r, c) a + b = b − a, , with its edges represented, , q), , π, 3, , r r r r, d) a × b = b × a, , by the vectors $i + $j, $i + 2 $j, , r), , π, 2, , (, , 5., , Column-I, , Column-II, , ), , (, , ) (, , is equal to, Column-I, , 9., , ), , Column-II, , 10., , and $i + $j + π k$, b) Angle between vectors, s) π, r r, r, r, r, a and b , where a, b and c, r r, r r, are unit vectors satisfying a + b + 3 c = 0, r, r, If a = ax $i + a y $j + az k$ , b = bx $i + by $j + bz k$ , we, have, Column-I, Column-II, rr, 11., a) a .b, p) a x = bx , a y = b y , a z = bz, rr, r r, b) a = b, q) a.b = 0, r, r, c) a ↑↑ b, r) a x bx + a y b y + a z b z, r, , r, , d) a ⊥ b, , a x a y az, s) b = b = b, x, y, z, , Column-I, r r, a) a × b = c$, r r, b) a × b = ab, r r, c) a × b = 0, , p) Not possible either, q) θ = 900, r) b=0, s) θ = 450, Column-II, , r r, p) If θ1 = θ 2 , b = c, q) Not possible, r) θ = 900, s) For any angle θ, between two vectors, Column-II, p) Not possible, r, r, q) a ↑↑ b, r) Both are unit vectors mutually, perpendicular, , r r, r r, d) a × b = −b × a s) Any angle θ between, r, r, a and b, Column-I, Column-II, rr, a) Parallel vectors, p) a.b = 0, r r, b) Perpendicular vectors q) a × b = 0, r, $a = a, c) Axial vector, r), a, d) Unit vector, , NARAYANAGROUP, , r, , p) a ⊥ b, r, r, q) a ↑↑ b, r, r, r) a ↓↓ b, r, r, s) a ↑↑ b & both are unit, vectors, Column-II, , a) a + b = a - b, r r, r r, b) a + b = a - b, , ( ), , b) The points (1,0,3),(-1,3,4), q) -1, (1,2,1) (t,2,5) are coplanar,, then t is equal to, c) The vectors (1,1,m),, r) 1, (1,1,m+1), (1,-1,m) are coplanar,, then value of m is, r r r r r r r r r, d) a × b × c + b × c × a + c a × b s) 2, , Column-I, rr, a) a .b = 1, rr, b) a .b = 0, rr, c) a .b = ab, rr, d) a .b = − ab, , s) a$ = b$ × c$, 215
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , VECTORS, 12., , Column-I, , Column-II, , 18., , a) $j.i$, , p) −k$, , Statements -1: If the magnitude of the sum of, two unit vectors is a unit vector, then magnitude, of their difference is 3, , b) $j + $i, , q), , Statements -2: a + b = a + b, , c) k$ + k$, , r) 2, , d) $j × $i, , s) 0, , r, , 2, 19., , 20., , 2.Statement I is true and statement II is false, 3.Statement I is false and statement II is true, 4.Statement I is false and statement II is false, 13., , 14., , 15., , 16., , Statements -1: A physical quantity cannot be, 21., called as a vector if its magnitude is zero., Statements -2: A vector has both magnitude and, direction, r r, Statements -1: a × b is perpendicular to both, r r, r r, a + b as well as a − b ., r r, r r, Statements -2: a + b as well as a − b lie in the 22., r, r, r r, plane containing a and b , but a × b lies, r, r, perpendicular to the plane containing a and b, Statements -1: Minimum number of non-equal, vectors in a plane required to give zero resultant, 23., is three., r r r, Statements -2: If a + b + c = 0 , then they must, lie in one plane, r r r r r, r, r, Let a + b + c = 0, a = 3, b = 5, c = 7, 24., , r, r 2π, Statements -1: Angle between a and b is, 3, r r2 r2 r2, rr, Statements -2: a + b = a + b + 2 a.b, 17., , Statements -1: The cross product of a vector, with itself is a null vector., 25., Statements -2: The cross product of two vectors, results in a vector quantity., , 216, , r r, , Statements -1:Angle between $i + $j and $i is 450 ., Statements -2: $i + $j is equally inclined to both, , Statement Type Questions, 1.Statement I is true and statement II is true, , r, , $i and $j and the angle between $i and $j is 900, rr rr, r, Statements -1: If a.b = b.c , then a may not, r, always be equal to c ., Statements -2: The dot product of two vectors, involves cosine of the angle between the two, vectors., Statements-1:, The, value, $i. $j × k$ + $j. k$ × $i + k., $ $i × $j is equal to 3, , ( ) ( ) ( ), , of, , $ b,, $ c$ are mutually, Statements-2: If a,, $ b,, $ c$ = 1, perpendicular unit vectors, then a,, , , uur uur uur, Statements -1: In ∆abc , ab + bc + ca = 0, uur r uur r, Statements-2: If oa = a,ob = b , then, uur r r, ab = a + b (triangle law of addition), r r r r, Statements -1 : If a + b = a − b , then angle, r, r, between a and b is 900, r r r r, Statements -2: a + b = b + a, r, r, Statements -1 : Let a = $i + $j and b = $i − $j be, two vectors. Angle between a$ + b$ and, a$ − b$ = 900, , r r, , r r, , Statements -2: Projection of a + b on a − b is, zero., r r, Statements -1 : a = b does not imply that, r r, a=b, r r, r r, Statements -2: If a = b , then a = b, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 26., , 27., , r, Statements -1 : If θ be the angle between a, r r, a×b, r, and b , then tan θ = r r, a.b, r r, rr, Statements -2: a × b is perpendicular to a.b, r r r, r r r, Statements -1 : τ = r × F and τ ≠ F × r, , VECTORS, 33., , r, Statements -1: The vector product of a force F, r, and displacement r is equal to the work done., , 34., , Statements -2: Work done is not a vector., r, r, Statements -1: If a is perpendicular to b and, r r r, r, c , then a × b × c = 0, , (, , r, r, Statements -2: If b is perpendicular to c , then, r r, b× c = 0, , Statements -2: Cross product of vectors is, commutative., 28., , r, Statements -1 :The scalar product of a force F, r, and displacement r is equal to the work done., , Statements -2: Work done is not a scalar., 29., , Statements -1:If dot product and cross product, r, r, of a and b are zero, it implies that one of the, r, r, vector a and b must be a null vector.., Statements -2: Null vector is a vector with zero, magnitude., , 30., , r, Statements-1: a = $i + pj$ + 2k$ and, , r, b = 2i$ + 3j$ + qk$ are parallel vectors if, 3, p = ,q = 4 ., 2, , r, Statements -2: If a = a1 $i + a 2 $j + a 3 k$ and, r, are parallel, then, b = b1 $i + b 2 $j + b 3 k$, a1 a 2 a 3, =, =, b1 b 2 b 3, , 31., , 32., , Statements -1: A null vector is a vector whose, magnitude is zero and direction is arbitrary., Statements -2: A null vector does not exist., r, r, Let a and b be two vectors inclined at an angle, π, θ , where 2 < θ < π ., r r r r, Statements -1: a + b < a − b, , ), , 35., , Statements -1: The sum and difference of two, vectors will be equal in magnitude, when the two, vectors are perpendicular to each other., Statements -2: If either of the two vectors is a, null vector., LEVEL-IV-KEY, , Matching Type Questions, 1) a → r,,, , b → p,, , c → q,, , d→p, , 2) a → p,, , b → r,,, , c → s,, , d→p, , 3) a → q,, , b → r,, , c → p, d → p, , 4) a → r,,, , b → q,, , c → p,, , d→p, , 5) a → s,, , b → q,, , 6) a → r,,, , b → p,, , c → s,, , d →q, , 7) a → s,, , b → p,, , c → q,, , d →r, , 8) a → r,,, , b → q,, , c → s,, , d→p, , 9) a → q,, , b → p,, , c → r,,, , d→s, , 10) a → r,,, , b → p,, , c → q,, , d→ s, , 11) a → q,, , b → p,, , c → s,, , d →r, , 12) a → s,, , b → q,, , c → r,,, , d→p, , Statement type questions, 13) 3, , 14) 1, , 15) 1, , 16) 1, , 17) 1, , 18) 2, 23) 1, , 19) 1, 24) 1, , 20) 1, 25) 1, , 21) 1, 26) 2, , 22) 2, 27) 2, , 28) 2, 33) 3, , 29) 1, 34) 2, , 30) 1, 35) 1, , 31) 2, , 32) 1, , r r2, rr, Statements -2: a + b = a 2 + b 2 + 2a.b, NARAYANAGROUP, , 217
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , MOTION IN A STRAIGHT LINE, SYNOPSIS, , , , , Motion in a straight line deals with the motion of, an object which changes its position with time, along a straight line., The study of the motion of objects without, considering the cause of motion is called, kinematics., , s, , If the position of a body does not change with, time with respect to the surroundings then it is , said to be at rest, if not it is said to be in motion., , Distance and Displacement:, , , Distance is the actual path covered by a moving, particle in a given interval of time while, displacement is the change in position vector,i.e., , a vector joining initial to final position. If a particle, moves from A to B as shown in Fig. the distance , travelled is s while displacement is, , , r rf ri, , R =A+B, B, , , A, , , , Distance is a scalar while displacement is a vector,, both having same dimensions[L] and SI unit is, metre., , h, , h, , ( c), , Rest and Motion:, , , , x, , x, , (d), Distance = s, distance = h + 2x, Displacement = s displacement = h, The magnitude of displacement is equal to minimum, possible distance between two positions; so, Distance Displacement, For motion between two points displacement is, single valued while distance depends on actual path, and so can have many values., For a moving particle distance can never decrease, with time while displacement can., Decrease in displacement with time means body, is moving towards the initial position., For a moving particle distance can never be, negative or zero while displacement can be, negative., In general, magnitude of displacement is not equal, to distance. However, it can be so if the motion is, along a straight line without change in direction., Magnitude of displacement is less than the distance, travelled in case of curvilinear motion., Ex : If an object turns through an angle along a, circular path of radius r from point A to point B, then, i) distance d r, , , h, A, , r, , O, , B, , (a), Distance = r, Displacement = 2r, , 38, , AB, (b), Distance = 2h, Displacement = 0, , , , x, , ii)displacement 2 x 2r sin / 2 sin 2 r , , , d, B x x, A, , r, , r, – –, , 22, O, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , , track of radius R in 40 s. What will be his, displacement at the end of 2 min 20 s? (2010E), Sol. The time = 2 min 20s = 140s, , WE-1: An athlete completes one round of a circular, , A, , R, , R, , B, , In 40 seconds athlete completes = 1 round, In 140 seconds athlete will completes, 140, =, round =3.5 rounds, 40, The displacement in 3 rounds = 0, So net displacement =2R, W.E-2 : If the position of a particle along Y axis is, represented as a function of time t by the, equation y(t)=t3 then find displacement of the, particle during the period t to t t, Sol. Position at time t is y t t 3, 3, , Position at time t t is y t t t t , displacement of the particle from t to t t is, 3, , y t t y t t t t 3, 2, , 2, , , , 3, , t 3 3t 2 t 3t t t t 3, 3t 2 t 3t t t , , For a given time interval average velocity is single, valued while average speed can have many values depending on path followed., During the motion if the body comes back to its, initial position., , , ( r 0 ), V avg = 0, but Vavg 0 and finite ( r 0 ), , For a moving body average speed can never be, negative or zero while average velocity can be, negative., , If a graph is plotted between distance (or displacement) and time, the slope of chord during a given, interval of time gives average speed (or ) average, velocity, Δr, Vavg =, = tan = slope of chord, Δt, Instantaneous speed and Instantaneous velocity :, , Instantaneous velocity is defined as rate of change, of position of the particle with time. If the posi, tion r of a particle at an instant t changes by, , r in a small time interval t, , , , r dr, V lim, , t o t, dt, , The magnitude of velocity is called speed,i.e, , 3, , , , Average Speed and Average Velocity:, , , , , , Average speed or velocity is a measure of overall , 'fastness' of motion during a specified interval of time., The average speed of a particle for a given 'interval, of time' is defined as the ratio of distance travelled to, the time taken while average velocity is defined as , the ratio of displacement to time taken., Thus, if a particle in time interval t after, , travelling distance r is displaced by r, Distance travelled, Average speed =, ,, Time taken, r, ........(i), t, Displacement, ,, Average velocity =, Time, , r, i.e, Vavg ............(ii), t, , i.e., Vavg , , , , , , Average speed is a scalar while average velocity, is a vector both having same units (m/s) and , dimensions LT -1 ., , NARAYANAGROUP, , Speed= velocity, i.e, V V, Velocity is a vector while speed is a scalar , both, having same units (m/s) and dimensions LT -1 ., If during motion velocity remains constant, throughout a given interval of time, then the, motion is said to be uniform and for uniform, , , motion,, V=constant=Vavg, If velocity is constant, speed will also be constant., However , the converse may or may not be true,i.e,, if speed=constant,velocity may or may not be, constant as velocity has a direction in addition to, magnitude which may or may not change,e.g, in, case of uniform rectilinear motion, , V constant and V constant, While in case of uniform circular motion, , V=constant but V is not constant , due to, change in direction., velocity can be positive or negative as it is a vector but speed can never be negative as it is magni, , tude of velocity, i.e, V V, , dr, , the slope of displacedt, ment-versus time graph gives velocity , i.e,, , As by definition V , , 39
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, Total time taken is, AB BC CA, T t1 t 2 t 3 , , , V1 V2 V3, r 2r 5r, , , (arc length = radius angle), 2S 6S 18S, 2 r, Vavg , 1.8 S, r 2 r 5 r, , , 2S 6S 18S, WE-4.For a man who walks 720 m at a uniform, speed of 2 m/s, then runs at a uniform speed, of 4 m/s for 5 minute and then again walks, at a speed of 1 m/s for 3 minutes. His average, speed is, s1, Sol. Where s1 = 720 m and t1 v 360s 6 min ., 1, s2 = (4)(5)(60) = 1200m, t2 = 300 s, s3 = (1)(3)(60) = 180 m, t3 = 180 s, s s s, vavg 1 2 3 = 720 1200 180, t1 t 2 t 3 360 300 180, , MOTION IN A STRAIGHT LINE, Sol., , A, , , , v avg 2.5m / s, , WE-5. A particle is at x = +5 m at t = 0, x = -7m at, t = 6s and x = +2m at t = 10s. Find the average, velocity of the particle during the interval (a), t = 0 to t = 6s; (b) t = 6s to t = 10s, (c) t = 0 to, t = 10s., Sol. x1 = +5m, t1 = 0, x2 = -7m; t2 = 6s, x3 =+2m, t3=, 10s, a) The average velocity between the times t = 0, to t = 6s is, , x x1, 7 5, v1 2, , 2m / s, t 2 t1, 60, b) The average velocity between the times t2 = 6s, to t3 = 10s is, , x x 2 2 7 9, v2 3, , 2.25m / s, t3 t2, 10 6, 4, c) The average velocity between times t1 = 0 to t3, = 10s is, , x x, 25, v3 3 1 , 0.3m / s, t 3 t1 10 0, WE-6. A particle traversed one third of the distance, with a velocity v0. The remaining part of the, distance was covered with velocity v1 for half, the time and with a velocity v 2 for the, remaining half of time. Assuming motion to, be rectilinear, find the mean velocity of the, particle averaged over the whole time of, motion., NARAYANAGROUP, , t, , t /2, , 1, , v0, , t /2, , 2, , C, , v1, S, , 2, , v2, , D, , B, , S, S, v0 t1 t1 , ---(1), 3v0, 3, 2S, CD DB, For CB;, 3, 4S, 2S, t , t , v1 2 v2 2 t 2 , 3 v1 v2 , 3, 2, 2, Since, average velocity is defined as, , For AC;, , v avg, , S 2S, , Total Displacement, , 3 3, t, t, Total Time, t1 2 2, 2 2, , vavg , , =, , S, t1 + t 2, , 3v0 v1 v2 , 4v0 v1 v2, , Acceleration, , , The rate of change of velocity is equal to, acceleration., , Average and Instantaneous acceleration, , , If the velocity of a particle changes ( either in magnitude or direction or both) with time the motion, is said to be accelerated or non-uniform. In case, , of non-uniform motion if change in velocity is V, in time interval t , then, , Δv v - v, a avg =, = 2 1, .......(1), Δt t 2 - t1, Instantaneous acceleration or simply acceleration, is defined as rate of change of velocity with time, , at a given instant. So if the velocity of a particle v, , at time t changes by V in a small time interval, t then, , , , ΔV dV, a = lim, =, .......... (2), Δt 0 Δt, dt, Regarding acceleration it is worth noting that:, , , , -2, It is a vector with dimensions LT and SI units, m / s 2 ., If acceleration is zero, velocity will be constant, and the motion will be uniform. However, if, acceleration is constant(uniform), motion is nonuniform and if acceleration is not constant then, both motion and acceleration are non-uniform., , , As by definition V ds / dt, , , , , , , , , , 41
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , , , , , , , , dv d ds d 2 s, So, a = dt = dt dt = dt 2 .........(3), , , i.e., if s is given as a function of time, second, derivative of displacement w.r.t. time gives acceleration., If velocity is given as a function of position, then, by chain rule, dx, , dv dv dx, dv, as, = v, a=, = ., or a = v, , dt dx dt, dx, dt, , , As acceleration a = dv / dt , the slope of velocity-time graph gives acceleration., , , , , , B, , , V1, , V, , , , , t, , 0, , 2, , , , 2 a , or 2as = 2uv - 2u 2 + v 2 + u 2 - 2uv, i.e., v 2 = u 2 + 2as .......(3), In scalar form, , , v.v = u.u + 2a.s or v 2 = u 2 + 2a.s, Distance travelled = average speed x time, uv, s, t .............(4), 2 , Distance travelled in nth second, 1, s n = u + a n - ..... (5), 2, If acceleration and velocity are not collinear, v can, be calculated using, a, , , , A, , 0, , t, , ds = u + at dt or, , v - u + 1 a v - u, s=u, , V, , V2, , s, , 1, s = ut + at 2, ........(2A), 2, 1 2, In vector form s = ut + at .............(2B), 2, From eqns. (1A) and (2A), we get, , , , Velocity - Time Graph, , , , or, , 2, , O, , t1, , t2, , v = u 2 + at + 2uat cosθ , , , at sinθ, with tan =, .......(6), u + atcosθ, , t, , , , V, a avg , tan , t, , , aavg = slope of the line joining two points in v-t graph., , , , , , , a=, , , , 42, , The slope of a versus t curve, i.e,, , , da, dt, , , v, , is a, , measure of rat e of non-uniformity of, acceleration(usually it is known as JERK)., Acceleration can be positive or negative. Positive, acceleration means velocity is increasing with time, while negative acceleration called retardation means , velocity is decreasing with time., , Equations of motion :, , If a particle starts with an initial velocity u,, acceleration a and it gains velocity v in time t then, v, t, dv, V, t, or dv = a dt or dv = a dt or Vu = a t 0, u, 0, dt, or v = u + at .......(1A), , In vector form v = u + a t ........(1B), Now again by definition of velocity, Eqn. (1A), ds, = u + at, reduces to, dt, , , , GRAPHS, , 1/ 2, , , at, , , u, , Characteristics of s-t and v-t graphs, Slope of displacement time graph gives velocity., Slope of velocity-time graph gives acceleration., Area under velocity-time graph gives displacement, Let us plot v-t and s-t graphs of some standard, results. To draw the following graphs assume, that the particle has got either a one-dimensional, motion with uniform velocity or with constant, acceleration., , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , S.No, , Situation, , 1, , Uniform motion, , MOTION IN A STRAIGHT LINE, , v-t graph, v-t graph, v, , s-t graph, s-t graph, s, s=vt, , Interpretation, i)Slope of s-t graph =, v = constant., ii) In s-t- graph s = 0 at, t=0, , v= constant, t, 2, , 3, , 4, , 5, , Uniformly, accelerated, motion with, u=0 and, s=0 at t = 0, Uniformly, accelerated, motion with, u 0 and, s s0 at t 0, , s, , v, v= at, , t, v, u, , Uniformly, accelerated, motion with, u 0 and, s s0 at t 0, , v, , Uniformly, retarded motion, till velocity be, comes zero, , v, , u, , u, , s, v=u+at, , 2, , s=ut+1/2 at, , I) s = s0 at t=0, , s, v=u+at, s0, , _, , s, v=u-at, t0, , 6, , Uniformly, retarded then, accelerated in, opposite, direction, , NARAYANAGROUP, , v, , t0, s, , u, t0, O, , I) u=0, i.e.,v=0 and t=0, ii) u=0, i.e., slope of s-t, graph at t=0, should, be zero, iii) a or slope of v-t, graph is constant, I) u 0 , i.e., v or, slope of s-t graph at, t=0 is not zero, ii) v or slope of s-t, graph gradually, goes on increasing, , t0, , I)Slope of s-t graph at, t=0 gives u., ii) Slope of s-t graph at, t=t0 becomes zero, iii) In this case u can’t, be zero., I)At time t=t0, v=0 or, slope of s-t graph is, zero., ii) In s-t graph slope or, velocity first decreases, then increases with, opposite sign., , 43
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , , If a particle starts from rest and moves with uniform, acceleration ‘a’ such that it travels distances, sm and sn in the mth and nth seconds, , u, , u u , n, , , , , a, 2n -1 1 (u=0), 2, a, Sm = 0 + 2m -1 2 , 2, sn sm, subtracting eq (2) from eq (1) a n m ., , , A particle starts from rest and moves along a, straight line with uniform acceleration. If ‘s’ is the , distance travelled by it in n seconds and sn is the, distance travelled in the nth second, then, , sn 2n 1, , (fraction of distance fallen in nth, 2, s, n, second during free fall ), , Moving with uniform acceleration, a body crosses, a point 'x' with a velocity 'u' and another point ‘y’, with a velocity ‘v’. Then it will cross the mid point, of, ‘x’ and ‘y’ with velocity, , v1, u, v, y, x, S, S, v 2 - u 2 v 2 - v12, =, In this case acceleration a = 1, 2S, 2S, v2 u 2, ., 2, If a bullet loses (1/n)th of its velocity while passing, through a plank, then the minimum no. of such, planks required to just stop the bullet is ., , 2, , m, , u, , u, , x, , u, n, , v=0, , x, , x, , Let m be the no of planks required to stop the, bullet, 2, , u, , 2, u - - u = 2ax, n, , , , , 0 2 - u 2 = 2amx, Dividing eq (2) with eq (1), , 02 - u 2, 2, , u, 2, u - n -u, , , 44, , =, , 2amx, 2ax, , 1, 2, , 1, 1 1 , n, , 1, n 1 , 1 , , n , , , m, , 2, , , , 2, , n2, n 2 n 2 1 2n, , n2, 2n 1, , 1, of the initial, n, velocity while penetrating a plank. The number of, such planks required to stop the bullet., u, u, u, v=0, n, , The velocity of a bullet becomes, , x, x, x, 2, u, 2, - u = 2ax 1, n, 2, 0 2 - u 2 = 2amx, From eq (1) and eq (2) ;, , , m, , n2, n2 1, , 1, of its velocity while penetrating, n, a distance x into the target. The further distance, travelled before coming to rest., , A bullet loses, , on solving, v1 , , , , 1, , , , 2, , Sn = 0 +, , then, , , , u2, , m, , u, , u, x, , u, n, , v=0, , y, , Let x is the distance covered by the bullet to lose, 1, the th of initial velocity, n, v 2 - u 2 = 2as, 2, , u, , 2, u - - u = 2ax 1, n, , Let y is the further distance covered by the bullet, to come to rest, 2, 2, 0 - u = 2a x + y 2 , From eq (1) and (2), n 1 2 , y x, , 2n , , , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, th, , , , 1, If the velocity of a body becomes of its initial, n, velocity after a displacement of ‘x’ then it will come, to rest after a further displacement of, u, n, , u, , 1 2, 2u, at = ut t , 2, a, 2u 2u 2, s bike ut u, , a, a, v car at 2u, , vcar w.r.t. bike at the t ime of meeting,, , , , v cb = vcar - v bike = 2u - u = u, x, y, , , 2, W.E-9: What does d v /dt and d v / dt repreu, 2, - u = 2ax 1, sent? can these be equal ? can:, n, , , 2, d v /dt=0 while d v / dt 0, (a), 2 u , , , 0 - = 2ay 2 , n, (b) d v / dt 0 while d v / dt 0 ?, x, , From eq. (1) and eq (2) ; y 2, d, v, / dt represents time rate of change of speed, Sol., n 1, , , as v u , while d v / dt represents magnitude, WE-7. A body covers 100cm in first 2 seconds and, 128cm in the next four seconds moving with, of acceleration., constant acceleration. Find the velocity of, If the motion of a particle has translational, the body at the end of 8sec?, acceleration (without change in direction), , Sol. distance in first two seconds is, dv d , v n , as v v n ,, 1, S1, S2, , s1 = ut1 + at12, dt dt , t1, t2, , 2, dv d , 1, n v [as n is constant], or, 100 = 2u + a 4, ......(1), dt, dt, 2, , distance in (2+4)sec from starting point is, dv d , or dt dt v 0 , 1, 2, s1 + s 2 = u t1 + t 2 + a t1 + t 2 , 2, In this case both these will be equal and not, 1, equal to zero., 228 = 6u + a 36, ......(2), (a) The given condition implies that:, 2, , , from eq (1) and (2), d v / dt 0, i.e., acc. 0 while d v / dt 0 ,, 2, We get a = -6 cm/s , sub a=-6 in eq - (1), i.e., speed =constant., 1, This actually is the case of uniform circular, 100 2u 6 4, 2, motion. In case of uniform circular motion, , 2u = 112, u = 56 cm/s, u2, d, v, v = u + at = 56 - 6 x 8 v 8cm / s, = a =, = constant, dt, r, WE-8. A car starts from rest and moves with, uniform acceleration ‘a’. At the same instant, , d , from the same point a bike crosses with a, v 0, while v =constant., i.e.,, dt, uniform velocity ‘u’. When and where will, , , , they meet ? What is the velocity of car with, (b) d v / dt 0 means a = 0 , i.e., a = 0, respect to the bike at the time of meeting ?, , , , d, v, /, dt, , 0, or, or v = constant, 1 2, s = at 1 , s = ut 2 , Sol. car, , bike, 2, And when velocity v is constant speed will be, if they meet at the same point, constant,, s = s, , v=0, , , , car, , , , bike, , NARAYANAGROUP, , 45
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , d , v 0, , v, i.e., speed, =constant or, dt, , dv, So, it is not possible to have dt 0, d , v 0., dt, WE-10:In a car race, car A takes time t less than, car B and passes the finishing point with a, velocity v more than the velocity with, which car B passes the point . Assuming, that the cars start from rest and travel with, , while, , constant accelerations a 1 and a 2 respectively , the value of, Sol., , V, t, , is, , WE-12 : An, , particle travels inside a straight, hollow tube 2m long of a particle accelerator, under uniform acceleration. How long is the, particle in the tube if it enters at a speed of, 1000 m/s and leaves at 9000 m/s. What is its, acceleration during this interval ?, Sol. Let ‘a’ be the uniform acceleration of -particle., According to given problem s = 2m,, v = 9000 m/s and u = 1000 m/s., Since v2 - u2 = 2aS,, 2, , 9000 1000 , 7, , 2, , 2a 2 , 2, , a = 2 x 10 m/s, Let the particle remains in the tube for time ‘t’,, then v u at, v u 9000 1000, t, , 4 10 4 s, 7, a, 2 10, , The distance covered by both cars is same, WE-13:A car starts from rest and moves with, Thus, s1=s2=s, uniform acceleration of 5 m/s2 for 8 sec. If the, If the cars take time t1 and t2 for the race and their, acceleration ceases after 8 seconds then find, velocities at the end of race be v1 and v2 , then it is, the distance covered in 12s starting from rest., given that, Sol. The velocity after 8 sec v = 0 + 5 x 8 = 40 m/s, v1 v2 v and t 2 t1 t, Distance covered in 8 sec, 1, 2a1s 2a 2s, v v1 v2, s 0 0 5 64 160m, , , 2, 2s, 2s, Now, t t 2 t1, , After 8s the body moves with uniform velocity, a2, a1, and distance covered in next 4s with uniform velocity., a1 a 2, v, s = vt = 40 x 4 = 160 m, , a 1a 2, t, 1, 1, The distance covered in12 s =160 +160 =320m., , a2, , a1, , WE-11 : A drunkard walking in a narrow lane takes, , WE-14 : A car is moving with a velocity of 40 m/s., , The driver sees a stationary truck ahead at a, 5 steps forward and 3 steps backward, followed, distance of 200 m. After some reaction time, again by 5 steps forward and 3 steps backward,, and so on. Each step is 1m long and requires, t the breaks are applied producing a, 1sec. How long the drunkard takes to fall in a, (reaction) retardation of 8 m / s 2 . What is the, pit 13m away from the start?, maximum reaction time to avoid collision ?, Sol. Distance of the pit from the start =13-5=8m, Sol. The car before coming to rest (v = 0), Time taken to move first 5m=5sec, 2, v 2 u 2 2as 0 40 2 8 s, 5 steps (i.e., 5m) forward and 3steps(i.e.,3m), backward means that net distance moved =5 s 100m, 3=2m, The distance travelled by the car is 100m, and time taken during this process =5+3=8sec, To avoid the clash, the remaining distance 200 8×8, 100 = 100m must be covered by the car with uni= 32sec, Time taken in moving 8m =, form velocity 40 m/s during the reaction time t ., 2, Total time taken to fall in the pit =32+5=37sec, 100, 40 t 2.5s, t, The maximum reaction time t 2.5s, 46, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , WE-15 : Two trains one travelling at 54 kmph and, , the other at 72 kmph are headed towards one, another along a straight track. When they, are 1/2 km apart, both drivers simultaneously, see the other train and apply their brakes. If, each train is decelerated at the rate of 1 m/s2,, will there be a collision ?, Sol. Velocity of the first train is 54 kmph = 15 m/s, Distance travelled by the first train before coming, to rest, , b) Displacement = area under the v-t graph, = area of OAB, 1, 1, 1, = base height t1 t 2 v max tv max, 2, 2, 2, 1 2, = , t, 2 , v avg , , v max, total displacement 1 , , t , total time, 2 , 2, , u 2 225, WE-17: A body starts from rest and travels a distance, , 112.5m, 2a, 2, S with uniform acceleration, then moves, Velocity of the second train is 72 kmph= 20 m/s, uniformly a distance 2S and finally comes to, Distance travelled by the second train before comrest after moving further 5S under uniform, ing to rest, retardation. Find the ratio of average, 2, velocity to maximum velocity, u, 400, s2 , , 200m, Sol., 2a, 2, Total distance travelled by the two trains before, v, coming to rest = s1 + s2 = 112.5 + 200 = 312.5m, Vmax, Because the initial distance of separation is 500m, which is greater than 312.5m, there will be no, collision between the two trains., t, O, WE-16. A bus accelerates from rest at a constant rate, t1, t2, t3, ‘ ’ for some time, after which it decelerates, at a constant rate ‘ ’ to come to rest. If the, area of ΔOAC S 1 Vmax t1 (or) t1 2S ;, total time elapsed is t seconds. Then evalu2, Vmax, ate following parameters from the given graph, 2S, a) the maximum velocity achieved, area of ABCD 2S Vmax t 2 (or) t 2 , b) the total distance travelled graphically and, Vmax, c) Average velocity, area of ΔBDE 5S 1 Vmax t 3 (or) t 3 10S, V, 2, Vmax, s1 , , Vavg , , A, , Vmax, , O, , t1, , , t2, , Vavg , , B, , t, , vmax, vmax, Sol. a) = Slope of line OA = t t1 , 1, vmax, vmax, = Slope of line AB = t t 2 , 2, , t t1 t 2 , , vmax vmax, , , v max , , , , , , V av g, V m ax, , Totaldisplacement, ;, Total time, S 2S 5S, 8S, , 2S, 2S 10S, 14S, , , Vmax Vmax Vmax, V m ax, , , 8S, 4, , 14S, 7, , WE-18: The acceleration-displacement (a - x) graph, , of a particle moving in a straight line is as, shown. If the particle starts from rest,then, find the velocity of the particle when displacement of the particle is 12m., , , v max , t, , NARAYANAGROUP, , 47
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , C, , 4, 2, , A, , B, , F, E, G D, O, 8 10 12, 2, x(m), 2, 2, Sol. v - u = 2ax, v2 - u 2 v 2 u 2, ax =, , ( u=0 ), 2, 2 2, , v 2 area under a x graph , Area under a-x graph =, Area of OAE + Area of rectangle ABEF, + Area of trapezium BFGC + Area of CGD, 1, 1, 1, Area 2 2 6 2 2 4 2 2 4 24, 2, 2, 2, v 2 24 4 3m / s, WE-19: Velocity-time graph for the motion of a, certain body is shown in Fig. Explain the, nature of this motion. Find the initial, velocity and acceleration and write the, equation for the variation of displacement, with time. What happens to the moving body, at point B ? How will the body move after, this moment ?, v, , 7, , A, , 11 15, t, B, C, Sol. The velocity -time graph is a straight line with, -ve slope, the motion is uniformly retarding one, upto point B and uniformly accelerated after, with-ve side of velocity axis ., At point B the body stops and then its direction of, velocity reversed., The initial velocity at point A is v 0 = 7ms -1, The acceleration, a=, , Δv vt - v0 0-7, =, =, = -0.6364ms-2 -0.64ms-2, Δt, Δt, 11, , WE-20: A particle starts from rest and accelerates, as shown in the graph. Determine, a) the particle’s speed at t = 10s and at t = 20s, b) the distance travelled in the first 20s., 2, a m/s, , 2, 1, 15 20, -1, , t sec, , 5 10, , -2, -3, Sol. a) Upto 10 sec the particle moves with uniform, acceleration, hence the velocity at t = 10s,, v1 = u + at = 0 + 2 x 10 = 20m/s, From t = 10s to t = 15s the acceleration is zero,so, The velocity of the particle at t=15s is 20m/s, Velocity at t = 20sec, v2 = v1 + at, = 20 + (-3)5 = 5 m/s, b) Distance travelled in first 10 sec, 1, 1, 2, s1 ut a1t 2 0 10 2 10 100m, 2, 2, Distance travelled when t = 10 sec to t = 15 sec, s 2 vt 20 5 100m, Distance travelled when t = 15 sec to t = 20 sec, 1, s 3 20 5 3 52 62.5m, 2, Total distance travelled in 20 sec = s1 + s2 + s3 =, 100 + 100 + 62.5 = 262.5 m, W.E-21: Velocity (v) versus displacement (x) plot, of a body moving along a straight line is as, shown in the graph. The corresponding plot, of acceleration (a) as a function of displacement (x) is, (EAM-2014), , v, Velocity, , a m / s2, , 0, , 100 200 x, Displacement, , The equation of motion for the variation of, displacement with time is, 1, s = 7t - 0.64t 2 7t 0.32t 2, 2, 48, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , a, , a, , 2), , 1), 0 100 200, Displacement, , x, , O, , 100 200, Displacement, , x, , a, , WE-22: A particle located at x = 0, at time t = 0,, starts moving along the positive x-direction, with a velocity v that varies as v = α x . The, displacement of particle varies with time as, dx, dx, x , dt, Sol., dt, x, x, t, dx, , On integrating, we get , dt, x 0, 0, [ at t = 0, x = 0 and let at any time t, particle be at x], x, , x1/ 2 , , 1/ 2, , t or x t x t 2, 2, 1 / 2 0, , 100 200 x, 4), , 3) O, , x, 100 200, WE-23 : The acceleration (a) of a particle moving, Displacement, Displacement, in a straight line varies with its displacement, Sol. From the given graph equation for velocity v = kx, (S) as a = 2S. The velocity of the particle is, on differentiation, zero, at zero displacement. Find the, dv, kv, ------ (i), corresponding velocity-displacement, dt, equation., dv, 2, = k(kx) = k x;, dv, dt, Sol. a = 2S v 2S vdv 2SdS, 2, ds, a=kx, and v = -kx + v0, O, , v, , dv, , kv k kx v 0 , on differentiation, dt, 2, a = k x - kv0, -------(ii), So, according to the eq. (ii) the shape of a-x graph, is as below, , v, , s, , 0, , S, , v2 , S2 , 2 , 2 0, 2 0, , vdv 2SdS, 0, , 2, , , , v, S2 v 2S, 2, , WE-24: An object moving with a speed of 6.25, m/s, is decelerated at a rate given by, , 0, , dv, = -2.5 v , where v is instantaneous, dt, speed. The time taken by the object, to come, to rest, would be, (AIEEE-2011), , x, , Equations of Motion for Variable, Acceleration :, , , 1, dv, dv 2.5dt, 2.5 v (or), Sol., When acceleration ‘a’ of the particle is a function, v, dt, of time, i.e., a = f(t), On integrating, within limits, dv, (v1 = 6.25 m/s to v2 = 0), , f t dv f t dt, dt, , Integrating both sides within suitable limits, we have, v, , , , t, , t, , t, , v, , dv f t dt v u f t dt, , v1 6.25m/s, , When acceleration ‘a’ of the particle is a function, of distance a = f(x), , 2 v1/2 , , u, , , , v2 0, , , v, , 0, , 0, , dv dx, dv, f x , f x, dt, dx dt, x, , 1/ 2, , dv 2.5 dt, 0, , 0, 6.25, , 2.5 t (or), 1/ 2, , t, x, , 2 6.25, 2.5, , 2s, , 2, 2, vdv f x dx v u 2 f x dx, u, , x0, , NARAYANAGROUP, , x0, , 49
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , WE-25:The velocity of a particle moving in the MOTION UNDER GRAVITY, positive direction of the X-axis varies as Equation of motion for a body projected, V K S where K is a positive constant. vertically downwards :, , When a body is projected vertically downwards, Draw V-t graph., with an initial velocity u from a height h, Sol. V K s, then a = g, s = h, S dS, t, 1 2, dS, K S , K dt, a) v = u + gt, b) h = ut + gt, 0, 0, 2, dt, S, g, 1, c) v2 - u2 = 2gh, d) Sn = u + (2n-1), 2 S Kt and S= K 2t 2, 2, 4, , In case of freely falling body u = 0, a = +g, dS 1 2, 1, 1 2, V , K 2t K 2t, a) v = gt, b) S gt, dt 4, 2, 2, V t, 1, , d) Sn g n , c) v 2 2gS, 2, , , For a freely falling body, the ratio of distances, travelled in 1 second, 2seconds, 3 seconds, .... =, V, 1 : 4 : 9 : 16...., , For a freely falling body, the ratio of distances, travelled in successive seconds = 1 : 3 : 5 : 9 ...., O, , A freely falling body passes through two points A, t, and B in time intervals of t1 and t2 from the start,, V t, then the distance between the two points A and, The V-t graph is a straight line passing through, g 2 2, the origin., B is = t 2 t1, 2, ACCELERATION DUE TO GRAVITY A freely falling body passes through two points A, and B at distances h1 and h2 from the start, then, , The uniform acceleration of a freely falling body, the time taken by it to move from A to B is, towards the centre of earth due to earth’s gravitational, force is called acceleration due to gravity, 2h 2, 2h 1, 2, , It is denoted by ‘g’, , , h 2 h1, t=, g, g, g, , Its value is constant for all bodies at a given place., Two bodies are dropped from heights h1 and h2, It is independent of size, shape, material, nature , simultaneously. Then after any time the distance, of the body., between them is equal to (h2~ h1)., , Its value changes from place to place on the, , A stone is dropped into a well of depth ‘h’, then, surface of the earth., the sound of splash is heard after a time of ‘t’., , It has maximum value at the poles of the earth., 2, The value is nearly 9.83 m/s ., 2h, t1 , , It has minimum value at equator of the earth. The, Time, taken, by, the, body, to, reach, water,, g, value is nearly 9.78 m/s2., Time taken by sound to travel a distance ‘h’,, g, , On the surface of the moon, g moon earth, 6, h, t2 , , The acceleration due to gravity of a body always, Vsound, directed downwards towards the centre of the, Time to hear splash of sound is, earth, whether a body is projected upwards or, downwards., 2h, h, +, , When a body is falling towards the earth, its, t t1 t2 =, g Vsound, velocity increases, g is positive., A stone is dropped into a river from the bridge, , The acceleration due to gravity at the centre of , and after ‘x’ seconds another stone is projected, earth is zero., , , , , , , , 50, , , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , down into the river from the same point with a, velocity of ‘u’. If both the stones reach the water , simultaneously, then S1 t S 2 t x , 1 2, 1, 2, gt u t x g t x , 2, 2, , , , A body is projected vertically up with a velocity, ‘u’ from ground in the presence of constant air, resistance ‘R’. If it reaches the ground with a, velocity ‘v’, then, a) Height of ascent = Height of descent, , mu, mg R, mv, c) Time of descent t d , mg R, d) t a t d, , A body dropped freely from a multistoried building, can reach the ground in t1 sec. It is stopped in its, path after t2 sec and again dropped freely from, the point. The further time taken by it to reach the, , b) Time of ascent t a , , ground is t 3 t12 t 2 2 ., , t2 H2, , H1, , e), , t1, , H3 t3=?, , We know that H1 = H2 + H3, , , 1 2 1 2 1 2, gt1 gt 2 gt 3, 2, 2, 2, , t12 t 2 2 t 32, , , , , t 3 t12 t 22, , Equations of motion of a body Projected, Vertically up :, , , , , , , , Acceleration (a) = -g, , , 1 2, a) v = u - gt, b) s = ut - gt, , 2, g, c) v2 - u2 = -2gh, d) sn = u - (2n-1), 2, Angle between velocity vector and acceleration , vector is 1800 until the body reaches the highest, point., At maximum height, v = 0 and a = g, , u2, H max , H max u 2 (independent of mass , 2g, of the body), A body is projected vertically up with a velocity, ‘u’ from ground in the absence of air resistance, ‘R’. then. t a t d , i) t a t d , , u, g, , 2u, ii) Time of flight T t a t d , g, NARAYANAGROUP, , v, , u, , mg R, v u, mg R, , f) For a body projected vertically up under air, resistance, retardation during its motion is > g, At any point of the journey, a body possess the, same speed while moving up and while moving, down., Irrespective of velocity of projection, all the, g, bodies pass through a height in the last second, 2, of ascent. Distance traveled in the last second of, g, its journey u ., 2, The change in velocity over the complete journey, is ‘2u’ (downwards), If a vertically projected body rises through a height, ‘h’ in nth second, then in (n-1)th second it will rise, through a height (h+g) and in (n+1)thsecond it, will rise through height (h-g)., If velocity of body in nth second is ‘v’ then in, , n 1, , th, , n 1, , th, , second it is (v + g) and that in, , second is (v - g) while ascending, A body is dropped from the top edge of a tower, of height ‘h’ and at the same time another body is, projected vertically up from the foot of the tower, with a velocity ‘u’., u=0, , h, a) The separation between them after ‘t’ seconds, is = h ut , , 51
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, b) The time after which they meet t , , h, u, , , , c) The height at which they meet above the, , , , a) If u2 u1 , the time after which both the bodies, will meet with each other is, h1 h2, 1, 1, 2, u1 x gx 2 u 2 x t g x t , 2, 2, 1, u 2 t gt 2, 2, for the first body.., x, u 2 u1 gt, , , gh 2 , h, , , , ground =, 2u 2 , , d) The time after which their velocities are equal, u, in magnitudes is t , 2g, e) If they meet at mid point, then the velocity of, thrown, body is u gh and its velocity of meeting is, zero, f) Ratio of the distances covered when the, magnitudes of their velocities are equal is 1 : 3., A body projected vertically up crosses a point P, at a height ‘h’ above the ground at time ‘ t1 ’, seconds and at time t2 seconds ( t1 and t2 are, measured from the instant of projection) to same, point while coming down., , b) If u1 u 2 u , the time after which they meet, , u t , for the first body and, g, 2, , , is , , 52, , u 2 gt 2, , 2g 8, A rocket moves up with a resultant acceleration, a. If its fuel exhausts completely after time ' t ', seconds, the maximum height reached by the, rocket above the ground is, h h1 h2, Height at which they meet =, , , , h t1, , 2, , u t , , g 2, , for the second body., , t2, , h ut 1/ 2 gt 2 ; gt 2 2ut 2h 0, This is quadratic equation in t, 2u, Sum of the roots, t1 t2 , (time of flight), g, 1, Velocity of projection , u g t1 t 2 , 2, 2h, Product of the roots, t1t2 , g, 1, Height of P is h gt1t 2, 2, Maximum height reached above the ground, 1, 2, H g t1 t 2 , 8, Magnitude of velocity while crossing P is, g t 2 t1 , , A body is projected vertically up with velocity u1, and after ‘t’ seconds another body is projected, vertically up with a velocity u2 ., , 1 2 1 2 2, a t, = at +, 2, 2g, , , , , v2 a 2t 2 , h, =, 2 2g = 2g , v at , , , , 1 a, h at 2 1 , 2 g, An elevator is accelerating upwards with an, acceleration a. If a person inside the elevator, throws a particle vertically up with a velocity u, relative to the elevator, time of flight is t , , , , In the above case if elevator accelerates down,, time of flight is t , , , , 2u, ga, , 2u, ga, , The zero velocity of a particle at any instant does, not necessarily imply zero acceleration at that, instant. A particle may be momentarily at rest and, yet have non-zero acceleration., For example, a particle thrown up has zero velocity, at its uppermost point but the acceleration at that, instant continues to be the acceleration due to, gravity., NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, WE-26: Drops of water fall at regular intervals from, the roof of a building of height h = 16m., The first drop striking the ground at the same, moment as the fifth drop is ready to leave from, the roof. Find the distance between the, successive drops., Sol. Step-I : Time taken by the first drop to touch the, ground = t , , 2h, g, , 16 2 , , 2, g, g, Time interval between two successive drops is, , For h = 16m, t , , 4, , 2, 1 1, t , t t , g, n 1 4 , Where n = number of drops, Step-II :, Distance travelled by 1st drop, , 2, 1, 1, 2, g 4t g 16 16m, 2, 2, g, nd, Distance travelled by 2 drop, S1 , , 2, 1, 1, 2, g 3 t g 9 9 m, 2, 2, g, Distance travelled by 3rd drop, S2 , , 2, 1, 1, 2, g 2 t g 4 4 m, 2, 2, g, th, Distance travelled by 4 drop, S3 , , MOTION IN A STRAIGHT LINE, V = gt = 10 2 = 20 m/s, Now at this instant gravity ceases to act,, there after velocity becomes constant., The remaining distance which is 125 - 20 = 105m, is covered by body with constant velocity20 m/s, Time taken to cover 105 m with constant velocity, is given by, S, 105, t1 t1 , 5.25s, V, 20, Total time taken = 2 + 5.25 = 7.25 s, WE-28: A parachutist drops freely from an, aeroplane for 10 seconds before the parachute, opens out. Then he descends with a net, retardation of 2 m/s2. His velocity when he, reaches the ground is 8 m/s. Find the height, at which he gets out of the aeroplane ?, Sol. Distance he falls before the parachute opens is, 1, S1 g 100 490m, 2, Then his velocity , u=gt = 98.0 m/s, Velocity on reaching ground = v=8 m / s, retardation = 2 m / s2, S2 2385m, v 2 u 2 2aS 2, Total distance S S1 S 2 = 2385 + 490, =2875 m( height of aeroplane), WE-29: If a freely falling body covers half of its, total distance in the last second of its journey., Find its time of fall, Sol. Suppose t is the time of free fall, g, g, S n 2n 1 and S n2 , 2, 2, g, 2n 1 , 2 1, Sn, 2 g, 2, n , S, 2, , 1, 1 2 , 2, g t g 1m, 2, 2 g , st, nd, Distance between 1 and 2 drops=, S1 S 2 16 9 7m, Distance between 2nd and 3rd drops=, S 2 S 3 9 4 5m, Distance between 3rd and 4th drops=, n 2 4n 2 0 and n 2 2 sec, S 3 S 4 4 1 3m, WE-30:A body is projected vertically up with a velocity, Distance between 4th and 5th drops=, u. Its velocity at half of its maximum height and, S 4 S 5 1 0 1m, at 3/4th of its maximum height are, WE-27: A body falls freely from a height of 125m, 2, Sol. From v2 - u2 = 2aS, here a = –g; when S = H/2, then, (g = 10 m/s ). After 2 sec gravity ceases to, act Find time taken by it to reach the, u2, u, ground ?, v 2 u 2 2 g , v, 4g, 2, Sol. 1) Distance covered in 2s under gravity, When S = 3H/4, then, 1, 1, 2, S1 gt 2 10 2 20m, 2, 2, 3u 2, u, 2, 2, v, , u, , 2, , g, v, , , Velocity at the end of 2s, 4 2g, 2, S4 , , , , , , , , NARAYANAGROUP, , 53
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , WE-31 : A stone is allowed to fall from the top of a Body Projected Vertically up from a Tower, tower 300m height and at the same time an- , A body projected vertically up from a tower of, other stone is projected vertically up from the, height ‘h’ with a velocity ‘u’ (or) a body dropped, from a rising balloon (or) a body dropped from, ground with a velocity 100 m/s. Find when, an helicopter rising up vertically with constant, and where the two stones meet ?, velocity ‘u’ reaches the ground exactly below the, point of projection after a time ‘t’. Then, 300-x, , 300 m, x, , h, , height of the tower, h= 300m, Suppose the two stones meet at a height x from, ground after t seconds, t, , -X, , +X, , Sol., , T, O, W, E, R, , -h, , 1, , (a) Height of the tower is h ut gt 2, 2, (b) Time taken by the body to reach the ground, , Sr, ur , u r u 0 u , S r h, , t=, , h 300, , 3sec, u 100, height of the stone from the point of projection is, t, , u u 2 2 gh, g, , (c) The velocity of the body at the foot of the tower, v = u 2 2gh, (d) Velocity of the body after ‘t’ sec. is, v u gt, (e) Distance between the body and balloon after, , 1 2, 1, g t 100 3 9.8 9 255.9m, 2, 2, WE-32: A stone is dropped from certain height, 1 2, above the ground. After 5s a ball passes, this time = gt, 2, through a pane of glass held horizontally and, WE-33:A ball is thrown vertically upwards from, instantaneously loses 20% of its velocity. If, the top of a tower. Velocity at a point ‘h’ m, the ball takes 2 more seconds to reach the, vertically below the point of projection is, ground, the height of the glass above the, twice the downward velocity at a point ‘h’ m, ground is, vertically above the point of projection. The, Sol. In 5s velocity gained v = gt = 50 m/s. Velocity, maximum height reached by the ball above, after passing through the glass pane, the top of the tower is, (MED-2012), 80, Sol. If AB is the tower then according to the problem,, 50 40m / s, 100, velocity at ‘P’ is given as twice the velocity at ‘Q’, X = u t-, , VP , , u, , 2, , 2 gh ; VQ , , u, , 2, , 2 gh ; VP 2 VQ , , C, H Q, h, , A, T, O, W, E, R, , Height of the glass pane above the ground is, 1, 1, 2, h ut gt 2 = 40 2 10 2 = 100m, 2, 2, 54, , h, P, , B, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , u, , 2, , 2gh , , 2, , , , , , 2, u 2gh u , , 2, , MOTION IN A STRAIGHT LINE, , , 10gh, 3, , Three bodies are projected from towers of same, height as shown. 1st one is projected vertically up, with a velocity ‘u’. The second one is thrown down, vertically with the same velocity and the third one, is dropped as a freely falling body. If t1, t2, t3 are, the times taken by them to reach ground, then,, , From the top of the tower maximum height reached, , 10gh , , , u, 3 5h, H, H, , 2g, 2g, 3, 2, , WE-34: From a tower of height H, a particle is, thrown vertically upwards with a speed u., The time taken by the particle to hit the, ground is n times that taken by it to reach, the highest point of its path. The relation, between H, u and n is ( jee main- 2014), Sol. Time taken to reach the maximum height t1 , , h, , u, g, , h, , 2, , For 1st body, h ut1 , , 3, 1 2, gt1, 2, , 1 2, For 2nd body, h ut2 gt2, 2, 1, 2, For 3rd body, h gt3, 2, from 1 t2 + 2 t1, , nu 1 n 2 u 2 , , , H, u, g 2 , But t2 = nt1 ; So,, g 2 g , nu 2 1 n 2 u 2 , H , , , 2 gH nu 2 n 2 , g 2 g , , 1, 2, 3, , 1, gt1t2 t1 t2 , 2, 1, i) Height of the tower h gt1t2 4 , 2, ii) From eq (3) & (4),, , WE-35 : A balloon starts from rest, moves vertically upwards with an acceleration g/8 ms-2., A stone falls from the balloon after 8 s from, the start. Further time taken by the stone to, reach the ground (g = 9.8 ms-2) is, Sol. The distance of the stone above the ground about, which it begins to fall from the balloon is, , h t1 t2 , , 5, t3 t1t2, iii) Equating R.H.S of (1) & (2),, 1, velocity of projection u g t1 t 2 6 , 2, iv) Time difference between first two bodies to, 2u, 7, reach the ground t , g, , h, , NARAYANAGROUP, , h, , 1, , If t2 is the time taken to hit ground, 1 2, i.e., H ut 2 gt, 2, , 1g 2, 8 4g, 28, The velocity of the balloon at this height can be, g, obtained from v = u + at; v 0 8 g, 8, This becomes the initial velocity (u) of the stone, as the stone falls from the balloon at the height h., u ' g, 1 2, 1, For the total motion of the stone h = -u t + gt, 2, 1 2, 4g gt gt and t 2 2 t 8 0, 2, Solving for ‘t’ we get t = 4 and -2s. Ignoring negative value of time, t = 4s., , u=0, , u=u, , Relative Motion in one dimension, , , , , , Velocity of one moving body with respect to other, moving body is called Relative velocity., A and B are two objects moving uniformly with, average velocities vA and vB in one dimension, say, along x-axis having the positions xA(0) and xB(0), at t = 0., If xA(t) and xB(t) are positions of objects A and, B at time t then, x A t x A 0 vA t ; x B t x B 0 vB t, 55
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MOTION IN A STRAIGHT LINE, , , JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , The displacement from object A to B is given by, , x BA t x B t x A t , , at 2 2vt 2d 0 t , , v v 2 2ad, a, , x B 0 x A 0 v B v A t, , for minimum velocity v 2 2ad 0 v 2ad, x BA 0 v B v A t, WE.37. Two trains, each travelling with a speed of, , , Velocity of A w.r.t B is v AB = v A - vB, 37.5kmh 1 , are approaching each other on, , the same straight track. A bird that can fly at, v AB = v 2A + v 2B - 2v A v B cosθ, 60kmph flies off from one train when they, , Two bodies are moving in a straight line in same, are 90 km apart and heads directly for the, , , , direction then, v AB = v A - v B ( 00 ), other train. On reaching the other train, it, flies back to the first and so on. Total, , Two bodies are moving in a straight line in opposite, , , , distance covered by the bird before trains, direction then, v AB = v A + v B , ( 1800 ), collide is, , If two bodies move with same velocity and in same, 1, direction, then position between them does not Sol. Relative speed of trains 37.5 37.5 75kmh, vary with time., S r 90 6, , If two bodies move with unequal velocity and in, Time taken by them to meet t u h, 75 5, same direction, then position between them first, r, decreases to minimum and then increases., Distance travelled by the bird, x Vbird t, , If the particles are located at the sides of n sided, symmetrical polygon with each side a and each, 6, particle moves towards the other, then time after, 60 72km ( Vbird 60kmh 1 ) ,, 5, which they meet is, WE.38:, On, a, two-lane, road, car A is travelling with, Initial separation, a, speed, of, 36, kmph., Two cars B and C, T, Re lative velocity of approach, approach car A in opposite directions with, speed of 54 kmph each. At a certain instant,, a, a, when the distance AB is equal to AC, both, T, T, , , , being 1 km, B decides to overtake A before, v 1 cos , v v cos , C does. What minimum acceleration of car, n , n , , B is required to avoid an accident?, a, and T , Sol. Velocity of a car A, VA 36km / h 10m / s, 2, 2v sin , Velocity of car B, VB 54km / h 15m / s, n, Shortcut to solve the problems, Velocity of car C, VC 54km / h 15m / s, 2a, Relative velocity of car B with respect to car A,, For Triangle n = 3 T , ;, 3v, VBA VB VA 15 10 5m / s, a, Relative velocity of car C with respect to car A,, For Square n = 4 T , v, VCA VC VA 15 10 25m / s, 2a, At a certain instance, both cars B and C are at the, For hexagon, n = 6 T , v, same distance from car A i.e., s 1km 1000m, WE.36: A passenger is at a distance ‘d’ from a bus,, Time taken (t) by car C to cover 1000m is, when the bus begins to move with a con1000, stant acceleration a. Then find the minimum, t, 40s, constant velocity with which the passenger, 25, should run towards the bus so as to catch is, The acceleration produced by car B is, Sol., 56, , 1 2, S passenger d Sbus ; vt d at, 2, , 1, 1600, 1000 5 40 at 2 a , 1m / s 2, 2, 1600, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , WE.39: A body is at rest at x = 0. At t = 0, it starts, moving in the positive x-direction with a, constant acceleration. At the same instant, another body passes through x = 0 moving in, the positive x-direction with a constant speed., , C.U.Q, DISTANCE AND DISPLACEMENT, 1., , The position of the first body is given by x1 t , after time ‘t’ and that of the second body by, , x2 t after the same time interval. Which of, the following graphs correctly describes 2., , x1 x2 as a function of time ‘t’ ?[AIEEE 2008], , x1 x2 , , x1 x2 , 3., , 2), , 1), O, , t, , O, , x1 x2 , , t, , x1 x2 , , The numerical ratio of displacement to, distance is, 1) always less than 1 2) always greater than 1, 3) always equal to 1, 4) may be less than 1 or equal to one, The location of a particle is changed. What, can we say about the displacement and, distance covered by the particle?, 1) Both cannot be zero, 2) One of the two may be zero, 3) Both must be zero 4) Both must be equal, Consider the motion of the tip of the minute, hand of a clock. In one hour, a) the displacement is zero, b) the distance covered is zero, c) the average speed is zero, d) the average velocity is zero, 1) a & b are correct 2) a,b & c are correct, 3) a & d are correct 4) b,c & d are correct, , SPEED AND VELOCITY, 4., , 4), , 3), t, , O, , O, , t, , 1 2, Sol. As, x1 t = at and x2 t vt, 2, , 5., , 1 2, at - vt (parabola), 2, Clearly, graph (2) represents it correctly., x1 - x 2 =, , WE.40.A particle has an initial velocity 3iˆ 4ˆj and 6., an acceleration of 0.4iˆ 0.3jˆ . It speed after, 10s is, (AIEEE-2009), , Sol. v = u + at, 3iˆ 4 ˆj 0.4iˆ 0.3 ˆj 10, , , , , , , , 3iˆ 4 ˆj 4iˆ 3 ˆj 7iˆ 7 ˆj, , v = 49 + 49 7 2, , NARAYANAGROUP, , The numerical value of the ratio of average, velocity to average speed is, 1) always less than one 2) always equal to one, 3) always more than one, 4) equal to or less than one., If a particle moves in a circle describing equal, angles in equal intervals of time, then the, velocity vector, 1) remains constant. 2) changes in magnitude., 3) changes in direction., 4) changes both in magnitude and direction., In which of the following examples of motion,, can the body be considered approximately a, point object, a) a railway carriage moving without jerks, between two stations., b) a monkey sitting on top of a man cycling, smoothly on a circular track, c) a spinning cricket ball that turns sharply, on hitting the ground, d) a trembling beaker that has slipped off the, edge of a table, 1) a,b, 2) b,c, 3)a,c, 4)b,d, , 57
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 7., , 8., , 13., An object may have, a) varying speed without having varying velocity, b) varying velocity without having varying, speed, c) non zero acceleration without having varying velocity, d) non zero acceleration without having varying speed., 1) a,b & c are correct 2) b & d are correct, 3) a,b & d are correct 4) a & d are correct, The distance travelled by a particle in a, straight line motion is directly proportional 14., to t1/2, where t = time elapsed. What is the, nature of motion ?, 1) Increasing acceleration, 2) Decreasing acceleration, 3) Increasing retardation, 4) Decreasing retardation, , ACCELERATION, 9., , 10., , 11., , 12., , 58, , If a body starts from rest, then the time in, which it covers a particular displacement with, uniform acceleration is, 1) inversely proportional to the square root of the, displacement, 2) inversely proportional to the displacement, 3) directly proportional to the displacement, 4) directly proportional to the square root of the, displacement, Check up only the correct statement in the, following., 1) A body has a constant velocity and still it can, have a varying speed, 2) A body has a constant speed but it can have a, varying velocity, 3) A body having constant speed cannot have any, acceleration., 4) None of these., When the speed of a car is u, the minimum, distance over which it can be stopped is s. If, the speed becomes nu, what will be the, minimum distance over which it can be, stopped during the same time ?, 1) s/n, 2) ns, 3) s/n2, 4) n2s., The distance covered by a moving body is, directly proportional to the square of the time., The acceleration of the body is, 1) increasing, 2) decreasing, 3) zero, 4) constant, , 15., , 16., , Mark the incorrect statement for a particle, going on a straight line., 1) If the velocity and acceleration have opposite, sign, then the object is slowing down., 2) If the position and velocity have opposite sign,, then the particle is moving towards the origin., 3) If the velocity is zero at an instant, then the, acceleration should also be zero at that instant., 4) If the velocity is zero for a time interval,then, the acceleration is zero at any instant within the, time interval., , MOTION UNDER GRAVITY, B 1, B2 and B3 are three balloons ascending, , with velocities v, 2v and 3v, respectively. If, a bomb is dropped from each when they are, at the same height, then, 1) bomb from B1 reaches ground first, 2) bomb from B2 reaches ground first, 3) bomb from B3 reaches ground first, 4) they reach the ground simultaneously, The distances moved by a freely falling body, during 1st, 2nd, 3rd,......nth second of its motion, are proportional to, 1) even numbers, 2) odd numbers, 3) all integral numbers, 4) squares of integral numbers, To reach the same height on the moon as on, the earth, a body must be projected up with, 1) higher velocity on the moon., 2) lower velocity on the moon., 3) same velocity on the moon and earth., 4) it depends on the mass of the body., , 17., , At the maximum height of a body thrown, vertically up, 1) velocity is not zero but acceleration is zero., 2) acceleration is not zero but velocity is zero., 3) both acceleration and velocity are zero., 4) both acceleration and velocity are not zero., , 18., , A ball is dropped freely while another is, thrown vertically downward with an initial, velocity ‘v’ from the same point, simultaneously. After ‘t’ second they are, separated by a distance of, 1), , vt, 2, , 2), , 1 2, gt, 2, , 3) vt, , 1 2, 4) vt gt, 2, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, 19., , 20., , 21., , 22., , MOTION IN A STRAIGHT LINE, , The average velocity of a freely falling body, is numerically equal to half of the acceleration, due to gravity. The velocity of the body as it, reaches the ground is, g, g, 4) 2g, 1) g, 2), 3), 26., 2, 2, Two bodies of different masses are dropped, simultaneously from the top of a tower. If air, resistance is proportional to the mass of the, body, then,, 1) the heavier body reaches the ground earlier., 2) the lighter body reaches the ground earlier., 3)both the bodies reach the ground simultaneously., 4) cannot be decided., A man standing in a lift falling under gravity 27., releases a ball from his hand. As seen by, him, the ball, 1) falls down, 2) remains stationary, 28., 3) goes up, 4) executes SHM, A particle is dropped from certain height. The, time taken by it to fall through successive, distances of 1 m each will be, 1) all equal, being equal to 2 / g second, 29., 2) in the ratio of the square roots of the integers, 1,2, 3, ..........., 3) in the ratio of the difference in the square roots, 30., of the integers, i.e.,, 1,, , 23., 24., , 25., , , , , , 2 1 ,, , , , 3 2 ,, , , , 4 3 ,......, , 4) in the ratio of the reciprocals of the square roots, 1 1 1, ,, ,, ,......, of the integers, i.e.,, 1 2 3, A body, freely falling under gravity will have, uniform, 1)speed 2)velocity 3)momentum 4)acceleration, A person standing near the edge of the top of, a building throws two balls A and B. The ball, A is thrown vertically upward and B is thrown, vertically downward with the same speed, The, ball A hits the ground with a speed VA and, the ball B hits the ground with a speed VB . 31., then, 1) VA VB, 2) VA VB, 3) VA VB, 4) the relation between VA and VB depends on, height of the building above the ground, A lift is coming from 8th floor and is just about, to reach 4th floor. Taking ground floor as, origin and positive direction upwards for all, , NARAYANAGROUP, , quantities, which one of the following is, correct?, 1) x<0, v<0, a>0, 2) x>0, v<0, a<0, 3) x>0, v<0, a>0, 4) x>0, v>0, a<0, , GRAPHS, Choose the correct statement :, 1) The area of displacement - time graph gives, velocity., 2) The slope of velocity - time graph gives, acceleration., 3) The slope of displacement - time graph gives, acceleration., 4) The area of velocity - time graph gives average, velocity., Velocity-time graph of a body thrown, vertically up is, 1) a straight line, 2) a parabola, 3) a hyperbola, 4) circle, Velocity - displacement graph of a freely, falling body is, 1) straight line passing through the origin, 2) straight line intersecting ‘x’ and ‘y’ axes, 3) parabola, 4) hyperbola, Displacement - time graph of a body projected, vertically up is, 1) a straight line, 2) a parabola, 3) a hyperbola, 4) a circle, The displacement - time graphs of two bodies, A and B are OP and OQ respectively. If, POX is 600 and QOX is 450, the ratio of, the velocity of A to that of B is, P, Y, Q, , d, , 0, , 60, , 0, , O, , 45, t, , X, , 1) 3 : 2 2) 3 : 1 3) 1: 3 4) 3:1, If the distance travelled by a particle and, corresponding time be laid off along y and x, axes respectively, then the correct statement, of the following is, 1) the curve may lie in fourth quadrant, 2) the curve lies in first quadrant, 3) the curve exhibits peaks corresponding to, maxima, 4) the curve may drop as time passes, 59
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , 33., , 34., , 35., , In relation to a velocity - time graph, 1) a,c,d, 2) b,c,d 3) a,b 4) c,d, 1) the curve can be a circle, 37. The displacement-time graph of a moving, 2) the area under the curve and above the time, particle is shown in Fig. The instantaneous, axis between any two instants gives the average, velocity of the particle is negative at the point, acceleration, 3) the slope at any instant gives the rate of change, of acceleration at that instant, 4) the area under the curve and above the time, axis gives the displacement, The displacement - time graph of a particle, D, moving with respect to a reference point is a, F, E, C, straight line, 1)the reference point is stationary with zero velocity, 2) the acceleration of the object is zero, O, Time, 3) body moves with uniform velocity, 1) D, 2) F, 3) C, 4) E, 4) all the above, 38. Which of the following option is correct for, For a uniform motion, having a straight line motion represented by, 1) the velocity - time graph is a straight line parallel, displacement-time graph., to time axis, t, 2) the position - time graph is a parabola, D, 3) the acceleration - time graph is a straight line, inclined with time axis, C, 4) the position - time graph is a straight line, Figure shows the displacement- time graph, of a particle moving on the x-axis, B, x, , Displacement, , 32., , A, , 36., , t, t0, 1) the particle is continuously going in positive X 39., direction, 2) the particle is at rest, 3) the velocity increases up to a time t0 and then, becomes constant., 4) the particle moves at constant velocity up to a, time t0 and then stops., The variation of quantity A with quantity B., plotted in the Fig. describes the motion of a, particle in a straight line., a) Quantity B may represent time., b) Quantity A is velocity if motion is uniform., c) QuantityA is displacement if motion is uniform., d) Quantity A is velocity if motion is uniformly, accelerated., v, A, O, , 60, , B, , t, , Displacement, , O, , s, O, 1) The object moves with constantly increasing, velocity from O to A then it moves with constant, velocity., 2) Velocity of the object increases uniformly., 3) Average velocity is zero., 4) The graph shown is impossible., The displacement of a particle as a function of, time is shown in the figure. The figure shows that, , 2, 1, , 1, , 3, 4, 2, Time in seconds, 1) the particle starts with certain velocity but the, motion is retarded and finally the particle stops, 2) the velocity of the particle is constant through out, 3) the acceleration of the particle is constant, throughout, 4) the particle starts with constant velocity, then, motion is accelerated and finally the particle moves, with another constant velocity., NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, 40., , MOTION IN A STRAIGHT LINE, , A uniform moving cricket ball is turned back, by hitting it with a bat for a very short time, interval.Show the variation of its acceleration, with time. (Take acceleration in the back ward, 1., direction as positive), a, a, , 2., , 1., , 2., , t, , t, a, , a, , 3., , 4., t, , t, , 3., , RELATIVE VELOCITY, 41., , 42., , A small body is dropped from a rising balloon., A person A stands on ground, while another, person B is on the balloon. Choose the correct, statement : Immediately, after the body is, released., 1) A and B, both feel that the body is coming, (going) down., 4., 2) A and B, both feel that body is coming up., 3) A feels that the body is coming down, while B, feels that the body is going up, 4) A feels that the body is going up, while B feels, that the body is going down., Seeta is moving due east with a velocity of, v1 m / s and Geeta is moving due west with a, velocity of v 2 m / s . The velocity of Seeta with 5., respect to Geeta is, 1) v1 v 2 due east, , 2) v1 v 2 due east, , 3) v1 v 2 due west, , 4) v1 v 2 due west, , C.U.Q - KEY, 1) 4, 7) 2, 13) 3, 19) 1, 25) 1, 31) 2, 37) 4, , 2) 1, 8) 4, 14) 1, 20) 3, 26) 2, 32) 4, 38) 3, , NARAYANAGROUP, , 3) 3, 9) 4, 15) 2, 21) 2, 27) 1, 33) 4, 39) 1, , 4) 4, 10) 2, 16) 2, 22) 3, 28) 3, 34) 1, 40) 1, , 5) 3, 11) 4, 17) 2, 23) 4, 29) 2, 35) 4, 41) 4, , 6) 1, 12) 4, 18) 3, 24) 3, 30) 2, 36) 1, 42) 1, , 6., , LEVEL - I (C.W), DISPLACEMENT AND DISTANCE, A body is moving along the circumference of, a circle of radius ‘R’ and completes half of, the revolution. Then, the ratio of its, displacement to distance is, 1) : 2 2) 2:1, 3) 2 : 4) 1:2, A body completes one round of a circle of, radius ‘R’ in 20 second. The displacement of, the body after 45 second is, R, 1), 2) 2 R 3) 2 R 4) 2R, 2, SPEED AND VELOCITY, If a body covers first half of its journey with, uniform speed v1 and the second half of the, journey with uniform speed v2, then the, average speed is, 2 v1v 2, 1) v1 v2, 2) v v, 1, 2, v1v 2, 3) v v, 4) v1v2, 1, 2, A car is moving along a straight line, say OP, in figure. It moves from O to P in 18 s and, return from P to Q in 6 s. What are the average velocity and average speed of the car in, going from O to P and back to Q?, , 1) 10 m s 1 , 20 m s 1 2) 20m s 1 ,10 m s 1, 3) 10 m s 1 ,10 m s 1 4) 20m s 1 , 20m s 1, For a body moving with uniform acceleration, ‘a’, initial and final velocities in a time interval ‘t’ are ‘u’ and ‘v’ respectively. Then, its, average velocity in the time interval ‘t’ is, at , at , , , 1) v at 2) v 3) v at 4) u , 2, 2, , , ACCELERATION, A body moves with a velocity of 3m/s due, east and then turns due north to travel with, the same velocity. If the total time of travel, is 6s, the acceleration of the body is, 1) 3 m/s2 towards north west, 1, 2), m/s2 towards north west, 2, 3) 2 m/s2 towards north east 4) all the above, 61
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , If a body travels 30m in an interval of 2s and 16., 50m in the next interval of 2s, then the, acceleration of the body is, 1) 10 m/s2 2) 5 m/s2 3) 20 m/s2 4) 25 m/s2, A bullet travelling horizontally looses 1/20th, of its velocity while piercing a wooden plank., Then the number of such planks required to, stop the bullet is, 1) 6, 2) 9, 3) 11, 4) 13, If S n 2 0.4n find initial velocity and, acceleration, 1) 2.2 units, 0.4 units 2) 2.1 units, 0.3 units, 3) 1.2 units, 0.4 units 4) 2.2 units, 0.3 units, A particle starts moving from rest under, uniform acceleration. It travels a distance ‘x’, in the first two seconds and a distance ‘y’ in, the next two seconds. If y = nx, then n=, (1993 E), 1) 1, 2) 2, 3) 3, 4) 4, A particle is moving in a straight line with initial velocity ‘u’ and uniform acceleration ‘a’., If the sum of the distances travelled in tth and, (t+1)th second is 100cm, then its velocity after ‘t’ seconds in cm/s is, 1) 20, 2) 30, 3) 80, 4) 50, A particle is moving with uniform acceleration along a straight line ABC. Its velocity at, ‘A’ and ‘B’ are 6 m/s and 9 m/s respectively., If AB : BC = 5 : 16 then its velocity at ‘C’ is, 1) 9.6 m/s 2) 12 m/s 3) 15 m/s 4) 21.5 m/s, A car moving on a straight road accelerates, from a speed of 4.1 m/s to a speed of 6.9 m/s, in 5.0 s. Then its average acceleration is, 1) 0.5m/s2 2) 0.6m/s2 3) 0.56m/s2 4) 0.65m/s2, MOTION UNDER GRAVITY, A body projected vertically upwards with a, velocity of 19.6 m/s reaches a height of 19.8, m on earth. If it is projected vertically up with, the same velocity on moon, then the maximum height reached by it is, 1) 19.18 m2) 3.3 m 3) 9.9 m 4) 118.8 m, A ball is thrown straight upward with a speed, v from a point h meter above the ground. The, time taken for the ball to strike the ground is, v, , 2hg , , v, , 2hg , , 1) g 1 1 v 2 , , , 3) g 1 1 v 2 , , , 62, , v, , 2hg , , 2) g 1 1 v2 , , , 4), , 17., , 18., , 19., , A ball is dropped on the floor from a height of, 10m. It rebounds to a height of 2.5m. If the, ball is in contact with the floor for 0.01 s, then, the average acceleration during contact is, nearly, 1) 500 2m / s 2 upwards, 2) 1800 2m / s2 downwards, 3) 1500 2m / s2 upwards, 4) 1500 2m / s2 downwards, A body falling from rest has a velocity ‘v’ after it falls through a distance ‘h’. The distance, it has to fall down further, for its velocity to, become double, is ..... times ‘h’., 1) 5, 2) 1, 3) 2, 4) 3, RELATIVE VELOCITY, A ball is dropped from a building of height, 45m. Simultaneously another ball is thrown, up with a speed 40m/s. The rate of change of, relative speed of the balls is, 1) 20 ms 1 2) 40 ms 1 3) 30 ms 1 4) 0 ms 1, Two cars 1 & 2 starting from rest are moving, with speeds v1 and v 2 m/s (v1 > v 2 ) . Car 2 is, ahead of car ‘1’ by s meter when the driver, of the car ‘1’ sees car ‘2’. What minimum retardation should be given to car ‘1’ to avoid, collision., (2002 A), v v, v v, 1) 1 2, 2) 1 2, s, s, 3), , 20., , v1 v 2 , , 2, , 4), , v1 v 2 , , 2, , 2s, 2s, Two cars are travelling towards each other, on a straight road at velocities 15 m/s and 16, m/s respectively. When they are 150m apart,, both the drivers apply the brakes and the cars, decelerate at 3 m/s2 and 4 m/s2 until they stop., Separation between the cars when they come, to rest is, 1) 86.5 m 2) 89.5 m 3) 85.5 m 4) 80.5 m, , LEVEL - I (C.W) - KEY, 01) 3, 07) 2, 13) 3, 19) 4, , 02) 2, 08) 3, 14) 4, 20) 4, , 03) 2 04) 1 05) 2, 09) 1 10) 3 11) 4, 15) 1 16) 3 17) 4, , 06) 2, 12) 3, 18) 4, , v, 2hg , 2 2 , , g, v , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , LEVEL - I (C.W) - HINTS, 11., , C, 1., , A, 2., , 3., 4., , 5., 6., , B, , 1, , Displacement : Distance R : 2R, In 40sec body completes two revolutions., In 5 sec it covers 1/4 th of the circle and angle, , , traced is . So displacement s 2 R sin, 2, 2, s1 s 2, 2v1 v 2, Average speed = t t v = v + v, 1, 2, 1, 2, total displacement s1 s 2, v avg =, = t t, total time, 1, 2, vavg, , 12., , v1 5x v2 16x v3, , 13., , a, , 14., 15., , 19., 20., , N, W, , -3i , 3i, , 8., 9., , 10., , v u ft, , ; a, , v22 v12 v32 v22, , 2s1, 2 s2, , , , 2gh 2 2gh1, , , , 17. v 2 2gh ; 4v 2 2gx, Relative acceleration is zero as ‘g’ is downwards, for the both the bodies., 2, 2, urel v1 v2 ; vrel 0 ; vrel, urel, 2as, u1, v1=0, v2=0 u2, t, , s1, , s, , s2, , v12 u12 2a1 s1 ; v22 u22 2a2 s2, , E, , s s s1 s2 ., S, , LEVEL - I (H.W), , ; v i = v1 i, vf = v 2 j, , Δv = vf - vi = v 2 j- v1 i, , 7., , t, 2t 1, 2, , s, , 4j, , s1, t1, , st 1 u , , vu, t, 2, hE gM, 1, u, h, ; h h g, g, 2g, M, E, 1 2, h vt gt ; gt 2 2vt 2h 0, 2, , 16. a=, 18., , v u v v at, , , 2, 2, , vu, a, t, , t, 2t 1 ;, 2, st st 1 100 ;, A s B s2 C, st u , , s2, t2, , ;, , v2, ; tan θ = v, 1, , 1, s1 ut1 t12, 2, , 1, 2, s1 s2 u t1 t2 a t1 t2 , 2, 1 1, n2, , n 20 ; no. of planks =, n 20, 2n 1, 1 , , 1, sn u an a u a an ..........(1), 2 , 2, , sn 2 0.4n, ..........(2), 1, from (1) and (2) 0.4n an ; u a 2, 2, 1, 1, 2, 2, x a 2 ; x y a 4, 2, 2, , NARAYANAGROUP, , 1., , 2., , 3., , DISPLACEMENT AND DISTANCE, A body moves from one corner of an, equilateral triangle of side 10 cm to the same, corner along the sides. Then the distance and, displacement are respectively, 1) 30 cm & 10 cm, 2) 30 cm & 0 cm, 3) 0 cm & 30 cm, 4) 30 cm & 30 cm., SPEED AND VELOCITY, For a train that travels from one station to, another at a uniform speed of 40 kmh–1 and, returns to final station at speed of 60 kmh–1,, then its average speed is, 1) 98 km/hr, 2) 0 km/hr, 3) 50 km/hr, 4) 48 km/hr, If the distance between the sun and the earth, is 1.5x1011 m and velocity of light is 3x108 m/, s, then the time taken by a light ray to reach, the earth from the sun is, 1) 500 s 2) 500 minute 3) 50 s 4) 5 103s, , 63
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, ACCELERATION, 4., , 5., , 6., , 7., , A body is moving with velocity 30ms 1, towards east. After 10s its velocity becomes, 40ms1 towards north. The average, acceleration of the body is [AIPMT 2011], 13., 1) 7 ms 2 2) 7 ms 2 3) 5 ms 2 4) 1ms 2, A body starting with a velocity ‘v’ returns to, its initial position after ‘t’ second with the, same speed, along the same line., Acceleration of the particle is, 2v, v, t, 2) zero, 3), 1), 4), t, 2t, 2v, A body starting from rest moving with uniform, acceleration has a displacement of 16 m in, first 4 s and 9 m in first 3 s. The acceleration, 14., of the body is, –2, –2, –2, –2, 1) 1 ms 2) 2 ms 3) 3 ms 4) 4 ms, A body starts from rest and moves with an, uniform acceleration. The ratio of distance, covered in the nth second to the distance, covered in ‘n’ second is, 2 1, 2 1 , 1 1, 2 1, 1) n n 2 2) n 2 n 3) n 2 n 4) 2, , , , , , , n n, , 8., , 9., , 10., , 11., , 64, , 12., , A bus accelerates uniformly from rest and, acquires a speed of 36kmph in 10s. The, acceleration is, 1) 1 m/s2 2) 2 m/s2 3) 1/2 m/s2 4) 3 m/s2, Speeds of two identical cars are U and 4U at, a specific instant. The ratio of the respective, distances in which the two cars are stopped, from that instant is, 1) 1:1, 2) 1:4, 3) 1:8, 4) 1:16, A car moving along a straight highway with, speed of 126Kmh 1 is brought to a stop with, in a distance of 200m. what is the retardation, of the car, 1) 3.06ms -2 2) 4ms -2 3) 5.06ms -2 4) 6ms -2, MOTION UNDER GRAVITY, Two balls are projected simultaneously with, the same velocity ‘u’ from the top of a tower,, one vertically upwards and the other, vertically downwards. Their respective times, of the journeys are t1 and t2. At the time of, reaching the ground, the ratio of their final, velocities is, 1) 1:1, 2) 1:2, 3) 2:3, 4) 2:1, , 15., , 16., , 17., , Two bodies are projected simultaneously with, the same velocity of 19.6 m/s from the top of, a tower, one vertically upwards and the other, vertically downwards. As they reach the, ground, the time gap is, 1) 0 s, 2) 2 s, 3) 4 s, 4) 6 s, Two bodies begin to fall freely from the same, height. The second one begins to fall s after the first. The time after which the 1st body, begins to fall, the distance between the bodies equals to l is, 1), , l , , g 2, , 2), , g, , l, , 2, g , , 4) , lg , l 2, A balloon is going upwards with velocity 12, m/sec. It releases a packet when it is at a, height of 65 m from the ground. How much, time the packet will take to reach the ground, , 3), , g 10m / sec , 2, , 1) 5 sec 2) 6 sec 3) 7 sec 4) 8 sec, A body thrown up with some initial velocity, reaches a maximum height of 50m. Another, body with double the mass thrown up with, double the initial velocity will reach a, maximum height of, 1) 100m 2) 200m 3) 400m 4) 50m, The distance moved by a freely falling body, (starting from rest) during the 1st, 2nd and, 3rd ... nth second of its motion, are, proportional to, 1) (n-1) 2) (2n-1) 3) (n2-1) 4) (2n-1)/n2, A ball released from a height ‘h’ touches the, ground in ‘t’s. After t/2s since dropping, the, height of the body from the ground, h, h, 3h, 3h, 2), 3), 4), 2, 4, 4, 2, A boy standing at the top of a tower of 20 m, , 1), , 18., , height drops a stone Assuming g 10 ms 2 ,, the velocity with which it hits the ground is, [AIPMT 2011], 19., , 1) 20 ms 1 2) 40 ms 1 3) 5 ms 1 4) 10 ms 1, A ball thrown vertically upwards with an initial velocity of 1.4 m/s returns in 2s. The toNARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , 20., , 21., , 22., , 23., , MOTION IN A STRAIGHT LINE, , tal displacement of the ball is, 1) 22.4 cm 2) zero 3) 44.8 m 4) 33.6m, A stone is dropped from a certain height which, can reach the ground in 5s. It is stopped momentarily after 3s and then it is again released. The total time taken by the stone to, reach the ground will be, 1) 6s, 2) 6.5s, 3) 7s, 4) 7.5s, RELATIVE VELOCITY, What are the speeds of two objects if, when, they move uniformly towards each other, they, get 4 m closer in each second and when they, move uniformly in the same direction with the, original speeds, they get 4 m closer each 10s?, 1) 2.8 m/s and 12 m/s, 2) 5.2 m/s and 4.6 m/s, 3) 3.2 m/s and 2.1 m/s, 4) 2.2 m/s and 1.8 m/s, Two trains are each 50m long moving, parallel towards each other at speeds, 10 m/s and 15 m/s respectively, at what time, will they pass each other?, 1) 8 s, 2) 4 s, 3) 2 s, 4) 6 s, A ball is dropped from the top of a building, 100 m high. At the same instant another ball, is thrown upwards with a velocity of 40 ms 1, form the bottom of the building. The two balls, will meet after., 1) 5 s, 2) 2.5 s 3) 2s, 4) 3 s, , LEVEL - I (H.W) - KEY, 01) 2, 07) 1, 13) 1, 19) 2, , 02) 4, 08) 1, 14) 1, 20) 3, , 03) 1, 09) 4, 15) 2, 21) 4, , 04) 3, 10) 1, 16) 2, 22) 2, , 05) 1, 11) 1, 17) 3, 23) 2, , 06) 2, 12) 3, 18) 1, , LEVEL - I (H.W) -HINTS, 1., 2., , vi = v1i ;, , vf = v 2 j ; Δv = vf - vi = v 2 j - v1 i, , Δv = v12 + v22 ;, , 5., 7., 8., 9., , a=, , Δv, Δt, , vu, 1, 1, , 2, ; 6. s n a n 2 ; s an, , , t, 2, a, 2n 1, a, s, sn 2n 1 : s n 2 ; n , 2, 2, s, n2, v = u + at ;, v 2 u 2 2as ; v = 0 both the cases, , a, , 2, , u2 s ;, , u1 s1 , , u2 s2 , , 10., , v 2 u 2 2as, , 11., , v u 2 2gh is same for both the bodies., , 12., , t , , 2u, g, 2, , gt 2, g t , ; H2 , 13., ; l H1 - H2, 2, 2, u2, 1 2, H, , h, , ut, , gt, 14., 15. max, ( independent of mass), 2g, 2, 1, g 3g 5 g, g, :, 16. sn g n ; Ratio = :, ..... 2n 1, 2 2 2, 2, 2, sn 2n 1, H1 , , 1 2, 17. h gt, 18. v 2gh, 2, 19. Since the ball returns back to its initial position,, the displacement is zero., 1 2 1 2 1 2, gt gt 1 gt 2 ; t tot t1 t 2, 2, 2, 2, , 20., , h h1 h 2 ,, , 21., , v A v B 4m /s ; v A v B , , 4, m/s, 10, , 22., , t, , l1 l2 l1 l2, , vr, v1 v2, d, h, t r , a r 0 ;, v r u1 u 2, , u1 0, , Displacement = shortest distance between initial, point and final point, 23., 2v1 v 2, s, v avg , 3.s = vt , t =, v1 v 2, v, , 4., , LEVEL - II (C.W), N, 40j, -30i , 30i, , NARAYANAGROUP, , W, , E, S, , 1., , DISPLACEMENT AND DISTANCE, A person moves 30m north and then 20m, towards east and finally 30 2 m in southwest direction. The displacement of the, person from the origin will be, 65
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , 2., , 1) 10m along north, 2) 10 m along south, 9., 3) 10m along west, 4) zero, SPEED AND VELOCITY, If a car covers 2/ 5th of the total distance with, , v1 speed and 3/ 5th distance with v2 then, average speed is, , 5v1v 2, 2v1v 2, 1, v v, v1v 2 2) 1 2 3), 4), v1 v 2 3v1 2v 2, 2, 2, Four persons A, B,C and D initially at the 10., corners of a square of side length ‘d’. If every, person starts moving with same speed v such, that each one faces the other always, the, person will meet after time, , 1), 3., , 1), 4., , 5., , 6., , 7., , 8., , 66, , d, v, , 2), , d, 2d, 3), 2v, v, , 4), , d, 2v, , A man walks on a straight road from his home, to a market 2.5 km away with a speed of 5, km/h. Finding the market closed, he instantly, turns and walks back home with a speed of, 7.5 km/h. What is the (a) magnitude of average velocity and (b) average speed of the man, over the time interval 0 to 50 min (in kmph)., 1) 0,4, 2) 0,6, 3) 0,8, 4)0,12, ACCELERATION, A starts from rest and moves with acceleration, a1. Two seconds later, B starts from rest and, moves with an acceleration a 2 . If the, displacement of A in the 5th second is the, same as that of B in the same interval, the, ratio of a1 to a2 is, 1) 9:5, 2) 5:9, 3) 1:1, 4) 1:3, A body travels 200cm in the first two seconds, and 220cm in the next 4 seconds with, deceleration. The velocity of the body at the, end of the 7th second is, 1) 20 cm/s 2) 15 cm/s 3) 10 cm/s 4) 0 cm/s, A bullet moving at 20 m/sec. It strikes a, wooden plank and penetrates 4 cm before, coming to stop. The time taken to stop is, 1) 0.008 sec, 2) 0.016 sec, 3) 0.004 sec, 4) 0.002 sec, An automobile travelling with a speed of, 60km/h can brake to stop within a distance, of 20m. If the car is going twice as fast i.e.,, 120km/h the stopping distance will be, 1) 20 m 2) 40 m 3) 60 m 4) 80 m, , 11., , 12., , A police party is moving in a jeep at a constant, speed v. They saw a thief at a distance x on a, motorcycle which is at rest. The moment the, police saw the thief, the thief started at, constant acceleration . Which of the, following relations is true if the police is able, to catch the thief? [2011-E], 1) v2 x, 2) v 2 2x, 3) v 2 2x, 4) v 2 x, Velocity of a body moving with uniform, acceleration of 3m/s2 is changed through, 30m/s in certain time. Average velocity of, body during this time is 30m/s. Distance, covered by it during this time is, 1) 300 m 2) 200 m 3) 400 m 4) 250 m, A person is running at his maximum speed of, 4 m/s to catch a train. When he is 6m from, the door of the compartment the train starts, to leave the station at a constant acceleration of 1 m / s 2 . Find how long it takes him to, catch up the train, 1. 2sec, 2. 3 sec 3. 4 sec 4. none, A body is moving along the +ve x-axis with, uniform acceleration of 4ms 2 . Its velocity, at x=0 is 10 ms 1. The time taken by the body, to reach a point at x=12m is, 1) 2s, 3s 2) 3s, 4s 3) 4s,8s 4) 1s, 2 s , , 13., , MOTION UNDER GRAVITY, A freely falling body takes ‘t’ second to travel, first (1/x)th distance. Then, time of descent is, , t, 1, x, 2) t x 3), 4), x, t x, t, The distance travelled by a body during last, second of its upward journey is ‘d’, when the, body is projected with certain velocity vertically, up. If the velocity of projection is doubled, the, distance travelled by the body during the last, second of its upward journey is, 1) 2d, 2) 4d, 3) d/2, 4) d, A rocket is fired and ascends with constant, vertical acceleration of 10m/s2 for 1 minute., Its fuel is exhausted and it continues as a free, particle. The maximum altitude reached is, (g=10m/s2), 1) 18 km 2) 36 km 3) 72 km 4) 108km, 1), , 14., , 15., , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , A parachutist after bailing out falls 50m, without friction. When parachute opens, it, decelerates at 2m / s 2 . He reaches the, ground with a speed of 3m / s . At what 24., height, did he bail out ?, 1) 91m, 2) 182m 3) 293m 4) 111m, A body is thrown vertically upwards with an, initial velocity ‘u’ reaches a maximum height, in 6s. The ratio of the distance travelled by, the body in the first second to the seventh, second is, 25., 1) 1:1, 2) 11:1, 3) 1:2, 4) 1:11, A body is thrown vertically up to reach its, maximum height in t seconds. The total time, from the time of projection to reach a point, at half of its maximum height while returning( in seconds ) is, (2008 E), , MOTION IN A STRAIGHT LINE, first second of its free fall, it passes through, ‘n’ stare then ‘n’ equal to, 1) 1, 2) 2, 3) 3, 4) 4, RELATIVE VELOCITY, Two particles P and Q simultaneously start, moving from point A with velocities 15m/s, and 20 m/s respectively. The two particles, move with accelerations equal in magnitude, but opposite in direction. When P overtakes, Q at B then its velocity is 30m/s. The velocity of Q at point B will be, 1) 30 m / s 2) 5 m / s 3) 10 m / s 4) 15 m / s, Two trains A and B, 100m and 60m long, are, moving in opposite directions on parallel, tracks. The velocity of the shorter train is 3, times that of the longer one. If the trains take, 4s to cross each other, the velocities of the, trains are, , 1) vA 10ms 1 , vB 30ms 1, 1 , t, , 3t, 1) 2 t, 2) 1 , 4), t 3), 2, 2, 2, , 2) v A 2.5ms 1 , vB 7.5ms 1, Water drops fall from a tap on to the floor 5.0m, 3) v A 20ms 1 , v B 60ms 1, below at regular intervals of time. The first drop, strikes the floor when the fifth drop beings to, 4) vA 5ms1 , v B 15ms1, fall. The height at which the third drop will be, from ground, at the instant when the first drop, LEVEL - II (C.W) - KEY, strikes the ground is (Take g = 10ms-2), 01) 3 02) 4 03) 1 04) 2 05) 2, 06) 3, 1) 1.25m 2) 2.15m 3) 2.75m 4) 3.75m, 07) 3 08) 4 09) 3 10) 1 11) 1, 12) 1, A boy throws n balls per second at regular, 13) 2 14) 4 15) 2 16) 3 17) 2, 18) 2, time intervals. When the first ball reaches the, 19) 4 20) 2 21) 3 22) 4 23) 1, 24) 2, maximum height he throws the second one, 25) 1, vertically up. The maximum height reached, LEVEL - II (C.W) - HINTS, by each ball is, 20 i, A, g, N, g, g, g, 1) 2 n 1 2 2), 4), 2 3), 2, , 2n, n, n, 30j, W, A body is thrown vertically upward from a, E, -30i O, point ‘A’ 125m above the ground. It goes up, C, to a maximum height of 250 m above the, S, ground and passes through ‘A’ on its 1., -30 j, downward journey. The velocity of the body, when it is at a height of 70m above the ground, B, is (g = 10 m/s2), (MED-2013), Total displacement from the origin,, 1) 20 m/s 2) 50 m/s 3) 60 m/s 4) 80 m/s, s 20i 30 j 30 j 30i =-10i, A body is released from the top of a tower of, s1 s 2, height H m. After 2s it is stopped and then, total distance s s, instantaneously released. What will be its 2., avg speed=, 1, 2, total time, height after next 2s (in metres)?, v1 v2, 1)H-5, 2) H-10 3)H-20 4)H-40, A ball dropped from 9th stair of a multistoried, building reaches the ground in 3 sec. In the, , NARAYANAGROUP, , 67
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, T, , 3., 4., 5., , 6., , 7., 8., , a, , , (n=4), 2v sin 2 ,, n, s1 s2, s1 s 2, (a) vavg = t t ;, (b) v avg = t t, 1, 2, 1, 2, a, a, s1 2n 1 ; s 2 2n 1, 2, 2, According to given prob. S1 S 2, s1, s2, t1, t2, 1 2, s1 ut1 t1, 2, 1, 2, s1 s2 u t1 t2 a t1 t2 , 2, u+v, v-u, v 2 - u 2 = 2as ; t = a ; s = 2 t, as, v=0, v 2 - u 2 = 2as ;, , 18., , 20., , 9. Distance travelled by the police party in ‘t’ sec. is vt., 21., 1 2, Distance travelled by thief x t, 2, 2, 1, αt, x t 2 vt, , - vt + x = 0, 2, 2, αt 2 - 2vt + 2x = 0 t=2v ± 4v 2 - 8αx, 4v 2 > 8αx ; v 2 > 2αx, 1 2, vu, ;s v avg t, 10. a , 11. d at vt, t, 2, 22., 1 2, h 1 2, 1 2, 12. s ut at, 13. gt h gt d, 2, x 2, 2, 14. distance covered in the last second of upward, journey = distance covered in 1st second of downward journey. This is independent of velocity of, projection., 1 2, v2, 15. h at , v at : max.height = H = 2g, 2, total distance from the ground = (H + h), 23., , , 1 2 a, at 1 , 2, g, , 16. u = 2g H, (g = 9.8 m/s2), v2 – u2 = 2as ; height at which he bails out= (H + s), u, 17. ta , u gta , u 6 g, g, 68, , 5th, x, 4th, 3x, h 3rd, 5x, 2nd, 7x, 1st, , 19., , 2, , u1 , s1, u 2 s ; u s, 2, 2, , g, height in the first second, h1 u , 2, height covered in the first second of downward, g, journey, h 2 , 2, 1 2 H 1, g(t ) 2 ; find t = ?, H = gt ;, 2, 2 2, Total time = t + t , , 24., 25., , h, 5, x, , h x 3x 5x 7 x 16 x ;, 16 16, Distance of 3rd drop from the ground = 12x, Time interval between two balls = Time of ascend, u2, 1 u, g, h, ;, = = u, 2g, n g, n, The body is freely falling from a height of 250m. Its, velocity at a height of 70m from the ground means, velocity of freely falling body after travelling 180m., v = 2gh ., , s1 s2, t1 t2, , 1 2, 1, gt1 t 2sec ; h2 gt22 t 2sec , 2, 2, 1, h H h1 h2 , , h1 , , nh , , 1 2, gt , h= height of each story (constant), 2, , n t2, v rel at ; For P ,, 30 = 15 + at, For Q,, v = 20 - at, 3v A v B ;, vrel = vA + vB, l +l, t= A B ;, s1 s 2 v A v B t, vrel, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , MOTION UNDER GRAVITY, LEVEL - II(H.W), 1., , 2., , 9., , Two particles move along x-axis in the same, direction with uniform velocities 8 m/s and, 4 m/s. Initially the first particle is 21m to the 10., left of the origin and the second one is 7m to, the right of the origin. The two particles meet, from the origin at a distance of, 1) 35 m 2) 32 m 3) 28 m 4) 56 m, A moving car possesses average velocities of, 5ms1 ,10ms1 and 15ms 1 in the first, second, 11., and third seconds respectively. What is the, total distance covered by the car in these 3s?, 1) 15m, 2) 30m, 3) 55m, 4) 45m, , The splash of sound was heard 5.35s after, dropping a stone into a well 122.5m deep., Velocity of sound in air is, 1) 350 cm/s 2) 350 m/s 3) 392 cm/s 4) 0 cm/s, Two stones are thrown vertically upwards with, the same velocity of 49m/s. If they are thrown, one after the other with a time lapse of 3, second, height at which they collide is, 1) 58.8 m 2) 111.5 m 3) 117.6 m 4) 122.5 m, A stone projected upwards with a velocity ‘u’, reaches two points ‘P’ and ‘Q’ separated by, a distance ‘h’ with velocities u/2 and u/3. The, maximum height reached by it is, , ACCELERATION, 3., , 4., , 5., , 6., , 7., , 8., , The average velocity of a body moving with, uniform acceleration after travelling a, distance of 3.06 m is 0.34 m/s. If the change, in velocity of the body is 0.18 ms-1 during this, time, its uniform acceleration is ( in ms-2 ), 1) 0.01, 2) 0.02, 3) 0.03, 4) 0.04, If a body looses half of its velocity on penetrating, 3cm in a wooden block, then how much will it, penetrate more before coming to rest, 1) 1 cm 2) 2cm, 3) 3cm, 4) 4cm, A car moving with a speed of 50km/hr can be, stopped by brakes after atleast 6m. If the, same car is moving at a speed of 100km/hr,, the minimum stopping distance is, 1) 12m, 2) 18m, 3) 24m, 4) 6m, A particle moving with a constant acceleration describes in the last second of its motion, 36% of the whole distance. If it starts from, rest,how long is the particle in motion and, through what distance does it moves if it describes 6 cm in the first sec.?, 2) 10 s;150 cm, 1) 5 s;150 cm, 3) 15 s;100 cm, 4) 20s ; 200cm, A bus starts from rest with a constant acceleration of 5 m / s 2 .At the same time a car, travelling with a constant velocity 50 m/s over, takes and passes the bus. How fast is the bus, travelling when they are side by side?, 1) 10 m/s 2) 50 m/s 3) 100 m/s 4) 150m/s, A particle moving with uniform retardation, covers distances 18m, 14m and 10m in, successive seconds . It comes to rest after, travelling a further distance of, 1)50 m, 2) 8 m, 3) 12 m 4) 42 m, , NARAYANAGROUP, , 1), 12., , 13., , 14., , 15., , 9h, 5, , 2), , 18h, 5, , 3), , 36 h, 5, , 4), , 72h, 5, , A ball is dropped from the top of a building., The ball takes 0.5s to fall past the 3m length, of a window at certain distance from the top, of the building. Speed of the ball as it crosses, the top edge of the window is (g=10m/s2), 1) 3.5 ms-1 2) 8.5 ms-1 3) 5 ms-1 4) 12 ms-1, A body thrown vertically up with a velocity ‘u’, reaches the maximum height ‘h’after‘T’ second., Correct statement among the following is, 1) at a height h/2 from the ground its velocity, is u/2, 2) at a time ‘T’ its velocity is ‘u’, 3) at a time ‘2T’ its velocity is ‘-u’, 4) at a time ‘2T’ its velocity is ‘-6u’, A ball is projected vertically upwards with a, velocity of 25 ms-1 from the bottom of a tower., A boy who is standing at the top of a tower is, unable to catch the ball when it passes him in, the upward direction. But the ball again, reaches him after 3 sec when it is falling. Now, the boy catches it Then the height of the tower, is (g=10ms-2), 1) 5 m, 2) 10 m 3) 15 m, 4) 20 m, A person sitting on the top of a tall building is, dropping balls at regular intervals of one, second. When the 6th ball is being dropped,, the positions of the 3rd, 4th, 5th balls from, the top of the building are respectively, 1) 4.9m, 19.6m, 44.1m 2) 4.9m, 14.7m, 24.5m, 3) 44.1m, 19.6m, 4.9m 4) 24.5m, 14.7m, 4.9m, , 69
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 16., , 17., , 18., , 19., , A stone projected vertically up from the 23., ground reaches a height y in its path at t1, seconds and after further t 2 seconds reaches, the ground. The height y is equal to, 1, 1, 2, 1) gt1 t 2 , 2) gt1 t 2 , 2, 2, 1, 24., 3) g t1t 2, 4) g t1t 2, 2, A person standing on the edge of a well throws, a stone vertically upwards with an initial, velocity 5 ms-1. The stone gone up, comes down, and falls in the well making a sound. If the, person hears the sound 3 second after, throwing, then the depth of water (neglect time 25., travel for the sound and take g = 10ms-2), 1) 1.25 m 2) 21.25 m3) 30m, 4) 32.5 m, A ball is thrown vertically upwards with a, speed of 10 m/s from the top of a tower 200m, height and another is thrown vertically, downwards with the same speed, simultaneously. The time difference between, them on reaching the ground is (g=10m/s2), 26., 1) 12s, 2) 6s, 3) 2s, 4) 1s, A body is projected vertically upwards with a, velocity ' u ' . It crosses a point in its journey, at a height ' h ' twice , just after 1 and 7 seconds .The value of u in ms 1is g 10 ms 2 , , 20., , 21., , 22., , 70, , 1) 50, 2)40, 3) 30, 4) 20, A stone thrown vertically up from the ground, reaches a maximum height of 50m in 10s., Time taken by the stone to reach the ground, from maximum height is, 1) 5s, 2) 10s, 3) 20s, 4) 25s, A freely falling body travels-- of total distance, 1., in 5th second, 2., 1) 8%, 2) 12% 3) 25%, 4) 36%, A body is projected with a velocity u. It passes, through a certain point above the ground in t1 3., sec. The time after which the body passes through, the same point during the return journey, 4., u, u, , 2, 2) 2 t1 , 1) t1 , g, , g, , 6., u2, , u2, , 3) 3 g t1 , 4) 3 g 2 t1 , , , , , , A boy throws a ball in air in such a manner, that when the ball is at its maximum height, he throws another ball. If the balls are thrown, with the time difference 1 second, the maximum height attained by each ball is, 1) 9.8 m 2) 19.6 m 3) 4.9 m 4) 2.45 m, , RELATIVE VELOCITY, Two cars are travelling in the same direction, with a velocity of 60 kmph. They are, separated by a distance of 5 km. A truck, moving in opposite direction meets the two, cars in a time interval of 3 minute. The, velocity of the truck is (in kmph), 1) 20, 2) 30, 3) 40, 4) 60, A police van moving on a highway with a, speed of 30 kmph fires a bullet at a thief’s, car speeding away in the same direction with, a speed of 192 kmph. If the muzzle speed of, the bullet is 150 m/s, with what speed does, the bullet hit the thief’s car? (Note: Obtain, that speed which is relevant for damaging the, thief’s car)., 1) 25m/s 2) 50m/s 3) 75m/s 4) 105m/s, Two cars are moving in same direction with, speed of 30kmph. They are separated by a, distance of 5km. What is the speed of a car, moving in opposite direction if it meets the, two cars at an interval of 4 min?, 1) 60 kmph, 2) 5 kmph, 3) 30 kmph, 4) 45 kmph, , LEVEL - II(H.W)-KEY, 01) 1, 07) 3, 13) 3, 19) 2, 25) 4, , 02) 2, 08) 2, 14) 4, 20) 2, 26) 4, , 03) 2, 09) 2, 15) 3, 21) 4, , 04) 1, 10) 2, 16) 3, 22) 2, , 05) 3, 11) 3, 17) 3, 23) 3, , 06) 1, 12) 1, 18) 3, 24) 3, , LEVEL - II(H.W)-HINTS, x 21 u1t ; x 7 u2t, s u1t1 u2t2 u3t3, distance, changein velocity v u, t=, , ,, a, =, v ave, total time, t, x n 1, y=, 2n 1, s, , 1 2, at ;, 2, , 2, , 5. v 2 u 2 2as , v=0 ; u 2 s, 2n 1 36, , n2, 100, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, 7., 8., 10., 12., 13., 14., , MOTION IN A STRAIGHT LINE, , 1 2, at , s = vt , v=at, 2, 1, 2h, h, , s u a n 9. t , , 2, g vsound, , 1., u 2 gt 2, u2, h, , 11. v 2 u 2 2as ; H , 2g 8, 2g, 1, 1, 2, S ut at 2 3 v 0.5 10 0.5 , 2, 2, After a time 2T, body returns to the point of projec- 2., tion with the same velocity in opposite direction., 2v, t, ; v 2 u 2 2gh, g, 6, s=, , g, 2, , 15., , 3, , 7g, 2, , 2, , 9g, 2, , 4., , 1, 16., 18., 20., , 1, h = y = gt1t 2, 2, 2u, t , g, , 17. h = –ut +, , 1 2, gt, 2, , 1, 19. u g(t 2 t1 ), 2, sn 2 n 1, 21. , h, n2, , ta td, , u, , t1 , g, , , , ta P, , 22., , t1, 23., , h, , u, 24., 25., , g, 2n 2, , u, ; ta , g, 5., , u, d, d, ; t v uv, rel, , d, , v, The relative velocity of bullet with respect to thief, v bt v b v p v t, u, u, , 26., , d, , NARAYANAGROUP, , v, , A motorist drives north for 35.0 minutes at, 85.0 km/h and then stops for 15.0 minutes., He next continues north, travelling 130 km, in 2.00 hours. What is his total displacement, 1) 85 km 2)179.6 km 3)20 km 4)140 km, The coordinates of a moving particle at any, time ‘t’ are given by x t 3 and y t 3 . The, speed of the particle at time ‘t’ is given by, 1), , 2 2, , 2) 3t 2 2, 4) t 2 2 2, , ACCELERATION, 3., , 4, , 5g, 2, , SPEED AND VELOCITY, , 3) 3t 2 2 2, , 5, , 3g, 2, , LEVEL-III, , 6., d, d, ; t = v = u+v, rel, , The relation between time t and distance x is, t ax 2 bx where a and b are constants. The, acceleration is, 1) 2av 3 2) 2av 2 3) 2abv 2 4) 2bv3, Two cars start in a race with velocities u1 and, u 2 and travel in a straight line with, accelerations ‘ ’ and . If both reach the, finish line at the same time, the range of the, race is, 1), , 2( u1 u 2 ), ( u1 u 2 ), ( ) 2, , 2), , 2( u1 u 2 ), ( u1 u 2), , , 2 u 1u 2, 2( u 1 u 2 ) 2, 3), 4), 2, , ( ), A point moves with uniform acceleration, v1 ,v2 and v3 denote the average velocities in, three successive intervals of time t1, t2 and, t3. Correct relation among the following is, 1) (v1-v2) : (v2-v3) = (t1-t2) : (t2-t3), 2) (v1-v2) : (v2-v3) = (t1+t2) : (t2+t3), 3) (v1-v2) : (v2-v3) = (t1-t2) : (t2+t3), 4) (v1-v2) : (v2-v3) = (t1+t2) : (t2-t3), A train starts from rest and moves with uniform, acceleration for some time and acquires a, velocity ‘v’. It then moves with constant, velocity for some time and then decelerates, 71
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , height of each storey is 4m, the number of, , at rate and finally comes to rest at the next, station. If ‘L’ is distance between two stations, then total time of travel is, 1), , L v 1 1, , v 2 , , 2), , L v 1 1, , v 2 , , 13., , L v 1 1, L v 1 1, 4) , v 2 , v 2 , A car, starting from rest, accelerates at the, rate f through a distance S, then continues at 14., constant speed for time t and then decelerate, at the rate (f/2) to come to rest. If the total, distance travelled is 15S, then, 1 2, 1) S ft, 2) S ft, 6, , 3), 7., , 3) S , 8., , 9., , 10., , 11., , 12., , 1 2, ft, 72, , 1, 4, , 2, 4) S ft, , 15., An express train moving at 30 m/s reduces, its speed to 10 m/s in a distance of 240 m. If, the breaking force is increased by 12.5% in, the beginning find the distance that it travels, before coming to rest, 1) 270 m 2) 240 m 3) 210 m 4) 195 m, For motion of an object along the x-axis, the, velocity v depends on the displacement x as, v 3x 2 2x, then what is the acceleration at 16., x 2m ., 1) 48ms2 2) 80ms 2 3) 18ms 2 4) 10ms 2, , The velocity of a particle is v v 0 gt ft 2 ., If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is (AIE-2007), g f, 1) v 0 2g 3f, 2) v 0 , 2 3, g, 3) v 0 g f, 4) v 0 f, 2, A ball A is dropped from the top of a building, and at the same time an identical ball B is, thrown vertically upward from the ground., When the balls collide the speed of A is twice, that of B. At what fraction of the height of the, building did the collision occur?, , MOTION UNDER GRAVITY, , 1, 2, 5, 7, 2), 3), 4), 3, 3, 3, 3, An object falls from a bridge that is 45m, above the water.It falls directly into a small, row-boat moving with constant velocity that, was 12m from the point of impact when the, object was released. What was the speed of, , The friction of the air causes a vertical, retardation equal to 10% of the acceleration, , the boat? g 10 ms 2 , , due to gravity. Take g 10m / s 2 . The, maximum height and time to reach the, maximum height will be decreased by, 17., 1) 9%, 9%, 2) 11%, 11%, 3) 9%, 10%, 4) 11%, 9%, A parachutist after bailing out falls for 10s, without friction. When the parachute opens, he descends with an acceleration of 2 m/s2, against his direction and reached the ground, with 4 m/s. From what height he has dropped, himself ? (g = 10m/s2), 1) 500m 2) 2496m 3) 2996m 4) 4296m, A body is dropped from the roof of a multistoried building. It passes the ceiling of the, 1, , 15th storey at a speed of 20 ms . If the, 72, , storeys in the building is (take g 10ms 2, and neglect air resistance), 1) 20, 2) 25, 3) 30, 4) 35, A body is projected vertically up with velocity 98ms 1 . After 2 s if the acceleration due, to gravity of earth disappears , the velocity, of the body at the end of next 3 s is, 1)49ms-1 2)49.6ms-1 3)78.4ms-1 4)94.7ms-1, , 1), , 1. 2 m / s 2. 3 m / s 3. 5 m / s 4. 4 m / s, , GRAPHS, An elevator is going up. The variation in the, velocity of the elevator is as given in the, graph. What is the height to which the elevator takes the passengers, , velocity, 3.6, O, 1) 3.6 m, , 2, , 10 12, Time, 2) 28.8 m 3) 36.0 m 4)72.0 m, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, The velocity time graph of a body moving in, a straight line is shown in the figure. The displacement and distance travelled by the body, in 6 sec are respectively (in metres) ., , v(m/s), , 18., , MOTION IN A STRAIGHT LINE, , 2, , 4, , x, , 6, , 0, , 5, , 1, , 10, , 5, , v, , 15, v, , 20, , t, , t(sec), 1) 8,16, 2) 16,8, 3)16,16 4) 8,8, The velocity-time graph of a stone thrown, vertically upward with an initial velocity of, 30ms 1 is shown in the figure. The velocity in, the upward direction is taken as positive and, that in the downward direction as negative., What is the maximum height to which the, stone rises?, , 20., , 30 A, 20, 10, 0, 10, 1, 20, 30, , 1) 0, , 3) 0, , 22., , 2, , B, 3, , 4, , 5, , Time (s), , 5 10 15 20, , t 2) 0, , 5 10 15 20, , t, , v, , v, , 5 10 15 20, , t 4) 0 5 10 15 20 t, , Velocity -time (v-t) graph for a moving object, is shown in the figure . Total displacement of, the object during the time interval when there, is non-zero acceleration and retardation is, , C, , 1) 30 m, 2) 45 m 3) 60 m 4) 90 m, The variation of velocity of particle moving, along a straight line is shown in the figure., The distance travelled by the particle in 4s is, , v(m/s), , -1, , velocity (ms ), , 19., , 4, 3, 2, 1, 0, , -1, , V(ms ), , 20, , 10 20, , 30, , 40 50 60, , t(sec), , 10, , 1) 60m, , 2) 50m, , 3) 30m, , 4) 40m, , 1, , 21., , 2, 3, 4, t(s), 1) 55 m 2) 30 m 3) 25 m 4) 60 m, Figure shows the displacement-time (x-t), graph of a body moving in a straight line., Which one of the graphs shown in figure, represents the velocity-time (v-t) graph of the, motion of the body., , NARAYANAGROUP, , 73
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , RELATIVE VELOCITY, 23., , 1 , 1 , , s t u1 t t u 2 t , 2 , 2 , , solve for t and substitute in the above equation., , An armored car 2m long and 3 m wide is, moving at 10 ms 1 when a bullet hits it in a, 1 3 , direction making an angle tan 4 with the, , , 24., , 25., , 5., , v, v, v, v, 1) sin 2) cos 3) tan 4) cot , a, a, a, a, A jet airplane travelling at the speed of 500 7., km/h ejects its products of combustion at the, speed of 1500 km/h relative to the jet plane., What is the speed of the later with respect to, an observer on ground?, 1) 100kmph, 2) 1000kmph, , 4) 11kmph, , LEVEL - III - KEY, 1) 2, 7) 3, 13) 3, 19) 2, 25) 2, , 2) 3, 8) 1, 14) 2, 20) 1, , 3) 1, 9) 2, 15) 2, 21) 4, , 4) 1, 10) 1, 16) 4, 22) 2, , 5) 2, 11) 3, 17) 3, 23) 1, , 6) 1, 12) 1, 18) 1, 24) 2, , LEVEL - III - HINTS, 1., , s1 v1t ; s s s, , 2., , vx , , 4., 74, , change in average velocity, average time, , length of the car as seen by a stationary, v, observer. The bullet enters one edge of the, v, v, car at the corner and passes out at the, =, =, t1, t3, diagonally opposite corner. Neglecting any, interaction between the car and the bullet and, , , effect of gravity, the time for the bullet to 6., v, cross the car is, O, t3, t1, t2, 1) 0.20 s 2) 0.15 s 3) 0.10 s 4) 0.50 s, Time, Two particles start simultaneously from the, same point and move along two straight lines., 1, 1, One with uniform velocity v and other with a T t1 t 2 t 3 ; weknow L = vt1 + vt 2 + vt 3, 2, 2, uniform acceleration a. If is the angle bevt, vt, L 1, tween the lines of motion of two particles then, L = 1 + vt 2 + 3 ; t 2 = - t1 + t 3 ......(1), the least value of relative velocity will be at, 2, 2, v 2, time given by, , 3) 10kmph, , 3., , a, , 1, , 2, , dx, dy, , v y ; v v2x v2y, dt, dt, dx, dv, v , a, dt, dt, Range = dist. covered before they meet, , s, v 2 - o 2 = 2fs, , , , f ® s3 = 2s, 2, 2, o - v = -2 .s3 , 2 , , s 2 = 12s ;, , 12s = vt, , also v = 2fs, , 1 2, ft, 12s = 2fs.t ;144s2 = 2fs.t ; s =, 72, , 8., , v 2 u 2 2as, , 9., , Diff. eq. (1) w.r.t x is, , 10., , h, , 1, ;, g, , g1 g , , 11., 12., , dv, dv, = 6x - 2 ; a v, dx, dx, , h2 g1 h2 h1, g g, , 100 1 2 100, ;, h1, g2, h1 g 2, g 11g, , 10 10, , 1 2, gt , v = gt ; v12 - v 2 = 2as 2 ; s = s1 + s 2, 2, Distance traveled when it gains velocity of 20m/s, s1 , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , v 2 400, h, =, ;, =, =20m, v - 0 = 2gh, 2g, 20, , second car is v = v + at = 0 + at, , 20, 5, 4, Total no.of stories 15 5 20, 25., v = u - gt . after g disappears body moves up, with uniform velocity., , vcosα, d 2, vr = 0 ; t =, , dt, a, Velocity of jet aeroplane = 500 j, , 2, , 2, , v rel = v 2 + at - 2vat cosα, , Number of stories =, , 13., 14., , 16., , LEVEL-IV, Matching Type Questions, 1., , 2h, g, , , v f = v p -1500 j, , 500 j 1500 j 1000 j, speed of fuel w.r.to ground is -1000km/hr.., , 1 2, ut, gt, , hB, 2, , 1, hA hB 1 2 , , gt ut gt 2 , 2, 2, , , t, , velocity of fuel w.r.to plane 1500 j, , v f - v p = -1500 j ;, , Given velocity of particle ; v = v0 + gt + ft 2, displacement s = vdt, , 15., , v rel is minimum if, , ; s = v×t, , Match the following, Column I Column II, , dv, a), dt, , 17. area of v t graph; 18.area of v t graph, 30, u2, 2, ; From graph u=30m/s; a 10m / s, 19. h , 3, 2a, 20. area of under v-t graph, 21. o to 5s - velocity is +ve and constant, 5 to 15s - slope is zero, 15 to 20s - velocity is -ve and constant, 22. Area under trapezium gives displacement., , , dv, b), dt, , dr, c), dt, , dr, d), dt, , p) Acceleration, q) Magnitude of acceleration, r) Velocity, s) magnitude of velocity, , RELATIVE VELOCITY, 2., , 23., 2m, , 0, , 3m vb sin 37, vb, , 10m/s, -1 3 , , =tan, =370, 0, v b cos 37, , , 4, , 2m, , d) 2 Area under a x graph s) Velocity, , Relative velocity along x-axis is v b cos37 -10 , 0, , Distance travelled by bullet along x- axis is, , v cos37, , 0, , b, , -10 t = 2 .........(1), , Distance travelled by bullet along y-axis is, , v sin37 t = 3 ..........(2), 0, , b, , Solving equation (1) and (2) we get t=0.20s, 24. At any time velocity of first car is V and that of, NARAYANAGROUP, , A particle moves in a straight line with zero, initial velocity. Then, match the following, terms :, Column I, Column II, 1, a) (slope of v 2 - x ) graph p) Speed, 2, q) Acceleration, b) slope of v - t graph, c) Slope of x t graph, r) Area under v-t, graph, , Statements Type Questions, Mark your answer as, 1) If Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I, 2)If Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I, 3) If Statement I is true, Statement II is false, 4) If Statement I is false, Statement II is true, , 75
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , Statement I : An object may have varying speed, without having varying velocity., Statement II : If the velocity is zero at an instant,, the acceleration may not be zero at that instant., Statement I : A body is momentarily at rest at, the instant it reverses the direction., Statement II : A body cannot have acceleration, if its velocity is zero at a given instant of time., Statement I : On a curved path average speed, of a particle can never be equal to average velocity., Statement II : Average speed is total distance, travelled divided by total time. Whereas average, velocity is, final velocity plus initial velocity divided, by two., Statement I : Particle A is moving eastwards and, particle B northwards with same speed. then,, velocity of A with respect to B is in south -east, direction., Statement II : Relative velocity between them is, zero as their speeds are same., Statement I : A lift is ascending with decreasing, speed means acceleration of lift is downwards., Statement II : A body always moves in the, direction of its acceleration., Statement I : Two balls of different masses are, thrown vertically upwards with the same speed., They will pass through their point of projection in, the downward direction with the same speed., Statement II : The height and the downward, velocity attained at the point of projection are, independent of the mass of ball., Statement I: The v-t graph perpendicular to time, axis is not possible in practice., Statement II: Infinite acceleration can’t be, realized in practice., Statement I: Magnitude of average velocity is, equal to average speed, if velocity is constant., Statement II: If velocity is constant , then there, in no change in the direction of motion., Statement I: The average velocity of a particle, having initial and final velocity v1 and v2 is, , 12., , 76, , Statement II: In last second distance travelled is, 4.9m. (take g 9.8m / s 2 ), 13., , Statement I: Distance between two particles, moving with constant velocities always remains, constant., Statement II: In the above case relative motion, between them is uniform., , 14., , Statement I: For one dimensional motion, the angle, between acceleration and velocity should be zero., Statement II: One dimensional motion is on a, straight line., , 15., , Statement I: If displacement is a linear function, of time, its average and instantaneous velocity will, be same., Statement II: If the acceleration of a moving, particle is zero, the particle moves linearly., , 16., , Statement I: If two particles are moving with, same speed but their velocity vectors are opposite, then the distance between the particles must, be changing., Statement II:If the velocity of one particle w.r.t., other is zero, then separation between the particle must be constant., , 17., , Statement I: If the magnitude of acceleration of, a particle is constant then speed must change., Statement II: In uniform circular motion speed, of the particle is constant but it has some acceleration., , 18., , Statement I: A bus moving due north takes a turn, and starts moving towards east with same speed., There will be no change in the velocity of bus., Statement II:Velocity is a vector quantity., , 19., , v1 v 2, 2, , Statement II: If r1 and r2 be the initial and final 20., r -r, displacement in time t, then vavg = 1 2 ., t, Statement I: If a particle is thrown upwards then, distance travelled in last second of upward journey, is independent of the velocity of projection., , Statement I:When a particle is moving on a, straight line, its velocity is constant., Statement II: The net acceleration of a moving, particle may change its direction of motion or, magnitude of velocity or both., Statement I:The direction of velocity vector is, always along the tangent to the path therefore its, magnitude may be given by its slope., Statement II:The slope of tangent to path always, measure the magnitude of velocity at that point., NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , LEVEL - IV - KEY, Matching type, , 12., , 1) a -p; b-q; c-r; d-s, 2) a-q; b-q; c-s; d-p, , Statement Type, 3) 4 4)3, 9) 1 10)1, 15)2 16)2, , 5)3, 11)4, 17)4, , 6)3, 12)1, 18)4, , 7)3, 13)4, 19)4, , 8) 1, 14) 4, 20)2, , body., , 4., , 5., , 13., , If speed varies, then velocity will definitely vary. 14., At highest point of a particle thrown upwards, a 0 but v=0., 15., Any body momentarily at rest must change its direction, when a particle is released from rest,, v 0 but a 0, 16., On a curved path, distance>displacement, , displacement, average velocity = Total time, , 17., , 7., , Ascending means velocity is upwards. Speed is decreasing. It means acceleration is downwards. Fur- 18., ther, the body moves in the direction of velocity., , 8., , Both maximum height and time of flight are independent on masses., , o, , t, , 10., , In one dimensional motion vavg is numerically equal, to vavg velocity, since there is no change in direction., , 11., , If acceleration is not constant v avg ¹, , v avg =, , v1 + v 2, 2, , If displacement is linear function of time then velocity is constant that is motion will be of uniform, nature., Correct speed same but direction of velocity opposite than distance between particle will be, changing. Statement-(1) correct statement-(2), correct but (2) is not the correct explanation of, Statement-(1)., acceleration = constant ; velocity changes, , speed = constant but it has some acceleration, statement-(2) correct., Velocity change either by changing its magnitude, or by direction, Statement-(1)-false. ; Statement-(2) - correct., , 19., , In st. line motion particle may move with constant, velocity but it is not necessary acceleration , velocity changes either by magnitude or by direction or both., , 20., , Statement -(1) correct, , v, , v, =a, t, , In one dimension motion angle between velocity, and acceleration becomes zero and 1800 in different situation. These are two possible value only., , In uniform circular motion, , tan 1 , 450 in S-E direction., , , Slope = =, , If distance between two particles moving with con-, , speed must changes statement-(1)-false, , v = u j, v = v - v = ui - u j, v A = ui,, B, AB, A, B, , 9., , g, = 4.9m, 2, , stant velocities always remains constant v r 0, , average speed > average velocity, Further,, , 6., , s=, , It is independent of velocity of projection., , LEVEL - IV - HINTS, 3., , If a particle is thrown upwards then distance travelled in last second of upward journey is equal to, distance travelled in first second of a freely falling, , Yes it may not have variable speed when have, constant velocity, Statement-(2) correct, But Statement-(2) is not the correct explanation, for Statement -(1)., , s r2 - r1, =, t, t, , NARAYANAGROUP, , 77
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , A. MOTION ALONG HORIZONTAL AXIS :, , 6., , LEVEL - V, , In 1.0 s, a particle goes from point A to point, B, moving in a semicircle (see figure). The, magnitude of the average velocity is, A, , SINGLEANSWER TYPE, 1., , A body moving along a straight line traversed, 1.0, one third of the total distance with a velocity, m, 4 m/sec in the first stretch. In the second, stretch the remaining distance is covered with, B, a velocity 2 m/sec for some time t0 and with, (A) 3.14 m/s (B) 2.0 m/s (C) 1.0 m/s (D) zero, 4m / s for the remaining time. if the average, MULTIPLE ANSWER QUESTIONS, velocity is 3 m/sec, find the time for which, body moves with velocity 4 m/sec in second 7. The velocity of a particle along a straight line, stretch:, increases according to the linear law v =, v0+kx, where k is a constant. Then, t0, 3, a) t 0, b) t0, c) 2t0, d), a) the acceleration of the particle is k v0 kx , 2, 2, , 2., , For motion of an object along the x-axis, the, velocity v depends on the displacement x as, v 3 x 2 2 x , then what is the acceleration at, x = 2 m?, , 3., , 4., , A) 48 ms 2 B) 80 ms 2 C) 18 ms 2 D) 10 ms 2, A police party is chasing a dacoit in a jeep 8., which is moving at a constant speed v . The, dacoit is on a motorcycle. When he is at a, distance x from the jeep, he accelerates from, rest at a constant rate ? Which of the, following relations is true if the police is able, to catch the dacoit ?, A) v 2 x, B) v 2 2 x, C) v 2 2 x, D) v 2 x, A point moves in a straight line so that its, displacement x metre time t second is given, by x 2 1 t 2 . Its acceleration in ms 2 at time, t second is, t, 1 1, 1 t2, A) 3/ 2, B) 3, C) 3 D) 2, x, x, x x, x x, A 2m wide truck is moving with a uniform, speed v0 8 ms 1 along a straight horizontal 9., road. A pedestrain strarts to cross the road, with a uniform speed v when the truck is 4 m, away from him. The minimum value of v so, that he can cross the road safely is, 1, , 5., , 78, , A) 2.62 ms 1, , B) 4.6 ms 1, , C) 3.57 ms 1, , D) 1.414 ms 1, , 1, , v , , 1, b) the particle takes a time k log e v to attain a, 0, velocity v1, c) velocity varies linearly with displacement with, slope of velocity displacement curve equal to k., d) data is insufficient to arrive at a conclusion., Two particles P and Q move in a straight line, AB towards each other. P starts from A with, velocity u1 and an acceleration a1 . Q starts, from B with velocity u2 and acceleration a2 ., They pass each other at the midpoint of AB, and arrive at the other ends of AB with equal, velocities., , 2u u, , , , 2, 1, a) They meet at midpoint at time t a a , 1, 2, b) The length of path specified i.e., AB is, 4 u 2 u 1 a 1 u 2 a 2 u 1 , , l, , a, , 1, , a2, , 2, , c) They reach the other ends of AB with equal, velocities if u 2 u1 a1 a 2 8 a 1u 2 a 2 u1 , d) They reach the other ends of AB with equal, velocities if, u 2 u1 a1 a 2 8 a 2u1 a1u 2 , Which of the following statements is/ are, correct ?, A) If the velocity of a body changes, it must have, some acceleration., B) If the speed of a body changes, it must have, some acceleration, C) If the body has acceleration, its speed must, change, D) If the body has acceleration, its speed may, change., NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , 10. A particle moves along a straight line so that, PASSAGE TYPE QUESTIONS, 2, its velocity depends on time as v 4t t . PASSAGE-1, Then for first 5s., A train starts from rest with constant, acceleration, a 1m / s 2 . A passenger at a, A) Average velocity is 25 / 3 ms 1, distance ‘S’ from the train runs at his, B) Average speed is 10 ms 1, maximum velocity of 10 m/s to catch the train, 1, at the same moment at which the train starts., C) Average velocity is 5 / 3 ms, 14. If S=25.5 m and passenger keeps running, find, D) Acceleration is 4 ms 2 at t 0, the time in which he will catch the train:, 11. A particle moves with an initial velocity v0 and, a) 5 sec b) 4 sec c) 3 sec, d) 2 2 sec., retardation v , where v is velocity at any time 15. Find the critical distance ‘S ’ for whcih passenger, c, t., will take the ten seconds time to catch the train:, a) 50m, b) 35m, c) 30m, d) 25m, v0, (A) The particle will cover a total distance, 16. Find the speed of the train when the passenger, , catches it for the critical distacne:, 1, a) 8 m/s b) 10 m/s c) 12 m/s d) 15m/s, (B) The particle will come to rest after time, PASSAGE-2, , A body is moving with uniform velocity of, (C) The particle will continue to move for a long, time., 8 ms 1 . When the body just crossed another, body, the second one starts and moves with, v, (D) The velocity of particle will become 0 after, uniform acceleration of 4 ms 2 ., e, 17. The time after which two bodies meet will be, 1, A) 2 s, B) 4 s, C) 6 s, D) 8 s, time, , 18. The distance covered by the second body, 12. A particle is moving along X–axis whose, when they meet is, A) 8 m, B) 16 m C) 24 m D) 32 m, t3, position is given by x 4 9t . Mark the, 3, , MARTIX MATCHING QUESTION, , correct statement(s) in relation to its motion. 19. A particle moves along a straight line such, (A) direction of motion is not changing at any of the, that its displacement S varies with time, instants, as S t t 2 ., (B) direction of motion is changing at t = 3 s, Column-I, (C) for 0 < t < 3 s, the particle is slowing down, i. Acceleration at t= 2 s, (D) for 0 < t < 3 s, the particle is speeding up., ii. Average velocity during 3rd sec, 13. A particle of mass m moves on the x-axis as, iii. Velocity at t 1 s, follows : it starts from rest at t = 0 from the, iv. Initial displacement, point x = 0, and comes to rest at t = 1 at the, Column-II, point x = 1. No other information is available, about its motion at intermediate times, a. 5, b. 2, 0 t 1 . If a denotes the instantaneous, c. , acceleration of the particle, then [1993], d. 2, (A) a cannot remain positive for all t in the interval, 0 t 1, INTEGER TYPE QUESTIONS, (B) |a| cannot exceed 2 at any point in its path, 20. In a car race, car A takes 4 s less than car B at, (C) |a| must be 4 at some point or points in its, the finish and passes the finishing point with a, path, velocity v more than the car B. Assuming that, (D) a must change sing during the motion, but no, the cars start from rest and travel with constant, other assertion can be made with the information, accelerations a1 4 ms 2 and a2 1 ms 2, given., respectively, find the velocity of v in ms 1 ., NARAYANAGROUP, , 79
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , 21. A police jeep is chasing a culprit going on a 26. A stone is dropped from the 25th storey of a, motorbike. The motorbike crosses a turning, multistoried building and it reaches the, at a speed of 72 kmh 1 . The jeep follows it at, ground in 5s. In the first second, it passes, through how many storeys of the building, a speed of 90 kmh 1 , crossing the turning 10, ( g= 10 ms 2 ), s later than the bike., A)1, B)2, C) 3, D) None, Assuming that they travel at constant, speeds, how far from the turning will the jeep 27. A body is projected upwards with a velocity, catch up with the bike ? ( in km), u , It passes through a certain point above, 22. A particle moves in a straight line such that, the ground after t1 . The time after which the, the displacement x at any time ‘t’ is given by, body passes through the same point during, 2, 3, x, is, in, m, and, t, is, in, second, the return journey is, x 6t t 3t 4., calculate the maximum velocity (In ms-1) of the, u, , u 2, particle., A) t1 , B) 2 t1 , g, , g, , B. MOTION UNDER GRAVITY, , u2, , u2, , 3, 3, , t, C) g 1 , D) g 2 t1 , 23. A ball is thrown upwards with speed v from, , , , , the top of a tower and it reaches the ground 28. A ball is dropped vertically from a height d, with speed 3v . What is the height of the, above the ground. It hits the ground and, tower ?, d, bounces up vertically to a height, ., 2, 2, 2, v, 2v, Neglecting subsequent motion and air, A), B), g, g, resistance, its velocity v varies with height h, above the ground as: [2004], 4v 2, 8v 2, C), D), V, g, g, , SINGLE ANSWER QUESTIONS, , V, , 24. An elevator in which a man is standing is, moving upwards with a speed of 10 ms 1 . If, the man drops a coin from a height of 2.45 m, from the floor of elavator, it reaches the floor, of the elavator after time ( g 9.8 ms 2 ), A), , 2 s B) 1/ 2 s, , C) 2 s, , d, , A), , h, , B), d, , h, , V, , D) 1/ 2 s, , V, , 25. A body is thrown vertically upwards from A,, d, d, the top of a tower. It reaches the ground in, h, h, C), D), time t1 . If it is thrown vertically downward, from A with the same speed, it reaches the, ground in time t2 . If it is allowed to fall freely, 29. A small block slides without friction down an, from A, then the time it takes to reach the, inclined plane starting from rest. Let sn be the, ground is given by, distance traveled from t =n – 1 to t = n. Then, t1 t2, t1 t2, sn, A) t , B) t , is :, [2005], sn 1, 2, 2, C) t t1t2, , t1, D) t t, 2, , (A), , 2n – 1, 2n, 2n – 1, , (C) 2n 1, 80, , 2n 1, , (B) 2n – 1, 2n, , (D) 2n 1, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , Column-I, Column-II, p) Acceleration at t= 5 s, S1,S2 and S3 are the different sizes of windows 1,2 a) 0, b), 2, q) Average speed from t =0 to t =6s, and 3 respectively, placed in avertical plane. A, r) Velocity at the point of reversal of, particle is thrown up in that vertical plane. Find the c) 3, correct options:, motion, a) average speed of the particle, d) 18, s) Total distance travelled from t =0 to, S3, t=6s, passing the windows may be equal if, t) Displacement from t =0 to t=6s, s1 s2 s3, b) average speed of the particle passing, S2, 34. For a body projected vertically up with a, the windows may be equal if, S1 S2 S3, velocity v0 from the ground, match the, c) If S1 S2 S3 , the change in speed, following, of the particle while crossing the, Column-I, , windows will satisfy V1 V2 V3 ., A. V av (Average velocity), d) If S1 S2 S3 , the time taken by, S1, B. U av ( Average speed), particle to cross the windows will, satisfy t1 t 2 t 3 ., C. Tascent, At t 0 , an bullet is fired vertically upward, D. Tdescent, with a speed of 100 ms 1 . A second bullet is, fired vertically upwards with the same speed, Column-II, at t 5 s. Then, i. Zero for round trip, A) The two bullets will be at the same height, , v1 v 2, above the ground at t 12.5 s, ii., over any t ime interval where, 2, B) The two bullets will reach back their strarting, , points at t 20 s and at t 25 s, v1 & v 2 are the intial and final velocities in the time, C) The ratio of the speeds of the first and second, interval, bulletsat t 20 s will be 2: 1, v0, over the total timeof its flight, iii., D) The maximum height attained by either bullet, 2, will be 1000 m, v0, From the top of a tower of height 200 m, a, iv., 1, ball A is projected up with 10 ms , and 2 s, g, later another ball B is projected vertically, INTEGER TYPE QUESTIONS, down with the same speed Then, 35. From a lift moving upwards with a uniform, A) Both A and B will reach the ground, acceleration a 2ms 2 , a man throws a ball, simultaneously, B) Ball A will hit the ground 2 s later than B hitting, vertically upwards with a velocity v 12 ms 1, the ground., relative to the lift. The ball comes back to the man, C) Both the balls will hit the ground with the same, after a time t. Find the value of t in second, velocity, (g=10ms-2)., D) Both the balls will hit the ground with the, 36. A body is thrown up with a velocity 100 ms 1 . It, different velocity, travels 5 m in the last second of upword journey., MATRIX MATCH QUESTIONS, If the same body is thrown up with a velocity, A particle moves such that , t x 3 , where, 200 ms 1 , how much distance ( in metre ) will it, x is in metre, t is in second. Based on this, travel in the last second of its opward journey, information, match the value in Column-I (in, , MULTIPLE ANSWER QUESTIONS, 30., , 31., , 32., , 33., , SI units) to their respective quantities for the, particles motion given in Column-II, NARAYANAGROUP, , g 10ms , 2, , 81
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , C. GRAPHS, 6, , SINGLE ANSWER QUESTIONS, , V, , 37. The following graph shows the variation of, velocity of a rocket with time. Then the, maximum height attained by the rocket is, , 2, 30, , 0, –2, –6, , 40, , t(s), , 10 18 24, , A) The maximum value of the position coordinate, of the particle is 54 m., B) The maximum value of the position coordinate, of the particle is 36 m., C) The particle is at the position of 36 m at, t 18 s ., D) The particle is at the position of 36 m at, t 30 s, , LEVEL - V - KEY, 1. D 2. B 3. C 4. C 5. C 6. B 7.A,B,C, 8.A,B,C 9. A,B,D 10. C,D 11. A,C,D 12. B,C, 13. A,C 14.C 15. A 16. B 17. B 18. D, 19. i - b, ii - a, iii - d, iv - c, 20. 8, 21. 1 22. 9 23. C 24. B 25. C, 26. A, 27. B 28. A 29. C 30. B,C,D, 31. A,B,C 32. A,C, 33. a- (r,t) b-(p) c-(q) d-(s), 34. A - i,ii B - iii, C - iv, D - iv, 35. 2, 36. 5 37. C 38. C 39. A,B,C,D, 40. A,C,D, , A) 1.1 km B) 5 km C) 55 km D) None, 38. The velocity-time graph of a particle moving, in a straight line is shown in figure. The, acceleration of the particle at t 9 is, v(ms–1), , 15, 10, 5, t(s), 2, , 4, , A) Zero B) 5 ms, , 6, , 2, , 8, , LEVEL - V - HINTS, , 10 12, , C) 5 ms, , 2, , D) 2 ms, , 2, , MULTIPLE ANSWER QUESTIONS, 39. Figure shows the velocity( v) of a particle, plotted against time ( t)., , 1., , s/3 s, , 4, 12, , 2s, 2s, 2 t 0 4 kt 0 t 0 2 4k or t 0 , 3 2 4k , 3, s, Average velocity = t t t k, 1, 0, 0, , v, T, O, , Sol: t1 , , 2T, t, , , A) The particle changes its direction of motion at, some point, B) The acceleration of the particle remains, constant ., C) The displacement of the particle is zero, D) The initial and final speeds of the particle are, the same., 40. A particle moves in astraight with the velocity, 2., as shown in figure. At t 0, x 16 m ., , 6 2 4k , s, , s 2s 1 k , 5 6k , , 12 3 2 4k , , v av 5 6k 12 24k gives k , , 1, 2, , t0, ., 2, Given v 3 x 2 2 x; differentiating v , we get, , Required time = kt 0 , , dv, dx, 6 x 2 6 x 2 v, dt, dt, 82, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , a 6 x 2 3 x 2 2 x Now put x= 2 m, , 6., , |Average velocity| =, , displacement, time, , 2, , 3., , a 6 2 2 (3 2 2 2) 80 ms 2, If police is able to catch the dacoit after time t, then, 1, , vt x t 2. This gives t 2 vt x 0, 2, 2, , AB, , 7., , , dx, , v v a kv k v kx , x , 0, dt, , , , v v 2 2 x, For t be real, v 2 2 x, or t , , 4.SOL: x 1 t or x 1 t, 2, , 2, , , , dv, dv, dv, kv , kv , kdt, dt, dt, v, , Further, a , , 2 1/ 2, , , , 1/ 2, 1/ 2, dx 1, 1 t 2 2t t 1 t 2 , dt 2, 1/ 2, d 2x, 1, a 2 1 t2, t 1 t2, dt, 2, , , , 2, , = time 1 = 2 m/s., dv , v 0 kx, Acceleration =, dt, , , , 1 t2, 3, x x, , v1, , t, , v , dv, 1, v v k 0 dt t k log e v10 , , 0, Since, v=v 0+kx. Hence slope of velocity, dv, k, displacement curve is, dx, u1a1, u2a2, , , , , 3/ 2, , 2t, , 8., , l/2, l/2, l, 1, l, 1, u1 t a1 t 2 ...(1) and u1t a 2 t 2, 2m, 8, 2, 2, 2, 2, 5., 4m, l, 1, , u 2 t a 2 t 2 ....(2), v cos, 2, 2, 2, u 2 u1 , Time of crosing v sin , subtracting (1) and (2) , we get t 2 a a ....(3), 1 2, Time in which truck just able to catch the man =, Substituting (3) in (1) or(2) and rearranging, we get., 4, 4 u 2 u1 , 8 v cos , l, a u a 2 u1 ...(4), 2 1 2, a1 a 2 , , 2, 4, Since the particle P & Q reach the other ends of, For safe crosing v sin = 8 v cos , A and B with equal velocities say v, 8, For particle P v 2 u12 2a 2 l ... (5), or 16 2 v cos 4v sin or v cos 2sin , For particle Q v 2 u 22 2a 2 l ...(6), For v minimum cos 2sin is maximum, Substracting and then substituting value of l and, d, rearranging, we get, so ,, cos 2sin 0, d, u 2 u1 a1 a 2 8 a1u 2 a 2 u1 , sin 2 cos 0, dv, 9. a , , If velocity changes, definiety there will, tan 2, dt, be acceleration. If speed changes, then velocity, 1, 2, cos , , sin , also changes, so definietely there will be, 5, 5, acceleration., Acceleration may be due to change in the, 8 5, 8, v min , , direction of velocity only and not magnitude., 5, 5, Now, If body has acceleration, its speed may changes, 3.57 m / s, if acceleration is due to change in magnitude of, velocity., v sin, , NARAYANAGROUP, , 83
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 5, , vdt, , 4t t dt, 2, , 0, 5, , , , 0, , 5, , dt, , dt, , 0, , 0, , 2 t3 , 125, 2t 3 , 50 , , 0, 3 25 5, , , 5, 5, 35 3, For average speed, let us put v= 0, which gives, t 0 amd t 4s, speed, =, average, 4, , 5, , 4, , 2, (4t t )dt vdt, , vdt vdt, 0, , 4, , , , 5, , 5, , 0, , 13. Since, the body is at rest at x = 0 and x = 1. Hence,, a cannot be positive for all time in the interval 0 t, 1, Therefore, first the particle is accelerated and then, retarded. Now, total time it = 1 s (given), S = Area under v – t graph, 2s, , \Height or vmax = t = 2 m/s is also fixed., 1, , , Area or S = 2 t vmax , , , V, Vmax = 2m/s, , 10. Average velocity, , , v, , 5, , 2, 1, , 3, , 4, , O, , 5, , dt, , 1, , t, , If height and base is fixed, area is also fixed, I n case 2 : A cceleration = Retardation = 4 m/s2, , 0, , 4, , 5, , 2 t3 , 2 t3 , 2t 3 2t 3 , 0 , 4, , 5, 4, , 5, , 2 t3 , 2 t3 , 2t 3 2t 3 , , 0 , 4 13 1, , ms, 5, 5, , For acceleration :, dv d, a, 4t t 2 4 2t, dt dt, dv, 11. v. dx v or, , At t 0, a 4 ms 2, , 0, , x0, , dv dx, , v0, , v, v0 x0 x0 0 ;, , v, t, dv, dv, v (or ) , dt, dt, v, v0, 0, v v0 e t or v 0 for t , , 0, , In case 1 : Acceleration > 4 m/s2 while Retardation, < 4 m/s2., Hence, a ³ 4 at some point or points in its path., 14. At time t, Xt and Xp are coordinates of train and, passenger respectively., 1, X t a1 t 2 and X p v P t S, 2, If passenger catches the train,, Xt = Xp, v P v P2 2a1S, 1 2, or a1t vP t S or t , a1, 2, , , 10 , , 10 , , 2, , 2 1 25.5 , , 3seconds, 1, 15. The critical distacne ‘Sc’ for which passenger will, take the ten seconds time to catch the train is given, , by Sc , , v 2P, 2a1, , The time is 10 seconds, if v 2P 2a1S 0, , v0, 1, 2, ., when t , v 2P 10 , Sc , , 50m, e, , 2a t 2 1, 12. The particle’s velocity is getting zero at t = 3 s,, 16. For critical distance, passenger catches the train in, where it changes its direction of motion., For 0 < t < 3 s, V is negative, a is positive, so, vP, particle is slowing down., time, t a So, required velocity of train = a t .t, t, For t < 3, both V and a are positive, so the particle, is speeding up., v , a t P VP / 2 10m / sec, at , 17. Let they meet after time t , then the distance, travelled by both in time t should be same, v, , 84, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , 26. Suppose h be the height of each storey, then, t1 t 2 t 3 if S3 S2 S1, 1, 1, 31. Let they meet at hieght h after time t., 25h 0 10 t 2 10 52 or h 5m, 2, 2, 1, h 100t gt 2 for first bullet, In first second, let the stone passes through n, 2, storey., 1, 2, 1, 2, 100 t 5 g t 5 for second bullet, So n 5 10 1 or n 1, 2, 2, t 12.5sec (after solving) So (a) is correct., 27. Suppose v be the velocity attained by body after, Time of flight of first bullet, time t1 . Then v u gt1, (i), 2u 2 100, T, , 20 s, Let the body reach the samepoint at time t2 . Now, g, 10, velocity will be downwards with same magnitude, Second bullet will reach after 5 s of reaching first, ( ii), v , then v u gt2, . So ( b) is correct., ( i) -( iii ) 2v g t2 t1 , , v1 100 10 20 100 ms 1, , v2 100 10 15 50 ms 1, u, , 2v 2, u gt1 2 t1 , or t2 t1 , v1, g g, g, , 2 :1, So ( c) is correct., ratio:, 28. (i) For uniformly accelerated/decelerated motion, v2, 2, 2, v = u 2gh, maximum height attained, i.e., v – h graph will be a parabola (because, 2, u 2 100 , equation is quadratic)., (ii) Initially velocity is downwards (–ve) and then H 2 g 2 10 500 m . Hence ( d ) is incorrect, after collision it reverses its direction with lesser 32. Ball, A will return to the top of tower after, magnitude. I.e., velocity is upwards (+ve). Graph, 2u 2 10, (A) satisfies both these conditions., T, , 2s, Therefore, correct answer is (A)., g, 10, Note that time t = 0 corresponds to the point on, With speed of 10ms 1 downward., the graph where h = d, And this time, B is also projected downwards, v, with 10ms 1 ., at t = 0, h = d, 2, So both reach ground simultaneously. Also they, 1 2: increases, d, h, will hit the ground with the same speed., downwards, 3 1, At 2 velocity changes, dx, 2, its direction, 2t 6, 33. x t 3 x t 2 6t 9 v , Collision takes, dt, 2 3V decreases upwards, place here 2, At the point of reversal, v=0, so 2t 6 0 t 3s, 29. Distance traveled in tth second is,, , st = u + at –, sn, , s, n 1, , 1, a, 2, , Given : u = 0, , 1, a, 2, 2n – 1, =, 1 = 2n 1, a n 1 – a, 2, an –, , Hence, the correct option is (C)., 30. As going up, speed of the particle is decreasing, and hence time taken in crossing the windows, (if S1 S2 S3 ) will be t1 t 2 t 3 . Since,, , v u at u t (as acceleration is same), So, v1 v 2 v3 (as for equal windows, t1 t 2 t 3 ) For unequal windows., 86, , t, x, v, a, , 0, 9, -6, 2, , 1, 4, -4, 2, , 2, 1, -2, 2, , 3, 0, 0, 2, , 4, 1, 2, 2, , 5, 4, 4, 2, , Nature, Decele Decele Decele Accele Accele Accele, of, rating rating rating rating rating rating, Motion, , Displacement of the particle from t =0 to t =6s is, zero Distance travelled by the particle from t =0 to, t=6 is 9+9=18m Average speed of the particle is, Total Distance Travelled, v av , Total Time Taken, vav , , 18, 3ms 1, 6, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , Average Velocity of the particle is, Displacement, , v av , 0, Time, Acceleration of the particle over the entire duration 37., -2., of motion is 2ms, , 200 5 20 2 1 5 m, In the last second of upward journey , the bodies, will travel same distance., Maximum height will be attained at 110 s., Because after 110 s, velocity becomes negative, and rocket will start coming down. Area from 0, , Displacement of the particle from t =0 to t =6s is, zero Distance travelled by the particle from t =0 to, 1, t=6 is 9+9=18m, to 110 s is 110 1000 55, 000 m 55 km, 2, Average speed of the particle is, 38., Acceleration, between 8 to 10 s( or at t= 9 s) ., Total Distance Travelled, v av , v v 5 15, Total Time Taken, a 2 1 , 5m / s 2, t2 t1 10 8, 18, 1, vav 3ms, 39. Particle changes direction of motion at t T ., 6, Acceleration remains constant, because the, Average Velocity of the particle is, velocity-time graph is a straight line. Displacement, Displacement, , is zero, because net area is zero. Initial and final, v av , 0, speeds are equal., Time, Acceleration of the particle over the entire duration 40. Maximum value of position coordinate = initial, of motion is 2ms-2., coordinate+area under graph up to t 24 s ( As, 34. For a round trip, displacement is zero; hence, up to t= 24 s, the displacement of the particle will, , be positive ), Also, v av 0., , , , , v1 v 2, 6, v av , , when v1 is initial, v2 is final., 2, 2, V 0, –2, , Hence i. a, b, Average speed, , lift = ( g+ a) downwards, so areal g a ,, , Maximum value of position coordinate, , 1 6, , 26, 16 2 10 , 18 10 2 24 18 , 2 , , , , LEVEL - VI, A. MOTION ALONG HORIZONTAL AXIS, SINGLE QUESTION TYPE, , initial velocity; urel v , final velocity vrel v as 1., the ball will reach the man with same speed w.r.t, lift., Apply vrel urel arel t v v g a t t 2s, 36. s u , , t(s), , –6, , 2, Total dis tan ce 2 v0 / 2 g v0, , , ( vav ) =, Time of flight, 2v0 / g, 2, , v, Tascent Tdescent 0 Henace iii d ., iv d, g, 35. Taking upward direction as positive, let us work in, the frame of lift. Acceleration of ball relative to, , 30 40, 10 18 24, , a, 2n 1, 2, , The decelaration experienced by a moving, motor boat, after its engine is cut-off is given, dv, kv 3 , where k is constant.If v0 is the, by, dt, magnitude of the velocity at cut-off, the, magnitude of the velocity at a time t after, the cut-off is, , u 100ms 1 , a 10 ms 2 and s 5m, , c) v0 e kt d), , v0, , 5 10 5 2n 1 gives n 10 s, , a) v0 / 2, , Body when thrown up with velocity 200 ms 1 2., will take 20 s to reach the highest point. Distance, travelled, in, second, is, 20th, , A jet plane starts from rest at S = 0 and is, subjected to the acceleration shown., Determine the speed of the plane when it has, travelled 60 m., , NARAYANAGROUP, , b) v, , 2v02 kt 1, , 87
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 6., , 2, , a(m/s), 22.5, , A train stops at two stations s distance apart, and takes time t on the journey from one station, to the other. Its motion is first of uniform, acceleration a and then immediately of uniform, retardation b, then, 1 1 t2, a) , a b s, , 150, , A) 46.47 m / s, 3., , S(m), , B) 36.47 m / s, , 7., C) 26.47 m / s, D) 16.47 m / s, Velocity versus displacement graph of a, particle moving in a straight line is shown in, figure. Corresponding acceleration versus, velocity graph will be :, v(m/s), , 10, , 1 1 t2, , a b s, , b), , 1 1 t2, 1 1 t2, d) , c) , a b 2s, a b 2s, Two stones are thrown up simultaneously with, initial speeds of u1and u2 (u2>u1). They hit the, ground after 6 s and 10 s respectively. Which, graph in fig.correctly represents the time, variation of x (x 2 x1 ) , the relative, position of the second stone with respect to, the first upto t=10 s? Assume that the stones, do not rebound after hitting the ground., A, , 10, , a(m/s2), , a(m/s2), , A), , B, 0, , 10, , 10, , A), , x, , x, , s(m), , B), , B, 0, , 2 4 6 8 10, t, , B), A, 10, , v(m/s), , 10, , a(m/s2), , v(m/s), , x, , C), 10, , C), 4., , 5., , 88, , A, , x, , a(m/s2), 10, , B, 0, , D), 8., 10, v(m/s), 10, v(m/s), The relation between time t and distance x, is t a x 2 b x . Where a and b are, constants. The retardation is, a) 2av 3, b) 2b v 3, c) 2ab v3, d) 2b 3 v 3, The motion of a body falling from rest in a, resisting medium is described by the equation, dv, = a -b v where a and b are constants., dt, The velocity at any time t is given by, a, b -bt, -b t, (a) v = 1 -e , (b) v= e , 9., b, a, a, b bt, -bt, (c) v = 1 + e , (d) v= e, b, a, , 2 4 6 8 10, t, , 2 4 6 8 10, t, , D), , B, 0, , 2 4 6 8 10, t, , A particle moving along x - axis has, acceleration a at time t given by, , t, a a0 1 where a0 and T are constants., T , The particle at t 0 has zero velocity. The, particles velocity when acceleration reduces, to zero., a), , 1, a 0T 2, 2, , b) a0T 2, , c), , 1, a0T, 2, , d) a0T, , A cone falling with a speed v0 strikes and, penetrates the block of a packing material. The, accelertation of the cone after impact is, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, a g cx 2 . Where c is a positive constant, and x is the penetration distance. If maximum, penetration depth is x m then c equals, 2 gx m v02, a), x m2, , 2 gx m v02, b), x m2, , 6 gx m 3v02, c), 2 x m3, , 6 gx m 3v02, d), 2 x m3, , MOTION IN A STRAIGHT LINE, s, , 5s t1, , t2, , t, , 15. The maximum hight reached by the stone is, A) 30m, B) 40m, C) 45m, D) 28m, MULTIPLE ANSWER QUESTIONS 16. t1 is, A) 4s, B) 6s, C) 2s, D) 1s, 10 If the velocity of the particle is given by v x, and intially particle was at x=4m, then which 17. t2 is, A) 4s, B) 4.45s C) 3.45s D) 9.45s, of the following are correct., (A) at t=2 sec, the position of the particle is x= 9 m, MATRIX MATCHING QUESTIONS, (B) Particle acceleration at t= 2 sec. is 1 m/s2., 18. Study the following., (C) Particle acceleration is ½ m/s2 through out the, List - I, motion, (D) Particle will never go in negative direction from a) A body covers first half of distance with a speed v1, its starting position., and second half of distance with a speed v2, 11. Starting from rest a particle is first accelerated, for time t1 with constant acceleration a1 and b) A body covers first half of a time with a speed v1, then stops in time t2 with constant retardation, and second half of time with a speed v2, a2. Let v1 be the average velocity in this case, and s1 the total displacment. In the second c) A body is projected vertically up from ground, case, it is accelerated for the same time t1 with, with a speed gh .Considering its total motion, constant acceleration 2a1 and comes to rest, with constant retardation a2 in time t3. If v2 is d) A body freely released from a height h, List - II, the average velocity in this case and s2 the, total displacement. Then, gh, (a) v2 = 2v1, (b) 2v1 < v2 < 4v1, i) Average speed is, 2, (c) s2 = 2s1, (d) 2s1 < s2 < 4s1, , PASSAGE TYPE QUESTIONS, , v v2, ii) Average speed is 1, Comprehension - 1, 2, Two particles A and B start from rest at the, 2v1v2, origin x=0 and move along a straight line such, iii), Average, speed, is, -2, 2, -2, that aA=(6t-3)ms and aB=(12t -8)ms , where, v1 v2, t is in seconds. Based on the above facts,, answer the following questions., gh, iv) Average speed is, 12. Total distance travelled by A at t=4 s is, 2, A) 40m B) 41m, C) 42m, D) 43m, 19. For a particle moving along X - axis if, 13. Total distance travelled by B at t=4 s is, acceleration (constant) is acting along -ve XA) 192m B) 184m C) 196m D) 200m, axis, then match the entires of Column I with, 14.. Total distance between them at t = 4 s is, entires of Column II., A) 144m B) 148m C) 152m D) 156m, Column -I, Comprehension - 2, (A) Initial velocity > 0, A balloon is start rising with constant, acceleration 2m/s2 from ground at t=0s. A stone, (B) Initial velocity < 0, is dropped at t=5s. s-t graph for the given, (C) x > 0, situation is shown in figure. Answer the following., (D) x < 0, NARAYANAGROUP, , 89
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MOTION IN A STRAIGHT LINE, , JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , shortest time in which the train can travel, Column -II, between two stations 8 km apart is x minutes, i) Particle may move in +ve X - direction with, and 10 s, if it stops at both stations. The value, increasing speed., of x is., ii) Particle may move in +ve X-direction with, 24. A body starts from rest with uniform, decreasing speed., acceleration. Its velocity after 2n second is, iii) Particle may move in -ve X - direction with, v0. the displacement of the body in last n, increasing speed., iv) Particle may move in -ve X - direction with, 3v0 n, second is, . Determine the value of ?, decreasing speed., , , INTEGER TYPE QUESTIONS, , B. MOTION ALONG VERTICAL AXIS :, , 20. A train starts from station A with uniform, SINGLE ANSWER QUESTIONS, acceleration a1 for some distance and then 25. A ball is thrown from the top of a tower in, vertically upward direction. Velocity at, goes with uniform retardation a2 for some, apoint h metre below the point of projection, more distance to come to rest at station B., is twice of the velocity at a point h metre, The distance between stations A and B is 4, above, the point of projection. find the, km and the train takes 1/15 h to complete, maximum height reached by the ball above, this journey. Acceleration are in km/ min 2, the top of tower., 1 1, a) 2 h, b) 3 h, c) (5/3) h d) (4/3) h, unit. If a a x , find the value of x., 26. A parachutist drops first freely from an, 1, 2, aeroplane for 10 s and then his parachute, 21. A cat, on seeing a rat a distance d 5 m,, opens out. Now he descends with a net, starts with velocity u 5 ms 1 and moves, retardation of 2.5 ms 2 . If he bails out of the, with acceleration 2.5 ms 2 in order to, plane at a height of 2495 m and g 10 ms 2 ,, catch it, while the rat with acceleration , his velocity on reaching the ground will be, starts from rest. For what value of will the, a) 5 ms 1 b) 10 ms 1 c) 15 ms 1 d) 20 ms 1, cat overtake the rat ? (in ms 2 ), MORE THAN ONE QUESTION TYPE, 22. On a two-lane road, car A is travelling with 27. A particle is projected vertically upward with, velocity u from a point A, when it returns to, a speed of 36 kmh 1 . Two cars B and C, point of projection, approach car A In opposite directions with a, a) Its average speed is u /2, 1, speed of 54 kmh each. At a certain instant,, b) Its average velocity is zero, when the distance AB is equal to AC, both, c) Its displacement is zero, being 1 km, B decides to overtakes A before, d) Its average speed is u, C does. What minimum acceleration of car B 28. A particle is thrown vertically in upward, direction and passes three equally spaced, (in m / s 2 ) is required to avoid an accident ?, windows of equal heights then, –1, –1, –1, vB = 15ms, vA = 10ms, vC = 15ms, (A) average speed of the particle while passing the, windows satisfies the relation u av u av u av, (B) the time taken by the particle to cross the, B, A, C, windows satisfies the relation t1 < t2 < t3, 1000 m, 1000 m, (C) the magnitude of the acceleration of the particle, 23. The accelerator of a train can produce a, while crossing the windows satisfies the relation a1, = a2 a3, uniform acceleration of 0.25 ms-2 and its brake, -2, can produce a retardation of 0.5 ms . The, 1, , 90, , 2, , 3, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, , (D) the change in the speed of the particle while, C. GRAPHS:, crossing the windows would satisfy the relation, SINGLE ANSWER TYPE, 37. The velocity-time graph of abody is given in, u1 u 2 u 3 ., figure. The maximum accceleration in ms 2, COMPREHENSION QUESTION, is, An elevator car whose floor to ceiling distance in, v(ms ), equal to 2.7 m starts ascending with constant, 2, acceleration 1.2m/s , 2 sec. after the starts a bolt, 60, begins falling from the ceiling of the car. Answer the, 2, following questions. (g=9.8 m/s ), 29. The bolt’s free fall time, 20, (A) 0.3 s (B) 0.5 s (C) 0.7 s (D) 0.9 s, t(s), 0, 30. The velocity of bolt at instant it loses contact is, 20 30 40, 70, a) 4, b) 3, c) 2, d) 1, (A) 1.2 m/s(B) 2.4 m/s (C) 4 m/s (D) 10 m/s, 38., The, velocity-time, graph, of, a, body, is shown, 31. Distance moved by elevator car w.r.t. ground, in, figure., The, ratio, of, magnitude, of, average, frame during the free fall time of the bolt., acceleration during the intervals OA and AB, (A) 1.44 m (B) 1.63 m (C) 1.68 m (D) 1.97 m, is, 32. Distance covered by the bolt during the free, v(ms ), fall time w.r.t. ground frame., (A) 0.7 m (B) 0.9 m (C) 1.1 m (D) 1.3m, D, C, 33. The displacement by the bolt during its free, 40, fall time w.r.t. ground frame, (A) 0.3 m (B) 0.7 m (C) 0.9 m (D) 1 m, 30°, 60°, E, –1, , –1, , INTEGER TYPE QUESTIONS, , O, , A, , B, , t(s), , a) 1, b) 1/2, c) 1/3, d) 3, 34. A stone is dropped from a height h., 39., The, displacement-time, graph, of, a moving, Simultaneously another stone is thrown up, particle with constant acceleration is shown, from the ground with such a velocity that it can, in the figure. The velocity time graph is given, reach a height of 4h. The time when two stones, by, cross each other is, , x(m), , h , , kg where k = ______, , 35. A particle moves along x-axis satisfying the, equation x t t 1t 2 (t is in seconds, and ‘x’ is in meters). Find the magnitude of, initial velocity of the particle in m/s., 36. The position vector of a particle varies with, , , , 0, , v, , , , 0, , 1, , t B), , 2, , v, , Kr0, . Determine the value of K?, 16, C), , NARAYANAGROUP, , t(s), , 2, , v, , time as r r0t 1 t where r 0 is a constant, vector and is a positive constant. The A), distance travelled by particle in a time interval, in which particle returns to its initial position, is, , 1, , 0, , t, 0, , 1, , 2, , v, , 1, , 2, , t D), , 0, , 1, , 2, , t, , 91
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 40. The displacement-time graph of a moving, particle is shown in figure. The instantaneous, velocity of the particle is negative at the, point., , x, , x, , displacement, , t, , t, , D, F, , C, , x, , x, , E, , a) D, , b) F, , c) C, , d) E, t, , MULTIPLE ANSWER QUESTIONS, , t, , 41. The velocity-time plot a particle moving on a, straight line is shown in figure., 43. The particle is moving with constant speed, a) In graphs( i ) and ( iii ), v(ms ), b) In graphs ( i ) and ( iv ), 10, c) In graphs ( i ) and ( ii ), d) In graphs ( i ), t(s), 0, 44. The particle has negative acceleration, 10, 20, 30, a) In graph ( i ), b) In graph ( ii ), –10, c) In graph ( iii ), d) In graph ( i v), –1, , –20, , MATRIX MATCHING QUESTION, a) The particle has a constant acceletration, 45. The displacement versus time is given, b) The particle has never tuned around, c) The particle has zero displacement, figure. Sections OA and BC are parabolic., d) The average speed in the interval 0 to 10 s is, CD is parallel to the time axis., the same as the average speed in the interval 10 s, C, D, to 20 s, 42. The displacement of a particle as a function, B, S, A, of time is shown in figure. It indicates, S, O, , t, , Column-I, A. OA, B. AB, C. BC, t, O, 1, 2, 4, 5, D. CD, a) The particle starts with a certain velocity, but, Column-II, the motion is retarded and finally the particle stops, i. Velocity increases with time linearly, b) The velocity of the particle decreases, ii. Velocity decreases with time, c) The acceleration of the particle is in opposite, iii. Velocity is independent of time, direction to the velocity, iv. Velocity is zero, d) The particle starts with a constant velocity, the, 46., Study the following v t graphs in Column I, motion is accelerated and finally the particle, carefully and match appropriately with the, moves with another constant velocity., statements given in Column II. Assume that, PASSAGE TYPE QUESTION, motion takes place from time 0 to T ., PASSAGE-1:, Study the following graphs:, , 92, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, Column-I, v, , MOTION IN A STRAIGHT LINE, v 75 22.5 45 13.5 46.47, 1, , 4., t x 2 x; 1 2 xv v v 2 x , 0 2 x.a v.v a 0 2 x a 2 v 2, , v, v0, , T, , A) O, , B) O, , t, , –v0, , T, , t, , 2 v 2, 2 v 2, , 2 v3, 1, 2 x , v, v, t, dv, dv, a bv , dt, 0 a bv, 0, dt, v, 2 , a bv, bt, ln a bv 0 t ln, a, b , a bv ae bt, a, v 1 e bt , b, 2, v, v2 v2 1 1 , s, , , 2a 2b 2 a b , , ab , Again t v t , , a b, ab, t2, 1 1, s , , 2, t2 1 1, 1 1 a b , 2 , 2s a b, a b, a, , v, , v, v0, C) O, , v0, T/2, , T, , D) O, , –v0, , T, , t, , i, , 5., , Column-II, i. Net displacement is positive, but not zero, ii. Net displacement is negative, but not zero, iii.Particle returns to its initial position again, iv. Acceleration is positive., , LEVEL - VI - KEY, , 6., , 1. D, 2. A 3. A 4. A 5. A 6. C, 7. A, 8. C 9. D 10. A,C,D, 1.A,D, 12) B, 13) D 14) C 15) A 16) B 17) D, 18. A-iii, B- ii, C-iv, D-i, 19. A-ii, B-iii, C-ii, D- ii,iii, 20. 2, 21. 5 22. 1 23. 5 24. 4 25. C, 26. A, 27. ABC 28. ABD 29. C 30. B 31. D, 32. D, 33. B 34. 8 35. 2 36. 8 37. A, 7., 38. C, 39. A 40. D 41. AD 42. ABC, 43. B, 44. C, 45. A - i, B - iii, C - ii, D - iv, 46. A - ii,iv, B - i,iv C - iii, D - i, , 1 2, gt, 2, 1, x2 u2t gt 2, 2, x1 u1t , , It is valid up t=6sec only, , LEVEL - VI - HINTS, 1., , dv, kv 3, dt, dv, or 3 kdt or, v, , 8., , Here, , v, t, dv, , 3, v0 v 0 kdt, v, , v02, v0, v , or v , 2, 1 2v0 kt, 2v02 kt 1, s, v, dv, a v a.ds v.dv v 2 u 2, 0, u, ds, Area, under, curve, as, , v2 u 2, 1, 1, 2, 2 150 22.5 2 90 13.5 v 0, , , 2, v 75 22.5 45 13.5, , NARAYANAGROUP, , T, , 0, , 1, 2, 1 , or 2v 2 kt or 2v 2 2v 2 kt , or, , V0, 0, 2, , a0T, a0t, a 0, at t a T, T, 0, T, , dv a0 dt , , v, , 2., , a a0 , , 9., , 0, , a0, t dt, T 0, , a0 T 2 a0T, , v a0T , 2, T 2, dv, v g cx 2, dx, 0, , , , x, , x, , v dv g dx c x2 dx, , v0, , , , 0, , v02, 2, , 0, , 3, , gx , , cx, 3, , 93
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A STRAIGHT LINE, 26. The velocity v acquired by the parachutist after, 10 s., , d rel 1000 m, vrel 10 15 25 ms 1, d rel 1000, Here t v 25 40 s, rel, , v u gt 0 10 10 100ms 1, 1 2, 1, gt 0 10 102 500m, 2, 2, The distance travelled by the parachutist under, retadation is, , Then, s1 ut , , Let acceleration of B be a for overtaking., d rel 1000 m; vrel 15 10 5 ms 1, , d rel a and t 40 s, Using d rel, , s2 2495 500 1995 m, , 1, urel t arel t 2, 2, , Let vg be the velocity on reaching the ground., Then vg2 v 2 2as2, , 1, 2, 100 5 40 a 40 a 1 ms 2, 2, , 23. tmin, , t m in , , 2 l , , , , , , , , 1, , 1/ 2, , vmax, , 0 .2 5 0 .5 8 1 0 3 2, 0 .2 5 0 .5, , 24. v 0 2na, 3 2, an, 2, , 25. H , , a, , , 2, , or vg2 100 2 2.5 1995 or vg 5 ms 1, , 2 l 2, , , , 3 10 s 5 m in 1 0 s, , v0, 1, 1, 2, s 2n an 2, 2n, 2, 2, , 3 v 0 2 3v 0 n, n , 2 2n, 4, , u2, ; given v2 2v1, 2g, , ., , 27. For vertically projected body, if it returns to the, starting point, displacement and average velocity, become zero. As acceleration is constant,, average speed during upward or downward, motion is u 0 / 2 u / 2 . The same will be the, average speed for the whole motion., 28. As the particle is going up, it is slowing down, i.e.,, speed is decreasing and hence we can say that time, taken by the particle to cover equal distances is, increasing as the particle is going up., Hence, t1 < t2 < t3., As u av , , Distance, 1, , we have u av , time, time, , (i) A to B: v12 u 2 2 gh, , Acceleration throughout the motion remains same, from equation,, , (ii) A to C v22 u 2 2 g h , , , v u at, u t . So, u1 u 2 u 3 ., , (iii) solving (i), (ii),(iii) we get the value of u 2 as, 1, 1, 10g/h/3 and then we get the value of H by using 29. yrel urel t arel t 2 2.7 0 9.8 1.2 t 2 t 0.7 sec, 2, 2, u2, 30. v = u + at = 0+(1.2)(2) = 2.4 m/s, H, 2g, 1 2, 1, 2, 31. y ut at 2.4 0.7 1.2 0.7 , 2, 2, v, H, , 1, , 2, , B, u, , h, , A, h, C, v2, , 2u 2, 2.4 1.3m, 0.7 , 32. Distance s | y | , 2g, 9.8, , , , , 33. Sb Sbe Se Sb 2.7 1.97 Sb 0.7m ., 34. S1 , , 1 2, 1, gt ; S 2 ut gt 2, 2, 2, , u2, u 8 gh, S1 S2 h; 4h , 2g, NARAYANAGROUP, , 95
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MOTION IN A STRAIGHT LINE, , ut h, 8gh t h, , t , , h, 8g, , , , 2, 35. x t t 3t 2, , , , t 3 3t 2 2t, v 3t 2 6t 2 v0 2, 36. r r0 t 1 t , , v r0 2 t r0, , JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, 41. Since the graph is astraight line, its slope is, constant, it means acceleration of the particle is, constant., Velocity of the particle changes from positive to, negative at t 10 s , so particle changes direction, at this time., The particle has zero displacement up to 20 s,, but not for the entire motion., The average speed in the interval of 0 to 10 s is, the same as the average speed in the internal of, 10 s to 20 s because distance covered in both, time interval is same., 42. Initially at orogin, slope is not zero, so the particle, has some initial velocity but with time we see that, slope is decreasing and finally the slope necomes, zero, so the particle stops finally., , PASSAGE TYPE, 1, 43,44., 2, distance travelled before coming to original, 1) For the graphs ( i ) and ( iv ), slope is constant, position, hence the velocity is constant, 2) For The graph ( iii ), the particle’s velocity first, r0 r, 0, decreases and then increases in negative directio., r 2 , 4, , It means negative acceleration is involved in this, 2, motion., k 8, dS, 2, 2t, 45. In OA, S t , v , dt, 37. Maximum acceleration will be from 30 to 40 s,, i. e., v t, because slope in this interval maximum, i.e., velocity increases with time, v2 v1 60 20, 2, a, , 4 ms, dS, t2 t1 40 30, 1, In AB, S t , v , dt, 1, 0, ms 2, 38. During OA , acceleration = tan 30 , 46. 1) Area of v t graph lies below the time axis, so, 3, displacement is negative, but slope is positive, so, 0, 2, accceleration is also positive., During AB , acceleration = tan 60 3 ms ., 2) Area of v t graph lies above the time axis, so, 1/ 3 1, displacement is positive, and slope is positive, so, , required ratio =, accceleration is also positive., v 0 at t , , 3, , 3, , 39. At t 0 , slope of the x-t graph is zero; hence,, velocity is zero at t 0 . As time increases, slope, increases in negative direction; hence, velocity, increases in negative deirection. At point’I’, slope, changes suddenly from negative to positive value:, hence, velocity changes suddenly from negative, to positive and then velocity starts decreasing and, becomes zero at’2’, option ( a) represents all, these clearly., 40. The slope of the graph is negative at this point., , 96, , 3) Displacement is zero, because half area is, above time axis and half below. Slope is negative,, so acceleration is negative, 4) Area of v t graph lies above the time axis, so, displacement is positive, and alope is negative,, so acceleration is also negative., , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , MOTION IN A PLANE, , , SYNOPSIS, Addition of Vectors, , , , , Two Vectors A and B of same kind are added, geometrically as shown., , Special Cases, , , , a) If P and Q are in the same direction i.e., , θ=00 then R = P + Q, , , b) If P and Q are in opposite direction i.e., , θ=1800 then R = P Q ; for P > Q, , , , c) If P and Q are perpendicular to each other, , 0, 2, 2, i.e. θ=90 then R = P +Q, , R = A+B, B, , A, , Resultant of number of vectors, , , , , , , θ, θ, d) If P Q then R = 2Pcos & α = = ., 2, 2, 0, i) If 60 then R 3P and 30, , ii) If 900 then R 2 P and 45, Resultant is a single vector that gives the total effect, of number of vectors., iii) If 1200 then R P and 60, Resultant can be found by using, Triangle Law of Vectors: If two given vectors of, a) Parallelogram law of vectors, same kind are represented both in magnitude and, b) Triangle law of vectors, direction by the two adjacent sides of a triangle,, c) Polygon law of vectors, taken in order then the closing side taken in the, Parallelogram Law of vectors, reverse order will give the resultant both in, magnitude and direction., P, B, , Q, , R, Q, , (Q sin ), , D, P, , , If P and Q are two vectors with angle between, , them, then the resultant vector R P Q ., Magnitude of resultant :, , R = P 2 +Q 2 + 2PQcosθ, , R PQ, , O, , O, , Q, , A, , P, , Law of equilibrium of forces:, If three forces represents the three sides of a, triangle taken in order then their resultant is zero., If such forces acts on a particle simultaneously, then they keep that particle in equilibrium., , Direction of resultant :, Qsinθ , , α = tan -1 , w.r.t P, P, +, Qcosθ, , , Psinθ , , β = tan -1 , w.r.t Q, Q + Pcosθ , , The resultant of two vectors always lie in the plane, containing the vectors, closer to vector of larger, magnitude., NARAYANAGROUP, , F2, , F3, , O, , , , , , , F1, , A, , , , F1 F2 F3 0, 97
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , E, , Lami’s theorem, , , If a body is in equilibrium under the action of three, coplanar concurrent forces P,Q,R at angles, , , as shown in the figure., Then the ratio of magnitude of one of the, force to sine of angle between the other two, vectors is always constant, P, Q, R, , , i.e:, sin sin sin , , P, , Q, , D, x, , 2x, , 3x, , 0, , 90, , F, , C, , 3x, 0, , 60, A, , B, AB x, AC 3 x, AD 2 x, AE 3 x, AF x, WE -1 : ABCDEF is a regular hexagon with point, ‘O’, as centre. Find the value of, , AB AC AD AE AF ., , E, , O, , F, , R, , D, , C, , Polygon Law of Vectors, , , , 3x, " If number of vectors of same kind acting at a, point in the same plane in different directions are, A, B, represented both in magnitude and direction by, , , , , , the adjacent sides of a polygon taken in order, Sol. From the diagram AB DE , , BC EF, then the closing side taken in the reverse order, , will give the resultant both in magnitude and, AB AC AD AE AF, , direction"., AB AB BC AD AD DE AD DE EF , , , , 3 AD 3 2 AO 6 AO, D, D, C, , , , , , C, A+, B+, C, , E, , A+B+C+D, , Applications of Polygon Law, E, , B, A+, , B, , B, , R, O, , A, , , , A, , OA + AB + BC + CD + DE = OE, If many vectors form a closed n sided polygon, with all the sides in the same order then the resultant, , is 0 ., Note :If x is the side of a regular hexagon, ABCDEF as shown in figure., , 98, , If 'n' equal forces act on a body such that each, 2, force makes an angle, with the previous one then, n, they form a closed polygon. So the resultant is zero., If each force of magnitude 'F' makes an angle , with previous one, then, a) the resultant is zero, if the number of forces is, 2, n=, , 2, 1 , then, b) If the number of forces are n =, , the resultant force is ‘F’., , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , Subtraction of Vectors:, , , , , , , Subtraction of a vector Q from vector P is the, , , addition of P and, , , -Q, , ., , , , , , S P Q P Q, , , , Q, , 0, , Q, Q, , P, 180-, S, , A, , B, C, here S P Q, , The magnitude of P Q is, , S P 2 Q 2 2 PQ cos 1800 , , tan , , Q sin 1800 , , P Q cos 180 , Q sin , , , w.r.t P, P Q cos , , , , , 0, , , , , Note: If P Q then P Q 2 P sin ., 2, Subtraction of vectors is used to find, i) change in velocity of a particle., ii) the relative velocity of one body with respect, to another body., iii) change in momentum., iv) acceleration of a particle etc., , Applications, , , , , d1, , Q, , , , d2, P, , P, , S P 2 Q 2 2 PQ cos and, , MOTION IN A PLANE, , , If two vectors P and Q represents adjacent sides, of a parallelogram both in magnitude and direction, then the two diagonals of parallelogram are, represented as, , , , , , d1 P Q , d2 P Q, , 1 , 1 , P d1 d 2 ; Q d1 d 2, 2, 2, , , P, Q are two sides and R , S are two diagonals, , , , , , , , , , , , of a parallelogram then R 2 S 2 2 P 2 Q 2 , W.E - 2: A particle is moving eastwards with a, velocity of 5 m/s. In 10s the velocity changes, to 5 m/s northwards.Find the average, acceleration in this time., , , Δv vf - vi, =, Sol . a avg =, Δt, Δt, , V, , v f 5m / s, 5 2m/s, , W, , 450, , V i 5m / s, , N, , V i 5m / s, , E, S, , , 5 ˆj 5iˆ 5 2, 1, a avg , , , m / s2, 10, 10, 2, , When a particle is performing uniform circular, motion with a constant speed v, then the, Along north-west direction, magnitude of change in velocity when it describes, θ, , , W.E -3: Two vectors A and B have precisely equal, an angle at the centre is Δv = 2vsin ., 2, , magnitudes. For the magnitude of A B to, , v, be larger than the magnitude of A B by a, factor of n, what must be the angle between, them?, 0, , , , v, A, , B, , n, A B, Sol, :, , If velocity of a particle changes from vi to vf in, , , time ‘t’ then, 2 A cos n2 A sin, A B , the, acceleration of the particle is given, 2, 2, v -v, by a = f i ., 1 , 1 1 , 1 1 , t, tan ; tan 2 tan , 2 n 2, n, n, , NARAYANAGROUP, , 99
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , W.E-4: The resultant of two forces whose, AB, or AB OB tan150, Clearly, tan150 , magnitudes are in the ratio 3:5 is 28 N. If, OB, Thus, distance travelled by the aircraft in 10s,, the angle of their inclination is 600 , then, AC=2AB=2(OB tan 150 ) = 2 3400 0.2679 1822m, find the magnitude of each force., speed of the aircraft, Sol . Let F1 and F2 be the two forces., 1822m, v=, = 182.2m / s ., 0, Then F1 3x; F2 5 x; R 28 N and 60, 10s, R F12 F22 2 F1 F2 cos , 2, , 3x 5x , , 28 , , 2, , 2 3x 5 x cos 600, , Position vector : Position vector of point A, with respect to O, Y, , 28 9 x 2 25 x 2 15x 2 7 x, A(x,y), 28, x, 4., 7, F1 3 4 12 N , F2 5 4 20 N., O, X, , W.E - 5: What is the displacement of the point of a, r x iˆ y ˆj, wheel initially in contact with the ground, when the wheel rolls forward half a Displacement vector in two dimensional Plane, revolution ? Take the radius of the wheel as, Y, R and the x-axis as the forward direction ?, Sol . From figure, during half revolution of the wheel,, A, the point A covers AC = R in horizontal, B, direction, and BC =2R in vertical direction, r1, B, , r2, , y, , , x, A, C, x R and y 2 R;, 2, AB x 2 y 2 R (2 R ) 2, and, , X, , O, , , , , If r1 , r2 are the initial and final position vectors, 2, , R 4, , 2R , y, Tan 1 Tan 1 , , x, R, , 2, Tan 1 with x- axis., , W.E -6: An aircraft is flying at a height of 3400 m, above the ground.If the angle subtended at a, point on the ground by the aircraft positions, 10 s apart is 300 , then what is the speed of, the aircraft ? (tan 150 0.2679), A t 0S B t 5S C t 10S , , , of a particle then AB, represents the displacement, vector of a particle., , , AB r 2 r1. AB x2 x1 iˆ y2 y1 ˆj, The magnitude of the displacement vector is, 2, , x2 x1 y2 y1 , , AB , , 2, , Application : Condition for collision, Two particles 1 and 2 move with constant velocities, , , v1 and v2 . At ‘t = 0’ their position vectors are, , , r1 and r2 .If particles collide at the point `P’ after, time `t’., Y, , 1, , 3400m, , r2, , S2 = V2 t, , P, S2 = V1 t, , r, , 2, , 0, , 15 150, , r1, , O, , 30, , From the diagram, , 0, , X, , ground, , Sol . Let O be observation point on the ground, and, A,B,C be the positions of the aircraft at t=0s, t=5s, and t =10s respectively., 100, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , r = r1 + s1 = r2 + s 2, , , r1 + v1t = r 2 + v 2 t, , , r1 - r 2 = v 2 - v1 t, , MOTION IN A PLANE, v 2B = 2gd , , Along OB, , vA, = cosα, vB, , W.E -9: Velocity and acceleration of a particle at, , tim e t=0 are u 2iˆ 3 ˆj m / s and, W.E 7 : A frictionless wire is fixed between A and, , B inside a circle of radius R. A small bead, a (4iˆ 2 ˆj ) m / s 2 respectively. Find the, slips along the wire.Find the time taken by, velocity and displacement of particle at t=2s., the bead to slip from A to B., , , , , , A, g, , gcos, d, 90°, B, O, , O, , A, , Sol.: S =, , , , B, , 1 2, at, 2, , i.e. AB , , 1, g cos t 2 ......... (1), 2, , From diagram AB 2 R cos ..........(2), From eq (1) and eq (2) t 2, , R, g, , Note: Time is independent of inclination of the wire., WE 8: Two particles 1 and 2 are allowed to descend, on the two frictionless chord OA and OB of, a vertical circle, at the same instant from point, O. The ratio of the velocities of the particle 1, and 2 respectively, when they reach on the, circumference will be (OB is the diameter)., O, , Sol :, , 1, , , , A, , R, B, , gc, os, , , O, , 2, , , d, g, , A, B, , OA = dcosα, a OA = gcosα, , , , 1 2, , Sol. From v = u + at and s = ut + at, 2, , v = 2iˆ + 3jˆ + 2 4iˆ + 2jˆ = 10iˆ + 7jˆ m / s, , , , , , , , , , , , ^, , 1, s = 2 2 ˆi + 3jˆ + (2 ) 2 4 i + 2ˆj = 1 2 ˆi + 1 0ˆj m, 2, , , , , , , , , , , , W.E -10: A particle starts from origin at t = 0 with, a velocity 5 iˆ m/s and moves in x-y plane, under action of a force which produce a, constant acceleration of 3i 2 j m / s 2 ., a) What is the y -coordinate of the particle, at the instant its x-coordinate is 84m?, b) what is speed of the particle at this time, Sol.: The position of the particle is given by, , 1 , r t = v0 t + at 2, 2, , , , , , 5 iˆ t 1/ 2 3 iˆ 2 ˆj t 2, 5t 1.5t 2 iˆ t 2 ˆj, 2, Therefore, x t 5 t 1.5 t 2 , y t t, , 5 t 1.5 t 2 84 t 6 s, 2, , A t t 6 s , y 6 36 m, , dr, v=, = 5 + 3 t ˆi + 2t ˆj, dt, , , At t = 6s, v = 23 ˆi + 12 ˆj, , speed = v = 23 2 + 12 2 = 26ms -1, , W.E -11: The coordinates of a body moving in a, plane at any instant of time t are x t 2 and, y t 2 . The speed of the body is., 2, Sol. x = αt v x =, , dx, = 2αt, dt, , Along OA v 2A = 2 gcosα dcosα , NARAYANAGROUP, , 101
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, 2), , dy, y = βt v y =, = 2βt, dt, 2, , 3), , speed v = v 2x + v 2y = 2t α 2 + β2, , W.E -12: Figure shows a rod of length l resting, against a wall and the floor. Its lower end A, is pulled towards left with a constant, velocity v. Find the velocity of the other end 4), B downward when the rod makes an angle, with the horizontal., B, 5), , l, , , , v, , 6), A, Sol. In such type of problems, when velocity of one, part of a body is given and that of other is required,, first find the relation between the two displacement,, then differentiate them with respect to time. Here, if the distance from the corner to the point A is x, 7), and upto B is y, then x 2 y 2 l 2, Differentiating with respect to time t, 8), 2x, , dx, dy, 2y, 0, dt, dt, , 9), , dx, dy, where vA = v = & vB = dt, dt, (- sign denotes that y is decreasing), x v = yv B, v B = v , , x, = v c o tθ, y, , Change in velocity, If Vi is the initial velocity of a particle, Vf is its, , final velocity, V is the change in velocity, and, , θ is the angle between Vi and Vf then, , V = Vf - Vi ., , V =, , Vf 2 +Vi 2 -2Vf Vi cosθ, , Relative Velocity, , , 1), 102, , , , If body A is moving with a velocity VA w.r.t., , ground and body B is moving with velocity VB, w.r.t. ground then, The relative velocity of body 'A' w.r.t. 'B' is given, , , by VAB = VA -VB, , The relative velocity of body 'B' w.r.t. 'A' is given, , , by VBA = VB -VA, , , Both VA -VB and VB -VA are equal in magnitude, but opposite in direction., , , V AB = -V BA and, , , V AB = VBA = VA 2 +VB 2 -2VA VB cosθ, For two bodies moving in same direction,, magnitude of relative velocity is equal to the, difference of magnitudes of their velocities., (θ = 00 , cos 0 = 1), , , VAB =VA -VB , VBA =VB VA, For two bodies moving in opposite directions,, magnitude of relative velocity is equal to the sum, of the magnitudes of their velocities., ( 1800 ; cos 1800 -1), , , V AB V BA = VB VA, Relative displacement of A w.r.t. B is, , , , X AB X AG X BG, , , , Where X AG displacement of ‘A’ w.r.t ground, , , , and X BG displacement of ‘B’ w.r.t ground, Relative velocity of A w.r.t. B is, , , , V AB V AG V BG, Relative acceleration of A w.r.t. B is, , , , a AB a AG a BG, Two trains of lengths l1 and l2 are moving on, parallel tracks with speeds v1 and v2 (v1 > v2 ), w.r.t ground. The time taken to cross each other, when they move in same direction is, S, l l, t1 = rel = 1 2, Vrel v1 - v2, when they move in opposite direction is, S, l l, t2 = rel = 1 2, Vrel v1 + v2, , Application:, Relative Motion on a moving train, , , If a boy in a train is running with velocity V BT, relative to train and train is moving with velocity, , V TG relative to ground, then the velocity of the, , boy relative to ground V BG will be given by, , , , V BG V BT V TG, So, if boy in a train is running along the direction, of train., , , , V BG V BT V TG, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , , Velocity of B relative to A is V BA V B V A, , If the boy in train is running in a direction opposite, to the motion of train, then, , , , , 2, 2, V BG V BT V TG, V BA 20 20 20 2km / h, W.E-13: When two objects move uniformly towards, , each other, they get 4 metres closer each, i.e., V BA is 20 2 km/h at an angle of 450 from, second and when they move uniformly in, east towards north., the same direction with original speed, they, , get 4 metres closer each 5s. Find their, A is at rest and B is moving with VBA in the, individual speeds., direction shown in Fig., Sol. Let their speeds be v1 and v2 and, Therefore the minimum distance between ships, let v1 > v2 ., 1 , In First case :, S min AC AB sin 450 10 , km 5 2km, 2, 4, Relative velocity, v1 + v 2 = = 4 m / s ....(1), and time taken is, 1, In Second case:, BC, 5 2 1, 4, t , h 15min, Relative velocity = v1 - v 2 = = 0.8 m / s...(2), V BA 20 2 4, 5, solving eqns.(1) and (2), we get, Rain umbrella Concept, v1 =2.4ms -1 ,v 2 =1.6ms -1, , W.E - 14 : A person walks up a stationary escalator If rain is falling with a velocity V R and man moves, in time t1 . If he remains stationary on the, , with a velocity V M relative to ground, he will, escalator, then it can take him up in time t2., How much time would it take for him to walk, observe the rain falling with a velocity, up the moving escalator?, , , Sol. Let L be the length of escalator ., V RM VR VM ., L, Case - I : If rain is falling vertically with a velocity, Speed of man w.r.t. escalator is v ME = t, , 1, V R and an observer is moving horizontally with, L, , velocity V M , then the velocity of rain relative to, Speed of escalator v E = t, 2, observer will be :, Speed of man with respect to ground would be, 1 1, v M = v ME + v E = L + , , VR, t1 t 2 , V RM, VR, L, t1 t 2, t, =, =, ., The desired time is, v M t1 + t 2, , V M, W.E -15: Two ships A and B are 10km apart on a, VM, VM, line running south to north. Ship A farther, north is streaming west at 20km/h and ship, V RM VR VM, B is streaming north at 20km/h. What is their, The magnitude of velocity of rain relative to man is, distance of closest approach and how long, do they take to reach it?, VRM VR2 VM2, Sol., If is the angle made by the umbrella with, VB 20km / h, A, , 10km, , C, V BA, , horizontal, then, tan VR, , VM, , 0, , V BA 20 2km / h, 450, VA 20 km / h, NARAYANAGROUP, , 45, , B, , If is the angle made by the umbrella with vertical,, then, tan V M, , VR, 103
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , Case - II : When the man is moving with a, W.E-16: Rain is falling vertically with a speed of, velocity VM1 relative to ground towards, 20ms 1 ., A person is running in the rain with, a velocity of 5 ms 1 and a wind is also, east(positive x-axis), and the rain is falling with a, , blowing with a speed of 15 ms 1 (both from, velocity V R relative to ground by making an angle, the west) The angle with the vertical at which, with vertical(negative z-axis). Then the velocity, the person should hold his umbrella so that, , he may not get drenched is :, of rain relative to man V RM is as shown in figure., , , , ˆ, , V, , V, Sol., Rain, R 20 -k, VR VRx iˆ VR y kˆ ; V M 1 V M 1 iˆ, , , , , ˆ , VWind VW 15iˆ, V, , V, , 5, i, M, an, M, VR VM, Resultant velocity of rain and wind is, and ta n , ...... ( 2 ), , VR, VRW 20kˆ 15 iˆ, N, Now, velocity of rain relative to man is, , , , E, W, VRW VM 20kˆ 15iˆ 5iˆ, , VR, VR, VRy, M1, S, 20kˆ 10iˆ, vertical, VRM, 1, , x, , , , 1, , y, , , , , , 1, , VM 1, VR, , VM 1, , VM 1, x, , Case - III : If the man speeds up, at a particular, , velocity V M 2 , the rain will appear to fall vertically, , , , with V RM 2 , then V RM 2 V R VM 2 as shown in, figure., , VRM, , 2, , VRM, VM, , , 20kˆ, , VR, 2, , 2, , VM, , VM, , 2, , , 10 iˆ, 1, 1, Tan Tan 1, 2, 2, W.E -17: To a man walking at the rate of 3km/h, the rain appears to fall vertically.When he, increases his speed to 6km/h it appears to, meet him at an angle of 450 with vertical., Find the angle made by the velocity of rain, with the vertical and its speed., Sol :, , 2, , Case - IV : If the man increases his speed further,, he will see the rain falling with a velocity as shown, in figure., , V RM, VRM, , 3, , VRM, VM, , 3, , , 3, , VR, , 450 , y, , VRM, VR, , 3, , 3, 6, , VM 3kmph, , VR y, , 0, From the diagram Tan45 , , VR, , VR x, , 3, , VM 3, VM, , 3, , VM VR, , , tan , ;, V RM 3 V R V M 3, VR, , x, , 3, , 3, .........(1), y, , and Tan y ..............(2), 3, , 1, , 3, , 0, From (1) and (2) 450 sin 45 V , 2 V, R, R, VR 3 2kmph, , y, , 104, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , W.E -18: Rain is falling, vertically with a speed, of 1m/s .Wind starts blowing after sometime, with a speed of 1.732 m/s in east to west, direction.In which direction should a boy, waiting at a bus stop hold his umbrella.?, N, vw, , W, , E, , S, Sol. If R is the resultant of velocity of rain ( Vr ) and, velocity of wind ( Vw ) then, 2, , R = v 2r + v 2w = 12 + 1.732 ms-1 = 2ms-1, The direction that R makes with the vertical is, vw, 3, =, = 600, vr, 1, Therefore, the boy should hold his umbrella in the, vertical plane at an angle of about 600 with the, vertical towards the east., W.E - 19 : Rain is falling vertically with a speed of, 1m/s . A woman rides a bicycle with a speed, of 1.732 m/s in east to west direction. What, is the direction in which she should hold her, umbrella ?, Sol. In Fig. vr represents the velocity of rain and vb ,, the velocity of the bicycle, the woman is riding., Both these velocities are with respect to the, ground.Since the woman is riding a bicycle, the, velocity of rain as experienced by, N, vb, , given by tanθ =, , , vr, , W, , Resultant velocity of the boat = VBR VR, The time taken for the boat to move a distance, `d’ along the direction of flow of water is., d, t1 , VBR VR .............(1), 2) Up stream, , ( 1800 ) :, , Resultant velocity of the boat = VBR VR, The time taken for the boat to move a distance, `d’ opposite to the direction of flow of water is., d, t2 , VBR VR .......................(2), t1 VBR VR, From equation (1) and (2) t V V, BR, R, 2, time taken by person to go down stream a, distance `d’ and come back is, d, d, , T t1 t2 , VBR VR VBR VR, 3) General approach :, Suppose the boat starts at point A on one bank with, velocity VBR and reaches the other bank at point D, C VR, , E, d, , v rb, , -v b, S, here is the velocity of rain relative to the velocity, of the bicycle she is riding. That is vrb = vr - vb, This relative velocity vector as shown in Fig., makes an angle with the vertical.It is given by, v, 3, Tan θ = b =, θ = 60o, , v, 1, r, Therefore,the woman should hold her umbrella at an, angle of about 600 with the vertical towards the west., NARAYANAGROUP, , 1) Down stream( 00 ):, , VBR, VR, , vr, , R, , Boat motion is classified into three categories, based on angle between VBR and VR they are, , B, , D, , VBRcos, , , , Motion of a Boat in the River, , V BR, , , , VBRsin A, , The component of velocity of boat anti parallel to, the flow of water is VBR sin , The component of velocity of boat perpendicular, to the flow of water is VBR cos , The time taken by the boat to cross the river is,, t, , d, VBR cos , 105
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, Along the flow of water, distance travelled by the, boat (or) drift is x (VR VBR Sin )t, , , , , , d, x (V R V B R sin ) , , V B R cos , (a) The boat reaches the other end of the river to, the right of B if VR >VBRsin , , , , (b) The boat reaches the other end of the river to, the left of B if VR < VBRsin , (c) The boat reaches the exactly opposite point, on the bank if VR = VBRsin , , V , θ = tan-1 R with the normal., VBR , iv)The distance (BC) travelled downstream, , d , =VR , is called drift, vBR , Motion of a Boat Crossing the River in, Shortest Distance, , C, , VR, , B, , Motion of a Boat Crossing the River in, Shortest Time, , d, , VBR, VR, , B, , C, , , VBR, , A, , d, , i) The boat is to be rowed upstream making some, angle ' θ with normal to the bank of the river which, , , , V , is given by θ = sin-1 R , VBR , , A, , If V BR , V R are the velocities of a boat and river, flow respectively then to cross the river in shortest, , time, the boat is to be rowed across the river i.e.,, along normal to the banks of the river., d, i) Time taken to cross the river, t= V where d, BR, , = width of the river. This time is independent of, velocity of the river flow, ii) Velocity of boat w.r.t. ground has a magnitude, , ii) The angle made by boat with the river flow (or), bank is = 900 +θ, iii) Velocity of boat w.r.t. ground has a magnitude, of VB= VBR2 -VR2, d, iv) The time taken to cross the river is t = V 2 -V 2, BR, R, , Note : VBR = Relative velocity of the boat w.r.t, river (or) velocity of boat in still water., , of VB= VBR2 +VR2, iii) The direction of the resultant velocity is, 106, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, W.E -20: A boat is moving with a velocity Sol :, v, = 5 km/hr relative to water. At time, bw, t = 0. the boat passes through a piece of cork, floating in water while moving down, stream.If it turns back at time t1 30 min., a) when the boat meet the cork again ?, b) The distance travelled by the boat during, this time., , t=0, , MOTION IN A PLANE, Let v be the river velocity and u be the velocity of, swimmer in still water. Then, , , , d, t1 2 , .....(i ), 2, 2, u v , d, d, 2ud, , 2 2 ........(ii ), u v u v u v, 2d, and t 3 , ..............(iii), u, from equation (i) ,(ii) and (iii), t2 , , t12 t2t3 t1 t2t3, W.E - 22: Two persons P and Q crosses the river, starting from point A on one side to exactly, Sol., opposite point B on the other bank of the, river.The person P crosses the river in the, C, VW, shortest path. The person Q crosses the river, VW, in shortest time and walks back to point B., Let AB =d is the distance travelled by boat along, Velocity of river is 3 kmph and speed of each, down stream in ` t1 ’ sec and it returns back and, person is 5kmph w.r.t river.If the two persons, reach the point B in the same time, then the, it meets the cork at point C after ` t2 ’ sec., speed of walk of Q is., Let AC=x is the distance travelled by the cork Sol :, during t1 t2 sec., For person(Q) :, For person (P) :, , d VB VW t1................(1), , C, , B, , B, , C, , x, , d x VB Vw t2 ............. 2 , and x Vw t1 t2 ............. 3, Substitute (1) and (3) in (2) we get t1 t2, The boat meets the cork again aft er, T 2t1 60 min and the distance (AB+BC), travelled by the boat before meets the cork is, , VB, , A, , D 2VBt1 2Vwt1 2Vwt1, 30, 5km, 60, W.E- 21: A swimmer crosses a flowing stream of, width `d’ to and fro normal to the flow of the, river in time t1 . The time taken to cover the, D 2VBt1 2 5 , , tQ , , VW, , d, , tP , , 2, , VB Vw, , D 2d x, D 2(VB Vw ) t1 Vw 2t1, , VB, , 2, , A, , , , d, 52 32, , , , d, 4, , d d, , tP tQ t, VB 5, , d d, x, , ,, 4 5 Vman, , d d VW d, , ,, 4 5 VBVman, , But x VW, , d, VB, , d d, 3d, , 4 5 5 Vman, , same distance up and down the stream is t2., , 1 1, 3, 1, 3, , ,, , 4 5 5Vman 20 5Vman, , If t3. is the time the swimmer would take to, swim a distance 2d in still water, then relation, between t1 , t2. & t3., , Vman , , NARAYANAGROUP, , VW, , 3 20 12kmph, 5, , 107
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , , When a body is moving in a plane, a) A body can have any angle between velocity, and acceleration, b) If the angle between velocity and, acceleration is acute, velocity increases., c) If the angle between velocity and acceleration, is obtuse, velocity decreases., d) If the angle between velocity and, acceleration is a right angle, velocity remains, constant., e) A body can have constant speed and, changing velocity, f) A body cannot have constant velocity and, changing speed., , Here vx = u cosθ and, v y = u y + a y t = usinθ - gt, , Hence v = ucosθ ˆi + ( usinθ - gt) ˆj, magnitude of velocity is given by, 2, , v = v2x +v2y = ucosθ + usinθ-gt , direction of velocity is given by, , vy , u sin - gt , tan -1 tan -1 , , u cos , vx , , Displacement vector ( s ), , , Projectiles :, , displacement s x i y j here, horizontal displacement during a time t, , Oblique Projectile :, , , Any body projected into air with some velocity, at an angle ‘ ’ [ (900 and 00)] with the, horizontal is called an oblique projectile., , y, , x u xt u cos t, vertical displacement during a time t, 1, 1, y u y t gt 2 u sin t gt 2, 2, 2, Equation of projectile, , vy, , g, , 2, 2, y tan x 2, x Ax Bx, 2, 2u cos , Where A and B are constants, g, A tan , B 2, 2u cos2 , Time of flight (T), , v, u sin , , , u, , vx H, , , u cos , , , x, R, , horizontal component of acceleration, ax 0, vertical component of acceleration , a y g, , At the Point of Projection, (a) Horizontalcomponent ofvelocity u x = u cos, (b)Vertical component of velocity u y u sin , , , , , 108, , Time of ascent ta = Time of descent td , , Horizontal component of velocity u x u cos ,, remains constant throughout the journey., Vertical component of velocity u y u sin ,, gradually decreases to zero and then gradually, increases to u sin . It varies at the rate ‘g’., , (c) velocity vector u u cos i u sin j, (d) Angle between velocity and acceleration, is 90 , At any instant ‘t’, , , , Velocity vector ( v ) is v = v x i + v y j, , 2, , , , uy, g, , , , Time of flight T = ta + td =, , , , u sin , g, 2u y, , , , 2u sin , g, , g, During time of flight, 1) angle between velocity and acceleration, vectors changes from 900 to 900 ., 2) change in momentum is 2 mu sin ., , (In general, change in momentum P mgT ), 3) vertical displacement is 0., 4) The angle between velocity and acceleration, during the rise of projectile is 1800> >900, 5) The angle between velocity and acceleration, during the fall of projectile is 00< <900, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , Maximum height (H), , , , 2, , are respectively x at and y bt ct 2 then, , 2, , u sin , 2g, 2g, At maximum height, 1) The vertical component of velocity becomes, zero., 2) The velocity of the projectile is minimum, at the highest point and is equal to u cos , and is horizontal., 3) Acceleration is equal to acceleration due to, gravity ‘g’, and it always acts vertically, downwards., 4) The angle between velocity and, acceleration is 900., Range (R):, 2u u, R u xT x y (or), g, H , , , , u, , 2, y, , , , 2u sin u 2 sin 2, R u cos T u cos , , g, g, 0, 1) Range is maximum when 45, u2, R, , 2) Maximum range, max, g, R, , u2, , 3) When ‘R’ is maximum, HMax= 4Max = 4g, 4) For given velocity of projection range is same, for complimentary angles of projection, , , R, , , , , , , If y Ax Bx 2 represents equation of a, projectile then, 1) Angle of projection = tan-1(A), 2) Initial velocity, , , u , , , , g 1 A2, , 1) Horizontal component of velocity = u cos , 2) Vertical component of velocity = -u sin , 3) Speed of projection is equal to striking, speed of projectile., 4) Angle of projection is equal to the striking, angle of projectile, 5) If the angle of projection with the horizontal is, then angle of deviation is 2, The projectile crosses the points A, D in time, interval t1 seconds and B,C in time interval, 8h, 2, 2, t2 seconds then t1 t2 , g, (h is the distance between BC and AD), , A, B, , A2, 4B, , 2A2, T, , , , 5) Time of flight, Bg, , If horizontal and vertical displacement of projectile, , NARAYANAGROUP, , iii) R 4 H1 H 2 iv) Rmax 2 H1 H 2 , If a man throws a body to a maximum distance ‘R’, then he can project the body to maximum vertical, height R/2., If a man throws a body to a maximum distance ‘R’, then maximum height attained by it in its path is R/4., , 2B, , 3) Range of the projectile R =, 4) Maximum height H =, , , , 2) velocity of projection u a 2 b 2, 3) acceleration of projectile = 2c, b2, 4) maximum height reached =, 4c, ab, 5) horizontal range =, c, In case of complimentary angles of projection, 1) If T1 and T2 are the times of flight then, T1, 2R, ii) T1T2 , T T R, i) T tan , g 1 2, 2, 2)If H1 and H 2 are maximum heights then, , At the point of striking the ground, , gT 2, and if 450 then, 2 tan , , gT 2, 2R, T , 2, g, , b, tan 1 , a, , H1, u2, 2, , tan, , H, , H, , i) H, ii) 1, 2, 2g, 2, , R, g, H g, H tan , (b), , (c) 2 , 2, T, 8, R, 4, T, 2 tan , , 2) R , , , , , , 0, i.e 1 2 90 , Relation between H, T and R, , 1), , , , 1) angle of projection , , B, , C, h, , A, , D, , 109
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , , A projectile is fired with a speed u at an angle , with the horizontal. Its speed when its direction, of motion makes an angle with the horizontal., v = u cos sec , , Then slope, , x, , v, , , , The path is a vertical straight line, , u cos , , v cos u cos , v u cos sec , If a body is projected with a velocity u making an, angle with the horizontal, the time after which, direction of velocity is perpendicular to the initial, u cosecθ, u, , , velocity is t , g, g sin , and its velocity at that instant is v = u cot , , The path of projectile as seen from, another projectile, , 1, 1, 2, 2, y u1 sin 1 u2 sin 2 t, , , , i) If u1 sin 1 u 2 sin 2 ( initial vertical, y, 0, x, , y, , , , 110, , of the maximum height is, , 1, nth, , u sin , n, , Resultant velocity at a height of, , 1, nth, , of maximum height, 2, y, , v= v +v =, , y u1 sin 1 u2 sin 2, slope = x u cos u cos , 1, 1, 2, 2, , x, The path is a horizontal straight line, , For a projectile, ‘y’ component of velocity at, , 2, x, , Suppose two bodies A and B are projected, simultaneously from the same point with initial, velocities u1 and u2 at angles 1 and 2 with, horizontal., The instantaneous positions of the two bodies, are given by, 1 2, Body A : x1 u1 cos 1t , y1 u1 sin 1t gt, , 2, 1 2 , Body B : x2 u2 cos 2t , y2 u2 sin 2t gt, 2, , x u cos u cos t, , components) then slope, , y, , x, , v cos , , , , , , y, , u, , u sin , , , , ii) If u1 cos 1 u 2 cos 2 ( initial horizontal, components), , ucosθ , , 2, , usinθ , +, , n , , 2, , (n 1) cos2 1, n, If n = 2, velocity of a projectile at half of maximum, u, , 1 cos 2 , 2, For a projectile, w.r.t stationary frame path (or), trajectory is a parabola., Path of projectile w.r.t frame of another projectile, is a straight line, Acceleration of a projectile relative to another, projectile is zero, A body is projected vertically up from a topless, car relative to the car which is moving horizontally, relative to earth, a) If car velocity is constant, ball will be caught by, the thrower., b) If car velocity is constant, path of ball relative, to the ground is a parabola and relative to this car, is straight up and then straight down, c) If the car accelerates, ball falls back relative to, the car, d) If the car retards ball falls forward relative to, the car, If a gun is aimed towards a target and the bullet is, fired, the moment when the target falls, the bullet, will always hit the target irrespective of the velocity, of the bullet if it is with in the range., height = u, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , Note : If air resistance is taken into consideration then, a) trajectory departs from parabola., b) time of flight may increase or decrease., c) the velocity with which the body strikes the, ground decreases, d) maximum height may decrease., e) striking angle increases, f) range decreases., , A particle is projected with a velocity u ai b j, , then the radius of curvature of the trajectory of, the particle at the, , a, (i) point of projection is r=, , 2, , +b2 , ga, , 2, , a, g, Expression for radius of curvature is, (ii) Highest point is r=, , , , r, , r, , velocity 2, , 3/2, , g 9.8ms 2 , ?, u02 sin 2, g, Rg, sin 2 2, u0, R, , using, T , , 300 or 600, , 2u0 sin, g, , 2 240 0.5, 24.5s, 9.8, 2 240 0.867, When 600 , T2 , 42.46s, 9.8, , When 300 , T1 , , W.E -25: The ceiling of a long hall is 20 m high., What is the maximum horizontal distance, that a ball thrown with a speed of 40 ms 1, can go without hitting the ceiling of the, 2, hall g 10ms ?, Sol. : Here, H =20 m, u 40ms 1., Suppose the ball is thrown at an angle with the, horizontal., , normalacceleration, u 2 cos2 , g cos3 , , 2, , , α is angle made by v with horizontal, , u 2 sin 2 , 40 sin 2 , Now H , 20 , 2g, 2 10, 0, , or, sin 0.5 30, W.E - 23 : A bullet fired at an angle of 300 with the, 2, 0, horizontal hits the ground 3.0 km away. By, u 2 sin 2 40 sin 60, Now, R, , , adjusting its angle of projection, can one, g, 10, hope to hit a target 5.0 km away? Assume, 2, the muzzle speed to be fixed, and neglect air, 40 0.866 138.56m, , resistance., 10, Sol . We are given that angle of projection with the W.E -26: A ball projected with a velocity of 10m/s, at angle of 300 with horizontal just clears two, horizontal, 300 , horizontal range R = 3km., vertical poles each of height 1m. Find, u 2 s in 2 , separation between the poles., R 0, , g, , 3, , 2, 0, , u s in 6 0, g, , Y, , 0, , , , 2, 0, , u, 3, , g, 2, , u02, 2 3km, or, g, Since the muzzle speed (u0 ) is fixed, 2, 0, , 30, O, , P, , R, , Q, , S, , 0, , X, , u, 1, 1, 2 3 2 1.732 3.464km, h uyt gt 2 10sin300 t 10 t 2, Sol., g, 2, 2, so, it is not possible to hit the target 5km away., the, 1 5t 5t 2 t 0.72 s, 2.76s are, W.E -24: A cannon and a target are 5.10 km, instants at which projectile crosses the poles., apart and located at the same level. How soon, separation between poles = OS - OQ, , will the shell launched with the initial, u cos t2 t1 , velocity 240 m/s reach the target in the, absence of air drag ?, = 10 cos300 2.76 0.72 17.7m, Sol . Here, u0 =240 ms 1 , R =5.10 km =5100m,, Rmax , , NARAYANAGROUP, , 111
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, W.E -27 : A body is projected with velocity u at an, angle of projection with the horizontal.The, body makes 300 with horizontal at t = 2, second and then after 1 second it reaches, the maximum height. Then find, a) angle of projection b) speed of projection., Sol. During the projectile motion, angle at any instant, t is such that, , x, But equation of trajectory is y x tan 1 , R, yR , tan , (ii ), xR x, , From Eqs. (i) and (ii), tan tan tan , , usinθ-gt, ucosθ, For t = 2 seconds, 300, 1 usinθ-2g, =, --------- (1), ucosθ, 3, , W.E -29: The velocity of a projectile at its greatest, , For t = 3 seconds, at the highest point 0o, , 2, 1 cos2 , u, Sol.: u cos , 5, 2, Squaring on both sides, , tanα=, , 0=, , 2, times its velocity, at half of itss, 5, greatest height, find the angle of projection., height is, , usinθ-3g, ucosθ, , usinθ=3g ------------(2), using eq. (1) and eq. (2), , 2 1 cos2 , u 2 cos 2 u 2 , , 5 , 2, , , ucosθ= 3g......................(3), , 10 cos 2 2 2 cos 2 , , Eq. (2) eq.(3) give 600 squaring and, adding equation (2) and (3), , 1, 8cos2 2 cos2 600, 4, , u 20 3 m / s., W.E-28: A particle is thrown over a triangle from W.E -30: A foot ball is kicked off with an initial, speed of 19.6 m/s to have maximum range. Goal, one end of horizontal base and grazing the, keeper standing on the goal line 67.4 m away, vertex falls on the other end of the base.If , in the direction of the kick starts running, and are the base angles and be the angle, opposite to the direction of kick to meet the ball, of projection, prove that, at that instant. What must his speed be if he is, tan tan tan ., to catch the ball before it hits the ground?, Sol.: The situation is shown in figure.From figure,we have, 2, u 2 sin 2 19.6 sin 90, Y, , Sol.: R , g, , or R= 39.2 metre., Man must run 67.4 m -39.2m=28.2m in the time, taken by the ball to come to ground Time taken, by the ball., , A(x,y), , O, , , R-x, , y, y, tan tan , x Rx, yR, tan tan , x R x, 112, , 2u sin 2 19.6 sin 450, 4, t, , , g, 9.8, 2, , y, , x, , 9.8, , x, , t 2 2 2 1.41 2.82sec., , Velocity of man , , 28.2m, 10m / sec., 2.82sec, , _______(1), NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, 2vy, , 25, , t , , 1s, W.E -31: A body projected from a point `0’ at an, g, 10, angle , just crosses a wall `y’ m high at a W.E - 33: A projectile of 2 kg has velocities, distance `x’ m from the point of projection, 3 m/s and 4 m/s at two points during its flight, and strikes the ground at `Q’ beyond the wall, in the uniform gravitational field of the, as shown, then find height of the wall, earth. If these two velocities are to each, Y, other then the minimum KE of the particle, during its flight is, , Sol. v1 cos v2 cos (90 ), -1, , 3 ms, , 3 cos 4 sin , , , 3, 4, 1, mv12 cos 2 , 2, , tan , , y, , , (R - x), , x, , O, , KE min, , X, , Q, R=range, Sol . We know that the equation of the trajectory is, y x tan , , gx 2, 2u 2 cos 2 , , , gx 2, y x tan 2, 2, 2u cos , , 4 ms, , -1, , 2, , , , 1, 4 9 16, 2 32 , 5.76 J, 2, 25, 5, , can be written as, , sin , , sin , , x 2 tan , y x tan 2, gx 2 tan , u sin 2, y x tan 2, u (2sin cos ), g, , W.E-34: In the absence of wind the range and, maximum height of a projectile were R and, H. If wind imparts a horizontal acceleration, a =g/4 to the projectile then find the, maximum range and maximum height., Sol : H 1 H ( u sin remains same ), , T1 T, , u 2 sin 2, x, y x tan 1 [ R , ], g, R, , 1, 1g 2, R1 u xT aT 2 R , T, 2, 24, , W.E - 32: A particle is projected with a velocity of 10 2, m/s at an angle of 450 with the horizontal . Find, the interval between the moments when speed is, 125 m/s g 10m / s, , 0, , 90 -, , 2, , 1, 8, , 2, = R gT, , H1 H, , R1 R H, , , , RH, , Sol., , , , u ai bj ck, , t, v 125 m / s, u x 10 2 cos 450 10m / s, u y 10 2 sin 450 10m / s, v 2 v x2 v 2y, 125 100 v y2 v y 5m / s, , If a body is projected with a velocity, , vx ux , , ( i east, , , j north k vertical ) then, , ux a 2 b2 ; u y c, , The required time interval is, , NARAYANAGROUP, , 113
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, 2, , T, , 2c, c, ;H , ,R , g, 2g, , 2, , , , , , a 2 b2 c, g, , W.E-35: A particle is projected from the ground, with an initial speed v at an angle with, horizontal.The average velocity of the, particle between its point of projection and, highest point of trajectory is [EAM 2013], , Y, , Sol., , e) The angle at which it strikes the ground, 2 gh , tan 1 , , u , f) If is angle of elevation of point of projection, from the point where body hits the ground then, , h gt 2 /2 gt, =, =, R, ut, 2u, tanθ, tanα =, 2, is, the, angle, with which body reaches the, , ground, Case (i) : If the body is projected at an angle , in upward direction from the top of the tower,then, tanα =, , u, , H, , , , , , X, , R/2, , v avg, , , ˆ, v + u ucosθiˆ + (ucosθiˆ + usinθj), =, =, 2, 2, , v, 1+ 3cos 2 θ, 2, Horizontal projectile, When a body is projected horizontally with a, velocity from a point above the ground level, it is, called a Horizontal Projectile., Path of the Horizontal Projectile is parabola, v av =, , , , , , u, , h, , R, , 2h, a) Time of descent t , (is independent ofu), g, b) The horizontal displacement (or) range, Ru, , 2h, g, , c) The velocity of projectile at any instant of time, is v = u 2 + g 2 t 2, 1 gt , The direction of velocity tan , u, d) The velocity with which it hits the ground, , v =, 114, , h, , x, , a) The time taken by projectile to reach same level, 2u sin , as point of projection is T , g, b) The time taken by projectile to reach ground is, 1 2, calculated from h u sin t gt, 2, c) The horizontal distance from foot of the tower, where t he projectile lands is given by, x u cos t, d) The velocity with which it strikes the ground, , v u 2 2 gh, e) The angle at which it strikes the ground, -usinθ + gt , α = tan -1 , ucosθ , , u 2 sin 2 θ + 2gh, -1, α, =, tan, , (or), ucosθ, , , , , , , Case (ii) : If the body is projected at angle , from top of the tower in the downward direction,, then, , u 2 + 2gh, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , , MOTION IN A PLANE, with each other is, , u, , 2 u1u2, , cot, g, 2, Two tall towers having heights h1 and h2 are, separated by a distance d. A person throws a ball, horizontally with velocity u from the top of the first, tower to reach the top of the second tower then, x u1 u2 , , h, , , x, , a) The time taken by projectile to reach ground, 1 2, is calculated from h u sin t gt, 2, b) The horizontal distance from foot of the, tower where the projectile lands is given by, x u cos t, c) The velocity with which it strikes the ground, , u, , h1 h2 , h1, h2, , v u 2 2 gh, , d, , d) The angle at which it strikes the ground, , , , , , u 2 sin 2 2 gh , tan 1 , , u cos , , , When an object is dropped from an aeroplane, moving horizontally with constant velocity, a) Path of the object relative to the earth is, parabola, b) Path of the object relative to pilot is a straight, line vertically down., Two bodies are projected horizontally from top, of the tower of height h in opposite directions, with velocities u1 and u2 then, a) The time after which their velocity vectors, are making an angle with each other, , 2 h1 h2 , , a) Time taken t , , , g, , b) Horizontal distance travelled d ut, A ball rolls off from the top of a stair case with a, horizontal velocity u. If each step has a height ‘h’, and width “b” then the ball will just hit the nth step,, directly if n equal to, nb = ut and nh =, , 1 2, gt, 2, u, , h, , b, 1, 2, , u1u2, , cot, g, 2, b) The distance between them when their, velocity vectors are making an angle with, , n, , t, , u1u 2, , cot, g, 2, c) The time after which their position vectors, are making an angle with each other, , R, , each other x u1 u2 , , 2 u1u2, , cot, g, 2, d) The distance between them when their, displacement vectors are making an angle , , , NARAYANAGROUP, , , , 2hu 2, n=, gb 2, From the top of the tower of height h , one stone, is thrown towards east with velocity u 1 and, another is thrown towards north with velocity u 2 ., The distance between them after striking the, ground,, , d t u12 u22 , t , , 2h, g, 115
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , W.E -36: A ball is thrown from the top of a tower of W.E - 38: A golfer standing on the ground hits a, ball with a velocity of 52 m/s at an angle , 61 m high with a velocity 24.4 ms 1 at an, elevation of 300 above the horizontal . What, 5, above the horizontal if tan , find the, is the distance from the foot of the tower to, 12, the point where the ball hits the ground?, time for which the ball is at least 15m above, Sol. :, the ground?, u sin , g 10m / s 2 , u, , , Sol. v y = u y 2 - 2gy , u y = usinθ, , u cos , , vy 52 52 , , h, , 5 5, 2 10 15, 13 13, , = 16 × 25 - 300 = 10, 2v y, , 2 ×10, = 2s, 10, 10, W.E - 39: Two paper screens A and B are separated, Also, d u cos t 105.65m, by a distance of 100m. A bullet pierces A and, W.E -37: A particle is projected from a tower as, B. The hole in B is 10 cm below the hole in, shown in figure, then find the distance from, A. If the bullet is travelling horizontally at, the foot of the tower where it will strike the, the time of hitting the screen A, calculate, ground. g 10m / s 2 , the velocity of the bullet when it hits the, screen A. Neglect resistance of paper and air., 37 0, Sol. : The situation is shown in Fig., 1, h gt 2 (u sin )t, 2, , t 5 sec onds, , 1500 m, , Δt =, , P, , 500, m/s, 3, , =, , u, , Q, 0.1 m, R, , Sol.:, u y u sin , , 500, sin 37 0, 3, , 1, s = ut + at 2, 2, 500, 1, 1500 , sin 37 t 10t 2, 3, 2, 1500 , , 500 3 , 2, t 5t, 3 5, , 300 20t t 2 t 20 s, horizontal distance = ( u cos ) t, , 116, , 500 4 , 4000, m, 10 , 3 5, 3, , A, d u, , 100 m, , B, , 2 h1 h2 , 2 0.1, 100 u 9.8 u 700m / s., g, , W.E -40: A boy aims a gun at a bird from a point,, at a horizontal distance of 100m. If the gun, can impart a velocity of 500m/sec to the, bullet, at what height above the bird must he, aim his gun in order to hit it?, Sol : x = vt or 100 = 500× t ; t 0.2sec., 1, 2, Now h 0 10 0.2 = 0.20m = 20cm., 2, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , W.E - 41:An enemy plane is flying horizontally at WE- 43: From points A and B, at the respective, heights of 2m and 6m, two bodies are thrown, an altitude of 2 km with a speed of 300 ms-1., simultaneously towards each other, one is, An army man with an anti - aircraft gun on, thrown horizontally with a velocity of 8m/s, the ground sights enemy plane when it is, and the other, downward at an angle 450 to, directly overhead and fires a shell with a, -1, the horizontal at an initial velocity v0 such, muzzle speed of 600ms . At what angle with, that the bodies collide in flight. The, the vertical should the gun be fired so as to, horizontal distance between points A and B, hit the plane?, equal to 8m . Then find, Sol. Let G be the position of the gun and E that of the, a) The initial velocity V0 of the body thrown, enemy plane flying horizontally with speed., at an angle 450, P, E, u, b) The time of flight t of the bodies before, colliding, v0, c) The coordinate (x,y) of the point of, collision (consider the bottom of the tower A, vy, as origin) is, B, , , , 450, , 4m, , V0, , A, , 0, , (90 -), , Sol :, , , , Ground, , G, , vx, u = 300ms , when the shell is fired with a speed, v0 , vx v0 cos , The shell will hit the plane, if the horizontal distance, EP travelled by the plane in time t = the distance, travelled by the shell in the horizontal direction in, the same time, i.e., u × t = v x × t or u = v x u = v0 cosθ, u 300, or cosθ = =, = 0.5 or 600, v0 600, Therefore, angle with the vertical 900 300., W.E -42: From the top of a tower, two balls are, thrown horizontally with velocities u1 and u2, in opposite directions. If their velocities are, perpendicular to each other just before they, strike the ground, find the height of tower., -1, , Sol. Time taken to reach ground t , , 2h, g, , at time of reaching ground respective velocities, , , are v1 = u1 i + gt j, v 2 = -u 2 i + gt j, , , u1u2, Given v1.v 2 = 0 , t , g, uu, uu, 2h, 1 2 h 1 2, g, g, 2g, is the height of the tower., , , NARAYANAGROUP, , 2m, , 2m, , 8m, a) From diagram, 4, 1, tan tan .................(1), 8, 2, , , , ˆ, v A = 8i, v B = -v 0cos450 ˆi - v0sin450 ˆj, , v, v, v BA = - 0 - 8 i - 0 j, 2, 2 , , Direction of vBA, tanθ =, , v0, , 2, , 2 v0 + 8 2 .................(2), From eq (1) and eq (2), , , , , , 2v0 = v0 + 8 2 , v0 = 11.28m / s, , v0, v, - 8 iˆ - 0 ˆj, b) v BA = 2, 2 , , v0 = 8 2 Þ v BA = -16iˆ - 8jˆ, , | v BA | t = SBA, , , , 16 , , t, , 2, , 8 , , 2, , t , , 82 42, , 80, 1, , t 0.5s, 320, 4, , c), x = v x t = 8 0.5 = 4, , 1 2 1, 1, gt = ×10× = 1.25, 2, 2, 4, y = 2 - y' = 0.75, , y' =, , 117
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , Motion of a Projected Body on an inclined, plane :, A body is projected up the inclined plane from, the point O with an initial velocity v0 at an angle, with horizontal., x, v, , , , Down the plane : Here, x and y-directions are, down the plane and perpendicular to plane, respectively, y, , x, , u, , 0, , x, , , , , A, , , g, , , , , , B, , O, x', , y', x, y, g sin , g cos, g, , O, , g cos, , , O, u x u cos( ), ax g sin , u y u sin( ), a y g cos , Proceeding in the similar manner , we get the, following results, 2u sin( ), T, ,, g cos , u2, sin 2 sin , g cos 2 , W.E -44: A particle is projected horizontally with a, speed “u” from the top of plane inclined at, an angle “ ” with the horizontal. How far, from the point of projection will the particle, strike the plane ?, u, , R, , , , , x1, , g sin, , B, , y1, , a) Acceleration along x axis , a x = -gsin α, b) Acceleration along y axis , a y = - g c o s α, c) Component of velocity along x axis, , u x = v0cos θ - α , d) Component of velocity along y axis, , y, R, , u y = v0sin θ- α, e) Time of flight T =, , 2v0sin θ - α , , gcosα, f) Range of projectile (OA), v 20, R=, sin 2θ - α - sinα . (or), gcos2α , 2v02sin θ - α cosθ, R=, gcos 2 α, For maximum range 2 , R max =, , , 2, , v 20 1- sinα , gcos 2 α, , x, y, , tan , x, , , R x2 y 2, , x, , 2, , x 2 x tan x 1 tan 2 x sec , 1 2, y 1 gt 2, gt ;, , 2, x 2 ut, gt, 2u, tan ; t , tan , 2u, g, 2u 2, x ut , tan ;, g, 2u 2, R , tan sec , g, x ut ;, , g) T 2 g 2 R max, horizontal range (OB) x R cos , 118, , , , y, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , W.E 45: A projectile has the maximum range of, 500m. If the projectile is now thrown up on, an inclined plane of 300 with the same, speed , what is the distance covered by it, along the inclined plane?, Sol:, , Average angular velocity av , , , t, , instantaneous angular velocity is, d, Lt , t 0 t, , dt, , SI unit rad S1, , 2, , R max =, 500 =, , u, g, , u2, g, , , , or u = 500g, , Note:, , v 2 - u 2 = 2as, , , 0 - 500g = 2× -gsin300 x, x = 500m., , , , Circular Motion, , , , , 1, Dimensional Formula T , Angular velocity is an axial vector., Its direction is given by right hand screw rule, Its direction is along axis of rotation, , When a body makes ‘N’ revolutions in ‘t’ sec then, 2πN, its average angular velocity is ω =, t, If a particle makes ‘n’ rotations per sec its angular, velocity is 2 n, 2, , 2 n, T, , Radius Vector : The line joining the centre, of rotation to position of particle in circular path is, , called radius vector, Angular displacement : The angle turned by the, Angular velocity of hands of a clock:, radius vector in a given time interval is called angular , Angular velocity of seconds hand, displacement , 2 2 , 1, , , , , d , , arc length, radius, , A, , , , , , , , , , , B, , , , SI unit : Radian, Small angular displacements are vectors, Large angular displacements are scalars as they , do not obey commutative law, The direction of angular displacement is along the axis, of rotation and it is given by right hand screw rule., When a particle completes one revolution the, angular displacement is 2 radian, When a particle completes N revolutions in a circle, , the angular displacement is N 2 , When an object moves in circular path at a, constant speed, the motion is uniform circular, motion, Angular Velocity , The time rate of change of angular displacement, of particle is called angular velocity, If is angular displacement in small interval of, time t then, , NARAYANAGROUP, , , , 60, , , , 30, , rad S, , Angular velocity of minutes hand, , , d, , T, , 2, , , rad S 1, 60 60 1800, , Angular velocity of hours hand, , , 2, , , rad S 1, 12 3600 21600, , In case of self rotation of earth about its own axis, , , 2, rad / sec, 24 60 60, , P, , , A, , , , , , O, , In the above fig, let the angular velocity of particle, (P) about the point ‘O’ is 0 , Let the angular, velocity of particle about A is A then 0 A, , 119
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, MAINS, - VOL, - -VI-I II, , MOTION IN A PLANE, , A, , of angular velocity., If a particle rotates with uniform angular velocity, then 0, B, If the particle has constant angular acceleration, , O, ( constant ), then we use following equations, of motion, In the above fig. if two particles A and B are, 1 2, 0 t , 0 t t, moving in same circular path in the same direction,, 2, for a person at the centre of the circle, 2, 2, 0 2, BA B A, WE, 46:, When a motor cyclist takes a U-turn in 4s, Time taken by one particle to complete one, what, is the average angular velocity of the, rotation with respect to another particle is, motor cyclist., 2, 2, TT, Sol. When the motor cyclist takes a U-turn, angular, T, , A B, rel B A TA TB, displacement, rad and t = 4 s., vB, The average angular velocity,, , vA, 0.7855rad s 1, t 4, rA, rB, Note:In general we can also use the following, , equations to solve the problems, , , , av , av , If two particles A and B are moving in concentric, t, t, circles as shown in the fig. , if they are nearer to, d, d d 2 , d, each other., , , 2 , dt, dt dt, d, rrel rB rA , vrel = vB - vA, d dt , d dt , d d, v, v -v, ωrel = rel = B A, Relation between linear and angular, rrel, rB - rA, variables, Angular acceleration ( ), Relation between linear and angular displacement, The time rate of change of angular velocity of a , particle is called angular acceleration, is ds rd, If be the change in angular velocity of the , Relation between linear and angular velocities is, , particle in time interval ‘ t ’ while moving, v r , v r, on a circular path, then, Relation between tangential and angular, , , , acceleration is a t = rα , a t = α × r, Average angular acceleration av , t, Linear acceleration of a particle moving in a circle., , Instantaneous angular acceleration, , We know v r diff. w.r.t time, we get, d , , , , inst Lt, , t 0 t, dt, d v d d r, , r , d d 2, d, dt, dt, dt, Note: 2 , , , , dt, dt, d, a = α× r + ω× v, SI units rad.sec-2, , 2, but, , r a t , it is tangential acceleration, , , T, Dimensional formula , , Its direction is in the direction of change in angular, v a c , it is centripetal acceleration, velocity and it is given by right hand screw rule., Due to change in direction of velocity there is an, When angular velocity increases the direction of , acceleration and is always directed towards the, angular acceleration is in the direction of angular, centre. This is called centripetal or radial, velocity, acceleration and the corresponding force acting, When angular velocity decreases the direction of, towards the centre is called centripetal force, angular acceleration is in the opposite direction, , , , , , , , , , , 120, , , , , , , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, JEE MAINS, - VOL - -VII, , MOTION IN A PLANE, , W.E -47: What is the linear velocity of a person at, equator of the earth due to its spinning, motion? ( Radius of the earth = 6400km)., Sol. The earth completes one rotation in 24 hour. Its, angular velocity., , , , 2 N, 2 1, , , , rad s 1, 24 60 60 43, 200, t, , The linear velocity ,, , , 465.5m / s, 43, 200, Centripetal Acceleration a r or a c , When a particle is moving along a circle of radius, r with a uniform speed v, then the centripetal, acceleration is a r ., v R 6.4 106 , , , , , , ar v r, , , , , , v2, 2 2, = 4 f r, r, , , The directions of a r , and v are mutually, perpendicular., Position vector is always perpendicular to velocity, , vector. i.e, r.v 0, Velocity vector is always perpendicular to the, , centripetal acceleration vector is v.a c 0, a r v r2 , , , , , , , , , , , , , Position vector r and centripetal acceleration, , a r are always antiparallel., , , , , Tangential acceleration, Due to change in magnitude of velocity (speed), of a particle in circular motion, it has tangential, acceleration and the corresponding force is called, tangential force, , dv, at =, also at r ( a t = α × r ), dt, dv, Ft = ma t = m, = mrα, dt, Net linear acceleration of particle in circular motion, a = a 2c + a 2t, , at, a, O, , NARAYANAGROUP, , , ac, , p, , If is the angle made by ‘a’ with a c, at, then Tan = a, c, Net force F Fc2 Ft 2, , Uniform circular motion:, In the above case if a c 0 , a t 0 , then the, particle under go uniform circular motion (or), When a particle moves in a circular path with, constant speed then it is said be in uniform circular, motion. in this case the acceleration of the particle, is a v v 2 / r r 2 ,, In uniform circular motion, (a) magnitude of velocity does not change, (b) direction of velocity changes, (c) velocity changes, (d) angular velocity is constant, (e) centripetal acceleration changes (only in, direction), (f) linear momentum changes, (g) angular momentum w.r. to centre does not, change, , Non Uniform Circular motion:, In a circular motion if a c 0 , a t 0 then the, particle undergo non uniform circular motion, in, this case the acceleration of particle is given by, , a = a c2 + a t2, If is the angle made by ‘a’ with ac then, a, Tan = t, ac, For a particle in non uniform circular motion, the, resultant force on the particle is F Fc2 Ft 2, In non uniform circular motion, (a) both magnitude and direction of velocity, changes, (b) angular velocity changes, (c) linear momentum and angular momentum are, not conserved, Note: In circular motion, a c is towards centre, v, a t are along tangential direction, d , , are along axis of rotation, , 121
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MOTION IN A PLANE, , JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , 1) North - East, 2) North - West, 3) South - West, 4) South - East, 9. A bus moves over a straight level road with a, r, r r r, If A + B = C and the angle between A and, constant acceleration a. A boy in the bus drops, r, r, a ball out side. The acceleration of the ball, B is 1200, then the magnitude of C, r r, w.r.t the bus and the earth are respectively, 1) must be equal to A − B, r r, 1) a 2 + g 2 ,g, 2) g, a 2 + g2, 2) must be less than A − B, r r, 3) a, g, 4) g, a, 3) must be greater than A − B, 10. A particle P moves with speed ‘v’ along AB and, r r, BC, sides of a square ABCD. Another particle, 4) may be equal to A − B, r, Q also starts at A and moves with the same, r, When two vectors A and B of magnitudes‘a’, speed but along AD and DC of the same square, and ‘b’ respectively are added, the magnitude, ABCD. Then their respective changes in, of resultant vector is always, velocities are, 1) Equal to (a+b), 2) Less than (a+b), 1) equal in magnitude but different in directions, 3) Greater than (a+b) 4) Not greater than (a+b), 2) different in magnitude but same in directions, r r r, If C = A + B then, 3) different both in magnitude and direction, r, r, 4) same both in magnitude and direction, 1) C is always greater than A, 11., A river is flowing from west to east at a speed, 2) C is always equal to A+B, of 5 m/s. A man on the south bank of the river, 3) C is never equal to A+B, r, r, r, r, capable of swimming at 10 m/s in a still water, 4) It is possible to have C < A and C < B, wants to swim, across the river in a shortest, Three forces start acting simultaneously on a, time. He should swim in a direction, ur, particle moving with velocity V . The forces, 1) Due north, 2) 300 east of west, 0, are represented in magnitude and direction by, 3) 30 west of north, 4) 600 east of north, the three sides of a triangle ABC (as shown). 12. A hunter aims his gun and fires a bullet directly, The particle will now move with velocity, at a monkey on a tree. At the instant the bullet, leaves the gun, the monkey drops. The bullet, B, A, 1) cannot hit the monkey, 2) may hit the monkey it its weight is more than 30, C, kg.wt, ur, ur, 3) may hit the monkey if its weight is less than 30, 1) less than V, 2) greater than V, kg.wt, ur, 4) hits the monkey irrespective of its weight., 3) V in the direction of largest force, 13., Keeping the speed of projection constant, the, ur, 4) V remaining unchanged, angle of projection is increased from 0º to 90º., The minimum number of forces of equal, Then the horizontal range of the projectile, magnitude in a plane that can keep a particle, 1) goes on increasing up to 900, in equilibrium is, 2) decreases up to 90º, 1) 4, 2) 2, 3) 3, 4) 5, 3) increases up to 450 and decreases afterwards, The minimum number of unequal forces in a, 4) decreases up to 450 and increases afterwards, plane that can keep a particle in equilibrium is, 14. Keeping the speed of projection constant, the, 1) 4, 2) 2, 3) 3, 4) 6, angle of projection is increased from 0° to 90º., The minimum number of non coplanar forces, Then the maximum height of the projectile, that can keep a particle in equilibrium is, 1) goes on increasing upto 900, 1) 1, 2) 2, 3) 3, 4) 4, 2) decreases upto 900, A train is moving due east and a car is moving, 3) increases upto 450 and decreases beyond it, due north with equal speeds. A passenger in, 4) decreases upto 450 and increases beyond it, the train finds that the car is moving towards, , C.U.Q, , 1., , 2., , 3., , 4., , 5., , 6., 7., 8., , 122, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 15. The path of one projectile as seen from another, projectile is a ( if horizontal components of, velocities are equal ), 1) straight line, 2) parabola, 3) hyperbola, 4) circle, 16. Two particles are projected with same speed, , 17., 18., , 19., , 20., , 21., , 1) equal horizontal and equal vertical ranges, 2) equal vertical ranges but different horizontal, ranges, 3) different vertical ranges but equal horizontal, ranges, 4) different vertical and different horizontal ranges, 0, but at angles of projection ( 45 − θ ) and 22. For body thrown horizontally from the top of a, tower,, ( 450 + θ ) . Then their horizontal ranges are in, 1 ) the time of flight depends both on h and v, the ratio of, 2 ) the horizontal Range depends only on v but not, 1) 1 : 2 2) 2 : 1 3) 1 : 1 4) none of the above, on h, The acceleration of a projectile relative to, 3) the time of flight and horizontal Range depend, another projectile is, on h but not on v, 1) -g, 2) g, 3) 2g, 4)0, 4) the horizontal Range depends on both v and h, A stone is just dropped from the window of a, 23. A body is projected from a point with different, train moving along a horizontal straight track, angles of projections 200, 350, 450, 600 with the, with uniform speed. The path of the stone is, horizontal but with same initial speed. Their, 1) a parabola for an observer standing by the side, respective horizontal ranges are R1, R2, R3 and, of the track, R4. Identify the correct order in which the, 2) a horizontal straight line for an observer inside, horizontal ranges are arranged in increasing order, the train, 1) R1, R4, R2, R 3, 2) R2, R1, R4, R 3, 3) both (1)&(2) are true, 3), R, ,, R, ,, R, ,, R, 4), R 4, R 1, R 2, R 3, 4) (1) is true but (2) is false, 1, 2, 4 3, A bomb is dropped from an aeroplane flying 24. Two particles are projected from the same, horizontally with uniform speed. The path of, point with the same speed at different angles, the bomb is, θ1 and θ2 to the horizontal. If their respective, 1) a vertical straight line for a stationary observer, times of flights are T1 and T2 and horizontal, on the ground, ranges are same then, 2) a parabola for the pilot of the aeroplane, 3) a vertical straight line for the pilot and parabola, a) θ1 + θ2 = 900 b) T1 = T2 tan θ1, for a stationary observer on the ground, c) T1 = T2 tan θ2 d) T1 sin θ2 = T2 sin θ1, 4) a horizontal straight line for the pilot and, 1) a, b, d are correct 2) a, c, d are correct, parabola for a stationary observer on the ground, 3) b, c, d are correct 4) a, b, c are correct, A and B are two trains moving parallel to each, other. If a ball is thrown vertically up from the 25. Two bodies are projected at angles 300 and 600, train A, the path of the ball is, to the horizontal from the ground such that the, 1) parabola for an observer standing on the ground, maximum heights reached by them are equal., 2) vertical straight line for an observer in B when, Then, B is moving with the same speed in the same, a) Their times of flight are equal, direction of A, b) Their horizontal ranges are equal, 3) a parabola for an observer in B when B is, c) The ratio of their initial speeds of projection, moving with same speed but in opposite, direction, is 3 : 1, 4) all the above are true, d) Both take same time to reach the maximum, A ball is thrown from rear end to the front end, height., of a compartment of a train which is moving at, 1) a, b, c and d are correct, constant horizontal velocity. An observer, 2) only a, b and c are correct, sitting in the compartment and another, 3) only a and c are correct, observer standing on the ground draw the, 4) only a, c and d are correct, trajectory of the ball. They will have, , NARAYANAGROUP, , 123
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MOTION IN A PLANE, , JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , 4) none of the above is correct, 26. A body is projected with an initial speed of, −1 at an angle of 60 0 above the, Answer 31 to 33 based on following information, 100 3ms, ur ur ur ur ur, -2, ur, horizontal. if g = 10ms then velocity of the, A, B , C , D , E and F are coplanar vectors, projectile, having the same magnitude each of 10 units, a) Is perpendicular to it's acceleration at the, and angle between successive vectors is 600, instant t = 15 sec., b) Is perpendicular to initial velocity of 31. The magnitude of resultant is, 1) 0 units 2) 1 units 3) 2 units 4) 3 units, projection at t = 20 sec., ur, 32. If A is reversed the magnitude of resultant is, c) Is minimum at the highest point, d) Changes both in magnitude and direction,, 1) 10 units 2) 20 units 3) 30 units 4) 40 units, ur, ur ur, during its flight., 33. If A , B & C are reversed magnitude of, 1) a, b, c and d are correct, resultant is, 2) only a, c and d are correct, 1) 10 units 2) 20 units 3) 30 units 4) 40 units, 3) only b, c and d are correct, Answer 34 to 36 based on following information, 4) only a, b and d are correct, On an open ground, a motorist follows a track, 27. A particle is moving along a circular path with, that turns to his left by an angle of 60° after, uniform speed . Through what angle does its, angular velocity change when it completes, every 500 m. Starting from a given turn, The, half of the circular path ?, path followed by the motorist is a regular, 1) 00, 2) 450, 3)1800, 4)3600, hexagon with side 500 m, as shown in the given, 28. A car of mass m moves in a horizontal circular, figure specify the displacement of the motorist, path of radius r metre. At an instant its, speed is V m / s and is increasing at a rate of, a m / sec2 . then the acceleration of the car is, 600 S, T, , 29., , 30., , 124, , 2, 600, V 2 , V2, V2, 2, 1), 2) a 3) a + 4) a +, r, r, r , 600, Consider the following two statements A and, U, V, B and identify the correct choice, R, 0, A)When a rigid body is rotating about its own, 60, axis, at a given instant all particles of body, 600, possess same angular velocity., 600, β, B)When a rigid body is rotating about its own, Q, P, axis, the linear velocity of a particle is directly 34. at the end of third turn., proportional to its perpendicular distance, 1) 500 m 2) 250 m 3) 1000 m 4) 1500 m, from axis, 35. at the end of sixth turn., 1) A is true but B is false, 1) 3000 m 2) 1500 m 3) 0m 4) 1000 m, 2) A is false but B is true, 36. at the end of eighth turn., 3) Both A and B are true, 1) 3000 m 2) 1500 m 3) 0, 4) 866 m, 4) Both A and B are false, Suppose a disc is rotating counter clockwise, C.U.Q-KEY, in the plane of the paper then, 1) It’s angular velocity vector will be perpendicu01)2 02) 4 03) 4 04) 4 05) 2 06) 3, lar to the page pointing up out of the page, 07)4 08) 2 09) 1 10) 1 11) 1 12) 4, 2) It’s angular velocity vector will be perpendicu13)3 14) 1 15) 1 16) 3 17) 4 18) 3, lar to the page pointing in wards, 19)3 20) 4 21) 2 22) 4 23) 1 24) 1, 3) It’s angular velocity vector acts along the tan25)4 26) 1 27) 1 28) 3 29) 3 30) 1, gent to the disc., 31)1 32) 2 33) 4 34) 3 35) 3 36) 4, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, 8., , LEVEL - I (C.W), PARALLELOGRAM LAW, 1., , 2., , 3., , 4., , 5., , The maximum resultant of two concurrent, forces is 10N and their minimum resultant is 9., 4N. The magnitude of large force is, 1) 5N, 2) 7N, 3) 3N, 4) 14N, The resultant of two vectors of magnitudes3, units and 5 units is perpendicular to 3 units., 10., The angle between the vectors is, 1) 127 0, 2) 1200, 3) 750, 4) 1500, The sum of two unit vectors is also a vector, of unit magnitude, then the magnitude of the, difference of the two unit vectors is, 1) 1 unit 2) 2 units 3) 3 units 4) zero, Which of the following sets of forces acting, 11., simultaneously on a particle keep it in, equilibrium?, 1) 3N, 5N, 10N, 2) 4N, 5N, 12N, 3) 2N, 6N, 5N, 4) 5N, 8N, 1N, r, r, The magnitude of two vectors P and Q differ by 1. The magnitude of their resultant, , Five equal forces each of 20N are acting at a, point in the same plane. If the angles between, them are same, the resultant of these forces, is, 1) 0, 2) 40N 3) 20N, 4) 20 2 N, A boy is hanging from a horizontal branch of, a tree. The tension in the arms will be maximum when the angle between the arms is, 1) 00, 2) 300, 3) 600, 4) 1200, A 10 kg body is suspended by a rope is pulled, by means of a horizontal force to make 600, by rope to vertical. The horizontal force is, 1) 10 kgwt, 2) 30 kgwt, 3) 10 3 kgwt, , 4) 30 3 kgwt, , T1, If ‘P’ is in equilibrium then T is, 2, , T2, , T1, , 600, , 300, , P, , ur, −1 3 , makes an angle of tan with P . The, 4, angle between, 6., , ur, , ur, P, , and Q is, 2) 00, 3) 1800 4) 900, 1) 450, Two vectors inclined at an angle θ have magnitude 3 N and 5 N and their resultant is of, magnitude 4 N. The angle θ is, 3, 4, −1 , −1 3, 2) cos, 3) cos, 4) cos − 5 , 5, 5, , , , 900 N, 1), , 12., , TRIANGLE LAW, POLYGON LAW, &LAMI’S THEOREM, 7., , _, , _, , vectors a = 3 i − 4 j + 2 k ,, _, , _, , _, , _, , _, , _, , b = 2 i − j − 6 k , c = 5 i − 5 j − 4 k is., 1) Quadrilateral, 3) Circle, , NARAYANAGROUP, , 2) Triangle, 4) Hyperbola, , 3), , 1, 3, , 4), , 1, 2, , A, , body, , starts, , with, , a, , velocity, , ( 2iˆ + 3 ˆj + 11kˆ ) m / s and moves with an acceleration ( 5iˆ + 5 ˆj − 5kˆ ) m / s . What is its ve2, , The plane which can be formed with the, _, , 2) 2, , MOTION IN A PLANE, , −1, , 1) 900, , 3, , 13., , locity after 0.2 sec?, ), ), 1) 7iˆ + 8 ˆj + 6k, 2) 2iˆ − 3 ˆj + 11k, ), ), 3) 3iˆ − 4 ˆj − 10k, 4) 3iˆ + 4 ˆj + 10k, The position vector of a particle is given by, r, r = 3t 2iˆ + 4t 2 ˆj + 7kˆ m at a given time t . The, , (, , ), , net displacement of the particle after 10 s is, 1) 500m 2) 400m 3) 300m 4) 700m, , 125
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, , CONCEPT OF RELATIVE, VELOCITY, , 14., , 15., , 20., A particle is moving eastwards with a velocity, 5 ms-1, changes its direction northwards in 10, seconds and moves with same magnitude of, velocity. The average acceleration is, 1, ms −2 towards N-E, 1) zero, 2), 21., 2, 3), , 1, ms −2 towards S-E, 2, , 4), , 1, ms −2 towards N-W, 2, , A man is going due east with a velocity of, 5 ms-1. It is vertically raining downwards with, a velocity of 4 ms-1. At what angle should he, hold the umbrella to the vertical so as to, protect him self from the rain?, 1) tan, 2) tan, 3), , −1, , −1, , 5, , 4, , in anti-clockwise direction, , 5, , 4, , in clock-wise direction, , 4, tan − 1 , 5, , OBLIQUE PROJECTILE, 22., , North of East, , 4, , 16., , 17., , 4) tan −1 5 East of North, Rain drops are falling down ward vertically, at 4kmph. For a person moving forward at, 3kmph feels the rain falling at, 23., 1)7 kmph 2)1 kmph 3)5 kmph 4)25 kmph, A man travelling at 10.8 kmph in topless car, on a rainy day. He holds an umbrella at angle, of 370 with the vertical so that he does not 24., wet.If rain drops falls vertically downwards, what is rain velocity., 1) 1 m/s 2) 2 m/s 3) 3 m/s 4) 4 m/s., , MOTION OF A BOAT IN A RIVER, 18., , 19., , 126, , stream. The ratio of times taken are, 1) 1 : 1, 2) 1 : 2, 3) 1 : 4, 4) 4 : 1, The velocity of water in a river is 2 kmph,, while width is 400 m. A boat is rowed from a, point rowing always aiming opposite point at, 8 kmph of still water velocity. On reaching, the opposite bank the drift obtained is, 1) 93 m 2) 100.8 m 3) 112.4 m 4) 100 m, A man can swim in still water at a speed of 4, kmph. He desires to cross a river flowing at, a speed of 3 kmph in the shortest time interval. If the width of the river is 3km time taken, to cross the river (in hours) and the horizontal distance travelled ( in km) are respectively, 3, 3 9, 3, 1 15, ,7, 2) ,3, 3) ,, 4), 1) ,, 7, 4 4, 5, 4 4, , A man can row a boat in still water with a 25., velocity of 8 kmph. Water is flowing in a river, with a velocity of 4 kmph. At what angle, should he row the boat so as to reach the, exact opposite point, 1) 1500 to flow of water, 2) 1200 to flow of water., 26., 3) 300 to flow of water., 0, 4) 90 to flow of water., A person can swim in still water at 5 m/s. He, moves in a river of velocity 3 m/s, first down, the stream and next same distance up the, , A particle is projected in xy plane with y-axis, along vertical, the point of projection is origin., g 2, The equation of the path is y = 3x − x ., 2, where y and x are in m. Then the speed of, projection in m s −1 is, 3, 2, If a body is thrown with a speed of 19.6m/s, making an angle of 300 with the horizontal,, then the time of flight is, 1) 1 s, 2) 2 s, 3) 2 3 s 4) 5 s, A particle is projected with an initial velocity, of 200 m/s in a direction making an angle of, 300 with the vertical. The horizontal distance, covered by the particle in 3s is, 1) 300 m 2) 150 m 3) 175 m 4) 125 m, A body is projected with an initial velocity, 20 m/s at 600 to the horizontal. Its initial, velocity vector is ____(g=10 m / s 2 ), , 1) 2, , 2), , 3, , 1) 10iˆ − 20 ˆj, , 3) 4, , 4), , 2) 10 3 iˆ + 10 ˆj, , 3) 10iˆ + 10 3 ˆj, 4) 5iˆ + 5 3 ˆj, A body is projected at an angle of 30° with the, horizontal with momentum P. At its highest, point the magnitude of the momentum is:, 1), , 3, P, 2, , 2), , 2, P, 3, , 3) P, , 4), , P, 2, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 27., , The potential energy of a projectile at its, maximum height is equal to its kinetic energy, 35., there. If the velocity of projection is 20 ms −1 ,, its time of flight is (g=10 ms −2 ), , 1, 1, s, s, 4), 2, 2, From a point on the ground a particle is 36., projected with initial velocity u , such that its, horizontal range is maximum. The magnitude, of average velocity during its ascent., 1) 2s, , 28., , MOTION IN A PLANE, , 2) 2 2 s 3), , 5u, 3, 5u, 37., 2), 3), 4) none, 4, 2 2, 2 2, 29. The horizontal and vertical displacements of, a projectile are given as x = at &, 1), , y = b t − ct 2 . Then velocity of projection is, 1), , a 2 + b2, , b 2 + c2, , 2), , 38., , 3) a 2 + c2, 4) b 2 − c 2, 30. Two bodies are thrown from the same point with, the same velocity of 50ms-1. If their angles of, projection are complimentary to each other and, the difference of maximum heights is 30m, the 39., minimum and maximum heights are(g=10 m/s2), 1) 50 m & 80 m, 2) 47.5 m & 77.5 m, 3) 30 m & 60 m, 4) 25 m & 55 m, 31. A missile is fired for maximum range with an, initial velocity of 20 m s −1 , the range of the 40., missile is ( g = 10 m / s 2 ) [AIPMT 2011], 1) 50m, 2) 60 m 3) 20m 4) 40 m, r, 32. If u = a $i + b $j + c k$ with $i, $j , k$ are in east,, north and vertical directions, horizontal 41., component of velocity of projectile is, 33., , 34, , 1) a, 2) b 3) a 2 + b 2 4) b 2 + c2, If the time of flight of a projectile is doubled,, what happens to the maximum height at42., tained?, 1) halved 2) remains unchanged, 3) doubled 4) become four times, r, If u = a $i + b $j + c k$ with $i, $j , k$ are in east,, north and vertical directions, the maximum 43., height of the projectile is, 1), , a2, 2g, , 2), , NARAYANAGROUP, , b2, 2g, , 3), , c2, 2g, , 4), , b 2 c2, 2g, , HORIZONTAL PROJECTILE, A body projected horizontally with a velocity ‘v’, from a height ‘h’ has a range ‘R’. With what, velocity a body is to be projected horizontally, from a height h/2 to have the same range ?, 1) 2 v 2) 2v 3) 6v, 4) 8v, A stone is thrown horizontally with velocity g, ms-1 from the top of a tower of height g metre., The velocity with which it hits the ground is, (in ms-1), 1) g, 2) 2g 3) 3g, 4) 4g, A body is thrown horizontally from the top of, a tower. It reaches the ground after 4s at an, angle 450 to the ground. The velocity of, projection is, 1)9.8ms-1 2)19.6ms-1 3)29.4ms-1 4) 39.2ms-1, Two cliff of heights 120 m and 100.4 m are, separated by a horizontal distance of 16 m if a, car has to reach from the first cliff to the second, the horizontal velocity of car should be, 1) 16 m/s 2) 4 m/s 3) 2 m/s 4) 8 m/s, , CIRCULAR MOTION, A circular disc is rotating about its own axis, at the rate of 200 revolutions per minute. Two, particles P,Q of disc are at distances 5cm,, 10cm from axis of rotation. The ratio of, angular velocities of P and Q is, 1) 1:2, 2) 1:1, 3) 2:1, 4) 4:1, A stationary wheel starts rotating about its, own axis at uniform angular acceleration 8rad/, s2. The time taken by it to complete 77, rotations is, 1) 5.5 sec 2) 7 sec, 3) 11 sec 4) 14 sec, A circular disc is rotating about its own axis at, uniform rate completes 30 rotations in one, minute.The angular velocity of disc in rad s–1 is, p, p, 2) p, 3), 4), 1) 2p, 2, 4, A particle is moving at uniform speed, 2 ms–1 along a circle of radius 0.5m. The, centripetal acceleration of particle is, 1) 1ms-2 2) 2ms-2 3) 4ms-2 4) 8ms-2, A particle P is moving in a circle of radius, ' a ' with a uniform speed v . C is the centre, of the circle and AB is a diameter. when, passing through B the angular velocity of P, about A and C are in the ratio, 1) 1 : 1, 2) 1 : 2, 3) 2 : 1, 4) 1 : 3, , 127
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 10., , LEVEL -I (C.W) - KEY, 01)2, 07)2, 13)1, 19)3, 25)3, 31)4, 37)4, 43)2, , 02)1, 08)1, 14)4, 20)4, 26)1, 32)3, 38)4, , 03)3, 09)4, 15)2, 21)1, 27)2, 33)4, 39)2, , 04)3, 10)3, 16)3, 22)1, 28)1, 34)3, 40)3, , 05)4, 11)3, 17)4, 23)2, 29)1, 35)1, 41)2, , ///////////////////////////////////, θ, , 06)4, 12)4, 18)2, 24)1, 30)2, 36)3, 42)4, , T, F, , LEVEL -I (C.W) - HINTS, 1. F1 + F2 = 10 , F1 − F2 = 4, 11., 13., , 5, 2., , 4, 1, , θ, , 3, θ = 53 ; θ = 180 − θ 1, θ, θ, R = 2 P cos , S = 2 P sin, 2, 2, 1, , 3., 4., 5., 6., 7., 8., , 9., , θ, 14., , 0, , f − f < f <f + f, 1, 2, 3 1, 2, , 15., 3, α = tan , 4, 16., 2, 2, 2, R = P + Q + 2 PQ cos θ, ) ) ), a + b + c = 0 (triangle law), If polygon is closed, resultant becomes zero. If, r, resultant of 5 forces is Q and magnitude of each 17., r, r, r r r, r, vector is P then Q + P = 0 ⇒ P = Q, 18., r r, P − Q = 1;, , −1, , mg, F = mgTanθ, ), ), T1 cos 300 = T2 cos 600 12. V¼, = u + a .t, r, r, At t = 0 , r1 = 7 k, r r r r, r, r, r, At t = 10, r2 = 300i + 400 j + 7k ; s = r2 − r1, N, aavg, vf, N-W, ) ), ) v f − vi, a=, t, v, -vi, O, i, O, -Vm, Vm, θ, v, Tanθ = m (clockwise), vr, Vrm, V, R, , VRM = VR − V M, O, -Vm, 0, 37, , Vrm, , Vm, , B, T cos θ, , θ, , Vb, , T, , VR, , θ, , θ, , T sin θ, , 90, A, , T sin θ, , VM, VR, , VR, , T cos θ, T, , Tan ( 370 ) =, , 0, , VR, , Vriver, Vboat ; α = 90 + θ with stream, d, d, t1 =, and t2 =, v+u, v−u, , sin θ =, , 19., , mg, 2T cos θ = mg ;, 128, , T=, , mg, 2 cosθ, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , x, , B, , MOTION IN A PLANE, , C, , LEVEL - I (H.W), , Vb, 20., , θ, , d, , x = VW ., , Vw, , A, 21., 22., , 25., 27., , 28., 29., 30., 31., 33., 35., , g, x2, 2, 2u cos2 θ, , 2u sin θ, T=, g, , 2., 24.x = ( u cos θ ) t, , r, u = u cos θ ˆi + usin θ ˆj 26. Px = P cos θ , Py = 0, 1 2 2, 1 2 2, P.E = K .E ⇒ 2 mu sin θ = 2 mu cos θ, 2u sin θ, use T =, g, u, 1 + 3cos2 θ, 2, dx, dy, ux = , u y =, and u = u x2 + u y2, dt, dt, avg velocity =, , u2, H, +, H, =, = 125, H 2 − H1 = 30 ; 2, 1, 2g, u2, Rmax =, 32.East and north are taken as ground., g, H=, , gT 2, 8, , R=u, , PARALLELOGRAM LAW,, TRIANGLE LAW & POLYGON LAW, 1., , d, d, t=, x = VW ., ;, VB, VB, Compare the equation with, y = x tan θ −, , 23., , d, VB, , 3., , 4., , 5., , Two forces each of 20N act on a body at 1200., The magnitude and direction of resultant is, 1) 20N;φ =60 0 2) 20 2N;φ =600, 3) 10 2N;φ =00 4) 10 2N;φ =1200, Two forces whose magnitudes are in the ratio, 3:5 give a resultant of 35N. If the angle, between them is 600, the magnitude of each, force is, 1) 3N, 5N, 2) 9N, 25N, 3) 15N, 25N, 4) 21N, 35N, The resultant of two forces 2P and 2P is, , 10P . The angle between the forces is, 3) 450, 4) 900, 1) 300 2) 600, Which one of the following cannot be, represented by the three sides of a triangle?, 1) 5,9,11 2) 5,7,13 3) 7,10,13 4) 3,8,9, r r, r, Figure shows three vectors a, b and c where, R is the mid point of PQ. then which of the, following relations is correct., P, , a, , 34. u x = a 2 + b 2 , u y = u sin θ = c, 2h, ; given R is same, g, , 36., , V = u 2 + 2 gh, , 38., 39., , R=, , 37. Tanα =, , O, , gt, u, , u. 2 ( h1 − h2 ), g, , Angular velocity ω = Constant, θ = 2π N ; t =, , 41., , ω = 2π n ;where n = no.of revolutions per sec, , 42., , v2, Centripetal acceleration a =, r, , 43., , Angular velocity of P about A , ω A =, , v, 2a, v, Angular velocity of P about C, ωC =, a, , NARAYANAGROUP, , 6., , 2θ, α, , 40., , R, c, , 7., , b, , Q, , r r, r, r r r, 1) a + b = 2c, 2) a + b = c, r r uur, r r r, 3) a − b = 2c, 4) a − b = c, Eleven forces each equal to 5N act on a particle simultaneously. If each force makes an, angle 300 with the next one, the resultant of, all forces is, 1) 15 N 2) 55 N 3) 5 N, 4) zero, A body of mass 3 kg is suspended by a, string to rigid support.The body is pulled, horizontally by a force F until the string, makes an angle of 300 with the vertical. The, value of F and tension in the string are, 1) 9.8 N,9.8N, 2) 9.8 N, 19.6 N, 3) 19.6 N, 19.6 N, 4) 19.6 N, 9.8 N, 129
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 8., , Two light strings of length 4cm and 3 cm are, tied to a bob of weight 500 gm. The free ends, of the strings are tied to pegs in the same, horizontal line and separated by 5 cm.The, ratio of tension in the longer string to that in, 15., the shorter string is, 1) 4:3, 2) 3:4, 3) 4:5, 4) 5:4, , MOTION IN A PLANE, 9., , 10., , A force 2iˆ + ˆj − kˆ newton acts on a body, which is initially at rest. If the velocity of the, body at the end of 20seconds is, 4iˆ + 2 ˆj + 2kˆ ms −1 , the mass of the body, 16., 1) 20kg, 2) 15kg, 3) 10kg 4) 5kg, The position vector of a moving particle at, t, r, 2, 3, seconds is given by r = 3iˆ + 4t ˆj − t kˆ. Its, displacement during an interval of 1s to 3s is, 2) 3iˆ + 4 ˆj − kˆ, 1) ĵ − kˆ, 3) 9iˆ + 36 ˆj − 27 kˆ, , which angle he should hold the umbrella with, vertical if he moves forward, if it is raining, vertically downwards, 2) 450, 3) 600, 4) 900, 1) 300, When it is raining vertically down, to a man, walking on road the velocity of rain appears, to be 1.5 times his velocity . To protect, himself from rain he should hold the umbrella, at an angle θ to vertical. Then tanθ =, , 2, 2, 3, 5, 2), 3), 4), 5, 3, 2, 2, A motor car A is travelling with a velocity of, 20m/s in the north - west direction and another, motor car B is travelling with a velocity of, 15 m/s in the north-east direction. The, magnitude of relative velocity of B with, respect to A is., 1) 25 m/s 2) 15 m/s 3) 20 m/s 4) 35 m/s, 1), , 4) 32 ˆj − 26kˆ, r, r r r, 11. If initial velocity of a body is u = 2i − 2 j + 3k, MOTION OF A BOAT IN A RIVER, r r r r, and the final velocity is v = 2i − 4 j + 5k and 17. A man can swim in still water at a speed of 6, kmph and he has to cross the river and reach, it is changed in time of 10 sec. Find the, just opposite point on the other bank. If the, acceleration vector?, river is flowing at a speed of 3 kmph, and the, r r, r, r r, r, width of the river is 2km, the time taken to, 3i − 2 j + 2k, −3i + j + 2k, cross the river is (in hours), 1), 2), 10, 10, 2, 2, 2, 2, r r, r, r r, 1), 2), 3), 4), −3i − 2 j + 2k, −j+k, 27, 45, 27, 3, `3), 4), 18. A boat moves perpendicular to the bank with, 10, 5, a velocity of 7.2 km/h. The current carries it, CONCEPT OF RELATIVE VELOCITY, 150m downstream, find the velocity of the, 12. A particle is moving eastwards with a, current(The width of the river is 0.5 km)., −1, velocity 15 ms . Suddenly it moves towards, 1) 0.4 m s − 1, 2)1.2 ms −1, north and moves with the same speed in a, 3) 0.5ms −1, 4) 0.6 ms −1, time 10 sec. The average acceleration dur-19. A swimmer is capable of swimming 1.65 ms −1, ing this time is, in still water. If she swims directly across a, 1) 3 / 2 NE, 2) 3 2 NE, 180m wide river whose current is 0.85 ms −1 ,, 3) 3 / 2 NW, 4) 3 2 NW, how far downstream(from a point opposite her, 13. A Person crossing a road with a certain, starting point) will she reach?, velocity due north, sees a car moving, 1) 92.7m 2) 40m 3) 48m 4)20m, towards east. The relative velocity of the car 20. A person swims at, 1350 to current of river, to, w.r.t the person is 2 times that of the vemeet target on reaching opposite point. The, locity of the person . The angle made by the, ratio of person’s velocity to river water verelative velocity with the east is, locity is, 0, 0, 0, 0, 1) 30, 2) 45, 3) 60, 4) 90, 1. 3 :1 2. 2 :1 3. 1: 2 4. 1: 3, 14. A Person is walking in rain feels the, velocity of rain as twice to his velocity . At, 130, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , OBLIQUE PROJECTILE, 21., , 29., , The parabolic path of a projectile is, , x2, in MKS units :, 3 60, Its angle of projection is (g = 10ms-2), 1) 300, 2) 450, 3) 600, 4) 900, 22. A body is projected at angle 300 to horizontal, with a velocity 50 ms −1 . Its time of flight is, 30., 1) 4 s, 2) 5 s, 3) 6 s, 4) 7 s, 23. A body is projected with velocity 60 m / s at, 300 to the horizontal. The velocity of the body, after 3 seconds is, 2) 30iˆ, 1) 20iˆ + 20 3 ˆj, represented by y =, , x, , −, , 24., , 31., 3) 10 3 ˆj, 4) 30 3iˆ, A body is projected with velocity u such that, in horizontal range and maximum vertical, heights are same.The maximum height is, , 25., , u2, 3u 2, 16u 2, 8u 2, 1), 2), 3), 4), 32., 2g, 4g, 17 g, 17 g, A cricket ball is hit for a six leaving the bat at, an angle of 600 to the horizontal with kinetic, energy ‘k’. At the top, K.E. of the ball is, [JEE MAIN-2007], , 26., , 27., , 28., , k, k, 33., 1) Zero, 2) k, 3), 4), 2, 4, A bomb at rest is exploded and the pieces, are scattered in all directions with a maximum, velocity of 20ms-1. Dangerous distance from, that spot is (g = 10 m/s2), 1) 10 m, 2) 20 m 3) 30 m 4) 40 m, 34., A boy can throw a stone up to a maximum, height of 10 m. The maximum horizontal, distance that the boy can throw the same, stone up to will be, [JEE MAIN-2012], 35., 1) 20 2 m 2) 10 m 3) 10 2 m 4) 20 m, A grass hopper can jump a maximum, horizontal distance of 0.3 m. If it spends, negligible time on the ground, its horizontal, component of velocity is (g=10 m/s2), 1) 3/2 m/s, , 2), , 3, 2, , 3) 1/2 m/s, , 4), , 2, m/s, 3, , NARAYANAGROUP, , m/s, , A stone is thrown with a velocity v at an, angle θ with the horizontal. Its speed when it, makes an angle β with the horizontal is, , v, cos β, , 1) v cos θ, , 2), , 3) v cos θ cos β, , v cos θ, 4) cos β, , A body is projected with a certain speed at, angles of projection of θ and 90– θ . The maximum heights attained in the two cases are 20, m and 10 m respectively. The maximum possible range is, 1) 60 m 2) 30 m 3) 20 m, 4) 80 m, The launching speed of a certain projectile is, five times the speed it has at its maximum, height. Its angle of projection is, 1) θ = cos-1(0.2), 2) θ = sin-1(0.2), 3) θ = tan-1(0.2), 4) θ = 00, A person throws a bottle into a dustbin at the, same height as he is 2m away at an angle of, 450 . The velocity of thrown is, 1) g, , 2), , g, , 3) 2g, , 4), , 2g, , HORIZONTAL PROJECTILE, A body projected horizontally from the top of, a tower follows y = 20 x 2 parabola equation, −2, where x , y are in m ( g = 10 m s ) .Then the, , velocity of the projectile is (ms-1), 1) 0.2 2) 0.3, 3) 0.4 4) 0.5, A bomb is dropped from an aeroplane flying, horizontally with a velocity of 720 kmph at, an altitude of 980m. Time taken by the bomb, to hit the ground is, 1) 1 s, 2) 7.2 s 3) 14.14 s 4) 0.15 s, A body is projected horizontally from a height, of 78.4 m with a velocity 10 ms −1 . Its velocity, after 3 seconds is (g=10 ms −2 )(Take direction, of projection as i and vertically upward, direction as j), 1) 10$i − 30 $j, 2) 10$i + 30 $j, 3) 20$i − 30 $j, , 4) 10$i + 10 3 $j, 131
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 36., , Two thin wood screens A and B are separated, by 200 m a bullet travelling horizontally at speed, of 600 m/s hits the screen A penetrates through, it and finally emerges out from B making holes 1., in A and B the resistance of air and wood are, negligible the difference of heights of the holes, 2., in A and B is., 1) 5 m, , 49, m, 2), 90, , 3), , 7, m, 90, , 4)zero, , CIRCULAR MOTION, 37., , 38., , 39., , 3., 4., , A fly wheel is rotating about its own axis at, an angular velocity 11 rad s–1, its angular, velocity in revolution per minute is, 1) 105, 2) 210, 3) 315, 4) 420, A stationary wheel starts rotating about its own, axis at constant angular acceleration. If the 5., wheel completes 50 rotations in first 2 seconds,, then the number of rotations made by it in next, two seconds is, 1) 75, 2) 100, 3) 125, 4) 150, A point size body is moving along a circle at an 6., angular velocity 2.8rads-1 . If centripetal, 1) 1.25ms-1, , 41., , θ, R = 2F cos 60 = F , α = 2, makes equal angle with both vectors, P 3, = ( given) ;Let P = 3x and Q = 5x, Q 5, , R = P 2 + Q 2 + 2 PQ cos θ, , acceleration of body is 7ms-2 then its speed is, , 40., , LEVEL -I (H.W)-HINTS, , 7., 2) 2.5ms-1, 3) 3.5ms-1, 4) 7ms-1, A circular plate is rotating about its own axis at, an angular velocity 100 revolutions per minute., The linear velocity of a particle P of plate at a, distance 4.2 cm from axis of rotation is, 1) 0.22 m/s 2) 0.44 m/s 3) 2.2 m/s 4) 4.4 m/s, An aircraft executes horizontal loop of radius 8., 1 km with steady speed of 900 kmph. The, ratio of centripetal acceleration with acceleration due to gravity is, 1) 6.0, 2) 6.4, 3) 5, 4) 7, , R 2 = P2 + Q 2 + 2 PQ cos θ, To get a closed ∆ le , the sum of any two vectors, in magnitude must be either equal or large in, magnitude of the third., , a, , c, , b, , r r, r, ∴ b + a = 2c, , 360, n=, = 12, No. of forces = 11;, θ, polygon formed with 1 side absent, resultant is, closing side., F = mgTanθ , T = F 2 + ( mg ) 2, , 5 cm, θ1, , θ2, , T1, , 4 cm, , 3 cm, T2, θ1, , θ2, , 500 g, θ2, 5, , LEVEL -I (H.W)-KEY, 01) 1, 07) 2, 13) 2, 19) 1, 25) 3, 31) 1, 37) 1, 132, , 02) 3, 08) 2, 14) 1, 20) 2, 26) 4, 32) 4, 38) 4, , 03) 3, 09) 3, 15) 1, 21) 1, 27) 4, 33) 4, 39) 2, , 04)2, 10)4, 16)1, 22)2, 28)2, 34)3, 40)2, , 05) 1, 11) 4, 17) 2, 23) 4, 29) 4, 35) 1, 41) 2, , 4, , 06) 3, 12) 3, 18) 4, 24) 4, 30) 1, 36) 2, , θ1, 3, T1, T2, =, sin(90 + θ1 ) sin(90 + θ ), 9., , →, , →, , →, , →, , v = ( a )t , and f = m a, , NARAYANAGROUP
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, , LEVEL - II (C.W), 7., , PARALLELOGRAM LAW, 1., , 2., , 3., , The greatest and least resultant of two forces, acting at a point are 29 Kg wt. and 5 Kg wt., respectively. If each force is increased by 3, Kg wt. the magnitude of the resultant of new, forces acting at right angles to each other is, 1) 45 kg wt., 2) 35 kg wt., 8., 3) 25 kg wt., 4) 15 kg wt., Two forces P and Q act at an angle of 1200, with each other. If the resultant is at right, angles to P and P is equal to 4 kg-wt, then the, value of Q is, 1) 4 kgwt 2) 8 kgwt 3) 6 kgwt 4) 3 kgwt, →, , →, , →, The resultant of two vectors P&Q is R, . If, , →, , the magnitude of Q is doubled the new, resultant becomes perpendicular to →, P then, →, is, magnitude of R, , 4., , 9., P+Q, P, P 2 − Q2, 1), 2), 3) Q 4), P−Q, Q, 2PQ, r r r r, P, Q, R, S are vector of equal magnitude. If, r, r r r, r, P + Q − R = 0 angle between P and Q is θ1 ., r r r, r, r, If P + Q − S = 0 angle between P and S is, 10., θ2 . The ratio of θ1 to θ2 is, 1) 1 : 2, , 4) 1: 3, TRIANGLE LAW, POLYGON LAW OF, 5., , 6., , 134, , 2) 2 : 1, , 3) 1 : 1, , VECTORS AND LAMI’S THEOREM, If ABCD is quadrilateral whose sides, uuur, represent vectors in cyclic order, AB is 11., equivalent is, uuur uuur, uuur, 1) CA + CB, 2) CD, uuur uuur uuur, uuur uuur, 3) AD + DC + CB, 4) AD + BD, An iron sphere of mass 100 kg is suspended, freely from a rigid support by means of a rope, of length 2m. The horizontal force required, to displace it horizontally through 50cm is, , nearly, 1) 980 N 2) 490 N 3) 245 N 4) 112.5 N, r, r, Three forces A = (iˆ + ˆj + kˆ ) , B = (2iˆ − ˆj + 3kˆ ), r, and C acting on a body to keep it in, r, ( 2008 - M), equilibrium. Then C is, 1) −(3iˆ + 4kˆ ), 2) −(4iˆ + 3kˆ ), 3) (3iˆ + 4 ˆj ), , 4) (2iˆ − 3kˆ ), , MOTION IN A PLANE, The displacement of the point of a wheel, initially in contact with the ground when the, wheel rolls forward quarter revolution where, perimeter of the wheel is 4π m, is (Assume, the forward direction as x-axis), −1 2, 1) (π + 2) 2 + 4 along tan, with x - axis, π, 2, −1, 2) (π − 2) 2 + 4 along tan, with x - axis, π −2, −1 2, 3) (π − 2) 2 + 4 along tan, with x - axis, π, 2, −1, 4) (π + 2)2 + 4 along tan, with x - axis, π −2, A particle starts from the origin at t = 0s with, a velocity of 10.0 ˆj m/s and moves in the, xy − plane with a constant acceleration of, , (8iˆ + 2 ˆj ) ms, , −2, , . Then y − coordinate of the, particle in 2 sec is, 1) 24 m 2) 16 m, 3) 8 m, 4) 12 m, A car moving at a constant speed of 36 kmph, moves north wards for 20 minutes then due, 1, to west with the same speed for 8 minutes., 3, what is the average velocity of the car during, this run in kmph, 1) 27.5 2) 40.5, 3)20.8, 4) 32.7, Velocity of a particle at time t = 0 is 2ms −1 ., A constant acceleration of 2 ms −2 acts on the, particle for 1 second at an angle of 600 with, its initial velocity . Find the magnitude of, velocity at the end of 1 second., 1) 3 m / s, 2) 2 3 m / s, 3) 4 m / s, , 4) 8 m / s, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , RELATIVE VELOCITY, 12., , 13., , 14., , 15., , 17., , 23., , A boat takes 2 hours to travel 8km and back, in still water lake.With water velocity of 4, kmph, the time taken for going upstream of, 8km and coming back is, 1) 160 minutes, 2) 80 minutes, 3) 320 minutes, 4) 180 minutes, 24., The velocity of water in a river is 2 kmph,, while width is 400 m. A boat is rowed from a, point rowing always aiming opposite point at, 8 kmph of still water velocity. On reaching, the opposite bank the drift obtained is, 1) 93 m 2) 100.8 m 3) 112.4 m 4) 100 m, , OBLIQUE PROJECTILE, 18., , 19., , An aeroplane moving in a circular path, with a speed 250 km/h. The change in, velocity in half of the revolution is., 1) 500km/h, 2) 250km/h, 3) 120 km/h, 4) zero, A car starting from a point travels towards, east with a velocity of 36 kmph. Another car 20., starting from the same point travels towards, north with a velocity of 24 kmph. The relative, velocity of one with respect to another is, 2)30 kmph, 1) 12 13 kmph, 3) 12 kmph, 4) 20 kmph, A ship is moving due east with a velocity of, 21., 12 m/sec, a truck is moving across on the ship, with velocity 4m/sec. A monkey is climbing, the vertical pole mounted on the truck with a, velocity of 3m/sec. Find the velocity of the, monkey as observed by the man on the shore, (m/sec), 1) 10, 2) 15, 3) 13, 4)20, 22., A man is walking due east at the rate of, 2kmph. The rain appears to him to come down, vertically at the rate of 2kmph. The actual, velocity and direction of rainfall with the, vertical respectively are, (2008 M), 1, kmph,30 0, 2), 1) 2 2kmph, 450, 2, 0, 3) 2 kmph, 0, 4) 1kmph, 900, , MOTION OF A BOAT IN A RIVER, 16., , MOTION IN A PLANE, , A particle is projected from ground with some, initial velocity making an angle of 450 with, the horizontal. It reaches a height of 7.5 m, above the ground while it travels a horizontal, distance of 10 m from the point of projection., The initial speed of the projection is, , NARAYANAGROUP, , 1) 5 m/s 2) 10 m/s 3) 20 m/s 4) 40 m/s, A particle is projected from ground at an, angle 45° with initial velocity 20 2 m s −1 ., The magnitude of average velocity in a, time interval from t = 0 to t = 3 s in m s −1 is, 1) 20.62 2) 10.31 3) 41.14 4) 5.15, A ball is thrown with a velocity of u making, an angle θ with the horizontal. Its velocity, vector normal to initial vector (u) after a time, interval of, u sin θ, u, u, u cos θ, 1), 2), 3), 4), g, g cos θ, g sin θ, g, A stone is projected with a velocity 20 2 m/, s at an angle of 450 to the horizontal. The average velocity of stone during its motion from, starting point to its maximum height is, 1) 10 5 m/s, , 2) 20 5 m/s, , 3) 5 5 m/s, 4) 20 m/s, A player kicks a foot ball obliquely at a speed, of 20 m/s so that its range is maximum., Another player at a distance of 24m away in, the direction of kick starts running at that, instant to catch the ball. Before the ball hits, the ground to catch it, the speed with which, the second player has to run is (g=10 ms-2), 1) 4 m/s, 2) 4 2 m/s, 4) 8 m/s, 3) 8 2 m/s, A particle is fired with velocity u making, angle θ with the horizontal . What is the, change in velocity when it is at the highest, point ?, 1) u cos θ 2) u 3) u sin θ 4) ( u cos θ − u ), Two projectiles A and B are thrown from the, v, respec2, tively. If B is thrown at an angle 45° with, horizontal. What is the inclination of A. when, their ranges are the same?, , same point with velocities v and, , −1 1 , 1) sin , 4, , 1 −1 1 , 2) sin , 2, 4, , −1 1 , 3) 2sin , 4, , 4), , 1 −1 1 , sin , 2, 8, , 135
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 25., , 26., , A particle is projected with a velocity v such 31., that its range on the horizontal plane is twice, the greatest height attained by it, The range, of the projectile is (when it is acceleration due, to gravity is ‘g’), [PMT 2010], 4v 2, 4v 2, v2, 4g, 1), 2) 2, 3), 4), 5g, 5g, g, 5v, A number of bullets are fired in all possible, directions with the same initial velocity u., The maximum area of ground covered by, bullets is, 2, , u , u, u , u2 , π, 1) g 2) π 2 g 3) π 4) π , , , , , g, 2g , 2, , 27., , 2, , 32., , 2, , 33., , A ball is projected from the ground with a, velocity ‘u’ such that its range is maximum., Then, 1) Its velocity at half the maximum height is, , 3, u, 2, , 34., , 2) Its velocity at the maximum height is ‘u’., 3) Change in its velocity when it returns to the, ground is ‘u’., 4) all the above are true., , HORIZONTAL PROJECTILE, 28., , 29., , A stair case contains ten steps each 10 cm, high and 20 cm wide. The minimum horizontal, velocity with which the ball has to be rolled, off the upper most step, so as to hit directly, the edge of the lowest step is (approximately) 35., 1) 42ms-1 2)4.2ms-1 3) 24ms-1 4)2.4ms-1, From certain height 'h' two bodies are projected horizontally each with velocity v. One, body is projected towards North and the other, body is projected towards east. Their separation on reaching the ground, 1), , 30., , 2), , 4v2 h, g, , 3), , v2 h, g, , 4), , 8v 2 h, g, , An object is projected horizontally from a top, of the tower of height h. The line joining the 36., point of projection and point of striking on, the ground makes an angle 450 with, ground,Then with what velocity the object, strikes the ground, 1), , 136, , 2v2h, g, , 11gh, 2), 2, , 9gh, 2, , 3), , 7gh, 4), 2, , 5gh, 2, , A ball is thrown horizontally from a cliff such, that it strikes the ground after 5s. The line of, sight makes an angle 37° with the horizontal., The initial velocity of projection in ms-1 is, 100, 100, 100, 1) 50, 2), 3), 4), 3, 2, 3, An object is launched from a cliff 20 m above, the ground at an angle of 300 above the horizontal with an initial speed of 30 m/s. How far, does the object travel before landing on the, ground? (in metre), 4. 60 3, 1) 20, 2) 20 3 3) 60, A bomber flying upward at an angle of 53°, with the vertical releases a bomb at an altitude, of 800 m. The bomb strikes the ground 20 s, after its release. If g = 10 m s −2 , the velocity, at the time of release of the bomb in ms-1 is, 1) 400 2) 800, 3) 100, 4) 200, Two particles move in a uniform gravitational, field with an acceleration g. At the initial, moment the particles were located at same, point and moved with velocities u1 = 9 m s −1, and u2 = 4 m s −1 horizontally in opposite, directions. The time between the particles at, the moment when their velocity vectors are, mutually perpendicular in s is (take, g = 10 m s−2 ), 1) 0.36 2) 3.6, 3) 0.6, 4) 6, An aeroplane is flying horizontally at a height, of 980 m with velocity 100 ms-1 drops a food, packet. A person on the ground is 414 m, ahead horizontally from the dropping point., At what velocity should he move so that he, can catch the food packet., 50 −1, ms, 2, , 1) 50 2ms −1, , 2), , 3) 100ms −1, , 4) 200ms −1, , CIRCULAR MOTION, A cyclist riding with a speed of 27kmph. As, he approaches a circular turn on the road of, radius 80m, he applies breaks and reduces, his speed at the constant rate of 0.50 m/s, every second. The net acceleration of, cyclist on the circular turn is, 1) 0.5m/s2 2)0.87m/s2 3)0.56m/s2 4)1m/s2, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 37., , MOTION IN A PLANE, , The length of minute hand in a pendulum clock, is 10cm the speed of tip of the hand is (in m/s), 1), , π, π, 2), 6000, 18000, , 3), , π, 3600, , 4), , 43., , π, 1200, , PREVIOUS EAMCET QUESTIONS, 38., , Equations of motion of a projectile are given 44., by x = 36t and 2 y = 96t − 98t 2 m .The angle of, projection is equal to, (2011 - M), −1, , 3, , 1) sin 4 , , 3), 39., , 4, sin −1 , 5, , −1, , 4, , 2) sin 3 , , , 45., , −1 3 , 4) sin 5 , , , A and B are two vectors of equal magnitude and, 'θ ' is the angle between them. The angle, between A or B with their resultant is(E - 2010), θ, θ, 2), 3) 2θ, 4)0, 4, 2, If a body is projected with an angle θ to the, horizontal then, (E - 2008), 1) it’s velocity always perpendicular to its, acceleration, 2) its velocity becomes zero at its maximum height, 3) it’s velocity makes zero angle with the horizontal, at its maximum height, 4) the body just before hitting the ground, the, direction of velocity coincides with the, acceleration., 1., A body is projected at an angle θ so that its, range is maximum. If T is the time of flight, then the value of maximum range is, (acceleration due to gravity = g) ( 2014 - E), , 1), , 40., , 41., , 2, , 2, , 2, , A body is projected horizontally from the top, of a tower with a velocity of 10 m/s. If it hits, the ground at an angle 450 , the vertical, component of velocity when it hits ground in, m/s is, (2014 - M), 3) 5 2 4)5, 1) 10, 2) 10 2, A body is projected with an angle θ . The, maximum height reached is h. If the time of, flight is 4 sec and g = 10m / s 2 , then the value, of h is, ( 2014 - M), 1) 10m, 2)40m, 3)20m, 4)5m, A person reaches a point directly opposite on, the other bank of a river. The velocity of the, water in the river is 4 m/s and the velocity of, the person in still water is 5 m/s. If the width, of the river is 84.6m, time taken to cross the, river in seconds is ( 2013 -M ), 1) 9.4, 2) 2, 3) 84.6, 4) 28.2, , LEVEL -II (C.W)-KEY, 01) 3, 07) 1, 13) 1, 19) 1, 25) 1, 31) 4, 37) 2, 43)1, , 02) 2, 08) 2, 14) 3, 20) 3, 26) 1, 32) 4, 38)3, 44)3, , 03) 3, 09) 1, 15) 1, 21) 1, 27) 1, 33) 3, 39) 2, 45) 4, , 04) 2, 10) 1, 16) 1, 22) 2, 28) 2, 34) 3, 40)3, , 05) 3, 11) 2, 17) 4, 23) 3, 29) 2, 35) 1, 41)3, , 06)3, 12) 1, 18) 3, 24) 2, 30) 4, 36) 3, 42)4, , LEVEL -II (C.W)-HINTS, a + b = 29; a − b = 5 ; a1 = a + 3, b1 = b + 3, , R ' = a1 + b1, 2, , 2, , R, , 2, , gT, gT, gT, gT, 2), 3), 4), 2, 2, 2, 2, The path of projectile is given by the equation, y = ax − bx 2 , where ‘a’ and ‘b’ are constants 2., and x and y are respectively horizontal and, vertical distances of projectile from the point, of projection. The maximum height attained, by the projectile and the angle of projection, 3., are respectively., ( 2014 - E ), , 1), 42., , 2a 2, , Tan −1 (a ), 1), b, 3), , a2, , Tan −1 (2b), b, , NARAYANAGROUP, , b2, −1, 2) , Tan (b), 2a, 4), , a2, , Tan −1 (a ), 4b, , 900, , Q sin 30 = P, , P, , R 2 = P 2 + Q2 + 2PQ cosθ when Q is doubled, tan 90 =, , 4., , 300, , Q, , 2Q sin θ, P + 2Q cos θ, , ur ur ur, ur ur ur, P+Q = R ; P+Q = S,, , θ2 = tan −1, , Q sin θ1, P + Q cos θ1, , θ1, ; θ =?, 2, 137
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , LEVEL - II (H.W), , MOTION IN A PLANE, 7., , The position of a particle is given by, , r, r = 3 t $i − 2 t 2 $j + 4 k$ m where t is in second, , and the co-efficients have proper units for r, to be in m. The magnitude and direction of, velocity of the particle at t = 2 s is, 1) 8.54 m s−1 , 20° with x - axis, , PARALLELOGRAM LAW, 1., , The greatest and least resultant of two, forces are 7 N and 3N respectively. If each, of the force is increased by 3 N and applied, at 600 .The magnitude of the resultant is, , 2) 10.54 m s −1 , 70° with x - axis, , 1) 7 N, 2) 3 N 3) 10 N, 4) 129N, TRIANGLE LAW, POLYGON LAW OF, 2., , 3., , 4., , 3) 8.54 m s−1 , 70° with x - axis, 4) 10.54 m s −1 , 20° with x - axis, A particle starts from origin at t = 0 with a, constant velocity 5 $i ms −1 and moves in xy, , VECTORS AND LAMI’S THEOREM, 8., In an equilateral triangle ABC, AL, BM and, CN are medians. Forces along BC and BA, represented by them will have a resultant, represented by, uuur, uuur, uuur, uuuur, 2) 2BM, 3) 2CN, 4) AC, 1) 2 AL, r r r, Given that A + B + C = 0 , out of three vectors two are equal in magnitude and the magnitude of third vector is 2 times that of either of two having equal magnitude. Then, 9., angle between vectors are given by, 1) 300 , 600 ,90 0, 2) 450 ,1350 ,150 0, , plane under action of a force which produces, , (, , y - coordinate of the particle at the instant its, x co-ordinate is 84 m in m is, 1) 6, 2) 36, 3) 18, 4) 9, , RELATIVE VELOCITY, When two bodies approach each other with, , 3) 900 ,1350 ,150 0, 4) 900 ,1350 ,1350, A and B are the two pegs separated by, 13 cm. A body of 169 kgwt is suspended by, thread of 17 cm connecting to A & B, such 10., that the two segments of strings are perpendicular. Then tensions in shorter and longer, parts of string are, 1) 100 kgwt, 69 kgwt 2) 65 kgwt, 156 kgwt, 3) 156 kgwt, 65 kgwt 4) 69 kgwt, 100 kgwt, , MOTION IN A PLANE, 5., , 6., , Two particles having position vectors, ur, r r, ur, r r, r1 = (3i + 5 j ) m and r2 = (−5i + 3 j )m are, ur, moving with velocities V1 = (4$i − 4 $j )ms −1, uur, and V2 = (ai$ − 3 $j )ms −1 . If they collide after, 2 seconds , the value of ‘ a’ is, 1) 2, 2) 4, 3) 6, 4) 8, A body is projected up such that its position, vector varies with time as, r, r = 3ti$ + ( 4t − 5t 2 ) $j m. Here t is in second., , {, , }, , The time when its y-coordinate is zero is, 1) 3 s, 2) 1 s, 3) 0.8 s, 4) 1.25 s, NARAYANAGROUP, , ), , −2, a constant acceleration of 3$i + 2 $j ms . The, , the different speeds, the distance between, them decreases by 120 m for every 1 min., The speeds of the bodies are, 1) 2 m/s and 0.5 m/s, 2) 3m/s & 2m/s, 3) 1.75m/s & 0.25 m/s 4) 2.5m/s&0.5m/s, An aeroplane is flying with the velocity of, V1 = 800kmph relative to the air towards, south. A wind with velocity of V2 = 15ms −1 is, blowing from west to east. What is the velocity of the aeroplane with respect to the earth., 1) 221.7ms −1 2) 150ms −1 3) 82ms −1 4) 40ms −1, , MOTION OF A BOAT IN A RIVER, 11., , 12., , A boat takes 4 hr upstream and 2 hr down, the stream for covering the same, distance.The ratio of velocity of boat to the, water in river is., 4) 3 :1, 1) 1:3, 2) 3:1 3) 1: 3, The width of a river is 2 3km. A boat is rowed, in direction perpendicular to the banks of, river. If the drift of the boat due to flow is 2, km, the displacement of the boat is., 1) 3 km 2) 6km 3) 5 km, 4) 4 km, , 139
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 13., , Person aiming to reach the exactly opposite, point on the bank of a stream is swimming, with a speed of 0.5ms −1 at an angle of 1200, with the direction of flow of water. The speed, 20., of water in the stream is, 1) 1ms −1 2) 0.25ms −1 3) 0.67ms −1 4) 3ms −1, , OBLIQUE PROJECTILE, 14., , A particle projected from the level ground just, clears in its ascent a wall 30 m high and 21., 120 3 away measured horizontally. The time, since projection to clear the wall is two, second. It will strike the ground in the same, horizontal plane from the wall on the other, side at a distance of (in metres), 1)150 3 2)180 3 3) 120 3 4) 210 3, , 15., , he can wet ( g = 10 m s −2 ) ( in metre2 ), 22., , A stone is projected with a velocity 20 2 m/, s at an angle of 450 to the horizontal. The average velocity of stone during its motion from, starting point to its maximum height is (g=10, m/s2), A ball is thrown with velocity 8 ms −1 making, an angle 60° with the horizontal. Its velocity 23., will be perpendicular to the direction of initial, velocity of projection after a time of, , 1.6, 4, s 2), s, 3) 0.6 s 4) 1.6 3 s, 3, 3, 24., The range of a projectile, when launched at, an angle of 150 with the horizontal is 1.5m., The additional horizontal distance the, projectile would cover when projected, with same velocity at 450 is, 1) 3 km 2) 4.5 km 3) 1.5 km 4) 2.5 km, A body is projected obliquely from the ground, such that its horizontal range is maximum.If the, change in its linear momentum, as it moves from, half the maximum height to maximum height, 25., is P, the change in its linear momentum as it, travels from the point of projection to the, landing point on the ground will be, 1), , 17., , 18., , 1) P, 19., , 140, , 2), , 2 P, , 3) 2 P, , 4) 2 2 P, , A projectile is thrown at angle β with vertical., It reaches a maximum height H. The time, , H, 2H, H, 2H, 2), 3), 4), g, g, 2g, g cos β, The maximum height attained by a projectile, is increased by 5%.Keeping the angle of projection constant, What is the percentage increase in horizontal range?, 1)5%, 2)10%, 3)15% 4)20%, A gardener wants to wet the garden without, moving from his place with a water jet whose, , 1), , velocity is 20 m s −1 the maximum area that, , 1)10 5 m/s 2)20 5 m/s 3)5 5 m/s 4)20m/s, 16., , taken to reach the highest point of its path is, , 1) 1600 π 2) 40 π 3) 40 0 π 4) 20 0 π, A particle is projected with speed u at angle, θ to the horizontal. find the radius of, curvature at highest point of its trajectory, u 2 cos2 θ, 1), 2g, , u 2 cos 2 θ, 3), g, , 2), , 3u 2 cos 2 θ, 2g, , 4), , 3u 2 cos 2 θ, g, , HORIZONTAL PROJECTILE, From the top of a tower of height 78.4 m two, stones are projected horizontally with 10 m/s, and 20 m/s in opposite directions. On reaching the ground, their separation is, 1) 120 m 2) 100 m 3) 200 m 4) 150 m, A body is projected vertically upwards. At its, highest point it explodes into two pieces of, masses in the ratio of 2:3 and the lighter piece, flies horizontally with a velocity of 6 m s−1 ., The time after which the lines joining the point, of explosion to the position of particles are, perpendicular to each other is, , 6, 12, 24, s 2), s 3), s 4) 2 s., 25, 15, 25, From the top of a building 80 m high, a ball is, thrown horizontally which hits the ground at, a distance. The line joining the top of the, building to the point where it hits the ground, makes an angle of 450 with the ground. Initial, velocity of projection of the ball is, (g =10 m/s2), 1) 10 m/s 2) 15 m/s 3) 20 m/s 4) 30 m/s, 1), , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 26., , 27., , 28., , MOTION IN A PLANE, , A stone is thrown from the top of a tower of, height 50 m with a velocity of 30 ms −1 at an, , 31., , angle of 300 above the horizontal . Find the, time during which the stone will be in air, 1) 2 sec 2) 3 sec, 3) 4 sec, 4) 5 sec, From the top of a tower 40 m high a ball is, projected upwards with a speed of 20 m s −1, at an angle 30° with the horizontal. The ratio, of the total time of flight to hit the ground to, the time taken by it to come back to the same, initial elevation is ( g = 10 m s −2 ), 1) 2:1, 2) 3:1, 3) 3:2 4) 4 :1, A body is thrown horizontally with a velocity, u from the top of a tower.The displacement, of the stone when the horizontal and vertical 1., velocities are equal is, , u2, u2, 2), 1), g, 2g, , u2 , 5, 3) 2g , , , An insect trapped in a circular groove of radius, 12 cm moves along the groove steadily and, complete 7 revolutions in 100 seconds.The, linear speed of the motion in cm/s, 1) 5.3, 2) 4, 3) 3, 4) 5, , LEVEL - II (H.W) - KEY, 01) 4, 07) 3, 13) 2, 19) 2, 25) 3, 31) 1, , 02) 2, 08) 2, 14) 2, 20) 1, 26) 4, , 03) 4, 09) 3, 15) 1, 21) 1, 27) 1, , 04) 2, 10) 1, 16) 1, 22) 3, 28) 3, , 05) 3, 11) 2, 17) 3, 23) 1, 29) 2, , 06) 3, 12) 4, 18) 4, 24) 3, 30) 3, , LEVEL - II (H.W) - HINTS, P+Q=7,P-Q=3 ; P1 =P+3, Q1 = Q + 3, , R = P1 + Q1 + 2 P1Q1 cos θ, 2, , 2u 2, 4), g, , 2, , 2., , A, , CIRCULAR MOTION, 29., , A ball is projected with 20 2 m/s at angle, , 30., , 45 with horizontal. The angular velocity of, the particle at highest point of its journey, about point of projection is, 1) 0.1 rad/s, 2) 1 rad/s, 3) 0.3 rad/s, 4) 0.4 rad/s, A particle is moving along a circular path in, xy − plane. When it crosses x − axis, it has, an acceleration along the path of 1.5m / s 2 ,, and is moving with a speed of 10 m/s in −ve, y − direction. The total acceleration is, y, , M, , N, , 0, , L, , B, →, , →, , C, , →, , from figure AB + BC = 2 BM, , θ2, C, , B, , 3., , θ1, , 2m, , A, , x, , →, , →, , →, , →, , →, , →, , →, , A+ B = − C , A = B , 2 A = C, →, , →, , anglebetween A and B is 90 0, , 1) 50iˆ − 1.5 ˆjm / s 2, 3) −50iˆ − 1.5 ˆjm / s 2, NARAYANAGROUP, , 10 m/s, 2) 10iˆ − 1.5 ˆjm / s 2, 4) 1.5iˆ − 50 ˆjm / s 2, , →, , →, , →, , →, , →, , B + C = A anglebetween B and C is 135 0, , 141
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 4., , MOTION IN A PLANE, , Two forces F1 and F2 are acting at a point, 8., whose resultant is F . If F2 is doubled F is, also doubled. If F2 is reversed then also F, is doubled. Then F1 : F2 : F is, 1), , 5., , 6., , 2), , 2: 2: 3, , 0, , T1, 0, , 150, O, , 60, , 20 N, , T2, , 1) 20N, 30N, , 0, , ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , 3) 3 : 2 : 3, 4) 2 : 3 : 2, Two vectors of equal magnitude P are, inclined at some angle such that the difference in magnitude of resultant and magnitude, of either of the vectors is 0.732 times either, of the magnitude of vectors. If the angle between them is increased by half of its initial, value then find the magnitude of difference 10., of the vectors, 1) 2P, 2) 2P, 3) 3P, 4) 3P, If ‘O’ is at equilibrium then the values of the, tension T1 and T2 are, (20 N is acting, vertically downwards at O)., 11., ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , 60, , 7., , 9., , 3: 3: 2, , 2) 20 3N,20N, , 3) 20 3N,20 3N, 4) 10N, 30N, A particle starts from the origin at t = 0 s with, a velocity of 10.0 ˆj m/s and moves in the xy − 12., plane with a constant acceleration of, , (8.0iˆ + 2.0 ˆj ) ms, , −2, , .What time is the x − coor--, , dinate of the particle 16m?, 1) t = 2s 2) t = 4s 3) t = 3s, NARAYANAGROUP, , 4) t = 1s, , Resultant of two vectors of magnitude P and, Q is of magnitude `Q’. If the magnitude of, ur, Q is doubled now the angle made by new reur, sultant with P is, 1) 300, 2) 900, 3) 600, 4) 1200, The two forces 2 2N and x N are acting at, a point their resultant is perpendicular to x̂N, and having magnitude of 6 N . The angle, between the two forces and, magnitude of x are., 1) θ = 1200 , X = 2 N 2) θ = 300 , X = 2 N, 3) θ = 1500 , X = 3 N 4) θ = 1500 , X = 2 N, The square of the resultant of two forces 4 N, and 3 N exceeds the square of the resultant, of the two forces by 12 when they are mutually perpendicular.The angle between the, vectors is., 2) 600, 3) 900, 4) 1200, 1) 300, An aeroplane flies along a straight line from, A to B with a speed v0 and back again with, the same speed v0 . A steady wind v is blowing., If AB = l then, 2v0l, a) total time for the trip is v 2 − v 2 if wind, 0, blows along the line AB, 2l, b) total time for the trip is v 2 − v 2 , if wind, 0, blows perpendicular to the line AB, c) total time for the trip decrease because of, the presence of wind, d) total time for the trip increase because of the, presence of wind, 1) a , b , d are correct, 2) a , b , c are correct, 3) only a , d are correct, 4) only b , d are correct, Two particles A and B move with constant, ur, uur, velocity v1 and v2 along two mutually, perpendicular straight lines towards, intersection point O as shown in figure. At, moment t = 0 particles were located at, distance l1 and l2 respectively from O. Then, 143
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, minimum distance between the particles and 15., time taken are respectively, , 1, A Vt, 1, S, , l1, 16., , 2 B, , O, , l2, , V 2t, 1), , 2), , l1v2 − l2 v1 l1v1 + l2 v2, ,, , v12 + v22, , l1v1 − l2v2 l1v2 + l2 v1, ,, , v +v, 2, 1, , 2, 2, , 13., , v +v, 2, 1, , 2, 2, , (, , v12 + v22, , l1v2 − l2 v1, 4), , 2, 2, , ), , (, , ), , The distance between two moving particles P 18., and Q at any time is a. If vr be their relative, velocity and if u and v be the components of, vr , along and perpendicular to PQ.The closest, distance between P and Q and time that, elapses before they arrive at their nearest, distance is, , av, 2) ( v + v, , r, , ), , , u, , a 1 + , vr , , 2, , 19., , 2, , av au, avr avr, , 2, 4) v , v 2, v u, r, r, 20., Two ships are 10 km apart on a line from south, to north. The one farther north is moving, towards west at 40 kmph and the other is, moving towards north at 40 kmph. Then, distance of their closest approach is, 10, km 4) 20 km, 1) 10 km 2) 10 2 km 3), 2, , 3), , 144, , 1) tan −1 ( 2 ) to west 2) tan −1 ( 2 ) to east, , l2 ( l1v1 + l2v2 ) l2, l1 , v12 + v22 l1, , a ( v + vr ) vr , 1), , a 1 + , u, v, , , 14., , 17., , v +v, 2, 1, , l1 ( l1v1 + l2 v2 ) l1, l2 , v12 + v22 l2, , l1v2 − l2v1, 3), , v12 + v22, , Two stones are projected from the top of a, tower in opposite direction, with the same, velocity V but at 300 & 600 with horizontal, respectively.The relative velocity of first, stone relative to second stone is, 2V, V, 3), 4), 1) 2v, 2) 2v, 3, 2, A motor boat going down stream comes over, a floating body at a point A. 60 minutes later, it turned back and after some time passed, the floating body at a distance of 12 km from, the point A. Find the velocity of the stream, assuming constant velocity for the motor boat, in still water., 1) 2 km/hr, 2) 3 km/hr, 3) 4 km/hr, 4) 6 km/hr, It is raining at a speed of 5 m / s −1 at an angle, 370 to vertical, towards east.A man is moving, to west with a velocity of 5 m / s −1 . The angle, with the vertical at which he has to hold the, umbrella to protect himself from rain is., −1 1 , −1 1 , 3) tan to south 4) tan to east, 2, 2, Rain, pouring down at an angle α with the, vertical has a speed of 10ms −1 . A girl runs, against the rain with a speed of 8ms −1 and, sees that the rain makes an angle β with the, vertical, then relation between α and β is, , 8 + 10sin β, 8 + 10sin α, 2) tan β =, 10 cos β, 10 cos α, 3) tan α = tan β, 4) tan α = cot β, A man can swim directly a stream of width, 100 m in 4 minutes, when there in no current, of water and in 5 minutes when there is, current of water. The velocity of the current, of water in the stream is, 1) 15ms −1 2) 5ms −1 3) 2.5ms −1 4) 0.25ms −1, The velocity of a boat in still water is 10 m/s. If, water flows in the river with a velocity of 6 m/s, what is the difference in times taken to cross, the river in the shortest path and the shortest, time. The width of the river is 80 m., 1) tan α =, , 1) 1s, , 2) 10s, , 3), , 3, s, 2, , 4) 2s, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 21., , 22., , MOTION IN A PLANE, , A boat takes 4 hrs to travel certain distance 26., in a river in down stream and it takes 6 hrs to, travel the same distance in upstream. Then, the time taken by the boat to travel, the same distance in still water is, , from rest with uniform acceleration of 3ms −1, for 0.5 minutes. If the maximum height, reached by it is 80m then the angle of, , 1) 4.8 hrs 2) 9.8 hrs, , projection is [ g = 10 ms −2 ], , 3) 24 hrs, , 4) 10 hrs., , A boats man finds that he can save 6 sec in, crossing a river by quicker path, then by, shortest path if the velocity of boat and river, be respectively 17 m/s and 8 m/s, then river, width is, 1) 675 m 2) 765 m 3) 567 m, , 23., , 4) 657 m, , −1 3, 2) tan ( ), 2, , −1 4 , 3) tan , 9, , −1 4 , 4) sin , 9, , A ball is thrown from a point with a speed V0 ,, , ing throws a ball with a speed of 10 m s −1 at an, , at an angle of projection θ . From the same, point and at the same instance a person starts, , 1) 5.2 m, , 2) 4.33 m, , running with a constant speed, , ^, , ^, , ^, , ^, , ity at that time is:, , 1., , ^, , 28., , ^, , (13i − 29 j − 31k ), 146, ^, , ^, , ^, , 2., , ^, , ^, , ^, , ^, , ^, , ^, , (13i− 29 j− 31k ), 146, , (13i + 29 j − 31k ), (13i + 29 j+ 31k ), 3., 4., 146, 146, A projectile is given an initial velocity, ^, , 1) yes, 60°, , 2) yes, 30°, , 3) No, , 4) yes, 45°, −1, , an initial speed of 20 m s at an angle of 45°, with the horizontal. At the moment the ball is, thrown, the player is 50 m from coach. The, speed and the direction that the player has, to run to catch the ball at the same height at, which it was released in m s −1 is, 1), , 5, away from coach, 2, , 2), , 5, towards the coach, 2, , 3), , 2, towards the coach, 5, , 4), , 2, away from the coach, 5, , ^, , (g = 10ms ), 1) y = 2x − 5x 2, 3) y = 9x − 5x 2, , NARAYANAGROUP, , to catch, , The coach throws a baseball to a player with, , i + 2 j .The cartesian equation of its path is, −2, , 2, , ^, , 3 i − 4 j + 5 k is i + 9 j − 8 k .Its angular veloc-, , ^, , V0, , the ball will the person be able to catch the, ball ? If yes, what should be the angle of, projection, , 3) 2.66 m 4) 8.66 m, , At a given instant of time the position vector, of a particle moving in a circle with a velocity, ^, , 25., , 1) tan −1 (3), , A boy playing on the roof of a 10 m high build- 27., angle of 300 with the horizontal. How far from, the throwing point will the ball be at a height of, 10 m from the ground?, , 24., , A projectile has initially the same horizontal, velocity as it would acquire if it had moved, , (JEE MAIN - 2013), 2) 9y = 12x − 5x 2, 4) 5y = x − 9x 2, , 145
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 29., , If a stone is to hit at a point which is at a, distance d away and at a height h (see fig), above the point from where the stone starts,, then what is the value of initial speed u if the, stone is launched at an angle θ ?, , u2 h, 1) d =, 2, 2, , 32., , 2u2h, 2) d =, g, 2, , 3) d = h, 4) gd2 = u2h, A ball is projected from the top of a tower, with a velocity 3i$ + 4 $j + 5 k$ m s −1 , where, , $i, $j & k$ are unit vectors along east, north, and vertical upwards respectively. If the, height of the tower is 30 m, its range is, , (g = 10ms ), −2, , h, θ, , 33., , with a velocity, , d, d, 1) sin θ, , g, d, 2), 2 ( d tan θ − h ) cosθ, , 30., , −2, , ( g = 10m / s ), 2, , 35., , t = 6s, , t = 2s, h, θ, 120 m, , 31., , 146, , m s −1 , where, , (g = 10ms ), , If a projectile crosses two walls of equal 34., height h symmetrically as shown in the fig., Choose the correct statement (s), , u, , 3i$ + 4 $j + 5 k$, , $i, $j & k$ are unit vectors along east, north and, vertical upwards respectively. If the height, of the tower is 30 m, its time of flight is, , g, 2(d tan θ − h), , gd 2, 4), ( d − h), , gd 2, 3), h cos 2 θ, , 1) 12 m 2) 9 m, 3) 15 m, 4) 25 m, A ball is projected from the top of a tower, , 36., 1) The time of flight is 8 sec, 2) The height of each wall is 60 m, 3) The maximum height of projectile is 80m, 4) All the above, A particle is dropped from a height h. Another, particle which is initially at a horizontal 37., distance ‘d’ from the first is simultaneously, projected with a horizontal velocity u and the, two particles just collide on the ground. Then, , 1) 5, 2) 3, 3) 0.3, 4) 0.5, A cricketer of height 2.5 m throws a ball at an, angle of 300 with the horizontal such that it is, received by another cricketer of same height, standing at distance of 50 m from the first one., Find the maximum height attained by the ball., 1) 10 m 2) 9 m, 3) 10.7 m 4) 9.7 m, A particle when fired at an angle, θ = 600 along the direction of the breadth of, a rectangular building of dimension, 9m × 8 m × 4 m so as to sweep the edges. Find, the range of the projectile., 8, 4, 1) 8 3, 2) 4 3 3), 4), 3, 3, A hiker stands on the edge of a cliff 490 m above, the ground and throws a stone horiozontally with, an initialspeed of15m s-1 neglecting air resistance., The time taken by the stone to reach the ground in, seconds is (g=9.8ms-2 ), 1) 5, 2) 10, 3) 1, 4) 15, A hiker stands on the edge of a cliff 490 m, above the ground and throws a stone, horizontally with an initial speed of 15ms-1, Neglecting air resistance, the speed with, which it hits the ground in ms-1 is (g=9.8ms-2 ), 1) 9.8, 2) 99, 3) 4.9, 4) 49, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 38., , The direction of projectile at certain instant, is inclined at angle α to the horizontal after t, sec.If it is inclined at an angle β then the, horizontal component of velocity is, 1), , 39., , MOTION IN A PLANE, , g, tan α − tan β, , 2), , the two fragments, when their displacement, vectors is inclined at 600 relative to each, other is ( g = 10ms −2 ), , gt, tan α − tan β, , t, gt, 4), 3), g(tan α − tan β ), (tan α + tan β ), Two bodies are projected from the same point, with same speed in the directions making an, , 1) 40 3 2) 80 3 3) 120 3 4) 480 3, 43., , An object in projected up the inclined at the, angle shown in the figure with an initial, velocity of 30ms −1 . The distance x up the, incline at which the object lands is, , angle α1 and α 2 with horizontal and strike, at the same point in the horizontal plane, , A, 30 ms-1, , through a point of projection. If t1 and t2, are their time of flights. Then, tan (α1 − α 2 ), , 2) sin α − α, ( 1 2), , sin (α1 − α 2 ), , 300, 1) 600 m 2) 104m 3) 60 m, 44., , sin 2 (α1 − α 2 ), 4) sin 2 α + α, ( 1 2), , 3) sin α + α, ( 1 2), , At a certain height a body at rest explodes, into two equal fragments with one fragment, receiving a horizontal velocity of 10 ms −1 The, time interval after the explosion for which the, velocity vectors of the two fragments become, , 41., , 4) 208 m, , A projectile fired with velocity u at right, angle to the slope which is inclined at an angle, θ with horizontal. The expression for R is, , 2u 2, tan θ, 1., g, , 2u 2, sec θ, 2., g, , u2, tan 2 θ, g, , perpendicular to each other is ( g = 10 ms −2 ), 1) 1s, 2) 2s 3) 1.5s, 4)1.7s, 45., At a certain height a shell at rest explodes into, two equal fragments one of the fragments, receives a horizontal velocity u.The time interval, after which the velocity vectors will be inclined, at 1200 to each other is, , In figure shown below, the time taken by the, projectile to reach from A to B is t then, the, distance AB is equal to, , u, 3g, , 2), , 3u, g, , 2u, , 3), , 3g, , B, , u, , 4), , 0, , 2 3g, , A bomb at rest at the summit of a cliff breaks, into two equal fragments.One of the, fragments attains a horizontal velocity of, 20 3ms −1 .The horizontal distance between, , NARAYANAGROUP, , 4., , 2u 2, tan θ sec θ, g, , 3., , 1), 42., , X, , 300, , 2, 2, 2, 2, , sin (α1 + α 2 ), , 1) tan α + α, ( 1 2), , 40., , t −t, t +t, 2, 1, 2, 1, , 60, , 0, , 30, , A, , 1), , C, , ut, 3, , 2), , 3ut, 2, , 3), , 3ut, , 4) 2ut, , 147
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, 46., , A particle moves on a circle of radius r with 2., centripetal acceleration as function of time, as a c = k 2 rt 2 where k is a positive constant., Find the resultant acceleration., 2) kr, 1) kt 2, , 47., , 3) kr k 2 t 4 + 1, 4) kr k 2 t 2 − 1, A particle moves in a circular path such that, , B, (0,b), , its speed v varies with distance as v = α s, where α is a positive constant. Find the 3., acceleration of particle after traversing a, distance S?, 2, 1) α, , 3) α, 48., , 1., , 148, , 1 S2, −, 4 R2, 1 S2, +, 2 R2, , 2, 2) α, , 1 S2, +, 4 R2, , 2, 4) α, , 1 S2, +, 2 R2, , A particle moves in a circle of radius 20 cm., Its linear speed is given by v = 2t where t is, in s and v in m/s. Then, a) the radial acceleration at t = 2s is 80ms–2, b) tangential acceleration at t = 2s is 2 ms–2, c) net acceleration at t = 2 s is greater than, 80 ms–2, d) tangential acceleration remains constant in, magnitude., 1) only a,b,c are correct 2) only a,b,d are correct, 3) only a,c,d are correct 4) all a,b,c,d are correct, LEVEL - III - KEY, 01) 2 02) 3 03) 4 04) 4 05) 2 06) 2, 07) 1 08) 2 09) 1 10) 2 11) 1 12) 1, 13) 4 14) 3 15) 2 16)4 17) 1 18) 2, 19)4 20) 4 21) 1 22) 2 23) 4 24) 2, 25) 1 26) 3 27) 4 28) 2 29) 2 30) 4, 31) 2 32) 3 33) 2 34) 4 35) 1 36) 2, 37) 2 38) 2 39) 3 40) 1 41) 1 42) 4, 43) 3 44) 1 45) 1 46) 3 47) 2 48) 4, LEVEL - III - HINTS, uuur uuur uuur, uuur uuur uuur, AC + CB = AB ;, AC + 4CR = AB, uuur, uuur uuur uuur uuur uuur uuur, AC + 4 AR − AC = AB ; 4 AR − 3 AC = AB, uuur uuur, uuur, ∴ AB + 3 AC = 4 AR, , (, , ), , r, p = 2 cos tiˆ + 2 sin tjˆ, r dpr, F=, = −2 sin tiˆ + 2 cos tjˆ, dt, r, r r, r, p.F = 0 ⇒ p ⊥ to F, , c c , ,, , , 2 2, C, 450, , A, (a, 0 ), , 0, , c, 2 = b−0, comparing the slopes, c, 0−a, 0−, 2, 2, 2, 2, F = F1 + F2 + 2 F1 F2 cos θ ...(1), b−, , 4., , 4F 2 = F12 + 4F22 + 4F1F2 cosθ ...(2), 4 F 2 = F12 + F22 − 2 F1 F2 cos θ ...(3), , 5., , 6., , 7., 8., , R − P = 0.732 P ;, R = 1.732 P = 3P, θ , θ, 1, = θ +, But 2 P.cos = 3P ; θ, 2, 2, 0, 0, T1 sin 30 = T2 cos 30 ; T1 = 3T2 ...(1), T1 cos 300 = 20 + T2 ...(2) ; solving (1) and (2), r, r, 1r, r, r, r, r ( t ) = v0t + at 2 ; r ( t ) = 4t 2 i + (10t + t 2 ) j, 2, 2, 4t = 16, Q 2 = P 2 + Q 2 + 2 PQ cos θ ;, P, 2Q sin θ, cosθ = −, ;Tan α =, 2Q, P + 2Q cos θ, , 9., , x, , 6, 2 2, , θ, , 900, , x, , NARAYANAGROUP
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, A–B=C, →, , →, , →, , b) A + B = C and, , f), , 2π, 3, , A–B=C, →, , →, , c) A − B = C and, A2 + B2 = C2, →, , →, , →, , d) A + B = C and, A= B =C, 2., , 5., , →, , g) 0, h) π /2, , The path of projectile is represented by y=PxQx2., List-I, List-II, (A) Range, (P) P/Q, 6., (B) Maximum height (Q) P, 2, (R)P /4Q, (C) Time of flight, (D) Tangent of angle (S), , 3., , 4., , 2 2, P, Qg, , of projection is, Angle between velocity and acceleration, vectors in the following cases., List - I, List - II, a) Vertically projected, e) 900, body, b) For freely dropped, f) changes from, 7., body, point to point, c) For projectile, g) zero, d) In uniform circular, h) 1800, motion, Acceleration a in case of circular motion is, , c) If ar ≠ 0, at = 0 r) motion is accelerated, translatory, d) If ar ≠ 0, at ≠ 0 s) no conclusion can be, drawn regarding motion, A projectile is launched at angle θ to, horizontal from A and it hits the target B on, level ground., Column-I, Column-II, a) Magnitude of radial, p) Increases, acceleration, b) Magnitude of tangential q) Decreases, acceleration, c) Power delivered by gravity r) First increases,, then decreases, d)Torque of gravity about B s)First decreases,, then increases, Considering a projectile motion.., Column-I, Column-II, a) Change in magnitude p) At highest point of, of momentum, the parabolic path, b) Maximum angular q)2mv sin θ between, momentum about, the point of throw, the point of throw, and target, 2mv3 sin2 θ cosθ, g, d)Magnitude of change s)Present along vertical, in momentum, direction, The correct match is, Motion is defined as rate of change of, position., Column-I, Column-II, Position r changes, a) Magnitude only, p) A coin dropped from, roof of house, b) Direction only, q) A coin thrown at any, , c) Minimum velocity, , angle with horizontal, , given by a = a r 2 + at 2 where ar , is radial, component of acceleration and a t the, tangential component of acceleration., , c) Both in magnitude, , r) A coin held in your, , and direction, , hand, , d) Remains invariant, , s) A coin is rotated in, circular path with, variable speed, , a t governs the magnitude of velocity v and, ar its direction of motion., Column-I, Column-II, a) If ar = 0, at = 0 p) motion is non-uniform, circular, b) If ar = 0, at ≠ 0 q) motion is uniform, circular, 152, , r) L =, , 8., , In uniform circular motion a body moves with, constant speed v on a circular path of constant, radius r. In this motion, Column-I, , Column-II, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , a) The acceleration of p) tangent to path at, body is, , every point, , b) The kinetic energy q) along axis of rotation, , 15., , angle with horizontal, c) The angular displac- r) constant in magnitude, ement of body at any, , but changing in direction, , 16., , instant is directed along, d) The velocity of body s) constant always, always directed along, , ASSERTION AND REASON, , 9., , 10., , 11., , 12., , Options :, 1) If both Assertion and Reason are true and, the Reason is correct explanation of the, Assertion., 2) If both Assertion and Reason are true, but, Reason is not correct explanation of the, Assertion., 3) If Assertion is true, but the Reason is false., 4)Both Assertion and Reason are false., Assertion(A): Two forces 7N and 5N are acting, at a point and their resultant can be 3N., Reason(R): If two vectors P and Q are acting at, a point, then magnitude of their resultant can have, a value between P–Q to P + Q., Assertion(A): Electric current and velocity of light, both have direction as well as magnitude but still, are not considered as vectors., Reason(R): Electric current and velocity of light, do not follow laws of vector addition., Assertion(A):A vector is not changed if it is slide, parallel to itself., Reason(R):Two parallel vectors of same, magnitude are said to be equal vectors., , 17., , 18., , 19., , 20., , F=, , 21., , Assertion(A):Angle between iˆ + ˆj and iˆ is 450, Reason(R): iˆ + ˆj is equally inclined to both iˆ and, ĵ and the angle between iˆ and ĵ is 900 ., , 13., , 14., , 22., , Assertion(A): Finite angular displacement is not, a vector quantity., 23., Reason(R): It does not obey the laws of vector, addition., Assertion(A): The minimum number of non, coplanar vectors whose resultant can be zero is four., , NARAYANAGROUP, , Reason(R): The resultant of two vectors of, unequal magnitude can be zero., Assertion(A): Two bodies thrown with same, speed from the same point at the same instants, but at different angles never collide in air., Reason(R): x and y co-ordinates of the two, projectiles always differ, Assertion(A): In projectile motion, the angle, between the instantaneous velocity and, acceleration at the highest point is 1800., Reason(R): At the highest point, velocity of, projectile will be vertically upward., Assertion(A): If a bomb is dropped from an, aeroplane moving horizontally with constant, velocity then the bomb appears to move along a, vertical straight line for the pilot of the plane., Reason(R): Horizontal component of velocity of, the bomb remains constant and same as the, velocity of the plane during the motion under, gravity., Assertion (A) :- Two particles of different mass,, projected with same velocity at same angles. The, maximum height attained by both the particle will, be same., Reason(R) :- The maximum height of projectile, is independent of particle mass., Assertion(A) :- The maximum horizontal range, of projectile is proportional to square of velocity., Reason(R):-The maximum horizontal range of, projectile is equal to maximum height attained by, projectile., Assertion(A) :- A body of mass 1kg is making 1, rps in a circle of radius 1m. Centrifugal force acting, on it is 4π 2 N, Reason(R) :- Centrifugal force is given by, mυ 2, r, , [AIIMS 2008], Assertion(A) :- A coin is placed on phonogram, turn table. The motor is started, coin moves, along the moving table., Reason (R) :- The rotating table is providing, necessary centripetal force to the coin., Assertion(A) :- The trajectory of projectile is, quadratic in y and linear in x., Reason (R) :- y component of trajectory is, dependent of x-component., Assertion(A) :- Two similar train are moving, along the equatorial line with the same speed but, in opposite direction. They will exert different, pressure on the rails., 153
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JEEJEE, MAINS, - C.W, - VOL, JEE-ADV, PHYSICS-VOL, --III, MAINS, - VOL, - VI, , MOTION IN A PLANE, , 24., , 25., , 26., , 27., , 28., , 29., , 30., , 31., , 32., , 154, , Reason (R) :- In uniform circular motion the, magnitude of velocity remains constant but the, direction continuously changes., Assertion(A) :- When a body is dropped or, thrown horizontally from the same height, it, would reach the ground at the same time., Reason (R) :- Horizontal velocity has no, effect on the vertical direction., Assertion(A) :- When the velocity of projection of a body is made n times, its time of flight, becomes n times., Reason (R) :- Range of projectile does not, depend on the initial velocity of a body., Assertion (A):- Improper banking of roads, causes wear and tear of tyres., Reason (R) :- The necessary centripetal, force is provided by the force of friction between the tyres and the road., Assertion(A) :- When range of a projectile is, maximum, its angle of projection may be 450 or, 1350., Reason (R) :- Whether θ is 450 or 1350,, value of range remains the same, only the sign, changes., Assertion(A) :- In order to hit a target, a man, should not point his rifle in the same direction as, target., Reason (R) :- The horizontal range of the, bullet is dependent on the angle of projectile, with horizontal direction., Assertion(A) :- When an automobile while, going too fast around a curve overturns, its inner, wheels leave the ground first., Reason (R) :- For a safe turn the velocity of, automobile should be less than the value of safe, limit velocity., Assertion(A) :- During a turn, the value of, centripetal force should be less than the limiting, frictional force., Reason(R) :- The centripetal force is, provided by the frictional force between the, tyres and the road., Assertion (A):- When a vehicle takes a turn on, the road, it travels along a nearly circular path., Reason (R) :- In circular motion, velocity of, vehicle remains same., Assertion(A) :- As the frictional force increases, the safe velocity limit for taking a turn, on an unbanked road also increases., Reason (R) :- Banking of roads will increase, , 33., , 34., , the value of limiting velocity., Assertion(A) :- If both the speed of a body, and radius of its circular path are doubled, then, centripetal force also gets doubled., Reason (R) :- Centripetal force is directly, proportional to both speed of a body and radius, of circular path., Assertion(A) :- In circular motion, the centripetal and centrifugal force acting in opposite, direction balance each other., Reason (R) :- Centripetal and centrifugal, forces don’t act at the same time., , LEVEL-IV KEY, Matching Type Questions, 1) a → g; b → e; c → h ; d → f, 2) A → P, B → R, C → S, D → Q, 3) a → h; b → g ; c → f ; d → e, 4) a → s; b → r ; c → q; d → p, 5) a → r ; b → s; c → p; d → q, 6) a → s; b → r; c → p; d → q, 7) a → p; b → s; c → q; d → r, 8) a → r ; b → s; c → q; d → p, , ASSERTION AND REASON, 09) 1, 15) 1, 21) 4, 27) 1, 33) 3, , 10) 1, 16) 4, 22) 4, 28) 2, 34) 4, , 11) 2, 17) 1, 23) 2, 29) 2, , 12) 1, 18) 1, 24) 1, 30) 1, , 13) 1, 19) 3, 25) 3, 31) 3, , 14) 3, 20) 1, 26) 1, 32) 2, , LEVEL - IV - HINTS, ASSERTION AND REASON, 11., 16., , Magnitude and direction of vector does not change, when it is displaced parallel to itself., At maximum height of projectile velocity is, horizontal and acceleration is vertically, downwards., , 18., , H=, , u 2 sin 2 θ, 2g, , i.e. it is independent of mass of, , projectile., 19., , R=, , u 2 sin 2θ, u2, ∴ Rmax = whenθ = 450, g, g, , ∴ Rmaxα u 2, H=, , u 2 sin 2 θ, u2, ⇒ H max =, when θ = 900, 2g, 2g, , It is clear that H max =, , Rmax, 2, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , mυ 2 m ( rω ), From relation F =, =, = mrω 2, r, r, 2, , 20., , = mr ( 2π v ) = 4π 2 mrv 2, Here, m = 1kg, v = 1rps, r = 1m, ∴ F = 4π 2 × 1× 1× 12 = 4π 2 N, Within a certain speed of the turn table the, frictional force between the coin and the turn, table supplies the necessary centripetal force, required for circular motion. On further increase, of speed, the frictional force cannot supply the, necessary centripetal force. Therefore the coin, flies off tangentially., gx 2, y = x tan θ − 2, 2u cos 2 θ, Due to earth’s axial rotation, the speed of the, trains relative to earth will be different and hence, the centripetal forces on them will be different., mυ 2, Thus their effective weights mg −, and, r, mυ 2, mg +, will be different. So they exert, r, different pressure on the rails., Both bodies will take same time to reach the, earth because vertical downward component of, , Rinner =, , 2, , 21., , 22., 23., , 24., , 2h, . Horizontal velocity has, g, no effect on the vertical direction., T α u and Rα u 2, When velocity of projection of a body is made n, times, then its time of flight becomes n times and, range becomes n2 times., When roads are not properly banked, force of, friction between tyres and road provides, partially the necessary centripetal force. This, cause wear and tear of tyres., u 2 sin 2θ, R, =, Range,, g, , 26., , 27., , u2, u2, 0, R, =, sin, 90, =, when θ = 45 , max, g, g, 2, u, −u 2, 0, 0, R, =, sin, 270, =, when θ = 135 , max, g, g, Negative sign shows opposite direction., The man should point his rifle at a point higher, than the target since the bullet suffers a vertically, 0, , 28., , , , 29., , 1, , , , downward deflection y = 2 gt due to gravity.., , , When automobile moves in circular path then, reaction on inner wheel and outer wheel will be, different., , NARAYANAGROUP, , 2, , υ 2h , M, υ 2h , g, −, R, =, g, +, outer, , ra and, ra , 2 , , , gra, h, If υ is equal or more than the critical value then, reaction on inner wheel becomes zero. So it, leaves the ground first., The body is able to move in a circular path due, to centripetal force. The centripetal force in case, of vehicle is provided by frictional force. Thus if, the value of firctional force µ mg is less than, centripetal force, then it is not possible for a, vehicle to take a turn and the body would, overturn., thus condition for safe turning of vehicle is,, mυ 2, µ mg ≥, r, In circular motion the frictional force acting, towards the centre of the horizontal circular path, provides the centripetal force and avoid overturning of vehicle. Due to the change in direction, of motion, velocity changes in circular motion., On an unbanked road, friction provides the, necessary centripetal force, mυ 2, = µ mg ∴υ = µ rg ., r, Thus with increase in friction, safe velocity limit, also increases., When the road is banked with angle of θ then, its limiting velocity is given by, In critical condition υ safe =, , 30., , 31., , 32., , velocity for t =, 25., , M, 2, , υ=, , 33., , 34., , rg ( tan θ + µ ), , 1 − µ tan θ, Thus limiting velocity increase with banking of, road., Centripetal force is defined from formula, mυ 2, υ2, F=, ⇒ Fα, r, r, If υ and r both are doubled then F also gets, doubled., While moving along a circle, the body has a constant tendency to regain its natural straight line path., This tendency gives rise to a force called centrifugal force. The centrifugal force does not act on, the body in motion, the only force acting on the, body in motion is centripetal force. The centrifugal force acts on the source of centripetal force to, displace it radially outward from centre of the path., , 155
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, 4., , LEVEL - V, , A boat moves relative to water with a velocity, which is 1/2 times the river flow velocity. At, what angle to the stream direction must the, boat move to minimize drifting., a) 450, b) 600, c) 1200, d) 900, Two bodies were thrown simultaneously from, the same point one straight up and the other, at an angle of θ =600 to the horizontal. The, initial velocity of each body is equal to, v 0 = 25m / s. Neglecting the air drag, find the, distance between the bodies t =1.70 s later., a) 20m, b) 18m, c) 22m, d) 24m, A sailor in a boat, which is going due east with, a speed of 8m/s observes that a submarine is, heading towards north at a speed of 12 m/s, and sinking at a rate of 2 m/s. The, commander of submarine observes a, helicopter ascending at a rate of 5 m/s and, heading towards west with 4 m/s. Find the, speed of the helicopter with respect to boat., a) 10m/s, b) 11m/s c) 12m/s d) 13m/s, Consider a collection of a large number of, particles each with speed v in a plane., The direction of velocity is randomly, distributed in the collection. The magnitude, of the average relative velocity of a particle, with velocities of all other particles is, a) > v, b) < v, c) = v d) none of these, A man in a row boat must get from point A to, point B on the opposite bank of the river (see, figure). The distance BC=a. The width of the, river AC=b. At what minimum speed u relative, to the still water should the boat travel to reach, the point B? The velocity of flow of the river is, v0., , SINGLE CORRECT ANSWERS QUESTIONS, A) RELATIVE MOTION, 1., , An aeroplane A is flying horizontally due east 5., at a speed of 400 km/hr. Passengers in A,, observe another aeroplane B moving, perpendicular to direction of motion at A., Aeroplane B is actually moving in a direction, 300 north of east in the same horizontal plane, as shown in the figure. Determine the, velocity of B, 6., N, B, E, , 30°, A, , 400 ˆ, j, B) 4 0 0 iˆ +, 7., 3, 400 $, i + 400 $j, D), C) 400 $i + 400 $j, 3, A boat moves relative to water with a velocity, which is n times less than the river flow, velocity. At what angle to the stream, direction must the boat move to minimize, 8., drifting? (u is velocity of water, v is velocity, of boat), A) 400 3 $i + 400 3 $j, , 2., , v, u, , −1, a) θ = sin from normal direction, v, u , , −1, b) θ = cos from normal direction, v, u, , −1, c) θ = tan from normal direction, , C, , u, v, , c) ( u − v cos θ ), 156, , b, , A man wishes to cross a river flowing with, velocity u swims at an angle θ with the river, flow. If the man swims with speed v and if the, width of the river is d, then the drift travelled 9., by him is, a) ( u + v cosθ ), , d, d, b) ( u − v cos θ ), v sin θ, v sin θ, d, v cos θ, , d) ( u + v cos θ ), , B, a), , −1, d) θ = sin from normal direction, , 3., , a, , d, v cos θ, , v0 b, a +b, 2, , 2, , b), , v 0a, a 2 + b2, , V0, c), , A, , v0 b, 2a, , d), , v 0a, 2b, , A man standing on a road has to hold his, umbrella at 530 with vertical to keep the rain, away. He throws the umbrella and starts, running at 12 km/h. He finds that rain drops, are falling on his head vertically. Find the speed, ( in km/hr) of raindrops w.r.t the moving man, a) 12km/hr, b) 14 km/hr, c) 16km/hr, d) 18 km/hr
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , B) BODY PROJECTED FROM THE GROUND, 10. A motor boat has a speed of 5 m/s. At time t 13. From a point on the ground at a distance a from, =0, its position vector relative to a origin is, the foot of a pole, a ball is thrown at an angle, of 450, which just touches the top of the pole, −11iˆ + 16ˆj m, having the aim of getting as, and strikes the ground at a distance of b, on, close as possible to a steamer. At time t =0,, the other side of it.Find the height of the pole., P (a,h), the steamer is at the point 4iˆ + 36jˆ m and is, , (, , ), , (, , (, , ), , ), , moving with constant velocity 10iˆ − 5jˆ m/s., Find the direction in which the motorboat, must steer, b) 3$i + 4 $j, a) 3$i + 3 $j, , Slope, , Slope, , Slope, , Slope, , h, α = 45°, x, O, a, b, ab, ab, 2ab, ab, a), b), c), d), a −b, a +b, a +b, a + 2b, c) 4$i + 3 $j, d) 4$i + 4 $j, 14. A heavy particle is projected with a velocity at, 11. A river 400 m wide is flowing at a rate 2.0 m/, an angle with the horizontal into the uniform, s. A boat is sailing at a velocity of 10.0 m/s, gravitional field. The slope of the trajectory of, repect to the water, in a direction, the particle varies as, perpendicular to the river., a), b), (a) Find the time taken by the boat ot reach, the opposite bank., (b) How far from the point directly opposite, O, O, t, t, to the starting point does the boat reach the, opposite bank?, a) 40 sec, 80 m, b) 30 sec, 40 m, c) 20 sec, 20 m, d) 35 sec, 80 m, c), d), 12. A block of mass m is floating in a river, flowing with a velocity of 2m/s. A boat is, O, O, moving behind the block with a velocity of 5, t, t, m/s with respect to the block as shown. From, the boat a stone is thrown with a velocity, 15. A fixed mortar fires a bomb at an angle of 530, v = v1 i$ − v2 $j + v3 k$ with respect to the river, above the horizontal with a muzzle velocity of, such that it hits the block. If v1 : v2 : v3 =, 80m/s–1. A tank is advancing directly towards, the mortar on level ground at a constant speed, 2 3 : 2 : 3 then the velocity of the stone with, of 5m/s. The initial separation ( at the instant, respect to the ground is(g=10m/s2 ), mortar is fired) between the mortar and tank,, y, so that the tank would hit is [Take g=10ms–2], Boat, 2m/sec, a) 662.4 m b) 526.3 m c) 486.6 m d) 678.4 m, 16. The angular elevation of an enemy’s position, x, on a hill ‘h’ ft high is α . What should be the, 10m, minimum velocity of the projectile in order to, hit the enemy?, 30°, Block, a) u = gh (cos α + 1) b) u = gh (sin α + 1), 10 $, 10 $, $, $, $, $, j + 5 3 k b) 12i −, j + 5 3k, a) 10i −, c) u = gh(cosecα + 1) d) u = gh (sec α + 1), 3, 3, 17. Two particles are projected simultaneously, 10 $, $, $, with the same speed v in the same vertical, j + 5 3k, c) 10$i − 10 $j + 5 3 k$ d) 10 3 i −, 3, plane with angles of elevation θ , and 2θ ,, where θ <450. At what time will their velocities, be parallel?, 157
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , v, θ, 3θ, v, θ, 3θ, a) t = tan cos ec b) t = cos cot, g, 2, 2, g, 2, 2, , v1, , P1, , v2, , a) 250 m, , 60°, A, , P2, , b) 500 m, , v, θ, 3θ, v, θ, 3θ, c) 750 m, B, cos tan, d) t = cos cos ec, 45°, 30°, 2, 2, g, 2, 2, g, d) 1000 m, O, 18. Figure shows a sphere moving in a steady flow, of air in the x-direction on a horizontal plane . 22. A particle is projected with a certain velocity, at an angle a above the horizontal from foot, The air stream exerts an essentially constant, 2, of an inclined plane of inclination 30o . If the, acceleration 1.8 m/sec on the sphere in the xdirection. If at t = 0 the sphere is moving as, particle strikes the plane normally then a is, shown in figure, determine the time t required, equal to, for the sphere to cross the y-axis again., 3, 1 , 3 0 o + tan − 1 , Air flow, 30o + tan −1 , , (b), (a), , y, a) 1/3 sec, 2 3, 2 , c) t =, , b) 2/3 sec, , 3 m/s, 30°, , c) 4/3 sec, x, 500 g, , d) 5/3 sec, , 19. A very broad elevator is going up vertically, with a constant acceleration of 2m / s 2 . At the, instant when the velocity is 4 m / s a ball is, projected from the floor of the lift with a speed, of 4 m / s relative to the floor at an elevation, of 30o . The time taken by the ball to return, the floor is (g = 10m / s 2 ), , 90°, , (, , ), , o, −1, (c) 60o, (d) 30 + tan 2 3, 23. A particle is projected at an angle 600 with, speed 10 3 m/s from the point A, as shown in, the figure. At the same time the wedge is made, to move with speed 10 3 m/s towards right as, shown in the figure. Then the time after which, particle will strike with wedge is, , 10 3m / s, 10 3m / s, 30°, , 60°, , A, 1, 1, 1, s, (b) s, (c) s, (d) 1s, 2, 3, 4, a) 2s, b) 2 3s, 20. A boy throws a ball with velocity v0 = 20m / s, 4, The wind impart horizontal acceleration of 4m/, s, c), d) None of these, 3, s2 to the left. The angle θ (with vertical )at, which the ball must be thrown so that the ball D) COLLISIONS BETWEEN PROJECTILES, returns to the boy’s hand is (g=10m/s2) :, 24. A particle is projected from a point A, a) tan −1 (1.2) b) tan −1 (0.2) c) tan −1 (2) d) tan −1 (0.4), withvelocity u 2 at an angle of 45o with, C) PROJECTILE MOTION ON INCLINED, horizontal as shown in figure. It strikes the, PLANE, plane BC at right angles. The velocity of the, 21. A particle is projected from an inclined plane, particle at the time of collision is, OP1 from A with velocity v1 =8ms–1 at an angle, C, u 2, 600 with horizontal. An another particle is, projected at the same instant from B with, velocity v2 =16ms–1 and perpendicular to the, 60°, 45°, plane OP2 as shown in the figure. After time, 2u, u, 3u, 10 3 sec there separation was minimim and, (a), (b), (c), (d) u, found to be 70m. Then find distance AB., 3, 2, 2, (a), , 158
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 25. A particle is dropped from point P at time t, = 0. At the same time another particle is, thrown from point O as shown in the figure and, it collides with the particle P. Acceleration due, to gravity is along the negative y-axis. If the, two particles collide 2s after they start, find, the initial v0 of the particle which was projected, from O. Point O is not necessarily on ground., y, , P, , 2m, , 10 m, , 10 m/s, , B, , 10 2m / s, , A, , 45°, , 20 m, , 10 m, d, , a) 10 m, b) 20 m, c) 30 m, d) 40 m, 28. In the figure shown, the two projectiles are fired, simultaneously. The minimum distance, between them during their flight is, , v0, , O, , 20 3 m / s, , θ, , x, , a), , 6 m / s −1 , θ = tan −1 (1) with X − axis, , b), , 26 m / s −1 , θ = tan −1 (5) with X − axis, , c), , 2 m / s −1 , θ = tan −1 (2) with X − axis, , 20 m/s, , 60°, , 30°, 20 3 m / s, , d) 13m / s −1 , θ = tan −1 (4) with X − axis, a) 20 m, b) 10 3 m c) 10 m, d) zero, 26. Shots are fired simultaneously from the top and 29. In the direction shown in figure with velocities, bottom of a vertical cliff at angles α and β and, v A = 20m / s and vB = 10m / s respectively.., they strike an object simultaneously at the, 1, same point. If the horizontal distance of the They collide in air after s . Find the distance x., 2, object from the cliff is a, the height of the cliff, vB = 10 m/s, is, U, o', , α, , vA = 20 m/s, , 1, , A, m, P, v, , a, 2, , n, , β, o, , (a), , a(cot α − cot β ), cot α cot β, , (b) a(sin β − tan α ), , θ, , B, x, , a) 2 3 m b) 3 3 m c) 4 3 m d) 5 3 m, E) NON UNIFORM TWO DIMENTIONAL, MOTION, 30. A particle starts from origin of co-ordinates at, time t = 0 and moves in the xy plane with a, constnat acceleration α in the y direction .It’ss, equation of motion is y = β x 2 . It’s velocity, component in the x direction is, , a tan α, (d) a(cot α − cot β ), tan β, 27. Two particles A and B are projected, α, 2α, α, (A) Variable (B), (C), (D), simultaneously from the two towers of height, 2β, β, 2β, 10m and 20m respectively. Particle A is, 31. Motion of a particle is governed by following, projected with an initial speed of 10 2m / s at, x, an angle of450 with horizontal, while particle relations y = ; V x = b − ct . (α , b , c are + ve const ), α, B is projected horizontally with speed 10m/s., The displacement (S) verson from (t) graph os, If they collide in air, what is, t h e, distane d between the towers?, (c), , 159
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , (A), , (B), s, , s, t, , t, (D), , s, , s, t, , t, , MULTIPLE ANSWER QUESTIONS, 32. At the instant a motor bike starts from rest in, a given direction, a car overtakes the motor, bike, both moving in the same direction. The, speed time graphs for motor bike and car are, represented by OAB and CD respectively., Then, , speed (in m/s), , A, 60, , bike, , B, , 40, , car, , D, , 20, , 34. A motor boat is to reach at a point 30°, upstream on other side of a river flowing with, velocity 5 m/s. Velocity of motor boat with, respect to water is 5 3 m/sec. The driver, should steer the boat at an angle of, (A) 30° up w.r.t. the line of destination from the, starting point, (B) 60° up w.r.t. normal to the bank, (C) 150° w.r.t. stream direction, (D) none of these, 35. A car is moving rectilinearly on a horizontal, path with acceleration a 0 . A person sitting, inside the car observes that an insect S is, crawling up the screen with an acceleration a., If θ is the inclination of the wind screen with, the horizontal, then the acceleration of the, insect., a) perpendicular to screen is a 0 tan θ, b) perpendicular to screen is a 0 sin θ, c) along the horizontal is a 0 − a cos θ, d) parallel to screen is a + a 0 cos θ, , O, , 36. Three particles A, B and C and situated at the, vertices of an equilateral triangle ABC of side, of length l at time t = 0 , Each of the particles, a) at t=18s the motor bike and car are 180m apart., m ove w ith constant speed u. A always has its, b) at t=18s the motor bike and car are 720m apart., velocity along AB, B along BC and C along, c) the relative distance between motor bike and, CA., car reduce to zero at t =27 s and both are 1080m, A, far from origin, d) the relative distance between motor bike and, car always remains same., l, 33. A man in a boat crosses a river from point A., l, If he rows perpendicular to the bank he, reaches point C(BC=120m) in 10 minutes. If, the man heads at a certain angle α to the, B, l, straight line AB (AB is perpendicular to the, banks)against the current he reaches point, 2l, a) The time after which they meet is, B in 12.5 minutes., 3u, B, C, b) Total distance travelled by each particle before, 2l, they, meet, is, v, W, α, 3, 3 6 9 12 15 18 21 24 27, time (in s), , A, a) The width of the river is 300m, b) The width of the river is 200m, c) The rowing velocity is 20m/min, d) The rowing velocity is 30m/min, 160, , 3u, 2, d) Relative velocity of approach between any two, 3u, particles is, 2, , c) Average velocity during the motion is
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 37. A man crosses a river in a boat. If he crosses, (b) x-t graph is a straight line passing through origin, the river in minimum time he takes 10 minutes, (c) y-t graph is a straight line passing through origin, with a drift 120 m. If he crosses the river taking, d) ? x − t graph is a straight line, shortest path, he takes 12.5 minute:(Assume, 41. Two particles are projected from ground with, vb/r > vr ), same intial velocities at angles 600 and 300, (a) width of the river is 200 m, (with horizontal). Let R1 and R2 be their, (b) velocity of the boat with respect to water 12 m/, horizontal ranges, H 1 and H 2 their maximum, min, heights and T1 and T2 are the time of flights., Then, (c) speed of the current 20 m/min, H1 H 2, H1 H 2, H1 H 2, H1 H 2, (d) velocity of the boat with respect to water 20 m/, >, <, >, <, (b), (c), (d), (a), min, R1 R 2, R1 R 2, T1 T2, T1 T2, MULTIPLE ANSWER QUESTIONS 42. A particle is projected from the ground with, velocity u at angle ? with horizontal. The, 38. The co-ordinates of a particle moving in a plane, horizontal range, maximum height and time of, are given by x ( t ) = a cos ( pt ) and, flight are R, H and T respectively. Now, keeping u as fixed, ? is varied from 30o to, y ( t ) = b sin ( pt ) where a, b ( < a ) and p are, 60o . Then, positive constant of appropriate dimensions., (a) R will first increase. H will increase and T will, Then, decrease, (A) The path of the particle is an ellipse., (b) R will first increase then dcrease while H and T, (B) The velocity and acceleration of the particle, both will increase, π, (c) R will decrease while H and T will increase, are normal to each other at t =, 2p, (d) R will increase while H and T will decrease, (C) The acceleration of the particle is always 43. Two projectiles A and B are fired, directed towards a fixed position, simultaneously as shown in figure. They collide, (D) The distance travelled by the particle in time, in air at point at time t. Then, π, y(m), internal t = 0 to t =, is ‘a’, 2p, θ2, B, 39. Trajectories of two projectiles are shown in, 20, figure. Let T1 and T2 be the time periods and, P, u1, u and u their speeds of projection. Then, 1, , 2, , y, 10, , u1, u2, , 1, , 2, , 0, , x, (a) T2 > T1 (b) T1 = T2 (c) u1 > u 2 (d) u1 < u 2, 40. In a projectile motion let ? x and ? y are the, horizontal and vertical components of velocity, at any time t and x and y are displacements, along horizontal and vertical from the point of, projection at any time t. Then, (a) ? y − t graph is a straight line with negative slope, and positive intercept, , θ1, A, 10, , 20, , x(m), , (a) t ( u1 cos ?1 − u 2 cos ? 2 ) = 20, (b) t ( u1 sin ? 1 − u 2 sin ? 2 ) = 10, (c) Both (a) and (b) are correct, (d) Both (a) and (b) are wrong, , 161
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MOTION IN A PLANE, , JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , 44. Suppose in the abscence of air resistance,, R = OB, H = AC, t1 = t OA and t 2 = t AB . If air, resisitance is taken into consideration and the, corresponding values are R ' , H ' , t1' and t 2 ', then, 47., y, , (a) t1 will decrease while t 2 will increase, (b) H will increase, (c) R 1 will decrease while R 2 will increase, (d) T may increase or decrease, A projectile is projected from the ground, making an angle of 30o with the horizontal. Air, exerts a drag which is proportional to the, velocity of the projectile, A, (a) at highest point velocity will be horizontal, (b) the time of ascent will be equal to the time of, x, descent, O, B, C, (c) the time of descent will be greater than the time, ', ', ', ', of ascent, (a) R < R,H < H, t1 > t1 and t 2 > t 2, (d) the time of ascent will be greater than the time, (b) R ' < R,H ' < H, t1' > t1 and t '2 < t 2, of descent, ', ', ', ', 48. A particle is fired from a point on the ground, (c) R < R,H > H, t1 > t1 and t 2 < t 2, with speed u making an angle ? with the, ', ', ', ', (d) R < R,H < H, t1 < t1 and t 2 > t 2, horizontal. Then, 45. From an inclined plane two particles P, Q are, (a) the radius of curvature of the projectile at the, projected with same speed at same angle ? ,, u 2 cos 2 ?, one up and other down the plane as shown in, highest point is, figure. Which of the following statement(s) is/, g, are correct ?, (b) the radius of curvature of the projectile at the, θ, u 2 sin 2 ?, Q, highest point is, g, θ, (c) at the point of projection magnitude of tangential, θ, acceleration is g sin ?, P, (a)The particles will collide the plane with same, (d) at the point of projection magnitude of tangential, speed, acceleration is g cos ?, (b) The times of flight of each particle are same, 49. A particle is projected from ground with velocity, (c) Both particles strike the plane perpendicularly, 40 2 m / s at 45o . At time t = 2 s, (d) The particles will collide in mid air if projected, simultaneuosly and time of flight of each, (a) displacement of particle is 100 m, particle is less than the time of collision, (b) vertical component of velocity is 20 m/s, 46. In a projectile motion let t OA = t1 and t AB = t 2 ., (c) velocity makes an angle of tan −1 ( 2 ) with, The horizontal displacement from O to A is R1, , vertical, d) particle is at height of 60 m from ground, , and from A to B is R 2 . Maximum height isH, COMPREHENSION TYPE QUESTIONS, and time of flight is T. If air drag is to be, Comprehension, -I, considered, then choose the correct, A motor cyclist is riding North in still air at, alternative(s)., y, 36 kmh−1 . The wind starts blowing West ward, with a velocity 18 kmh−1, 50. The direction of apparent velocity is, A, a) tan−1 (1/2) West of North, H, B x, b) tan−1 (1/2) North of West, O, c) tan−1 (1/2) East of North, R1, R2, d) tan−1 (1/2) North of East, 162
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 51. If the wind velocity becomes 36 kmh−1 due, West, then how much more distance the motor, cyclist would cover in 10 min, a) 10 km, b) 1.8 km c) 3.6 km d) 8.5 km, , P, , 45°, , 45°, , Q, , 12.5 m, , Comprehension-II, , A, , 15 m, , B, , 20 m, A river of width ‘a’ with straight parallel banks, flows due north with speed u. The points O and A, are on opposite banks and A is due east of O. 57. The speed of the particle at point P will be :, Coordinate axes OX and OY are taken in the east, a) 5 10m / s, b) 10 5m / s, and north directions respectively. A boat, whose, c) 5 15m / s, d) 5 5m / s, speed is v relative to water, starts from O and, crosses the river. If the boat is steered due east 58. The speed of projection of the particle at A will, be :, v, and u varies with x as u = x ( a − x ) 2 find, a) 5 10m / s, b) 10 5m / s, a, 52. equation of trajectory of the boat, c) 5 15m / s, d) 5 5m / s, 2, 2, 3, 59. The range that is AB will be :, x x, x, x, − 2, a) y = −, b) y =, a) 5 10m b) 25 3m c) 5 15m d) 25 5m, a 2a, 2a 3a, , Passage: 3, x 2 x3, x 2 x3, Two projectiles are projected simultaneously from, −, −, c) y =, d) y =, a a2, a 3a 2, the top and bottom of a vertical tower of height h at, 53. Time taken to cross the river, angles 450 and 600 above horizontal respectively., Body strike at the same point on ground at distance, a, v, 2a, 2v, 20m from the foot of the tower after same time., a), b), c), d), v, a, v, a, 60. The speed of projectile projected from the, 54. The direction of absolute velocity of boat, bottom is, man when he reaches the opposite bank, 20, 20, m/ s, a) west, b)south, c)east, d) north, m, /, s, a) 40 m/s b), c) 40 3m / s d), 3, 3, Comprehension-III, A car is moving towards south with a speed of 20 61. The ratio of the speed of the projectile projected, from the top and the speed of the, projectile, ms-1. A motorcyclist is moving towards east with, projected from the bottom of tower is, -1, a speed of 15 ms . At a certain instant, the, a) 1: 2, b) 1: 3, c) 5 :1, d) 7 :1, motorcyclist is due south of the car and is at a, 62., The, time, of, flight, of, projectiles, is, distance of 50m from the car., 1, 1, 1, 1, 55. The shortest distance between the, a) (3) 4, b) 2(3) 4, c) 3(3) 4, d) 4(3) 4, motorcyclist and the car is, Passage:4 (IIT JEE 1996), a) 10m, b) 20m, c) 30m, d) 40m, Two guns situated on top of a hill of height 10m fire, 56. The time after which they are closest to each, other, one shot each with the same speed 5 3 m/s at some, a) 1/3s, b) 8/3s, c) 1/5s, d) 8/5s, interval of time. One gun fires horizontally and the, other fires upwards at an angle of 600 with the, COMPREHENSION TYPE QUESTIONS, horizontal. The shots collide in air at a point P., Passage:1, Find, A particle is fired from ‘A’ in the diagonal plane of, 63. The time interval between the firings and, a building of dimension 20m (length) x 15m(breadth), a) 1 s, b) 2s, c) 3 s, d) 4s, x 12.5m (height), just clears the roof diagonally & 64. the coordinates of point P. Take the origin of, falls on the other side of the building at B. It is, coordinate system at the foot of the hill right, observed that the particle is traveling at an angle, below the muzzle and trajectories in the xyplane., 450 with the horizontal when it clears the edges P, and Q of the diagonal. Take g=10m/s2., a) ( 5 m , 5 m ), b) 5 3 m,5 3m, , (, , c) 5 3 m ,5m, , ), , (, d) ( 5 m ,5, , ), 3m ), , 163
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , MATRIX MATCHING TYPE, , its horizontal velocity component reverses the, direction without change in magnitude and the, 65. Vr ,Vw ,Vm are the velocities of rain, wind and, vertical velocity component remains same., Ball stops after hitting the ground.Match the, man are given in Column-I Vrm in Column-II, statement of column I with the distance of the, match Column-I withColumn-II, wall from the point of throw in column II., Column-I, Column-II, Column-I, Column-II, $, $, $, a), Ball, strikes, the, wall, directly p) d = 8m, a) Vr = −5 j , Vw = 5i, Vm = 0, p) Vm = 5i, b) Ball strikes the ground, q) d = 10 m, b) Vr = −5 $j, Vw = 5$i, Vm = 5$i, q) Vrm = 5i$ − 5$j, at x = 12 m from the wall, c) Ball strikes the ground at, r) d = 15 m, c) Vr = −5 $j , Vw = 5$i − 5 $j, r) Vrm = −5$j, x = 10 m from the wall, d) Ball strikes the ground, s) d = 25 m, Vm = 5$i, at x = 5 m from the wall, 68. Trajectories are shown in figure for three, d) Vr = −5 $j , Vw = −5$i − 5 $j,, s) Vrm = − 10i$ − 10 $j, kicked footballs. Initial vertical and horizontal, velocity components are u y and u x, Vm = 5$i, respectively. Ignoring air resistance, choose, the correct statement from column-2 for the, MATRIX MATCHING TYPE QUESTIONS, value of variable in column-1., 66. A particle is projected on an inclined plane which, is inclined at 300 with the horizontal as shown in, fig. Initial speed of the particle isv0,and inclined, B, A, plane is sufficiently large. Match the Column –, I and Column – II, C, O, , v0, , 30°, 30°, , Column – I, , Column – II, , a) Range on the inclined plane, b) Velocity of the particle is, , p), q), , v0, g 3, 2v0, g 3, , parallel to the inclined plane at time, c) Time after which particle, , r), , 4v0, 3g, , strikes the plane is, 2v02, d) For the given velocity, s), 3g, maximum range on the inclined plane, (angle of projection changing), 67. A ball is thrown at an angle 750 with the, horizontal at a speed of 20 m/s towards a high, wall at a distance d. If the ball strikes the wall,, 164, , Column-1, A) Time of flight, B) u y / u x, , Column-2, P) greatest for A only, Q) greatest for C only, , C) ux, , R) equal for A and B, , D) u x u y, , S) equal for B and C, , INTEGER ANSWER TYPE QUESTIONS, 69. Three points are located at the vertices of, an equilateral triangle whose sides equal to, a =3m. They all start moving simultaneously, with speed v =1 m/s, with the first point, heading continually for the second, the, second for the third, and the third for the, first. How soon will the ponits meet?, 70. The slopes of wind screen of two cars are, α1 = 300 and α 2 = 150 respectively. At what, v1, ratio of v of the velocities of the cars will, 2, their drivers see the hail stones bounced back, by the wind screen on their cars in vertical, direction assume hail stones fall vertically, downwards and collisions to be elastic
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 71. A heavy particle is projected from a point at, the foot of a fixed plane, inclined at an angle, 450 to the horizontal, in the vertical plane, containing the line of greatest slope through, the point. If φ (> 450 ) is the inclination to the, horizontal of the initial direction of projection,, for what value of tan φ will the particle strike, the plane horizontal., 72. A projectile is fired from the base of coneshaped hill. The projectile grazes the vertex, and strikes the hill again at the base. If α be, the half - angle of the cone, h its height, u the, initial velocity of projection and θ angle of, projection, then tan θ tan α is, , u, , α, , h, , 90°, H, 90°, , 30°, , LEVEL-V - KEY, SINGLE ANSWER TYPE, 1)b, 8)A, 15) d, 22) b, 29) d, , 2)a, 9) C, 16)c, 23) a, 30) d, , 3)A, 10) C, 17) d, 24) c, 31) d, , 4)C, 11)A, 18) d, 25) b, , 5)C, 12) B, 19) b, 26) a, , 6)D, 13) b, 20) d, 27) b, , 7)A, 14) a, 21) a, 28) b, , MULTI ANSWER TYPE, 32)AC, 36)A,B,C,D, 40) a,b,d, 44) d, 48) a,c, , 33)BC, 37)A,D, 41) a,c, 45) b,d, 49) a,b,c,d, , 34) ABC, 38) a,b,c, 42) b, 46) a,d, , 35)BC, 39) b,d, 43) b, 47) a,d, , COMPREHENSION TYPE, 50) A 51) B 52)B 53)A 54)C 55)C 56)D, 57) a 58) b 59) b 60) d 61) a 62) b 63) a 64)c, , θ, 2h tanα, , MATRIX MATCH TYPE, , 73. Three balls A,B and C are projected from, ground with same speed at same angle with, the horizontal. The balls A, B and C collide with, the wall during a flight in air and all three collide, perpendicularly and elastically with the wall as, shown in, figure. If the time taken by the, ball A to fall back on ground is 4 seconds and, that by ball B is 2 seconds. Then the time taken, by the ball C to reach the ground after, projection will be ...., , 65) a-q,b-r,c-p,d-s, 66) a → s, b → p, c → q, d → s, 67) A → p, q, r ; B → p ; C → q ; D → r , s, 68) a-r b-p c-q d-s, , INTEGER TYPE, 69) 2 70)3, , B, , θ, , C, θ, , 74. In the given figure, the angle of inclination of, the inclined plane is 300. A particle is projected, with horizontal velocity V0 from height H. Find, the horizontal velocity V0(in m/s) so that the, particle hits the inclined plane perpendicularly., Given, H = 4m, g =10 m/s2, , 72) 2 73) 6 74) 4, , LEVEL - V - HINTS, , 1., A, , 71) 2, , r, VA = 400iˆ ;, r, VB = υ B cos 300 iˆ + υ B sin 300 ˆj, r υ 3 υB, ˆj ;, VB = B, iˆ +, 2, 2, r, r r, VB / A = VB − VA, r, υ 3 υb, VB / A = B, iˆ +, 2, 2, , , , ˆj − 400iˆ, , , , ur, υ 3, υ, V B / A = B, − 400 $i + B $j, 2, 2, , 800 r, 400 ˆ, υ, =, j, ; VB = 400iˆ +, 3 υB − 800 = 0, B, 3, 3, 165
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , magnitude of average velocity when averaged over, all such pairs, Thus, 2π, , v 21 =, , ∫v, 0, , dθ, , 21, , 2π, , ∫ dθ, , =, , 4v, = 1.273v ⇒ > v ., π, , 0, , 8., , Suppose u is the speed of the boat relative to water,, then velocity of the flow (w.r.t. bank) V0, , v x = ( u cos θ + v0 ) , and perpendicular to flow will, be v y = u sin θ Time to cross the river,,, b, . In the time the distance travelled by, u sin θ, the boat in the direction of flow, t=, , C, , a, , r, υS = 10iˆ − 5jˆ, r, r, r, υBS = υB − υS = ( a − 10 ) ˆi + ( b + 5 ) ˆj, r, r, r r, Now υB ⊥ υBS or υB .υBS = 0, or a ( a − 10 ) + b ( b + 5 ) = 0 ..(1), But we know that speed of the motorboat is, 5 m/s, so a 2 + b2 = 25 ...(2), Solving (1) and (2), we get a = 0 or 4, when a = 0, b=-5 and a=4, b=3 Hence either, r, r, υ = −5jˆ or υ = 4iˆ + 3jˆ, B, , B, , r, However a diagram shows that when υB = −5jˆ ,, the motorboat is moving further away from the, r, steamer. So υ = 4iˆ + 3jˆ, B, , B, , Y, b, , u, , v0, , θ, A, , b, u sin θ, or au sin θ = ub cos θ + v0 b, a = v x t = ( u cos θ + v 0 ), , ∴u =, , v0 b, ( a sin θ − b cos θ ) ..(i) u to be minimum, duld, , v0 b, d , , θ = 0 or dθ a sin θ − b cos θ = 0, , , , (4,36), , (−11,16 ), , 10iˆ − 5 ˆj, 4iˆ + 3 ˆj, , −5iˆ, , X, , 11. As it is given that boat is sailing in a direction, normal to current. Crosing velocity of boat is =, 10 m/s. So time taken by the boat to reach the, other bank is, , 400, = 40 s. Drift due to flow of river, 10, , is = Drift velocity x time to cross the river Here, boat is sailing in normal direction so direction so, drift velocity is the river current velocity. Thus,, dirft is x = 2.0 × 40 = 80 m, , b, a, ∴ cos θ =, 2, b, a + b 2 12. Let v1 = 2 3k v2 = 2k v3 = 3k ; (2k)t = 10, On substituting these values in equation (i), we get, 3k, 5, 1, 3k ) t − × 10 × t 2 = 0 ; t = , k =, (, 5, v0 b, 3, 2, u min =, uur uuur uuv, 2, 2 ., a +b, vb = vb / r + vr, 13. Let h be the height of the pole. We have, v RM, 4, 0, 9. v = tan 53 = 3 When the man is running, x, MG, y = x tan α 1 − Since top of pole, lie on, r, 4, 4, ⇒ v RM = v MG = × 12 = 16km / hr, curve(1),, 3, 3, y, 10. In order to approach as close as possible to, steamer, the direction of motion of motorboat, P (a,h), should be perpendicular to the relative motion., Let the optimum velocity of motorboat is, h, r, ˆ, ˆ, α = 45°, υB = ai + bj Velocity of steamer is, x, O, a, b, 167, , or tan θ = −
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, a , ab, , a + b − a, h = a tan 450 1 −, =, =a, , a +b, a+b a+b, Vy = U sin θ − gt, 14. Vx = U cosθ, sin θ g, U sin θ − gt, −, ; Tanα =, cos θ U cos θ, U cos θ, Comparing with y = c − mx, , Tanα =, , , t, , , 15., , 18. The motion of the sphere is similar to projection, motion. The components of its acceleration are, , ax = 1.80m / s 2 , a y = 0 . When the sphere, crosses the y-axis, its displacement component, 1, along x-axis is zero. 0 = vixt + ax t2 ;, 2, 1, 0 = 3sin 300 t − (1.8)t 2 or t = 1.667 s, 2, 19. Components of velocity of ball relative to lift are:, u x = 4 cos 30o = 2 3 m / s, and u y = 4sin 30o = 2 m / s, , O, , R, , A, d, , x, , y, , B, , u = 4m/s, , d = R+x, R = 614.4, T = 12.8 sec, , 30°, , A B = ( 5 )12.8 sec = 64 m, , x, , and acceleration of ball relative to lift is 12 m / s 2 in, negative y-direction or vertically downwards. Hence, time of flight is 1/3 sec., y, , d = 614.4 + 64, = 678.4 m ., , 16. ‘O’ is the point of projection of the shell and ‘A’ is, the position of enemy at a height ‘h’ above the level, A, u, , a, , h, , of ‘O’., O, , 20., , α, B, , If ‘u’ is the minimum initial velocity of the projectile, to shell the enemy, then ‘OA’ must be the maximum, range up the inclined plane of angle α ., 21., So OA =, , u2, g (1 + sin α ), , --------- (i), , From ∆OAB , OA = h cosecα ---------- (ii), , tan θ =, , u, θ, , g, x, , (VB. A )horizontal = v2 cos 60 + v1 cos 60 = 12m / s, (VB. A )vertical = v2 sin 60 − v1 sin 60 = 4 3m / s, VB. A =, , (12 ), , 2, , (, , + 4 3, , 70, m, , From eqn. (i) and (ii), u = gh(cosecα + 1), 17. Velocity of particle after time t is, ∧, ∧, ∧, ur, V1 = (v cos θ i + v sin θ j) − ( g j)t, ∧, ∧, ∧, uur, V2 = (v cos2θ i + v sin2θ j) − ( g j)t, To be parallel of V 1 and V 2, v cos θ, v sinθ − gt, ⇒, =, v cos 2θ v sin 2θ − gt, Solving the above equation, we get,, v, θ, 3θ, 168 t = cos cos ec, g, 2, 2, , a, = 0.4, g, , ), , 2, , = 192, , X, , A, , B, , 45°, , 30°
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , x = v21t = 192 × 10 3 = 240m, from Vle ABC, , vsin 60°, 60°, , AB 2 = Ac 2 + Bc 2, AB = 250m., , vcos 60°, 60°, , u, , ( u 2 ) , , B, , 22., , 30°, , 1 , 3?, 2, u, = 2 ∴ ?=, 2, 3, , (or), , u, 0, At the moment of collision tan 60 = ( gt − u ) = 3, , α, , v = u 2 + ( gt − u ) =, 2, , t AB = time of flight of the projectile, , 2u, 3, , 25. x = ( U cos θ ) t = 2 ; U cos θ = 1, 1, g cos 30o, U sin θ ) t − gt 2 = −10 U sin θ = 5, (, Now component of velocity along the plane, 2, becomes zero at point B., 2, 2, V = ( U cosθ ) + ( U sin θ ) = 26 ; Tanθ = 5, 26. Coordinates of P w.r.t O and O' as origins are (a,, 0 = ucos( a −30o ) − ( gsin30o ) T, ∴, n) and (a, -m) The height of the cliff = h = m + n, , =, , or, or, ∴, , 23., , 2u sin ( a − 30o ), , ucos ( a −30 ) = ( gsin30, o, , o, , ), , 2usin( a −30o ), gcos30o, , tan ( a − 30o ) =, , 1, 3, =, o, 2tan30 2, 3, a = 30o + tan−1 , 2, , , θ=, , 30, α = 30, , In the frame of wedge the particle appears to be, projected up the plane as shown in figure., 2Usin θ, Its time of flight is, g cos α = 2s, 24. Let ? be the velocity at time of collision. Then,, , u 2 cos 45o = ? sin 60o, , gx 2, 2u 2 cos 2 α, ga 2, ∴ −m = a tan α − 2, 2u cos 2 α, y = x tan α −, , ga 2, 2 gh, −, =, −, n, a, tan, β, v0 =, ;, 2v 2 cos 2 β, 2 + cot 2 θ, ∴ ( m + n) = a (tan β − tan α ), because u cos α = v cos β because both the shots, reach P simultaneously. This is possible only If the, horizontal components of the velocities are equal., ∴ Ans. (a),, 27. In the frame of B, rest, 10m, , 10m/s, 20m/s, 10m, , 10m, d, , 10, = 1s d = 20 ×1 = 20m., 10, 28. In a reference frame, take a differentiation of, distance between projectiles is zero., t=, , 169
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, VAB = u + u cos 60 =, , MOTION IN A PLANE, H1 = H 2 ⇒ u y1 = u y2 ;Hence,, , 3u, 2, , T1 = T2, , u 2 sin 2÷ 2 ( u sin ÷ )( u cos ÷ ) 2u x u y, =, =, g, g, g, R 2 > R1 ∴ u x2 > u x1 or u 2 > u1, Range R =, , 37., , θ, , W, , Shortest time Shortest path, w, vr, 10m =, 12.5 =, vbr, vbr cos θ, 120 = 10 × vr 12.5 = 10 secθ, vr = 12 m / min, , 38., , 40. If u is the initial speed and ? the angle of projection., Then ? y = u sin ? − gt i.e., ? y -t graph is a straight, line with negative slope and positive intercept., , x = ( ? cos ? ) t i.e., x-t graph is a straight line, , 4, cos θ =, 5, , 1 2, passing through the origin. y = ( u sin ? ) t − gt, 2, i.e., y-t graph is parabola i.e., ? x -t graph is a straight, line parallel to t-axis., , x2 y2, +, = 1 ; So the path is an ellipse, a 2 b2, Vx = − ap sin pt , V y = bp cos pt, ax = − ap 2 cos pt a y = −bp 2 sin pt, ur r, ur r, So V − a = 0 when V ⊥ a, So a 2 p3 sin pt.cos pt − b 2 p 3 sin pt.cos pt, , ⇒ a 2 p3 sin pt.cos pt = b2 p3 sin pt.cos pt, as a ≠ b, So sin pt.cos pt = 0, , 41. At 30o and 60o , R 1 = R 2, Further, H ∝ sin 2 ? and T ∝ sin ?, ∴, , H, H, ∝ sin 2 ? and ∝ sin ?, R, T, o, o, sin 60 > sin 30, , ∴, , H1 H 2, H, H, >, and 1 > 2, R1 R 2, T1, T2, , ∴, , 10 + ( u1 cos ?1 ) t = 30 − ( u 2 cos? 2 ) t, , π, −−−−−, 2P, 2, The motion is similar to motion of earth around sun 42. R ∝ sin 2?, H ∝ sin ? and T ∝ sin ? , sin 2? will, So force always towards focus and hence, first increase, then decrease. While sin ? will only, increase., acceleration. At t = 0 particle is at ( a, o ), 43., π, particle is at ( o, b ), At t =, XA = XB, 2p, , ⇒sin p2t = 0 ⇒ p2t = π ,2π ⇒ t =, , (o,b), , or, , t (u 1 cos ?1 + u 2 cos ? 2 ) = 20, yA = yB, , (a,o), , ∴, , So distance travelled along X axis is a not the actual, distance, which is the length of the part of the ellipse, between ( a, o ) to ( o, b ) you can try out for, distance by following method ds = dx 2 + dy 2, 2, , 2, , s, 0, dy , dy , ⇒ ds = 1 + ⇒ ∫ ds = ∫ 1 + .dx, 0, a, dx , dx , 39. Maximum height and time of flight depend on the, vertical component of intial velocity., , ∴, , 1, 10 + ( u1 sin ? 1 ) t − gt 2, 2, 1, = 20 + ( u 2 sin ? 2 ) t − gt 2, 2, u, sin, ?, −, u, sin, ?, ( 1 1 2, 2 ) t = 10, , 44. During upward journey gravity and air fraction both, will oppose the motion. Hence, t1' < t1, 45., , 171
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , θ, θ, P, , Q, , θ, u, , a = an =, , but, , a =g, , ∴, , ( u cos ? ), g=, , β α, , u, , β, Time of flight of P is, 2u sin ( 2÷ − ÷ ), , 2u sin (÷ ), , but, , a =g, , 2, , v2, R, , ( u cos ? ), g=, , =, , 2u tan ÷, ∴, T1 = T2, g cos ÷, g, Further acceleration of both the particles is g, downwards. Therefore, relative acceleration, between the two is zero or relative motion between, the two is uniform. Now relative velocity of P with, respect to Q is towards PQ. Therefore, collision, will take place between the two in mid air., 46. In the upward journey resistance due to air will be, in downward direction. While in downward motion, resistance will be upwards. Therefore, t1 will, , R, u cos 2 ?, R=, g, 2, , =, , θ, , g, At point of projection component of acceleration, , ( = g ) along velocity vector is −g cos ( 90o − ? ) or, , −g sin ? ., 49. u x = 40 m / s, u y = 40 m / s ; At t = 2s., , ? x = 40 m / s and ? y = 40 − 10 × 2 = 20 m / s, x = ? x t = 80 m, 1, y = u x t − gt 2 = 60 m, 2, ∴, , s = x 2 + y 2 = 100m, ?, ? = tan −1 x, ?, y, , , −1, = tan ( 2 ), , , COMPREHENSION TYPE, decrease, while t 2 will increase. Hence, T = t1 + t 2, AC 5 1, may increase or decrease. H will decrease. In, =, =, 50. From figure tan β =, horizontal direction air resistance is opposite to the, OA 10 2, motion.Therefore, R 1 will decrease while R 2 may, 1, β = tan−1 West of North, So, decrease or increase., 2, 47. During upward journey acceleration due to gravity, N, and vertical component of retardation are in same, direction. While during downward journey they are, C, in opposite directions. Hence, time of descent will, A, be greater than time of ascent, r, r, R β 10ms-1, 48. At highest point angle between a and v is zero., B, Hence, total acceleration is only normal or radial, -1, W, , 5ms, , O, , v = ucosθ, , acceleration., , a=g, , S, 172, , 2, , θ, 90°, , 2u tan ÷, g cos ÷, g, and time of flight of Q is, T2 =, , a = an =, , u, , Q, , T1 =, , ∴, , ∴, R, u 2 cos 2 ?, R=, or, g, , or, , α, , v2, R, , ∴, , E
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 51. New apparent velocity, R ' = 102 + 102, = 10 2ms −1, , (, , North, , ), , C, , Distance covered, s ' = 10 2 × (10 × 60 ) m, , N, W, , Vr, , O, , A, , Vbr, , a, , uur uur uur, v b = vbr + vr, , X, , j, E, i, , S, , uur, uur, Given; v br = v and vr = u, , dy, v, = x ( a − x ) 2 ..(1), dx, a, dx, = v ..(2), and v = v x =, dt, Dividing (1) by (2), we get, , Now u = v y =, , y, , dy = ∫, , y, , 53°, , M, , East, Vm, , −1 20 , 0, v mC = 152 + 20 2 = 25ms −1 ; θ = tan 15 = 53, , with x-axis, The motorcyclist appears to move along the line, MP with speed 25 ms-1 . The shortest distance=, perpendicular distance of MP from, C = d ⇒ d = 50 cos 530 ⇒ d = 30m . Time taken, to come closest = time taken by motorcyclist to, , reach B. t =, , MB 50sin 530, =, ⇒ t = 1.6s, v mC, 25, , 125 − ( U sin θ ) = −2 (10 ) 125, , → (1), , 5 10 cos 45 = V cos θ, , → ( 2), , 2, , 2, , 3, , a, , x-axis, , 58. Vy2 − U2y = 2a ys y, , x (a − x ), x, x, dx or y =, − 2 ..(3), 2, 0, 0, a, 2a 3a, This is the desired equation of trajectory., , ∫, , −Vc, , B, , U 2 sin 2θ, = 25 ; U = 5 10 m / s., 57. PQ length ;, g, , x (a − x ), dy x ( a − x ), =, or dy =, dx, 2, dx, a, a2, or, , V mc, , 50m, , Extra distance covered = 8.5–6.7 = 1.8 km, uur, 52. Let v br be the velocity of boatman relative to, uur, uur, river, v r the velocity of river and v b is the, absolute velocity of boatman. Then, Y, , d, 53°, , = 6 2 km = 8.5km, , P, , a, , On solving U = 500m / s ; θ = Tan −1 3 = 60o, 59. Range =, , U 2 sin 2θ, g, , 53. Time taken to cross the river is t = v = v, (60, 61, 62), x, 54. When the boatman reaches the opposite side,, U, 2 3, R1 = ( U cos θ ) t =, x=a or vy = 0 (from equation 1), 2, 55,56Taking North as +y-axis and East as +x Axis, Imagine yourself as an oberver sitting inside the, U, 1, 10 2 U, 2 3 = 20 ; U =, =, car.You will regard the car as being at rest (at C)., 4, V, 2, 2, 3, Relative to you, the speed of the motorcyclist is, obtained by imposing the reversed velocity of the (63,64), Let gun 1 and gun 2 to be fired at an interval ∆t ,, car on motorcyclist as shown in figure., such that t1 = t2 + ∆t ---------- (1), v m = 15ms −1 , v C = 20ms −1, Where t1 and t2 are the respective times taken by, the two shots to reach point P., For gun 1 :, x-component, y-component, 1, y − yi = vi sin 600 t1 − gt12, x − xi = vi cos 600 t1, 2, 173
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , P, , a) cot α =, , u cos φ, , h, , u, , 2., , φ, 45°, O, , N, , 1 2, gt ---------(3), 2, Using eqns. (1) and (2) in (3), 1, (u cos φ )(t ) = (u sin φ )(t ) − (u sin φ )t, 2, tan φ = 2, , and, , h = u sin φ t −, , 72. Here range = 2h tan α =, , h=, , u 2 sin 2 θ, Dividing, 2g, , 2 gh, =, sin 2 θ , , , , , ;, , u 2 sin 2θ, and, g, 2 tan α =, , 2 sin 2θ, ⇒, sin 2 θ, , tan α = 2 cot θ, ∴u 2, , 3., , 4., , tan θ = 2cot α, 73. S = 10 × 4 = 40m ., v0 cos300, 74. 0 = v0 cos 30 − g sin30 t ⇒ t =, .....(1), g sin 300, 0, , − H cos 30 0 = −v0 sin 300 t −, , 1, g cos 30 0 t 2 ..(2), 2, , By equation (1) and (2), we get, H=, , v02 cot2 α , 2gH, 1+, ⇒ v0 =, = 4m / s (α = 300 ), , , g, 2 , 5, , 5., , LEVEL - VI, SINGLE ANSWER QUESTIONS, A) RELATIVE MOTION, 1., , v0, rω, v0, d) tan α =, rω, b) tan α =, , rω, c) cot α = v, 0, A man standing, observes rain falling with, velocity of 20 m/s at an angle of 30 0 with the, vertical. Find out velocity of man so that rain, again appears to fall at 300 with the vertical., a) 20 m/s b) 30 m/s c) 40 m/s d) 10 m/s, A person standing on a road has to hold his, umbrella at 600 with the vertical to keep the, rain away. He throws the umbrella and starts, running at 20 ms-1. He finds that rain drops, are falling on him vertically. Find the speed, of the rain drops with respect to, 1. The road 2. The moving person, 40, 20, 40, 22, m/s,, m/s, b), m/s,, m/s, a), 3, 3, 3, 3, 20, 40 3, 20 3, 40 3, m/s,, m/s d), m/s,, m/s, 3, 3, 3, 3, Two swimmers leave point A on one bank of, the river to reach point B lying right across, on the other bank. One of them crosses the, river along the straight line AB while the, other swims at right angles to the stream and, the walks the distance that he has been, carried away by the stream to get to point B., What was the velocity u of his walking if both, swimmers reached the destination, sumultaneously The stream velocity v 0 = 2.0, , c), , 2 gh, 2 gh (1+ 4cot 2 α ), 1, , =, = gh tan 2 α + 2 , , 2, , 4cot 2 α, 2cot α , , 2, 1+ 4cot α , , 0, , v0, rω, , An open merry go round rotates at an angular, velocity ω . A person stands in it at a distance, r from the rotational axis. It is raining and, the rain drops falls vertically at a velocity v 0. 6., How should the person hold an umbrella to, prorect himself from the rain in the best way., Angle made by umbrella with the vertical is, , km/hour and the velocity v1 of each swimmer, with respect to water equals 2.5km per hour., a) 3km/hr b) 3.5km/hr c) 4km/hr d) 5km/hr, A ball is thrown vertically upward from the, 12m level in an elevator shaft with an initial, velocity of 18m/s. At the same instant an open, platform elevator passes the 5m level,, moving upward with a constant velocity of 2, m/s. Determine (a) when and where the ball, will hit the elevator, (b) the relative velocity, of the ball with respect to the elevator when, the ball hits the elevator., a) 10.2m 9.8m/s, b) 12.3m 19.8m/s, c) 12m 10.2m/s, d) 12.5m 22m/s, From a point A on bank of a channel with still, water a person must get to a point B on the, opposite bank. All the distances are shown, in figure. The person uses a boat to travel, 175
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , across the channel and then walks along the 10. A stone is projected from the point of a ground, bank to point B. The velocity of the boat is, in such a direction so as to hit a bird on the top, of a telegraph post of height h and then attain, v1 and the velocity of the walking person is, the maximum height 2h above the ground. If, v 2 . Prove that the fastest way for the person, at the instant of projection, the bird were to, fly away horizontally with a uniform speed. Find, to get from A to B is to select the angles α1, the ratio between the horizontal velocities of, and α 2 in such a manner that, the bird and the stone, if the stone still hits, A, bird while decending., sin α1 v 2, y, a) sin α = v, 2, 1, sin α1 v1, a, α1, M, b) sin α = v, 2, 2, cos α1 v 2, Q, uQ, =, c), cos α 2 v1, B b, x, α2, cos α 2 v1, x, O, d) cos α = v, N, 1, 2, d, 2, 1, 7. On morning, Joy was walking on a grass-way, a), b), in a garden. Wind was also blowing in the, 2 +1, 2 +1, direction of his walking with speed u. He, 2, 1, c), d), suddenly saw his friend Kim walking on the, 2 −1, 2 −1, parallel grass-way at a distance x away. Both 11. The benches of a gallery in a cricket stadium, stopped as they saw each other when they, are 1 m high and 1 m wide. A batsman strikes, were directly opposite on their ways at a, the ball at a level 1 m about the ground and, distance x. Joy shouted “Hi Kim”. Find the, hits a ball. The ball starts at 35 m/s at an angle, time after which Kim would have heard his, of 53o with the horizontal. The benches are, greeting. Sound speed in still air is v., perpendicular to the plane of motion and the, x, 2x, x, x, first bench is 110 m from the batsman. On which, a) 2 2 b), bench will the ball hit., 2, 2 c), 2, 2 d), 2, 2, v −u, v −u, 2 v −u, 4 v −u, a) 4th step b) 5th step c) 6th step d) 7 th step, B) BODY PROJECTED FROM THE GROUND 12. If R is the horizontal range for inclination, θ, and, h, is, the, maximum, height, reached, by the, 8. A projectile is fired with velocity v0 from a gun, projectile, Then maximum range is, adjusted for a maximum range. It passes, through two points P and Q whose heights, R2, R2, R2, R2, +, 2, h, +, 2, h, +, 8, h, +h, a), b), c), d), above the horizontal are h each. The, h, 8h, 8h, h, separation of the two points is, 13. The acceleration of gravity can be measured, by projecting a body upward and measuring, v0 2, v0 2, v0 − 4 gh, v0 + 4 gh, a), b), the time it takes to pass two given points in, g, g, both directions. Show that if the time the body, takes to pass a horizontal line a in both, v0 2, v0, v0 − 4gh, v 02 − gh, c) 2, d), directions is t A antime to go by a second line B, g, g, in both direction is t B , then assuming that the, 9. A shot is fired with a velocity u at a very high, acceleration is constant, its magnitude is, vertical wall whose distance from the point of, g = (where h is the height of the line B above, projection is x. The greatest height above the, line A.), level of the point of projection at which the, h, 8 h, bullet can hit the wall is ., a), b), 2, 2, 2, tA − tB, tA − tB 2, u 4 + g 2 x2 u 4 − g 2 x 2 u 4 − g 2 x2, u 4 − g 2 x2, a), b), c), d), 8 h, 4 h, 2 gu 2, gu 2, 4 gu 2, 2 gu 2, c) t 2 + t 2, d) t 2 + t 2, A, , 176, , B, , A, , B
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , C) BODY PROJECTED FROM TOP OF A D) PROJECTILE MOTION ON INCLINED, TOWER, PLANE, 14. A particle is released from a certain height H, = 400 m. Due to the wind the particle gathers 16. A body has maximum range R1 when projected, up the inclined plane. The same boby when, the horizontal velocity component, projected down the inclined plane. it has, ν x = ay where a = 5 s −1 and y is the vertical, maximum range R2 . Find its maximum, displacement of the particle from point of, horizontal range. Assume the equal speed of, release, then find the horizontal drift of the, particle when it strikes the ground, projection in each case and the boby is, a) 2.67 km b) 5.67 km c)12.67 km d) 4.97 km, projected onto the e greatest slope., v0, 2R1 R2, 14(a) A fighter plane enters inside the enemy, a) R = R − R, territory, at time t = 0, with velocity, v0, 1, 2, 2R1 R2, O, υo = 250 m / s a moves horizontally with P, R, =, b), R1 + R2, constant acceleration a = 20 m / s2 (see figure)., R1 R2, Q, An enemy tank at theborder, spot the plane, c) R = R − R, 1, 2, and fire shots at an angle θ = 60o with the, 4R1 R2, β, horizontal and with velocity u = 600 m/s. At, d) R = R + R, 1, 2, what altitude H of the plane it can be hit by the 17. A particle P is projected from a point, on the, shot?, surface of smooth inclined plane., Simultaneously another particle Q is released, on the smooth inclined plane from the same, position. P and Q collide on the inclined plane, after t = 4 second. The speed of projection of, 600 m/s, P is (Take g = 10 m/s2), P, H, Q, , θ = 60°, 60°, a) 1500 m b) 2473 m c) 1650 m d) 1800 m, 15. A bomber plane moving at a horizontal speed, a) 5 m/s, b) 10 m/s c) 15 m/s d) 20 m/s, of 20 m/s releases a bomb at a height of 80 m 18. A particle is projected from surface of the in, above ground as shown. At the same instant a, clined plane with speed u and at an angle ?, Hunter starts running from a point below it, to, with the horizontal. After some time the particle, catch the bomb at 10 m/s. After two seconds, collides elastically with the smooth fixed, he realized that he cannot make it, he stops, inclined plane for the first time and, subsequently moves in vertical direction., running and immediately hold, his gun, Starting from projection, find the time taken, and fires in such direction so that just before, by the particle to reach maximum height., bomb hits the ground, bullet will hit it. What, (Neglect time of collision), should be the firing speed of bullet. (Take g =, 2, 10 m/s ), θ, 20 m/s, , 45°, 80 m, Ground, , 10 m/s, , a) 10 m/s, c) 10 10m / s, , 2u sin? u ( sin ?+cos ? ) 2u, 2ucos?, b), c), d), g, g, g, g, 19. A perfectly elastic particle is projected with a, velocity v on a vertical plane through the line, of greatest slope of an inclined plane of, elevation α . If after striking the plane, the, particle rebounds vertically, show that it will, a), , b) 20 10m / s, d) None of these, , 177
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , return to the point of projection at the end of 23. Two particles are projected from the same, time equal to, point on ground simultaneously with speeds, 6v, 6v, and 20 m / s and 20 / 3 m / s at angles 30o and, b), a), g 1 + 8sin 2 α, g 1 + sin 2 α, 60 o with the horizontal in the same direction., The maximum distance between them till both, v, v, of them strike the ground is approximately (g, c), d), g 1 + 8sin 2 α, g 1 + sin 2 α, = 10m/s2), 20. Two bodies A and B are projected from the, (a) 23.1 m (b) 16.4 m (c) 30.2 m (d) 10.4 m, same place in same vertical plane with veloci 24. Two particles A and B are projected, ties v1 and v2 .From a long inclined plane as, simultaneously from a point situated on a, shown Find the ratio of their times of flight, horizontal plane.The particle A is projected, v2, vertically up with a velocity ? A while the, particle B is projected up at an angle of 30o, v1, θ, , a), , v1 sin θ, 2v1 sinθ, b), v2, v2, , A B, , c), , v1 sin θ, v1 cos θ, d), 2v2, v2, , E) COLLISIONS BETWEENPROJECTILES, 21. A particle A is projected from the ground with, an initial velocity of 10 m/s at an angle of 60°, with horizontal. From what height should an, another particle B be projected horizontally with, velocity 5 m/s so that both the particles collide, in ground at point C if both are projected, 25., simultaneously g = 10 m/s2., B, , h, , 5m/s, , (b) 100ms −1 ,50 ms−1, (c) ? A can have any value grater than, 25ms −1 ,100 ms −1 (d) 20ms −1 , 25ms −1, An aircraft moving with a speed of 250 m/s is, at a height of 6000m, just overhead of an, antiaircraft gun. If the muzzle velocity is 500, m/s, the firing angle θ should be:, 250m/s, , 10m/s, , 6000m, , 60°, A, , with horizontal with a velocity ? B . After 5 s, the particles were observed moving mutually, perpendicular to each other. The velocity of, projection of the particle ? A and ? B, respectively are, (a) 5ms −1 ,100ms −1, , C, , 500m/s, , θ, (A) 10 m, (B) 15 m (C) 20 m (D) 30 m, 0, 22. A smooth square platform ABCD is moving, (a) 30, (b) 450 (c) 600 (d) none of these., towards right with a uniform speed ?. At what 26. A cannon fires successively two shells with, velocity v0 =250 m/s, the first at an angle, angle ? must a particle be projected from A, with speed u so that it strikes the point B?, θ1 = 600 and the second at an angle θ 2 = 450 to, B, C, the horizontal, the azimuth being the same., Neglecting the air drag, find the time interval, u, V, between firings leading to the collision of the, θ, shells, A, D, a) 4 sec, b) 7 sec, c) 17 sec d) 11 sec, −1 u , −1 ? , 27., A, shell, is, projected, from, a gun with a muzzle, sin, cos, (a), (b), , , ?, u, velocity v. The gun is fitted with a trolley car, at an angle θ as shown in the fig. if the trolley, −1 u , −1 ? , (c) cos , (d) sin , car is made to move with constant velocity v, ?, u, towards right, find the horizontal range of the, shell relative to ground., 178
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , MULTI ANSWER QUESTIONS, , u, θ, , a) R =, b) R =, c) R =, , 30. A child in danger of drowing in a river is being, carried downstream by a current that flows, uniformly at a speed of 2.5km/h. The child is, 0.6km from shore and 0.8km upstream of a, boat landing when a rescue boat sets out. If, the boat proceeds at its maximum speed of, 20km/h with respect to the water, what angle, does the boat velocity v make with the, shore? How long will it take boat to reach, the child., , V, , 2u sin θ ( u cos θ + v ), g, 2u sin θ ( u cos θ − v ), g, u sin θ ( u cos θ + v ), , a) The angle made by the boat with the shore is 530, , b) The angle made by the boat with the shore is, 370, , 2g, u sin θ ( u cos θ + v ), , c) The time taken by boat to reach the child is 4, min, g, d) The time taken by boat to reach the child is 3, 28. Two guns are projected at each other, one, min, upward at an angle of 300and the other at the 31. A launch plies between two points A and B on, same of depression, the muzzles being 30m, the opposite banks of a river always, apart as shown in the figure. If the guns are, following the line AB. The distance S, shot with velocities of 350m/s upward and 300, between points A and B is 1,200m. The, m/s downward respectively. where the bullets, velocity of the river current v =1.9m/s is, may meet., constant over the entire width of the river., The line AB makes an angle α = 600 with the, direction of the current. With what velocity, B, u and at what angle β to the line AB should, 30 m, P, the launch move to cover the distance AB, y, and back in a time t =5 min? The angle β, 30°, remains the same during the passage from A, A, to B and from B to A., X, d) R =, , a) x = 14m,y = 8.07m, b) x = 4m,y = 4.07m, c) x = 10m,y = 10.07m d) x=5m,y = 1 8.07m, 29. Two particles A and B are projected in same, vertical plane as shown in the figure. Their, initial positions (t = 0), initial speed and angle, of projections are indicated in the diagram. If, initial angle of projection qB = 370 , what, should be initial speed of projection of particle, B, so that it hits particle A. U A = 60m / s, 32., , 53°, , θB, , A, 1) 80 m/s, , B, 100m, 2) 75 m/s 3) 40 m/s, , 4) 45 m/s, , B, u, , β, , v, , α, A, , a) The velocity of the boat is 8m/s, b) The velocity of the boat is 6m/s, c) The angle made by u with the line AB is 120, d) The angle made by u with line AB is 100, The current velocity of river grows in, proportion to the distance from its bank and, reaches the maximum value v 0 in the middle., Near the banks the velocity is zero. A boat is, moving along the river in such a manner that, it always perpendicular to the current. The, speed of the boat in still water is u. Find the, distance through which the boat crossing the, river will be carried away by the current if the, width of the river is c. Also determine the, trajectory of the boat., 179
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, 2cu, a) The distance carried by the boat is X max = v, 0, , b) The distance carried by the boat is x max =, , cv 0, 2u, , relative velocity. River is flowing with speed, of 2km/hr and is 100m wide. speeds of A and B, on the ground are 8km/hr and 6km/hr, respectively., C, , v0 c, 2km/hr, x, u, 100m, uc, 2, d) The trajectary of the boat is y = v x, 0, 127°, 37°, B, 33 Two swimmers A and B start swimming from, A, different positions on the same bank as shown, 300m, 125m, in figure. The swimmer A swims at angle 900, (A), A, will, win, the, race, (B) B will win the race, with respect to the river to reach point P. He, (C), the, time, taken, by, A, to, reach, the point C is 165 sec, takes 120 seconds to cross the river of width, (D) the time taken by B to reach the point C is 150, 10m. The swimmer B also takes the same time, sec, to reach the point P, 36. Two trains A and B are moving with same speed, y, of 100km/hr. Train ‘A’ moves towards east and, 30m, P, train B moves towards west. At an instant when, the trains are moving side by side, an aeroplane, x, files above the trains horizontally. For the, passengers in the train A, the plane appears, 10m, to fly from North to South direction. For the, passengers in the train B, the plane appears, to fly in a direction making an angle 600 to, A, B, North – South direction., (A) The speed of the plane with respect to ground, 5m, a) velocity of A with respect to river is 1/6 m/s, 7, is 100 km / hr, b) river flow velocity is ¼ m/s., 3, c) Velocity of B along y-axis with respect to earth, (B) The speed of the plane with respect to ground, is 1/3 m/s., is 100 3 km / hr, d) velocity of B along x-axis with respect to earth, (C) The plane moves in a direction at an angle of, is 5/24 m/s., 2, c) The trajectary of the boat is y =, , 34. Two frames of reference P and Q are moving, 3, tan −1, to North-South direction (with respect, relative to each other at constant velocity. Let, 2, r, r, to ground), v OP and a OP represent the velocity and the, (D) The plane moves in a direction at an angle of, acceleration respectively of a moving particle, O as measured by an observer in frame P and, 5, r, r, tan −1, to North-South direction (with respect, vOQ and a OQ represent the velocity and the, 2, to ground), acceleration respectively of the moving, 37., Two shells are fired from cannon with speed u, particle O as measured by an observer in frame, each, at angles of α and β respectively with, Q, then, r, r, r, r, r, the, horizontal. The time interval between the, (A) v OP = vOQ, (B) v OP = vOQ + v QP, shots, is T. They collide in mid air after time t, r, r, r, r, r, from the first shot. Which of the following, (C) a OP = a OQ, (D) a OP = a OQ + a QP, conditions must be satisfied?, 35. Two swimmers start a race. One who reaches, a) α > β, b) t cos α = (t − T ) cos β, the point C first on the other bank wins the, race. A makes his strokes in a direction of, c) (t − T ) cos α = t cos β, 370 to the river flow with velocity 5km/hr, 1 2, 1, 2, d) ( u sinα ) t − gt = (u sin β )(t −T) − g(t −T ), relative to water. B makes his strokes in a, 2, 2, direction 1270 to the river flow with same, 180
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , 38. Two inclined panes OA and OB having inclina, tion 300 and 600 with the horizontal respectively, intersect each other at O, as shown in figure. A, particle is projected from point P with a velocity, u = 10 3m / s along a direction perpendicular, to plane OA. If the particle strikes plane OB, perpendicular at Q, , 4m/s, 3m/s, 53°, 37°, , x, , y, v, u, , B, Q, , A, h P, , 30°, , 60°, O, , Which of the following is/are correct, (a) The time of flight 2s, (b) The velocity with which the particle strikes the, plane OB=10 m/s, (c) The height of the point P from point O is 5m, (d) The distance PQ = 20m, 39. Two balls are thrown from an inclined plane at, angle of projection α with the plane, one up, the incline and other down the incline as shown, in figure (T stands for total time of flight) :, v1, , v0, , v0, h2, , h1, α, , θ, , R1, , v1, , v02 sin 2 α, a) h1 = h2 =, 2 g cos θ, , θ, , α, , R1, , b) T1 = T2 =, , 2v0 sin α, g cos θ, , a) Their relative velocity is along vertical direction, b) Their relative acceleration is non-zero and it is, along vertical direction, c) They will hit the surface simultaneously, d) Their relative velocity is constant and has, magnitude 1.4 m/s, 42. A particle moves along x-axis with constant, acceleration and its x-position depend on time, ‘t’ as shown in the following graph (parabola);, then in interval 0 to 4 sec., , x, (m), , 45°, , 45°, 4, , t(sec), , a) relation between x- coordinate & time is, x = t − t2 / 4 ., b) maximum x-coordinate is 1m, c) total distance traveled is 2m, d) average speed is 0.5 m/s, 43. A railway compartment is 16 m long, 2.4 m wide, and 3.2 m high. It is moving with a velocity ‘v’., A particle moving horizontally with a speed ‘u’,, perpendicular to the direction of ‘v’ enters, through a hole at an upper corner A and strikes, the diagonally opposite corner B. Assume g =, 10 m/s2., , c) R2 = R1 = g (sin θ )T12 d) vt2 = vt1, 2.4 m, 40. An aeroplane at a constant speed releases a, v, bomb. As the bomb drops away from the, A, O, aeroplane,, u, B, a) It will always be vertically below the aeroplane, 3.2 m, b) It will always be vertically below the aeroplane, only if the aeroplane was flying horizontallly., 16 m, c) It will always be vertically below the aeroplane, a) v = 20 m/s, b) u = 3 m/s, only if the aeroplane was flying at an angle of, c) To an observer inside the compartment, the path, 450 to the horizontal, of the particle is a parabola, d) It will gradually fall behind the aeroplane if the, d) To a stationary observer outside the, compartment,the path of the particle is parabola, aeroplane was flying horizontally., 41. Two particles are projected with speed 4 m/s 44. Two particles A and B are projected from the, same point with the same speed but at different, and 3 m/s simultaneously from same point as, shown in the figure. Then :, angles α and β with the horizontal, such that, 181
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, the maximum height of A is two–third of the, horizontal range of B. Then which of the, following relations are true ?, a) range of A = maximum height of B, b) 3(1 – cos 2 α ) = 8 sin 2 β, c) maximum value of β is sin–1 (3/4), d) maximum horizontal range of A = u2/g and this, 1 −1 3 , occurs when β = sin , 2, 8, 45. Two particles are projected from the same, point , with the same speed, in the same, vertical plane, at different angles with the, horizontal. A frame of reference is fixed to one, particle. The position vector of the other, r, particle as observed from this frame is r, Which of the follwing satements are correct?, r, a) direction of r does not change, r, b) r changes in magnitude and direction with time, r, c) The magnitude of r increases linearly with time, r, d) The direction of r changes with time; its, magnitude may or may not change, depending on, the angles of projection, , Passage-2, A man is riding on a flat car travelling with a constant, speed of 10m/s. He wishes to throw a ball through, a stationary hoop 15 m above the height of his, hands in such a manner that the ball will move, horizontally as it passes through the hoop. He throws, the ball with a speed of 12.5 m/s w.r.t. himself., 48. How many seconds after he release the ball, will it pass through the hoop ?, a) 1 sec, b) 2 sec, c) 3 sec, d) 4 sec, 49. At what horizontal distance in front of the hoop, must he release the ball ?, a) 12.5 m, b) 15.5 m c) 17.5 m d) 20 m, , Passage-3, A cannon is fixed with a smooth massive trolley car, at an angle θ as shown in the figure. The trolley car, slides from rest down the inclined plane of angle of, inclination β ., The muzzle velocity of the shell fired at t = t0 from, the cannon is u, such that the shell moves, perpendicular to the inclined plane just after the, t=0, θ, , COMPREHENSION TYPE QUESTIONS, firing., , Passage-1, A river of width w is flowing such that the stream, , β, , , 3 −1 , velocity varies with y as v R = v 0 1 + w y ; 50. The value of t0 is:, , , , u cos θ, u cos θ, u cos θ, u sin θ, where y is the perpendicular distance from one, (a), (b), (c), (d), g, g cos β, g sin β, g cos β, bank. A boat starts rowing from the bank with, constant velocity v =2v0 in such a way that it 51. the time of flight of the shell is:, always moves along a straight line perpendicular, u cos θ, 2u sin θ, u, u sin θ, to the banks., (a), (b), (c), (d), g sin β, g cos β, g, g sin β, 46. At what time will he reach the other bank, 52. the difference in range of the shell relative to, wπ, wπ, the trolley car and ground is:, a) t = 6v, b) 6 2 − 1 v, 0, 0, u 2 sin 2θ, u 2 cos 2 θ, (a), (b), g cos β, 2 g sin β, wπ, wπ, c) 6 3 − 1 v, d), 3 − 1 v0, u 2 sin θ sin β, 2U 2 sin θ cos(θ − β ), 0, (c), (d), 2g, g cos 2 β, 47. What will be the velocity of the boat along, the straight line when he reaches the other 53. after what time should the shell be fired such, that it will go vertically up?, bank, , (, , (, , a) v 0, , ), , (, , b), , 2v0, , c), , ), , ), , v0, 2, , d) 2c0, , (a), (c), , 182, , u cos θ, g sin β, , u cos (θ + β ), g cos β, , (b), (d), , u sin (θ + β ), g cos θ sin β, , u cos (θ + β ), g sin β cos β
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , Passage: 4, , 58. The magnitude of the velocity of the point as a, function of time is, When we analyse the projectile motion from any, uur uur, uur, 2, 2, acccelerated frame O as rO , u O and aO, (A) a 1 + (1 − α t ), (B) a 1 + (1 − 2α t ), respectively, express the following terms;, r, r r r, r r, r, r r, 2, 2, (C) 2a 1 + (1 − α t ) (D) 2a 1 + (1 − 2α t ), rpO = rp − rO , u pO = u p − uO and apO = ap − aO ,, where P stands for projectile. Then using the Passage - 6:, following kinematical equations of the projectile (For, At time t = 0 , the position vector of a particle, constant acceleration) relative to the accelerating, moving in the x − y plane is 5i$ m. By time t = 0 ., 1r 2 r, r, r, frame, we have s pO = u pOt + a pO t , v pO, 62 sec , it’s position vector has become, 2, r, r, rr, 5.1 $i + 0.4 $j m. with this data answer the, = u pO + a pOt and v 2pO = u 2pO + 2a.s pO, Using the above expressions, answer the following, following questions., question: A projectile has initial velocity v0 relative 59. The magnitude of the average velocity during, the above time interval., to the large plate which is moving with a constant, (B) 0.206 m / sec, (A) . 0206 m / sec, upward accelertion a., v0, (C) 20.6 m / sec, (D) 2.06 m / sec, 60. The angle θ made by the average velocity with, rest, the positive x axis, B, A, θ0, a, (B) tan −1 ( 3), (A) tan −1 ( 2 ), 54. Which of the following remain/s equal for the, (C) tan −1 (1), (D) tan −1 ( 4 ), observers A and B ?, (a) Maximum height, (b) Range, Passage - 7:, (c) Time of flight, (d) Angle of projection, The position vector of a particle at time t is given, r, 55. Refering to Q.1, velocity of the projectile, by, r = 2t $i + 5t $j + 4sin ωt k$ where ω is a, relative to B ofter some time, constant . Answer the following questions, (a) < v0 at an angle θ < θ0, 61. Velocity vector of the particle is, (A) Constant in magnitude but with variable direction, (b) > v0 at an angle θ > θ 0, (B) constant in direction must variable with, (c) > v0 at an angle θ = θ 0, magnitude, (C) constant, (d) v0 at an angle θ = θ 0, (D) varying with magnitude an well as direction, Passage - 5:, 62. Velocity vector is perpendicular to ...... vector, A point moves in the plane xy according to the law,, (A) 2 $i + 4 $j (B) 3 $i + 2 $j, x = a sin ωt , y = a (1 − cos ωt ) Answer the, (C) 5 $i − 2 $j (D) None, following question taking a and ω as positive, 63. Acceleration of the particle is, constant, (A) Constant in magnitude but variable with direction, 56. The distance travelled by the point during the, (B) constant, time T is, (C) Constant in direction but variable with magnitude, (A) 2aωT (B) 3aωT, (C) 4aωT (D) aωT, (D) Varying with magnitude as well as direction, 57. The equation of the trajectory of the particle is, , (, , (A) y = x −, , x 2α, a, , (B) y = 2 x −, , (C) y = x −, , x 2α, 2a, , (D) y = x −, , x 2α, a, , 2x 2α, a, , ), , MATRIX MATCHING TYPE QUESTIONS, 64. Two particles A and B moving in x-y plane, are at origin at t=0sec. The initial velocity, vectors of A and B are vA = 8i m / s and, , vB = 8j m / s . The acceleration of A and B aree, 183
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, a A = −2i m / s 2, , constant and are, , and, , V R / M = 4m / s, , a B = −2 j m / s . Column-I given certain, 2, , M, , r), , 4m/s, 30°, 30, °, , statements regarding particle A and B, Column-II given corresponding results., Match the statements in Column-I with, corresponding results in Column-II., Column-I Column-II, (a) The time ( in secs) at which, (p) 16 2, velocity of A relative to B is zero, (b) The distance (in m) between A (q) 8 2, and B when their relative velocity is zero, (c) The time (in sec) after t =0 at, (r) 8, which A and B are at same position, (d) The magnitude of relative, velocity of A and B at the instant, they are at same position, (s) 4, 65. Consider 5 different situations a man M, ur, moving and rain as observed by him. V R →, ur, velocity of rain, V R / M → velocity of rain relative 66., ur, to man, V M → velocity of man The situations, are shown on right hand column, Column - I, a) VR lies in which of the following ranges, , V R / M = 5m / s, , M, , s), , 2m/s, 60°, V R / M = 5m / s, , M, , t), , 3m/s, 60°, , A particle projected onto an inclined plane:, , vx'', , v′x, , C, B, vy', , x, , 3.3 m/s ≤ VR ≤ 4.3 m/s, b) 4.3 m/s < VR ≤ 5.3 m/s, c) 5.3 m/s < VR ≤ 6.3 m/s, d) 6.3 m/s < VR ≤ 7.3 m/s, Column - II, , vx, , vy, A, Column-I, , V R / M = 4m / s, , v 'y, , θ x, Column-II, , (a) v, y, , (p) > 1, , (b) t AC, , (q), , g cos θ, , x, x', , (r), , vx − v ' x, g sin θ, , 30°, , p), , M, , 3m/s, , V R / M = 4m / s, , (c), , t AB, (d) t, BC, , q), , M, , 2m/s, , 2v y, , (s) 1, , 67. A projectile is thrown at an angle θ with the, horizontal with a initial velocity v0. If the, magnitude of velocity of the projectile and time, 2, , v2 b , c2, −, t, −, =, , then, are related as 2 , , a a, a2, 184
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, Column-I, (a) Range is, , MOTION IN A PLANE, , Column-II, (p) c, , 72. A body is projected up from the bottom an, inclined plane with a velocity 3 3 m/sec which, makes an angle 600 if the horizontal. The angle, 2b, (b) Height, (q), of projection is 300 with the plane then the time, a, of flight when it strikes the same plane is 0.1x., Then the value of x is, 2bc, (c) Time of flight is, (r), 73., A ball is thrown with a velocity whose horizontal, a, component is 12ms -1 from a vertical wall, 2, b, 18.75m high in such away that it just clears, (d) Velocity at highest point (s), the wall. At what time will it reach the ground, 2a, ? (g = 10ms-2), INTEGER ANSWER TYPE QUESTIONS, 74. A golfer standing on level ground hits a ball, 68. The distance between two moving particles, with a velocity of u = 50ms −1 at an angle α, at any time is a= 32m. If v be their relative, 5, velocity and v1 = 4 m/s and v 2 =8m/s be the, above the horizontal. If tan α = , then the, 12, components of v along and perpendicular, time, for, which, the, ball, is, at, least, 15 m above, to a. The time when they are closest to each, other is ( in meter), the ground will be (take g g = 10ms −2 ), 69. Airplanes A and B are flying with constant 75. A particle is projected from a stationary trolley., velocity in the same vertical plane at angles, After projection, the trolly moves with velocity, 0 and, 0 with respect to the horizontal, 30, 60, 2 15m / s. For an observer on the trolley, the, respectively as shown in figure. The speed, direction of the particle is as shown in the figure, while for the observer on the ground, the ball, of A is 100 3m / s. At time t=0 s, an observer, rises vertically. The maximum height reached, in A finds B at a distance of 500 m. The, by the ball from the trolley is h metre. The, observer sees B moving with a constant, value of h will be, velocity perpendicular to the line of motion, of A. If at t = t0 , A just escapes being hit by, B, t0 in seconds is, , (adv 2014), , V, , (W.r.t Trolley), 60°, 10m/s, , A, 76. A projectile is launched at time t = 0 from point, B, A which is at height 1m above the floorw i t h, speed v m/sec and at an angle θ = 450 with, the floor. It passes through a hoop at B which, is 1 m above A and B is the highest point of, the trajectory. The horizontal distance, 60°, 30°, between A and B is d metres. The projectile, 0, then, falls into a basket, hitting the floor at C a, 70. A rock is launched upward at 45 . A bee moves, horizontal distance 3d metres from A. Find l, along the trajectory of the rock at a constant, (in, m)., speed equal to the initial speed of the rock., B, The magnitude of acceleration of the bee at, the top point of the trajectory is xg ? For the, 45°, rock, neglect the air resistance. Find the value, A, of x ., l, C, 71. A balll is thrown horizontally from a height of, d, O, 3d, 20 m. If hits the ground with a velocity of ‘3’, times the velocity of projection. The velocity, of projection is 3.5x m/s, then x is, 185
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , uuur, r, v r/p = OB = velocity of rain w.r.t. person., , LEVEL-VI - KEY, SINGLE ANSWER TYPE, 1) A 2)A, 8) a, 9) d, 14(a) b 15) c, 21) b 22) b, 27) a 28) a, , 3) C, 10) a, 16) b, 23) a, 29) a, , 4) A, 11) c, 17) b, 24) c, , 5) B, 12) b, 18) c, 25) c, , 6) A 7)A, 13) b 14) a, 19) a 20) a, 26) d, , 60°, D −V p, , 31)AC, 35) A,D, 39) a,b,c,d, 43) a,b,c,d, , 32)BD 33)B,D, 36) A,C 37) a,b,d, 40) a 41) a,d, 44) b,d 45) a,c, , B, , 46)C 47)A 48) b 49) c 50) c 51) b 52) d, 53) d 54) d 55) b 56) d 57) a 58) b 59) c, 60 ) d 61) d 62) c 63) d, , MATRIX MATCH, , ∆OCB,, , 64)a-s,b-p,c-r,d-q, 65) A – p, B – q,t, C – s, D – r, 66) a-s b-q,r c- p d-s 67) a-r b-s c- q d-p, , INTEGER TYPE, , CB, 20, 40 40 3 -1, =, =, =, ms ., 0, sin 60, 3, 3/2, 3, r, r, 2. Speed of rain w.r.t.the person v r/p = OB, , Let the velocity of the drops above the person, rel. to the merry go round be at an angle α to the, vertical. This angle can be determined from the, v 0 = vrel + v m.g.r ,, velocity, triangle, , v0, ., rω, , ⇒ OB = CBcot 600 =, 4., , V RM, , vx, , 60°, , d, v 2 − v r2, vA × A, vA, , Vm, ∧, ∧, ∧, r, r, v R = 10 i − 10 3 j ;, vm = vx i ;, ∧, ∧, r, v RM = (10 − v x ) i − 10 3 j, , Angle with the vertical = 30r, tan 300 =, , 3., , 186, , 10 − vx, 10 3, , ⇒ v x = 20 m / s, , Given θ = 600 and velocity of person, uuur, r, v P = OA =20ms -1. This velocity is same as the, velocity of person w.r.t.ground. First of all let’s, see how the diagram works out., , CB, vt, = tB + r B, walking speed, u, , d, d vr t B, d, vd, =, +, =, + r, v A cos θ vB, u, vB u vB, , 10 3, 60°, , 20 20 3 −1, =, ms, 3, 3, , According to problem,, t A = t B + time taken in walking from C to B., t A = tB +, , 30° 30°, 2., , OB, = cot 600, CB, , From ∆OCM,, , 10, V R = 20m / s, , CB, = sin 60 0, OC, , ⇒ OC =, , 75) 9, , LEVEL-VI - HINTS, , v rel = v0 − v m.g.r vm.g.r = rω , cot α =, , C, , r uuur, v r = OC = velocity of rain w.r.t .earth, r, r, Values of v r and v r/p can be obtained by using, simple trignometric relations, r uuur, 1. Speed of rain drops w.r.t.earth v r = OC from, , COMPREHENSION TYPE, , 1., , Vr, , Vr / p, , 68)3 69)5 70) 2 71) 4 72) 6 73) 3 74) 5, 76) 3, , A, , O, , MULTIPLE TYPE, 30)BD, 34)B,C,D, 38) a,b,c,d, 42)a,b,c,d, , Vp, , u=, , d vr d, +, v B uv B, , v0, v , 1 − , v , 2, 0, 2, , 5., , =, , −, , 1, 2, , = 3.0km / hr, −1, , a) Let the two meet at a distance s from ground., 1, 2, Then s − 12 = 18t − x9.8xt ...(i), 2, and s − 5 = 2t ...(ii) Solving these two equations,, we get t = 3.65s and s= 12.30m, b) vb = 18-(9.8)(3.65) = - 17.8 m/s i.e, velocity, of ball is 17.8 m/s(downward) at the time of, impact or relative velocity= 19.8 m/s (downward)
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, , t=, , 2 gh ± (2 gh ) 2 − 4 ×, g, 2×, 2, , g, h, h, , & t2 =, , 2h, ( 2 + 1), g, , Where t 1 and t 2 correspond to P and Q in the figure., Suppose v is the horizontal velocity of the bird., Then PQ = vt2 ., 11. Equation of ball,, y, N, , u2, u2, 2, 2, =, θ, +, sin, θ, cos, (, ) g = Rmax, g, 13. Suppose height of line A is y from the ground.For, the motion of body along vertical direction we have,, =, , Height, B, , TB, , h, , tA, , A, , p, , time, , y = ut −, , O, ground, level, , 53° M, Ist, , (110, 0), , X, t=, , 100m, , y = x tan θ −, , u ± u2 − 4 ×, , 2, , gx, Substituting the values,, 2u cos2 θ, 2, , y = 1.33x − 0.0113x, ..... (1), Slope of line MN is 1 and it passes through point, (110 m, 0). Hence the equation of this line can be, 2, , 1 2, g t2, g t or, − u t + y = 0 ...(i), 2, 2, , 2×, , g, ×y, 2, , g, 2, , ...(ii), , Let t1 and t2 are time the body passes the two, points of the same horizontal line,, Then t A = t2 − t1 From (ii),, , u − u 2 − 2 gy, u + u 2 − 2 gy, ..... ( 2 ), written as, y = x − 110, t1 =, t, =, and 2, g, g, Point of interrsection of two curves is say P. Solving, 2, (1) and (2) we get positive value of y equal to 4.5, u + u 2 − 2 gy u − u 2 − 2 gy , 2, 2, ∴ t A = ( t2 − t1 ) = , −, , m. i.e.,, y P = 4.5, g, g, , , Height of one step is 1 m. Hence, the ball will collide, 2, somewhere between y = 4 m and y = 5m. which, 2 2, , 4, comes out to be 6th step, u − 2 gy = 2 ( u 2 − 2 gy ), =, g, g, , u 2 sin 2θ, 12. We know that horizontal range, R =, 4 2, , g, 2, Similarly, t B = 2 u − 2 g ( h + y ) , g, , u 2 sin 2 θ, and maximum height h =, 8h, 2, 2, 2g, Now t A − tB = −, g, 2, 2, u sin 2θ , , , g, u 2 sin 2 θ , R, ∴ + 2h = 2 2 + 2 , 8h, 2 g , u sin θ , , 8, , 2g , 2, , u 4 ( 2sin θ cos θ ) u 2 sin 2 θ, =, +, u 2 sin 2 θ, g, 2, g ×8, 2g, , 8h, ∴ Magnitude of g = t 2 − t 2, B, A, , 14. ν y =, , 2, , 188, , νx =, , dy, = 2gy, dt, , .... (1), , dx, = ay, dt, , .... ( 2 )
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, 1, 2, travelled by bomb = ´10´ 2 = 20m., 2, In 2 sec. horizontal distance travelled by Hunter, = 10 ´ 2 = 20 m., Time remaining for bomb to hit ground, , Dividing (1) by ( 2 ) ,weget, 2gy, 2, =, ay, y, , dy, =, dx, , ( g = 10 m / s, , 2, , and a = 5 s −1, , ), , 2´80, - 2 = 2 sec., 10, Let V x and Vy be the velocity components of bullet, along horizontal and vertical direction. Thus we use., 2Vy, = 2 Þ Vy = 10 m/s and, g, =, , y.dy = 2dx, , or, ∫ 0400m, , or, , y .dy = 2 dx, , 2 3 / 2 400m, ( y )0 = 2x, 3, 1, 3/2, x = ( 400 ), 3, 8, x = × 10 3 m, 3, x = 2.67 km, , or, or, or, or, , 20, = 2 Þ Vx = 30 m/s, Vx - 20, Thus velocity of firing is, , X, y, H = 400m, , υx, υy, Y, , 1 2, 14(a). If it is being hit then d = ν o t + at = ( u cos θ ) t, 2, , V = Vx2 + Vy2 = 10 10 m/s., C) projectile motion on inclined plane, 16. As derive earlier,, v02, R, =, for upward projection, max g 1 − sin β = R1, (, ), ...(i), For downward projection,, v02, = R2 ...(ii), g (1 + sin β ), For a projection of horizontal surface substituting, Rmin =, , β = 0, Then, we have Rmax, , Q, , To establish a relation between R, R1 and R2 , we, , 600m/s, H, , need to eliminate sin β ., , θ = 60°, d, t=, , or, ∴, , v02, = = R ( say ) ...(iii), g, , u cos θ − νo, a/2, t = 5s, 1, H = ( u sin θ ) t − g t 2, 2, H = 1500 3 − 125, H = 2473 m, , 1, Adding R form eq.(i), 1, , 1, 2 1 1, with R from eq.(ii) we have R = R + R Then,, 2, 1, 2, v0, , P, 2R1 R2, ans. R = R + R, 1, 2, , R1, , v0, , O, R2, , Q, β, , 17. It can be observed from figure that P and Q shall, collide if the initial component of velocity ‘P’ on, 15. In 2 sec. horizontal distance travelled by bomb, inclined plane i.e. along incline u = 0 that is particle, = 20´ 2 = 40m. In 2 sec. vertical distance, is projected perpendicular to incline., ⇒, , 189
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, gT cos q, = 10m / s, 2, 18. After the elastic collision with inclined plane the, projectile moves in vertical direction. The inclination, of plane with horizontal is 45o. Hence velocity of, particle just before collision should be horizontal., Time required to reach maximum height, 2u, , \ Time of light T = g cos q \ u =, , C, 45°, 45°, , uy = ucosθ, , B, u, A, , 45°, , θ, , cos θ 2sin θ tan α, = cot θ − 2 tan α, sin θ, 3 tan α, ∴ cos θ =, 1 + 9 tan 2 α, , tan α =, , 1 + 9 tan 2 α, , 1, θ, 3tanα, Total time taken by paticle is equal to the sum of, time taken from O to P and P to Q and then from Q, to P to Q. Thus total time, = 2 T + 2t=2(T+t) For t, we hae 0 = v, v, - gt or t =, g, , 2v sin θ v , = 2, + , u sin q u cos q, Total, time, ∴, = t AB + t BC =, +, g cos α g , g, g, 2v sin θ 1 v cos θ 2v sin θ , 19. The path of the motion of particle is as shown in, = 2, + , −, , figure., cos α , g cos α g sin α, Q, 3tan α , v, 2v , y, , v, α, 2v cos θ, 1 + 9 tan 2 α , , x, =, =, g sin α, g sin α, p, g sinα, g cosα, 6v, θ α, o, After solving, we get Total times =, g 1 + 8sin 2 α, Let particle is projected at an angle θ with the plane., Its displacement along y-axis becomes zero in time, 2vy 2v1 sinθ, 2v, , T2 = 2, T, =, =, 1, T. Then we have, ay, g cosθ, g cosθ, 2, v, sin, θ, 1, 20. T1 v1 sinθ, y = uy T + ayT 2 or T =, . .. (i), =, g cos α, 2, T2, v2, Let v is the velocity with which particle strikes the, 2Usin θ, 1, plane α is the angle which it makes with the vertical., = 3, S = gt 2 h = 15m, 21. t =, g, Then we have, v sin α = v cosθ − g sin θ T ...(ii), 2, 22. Particle will strike the point B if velocity of particle, and v cos α = v sin θ − g cos θ T ...(iii), with respect to platform is along AB or component, 2v sin θ, of its relative velocity along AD is zero. i.e.,, From(ii), v sin α = v cosθ − g sin α ×, or, g cos α, u cos ? = ?, v cos θ 2v sin θ, ?, or, ? = cos −1 , −, ...(iv), u, sin α, cos α, Substituting the value of T in equation (ii) and (iii), 23. Components of velocities of both the particles in, vertical direction are equal. Therefore, their time, 2v sin θ , of flights are equal and their relative motion is in, v cos α = v sin θ − g cos α , ...(vi), g, cos, α, , , horizontal direction only. Thus, the maximum, Dividing equation (v) by (vi), we have, distance between them is equal to the difference, in their horizontal ranges., v=, , 190
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, 42., , dx, t, t, = 1−, 1− = 0, dt, 2, 2, 2, t, x=t−, x = 1m, 4, total distance = 2m., , MOTION IN A PLANE, , B, , t = 2s, , v, , w, , 2, = 0.5 m / s ., Average speed Vav =, 24, 43. (a, b, c, d) consider the vertical motion of the particle, after entering the compartment. Let it reach the floor, in time ‘t’, 1, 3.2 = (10) t 2, or t = 0.8 s, 2, Due to the velocity component ‘u’, which remain, constant, it covers a distance of 2.4 m in 0.8 s, 2.4m, 16 m, u=, = 3m / s Also v =, = 20 m / s, 0.8 s, 0.8, v 2 sin 2 α 2 2 sin 2 β, 4, = U, sin 2 α = sin 2 β, 44., 2g, 3, 3, g, 3, 0 ≤ sin 2 α ≤ 1 0 ≤ sin 2 β ≤ 1 β ≤ sin −1 , 4, s i n 2α, RA =, g, T, R max a t α =, 4, U, , 3sin2 α = 4sin2β, 3(1− cos2α ) = 8sin2β, , β =, , 2, , 1, 3, sin − 1 , 2, 8, , RA ≠ H max.B, 45. Let u = the speed of projection, θ1 and θ2 = the, ur, ur, angles of projection. Let r1 and r2 be the position, ur, vectors of the two particles in a ground frame . r1, , Y, α, , vR, , X, , A, Resultant velocity of boatman should be along AB, uur, r, or perpendicular to AB components of v and v R, should be zero. Hence v cos α = v R, , , 3 −1 , , or ( 2v 0 ) cos α = v0 1 + w y , , , Therefore, resultant velocity along AB is, v y = v sin α or, dy, = ( 2v 0 ) sin α =, dt, , ( 2v0 ), , { (, , 4w 2 − w +, , { (, , )}, , v, = 0 4w 2 − w + 3 − 1 y, w, w, dy, , ∫, or, , 0, , { (, , 4w 2 − w +, , t=, , (, , 2, , 2w, , 2, , )}, , 3 −1 y, , )}, , 3 −1 y, , =, , 2, , wπ, , v0, w, , ∫ dt, t, , 0, , ), , 6 3 − 1 v0, 47. When the boatman reaches the opposite bank, y=w or v R = 3v 0 or v cos α = v R, , Solving this,we get, , Hence, ( 2v 0 ) cos α = 3v0, Hence resultant velocity will be, v = vsin α = ( 2v 0 ) sin 300 , v y = v 0 ., 1 2 $, Passage-2. (48 & 49), $ , = (u cos θ1 ) ; t i + ( u sinθ1 ) t − gt j, Two important aspects to be noticed in this problem, 2 , , are : (1) Velocity of projection of ball is relative to, ur, $ + ( u sinθ ) t − 1 gt 2 $j, man in motion, t, i, 2, r2 = (u cos θ2 ), , 2 , (2) Ball clears the hoop when it is at the topmost, , point, The position vector of one particle with respect to, ur, ur, ur, ur, ur, ur, r ur ur, $, V ball , man = V ball − V man ; V ball = V ball , man + V man, another is r = r1 − r2 = u ( cos θ1 − cos θ 2 ) t i, Now we apply the above relation to x- as well as, $, $, $, $, $, , , y-component of velocity. If ball is projected with, =, at, i, +, bt, j, =, a, i, +, b, j, t, ,, + u ( sin θ1 − sin θ 2 ) t j, , , velocity v0 and angle θ , then x-component of, where a and b are constants., ur, COMPREHENSION TYPE, V ball = (v0 cos θ + 10)m / s y-component of, ur, uur, , r, 3 −1 , V, ball = ( v0 sin θ ) m / s . Since vertical component, v, =, v, 1, +, y, and v = 2v0, 0 , 46. Given R, w, of ball’s velocity is unaffected by horizontal motion, , , of car, we can use formula for time of flight., (12.5sin θ )2, 5 × (2 × 10), = 5m ; or sin 2 θ =, i.e.,, 2g, 12.5 × 12.5, 193
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JEE-ADV, PHYSICS-VOL, -I, JEE MAINS, - VOL - VI, , MOTION IN A PLANE, 4, 3, and cos θ =, 5, 5, 4, , and v0 sin θ = (12.5) × = 10m / s, 5, , or sin θ =, , 2v0 sin θ, Time taken to reach maximum height =, g, 2 × 10, =, = 2 seconds, 10, Horizontal distance of loop from point of projection, = (12.5 cos θ + 10) × 1 = 17.5 m, Passage-3., 50. Velocity along tralley is zero i.e., Vx = U x + a x t, U cos θ, 0 = U cos θ − ( g sin β ) t ; t =, ., g sin β, y, , 51., β, , x x , x 2α, y, =, a, 1, −, ., α, Again, , ⇒ y = x−, a a , a, ur, r, If V and a makes an angle 450 with each other, ur, V must be making an angle 450 with X- a x i s, 1, So Vx = V y ⇒ a = a − 2 aα t0 ⇒ t0 =, α, uur, Displacement ∆x = .1$i + .4 $j m, ur, V avg = Average velocity, .1$i + .4 $j , , m / sec = 5 $i + 20 $j m / sec, .02, , , uuuur, So Vavg = 425 m / sec = 20.6 m, , (, , θ = tan −1, , U x = U cos θ, Vy = V sin θ, , 2Vy, T, =, Time of flight, ay, , =, , 2Vsin θ, g cos β, , 1, 2, 52. Distance moved by shell = ( Vo cos θ ) t + g sin β t, 2, 2, 2Usin θ, 2U sin θ, t=, cos (θ − β ), on solving x =, g cos β, g cos 2 β, 53. Vx = U x ; ( ( a x ) cos β ) t = U cos (θ + β ), , g sin β cos β t = U cos (θ + β ) ; t =, , U cos (θ + β ), g sin β cos β, , (54,55) Conceptual, (56,57,58), , ), , ), , 20, = tan −1 4, 5, , ( 61,62,63 ), r, r = 2t $i + 5t $j + 4ω sin ωt k$, ur, r, ⇒ V = 2 i$ + 5 $j + 4sin ωt k$ ⇒ a = −4ω2 sinωtk$, We can earliy acc/ is varify with magnitude but along, a constant direction. As velocity is not changing, along x axis and y axis , we can early realise a vector, parallel to XY plane which is ⊥r to velocity . say, ur ur, ur ur, ur, x . $i + y $j = A Hence V . A = 0 for V ⊥ A, The above equation gives are solution, x = 5 and y = −2 or x = 10 and y = −4 and so, on ( Realise the above solution in 3D and you will, get a clear idea)., , (, , ), , MATRIX MATCH, , x = a sin ωt ⇒ Vx = aω cos ωt ⇒ a x = aω 2 sin ωt, y = a (1 − cos ωt ) ⇒ Vy = aω sin ωt, , ⇒ a y = aω 2 cos ωt, Again x 2 + ( y − a ) = a 2, So the motion is a circular motion with centre at, ( 0, a) and radius of ‘a’. We can also realise that, 2, , speed at any instant V = V x2 + V y2 = aω =, Constant. So distance travelled in a time T = aωT, 194, , y = at (1 − α t ) ⇒ Vy = a − a 2α t ⇒ a y = −2α a, , (, , x, , a x = −g sin β, a y = −g cos β, , As motion is uniform circular motion velocity is, always perpendicular to accelaration., (59 , 60), x = at ⇒ Vx = a ⇒ ax = 0, , 64. The initial relative velocity of A w.r.t B is, u AB = 8j − 8i m / s , u AB = 8 2m / s, , a AB = ( −2i + 2j) m / s , a AB = 2 2m / s 2, Since B observes initial velocity and constant, acceleration of A in opposite directions. Hence B, observes A moving along a straight line From, frame of B, u AB, Hence time when u AB = 0 in t = a = 4 sec ., AB, The distance between A & B when u AB = 0, u 2AB, g=, = 16 2m, 2a AB
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , NEWTONS LAWS OF MOTION, FRICTION, & UNIFORM CIRCULAR MOTION, SYNOPSIS, Inertia :, Ø, Ø, Ø, , Ø, , Ø, , It is the inability of a body to change its state of, rest or of uniform motion or its direction by itself., Mass is a measure of inertia in translatory motion, Heavier the mass, larger the inertia & vice–versa., Types of inertia: There are three types of inertia., (i) Inertia of rest (ii) Inertia of motion and (iii)Inertia, of direction., Inertia of rest: It is the inability of a body to change, its state of rest by itself ., Ex:When a bus is at rest and starts suddenly moving, forward the passengers inside it will fall back., Inertia of motion: It is the inability of a body to, change its state of uniform motion by itself., Ex: Passengers in a moving bus fall forward, when, brakes are applied suddenly., Inertia of direction: It is the inability of a body to, change its direction of motion by itself ., Ex:When a bus takes a turn, passengers in it experience, an outward force., A person sitting in a moving train, throws a coin, vertically upwards, then, i) it falls behind him, if the train is accelerating, ii) it falls infront of him, if the train is retarding, iii) it falls into the hand of the person, if the train is, moving with uniform velocity., iv) It falls into the hand of the person if the train is at, rest, , Ø, Ø, , Change in momentum of a body in, different cases, Ø, , Ø, Ø, , small time interval Dt ., , r r, r, Change in momentum of body = DP = Pf - Pi, , Where, , Consider a body of mass ‘m’ moving, r, with velocity ‘ v ’along a straight line, Ø, , NARAYANAGROUP, , Case (i) : If it hits a wall and comes to rest,, Change in momentum of the body, , m, v, , Every body continues to be in its state of rest (or), uniform motion in a straight line unless it is acted, upon by a net external force to change its state, It defines inertia,force and mechanical equilibrium., If the net external force on an object is zero , then, acceleration of object is zero., Linear momentum is the product of the mass of a, r, r, body and its velocity. p = mv, , Pi = initial momentum, , Pf = final momentum, uur, ur, r, DP = mv f - mvi, r, r, r, DP = Pf - Pi = Pf2 + Pi 2 - 2 Pf Pi cos q, r, r, where q = angle between Pf and Pi, , ur r, r, r, D P = Pf - Pi = 0 - (mv ) iˆ, ur, r, = -mviˆ ; DP = mv , along the normal and, , Linear momentum :, Ø, , Consider a body of mass m moving with velocity, ur, r, vi and momentum Pi . Due to a collision (or) due, to the action of a force on it suppose its velocity, uur, r, changes to v f and momentum changes to Pf in a, , Newton's First Law ( law of Inertia), Ø, , Linear momentum is a vector.It has the same, direction as the direction of velocity of the body., SI unit: kg m s-1 , CGS unit: g cm s-1, D.F: MLT-1, , away from the wall., Ø, , Case(ii) : If the body rebounds with same speed, ‘ v ’ then q = 1800, 1
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, Ø, Ø, , Ø, , Ø, Ø, Ø, Ø, , Ø, , r, ur, ur, d mv, ur, dp, F net =, (or) F net =, dt, dt, In a system if only velocity changes and mass remain, r, ur, r, dv, = ma, constant , F net = m, dt, In a system, if only mass changes and velocity, ur, r dm, remians constant F net = v, dt, Force is a vector and the acceleration produced in, the body is in the direction of net force,, SI unit : newton (N). CGS unit :dyne., One newton = 105 dyne., D.F=MLT-2, Gravitational units of force: Kilogram weight, (kg wt) and gram weight (gm wt); 1 kg.wt = 9.8, N, 1 gm.wt= 980 dyne., A metallic plate of mass ‘M’ is kept held in mid air, by firing ‘n’ bullets in ‘t’ seconds each of mass ‘m’, with a velocity ‘v’ from below., (a) If the bullet falls dead after hitting the plate then, , UNIFORM CIRCULAR MOTION, , ( ), , mnv, = Mg, t, (b) If the bullet rebounds after hitting the plate with, 2m n v, = Mg, t, W.E-1: A force produces an acceleration16 ms −2, in a mass 0.5 kg and an acceleration 4 ms −2, in an unknown mass when applied separately., If both the masses are tied together, what will, be the acceleration under same force?, Sol. Force F=ma=0.5×16=8Ν when both masses are, joined and same force acts, acceleration is given, , same velocity then, , 1, by a =, , F, 8, =, = 3.2ms −2, 1, m + m 0.5 + (8 / 4), , W.E-2:When the forces F1 , F2 , F3 are acting on a, particle of mass m such that F2 and F3 are, mutually perpendicular, then the particle remain stationary. If the force F1 is now removed,, then find the acceleration of the particle ., Sol. If mass 'm' is stationary under three forces,, , NARAYANAGROUP, , Pf, , r r r, F1 + F2 + F3 = 0, r, r r, F1 = − F2 + F3, , B, r, , (, , θ, , ), , F + F = F1, 2, 2, , A, , 2, 3, , Pi, , Obviously if F1 is removed then the mass will have, acceleration, a =, , F22 + F32, m, , ( or ) a =, , F1, m, , W.E-3:A body of mass m=3.513 kg is moving along, the x-axis with a speed of 5ms-1.The magnitude, of its momentum is recorded as (AIEEE - 2008), Sol. m=3.513kg,v=5ms-1 momentum,, p = mv=3.513×5 =17.565kgms-1, W.E-4:A very flexible chain of length L and mass, M is vertically suspended with its lower end, just touching the table. If it is released so that, each link strikes the table and comes to rest., What force the chain will exert on the table, at the moment ‘y’ part of length falls on the, table ?, Sol. Since chain is uniform , the mass of ‘y’ part of the, M , , chain will be L y . When this part reaches the, table, its total force exerted must be equal to the, weight of y part resting on table + Force due to the, momentum imparted, M , dy 2 gy, M, F = yg + L , L, dt, , =, , Mg, M, y + v. 2 gy, L, L, , dy, , Mg, M, My, y+, 2 gy . 2 gy = 3, g, Q = v =, L, L, L, dt, , , W.E-5: A body of mass 8kg is moved by a force, F =(3x)N, where x is the distance covered., Initial position is x = 2 m and final position, is x =10m. If initially the body is at rest, find, the final speed., [2014E], Sol: F=ma ⇒ F=m, 3x = 8, 10, , dv, dv dx, ⇒ 3 x =m, dt, dx dt, , dv, v ⇒ 3xdx = 8vdv, dx, v, , 10, , v, , 2, 2, 3 xdx = 8 vdv ⇒ 3 x = 8 v , 2 2, 2 0, 2, 0, , ∫, , ∫, , 2, 2, 3[100-4]=8 v ⇒ v =, , 3 × 96, = 36 ⇒ v =6ms-1, 8, 3
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , W.E-6:Sum of magnitudes of the two forces acting, at a point is 16 N. If their resultant is normal, to the smaller force, and has a magnitude 8N,, then the forces are, ( 2012E), Sol. F1 + F2 = 16 ——(1) Resultant force is perpen-, , time., v, , a=, , t, , F 1 , , F 1 , , ⇒ v = 0 e −bt = 0 e− bt , m −b , 0 m b, , F0 0 − bt, F0, =, e −e, =, 1 − e −bt, mb, mb, , dicular to F1, then F22 − F12 = F 2, , (, , F2 2 − F12 = 82 ⇒ ( F2 − F1 )( F2 + F1 ) = 64, , ( F2 − F1 ) ×16 = 64 ————(2), , Solving(1) &(2),we get F1 = 6 N , F2 = 10 N, ur, br, W.E-7:A particle is at rest at x=a. A force F = − 2 i, x, begins to act on the particle. The particle starts, its motion, towards the origin, along X–axis., Find the velocity of the particle, when it, reaches a distance x from the origin., b, d, b, ⇒ ( p) = − 2, 2, x, dt, x, b, d, b, dv dx, = − 2, ( mv) = − 2 ⇒ m, x, dt, dx dt, x, , Sol. F = −, , mv dv = −, v, , x, , ∫ v dv =∫ −, 0, , a, , b, b, dx ⇒ vdv = − 2 dx, 2, x, mx, x, , b, v2 b 1 , dx, ⇒, =, mx 2, 2 m x a, , v, b 1 1, =, −, 2 m x a , 2, , ∴v =, , 2b a − x , , , m xa , , W.E-8:A particle of mass m is at rest at the origin, at time t=0. It is subjected to a force, F(t)= F0 e − bt in the X-direction. Its speed V(t), is depicted by which of the following curves., (AIEEE-2012), , 1) v(t), , Ø, , Ø, , F0, mb, , 3) v(t), , 4) v(t), t, , Ø, t, , Sol: As the force is exponentially decreasing, its, acceleration, i.e,rate of increase of velocity will, decrease with time.Thus, the graph of velocity will, be an increasing curve with decreasing slope with, 4, , Ø, , t, , F0b, m, , (, , t, , ), , So, velocity increases continuously and attains a, F, maximum value, vmax = 0, Ans: 3, mb, W.E-9:A bus moving on a level road with a velocity, v can be stopped at a distance of x , by the, application of a retarding force F. The load, on the bus is increased by 25% by boarding, the passengers. Now, if the bus is moving with, the same speed and if the same retarding force, is applied, the distance travelled by the bus, before it stops is, [2014E], 2, Sol :By using equations of motion v - u2 = 2as, F, F, v 2 − u 2 = −2 s −u 2 = −2 s, m, m, m1 s1, Fs, 2Fs, u2 = 2, ⇒ m= 2 ⇒m∝s ⇒ m = s, m, u, 2, 2, Given s1 = x ,m1 = m, and, 25, m, 5m, ( m) = m +, =, m2 = m +, 100, 4, 4, m, x, 5x, ⇒ 5m / 4 = s ⇒ s2 = = (1.25x ) m, 4, 2, , 2) v(t), t, , ), , 0, , Applications of variable mass :, When a machine gun fires ‘n’ bullets each of mass, ‘m’ with a velocity v in a time interval ‘t’ then, force needed to hold the gun steadily is F =, , F0b, m, , F0, mb, , t, , F F0 − bt, dv F0 − bt ⇒ dv = F0 e −bt dt, = e ⇒, = e, ∫0 ∫0 m, m m, dt m, , Ø, , nmv, t, , When a jet of liquid coming out of a pipe strikes a, wall normally and falls dead , then force exerted, by the jet of liquid on the wall is F=Adv2 A = Area, of cross section of the pipe v = Velocity of jet d =, density of the liquid, If the liquid bounces back with the same velocity, then the force exerted by the liquid on the wall is, F = 2 Adv 2, If the liquid bounces back with velocity v ' then, the force exerted on the wall is F = Adv(v + v′), When a jet of liquid strikes a wall by making an, angle 'θ ' with the wall with a velocity ‘ v ’ and, rebounds with same velocity then force exerted by, the water jet on wall is F = 2 Adv 2 sin θ, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , Ø, , If gravel is dropped on a conveyor belt at the rate, dm, of, ,extra force required to keep the belt moving, dt, dm , with constant velocity ' u ' is F = u , , dt , W.E-10:A gardener is watering plants at the rate, 0.1litre/sec using a pipe of cross- sectional area, 1 cm2. What additional force he has to exert, if he desires to increase the rate of watering, two times?, Sol :F = Ad v 2 =, , ( Av ), , 2, , d, , A, , . If rate of watering of, , plant (A v ) is doubled, it means that the amount of, water poured/sec is doubled which is possible only, if velocity is doubled. Hence, force is to be made 4, times., ∴ Additional force = 3 times initial force, = 3 Adv, , 2, , ( Av ), =3, , 2, , A, , d, , 3 × 0.1× 0.1×103, =, = 3 ×105 N, −4, 10, W.E-11: A liquid of density ρ flows along a, horizontal pipe of uniform cross – section A, with a velocity v through a right angled bend, as shown in Fig. What force has to be exerted, at the bend to hold the pipe in equilibrium?, Sol :Change in momentum of mass ∆m of liquid as it, passes through the bend, , W.E-12:A flat plate moves normally with a speed, v1 towards a horizontal jet of water of uniform, area of cross section. The jet discharges water, at the rate of volume V per second at a speed of, v2 . The density of water is r . Assume that water, splashes along the surface of the plate at right, angles to the original motion. The magnitude, of the force acting on the plate due to the jet is, dp, dm, = ur, Sol. Force acting on the plate F =, dt, dt, V, dm, = A(v1 + v2 )ρ = (v1 + v2 ) ρ, Since Av2 = V ⇒, v2, dt, ( ur = v1 + v2 = velocity of water coming out of jet, w.r.t plate), V, V, F = (v1 + v2 ). (v1 + v2 ) ρ = (v1 + v2 )2 ρ N, v2, v2, ur, Impulse ( J ) :, Ø It is the product of impulsive force and time of act, ion that produces a finite change in momentum of, body., Ø J=Ft = m(v-u) = change in momentum. SI unit: Ns, (or)K g -m s-1; DF: MLT-1, Ø It is a vector directed along the force, Ø change in momentum and Impulse are always in, the same direction., Ø For constant force, J=Ft,, Ø Impulsive force is a variable, then, ur, t2, ur d p, J = ∫ Fdt, F=, dt ,, t1, Ø The area bounded by the force-time graph, measures Impulse., , v, , F, , 45°, , v, , dP = Pf − Pi = 2d ∆mv, dP, dm, F=, = 2 v, ; [ as dm = ρ AdL ], dt, dt, , ( ), , F = 2v, , ( ρ . AdL ) ;, dt, , [ as dL / dt = v], , t1, , ∆t, , t2, , Application of Impulse :, a) shock absorbers are used in vehicles to reduce, the magnitude of impulsive force., b) A cricketer lowers his hands, while catching the, ball to reduce the impulsive force., , F = 2 ρ Av 2, NARAYANAGROUP, , 5
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , W.E-13:Find the impulse due to the force, r, F = ai$ + bt $j , where a=2 N and b=4 Ns-1 if this, force acts from ti=0 to tf=0.3s, tf, 0.3, ur, Sol: J = Fdt = (ai$ + bt $j )dt, ∫, ∫, ti, , 0, , 0.3, , t2 , 0.3, J = a ∫ dt $i + b ∫ t dt $j = a [ t ]0 $i + b $j, 0, 0, 2 0, 0.3, , 0.3, , W.E-16:A bullet is fired from a gun. The force on, a bullet is, F = 600 − 2 × 105 t newton. The, force reduces to zero just when the bullet, leaves barrel.Find the impulse imparted to the, bullet., Sol. F = 600 – 2 × 105 t , F becomes zero as soon as, the bullet leaves the barrel., 0 = 600 – 2 × 105 t ⇒ 600 = 2 × 105 t, t, , t = 3 × 10–3 s ⇒ Impulse = ∫ Fdt, , ( 0.3) × $j = $, = 2 × 0.3 × $i + 4 ×, 0.6i + 0.18 $j NSec, 2, W.E-14:A ball falling with velocity, r, v i = (−0.65$i − 0.35 $j ) ms-1 is subjected to a net, 2, , 0, , t, , =, , ), , a mass of 0.275kg, calculate its velocity, immediately following the impulse, r, r, r, r I, r, r, Sol: mv f − mvi = I ; v f = vi +, m, r, 0.6iˆ + 0.18 ˆj, v f = −0.65iˆ − 0.35 ˆj +, 0.275, , r, v f = −0.65iˆ − 0.35 ˆj + 2.18iˆ + 0.655 ˆj, r, v f = 1.53iˆ + 0.305 ˆj m s − 1, , (, , ), , W.E-15:A body of mass 2kg has an initial speed 5, ms −1 . A force acts on it for some time in the, direction of motion. The force–time graph is, shown in figure. Find the final speed of the body, , dt, , 0, , r, impulse I = 0.6$i + 0.18 $j Ns. If the ball has, , (, , ∫ (600 − 2 × 10 t ), 5, , Ø, , , t2 , = 600t − 2 × 105 , 2 0, , , = 600 × 3 × 10–3 – 105 × 9 × 10–6 = 0.9Ns, Equilibrium: The necessary and sufficient, conditions for the translational equilibrium of the, rigid body., ∑ F = 0 ; ∑ Fx = 0 , ∑ Fy = 0, ∑ Fz = 0 For, rotational equilibrium, ∑ τ = 0 ; ∑ τ x = 0 , ∑ τ y = 0, ∑ τ z = 0, r, As for, F = 0 mar = 0 (or) m ( dv / dt ) = 0, , dv, = 0 (or) vr =constant or zero, dt, If a body is in translatory equilibrium it will be either, at rest or in uniform motion.If it is at rest, the, equilibrium is called static,otherwise dynamic., If ‘ n ’coplanar forces of equal magnitudes acting, simultaneously on a particle at a point, with the angle, between any two adjacent forces is ‘ θ ’ and keep, as m ± 0,, , Ø, Ø, , F(N), , it in equilibrium, then θ =, A, , 4, , B, C, , 2.5, F, , G, , H E, 4 4.5 6.5, , 2, , O, , D, , t → (sec), , 1, Sol. Area of OAF = × 2 × 4 = 4, 2, , 3×10 −3, , 360, n, , Lami’s Theorem :, Ø, , If an object O is in equilibrium under three, uur uur, uur, concurrent forces F1 , F2 and F3 as shown in, F1, F, F, = 2 = 3, sin α sin β sin γ, , figure. Then,, , Area of ABG F = 2 × 4 = 8, , F2, , 1, 2, , Area of BGHC = ( 4 + 2.5) × 0.5 = 1.625, Area of CDEH = 2 × 2.5 = 5, Total area under F-t graph = Change in momentum, ⇒ m(v – u) = 18.625, 18.625, + 5 = 14.25ms −1, ⇒ v=, 2, 6, , γ, , α, , F1, , β, F3, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, Ø, , UNIFORM CIRCULAR MOTION, , If the bob of simple pendulum is held at rest by, applying a horizontal force ‘F’ as shown in fig, , θ, , l, T, , WE-18:A mass M is suspended by a weightless, string. The horizontal force required to hold, the mass at 600 with the vertical is (2013E), Sol :, , T cos θ, , θ, , T sin θ, , l T cosθ, , θ, , T θ, T sin θ, , F, mg, , Mg, , - - - - - (1), F = T sin θ, - - - - - (2), Mg = T cos θ, Dividing Eq.(1) and Eq.(2), , If body is in equilibrium, T cosθ = mg ,, T sin θ = F ,, F = mg tan θ ,, x l, l −x, = =, F T, mg, 2, , F 2 + (m g ) = T, 2, , F, T sin θ, =, ; F = Mg tan θ, Mg T cos θ, , 2, , WE-17:A mass of 3kg is suspended by a rope of, length 2m from the ceiling. A force of 40N in, the horizontal direction is applied at midpoint, P of the rope as shown. What is the angle the, rope makes with the vertical in equilibrium, and the tension in part of string attached to, the ceiling? (Neglect the mass of the rope,, g = 10m/s2), Sol :Resolving the tension T1 into two mutually, perpendicular components, we have, T1 cos θ = W = 30 N, , T1 sin θ = 40 N, , F = Mg tan 600 ; F = 3Mg, W.E.19:A chain of mass 'm' is attached at two, points A and B of two fixed walls as shown in, the figure. Find the tension in the chain near, the walls at point A and at the mid point C., A, , B, , T1, θ, , 1m, , C, , Sol., i) At point A, , T sin θ, 2T sin θ, , θ, T cos θ, , θ, T cos θ, , θ, , θ T cos θ, , T, θ, T cos θ, , 40 N, , w, , 30 N, , The tension in part of string attached to the ceiling, T1 = W 2 + F 2 = 302 + 402 = 50 N, NARAYANAGROUP, , T sin θ, , T, , T1cos?, , P, T, 1m 2, , θ, , θ, , 4, 4, ∴ tan θ = (or) θ = tan −1 = 530, 3, 3, , T1sin?, , F, , X, , X, , 1, mg cos ecθ, 2, ii) Tension along horizontal direction is same everywhere, Q(no external force is acting on it in horizontal direction.), At point C, mg cos θ mg cot θ, T 1 = T cosθ =, =, 2sin θ, 2, 2T sin θ = mg ⇒ T =, , 7
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , Ø, , Newton's third law:, Ø, Ø, Ø, , Ø, Ø, Ø, , Ø, Ø, Ø, Ø, , Ø, Ø, Ø, , For every action there is always an equal and, opposite reaction, Action and reaction do not act on the same body, and they act on different bodies at same instant of, time, Action and reaction, known as pair of forces, are, equal in magnitude and opposite in directions acting, on different bodies in interaction. So they never, cancel each other, Newton’s third law is not applicable to pseudo, forces., Newton’s third law defines nature of force and gives, the law of conservation of linear momentum., Examples:, When we walk on a road we push the road, backwards and road applies equal (in magnitude), and opposite force on us, so that we can move, forward., When we swim on water we push water backward, and water applies equal (in magnitude) and opposite, force on us,so that we can move forward., A bird is in a wire cage hanging from a spring, balance. When the bird starts flying in the cage,, the reading of the balance decreases., If the bird is in a closed cage (or) air tight cage and, it hovers in the cage the reading of the spring, balance does not change., In the closed cage if the bird accelerates upward, the reading of the balance is R = Wbird + ma, Limitations of newton’s third law:Newton’s third law is not strictly applicable for the, interaction between two bodies separated by large, distances, of the order of astronomical units., It does not apply strictly when the objects move, with velocity nearer to that of light, It does not apply where the gravitational field is, strong., Normal reaction/force : Normal force acts, perpendicular to the surfaces in contact when one, body tries to press on the surface of the second, body.In this way second body tries to push, away the first body., , When the body lies on a horizontal surface N = mg, N, , mg, , Ø, , When the body lies on an inclined surface, N = mg cos ?, ., N, , mg cosθ, , A, , A, N, , A, , 8, , θ, , are connected by strings, springs, surfaces of, contact, then all the forces acting on a body are, considered and sketched on the body under, consideration by just isolating it . Then the diagram, so formed is called Free Body Diagram (FBD)., , Some examples:, i) A block is placed on a table and the table is kept, on earth.Assuming no other body in the universe, exerts any force on the system,make the FBD of, block and table., m1, m2, , FBD of block,, FBD of table, , N1 = m1 g, , N 2 = N1 + m2 g = m1 g + m2 g = ( m1 + m2 ) g, N1, m1, N1, m2g, , A, , B, , mg sinθ, , mg, , Free Body Diagram:- When several bodies, , m1g, , B, , θ, , B, , B, , N2, , ii) A block of mass M is suspended from the ceiling, by means of a uniform string of mass m.Find the, tension in the string at points A,B and C. B is the, mid point of string. Also find the tensions at A,B, and C if the mass of string is negligible or it is, massless., NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , d) Equation of motion of mass ‘m’ moving with, acceleration ‘a’ relative to an observer in an inertial, , A, , r, , frame is ∑ Freal = ma, , m, , B, , Examples:, , C, M, , Tension at any point will be weight of the part, below it., , m, , So, TA = ( M + m) g,TB = + M g,TC = Mg ., 2, , Now if the string is massless: m=0 then, TA = TB = TC = Mg . So in a massless string,, tension is the same at every point., (iii) Find the tension in the massless string, connected to the block accelerating upward., T, T, m, , m, , mg, , mg, , a, , Net force :, Fnet = T − mg, , Now apply Fnet = ma, , ⇒ T − mg = ma ⇒ T = mg + ma = m ( g + a ), , Note: If ‘a’ is downward, then replace a with, -a; we get T = m ( g − a ), In free fall a=g then T=0., , Frames of Reference:, Ø, , A system of coordinate axes which defines the, position of a particle or an event in two or three, dimensional space is called a frame of reference., There are two types of frames of reference, a) inertial or unaccelerated frames of reference, b) non-inertial or accelerated frames of reference, , Inertial frames of reference :, a) Frames of reference in which Newton’s Laws, of motion are applicable are called inertial frame., b) Inertial frames of reference are either at rest or, move with uniform velocity with respect to a fixed, imaginary axis., c) In inertial frame, acceleration of a body is caused, by real forces., NARAYANAGROUP, , 1) A lift at rest,, 2) Lift moving up(or)down with constant velocity,, 3) Car moving with constant velocity on a straight, road., Real Force : Force acting on an object due to, its interaction with another object is called a real, force., Ex: Normal force, Tension, weight, spring force,, muscular force etc., a) All fundamental forces of nature are real., b) Real forces form action, reaction pair., , Non-Inertial frames :, a) Frames of reference in which Newton Laws are, not applicable are called non-inertial frames., b) Accelerated frames move with either uniform, acceleration or non uniform acceleration., c) All the accelerated and rotating frames are noninertial frames of reference., , d) Examples:, 1) Accelerating car on a road., 2) Merry go round., 3) Artificial satellite around the earth., , Pseudo force :, a) In non-inertial frame Newton’s second law is, not applicable. In order to make Newton’s second, law applicable in non-inertial frame a pseudo force, is introduced., r, b) If a is the acceleration of a non-inertial frame,, the pseudo force acting on an object of mass m, as, measured by an observer in the given non-inertial, ur, r, frame is F Pseudo = −ma, i.e. Pseudo force acts on an object opposite to the, direction of acceleration of the non-inertial frame., c) Pseudo forces exist for observers only in noninertial frames, such forces have no existence relative, to an inertial frame., d) Equation of motion relative to non-inertial frame, ur, , ur, , ur, , is ∑ ( F real + F Pseudo ) = ma′, Where a′ is the acceleration of body as measured, in non-inertial frame., e) Earth is an inertial frame for an observer on the, earth but it is an accelerated frame for an observer, at centre of earth (or) in a satellite., Examples : (i) Centrifugal force and deflection, of pendulum relative to accelerating car.(ii) Gain or, loss of weight experienced in an accelerating, elevator., 9
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , Apparent weight of a body in a moving elevator, Weight of a body on a surface comes due to the, reaction of a supporting surface, i.e.,Apparent, weight of a body in a lift, Wapp = Reaction of supporting surface. Consider, a person standing on a spring balance , or in a lift., The following situations are possible:, Case(i) :If lift is at rest or moving with constant, velocity then the person will be in translatory, equilibrium. So, R = mg, [as Wapp = R ], ∴ Wapp = mg, or Wapp = W0 [as W0 = mg = true weight], R, , R, , R, a, , 70 kg, mg, (a), , 80 kg, mg, (b), , a, , 60 kg, mg, (c), , i.e., apparent weight (reading of balance) will be, equal to true weight., Case(ii) : If lift is accelerated up or retarding down, with acceleration a from Newton's II law we have, R − mg = ma or R = m ( g + a ), or Wapp = m ( g + a ), , a, a, = mg 1 + = W0 1 + or Wapp >W0, g, g, i.e., apparent weight (reading of balance) will be, more than true weight., Case (iii) : If lift is accelerated down or retarding, up with acceleration ‘a’ mg − R = ma i.e.,, , R = m ( g − a), , a, or Wapp = m ( g − a ) [as Wapp = R ] = mg 1 − g , , , a, i.e., Wapp = W0 1 − Wapp <W0, g, i.e., apparent weight (reading of balance) will be, lesser than true weight., Note: If a > g ,Wapp will be negative; negative, weight will mean that the body is pressed against, the roof of the lift instead of floor (as lift falls more, faster than the body) and so the reaction will be, downwards, the direction of apparent weight will, be upwards., 10, , Case (iv) : If lift is in freely falling, Then a=g ,, So mg − R = mg i.e., R = 0 . So, Wapp = 0, v, , v, Satellite, a=g, , a=g, , Planet, a=g, , ( a), , (b), , (c), , ( a ) Freely falling lift, ( b ) Satellite motion, ( c ) Projectile motion, i.e., apparent weight of a freely falling body is zero., Ø This is why the apparent weight of a body is zero,, or body is weightless if it is in a (i) lift whose cable, has broken, (ii) orbiting satellite., W.E.20-: A mass of 1kg attached to one end of a, string is first lifted up with an acceleration, 4.9m/s 2 and then lowered with same, acceleration. What is the ratio of tension in, string in two cases., Sol :When mass is lifted up with acceleration 4.9m/s2, T1 = m( g + a) =1 (9.8 + 4.9)=14.7N, When mass is lowered with same acceleration, T2 = m( g − a) =1(9.8 – 4.9)=4.9 N, , ∴, , T1 14.7, =, = 3 :1, T2 4.9, , W.E.21:The apparent weight of a man in a lift is, W1 when lift moves upwards with some, acceleration and is W2, when it is accelerating down with same, acceleration. Find the true weight of the, man and acceleration of lift., Sol :(a) W1 = m( g + a), W2 = m( g − a), W1 + W2 = 2mg ⇒ W1 + W2 = 2W (QW = mg ), ⇒, , W1 + W2, =W, 2, , W1 m( g + a) g + a, (b) W = m( g − a ) = g − a, 2, , W − W2 , g W1 + W2, =, ⇒a = g 1, , a W1 − W2, W1 + W2 , , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , Connecting Bodies:, Ø, , T2, , If masses are connected by strings then acceleration, of system and tension in the strings on smooth, horizontal surface are, , M1, , M2, , T, , M1g, , Free body diagram for M2, , T, , From (1) and (2), T2 = F + (M 1 + M 2 ) g, WE.22: Three blocks connected together by strings, are pulled along a horizontal surface by, applying a force F. If F= 36N, What is the, tension T2?, , T=M2a .....(1), , M2, , Free body diagram for M1, , F, , M1, , T, , F-T=M1a ....(2), , from (1) and (2), F, M 2F, a=, and T =, ( M1 + M 2 ), ( M1 + M 2 ), T1 T1, T2 T2, b) F, M, M1, 2, , T1, , Sol :Suppose the system slides with acceleration ‘a’., m1 = 1kg , m2 = 8kg , m3 = 27 kg, , M3, , F − T2 = m3 a , T2 − T1 = m2 a , T1 = m1a, Solving the above equations ,we get, a=, =, , M1, T1, M2, , F, , Free body diagram for M2, T1, , T1 = F + M 2 g ............(1), , M2g, , Free body diagram for M1, NARAYANAGROUP, , F, m1 + m2 + m3, , 36, 36, =, = 1 ms −2, 1 + 8 + 27 36, , From the above equation, T2 = F − m3 a, , T2, , F, , T2, F, , M3F, T2 =, ( M1 + M 2 + M 3 ), If masses are connected by a string and suspended, from a support then tension in the string when force, F is applied downwards as shown in the figure, , M2, , 27kg, , 8kg, , 1kg, , F, (M 2 + M3 ) F, a=, ; T1 =, M1 + M2 + M 3, (M1 + M2 + M3 ), , Ø, , T2 = T1 + M 1 g ............(2), , T1, , Ø, , T2 = 36 − 27 × 1 =9 N, Contact Forces : When two objects are in contact, with each other, the molecules at the interface, interact with each other. This interaction results in, a net force called contact force. The contact force, can be resolved into two components., (a) Normal force (N): Component of the contact, force along the normal to the interface. Normal, force is independent of nature of the surfaces in, contact., (b) Friction (f): Component of the contact force, along the tangent at the interface. Friction depends, on the roughness of the surfaces in contact. This, component can be minimised by polishing the, surfaces., The tension and contact forces are self adjustable, forces. Their magnitude and direction change when, other forces involved in a physical arrangement, change., 11
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , W.E-24:Figure shows three blocks of mass ‘m’ each, hanging on a string passing over a pulley., Calculate the tension in the string connecting A to B and B to C?, Sol. Net pulling force = 2 mg – mg = mg, Total mass = m+ m + m = 3m, , T, , T1, , T1, , a, , B, a, , T, , A, mg, , Sol. we know that Normal reaction = scale reading, For man , T = Mg − R, , T2, T2, a, , C, , T, , mg, , Acceleration, a =, , T, R, , m, , mg g, =, 3m 3, , R, T2, , a, , a, , C, , Mg, , T1, , mg, , mg, , For box : T = mg + R, Mg − R = mg + R ;, , A, , mg, , Considering block A,, T1 − mg = ma ; T1 = mg + ma, g, 4, T1 = mg + m ⇒ T1 = mg, 3, 3, , R=, , (60 - 30)´10, 2, , 2 R =(M –m)g, , = 150 N, , W.E.26:Two unequal masses are connected on two, sides of a light string passing over a light and, smooth pulley as shown in figure. The system, is released from rest. The larger mass is, stopped for a moment, 1sec after the system is, set into motion. Find the time elapsed before, the string is tight again . (g = 10 m/s2), , Considering block C,, mg − T2 = ma ⇒ T2 = mg − ma, mg 2, = mg ., 3, 3, W.E-25:A man of mass 60 kg is standing on a, weighing machine kept in a box of mass 30, kg as shown in the diagram. If the man manages to keep the box stationary, find the reading of the weighing machine., ⇒ T2 = mg −, , NARAYANAGROUP, , 1kg, 2kg, 13
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , General Constraints:, , mA g sin 60 − T1 = m Aa, , i) A body placed on floor : The floor acting as, , T1 = mA g sin 60 − m Aa = mA ( g sin 60 − a ), , a constraint restricts the kinematical quantities in, the downward direction such that, , , , 3, T1 = (1) 10 ×, − 4.1 = 4.56 N, 2, , , , m, , x, , For the tension in the string between B and C, FBD of body C, , y, , y = 0 ; v y = 0 and a y = 0 for the body placed, , T2, a, , on the floor., ii) Two bodies connected with a string or rod., , C, , inextensible string, , 0, T2 − mC g sin 300 = mC a ; T2 = mC ( g sin 30 + a ), , a, , A force F is applied on the massless pulley as shown, in the figure and string is connected to the block on, smooth horizontal surface. Then, T, , T, , T, , T, , F, , m, , Ø, , Ø, , (a) Constraint : Restriction to the free motion, of body in any direction is called constraint., (b) Constrained Body : A body, whose, displacement in space is restricted by other bodies,, either connected to or in contact with it, is called a, constrained body., (c) Kinematic Constraints : These are, equations that relate the motion of two or more, particles., , (d) Types of Constraints :, i) General constraints, iii) Wedge constraints, 16, , ii) Pulley constraints, iv) Mixed constraints, , b, , ds A dsB, =, ⇒ v A = vB, dt, dt, , Again differentiating, , dv A dvB, =, ⇒ a A = aB, dt, dt, iii)Two bodies in contact with each other, Ø, , F= 2T and T = mablock, If the block moves a distance ‘x’ the pulley moves, x/2 (Total length of the string remains constant), a, Therefore acceleration of the pulley = block, 2, T, F /2 F, =, =, ., =, 2m, 2m, 4m, , Constrained Motion:, , B, , A, , The string / rod is inextensible., ∴ Displacements of A and B are equal in horizontal, direction ⇒ s A = s B, Differentiating w.r.t time,, , 1, , T2 = 2 10 + 4.1 = 18.2 N, 2, , , Ø, , B, , A, , mcg sin 30, , inextensible rod, , Displacement of A and B are equal in horizontal, direction., A, , B, , ⇒ s A = sB, By differentiating, we will get, , v A = vB and a A = aB in horizontal direction, Pulley Constraints:, Ø, , For example, the motion of block A is downwards, along the inclined plane in fig. will cause a, corresponding motion of block B up the other, inclined plane.Assuming string AB length is, inextensible, i.e., length of AB is constant., , xA, , A, , B, , xB, , A, , B, β, α, |||||||||||||||||||||||||||||||||||||||||||||||||||||||, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , ∴ The displacements of A ( x A ) and B ( xB ) are, equal ∴ x A = xB, Differentiating w.r.t. time, ⇒ v A = vB, Once again differentiating w.r.t. time, ⇒ a A = aB, i.e., if one body (A) moves down the inclined plane, with certain acceleration, then the other body will, move up inclined plane with an equal acceleration, (magnitude)., Alternate Method : First specify the location, of the blocks using position co-ordinates S A and, SB., C, , D, , SB, , SA, O, , α, , A, , B, , x A = xB ⇒ v A = v B ⇒ a A = a B, Mixed constraints :, Ring sliding on a smooth rod :, Ø, , Consider a ring of mass m connected through a, string of length L with a block of mass M. If the, ring is moving up with acceleration am and aM is the, acceleration of block. As the length of the string is, constant,, , L = d 2 + y2 + x, , B, , A, , 2), , β, , d, , From the fig. the position co-ordinates are related, by the equation s A + lCD + sB = L, , y, , θ, , where lCD = the length of the string over arc, CD = constant L = total length of the string =, constant Differentiating w.r.t. time, we get, , x, , m, , ds A dsB, +, = 0 ⇒ vB = −v A, dt, dt, , M, , The negative sign indicates that when block A has, a velocity downward, i.e., in the direction of positive, s A , it causes a corresponding upward velocity of, , Since, L is constant, differentiating with respect to, time t, we get, , block B, i.e., B moves in the negative s B direction., Again differentiating w.r.t. time,, , dL 1, =, dt 2, , dvB, dv, =− A, dt, dt, , ⇒ aB = −a A, Since, , Similarly, , a, , 1), a, B, , x A = xB ⇒ v A = v B ⇒ a A = a B, NARAYANAGROUP, , (d, , 2, , +y, , ), , 1, 2 2, , dy dx, =0, +, dt dt, , dy, dx, = vm and, = vM and, dt, dt, , cosθ =, , A, , 2y, , y, , so vM = −vm cosθ ., d 2 + y2, By differentiating, relation between am and aM can, be obtained, however, while doing so remember, that cosθ is not constant, but it is variable., Two blocks connected with pulley : If the, blocks are connected as shown in fig, then the, length of the string is, 17
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , W.E-33:In the fig, find the acceleration of m1 and m2, m1, , a1, , ||||||||||||||, T, , l1, , T, , T, , Sol : T sin θ = ma0 ...... ( i ) ; T cos θ = mg ...... ( ii ), a0, dividing (i)&(ii) tan θ =, g, a , , −1, ∴The string is making an angle θ = tan g0 with, , the vertical at equilibrium, Squaring and adding (i) and (ii), T 2 sin 2 θ + T 2 cos 2 θ = m 2 ( a02 + g 2 ), , l2, , 2T, m2, , T = m a02 + g 2, , a2, m2g, , W.E.35: For what value of ‘a’ the block falls freely?, , a1, Sol. l 1 + 2l 2 = constant, , m1, , T, , a1 = 2a2 ; T = m1a1, m2 g − 2T = m2 a2 ; m2 g = 2m1a1 + m2 a2, 2T, m2 g = 2m1 ( 2a2 ) + m2 a2, , θ, , m2, a2, , m2g, m2 g, 2m2 g, a2 =, a1 =, ,, 4m1 + m2, 4m1 + m2, W.E.34:A pendulum is hanging from the ceiling, of a car having an acceleration a 0 with respect, to the road. Find the angle made by the string, with vertical at equilibrium. Also find the, tension in the string in this position., , θ, , h, , a, , a0, , x, Sol :In the time the wedge moves a distance ‘x’ towards, left with an acceleration a the block falls from a, height ‘h’ with acceleration ‘g’, x=, , a, x a, 1 2, 1, at , h = gt 2 ⇒ = , ⇒ cot θ = ⇒ a = g cot θ, h, g, g, 2, 2, , W.E.36:A block of mass m is placed on a smooth, wedge of inclination q . The whole system is, accelerated horizontally so that the block does, not slip on the wedge.Find the i) Acceleration, of the wedge ii) Force to be applied on the wedge, iii) Force exerted by the wedge on the block., Sol. (i). For an observer on the ground :, R cos θ, θ, , R sin θ, , F, , T cos θ, , mg, M, , θ, ma0, , NARAYANAGROUP, , T, T sin θ, , mg, , R, , a, θ, , R sin θ = ma, R cos θ = mg, ⇒ a = g tan θ, ii) F = (M + m)a = (M + m) g tan q, iii) Force exerted by the wedge on the block, mg, or R = mg sec θ, ⇒ R=, cos θ, , Note :If inclination is given as 1 in x, sin θ =, , 1, x, 19
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , tan q =, , ma cos θ, , xq 1, , 1, x 2 -1, , a, ma, , g, , Þ Acceleration a = g tan q =, , x -1, W.E.37:Two fixed frictionless inclined planes, making an angles 300 and 600 with the vertical are shown in the figure. Two blocks A and, B are placed on the two planes. What is the, relative vertical acceleration of A with respect, to B?, (AIEEE-2010), B, , mg sin θ, , ma sin θ θ, mg, mg cos θ, , 2, , A, , θ, , W.E.39:A solid sphere of mass 2kg rests inside a, cube as shown. The cube is moving with ver, , −1, ˆ, locity υ = (5ti$ + 2tj)ms, where 't' is in sec, and ' u ' is in m/s. What force does sphere exert, on cube?, , Y, , 60°, , 30°, , Sol: mg sin θ = ma ⇒ a = g sin θ, where a is along the inclined plane, ∴ vertical component of acceleration is g sin 2 θ, ar = a AB = a A − aB, ∴ relative vertical acceleration of A with respect, g, 2, 0, 2, 0, −2, to B is g ( sin 60 − sin 30 ) = = 4.9 ms, 2, (in vertical direction), , O, , X, , r, , Sol. As given, υ = 5ti$ + 2t $j ;, ∴ ax =, , dυ y, dυ x, = 5, a y =, =2, dt, dt, , W.E.38:For what value of 'a' block slides up the, plane with an acceleration 'g' relative to the, inclined plane., , ay, , ax, mg, , a, Sol., Fnet = ma cos θ − mg sin θ, ma′ = ma cos θ − mg sin θ, If a′ = g , mg = ma cos θ − mg sin θ, , 1 + sin θ , a cos θ = g + g sin θ ⇒ a = g , , cos θ , ⇒ a = g (sec θ + tan θ ), , 20, , When cube is moving with above accelerations, along x and y-axes, the forces that exert on cube, are, Fx = − max = −2 × 5 = −10 N, , (, , ), , Fy = − mg + ma y = − ( 20 + 2 × 2 ) = −24 N, Net force,, , F=, , ( Fx )2 + ( Fy ), , 2, , = (10) 2 + ( 24) 2 = 26N, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , W.E.40:A block is placed on an inclined plane, moving towards right with an acceleration, a0 = g. The length of the inclined plane is l0 ., All the surfaces are smooth. Find the time, taken by the block to reach from bottom to, top., ma cos θ, , ma0, , a0, θ, L, , Sol. Let us solve the problem in the elevator frame. The, free body diagram is shown. The forces are, N, , 30°, mg sin θ, 30°, , ma0sinθ, , Sol. ma = ma0 cos300 − mg sin 300, , mgsinθ, , ma cos300 − mg sin300, a= 0, m, 3, 1, æ 3 -1÷ö, ma0, - mg, ÷, = g ççç, 2, 2, a=, çè 2 ÷÷ø, m, 1 2, 1, l 0 = at 2, from s = ut + at ;, 2, 2, 1 æ 3 -1÷ö 2, 4l 0, ÷÷ t, l 0 = g ççç, 2 çè 2 ÷ø ⇒ t =, sec, g 3 -1, , (, , Sol., , mg sin α, ma0, , mg sin θ + ma0 sin θ = ma ∴ a = ( g + a0 ) sin θ, , ), , This is the acceleration with respect to the elevator, The distance travelled is, , T sin q = ma0 + mg sin a -------------(1), T cos q = mg cos a -------------(2), a + g sin a, 1, tan q = 0, 2 ⇒, g cos a, é a + g sin a ù, ú, q = tan -1 ê 0, êë g cos a úû, , W.E.42:A block slides down from top of a smooth, inclined plane of elevation q fixed in an, elevator going up with an acceleration a 0.The, base of incline has length L. Find the time, taken by the block to reach the bottom., NARAYANAGROUP, , L, cos θ, , . If ‘t’ is the time for, , reaching the bottom of inclined plane, L, 1, = 0 + ( g + a0 ) sin θ .t 2, cos θ, 2, 1, , θ, T sin θ, , α, cos, mg α, , θ, , (iii) ma0 ( pseudo force).acting vertically down, If a is acceleration of the body with respect to, inclined plane, taking components of forces, parallel to the inclined plane., , a0, T, , ma0, , (i) N normal reaction to the plane,, (ii) mg acting vertically downwards,, , W.E.41:A pendulum of mass m hangs from a support fixed to a trolley. The direction of the, string when the trolley rolls up a plane of, inclination a with acceleration a0 is, T cos θ, , mg, , a0, , θ, , 1, , , 2 , 2, 2L, 4L, t=, =, , ( g + a0 ) sin θ cos θ , ( g + a0 ) sin 2θ , , Law of conservation of momentum:, Ø, , Ø, Ø, Ø, , When the resultant external force acting on a system, is zero, the total momentum (vector sum) of the, system remains constant. This is called “law of, conservation of linear momentum”., Newton’s third law of motion leads to the law of, conservation of linear momentum., Walking, running, swimming, jet propulsion, motion, of rockets, rowing of a boat, recoil of a gun etc.,, can be explained by Newton’s third law of motion., Explosions, disintegration of nuclei, recoil of gun,, collisions etc., can be explained on the basis of the, law of conservation of linear momentum., , Applications:, 21
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, W.E.44:A particle of mass 4 m explodes into three, pieces of masses m,m and 2m. The equal, masses move along X-axis and Y- axis with, velocities 4ms-1 and 6 ms-1 respectively . The, magnitude of the velocity of the heavier mass, is, (E - 2009), Sol: M=4 m ,u=0, m1 = m, m2 = m, m3 = 2m, v1 = 4ms −1 , v2 = 6ms −1 , v3 = ?, According to law of conservation of momentum,, ur ur ur, P1 + P2 + P3 = 0, ur, ur ur, ur, ur ur, P 3 = − P1 + P 2 , P 3 = P1 + P 2, , (, , ), , P3 = P12 + P2 2 + 2 P1 P2Cosθ, P1 and P2 are perpendicular to each other, , P3 = P12 + P22 , m3v3 = (m1v1 ) 2 + (m2 v2 ) 2, 2mv3 = (m × 4)2 + ( m × 6 ), , 2, , 2v3 = 16 + 36 ⇒ v3 = 13ms −1, W.E.45:A rifle of 20kg mass can fire 4 bullets/s., The mass of each bullet is 35 ×10−3 kg and its, final velocity is 400ms −1 . Then, what force, must be applied on the rifle so that it does not, move backwards while firing the bullets?, (2007E), Sol :Law of conservation of momentum MV + 4mv = 0, 4mv, , 4 × 35 × 10 × 400, , m, l, θ, , M, L, , Sol : When the block slides down on the smooth wedge,, the wedge moves backwards. In the horizontal, r, direction there is no external force ; Fx = 0, r, ∴ Px =constant, NARAYANAGROUP, , r, r, Pf = Pi (along x-axis) ;, , r r, r, mu + MV = 0, x1 = forward distance moved by the block along, X-axis., x2 = backward distance moved by the wedge along, X-axis., r, r, mu = − MV ;, x, x, m 1 =M 2, t, t, ML, M l cos θ, =, mx1 = Mx2 , x1 =, M +m, M +m, mL, ml cos θ, =, x2 =, M +m, M +m, W.E.47: A bomb of 1 kg is thrown vertically up, with speed 100 m/s. After 5 seconds, it, explodes into two parts. One part of mass, 400gm goes down with speed 25m/s. What will, happen to the other part just after explosion, Sol :After 5 sec, velocity of the bomb,, v = u + at, r, v = u $j – gt $j = (100 – 10 x 5) $j = 50 $j m/s, , m = 1kg , m1 = 0.4kg , m2 = 0.6kg , v1 = 25ms −1, According to law of conservation of, momentum mv = m1v1 + m2 v2, r, , 1 × 50 $j = −0.4 × 25 $j + 0.6v2, ⇒ v = 100 $j = v = 100ms −1 ,vertically upwards, 2, , 2, , −3, , = -2.8 ms −1, ⇒V = − M = −, 20, Force applied on the rifle, MV, 20 × 2.8, F=, =−, = -56 N, t, 1, W.E.46:All surfaces are smooth.Find the, horizontal displacements of the block and the, wedge when the block slides down from top to, bottom., , h, , UNIFORM CIRCULAR MOTION, , W.E-48:A particle of mass 2m is projected at an, angle 450 with horizontal with a velocity of, 20 2 m/s. After 1sec, explosion takes place, and the particle is broken into two equal, pieces. As a result of explosion one part comes, to rest. The maximum height attained by the, other part from the ground is (g = 10m/s2), Sol : M = 2m, θ = 450 , u = 20 2ms −1, 1, = 20 ms − 1, 2, 1, u y = u sin θ = 20 2 ×, = 20 ms −1, 2, u x = u cos θ = 20 2 ×, , But height attained before explosion , H1, = ut −, , 1 2, 1, gt = 20 × 1 − × 10 × 12 = 15 m, 2, 2, , After 1sec, v x = 20 ms −1, v y = u y − gt = 20 − 10 = 10ms −1, , Due to explosion one part comes to rest,, 23
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , frictional force on the rear wheels will be in the, direction of motion and on the front wheels in the, opposite direction of motion, Note-iii:In cycling ,the force exerted by rear wheel, on the ground makes the force of friction to act on, it in the forward direction. Front wheel moving by, itself experience force of friction in backward, direction., Note-iv: If the pedaling cycle is accelarating on, the horizontal surface, then the forward friction on, the rear wheel is greater than the backward friction, on the front wheel., Note-v:When pedaling is stopped,the frictional, force is in backward direction for both the wheels., , m1 = m2 = m, v1 = 0, M ( v x i + v y j ) = m1 v1 + m 2 v 2, 2 m (20 i + 10 j ) = m (0) + m v 2, v 2 = 40 i + 20 j, , v y1 = 20ms −1, Height attained after explosion =, , (v ), =, , 1 2, , H2, , Ø, Ø, Ø, , Ø, Ø, Ø, , 2g, , =, , 20 × 20, = 20m, 2 ×10, , H TOT = H1 + H 2 = 15 + 20 = 35m, Friction: If we slide or try to slide a body over, another surface, the motion of the body is resisted, by bonding between the body and the surface.This, resistance is called friction., The force of friction is parallel to the contact surfaces, and opposite to the direction of intended or relative, motion., There are three types of frictional forces, i. Static friction, ii. Dynamic friction, iii. Rolling friction, If a body is at rest and no pulling force is acting on, it,force of friction on it is zero., If a force is applied to move the body and it does, not move,the friction developed is called static, friction, which is equal in magnitude and opposite, in direction to the applied force (static friction is, self adjusting force)., If a force is applied to move the body and it, moves,then the friction developed is called dynamic, or kinetic friction., When a body rolls on the surface of another, body friction developed is called as rolling, friction., It is due to the deformation at the point of contact, and depends on area of contact., F, , B, Static friction, , t, es, St, at, eo, fr, , Frictional force, , A, , Dynamic, friction, , Ø, , y, , State of, motion, , Pulling, O, D, C, force, Note-i: If you are walking due east, then the friction, on the feet is due east and the friction on the surface, is due west., Note-ii: Engine is connected to rear wheels of a, car. When the car is accelerated, direction of, 24, , Laws of Friction:, Ø, Ø, , Friction is directly proportional to the normal, reaction acting on the body., The law of static friction may thus be written as, f s α N . ⇒ ( f s )max = µ s N = fl, Generally 0 ≤ static friction ≤ f l, Where the dimensionless constant µ s is called the, coefficient of static friction and N is the magnitude, of the normal force., , ( fs )max = fl = µsN ;, , Ø, Ø, Ø, , f l = Limiting friction, Coefficient of static friction ( µ s) depends on the, nature of the two surfaces in contact and is, independent of the area of contact., Static friction is independent of the area of contact, between the two surfaces, Coefficient of kinetic friction ( µ k) =, , fk, ., N, , It is independent of velocity of the body., fR, N, , Ø, , Coefficient of rolling friction ( µ R ) =, , Ø, , Rolling friction depends on the area of the, surfaces in contact., Note : µ S > µ K > µ R, Friction depends on the nature of the two surfaces, in contact i.e., nature of materials, surface finish,, temperature of the two surfaces etc., , Ø, Ø, , Angle of Friction:, Ø, Ø, Ø, , Angle made by the resultant of f and N with the, normal reaction N is called angle of friction., Friction is parallel component of contact force to, the surfaces., Normal force is perpendicular component of, contact force to the surfaces., NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , N, , R, , a=, , Ø, , φ, , mg, , f, , R=, , Ø, Ø, , Ø, , f, ; φ ≤ φs, N, When the block is in impending state,, µN, tan φs = s = µs, N, Where φs → maximum angle of friction., µk N, = µk, When block is sliding, tan φk =, N, Since µ s > µk , it follows that φs > φk ., , Here F 1ext > f l, , If the block slides with an acceleration ‘ a ’under, the influence of applied force ‘F’,, FR = F − f k ; ma = F − fk, , ∴a =, , f 2 + N2, , F1ext − fk, m, , F − f k F − µk mg, =, ( f k = µ K N = µ K mg ), m, m, N, a, , When the block is static tan φ =, , FR =, , FR = mg, , 2, , 2, , 2, , 2, s, , |||||||||||||||||||||||||||||||||||||||||||||, mg, , Bodies in contact with vertical surfaces:, Ø, , A block of mass m is pressed against a wall, without falling, by applying minimum horizontal, mg, force F. Then F = µ, s, , +1, , fs, |||||||||||||||||||||||||||||||||||||||||||||||, , ( µs N ) + N = N µ, (∴ µs = tan φs ), tan 2 φs + 1, , fl + N =, 2, , F, , fk, , FR = mg sec φs, , Motion on a horizontal rough surface:, N, , Consider a block of mass ‘m’ placed on a, horizontal surface with normal reaction N., Case (i) :If applied force F = 0, then the force of, friction is also zero., F(Applied, force), m, f, |||||||||||||||||||||||||||||||||||||||||||||||||||||||, Case (ii) :If applied force F < ( f s ) max , the block, Case (iii) : If applied force F = ( f s ) max block just, ready to slide and frictional force, , ( f s ) max =, , fl = µ s N, , NARAYANAGROUP, , Ø, , mg, ‘F’ by each hand. Then F = 2 µ, s, fs, fs, , F, , ||||||||||||||||||||||||||||||||||||||||||, , F = µ s mg, (Q N = mg ) ; (at time t=0 ), Case(iv) : If the above applied force continues to, act ( t > 0 ) the body gets motion, static friction, converts as kinetic friction and body possesses, acceleration, F − fk, f − fk, a = ext, = l, = ( µs − µk ) g, m, m, Case (v) :If the applied force is greater than limiting, friction the body starts moving and gets, acceleration, , mg, , As the body is in vertical equilibrium, f s = mg ; µ s N = mg, mg, µ s F = mg (Q N = F ) ⇒ F = µ, s, A block is pressed between two hands without, falling, by applying minimum horizontal force, , ||||||||||||||||||||||||||||||||||||||||||, , does not move and the force of friction is fs = F, , F, , F, , mg, 25
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 1) The maximum acceleration of the truck for which, block does not slide on the floor of the truck is, a = µs g, 2) If a < µs g block does not slide and frictional, force on the block is f=ma., 3) If a > µs g block slips or slides on the floor the, acceleration of the block relative to the truck is, , mg, W=2 fs, ; mg = 2 µs F ⇒ F = 2µ, s, Note : Here in the above two cases, by applying, any amount of horizontal force ‘F’, the frictional, force f s can never be greater than ' mg ', , Sliding block on a horizontal rough surface, coming to rest :, , a′ = a − µk g, 4) If l is the distance of the block from rear side, , N, , of the truck, time taken by the block to cover a, , v=0, , 2l, a - µkg, 4) Acceleration of the block relative to ground is, a′′ = µk g, , u, fk, , distance l ., , |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, S, mg, , a) The acceleration of the block is a = − µ k g, b) Distance travelled by the block before coming, u2, , to rest is S = 2µ g, k, c) Time taken by the block to come to rest is, , Ø, , t=, , Body placed in contact with the front, surface of accelerated truck:, When a block of mass ‘m’ is placed in contact with, the front face of the vehicle moving with acceleration, ‘a’ then a pseudo force ‘Fpf ‘ acts on the block in a, direction opposite to the direction of motion of the, vehicle, fl, , u, mk g, An insect is crawling in a hemispherical bowl of, radius ‘r’. Maximum height upto which it can, crawl is, t=, , Ø, , θ, r sin θ, h, , N, , mg, |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , N, P, θ, , Under equilibrium, fl = mg; N = ma, mg cos θ, , mg sin θ, mg, , Ø, , ma, , fs=µs N, , O, r, , a, , , 1 , , , =, −, r, 1, h = r (1- cosθ ), 2, , µs + 1 , , Maximum angular displacement upto which it can, crawl is ‘ ? ’. Then µ s = tan θ, A block is placed on rear horizontal surface of a, truck moving along the horizontal with an, acceleration ‘a’. Then, , g, µ S N = mg ⇒ µ S ma = mg ⇒ amin = µ, s, W.E.49:A man of mass 40 kg is at rest between the, walls as shown in the figure. If ‘ m ’ between, the man and the walls is 0.8, find the normal, reactions exerted by the walls on the man., Sol. Since man is at rest,, µN1, , µN2, , F2, mg, F1, , N2, , N1, , ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, 26, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , W.E.52:A block on table shown in figure is just on, the edge of slipping. Find the coefficient of, static friction between the block and table, , N1 − N 2 = 0 ( horizontal equilibrium ), \ N1 = N2 = N , F1 = F2 = F (say), , T cos 30, , \ 2m N = mg ( vertical equilibrium ), , O, , = 2´ 0.8´ N = 400 \ N = 250N, W.E.50:A 2 kg block is in contact with a vertical, wall having coefficient of friction 0.5, between the surfaces. A horizontal force of, 40N is applied on the block at right angles, to the wall. Another force of 15N is applied,, on the plane of the wall and at right angles, to 40N force. Find the acceleration of the, block., Sol., , 30 T, 40N, T sin 30, , f, , 80N, , Sol., fl = T sin θ, µ mg = T sin θ, , 80 = T cos θ ...........(2), , f=20N, , F=15N, , 15N, F=40N, , F=40N, , 20N, , 25N=FR, , W=20N, , = 25N, , frictional force f = m N = 0.5 x 40 = 20N, This acts in a direction, opposite to 25N force., \ Net force acting on the block, Fnet = 25–20=5, 5, 2, , \ acceleration of the block a = = 2.5ms–2, W.E.51:A block of mass 4 kg is placed on a rough, horizontal plane. A time dependent, horizontal force F = kt acts on the block (k =, 2 N/s). Find the frictional force between the, block and the plane at t = 2 seconds and t = 5, seconds (µ = 0.2), Sol. Given F = kt, When t = 2sec ; F = 2(2)=4N ..... case (i), f ms = ms mg = 0.2 × 4 × 10 = 8N, Here F < f ms ∴ friction = applied force=4N, When t = 5 sec ; F = 2 (5) = 10N......case(ii), F > f ∴ frictional force < 8N, NARAYANAGROUP, , T sin 30 0 µ mg, =, ;, 80, T cos30 0, , Tan 300 =, , 1, µ, µ 40, =, ;, 80, 3 2, , ⇒µ=, , 2, 3, , = 1.15, , WE.53:When a car of mass 1000 kg is moving with, a velocity of 20ms-1 on a rough horizontal road,, its engine is switched off. How far does the, car move before it comes to rest if the, coefficient of kinetic friction between the road, and tyres of the car is 0.75 ?, , Resultant of W=20N and 15N, FR = 20 2 + 152, , .........(1), , Sol. Here v = 20ms −1 , µk = 0.75, g = 10ms −2, v2, = 26.67m, 2 µk g, W.E.54:A horizontal conveyor belt moves with a, constant velocity V. A small block is projected, with a velocity of 6 m/s on it in a direction, opposite to the direction of motion of the belt., The block comes to rest relative to the belt in, a time 4s. µ = 0.3 , g = 10 m/s2. Find V, Stopping distance S =, , ur, , Sol. V b,c = Vb + Vc = 6 + V, f = µmg = 0.3 × m × 10 = 3m, , Retardation a =, , F 3m, = 3m / s 2, =, m, m, , ur = 6 + V,Vr = 0, t = 4 sec , ar = - 3ms -2, , Vr = ur + ar t , 0 = (6+V) – 3 x 4,V = 6 m/s, 27
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , N, , WE.55:The rear side of a truck is open. A box of, 40 kg mass is placed 5m away from the open, end as shown in figure. The coefficient of, friction between the box and the surface is, 0.15. On a straight road, the truck starts from, rest and accelerating with 2 m/s2. At what, distance from the starting point does the box, fall off the truck? (Ignore the size of the box)., , L−, , fs, , |||||||||||||||||||||||||||||||, m, , m− g, n, , , m, g, n, , N, Fpf, , fk, ||||||||||||||||||||||||||||||||||||||||, 5m, , Weight of the chain lying on the table =, , a = 2m/s 2, , mg, 1, = mg 1 − , n, n, When the chain is about to slide from edge of the, table,, The weight of the hanging part of the chain =, frictional force between the chain and the table, surface., mg −, , mg, , |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , Sol :Because of the acceleration of the truck the pseudo, force on the box = m x a = 40 x 2 = 80N., This force acts opposite to the acceleration of the, truck. The frictional force on the truck which acts, , mg, 1, = µ s mg 1 − , n, n, mg, n −1 , ⇒, = µ s mg , , n, n , , in the forward direction f k = µ N = 0.15 x 40g, = 58.8N Since pseudo force is greater than frictional, force, the block will accelerate in backward, direction relative to truck with a magnitude, , n=, , ∴The maximum fractional length of chain, hanging from the edge of the table in, l, µs, equilibrium is L = µ + 1, s, , Ø, , friction “ µ s ”. When 1/nth of its length is hanging, from the edge of the table, the chain is found to be, about to slide from the table. Weight of the hanging, part of the chain =, , 28, , mg, n, , Fractional length of chain on the table, 1, L−l, =, L, µs + 1, , Sliding of a chain on a horizontal table:, Consider a uniform chain of mass “m” and length, “L” lying on a horizontal table of coefficient of, , L, Substituting this in the above expression we, l, , l, L µs + 1, get, µ s = L − l ( or ) n = l = µ, s, , 1, 1, s′ = a′t 2 = × 2 × (4.34)2 = 18.87 m, 2, 2, , Ø, , 1, ( n − 1), , If l is the length of the hanging part, then, , 80 − 58.8, a=, = 0.53 m / s 2, 40, The time taken by box to cover the distance 5m is, given by, , 1, 2s, s = 0 + at 2 ⇒ t =, = 4.34sec, 2, a, The distance travelled by truck in this time, is , a ′ = 2 ms − 2, , ∴ µs =, , Connected Bodies :, Ø, , A block of mass m1 placed on a rough horizontal, surface, is connected to a block of mass m2 by a, string which passes over a smooth pulley.The, coefficient of friction between m1 and the table is, µ., , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , a, N, , UNIFORM CIRCULAR MOTION, , W.E.57:A block of mass ‘m’ is placed on a rough, x3, , surface with a vertical cross section of y = . If, 6, the coefficient of friction is 0.5, the maximum, height above the ground at which the block, can be placed without slipping is(JEE MAIN -2014), Y, , T, T a, , m1g, , m2g, For body of mass m2, , θy, x, , m2 g − T = m2 a ————— (i), , X, , Sol:, , For body of mass m1, T − f k = m1a ⇒ T − µk N = m1a —— (ii), Solving Eqs (i) and (ii), we get, , m − µk m1 , m1m2 g, (1 + µ ), a= 2, g; T =, m, +, m, m, +, m, 1, 2, 1, 2 , WE.56:A block of mass 10kg is pushed by a force F, on a horizontal rough plane is moving with, acceleration 5ms −2 . When force is doubled,, , Tanθ =, , x2, dy d x3, = ( ) ⇒ Tanθ =, dx dx 6, 2, , At limiting equilibrium,we get µ = Tanθ ⇒ 0.5 =, , x2, 2, , x 2 = 1 ⇒ x = ±1, x3, , we get, 6, , its acceleration becomes 18ms −2 .Find the, coefficient of friction between the block and, , Now putting the values of ‘ x ’in y =, , −2, rough horizontal plane. ( g = 10ms ) ., , When x = 1 ⇒ y = ; x = −1 ⇒ y = −, 6, 6, So the maximum height above the ground at which, 1, the block can be placed without slipping is y = m, 6, , 1, , Sol :, , N, , a, , 1, , Motion of a body on an inclined plane :, F, , fk, , Case (i) :Body sliding down on a smooth, inclined plane :, , |||||||||||||||||||||||||||||||||||||||||||||, mg, On a rough horizontal plane, acceleration of a block, F, of mass ‘m’ is given by a = − µ k g ..........(i), m, , Initially, a = 5 ms −2, 5=, , F, − µk (10 ) ............ ( ii ), 10, , Ø, Ø, Ø, , (Q m = 10kg ), , 2F, − µ k (10 ) .......... ( iii ), 10, , Multiplying Eq(ii) with 2 and subtracting from Eq.(iii), 8, 8 = µ k (10 ) ⇒ µ k =, = 0.8, 10, NARAYANAGROUP, , Acceleration of sliding block a = g sin θ, If l is the length of the inclined plane and h is the, height. The time taken to slide down starting from, rest from the top is t =, , When force is doubled a = 18 ms −2 ., 18 =, , Let us consider a body of mass ‘m’ kept on a, smooth inclined plane. ., Normal reaction N = mg cos θ, , Ø, , 2l, 1 2h, =, g sinθ sinθ g, , h , , Q l =, , sin θ , , Sliding block takes more time to reach the bottom, than to fall freely in air from the top of the inclined, plane to the ground., , 29
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , Ø, , Velocity of the block at the bottom of the inclined, plane is same as the speed attained if block falls, freely from the top of the inclined plane., , V =, , 2 gl sin θ =, , 3., , 2 gh, , mg sin θ 2 > µk mg cos θ 2 acceleration, , Case(ii) :Body projected up on a, smooth inclined plane :, v=0, B, u, A, , Ø, Ø, , s, , h, , θ, , C, , If a block is projected up the plane with a velocity, u, the acceleration of the block is a = − g sin θ, Distance travelled by the block up the plane before, , a=, , mg sin θ 2 − µ k mg cos θ 2, m, , a=, , µ s mg cos θ 2 − µ k mg cos θ 2, m, , = g cos θ 2 ( µs − µk ), , 4. When θ > α , the body slides f k = µk mg cos θ, The resultant force acting on the body down the, plane is FR = mgsinθ -f K ,, FR = mg ( sin θ − µk cos θ ), , The acceleration of the body a = g ( sin θ − µk cos θ ), , u2, , its velocity becomes zero is S = 2 g sin θ, Ø, , When θ 2 = α and ( t > 0 ) the same inclination is, continued the block moves downwards with, accleration a., , N A, , u, , Time of ascent t = g sin θ, , mg sinθ, , Case (iii) Motion of a body down the, rough inclined plane:, Ø, , Let a body of mass ‘m’ be sliding down a rough, inclined plane of angle of inclination ? and coefficient, of kinetic friction µ k., N, , mg sinθ, , Ø, , θ, , θ, , Ø, , v, , Velocity of the body at the bottom of the plane, V = 2 g ( sin θ − µk cos θ ) l, , mg mg cosθ, , If ‘t’is the time taken to travel the distance ‘l’ with, initial velocity u = 0 at the top of the plane,, 2l, g (sin θ − µk cosθ ), , t=, , Ø, , The time taken by a body to slide down on a rough, inclined plane is ‘n’ times the time taken by it to, slide down on a smooth inclined plane of same, inclination and length, then coefficient of friction is...., n=, , trough, tsmooth, , 1. When θ1 < α ; the block remains at rest on the, inclined plane. Frictional force mg sin θ1 (self, adjusting), acceleration a=0, 2. When θ 2 = α ; the block remains at rest on inclined, plane or impending state of motion is achieved., , mg sin θ 2 = µ s mg cos θ 2, , (at time t=0), , Here θ 2 > θ1 and f s = fl acceleration a=0, 30, , h, , B, , Ø, , Angle of Repose ( α ) : Angle of repose is the, minimum angle of the rough inclined plane for, which body placed on it may just start sliding, down. It is numerically equal to the angle of friction., Let ? be the angle of inclination of a rough inclined, plane, α be the angle of repose, m be the mass, of the body and µ be the coefficient of friction., At limiting equilibrium (about to slide), mg sin α = µs mg cosα ⇒ tan α = µs ⇒ α = tan −1 ( µs ), , θ mg cosθ, mg, , fk, θ, , u=0 fk, , n2 =, , =, , 2l, g (sin θ − µ k cos θ ), 2l, g sin θ, , sin θ, sin θ − µ k cos θ, , ⇒ n 2 sin θ − n 2 µ k cos θ = sin θ, 1, , ⇒ µk = Tanθ 1 − 2 , n , , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , Body projected up a rough inclined plane:, , 2l, g (sin θ + µ k cos θ ), , the plane, t =, Ø, , Force required to drag with an acceleration ‘a’ is, F = ( µ k mg cos ? + mg sin ? + m a ), WE.58:A body is moving down a long inclined, plane of angle of inclination ‘ θ ’for which the, coefficient of friction varies with distance x, as µ ( x ) = kx , where k is a constant. Here x is, the distance moved by the body down the, plane. The net force on the body will be zero, at a distance x0 is given by, , θ, , F = mg ( sin θ − kx cos θ ), tan θ, k, WE.59:A body of mass ‘m’ slides down a smooth, inclined plane having an inclination of 450, with the horizontal. It takes 2s to reach the, bottom. If the body is placed on a similar plane, having coefficient of friction 0.5 Then what is, the time taken for it to reach the bottom?, Sol : Mass = m, θ =450, µ =0.5 Time taken by the body, to reach the bottom without friction is, T1 =, , sin θ − kx0 cos θ = 0 ⇒ x0 =, , 2l, g sin θ = 2sec, , NARAYANAGROUP, , =2, , sin θ, sin θ − µ cos θ, , sin 450, sin 450 − (0.5) cos 450, , (1/ 2), = 2 × 2 = 2.828s, (1/ 2) − (0.5)(1/ 2), WE.60:Two blocks of masses 4 kg and 2 kg are in, contact with each other on an inclined plane, =2, , of inclination 300 as shown in the figure., The coefficient of friction between 4 kg mass, and the inclined plane is 0.3, where as, between 2 kg mass and the plane is 0.2. Find, the contact force between the blocks., , g, , f = µ N = µ mg cos θ, , F = mg sin θ − µ mg cos θ, , If F = 0 ;, , T2 = T1, , sin θ − µ cos θ, sin θ, , 2k, , mg cosθ, , mg, , F = mg sin θ − f, N = mg cos θ ;, , 2l, T1, =, ⇒, g ( sin θ − µ cos θ ), T2, , g, , θ, , mg sinθ, , T2 =, , f, , N, , Sol :, , Time taken with friction is, , 4k, , Ø, Ø, , If a body is projected with an initial velocity ‘u’to, slide up the plane, the kinetic frictional force acts, down the plane and the body suffers retardation, due to a resultant force, FR =(mg sin θ +fk ), acceleration a = - g(sin ? + µ k cos ? ), Time taken to stop after travelling a distance l along, , UNIFORM CIRCULAR MOTION, , 300, , Sol :The acceleration of 4 kg mass,, If θ = 300 , µk = 0.3, 1, 3, −2, a4 = g (sin θ − µ k cos θ ) = 10 − 0.3 ×, = 2.6ms, 2, 2, , , , The acceleration of 2 kg mas, 1, 3, −2, a2 = 10 − 0.2 ×, = 3.27 ms, 2 , 2, ∴ a2 > a4, , Thus, there will be contact force between the blocks, and they move together. If ‘a’ is the common, acceleration,, (m1 + m2 )a =, (m1 + m2 ) gsin θ − ( µ1 m1 + µ2 m2 ) g cos θ, 31
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 1, 3, 6a = 6 × 10 × − (0.3 × 4 + 0.2 × 2) ×10 ×, 2, 2, 6a = 30 − 13.856 ⇒ a = 2.7 ms −2, For, 4 kg mass; mg sin θ + f contact − f friction = ma, 1, 3, 4´10´ + f c - 0.3´10´10 ´, = 4 x 2.7, 2, 2, , fc = 10.8 + 10.4 – 20 Þ f c = 1.2N, WE.61:A 30kg block is to be moved up an inclined, plane at an angle 300 to the horizontal with a, velocity of 5ms–1 . If the frictional force, retarding the motion is 150N, find the, horizontal force required to move the block up, the plane. (g=10ms–2.), Sol., , WE.63:In the given figure, the wedge is acted upon by, a constant horizontal force 'F'. The wedge is, moving on a smooth horizontal surface. A ball of, mass 'm' is at rest relative to the wedge. The ratio, of forces exerted on 'm' by the wedge when 'F' is, acting and 'F' is withdrawn assuming no friction, between the edge and the ball,is equal to, F, , Sol., θ, |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , When Force F is applied, ma cosθ, , P, , N, , θ, ma, mg, , ma sinθ, , 300, , The force required to move a body up an inclined plane is, F = mg sin q + f k, , N1 = mgcos θ + Fsin θ, ( F cos θ = mg sin θ ⇒ F = mg tan θ ), , f k = µk ( mg cos θ + P sin θ ) = 150 N, , = 30(10) sin 300 +150 = 300N., If P is the horizontal force, F = P cos q, P=, , F, 300, 300´ 2, =, =, = 200 3 = 346 N, cos q cos300, 3, , WE.62: A body is sliding down an inclined plane, having coefficient of friction 0.5. If the normal, reaction is twice that of resultant downward force, along the inclined plane, then find the angle, between the inclined plane and the horizontal, Sol : µ = 0.5 , N = mg cos θ, N = 2 F , F = mg (sin θ − µ cos θ ), , N, , c os θ = 2 cos θ ( tan θ − µ ), 1, 1, = tan θ − ⇒ tan θ = 1 ⇒ θ = 450, 2, 2, 32, , F sin θ, , 1, If F=0 ; N2 = mgcos θ , N = 1 + mg cos θ, 2, , Ø, , N1, mg tan θ sin θ, =1+, 2, 2, N2, mg cos θ = 1 + tan q = sec q, , Two blocks of mass m and M are placed on a rough, inclined plane as shown, when (θ > α ), i)Minimum value of M for which m slides upwards is, M = m ( sin θ + µs cosθ ), , m, , N = 2 mg ( sin θ − µ cos θ ), mg cos θ = 2 mg ( sin θ − µ cos θ ), , mg sinθ, , mg cosθ, , mg sinθ, , M, θ, , Mg, mg, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , Ø, , UNIFORM CIRCULAR MOTION, , ii) Maximum value of M for which m slides, downwards: M = m ( sin θ − µs cosθ ), A body is released from rest from the top of an, inclined plane of length ‘L’ and angle of inclination, L, ( n > 1) is, ‘ θ ’. The top of plane of length, n, smooth and the remaining part is rough. If the, body comes to rest on reaching the bottom of the, plane then find the value of coefficient of friction of, rough surface, N, u=0, mg sinθ, θ, L mg, n, , v, , θ, , mg cosθ, h, , n −1, n L, , , , N, F, , fr, , mg, , Ø When a lawn roller is pushed by a force ‘F’,which, makes an angle θ with the horizontal, then, Ø Normal reaction N = mg + F sin θ ., Ø Frictional force f r = µr N = µr ( mg + F sin θ ), ∴The net horizontal pushing force is given by, F1= F(cos θ – sin θ ) – µ rmg, , ii) A Lawn Roller on a Horizontal Surface, Pulled by an Inclined Force, F sin θ, , L, 2, Using v 2 − u 2 = 2as ; V = 2a1 ,, n, a1 = g sin θ , a2 = g (sin θ − µ cos θ ), , Ø, , n , µ = Tanθ , , n − 1, A body is pushed down with velocity ‘u’ from the, top of an inclined plane of length ‘L’ and angle of, inclination ‘ θ ’. The top of plane of length, L, ( n > 1) is rough and the remaining part is smooth., n, , If the body reaches the bottom of the plane with a, velocity equal to the initial velocity ‘u’, then the, value of coefficient of friction of rough plane, is µ K = n ( tan θ ), Note : If the top surface is smooth and the remaining, n−2, is rough, then µ k = tan θ , , n −1 , , Pushing & Pulling of a Lawn Roller :, i) A Roller on Horizontal Surface Pushed by, an Inclined Force :, NARAYANAGROUP, , N, F, θ, , Ø, Ø, Ø, Ø, , F cos θ, , |||||||||||||||||||||||||||||||||||||||||||||||||||||||, , fr, , For rough part, , g sin θ = − g [sin θ − µ cosθ ] ( n − 1), , F, ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, F sin θ, , For smooth part :, , n −1 , 0 − V 2 = 2a2 , L, n , L, n −1 , 2a1 = −2a2 , L, n, n , , F cos θ, , θ, , mg, Let a lawn roller be pulled on a horizontal road by, a force ‘F’, which makes an angle θ with the, horizontal., Normal reaction N =mg – Fsin θ, Frictional force fr = µrN = µr(mg – Fsin θ ), The net horizontal pulling force is, F2 = F (cos θ + sin θ )-µ r mg Pulling is easier than, Pushing., , Applying an Inclined Pulling Force :, Let an inclined force F be applied on the body so, as to pull it on the horizontal surface as shown in, the figure., N, F, F sin θ, θ, , fr, , F cos θ, , |||||||||||||||||||||||||||||||||||||||||||||||||||||||, mg, 33
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, Ø, , UNIFORM CIRCULAR MOTION, , accelerated by the force of friction acting upon it., The maximum acceleration of the system of two, blocks to move together without slipping is, amax = µ s g , where µs is the coefficient of static, friction between the two blocks. The maximum, applied force for which both the blocks move, together is Fmax = µs g ( mu + mB ), , Ø, , If a < µs g blocks move together and applied force, is F = (m B + mu )a In this case frictional force, between the two block f = mu a ., , Ø, , If a > µs g , blocks slip relative to each other and, have different accelerations. The acceleration of, the upper block is au = µ k g and that for the bottom, block is aB =, , Ø, , F − µk mu g, mB, , Case - II:Upper block pulled and there is no, friction between bottom block and the horizontal, surface., , mu, , f, , F, , mB, |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, Ø, Ø, , When the upper block is pulled, bottom block is, accelerated by the force of friction acting on it., The maximum acceleration of the system of two, blocks to move together without slipping is, amax = µs, , mu, g, mB where , µs =coefficient of static, , friction between the two blocks The maximum force, for which both blocks move together is, Fmax = µs, , Ø, , mu, g ( mu + m B ), mB, , If a < a , blocks move together and frictional, max, force between the two blocks is f = mB a The, applied force on the upper block is, F = (m B + mu )a, , Ø, , hence they have different accelerations.The, mu, acceleration of the bottom block is aB = µ K m g, B, and the acceleration of the upper block is, F − µ k mu g, au =, mu, Ø A number of blocks of identical masses m each, are placed one above the other. Force required, to pull out N th block from the top is, F = (2N–1) µ mg, WE.65:A block of mass 4kg is placed on another, block of mass 5kg, and the block B rests on a, smooth horizontal table, for sliding the block, A on B, a horizontal force 12N is required to, be applied on it .How much maximum, horizontal force can be applied on ‘B’ so that, both A and B move together? Also find out, the acceleration produced by this force., , M1, F, , M2, , A, B, , Sol: Here M 1 =4kg and M 2 =5kg, Limiting friction between the blocks is flim, Acceleration of system is, F, F, F, =, = m / s2, a=, M1 + M 2 4 + 5 9, Because of this acceleration the block A, experiences a pseudo force of magnitude, F, Fpseudo = M1a = 4 ×, 9, M1a, M1, A, 12N, F, B, M2, As block A move together with B, Fpseudo ≤ flim For, maximum value of applied force, 4F, = 12 ⇒ F = 27 N, Fpseudo = f lim ;, 9, 27, = 3m / s 2, The acceleration of blocks =, 9, , If a > amax blocks slide relative to each other and, , NARAYANAGROUP, , 35
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , WE.66:Two blocks of masses ‘m’ and ‘M’ are, arranged as shown in the figure. The, coefficient of friction between the two blocks, is ‘ µ ’, where as between the lower block and, the horizontal surface is zero. Find the force, ‘F’ to be applied on the upper block, for the, system to be under equilibrium?, F, m, , F = µ R1 + µ R2 = 3 + 0.25 ( 4 ) = 4 N, (c) In this situation for dynamic equilibrium of B, F = µ R1 + µ R2 + T ...........(i), While for the uniform motion of A,, T = µ R2 ............(ii), Substituting T from eqn (ii) in (i) we, get F = µ R1 + 2 µ R2 = 3 + 2 × 1 = 5 N, , M, , F, , Sol :, , Uniform circular motion, , m, , T, µ mg, , T, m, µ mg, On the upper block,, F=T+f=T+ µ N;, F = T + µ mg ……(1), On the lower block, T = µ mg ……… (2), from (1) and (2), we get, F = 2 µ mg, W.E-67 : Block A weighs 4N and block B weighs, 8N.The coefficient of kinetic friction is 0.25, for all surfaces.Find the force F to slide B at, a constant speed when (a) A rests on B and, moves with it (b) A is held at rest and (c) A and, B are connected by a light cord passing over a, smooth pulley as shown in fig (a),(b) and (c), respectively, A, R2, , A, R2, F, , F, , R1 f1, , B, , R1, , A, R2, , f2, B, , f2, R1, , T, , S, (C), , f1, , Sol :(a) When A moves with B the force opposing the, motion is the only force of friction between B and, S the horizontal and as velocity of system is, constant,, , F = f1 = µ R1 = 0.2 ( 4 + 8 ) = 3 N, (b) When A is held stationary, the friction opposing the, motion is between A and B. So, 36, , v2, = rω 2 ,Where v, r & ω are linear, r, velocity, radius and angular velocity of the particle., In uniform circular motion, (a) magnitude of linear velocity does not change, (b) direction of linear velocity changes, (c) Linear velocity changes, (d) Angular velocity is constant, (e) Linear momentum changes, (f) Angular momentum w.r.t to centre does not, change, , Centripetal force ( FC ):, Ø, , It is the force required to keep the body in uniform, circular motion. This force changes the direction of, linear velocity but not its magnitude, , mv 2, = mrω 2 = mvω, ( v = rω ), r, Direction of centripetal force is always perpendicular to the direction of linear velocity. The work done, by centripetal force is always zero, as it is perpendicular to velocity and instantaneous displacement, Ex1: When an electron moves around the nucleus, in a circular orbit, the electrostatic force of attraction between the electron and nucleus is the centripetal force., Ex2: If an electron of mass m and charge e moves, around the nucleus of atomic number Z in a circular orbit of radius r , centripetal force on it is, FC =, , Ø, , T, , When a particle moves in a circular path with constant speed then it is said to be in uniform circular, motion. In this case the acceleration of the particle, is a = v ( ω ) =, , f2, f1, , S, (b), , S, (a), , F, , B, , Ø, , mv 2, 1 Ze 2, =, r, 4πε 0 r 2, Ex3: When planet of mass m moves around the, sun in a circular orbit of radius r , centripetal force, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , mv 2 GMm, = 2, on it is, r, r, Ex4: When a stone is whirled round in horizontal, circle by attaching it at the end of string, tension in, the string provides the centripetal force., , Ø, , mv 2, =T, r, , Centrifugal force:, Ø, , Ø, , Ø, , The Pseudo force which acts radially outward on, the body moving along a circle is called centrifugal, force., Even though centripetal and centrifugal forces are, equal in magnitude and opposite in direction, they, do not form action - reaction pair because they act, on the same body in two different frames., Circular turning on roads:, The necessary centripetal force while taking a circular turn is being provided to a vehicle by following three ways, 1) by friction only, 2) by banking of roads only, 3) by both friction and banking of roads, , Friction only:, In this case the necessary centripetal force is provided by static friction, v, , Let θ be the angle through which the outer edge is, raised relative to the inner edge. This angle is also, called “angle of banking”. The normal reaction N, exerted by the road on the vehicle is directed normal to the surface as shown in the figure., N cos θ balances the weight of the vehicle., N cos θ = mg, N sin θ is directed towards the centre of the circular path. Which provides the centripetal force., , mv 2, r, From the above equations, we get, N sin θ =, , tan θ =, , v2, (or) v = rg tan θ ., rg, , Motion of a vehicle on a rough banked road:, Ø, , If friction is present between the road and the, tyres,the components of friction and normal reaction provide the centripetal force., , mv 2, ,the vehicle possesses, r, the tendency to slip down the plane. The minimum, speed for avoiding slipping down the plane can be, obtained by taking friction up the plane., , Case-I:If N sin θ >, , r, o, , N, , N cos θ, f sin θ, , f, θ, , Ø, , v ≤ µs gr ;, vmax = µ s gr, For a given radius of curvature and coefficient of, friction, the safe maximum velocity of the vehicle is, , N sin θ, , Ø, , θ, , f cos θ, , mg, , θ, , given by vmax = µ s rg, , f, , fig - (i), , Banking of roads only:, N, , N cos θ, , To find minimum speed we can use fig (i),, , θ, , N sin θ − f cos θ =, , N cos θ + f sin θ = mg.... ( 2 ), , N sin θ, A, O, NARAYANAGROUP, , 2, mvmin, .... (1), r, , θ, , Mg, , X, , (f, , = µN ), , From (1) and (2) we get, , vmin =, , rg ( sin θ − µ cos θ ), rg (tan θ − µ ), =, (1 + µ tan θ ), ( cos θ + µ sin θ ), 37
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , radius 18m without the danger of, skidding.The coefficient of friction between, the tyres of the car and the surface of the, curved path is 0.2. What is the maximum, speed in kmph of the car for safe driving? (g=10, ms −2 ), , mv 2, Case-II:If N sin θ <, ,the vehicle possesses, r, the tendency to skid up the plane. The safe maximum speed for avoiding skidding can be obtained, by taking friction acting down the plane., N, , N cos θ, θ, , N sin θ, f cos θ θ, f, , f sin θ, mg, , θ, , fig - (ii), To find maximum safe speed, we have to consider, figure (ii)., , Sol :, , Maximum speed. v = µ s gr, , v = 0.2 ×10 ×18 = 36 = 6ms −1, W.E.70:A point P moves in a counter clock wise, direction on a circular path as shown in fig., The movement of ‘P’ is such that it sweeps, out a length S=t3+5, where ‘S’ is in metres, and ‘t’ is in seconds. The radius of the path, is 20m. The acceleration of ‘P’ when t=2s is, nearly., (AIEEE - 2010), , B, , mv 2, N sin θ + f cos θ = max .... (1), r, , P(x, y), , N cos θ − f sin θ = mg .........(2), , From (1) and (2) we get, , vmax =, Ø, , O, , rg ( sin θ + µ cos θ ), rg (tan θ + µ ), =, (1 − µ tan θ ), ( cosθ − µ sin θ ), , A cyclist is taking a turn of radius r with speed v, then he should bend through an angle θ with the, vertical such that tan θ =, , 2, , v, rg, , W.E.68:Two cars of masses m 1 and m 2 are moving, in circles of radii r1 and r2 respectively. Their, speeds are such that they make complete circle, in the same time t. The ratio of their, centripetal acceleration is (AIEEE - 2012), Sol :As their time period of revolution is same,angular, speed is also same.centripetal acceleration is, a = ω 2r ;, a1 ω 2 r1 r1, =, =, a2 ω 2 r2 r2, WE.69: A car is driven round a curved path of, , 38, , 3, Sol: S = t + 5 ⇒ v =, , A, dS, = 3t 2, dt, , For t=2s, v = 3 × 4 = 12ms −1, Tangential acceleration, at =, , dv, = 6t, dt, , For t=2s, at = 12ms −2, Centripetal acceleration, , ac =, , v 2 144, =, = 7.2ms −2, R 20, , net acceleration = at 2 + ac 2 ≈ 14ms −2, WE.71:A turn of radius 20m is banked for the, vehicle of mass 200kg going at a speed of, 10m/s. Find the direction and magnitude, of frictional force acting on a vehicle if it, moves with a speed a) 5 m/s b) 15 m/s, Assume that friction is sufficient to, prevent slipping (g=10m/s2), NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , Sol : v = 10m/s, , ω=, , (10 ) = 1, v, 1, =, ⇒ θ = tan−1 , rg ( 20 )(10 ) 2, 2, Now, as speed is decreased, force of friction, f acts upwards. Using the equations, tan θ =, , 2, , 2, , N sin θ − f cos θ =, , g tan θ, r, 2π, , but r = l sin θ and ω = T ( T0 is the time period, 0, of pendulum), , 2, , mv, ; N cos θ + f sin θ = mg, r, , −1 1 , Substituting θ = tan , v = 5m/s, m=200kg, , 2, , Time period of the pendulum is T0 = 2π, Ø, , l cos θ, g, , For the coin not to fly off on the turn table, the, condition is, , and r = 20m, in the above equations, we get, f = 300 5 N, b) In the second case force of friction f will act, downwards, , ω, , mv 2, N sin θ + f cos θ =, ; N cos θ − f sin θ = mg, r, −1, , 1, , Substituting θ = tan , v = 15m/s,, 2, m=200kg and r = 20m, in the above, equations, we get f = 500 5 N, , Conical pendulum:, A bob of mass M is given a horizontal push a little, through angular displacement θ and arranged such, that the bob describes a horizontal circle of radius, ‘ r’ with uniform angular velocity ω in such a way, that the string always makes an angle θ with the, vertical and T is the tension in the string., Suppose the body is in rotational equilibrium,then, T cos θ = Mg ------(1), T sin θ = Mrω 2 ----(2), rω 2, tan θ =, g, , From (1) and (2), , O, , T, T, , T cos θ, , A, , T sin θ, Mg, , NARAYANAGROUP, , θ, , r, , ⇒, Ø, , mv 2, ≤ µ s mg, r, , Motion of a Cyclist in a Death Well:, For equilibrium of cyclist in a death well, as shown, in the figure, the normal reaction N provides the, centripetal force needed and the force of friction, balances his weight mg., , mv 2, r, , Thus,, , N=, , and, , f = µ N = mg, , ⇒, , Vmax =, , mv 2, r, , rg, µ, , f, N, mg, r, 39
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LAWS OF MOTION, FRICTION, , C.U.Q, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 40, , The behaviour of a body under zero resultant, force is given by, 1) first law of motion 2) second law of motion, 3) third law of motion 4) law of gravitation, Which law of Newton defines an ‘inertial, frame of reference’?, 1) First law of motion 2) Second law of motion, 3) Third law of motion 4) Law of gravitation, The statement “acceleration is zero if and only, if the net force is zero” is valid in, 1) non-inertial frames 2) inertial frames, 3) both in inertial frames and non-inertial frames, 4) neither inertial frames nor non-inertial frames, You move forward when your car suddenly, comes to a halt and you are thrown backward, when your car rapidly accelerates. Which law, of Newtons is involved in these?, 1) third law, 2) second law, 3) first law, 4) law of gravitation, You are thrown outer side when your car suddenly, takes a turn. Which law of Newton is involved in, this?, 1) Third law, 2) Second law, 3) First law, 4) Law of gravitation, An object is thrown vertically upward with, some velocity. If gravity is turned off at the, instant the object reaches the maximum height,, what happens?, 1) The object continues to move in a straight line, 2) The object will be at rest, 3) The object falls back with uniform velocity, 4) The object falls back with uniform acceleration, Which of the following is the most significant, law of motion given by Newton?, 1) First law of motion 2) Second law of motion, 3) Third law of motion 4) Zeroth law of motion, The quantity of motion of a body is best, represented by, 1) its mass, 2) its speed, 3) its velocity, 4) its linear momentum, A certain particle undergoes erratic motion., At every point in its motion, the direction of, the particle’s momentum is always, , JEE-ADV PHYSICS-VOL - II, 1) the same as the direction of its velocity, 2) the same as the direction of its acceleration, 3) the same as the direction of its net force, 4) the same as the direction of its kinetic energy, 10. Inside a railway car a plumb bob is suspended, from the roof and a helium filled balloon is tied, by a string to the floor of the car. When the, railway car accelerates to the right, then, 1) both the plumb bob and balloon move to the left, 2) both the plumb bob and balloon move to the right, 3) plumb bob moves to the left and the balloon, moves to the right, 4) plumb bob moves to the right and the balloon, moves to the left, 11. A constant force (F) is applied on a stationary, particle of mass ‘m’. The velocity attained by, the particle in a certain displacement will be, proportional to, 1, 1) m, 2) 1/m, 3) m, 4), m, 12. A constant force (F) is applied on a stationary, particle of mass ‘m’. The velocity attained by, the particle in a certain interval of time will be, proportional to, 1, 1) m, 2) 1/m, 3) m, 4), m, 13. A force produces an acceleration of a1 in a, body and the same force produces an, acceleration of a2 in another body. If the two, bodies are combined and the same force is, applied on the combination, the acceleration, produced in it is, a1 + a2, a1a2, 1) a1 + a2 2) a a, 3) a + a 4) a1a2, 1 2, 1, 2, 14. To keep a particle moving with constant, velocity on a frictionless surface, an external, force, 1) should act continuously, 2) should be a variable force, 3) not necessary, 4) should act opposite to the direction of motion, 15. If action force acting on a body is gravitational, in nature, then reaction force, 1) may be a contact force, 2) may be gravitational force, 3) may be a gravitational or contact force, 4) may be a force of any origin, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 16. Action and reaction can never balance out, because, 1) they are equal but not opposite always, 2) they are unequal in magnitude even though, opposite in direction, 3) though they are equal in magnitude and opposite, in direction they act on different bodies, 4) they are unequal in magnitudes, 17. The propulsion of a rocket is based on the, principle of conservation of, 1) linear momentum, 2) energy, 3) angular momentum, 4) mass, 18. An automobile that is towing a trailer is, accelerating on a level road. The force that, the automobile exerts on the trailer is, 1) equal to the force the trailer exerts on the, automobile, 2) greater than the force the trailer exerts on the, automobile, 3) equal to the force the trailer exerts on the road, 4) equal to the force the road exerts on the trailer, 19. A man is standing in the middle of a perfectly, smooth ‘island of ice’ where there is no friction, between the ground and his feet. Under these, circumstances, 1) he can reach the desired corner by throwing, any object in the same direction, 2) he can reach the desired corner by throwing, any object in the opposite direction, 3) he has no chance of reaching any corner of the island, 4) he can reach the desired corner by pursuing on, the ground in that direction, 20. Which law of Newton reveals the underlying, symmetry in the forces that occur in nature?, 1) First law, 2) Second law, 3) Third law, 4) Law of gravitation, 21. You hold a rubber ball in your hand. The, Newton’s third law companion force to the, force of gravity on the ball is the force exerted, by the, 1) ball on the earth, 2) ball on the hand, 3) hand on the ball, 4) earth on the ball, 22. A lift is going up with uniform velocity. When, brakes are applied, it slows down. A person in, that lift, experiences, 1) more weight, 2) less weight, 3) normal weight, 4) zero weight, , NARAYANAGROUP, , UNIFORM CIRCULAR MOTION, , 23. While we catch a cricket ball, we catch it at, the front and make the hands move with the, ball backwards. Why is that?, 1) To reduce the impulse, 2) To increase the time of contact, there by increase, the force, 3) To increase the impulse, 4) To increase the time of contact, there by decrease, the force, 24. The change in momentum per unit time of a, body represents, 1) impulse, 2) force, 3) kinetic energy, 4) resultant force, 25. A father and his seven years old son are facing, each other on ice skates. With their hands,, they push off against one another. Regarding, the forces that act on them as a result of this, and the acceleration they experience, which, of the following is correct?, 1) Father exerts more force on the son and, experiences less acceleration, 2) Son exerts less force on the father and, experiences more acceleration, 3) Father exerts as much force on the son as the, son exerts on the father, but the father experiences, less acceleration, 4) Father exerts as much force on the son as the, son exerts on the father, but the father experiences, more acceleration, 26. A student initially at rest on a frictionless, frozen pond throws a 2 kg hammer in one, direction. After the throw, the hammer moves, off in one direction while the student moves, off in the other direction. Which of the, following correctly describes the above, situation?, 1) The hammer will have the momentum with, greater magnitude, 2) The student will have the momentum with greater, magnitude, 3) The hammer will have the greater kinetic energy, 4) The student will have the greater kinetic energy, 27. A ball falls towards the earth. Which of the, following is correct?, 1) If the system contains ball, the momentum is, conserved, 2) If the system contains earth, the momentum is, conserved, 3) If the system contains the ball and the earth, the, momentum is conserved, 4) If the system contains the ball and the earth and, the sun, the momentum is conserved
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LAWS OF MOTION, FRICTION, , 28. A block moving in air breaks into two parts, and the parts separate, 1) the total momentum must be conserved, 2) the total kinetic energy must be conserved, 3) the total momentum must change, 4) the potential energy must be conserved, 29. Regarding linear momentum of a body, a) It is a measure of quantity of motion, contained by the body, b) Change in momentum is the measure of, impulse, c) Impulse and acceleration act in opposite, direction to the change in momentum, d) In the case of uniform circular motion the, linear momentum is conserved., 1) a& b are true, 2) b & c are true, 3) c & d are true, 4) a,b & c are true, 30. Compare the impulses exerted on a wall by, the two objects, a golf ball and a lump of mud,, both having the same mass and the velocity., 1) the golf ball imparts greater impulse, 2) the lump of mud imparts the greater impulse, 3) both impart equal impulse, 4) nothing can be said, 31. Two objects X and Y are thrown upwards, simultaneously with the same speed. The mass, of X is greater than that of Y. The air exerts, equal resistive force on two objects, then, 1) X reaches maximum height than Y, 2) Y reaches maximum height than X, 3) the two objects will reach the same height, 4) cannot say, 32. A man drops an apple in the lift. He finds that, the apple remains stationary and does not fall., The lift is, 1) going down with constant speed, 2) going up with constant speed, 3) going down with constant acceleration, 4) going up with constant acceleration, 33. Internal force can change, 1) linear momentum as well as kinetic energy, 2) linear momentum but not the Kinetic energy, 3) the kinetic energy but not linear momentum, 4) Neither the linear momentum nor the kinetic, energy, 34. A man is standing on a spring platform., Reading of spring balance is 60 kg wt. If man, jumps outside the platform, then the reading, of the spring balance, 1) remains same, 2) decreases, 3) increases, 4) first increases and then decreases to zero, 42, , JEE-ADV PHYSICS-VOL - II, 35. A stretching force of 10N is applied at one end, of a spring balance and an equal force is, applied at the other end at the same time. The, reading of the balance is, 1) 5 N, 2) 10 N, 3) 20 N, 4) 0, 36. A ball is dropped from a spacecraft revolving, around the earth at a height of 1200 km. What, will happen to the ball?, 1) It will continue to move with velocity V along, the original orbit of spacecraft, 2) It will move with the same speed tangential to, the space craft, 3) It will fall down to the earth gradually, 4) It will go far in space, 37. A body is under the action of three forces, uur, uur uur, F1 , F2 and F3 . In which case the body cannot, undergo angular acceleration?, uur, uur uur, 1) F1 , F2 and F3 are concurrent, uur uur uur, 2) F1 + F2 + F3 = 0, uur, uur uur, 3) F1 , F2 is parallel to F3 but the three forces are, not concurrent, uur, uur, uur, 4) F1 and F2 act at the same point but F3 acts at, different point., 38. In the system shown in figure m1 > m2 ., System is held at rest by thread BC. Just, after the thread BC is burnt., , 1) acceleration of m1 will be equal to zero, 2) acceleration of m2 will be downwards, 3) magnitude of acceleration of two blocks will be, non–zero and unequal, 4) magnitude of acceleration of both the blocks, m1 − m 2 , , will be m + m g, 1, 2 , , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 39. A lift is ascending with a constant speed “V”. A, passenger in the lift drops a coin. The, acceleration of the coin towards the floor will be, 1) Zero, 2) g, 3) <g, 4) >g, 40. A reference frame attached to the earth with, respective to an observer in space, 1) is an inertial frame because Newton’s laws of, motion are applicable in it, 2) is an inertial frame by definition, 3) cannot be an inertial frame because earth is, rotating about its axis, 4) can be an inertial frame because earth is, revolving around the sun., 41. A Stationary railway platform on earth is, 1) an inertial frame of reference for an observer on, earth., 2) a Non inertial frame of reference for an observer, on moon, 3) both are true, 4) both are false, 42. A rotating platform for a stationary observer, outside it is, 1) inertial frame of reference, 2) non inertial frame of reference, 3) both, 4) some times inertial (or) some times non inertial, 43. The acceleration of a particle is found to be, non zero when no force acts on the particle., This is possible if the measurement is made, from, 1) inertial frame 2) non inertial frame, 3) both, 4) some times inertial (or) some times non inertial, 44 Frictional force between two bodies, 1) increases the motion between the bodies, 2) destroys the relative motion between the, bodies, 3) sometimes helps and sometimes opposes, the motion, 4) increases the relative velocity between the, bodies, 45. Maximum value of static friction is, 1) limiting friction, 2) rolling friction, 3) static friction, 4) normal reaction, 46. A good lubricant should be highly, 1) viscous, 2) non-volatile, 3) both (1and 2), 4) transparent, 47. Theoretically which of the following are best, lubricants?, 1) Solids, 2) Liquids, 3)Gases, 4) Both 2 and 3, , NARAYANAGROUP, , UNIFORM CIRCULAR MOTION, , 48. A block ‘B’ rests on ‘A’. A rests on a, horizontal surface ‘C’ which is frictionless., There is friction between A and B. If ‘B’ is, pulled to the right, 1) B moves forward and A to the left, 2) ‘B’ only moves to the left, 3) ‘B’ does not move, 4) ‘A’ and ‘B’ move together to the right, 49. Sand is dusted to the railway tracks during, rainy season to, 1) make it always wet, 2) increase friction, 3) to reduce consumption of fuel, 4) make it always dry, 50. With increase of temperature, the frictional, force acting between two surfaces, 1) increases, 2) decreases, 3) remains same 4) may increase or decrease, 51. If we imagine ideally smooth surfaces and if, they are kept in contact, the frictional force, acting between them is, 1) zero, 2) a finite value but not zero, 3) very large, 4) we can’t predict, 52. If man is walking, direction of friction is, 1) opposite to the direction of motion, 2) same as that of direction of motion, 3) perpendicular to that of direction of motion, 4) 45º to the direction of motion, 53. Aeroplanes are streamlined to reduce, 1) fluid friction, 2) sliding friction, 3) kinetic friction, 4) limiting friction, 54. The limiting friction between two surfaces, does not depend, 1) on the nature of two surfaces, 2) on normal reaction, 3) on the weight of the body, 4) on volume of the body, 55. While walking on ice one should take small, steps to avoid slipping. This is because, smaller steps ensure, 1) larger friction, 2) smaller friction, 3) larger normal force 4)smaller normal force, 56. In order to stop a car in shortest distance, on a horizontal road, one should, 1) apply the brakes very hard so that the wheels, stop rotating, 2) apply the brakes hard enough to just prevent, slipping, 3) pump the brakes (press and release), 4) shut the engine off and not apply brakes
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 57. A body rests on a rough horizontal plane. A, force is applied to the body directed towards, the plane at an angle ? with the vertical., The body can be moved along the plane, 1) only if ? is greater than the angle of friction, 2) only if ? is lesser than the angle of friction, 3) only if ? is equal to the angle of friction, 4) for all values of ?, 58. A lift is moving down with an acceleration, equal to the acceleration due to gravity. A, body of mass M kept on the floor of the lift, is pulled horizontally. If the coefficient of, friction is µ , then the frictional resistance, offered by the body is, 1) µ k Mg, 2) Mg 3) Zero 4) µ kMg2, 59. A body is struck to the front part of the truck., The coefficient of friction between the body, and truck is µ . The minimum acceleration, with which the truck should travel so that, the body does not fall down is, 1) µ / g 2) µ g, 3) g / µ 4) µ 2 g, 60. When a bicycle is in motion, the force of, friction exerted by the ground on the two, wheels is such that it acts, 1) in the backward direction on the front wheel, and in the forward direction on the rear wheel, 2) in the forward direction on the front wheel, and in the backward direction on the rear wheel, 3) in the backward direction on both the front, and rear wheels, 4) in the forward direction on both the front and, rear wheels, 61. A boy of mass M is applying a horizontal, force to slide a box of mass M 1 on a rough, horizontal surface. The coefficient of friction, between the shoe of the boy and the floor is, µ and that between the box and the floor is, ‘ µ 1’In which of the following cases is it, certainly not possible to slide the box?, 1) µ < µ 1; M < M1, 2) µ > µ 1; M > M1, 3) µ < µ 1; M > M1, 4) µ > µ 1; M < M1, 62. When a person walks on a rough surface, 1) the frictional force exerted by the surface, keeps him moving, 2) reaction of the force applied by the man, on the surface keeps him moving, 3) the force applied by the man keep him, moving, 4) weight of the man keeps him moving, 44, , 63. The maximum speed of a car on a curved, path of radius ‘r’ and the coefficient of friction, µ k is, 1) v =, , µk, gr, , 2) v = µ k gr, , 3) v =, , gr, µk, , 4) v =, , 1, µ k gr, , 64. The angle which the rough inclined plane, makes with the horizontal when the body, placed on it just starts sliding down is called, 1) angle of Friction, 2) angle of repose, 3) critical angle, 4) brewster’s angle, 65. A body of mass M is placed on a rough, inclined plane of inclination ? and coefficient, of friction µ k. A force of (mg sin ? + µ k mg, cos ? ) is applied in the upward direction, the, acceleration of the body is, 1) g sin ?, 2) g (sin ? + µ kcos ? ) 3) g, (sin ? – µ k cos ? ) 4) Zero, 66. It is easier to pull a lawn roller than to push it, because pulling, 1) involves sliding friction, 2) involves dry friction, 3) increases the effective weight, 4) decreases normal reaction, 67. A block of mass m and surface area A just, begins to slide down an inclined plane when, the angle of inclination is p /5. Keeping the, mass of the block same, if the surface area, is doubled, the inclination of the plane at which, the block starts sliding will be, 2) p /10, 3) 2 p /5 4) p / 5 2, 1) p /5, 68. A block X kept on an inclined surface just, begins to slide if the inclination is θ 1. The, block is replaced by another block Y and it, is found that it just begins to slide if the, inclination is θ 2 ( θ 2 > θ 1). Then, 1) Mass of X = mass of Y, 2) Mass of X < mass of Y, 3) Mass of X > mass of Y, 4) All the three are possible, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 69. A Particle is acted upon by a force of constant, magnitude which is always perpendicular to the, velocity of the particle. The motion of the, particle takes place in a plane. It follows that, 1) the Kinetic energy of the particle changes with, time., 2) the acceleration of the particle is constant., 3) the velocity of the particle is constant, 4) the speed of the particle is constant, 70. The direction of angular acceleration of a body, moving in a circle in the plane of the paper is, 1) along the tangent, 2) along the radius inward, 3) along the radius outward, 4) perpendicular to the plane of the paper, 71. Suppose a disc is rotating counter clockwise, in the plane of the paper then, 1) Its angular velocity vector will be perpendicular, to the page pointing up out of the page, 2) Its angular velocity vector will be perpendicular, to the page pointing inwards, 3) Its angular velocity vector acts along the tangent, to the disc, 4) None of the above, 72. A Particle of mass’M’ moves in a uniform, circular path of radius ‘r’ with a constant speed, ‘ v ’, then its centripetal acceleration is, 1), , v2, r, , 2), , v2, r2, , 3) v 2 r, , 4) Zero, , 73. A vehicle moves safe on rough , curved and, unbanked road. Then, a) The direction of static friction is radially, out wards, b) The direction of static friction is radially, inwards, c) The direction of kinetic friction is tangential, to curved path, d) Static friction does not exist, 1) a & b are correct, 2) c & d are correct, 3) b & c are correct, 4) a & c are correct, , NARAYANAGROUP, , C.U.Q - KEY, 01) 1, 07) 2, 13) 3, 19) 2, 25) 3, 31) 1, 37) 1, 43) 2, 49) 2, 55) 3, 61) 1, 67) 1, 73) 3, , 02) 1, 08) 4, 14) 3, 20) 3, 26) 3, 32) 3, 38) 1, 44) 3, 50) 2, 56) 2, 62) 1, 68) 4, , 03) 2, 09) 1, 15) 2, 21) 1, 27) 4, 33) 3, 39) 2, 45) 1, 51) 3, 57) 1, 63) 2, 69) 4, , 04) 3 05) 3, 10) 3 11) 4, 16) 3 17) 1, 22) 2 23) 4, 28) 1 29) 1, 34) 4 35) 2, 40) 3 41) 3, 46) 3 47) 3, 52) 2 53) 1, 58) 3 59) 3, 64) 2 65) 4, 70) 4 71) 1, , 06) 2, 12) 2, 18) 1, 24) 4, 30) 1, 36) 1, 42) 2, 48) 4, 54) 4, 60) 1, 66) 4, 72) 1, , LEVEL–I (C.W), NEWTON’S LAWS OF MOTION, 1., , n balls each of mass m impinge elastically in, each second on a surface with velocity u. The, average force experienced by the surface will, be, 1) mnu 2) 2mnu, 3) 4mnu, 4) mnu/2, 2. A ball reaches a racket at 60 m/s along + X, direction, and leaves the racket in the opposite, direction with the same speed. Assuming that, the mass of the ball as 50gm and the contact, time is 0.02 second, the force exerted by the, racket on the ball is, 1) 300 N along + X direction, 2) 300 N along - X direction, 3) 3,00,000 N along + X direction, 4) 3,00,000 N along - X direction, 3. 'P' and 'Q' horizontally push in the same, direction a 1200 kg crate. 'P' pushes with a, force of 500 newton and 'Q' pushes with a force, of 300 newton. If a frictional force provides, 200 newton of resistance, what is the, acceleration of the crate?, 1) 1.3 m / s 2, , 2) 1.0 m / s 2, , 3) 0.75 m / s 2, , 4) 0.5 m / s 2
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 4., , A ball of mass 'm' moves normal to a wall, with a velocity 'u' and rebounds with a, velocity 'v'. The change in momentum of the, ball during the rebounding is, 1) m ( u + v ) towards the wall, 2) m ( u − v ) towards the wall, 3) m ( u + v ) away from the wall, , 5., , 6., , 7., , 4) m ( u − v ) away from the wall., If a force of 250N acts on a body,the, momentum required is 125kgms −1 .The period, for which the force acts on the body is, 1) 0.1 s, 2) 0.3 s, 3) 0.5 s, 4) 0.2 s, A machine gun fires a bullet of mass 40g with, a velocity 1200 ms–1 . The man holding it can, exert a maximum force of 144 N on the gun., How many bullets can he fire per second at, the most?, 1) One 2) Three, 3) Two 4) Four, A truck of mass 500kg is moving with, constant speed 10 ms −1 .If sand is dropped into, the truck at the constant rate, 10kg/min,the force required to maintain the, motion with constant velocity is, , 1) 5kg ms −1, , 2) 10 5 kgms −1, , 3) 20 kg ms −1, 4) 15 kg ms −1, 10. A ball of mass’m’ is thrown at an angle is ‘ θ ’, with the horizontal with an initial velocity, ‘u’.The change in its momentum during its, flight in a time interval of ‘t’ is, 1) mgt, 2) mgt cosθ, 3) mgt sin θ, 4) ½ mgt., 11. A body of mass 2kg has an initial speed, 5ms–1. A force acts on it for 4 seconds in the, direction of motion. The force time graph is, shown in figure. The final speed of the body, is, , 1) 8.5 ms–1, 2) 11 ms–1, –1, 3) 14.31 ms, 4) 4.31 ms–1, 12. A force time graph for the motion of a body is, as shown in figure. Change in linear, momentum between 0 and 6s is, , 3, 5, 7, 5, N 2) N, 3) N, 4) N, 2, 4, 5, 3, A 5000 kg rocket is set for vertical firing. The, exhaust speed is 800ms −1 . To give an upward, , 1), , 8., , acceleration of 20ms −2 , the amount of gas, ejected per second to supply the needed, thrust is ( g = 10ms −2 ), 1) 127.5 kg s −1, , 2) 137.5 kg s −1, , 3) 187.5 kg s −1, , 4) 185.5 kg s −1, , IMPULSE, 9., , A small sphere of mass m = 2kg moving with, a velocity u = 4$i − 7 $j m / s collides with a, smooth wall and returns with a velocity, v = −$i + 3$j m / s . The magnitude of the impulse, received by the ball is, , 46, , 1) zero, 2) 8 Ns, 3) 4 Ns 4) 2 Ns, 13. An object of mass 3 kg is at rest. Now a force, F = 6t 2 iˆ + 2tjˆ N is applied on the object. Find the, velocity of the object at t = 3 sec., r r, r r, 1)18i + 3 j, 2) 18i − 3 j, r, r, r, r, 3) 3i − 18 j, 4) 3i + 18 j, 14. An impulse "I" given to a body changes its, velocity from " v1 to v2 ". The increase in the, kinetic energy of the body is given by, 1) I ( v1 + v2 ), , 2) I ( v1 + v2 ) / 2, , 3) I ( v1 − v2 ), , 4) I ( v1 − v2 ) / 2, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , OBJECTS SUSPENDED BY, STRINGS & APPARENT WEIGHT, 15. A 60kg man is inside a lift which is moving up, , with an acceleration of 2.45 ms −2 . The appar-ent percentage change in his weight is,, 1) 20% 2) 25%, 3) 50%, 4) 75%, 16. The apparent weight of a person inside a lift, is W 1 when lift moves up with certain, acceleration and is W 2 when lift moves down, with same acceleration. The weight of person, when lift moves up with constant speed is, , 1) 6 N to the right 2) 12 N to the right, 3) 6 N to the left 4) 12 N to the left, 21. Two masses m1 and m2 are attached to a spring, balance S as shown in Figure. If m1 > m2 then, the reading of spring balance will be, , W1 + W2, W1 − W2, 2), 3) 2W1 4) 2 W2, 2, 2, 17. A person of mass 60 kg is in a lift. The change, in the apparent weight of the person, when the, lift moves up with an acceleration of 2ms −2 and, , 1), , then down with an acceleration of 2ms −2 , is, (take g = 10 m/sec2), 1) 120 N 2) 240 N 3) 480 N 4) 720 N, 18. A rope of length 10m and linear density 0.5kg/, m is lying length wise on a smooth horizontal, floor. It is pulled by a force of, 25 N., The tension in the rope at a point 6m away, from the point of application is, 1) 20 N 2) 15 N, 3) 10 N 4) 5 N, 19. Three blocks of masses m1, m2 and m3 are, connected by a massless string as shown in, figure on a frictionless table. They are pulled, with a force T3 = 40N. If m1 = 10kg , m2 = 6kg, and m3 = 4kg , then tension T 2 will be, , 1) ( m1 − m2 ), , 2) ( m1 + m2 ), , 2m1m2, 3) m + m, 1, 2, , m1m2, 4) m + m, 1, 2, , 22. Two masses (M+m) and (M-m) are attached, to the ends of a light inextensible string and, the string is made to pass over the surface of, a smooth fixed pulley. When the masses are, released from rest, the acceleration of the, system is, 1) gm/M, 2) 2gM/m, 3)gm/2M, , 4) g ( M 2 -m 2 ) /2M, , 23. Two bodies of masses 5kg and 4kg are tied to, a string as shown. If the table and pulley are, smooth, then acceleration of 5kg mass will be, , 1) 10 N, 2) 20 N, 3) 32 N 4) 40 N, 20. A horizontal force F pushes a 4 kg block (A), which pushes against a 2 kg block (B) as, shown. The blocks have an acceleration of, 3m/s 2 to the right. There is no friction between, the blocks and the surfaces on which they slide., What is the net force B exerts on A?, NARAYANAGROUP, , 1) 19.5m/s2, 3) 2.72 m/s2, , 2) 0.55 m/s2, 4) 5.45m/s2
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , LAW OF CONSERVATION OF, MOMENTUM, 24. The object at rest suddenly explodes into three, parts with the mass ratio 2:1:1.The parts of, equal masses move at right angles to each, other with equal speed ‘ v ’.the speed of the, third part after explosion will be, v, v, 2) 2v, 3), 4), 1) v, 2, 2, 25. A man and a cart move towards each other., The man weighs 64 kg and the cart weighs, 32kg. The velocity of the man is 5.4 km/hr and, that of the cart is 1.8 km/hr. When the man, approaches the cart, he jumps on to it. The, velocity of the cart carrying the man will be, 1) 3 km/hr, 2) 30 km/hr, 3) 1.8 km/ hr, 4) zero, 26. A bomb of mass 6 kg initially at rest explodes, in to three identical fragments. One of the, fragments moves with a velocity of, ∧, , 10 3 i m / s , another fragment moves with a, ∧, , velocity of 10 j m / s , then the third fragment, moves with a velocity of magnitude., 1) 30 m/s 2) 20 m/s 3) 15 m/s 4) 5 m/s, , EQUILIBRIUM OF A PARTICLE, 27. A mass of 10 kg is suspended by a rope of, length 2.8m from a ceiling. A force of 98 N is, applied at the midpoint of the rope as shown, in figure. The angle which the rope makes with, the vertical in equilibrium is, , 1) 300, , 48, , 2) 600, , 3) 450, , 4) 900, , 28. A mass of M kg is suspended by a weightless, string.The horizontal force that is required to, displace it until the string makes an angle 450, with the initial vertical direction is, 1) Mg, , 2), , 3) Mg ( 2 + 1), , 4), , Mg, 2, 2Mg, , LAWS OF FRICTION, 29. The coefficients of static and dynamic friction, are 0.7 and 0.4. The minimum force required, to create motion is applied on a body and if it, is further continued, the acceleration attained, by the body in ms–2 is (g = 10m/s2), 1) 7, 2) 4 3) 3, 4) Zero, 30. The coefficient of static friction between, contact surfaces of two bodies is 1. The, contact surfaces of one body support the, other till the inclination is less than, 1) 30º 2) 45º, 3) 60º 4) 90º, , MOTION ON A HORIZONTAL, ROUGH SURFACE, 31. Brakes are applied to a car moving with, disengaged engine, bringing it to a halt after, 2s. Its velocity at the moment when the, breaks are applied if the coefficient of, friction between the road and the tyres is, 0.4 is, 1) 3.92 ms–1, 2) 7.84 ms–1, 3) 11.2ms–1, 4) 19.6 ms–1, 32. A book of weight 20N is pressed between, two hands and each hand exerts a force of, 40N. If the book just starts to slide down., Coefficient of friction is, 1) 0.25 2) 0.2, 3) 0.5 4) 0.1, 33. A car running with a velocity 72 kmph on a, level road, is stopped after travelling a, distance of 30m after disengaging its engine, (g = 10ms–2). The coefficient of friction, between the road and the tyres is, 1) 0.33 2) 4.5, 3) 0.67, 4) 0.8, 34. In the above problem car got a stopping, distance of 80m on cement road then µ k is, (g = 10 m/sec2), 1) 0.2, 2) 0.25 3) 0.3, 4) 0.35, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 35. A 10kg mass is resting on a horizontal, surface and horizontal force of 80N is applied., If µ = 0.2, the ratio of acceleration without, and with friction is (g = 10ms–2), 1) 3/4, 2) 4/3, 3) 1/2 4) 2, 36. A block of mass 20 kg is pushed with a, horizontal force of 90N. If the coefficient of, static and kinetic friction are 0.4 and 0.3, the, frictional force acting on the block is, ( g = 10ms −2 ), 1) 90N 2) 80N, 3) 60N 4) 30N, 37. A force of 150N produces an acceleration, of 2ms–2 in a body and a force of 200N, produces an acceleration of 3ms–2. The mass, of the body and the coefficient of kinetic, friction are, 1)50kg; 0.1, 2)25kg; 0.1, 3)50kg; 0.5, 4)50kg; 0.2, 38. A heavy uniform chain lies on horizontal table, top. If the coefficient of friction between the, chain and the table surface is 0.25, the, maximum percentage of the length of the chain, that can hang over one edge of the table is, 1) 20% 2) 25%, 3) 35%, 4) 15%, , MOTION OF A BODY ON THE, INCLINED PLANE, 39. The angle of inclination of an inclined plane is, 60º. Coefficient of friction between 10kg body, on it and its surface is 0.2, g = 10 ms–2. The, acceleration of the body down the plane in ms–2 is, 1) 5.667 2) 6.66 3) 7.66, 4) Zero, 40. In the above problem the resultant force on, the body is, 1) 56.6 N 2) 66.6 N 3) 76.6 N 4) 86.6 N, 41. In the above problem, the frictional force on, the body is, 1) Zero 2) 5 N, 3) 7.5 N, 4) 10 N, 42. In the above problem, the minimum force, required to pull the body up the inclined plane, 1) 66.6 N 2) 86.6 N 3)96.6 N 4)76.6 N, 43. When a body slides down an inclined plane, with coefficient of friction as µ k, then its, acceleration is given by, 1) g( µ k sin ? + cos ? ) 2) g( µ k sin ? – cos ? ), 3) g(sin ? + µ k cos ? ) 4) g(sin ? – µ k cos ? ), NARAYANAGROUP, , UNIFORM CIRCULAR MOTION, , 44. A brick of mass 2kg just begins to slide down, on inclined plane at an angle of 45º with the, horizontal. The force of friction will be, 1) 19.6 sin 45º, 2) 9.8 sin 45º, 3) 19.6 cos 45º, 4) 9.8 cos 45º, 45. The lengths of smooth & rough inclined planes, of inclination 450 is same. Times of sliding of, a body on two surfaces is t1 , t2 and µ = 0.75 ,, then t1 : t2 =, 1) 2 : 1 2) 2 : 3, , 3) 1 : 2, , 4) 3 : 2, , PULLING / PUSHING A BODY, 46. A block of weight 200N is pulled along a, rough horizontal surface at constant speed, by a force of 100N acting at an angle 30º, above the horizontal. The coefficient of, kinetic friction between the block and the, surface is, 1) 0.43 2) 0.58, 3) 0.75, 4) 0.83, , UNIFORM CIRCULAR MOTION, 47. The centripetal force required by a 1000, kg car that takes a turn of radius 50 m at a, speed of 36 kmph is, 1) 1000 N, 2) 3500 N, 3) 1600 N, 4) 2000 N, 48. A stone of mass 0.5 kg is attached to a string, of length 2m and is whirled in a horizontal, circle. If the string can withstand a tension of, 9N, the maximum velocity with which the stone, can be whirled is, 1) 6ms −1 2) 8ms −1 3) 4ms −1 4) 12ms −1, , LEVEL- I (C.W) - KEY, 1) 2, 7) 4, 13) 1, 19) 3, 25) 1, 31) 2, 37) 1, 43) 4, , 2) 2, 8) 3, 14) 2, 20) 3, 26) 2, 32) 1, 38) 1, 44) 1, , 3) 4, 9) 2, 15) 2, 21) 3, 27) 3, 33) 3, 39) 3, 45) 3, , 4) 3, 10) 1, 16) 1, 22) 1, 28) 1, 34) 2, 40) 3, 46) 2, , 5) 3, 11) 3, 17) 2, 23) 4, 29) 3, 35) 2, 41) 4, 47) 4, , 6) 2, 12) 1, 18) 3, 24) 4, 30) 2, 36) 3, 42) 3, 48) 1, , LEVEL- I (C.W) - KEY, 1., , Favg = ma, a =, , 3., , Fnet = ma , a =, , v−u, t, , 2. Favg = ma, a =, , Fnet ( FP + FQ ) − f, =, m, m, , 4. ∆P = Pf − Pi = m (V − u ) 5. F =, , v−u, t, , dp, dp, ⇒ dt =, dt, F
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 2., , A 0.2 kg object at rest is subjected to a force, , 1) 2mu towards the wall, 2) 2mu away from the wall, 0.3iˆ − 0.4 ˆj N . What is its velocity vector, 3) zero, after 6 sec, 4) mu away from the wall, 9. Bullets of 0.03 kg mass each hit a plate at the, 1) 9iˆ − 12 ˆj, 2) 8iˆ − 16 ˆj, rate of 200 bullets per second with a velocity, of 30 m/s. The average force acting on the, 3) 12iˆ − 9 ˆj, 4) 16iˆ − 8 ˆj, plate in newton is, A body of mass 2 kg is moving with a velocity, 1) 120, 2) 180, 3) 300, 4) 480, r, of u = 3iˆ + 4 ˆj m / s ., A steady force 10. A vehicle of mass 10kg is moving with a, −1, ur, velocity of 5 ms .To stop it in 1/10 sec the, F = iˆ − 2 ˆj N begins to act on it. After four, required force in opposite direction is, seconds , the body will be moving along., 1) 5000N 2) 500N, 3) 50N, 4) 1000N, 1) X-axis with a velocity of 2 m/s, , (, , ), , (, (, , 3., , UNIFORM CIRCULAR MOTION, , ), ), , (, (, , ), ), , 2) Y-axis with a velocity of 5 m/s, 3) X-axis with a velocity of 5 m/s, 4) Y-axis with a velocity of 2 m/s, 4., , 5., , 6., , 7., , 8., , Three forces F 1 , F2 and F 3 are, simultaneously acting on a particle of mass 'm', and keep it in equilibrium. If F 1 force, is reversed in direction only, the acceleration, of the particle will be., 1) F1 / m, 2) 2 F1 / m, , IMPULSE, , 11. An impulse is supplied to a moving object with, the force at an angle1200 with the velocity, vector. The angle between the impulse vector, and the change in momentum vector is, 1) 1200 2) 00, 3) 600 4) 2400, 12. A 20 kg body is pushed with a force of 7N for, 1.5 sec then with a force of 5N for 1.7 sec and, finally with a force of 10N for 3 sec, the total, impulse applied to the body and change in, velocity will be, 1) 49 Ns,12.5ms −1, 2) 49 Ns, 2.45ms −1, 3) 98 Ns, 4.9ms −1, 4) 4.9 Ns , 2.45ms −1, 13. A body of mass 5 kg is acted upon by a net, force F which varies with time t as shown in, graph, then the net momentum in SI units, gained by the body at the end of 10 seconds is, , 4) −2 F1 / m, 3) − F1 / m, A block of metal weighing 2kg is resting on a, frictionless plane. It is struck by a jet releasing, water at a rate of 1 kg/s and at a speed of 5 m/, s. The initial acceleration of the block will be, 1) 2.5 m/s2 2) 5 m/s2 3) 10 m/s2 4)20 m/s2, A body of mass 2kg moving on a horizontal, surface with an initial velocity of 4 ms–1, comes, to rest after 2 second. If one wants to keep, this body moving on the same surface with a, velocity of 4ms–1, the force required is, m1, 1) zero 2) 2 N, 3) 4 N, 4) 8 N, Ten coins are placed on top of each other on a, 8kg, horizontal table. If the mass of each coin is, 10g and acceleration due to gravity is 10 ms −2 ,, 30°, m2, what is the magnitude and direction of the, 1. 0, 2. 100, th, force on the 7 coin (counting from the bottom), 3. 140, 4. 200, due to all the coins above it?, 14. A body is acted on by a force given by, 1) 0.3 N downwards, 2) 0.3 N upwards, F = (10 + 2t ) N . The impulse received by the, 3) 0.7 N downwards, 4) 0.7 N upwards, body during the first four seconds is, A ball of mass ’m’ moves normal to a wall with, 1) 40 N s, 2) 56 N s, a velocity ‘u’ and rebounds with the same, 3), 72, N, s, 4) 32 N s, speed. The change in momentum of the ball, during the rebounding is, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 15. A unidirectional force F varying with time t as, shown in the Fig. acts on a body initially, at rest for a short duration 2T. Then the, velocity acquired by the body is, , 21. Three equal masses A, B and C are pulled with, a constant force F. They are connected to each, other with strings. The ratio of the tension, between AB and BC is, , 1) 1 : 2, π F0T, π F0T, FT, 2), 3) 0, 4) zero, 4m, 2m, 4m, 16. If the average velocity of a body moving with, uniform acceleration under the action of a, force is “v” and the impulse it receives during, a displacement of “s” is “I”, the constant force, acting on the body is given by, I ×v, 2I × v, I ×v, I ×s, 2), 3), 4), 1), 2s, s, s, v, , 1), , OBJECTS SUSPENDED BY, STRINGS AND APPARENT WEIGHT, 17. A 6.0kg object is suspended by a vertical, string from the ceiling of an elevator which is, accelerating upward at a rate of 2.2 ms −2 .the, tension in the string is, 1) 11N 2) 72N, 3) 48N, 4)59N, 18. A young man of mass 60 kg stands on the floor, of a lift which is accelerating downwards at, 1 m / s 2 then the reaction of the floor of the lift, on the man is (Take g = 9.8 m / s 2 ), 1) 528 N 2) 540 N 3) 546N 4) none, 19. Three masses of 16 kg, 8 kg and 4kg are, placed in contact as shown in Figure. If a force, of 140 N is applied on 4kg mass, then the force, on 16kg will be, , 1) 140 N 2) 120 N, 3) 100 N, 4) 80 N, 20. A body of mass M is being pulled by a string, of mass m with a force P applied at one end., The force exerted by the string on the body is, Pm, PM, 1) ( M + m ), 2) ( M + m ), 3) Pm ( M + m ), 52, , P, 4) ( M − m ), , 2) 2 : 1, , 3) 3 : 1 4) 1 : 1, , 22. A coin is dropped in a lift. It takes time t1 to, reach the floor when lift is stationary. It takes, time t2 when lift is moving up with constant, acceleration. Then, 1) t1 > t2 2) t2 > t1, , 3) t1 = t2, , 4) t1 ≥ t 2, , 23. A light string passing over a smooth light, pulley connects two blocks of masses m1 and, m2 (vertically). If the acceleration of the, system is g/8, then the ratio of masses is, 1) 8:1, , 2) 4:3, , 3) 5:3, , 4) 9:7, , 24. A pendulum bob is hanging from the roof of, an elevator with the help of a light string. When, the elevator moves up with uniform, acceleration ‘a’ the tension in the string is, T1 .When the elevator moves down with the, same acceleration, the tension in the string, is T2 .If the elevator were stationary, the, tension in the string would be, 1), , T1T2, T1 + T2, 2) T1 + T2 3) T + T, 2, 1, 2, , 2T1T2, 4) T + T, 1, 2, , 25. Three bodies are lying on a frictionless, horizontal table and these are connected as, shown in the figure. They are pulled towards, right with a force T3 = 60 N If m1, m2 and m3 are, equal to 10 kg,20kg and 30kg respectively,, then the values of T1 and T2 will be, , 1)10N,10N, 3)10N,30N, , 2)30N,10N, 4)30N,30N, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , LAW OF CONSERVATION OF, MOMENTUM, 26. A bullet of mass 20gm is fired from a riffle of, mass 8 kg with a velocity of 100m/s. The, velocity of recoil of the rifle is, 1) 0.25 m/s 2) 25 m/s, 3) 2.5 m/s 4) 250 m/s, 27. A space craft of mass 2000 kg moving with a, velocity of 600m/s suddenly explodes into two, pieces. One piece of mass 500 kg is left, stationary. The velocity of the other part must, be ( in m/s), 1) 600, 2) 800 3)1500, 4) 1000, 28. A person weighing 60 kg in a small boat of, mass 140 kg which is at rest, throws a 5 kg, stone in the horizontal direction with a velocity, of 14 ms -1 . The velocity of the boat, immediately after the throw is ( in m/s), 1) 1.2, 2) 0.5, 3) 0.35 4) 0.65, , LAWS OF FRICTION, 29. A body of mass 60kg is pushed with just, enough force to start it moving on a rough, surface with µ s = 0.5 and µ k = 0.4 and the, force continues to act afterwards. The, acceleration of the body is ( in m/sec2 ), 1) 0.98 2) 3.92 3) 4.90, 4) Zero, 30. If the coefficient of friction is 3 , the angle of, friction is, 2) 60 0, 3) 450, 4) 37 0, 1) 300, , MOTION ON A HORIZONTAL ROUGH, SURFACE, 31. The coefficient of friction between a car wheels, and a roadway is 0.5 The least distance in, which the car can accelerate from rest to a, speed of 72 kmph is (g=10 ms −2 ), 1)0m, 2)20m, 3)30m, 4)40m, 32. An eraser weighing 2N is pressed against the, black board with a force of 5N. The coefficient, of friction is 0.4. How much force parallel to, the black board is required to slide the eraser, upwards, 1) 2N, 2)2.8N 3)4N, 4) 4.8N, 33. A marble block of mass 2 kg lying on ice when, given a velocity of 6ms −1 is stopped by friction, in 10 s. Then the coefficient of friction is, ( g = 10ms −2 ), 1) 0.02 2)0.03, , NARAYANAGROUP, , 3)0.06, , 4)0.01, , 34. A block of weight 100N is pushed by a force, F on a horizontal rough plane moving with, an acceleration 1 m/s2, when force is doubled, its acceleration becomes 10m/s2. The, coefficient of friction is (g=10ms–2), 1) 0.4, 2) 0.6, 3) 0.5, 4) 0.8, 35. A block of mass 5kg is lying on a rough, horizontal surface. The coefficient of static and, kinetic friction are 0.3 and 0.1 and g = 10ms −2 ., If a horizontal force of 50N is applied on the, block, the frictional force is, 1)25N, 2)5N, 3)10N, 4)Zero, 36. A heavy uniform chain lies on horizontal table, top. If the coefficient of friction between the, chain and the table surface is 0.5, the maximum percentage of the length of the chain that, can hang over one edge of the table is, 1) 20% 2) 33.3%, 3)76%, 4) 50%, , MOTION OF A BODY ON THE INCLINED, PLANE, 37. A body is sliding down an inclined plane, forming an angle 300 with the horizontal. If, the coefficient of friction is0.3 then acceleration, of the body is, 1)1.25. ms −2 2) 2.35ms −2 3) 3.4ms −2 4) 4.9ms −2, 38. In the above problem its velocity after 3, seconds in ms −1 is, 1)7.05, 2)14.7, 3)29.4, 4) zero, 39. In the above problem its displacement after 3, seconds is, 1)78.4m 2)44.15m, 3)10.57m, 4)Zero, 40. A block sliding down on a rough 450 inclined, plane has half the velocity it would have been,, the inclined plane is smooth.The coefficient, of sliding friction between the block and the, inclined plane is, , 1, 1, 1, 3, 2), 3), 4), 2 2, 2, 4, 4, 41. A cube of weight 10N rests on a rough inclined, plane of slope 3 in 5. The coefficient of friction, is 0.6. The minimum force necessary to start, the cube moving up the plane is, 1)5.4N 2)10.8N 3) 2.7N, 4) 18N, 1)
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , CIRCULAR MOTION, 42. A body moves along a circular path of radius, 5 m.The coefficient of friction between the, surface of the path and the body is 0.5.The, angular velocity in rad/s with which the body, should move so that it does not leave the path, is (g=10 ms–2), 1) 4, 2) 3, 3) 2, 4) 1, 43. A van is moving with a speed of 72 Kmph on a, level road,where the coefficient of friction, between tyres and road is 0.5.The minimum, radius of curvature,the road must have,for safe, driving of van is, 1) 80 m 2) 40 m, 3) 20 m, 4) 4 m, 44. What is the smallest radius of a circle at which, a bicyclist can travel if his speed is, 7 m/s and the coefficient of static friction the, between tyres and road is 0.25, 1) 10 m, 2)20 m, 3)5 m, 4)15 m, , LEVEL - I (H.W) - KEY, 1) 2, 7) 1, 13) 3, 19)4, 25) 3, 31) 4, 37) 2, 43) 1, , 2) 1, 8) 2, 14) 2, 20)2, 26) 1, 32) 3, 38) 1, 44) 2, , 3) 3, 9) 2, 15) 4, 21)2, 27) 2, 33) 3, 39) 3, , 4) 4, 10) 1, 16) 3, 22) 1, 28) 3, 34) 4, 40) 2, , 5) 1, 11) 2, 17) 2, 23) 4, 29) 1, 35) 2, 41) 2, , LEVEL - I (H.W) - HINTS, , 14. J = ∫ F .dt, , 15. From O to T, area is ( + ) ve and from T to 2T,,, area is ( − ) ve , net area is zero, hence, no chang in, momentum., s, 16. J = F × t , t =, 17.T=m(g+a), v, F, 18. R = m( g − a ) 19. F = ma , a =, m1 + m2 + m3, , F, F, =, 20. F = ma , a =, ;T = Ma, M M +m, 21. F − T1 = m1a, T1 − T2 = m2 a, T2 = m3 a, , 22. For stationary lift t1 =, , 2h, and when the lift is movg, 2h, , ing up with constant acceleration t2 = g + a, m −m , , 6) 3, 12) 2, 18) 1, 24) 1, 30) 2, 36) 2, 42) 4, , ur, 1 1 1, r F r r r, 1. F = ma , a = a + a 2. a = , v = u + at ., 1, 2, m, 3. v = u + at , F = ma, r r, r, 4. Under equilibrium condition F1 + F2 + F3 = 0, − F1 + F2 + F3, F1 = −( F1 + F2 ) , a =, m, dp, dm, F, 5. F =, ,F =v, ,a=, dt, dt, m, 6. v = u + at , F = ma, 7. Force on the seventh coin = weight of the three, coins above it. F = 3mg, dp, mnv, F=, 8. ∆P = m ( v − u ) 9., ,F =, dt, t, dp, m (v − u ), 10. F =, ,F =, dt, t, 11. Both the impulse and change in momentum are in, same direction., 54, , 12. J = F × t , J = F1t1 + F2 t2 + F3t3 , J = m(v − u ), 13. Impulse=Area under F-t curve ; J = m(v − u ), , 1, 2, 23. a = m + m g, 1, 2 , 24. T1 = m( g + a), T2 = m( g − a ), T = mg, 25. T3 − T2 = m3 a ,, , T3, , T2 − T1 = m2 a , T1 = m1a , a = m + m + m, 1, 2, 3, 26. MU = m1v1 + m2v2 27. MU = m1v1 + m2v2, 28. m1u1+m2u2=m1v1+m2v2, 29. FR = f s − f k ,a = ( µ s – µ k) g 30. µ = tan θ, 31. v 2 − u 2 = 2as , a = µk g, 32. F 1 = f s + mg , f s = µ s mg 33. v = u + at , a = µk g, 34. FR = F − f , f = µ k mg, 35. f s = µ S N , f k = µ k N , N = mg, 36., , l, µ, × 100 =, × 100, L, µ +1, , 37. a = g (sin θ − µk cosθ ), , 38. v = u + at , a = g (sin θ − µk cos θ ), 1, 2, , 2, 39. S = ut + at , a = g (sin θ − µk cosθ ), , 40. v 2 − u 2 = 2as, VR = 2 gl (sin θ − µk cos θ ), VS = 2 gl sin θ, 41. F = ma , F = mg ( sin θ + µk cos θ ), 42. F = µ s N ,, , mv 2, = µ mg, r, , 2, , mv, = µ mg, 43. F = µ s N ,, r, 2, , v, , 2, ⇒r=, 44. mv = µ mg , v = µ rg, µs g, , r, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 5., , LEVEL-II (C.W), , If F = F0 (1 − e −t / λ ) , the F-t graph is, , NEWTON’S LAWS OF MOTION, 1., , The momenta of a body in two perpendicular, directions at any time't' are given by, , 3t 2, + 3 .The force acting, 2, on the body at t = 2 sec is, 1) 5 units 2) 2 units 3) 10 units 4) 15 units, When a force F acts on a body of mass m,, the acceleration produced in the body is a., If three equal forces F1 = F2 = F3 = F act on, the same body as shown in figure the, acceleration produced is, PX = 2t 2 + 6 and PY =, , 2., , 1), 3., , ), , 2 − 1 a 2), , (, , ), , 2 + 1 a 3), , 2a 4) a, , Two blocks of masses m and M are placed on, a horizontal frictionless table connected by, light spring as shown in the figure . Mass M, is pulled to the right with a force F. If the, acceleration of mass m is a ,then the, acceleration of mass M will be (AIEEE-2007), , ( F − ma ), , ( F + ma ), , F, am, 4), M, M, M, M, The displacement of a body moving along a, straight line is given by : S = bt n , where 'b' is, a constant and 't' is time. For what value of 'n', the body moves under the action of constant, force?, 1) 3/2, 2) 1, 3) 2, 4) 1/2, , 1), , 4., , (, , 2), , NARAYANAGROUP, , 3), , 6., , Three forces 2 0 2 N , 2 0 2 N and 40N are, acting along X, Y and Z – axes respectively, on a 5 2 k g mass at rest at the origin. The, magnitude of its displacement after 5s is,, 1) 50m 2) 25m, 3) 60m 4) 100m, 7. A horizontal jet of water coming out of a pipe, of area of cross-section 20cm 2 hits a vertical, wall with a velocity of 10ms −1 and rebounds, with the same speed. The force exerted by, water on the wall is,, 1) 0.2 N 2) 10 N, 3) 400 N 4) 200 N, 8. A rocket of mass 40 kg has 160 kg fuel.The, exhaust velocity of the fuel is 2 kms −1 .The rate, of consumption of fuel is 4 kgs −1 .Calculate the, ultimate vertical speed gained by the rocket., −1, −1, 1)2.82 kms −1 2)4.82 kms 3)3.61 kms −1 4)5.62 kms, 9. A body of mass 5kg starts from the origin with, an initial velocity ur = 30iˆ + 40 ˆjms −1 . If a conr, stant force F = − iˆ + 5 ˆj N acts on the body,,, the time in which the y − component of the, velocity becomes zero is, 1) 5s, 2) 20s, 3) 40s, 4) 80s, 10. A professional diver of mass 60 kg performs, a dive from a platform 10 m above the water, surface. Find the magnitude of the average, impact force experienced by him if the impact, time is 1s on collision with water, surface.Assume that the velocity of the diver, just after entering the water surface is, 4 ms −1 .(g=10 ms −2 ), 1) 240N 2) 600N 3) 300N 4) 60N, , (, , )
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 11. An open knife edge of mass 200 g is dropped, from height 5m on a cardboard. If the knife, edge penetrates distance 2m into the card, board, the average resistance offered by the, cardboard to the knife edge is (g = 10m/s2), 1) 7 N, 2) 25 N, 3) 35 N 4) None, 12. six forces lying in a plane and forming angles, of 600 relative to one another are applied to, the centre of a homogeneous sphere with a, mass m=6kg. These forces are radially, outward and consecutively 1N,2N,3N,4N,5N, and 6N. The acceleration of the sphere is, 1) 0, 2) 1/2 m/s2, 3) 1m/s2 4) 2 m/s2, , IMPULSE, 13. A particle of mass m, initially at rest is acted, upon by a variable force F for a brief interval, of time T. It begins to move with a velocity u, after the force stops acting. F is shown in the, graph as a function of time. The curve is a, semicircle. Then, , m, , + M .F, , 1) 2, m+M, , M, , + m .F, , 2) 2, m+M, , M .F, 3) zero, 4), m+M, 16. A ball is suspended by a thread from the ceiling, of a tram car. The brakes are applied and the, speed of the car changes uniformly from 36, kmh–1 to zero in 5 s. The angle by which the, ball deviates from the vertical is (g = 10 ms–2), , −1 , 1) tan 3 , 1, , , , −1, , −1 , 2) sin 5 , , , 1, , 1, , −1, , 1, , 3) tan 5 , 4) cot 3 , , , 17. A block is kept on a frictionless inclined, surface with angle of inclination α . The, incline is given an acceleration ‘a’ to keep the, block stationary. Then ‘a’ is equal to, , g, 2) g cos ecα, tan α, 3) g, 4) g tan α, 18. A man sits on a chair supported by a rope, passing over a frictionless fixed pulley. The, man who weighs 1,000 N exerts a force of 450, N on the chair downwards while pulling the, rope on the other side. If the chair weighs, 250N, then the acceleration of the chair is, 1) 0.45 m / s 2, 2) 0, , 1), , π F02, πT 2, u, =, 2), 2m, 8m, π F0T, π F0T, 3) u =, 4) u =, 4m, 2m, 14. A ball of mass 0.2kg strikes an obstacle and, moves at 600 to its original direction. If its speed, also changes from 20m/s to 10m/s, the magnitude, of the impulse received by the ball is, 1) 2 7 N s 2) 2 3 N s 3) 2 5 N s 4) 3 2 N s, 1) u =, , OBJECTS SUSPENDED BY, STRINGS AND APPARENT WEIGHT, 15. The block is placed on a frictionless surface in, gravity free space. A heavy string of a mass m, is connected and force F is applied on the string,, then the tension at the middle of rope is, , 56, , 3) 2 m / s 2, 4) 9 / 25 m / s 2, 19. A balloon of mass M is descending at a constant, acceleration α . When a mass m is released from, the balloon it starts rising with the same, acceleration α . Assuming that its volume does, not change, what is the value of m?, 1), , α, M, α+g, , 2), , 2α, M, α+g, , 3), , α+g, M, α, , 4), , α+g, M, 2α, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 20. A monkey of mass 40 kg climbs on a massless, rope of breaking strength 600 N. The rope will, break ifthe m onkey. (Take g = 10 m /s2), 1) climbs up with a uniform speed of 6m / s, 2) climbs up with an acceleration of 6 m / s 2, , 25. In the following figure, the pulley is massless, and frictionless. There is no friction between, the body and the floor. The acceleration, produced in the body when it is displaced, through a certain distance with force ‘P’ will be, , 3) climbs down with an acceleration of 4 m / s 2, 4) climbs down with a uniform speed of 5 m / s, 21. Two persons are holding a rope of negligible, weight tightly at its ends so that it is horizontal., A 15 kg weight is attached to rope at the, midpoint which now no more remains, horizontal. The minimum tension required to, completely straighten the rope is, 1) 150 N 2) 75 N 3) 50 N 4) Infinitely large, 22. A straight rope of length 'L' is kept on a, frictionless horizontal surface and a force 'F', is applied to one end of the rope in the direction, of its length and away from that end. The, tension in the rope at a distance 'l' from that, end is, 1), , F, l, , 2), , l, , LF, 3) 1 − F, l, L, , l , , 4) 1 + F, L, , 23. Consider three blocks of masses m1 , m2 , m3, interconnected by strings which are pulled by, a common force F on a frictionless horizontal, table as in the figure. The tension T1 and T2, are also indicated, , P, P, P, P, 2), 3), 4), M, 2M, 3M, 4M, 26. Two identical blocks each of mass “M” are, tied to the ends of a string and the string is, laid over a smooth fixed pulley. Initially the, masses are held at rest at the same level. What, fraction of mass must be removed from one, block and added to the other , so that it has an, , 1), , acceleration of 1/ 5th of the acceleration due, to gravity, 1.1/10, 2.1/5, 3.2/5, 4.1/20, 27. In the given arrangement, n number of equal, masses are connected by strings of negligible, masses. The tension in the string connected, to nth mass is :, , a) T2 > T1 if m2 > m1, b) T2 = T1 if m2 = m1, , c) T2 > T1 always, , F, d) acceleration of the system = m + m + m, 1, 2, 3, , 1) a, b, , 2) b, d, , 3) a, d, , 4) c, d, , 24. A railway engine of mass 50 tons is pulling a, , wagon of mass 40 tons with a force of 4500N., The resistance force acting is 1N per ton. The, tension in the coupling between the engine, and the wagon is, 1) 1600 N 2) 2000 N 3) 200 N 4) 1500N, NARAYANAGROUP, , mMg, mMg, 2), 3) mg, 4) mng, nm + M, nmM, 28. A 40 N block is supported by two ropes. One, rope is horizontal and the other makes an angle, of 300 with the ceiling. The tension in the rope, attached to the ceiling is approximately :, 1) 80 N 2) 40 N 3) 34.6 N 4) 46.2 N, , 1)
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 29. The pulley arrangements shown in figure are, identical, the mass of the rope being negligible., In case I,the mass m is lifted by attaching a, mass 2m to the other end of rope with a, constant downward force F = 2mg, where g is, acceleration due to gravity. The acceleration, of mass m in case I is, , 32. A block of mass 3kg which is on a smooth, inclined plane making an angie of 300 to the, horizontal is connected by cord passing over, light frictionless pulley to second block of, mass 2kg hanging vertically. What is the, acceleration of each block and what is the, tension of the cord?, 1) 0.98 m/s2; 17.6N, 2) 1.98 m/s2; 19.6N, 3) 0.49 m/s2; 9.8N, 4) 1.47 m/s2; 4.9N, 33. If m 1 = 10kg,m 2 = 4kg,m3 = 2kg, the, acceleration of system is, , F = 2mg, m, , m, 2m, , 1) zero, 2) more than that in case II, 3)less than that in case II 4)equal to that in case II, 30. Two masses of 10 kg and 5 kg are suspended, from a rigid support as shown in figure. The, system is pulled down with a force of 150 N, attached to the lower mass. The string, attached to the support breaks and the system, accelerates downwards., , In case the force continues to act.what will be, the tension acting between the two masses?, 1) 300 N 2) 200 N, 3) 100 N, 4) zero, 31. Two bodies of masses 3kg and 2kg are, connected by a long string and the string is, made to pass over a smooth fixed pulley., Initially the bodies are held at the same level, and released from rest. The velocity of the 3kg, body after one second is (g=10m/ s 2 ), 1)2m/s 2) 1m/s 3) 0.4m/s, 4) 4m/s, 58, , 1) 5g/2 2) 5g/3, 3) 5g/8, 4)5g/14, 34. The string between blocks of masses ‘m’ and, ‘2m’ is massless and inextensible.The system, is suspended by a massless spring as shown. If, the string is cut, the magnitudes of accelerations, of masses 2m and m (immediately after cutting), , 1) g , g, , 2) g ,, , g, 2, , 3), , g, ,g, 2, , 4), , g g, ,, 2 2, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 35. All surfaces are smooth. The acceleration of, mass m relative to the wedge is, , UNIFORM CIRCULAR MOTION, , while still moving in air. The value of, ( m1vr11 + m2vr21 ) − ( m1vr1 + m2vr2 ) is, 2) ( m1 + m2 ) gt0, , 1) zero, , 3) 2 ( m1 + m2 ) gt0 4), , 1, ( m1 + m2 ) gt0, 2, , EQUILIBRIUM OF A PARTICLE, 40. Two masses M 1 and M 2 connected by means, , 1) g sin θ, , 2) g sin θ + a cos θ, , 3) g sin θ − a cosθ, , 4) a cos θ, , LAW OF CONSERVATION OF, MOMENTUM, 36. A bullet of mass 10 gm moving with a, horizontal velocity 100m/s passes through a, wooden block of mass 100 gm. The block is, resting on a smooth horizontal floor. After, passing through the block the velocity of the, bullet is 10m/s. the velocity of the emerging, bullet with respect to the block is, 1) 10 m/s 2) 9 m/s 3) 1 m/s., 4) 5 m/s, 37. A shell is fired from the ground at an angle θ, with horizontal with a velocity 'v'. At its highest, point it breaks into two equal fragments. If one, fragment comes back through its initial line of, motion with same speed, then the speed of the, second fragment will be, 1) 3v cos θ, 2) 3v cos θ / 2, , of a string which is made to pass over light,, smooth pulley are in equilibrium on a fixed, smooth wedge as shown in figure. If θ = 600, and α = 300 ,then the ratio of M 1 to M 2 is, , 1) 1 : 2, , 2) 2 : 3, , 3) 1: 3, 4) 3 :1, 41. If ‘O’ is at equilibrium then the values of the, tension T1 and T2 respectively., , T1, , 3) 2v cos θ, 4) 3v cos θ / 2, 38. Two trolleys of masses m and 3m are connected by a spring. They are compressed and, released, they move off in opposite direction, and come to rest after covering distances s1, and s2 respectively. If the frictional force between trolley and surface is same in both the, cases then the ratio of distances s1 : s2 is, 1) 1:9, 2) 1:3, 3) 3:1, 4) 9:1, 39. Two particles of masses m1 and m2 in projecr, r, tile motion have velocities v1 and v2 respectively at time t = 0 . They collide at time t0 ., r, r, Their velocities become v11 and v21 at time 2t0, NARAYANAGROUP, , O, , 1500, 600, T2, , 20N, 1) 20N, 30N, , 2) 20 3 N,20N, , 3) 20 3 N,20 3 N, , 4) 10N, 30N
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 42. A 1N pendulum bob is held at an angle θ, from the vertical by a 2 N horizontal force F, as shown in the figure. The tension in the string, supporting the pendulum bob (in newton) is, (2011E), , 1) cosθ 2), , 2, cosθ, , 3), , 5, , 4)1, , MOTION ON A HORIZONTAL, ROUGH SURFACE, 43. The coefficient of friction between a, hemispherical bowl and an insect is 0.44, and the radius of the bowl is 0.6m. The, maximum height to which an insect can crawl, in the bowl will be, 1) 0.4m 2) 0.2m, 3) 0.3m 4) 0.1m, 44. A 500 kg horse pulls a cart of mass 1500 kg, along a level road with an acceleration of, 1 m/s2. If coefficient of sliding friction is 0.2,, then force exerted by the earth on horse is, 1)3000N 2)4000N 3)5000 N 4)6000N, 45. An aeroplane requires for take off a speed, of 108 kmph the run on the ground being, 100m. Mass of the plane is 104kg and the, coefficient of friction between the plane and, the ground is 0.2. Assuming the plane, accelerates uniformly the minimum force, , required is ( g = 10ms −2 ), 1) 2 x 104 N, 2) 2.43 x 104 N, 4, 3) 6.5 x 10 N 4) 8.86 x 104 N, 46. A duster weighs 0.5N. It is pressed against a, vertical board with a horizontal force of 11N. If, the co-efficient of friction is 0.5 the minimum, force that must be applied on the duster parallel, to the board to move it upwards is, 1) 0.4 N 2) 0.7 N, 3) 6 N 4) 7 N, 47. A man of mass 65 kg. is standing stationary, with respect to a conveyor belt which is accel-, , 60, , erating with 1m / s 2 . If µ s is 0.2, the net force, on the man and the maximum acceleration of, the belt so that the man is stationary relative, 2, to the belt are ( g = 10m / s ), , 1) zero, 2m / s 2 2) 65 N , 2m / s 2, 3) zero, 1m / s 2 4) 65 N ,1m / s 2, 48. A man of mass 60kg sitting on ice pushes a, block of mass of 12kg on ice horizontally with, a speed of 5ms −1 .The coefficient of friction, between the man and ice and between block, and ice is 0.2. If g = 10ms −2 ,the distance, between man and the block, when they come, to rest is, 1)6m, 2)6.5m, 3) 3m, 4)7m, 49. A vehicle of mass M is moving on a rough, horizontal road with a momentum P. If the, coefficient of friction between the tyres and, the road is µ , then the stopping distance is, (EAM-2012), P, P, P2, P2, 1), 2), 3), 2 4), 2µ M 2 g, 2µ Mg, 2µ M g, 2µ Mg, 50. The rear side of a truck is open and a box of, 40 kg mass is placed 5 m away from the open, end as shown in figure. The coefficient of friction between the box the surface below it is, 0.15. On a straight road, the truck starts from, rest and accelerates with 2ms −2 . At what distance from the starting point does the box fall, from the truck? (Ignore the size of the box.), , 1) 20m 2) 10m, 3) 20m 4) 5m, 51. A grinding machine whose wheel has a radius, 1, of, is rotating at 2.5 rev/sec. A tool to be, π, sharpened is held against the wheel with a, force of 40N. If the coefficient of friction between the tool and the wheel is 0.2, power required is, 1) 40 W 2) 4 W, 3) 8 W, 4) 10 W, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 52. A block A of mass 3kg and another block B of, , mass 2 kg are connected by a light inextensible, string as shown in figure. If the coefficient of, friction between the surface of the table and A, is 0.5. What maximum mass C is to be placed, on A so that the system is to be in equlibrium?, , tion is doubled for the same length of the plane,, what will be the velocity of the body on reaching the ground, 1, , 1)v, , 2) 2v, , 1, , 3) ( 2cos θ ) 2 v 4) ( 2sin θ ) 2 v, , 57. The force required to move a body up a rough, inclined plane is double the force required to, prevent the body from sliding down the plane., The coefficient of friction when the angle of, inclination of the plane is 600 is (EAM - 2014), , 1, 1, 1, 1, 2), 3), 4), 2, 3, 2, 3, 58. A smooth block is released from rest on a 450, inclined plane and it slides a distance ‘d’. The, time taken to slide is n times that on a smooth, inclined plane. The coefficient of friction, (2010E), 1), , 1) 3kg, , 2) 2kg, , 3) 1kg, , 4) 4kg, , Motion Of A Body On The Inclined Plane, 53. A block slides down a rough inclined plane of, slope angle θ with a constant velocity. It is, then projected up the same plane with an initial, velocity v. The distance travelled by the, block up the plane before coming to rest is, , 1) µ k = 1 −, , 1, n2, , 2) µ k = 1 −, , 1, n2, , 2, , 4 gv 2, sin?, 54. The minimum force required to start pushing, a body up a rough (frictional coefficient µ ), 1), , v, v2, v2, 2), 3), 4gsin?, 2gsin?, gsin?, , 4), , inclined plane is F1 while the minimum force, needed to prevent it from sliding down is F2 ., If the inclined plane makes an angle θ with, the horizontal such that tan θ = 2 µ , then the, F1, ratio F is, (AIEEE-2011), 2, 1) 4, 2) 1, 3) 2, 4) 3, 55. The horizontal acceleration that should be, given to a smooth inclined plane of angle, , sin–1, , 1, to keep an object stationary on the, l, , plane, relative to the inclined plane is, 1), , g, , 2) g l 2 − 1 3), l 2 −1, , l2 −1, 4), g, , g, l +1, , 1) µ = 2 tan θ, , 2) µ =, , 2, tan θ, , 3) µ = tan θ, , 4) µ =, , 1, tan θ, , 60. A 30 kg box has to move up an inclined plane, , of slope 300 to the horizontal with a uniform, velocity of 5 ms-1.If the frictional force retarding, the motion is 150N, the horizontal force required, to move the box up is(g=10ms-2), , 2, , 56. A body is released from the top of a smooth, inclined plane of inclination θ . It reaches the, bottom with velocity v . If the angle of inclina-, , NARAYANAGROUP, , 1, 1, 4) µ k =, 2, 1− n, 1 − n2, 59. The upper half of an inclined plane of, inclination ‘ θ ’ is perfectly smooth while the, lower half is rough. A block starting from rest, at the top of the plane will again come to rest, at the bottom. The coefficient of friction, between the block and the lower half of the, plane is given by, (2013E), , 3) µ k =, , 1) 300×, 3) 300N, , 2, N, 3, , 2) 300×, 4) 150N, , 3, N, 2
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , PULLING / PUSHING A BODY, 61. A block weighing 10kg is at rest on a, horizontal table. The coefficient of static, friction between the block and the table is, 0.5. If a force acts downward at 60º with the, horizontal, how large can it be without, causing the block to move? (g = 10ms–2), 1) 346 N 2)446 N 3) 746 N 4)846 N, 62. A pulling force making an angle θ with the, horizontal is applied on a block of weight W, placed on a horizontal table. If the angle of, friction is φ , the magnitude of the force, required to move the body is equal to, WCosφ, , 1) Cos(θ − φ ), , W sin φ, , 2) Cos(θ − φ ), , WTanφ, , WSinφ, , 3) Sin (θ − φ ), 4) Tan(θ − φ ), 63. A block of mass 3 kg is kept on a frictional, 1, , surface with µ =, . The minimum force to, 2 3, be applied as shown to move the block is, , rb, vb, v 2b, vb 2, 2), 3), 4) 2, Rg, R g, Rg, Rg, 66. The centripetal force required for a 1000 kg, car travelling at 36 kmph to take a turn by 900, in travelling along an arc of length 628 m is, 1) 250 N 2) 500 N 3) 1000 N 4) 125 N, 67. A small coin is placed on a flat horizontal turn, table. The turn table is observed to make three, revolutions in 3.14 sec. What is the coefficient, of static friction between the coin and turn, table if the coin is observed to slide off the, turn table when it is greater than 10cm from, the centre of turn table, 1) 0.4 2) 0.36, 3) 4, 4) 0.004, 68. A particle of mass m is suspended from the, ceiling through a string of length L. The particle moves in a horizontal circle of radius r ., The speed of the particle is, 1), , rg, , r g, , 2), , 1, 2 4, , 3), , 1) maω 2 2) 3maω 2 3), 1) 5N, , 2) 20 N 3) 10 N, , 4) 20/3 N, , CIRCULAR MOTION, 64. A car is moving in a circular horizontal track of, radius 10 m with a constant speed of 10ms −1 .A, plumb bob is suspended from the roof of the, car by a string of length 1m. The angle made, by the string with vertical is ( g = 10ms −2 ), 1) 00, 2) 300, 3) 450, 4) 600, 65. A vehicle is moving with a velocity v on a, curved road of width b and radius of curvature R . For counteracting the centrifugal force, on the vehicle the difference in elevation required in between the outer and inner edges, of the road is, , 62, , mgL, , r g, , 1, L2 − r 2, ( L2 − r ) ( L2 − r ) 4) ( L2 − r 2 ) 2, 69. Three point masses each of mass m are joined, together using a string to form an equilateral, triangle of side a . The system is placed on a, smooth horizontal surface and rotated with a, constant angular velocity ω about a vertical, axis passing through the centroid. Then the, tension in each string is, , 1), , 1, 2 2, , maω 2, 3, , 4), , maω 2, 3, , PREVIOUS EAMCET QUESTIONS, 70. A steel wire can withstand a load up to 2940N., A load of 150kg is suspended from a rigid, support. The maximum angle which the wire, can be displaced from the mean position , so, that the wire does not break when the load, passes through the position of equilibrium is, (EAM - 2008), 0, 0, 0, 1) 30, 2) 60, 3) 80 4) 850, 71. A car is travelling along a curved road of, radius r. If the coefficient of friction between, the tyres and the road is µ , the car will skid, if its speed exceeds, (2010M), 1) 2 µ rg 2) 3µ rg 3) 2 µ rg 4) µ rg, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 72. A boy of mass 50 kg is standing on a weighing, machine placed on the floor of a lift. The, machine reads his weight in newtons. The, reading of the machine if the lift is moving, upwards with uniform speed of 10 ms-1, (2009M), 1) 510 N 2) 480 N 3) 490 N 4) 500 N, , LEVEL-II (C.W) - KEY, 01) 3, 07) 3, 13) 3, 19) 2, 25) 2, 31) 1, 37) 1, 43) 4, 49) 3, 55) 1, 61) 3, 67) 2, , 02) 1, 08) 1, 14) 2, 20) 2, 26) 2, 32) 1, 38) 4, 44) 4, 50) 1, 56) 3, 62) 2, 68) 2, , 03) 1, 09) 3, 15) 1, 21) 4, 27) 1, 33) 3, 39) 3, 45) 3, 51) 1, 57) 2, 63)2, 69) 3, , 04) 3, 10) 2, 16) 3, 22) 3, 28) 1, 34) 3, 40)3, 46)3, 52)3, 58) 1, 64) 3, 70) 2, , 05) 3, 11) 1, 17) 4, 23) 4, 29) 3, 35) 2, 41) 2, 47) 1, 53) 1, 59) 1, 65) 1, 71) 4, , 06)4, 12)3, 18)3, 24)2, 30)3, 36)3, 42)3, 48)2, 54)4, 60)1, 66)1, 72)3, , LEVEL-II (C.W) - HINTS, 1., 2., 3., , dP, dPx, , Fy = y , F = Fx 2 + Fy 2, dt, dt, F, F, a = , FR = F12 + F22 − F3 , a1 = R, m, m, For mass ‘m’ , F 1 = ma, For mass ‘M’ F − F 1 = Ma1, Fx =, , (, , ), , 14. ∆Px = m(vx − ux ) , ∆Py = mv y ,, J = ∆P = ∆px2 + ∆p y2, m, , F, 15. F = ma , a =, , T = + M a, M +m, 2, , a, 16. v = u + at , tan θ = 17. ma cos α = mg sin α, g, 18. For man : T + F − W p = m p a, , For chair : T − F − Wc = mc a, 19. While descending, Mg − FB = M α, , While ascending FB − ( M − m ) g = ( M − m ) α, , 20., , 21., 22., 23., 24., , 25., , Where ‘ FB ’is the buoyant force, To move up with an acceleration a the monkey, will push the rope downwards with a force of ma., Tmax = mg + mamax, T = F , 2T cos θ = mg, To make rope straight, θ = 900, F − T = m1a , T = m2 a, Fnet, F = ma a = M , T1 = m1a , T2 = ( m1 + m2 ) a, Fnet, F = ma , a = m + m ,, 1, 2, For engine F − f − T = m1a, T, F = ma , 2T = p, a = m, m1 − m2 , a, =, , g, ,, F = ma, m1 + m2 , , d 2S, 4. F = ma , F = m 2 = constant, dt, 5. For t=0,F=0 ;, For t = λ , F = 0.63F0, F, 1 2, dp, 6. a = R , S = ut + at 7. F =, , F = 2 ρ AV 2, m, 2, dt, m0 , 8. u0 = − gt , V = uo + u log e , m, F, 9. V = u + at , a =, m, dp m(v − u ), 2, 2, =, 10. v − u = 2as , F =, dt, t, 2, 2, 11. u = 2 gh ,using v − u = 2as retardation due to air, , 26., , resistance a1 = g + a, Force due to air resistance = Ma ', 12. Find net force and hence, a=Fnet/m, 13. Impulse = Area of semi circle, , m1 − m2 , 29. F = ma , a = g m + m ,F –T = 0 and, 1, 2 , T = 2mg also T–mg = ma1 Finally a < a1, , NARAYANAGROUP, , 27., , from diagram mg − T = ma → (1) , T = nma → (2), for nth block Tn = ma, 28. T1 = F cos θ , T2 = F sin θ , T = T12 + T2 2
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 3., , 4., , 5., , 6., , 7., , A ball of mass 10 gm dropped from a height of, 5m hits the floor and rebounds to a height of, 1.25m. If the ball is in contact with the ground, for 0.1s, the force exerted by the ground on, the ball is (g=10 m/s2), 1) 0.5 N 2) 1.5 N 3) 0.15N 4) 2.5 N, A stream of water flowing horizontally with a, speed of 15 ms-1 pushes out of a tube of cross, sectional area 10-2m2 and hits a vertical wall, near by what is the force exerted on the wall, by the impact of water assuming. that it does, not rebound? (Density of water=1000 kg m -3), 1)1250N 2)2250N 3)4500N 4) 2550N, What is the magnitude of the total force on a, driver by the racing car he operates as it, accelerates horizontally along a straight line, from rest to 60m/s in 8.0s (mass of the, driver=80kg), 1) 0.06KN 2)0.78KN 3)1.4KN 4) 1.0KN, A base ball of mass 150 gm travelling at speed, of 20 m/s is caught by a fielder and brought to, rest in 0.04 s.The force applied to the ball and, the distance over which this force acts are, respectively, 1) 75 N, 0.8 m 2) 37.5 N,0.4 m, 3) 75 N,0.4 m, 4) 37.5 N, 0.8m, A dynamometer D is attached to two blocks, of masses 6 kg and 4 kg. Forces of 20 N and, 10N are applied on the blocks as shown in Fig., The dynamometer reads, , 1) 10N, , 2) 20N, , 3) 6N 4) 14N, , Impulse, 8., , A particle of mass m moving with velocity u, makes an elastic one-dimensional collision, with a stationary particle of mass m. They are, in contact for a very brief time T. Their force, of interaction increases from zero to F0 linearly, 1, in time T ., The magnitude of F 0 is, 2, , UNIFORM CIRCULAR MOTION, , mu, 2mu, mu, 3mu, 2), 3), 4), T, T, 2T, 2T, The position-time graph of a body of mass, 0.04kg is shown in the figure. The time between two consecutive impulses received by, the body and the magnitude of each impulse, is, (AIEEE-2010), 1), , 9., , 1) 4 sec, 4 × 10−4 kgm / s, 2) 2 sec, 8 × 10 −4 kgm / s, 3) 6 sec, 4 × 10−4 kgm / s, 4) 8 sec, 8 × 10 −4 kgm / s, , OBJECTS SUSPENDED BY STRINGS, AND APPARENT WEIGHT, 10. The elevator shown in figure is descending, with an acceleration of 2m/s2. The mass of, the block A = 0.5 kg. The force exerted by, the block A on block B is, , 1) 2 N, 2) 4 N 3) 6 N, 4) 8 N, 11. A block of mass m is pulled by a uniform chain, of mass m tied to it by applying a force F at, the other end of the chain. The tension at a, point P which is at a distance of quarter of the, length of the chain from the free end, will be, , 1), , NARAYANAGROUP, , 3F, 4, , 2), , 7F, 8, , 3), , 6F, 7, , 4), , 4F, 5
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 12. Two masses of 8 kg and 4 kg are connected, by a string as shown in figure over a, frictionless pulley. The acceleration of the, system is, , 1) 6.15 N 2) 4.92 N 3) 9.84N 4) 2.46N, 16. Three blocks of equal masses (each 3kg) are, suspended by weightless strings as shown. If, applied force is 100N,then T1 is equal to, ( g = 10m / s 2 ), , m1, 8kg, , 30°, , m2, 1) 4m / s 2) 2m / s 3) zero 4) 9.8m / s 2, 13. Consider the system shown in figure. The, pulley and the string are light and all the, surface are frictionless. The tension in the, string is, (Take g = 10 m/s2), 2, , 2, , 1) 0 N, 2) 1 N, 3) 2 N, 4) 5 N, 14. In the figure, a smooth pulley of negligible, weight is suspended by a spring balance., Weights of 1 kg and 5kg are attached to the, opposite ends of a string passing over the, pulley and move with an acceleration due of, gravity. During their motion, the spring, balance reads a weight of, , 1) 6 kg, 2) less than 6 kg, 3)more than 6 kg 4)may be more or less than 6kg, 15. A chain consisting of 5 links each of mass 0.1, kg is lifted vertically up with a constant, acceleration of 2.5m/s 2 . The force of, interaction between 1st and 2nd links as shown, 66, , 1)130N 2)190N 3)100N, 4)160N, 17. Pulleys and strings are massless. The, horizontal surface is smooth.What is the, acceleration of the block, , F, F, 2F, m, 2), 3), 4), 2m, m, m, 2F, 18. When a train starting from rest is uniformly, accelerating, a plumb bob hanging from the roof, of a compartment is found to be inclined at an, angle of 450 with the vertical . The time taken by, the train to travel a distance of ½km will be nearly, 1) 7s, 2)10s, 3)15s, 4) 25s, 19. The pulley and strings shown in the figure are, smooth and of negligible mass. For the system, to remain in equilibrium, the angle ‘ θ ’ should be, , 1), , 1)00, , 2) 300, , 3) 450, , 4) 600, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 20. Two bodies of masses 4kg and 6kg are, attached to the ends of a string which passes, over a pulley,the 4kg mass is attached to the, table top by another string.The tension in this, string T1 is equal to, , 2 iˆ m/s and the other with a velocity 3 ĵ m/s., If the explosion takes place in 10-5 sec, the, average force acting on the third piece in, Newtons is:, ^ ^ −5, 2 i + 3 j 10, 1) , , , , T1, , , , T, , 4kg, , 6kg, , 2) 10.6N, , 3) 25N, , 4) 20N, , 0, 21. Acceleration of block m is (θ < 45 ), , A, m, , θ, , B, , C, , 1) g sin θ, , 2) g cos θ, , 3) g ( cos θ + sin θ ), , 4) g ( cos θ − sin θ ), , LAW OF CONSERVATION OF, MOMENTUM, 22. A stationary shell breaks into three fragments., The momentum of two of the fragments is P, each and m ove at 600 to each other. The, momentum of the third fragment is, 3), , P, 3, , 4), , 3P, 23. An object initially at rest explodes,, disintegrating into 3 parts of equal mass. Parts, 1 and 2 have the same initial speed 'v', the, velocity vectors being perpendicular to each, other. Part 3 will have an initial speed of, 1) 2 v 2) v / 2 3) v / 2 4) 2v, 24. A man of 50 kg is standing at one end on a, boat of length 25m and mass 200 kg. If he, starts running and when he reaches the other, end, has a velocity 2ms-1 with respect to the, boat. The final velocity of the boat is, 2, 2, 8, 8, 1) ms-1 2) ms-1 3) ms-1 4) ms-1, 3, 5, 5, 3, 25. A stationary body of mass 3 kg explodes into, three equal pieces. Two of the pieces fly off, at right angles to each other, one with a velocity, NARAYANAGROUP, , , , , , , , ^, , ^, , −5, 4) 2 j − 2 i 10, , , , , , MOTION ON A HORIZONTAL ROUGH SURFACE, 26. A particle is placed at rest inside a hollow, hemisphere of radius R. The coefficient of, friction between the particle and the hemisphere, is µ =, , 1, . The maximum height up to which, 3, , , 3, R, 3R, 3, R 4), 2) 1 − 2 R, 3), 8, 2, , , 2, 27. A horizontal force is applied on a body on a, rough horizontal surface produces an acceleration ‘a’. If coefficient of friction between, the body and surface which is µ is reduced, to µ /3, the acceleration increases by 2 units., The value of ‘ µ ’ is, 1) 2/3g 2) 3/2g, 3) 3/g, 4) 1/g, 28. A block of mass 4kg is placed in contact with, the front vertical surface of a lorry. The co, efficient of friction between the vertical surface and block is 0.8.The lorry is moving with, an acceleration of 15m / s 2 .The force of fric-, , 1), , M, , 2) 2P, , , , ^, , the particle can remain stationary is, , a0=g, , 1) P, , ^, , −5, 3) 3 j − 2 i 10, , T, , 1) 10N, , ^, , ^, , +5, 2) − 2 i + 3 j 10, , tion between lorry and block is ( g = 10ms −2 ), 1) 48N 2) 24N 3) 40N, 4) Zero, 29. A person of mass 72kg sitting on ice pushes a, block of mass of 30kg on ice horizontally with a, speed of 12ms −1 . The coefficient of friction between the man and ice and between block and, ice is 0.02. If g = 10ms −1 ,the distance between, man and the block,when they come to rest is, 1) 360m 2) 10m 3) 350m 4) 422.5m, 30. Consider a 14- tyre truck, whose only rear 8, wheels are power driven (means only these 8, wheels can produce an acceleration). These 8, wheels are supporting approximately half of the, load. If coefficient of friction between road and, each tyre is 0.6, then what could be the maximum attainable acceleration by this truck is, 1) 6ms-2 2) 24ms-2, 3) 3ms-2 4)10ms-2
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 31. A block is sliding on a rough horizontal surface. If the contact force on the block is 2, times the frictional force, the coefficient of friction is, , 1, 3) 2, 4) 1, 2, 32. A block is in limiting equilibrium on a rough, horizontal surface. If the net contact force is, 1) 0.25, , 2), , 3 times the normal force, the coefficient of, static friction is, , 1, 1, 3), 0.5, 4), 2, 2, 3, 33. A block of mass 2kg is placed on the surface, of trolley of mass 20kg which is on a smooth, surface. The coefficient of friction between, the block and the surface of the trolley is 0.25., If a horizontal force of 2 N acts on the block,, the acceleration of the system in ms −2 is, 1), , 2), , ( g = 10ms ), −2, , 1) 1.8, 2) 1.0, 3) 0.9, 4) 0.09, 34. A man slides down on a telegraphic pole with, an acceleration equal to one-fourth of, acceleration due to gravity. The frictional, force between man and pole is equal to (in, terms of man’s weight W), , 37. Sand is piled up on a horizontal ground in the, form of a regular cone of a fixed base of radius R. The coefficient of static friction between sand layers is µ . The maximum volume of sand that can be piled up, without the, sand slipping on the surface is, µ R3, 1), 3π, , πR, µ R3, 2), 3), 3µ, 3, , 3, , 4), , µπ R3, 3, , MOTION OF A BODY ON THE, INCLINED PLANE, 38. A body is allowed to slide from the top along a, smooth inclined plane of length 5m at an angle, of inclination 300. If g=10ms-2, time taken by, the body to reach the bottom of the plane is, , 1, 3, s 4) 2s, s 2) 1.414s 3), 2, 2, 39. A body slides down a smooth inclined plane, of height h and angle of inclination 30º, reaching the bottom with a velocity v. Without, changing the height, if the angle of inclination, is doubled, the velocity with which it reaches, the bottom of the plane is, 1) v, 2) v/2, 3) 2v, 4) 2 v, 40. A body is projected up along an inclined plane, from the bottom with speed is 2v. If it reaches, the bottom of the plane with a velocity v, if θ, is the angle of inclination with the horizontal, and µ be the coefficient of friction., 1), , W, 3W, W, 2), 3), 4) W, 4, 4, 2, 35. A box is placed on the floor of a truck moving, with an acceleration of 7 ms −2 . If the, coefficient of kinetic friction between the box, and surface of the truck is 0.5,find the, acceleration of the box relative to the truck, , 5, 3, 1, 2, tan θ 2) tan θ 3) tan θ 4) tan θ, 3, 5, 5, 5, 41. The minimum force required to move a body, up on an inclined plane is three times the, minimum force required to prevent it from sliding, down the plane. If the coefficient of friction, between the body and the inclined plane is, , 1)1.7 ms −2 2)2.1 ms −2 3)3.5 ms −2 4)4.5 ms −2, 36. A block is placed at a distance of 2m from the, rear on the floor of a truck (g=10ms-2). When, the truck moves with an acceleration of 8ms-2,, the block takes 2 sec to fall off from the rear of, the truck. The coefficient of sliding friction, between truck and the block is, 1) 0.5, 2) 0.1, 3) 0.8, 4) 0.7, , , the angle of the inclined plane is, 2 3, 1) 60º 2) 45º, 3) 30º, 4) 15º, 42. Starting from rest, the time taken by a body, sliding down on a rough inclined plane at 450, with the horizontal is, twice the time taken to, travel on a smooth plane of same inclination, and same distance. Then the coefficient of, kinetic friction is, (2008 E), 1) 0.25 2) 0.33, 3) 0.50 4) 0.75, , 1), , 68, , 1), , 1, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 43. A body is sliding down a rough inclined plane., The coefficient of friction between the body, and the plane is 0.5. The ratio of the net force, required for the body to slide down and the, normal reaction on the body is 1 : 2. Then the, angle of the inclined plane is, 2) 300, 3) 450, 4) 600, 1) 150, 1, 44. A body takes 1 times as much time to slide, 3, down a rough inclined plane as it takes to slide, down an identical but smooth inclined plane., If the angle of inclination is 450, find the, coefficient of friction., 1, 3, 5, 7, 1), 2), 3), 4), 16, 16, 16, 16, 45. A body is sliding down an inclined plane having, coefficient of friction 1/3. If the normal reaction, is three times that of the resultant downward, force along the inclined plane, the angle between, the inclined plane and the horizontal is, −1 1 , 1) tan , 2) tan −1 ( 2 ), 2, , −1 2 , −1 3 , 3) tan , 4) tan , 3, 2, 46. A box of mass 4 kg is placed on a rough inclined, plane of inclination 600. Its downward motion, can be prevented by applying an upward pull, is F and it can be made to slide upwards by, applying a force 3F. The coefficient of friction, between the box and inclined plane is, 2, 1, 1, 3, 1), 2), 3), 4), 3, 2, 2, 2, , UNIFORM CIRCULAR MOTION, , CIRCULAR MOTION, 49. A ball of mass 0.5kg is attached to the end of, a string having length 0.5m. The ball is rotated, on a horizontal circular path about vertical, axis. The maximum tension that the string can, bear is 324N. The maximum possible value of, angular velocity of ball in (rad/s), 1) 9, 2) 18, 3) 27, 4) 36, 50. A disc rotates at 60 rev/min around a vertical, axis. A body lies on the disc at the distance of, 20cm from the axis of rotation. What should, be the minimum value of coefficient of friction, between the body and the disc, so that the body, will not slide off the disc, 1) 8π 2 2) 0.8π 2 3) 0.08π 2 4) 0.008π 2, 51. A car is moving on a circular level road of, radius of curvature 300m. If the coefficient of, friction is 0.3 and acceleration due to gravity, is, 10m/s2. The maximum speed the car, can have is, 1) 30km/h, 2) 81km/h, 3) 108km/h, 4)162km/h, , KEY-LEVEL-II (H.W), 1) 2, 7)4, 13) 4, 19) 3, 25) 2, 31) 4, 37) 4, 43) 3, 49) 4, , PULLING / PUSHING A BODY, 47. A block of weight 100N is lying on a rough, horizontal surface. If coefficient of friction, 1, . The least possible force that can move, 3, the block is, 100, 1), N, 2) 100 3 N, 3, 3) 50 3 N, 4) 50N, 48. A weight W rests on a rough horizontal plane., If the angle of friction is θ , the least force, that can move the body along the plane will be, 1) W cos θ, 2) W tan θ, 3) W cot θ, 4) W sin θ, NARAYANAGROUP, , 2)2, 8) 2, 14) 2, 20) 4, 26)2, 32) 1, 38) 2, 44) 4, 50) 3, , 3) 2, 9) 2, 15) 2, 21) 4, 27) 3, 33) 4, 39) 1, 45) 3, 51) 3, , 4) 2, 10) 2, 16) 1, 22) 4, 28) 3, 34) 2, 40) 2, 46) 2, , 5)3, 11) 2, 17) 1, 23) 1, 29) 4, 35)2, 41) 3, 47) 4, , 6)3, 12)3, 18)2, 24)2, 30)3, 36)4, 42)4, 48)4, , HINTS - LEVEL – II (H.W), 1., 2., , 3., 4., 5., , v − u 2 = 2as ; F − mg = ma, When it reaches the point ‘P’again S=0,, F, 1 2, Using a = ; S = ut + at, m, 2, dp m( v−u), F= =, where v = 2 gh2 , u = 2 gh1, dt, t, dp, F=, = Aρ v 2, dt, m (v − u ), ; F2 = mg , F = F1 + F2, F = ma , F1 =, t, 2
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , LEVEL-III, NEWTON’S LAWS OF MOTION, 1., , 2., , A rope is stretched between two boats at rest. A, sailor in the first boat pulls the rope with a, constant force of 100N. First boat with the sailor, has a mass of 250kg where as the mass of second, boat is double of this mass. If the initial distance, between the boats was 100m, the time taken for, two boats to meet each other in seconds (neglect, water resistance between boats and water), 1) 13.8 2) 18.3, 3) 3.18, 4) 31.8, In order to raise a block of mass 100kg a man, of mass 60kg fastens a rope to it and passes, the rope over a smooth pulley. He climbs the, rope with an acceleration, , 3., , A, B, , 5., , C, 1) zero 2) 13 N 3) 3.3 N, 4) 19.6 N, In the figure show n a3 = 6m/s2 (downwards), and a2 = 4m/s2 (upwards). Find acceleration, of 1., , 5g, relative to rope., 4, , −2, The tension in the rope is ( g = 10ms ), 1) 1432N 2) 928 N 3) 1218N 4) 642N, In the pulley-block arrangement shown in, figure.Find the relation between acceleration, of block A and B., , 1, , 2, , 6., , B, A, , 4., , 3, 1) 1m/sec2 upwards, 2) 2m/sec2 upwards, 3) 1m/sec2 downwards 42m/sec2 downwards, A man of mass m stands on a platform of equal, mass m and pulls himself by two ropes passing, over pulleys as shown in figure. If he pulls, each rope with a force equal to half his weight,, his upward acceleration would be, , 1) aB=-3aA 2) aB=-aA 3) aB=-2aA 4) aB=-4aA, Three equal weights A, B and C of mass, 2 kg each are hanging on a string passing over, a fixed frictionless pulley as shown in the fig., The tension in the string connecting weights, B and C is, , 1), NARAYANAGROUP, , g, 2, , 2), , g, 4, , 3) g, , 4) zero
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 7., , A block is sliding along inclined plane as shown, in figure. Ifthe acceleration ofcham beris‘a’, as shown in the figure. The time required to, cover a distance L along inclined plane is, a, , M, , 8., , 2L, g sin θ − a cos θ, , 2), , 2L, g sin θ + a sin θ, , 3), , 2L, g sin θ + a cos θ, , 4), , 2L, g sin θ, , 30°, , 9., , 2F, 2) m, , x, x, 2, , F, 4) m, , − x2 ), , (a, , (a, , x, 2, , − x2 ), , x, 2, , − x2 ), , m2=2kg, , F = 2t, , A, , O, , (a, , − x2 ), , 2, , m1=1kg, , An inclined plane makes an angle 300 with the, horizontal. A groove (OA) of length 5 m cut, in, the plane makes an angle 300 with OX. A short, smooth cylinder is free to slide down under, the influence of gravity. The time taken by the, cylinder to reach from A to O is (g = 10 m/s2), Cylinder, , (a, , 10. A lift is going up, the total mass of the lift and, the passengers is 1500 kg. The variation in, the speed of lift is shown in fig. Then the, tension in the rope at t = 1 s will be, , θ, , 1), , 2F, 1), m, F, 3) 2m, , X, , 1) 4 s, 2) 2 s, 3) 3 s, 4) 1 s, Two masses each equal to m are lying on Xaxis at (-a, 0) and (+a, 0), respectively, as, shown in fig. They are connected by a light, string. A force F is applied at the origin along, vertical direction. As a result, the masses move, towards each other without loosing contact, with ground. What is the acceleration of each, mass? Assume the instantaneous position of, the masses as (-x, 0) and (x, 0), respectively, y, , F, , F = 15N, µ1=0.6, , µ2=0.5, , 1)17400N 2)14700N 3)12000N 4)10000 N, 11. In the above problem the tension in the rope, will be least at, 1) t = 1 s 2) t = 4 s 3) t = 9 s 4) t = 11 s, 12. A piece of wire is bent in the shape of a, parabola y = kx 2 (y-axis vertical) with a bead, of mass m on it. The bead can slide on the, wire without friction. It stays at the lowest, point of the parabola when the wire is at rest., The wire is now accelerated parallel to the xaxis with a constant acceleration a. The, distance of the new equilibrium position of the, bead, where the bead can stay at rest with, respect to the wire, from the y-axis is :, a, a, 2a, a, 2), 3), 4), 1), gk, 2 gk, gk, 4 gk, , Friction, 13. A block of mass m = 4kg is placed over a rough, inclined plane having coefficient of friction, µ = 0.6 as shown in fig. A force F = 10N is, applied on the block at an angle 300 . The contact, force between the block and the plane is, F, 30°, , -a, 0, -x, , a, 0, , m, m, ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||, , x, 45°, , 1)10.65N 2)16.32N 3) 27.15 N 4) 32.16 N, 72, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 14. A block of mass m slides down an inclined, plane of inclination θ with uniform speed. The, coefficient of friction between the block and, the plane is µ . The contact force between, the block and the plane is, 1) mg sin θ 1 + µ 2 2) ( mg sin θ ) + ( µ mg cosθ ), 3) mg sin θ, 4) mg, 15. In the pulley arrangement shown, the pulley, P2 is movable.Assuming coefficient of friction, between m and surface to be µ , the minimum, value of M for which m is at rest is, 2, , 2, , P1, , m, , F = 15N, µ1=0.6, , F, , A, 2m, , B, m, , Coefficient of friction between two blocks, shown in figure is µ = 0.4 . the blocks are, given velocities of 2 m/s and 8 m/s in the, directions shown in figure. Find., 2 m/s, , 1 kg, , 8 m/s, , 2 kg, , 20. The time when relative motion between them, will stop, 1) 1 sec 2) 2 sec 3) 3 sec, 4) 4 sec, 21. The common velocities of blocks upto that, instant., 1) 4m/sec 2) 6m/sec 3) 8 m/sec 4) 10 m/sec, 22. Displacements of 1 kg and 2 kg blocks upto, that instant (g = 10 m/s2), 1) 4 m towards right, 7 m towards right, 2) 4 m towards left, 7 m towards right, 3) 4 m towards left, 7 m towards left, 4) 4 m towards right, 7 m towards left, 23. A 2kg block is pressed against a rough wall, by a force F = 20N as shown in figure. find, acceleration of the block and force of friction, acting on it. (Take g = 10 m/s2), , C, 2m, , 3 , 5 , 1) , 2) , mg, mg, 2µ , 2µ , 5, 3, 3) µ mg, 4) µ mg, 2, 2, 19. Two blocks A and B are separated by some, distance and tied by a string as shown in the, figure. The force of friction in both the blocks, at t = 2s is., NARAYANAGROUP, , 2) 2N( → ), 5N( ← ), 4) 1N( ← ), 10N( ← ), , Passage:, , M, , m, M, µm, µM, 2) m =, 3) M =, 4) m =, 2µ, 2µ, 2, 2, 16. On an inclined plane of inclination angle 300,, a block is placed. It is observed that the force, to drag the block along the plane upwards is, smaller than the force required to lift it. The, maximum value of coefficient of friction is, 1, 1, 2, 3, 2), 3), 4), 1), 3, 2, 3, 2, 17. A body slides over an inclined plane forming, an angle of 450 with the horizontal. The, distance x travelled by the body in time t is, described by the equaiton x = kt2, where k =, 1.732. The coefficient of friction between the, body and the plane has a value, 1) µ = 0.5 2) µ = 1 3) µ = 0.25 4) µ = 0.75, 18. The system is pushed by a force F as shown in, the figure. All surfaces are smooth except, between B and C. Coefficient of friction, between B and C is µ . Minimum value of F, to prevent block B from downward slipping is, , µ2=0.5, , 1) 4N( → ), 5N( ← ), 3) 0N( → ), 10N( ← ), , P2, , 1) M =, , m2=2kg, , m1=1kg, F = 2t, , 20N, , µs= 0.8, , 2 kg, , µk= 0.6, , wall, , 1) 4 m/sec2 downward, 12N upward, 2) 2 m/sec2 downward, 6N upward, 3) 12 m/sec2 downward, 4N upward, 4) 8 m/sec2 downward, 12N upward
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 24. Three blocks are kept as shown in figure., Acceleration of 20 kg block with respect to, ground is, , 3 kg, B, 6 kg, A, , 100 N, 10 kg, , 20 kg, , F, , µ = 0.5, µ = 0.25, µ=0, , 30 kg, , 1) 5ms-2 2) 2 ms-2, 3) 1 ms-2, 4) 0, 25. A suitcase is gently dropped on a conveyor, belt moving at a velocity of 3 m/s. If the, coefficient of friction between the belt and the, suitcase is 0.5, find the displacement of the, suitcase relative to conveyor belt before the, slipping between the two is stopped, , 1) 72 N 2) 40 N, 3) 36 N, 4) 20 N, 28.Find the least horizontal force P to start motion, of any part of the system of the three blocks, resting upon one another as shown in fig. The, weights of blocks are A = 300 N, B = 100 N and, C = 200 N. Between A and B, coefficient of, friction is 0.3, between B and C is 0.2 and, between C and the ground is 0.1., A, , P, , B, C, , ( g = 10m / s ), 2, , 1) 2.7 m 2) 1.8 m 3) 0.9 m 4) 1.2 m, 26. Blocks A and B in the fig, are connected by a, bar of negligible weight. Mass of each block, , 1) 60 N, 2) 90 N 3) 80 N, 4) 70 N, 29.Determine time in which the smaller block reaches, other end of bigger block as shown in the fig., , is 170 kg and µ A = 0.2 and µ B = 0.4,, where µ A and µ B are the coefficients of, limiting friction between blocks and plane., Calculate the force developed in the bar, , ( g = 10m / sec ) ., 2, , B, , µ=0.3, 2 kg, , 10N, , µ=0, 8 kg, L=3.0m, , 1) 4 s, 2) 8 s 3) 2.19 s, 4) 2.13 s, 30.A block of weight W is kept on a rough horizontal, surface (friction coefficient µ ). Two forces, W/2 each are applied as shown in the figure., choose the correct statement., w/2, w/2, , 8, , 30°, , µ, , A, , w, , 15, , 1) For µ >, 1) 150 N 2) 75 N 3) 200 N, 4) 250 N, 27. Two blocks A and B of masses 6 kg and 3 kg, rest on a smooth horizontal surface as shown, in the fig. If coefficient of friction between A, and B is 0.4, the maximum horizontal force, which can make them without separation is, 74, , 3, 5, , block will move, , 3, 2) For µ < , work done by frictional force is zero, 5, , (in ground frame), 3) For µ >, , 3, , frictional force will do positive work, 5, , (in ground frame), , 4) For µ <, , 3, block will move, 5, , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , UNIFORM CIRCULAR MOTION, , 31. A 2kg block is placed over a 4kg block and, , both are placed on a smooth horizontal surface. The coefficient of friction between the, blocks is 0.20. The acceleration of the two, blocks if a horizontal force of 12N is applied, to the lower block is (g= 10ms–2), 1) 2 ms–2, 2 ms–2, 2) 2 ms–2, 1 ms–2, 3) 3 ms–2, 1 ms–2, 4) 4 ms–2, 1 ms–2, 32. Blocks A and B shown in the figure are, connected with a bar of negligible weight. A, and B each has mass 170Kg, the coefficient, of friction between A and the plane is 0.2 and, that between B and the plane is 0.4. What is, the total force of friction between the blocks, and the plane (g = 10 ms-2), , LEVEL-III-kKEY, 1) 2, 7) 3, 13) 3, 19) 4, 25) 3, 31) 1, , 2) 3, 8) 2, 14) 4, 20) 1, 26) 1, 32) 1, , 3) 1, 9) 3, 15) 1, 21) 2, 27) 3, 33) 1, , 4) 2, 10) 1, 16) 3, 22) 1, 28) 1, 34) 1, , 5) 1, 11) 4, 17) 1, 23) 1, 29) 3, , 6) 4, 12) 2, 18) 2, 24) 3, 30) 4, , HINTS - LEVEL-III, 1., , F = 100N, m1 = 250kg, m2 = 500kg,S = 100m, Force acts on both the boats is same., using F = ma; F = m1a1 ; F = m2 a2, F 100, F 100, a1 =, =, a2 =, =, m1 250, m2 500, a1 = 0.4 ms −2, , a2 = 0.2 ms −2, , Relative acceleration arel = a1 + a2, arel = 0.4 + 0.2 = 0.6 ms −2, 1, 2, using s = ut + arel t ,u=0, 2, t=, , 2., 1) 900N 2) 700N, 3) 600N 4) 300N, 33. From the above problem what is the force, acting on the connecting bar ?, 1) 150N 2) 100N, 3) 75N 4) 125N, 34. A block of mass m, lying on a rough horizontal, plane is acted upon by a horizontal force P and, another force Q, inclined at an angle θ with ver-tical. The block will remain in equilibrium, if coefficient of friction between it and surface is, , 2s, 2 ×100, =, = 18.3s, arel, 0.6, , mm = 60kg , mB = 100kg, ‘a’ be acceleration of rope., 5g, arel =, , arel = am + a ; am = arel − a, 4, 5g, am =, −a, 4, T − mB g = mB a → (1), T, T, , mB, , 1., , (P + Q sin θ ), (P cos θ + Q), 2., (mg + Q cos θ ) (mg − Q sin θ ), , (P+ Q cos θ ), (P sin θ − Q ), 3., 4., (mg + Q sin θ ) (mg − Q cos θ ), , mm, , mmg, mBg, T − mm g = m2 am → ( 2 ), solving (1) & (2), we get, T - 100g = 100a -------(3), , T - 60g =, , 5g, , 60 , − a ------(4), 4, , , (3 ) - (4) , −40 g = 100a − 75 g + 60 a, 35, 160a = 350 ⇒ a = ms −2, 16, T = 100g + 100a = 1000 + 100 ×, , NARAYANAGROUP, , 35, = 1218 N, 16
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 2., , A man of mass m is on the floor of a lift then, match the following, LIST – I, LIST –II, a) lift is moving up with e) apparent weight, acceleration a, is greater than, b) lift is moving down, f) apparent weight is, with acceleration a, zero, c) lift is moving with, g) apparent weight is, uniform velocity, equal to true weight, d) lift is freely falling, h) apparent weight is, less than true weight, , Assertion & Reason, , 3., , 4., , 5., , 6., , 7., , 8., , 78, , (1) A and R are true and R is the correct, explanation of A, (2) A and R are true and R is not the correct, explanation A, (3) A is true and R is false, (4) A is false and R is true, Assertion (A): A body in equilibrium has to be at, rest only, Reason (R): A body in equilibrium may be moving, with a constant speed along a straight line path., Assertion (A): If net force on a rigid body is zero,, it is either at rest or moving with a constant linear, velocity., Reason (R): Constant velocity means linear, acceleration is zero, Assertion (A): A cricket player while catching a, ball lowers his hands to save himself from getting, hurt., Reason (R) : The impulsive force on hands is, reduced by increasing the time of action, Assertion (A): When a ball of mass m hits, normally a wall with a velocity ' v ' and rebounds, with same velocity v , impulse imparted to the wall, is 2mv ., Reason (R) : Impulse = change in linear, momentum, Assertion (A):A concept of pseudo forces is valid, both for inertial as well as non-inertial frame of, reference, Reason (R) : A frame accelerated with respect to, an inertial frame is a non-inertial frame., Assertion (A):When a person walks on a rough, surface,the frictional force exerted by surface on, the person is opposite to the direction of his motion., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , Reason (R) :It is the force exerted by the road, on the person that causes the motion., Assertion (A):Coefficient of friction can be greater, than unity., Reason (R) :Force of friction is dependent on, normal reaction and ratio of friction and normal, reaction cannot exceed unity., Assertion (A): Inertia is the property by virtue of, which the body is unable to change by itself the, state of rest or uniform motion., Reason (R): The bodies do not change their state, unless acted upon by an unbalanced external force., Assertion (A): If the net external force on the body, is zero, then its acceleration is zero, Reason (R): Acceleration does not depend on, force, Assertion (A): A man in a closed cabin falling, freely does not experience gravity, Reason (R): Inertial and gravitational mass have, equivalence, Assertion (A): Force is required to move a body, uniformly along a circle, Reason (R): When the motion is uniform, acceleration is zero, Assertion (A): A body subjected to three concurrent forces can be in equilibrium., Reason (R): If large number of concurrent forces, acting on the same point, then the point will be in, equilibrium, if sum of all the forces is equal to zero., Assertion (A): Aeroplanes always fly at low altitudes., Reason (R): According to Newton's third law of, motion, for every action there is an equal and opposite reaction., Assertion (A): A table cloth can be pulled from a, table without dislodging the dishes., Reason (R): To every action there is an equal, and opposite reaction., Assertion (A): Mass is a measure of inertia of the, body in linear motion., Reason (R): Greater the mass, greater is the force, required to change its state of rest or of uniform, motion, Assertion (A): The slope of momentum versus, time curve give us the acceleration., Reason (R): Acceleration is given by the rate of, change of momentum., , NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 19. Assertion (A): A cyclist always bends inwards, while negotiating a curve., Reason (R): By bending, cyclist lowers his centre of gravity., 20. Assertion (A): The work done in bringing a body, down from the top to the base along a frictionless, incline plane is the same as the work done in bringing it down the vertical side, Reason (R): The gravitational force on the body, along the inclined plane is the same as that along, the vertical side., 21. Assertion (A): Linear momentum of a body, changes even when it is moving uniformly in a circle., Reason (R): Force required to move a body uniformly along a straight line is zero., 22. Assertion (A): A bullet is fired from a rifle. If the, rifle recoils freely, the kinetic energy of rifle is more, than that of the bullet., Reason (R): In the case of rifle bullet system the, law of conservation of momentum violates., 23. Assertion (A): A reference frame attached to earth, is an inertial frame of reference., Reason (R): The reference frame which has zero, acceleration is called a non inertial frame of reference., 24. Assertion (A): The apparent weight of a body in, an elevator moving with some downward acceleration is less than the actual weight of body., Reason (R): The part of the weight is spent in, producing downward acceleration, when body is, in elevator., 25. Assertion (A): When the lift moves with uniform, velocity the man in the lift will feel weightlessness., Reason (R): In downward accelerated motion of, lift, apparent weight of a body increases., 26. Assertion (A): Newton's third law of motion is, applicable only when bodies are in motion or rest., Reason (R): Newton's third law applies to all, types of force, e.g. gravitational, electric or magnetic forces etc., 27. Assertion (A): Linear momentum of a body, changes even when it is moving uniformly in a circle., Reason (R): In uniform circular motion velocity, remain constant., , UNIFORM CIRCULAR MOTION, , 28. Statement A: When a person is on the floor of a, lift which is at rest, the resultant force on him is, equal to his weight, Statement B: When the lift is moving with uniform, velocity, then the apparent weight of the man is zero, 29. Statement A: If the force varies with time in a, complicated way then the net force is measured by, the total change in momentum of the body, Statement B: Change in momentum and impulsive, force are numerically equal, 30. Statement A: Shock absorbers reduce the, magnitude of change in momentum, Statement B: Shock absorbers increase the time, of action of impulsive force, 31. Statement A:For a body resting on a rough, horizontal table, it is easier to pull at an angle, than push at the same angle to cause motion, Statement B:A body sliding down a rough, inclined plane of inclination equal to angle of, friction has non–zero acceleration, , Other models, 32. If µ s, µ k and µ R are the coefficients of, limiting, kinetic and rolling frictions between, two given surfaces. Arrange them in ascending, order, 1) µ R, µ S , µ K 2) µ R, µ K , µ s, 3) µ S, µ K , µ R 4) µ K, µ R , µ S, 33. Let F, FN and f denote the magnitude of the, contact force, normal force and the friction, exerted by one surface on the other kept in, contact. If none of these is zero, a. F > FN, b. F > f, c. FN > f, d. FN– f < F < FN+ f, 1) a & c are correct, 2) b & c are correct, 3) a, b & d are correct 4) a & b are correct, 34. Two blocks A and B are pressed against a, rough vertical wall by applying a horizontal, force ‘F’. There is no friction between A, and B. Then, a. both blocks A and B can be at rest for any, magnitude of F, b. B can be at rest A moves down for smaller, magnitude of F, c. both A and B will move down for smaller, magnitude of F, d. A can be at rest and B moves down for larger, magnitude of F, , Statement Type Questions, Options :, 1. Statement-A is true and statement-B is true, 2. Statement-A is true and statement-B is false, 3. Statement-A is false and statement-B is true, 4. Statement-A is false and statement-B is false, NARAYANAGROUP, , AB, 1) a & b are correct, 3) a & d are correct, , F, 2) c & d are correct, 4) b & c are correct
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JEE-ADV PHYSICS-VOL - II, , LAWS OF MOTION, FRICTION, , 35. A block is thrown up an inclined plane with, certain velocity to reach the top of it. If the, length of the inclined plane is l and its angle, of inclined plane is equal to angle of repose,, then, a. the retardation is 2g sin ?, b. the retardation is 2g tan ?, c. the time of ascent is, d. the time of ascent is, , l, g sin θ, , l, 2 g tan θ, , 1) a & c are correct, 2) a & d are correct, 3) b & c are correct, 4) b & d are correct, 36. Two blocks of masses m and 4m connected, by a spring of negligible mass are, compressed and released. They move off in, opposite direction with velocities V 1 and V 2, immediately after release and come to rest, after covering distances S 1 and S 2. Then, V1, 4, a. V = 1, 2, , S1, , 16, , S, , 4, , 2, c. S = 1 (if friction is same for both blocks), 1, , 1) a → h, b → g , c → e and d → f, 2) a − e, b − h , c − g , d − f, Assertion & Reason Type, 04) 2, 10) 1, 16)2, 22)4, , 05) 1, 11)3, 17)1, 23)4, , 06) 1, 12)1, 18)4, 24)3, , 07) 4, 13)2, 19)3, 25)4, , Statement Type Questions, 29)3, , 30)4, , 31)2, , Other Models, 32)2, 80, , 4., , If net force is zero,rotational motion takes place., , 5., , F=, , 6., , J = m(v− u), u = v, v = −v ⇒ J = 2mv, , 7., , Pseudo force is applied only for non inertial frames., , 8., , Frictional force is in the direction of motion., , 9., , Coefficient of friction µ = Tanθ .The value of, , dp, 1, , Fα, If ‘ dt ’is more then F will be less., dt, dt, , 10. Inertia is the property by virtue of which the body, is unable to change by itself not only the state of, rest, also the state of motion., 11. According to Newton's second law, , 33)3, , 34)2, , 35)1, , Acceleration, , =, , Force, M ass, , i.e. if net external force on, , the body is zero then acceleration will be zero, 12. m grav . g − N = minertial .a For freely falling a=g., Since m grav = minertial ⇒ N = 0, , KEY - LEVEL-IV, Matching Type, , 28)2, , In equilibrium,net force on the body is, zero.Therefore,its acceleration ‘a’ is zero.If the body, is at rest ,it will remain at rest.If the body is moving, with a constant speed along a stright line path,it, will continue to do so., , 16, , d. S = 1 ( if coefficient of friction is same), 1, for both blocks, 1)only a & c are correct 2)only a & b are correct, 3)only b & c are correct 4) c & d are correct, , 03) 4, 09) 3, 15)1, 21)2, 27)3, , 3., , Tanθ may exceed unity.., , b. S = 1 (if coefficient of friction is same, 2, for both blocks), , S2, , HINTS - LEVEL-IV, , 36)2, , 08) 4, 14)1, 20)3, 26)2, , 13. When a body is moving in a circle, its speed remains same but velocity changes due to change in, the direction of motion of body. According to first, law of motion, force is required to change the state, of a body. As in circular motion the direction of, velocity of body is changing so the acceleration, cannot be zero.But for a uniform motion acceleration is zero (for rectilinear motion)., 14. A body subjected to three concurrent forces is, found to in equilibrium if sum of these forces is equal, to zero., , uur uur uur, i.e. F1 + F2 + F3 + ........ = 0, NARAYANAGROUP
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 15. The wings of the aeroplane pushes the external air, backward and the aeroplane move forward by reaction of pushed air. At low altitudes density of air, is high and of pushed air. At low altitudes density, of air is high and so the aeroplane gets sufficient, force to move forward., 16. According to law of inertia (Newton's first law),, when cloth is pulled from a table, the cloth come in, state of motion but dishes remains stationary due, to inertia. Therefore when we pull the cloth from, table the dishes remains stationary., 17. According to Newton's second law of motion, a=F/m i.e. magnitude of the acceleration produced, by a given force is inversely proportional to the, mass of the body. Higher is the mass of the body,, lesser will be the acceleration produced i.e. mass, of the body is a measure of the opposition offered, by the body to change a state, when the force is, applied i.e. mass of a body is the measure of its, inertia., 18. F =, , dp, =Slope of momentum-time graph, dt, , i.e. Rate of change of momentum = Slope of momentum-time graph =force., 19. The purpose of bending is to acquire centripetal, force for circular motion. By doing so component, of normal reaction will counter balance the centrifugal force., 20. Work done in moving an object against gravitational, force (conservative force) depends only on the initial, and final position of the object, not upon the path taken., But gravitational force on the body along the inclined, plane is not same as that along the vertical and it varies with the angle of inclination., 21. In uniform circular motion of a body the speed remains constant but velocity changes as direction of, motion changes., As linear momentum=mass × velocity, therefore linear momentum of a body changes in a circle., NARAYANAGROUP, , UNIFORM CIRCULAR MOTION, , On the other hand, if the body is moving uniformly, along a straight line then its velocity remains constant and hence acceleration is equal to zero. So, force is equal to zero., 22. Law of conservation of linear momentum is correct when no external force acts. When bullet is, fired from a rifle then both should possess equal, , p2, 2m, ∴ Kinetic energy of the rifle is less than that of, bullet because E ∝ 1/ m, momentum but different kinetic energy. E =, , 23. An inertial frame of reference is one which has zero, acceleration and in which law of inertia hold good, i.e. Newton's law of motion are applicable equally., Since earth is revolving around the sun and earth is, rotating about its own axis also, the force are acting on the earth and hence there will be acceleration of earth due to these factors. That is why earth, cannot be taken as inertial frame of reference., 24. The apparent weight of a body in an elevator moving with downward acceleration a is given by, , W = m( g − a) ., 25. For uniform motion apparent weight = Actual weight, For downward acceleration motion,, Apparent weight < Actual weight, 26. According to third law of motion it is impossible, to have a single force out of mutual interaction between two bodies, whether they are moving or at, rest. While, Newton's third law is applicable for all, types of force., 27. In uniform circular motion, the direction of motion, changes, therefore velocity changes., As P = mv therefore momentum of a body also, changes in uniform circular motion.0
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, , LEVEL - V, , F, , θ, , SINGLE ANSWER QUESTIONS, 1., , The car A is used to pull a load B with the, pulley arrangement shown. If A has a forward, velocity v A determine an expression for the, , mg 2 cos θ, A) , 2, 2c sin θ, , upward velocity vB , of the load in terms of, , mg 2 cos θ, B) , 2, 2c sin θ, , VA and θ. θ is angle between string and, horizontal, A), , B) VA sin θ, , l, , h, , 1, VA cos θ, 2, , 4., C) VA cos θ, , B, , 2., , , m/s ,, , , , , m, , , , m2 g 3 cos θ, , 6c2 sin 3 θ, , , , m, , , , mg cos θ , m / s,, C) , 2θ , 2, c, sin, , , , m2 g 3 sin θ , , m, 6c2 cos3 θ , , , , mg 2 cos θ, D) , 2, 2c sin θ, , m2 g 3 sin θ , , m, 6c 2 cos3 θ , , , , , m / s,, , , , The vertical displacement of block A in meter, , t2, where t is in second., 4, Calculate the downward acceleration aB of, block B., is given by y =, , A, , 1, VA tan θ, 2, Identify the relationship which governs the, velocities of the four cylinders. Assume all, velocities as positive downward., , D), , X, , , mg 3conθ, m / s ,, , 6c 2 sin3 θ, , , , A, D, , C, , B, , A) 3v A + 6vB + 4vC + vD = 0, B) 4vA + 8vB + 4vC + vD = 0, C) 3v A + 6vB + 2vC + vD = 0, 3., , y, , A, , 5., , A)2ms2, B)1ms2, 2, C)4ms, D) 9ms2, Find the acceleration of block B relative to, the block A and realtive to the ground, if the, blcok A moves to the left with an acceleration, a0, , D) 3v A + 10vB + 2vC + vD = 0, At t = 0, force F = ct is applied to a small body, of mass m resting on a smooth horizontal, plane (c is a constant). The force is at an angle, θ with the horizontal .The velocity of the body, at the moment of its breaking off the plane, and the distance travelled by the body up to, this moment are, , A, , B, 60°, A), , 82, , B, , 31a0 B), , 25a0 C), , 30a0 D) 30a0
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 6., , NEWTON LAWS OF MOTION, , Under the action of force P, the constant, acceleration of block B is 3ms-2 to the right., At the instant when the velocity of B is 2ms-1, to the right, determine the velocity of B, relative to A, the acceleration of B relative to, A and the absolute velocity of point C of the, cable, , 9., , If A and B moves with acceleration a. block c, moves up with acceleration b. calculate, acceleration of D with respective A., a, a, b, , A, , B, θ θ, , C, A, , 7., , B, , D, , P, , A) 2, B) 1, C) 3, D) 4, Block B has a mass m and is released from, rest when it is on top of wedge A, which has a, mass 3m. Determine the tension in cord CD, required to hold the wedge from moving while, B is sliding down A. Neglect friction, , C, , A) 2a+b, B) 2a + b cos θ, C) b cos θ + a sin θ, D) bsin θ + a cos θ, 10. Three identical rigid circular cylinders A,B and, C arranged on smooth inclined surfaces as, shown in figure. The least value of θ that, prevents the arrangement from collapsing is, , B, D, , A), , 8., , C, , θ, , mg, sin ( 2θ ), 2, , C, A, , A, , B), , B), , m1, C), , θ, , m2, D), , B, , θ, , θ, , mg, sin ( 3θ ), 2, , mg, mg, sin ( 3θ ), sin ( 2θ ), C), D), 3, 2, Find the acceleration of the body of mass m2, in teh arrangement shown in figure. If the mass, m2 is η times great as the mass m1 and the, angle that the inclined plane forms with the, horizontal is equal to θ . The masses of the, pulleys and threads, as well as the friction, are, assumed to be negligible., , A), , E, , −1 1 , B) tan , , 2 3, , −1 1 , A) tan , 2, , −1 1 , −1 1 , C) tan , D) tan , , , 3 3, 4 3, 11. In the arrangement shown, blocks A and B, connected with an inextensible string move, with velocities v1 and v2 along horizontal, , v2, direction. The ratio of v is, 1, , 2 g ( 2η − sin θ ), 2η + 1, 2 g ( 2η − sin θ ), 4η + 1, , α, , 2 g ( 2η − sin θ ), , A, , 3η + 1, 4 g ( 2η − sin θ ), 3η + 1, , A), , sin α, sin β, , β, , V1, , B), , B V, 2, , cos α, sin β, cos β, C), D), cos β, sin α, cos α, 83
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 12. In the arrangements shown, the pulleys,, strings and springs are weightless and the, systems can move freely without friction. The, extension of spring in figure 1 is x1 and that in, figure 2 is x2. Then, , MULTIPLE ANSWER QUESTIONS, 15. In the figure small block m is kept on planck, of mass M and a force F is applied on planck, as shown in diagram then which of the following, statements is /are correct., , µ=0, , µ=0, , m, M, A, , k, , 5 kg, , 10 kg, , Fig 1, , l, , B, , F, ., m, (B) the acceleration of m w.r.t. ground is zero, , k, , 5 kg, , F, , (A) the acceleration of m w.r.t. ground is, , 5 kg, , Fig 2, , A) x1=x2 B) x2>x1>0 C) x1>x2=0 D) x1>x2>0, 13. Figure shows a system of four pulleys with two, masses m A = 3 kg and mB = 4 kg. At an instant,, force acting on block A, if block B is going up, at an acceleration of 3 m/s2 and pulley Q is, going down at an acceleration of 1 m/s2 is, (A) 7 N acting upward, , (C) the time taken by m to separate from M is, , 2lm, F, , 2lM, F, 16. A particle of mass m starts moving at t = 0, due to a force F = F0 sin ωt where F0 and ω are, constant. Then correct statement is/are, (D) the time taken by m to separate from M is, , (A) it will stop first time at, , π, ω, , (B) It will travel distance S =, , F0, during this time, mω2, , (C) During this distance maximum velocity of, , Q, , (B) 7 N acting downward, (C) 10.5 N acting upward, , R, , P, B, , particle is v max =, , F0, mω, , (D) it will stop for first time at 2π / ω, 17. From the given diagram, choose the correct, option, , (D) 10.5 N acting downward., , A, , 14. If A and B moves with acceleration a as shown, in diagram calculate acceleration of C with, respect to B, a, a, , A, , P1, P3, P2, , A, , B, , B, , C, , (A) 2a, 84, , (B) a 2, , (C) 3a, , (D) 4a, , (A) acceleration of block A is zero, (B) acceleration of B is g, (C) acceleration of block A is non zero, (D) tension in the string connecting A is zero
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, , 18. In the diagram shown, the acceleration of the, block B as shown in figure relative to the block, A and relative to ground is aBA and a BG, respectively. If the block A is moving towards, left with an acceleration a 0, then, , 20. A block of mass m is placed on a wedge. The, wedge can be accelerated in four manners, marked as (1), (2), (3) and (4) as shown. If the, normal reactions in situation (1), (2), (3) and, (4) are N1, N2, N3 and N4 respectively and, acceleration with which the block slides on the, wedge in situation are b 1, b2, b3 and b 4, respectively then :, , (2), , (1), B, , A, , A) aBA = 2a0, , B) aBG = 3a0, , D, , a, , a, , 37°, , 37°, , D) aBG = a0 10 + 6cos θ, C) aBA = 3a0, 19. In the pulley system shown the movable, pulleys A,B and C have mass m each, D and E, are fixed pulleys. The strings are vertical, light, and inextensible. Then,, , a, , (4), , (3), , m, m, , E, , 37°, a, , 37°, , (A) N3 > N1 > N2 > N4 (B) N4 > N3 > N1> N2, (C) b2 > b3 > b4 > b1 (D) b2 > b3 > b1 > b4, , A, B, , PASSAGE TYPE QUESTIONS, PASSAGE : 1, A body of mass m = 1.8 kg is placed on an inclined, , C, , plane, the angle of inclination is α = 370 , and is, attached to the top end of the slope with a thread, which is parallel to the slope. Then the plane slope, is moved with a horizontal acceleration of a. Friction, is negligible., , A) the tension throughout the string is the same and, equals T =, , m, , m, , θ, , 2mg, 3, , g, each in, 3, downward direction and pulley C has acceleration, , B) pulleys A and B have acceleration, g, in upward direction, 3, , C) pulleys A,B and C all have acceleration, , a, m, , g, in, 3, , downward direction, , 21. The acceleration, if the body pushes the plane, , g, D) pulley A has acceleration, in downward, 3, direction and pulleys B and C have acceleration, g, each in upward direction., 3, , α, , with a force of, A), , 3, mg is :, 4, , 5, 5, m / s 2 B) 0.5m / s 2 C) 0.75 m / s 2 D) m / s 2, 43, 6, 85
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 22. The tension in thread in the above question is, :, A) 12 N B) 10 N, C) 8 N, D) 4 N, 23. At what acceleration will the body lose contact, with plane :, A), , 40, m / s2, 3, , MATRIX MATCHING QUESTIONS, 27. In the diagram strings, springs and the pulley, are light and ideal. The system is in equilibrium, with the strings taut (T>0), match the column., Masses are equal., , B) 7.5 m / s 2, , C) 10 m / s 2, , D) 5 m / s 2, , PASSAGE-2, Two smooth blocks are placed at a smooth corner, as shown. Both the blocks are having mass m. We, apply a force F on the small block m. Block A, presses the block B in the normal direction, due to, which pressing force on vertical wall will increase,, and pressing force on the horizontal wall decrease,, as we increase F. ( θ = 37° with horizontal). As, soon as the pressing force on the horizontal wall, by block B becomes zero, it will lookse the contact, with the ground. If the value of F is further increase,, the block B will accelerate in upward direction and, simultaneously the block A will move toward right, , Y, B, A, , m, , smooth, , m 0=37°, F, X, 24. What is minimum value of F, to lift block B, from ground :, 25, 5, 3, 4, mg (B) mg (C) mg (D) mg, 12, 4, 4, 3, 25. If both the blocks are stationary, the force, exerted by ground on block A is :, , m B, X, m A, W, , 3F, 4, , (B) mg −, , 3F, 4, , 4F, 4F, (D) mg −, 3, 3, 26. If acceleration of block A is a rightward, then, acceleration of block B will be :, , (C) mg +, , (A), , 3a, upwards, 4, , 3a, (C), upwards, 5, 86, , (B), , 4a, upwards, 3, , 4a, (D), upwards, 5, , z, , Column - 1, Column- 2, A) Just after string W breaks P) a A = 0, B) Just after spring X breaks Q) aB = 0, C) Just after string Y breaks R) aC = 0, D) Just after spring Z breaks S) aB = aC, 28. In the situation shown, all surfaces are, frictionless and triangular wedge is free to, move, In column-2, the direction of certain, vectors are shown. Match the direction of, quantities in Column-1 with possible vector in, Column-2., , (A), , (A) mg +, , y, , θ, , Column - 1, Column-2, A) acceleration of the block, X relative to ground, , θ, , (P), , B) acceleration of block X, relative to wedge, , (Q), , C) normal force by block, on wedge, , (R), , D) net force on the wedge, , (S), , θ
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 29. See the diagrams carefully in Column-1 and, match each with the obeying relation(S) in, column-2, The string is massless, inextensible, and pulley is frictionless in each case., a = g / 3, m = mass of block T = tension in a, given string, a pulley = acceleration of movable, pulley in each case, acceleration due to gravity, is g., Column -1, Column -2, , (A), , NEWTON LAWS OF MOTION, 30. In the system shown in figure, masses of the, blocks are such that when system is released,, acceleration of pulley P1 is a upwards and, acceleration of block 1 is a1 upwards. It is, found that acceleration of block 3 is same as, that of 1 both in magnitude and direction., , a, P1, , T1, , T, , P) ablock ≤ a, , m, , a1, , 3, , 1, 2, , (B), , a, , T1, , Q) apulley ≤ a, , T, , P2, , a, , Column - I, A) Acceleration of 2, B) Acceleration of 4, C) Acceleration of 2 w.r.t. 3, D) Acceleration of 2 w.r.t. 4, , 4, Column - II, P)2a+ a1, Q) 2a – a1, R) upwards, S) downwards, , INTEGER TYPE QUESTIONS, m, (C), T1, , R) T > mg, , T, m, a, , (D), , 31. Under the action of a constant force F = 10, N , a body moves in a straight line so that the, relation between the distance S moved by, the body and the time t is described by the, equation S = A- Bt +Ct2. Find the mass of the, body if C = 1m/s2., 32. In the arrangement shown, by what, acceleration (in ms–2) the boy of mass 50 kg, must go up so that 100 kg block remains, stationary on the wedge. The wedge is fixed, and friction is absent everywhere. Take g =, 10 m/s2., , T1, , S)Force on fixed, , T, m, , a, , 100 kg, , m = 50 kg, , 53°, support T1 > ( 3/ 2 ) mg, 87
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 33. Two blocks of mass 2.9 kg and 1.9 kg are, suspended from a rigid support S by two, inextensible wires each of length 1 m (see Fig.), The upper wire has negligible mass and the, lower wire has a uniform mass of 0.2 kg/m., The whole system of blocks, wires and support, have an upward acceleration of 0.2 m/s2. The, acceleration due to gravity is 9.8 m/s2. If, tension at the midpoint of upper wire is 10x ., Find x., , 36. The pull P is just sufficient to keep the 14 N, block in equilibrium as shown. Pulleys are, ideal. Find the tension (in N) in the cable, connected with ceiling., , S, , 2.8 kg, , m = 1.4 kg, P, , 37. In the given figure find the and acceleration, of B, if instantaneous velocity and acceleration, of A are as shown in the Fig., , 1.9 kg, , 34. If, , the tension T needed to hold the cart, , equilibrium is, , 3W, , there is no friction. Find, x, , A, 1m/s, , value of x, , 2m/s2, W, T, , B, , 30°, , 35. The elevator is going up with an acceleration, of g/10, the pulley and the sting are light and, the pulley is smooth.If reading of spring, balance shown is 0.8x. Calculate x, , ( take g = 10m/s ), , 38. In Fig. shown, both blocks are released room, rest. Length of 4 kg block is 2 m and of 1 kg is, 4 m. Find the time they take to cross each other, ? Assume pulley to be light and string to be, light and inelastic., , 2, , g, 10, , 2m, 4 kg, 1.5 kg, , 4m, 3 kg, , 88, , 1 kg
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, , LEVEL -V - KEY, , So, the distance traversed till the time of break off,, , SINGLE ANSWER, 1) A 2) B 3)B, 4) C 5) A 6) B 7) A, 8) B 9) C 10)C 11)D 12) D 13)D 14) C, MULTI ANSWER, 15) A,C, 16) C,D, 17)A,B,D 18) C,D, 19) A,B, 20) A,C, PASSAGE TYPE, 21) D 22) A 23) A 24) C 25) C 26) A, MATCHING, 27. A-q,r,s, B-s, C-p, D-p,s, 28. A-q, B-p, C-r, D-s, 29. A-q,r,s; B-p, q, r; C-p,q,r,s; D-p,q, 30. A - qr; B - ps; C - s; D -r, INTEGER, 31) 5 32) 6 33) 5 34) 4 35) 6 36) 2 37) 1 38) 1, , i.e., in the time interval t =, , Let θ angle between string and horizontal T is, tension in string, , T, θ, , 2T, , 3, , 4., , 5., , 2, , 6., , 7., , 3., , VA =, , 3, 3, VB =, 4, 2, , VB + VC, = VA ; VC = 2VA − VB ; VC = 3 − 2 = 1, 2, mg, sin 2θ, T = mg sin θ cos θ , T =, 2, Ta 2 − 2Ta1 = 0, a 2 = 2a1 ....(i), , 2T − m1g sin θ = m1a1 ...(iii), solving three equations, , 1, VB = VA cos θ, 2, Let T be the tension, TVD + 4TVC + 8TVB + 4TVA = 0, , 9., , VD + 4VC + 8VB + 4VA = 0, , 10., , Since F cos θ =ma or ct cos θ = m, , −3T VB + 4T VA = 0 ;, , m 2 g − T = m 2a 2 ...(ii), , VA, , ∑ T.V = 0 ; −T cos θVA + 2T VB = 0, 2., , d2y 1, = m / sec 2 ; −TVB + 8TVA = 0, dt 2 2, 1, VB = 8VA = 8x = 4ms 2 ;, 2, 5T a 0 − Ta BA = 0 ;, a BA = 5a 0, , a B = a 20 + ( 5a 0 ) + 2a 0 5a 0 cos 600 ; a B = 31a 0, , 8., , VB, , c cos θ m g , m 2 g 3 cos θ, =, , , 6 m c sin θ , 6 c 2 sin 3 θ, , S =, , LEVEL-V - HINTS, 1., , mg, is given by, c sin θ, , a2 =, , 2g ( 2η − sin θ ), 4η + 1, , Let T be the tension in string, T cos θb + T sin θa − Ta D = 0, a D = b cos θ + a sin θ ., , dv, dt, , N, , N, A, , c cos θ t, c cos θ 2, t dt or v = , t, ∫0, ∫, 0, m, 2m , But velocity at the time of breaking off or at, v, , dv =, , t =, , ∴ v=, , c cos θ m g , m g 2 cos θ, x, =, 2m, 2 c sin 2 θ, c sin θ , , Now, v =, , ∫, , 0, , dS =, , B, , 60, , θ, 60, , 60, C, mg, , mg, c sin θ, 2, , s, , 60, , dS c cos θ 2, =, t, dt, 2m, , c cos θ, 2m, , c cos θ 3, ∫0 t dt or S = 6m t, t, , N, , F.B.D of A, 2N cos 300 = mg, F.B.D. of C, net force along the inclined plane =0, , N cos ( 60 + θ ) = mg sin θ, solving tan θ =, , 1, 3 3, 89
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, Similarly for this case get N 4 =, 3, 3, g− a, 5, 5, PASSAGE: 1 Question no: 21-23, N, T, , and b 4 =, , masin37° N2, , (2), , 4, 4, mg − ma, 5, 5, , a, , (Pseudo force), ma, macos37°, mgsin37°, , mgcos37°, , mg, , ma, mg, , 4, 3, mg − ma, 5, 5, 4, 3, N 2 = mg − ma, 5, 5, , α, , Similary for this case get N 2 =, and b 2 =, , 3, 4, g+ a ;, 5, 5, , T, , N, , a, , a, , α, F.B.D with respect to plane, with respective to ground, N + ma sin α = mg cos α, T = ma cos α + mg sin α, , N3, , (3), m(g+a)cos37°, , m(g+a)sin37°, mg, , ma, , ---(1), ---(2), 3, mg, 4, , g, 5, or a = m / sec 2 and T=12 N, 12, 6, The body will loose contact for N=0 For N=0, 40, m / sec 2, a = g cot α [From eq(i)] a =, 3, 24. Solving equations acceleration can be calculated, ⇒a =, , Similarly for this case get N 3 =, and b 3 =, , solving equation (1) and (2) for N =, , F.B,D, , 4, 4, mg + ma, 5, 5, , 3, 3, g+ a, 5, 5, , m, ma, N4, , (4), , m, , F, , m(g–a)cos37°, , m(g–a)sin37°, , N cos 370 = mg ...(1), , tan 370 =, , mg, a, , F=, , F, mg, , N, N sin 37 0 = F, 3, F, =, 4, mg, , 3, mg, 4, , 91
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 4, 25. F = N sin 37 0 ; N A = mg + N cos 37 0 = mg + F, 3, , ;T =, , 7mg, 7 mg, ; T1 =, ; B − P, Q, R, 6, 12, , 26. As bodies remain in contact a sin 37 0 = b cos 370, T1, , 3, 4, 3, =b ; b= a., 5, 5, 4, 27. Following F B D explains the options, a, , Kx, , 2, , T1, , T', , T1, , T', T' ;, , a/2, , T=, , 7mg, 7 mg, ; T1 =, ; C − P, Q, R, S, 6, 4, , T, , B, , A, , C, , a/2, , a, T1, , mg, , T2, , mg, , Kx2, , mg, , Kx1, , a, a, , T, , ; T1 =, , 28., b, , a is acceleration of wedge with respect ground. b, is acceleration of block with respect wedge., r r, a + b acceleration of block with respect ground., , (, , ), , a, , a/3, , m, , 14mg, 7 mg, ; T1 =, ; D − P,Q, 9, 9, , a, , 30. Let accelerations of various blocks are as shown, in figure. Pullet P2 will have downward acceleration, a +a, a. Now a = 1 2 ⇒ a2 = 2a − a1 > 0, 2, , T, , 29., , T1, , P1, , P2, , a, , 2a, , a, 1, , T1 = 5mg / 3 T = 5mg / 3 A − Q, R , S, , 3, , a1, , 4, 2, , a1, , a4, , a2, , a/2, , a/2, T, m, , 92, , So acceleration of 2 is upwards. Hence (a) → (q,r), −a + a, and a = 1 4 ⇒ a4 = 2a + a1 > 0, 2, So acceleration of 4 is downwards, Hence (b) → (p, s), Acceleration of 2 w.r.t. 3:, a2/3 = a2 – a3 = a2 –a1 = 2(a – a1) < 0, This is downwards. Hence (c) → (s), Acceleration of 2 w.r.t. 4a2/4 = a2 – (–a4) = 4a >0, This is upwards Hence (d) → (r)
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, or vC = v A ⇒ vC = 1m / s ↑, , LEVEL - VI, , where vC is velocity of pulley C., , SINGLE ANSWER QUESTIONS, , aC = a A, , Similarly, , 1., , 2, ∴ aC = 2m / s ↓, , Now l4 + l5 = l6, ∴, , dl4 dl5 dl6, +, =, dt dt, dt, , ∴ −vC + ( vB ) = −vB, , ' Tanβ ' is, , ∴ vB = 0.5m / s ↑, , Similary, aB =, , In the figure the heavy mass m moves down, the smooth surface of a wedge making an angle, α with the horizontal. The wedge at rest t = 0, is on a smooth surface. The mass of the wedge, is M. The direction of motion of the mass m, makes an angle β with the horizontal, then, , m, , aC, 2, , ∴ aB = 1m / s ↓, 2, , M, , 38. Solution :, A), 2., , T, T, a 2m, , T, , B, 4 kg, , T, , a, , a, , m, m, M, , M, tan α B), tan α C) 1 + tan α D) 1 + tan α, M, m, M, m, , , , , A weightless inextensible rope rests on a, stationary wedge forming an angle α with a, horizontal. One end of the rope is fixed to the, wall at point A. A small load is attached to the, rope at point B. The wedge starts moving to, the right with a constant acceleration a. The, acceleration of the load is given by :, , a, A, , 4m, , α, , A, , 4g, , B, , 1g, , 1 kg, , a, , From FBD of blocks A and B solve acceleration of, each block, 4 g − T = 4a, .......(1), .......(2), T − 1g = 1× a, After solving eqns. (1) and (2), a =, , 3g, 5, , ( ), , 3., , ( ), , A) a B) 2a sin α 2 C) a sin α D) sin α 2, Block is attached to system of springs., Calculate equivalent spring constant., , 6g, = 12m / s 2, 5, If A will cross B then distance travelled by A w.r.t., B is 6m, 1, 6 = 0 + × 12 × t 2 , t = 1sec, 2, For, F0 = 4m2 g, , acceleration of A w.r.t B aA / B =, , K, K, , m, , A) K, 94, , B) 2K, , C) 3K, , D) 4K
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 4., , NEWTON LAWS OF MOTION, , Block A and C start from rest and move to, , 7., , the right with acceleration a A = 12t m/s and, 2, , acceleration of 10 3 m / s 2 . It is seen that the, block starts climbing upon the smooth inclined, face of wedge. What will be the time taken by, the block to reach the top?, , a C = 3m / s 2 . Here t is in seconds. The t i m e, when block B attain comes to rest is, , A, , In the figure , the wedge is pushed with an, , B, m, , C, (A) 2 s, 5., , (B) 1 s, , (A), (C) 3/2 s, , (D) 1/2 s, , In the arrangement shown in fig. m1 = 1kg ,, m 2 = 2kg . Pulleys are massless and string are, light. For what value of M the mass m 1 moves, with constant velocity (neglect friction), , M, , a = 10 3m / s 2, , 1m, , 2, s, 5, , (B), (D), , 5s, , 1, , K, massless, block, , 2, B, m2, , 6., , 1, s, 5, , 5, s, 2, In the above diagram system is in equilibrium., If applied force F is doubled how much mass, less block will more towards right before new, equilibrium is achieved., (C), , 8., , 30°, , B, m1, , A), , (A) 6 kg (B) 4 kg, (C) 8 kg (D) 10 kg, Find equivalent spring constant for the system, , K, , F, K, , B), , F, , 2F, K, , F, F, D), 3K, 9K, In the above diagram all surface friction less, what horizontal force has to be applied on, wedge such that in equilibrium steady state, , C), 9., , sping is compressed by, , mg sin θ, K, , k, m, , m, θ, , A) k, , B) 2K, , C) 64K, , D) 8K, , A) 2mg tan θ, C) 4mg tan θ, , B) 2mg sin θ, D) 2mg tan θ, 95
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 10. If the above diagram initially there is not, elongation in spring if the block is displaced, towards right by x0. Calculate the elongation, of spring A., , 13. In the above situation all surface are, frictionless system is released from rest. Then, which of the following statements is/are, correct., , C, K, , D, , m, , m, , A, 60°, M, , B, 60°, M, , 3K, A, , B, x0, , 3, A) x 0, 7, , x, B) 0, 4, , x, C) 0, 7, , x, D) 0, 3, , MULTIPLE ANSWER QUESTIONS, 11. A book leans against a crate on a table., Neither is moving. Which of the following, statements concerning this situation is/are, incorrect ?, Crate, , Book, , A) The force of the book on the crate is lessthan, that of crate on the book, B) Although there is no friction acting on the crate,, there must be friction acting on the book or else it, will fall, C) The net force acting on the book is zero, D) The direction of the frictional force acting on, the book is in the same direction as the frictional, acting on the crate., 12. An iron sphere weighing 10 N rests in a V, shaped smooth trough whose sides form an, angle of 600 as shown in the figure. Then the, reaction forces are :, A, , B, , B, , B, , A) acceleration of wedges are zero, B) wedges accelerate towards right, C) Normal force exerted by ground on A is more, than normal force exerted by ground on B, D) Tension in connecting string is nonzero., 14. Two blocks of masses m 1 and m -2 (m1 > m2) are, connected by a massless threads, that passes, over a massless smooth pulley. The pulley is, suspended from the ceiling of an elevator. Now, the elevator moves up with uniform velocity, V0. Now, select the correct options., , V0, , m1, , A) Magnitude of acceleration of m1 with respect to, ground is greater than, , 60°, A, , (I), , 60°, , 60°, , (II), , (III), , A) RA = 10 Nand RB = 0 in case ( i ), B) RA = 10N and R3 = 10 N in case (ii ), C) RA =, , 20, 10, N and RB =, N in case ( iii ), 3, 3, , D) RA = 10 N and RB = 10 N in all the 3 cases, 96, , ( m1 − m2 ) g, , m1 + m2, B) Magnitude of acceleration of m1 with respect to, , A, 60°, , m2, , ground is qual to, , ( m1 − m2 ) g, , m1 + m2, C) Tension in the thread that connects m1 and m2 is, , 2m1 m2 g, equal to m + m, 1, 2, D) Tension in the thread that connects m1 and m2 is, 2m1m2 g, greater than m + m, 1, 2
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, , 15. A horizontal bar of mass m1 and Prism of mass, m2 can move as shown. There is no friction at, any contact point. During the motion, the, length of the rod is always horizontal. Now,, magnitude values of, m1, , Smooth, wall, , m2, θ, , 4m1 m2 g, A) Tension on the string is m + m, 1, 2, 2m1 m2 g, B) Tension on the string is m + m, 1, 2, C) The acceleration of mass m1 with respect to, 3m 2 − m1, ground is m + m g, 1, 2, D) The acceleration of mass m1 with respect to, , Smooth surface, , A) Acceleration of m1 is g / (1 + η cot 2 θ ) , where, , η = m2 / m1, g tan θ, B) Acceleration of m1 is η 1 + tan 2 θ , where, , , , η = m2 / m1, , 2 ( m 2 − m1 ), , g, m1 + m 2, 18. In the arrangement shown in the figure all, contact surfaces are smooth strings and, pulleys are massless., ground is, , Given M1 = 1kg, M 2 = 2kg, M 3 = 4kg and, g=10ms-2, , C) Acceleration of m2 is g / ( tan θ + η cot θ ) , where, , η = m2 / m1, , T', , g tan θ, D) Acceleration of m2 is η 1 + tan 2 θ , where, , , 2, , η = m2 / m1, 16. Which of the following regarding frame of, reference is correct, A) Newton’s third law is valid from both inertial, and non inertial frame., B) Natural forces like tension, normal force are, same from all inertial frame, C) sun can be considered perfectly inertial frame, D) Acceleration of a body measured from different, inerital frames are different., 17. Two masses m1 and m 2 are connected by light, inextensible string passing over a smooth, pulley. P. If the pulley moves vertically, upwards with an acceleration equal to g then., P, , M1, , T', T', , T, M2, , a↓, , T', T, M3, , A) The acceleration of block of mass M3 is 4ms-2, B) The acceleration of block os mass M1 is 4ms-2, C) The tension (T) in the string connecting blocks, of masses M3 and M2 is 24N., D) The tension (T) in the string connecting block, of mass M1 and M2 is 24N, 19. In the figure shown, two blocks, one of mass, 5kg and the other of mass 2kg are connected, by light and inextensible string. Pulleys are, light an d frictionless. Choose the correct, statement, , ↑g, , m1, , a↑, , 5kg, , m2, , Y, ground, , 2kg ↓a2, , O, 97
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, A) The acceleration of 5kg mass is, , 5g −2, ms, 11, , 5g −2, ms, B) The acceleration of 2kg mass is, 11, , C) Tension in the string is, D) Tension in the string is, , 12g, N, 11, , 23. The acceleration of m 1 with respect to ground, is, A), , C), , 10g, 11, , PASSAGE : I, A shot putter with a mass of 80 kg pushes the iron, ball of mass of 6 kg from a standing position, accelerating it uniformly form rest at an angle of, 450 with the horizontal during a time interval of, 0.1 seconds. The ball leaves his hand when it is 2m, high above the level ground and hits the ground 2, seconds later. ( g = 10m / s ), 20. The accleration of the ball in shot putter’s, hand:, 2, , A) 11 2 m / s, , m2 + m1 sin 2 θ, , ( m1 + m2 ) g sin 2 θ, m2 − m1 sin 2 θ, , ( m1 + m2 ) g sin 2 θ, , B), , m1 + m2 sin 2 θ, , ( m1 + m2 ) g sin 2 θ, , D), , m1 − m2 sin 2 θ, , 24. The acceleration of m2 with respect to ground, is, , PASSAGE TYPE QUESTION, , 2, , ( m1 + m2 ) g sin 2 θ, , B) 100 2 m / s, , 2, , 2, , 2, , C) 90 2 m / s, D) 9 2 m / s, 21 The horizontal distance between the point of, release and the point where the ball hits the, ground:, A) 16 m B) 18 m, C) 20 m D) 22m, 22. The minimum value of the static coefficient of, friction if the shot putter does not slip during, the shot is closest to :, A) 0.28, B) 0.40, C) 0.48, D) 0.58, PASSAGE : II, , A), , C), , ( m1 + m2 ) g sin 2 θ, m2 + m1 sin θ, 2, , ( m1 + m2 ) g sin 2 θ, , ( m1 + m2 ) g sin θ, , B), , m2 + m1 sin 2 θ, , ( m1 + m2 ) g sin 2 θ, , D), m2 − m1 sin 2 θ, m1 − m2 sin 2 θ, 25. Normal reaction on m1- is:, B) ( m1 + m2 ) g, , A) m1 g, m1m2 g cos 2 θ, C), m2 + m1 sin 2 θ, D), , m1 g 1 − ( m1 + m2 ) sin 2 θ , m1 + m2 sin 2 θ, , MATRIX MATCHING QUESTIONS, 26. Two blocks of masses m1 =5kg and m2 = 2kg, are connected by threads which pass over the, pulleys as shown in the figure. The threads, are mass less and the pulleys are mass less, and smooth. The blocks can move only along, the vertical direction. T1 and T2 are the, tensions in the string as shown. Now match, the following:, [take g = 10m/s2], , m1, FIXED, , m2, θ, , Two blocks m1 and m2 are allowed to move without, friction. Block m1 is on block m-2 and m2 slides on, smooth fixed incline as shown. The angle of, inclination of inclined plane is θ ., 98, , T2, , T1, m1, m2
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, , INTEGER TYPE QUESTIONS, Column-I, , Column-II, , A) Magnitude value of, acceleration of m1 with, respect to ground., , p), , 500, SI units, 19, , B) Magnitude value of, acceleration of m2 with, respect to ground., , q), , 250, SI units, 19, , C) The value of tension T 1, , r), , 60, SI units, 19, , D) The value of tension T 2, , r), , 40, SI units, 19, None of these, , t), , 29. Two smooth blocks of same mass are, connected by an inextensible and massless, string which is passing over a smooth pulley, are kept in a lift. The lift is going down with, acceleration ‘a’ as shown in the figure. What, should be the value of a (in m/s2) so that, acceleration of block A w.r.t. ground will be, minimum ? (g = 10 m/s2), , A, , 27. In thediagram show n,m atch the follow ing (g, = 10m/s2), , m, a, , F2 = 18N, , B, , m, , 3kg, F1 = 60N, , 2kg, , 1kg, Smooth, , 30°, Blocks are on smooth incline. F1 and F2 are, parallel to the inclined plane. The motion of, the blocks is along the incline the surface., Column I, Column II, SI UNITS, A) Acceleration of 2kg block, (p)39, B) Net force on 3kg block, (q)25, C) Normal reaction between, 2kg and 1kg, (r) 2, D) Normal reaction between 3kg, and 2kg, (s) 6, 28. Column - I, (A) When lift is accelerated up then apparent weight, (B) When lift is accelerated down, then apparent, weight, (C) When lift is moving up or down constant velocity,, (D) When lift is free falling then apparent weight, Column - II, (P) Less than actual weight, (Q) Greater than actual weight, (R) Zero, (S) Equal to actual weight., (T) Negative, , 30. Fig. shows a block of mass 0.1 kg placed on a, , 1, , kg . If the block, 5 3, of mass m will move vertically downward with, acceleration 10 m/s2 . Then the value of, tension, (in newton) in the string is, smooth wedge of mass, , (q = 300 )., , m, M, θ, Smooth, , 31. Two blocks of masses 10 kg and 20 kg are, connected by a massless spring and are placed, on a smooth horizontal surface. A force of 200, N is applied on 20 kg mass as shown in the, diagram. At the instant, the acceleration of 10, kg mass is 12 ms–2, the acceleration of 20kg, mass is., , 200N, 10kg, , 20kg, 99
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 32. Two block A and B having masses, m1 = 1kg , m2 = 4kg are arranged as shown in, the figure. The pulleys P and Q are light and, frictionless. All the blocks are resting on a, horizontal floor and the pulleys are held such, that strings remains just taut . At moment, , 34. A bead C can move freely on a horizontal rod., The bead is connected by blocks B and D by a, string as shown in the figure. If the velocity of, B is v. The velocity of block D is 4v/x, find, the value of x, , t = 0 a force F = 30t ( N ) starts acting on the, , pulley P along vertically upward direction as, shown in the figure. The time when the blocks, A and B loose contact with ground is 4/x sec, then x is, , F – 30t(N), , 53°, , D, , B, , 35. A lift goes up with 10m / s. A pulley P is fixed, to the celling of the lift. To this pulley other, two pulley P1 and P2 are attached. P1 moves, up with velocity 30m / s. A moves up with, velocity 10m / s. D is moving downwards with, velocity 10m / s. at same instant of time., Assume that all velocities are relative to the, ground. If velocity of v is 10x, calculate x, , p, , Q, , B, , 37°, , A, , 10m/s, , 33. In the figure shown, friction force between the, nl, mg, . If t =, 7g, 4, where t is the time in which the bead loose, contact with the string after the system is, released from rest, find n, , P, , bead and the light string is, , P1, , P2, D, , B, A, , m, ↑, l, ↓, m, , 100, , C, , 36. In the situation given, all surfaces are, frictionless. pulley is ideal and string is light if, F = Mg / 2, the acceleration of the big block, is g/x then x is, , x, M, M, , y
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, , NEWTON LAWS OF MOTION, , 37. Three blocks shown in figure, move vertically, with constant velocities, The relative velocity, of A w.r.t C is 100m / s upward and the relative, velocity of B w.r.t A is 50m / s downward. All, the string are ideal. The velocity of C with, respect to ground is 125/x calcualte x, , 40. A system is shown in the figure. End B of, string is moving upwards with 3 m/s. Pulley, is moving with speed 2 3 m/s in direction, shown in the figure. The velocity of the block, A is x + 2 3 (m/s) find x, , B, u = 3m / s, 60°, 30°, , A, , 2 3m / s, , A, , 41. If at t = 0 right spring in (A) and right string in, (B) breaks. The ratio of magnitudes of, instantaneous acceleration of blocks A & B is, C, , 38. Block A of mass m is placed over a wedge of, same mass m. Both the block and wedge are, placed on a fixed inclined plane. Assuming all, surfaces to be smooth, the displacement of the, , g sin 2 θ, x + sin 2 θ, , block A in ground frame in 1s is, then the value of x is, N, Block, , y, , wedge, , mg, , A, , x, B, , N, N1, , Fixed, incline, , θ, (a), , B, , mg, (b), , 39. A small , light pulley is attached with a block, C of mass 4 kg as shown in Fig . A block B of, mass 1.5 kg is placed on the top horizontal, surface of C.Another block A of mass 2kg is, hanging from a string , attached with B and, passing over the pulley. Taking g = 10 ms-2, and neglecting friction, acceleration of block, C when the system is released from rest is x/, 4 calculate x., , 5x, , calculate x, 24, 37° 37°, , 37°, , 2kg, , 37°, , 2kg, , (A), (B), 42. In the figure shown P1 and P2 are massless, pulleys. P 1 is fixed and P 2 can move. Masses, 9m, , 2m and m, of A,B and C are, 64, respectively. All contacts are smooth and, −1 3 , the string is massless. θ = tan . (Take, 4, 2, g = 10m/s ), , P1, , B, A, , θ, , vertical wall, , B, , P2, , C, , horizontal floor, 12.5cm, C, , the tension in string connecting pulley P 2 and, block C is, , 13, , Calculate x (Take m = 1 kg), x, 101
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, 43. In the arrangement shown in the figure,, pulleys are light, small and smooth. Mass, of blocks A, B and C is m1 = 14 kg, m2 = 11, kg and M = 52 kg respectively. The block A, can slide freely along a vertical rail, fixed, to left vertical face of block C. Assuming, all the surfaces to be smooth, magnitude of, acceleration of block A is, , N, m, , mA sinα, a, mg sinα, , α, , 10, , Calculate x., x, , mA cosα, , mg cosα, From free body diagrams, ----- (1), N sin α = Ma, N + mA sin α = mg cos α ----- (2), mg sin α + mA cos α = ma ------(3), on solving equations (1) (2) (3), mg cos α sin α, ( M + m ) g sin α, ; a=, 2, M + m sin α, M + m sin 2 α, uuuur uuuuuuuuur uuuuur, Now ablock = ablock / wedge + awegde, r, $, $, $, a, ∴ block = a cos α i − a sin α j − Ai, A=, , (, , LEVEL -VI - KEY, 1.C, 8.D, , SINGLE ANSWER, 3.B, 4.D, 5.C, 6.C, 10.A, MULTI ANSWER, , 2.B, 9.C, 11.A,B,D, 15. A,C, 19. B,C, 24.B, , 12.A,B,C, 16. ABD, 20.C, 25.C, , 13.A,C,D, 17.A,D, 21.B, , ), , ( a cos α − A ) $i − a sin α $j, 7.B, , 14. B,C, 18.A,C, 22.B 23.A, , 2., , a sin α, m, , ∴ tan β = a cos α − A = 1 + M, , Resultant acc. of B, , = a 2 + a 2 + 2aa cos (180 − α) = 2a sin ( α / 2 ), a, , MATCHING, , (180 – α), , 26. A – s;B – r;C – q;D – p, , 27. A - r ; B - s; C - q ;D – p, 28. A-q, B-p, C-s, INTEGER TYPE, 29.2 30. 2 31.4 32.2 33.8, 36.4 37.2 38. 1 39.5 40. 3 41.5, , 34. 3, 42.2, , 102, , 35.5, 43.5, , 3., 4., , 1, 1, 1, 1, =, +, =, K eq 4K 4K 2K ; K eq = 2K ., From constraint relation we can see that, , 3 − 12t , proper signs. Hence a B = , = 1.5 − 6t or, 2 , , 90 – α, M, α, , N, , α, , D-r, , a A + aC , acceleration of block B is a B = , with, 2 , , A, , 1., , a, , α, , LEVEL -VI - HINTS, , N sinα, , , tan α, , , dv B, = 1.5 − 6t or, dt, , vB, , t, , 0, , 0, , ∫ dvB = ∫ (1.5 − 6t ) dt o r, , 1, v B = 1.5t − 3t 2 or v B = 0 at t = s, 2
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 5., , NEWTON LAWS OF MOTION, , Mass m1 moves with constant velocity if tension in, , the lower string T1 = m1g = (1)(10 ) = 10N ….(i), ∴ Tension in the upper string is, T2 = 2T1 = 20N, ….(ii), Acceleration of block M is therefore,, T 20, a= 2 =, …..(iii), M M, This is also the acceleration of pulley 2., T = 10N, 40, Pseudo force = m 2 a =, M, a, , kx, N, θ, , a, , mg, , kx + mg sin θ = ma cos θ, , F, cos θ ; 4mg tan θ = F ., 2, 1, 1, 1, 7, 12K, x0, 10. K = 3K + 4K = 12K ; T =, 7, eq, , B, , Absolute acceleration of mass m1 is zero. Thus,, acceleration of m1 relative to pulley 2 is a upwards, or acceleration of m2 with respect to pulley 2 is a, downwards. Drawing free body diagram of m2 with, respect to pulley 2. Equation of motion gives, 40, 40, 20 − − 10 = 2a =, solving this we get M=8kg, M, M, 6. Refering the method to solve eqivalent spring, T, T T, T, consant Tx = x1 =, ; 64K =, 8, 8 8K, x, , 3, 10 3 cos 30° = 10 3, = 15 m / s 2, 2, , 12, K / x 0 = 4K / x1 ;, 7, , T = 4K x1 ;, , x1 =, , 3, x0, 7, , MULTI ANSWER TYPE, 11. Application of newton’s first law and second law, 12. A,B,C, RA, , RB sin60, 60°, , W, , RB, , RB cos60, , Since the sphere is not moving ∑ FH = 0, RB sin 60 = 0, , 10 3m / s 2, , ∴ RB = 0, , 30°, , & RA = W = 10 N, , g sin30° = 5m/s2, , ∴ a = 15 − 5 = 10 m / s 2 ∴ S =, , 8., , applying newtons law among incline, , 2mg sin θ =, , m2g = 20N, , 7., , 9., , 1 2, at, 2, , Fig-2, , 1, 1, S, 1 = (10)t 2 or t =, 5, 2, equivalent spring constant is 9K., , RA, 60°, RAsin60, , 9K, , displacement is, , 2F, F, F, −, =, 9K 9K 9K, , ∑F, , H, , =0, 103
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JEE, MAIN-JR-VOL, -I, JEE-ADV, PHYSICS-VOL, - II, 34. Concept : along the length of string velocity, component is same for all the points on string.., , uB, , 53°, , Y, , uC, , vp, , uC, , NEWTON LAWS OF MOTION, r r, 37. Let velocity of blocks, A, B and C are v A , v B and, r r, r r r, vC ; v A − v C = 100 ˆj ...(1) v B − v A = 50 ˆj .(2), , 37°, , uB, , uCcos37°, , (a), , X, VA, , A, , VP, , vB = vC cos 530 ........ (1) vC cos 37 0 = vD .......(2), from eqns. (1) and (2) we get, , vB cos 37 0 vB ( 4 / 5 ), 4, =, vD = v, ;, 0, cos 53, ( 3 / 5), 3, 35. Apply constraint on pulley P, r, r, r, r, r, r, v p1 / p = −v p2 / p ; v p1 − v p = − v p2 − v p, r r r, v p1 , v p , v p2 are respective velocity w.r.t. ground,, r, r r, v p2 = 2v p − v p1 ; = 2 10 ˆj − 30 ˆj = −10 ˆj, Now apply constraint eqn. on pulley P2, r r, r, v C − v p2 = − vp − vp2, r, r, r, r, v C = 2v p2 − v D = 2 −10 ˆj − −10 ˆj v C = 10 ˆj, vD =, , (, , (, , ), , ), , Apply constraint eqn. on pulley P1 to get, r r, r r, r, r, r, v A − v p1 = − v B − v p1 ; v B = 2v p1 − v A, , (, , ), , = 2 30 ˆj − 10 ˆj = 50 ˆj, , 36. First consider both the blocks as system force that, we apply at one end of string is tension in the string., System, T, N, mg, , F = Mg/2, , aB/g, ar, aA /B, , For system block ( A + B ), , B, , A, , Mg, = 2 Ma a = g / 4, 2, , Thus, a = g / 4iˆ, Mg, = Mar, For system block A : Mg −, 2, r, a r = g / 2 jˆ, r, Thus,, a A = g / 4iˆ + g / 2 ˆj, r, a3 = g / 4 ˆj, , VA, , B, C, , Now write constraint equation for pulley to get, r r, r r, r, r, r, v B − v p = − v C − v p or,, 2v p = v B + v C, r, r, r, also v p = −v A Thus we get v A = 37.5 ˆj m / s, r, r, v B = ( −12.5 g ) m / s ; v C = −62.5 ˆj m / s, 38. Let acceleration of wedge in ground frame is a down, the plane. The acceleration of block A will be a, r, r, r, sinθ vertically downward a A / g = a A / B + a B / g .(1), , (, , ), , (, , (, , ), , ), , a A / g = [a A/ B ]x + a B / g .......(2), x, x, From FBD of A it is clear that Block A cannot, accelerate horizontally. i.e., in x-direction because, there is no force in x-direction. Block A can, accelerate in y-direction only. a A / g x = 0 There, fore a A / g x = − aB / g x That means for an, observation on wedge block moves only x > 0., For block A ; mg − N = m ( a sin θ ) .......(3), For block B; ( N + mg ) sin θ = ma .......(4), 2 g sin θ , On solving eqns. (3) and (4), we get a = , 2, , 1 + sin θ , The acceleration of block A, aA = a sin θ, , 2 g sin 2 θ , 2 g sin θ , =, sin, θ, =, 1 + sin 2 θ , 1 + sin θ , , , Displacement of block A in 1 s is, 2, 1 2 g sin 2 θ , 2 g sin θ , 1, 2, × (1) , S = 0 + a At = × , , 2, 2 1 + sin 2 θ , 2, 1 + sin θ , 39. Let vertically downward component of acceleration, of A be a and let acceleration of C be c (right ward), , 107
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JEE MAINS, - C.W - VOL- -IIII, JEE-ADV, PHYSICS-VOL, , NEWTON LAWS OF MOTION, Then horizontal component of acceleration of A i c, (right ward) and acceleration of B, relative to C is, a leftward. Hence resultant acceleration of B is (a, -c) left ward. Now considering free body diagram, (Fig ), 1.5(a–c), , T, , T, H, , T, , ⇒ a2 = 8, 2, , 1.5g, , 25, a1, 25, = 3 =, a2, 8 24, , ∴, , B, N, N, , T, , A, , 4, ma2 = 20cos37 0 = 20 = 16 N, 5, , 20 N, , 2a, 2c, , 4g, , T, , 2g, , ←H, , T, aB↓, , 42., , aB = 2aC → ( 5) after solving above equations we get, , a A = 8 m / sec 2 , aB = 6 m / sec 2 ; aC = 3 m / sec 2 ;T =, 2T, , 43., , mA, , ), , Fr, , Fr, , A, , NA, , aB, mBA, , B, , NB, , 110, , T, NB, , C, A, , Equations are : 2T − 140 = 14a A → (1), N A = 14 A → ( 2 ) 110 − T = 11aB → ( 3), , 2, , N B = 11A → ( 4 ) T − N A − N B = 52 A → ( 5 ), , 37°, , constraint relation is : aA + aA − aB = 0, ⇒ aB = 2aA → ( 6 ) after solving equations 1 to 6, , 50, 3, , 37°, 2, , 50, FR = N ;, 3, , acceleration, , we get resultant acceleration of A =, 1 + 1 = 2 m / s 2 resultant acceleration of B =, 9 + 1 = 10 m / s 2 resultant acceleration of C =, 1 m / sec 2 ∴, , 20, , 108, , 50, N, 3, , 2T, NA, , 13, N, 2, , T, , 140, , ⇒ Fr =, , 2T, ↑aC, , 3, 2T − mC g = mC aC → ( 3) aB = a A → ( 4 ), 4, , ∴ VA = 3 + 2 3 m / sec, 37° 37°, , N, , aA, , where: VP = 2 3 , VB = 3 m / sec, , 41., , N, , 3, 4, N1 = m A .a A → (1) mB g − T − N1 = mBα B → ( 2 ), 5, 5, , V, 3, VP + VB + P = 0, 2, 2, , (, , A, , B, , T, , C, , For block A, H = 2c ....(1) 2g-T=2a .....(2), For horizontal forces on block B,, T= 1.5 (a-c) ....(3) from above equations,, a = 6.25ms-2 and c = 1.25ms-2, ∴ Vertical acceleration of A=a=6.25ms-2 (downward), Horizontal acceleration of A, = c = 1.25 ms-2 (right ward), Acceleration of B = ( a - c) = 5ms-2 (left ward), Acceleration of C = C = 1.25 ms-2 (right ward), 40. Use pulley constant equation:, −VA +, , aA, , 4c, , V, , T, , 25, 25, ; a1 =, 3, 3, , slipping between A & B both move, , with common acceleration a =, , 20, = 4m / s 2, 5
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Ø, , Ø, , 1. Force and displacement are perpendicular., 2. Displacement of point of application of, Ø, force is zero., 3. Net force acting on the body is zero., ur r, r, As W = ∫ F .d s so, if d s = 0 , W=0 i.e., if the, displacement of a particle or body is zero whatever, be the force , work done is zero (except nonconservative force), (a) When a person tries to displace a wall or stone, by applying a force and it (actually its centre of mass, ) does not move, the work done is zero., (b) A weight lifter does work in lifting the weight from, Ø, the ground but does not work in holding it up., As W = ∫ F ds cos θ , so W = 0 , if θ = 900 , i.e.,, if force is always perpendicular to motion, work, done by the force will be zero though neither force, nor displacement is zero. This is why:, (a) When a porter moves with a suitcase on his, head on a horizontal level road, the work done by, Ø, the lifting force or force of gravity is zero., (b) When a body moves in a circle the work done, by the centripetal force is always zero., (c) When the bob of a simple pendulum swings,, the work done by tension in the string is zero., Ø, , WORK DONE BY VARIABLE FORCE:, , Graphical representation of work done:, The area enclosed by the F-S graph and, displacement axis gives the amount of work done, by the force., F, P, , Q, , O, , S, , R, , Work = FS = Area of OPQR, Work done by variable force., F, , xi, , xf S, , dx, , For a small displacement dx the work done will be, the area of the strip of width dx, W=, , xf, , xf, , xi, , xi, , ∫ dW = ∫ F dx, , If area enclosed above X-axis, work done is +ve, and if the area enclosed below X-axis, work done, is –ve., F, O, , xi, , xf, , Negative work, , Applications on work, Ø, , Ø, When the magnitude and direction of a force varies, with position, then the work done by such a force, for an infinitesimal displacement ds is given by, ur uur, dW = F .ds, Ø, The total work done in going from A to B is, B ur uu, r B, WAB = ∫ F .ds = ∫ ( F cos θ ) ds, A, , A, , In terms of rectangular components, r, uur, F = Fx iˆ + Fy ˆj + Fz kˆ ; ds = dxiˆ + dyjˆ + dzkˆ, W =, , x2, , y2, , If a force is changing linearly from F1 to F2 over a, displacement S then work done is, F + F2 , W = 1, S, 2 , If a force displaces the particle from its initial, r, uur, position ri to final position rf then displacement, ur uur ur, vector is S = rf − ri ., y, , rf, , x, , y, , y1, , z, , z1, , ri, , ur ur ur uur ur, W = F .S = F . rf − ri, , (, , 146, , F, , z2, , ∫ F dx + ∫ F dy + ∫ F dz, x1, , S, , ), , X, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, Ø, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Work done in pulling the bob of mass m of a simple Ø, pendulum of length L through an angle θ to vertical, by means of a horizontal force F., , θ, , L-h, , T, , Ø, , L, C, A, , F, mg, , h, B, , cos θ =, , L−h, h h, = 1− ;, = 1 − cos θ, L, L L, , A block of mass m is suspended vertically using a, rope of negligible mass. If the rope is used to lift the, block vertically up with uniform acceleration ‘a’,, work done by tension in the rope is, W = m ( g + a ) h (h= height), If block is lowered with acceleration ‘a’, then, W = − m( g − a ) h, A uniform chain of mass M and length L is kept on, 1, smooth horizontal table such that th of its length, n, is hanging over the edge of the table., The work done by the pulling force to bring the, hanging part onto the table is, L, 2n L, n, , h = L (1 − cos θ ), , Ø, , Work done by gravitational force, W = −mgh = −mgL (1 − cos θ ), Work done by horizontal force F is W = FL sin θ, Work done by tension T in the string is zero., Work done by gravitational force in pulling a uniform, rod of mass m and length l through an angle θ is, given by, , C.G, , , , W = n gh = n g 2n = 2n 2, , , M, , C.G, h, , C.G, , l, (1 − cos θ ) , Where l is the distance, 2, 2, of centre of mass from the support., Ø, A ladder of mass ‘m’ and length ' L' resting on a, level floor is lifted and held against a wall at an, angle θ with the floor, Work done by gravitational force is, W = − mg, , Ø, , L, , MgL, , M, , Ø, , θ, , M, , Mass of hanging part is, n, A uniform chain of mass M and length L rests on a, 1, smooth horizontal table with n th part of its length, 1, is hanging from the edge of the table. Work done in, pulling the chain partially such that, 1, n2 th part is hanging from the edge of the table is, , given by, , W=, , MgL 1, 1, − 2, , 2, 2 n1 n2 , , A uniform chain of mass ‘M’ and length L is, suspended vertically. The lower end of the chain is, lifted upto point of suspension, , L, , , G2, , Wg = − mgh = −mg sin θ, 2, , l/4, , G1, G1, , l/4, G2, , L/ 2, θ, , Ø, , A bucket full of water of total mass M is pulled by, using a uniform rope of mass m and length l. Work, done by pulling force, , NARAYANAGROUP, , L L L, + = = raise in centre of mass of, 4 4 2, lower half of the chain ., Work done by gravitational force is, MgL, M L, Wg = − g = −, 2 2, 4, h=, , W = Mgl + mg, , l, 2, , 147
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, Ø, , N, , The Work done in lifting a body of mass ‘m’ having, , F cos θ, , density ' d1 ' inside a liquid of density ' d 2 ' through, , F, , a height ‘ h ’ is, , θ mg cos θ, , mg sin θ, , W = F S = ma S, Ø, , F sin θ, , θ, , mg, d2 , W = mg |h = mgh 1 − d , , 1, A body of mass ‘m’ is placed on a frictionless, horizontal surface. A force F acts on the body, parallel to the surface such that it moves with an, acceleration ‘a’, through a displacement ‘S’. The, work done by the force is, , Ø, , mg, , h, , FB, , Ø A body of mass ‘m’ is sliding down on rough inclined, plane of inclination θ . If L is the length of incline and, µK is the coefficient of kinetic friction then work done, by the resultant force on the body is, N, fk, mg sin θ, , θ, , mg cos θ, , mg, , (Qθ = 0 ), 0, , θ, , A body of mass ‘m’ is placed on a rough horizontal, W = ( mg sinθ − fk ) L = ( mg sin θ − µ k mg cos θ ) L, surface of coefficient of friction µ . A force F acts, = mgL ( sin θ − µk cos θ ), on the body parallel to the surface such that it moves, with an acceleration ‘a’, through a displacement Ø A uniform solid cylinder of mass m, length l and, radius r is lying on ground with curved surface in, ‘S’. The work done by the frictional force is, contact with ground. If it is turned such that its, f = µ mg cos θ ; but θ = 00, circular face is in contact with ground then work, 0, done by applied force is, ∴ f = µ mg cos 0 = µ mg ⇒ W f = µ mgs, , Wnet = ( f + ma ) S = ( µ mg + ma ) S = m ( µ g + a ) S, , Ø, , If the body moves with uniform velocity then, W = f S = µ mg S, A body of mass m is sliding down on a smooth, inclined plane of inclination θ . If L is length of, inclined plane then work done by gravitational force, is, N, L, , l, l/2, r, , Ø, mg sin θ, , θ, , mg cos θ, , l, , l , , W = mgh = mg − r , Qh = −r, 2 , , 2, , A gas at a pressure P is enclosed in a cylinder with, a movable piston. Work done by the gas in, producing small displacement dx of the piston is, , mg, θ, , Ø, , Wg = F S = mg sin θ L, A body of mass ‘m’ is moved up the smooth inclined, plane of inclination θ and length L by a constant, horizontal force F then work done by the resultant, force is, W = ( F cos θ − mg sin θ ) L, , 148, , NARAYANAGROUP
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, L, , L, , 0, , 0, , L, , L, , ∫, , ∫, , 2, W = ∫ Fdx = ∫ ( ax + bx 2 )dx = axdx + bx dx, , x2 , W=a, , 2, , L, , 0, , 0, , 0, , L, , x3 aL2 bL3, +b, +, =, 2, 3, 3, , Y, , H, , Sol:, α, , X, x, , 2, 2, ur r, Fy = − mg $j; ry = H max = u sin α ˆj, , , , 2g, , r r, u sin α ˆ, W = Fy .ry = −mgjˆ . , j, 2g, , (, , ), , , , , , Work done W = mg = l g, 2 L 2, 0.6, 4, W = × 0.6 × 10 ×, = 3.6 J, 2, 2, M, , l, , WE-8: Find the work done in lifting a body of mass, 20 kg and specific gravity 3.2 to a height of 8, m in water? (g = 10 m/s 2 ), ρb, Sol:Given specific gravity ρ = 3.2, w, ρ b = 3.2 × ρ w = 3.2 × 1000 = 3200, , r, , 2, , WE-7: A uniform chain of length 2 m is kept on a, table such that a length of 60 cm hangs freely, from the edge of the table. The total mass of, the chain is 4 kg. What is the work done in, pulling the entire chain back onto the table?, Sol:M = 4 kg, L = 2 m, l = 0.6 m, g = 10 m/s2, l, , 0, , WE-5: A particle of mass ‘m’ is projected at an angle, α to the horizontal with an initial velocity u., Find the workdone by gravity during the time it, reaches the highest point., , O, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , , , 2, , , , , , , , Workdone W = mgh 1 − ρw = 20×10 × 81−, ρb , 2200 , W = 20 × 10 × 8 , = 1100 J, 3200 , , , , 1000 , , 3200 , , WE-9: A block of mass ‘m’ is lowered with the help, of a rope of negligible mass through a distance, ‘d’ with an acceleration of g/3. Find the work, done by the rope on the block?, Sol:During lowering a block, tension in rope is, T = m ( g − a ) and S = d, , 1, W = − mu 2 sin 2 (α ), 2, WE-6: A 10 kg block is pulled along a frictionless, W = − m( g − a) d, work done, surface in the form of an arc of a circle of, radius10 m. The applied force is 200 N. Find, g, 2 mgd, , the work done by (a) applied force and, W = −m g − d =−, 3, 3, , , (b) gravitational force in displacing through, 0, an angle 60, WE-10: If the system shown is released from rest., Find the net workdone by tension in first one, second (g=10m/s2), r, 0, 60, F, r, , Sol: Work done by applied force W = Fr sin θ, , 3, W = 200 ×10 × sin60 = 200 ×10 × = 1732 J, 2, work done by gravitational force, W = − mgr (1 − cosθ ), 0, , W = − 10 × 9.8 × 10 (1 − cos 600 ), 1, , W = − 98 × 10 1 − = − 490 J, 2, , , 150, , 3 kg B, , A 2 kg, m − m1 , 3− 2 , Sol. a = 2, g =, 10 = 2m/s2, m, +, m, 2, +, 3, , , 1, 2 , , 2 m1m 2 g 2 × 2 × 3 × 10, =, = 24N, m1 + m 2, 2+3, 1 2 1, for each block S = at = × 2 × 1 = 1m, 2, 2, ∴ Wnet = W1 +W2 = TS − TS = 0, T =, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Energy:, Ø, Ø, Ø, Ø, Ø, , Ø, Ø, , Examples for bodies having K.E, 1) A vehicle in motion., 2) Water flowing in a river., 3) A bullet fired from a gun., Kinetic energy depends on frame of reference., Ex: kinetic energy of a person of mass m sitting in, a train moving with speed v is zero in the frame of, 1 2, train but mv in the frame of earth., 2, , Energy is the ability or capacity to do work. Greater, the amount of energy possessed by the body, greater, the work it will be able to do., Ø, Energy is cause for doing work and work is, effect of energy., Energy is a scalar. Energy and work have same, units and dimensions., The different forms of energy are Mechanical energy,, Light energy, Heat energy, Sound energy, Electrical Relation between K.E. and linear momentum, energy, Nuclear energy etc., 1 2 P2 1, Mechanical energy is of two types, KE, =, mv =, = Pv (Q P = mv ), Ø, a) Potential Energy, b) Kinetic Energy, 2, 2m 2, Ø If two bodies of different masses have same, Potential energy (U), momentum then lighter body will have greater KE, Potential energy of a body is the energy possessed, 1 , , by a body by virtue of its position or configuration, Q K E α , m, , in the field., Ø, When, a, bullet, is fired from a gun the momentum of, Potential energy is defined only for conservative, the bullet and gun are equal and opposite., forces. It does not exist for non-conservative, M gun, K E bullet, forces. In case of conservative forces., , i.e, , r2, ur uur, dU , ur uur U2, F = −, ∴ dU = − F .dr ⇒ ∫ dU = −∫ F .dr, dr , U, r, 1, , 1, , ur uur, U 2 − U1 = − ∫ F . dr = −W, r2, , Ø, , r1, , If r1 = ∞ , U 1 = 0, Ø, Ø, Ø, Ø, Ø, Ø, , Ø, Ø, , Ø, Ø, , r ur uu, r, ∴ U = ∫ F .dr = −W, ∞, , P.E can be +ve or -ve or can be zero., P.E depends on frame of reference., Ex: Water stored in a dam , A stretched bow,, A loaded spring etc., possesses P.E, In case of conservative force (field) potential energy, is equal to negative of work done in shifting the, body from some reference position to given position, A moving body may or may not have potential, energy., Potential energy should be considered to be a, property of the entire system, rather than assigning, it to any specific particle., Kinetic energy, Kinetic energy is the energy possessed by a body, by virtue of its motion., Kinetic energy of a body of mass ‘m’ moving with, 1 2, a velocity ‘v’, KE = mv, 2, Kinetic energy is a scalar quantity., The kinetic energy of an object is a measure of the, work an object can do by the virtue of its motion., , NARAYANAGROUP, , Ø, , Ø, 1), , K E gun, , =, , M, , bullet, , Hence, the KE of the bullet is greater than that of, the gun, A body can have energy without momentum. But it, can not have momentum without energy., A bullet of mass ‘m’ moving with velocity ‘v’ stops, in wooden block after penetrating through a distance, ‘x’. If F is resistance offered by the block to the, bullet, (Assuming F is constant inside the block), 1 2, mv 2, mv = Fx ; F =, ∴v 2α x, 2, 2x, For a given body, The graph between KE and P is a parabola., KE, , P, , 2), , The graph between KE and P is a straight line, 1, passing through the origin.Its slope =, 2m, K, , P, , 3), , 1, , The graph between KE and is a rectangular, P, hyperbola., 151
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , K, , and slope =, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , 1, 2m, , P 2 = 2mK, , K, , 1, p, Ø, , A particle is projected up from a point at an angle, ' θ ' with the horizontal. At any time ‘t’ if ‘P’ is linear, momentum, ‘y’ is vertical displacement and ‘x’ is, horizontal displacement, then nature of the curves, drawn for KE of the particle (K) against these, parameters are, , i) K - y graph:, K = Ki − mgy ; It is a straight line, , P, P 2 αK, , Conservative and non - Conservative forces, Ø, , Ø, y, , ii) K - t graph, 1, , , K = K i − mg uy t − gt 2 , 2, , , , 1, Q y = u yt − gt 2 ; It is a parabola, 2, K, , Ø, t, , Ø, , iii) K - x graph, , gx , K = K i − mg x tan θ −, , 2u2x , , 2, , Ø, , g , Q y = ( tan θ ) x − 2 x 2 ; It is also a parabola, 2u x , , Ø, , 2, , X, , It is a straight line passing through origin, 152, , If work done by a force around a closed path is, zero and is independent of path then the force is, said to be conservative force., dU, Under conservative force F = −, where U is, dr, ur uur, Potential Energy., U = ∫ dU = − ∫ F .dr, ur, ( F = Fx $i + Fy $j + Fz k$ and, uur, dr = dxi$ + dy $j + dzk$ ), ur, ∂u ∂u $ ∂u $ , F = − i$ +, j+ k, ∂y, ∂z , ∂x, Ex1: Gravitational force is a conservative force, Ex2: Elastic force in a stretched spring is a, conservative force, , Non-Conservative Forces:, If the work done by a force around a closed path is, not equal to zero and is dependent on the path, then the force is non-conservative force, Ex:-Force of friction , Viscous force., Work done by the non-conservative force will not, be stored in the form of Potential energy., Potential energy is defined only for conservative, forces., , Spring force, Ø, , K, , iv) K - P graph, , 2, , Ø, , Spring force is an example of a variable force which, is conservative., In an ideal spring, the spring force Fs is directly, proportional to ‘x’. Where x is the displacement of, the block from equilibrium position. i.e.,, Fs = − Kx .The constant K is called spring, constant., The work done on the block by the spring force as, the block moves from undeformed position, x = 0 to x = x1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , ur uur, dW = F .dx = − Kxdx, , WE-13: An athlete in the Olympic games covers a, distance of 100 m in 10s. His kinetic energy can, x, W = ∫ dW = ∫ − Kxdx = − 1 K ( x2 ) 1 = − 1 Kx12, 0, be estimated to be in the range(JEE MAIN 2008), 2, 2, 0, 1) 200 J - 500 J 2) 2 × 105 J − 3 × 105 J, If the block moves from x = x1 to x = x2 the work, x2, 3) 20,000J - 50,000J 4) 2,000 J - 5, 000 J, Sol:Approximate mass of the athlete = 60 kg, done by spring force is W = ∫ −Kxdx, x1, Average velocity = 10 m/s., 1, 1 2 1 2, 2, 2, 1 2 1, 2, W = K x1 − x2 = Kx1 − Kx2, Approximate K.E. = mv = × 60 ×10 = 3000 J, 2, 2, 2, 2, 2, Potential energy stored in a spring:, Range of KE = 2000 J to 5000J, The change in potential energy of a system W.E-14:Kinetic energy of a particle moving along, corresponding to a conservative internal force is, a circle of radius ‘r’ depends on the distance, x, r uur, as KE = cs2, (c is constant,s is displacement)., dU = −∫ F . dx ,, 0, Find the force acting on the particle, dU = - (work done by the spring force), 2c , 1 2, 2, mv, =, cs, ⇒, v, =, − Kx 2 , , s, 1 2, Sol. KE = 2, dU = − , ; U f − U i = Kx, m , x1, , Ø, , (, , Ø, , , , ), , , , 2, , 2, , since U i is zero when spring is at its natural length, 1 2, Kx, 2, W.E-11:Two spheres whose radii are in the ratio, 1 : 2 are moving with velocities in the ratio, 3 : 4. If their densities are in the ratio 3 : 2,, then find the ratio of their kinetic energies., r1 1 v1 3 ρ1 3, Sol. r = 2 , v = 4 , ρ = 2, 2, 2, 2, ∴U f =, , K .E =, , 1 2 1, 14, , mv = (V ρ ) v 2 = π r 3 ρ v 2, 2, 2, 23, , 3, , 2, , KE1 ρ1 r1 v1 3 1 , = × × = × , KE2 ρ2 r2 v2 2 2 , KE1 3 1 9, 27, = × × =, KE2 2 8 16 256, , 3, , 2, , 3, × , 4, , at =, , dv, 2c ds, 2c, =, × =v, dt, m dt, m, , Ft = mat = mv, , 2c , 2c 2c, = m, s, = 2cs, m , m m, , 2, 2, Total force F = Ft + Fc =, , F = 2cs 1 +, , mv 2 , 2, ( 2cs ) + , , r , , 2, , s2, r2, , W.E-15: A rectangular plank of mass m1 and height, ‘a’ is on a horizontal surface. On the top of it, another rectangular plank of mass m2 and, height ‘b’ is placed. Find the potential energy, of the system?, , m2, , b, W.E-12:A particle is projected at 600 to the horizontal, m1, a, with a kinetic energy ‘K’.Find the kinetic, energy at the highest point ?(JEE MAIN 2007), 1 2, Sol: Total potential energy of system U = U1 + U2, Sol.Initial kinetic energy is K = mu, 2, a, b m1, , b , , = m1 g + m2 g a + = 2 + m2 a + m2 2 g, The velocity at highest point vx = u cos θ ., 2, 2 , , , , kinetic energy of a particle at highest point, WE-16: A rod of mass m and length L is held, vertical. Find its gravitational potential, 1, 1, K, KH = mvx 2 = mu 2 cos2 θ = K cos 2 60 0 =, energy with respect to zero potential energy at, 2, 2, 4, the lower end?, NARAYANAGROUP, , 153
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , dy, y, Sol.Choose a small element of length dy, then, m, mass of the element dm = dy ., L, The potential energy of the element dU= (dm)g(y), Potential energy of the entire rod, , 1 2, Sol: Initial potential energy U i = kx, 2, 1, 2, Final potential energy U f = k ( x + y ), 2, 1, 1 2, 2, Work done W= Uf - Ui = k ( x + y ) − kx, 2, 2, 1, W = ky ( 2 x + y ), 2, , Work - energy theorem, Ø, , m, m, U = ∫ ( dm ) gy = ∫ ( dy ) . gy = g ∫ ydy, L 0, 0, 0 L , L, , L, , L, , L, , m y2 , mgL, U = g =, L 2 0, 2, Ø, WE-17: A chain of length l and mass ‘m’ lies on, the surface of a smooth hemisphere of radius Ø, R > l with one end tied to top of the, hemisphere.Find the gravitational potential, energy of the chain?, Ø, Sol., (Rd θ), , R, , θ dθ, , y = Rcosθ, , m, The mass of the element dm = Rdθ, l, The gravitational potential energy of the, element du= (dm)gy, The gravitational potential energy of total chain, , m, , U = ∫ ( dm )gy = ∫ Rdθ g ( R cos θ ), l, , 0, l, R, 0, , Ø, , Ø, , l, mgR 2, mgR 2, l , sin , [sin θ ]0R =, l, l, R, WE-18: A spring of force constant ‘k’ is stretched, by a small length ‘x’. Find work done in, Ø, stretching it further by a small length ‘y’?, , U=, , 154, , Work done by all forces acting on a body is equal, to change in its kinetic energy., 1 2 1, 2, i.e., W = K f − K i = mv − mu, 2, 2, Where Kf and Ki are the final and initial kinetic, energies of the body., Work energy theorem is applicable not only for a, single particle but also for a system of particles., When it is applied to a system of two or more, particles change in kinetic energy of the system is, equal to work done on the system by the external, as well as internal forces., Work-energy theorem can also be applied to a, system under the action of variable forces, pseudo, forces, conservative as well as non-conservative, forces., , Applications of work-energy theorem:, Ø, , Consider a small element of chain of width, dθ at an angle θ from the vertical, , l, R, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , A body of mass m starting from rest acquire a, velocity ‘v’ due to constant force F. Neglecting air, resistance., 1 2, Work done =change in Kinetic energy = mv, 2, A particle of mass ‘m’ is thrown vertically up with a, speed ‘u’. Neglecting the air friction, the work done, by gravitational force, as particle reaches maximum, height is Wg = ∆K = K f − K i, 1, 1, 1, 2, Wg = m ( 0) − m × u 2 = − mu 2, 2, 2, 2, A particle of mass ‘m’ falls freely from a height ‘h’, in air medium onto the ground. If ‘v’ is the velocity, with which it reaches the ground, the work done, by air friction is Wf and work done by gravitational, 1, 1, force Wg then , Wg + W f = mv 2 − 0 = mv 2, 2, , 2, , A block of mass ‘m’ slides down a frictionless, incline of inclination ‘ θ ’ to the horizontal. If h is, the height of incline, the velocity with which body, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , reaches the bottom of incline is, , h, mg ( h + s ) + ( − Rs ) = 0 ; R = mg 1 + , s, , 1 2, Wg = ∆K ; mgh = mv − 0, 2, , u=0, h, , 1 2, mv ; v = 2 gh, 2, A body of mass ‘m’ starts from rest from the top of, a rough inclined plane of inclination ‘ θ ’ and length, ‘l’. The velocity ‘v’ with which it reaches the bottom, of incline if µk is the coefficient of kinetic friction is, Wg + W f = ∆k, mgh =, , Ø, , ( mg sin θ ) l + ( − µk mg cosθ ) l =, v = 2 gl ( sin θ − µ k cosθ ), , Ø, , Ø, , A bob of mass m suspended from a string of length, l is given a speed u at its lowest position then the, speed of the bob v when it makes an angle θ with, the vertical is, 1, Wg + WT = ∆K ⇒ −mgl (1 − cosθ ) + 0 = m ( v2 − u2 ), 2, , v=0, , Here time of penetration is given by impulse, equation ( R − mg ) t = 0 − m 2 gh, A body of mass ‘m’ is initially at rest. By the, application of a constant force, its velocity changes, to v0 in time to the kinetic energy of the body at, time ‘t’ is, W = ∆K = K f − K i = K f − 0, , 1, 2, K f = W = mas = ma at = ma 2t 2, 2, 1, , , , , , 2, , 2, , 1 v , o, Since a = t ; K f = m 0 t 2, 2 t0 , o, , v, , WE-19: Under the action of force 2kg body moves, v = u 2 − 2 gl (1 − cos θ ), such that its position ‘x’ varies as a function, t3, A bullet of mass ‘m’ moving with velocity ‘v’ stops, of time t given by x = , x is in metre and t in, in a wooden block after penetrating through a, 3, second. Calculate the workdone by the force, distance x. If ‘f ’ is the resistance offered by the, block to the bullet., in first two seconds., W f = K f − K i ; − fx = 0 − KEi, Sol. From work-energy theorem W = ∆KE, KE, , Ø, , soil, , s, , Ø, , 1 2, mv − 0, 2, , air, v1 = 2 gh, , mv 2, , P2, , 3, , t, dx, i.e., stopping distance x= f i = 2 f = 2mf, x = , Velocity v = = t 2, dt, 3, A block of mass ‘m’ attached to a spring of spring, At t = 0, v1 =0, At t = 2s, v2 = 4m/s, constant ‘K’ oscillates on a smooth horizontal table., 1, 1, W = m ( v22 − v12 ) = × 2 ( 4 2 − 0 ) = 16 J, The other end of the spring is fixed to a wall. It has, 2, 2, a speed ‘v’ when the spring is at natural length. The, WE-20: A uniform chain of length ‘l’ and mass’M’, distance it moves on a table before it comes to rest, is on a smooth horizontal table, with (1/n)th, is calculated as below, part of its length hanging from the edge of the, table. Find the kinetic energy of the chain as, WS .F + Wg + WN = ∆K, (S.F=spring force), it completely slips off the table., Let the mass be oscillating with amplitude ‘x’., , On compressing the spring WS . F = − 1 Kx 2, 2, , Wg = FS cos 900 = 0 ;, , 1, , 1, , m, , ⇒ − 2 Kx = 0 − 2 mv ⇒ x = v K, A pile driver of mass ‘m’ is dropped from a height ‘h’ Sol: Work done ∆W = Ui − U f = K f − K i, above the ground. On reaching the ground it pierces, 1, Mgl Mgl 1, , through a distance ‘s’and then stops finally. If R is the, − 2 = Mv 2 ; v = gl 1 − 2 , average resistance offered by ground then, 2, 2n, 2, n , , WS .F = K f − K i, , Ø, , L, n, , WN = NS cos 900 = 0, , Wg + WR = K f − Ki =, , NARAYANAGROUP, , 2, , 2, , 1, 1, mu 2 − mv 2, 2, 2, , 155
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , WE-21:Two blocks having masses 8 kg and 16kg WE-23: In the below figure, what constant force, ‘P’ is required to bring the 50kg body, which, are connected to the two ends of a light spring., starts from rest to a velocity of 10m/s in moving, The system is placed on a smooth horizontal, 7m along the plane? (Neglect friction), floor. An inextensible string also connects B, with ceiling as shown in figure at the initial, moment. Initially the spring has its natural, p 0, length.A constant horizontal force F is applied, 30, 50, kg, to the heavier block as shown. What is the, maximum possible value of F so that lighter, block doesn’t loose contact with ground., 0, 30, , 4m, , 5m, B, , Sol. Work done by force P in displacing the, , A, , 8kg, , 16kg, , F, , Sol: Draw FBD of B to get extension in spring., When block B just looses contact with ground, resultant force on it is zero., N, , block by 7m, W1 = ( F cos θ )( S ), W1 = ( P cos 300 ) 7 =, , 7 3, PJ, 2, , W2 = − mgh = −50 × 9.8 × 7 sin 300 = −1715 J, , According to work energy theorem, , T, θ, , W1 + W2 =, , kx, , (, , 1, m v22 − v12, 2, , ), , 7 3, 1, P − 1715 = × 50 × (10 2 − 0 2 ) ⇒ P = 607 N, 2, 2, , mg, Kx, Kx − T cos θ = 0 ⇒ T =, cos θ, , ; T sin θ + N − mg = 0, Kx, , When N = 0 then T sin θ = mg ⇒ cos θ sin θ = mg, x=, , mg, 80, 60, =, =, K tan θ K × (4 / 3) K, , If spring has to just extend till this value then, from work energy theorem we get, Fx =, , 1 2, Kx ⇒ F = 30 N, 2, , WE-22: A 2kg block slides on a horizontal floor, with a speed of 4 m/s. It strikes an, uncompressed spring and compresses it, till the block is motionless.The kinetic, frictional force is 15 N and spring constant is, 10,000 Nm -1. Find the compression in the, spring?(JEE MAIN 2007), Sol:, , 1, 1, KE = mv 2 =W friction + Kx 2, 2, 2, , 1, 1, ⇒ × 2 × 4 2 = 15 x + × 10000 × x 2, 2, 2, , ⇒ 5000 x 2 + 15 x − 16 = 0, ⇒ x = 0.055 m or x = 5.5 cm, 156, , WE-24: Figure shows a spring fixed at the bottom, end of an incline of inclination 370. A small, block of mass 2 kg starts slipping down the, incline from a point 4.8 m away from the, spring. The block compresses the spring by 20, cm, stops momentarily and then rebounds, through a distance 1 m up the incline. Find, (i) the friction coefficient between the plane, and the block and (ii) the spring constant of, the spring. (g = 10 ms-2 ), , 0, , 37, , Sol: Applying work energy theorem for, downward motion of the body W = ∆KE, mg sin θ ( x + d ) − f × l1 −, , 1 2, Kx = ∆KE, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 1, 2, 20sin370 ( 5) − µ × 20 cos370 ×5 − K ( 0.2) = 0, 2, 80 µ + 0.02 K = 60 → (1), For the upward motion of the body, 1, −mg sin θ l2 + ( f × l2 ) + Kx 2 = ∆KE, 2, 1, 2, −2 ×10sin370 ×1 − µ × 20 cos370 ×1 + K ( 0.2) = 0, 2, 16µ − 0.02 K = − 12 → ( 2 ), Adding equations (1) and (2), we get, 96µ = 48 ⇒ µ = 0.5, Now, use the value of µ in equation (1),, we get K = 1000 N/m., , 2., 3., , 4., , dU, = 0 (or) slope of U-r graph is zero, dr, , When displaced from its equilibrium position, a net, force starts acting on the body which moves the, body in the direction of displacement or away from, the equilibrium position, PE in equilibrium position is maximum as compared, to other positions as, , 5., , d 2U, is negative, dr 2, , When displaced from equilibrium position the centre, of gravity of the body goes up, , Neutral equilibrium, , Types of Equilibrium, , 1., A body is said to be in translatory equilibrium, if net, ur, 2., force acting on the body is zero i.e., F net = 0, 3., dU, If the forces are conservative F = −, dr, 4., and for equilibrium F = 0 ,, dU, dU, = 0 or, = 0 , ∴ At equilibrium position, so −, dr, dr, 5., slope ofU -r graph is zero or the potential energy is, optimum (maximum or minimum or constant), There are three types of equilibrium, (i) Stable equilibrium, (ii) Unstable equilibrium, (iii) Neutral equilibrium., , Net force is zero, dU, = 0 or slope of U-r graph is zero, dr, , When displaced from its equilibrium position the, body has neither the tendency to come back nor, move away from the original position., PE remains constant even if the body is moving to, , d 2U, =0, neighbouring points, dr 2, When displaced from equilibrium position the centre, of gravity of the body remains constant, , Potential energy and Equilibrium, U, B, , Stable equilibrium, 1., 2., 3., , 4., , 5., , Net force is Zero, 1., dU, = 0 or slope of U-r graph is zero, dr, When displaced from its equilibrium position, a net, retarding forces starts acting on the body, which, has a tendency to bring the body back to its, equilibrium position, PE in equilibrium position is minimum as compared, to its neighbouring points as, , d 2U, is positive, dr 2, When displaced from equilibrium position the centre, of gravity of the body comes down, , Unstable equilibrium, 1., , Net force is zero, , NARAYANAGROUP, , A, x, , In the figure, at A :, dU, d 2U, = 0 , and, is positive, dx, dx 2, , Thus at A the particle is in stable equilibrium., dU, d 2U, = 0 , and, At B;, is negative, dx, dx 2, Thus at B the particle is in unstable equilibrium, WE-25: In a molecule, the potential energy between, two atoms is given by U(x) =, , a, b, − 6 . Where, 12, x, x, , ‘a’ and ‘b’ are positive constants and ‘x’ is the, distance between atoms. Find the value of ‘x’, at which force is zero and minimum P.E at that, point., (JEE MAIN 2010), , 157
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, Sol: Force is zero ⇒, , dU, =0, dx, , i.e., a ( −12) x −13 − b( −6) x −7 = 0, −12a 6b, 12a 6b, + 7 = 0 ⇒ 13 = 7, 13, x, x, x, x, 1, , 2a, 2a 6, ⇒x =, ∴x = , b, b , Substituting the value of x, , 2, , h1 + y1 y1 0.24 + 0.01 0.01 , = ;, =, ; h2 = 3.96 m, h2 + y2 y2 h2 + 0.04 0.04 , , 6, , 12, , 6, , b 6, b 6, ⇒ U min = a , − b, , 2a , 2a , b2 b2 , − b2, U min = , −, ⇒, U, =, , , min, 4a 2a , 4a, , Law of conservation of Mechanical, energy:, Ø, , Total mechanical energy of a system remains, constant, if only conservative forces are acting on a, system of particles and the work done by all other, forces is zero., ∴ U f − U i = −W, From work energy theorem W = k f − ki, ∴ U f − U i = − ( k f − ki ), , Sol. By conservation of mechanical energy, 1, mg ( h + y ) = Ky 2, 2, h = height of particle, y = compression of the spring, As here particle and spring remain same, 2, , WE-27: A small mass ‘m’ is sliding down on a, smooth curved incline from a height ‘h’ and, finally moves through a horizontal, smooth surface. A light spring of force, constant K is fixed with a vertical rigid stand, on the horizontal surface, as shown in the, figure.Find the value for the maximum, compression in the spring if mass ‘m’ is, released from rest from height ‘h’ and hits the, spring on the horizontal surface., A, m, h, C, , K, , ∴U f + k f = U i + ki ⇒ U + K = constant, The sum of potential energy and kinetic energy Sol. Conservation of energy b/w positions A and C, remains constant in any state., ( PE A )block + KE A = ( PEC )spring + KEC, Ø A body is projected vertically up from the ground., When it is at height ‘h’ above the ground, its, 1 2, 1 2, 2mgh, potential and kinetic energies are in the ratio x : y. If, mgh + 0 = Kx + 0 ;mgh = Kx ; x =, H is the maximum height reached by the body, then, 2, 2, K, x, h, h, x, WE-28:A vehicle of mass 15 quintal climbs up a, =, =, or, y H −h, H x+ y, hill 200m high. It then moves on a level road, WE-26: A massless platform is kept on a light, with a speed of 30ms −1 .Calculate the potential, elastic spring as shown in figure. When a sand, energy gained by it and its total mechanical, particle of 0.1kg mass is dropped on the pan, from a height of 0.24m, the particle strikes the, energy while running on the top of the hill, pan and the spring compresses by 0.01m. From, −2, what height should particle be dropped to Sol. m = 15 quintal = 1500kg, g= 9.8ms ,h = 200m, cause a compression of 0.04m., P.E.gained,U=mgh =1500 x 9.8 x 200=2.94 x 106 J, 0.1 kg, 1 2 1, K.E. = mv = x 1500 x (30)2= 0.675 x 106J, 2, 2, Total mechanical energy, E = K + U = (0.675 + 2.94) x 106= 3.615 x 106J, WE-29: A particle is released from height H.At, certain height from the ground its kinetic, energy is twice its gravitational potential, energy. Find the height and speed of particle, at that height, 158, , NARAYANAGROUP
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2, , JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, WE-31: The potential energy of 1 kg particle free, to move along X - axis is given by, , Sol. K.E = 2PE But KE = TE – PE, mg(H – h) = 2mgh; mgH = 3mgh, H, Also K.E = 2P.E ,, ⇒h= ;, 3, 1 2, H, gH, mv = 2mg ⇒ v = 2, 2, 3, 3, , x4 x2 , U (x) =, − J. The total mechanical, 4 2 , , WE-30: A heavy flexible uniform chain of length, π r and mass λπ r lies in a smooth, semicircular tube AB of radius ‘r’. Assuming, a slight disturbance to start the chain in, motion, find the velocity v with which it will, emerge from the end of the tube?, , C.G, , 2r/π, A, , B, Reference level, , Sol:, , B, , A, __, πr, 2, C.G, , energy of the particle is 2 J. Find the, maximum speed of the particle., dU, = 0., Sol:For maximum value of U,, dx, 4 x3 2 x, ∴, −, = 0 or x = 0, x = ± 1., 4, 2, d 2U, d 2U, =, −, 1, =2, At x = 0,, and, At, x, =, 1,, ±, dx 2, dx 2, Hence U is minimum at x = ± 1 with value, 1 1, 1, U min = − = − J, 4 2, 4, 1, 9, Kmax + Umin = E or K max − = 2 or K max =, 4, 4, 3, 1, 9, ⇒ vmax =, ms −1, ⇒ mv 2 =, 2, 4, 2, WE-32:Figure shows a particle sliding on a, frictionless track which terminates in a straight, horizontal section. If the particle starts slipping, from the point A, how far away from the track, will the particle hit the ground?, , A, B, 1.0 m, 0.5 m, , Sol: Applying the law of conservation of, mechanical energy for the points A and B,, Centre of gravity of a semicircular arc is at a, 1 2, 2r, mgH, =, mv + mgh, distance, from the centre ., 2, π, v2 g, Initial potential energy U i = ( λπ r ) g 2r , g− =, or v 2 = g ⇒ v = g = 3.1 ms −1, π, 2 2, , After, point B the particle exhibits projectile motion, −π r , with θ = 00 and y = − 0.5 m, Final potential energy U f = ( λπ r ) g , , 2 , Horizontal distance travelled by the body, When the chain is completely slipped off the tube,, 2h, 2 × 0.5, all the links of the chain have the same velocity v., R=u, = 3.1×, = 1m, g, 9.8, 1 2, 1, 2, kinetic energy of chain k = mv = ( λπ r ) v, POWER, 2, 2, Ø The rate of doing work is called power., From conservation of energy ,, Power or average power is given by, 2r , −π r 1, 2, work done, λπ rg = ( λπ r ) g , + ( λπ r ) v, Pavg =, , Power is a scalar, π , 2 2, time, SI Unit: watt(W) (or) J/s, CGS Unit : erg/sec, 2 π, Other Units : kilo watt, mega watt and horse power, v, =, 2, rg, +, 0n solving we get,, , , One horse power (H.P)=746 watt, π 2 , NARAYANAGROUP, , 159
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, Ø, , Instantaneous Power:, , Ø, , ∆W , P = Lt , , ∆t →.0, ∆t , , ur ur, It is also calculated by P = FV cos θ = F .V, Ø, , Ø, 1 ur ur, W mv 2 = 1 mv v 1, Pavg = =, = mav = F .V, 2 t 2, 2, t, 2t, 1, Pavg = Pinst, 2, Ø, Ø The area under P − t graph gives work done, dW, P=, ∴W = ∫ P.dt, Ø, dt, The slope of W-t curve gives instantaneous power, dW, P=, = tan θ, dt, , power, , Work, , Ø, , O, Ø, , Ø, Ø, , time, , O, , The power of a machine gun firing ‘n’ bullets each, of mass ‘m’ with a velocity ‘v’ in a time interval ‘t’, 1, , n mv 2 , 2, = nmv, is given by P= 2, t, 2t, , A crane lifts a body of mass ‘m’ with a constant, velocity v from the ground, its power is, P=Fv=mgv, Power of lungs of a boy blowing a whistle is, 1, ( mass of air blown per sec) (velocity) 2, 2, , Power of a heart pumping blood, = (pressure) (volume of blood pumped per sec), Ø A conveyor belt is moving with a constant speed, ‘v’ horizontally and gravel is falling on it at a rate of, dm, . Then additional force required to maintain, dt, dm, speed v is F = v, and additional power, dt, 2 dm, required to drive the belt is, P = F v = v, dt, , 160, , 1, mgh+ mv 2, 2, the motor P=, t, , If a body of mass ‘m’ starts from rest and accelerated, uniformly to a velocity v0 in a time t0 , then the work, done on the body in a time ‘t’ is given by, , v , v=at= 0 t, to , Instantaneous power, P = F v = m a v, v 02, v0 v0 , ∴ P = m t = m 2 t, to to , t0, A motor pump is used to deliver water at a certain, rate from a given pipe. To obtain ‘n’ times water, from the same pipe in the same time by what amount, of (a) force and (b) power of the motor should be, increased., If a liquid of density ‘ ρ ’ is flowing through a pipe, of cross section ‘A’ at speed ‘v’ the mass coming, dm, = Av ρ ., out per second will be, dt, To get ‘n’ times water in the same time, 2, , x, , Ø, , 1 mv 2 1, = ?Av 3, 2 t, 2, , (Q mass=density x volume = m = ρ × A × l ), A vehicle of mass ‘m’ is driven with constant, acceleration along a straight level road against a, constant external resistance ‘R’ when the velocity, is ‘v’, power of engine is P = F v = ( R + m a ) v, If P is a rated power of a device and if its efficiency, x, 1, P, is x% , useful power is (output power) P =, 100, If a motor lifts water from a well of depth ‘h’ and, delivers with a velocity ‘v’ in a time t then power of, , W=, , Applications on power, , P=, , Ø, , x, , When a liquid of density ‘ ρ ’ coming out of a hose, pipe of area of cross section ‘A’ with a velocity ‘v’, strikes the wall normally and stops dead. Then, power exerted by the liquid is P=, , Relation Between Pavg and Pins :, , θ, time, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , 1 2 1 vo t , mv = m ;, 2, 2 to , , 1, , dm , dm , , = n, , dt, , , dt , , a=, , v0, ;, to, , ⇒ A' v ' ρ ' = n ( Avρ ), , As the pipe and liquid are not changed,, ρ | = ρ ; A' = A &v ' = nv, 1, , dm , dm , v| , nv ) n, (, , , F', dt =, dt = n 2, dm, =, as F = v, ⇒ F, dm , dm , dt, v, v, , , dt, , , dt , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , as P = Fv ⇒, , WE-34: The 50 N collar starts from rest at A and is, lifted with a constant speed of 0.6 m/s along, the smooth rod. Determine the power, developed by the force F at the instant shown., , P ' F ' v ' ( n F ) ( nv ), =, =, = n3, P, Fv, Fv, 2, , ∴ F ' = n2 F, ∴ P ' = n3 P, To get ‘n’ times of water force must be increased, n 2 times while power n3 times., , Position and velocity of an automobile, w.r.t.time:, , 0.9 m, F, T, θ T, , 1.2 m, A, , An automobile of mass ‘m’ accelerates starting from, rest, while the engine supplies constant power,its Sol: Since the collar is lifted with a constant speed, position and velocity changes w.r.t time as, T cos θ − mg = 0 ⇒ T cos θ = mg = 5 × 10, Velocity : As F v = P = constant, ur r, Now, P = F .v = T cos θ × v ; Here T = F, dv, dv , , v=P, i.e. m, F=m , P = 50 × v = 50 × 0.6 = 30W, dt, dt , , WE-35: A machine delivers power to a body which is, P, directly proportional to velocity of the body. If, or ∫ v dv= ∫ dt on integrating we get, m, the body starts with a velocity which is almost, negligible, find the distance covered by the body, v2 P, = t+C 1, in attaining a velocity v., 2 m, As initially the body is at rest,, dv , Sol. Power P = Fv cos 0 = Fv = m v ∝ v, ie. v = 0 at t = 0 ⇒ C 1 = 0 ;, dt , 1/2, , 2Pt , v= , , m , , ⇒ v a t1/2, , Position: From the above expression, 1/2, , 2Pt , v= , , m , , 1/2, , (or), 1/2, , 2Pt , ∫ ds = ∫ m , , ds 2 Pt , =, , dt m , , 1, , 2, dt = 2 p ∫ t 2 dt, m, 1, , integrating on both sides we get, 1/2, , dv, = K 0 v , Where K0 =constant, dt, dv, dv dx, m, = K 0 ; m = K0, dx dt, dt, dv, K , mv, = K 0 ; vdv = 0 dx, m, dx, mv, , Integrating, , ∫, , v, , 0, , x K, vdv = ∫ 0, 0, m, , , dx ;, , , 1 mv 2, v2 K0 , ⇒, x, =, =, x, , 2 K0, 2 m, Now at t = 0, S = 0 ⇒ C2 = 0, WE-36: Find the power of an engine which can, 1/2, 8P 3/2, 3/2, draw a train of 400 metric ton up the inclined, S =, , ∴S α t, t, 9m , plane of 1 in 98 at the rate 10 ms-1.The, WE-33: An automobile is moving at 100 kmph and, resistance due to friction acting on the train, is exerting attractive force of 3920 N. What, is 10 N per ton., 1, horse power must the engine develop, if 20 %, Given sin θ =, ; m = 400 × 103 kg, So1., of the power developed is wasted?, 98, 5, frictional force f = 10 × 400 = 4000 N ;, Sol :Velocity = 100 kmph = 100 × m/s, 18, velocity v = 10 ms-1, Force = 3920 N;;Useful power = 80%, ∴ Power P = (mg sinθ + f) v, 80, 5, W F.S, Power = =, =F.v ⇒ 100 P = 3920 × 100 × 18, , 1 , , 3, 2P , S =, , m , , t, , P=, , 2 3/2, . t + C2, 3, , t, , 100, 5, × 3920 × 100 ×, = 13.16 × 104 W = 182.5hp, 80, 18, , NARAYANAGROUP, , ∴ P = 400 × 10 × 9.8 ×, + 4000 × 10, 98 , , , , = 440000W=440KW, 161
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, WE-37: A hose pipe has a diameter of 2.5cm and is Ø, required to direct a jet of water to a height of, atleast 40m. Find the minimum power of the, pump needed for this hose., Sol.Volume of water ejected per sec, 2, , d , Av = π × 2 gh m3 / s; ∴ v = 2 gh, 2, , TE A = TE P, , 3, 2, , Ø, , WE-38: A body of mass m accelerates uniformly, from rest to velocity v0 in time t0 , find the, instantaneous power delivered to body when, v0, Ø, ., velocity is, 2, Ø, 2, 0, , v0 mv0 v0 mv, Instantaneous power P= F . 2 = t 2 =, 2t0, 0 , , Vertical circular motion with variable speed:, , = V12 - 4 g r, , makes an angle q with vertical., Tq =, , Ø, , O, r, V1, A, , Consider a body of mass ‘m’ tied at one end of a, string of length ‘r’ and is whirled in a vertical circle, by fixing the other end at ‘O’. Let V1 be the velocity Ø, of the body at the lowest point., , 1) At the lowest point q = 00 tension in the string is, , mg, , O Thor, Tθ, P, θ, mg, , mVθ2, r, mg cos θ, , Ø, , mV 2 2, - mg (minimum), r, , 3) When the string is horizontal, q = 900 , tension, 2, mV horz, in the string at this position is T(hor ) =, r, 4) The difference in maximum and minimum tension, in the string is, mV12, mV 2, + mg - 2 + mg, Tmax–Tmin=, r, r, m 2, 2, = (V1 -V2 ) + 2mg, r, m, = ( 4 gr ) + 2mg = 4mg + 2 mg = 6 mg, r, 5) Ratio of maximum tension to minimum tension in, the string is, Tmax, Tmin, , Vhor, , mV12, + mg (maximum)., r, , 2) At the highest point q = 180 0 ., , V2, , TH, , mVq 2, + mg cos q, r, , The tension in the string is TH =, , Ø, , 162, , 2, , Let Tq be the tension in the string when the string, , TL=, , mv0, v0, Sol.Acceleration a = t ; Force F = t, 0, 0, , V1, , V 1 - 2 g r (1 + 1 ), , Tension in the string at any point :, , = × 3.14 × ( 2.5 ×10−2 ) × ( 2 × 9.8 × 40) ×1000 =21.5KJ, , A, mg, , 2, , V2 =, , 3, 1, 1, K .E = mv2 = π d 2 × ( 2 gh ) 2 × ρ, 2, 8, , TL, , 1, 1, mV12 + 0 = mVθ2 + mgh, 2, 2, , If V2 is the velocity of the body at highest point, ( q = 180 0 ), , Kinetic energy of water leaving hose / sec, , θ, , ;, , 2, 2, Vq = V1 - 2 gh , but h = r (1- cos q ), 2, , 1, 4, , 2, , Velocity of the body at any point on the vertical, circle:, , Vq = V1 - 2 gr (1 - cos q ) ; Vq = V12 - 2 gr (1- cos q ), , Mass ejected per sec is M = π d 2 × 2 gh ρ Kg/s, , 1, 8, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , mV12, + mg, = r 2, mV 2, - mg, r, , =, , V 12 + r g, V 22 - r g, , When the particle is at ‘P’, a) Tangential force acting on the particle is, Ft = mg sin q ., Tangential acceleration a t = g sin q, b) Centripetal force acting on the particle is, æ mV 2 ö÷, q ÷, Fc = ççç, = Tq - mg cos q ., çè r ø÷÷, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , V2, Centripetal acceleration ac = q, r, c) Net acceleration of the particle at the point, , From the Law of conservation of energy, total energy at point ‘A’ = total energy at point P, U A + KEA = UP + KEP, , ‘P’ is a = at2 + ac2 ., , Ø, , 1, 1, O + mV12 = mgh + mVθ2, 2, 2, 1, 1, m ( 5 gR ) = mgR (1 − cos θ ) + mVθ2, 2, 2, 5gmR, 1, = mgR − mgR cosθ + mVθ2, 2, 2, 5gmR, 1, − mgR + mgR cos θ = mVθ2, 2, 2, mgR, 1, [3 + 2cos θ ] = mVθ2, 2, 2, , d) The net force acting on the particle at point, ‘P’ is F = Ft 2 + Fc2, Angle made by net force or net acceleration with, at, , centripetal component is f and tan f = F = a, c, c, Ft, , Condition for vertical circular motion of, a body, B, , V2, , mg, , T2, O, , Vθ = gR ( 3 + 2cosθ ), , T1, V1, , A mg, mV2 2, Ø, T, =, − mg, We know that 2, r, The body will complete the vertical circular path, when tension at highest point is such that, , mV2 2, − mg ≥ 0 ; V2 min = gr, T2 ≥ 0 ,, r, Hence the minimum speed at highest point to just, , θ, , Tθ = mg cos θ +, , mg cos θ, , m, mvθ2, = mg cosθ + gR ( 3 + 2 cosθ ), R, R, , = mg cos θ + 3mg + 2mg cosθ, , Ø, , 5, mgr ⇒ V1 = 5 gr, 2, For the body to continue along a circular path the, , = 3mg cos θ + 3mg = 3mg (1 + cos θ ), In case of non uniform circular motion in a vertical, plane if velocity of the body at the lowest point is, less than 5gr , the particle will not complete the, circle in vertical plane, the particle may either, oscillate about the lowest point or it leaves the circle, with out looping., , critical velocity at lowest point is 5gr, Critical velocity at any point on the vertical circle:, , NARAYANAGROUP, , P, , mg sinθ mg, , =, , O, Vθ, R-h θ Tθ, P, B, θ, AV, 1 mg, , Vθ, , Tθ, θ, , 1, 1, O + mV12 = mg ( 2r ) + mV22, 2, 2, Q V2 = gr , , , , Let Tθ be the tension in the string when the string is, making an angle θ from lowest point, , O, , complete the vertical circle is gr, From the law of conservation of mechanical energy, total energy at lowest point A = total energy at, highest point B, U A + KE A = U B + KE B, , 1, 1, mV12 = 2mgr + mgr, 2, 2, , Minimum tension in the string to just, complete vertical circle:, , Condition for oscillating about the, lowest position:, 1), , If 0 < VL < 2 gr , in this case, velocity becomes, zero before tension vanishes and the particle, oscillates about its lowest position with angular, amplitude 0 0 < θ < 90 0, 163
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 2), , If velocity of the body at the lowest point, VL < 2gr , then the maximum height reached by, Ø, VL2, given by h = ., 2g, The angle made by the string with the vertical when, its velocity becomes zero is given by, , cos q = 1-, , mv 2, r, mv 2, and normal reaction N=mgcos? r, , Centripetal force = mgcos? - N=, , the body just before its velocity becomes zero is, , 3), , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , A ball of mass ‘M’ is suspended vertically by a string, of length ‘L’. A bullet of mass ‘m’ is fired horizontally, with a velocity ‘u’ onto the ball, sticks to it. For the, system to complete the vertical circle, the minimum, value of ‘u’ is given by u =, , ( M + m), m, , VL2, 2gr, , 5 gL, , Note: If 0 < VL ≤ 2 gr then the particle oscillates, 0, , such that 00 < θ ≤ 900, , Condition for leaving the circular path, without looping:, Ø, , Ø, , L, , If 2 gr < VL < 5 gr . the particle is not able to, complete the vertical circle, it goes to certain height Ø, and leaves the circular path (90o< θ <180o ), while leaving the circular path T = 0 but V ≠ 0, The angle made by the string with downward vertical, when the tension in the string becomes zero is given, 2, , 2 VL, 3 3 gr, , by cos θ = −, Ø, , The height at which the tension in the string becomes, VL + gr, 3g, 2, , zero is given by h =, , u, , m, , M, , A nail is fixed at a certain distance ‘x’ vertically, below the point of suspension of a simple pendulum, of length L. The bob is released when the string, makes an angle θ with vertical. The bob reaches, the lowest position then describes a vertical circle, whose centre coincides with the nail. Then, L ( 3 + 2 cosθ ), xmin =, 5, , Ø When car moves on a concave bridge of radius, N, , x, , θ, , L, , mg cos θ, , v, Concave Bridge, , θ, , mg, , Ø, , mv 2, Centripetal force = N - mgcos?=, r, mv 2, and normal reaction N=mgcos?+, r, , L-x, , Ø, , A body of mass ‘m’ is allowed to slide down from, rest, from the top of a smooth incline of height ‘h’., For the body to move in a loop of radius ‘r’ on, arriving at the bottom., , When car moves on a convex bridge of radius r, , h, , N, V, θ, , mg, 164, , o, nail, , r, , 5r , a) Minimum height of smooth incline h = , 2, b) ‘h’ is independent of mass of the body, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, Ø, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , A small body is freely sliding down from the top of, a smooth convex hemisphere of radius r, placed on, a table with its flat face on the table then, , To complete vertical circle, , Vmin = 5 gr = 5 g ( l − d ), , — (2), , Equating ,equations (1) and (2), we get, 4l 4, gl = 5 g ( l − d ) ⇒ d = = = 0.80m, 5 5, WE-40: A body slides without friction from a height, h, θ r, H = 60cm and then loops the loop of radius R =, 20cm at the bottom of an incline. Find the ratio, a) Normal reaction on the body is zero at the instant, of forces exerted on the body by the track at the, the body leaves the hemisphere., −2, positions A, B and C ( g = 10ms ), b) the vertical height from table at which the body, C, leaves the hemisphere is h=2r/3, c) If the position vector of the body with respect to, the centre of curvature makes an angle θ with, vertical when the body leaves the hemisphere, then, B, cos θ = 2 / 3, H, R, 2 gr, d) velocity of block at that instant is V =, 3, e) If the block is given a horizontal velocity ‘u’ from, A, the top of the smooth convex hemisphere then the, Sol. From data H = 3R, angle θ with vertical at which the block leaves, Velocity at A, VA = 2 gH = 2 g ( 3R ) = 6 gR, 2 u2, cos, θ, =, +, hemisphere is, Velocity at B, V B = 4gR, 3 3 gr, WE-39: A nail is located at certain distance vertically, Velocity at C, VC = 2 gR, below the point of suspension of a simple, mVA2, pendulum. The pendulum bob is released from, + mg cos ( 00 ), Reaction, force, at, A, =, R, =, 0, 1, the position where the string makes an angle 60, R, from the vertical. Calculate the distance of the, m × 6 gR, + mg =7mg, =, nail from the point of suspension such that the, R, bob will just perform revolutions with the nail, mVB2, as the centre. Assume the length of the pendulum, + mg cos ( 900 ), Reaction, force, at, B, =, R, =, 2, to be 1m., R, m × 4 gR, + 0 = 4mg, =, R, 600, mVC2, + mg cos (1800 ), Reaction force at C = R3 =, R, P, m × 2 gR, − mg =mg, =, R, ∴ R1 : R2 : R3 = 7 : 4 :1, Sol.Velocity of bob at lowest position, , V = 2 g l (1 − cos θ ), l, = 2 g × l (1 − cos60 ) = 2 g = g l ..(1), 2, Let ‘d’ be the distance of nail from the, point of suspension. The bob will have to, complete the circle of radius r = l − d, 0, , NARAYANAGROUP, , WE-41: A heavy particle hanging from a fixed point by, a light inextensible string of length l is projected, horizontally with a speed of gl .Find the speed, of the particle and the inclination of the string to, the vertical at the instant of the motion, when the, tension in the string is equal to the weight of the, particle., 165
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, Ø, Sol:, , T, θ, h u, , B, , mg mg cos θ, A, u = gl, mg sin θ, Let T = mg at an angle ‘ θ ’as shown in the, , Ø, Ø, , figure and h = l (1 − cos θ ) à (i), Applying law of conservation of mechanical energy, between the points A and B,, 1, 2, 2, we get m ( u − v ) = mgh, 2, 2, Here u = g l à (ii), and v 2 = u 2 − 2 gh à (iii), , mv 2, (T = mg), l, v 2 = g l (1 − cos θ ) à(iv), From equations (i), (ii), (iii) and (iv), we have g l (1 − cos θ ) = g l − 2 g l (1 − cos θ ), , Further T − mg cos θ =, , 2, θ = cos , 3, gl, From equation (iv) v =, 3, 2, cosθ = ;, 3, , −1, , Collisions, Ø, Ø, , Ø, , The strong interaction among bodies involving, exchange of momentum in a short interval of time, is called collision., During collision bodies may or may not come into, physical contact., Ex: In the collision of α particle with nucleus, due, to coulombic repulsive forces α particle is scattered, away without any physical contact., Based on the direction of motion of colliding bodies,, collisions are classified into, (i) Head on or one dimensional collision, (ii) oblique collision, , Head on (or) one dimensional collision, A, C, , Ø, , 166, , Line of impact and, line of motion, , Ø, , It is the collision in which the velocities of the, colliding bodies are not confined to same straight, line before and after collision., Oblique collision may be two dimensional or three, dimensional., When a particle hits elastically and obliquely another, stationary particle of same mass, then they move, perpendicular to each other after collision., Types of Collision:Based on conservation of, kinetic energy collisions are classified into, (i) Elastic Collision (ii) Inelastic collision, Elastic Collision:It is the collision in which both, momentum and kinetic energy are conserved., Forces involved during collision are conservative, in nature, Ex.1. Collision between atomic particles., 2. Collision between two smooth billiard balls., 3. Collision of α particle with nucleus., Inelastic collision:It is the collision in which, momentum is conserved but not kinetic energy. Some, or all the forces involved during collision are non, conservative., Ex: Collision between two vehicles., , Perfectly inelastic collision:, It is the collision in which the colliding bodies stick, together and move as a single body after collision., In perfectly inelastic collision the momentum, remains conserved but the loss of kinetic energy is, maximum., Ex: A bullet is fired into a wooden block and, remains embedded in it., Line of impact:The line passing through the, common normal to the surfaces in contact during, impact is called line of impact. The force during, collision acts along this line on both bodies., Ex 1: Two balls A and B are approaching each, other such that their centres are moving along line, CD., A, C, , Line of impact and, line of motion, B, D, , A, , B, , Head on Collision, Ex 2: Two balls A and B are approaching each, other such that their centres are moving along dotted, lines as shown in figure., B, , B, D, , A, , Ø, , Oblique Collision:, , B, , It is the collision in which the velocities of the, colliding bodies are confined to same straight line, before and after collision., , Line of motion, of ball B, A, Oblique collision, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Ex 3: Ball is falling on a stationary wedge, Line of motion, of ball, Line of impact, , Applications:, Ø, , collides elastically with a stationary mass m2, , Elastic collision in one dimension:, , m1, , u2, , v1, , m2, , v2, , m -m r, , 1, 2, Velocity of first body after collision v1 = m +m u1, , 2), , Velocity of second body after collision, , , , 1, , 2, , , , r 2m1 r, v2 = , u1, m1 + m2 , , 3), , When two particles of masses m1 and m2 are, moving along the line joining their centers with, velocities u1 and u2 ( u1 > u2 ) before collision., Then v1 and v2 are their velocities after collision, , r, , 1), , Oblique Collision, , u1, , A body of mass m1 moving with a velocity v1, , KE of first body after collision (or) KE retained by, 2, , first body, , 1, 1 m − m2 2, K .E1 = m1v12 = m1 1, u1, 2, 2 m1 + m2 , 2, , K.E ret =, , 4), , m − m2 , m1 − m 2 , 1, m1u12 1, = KE i , , +, 2, m, m, 1, m1 + m 2 , 2 , , 2, , Fraction of KE retained by 1st body, 2, , m1, , Ø, , m2, Before collision, , During, Collision, , After collision, , From the conservation of linear momentum, 5), r r, r r, m1 ( u1 - v1 ) = m 2 ( v 2 - u 2 ), From Law of conservation of K.E, 1, 1, 1, 1, m1u12 + m 2 u 22 = m1v12 + m 2 v22, 2, 2, 2, 2, r r, r r, ∴ u1 - u 2 = v 2 - v1, i.e Relative velocity of approach before collision, = Relative velocity of separation after collision, Velocities after collision are, r m -m r 2m 2 r, v1 = 1 2 u 1 + , u2, m1 +m2 , m1 +m2 , r 2m1 r m 2 -m1 r, v2 = , u1 + , u2, m1 +m 2 , m1 +m 2 , , 6), , 2), , If colliding particles have equal masses, , r r, r r, i.e m1 = m2 = m ; v1 =u 2 , v 2 =u1, , 3), , 4), , 7), , If two bodies are of equal masses and the second, r r, body is at rest ie., m = m =m and u2 = 0 then 8), r r, v1 =0, , r r, v 2 =u 1, , 1, , 2, , 2, , 2 m1 2, 1, 1, 2, KE 2 = m 2 v 2 = m 2 , u1, 2, 2, m1 + m 2 , , 4m m 1, , 1 2, KE2 = , m u2, ( m + m ) 2 2 1 1 , 2, 1, , 4m m, 1 2, KEtra = , ( m + m )2, 2, 1, , , KEi, , , , Fraction of KE transferred from 1st body to second, body (or) Fraction of KE lost by 1st body is, KEtra, 4m1m2, =, 2, KEi, ( m1 + m2 ), , Special cases:, 1), , K .Eret m1 − m2 , =, , K .Ei m1 + m2 , KE of second body after collision (or) KE, transferred to the second body, , Fraction of momentum retained by m1, P1 m1v1 m1 − m2, =, =, Pi m1u1 m1 + m2, Fraction of momentum transferred from 1st body, to second body, , ;, m − m2 , P2 Pi − P1, P, 2m2, =, = 1− 1 = 1− 1, =, A lighter particle collides with heavier particle which, Pi, Pi, Pi, m1 + m2 m1 + m2, r, r, is at rest m1 <<< m 2 , u 2 = 0, WE-42: A bullet of mass ‘m’ moving at a speed ‘v’, r, r, r, hits a ball of mass ‘M’ kept at rest. A small, v 1 =-u 1 ,, v2 =0, part having mass m1 breaks from the ball and, A heavier body collides with lighter body at rest, r, r, sticks to the bullet. The remaining ball is found, m1 >>> m 2 , u 2 = 0 ;, to move at a speed v2 in the direction of the, r r, r, r, v1 = u 1 ,, v 2 =2u1, bullet.Find the velocity of the bullet after the, collision., , NARAYANAGROUP, , 167
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , Sol:Mass of bullet = m and speed = v., WE-45: n elastic balls are placed at rest on a, Mass of the ball M and fractional mass of the ball m1, smooth horizontal plane which is circular at, According to law of conservation of linear, the end with radius ‘r’ as shown in the figure., momentum, m m, m, m, ,, ,, ,........, The, masses, of, the, balls, are, 2, mv+0 = ( m+m 1 ) v1 + ( M -m 1 ) v 2, 2 2, 2n −1, respectively. Find the minimum velocity that, Where v1 = final velocity of the, should be imparted to the first ball of mass, (bullet + fractional mass), ‘m’ such that the ‘ nth ’ ball will complete the, mv- ( M-m1 ) v 2, vertical circle., v1 =, , ( m+m1 ), , r, WE-43: Two bodies of masses m1 and m2 are, n, 1, 2, moving with velocities 1ms −1 and 3ms −1, respectively in opposite directions. If the, bodies undergo one dimensional elastic Sol:Let speed to be imparted to the first ball be v0 ., Consider the impact between the first two balls and, collision, the body of mass m1 comes to, v1 and v 2 be the velocities of balls 1 and 2 after, the impact respectively., rest.Find the ratio of m1 and m 2, According to law of conservation of linear, Sol. u1 = 1m / s, u 2 = -3m / s, v1 = 0, m, momentum mv0 =mv1 + v 2 → (1), m1 -m 2 , 2m 2 , v1 = , 2, u1+ , u2, According to law of conservation of kinetic energy, m1 +m2 , m1 +m 2 , 1, 1, 1m, m − m2 2 m2 , mv0 2 = mv12 + v 2 2 → ( 2 ), 0= 1, 1 + , ( −3), 2, 2, 2 2 , m1 + m2 m1 + m2 , 4, m1 7, Solving equations (1) and (2), we get v2 = v0, m1 − m2 = 6m2 ; m1 = 7m2 ; m = 1, 3, 2, n-1, WE-44: Two identical balls A and B are released, 4, Similarly, for nth ball v n = v 0 → ( 3 ), from the positions as shown in the figure. They, 3, collide elastically on the horizontal portion., th, For the n ball to complete the vertical circular, The ratio of heights attained by A and B after, motion vn = 5gr → ( 4 ), From equations (3) and (4), we have, , collision (neglect friction), A, , 4h, 0, , 45, , 600, , h, Ø, , Sol.As mass of two balls are equal, they exchange their, velocities after collision., u A2, = h;, u A = 2 gh , u B = 2 g ( 4h ) = 8 gh ; hA =, 2g, , hB = h +, , v B 2 sin 2 600, 9h 13h, =h+, =, 2g, 4, 4, , (Q( vB ) − uB2 = −2gh ⇒ vB 2 = uB2 − 2 gh ⇒ vB2 = 6gh ), 2, , hA, 4, =, hB 13, , 168, , 4, , 3, , B, , n-1, , 3, v 0 = 5gr ; v 0 = , 4, , n-1, , 5gr, , Coefficient of restitution, , Newton introduced a dimensionless parameter, called the coefficient of restitution (e) to measure, the elasticity of collision. It is defined as the ratio of, the relative velocity of separation to the relative, velocity of approach of the two colliding bodies, r r, Relative velocity of separation v 2 -v1, e=, = r r, Relative velocity of approach, u1 -u 2, Ø This formula is applied along the line of impact. Here, the velocities mentioned in the expression should, be taken along the line of impact., For a perfectly elastic collision e = 1, For an inelastic collision o < e < 1, For completely inelastic collision e = 0, NARAYANAGROUP
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, Ø, , 1), 2), , 3), , A ball is projected with an initial velocity u at an angle 11) The sum of maximum heights reached by the ball is, θ to the horizontal surface. If ‘e’ is the coefficient of, H 1 = H + H1 + H 2 + ........, restitution between the ball and the surface then, = H + e 2 H + e 4 H + ........, Y, H, u, 1, = H 1 + e2 + e 4 + ....... , H =, 1 − e2, If the collision is elastic e = 1 and H ′ = ∞, θ, x, Head on inelastic collision, 3rd, 2nd, 1st, Two bodies of masses m1 and m2 moving with, 2u sin θ, r, r, Time taken for 1st collision, T = g, initial velocities u1 and u2 ( u1 > u2 ) collide. After, r, Time interval between 1st and 2nd collisions,, collision two bodies will move with velocities v1 and, 2v sin θ, r, (Q v1 = eu ), T1 = 1, v2 ., g, From Law of conservation of linear momentum, 2 ( eu ) sin θ, r r, r r, T1 =, = eT, m 1 ( u 1 -v 1 ) = m 2 ( v 2 -u 2 ), g, By the definition of coefficient of restitution, Time interval between 2nd and 3 rd collisions,, r r, r r, 2, v 2 -v1 = e ( u1 − u 2 ), 2v sin θ 2 ( e u ) sin θ, 2, 2, T2 =, , 4), , 2, , g, , =, , =e T, , g, , (Q v, , 2, , = e u), , r m − em2 r (1 + e ) m2 r, v1 = 1, u2, u1 + , m1 + m2 , m1 + m2 , r (1 + e ) m1 r m2 − em1 r, v2 = , u1 + , u2, m1 + m2 , m1 + m2 , , The total time of flight is, T 1 = T + T1 + T2 + ........., , = T + eT + e2T + e3T + ......, = T [1 + e + e 2 + e 3 + ........], , Ø, , T, T1 =, 1− e, , 5), , If collision is elastic, e = 1 then T 1 = ∞, The horizontal distance covered by the ball before, 1st collision is, R=, , 6), 7), 8), , JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , u 2 sin 2θ, = u cos θ × T, g, , If m1 = m2 = m, u2 = 0 then, u, u, v1 = (1 − e ) 1 ; v 2 = (1 + e ) 1, 2, 2, v1 1 − e, =, v2 1 + e, , Loss of kinetic energy of the system:, , The horizontal distance covered by it between 1 st, and 2nd collisions, R1 = u cos θ × eT = eR, horizontal distance covered between 2nd and 3rd, collisions, R2 = u cos θ × e 2T = e 2 R, Total horizontal distance covered by the ball is, R1 = R0 + R1 + R2 + R3 + ........., = R + eR + e 2 R + ......, Ø, = R 1 + e + e 2 + ......., , ∆ KE = KE I − KE F, , 1 m1m2 r r 2, 2, , ( u1 − u2 ) (1 − e ), 2 m1 + m2 , In case of perfectly in-elastic collision, e = 0, ∴ loss in KE of system is, 1 mm r r 2, ∆KE = 1 2 ( u1 − u2 ), 2 m1 + m2 , If two bodies are approaching each other then, loss in KE of the system is maximum, ∆KE =, , R, 1 mm , 2, ∆KEmax = 1 2 ( u1 + u2 ), 1− e, 2 m1 + m2 , WE-46: Ball 1 collides with an another identical, For perfectly elastic collision e = 1 and R1 = ∞, st, ball 2 at rest as shown in the figure. For what, 9) The maximum height reached by the ball before 1, value of coefficient of restitution e, the velocity, 2, u 2 sin 2 θ ( u sin θ ), of second ball becomes two times that of first, =, collision H =, ball after collision?, 2g, 2g, st, nd, 10) Maximum height it reaches between 1 and 2, R1 =, , collisions is H, 170, , 1, , ( eu sin θ ), =, 2g, , 1, , 2, , 2, , =e H, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Sol.Here m1 = m2 and u2 = 0, , Sol : Let ‘u’ be the velocity of ball before collision., Speed of the ball after collision will become, , 1+ e , 1− e , v = u 2 sin 2 θ + e 2 u2 cos2 θ, After collision, v 2 = , u & v1 = , u, 2, 2, , , , , 2, 2, 5, u u , 1+ e , 1− e , +, =, .u, = , , u, =, u, 2, Given v 2 =2v1 ; , 8, , , , 2 2 2, 2 , 2 , \ Fraction of KE lost in collision, 1, 1, 1, 1+e = 2 – 2e ; 3e = 1; e =, mu 2 − mv 2, 3, 2, 2, 5 3, v, =2, WE-47: A body ‘A’ with a momentum ‘P’ collides, =, 1, −, = 1− =, 1, , , 2, u, with another identical stationary body ‘B’ one, 8 8, mu, 2, dimensionally. During the collision, ‘B’ gives, an impulse ‘J’ to the body ‘A’ . Then the WE-50: Two equal spheres A and B lie on a smooth, horizontal circular groove at opposite ends of a, coefficient of restitution is, diameter. At time t = 0 , A is projected along the, Sol : From the law of conservation of linear momentum,, groove and it first impinges on B at time t =T1, m1u1 +m2u2 = m1v1 + m2v2, and again at time t = T2 . If ‘e’ is the coefficient of, mu + m(0) = mv1 + mv2, T2, Þ P – P1 = P2 where P 2 = J, (given), restitution, find the ratio of T, 1, v 2 -v1 mv 2 -mv1 P2 -P1, e=, =, =, \ u -u, v, t=T, mu-0, P, 1, 2, 1, , 1, , P2 -( P - P2 ), , 2P2 - P 2 J - P 2 J, u, =, =, -1, A, P, P, P, P, WE-48: A ball of mass m collides with the ground, at an angle a with the vertical . If the collision, πR, lasts for time t, the average force exerted by, T1 =, Sol, :, the ground on the ball is : (e =coefficient of, u1, restitution between the ball and the ground), =, , =, , 1, , A, , u2 =0, , B, , B, , v2, , ...... (1), , v2 − v1, = e ⇒ v 2 − v1 = eu1, u1, , Time taken for A to collide with B again is, T2 − T1 =, u, α, , T2, , Sol : Impulse = change in linear momentum., eu cos α, , u sin α, u cos α, (Before Collision), , u sin α, u cos α, (After Collision), , mucos α (1 + e ), , \ Ft = m ( eu cos α + u cos α ) or F =, t, WE-49: A ball strikes a horizontal floor at an angle, θ = 45 0 with the normal to floor.The, coefficient of restitution between the ball and, the floor is e = 1/2 . The fraction of its kinetic, energy lost in the collision is, NARAYANAGROUP, , 2πR, 2 πR, ⇒ T2 − T1 =, v 2 − v1, eu1, , .... (2), , 2+e, , from (1) and (2), T = e, 1, WE-51: After perfectly inelastic collision between, two identical particles moving with same speed, in different directions, the speed of the, combined particle becomes half the initial, speed of either particle . The angle between, the velocities of the two before collision is, Sol : In perfectly inelastic collision between two particles,, linear momentum is conserved . Let θ be the angle, between the velocities of the two particles before, collision. Then, P 2 = P12 + P22 + 2 P1 P2 cos θ, , or, , 2, , v, 2, 2, , 2 m = ( mv ) + ( mv ) + 2 ( mv )( mv ) cos θ, 2, , 1, or 1 = 1 + 1 + 2 cosq or cos θ = − ; (or) θ = 1200, 2, 171
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JEE, MAINS, - C.W - VOL - -IIII, JEE-ADV, PHYSICS-VOL, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, WE-52: A bullet of mass ‘m’ moving with velocity, ‘u’ passes through a wooden block of mass, V, M = nm as shown in figure. The block is resting, on a smooth horizontal floor. After passing, V1, through the block, velocity of the bullet, becomes ‘v’ . Its velocity relative to the block is 3), , Y, h, , O, , m, , u, , t1 2t1, , t, , 3t1, , t, , -V1, , M=nm, Y, , V, , Sol : Let v' be the velocity of block. Then from, conservation of linear momentum., , V1, , u − v, v' = , n , , mu = mv + mnv' (or), , 4) O, , t1, , h, , 3t1, 2t1, , t, , 4t1, , t, , -V1, , \ velocity of bullet relative to block will be, u − v (1 + n ) v − u, v r = v − v' = v − , =, n , n, , Sol: When ball strikes the surface its velocity, will be reversed so correct option is (3)., , WE-53: A block of mass 0.50Kg is moving with a, Ballistic pendulum :, speed of 2.00 m/s on a smooth surface. It strikes, It is an arrangement used to determine the velocities, another mass of 1.00 kg and then they move, of bullets .A log of wood of mass ‘M’ is suspended, together as a single body. Find the energy loss, by a string of length ‘l’ as shown in the figure. A, during the collision (JEE MAIN 2008), bullet of mass ‘m’ is fired horizontally into the, Sol: From LCLM, m1u1 + m2 u2 = ( m1 + m2 ) v, wooden block with a velocity ‘u’, Case, I : Let the bullet gets embedded in the block and, 2, −1, 0.50 × 2 + 1× 0 = ( 0.5 + 1) v ⇒ v = ms, system rises to a height ‘h’ as shown in the figure., 3, 1, , 1, , 2, 2, ∴ Energy loss ∆KE = 2 m1u1 − 2 ( m1 + m2 ) v, , ∆KE =, , 1, 1, 2, 2, ( 0.5)( 2 ) − (1.5) , 2, 2, 3, , θ, , 2, , = 0.67 J, , WE-54: Consider a rubber ball freely falling from, a height h = 4.9 m on a horizontal elastic plate., Assume that the duration of collision is, negligible and the collision with the plate is, totally elastic. Then the velocity as a function, of time and the height as a function of time, will be; (JEE MAIN 2009), Y, V, , h, , V1, 1), , O, , t, , t1, , O, Y, , V, V1, , 2), , 172, , O, -V1, , h, t, t, , M, m u, , m, h, , M, , From the law of conservation of linear momentum, m1u1 + m2u2 = ( m1 + m2 ) v, mu, mu + 0 = ( m + M ) v ⇒ v =, .......(1), m+M, KE of the system after collision is given by, 1, KE = ( m + M ) v 2, 2, PE at highest point = ( m + M ) gh, 1, 2, From LCE, ( m + M ) v = ( m + M ) gh, 2, v 2 = 2 gh ( or )v = 2 gh ....(2), From (1) and (2) velocity of the bullet, M +m, M +m, u=, 2 gh =, 2 gl (1 − cos θ ), m, m, Loss in KE of the system = K E 1 − KE 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, ∆KE =, , ∆KE =, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 1, 1, mu 2 − ( m + M ) v 2, 2, 2, , and it emerges out of the bob with a speed v1 ., 3, Find the initial speed of the bullet if the bob, just completes the vertical circle., Sol: From the Law of conservation of momentum, v , m 2v, , mv = m1 v1 − 1 or v = 1 × 1, 3, m, 3, , , To describe a vertical circle v = 5gl, , m 3 5 gl, m1 2v1, ×, hence 5 gl =, ⇒ v1 = m × 2, m 3, 1, , Collisions in two dimensions (oblique, collisions), , 2., , 3., , 4., , Tangential axis, , 1 2, m 2u 2 , mu − ( m + M ), 2, 2 , ( m + M ) , , 1 mM 2, ∆ KE = , u, 2 m + M , Case II :, If the bullet emerges out of the block with, velocity ‘v’ then, mu = mv + MV Where V = 2 gh, WE-55: A pendulum consists of a wooden bob of, mass ‘m’ and of length l . A bullet of mass m 1 is, fired towards the pendulum with a speed v1, , 1., , Y, , X, m1, , m2 Normal, axis, , v2, , v1, β1, m1, , β2, m2, , From law of conservation of linear, momentum along x-axis:, m1u1 cos θ1 + m2u2 cos θ 2 = m1v1 cos β1 + m2 v 2 cos β 2, Along y-axis:, m1u1 sin θ1 + m2u2 sin θ 2 = m1v1 sin β1 + m2 v 2 sin β 2, Coefficient of restitution, v cos β1 − v 2 cos β 2, e=− 1, u1 cos θ1 − u2 cos θ2, WE-56:Two billiard balls of same size (radius r), and same mass are in contact on a billiard, table. A third ball also of the same size and, mass strikes them symmetrically and remains, at rest after the impact. The coefficient of, restitution between the balls is, v, , A pair of equal and opposite impulses act along, common normal direction.Hence,linear momentum, θ, of individual particles changes along common, u, normal direction., No component of impulse acts along common Sol: :, tangent direction. Hence, linear momentum (or), v, linear velocity of individual particles remains, unchanged along this direction., r 1, sin θ = = ; \ θ = 30 0, Net impulse on both the particles is zero during, 2r 2, From conservation of linear momentum, collision. Hence, net momentum of both the, particles remain conserved before and after collision, u, 0, v, =, or, mu, =, 2mv, cos30, in any direction., 3, Definition of coefficient of restitution can be applied, relative velocity of separation, along common normal direction., Now e =, relative velocity of approach, u2, u1, in common normal direction, θ1, θ2, v, u/ 3 2, =, =, Hence, e =, 0, u, cos30, 3, u, 3, /, 2, m1, m, 1, , NARAYANAGROUP, , 173
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , CONSERVATIVE AND, NON-CONSERVATIVE FORCES, , C.U.Q, , Potential energy is defined for, 1) non-conservative forces only, In which of the following, the work done by, 2) conservative forces only, the mentioned force is negative?, 3) both conservative & non-conservative forces, The work done by, 4) neither conservative nor non-conservative, 1) the tension in the cable while the lift is, forces, 9. Which of the following forces is called a, ascending, conservative force?, 2) the gravitational force when a body slides, 1) Frictional force, 2) Air resistance, down an inclined plane, 3), Electrostatic, force, 4) Viscous force, 3) the applied force to maintain uniform motion, 10., Identify, the, non-conservative, force in the, of a block on a rough horizontal surface, following, 4) the gravitational force when a body is thrown, 1) Weight of a body 2) Force between two ions, up, 3) Magnetic force 4) Air resistance, A man pushes a wall and fails to displace it. 11. If x , F and U denote the displacement, force, He does, acting on and potential energy of a particle,, 1) negative work, then, 2) positive but not maximum work, 1 dU , dU, dU, 3) F = −, 4) F = x dx , 1) U = F 2) F = +, 3) maximum work 4) no work at all, dx, dx, , , A bucket full of water is drawn up by a person. 12. In the case of conservative force, In this case the work done by the gravitational, 1) work done is independent of the path, 2) work done in a closed loop is zero, force is, 3) work done against conservative force is stored, 1) negative because the force and displacement are, is the form of potential energy, in opposite directions, 4) all the above, 2) positive because the force and displacement are, KINETIC ENERGY, in the same direction, 13., The, change, in kinetic energy per unit ‘space’, 3) negative because the force and displacement are, (distance), is, equal to, in the same direction, 1), power, 2), momentum, 3) force 4) pressure, 4) positive because the force and displacement are, 14. When the momentum of a body is doubled, the, in opposite directions, kinetic energy is, A man is rowing a boat upstream and inspite, 1) doubled, 2) halved, of that the boat is found to be not moving with, 3) becomes four times 4) becomes three times, respect to the bank. The work done by the man 15. For the same kinetic energy, the momentum, is, shall be maximum for which of the following, particle?, 1) zero, 2) positive, 1) Electron 2) Proton 3) Deuteron 4) Alpha particle, 3) negative, 4) may be +ve or –ve, A ball is thrown vertically upwards from the 16. If the momentum of a particle is plotted on, X-axis and its kinetic energy on the Y-axis,, ground. Work done by air resistance during, the graph is a, its time of flight is, 1) straight line, 2) parabola, 1) positive during ascent and negative during, 3) rectangular hyperbola 4) circle, descent, 17. When two identical balls are moving with equal, 2) positive during ascent and descent, speeds in opposite direction, which of the, 3) negative during ascent and positive during, following is true? For the system of two bodies, 1) momentum is zero, kinetic energy is zero, descent, 2) momentum is not zero, kinetic energy is zero, 4) negative during ascent and descent, 3) momentum is zero, kinetic energy is not zero, An agent is moving a positively charged body, 4) momentum is not zero, kinetic energy is not, towards another fixed positive charge. The, zero, work done by the agent is, 18. The product of linear momentum and velocity, 1) positive, 2) negative, of a body represents, 3) zero, 4) may be positive or negative, 1) half of the kinetic energy of the body, Workdone by force of friction, 2) kinetic energy of the body, 3) twice the kinetic energy of the body, 1) can be zero, 2) can be positive, 4) mass of the body, 3) can be negative 4) any of the above, , WORK, , 1., , 2., , 3., , 4., , 5., , 6., , 7., , 174, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 8., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 19. The KE of a freely falling body, 1) is directly proportional to height of its fall, 2) is inversely proportional to height of its fall, 3) is directly proportional to square of time of its fall, 4) 1 and 3 are true, 20. Consider the following statements, A) Linear momentum of a system of particles, is zero, B) Kinetic energy of a system of particles is, zero then, 1) A does not imply B & B does not imply A, 2) A implies B and B does not imply A, 3) A does not imply B but B implies A, 4) A implies B and B implies A, 21. Internal forces can change, 1) Linear momentum as well as kinetic energy, 2) Linear momentum but not the kinetic energy, 3) Kinetic energy but not linear momentum, 4) neither the linear momentum nor the kinetic, energy, 22. If the force acting on a body is inversely, proportional to its speed, then its kinetic, energy is, 1) linearly related to time, 2) inversely proportional to time, 3) inversely proportional to the square of time, 4) a constant, 23. Which of the following graphs depicts the, variation of KE of a ball bouncing on a, horizontal floor with height? (Neglect air, resistances), , 26. Two bodies of masses m1 and m2 have equal, KE. Their momenta is in the ratio, 1) m2 : m1 2) m1:m23) m1 : m2 4) m12 : m22, 27. A body can have, 1) changing momentum and finite kinetic energy, 2) zero kinetic energy and finite momentum, 3) zero acceleration and increasing kinetic energy, 4) finite acceleration and zero kinetic energy, 28. A rock of mass m is dropped to the ground, from a height h. A second rock with mass 2m, is dropped from the same height. When, second rock strikes the ground, its kinetic, energy wii be, 1) twice that of the first rock, 2) four times that of the first rock, 3) the same as that of the first rock, 4) half that of the first rock, , POTENTIAL ENERGY, 29. These diagrams represent the potential, energy U of a diatomic molecule as a function, of the inter-atomic distance r. The diagram, corresponds to stable molecule found in, nature is, U, , U, 1), , 2), , r, , r, , K, , K, , U, , U, , 1), , 3), , 2), h, , h, , r, , 30. In the fig. the potential energy U of a particle, plotted against its position x from origin., Which of the following statement is correct?, U, , K, 3), , 4) None of these, h, , 24. Which of the following statement is correct?, 1) KE of a system cannot be changed without, changing its momentum, 2) KE of a system can be changed without changing, its momentum, 3) Momentum of a system cannot be changed, without changing its KE, 4) A system cannot have energy without having, momentum, 25. Two bodies of different masses have same linear, momentum. The one having more KE is, 1) lighter body, 2) heavier body, 3) both, 4) none, NARAYANAGROUP, , 4), r, , O, , x1, , x2, , x3, , x, , 1) at x1 particle is in stable equilibrium, 2) at x2 particle is in stable equilibrium, 3) at x3 particle is in stable equilibrium, 4) at x1 ,x2 and x3particle is in unstable equilibrium, , POTENTIAL ENERGY OF A SPRING, , 31. When a spring is wound, a certain amount of, PE is stored in it. If this wound spring is, dissolved in acid, the stored energy, 1) is completely lost, 2) appears in the form of electromagnetic waves, 3)appears in the form of heat raising the, temperature of the acid, 4) appears in the form of KE by splashing acid, drops, 175
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 32. Two springs have their force constants K1, and K2 and they are stretched to the same, extension. If K 2>K1 work done is, 1) same in both the springs, 2) more in spring K1, 3) more in spring K2, 4) independent of spring constant K, 33. Two springs have their force constants K1, and K 2 (K2>K1). When they are stretched by, the same force, work done is, 1) same in both the springs 2) more in spring K1, 3) more in spring K2, 4) independent of spring constant K, , WORK ENERGY THEOREM, 34. A lorry and a car moving with the same KE, are brought to rest by applying the same, retarding force. Then, 1) lorry will come to rest in a shorter distance, 2) car will come to rest in a shorter distance, 3) both come to rest in same distance, 4) any of above, CONSERVATION OF MECHANICAL, ENERGY, 35. A shell is fired into air at an angle θ with the, horizontal from the ground. On reaching the, maximum height,, 1) its kinetic energy is not equal to zero, 2) its kinetic energy is equal to zero, 3) its potential energy is equal to zero, 4) both its potential and kinetic energies are zero, 36. A cricket ball and a ping-pong ball are, dropped in a vacuum chamber from same, height. When they have fallen half way, down, they have the same, 1) velocity, 2) potential energy, 3) kinetic energy 4) mechanical energy, 37. A cyclist free-wheels from the top of a hill,, gathers speed going down the hill, applies his, brakes and eventually comes to rest at the, bottom of the hill. Which one of the following, energy changes take place., 1) Potential to kinetic and to heat energy, 2) Kinetic to potential and to heat energy, 3) chemical to heat and to potential energy, 4) Kinetic to heat and to chemical energy, 38. If ‘E’ represents total mechanical energy of, a system while ‘U’ represents the potential, energy, then E - U is, 1) always zero, 2) negative, 3) either positive or negative 4) positive, 39. For a body thrown vertically upwards, its, direction of motion changes at the point, where its total mechanical energy is, 1) greater than the potential energy, 2) less than the potential energy, 3) equal to the potential energy 4) zero, 176, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 40. Internal forces can change, 1) Kinetic energy 2) mechanical energy, 3) Momentum, 4) 1 and 2, 41. Negative of work done by the conservation, forces on a system is equal to, 1) the change in kinetic energy of the system, 2) the change in potential energy of the system, 3) the change in total mechanical energy of the, system, 4) the change in the momentum of the system, 42. Which of the following statement is wrong?, 1) KE of a body is independent of the direction of, motion, 2)In an elastic collision of two bodies, the, momentum and energy of each body is conserved, 3) If two protons are brought towards each other,, the PE of the system increases, 4) A body can have energy without momentum, 43. When a body falls from an aeroplane there is, increase in its:, 1) acceleration, 2) potential energy, 3) kinetic energy 4) mass, , POWER, 44. A body is moved along a straight line by a, machine delivering constant power. The, distance moved by the body in time t is, proportional to, 1) t1/2, 2) t3/4 3) t3/2 4) t 2, 45. A particle is projected at t = 0 from a point on, the ground with certain velocity at an angle, with the horizontal. The power of, gravitational force is plotted against time., Which of the following is the best, representation?, P, , 2) P, , 1), t, , t, , 3), , P, , P, t, , 4), , t, , 46. A body starts from rest and acquires a, velocity V in time T. The instantaneous power, delivered to the body in time ‘t’ is, proportional to, V, V2 2, V2, V2 2, t, 1) t, 2), 3) 2 t 4) 2 t, T, T, T, T, 47. A car drives along a straight level frictionless, road by an engine delivering constant power., Then velocity is directly proportional to, 1, 1) t, 2), 3) t, 4) t 2, t, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 48. A particle is projected with a velocity u making, an angle θ with the horizontal. The, instantaneous power of the gravitational force, 1) varies linearly with time, 2) is constant throughout the path, 3) is negative for complete path, 4) varies inversly with time, , MOTION IN A VERTICAL CIRCLE, 49. A motor car of mass m travels with a uniform, speed v on a convex bridge of radius r ., When the car is at the middle point of the, bridge, then the force exterted by the car on, the bridge is, mv 2, mv 2, mv 2, 1) mg 2) mg +, 3) mg −, 4) mg ±, r, r, r, 50. A gramphone record is revolving with an, angular velocity ω . A coin is placed at a, distance R from the centre of the record., The static coefficient of friction is µ . The, coin will revolve with the record if, µg, µg, µg, µg, 1) R > 2 2) R = 2 only 3) R < 2 4) R ≤ 2, ω, ω, ω, ω, 51. A small sphere of mass ‘m’ is attached to a, cord and rotates in a vertical plane about a, point O . If the average speed of the sphere, is increased, the cord is most likely to break, at the orientation when the mass is at :, A, , m, l, C, , O, , B, , D, , 1) bottom point B 2) the point C, 3) the point D, 4) top point A, 52. A car is moving up with uniform speed along a, fly over bridge which is part of a vertical circle., The true statement from the following is, 1)Normal reaction on the car gradually decreases, and becomes minimum at highest position of, bridge, 2)Normal reaction on the car gradually increases, and becomes maximum at highest position, 3) Normal reaction on car does not change, 4)Normal reaction on the car gradually decreases, and becomes zero at highest position, 53. A bottle of soda water is rotated in a vertical, circle with the neck held in hand. The air, bubbles are collected, 1) near the neck, 2) near the bottom, 3) at the middle, 4) uniformly in the bottle, NARAYANAGROUP, , 54. A vehicle is moving with uniform speed along, horizontal, concave and convex surface, roads. The surface on which, the normal, reaction on the vehicle is maximum is, 1) concave, 2) convex, 3) horizontal, 4) same at all surfaces, , COLLISIONS, , ur, 55. A ball with initial momentum P collides with, uur, rigid wall elastically. If P1 be its momentum, after collision then, uur ur uur, ur uur, ur uur, ur, 1) P1 = P 2) P1 = − P 3) P1 = 2 P 4) P1 = −2 P, 56. Choose the false statement, 1) In a perfect elastic collision the relative velocity, of approach is equal to the relative velocity of, separation, 2) In an inelastic collision the relative velocity of, approach is less than the relative velocity of, separation, 3) In an inelastic collision the relative velocity of, separation is less than the relative velocity of, approach, 4) In perfect inelastic collision relative velocity of, separation is zero, 57. Two particles of different masses collide head, on. Then for the system, 1) loss of KE is zero, if it was perfect elastic, collision, 2) If it was perfect inelastic collision, the loss of, KE of the bodies moving in opposite directions is, more than that of the bodies moving in the same, direction, 3) loss of momentum is zero for both elastic and, inelastic collision, 4) 1, 2 and 3 are correct, 58. A 2 kg mass moving on a smooth frictionless, surface with a velocity of 10ms −1 hits another, 2kg mass kept at rest, in a perfect inelastic, collision. After collision, if they move, together, 1) they travel with a velocity of 5ms −1 in the same, direction, 2) they travel with a velocity of 10ms −1 in the, same direction, 3) they travel with a velocity of 10ms −1 in, opposite direction, 4) they travel with a velocity of 5ms −1 in opposite, direction, 59. A body of mass ‘m’ moving with a constant, velocity v hits another body of the same mass, moving with the same velocity v but in, opposite direction and sticks to it. The, velocity of the compound body after the, collision is, 1) 2v, 2) v, 3) v/2 4) zero, 177
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 60. In an inelastic collision, the kinetic energy, after collision, 1) is same as before collision, 2) is always less than before collision, 3) is always greater than before collision, 4) may be less or greater than before collision, 61. A ball hits the floor and rebounds after an, inelastic collision. In this case, 1) the momentum of the ball just after the collision, is same as that just before the collision, 2) The mechanical energy of the ball remains the, same in the collision, 3) The total momentum of the ball and the earth is, conserved, 4) the total kinetic energy of the ball and the earth, is conserved, 62. About a collision which is not correct, 1) physical contact is must, 2) colliding particles can change their direction of, motion, 3) the effect of the external force is not considered, 4) linear momentum is conserved, 63. In one– dimensional elastic collision, the, relative velocity of approach before collision, is equal to, 1) relative velocity of separation after collision, 2) ‘e’ times relative velocity of separation after, collision, 3) ‘1/e’ times relative velocity of separation after, collision, 4) sum of the velocities after collision, 64. Two identical bodies moving in opposite, direction with same speed, collide with each, other. If the collision is perfectly elastic then, 1) after the collision both comes to rest, 2) after the collision first comes to rest and, second moves in the opposite direction with same, speed., 3) after collision they recoil with same speed, 4) both and 1 and 2, 65. A body of mass ‘m’ moving with certain, velocity collides with another identical body, at rest. If the collision is perfectly elastic and, after the collision both the bodies moves, 1) in the same direction, 2) in opposite direction, 3) in perpendicular direction 4) at 45° to each other, 66. Six steel balls of identical size are lined up, along a straight frictionless groove. Two, similar balls moving with speed v along the, groove collide with this row on the extreme, left end. Then, 1) one ball from the right end will move on with, speed v, 2) two balls from the extreme right end will move, on with speed v and the remaining balls will be at, rest, 178, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 3) all the balls will start moving to the right with, speed v/8, 4) all the six balls originally at rest will move on with, speed v/6 and the incident balls will come to rest, 67. A lighter body moving with a velocity v, collides with a heavier body at rest. Then, 1) the lighter body rebounced with twice the, velocity of bigger body, 2) the lighter body retraces its path with the same, velocity in magnitude, 3) the heavier body does not move practically, 4) both (2) and (3), 68. A heavier body moving with certain velocity, collides head on elastically with a lighter body, at rest, then, 1) smaller body continues to be in the same state, of rest, 2) smaller body starts to move in the same, direction with same velocity as that of bigger body, 3) the smaller body start to move with twice the, velocity of the bigger body in the same direction, 4) the bigger body comes to rest, 69. A perfectly elastic ball P1 of mass ‘m’ moving, with velocity v collides elastically with three, exactly similar balls P2 , P3 , P4 lying on a, smooth table. Velocity of the four balls after, collision are, , P1, , P2, , P3, , P4, , 1) 0,0,0,0 2) v, v, v, v 3) v, v, v,0 4) 0, 0, 0, v, 70. Two bodies P and Q of masses m1 and m2, , ( m2 > m1 ) are moving with velocity v1 and v2, respectively, collide with each other. Then the, force exerted by P on Q during the collision is, 1) greater that the force exerted by Q on P, 2) less than the force exerted by Q on P, 3) same as the force exerted by Q on P, 4) same as the force exerted by Q on P but, opposite in direction, 71. The coefficient of restitution (e) for a, perfectly elastic collision is, 1) −1, 2) 0 3) ∞ 4)1, 72. A ball of mass M moving with a velocity v, collides perfectly inelastically with another, ball of same mass but moving with a velocity, v in the opposite direction. After collision, 1) both the balls come to rest, 2) the velocities are exchanged between the two, balls, 3) both of them move at right angles to the original, line of motion, 4) one ball comes to rest and another ball travels, back with velocity 2v, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 73. A ball of mass ‘m’ moving with speed ‘u’, undergoes a head-on elastic collision with a, ball of mass ‘nm’ initially at rest. Find the, fraction of the incident energy transferred to, the second ball., n, 2n, 4n, n, 2, 2, 1), 2) n + 1 3) 1 + n 4) 1 + n 2, ( ), ( ) ( ), n +1, 74. A small bob of a simple pendulum released, from 30° to the vertical hits another bob of, the same mass and size lying at rest on the, table vertically below the point of suspension., After elastic collision, the angular amplitude, of the pendulum will be, 1) 30°, 2) 60° 3) 15° 4) zero, 75. Two spheres ‘X’ and ‘Y’ collide. After, collision, the momentum of X is doubled. Then, 1) the initial momentum of X and Y are equal, 2) the initial momentum of X is greater then that, of Y, 3) the initial momentum of Y is double that of X, 4) the loss in momentum of Y is equal to the initial, momentum of X, 76. A bullet is fired into a wooden block. If the, bullet gets embedded in wooden block, then, 1) momentum alone is conserved, 2) kinetic energy alone is conserved, 3) both momentum and kinetic energy are, conserved, 4) neither momentum nor kinetic energy are, conserved, 77. During collision, which of the following, statement is wrong?, 1) there is a change in momentum of individual, bodies, 2) the change in total momentum of the system, of colliding particle is zero, 3) the change in total energy is zero, 4) law of conservation of momentum is not valid, , LEVEL - I (C.W), 1., , 2., , 3., , 4., , 5., , 6., , C.U.Q-KEY, 01) 4, 07) 4, 13) 3, 19) 4, 25) 1, 31) 3, 37) 1, 43) 3, 49) 3, 55) 2, 61) 3, 67) 4, 73) 4, , 02) 4, 08) 2, 14) 3, 20) 3, 26) 3, 32) 3, 38) 4, 44) 3, 50) 4, 56) 2, 62) 1, 68) 3, 74) 4, , NARAYANAGROUP, , 03) 1, 09) 3, 15) 4, 21) 3, 27) 1, 33) 2, 39) 3, 45) 3, 51) 1, 57) 4, 63) 1, 69) 4, 75) 4, , 04) 1, 10) 4, 16) 2, 22) 1, 28) 1, 34) 3, 40) 4, 46) 3, 52) 2, 58) 1, 64) 3, 70) 4, 76) 1, , 05) 4, 11) 3, 17) 3, 23) 1, 29) 1, 35) 1, 41) 2, 47) 3, 53) 1, 59) 4, 65) 3, 71) 4, 77) 4, , 06) 1, 12) 4, 18) 3, 24) 1, 30) 4, 36) 1, 42) 2, 48) 1, 54) 1, 60) 2, 66) 2, 72) 1, , 7., , WORK DONE BY CONSTANT FORCE, ^, ^, ^, ur, If F = 2 i + 3 j + 4 k acts on a body and, ^, ^, ^, ur, displaces it by S = 3 i + 2 j + 5 k , then the work, done by the force is, 1) 12 J 2) 20 J 3) 32 J 4) 64 J, A force of 1200 N acting on a stone by means of, a rope slides the stone through a distance of, 10m in a direction inclined at 600 to the force., The work done by the force is, 1) 6000 3J 2) 6000J 3) 12000J 4) 8000J, A man weighing 80 kg climbs a staircase, carrying a 20 kg load. The staircase has 40, steps, each of 25 cm height. If he takes 20, seconds to climb, the work done is, 1) 9800J 2) 490 J 3) 98x105J 4) 7840J, ur, The work done by a force F = 3$i − 4$j + 5k$, displaces the body from a point (3,4,6) to a, point (7,2,5) is, 1) 15 units 2) 25 units, 3) 20 units 4) 10 units, ur, A force F = (6$i − 8 $j ) N , acts on a particle and, displaces it over 4 m along the X-axis and 6m, along the Y-axis. The work done during the, total displacement is, 1) 72 J, 2) 24 J 3) - 24 J 4) zero, A lawn roller is pulled along a horizontal, surface through a distance of 20 m by a rope, with a force of 200 N. If the rope makes an, angle of 60° with the vertical while pulling, the, amount of work done by pulling force is, 1) 4000 J 2) 1000 J 3) 2000 3 J4) 2000 J, WORK DONE BY VARIABLE FORCE, An object has a displacement from position, r, r, vector r1 = 2$i + 3 $j m to r 2 = 4$i + 6 $j m, , (, , ), , (, , ), , ur, 2, under a force F = 3x $i + 2 y $j N, then work, , (, , done by the force is, 1) 24J, 2) 33J, 3) 83J, , ), , 4) 45J, , KINETIC ENERGY, 8., , A shot is fired at 30° with the vertical from a, point on the ground with kinetic energy K. If, air resistance is ignored, the kinetic energy, at the top of the trajectory is, 1) 3K/4 2) K/2, 3) K, 4) K/4, 179
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 9., , 10., , 11., , 12., , 13., , 14., , 15., , A body starts from rest and is acted on by a, constant force. The ratio of kinetic energy, gained by it in the first five seconds to that, gained in the next five seconds is, 1) 2 : 1 2) 1 : 1, 3) 3 : 1 4) 1 : 3, The mass of a simple pendulum bob is 100, gm. The length of the pendulum is 1 m. The, bob is drawn aside from the equilibrium, position so that the string makes an angle of, 60° with the vertical and let go. The kinetic, energy of the bob while crossing its, equilibrium position will be, 1) 0.49 J 2) 0.94 J 3) 1 J, 4) 1.2 J, A body starts from rest and moves with, uniform acceleration. What is the ratio of, kinetic energies at the end of 1st, 2nd and, 3rd seconds of its journey?, 1)1 : 8 : 27, 2)1 : 2 : 3 3)1 : 4 : 9 4)3 : 2, :1, A liquid of specific gravity 0.8 is flowing in a, pipe line with a speed of 2 m/s. The K.E. per, cubic meter of it is, 1) 160 J 2) 1600 J 3) 160.5 J 4) 1.6 J, A 60 kg boy lying on a surface of negligible, friction throws horizontally a stone of mass 1, kg with a speed of 12 m/s away from him. As, a result with what kinetic energy he moves, back?, 1) 2.4 J 2) 72 J 3) 1.2 J4) 36 J, Two stones of masses m and 2 m are projected, vertically upwards so as to reach the same, height. The ratio of the kinetic energies of their, projection is, 1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4, A neutron, one of the constituents of a, nucleus, is found to pass two points, 60 metres apart in a time interval of, 1.8 × 10-4 sec. The mass of the neutron is, 1.67 × 10-27 kg. Assuming that the speed is, constant, its kinetic energy is, 1) 9.3 × 10-17 joule, 2) 9.3 × 10-14 joule, 3) 9.3 × 10-21 joule, 4) 9.3 × 10-11 joule, , POTENTIAL ENERGY, , 16. A tank of size 10 m × 10 m × 10 m is full of, water and built on the ground. If g = 10 ms-2, the, potential energy of the water in the tank is, 1)5 × 107J 2)1 × 108 J 3)5 × 104J 4)5 × 105 J, 17. A bolt of mass 0.3kg falls from the ceiling of, an elevator moving down with an uniform, speed of 7m/s. It hits the floor of the elevator, (length of the elevator = 3m) and does not, rebound. What is the heat produced by, impact?, 1)8.82J 2)7.72J 3)6.62J 4)5.52J, 180, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , POTENTIAL ENERGY OF A SPRING, 18. A spring when compressed by 4 cm has 2 J, energy stored in it. The force required to, extend it by 8 cm will be, 1) 20 N 2) 2 N, 3) 200 N 4) 2000 N, 19. The elastic potential energy of a stretched, spring is given by E = 50x2. Where x is the, displacement in meter and E is in joule, then, the force constant of the spring is, 1)50Nm 2)100N m-1 3)100 N/m²4) 100 Nm, , WORK ENERGY THEOREM BY, CONSTANT FORCE, 20. A body of mass 2 kg is projected with an, initial velocity of 5 ms -1 along a rough, horizontal table. The work done on the body, by the frictional forces before it is brought to, rest is, 1) 250 J 2) 25 J 3) -250 J 4) -25 J, 21. An object is acted on by a retarding force of, 10 N and at a particular instant its kinetic, energy is 6 J. The object will come to rest, after it has travelled a distance of, 1) 3/5 m 2) 5/3 m, 3) 4 m 4) 16 m, 22. By applying the brakes without causing a, skid, the driver of a car is able to stop his car, with in a distance of 5 m, if it is going at, 36 kmph. If the car were going at 72 kmph,, using the same brakes, he can stop the car, over a distance of, 1) 10 m 2) 2.5 m 3) 20 m, 4) 40 m, 23. A bullet fired into a trunk of a tree loses 1/4 of, its kinetic energy in travelling a distance of 5, cm. Before stopping it travels a further, distance of, 1) 150 cm 2) 1.5 cm 3) 1.25 cm 4) 15 cm, , WORK ENERGY THEOREM FOR, VARIABLE FORCE, 1, 24. A bead of mass kg starts from rest from, 2, “A” to move in a vertical plane along a, smooth fixed quarter ring of radius 5m, under, the action of a constant horizontal force F = 5, N as shown. The speed of bead as it reaches, point “B” is, F, , A, , R=5m, , B, , 1) 14.14 m/s 2) 7.07 m/s 3) 5 m/s 4) 25 m/s, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , CONSERVATION OF, MECHANICAL ENERGY, , 33. An electric motor creates a tension of 4500, newton in a hoisting cable and reels it at the, rate of 2m/s. What is the power of the motor?, 1) 15 kW 2) 9 kW 3) 225 W 4) 9000 kW, 34. A juggler throws continuously balls at the rate, of three in each second, each with a velocity, of 10 ms-1. If the mass of each ball is 0.05 kg, his power is, 1) 2 W, 2) 50 W 3) 0.5 W 4) 7.5 W, , (, , 3.8 m, , 6.3m, , 25. A cradle is ‘h’ meters above the ground at the, lowest position and ‘H’ meters when it is at, the highest point. If ‘v’ is the maximum speed, of the swing of total mass ‘m’ the relation, between ‘h’ and ‘H’ is, 1) ½ mv2 + h = H 2) (v2/2g) + h = H, 3) (v2/g) + 2h = H 4) (v2/2g) + H = h, 26. AB is a frictionless inclined surface making an, MOTION IN A VERTICAL CIRCLE, angle of 300 with horizontal. A is 6.3 m above the, 35., A, body of mass 2 kg attached at one end of light, ground while B is 3.8 m above the ground. A, string, is rotated along a vertical circle of radius, block slides down from A, initially starting from, 2 m. If the string can withstand a maximum, rest. Its velocity on reaching B is, A, tension of 140.6 N, the maximum speed with, which the stone can be rotated is, 1) 22 m/s 2) 44 m/s 3) 33 m/s 4) 11m/s, 300, 36., A pilot of mass m can bear a maximum, B, apparent weight 7 times of mg. The, aeroplane is moving in a vertical circle. If the, 1) 7 m s-1 2)14 m s-1 3)7.4 m s-1 4) 4.9 m s-1, velocity of aeroplane is 210 m/s while diving, 27. A stone of mass “m” initially at rest and, up from the lowest point of vertical circle,, dropped from a height “h” strikes the surface, then the minimum radius of vertical circle, of the earth with a velocity “v”. If the, should be, gravitational force acting on the stone is W,, 1) 375 m 2) 420 m 3) 750 m 4) 840 m, then which of the following identities is, correct?, 37. The length of a ballistic pendulum is 1 m and, 1) mv - mh = 0 2) ½ mv2 - Wh2 = 0, mass of its block is 0.98 kg. A bullet of mass 20, 3) ½ mv2 - Wh = 0, 4) ½ mv2 - mh = 0, gram strikes the block along horizontal, POWER, direction and gets embedded in the block. If, 28. A motor boat is going in a river with a velocity, block + bullet completes vertical circle of radius, ur, ˆ ˆ ˆ ms-1 . If the resisting force due, 1m, then the striking velocity of bullet is, V= 4i-2j+k, 1) 280m/s 2) 350m/s 3) 420m/s 4) 490m/s, r, ˆ, ˆ, ˆ, 38., A simple pendulum is oscillating with an angular, to stream is F= 5i-10j+6k N, then the power, amplitude 60o . If mass of bob is 50 gram, then, of the motor boat is, 1) 100 W 2) 50 W, 3) 46 W 4) 23 W, the tension in the string at mean position is (g =, 29. Two riffles fire the same number of bullets in, 10ms–2), a given interval of time. The second fires, 1) 0.5 N 2) 1 N, 3) 1.5 N, 4) 2N, bullets of mass twice that fired by the first 39. A body is moving in a vertical circle such that, and with a velocity that is half that of the first., the velocities of body at different points are, The ratio of their powers is, critical. The ratio of velocities of body at, 1) 1 : 4 2) 4 : 1 3) 1 : 2 4) 2 : 1, 30. A car weighing 1000 kg is going up an incline, angular displacements 60o and120o from, with a slope of 2 in 25 at a steady speed of 18, lowest point is, kmph. If g= 10 ms-2, the power of its engine is, 1) 5: 2 2) 3: 2 3) 3:1 4) 2 : 1, 1) 4 kW 2) 50 kW 3) 625 kW 4) 25 kW, 31. A crane can lift up 10,000 kg of coal in 1 hour 40. A ball of mass 0.6kg attached to a light, from a mine of 180 m depth. If the efficiency, inextensible string rotates in a vertical circle, of the crane is 80 %, its input power must be, of radius 0.75m such that it has speed of 5ms–, (g = 10 ms-2), 1, when the string is horizontal. Tension in the, 1) 5 kW 2) 6.25 kW 3) 50 kW 4) 62.5 kW, string when it is horizontal on other side is, 32. A man carries a load of 50 kg through a height, (g =10ms–2), [2007M], of 40 m in 25 seconds. If the power of the man, 1), 30N, 2), 26N, 3), 20N, 4), 6N, is 1568 W, his mass is, 1) 5 kg 2) 1000 kg, 3) 200 kg 4) 50 kg, , ), , NARAYANAGROUP, , (, , ), , 181
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , ELASTIC AND INELASTIC, COLLISIONS IN ONE DIMENSION, 41. A 6 kg mass travelling at 2.5 ms −1 collides, head on with a stationary 4 kg mass. After the, collision the 6 kg mass travels in its original, direction with a speed of 1 ms −1 . The final, velocity of 4 kg mass is, 1) 1 ms − 1 2) 2.25ms −1 3) 2 ms −1 4) 0 ms −1, 42. A body of mass 10 kg moving with a velocity, of 5 ms −1 hits a body of 1 gm at rest. The, velocity of the second body after collision,, assuming it to be perfectly elastic is, 1) 10 ms −1, 2) 5 ms −1 3) 15 ms −1 4), 0.10 ms −1, , 43. A block of mass 1 kg moving with a speed of, 4 ms −1 , collides with another block of mass 2, kg which is at rest. The lighter block comes, to rest after collision. The loss in KE of the, system is, 1) 8 J, 2) 4 × 10 −7 J 3) 4 J 4) 0 J, 44. A marble going at a speed of 2 ms −1 hits, another marble of equal mass at rest. If the, collision is perfectly elastic, then the velocity, of the first marble after collision is, 1) 4 ms −1 2) 0 ms −1 3) 2 ms −1 4) 3 ms −1, 45. A massive ball moving with a speed v collides, head on with a fine ball having mass very much, smaller than the mass of the first ball at rest., The collision is elastic and then immediately, after the impact, the second ball will move with a, speed approximately equal to, 1) v, 2) 2v 3) v/3 4) infinite, 46. A 1 kg ball moving at 12 m/s collides head on, with a 2 kg ball moving in the opposite, direction at 24 m/s. The velocity of each ball, after the impact, if the coefficient of, restitution is 2/3 is, 1) -28 m/s ; -4 m/s, 2) 28 m/s ; -4 m/s, 3) 20 m/s ; 24 m/s 4) -20 m/s ; -4 m/s, 47. A 6 kg mass collides with a body at rest. After, the collision, they travel together with a, velocity one third the velocity of 6 kg mass., The mass of the second body is, 1) 6 kg 2) 3 kg 3) 12 kg, 4) 18 kg, 48. A body of mass m moving at a constant, velocity v hits another body of the same mass, moving with a velocity v/2 but in the opposite, direction and sticks to it. The common, velocity after collision is, 1) v, 2) v/4 3) 2v 4) v/2, 182, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 49. An 8 gm bullet is fired horizontally into a, 9 kg block of wood and sticks in it. The block, which is free to move, has a velocity of 40 cm/, s after impact. The initial velocity of the, bullet is, 1)450 m/s 2)450 cm/s 3)220 m/s 4) 220 cm/s, 50. A block of wood of mass 9.8 kg is suspended, by a string. A bullet of mass 200 gm strikes, horizontally with a velocity of 100 ms −1 and, gets embedded in it. The maximum height, attained by the block is g = 10 ms −2, 1) 0.1 m 2) 0.2 m, 3) 0.3 m, 4) 0 m, 51. A 15 gm bullet is fired horizontally into a 3 kg, block of wood suspended by a string. The bullet, sticks in the block, and the impact causes the, block to swing 10 cm above the initial level. The, velocity of the bullet nearly is ( in ms-1), 1)281, 2)326, 3)184, 4)58, 52. A body of mass 20 gm is moving with a certain, velocity. It collides with another body of, mass 80 gm at rest. The collision is perfectly, inelastic. The ratio of the kinetic energies, before and after collision of the system is, 1) 2 : 1 2) 4 : 1, 3) 5 : 1, 4) 3 : 2, , (, , ), , COEFFICIENT OF RESTITUTION, 53. A rubber ball drops from a height ‘h’. After, rebounding twice from the ground, it rises to, h/2. The co - efficient of restitution is, 1), , 1, 2, , 1/ 2, , 1, 2, , 2) , , 1/ 4, , 1, 2, , 1/ 6, , 1, 2, , 3) 4) , , 54. A body dropped freely from a height h o n t o, a horizontal plane, bounces up and down and, finally comes to rest. The coefficient of, restitution is e. The ratio of velocities at the, beginning and after two rebounds is, 1) 1 : e 2) e : 1 3) 1 : e², 4) e² : 1, 55. In the above problem, the ratio of times of, two consecutive rebounds is, 1) 1 : e 2) e : 1 3) 1 : e², 4) e² : 1, 56. In the above problem the ratio of distances, travelled in two consecutive rebounds is, 1) 1 : e 2) e : 1 3) 1 : e², 4) e² : 1, 57. A ball is dropped onto a horizontal floor. It, reaches a height of 144 cm on the first bounce, and 81 cm on the second bounce. The, coefficient of restitution is, 1) 0, 2) 0.75 3) 81/144, 4) 1, 58. A ball is dropped onto a horizontal floor. It, reaches a height of 144 cm on the first bounce, and 81 cm on the second bounce. The height it, attains on the third bounce is, 1) 45.6 cm 2) 81 cm 3) 144 cm 4) 0 cm, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 59. A ball is dropped from height 'H' onto a, horizontal surface. If the coefficient of, restitution is 'e' then the total time after which, it comes to rest is, 1), , 2H 1 − e , , , g 1+ e , , 3), , 2H 1 + e 2 , , , g 1 − e 2 , , 2), , 2H 1 + e , , , g 1− e , , 4), , 2H 1 − e 2 , , , g 1 + e 2 , , 60. A stationary body explodes into two, fragments of masses m1 and m 2. If momentum, of one fragment is p, the energy of explosion, is, , p2, 2) 2 m m, 1 2, , p2, 1) 2 m + m, ( 1 2), , 3), , p 2 ( m1 + m2 ), , 1., 3, 4., 5., 6., 7., 8., , 9., , 02)2, 08)4, 14)2, 20)4, 26)1, 32)4, 38)2, 44)2, 50)2, 56) 3, , 03)1, 09)4, 15)1, 21)1, 27) 3, 33)2, 39) 4, 45)2, 51) 1, 57) 2, , 04)1, 10)1, 16)1, 22)3, 28)3, 34)4, 40)3, 46)1, 52) 3, 58) 1, , 05)3, 11)3, 17)1, 23)4, 29)4, 35)4, 41)2, 47)3, 53)3, 59)2, , 06)3, 12)2, 18)3, 24)1, 30)1, 36) 3, 42)1, 48)2, 54)3, 60) 3, , LEVEL - I (C.W) - HINTS, ur ur, W = F .S 2. W = FS cos θ, ur ur, , W = F .S = FS = ( M + m ) g ( n × heach step ), ur ur ur ur ur, W = F .S = F . r2 − r1, , (, , W=Wx+Wy, , ), , ur, , ur, , Wx= F . x $i , Wy = F . y $j, , ur ur, W = F .S = FS cos θ, ;, , x2, , y2, , x1, , y1, , W = ∫ dw = ∫ Fx dx + ∫ Fy dy, 1, mu2, 2, 1, 1, 2, 2, At maximum height, K = mu cos θ, 2, 1, 1, 2, KE = mv 2 = m ( gt ) (Q v = gt ), 2, 2, , At projection, K =, , NARAYANAGROUP, , 1, 1, 2, 2, 1 2 1, 1, 2 K .E, = ρv2, 12. KE = mv = ( ρV ) v ;, V, 2, 2, 2, 2, 2 2, 11. KE = mv = mg t (Q v = gt ), , 13. m1v1 = m2 v2 ; KE =, , 1, m2 v22, 2, , 1, 2, , 2, 14. KE = mv , when two bodies reach the same, , KE, , LEVEL - I (C.W) - KEY, 01)3, 07)3, 13)3, 19)2, 25)2, 31)2, 37) 2, 43)3, 49)1, 55)1, , 10. K .Emean = P.Eextreme = mgl (1 − cosθ ), , m, , (, , 1, 1, height, v1=v2; KE = m Q v = 2 gh, 2, 2, , p2, 4) 2 m − m, ( 1 2), , 2m1m2, , K1, t2, = 2 1 2 where t =5sec and t =10sec, 1, 2, K 2 t2 − t1, , 15., , KE =, , 1, 1 s, mv 2 = m , 2, 2 t , , ), , 2, , h, 16. P.E=mgh1; here h1 = and m = ρ × V, 2, 17 Heat produced = loss of potential energy, = mgh, 1 2, 2U, 18. U = 2 Kx1 ⇒ K = x 2 and F = Kx2, 1, 1, , 2, 19. U = K x --(1),U=50x 2--(2),, 2, equation (1) and (2) to find K, , compare, , , 2, 2 , 20. W f = m v − m v i , 2, 2, , , 21. According to work energy theorem, W= ∆ KE= -FS, ∆KE1 W1 FS1, 22. W = ∆KE ; ∆KE = W = FS, 2, 2, 2, S1, ∆ KE 1, 23. W= F .S = ∆ K E ; S = ∆ KE, 2, 2, 24. Applying the work - energy theorem, we get, 1, × mv 2 − 0 = W1 + W2, 2, = Horizontal force × displacement + Vertical force, × displacement., = F × R + mg × R, 25. K.E at mean= P.E at extreme position, 1, , 1, , f, , 26. gain in K.E=Loss of P.E = mg ( h1 − h2 ), r r, 27. gain in K.E=Loss of P.E, 28. P = F .V, P1 m1 v12 , ur r, 29. P = m v 2 30. P = F.v = Fv = mg sin θ v, 2, , , , 2, , , , 2, , , , 31. η = Pout , where P = W = mgh, out, Pin, t, t, 183
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, ur ur, W ( m + M ) gh, =, 33. Pinst = F .V = FV cos θ, t, t, 1, , v2, , n mv2 , +g, 34 P = W = 2, 35. Tmax = m , r, , t, t, 36. At lowest point of vertical circle,, mv 2, Tmax =, + mg, rmin, 37. According to law of conservation of linear, momentum mu=(M+m)v, , KEi =, , 32. P =, , ( M + m), u=, , 5 gr, , m, mv 2, m, T, =, mg, +, = mg + 2 gl (1 − cos θ ) , 38., r, l, v1, 3 + 2 cos θ1, 39. v = gR ( 3 + 2 cos θ ) ⇒ v = 3 + 2 cos θ, 2, 2, , 2m1 , 42. v2 = m + m u1, 1, 2 , , 43. m1u1 + m2u2 = m1v1 + m2v2, 1, 1, ∆KE = m1u12 − m2 v22, 2, 2, r m1 − m 2 r 2m 2 r, u1 + , u2, 44. v1 = , m1 + m 2 , m1 + m 2 , , 59., , t=, , 184, , 57. e =, , h, 2, h, 1, , 2h1, 2h 2, 2h, +2, +2, + − − and h = e2n h, n, g, g, g, , 60. E = E1 + E 2 =, , p12, p2, + 2, 2 m1 2 m 2, , LEVEL-I (H.W), , 1., , WORK DONE BY CONSTANT FORCE, ur r r r, If a force F = i + 2 j + k N acts on a body, produces a displacement of, ur, r r, r, S = 4i + j + 7 k m, then the work done is, 1) 9J, 2) 13J, 3) 5J, 4) 1J, Work done by the gravitational force on a, body of mass “m” moving on a smooth, horizontal surface through a distance ‘s’ is, 1) mgs, 2 -mgs 3) 0, 4) 2mgs, A body of mass 1 kg is made to travel with a, uniform acceleration of 30 cm/s2 over a, distance of 2m, then work to be done is, 1) 6J, 2) 60J 3) 0.6J, 4) 0.3J, A uniform cylinder of radius ‘r’ length ‘L’ and, mass ‘m’ is lying on the ground with the, curved surface touching the ground. If it is to, be oriented on the ground with the flat, circular end in contact with the ground, the, work to be done is, 1)mg[(L/2)-r] 2) mL[(g/2)-r] 3) mr(gL-1) 4)mgLr, A meter scale of mass 400 gm is lying, horizontally on the floor. If it is to be held, vertically with one end touching the floor, the, work to be done is, 1) 6 J, 2) 4 J 3) 40J 4) 2 J, A force F is applied on a lawn mover at an angle, of 600 with the horizontal. If it moves through a, distance x, the work done by the force is, 1) Fx/2 2) F/2x 3) 2Fx 4) 2x/F, A weight lifter jerks 220 kg vertically through, 1.5 metre and holds still at that height for two, minutes. The work done by him in lifting and, in holding it still are respectively, 1) 220J, 330J, 2) 3234 J, 0 J, 3) 2334 J, 10 J 4) 0 J, 3234 J, , (, , (, , 2., , 3., , 4., , 5., , 6., , 48. m1u1 − m2u2 = ( m1 + m2 ) v 49. mu = ( m + M ) v, , v2, 50. mu = ( m + M ) v and h =, 2g, v2, mu, =, m, +, M, v, and, h, =, (, ), 51., 2g, 52. m1u1=(m1+m2)v, , n, , hn = e 2nh, , r m (1 + e ) r m 2 − em1 r, v2 = 1, u1 + , u2, m1 + m 2 , m1 + m2 , , 47. m1u1 + m2u2 = ( m1 + m2 ) v, , h = e 2 n h 54. vn = env 55. tn = e nt, , 58., , 45 m2 <<< m1 and, , r 2m1 r m 2 − m1 r, v2 = , u1 + , u2, m1 + m2 , m1 + m 2 , r m1 − em2 r m 2 (1 + e ) r, 46. v1 = m + m u 1 + m + m u 2, 1, 2 , 2 , 1, , 1, 1, m1u12 ; KE f = ( m1 + m2 ) v 2, 2, 2, , 2n, 56. hn = e h, , 2, H, , mv, where vH = 3gr, r, 41. According to law of conservation of linear, momentum, m1u1+m2u2=m1v1+m2v2, , 40. T =, , 53., , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 7., , ), , ), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , A tennis ball has a mass of 56.7 gm and is, served by a player with a speed of 180 kmph., The work done in serving the ball is nearly, 1) 710J 2) 71J, 3) 918J 4) 91.8J, A body of mass 2kg is projected vertically up, with velocity 5ms-1.The work done on the, body by gravitational force before it is, brought to rest momentarily is, 1) 250J 2) 25J, 3) 0J, 4) -25J, WORK DONE BY VARIABLE FORCE, A force F = (2+x) N acts on a particle in x direction where ‘x’ is in metre. The work, done by this force during a displacement, from x = 1 m to x = 2 m is, 1) 2 J, 2) 3.5 J, 3) 4.5 J 4) 5 J, KINETIC ENERGY, On increasing the speed of a body to 2 ms-1,, its kinetic energy is quadrupled. Then its, original speed must be, 1) 0.25ms-1 2) 1ms-1 3) 4ms-1 4) 2ms-1, A bullet of mass 10 gm strikes a target at 400, m/s velocity and loses half of its initial velocity., The loss of kinetic energy in joules is, 1) 800, 2) 200, 3) 400 4) 600, An object is acted on by a retarding force of, 10 N and at a particular instant its kinetic, energy is 6J. The object will come to rest, after it has travelled a distance of, 1) 3/5m 2) 5/3m, 3) 4m, 4) 16m, A man standing on the edge of the roof of a, 20 m tall building projects a ball of mass 100, gm vertically up with a speed of 10ms-1. The, kinetic energy of the ball when it reaches the, ground will be [g=10 ms-2], 1)5J, 2) 20J, 3) 25J, 4) Zero, A river of salty water is flowing with a, velocity 2 m/sec. If the density of water is 1.2, gm/cc, the kinetic energy of each of cubic, metre of water is, 1) 2.4 J 2) 24 J, 3) 4.8 KJ 4) 2.4 KJ, If the kinetic energy of a body increases by, 125% , the percentage increase in its, momentum is, 1) 50% 2) 62.5% 3) 250% 4) 200%, The kinetic energy of a body is ‘K’. If onefourth of its mass is removed and velocity is, doubled, its new kinetic energy is, 1) K, 2) 3K, 3) 4K, 4) 9K/4, POTENTIAL ENENGY, An inelastic ball falls from a height of 100, meters. It loses 20% of its total energy due to, impact. The ball will now rise to a height of, 1) 80 m 2) 120m 3) 60m 4) 9.8m, A woman weighing 63 kg eats plum cake, whose energy content is 9800 calories. If all, , NARAYANAGROUP, , 20., , 21., , 22., , 23., , 24., , 25., , 26., , this energy could be utilized by her, she can, ascend a height of, 1) 1m, 2) 67m, 3) 100m 4) 42m, POTENTIAL ENERGY OF A SPRING, A spring of spring constant 5x103 N/m is, stretched initially by 5cm from the unstretched, position. Then the work required to stretch it, further by another 5cm is., 1) 6.25Nm 2) 12.50Nm 3) 18.75Nm 4) 25Nm, A spring with spring constant K when, stretched through 1cm, the potential energy, is U. If it is stretched by 4cm, the potential, energy will be, 1) 4U, 2) 8U, 3) 16U 4) 2U, WORK ENERGY THEOREM, BY CONSTANT FORCE, A body moving with a kinetic energy of 6J, comes to rest at a distance of 1m due to a, retarding force of, 1) 4 N, 2) 6 N, 3) 5 N 4) 8 N, A ship of mass 3x107 kg initially at rest is, pulled by a force of 5x104 N through a, distance of 3 meters. Assuming that the, resistance due to water is negligible, the, speed of the ship is, 1) 0.1m/s 2) 1.5 m/s 3) 5m/s 4) 60 m/s, A vehicle of mass 1000 kg is moving with a, velocity of 15 ms-1 .It is brought to rest by, applying brakes and locking the wheels. If, the sliding friction between the tyres and the, road is 6000N, then the distance moved by, the vehicle before coming to rest is, 1) 37.5 m 2) 18.75 m 3) 75 m 4) 15 m, The workdone to accelerate a body from 30, ms-1 to 60 ms-1 is three times the work done to, accelerate it from 10 ms-1 to ‘v’ . The value of, ‘v’in ms-1 is, 1) 30, 2) 20 2 3) 30 3 4) 10 10, WORK ENERGY THEOREM, FOR VARIABLE FORCE, A block of mass 4 kg is initially at rest on a, horizontal frictionless surface. A horizontal, ur, force F = ( 3 + x ) $i newtons acts on it, when the, , block is at x=0. The maximum kinetic energy of, the block between x=0 and x=2m is, 1) 6J, 2) 8J, 3) 9J, 4) 10J, CONSERVATION OF, MECHANICAL ENERGY, 27. A block of mass 4 kg slides on a horizontal, frictionless surface with a speed of 2m/s. It is, brought to rest in compressing a spring in its, path. If the force constant of the spring is 400, N/m, by how much the spring will be, compressed, 1) 2x10-2m 2) 0.2m 3) 20 m 4) 200m, 185
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 28. At what height above the ground must a mass, of 5 kg be to have its PE equal in value to the, KE possessed by it when it moves with a, velocity of 10 m/s? (g=10 m/s2), 1)1m, 2) 5m, 3) 10m 4) 50m, 29. A body slides down a fixed curved track that, is one quadrant of a circle of radius R, as in, the figure. If there is no friction and the body, starts from rest, its speed at the bottom of the, track is, , R, , 1) 5gR, , 2), , 3), , 5gR, , 2gR 4) gR, , POWER, 30. An electric motor in a crane while lifting a, load produces a tension of 4000 N in the cable, attached to the load. If the motor is winding, the cable at the rate of 3ms-1, the power of the, motor expressed in kilo watt units must be, 1) 4, 2) 3, 3) 12, 4) 6, 31. An electric motor operates with an efficiency, of 90%. A pump operated by the motor has an, efficiency of 80%. The overall efficiency of, the system is, 1) 85% 2) 100% 3) 72% 4) 60%, 32. A machine gun fires 420 bullets per minute., The velocity of each bullet is 300ms-1 and the, mass of each bullet is 1gm. The power of the, machine gun is, 1) 315W, 2) 315000W, 3) 630W, 4) 3150W, 33. A 1 kg mass at rest is subjected to an, acceleration of 5 m/s2 and travels 40m. The, average power during the motion is, 1) 40W 2) 8W, 3) 50W 4) 200W, 34. If the power of the motor of a water pump is, 3kW, then the volume of water in litres that, can be lifted to a height of 10m in one minute, by the pump is (g=10 ms-2), 1) 1800 2) 180, 3) 18000 4) 18, 35. A particle moves with a velocity, 5$i + 3 $j + 6k$ m/s under the influence of a, , (, , ), , (, , ), , constant force 5$i + 5 $j + 10k$ N. The, instantaneous power applied to the particle is, 1) 100W 2) 40W 3) 140W 4) 170W, 186, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , MOTION INA VERTICAL CIRCLE, 36. A body of mass m is rotated in a vertical circle, of radius R by means of light string. If the, velocity of body is gR while it is crossing, highest point of vertical circle then the, tension in the string at that instant is, mg, 1) 2mg, 2) mg, 3), 4) Zero, 2, 37. A body of mass m is rotated in a vertical circle, with help of light string such that velocity of, body at a point is equal to critical velocity at, that point. If T1, T2 be the tensions in the, string when the body is crossing the highest, and the lowest positions then the following, relation is correct, 1) T 2-T1=6mg, 2) T 2-T1=4mg, 3) T 2-T1=3mg, 4) T 2-T1=2mg, 38. A vehicle is travelling with uniform speed, along a concave road of radius of curvature, 19.6m. At lowest point of concave road if the, normal reaction on the vehicle is three times, its weight, the speed of vehicle is, 1) 4.9m/s 2) 9.8m/s 3) 14.7m/s 4) 19.6m/s, 39. A car is travelling along a flyover bridge, which is a part of vertical circle of radius 10m., At the highest point of it the normal reaction, on the car is half of its weight, the speed of, car is, 1) 7m/s 2) 10m/s 3) 14m/s 4) 20m/s, 40. A very small particle rests on the top of a, hemisphere of radius 20 cm. The smallest, horizontal velocity to be given to it, if it has to, leave the hemisphere without sliding down its, surface (g=9.8 ms-2 )is, 1) 9.8 m/s2) 4.9 m/s3) 1.96 m/s4) 3.92 m/s, ELASTIC AND INELASTIC COLLISIONS, 41. A ball of 4 kg mass moving with a speed of, 3 ms −1 has a head on elastic collision with a 6, kg mass initially at rest. The speeds of both, the bodies after collision are respectively, 1) 0.6 ms −1 , 2.4 ms −1 2) −0.6 ms −1 , − 2.4 ms −1, 3) −0.6 ms −1 , 2.4 ms −1 4) −0.6 ms −1 , − 2.4 ms −1, 42. A ping - pong ball strikes a wall with a, velocity of 10 ms −1 . If the collision is, perfectly elastic, find the velocity of ball after, impact, 1) − 20 ms −1 2) − 5 ms −1 3) 1.0 ms −1 4) − 10 ms −1, 43. Two identical balls collide head on. The initial, velocity of one is 0.75 ms-1, while that of the, other is -0.43ms-1 . If the collision is perfectly, elastic, then their respective final velocities, are, 1) 0.75ms-1;-0.43ms-1 2) -0.43ms-1;0.75ms-1, 3) -0.75ms-1;0.43ms-1 4) 0.43ms-1;0.75ms-1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 44. A truck of mass 15 tons moving with 1 ms-1, collides with a stationary truck of mass 10, tons and automatically connected to move, together. The common velocity is, 1) 1ms-1 2) 0ms-1, 3) 0.4ms-1 4) 0.6ms-1, 45. In the above problem the total KE before, collision is, 1) 4500J 2) 7500J 3) 3000J 4) 0J, 46. In the above problem loss of KE during, collision is, 1) 4500J 2) 7500J 3) 3000J 4) 0J, 47. A bullet of mass ‘x’ moves with a velocity y,, hits a wooden block of mass z at rest and, gets embedded in it. After collision, the, wooden block and bullet moves with a, velocity, z, x, x+ y, y 4) x + y y, y 2), y 3), 1), x+ y, x+z, x, z, 48. A railway truck of mass 16000kg moving, with a velocity of 5ms-1 strikes another truck, of mass 4000kg at rest. If they move, together after impact, their common velocity, is, 1) 2ms-1 2) 4ms-1 3) 6ms-1 4) 8ms-1, COEFFICIENT OF RESTITUTION, 49. A ball falls from a height of 10m on to a, horizontal plane. If the coefficient of, restitution is 0.4, then the velocity with which, it rebounds from the plane after second, collision is, 1) 2.24ms-1, 2) 5.6ms-1 3) 2.8ms-1 4), -1, 0.9ms, 50. A ball is dropped from a height of 3m. If, coefficient of restitution between the surface, and ball is 0.5, then the total distance covered, by the ball before it comes to rest is, 1) 3m, 2) 4m, 3) 5m, 4) 6m, 51. A glass sphere of mass 5mg, falls from a, height of 3 meters on to a horizontal surface., If the coefficient of restitution is 0.5, then, after the impact the sphere will rise to a, height of, 1) 0.075m 2) 0.75m 3) 7.5m 4) 75m, 52. A particle falls from a height ‘h’ upon a fixed, horizontal plane and rebounds. If ‘e’ is the, coefficient of restitution, then the total, distance travelled before it comes to rest is, , 1+ e , 1) h 1 − e 2 , , , , 1− e , 2) h 1 + e 2 , , , , h 1 − e2 , 3) 2 1 + e 2 , , , , h 1 + e2 , 4) 2 1 − e 2 , , , , 2, , NARAYANAGROUP, , 2, , LEVEL-I ( H.W)-KEY, 01) 2, 07) 2, 13) 1, 19) 2, 25) 4, 31) 3, 37) 1, 43) 2, 49) 1, , 02) 3, 08) 2, 14) 3, 20) 3, 26) 2, 32) 1, 38) 4, 44) 4, 50) 3, , 03) 3, 09) 4, 15) 4, 21) 3, 27) 2, 33) 3, 39) 1, 45) 2, 51) 2, , 04) 1, 10) 2, 16) 1, 22) 2, 28) 2, 34) 1, 40) 3, 46) 3, 52) 1, , 05) 4, 11) 4, 17) 2, 23) 1, 29) 3, 35) 1, 41) 3, 47) 1, , 06) 1, 12) 4, 18) 1, 24) 2, 30) 3, 36) 4, 42) 4, 48) 2, , LEVEL-I ( H.W)-HINTS, , 1., 3, 4., , 5., 6., 7., , 8., , ur ur, ur ur, W = F .S 2. W = F . S = FS cosθ and θ = 900, ur ur, W = F .S = FS = maS, W = U i − U f ; where U i = mgh1 ; U f = mgh2 ;, , ∴W =mg ( h1 − h2 ), ur ur, L, W = F .S = FScom ;where F = mg and S com =, 2, ur ur, =FScos, θ, W = F .S, ur ur, W= F .S = FS cos θ, In lifting the weight F=mg,θ =00 ;, in holding the weight, S=0, 1, 1, W = mvi2 ; 9. W = − mgh = − mu 2, 2, 2, 2, , 10. W =, , ∫, , F dx, , 1, , 11. K = 1 Mv 2 and 4 K = 1 M ( v + 2 )2, 2, 2, , (, , 12. ∆KE = 1 m v12 − v2 2, , 2, , ), , , v2 =, , 13. ∆ KE = W ⇒ KE = FS, i, 14. K .E = K .E + P.E, f, i, i, , v1, 2, , 1, mv2, K, E, 1, 2, 15., =, = ρv2, V, V, 2, , 16. P = 2 mKE, KE2 − KE1, P2 − P1 , , × 100 = , KE1, P1 , , , , × 100, , , , 2, 1, 1, m v 2 , K E ' = m ' (v ' ), 2, 2, 3, m, m' =, , v' = 2v, 4, 18. with 80% of available energy it can rise to height, h' = 0.8h, , 17. K E =, , 187
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 19. W = JQ ⇒ mgh = JQ ( 1cal= 4.2J ), 1, 1 2, U1 x12, 2, 2, 20. W = K ( x2 − x1 ) 21. U = 2 Kx ⇒ U = x 2, 2, 2, 2, 22. W = ∆KE ⇒ KEi = Fret S, 23. W= FS = ∆KE, 24. W= FS = ∆KE, x2, ur uur, 1 2 1 2, ∆, KE, =, W, =, F, W, =, mv, −, mu, 25., 26., ∫x .dx, 2, 2, 1, 27. K.E of block is converted into elastic P.E in spring, 1 2 1 2, i.e mv = kx, 2, 2, 1 2, 28. mgh = mv, 2, 1 2, 29. Loss of P.E= gain in K.E ; mgh = mv, 2, ur ur, 30. P = F .V 31.η =η1 ×η 2, 32., , W, P =, =, t, , 33. Pavg =, , 1, , n mv2 , 2, , t, , W FS maS, =, =, ;v=, t, t, t, , W mgh, =, and m = V ρ, t, t, ur ur, P = F .V, , 34. P =, , 2 as , t =, , 36. Tension at the highest point, mv 2, T =, − mg where v = gr, r, , m v 22, m v12, + mg, − m g , T2 =, 37. T1 =, r, r, v22 − v12 = 4 gr , v1 = gr, , r m1 − m 2 r 2m 2 r, 41. v1 = m + m u1 + m + m u 2, 1, 2 , 1, 2 , r 2m1 r m2 − m1 r, v2 = , u1 + , u2, m1 + m 2 , m1 + m 2 , , 42. m1u1 + m2u2 = m1v1 + m2v2, 188, , r 2m1 r m2 − m1 r, v2 = , u1 + , u2, m1 + m2 , m1 + m2 , , 44., , 46., 47., 48., 49., , r, r, r m1u1 + m2u2, 1, 1, 2, 2, v=, ; 45.KEbefore= m1u1 + m2u2, m1 + m2, 2, 2, Loss in KE=KEbefore -KEafter, r, r, r m u + m2u2, v= 1 1, mr1 + m2 r, r m1u1 + m2u2, v=, m1 + m2, v n = e n v where v= 2gh, , 1 + e2 , d, =, h, , 50., 2 , 1− e , 51. hn = e2n h, 52. d=h+2h1+2h2+------and hn=e2nh, , LEVEL-II (C.W), WORK DONE BY CONSTANT FORCE, 1. A body of mass 5 kg is moved up over 10 m, along the line of greatest slope of a smooth, inclined plane of inclination 30° with the, horizontal. If g = 10 m/s2 , the work done will, be, 1) 500 J 2) 2500 J 3) 250 J 4) 25 J, 2. A particle of mass 0.5 kg is displaced from, ur, ur, position r1 (2,3,1) to r2 (4,3, 2) by applying a, force of magnitude 30 N which is acting along, , 2, , mv, ;Given N=3mg, r, mv 2, 39. Normal reaction, N = mg −, r, mg, Given that N =, 2, 40. Vmin = rg, 38. Normal reaction, N = mg +, , r m1 − m2 r 2m2 r, 43. v1 = m + m u1 + m + m u 2, 1, 2 , 1, 2 , , v, a, , 35., , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 3., , 4., , ^ ^ ^ , i + j + k . The work done by the force is, , , 1) 10 3 J 2) 30 3 J 3) 30 J 4) 40 J, Kinetic energy of a particle moving in a, straight line varies with time t as K = 4t 2 ., The force acting on the particle, 1) is constant, 2) is increasing, 3) is decreasing, 4) first increases and then decreases, A block of mass 5 kg initially at rest at the, origin is acted upon by a force along the, positive X - direction represented by, F=(20 +5x)N. Calculate the work done by the, force during the displacement of the block, from x = 0 to x = 4m., 1) 100 J 2) 150 J 3) 120 J 4) 75 J, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, 5., , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , A force F acting on a particle varies with the, position x as shown in the graph. Find the, work done by the force in displacing the, particle from x = -a to x=+2a, +2b, (F), -a, (x), , +2a, , -b, , 3ab, 1), 2, , 6., , 7., , 8., , 9., , 4ab, 2, 2, 2), 3), 4), 2, 3ab, 4ab, ur, A force F = 2$i + 3 $j − 4k$ N acts on a particle, which is constrained to move in the XOY plane, along the line x = y. If the particle moves 5 2m ,, the work done by force in joule is, 1) 25 2, 2) 5 58 3) 25 4) 10, Two forces each of magnitude 10 N act, simultaneously on a body with their directions, inclined to each other at an angle of 120° and, displaces the body over 10 m along the bisector, of the angle between the two forces. Then the, work done by each force is, 1) 5 J, 2) 1 J 3) 50 J, 4) 100 J, ‘n’ identical cubes each of mass ‘m’ and edge, ‘L’ are on a floor. If the cubes are to be, arranged one over the other in a vertical, stack, the work to be done is, 1) Lmng (n-1)/2 2) Lg(n-1)/mn, 3) (n-1)/Lmng, 4) Lmng/2(n-1), A chain of mass m and length ‘L’ is over, hanging from the edge of a smooth horizontal, table such that 3/4 th of its length is lying on, the table. The work done in pulling the chain, completely on to the table is, 1) mgL/16 2) mgL/32 3) 3mgL/32 4) mgL/8, , (, , ), , WORK DONE BY VARIABLE FORCE, 10. A body is displaced from (0,0) to (1m, 1m), along the path x = y by a force, ^, ^, , F = x 2 j + y i N. The work done by this, , , force will be, 4, 5, 3, 7, 1) J, 2) J 3) J 4) J, 3, 6, 2, 5, 11. A particle moves under the effect of a force, F = C x from x = 0 to x = x1. The work done in, the process is (treat C as a constant), 1) C2 / x12 2) Cx12 3) ½ Cx12 4) ½ C2/x12, , NARAYANAGROUP, , 12. Under the action of a force, a 2 kg body, moves such that its position ‘x’ in meters as a, function of time ‘t’ in seconds given by:, x = t2/2. The work done by the force in the, first 5 seconds is, 1) 2.5 J 2) 0.25 J, 3) 25 J 4) 250 J, 13. A body of mass 5 kg at rest under the action, of a force which gives its velocity given by, v = 3×t m/s, here ‘t’ is time in seconds. The, work done by the force in two seconds will be, 1) 90 J 2) 45 J, 3) 180 J 4) 30 J, , KINETIC ENERGY, 14. A body freely falls from a certain height onto, the ground in a time ‘t’. During the first one, third of the interval it gains a kinetic energy, , ∆K1 and during the last one third of the, interval, it gains a kinetic energy ∆ K 2 . The, ratio ∆K1 : ∆K 2 is, 1) 1 : 1 2) 1 : 3 3) 1 : 4 4) 1 : 5, 15. A man has twice the mass of a boy and has half, the kinetic energy of the boy. The ratio of the, speeds of the man and the boy must be, 1) 2 : 1 2) 4 : 1 3) 1 : 4 4) 1 : 2, 16. The speed of a car changes from 0 to 5 ms-1 in, the first phase and from 5 ms-1 to 10 ms-1 in the, second phase and from 10 ms-1 to 15 ms-1 during, the third phase. In which phase the increase in, kinetic energy is more?, 1) first phase 2) second phase, 3) third phase 4) same in all the three phases, , POTENTIAL ENERGY, 17. A rubber ball falling from a height of 5m, rebounds from a hard floor to a height of, 3.5m. The % loss of energy during the impact, is, 1) 20% 2) 30%, 3) 43% 4) 50%, 18. When a long spring is stretched by x cm, its, P.E is U . If the spring is stretched by Nx cm,, the P.E stored in it will be, U, U, 1), 2) NU 3) N 2U, 4) 3, N, N, 19. An elastic spring is compressed between two, blocks of masses 1 kg and 2 kg resting on a, smooth horizontal table as shown. If the, spring has 12J of energy and suddenly, released, the velocity with which the larger, block of 2 kg moves will be, 2kg, 1kg compressed spring, B, A, 1) 2 m/s, 2) 4 m/s, 3) 1 m/s 4) 8 m/s, 189
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 20. A block of mass 2 kg is on a smooth horizontal, surface. A light spring of force constant 800 N/, m has one end rigidly attached to a vertical, wall and lying on that horizontal surface. Now, the block is moved towards the wall, compressing the spring over a distance of, 5 cm and then suddenly released. By the time, the spring regains its natural length and looses, contact with the block, the velocity acquired by, the block will be, 1) 200m/s 2) 100 m/s 3) 2m/s 4) 1m/s, , WORK ENERGY THEOREM FOR, CONSTANT FORCE, 21. A bullet of mass 10gm is fired horizontally, with a velocity 1000ms -1 from a riffle, situated at a height 50 m above the ground. If, the bullet reaches the ground with a velocity, 500ms −1 , the work done against air, resistance in the trajectory of the bullet is (in, , joule) ( g = 10ms −2 ), 1) 5005 2) 3755 3) 3750 4) 17.5, 22. A drop of mass 1.00 g falling from a height, 1.00 km. It hits the ground with a speed of, 50.0ms -1 . What is the work done by the, unknown resistive force?, 1) -8.75J 2) 8.75J 3) -4.75J 4) 4.75J, 23. A block of mass 5 kg is initially at rest on a, rough horizontal surface. A force of 45 N acts, on it in a horizontal direction and pushes it, over a distance of 2 m. The force of friction, acting on the block is 25 N. The final kinetic, energy of the block is, 1) 40 J 2) 90 J 3) 50J, 4) 140 J, , WORK ENERGY THEOREM FOR, VARIABLE FORCE, 24. A block of mass 2 kg is initially at rest on a, horizontal frictionless surface. A horizontal, force F = (9 − x 2 ) i newton acts on it, when, the block is at x = 0. The maximum kinetic, energy of the block between x = 0m and, x = 3 m in joule is, 1) 24, 2) 20, 3) 18 4) 15, , Conservation of mechanical energy, 25. A freely falling body takes 4s to reach the, ground. One second after release, the, percentage of its potential energy, that is still, retained is, 1) 6.25% 2) 25% 3) 37.5% 4) 93.75%, 190, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 26. A vertically projected body attains the, maximum height in 6s. The ratio of kinetic, energy at the end of 3rd second to decrease in, kinetic energy in the next three seconds is, 1) 1: 1, 2) 1 : 3 3) 3 : 1 4) 9 : 1, 27. Two identical blocks A and B, each of mass, ‘m’ resting on smooth floor are connected by, a light spring of natural length L and the, spring constant k, with the spring at its, natural length. A third identical block C, (mass m) moving with a speed v along the line, joining A and B collides with A. The maximum, compression in the spring is:, 1) v, , m, 2k, , 2) m, , v, 3), 2k, , mv, 2k, , 4), , mv, 2k, , 28. A block of mass m=25kg on a smooth, r, horizontal surface with a velocity v = 3ms −1, meets the spring of spring constant, k = 100 N/m fixed at one end as shown in, figure. The maximum compression of the, spring and velocity of block as it returns to, the original position respectively are, , K, V, m, , 1) 1.5m, −3ms −1 2) 1.5m, 0ms −1, 3) 1.0m,3ms −1 4) 0.5m, 2ms −1, 29. A body is thrown vertically up with certain, initial velocity, the potential and kinetic, energies of the body are equal at a point P in, its path. If the same body is thrown with, double the velocity upwards, the ratio of, potential and kinetic energies of the body, when it crosses the same point, is, 1) 1:1, 2) 1: 4, 3) 1 : 7 4) 1:8, , POWER, 30. A machine rated as 150W, changes the, velocity of a 10kg mass from 4 ms −1 to, 10 ms −1 in 4s. The efficiency of the machine, is nearly, 1) 70% 2) 30% 3) 50%, 4) 40%, 31. A pump is required to lift 600 kg of water per, minute from a well 25m deep and to eject it, with a speed of 50 ms −1 . The power required, to perform the above task is, 1) 10 kW 2) 15 kW 3) 20 kW 4) 25 kW, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 32. A tank on the roof of a 20 m high building can, ELASTIC AND INELASTIC, hold 10 m3 of water. The tank is to be filled, COLLISIONS, from a pond on the ground in 20 minutes. If, the pump has an efficiency of 60 %, then 40. A body of mass 5kg moving with a speed of, 3ms-1 collides head on with a body of mass, the input power in kW is, 3kg moving in the opposite direction at a, 1) 1.1, 2) 2.74, 3) 5.48, 4) 7.0, speed of 2ms-1. The first body stops after the, 33. An electric fan, with effective area of crosscollision. The final velocity of the second, section ‘A’, accelerates air of density ‘ ρ ’ to a, body is, speed ‘v’. What is the power needed for this, 1) 3ms-1 2) 5ms-1 3) -9ms-1 4)30ms-1, process?, 41. Three identical particles moving with velocities, v2 4)½ ρ Av, v3, 1) ρ Av 2) ½ ρ Av 3) ρ Av, v0 i$, − 3v0 $j and 5v0 k$ collide successively with, MOTION IN A VERTICAL CIRCLE, each other in such a way that they form a single, 34. A point size mass 100 gm is rotated in a, particle. The velocity of resultant particle in i, j,, vertical circle using a cord of length 20cm., k form is, When the string makes an angle 60° with the, v, vertical , the speed of the mass is, 1) v0 $i − 3 $j + 5k$, 2) 0 $i − 3 $j + 5k$, 3, 1.5m/s. The tangential acceleration of the, –2, v, v, o $, mass in that position is ( in ms ), i − 3 $j + 5k$, 4) 0 $i + 3 $j + 5k$, 3), 2, 3, 1) 4.9 2) 4.9 2 3) 4.9 3 4) 9.8, 42. From the top of a tower of height 100m a 10, 35. A vehicle is travelling along concave road, gm block is dropped freely and a 6gm bullet, then along convex road of same radius of, is fired vertically upwards from the foot of, curvatures at uniform speed. If the normal, the tower with velocity 100ms -1, reactions on the vehicle as it crosses the, simultaneously. They collide and stick, lowest point of concave surface, highest point, together. The common velocity after, collision is (g=10ms-2), of convex surface are 1.5 × 104 N , 3 × 103 N, 1) 27.5ms-1 2) 150ms-1 3) 40ms-1 4) 100ms-1, respectively, then the mass of vehicle is(g=10, 43. A steel ball of radius 2cm is initially at rest. It, m/s–2), is struck head on by another steel ball of, 1) 400 kg 2) 450 kg 3) 800 kg 4) 900kg, radius 4cm travelling with a velocity of 81 cm/, 36. The length of a simple pendulum is 1 m. The, s. If the collision is elastic their respective, bob is given a velocity 7 ms-1 in horizontal, final velocities are, direction from mean position. During upward, 1) 63 cm/s, 144 cm/s, 2) 144 cm/s, 63 cm/s, motion of bob, if the string breaks when the, 3) 19 cm/s, 100 cm/s, 4) 100 cm/s, 19 cm/s, bob is horizontal, then the maximum vertical 44. A steel ball of radius 2cm is initially at rest. It, height of ascent of bob from rest position is, is struck head on by another steel ball of, radius 4cm travelling with a velocity of, 1. 2.5 m 2. 2 m 3. 3 m 4. 3.5 m, 81cm/s. The common velocity if it is perfectly, 37. A body is allowed to slide down a frictionless, inelastic collision, track from rest position at its top under, 1)144 cm/s 2)61 cm/s 3)81 cm/s 4) 72 cm/s, gravity. The track ends in a circular loop of, diameter D. Then, the minimum height of the, COEFFICIENT OF RESTITUTION, inclined track ( in terms of D ) so that it may 45. A tennis ball bounces down a flight of stairs,, complete successfully the loop is, striking each step in turn and rebounding to, 1) 7D/4 2) 9D/4 3) 5D/4, 4) 3D/4, half of height of the step. The coefficient of, 38. A body of mass m is rotating in a vertical, restitution is, 1/ 2, 1/ 4, circle of radius 'r' with critical speed. The, 1, 1 , 1 , difference in its K.E. at the top and at the, , , 1) 1/2 2), 3) , 4) , 2, 2, 2, bottom is, 46. A ball hits the ground and loses 20% of its, 1) 2mgr 2) 4 mgr 3) 6 mgr 4) 3 mgr, momentum. Coefficient of restitution is, 39. A simple pendulum of length 'l' carries a bob, 1) 0.2, 2) 0.4 3) 0.6 4) 0.8, of mass 'm'. If the breaking strength of the, 47. A plastic ball falling from a height 4.9m, string is 2 mg, the maximum angular, rebounds number of times. If total time for, amplitude from the vertical can be, second collision is2.4 sec, then coefficient of, 1) 0°, 2) 30°, 3) 60°, 4) 90°, restitution is, 1) 0.3, 2) 0.4 3) 0.7 4) 0.6, , (, (, , NARAYANAGROUP, , ), ), , (, (, , ), ), , 191
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 48. A ball is dropped from a height ‘h’ on to a 56. A body is thrown vertically upward from a, floor of coefficient of restitution ‘e’. The total, point ‘A’ 125 m above the ground. It goes up, distance covered by the ball just before, to a maximum height of 250 m above the, second hit is, (2008 E), ground and passes through ‘A’ on its, 2, 2, downward journey. The velocity of the body, 2) h (1 + 2e ), 1) h (1 − 2e ), when it is at a height of 70 m above the, 3) h (1 + e2 ), 4) he 2, ground is ( g = 10 ms-2 ), (2013 M), 1) 50 ms-1 2) 60 ms-1 3) 80 ms-14) 20 ms-1, 49. In two separate collisions, the coefficient of, 57. A body of mass 300 kg is moved through 10 m, restitutions e1 and e2 are in the ratio 3:1. In the, along a smooth inclined plane of inclination, first collision the relative velocity of approach is, angle 300 . The work done in moving (in joules), twice the relative velocity of separation. Then,, is ( g = 9.8 ms-2 ), (2013 M), the ratio between relative velocity of approach, 1) 4900 2) 9800 3) 14,700 4) 2450, and relative velocity of separation in the second 58. A ball of mass ‘m’ moving with a horizontal, collision is (2007 E), velocity ‘v’ strikes the bob of mass ‘m’ of a, 1) 1:6, 2) 2:3, 3) 3:2, 4) 6:1, pendulum at rest. During this collision, the, 50. A sphere of mass m moving with constant, ball sticks with the bob of the pendulum. The, velocity u, collides with another stationary, height to which the combined mass raises is, sphere of same mass. If e is the coefficient of, (g = acceleration due to gravity) (2013 M), restitution, the ratio of the final velocities of, v2, v2, v2, v2, the first and second sphere is (2007 M), 2), 3), 4), 1), 4g, 8g, g, 2g, 1+ e, 1− e, e, 1+ e, 2), 3), 4), 1), 59., The, velocity, ‘v’, reached, by, a, car, of, mass ‘m’, 1− e, 1+ e, 1− e, e, on, moving, a, certain, distance, from, the, PREVIOUS EAMCET QUESTIONS, starting, point, when, driven, by, a, motor, with, 51. A canon shell fired breaks into two equal parts, constant power ‘P’ is such that (2012 E), at its highest point. One part retraces the path, 2, to the canon with kinetic energy E1 and the, 3P, 3P, 3P, 3P , 2, 3, 1) v ∝, 2) v ∝, 3) v ∝, 4) v ∝ , kinetic energy of the second part is E2. Relation, , m, m, m, m, between E1 and E2 is, (2014 E), 60. A ball ‘A’ of mass ‘m’ moving along positive, 1) E2 = E1, 2) E2 = 4 E1, x- direction with kinetic energy K and linear, momentum “P” undergoes elastic head on, 3) E2 = 9 E1, 4) E2 = 15E1, collision with a stationary ball ‘B’ of mass M., 52. A body of 200 g begins to fall from a height, After the collision, the ball A moves along, where its potential energy is 80 J. Its velocity at, negative x- direction with kinetic energy K/9,, a point where its kinetic and potential energies, the final linear momentum of the ball B is, are equal(in m/s) ( 2014 M), ( 2012E), 1) 10 8 2) 4 3)400 4)20, 1) P, 2) P/3 3) 4P/3 4) 4P, 61., Displacement, of a body is 5i +3j - 4k m due to, 53. The work done by a force F = 2i- j-k in, the, action, of, a, force 6i + 6j + 4k N on it for 5 s., moving an object from origin to a point whose, The power in watt is, ( 2012 M ), position vector is r = 3i + 2j - 5k ( 2013 E), 1) 16, 2) 9.6 3) 6.4 4) 3.2, 1) 1 unit 2)9 units 3)13 units 4)60 units, 62. A ball at rest is dropped freely from a height of, 54. A ball at rest is dropped from a height of 12 m. If, 20 m. It loses 30% of its energy on striking the, it looses 25 % of its kinetic energy on striking, ground and bounces back. The height to which it, the ground and bounces back to a height ‘h’. the, bounces back is, ( 2012 M), value of ‘h’ is equal to ( 2013 E), 1) 14 m 2) 12 m, 3) 9 m 4) 6 m, 1) 3 m, 2)6 m 3) 9 m 4) 12 m, 63. A 3 kg sphere makes an inelastic collision, 55. A block of mass 2.9 kg is suspended from a, with another sphere at rest and they stick, string of length 50 cm and is at rest. Another, after the collision. If the composite mass, block of mass 100 g, which is moving with a, th, 1, speed of 150 m/s strikes and sticks to the first, of the initial speed, moves with a speed of, block. Subsequently when the string makes an, 4, angle of 60 0. With the vertical, the tension in, of 3 kg sphere, the mass of second sphere is, the string will be ( g = 10 ms-2 ) (2013 E ), (2012 M ), 1) 140 N 2) 135 N 3) 125 N 4) 90 N, 1) 12 kg, 2) 9 kg, 3) 6 kg, 4) 3 kg, 192, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 64. A ball is let to fall from a height h 0 . It makes, ‘n’ collisions with the horizontal ground. If, after ‘n’ collisions it rebounds with a velocity, v n and the ball rises to a height hn, then the, coefficient of restitution for the collision is, ( 2011 E), 1, , n, , n, , 67. A motor of power P0 is used to deliver water, at a certain rate through a given horizontal, pipe. To increase the rate of flow of water, through the same pipe ‘n’ times, the power of, the motor is increased to P1 . The ratio of P1, to P0 is, (2009E), 2, 3, 1) n : 1 2) n :1, 3) n :1 4) 1: n 3, 68. A body of mass 5 kg makes an elastic, collision with another body at rest and, continues to move in the original direction, after the collision with a velocity equal to 1/, 10th of its original velocity. The mass of the, second body is, (2009 E), 1) 4.09 kg 2) 0.5 kg, 3) 5 kg 4) 5.09 kg, , LEVEL - II (C.W) - KEY, 02) 2, 08)1, 14)4, 20)4, 26)1, 32)2, 38)1, 44)4, 50)2, 56)2, 62)1, 68)1, , NARAYANAGROUP, , 03) 1, 09)2, 15)4, 21)2, 27)1, 33)4, 39)3, 45)2, 51)3, 57) 3, 63) 2, , 04) 3, 10)2, 16)3, 22)1, 28)1, 34)3, 40)1, 46)4, 52) 4, 58) 2, 64) 4, , 05) 1, 11) 3, 17)2, 23)1, 29)3, 35)4, 41)2, 47)3, 53)2, 59)3, 65)2, , 1., , ur ur, W = F .S =FScos θ ; here F=mgsin α and, θ = 00 , α = 300, , 2., , Fx = F cos α , Fy = F cos β , Fz = F cos γ ,, , 3., , W = F .S, 1 2, 8, mv = 4t 2 ;, ∴v =, t, 2, m, comparing with v=at, a = constant, i.e., the force acting on the particle is constant, , 1, , h 2n, h 2, h 2, h 2n, 1) e = n 2) e = n 3) e = 0 4) e = 0 , ho , ho , hn , hn , 65. A bullet is fired normally towards an, immovable wooden block. If it loses 25% of, its kinetic energy in penetrating through the, block at thickness x, the further distance, penetrated by the bullet into the block is, ( 2011 M), 1) 2x, 2) 4x 3) 6x 4) 8x, 66. A ball is falling freely from a certain height., When it reaches 10 m height from the ground, its velocity is v0 . It collides with the, horizontal ground and loses 50% of its, energy and rises back to height of 10 m. The, value of velocity v0 is, ( 2010 E), −, 1, −, 1, −, 1, 1) 7 ms 2) 10 ms 3) 14 ms 4)16 ms −1, , 01) 3, 07)3, 13)1, 19)1, 25)4, 31)2, 37)3, 43)2, 49)4, 55) 2, 61) 3, 67) 3, , LEVEL - II (C.W) - HINTS, , 06)3, 12)3, 18)3, 24)3, 30)1, 36)1, 42)1, 48)2, 54)3, 60)3, 66)3, , 5. W = area under F - S curve., W= ∫ Fdx, 0, ur ∧ ∧, ur ur, 6., 7.W= FS cos θ, S = 5 i + 5 j, W = F.S, L, nL , 8. U i = nmg , U f = nmg , W = Ui − U f, 2, 2 , mgl, 9. W = 2, 2n, (1,1) ur uur, ^, ^, ^, W, =, F . d s ; Here uur, 10., ∫, ds, =, dx, i, +, dy, j, +, dz, k, (0,0), 4, , 4., , (1,1), , ∫ (x, , ∴W =, , 2, , dy + ydx ) =, , (0,0), , (1,1), , ∫ ( y dy + x.dx ), 2, , (0,0), , (as x = y), 11. W =, , X1, , ∫ Fdx 12., 0, , W=, , 1 2, dx, mv where v =, 2, dt, , 1 2, mv, 2, 2, 1, 1 2, , 2t , ∆, k, =, mg, Q ∆k = mv ; v = gt , 1, 14., 2, 9, 2, , , , 13. W =, , 1, 4t 2 , 2 2, ∆k 2 = mg t −, , 2, 9 , , 1, ( K .Eboy ), 2, 1 2 1 1 2 , Mu = mv (QM = 2m), 2, 2 2, , 1, ∆KE = m ( v 2 − u 2 ), 2, h1 − h2, percentage loss of energy = h ×100%, 1, 1, 1, 2, U = kx 2 &U ' = K ( Nx ), 2, 2, 1 2, Kx = KE1 + KE2 and m1v1 = m2 v2, 2, , 15. K .Eman =, , 16., 17., 18., 19., , 193
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 52. T.E at top = T.E at given point ; P.E=P.E1+K.E1, 80=2K.E1 (Q P.E 1 = K .E 1 ), ur ur ur ur ur, 53 W = F .S = F .(r2 − r1 ), 54. percentage loss of PE, 55., , =, , Q, 5m, 5N, , mu, mv 2, and T = mg cos θ +, M +m, r, 2, Where v = vbottom − 2 gr (1 − cos θ ), , P, , vcommon =, , dv, , P, , 2, 2, 59. P = F .v; P = mv, ; v dv = ∫ ds, m, ds ∫, 60. Law of conservation of linear momentum., mu1 = − mv1 + PB, , 4., , 11, , 2, m v 0 + m g h1 = m g h 2, 22, , , 5., , 6., , m −m , , 1, 2, 67. P = Aρ v3 ⇒ Pα v 3 68. v1 = m + m u1, 1, 2 , , 7., , LEVEL-II (H.W), WORK DONE BY CONSTANT FORCE, 1. A bicycle chain of length 1.6 m and of mass 1, kg is lying on a horizontal floor. If g=10ms-2,, the work done in lifting it with one end, touching the floor and the other end 1.6 m, above the floor is, 1) 10 J 2) 3.2 J, 3) 8 J 4) 16 J, 2. A bucket of mass ‘m’ tied to a light rope is, lowered at a constant acceleration of g/4. If, the bucket is lowered by a distance ‘d’, the, work done by the rope will be (neglect the, mass of the rope), 1, 3, 3, 5, 1) mgd 2) mgd 3) − mgd 4) − mgd, 4, 4, 4, 4, NARAYANAGROUP, , 4m, , 8., , R, , What is the work done in joules?, 1) 15, 2) 20, 3) 25, 4)35, A 5 kg stone of relative density 3 is resting at, the bed of a lake. It is raised through a height, of 5 m in the lake. If g = 10 m/s2, then work, done is, 350, , s, P = F .v = F . , t , , h1 − h2, 62. percentage loss of PE = h ×100%, 1, 63. m u = ( M + m ) v, 64., hn = e 2n h0, 65. Work - energy theorem; W = FS cos θ = ∆K .E, Sα K f − K i, , 66., , A weight of 5 N is moved up a frictionless, inclined plane from R to Q as shown., , h1 − h2, × 100%, h1, , 56. v A = 2 gh here h = 125m, Velocity at a distance 55m below ‘A’ is, v 2 = v 2A + 2 gh1 here h1 = 55m, 57. W = mgl sin θ, v2 2, h, =, 58. Mv = ( M + m ) v2 ;, 2g, , 61., , 3., , 750, , 550, , J, J 4), J, 1) 500 J 2), 3), 3, 3, 3, 3, Water is drawn from a well in a 5 kg drum of, capacity 55 L by two ropes connected to the, top of the drum. The linear density of each, rope is 0.5 kgm-1. The work done in lifting, water to the ground from the surface of water, in the well 20 m below is (g = 10 ms-2), 1)1.4×104J, 2)1.5×104J, 3)9.8×6×10 J, 4) 18 J, A ball is dropped from the top of a tower.The, ratio of work done by force of gravity in 1st,, 2nd, and 3rd second of the motion of ball is, 1) 1:2:3 2)1:4:16 3)1:3:5 4) 1:9:25, A plate of mass m, breadth ‘a’ and length ‘b’is, initially lying on a horizontal floor with length, parallel to the floor and breadth, perpendicular to the floor. The work done to, erect it on its breadth is, b, b, , 1) mg, 2) mg a + 2 , , , 2, b−a, b+a, 3) mg , 4) mg 2 , , , , 2 , A block of mass 10 kg slides down a rough, slope which is inclined at 450 to the horizontal., The coefficient of sliding friction is 0.30., When the block has to slide 5 m, the work, done on the block by the force of friction is, nearly, 1)115J 2)-75 2 J 3) 321.4J 4) -321.4 J, 195
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 9., , A uniform rope of length ‘L and linear density 17. The kinetic energy of a projectile at the, highest point of its path is found to be 3/4th of, ‘ µ ’ is on a smooth horizontal table with a, its initial kinetic energy. If the body is, length ‘l’ lying on the table. The work done in, projected from the ground, the angle of, pulling the hanging part on to the table is, projection is, 1)00, 2)300, 3)600, 4) 400, 2, 2, µg ( L − l), 18. The kinetic energy of a moving body is given, 1) µ g ( L − l ), 2), 2l 2, by k = 2v2, k being in joules and v in m/s. It’s, 2, momentum when travelling with a velocity of, 2, µ gL, µg (L − l), 2 m/s will be (in kgms-1 ), 3), 4), 2(L − l), 1)16, 2) 4, 3) 8, 4) 2, 2 L2, POTENTIAL, ENERGY, 10. A uniform rod of mass 2 kg and length l is, lying on a horizontal surface. If the work done 19. A simple pendulum is swinging in vertical, plane. The ratio of its potential energy when, in raising one end of the rod through an angle, it is making 450 and 900 with the, 450 is ‘W’, then the work done in raising it, vertical is, further 450 is, 1)1:1 2)1: 2 + 1 3) 2 :1 4) 2 − 1 : 2, W, 1) W, 2) 2 W, 3), 4) 2 − 1 W, POTENTIAL ENERGY OF A SPRING, 2, 20. A spring of force constant 800 Nm-1 is, WORK DONE BY VARIABLE FORCE, stretched initially by 5 cm. The work, 11. A block is constrained to move along x-axis, done in stretching from 5 cm to 15 cm is, under a force F= - (2x)N. Find the work done, 1) 12.50 N-m, 2) 18.75 N-m, by the force when the block is displaced from, 3) 25.00 N-m, 4) 6.25 N-m, x = 2m to x = 4m, 21. When a spring is compressed by a distance, 1)12J, 2)8J 3)-12J, 4)-8J, ‘x’, the potential energy stored is U1. It is, 2, 12. A force of (4x +3x)N acts on a particle which, further compressed by a distance ‘2x’, the, displaces it from x=2m to x= 3m. The work, increase in potential energy is U2. The ratio, done by the force is, of U1:U2 is, 1)32.8J 2)3.28J 3)0.328J 4)Zero, 1)1:7, 2) 1:4, 3)1:8, 4)1:3, 13. A body of mass 6kg is under a force which, causes a displacement in it which is given by 22. A massless spring with a force constant, K=40N/m hangs vertically from the ceiling., t2, A 0.2kg block is attached to the end of the, S = m, where ‘t’ is time. The work done by, 4, spring and held in such a position that the, the force in 2sec is, spring has its natural length and suddenly, 1)12J, 2)9J, 3)6J, 4)3J, released.The maximum elastic strain energy, KINETIC ENERGY, stored in the spring is(g=10m/s2), 1)0.1J, 2)0.2J, 3)0.05J 4) 0.4J, 14. Two spheres of same material are moving, with kinetic energies in the ratio 108:576. If, WORK ENERGY THEOREM BY, the ratio of their velocities is 2:3 , then the, CONSTANT FORCE, ratio of their radii is, 23. A bulletofm ass‘m ’isfired w ith a velocity‘v’, 1)1:1, 2)2:3 3) 3:4, 4) 4:3, into a fixed log of wood and penetrates a, 15. If the momentum of a body decreases by, distance ‘s’ before coming to rest. Assuming, 30%, then kinetic energy decreases by, that the path of the bullet in the log of wood is, 1) 60%, 2) 51% 3) 69% 4) 90%, horizontal, the average resistance offered by, 16. If the mass of a moving body decreased by, the log of wood is, one third of its initial mass and velocity is, tripled, then the percentage change in its, mv, 2s, mv 2, ms 2, kinetic energy is, 1) 2, 2), 3), 4), 2s, mv 2, 2s, 2v, 1)500% 2) 600% 3) 300%, 4) 200%, , (, , 196, , ), , (, , ), , (, , ), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 24. A ball of mass ‘m’ is thrown in air with speed, v1 from a height h 1 and it is caught at a height, h2> h1 when its speed becomes v2. Find the, work done on the ball by air resistance., 1, 2, 2, 1) mg ( h2 − h1 ) + m ( v2 − v1 ), 2, 1, 2, 2, 2) mg ( h2 − h1 ) 3) m ( v2 − v1 ), 2, 1, 2, 2, 4) mg ( h2 − h1 ) − m ( v2 − v1 ), 2, 25. An object of mass 5 kg falls from rest, through a vertical distance of 20 m and, attains a velocity of 10 m/s. How much work, is done by the resistance of the air on the, object (g = 10 m/s2), 1)750 J, 2)– 750 J 3) 850 J 4)- 650 J, 26. The velocity of a 2 kg body is changed from, ^, , 30., , 31., , 32., , ^, , -1, -1, (4 i + 3 j ) ms to 6 ms . The work done on, the body is, 1)9 J, 2)11 J 3)1 J, 4) 5 J, 27. An out fielder throws a cricket ball with, an initial kinetic energy of 800J and an, infielder catches the ball when its kinetic, energy is 600J. If the path of the ball, between them is assumed straight and is, 20m long, the air resistance acting on, the ball is, 1)26.6N 2)1.33N, 3)100N 4)10N, 28. Velocity -time graph of a particle of, mass 2kg moving in a straight line is as, shown in the figure. Work done by all, the forces on the particle is, , V(m/s), , 33., , 34., , 35., , 20, , 36., O, , 2, , t(s), , 1)400J, 2)-400J, 3)-200J 4)200J, 29. A block of mass of 1kg slides down a curved, track that is one -quadrant of circle of radius, 1m. Its speed at the bottom is 2m/s. The, workdone by the frictional force is, , 1)8J, , 2)-8J, , NARAYANAGROUP, , 3)4J, , 4)-4J, , WORK ENERGY THEOREM FOR, VARIABLE FORCE:, A block of mass 4kg is initially at rest on, a horizontal frictionless surface.A force, ur, F = ( 2 x + 3x2 ) i$ N acts horizontally on it. The, maximum kinetic energy of the block, between x=2m and x=4m in joules is, 1)40, 2) 36, 3)68, 4) 52, A force F=Ay2+By+C acts on a body at, rest in the Y-direction. The kinetic, energy of the body during a displacement, y = - a to y = a is, 2 Aa 3, 2 Aa 3, + 2ca, 1), 2), 3, 3, 2 Aa 3 Ba 2, 2 Aa3 Ba 2, +, +, ca, +, 3), 4), 3, 2, 3, 2, LAW OF CONSERVATION OF, MECHANICAL ENERGY, A 3kg model rocket is launched straight up, with sufficient initial speed to reach a, maximum height of 100 m, even though air, resistance (a non-conservative force), performs - 900 J of work on the rocket. The, height the rocket would have gone without air, resistance will be, 1) 70 m 2) 130 m 3) 180 m 4) 230 m, A body of mass 2kg is thrown up vertically with, kinetic energy of 490J. If g = 9.8m / s 2 , the, height at which the kinetic energy of the body, becomes half of the original value, is (2007 M), 1) 50m, 2) 25m, 3) 12.5m 4) 19.6 m, A simple pendulum bob has a mass “m” and, length “L”. The bob is drawn aside such that, the string is horizontal and then it is released., The velocity of the bob while it crosses the, equilibrium position is, 3) 5 gL 4) 3gL, 1) gL 2) 2 gL, A 100 gm light bulb dropped from a tower, reaches a velocity of 20 m/s after falling, through 100 m. The energy transferred to the, air due to viscous force is, 1) 98 J 2) 20 J 3) 118 J, 4) 78 J, In the arrangement shown in figure , string, is light and inextensible and friction is, absent every where .The speed of both, blocks after the block ‘A’ has ascend a, height of 1m will be (g=10m/s2), , A, 1kg, , B, 2kg, , 1)2m/s 2) 2.58m/s 3) 3m/s 4) 3.58 m/s, 197
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , POWER, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 44. A pilot of mass m can withstand a maximum, apparent weight 6 times of mg. The minimum, radius of curvature of vertical circle in which the, aeroplane dives up from lowest point with a, speed 504 kmph is, 1) 200 m 2) 300 m, 3) 400 m 4) 500m, 45. A simple pendulum is oscillating with an, angular amplitude 900. If the direction of, resultant acceleration of the bob is horizontal, at a point where angle made by the string, with vertical is, −1 1 , −1 1 , 2) cos , 1) sin , 3, 3, , 37. A car drives along a straight level frictionless, road by an engine delivering constant power., Then velocity is directly proportional to, 1, 3) t, 4) t 2, 1) t, 2), t, 38. The input power to an electric motor is 200KW., Its efficiency is 80%. It operates a crane of, efficiency 90%. If the crane is lifting a load of, 3.6 tonnes, the velocity with which the load, moves is, 1) 8 ms-1 2) 4 ms-1 3) 2 ms-1 4) 40 ms-1, 39. The human heart discharges 75 cm3 of blood, per beat against an average pressure of, −1 1 , −1 1 , 10 cm of Hg. Assuming that the pulse, 3) sin , 4) cos , 3, 3, frequency is 75 per minute, the power of the, 46. A steel wire can withstand a load up to 2940, heart is (density of Hg = 13.6 gm cm-3), N. A load of 150 kg is suspended from a rigid, 1) 1.25W 2) 12.5W, 3) 0.125W, support. The maximum angle through which, 4)125W, the wire can be displaced from the mean, 40. An elevator can carry a maximum load of, position, so that the wire does not break when, 1800kg(elevator+passengers) is moving up, the load passes through the position of, with a constant speed of 2m/s. The frictional, equilibrium, is (2008 E), force opposing the motion is 400N., 1) 30°, 2) 60° 3) 80° 4) 85°, Determine the minimum power delivered by, 47., A, small, block, is freely sliding down from top, the motor to the elevator( in horse power)., of, a, smooth, inclined, plane. The block reaches, 1) 59, 2)8, 3)22, 4)20, bottom, of, inclined, plane then the block, 41. A body is initially at rest. It undergoes, describes, a, vertical, circle of radius 0.5m, one - dimensional motion with constant, along, smooth, track., The, minimum vertical, acceleration. The power delivered to it at, height of inclined plane should be, time t is proportional to, 1) 1m, 2) 1.25 m, 3) 3m 4) 2.5 m, 1, 2, 48. A stone of mass 6 kg is revolved in a vertical, 2) t, 3) t 3, 4) t 2, 1) t 2, circle of diameter 6m., such that its speed is, 42. A dam is situated at a height of 550 m above, minimum at a point. If the K.E at the same point, sea level and supplies water to a power house, is 250 J , then minimum PE at that point is, which is at a height of 50 m above sea level., 1) 200J 2) 150J, 3) 100J, 4, ), 2000 kg of water passes through the turbines, per second. What would be the maximum, 450J, electrical power output of the power house if 49. The breaking strength of a string is 55 kg wt., the whole system were 80% efficient, The maximum permissible speed of a stone of, 1)8MW 2)10MW 3)12.5MW 4)16MW, mass 5 kg which is revolved in a vertical, MOTION IN A VERTICAL CIRCLE, circle of radius 4m with the help of, 43. A stone tied to a string of length ‘L’ is whirled in, this string is (g = 10m/s2), a vertical circle with the other end of the string, 1) 10m/sec, 2)15m/sec 3) 20 m/sec, at the centre. At a certain instant of time, the, 4)25m/sec, stone is at its lowest position and has a speed u., ELASTIC AND INELASTIC COLLISIONS, The magnitude of the change in its velocity as it 50. A 16 gm mass is moving in the +x direction at, reaches a position where the string is horizontal, 30 cm/s while a 4 gm is moving in the -x, direction at 50 cm/s. They collide head - on, is, and stick together. Their common velocity, after impact is, 1) 2, 2) 2gL, u − 2 gL, 1) 0.14cm/s 2)0.14 m/s 3) 0 ms −1 4) 0.3 m/s, 3) u 2 − gL, 4) 2( u 2 − gL ), 198, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 51. A bullet of mass 50 grams going at a speed of, 200 ms −1 strikes a wood block of mass 950, gm and gets embedded in it. The velocity of, the block after the impact is, 1) 5 ms −1 2) 10 ms −1 3) 20 ms −1 4) 50 ms −1, 52. A block of mass 1 kg moving with a speed of, 4 ms −1 , collides with another block of mass 2, kg which is at rest. If the lighter block comes, to rest after collision, then the speed of the, heavier body is, −1, 1) 2 ms −1 2) 1 ms 3) 1.5 ms −1 4) 0.5 ms −1, COEFFICIENT OF RESTITUTION, 53. A neutron travelling with a velocity v and, kinetic energy E collides perfectly elastically, head on with the nucleus of an atom of mass, number A at rest. The fraction of the total, kinetic energy retained by the neutron is, 2, , 2, , 2, , A +1, A −1 , A +1, A −1 , 2) , 3) , 4) , , 1) , A, +, 1, A, −, 1, A, , , , , , , A , , 2, , 54. Two balls each of mass ‘m’ are moving with, same velocity v on a smooth surface as shown, in figure. If all collisions between the balls, and balls with the wall are perfectly elastic,, the possible number of collisions between the, balls and wall together is, (2008 M), , v, , v, , m, , 1) 1, , m, , 2) 2, , 3) 3, , 4) Infinity, , LEVEL - II(H.W)-KEY, 1) 3, 7) 3, 13)4, 19)4, 25)2, 31)2, 37)3, 43)4, 49)3, , 2) 3, 8) 2, 14)3, 20)3, 26)2, 32)2, 38)2, 44)3, 50)2, , 3) 1, 9) 1, 15)2, 21)3, 27)4, 33)3, 39)1, 45)4, 51)2, , 4) 1, 10)4, 16)1, 22)3, 28)2, 34)2, 40)1, 46)2, 52)1, , 5) 1, 11)3, 17)2, 23)2, 29)2, 35)4, 41) 2, 47) 2, 53)1, , 6) 3, 12)1, 18)3, 24)1, 30)3, 36)2, 42)1, 48)1, 54)3, , LEVEL-II-(H.W) - HINTS, 1., 2., , l, ur ur, 0, W= F .S = FS cos θ ; here F=mg; S = , θ = 0, 2, W = FS cos θ here F=m(g-a),S=h, θ = 180 0, , 3., 5., 6., , 7., 8., , dL , 4. W = mgh 1 − d , , B , h, , W =Mgh + ρ wVgh + mg 2, 2, , ( 2n − 1)1, W, S, W = FS ⇒ 1 = 1 =, W2 S2 ( 2n − 1)2, , W = mgh, , , 1 , , Q S = u + a n − 2 , , , , W = mg ( h2 − h1 ), , W = − f × S = − µ mg cos θ × S, (L − l), 1, 9. W = m gh = µ ( L − l ) g, 2, Where m1 is mass of hanging part, l, W1 = W = mg sin θ1 here θ1 = 450, 10., 2, l, W 2 = m g sin θ 2 here θ 2 = 90 0, 2, required work = W 2 − W 1, xf, , 11. W =, , ∫ f ( x )dx, , W=, , 12., , xi, , ∫ f ( x )dx, , xi, , 1 2, ds, mv ; v =, 2, dt, 1 2 1 4 3 2, 14. K = mv = ρ π r v, 2, 2 3, , 2, 2, P, K, P, ⇒ 1 = 12, 15. K =, 2m, K2 P2, 13. W =, , 1 2, K1 m1v12, =, 16. K = mv ⇒, 2, K 2 m2v22, 1, 3 1, 2, , 3, ( K .Ei ) ; 2 m ( u cos θ ) = 4 2 mu 2 , 4, , , 1 2, 2, 18. K = 2v = mv and P = mv, 2, 1, 2, 2, 19. U = mgl (1 − cosθ ) 20. W = K ( x2 − x1 ), 2, 1 2, 1, mg, PE = kx 2 ; k =, 21. U = kx, 22., 2, 2, x, 23. W= F × S = ∆K . E 24. Wmg + Wair = ∆K . E ., , 17. K .Etop =, , 25. Wmg + Wair = ∆K .E ., 26. Work done = change in kinetic energy, 27. W= − FS = K 2 − K1, , NARAYANAGROUP, , xf, , 28. W = K 2 − K1, 199
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JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, xf, , 29. W friction = K .E f − P.Ei 30. K .E = W =, , ∫ f ( x)dx, , xi, , yf, , 31. K .E = W =, , ∫ F ( y )dy, , 50. v =, , 32. mgh + W f = mgH, , yi, , 34. Loss of P.E= Gain in K.E ; mgL =, , m u + m u2, , 1 1, 2, 51 v = m + m, 1, 2, , v, , WORK DONE BY A FORCE, , ur r, 40. F = mg+frictional force ;, P = F .v, dv, = cons tan t, 41. F = ma = constant; a =, dt, P = Fv and v=u+at, 80 dm ( gh ), 42. Power output =, 100 dt, 43. Using K.EB=P.EH+K.EH find v, ^, ^, uuur r r, ∆V = v − u = v j − u i ;, , A long rod ABC of mass “m” and length “L”, has two particles of masses “m” and “2m”, attached to it as shown in the figure. The, system is initially in the horizontal position., The work to be done to keep it vertical with A, hinged at the bottom is, , A, , B, L/2, , 2., , ∆V = v 2 + ( − u ) = v 2 + u 2 ;, 2, , mv 2, + mg = 6mg, r, ac, v2 / r, tan, θ, =, =, 45., at g sin θ, 2, but v = 2 gh ;, h = r cos θ, , 44. Tmax =, , 3., , m L/2, , C, 2m, , 1) 2mgL 2) 3mgL/2, 3) 5mgL/2 4) 3mgL, A particle of mass 100g is thrown vertically, upwards with a speed of 5m/s. The work done, by the force of gravity during the time the, particle goes up is, 3) 1.25J 4) 0.5J, 1) −0.5J 2) −1.25J, A large slab of mass 5 kg lies on a smooth, horizontal surface, with a block of mass 4 kg, lying on the top of it. The coefficient of, friction between the block and the slab is, 0.25. If the block is pulled horizontally by a, force of F = 6 N, then the work done by the, force of friction on the slab, between the, instants t=2s to t=3s is ( g = 10 ms −2 ), , 4kg, , ), , 48. T .Ebottom = T .Egiven po int ; m ( 5 gr ) = P.E + K .E, 2, , (Ball 2 rebounds), (Both balls rebounds), (Ball 2 rebounds), , LEVEL-III, 1., , (, , v, m 2, , Ball 2 with wall - 1st, Ball 2 with ball 1 - 2nd, Ball 2 with wall - 3rd, , W Pr essure× volume, P= =, t, t, , 46. While crossing the equilibrium position, mv 2, m, T = mg +, = mg + 2 gl (1 − cos θ ) , r, l, = mg + 2mg − 2mg cos θ = mg [3 − 2cos θ ], 1, 47. Loss of P.E =Gain in K.E ; mgh = m 5rg, 2, , m 1, , 54., , 1, 1, 2, 2, 36. mB gh = m A gh + m Av + mB v, 2, 2, dv, P, 37. P = Fv = m v ; Hence, vdv = dt, dt, m, On integration, we find v ∝ t, 72, P = FV here F = mg, 38. η = η1 ×η 2 ;, 100, , 1, 2, , 2, , 1, MV2, 2, , 35. W=P.Ei-K.Ef, , 200, , m1u1 + m2u2, m1 + m2, , KE1 m1 − m2 , 52. m1u1 + m2u2 = m1v1 + m2v2 53. KE = m + m , 1, 2 , , 33. At projection, T.E=K.E=490, at given point, P.E1+K.E1=T.E1, K .E, mgh +, = T .E, 2, , 39., , 2, 49. Tmax = m v + g , , , r, , , 2, , 5kg, 1) 2.4 J, , 2) 5.55 J 3) 4.44J, , 4) 10 J, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, 4., , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , In the pulley – block system shown in figure,, strings are light. Pulleys are massless and, smooth. System is released from rest. In 0.3, seconds, 5kg, , 2kg, , F, , F, , 2), , 1), , 2kg, , time, , time, F, , 1kg, , 5., , a) work done on 2 kg block by gravity is 6J, b) work done on 2 kg block by string is -2J, c) work done on 1 kg block by gravity, is –1.5J, d) work done on 1 kg block by string is 2 J, 1) only a, d are correct, 2) only b, d are correct, 3) only a, b, c are correct, 4) All are correct, A body of mass 0.5 kg travels in a straight, 3, , line with a velocity v = ax 2 where, −, , 1, 2 −1, , a = 5m s . What is the work done by the, net force during its displacement from, x = 0 to x = 2m?, 1) 50 J 2) 20 J 3) 80 J 4) 45.5 J, , F, , 4), , 3), time, , time, , WORK - ENERGY THEOREM, 9., , A lifting machine, having an efficiency of 80%, uses 2500 J of energy in lifting a 10 kg load over, a certain height. If the load is now allowed to fall, through that height freely, its velocity at the end, of the fall will be (g=10 m/s2 ), 1) 10 m s-1 2) 15 m s-1 3) 20 m s-1 4) 25 m s-1, 10. A chain AB of length L is lying in a smooth, horizontal tube so that a fraction h of its length, L, hangs freely and touches the surface of the, table with its end B. At a certain moment, the, end A of the chain is set free. The velocity of, end A of the chain, when it slips out of tube , is, , P.E AND K.E, 6., , 7., , 8., , A particle of mass 2 kg starts moving in a, straight line with an initial velocity of 2 m/s at, a constant acceleration of 2m/ s 2 . Then rate, of change of kinetic energy, 1) is four times the velocity at any moment 2) is, two times the displacement at any moment, 3) is four times the rate of change of velocity at any, moment, 4) is constant throughout, A running man has half the kinetic energy that a, running boy of half his mass has. The man, speeds up by 1.0 m/s and now has the same, kinetic energy as the boy. The original speed of, the man expressed in m/s units must be, 1) 2 + 1 2) 2 − 1 3) 2 + 2 4) 2 − 2, The kinetic energy (KE) versus time graph, for a particle moving along a straight line is, shown in the figure. The force vs time graph, for the particle may be, , NARAYANAGROUP, , A, , h, B, , 1) h, , 2g, L, 2) 2 gh log e h 3), , Lh, , 1, L, 2 gl loge 4), h, hL, , 2g, , 11. A block of mass m = 1 kg moving on a, horizontal surface with speed υ i = 2 ms −1, enters a rough patch ranging from x = 0 .10m, −k, , to x = 2.01m. The retarding force Fr =, for, x, 0.1m < x < 2.01 m,, Fr =0 for x < 0 .1m, and x > 2.01 m where k = 0.5 J, the final, speed υ f of the block as it crosses this patch, 1) 2ms −1 2) 1ms −1 3) 3ms −1 4) 0.5ms −1, 201
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, 12. A 1.5kg block is initially at rest on a, horizontal frictionless surface. A horizontal, ^, ur, force F = ( 4 − x 2 ) i is applied on the block ., Initial position of the block is at x = 0. The, maximum kinetic energy of the block, between x = 0 and x = 2m is, 1) 2.33J 2) 8.67 J 3) 5.33 J 4) 6.67 J, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , negligible mass and the pulley is frictionless., ( g = 10m / s 2 ), , K=100N/m, , 1KG, , CONSERVATION OF, MECHANICAL ENERGY, , 37O, , 13. The bob of a pendulum is released from a, 1) 0.125 2) 1.25 3) 5.2 4) 4.5, horizontal position. If the length of the 17. A light spring of force constant 'K' is held, pendulum is 1.5 m, the speed with which the, between two blocks of masses 'm' and '2m'., bob arrives at the lowest point, given that it, The two blocks and the spring system rests, dissipated 5% of its initial energy against air, on a smooth horizontal floor. Now the blocks, resistance is (m/s), are moved towards each other compressing, 1)3.14, 2) 5.28 3) 1.54, 4) 8.26, the spring by 'x' and suddenly released. The, 14. System shown in fig is released from rest, relative velocity between the blocks when the, with mass 2kg in contact with the ground., spring attains its natural length will be, Pulley and spring are massless and the, 3K , 2K , K , friction is absent everywhere. The speed of, 2) 3m x, 3) 3m x, 1) 2 m x, , , , , , , 5kg block when 2kg block leaves the contact, K , with the ground is ( force constant of the, 4) 2m x, 2, , , spring k=40 N/m and g = 10m / s ), 18. A ball of mass m is released from A inside a, smooth wedge of mass m as shown in fig., What is the speed of the wedge when the ball, reaches point B?, A, R, 45, , 0, , B, , 5kg, , 2kg, smooth, , 1) 2 m/s 2) 2 2 m/s 3) 2 m/s 4) 2 m/s, 1, 1, 2, 2, 3, 5, gR, , , gR, , , 15. The potential energy of a particle of mass m, 1) , 2) 2gR 3) 2 3 4) 2 gR, 1 2, , , 3 2 , is given by U = kx for x < 0 and U=0 for, POWER, 2, ., If, total, mechanical, energy of the 19. Power supplied to a particle of mass 2kg, x≥0, varies with time as P = 3t 2 / 2 W. Here t is in, 2E, second. If velocity of particle at t=0 is v=0,, is, particle is E. Then its speed at x =, k, the velocity of particle at time t= 2s will be, 2E, E, E, 1) 1 m/s 2) 4 m/s, 3) 2 m/s 4) 2 2 m/s, 1) zero 2), 3), 4), m, m, 2m, 20. A particle of mass m is moving in a circular, 16. A 1 kg block situated on a rough inclined, path of constant radius r such that its, plane is connected to spring of a spring, centripetal acceleration ac is varying with, constant 100 Nm-1 as shown in fig. The block, time t as ac = k 2 rt 2 where k is a constant. The, is released from rest with the spring in the, unstretched position. The block moves 10 cm, power delivered to the particle by the forces, down the incline before coming to rest. Find, acting on it, is, the coefficient of friction between the block, 1) zero 2) mk 2r 2t 2, 3) mk 2r 2t 4) mk 2 rt, and the incline. Assume that the spring has a, 202, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 21. A constant power P is applied to a particle of, mass m. The distance travelled by the, particle when its velocity increases from, v1 to v2 is (neglect friction), 3P 2 2, m, v2 − v1 ), ( v2 − v1 ), (, 2), m, 3P, m 3 3, m 2 2, v2 − v1 ), (, ( v2 − v1 ), 4), 3), 3P, 3P, 22. A body is moved from rest along a straight, line by a machine delivering constant power., The ratio of displacement and velocity (s/v), varies with time t as, , 1), , t, , t, , 2), , 1), , s/v, , s/v, t, , stopper is ejected out horizontally while test, tube completes a vertical circle of radius 25, cm. The minimum velocity with which stopper, should be ejected out is, 1) 72 kmph, 2) 90 kmph, 3) 180 kmph, 4) 360 kmph, 26. A nail is fixed at a point P vertically below the, point of suspension 'O' of a simple pendulum, of length 1m. The bob is released when the, string of pendulum makes an angle 300 with, horizontal. The bob reaches lowest point and, then describes vertical circle whose centre, coincides with P. The least distance of P from, O is, 1) 0.4 m 2) 0.5 m 3) 0.6 m 4) 0.8 m, 27. A simple pendulum with a bob of mass 'm', swings with angular amplitude of 60°. When its, angular displacement is 30°, the tension in the, string would be, , t, 4), , 28., , s/v, , s/v, , 23. Power applied to a particle varies with time as, P=(3t2-2t+1) watt, where ‘t’ is in second. Find, the change in kinetic energy between t=2s, and t=4s, 1)32J, 2)46J 3)61J 4)100J, 24. A car of mass M accelerates starting from, rest. Velocity of the car is given by, , 29., , 1, , 2Pt 2, v=, where P is the constant power, M , supplied by the engine. The position of car as, a function of time is given as, 1, , 3, 8P 2 2, t, 1) , , 9M , , 1, , 8P 2 3, 3) , t, 9M , 2, , 1, , 3, 9P 2 2, t, 2) , , 8M , , 9P, 4) , 8M, , 30., , 3, t, , , MOTION IN VERTICAL CIRCLE, 25. Dry gas of negligible mass is sealed in a test, tube of mass 50 gm with the help of a stopper, of mass 3.5 gm. The test tube is suspended, from a fixed point with help of massless string, such that the test tube is horizontal and, distance between point of suspension and, centre of mass of test tube is 25 cm. The test, tube is heated to a temperature due to which, NARAYANAGROUP, , 31., , mg, (3 3 − 2), 2, , 3) 2 mg 3 + 2 , 4) 2 mg ( 3 − 2 ), A block is freely sliding down from a vertical, height 4 m on smooth inclined plane. The block, reaches bottom of inclined plane and then it, describes vertical circle of radius 1 m along, smooth track. The ratio of normal reactions on, the block while it is crossing lowest point and, highest point of vertical circle is, 1) 6 : 1 2) 5 : 1, 3) 3 : 1 4) 5 : 2, The length of a ballistic pendulum is 1 m and, mass of its block is 1.9 kg. A bullet of mass 0.1, kg strikes the block in horizontal direction with, a velocity 100 ms–1 and got embedded in the, block. After collision the combined mass, swings away from lowest point. The tension in, the string when it makes an angle 60° with, vertical is ( g = 10 ms–2 ), 1) 20 N 2) 30 N, 3) 40 N 4) 50 N, A stone attached to a string is rotated in a, vertical circle such that when it is at the top of, the circle its speed is V and there is neither, tension nor slacking in the string. The speed, of stone when its angular displacement is 120°, from the lowest point is, 3, 3, V, 2) 2 V 3) 3 V 4), 1) V, 2, 2, Mass of the bob of a simple pendulum of, length L is m. If the bob is projected, horizontally from its mean position with, velocity 4gL , then the tension in the string, becomes zero after a vertical displacement of, 1) L/3, 2) 3L/4, 3) 4L/3 4) 5L/3, 1, , 3), , 2), , 1) 3 3mg, , , 3, , , , 1, , 203
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 32. A bob of mass M is suspended by a massless 36. A particle of mass m has a velocity –v0i, while, a second particle of same mass has a velocity, string of length L . The horizontal velocity v, v0 j. After the particles collide, first particle is, at position A is just sufficient to make it, −1, reach the point B . The angle θ at which the, v 0 i then the, found, to, have, a, velocity, speed of the bob is half of that at A , satisfies, 2, B’, O, velocity of other particle is, B, r, r, −1 r, 1 r, α, v0 i + v0 j, 1), 2) v 0 i + v 0 j, 2, 2, r, r, r, r, 4) − v 0 i + v 0 j, 3) v 0 i + v 0 j, C, 37. At high altitude a body at rest explodes into, two fragments of equal masses with one, A, π, 3π, π π, fragment receiving horizontal velocity of, 3π, π, <θ < π, 1) θ = 2) < θ < 3) < θ <, 4), 10 ms-1. Time taken by the two radius vectors, 2, 4, 4 4, 2, 4, connecting point of explosion to fragment to, 33. A simple pendulum is vibrating with an, make 900 is (g=10ms-2), angular amplitude of 900 as shown in the, 1) 10 s 2) 4 s 3) 2 s, 4) 1 s, figure. For what value of α is the, 38. A test tube of mass 20 gm is filled with a gas, acceleration directed, and fitted with a stopper of 2gm. It is, B’, , O, , B, , suspended horizontally by means of a thread, of 1m length and heated. When the stopper, kicks out, the tube just completes a circle in, vertical plane. The velocity with which the, C, stopper kicked out is, A, 1) 7ms-1 2) 10ms-1, 3) 70ms-1 4)0.1ms-1, i) vertically upwards, ii) horizontally, 39. Two bodies move towards each other and, iii) vertically downwards, collide inelastically. The velocity of the first, 0, 0, 0, 0, −1 1 , −1 1 , 0, ,cos, ,90, 90, ,cos, ,0, body is 2m/s and that of the second is 4m/sec, 1), 2), , , 3, 3, before impact.The common velocity after, collision is 1m/s in the direction of the first, 0, 0, 0, 0, −1 1 , −1 1 , cos, ,0, ,90, cos, ,90, ,0, body. The number of times did the KE of the, 3), 4), , , 3, 3, first body exceed that of the second body, COLLISIONS, before collision., 34. A block of wood of mass 3M is suspended by, 1) 4.25 2) 3.25 3) 2.25 4) 1.25, 10, 40. Three particles A, B and C of equal masses,, m. A bullet of mass M, a string of length, moving with the same speed ‘v’ along the, 3, medians of an equilateral triangle, collide at, hits it with a certain velocity and gets, the centroid G of the triangle. After collision,, embedded in it. The block and the bullet, A comes to rest and B retraces its path with a, swing to one side till the string makes 1200, with the initial position. The velocity of the, speed ‘v’ . The speed of C after the collision is, y-axis, v, bullet is (g=10ms–2), 40 −1, ms 2) 20ms–1, 1), 3) 30ms–1 4) 40ms–1, 3, 35. A wooden block of mass 10gm is dropped, θ, from the top of a cliff 100m high., x-axis, θ, Simultaneously a bullet of same mass is fired, m, m, from the foot of the cliff vertically upwards, with a velocity of 100ms-1. If the bullet after, collision gets embedded in the block, the, common velocity of the bullet and the block, v, immediately after collision is (g=10 ms-2), v, 1) v along BG, 2) along GB, 1) 40 ms-1 downward, 2) 40ms-1 upward, 2, 3) 80ms-1 upward, 4) zero, 3) Zero, 4) v along CG, α, , 1, , 1, , 1, , 2, , 2, , 2, , 204, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 41. A moving sphere P collides another sphere Q, at rest. If the collision takes place along the, line joining their centers of mass such that, their total kinetic energy is conserved and, the fraction of K.E. transferred by the, 8, colliding particle is , then the mass of P and, 9, , the mass of Q bears a ratio, 1) 8 : 3 2) 9 : 8 3) 2 : 3 4) 2 : 1, 42. A particle strikes a horizontal frictionless, floor with a speed 'u' at an angle ' θ ' with the, vertical and rebounds with a speed 'v' at an, angle ' α ' with the vertical. Find the value of, 'v' if 'e' is the coefficient of restitution., 1) v = u e2 sin2 θ+cos2 θ 2) v = u e2 cos2 θ +sin2 θ, 3) v =u e2 cos2 θ+tan2 θ 4) v = u cot2 θ + e2 cos2 θ, 43. Two spheres A and B of equal masses lie on, the smooth horizontal circular groove at, opposite ends of diameter and at the end of, time 't', ‘A’ impinges on ‘B’. If 'e' is the, coefficient of restitution, the second impinge, will occur after a time, 2t, t, πt, 2πt, 1), 4), 2), 3), e, e, e, e, 44. A ball is thrown at an angle of incidence ' θ ', on a horizontal plane such that the incident, direction and the reflected direction are at, right angles to each other. If the coefficient of, restitution is 'e' then ' θ ' is equal to, 1) tan-1 (e), 2) tan-1 (2e), 3) tan-1 ( 2 e), 4) tan-1 e, 45. Consider the collision depicted in fig to be, between two billiard balls with equal masses, m1 = m2 . The first ball is called the target., The billiard player wants to ‘sink’ the target, ball in a corner pocket, which is at an angle, θ 2 = 37o . Assume that the collision is elastic, and that friction and rotational motion are, not important, then θ1 is, , ( ), , y-axis, , v1, , 1) 37o, , m2, , 2)90o, , NARAYANAGROUP, , T1 , , be the ratio of time of flight T , maximum, 2, H1, , x-axis, , θ2, , 3) 45o, , v2, , 4) 53o, , R , , height H and horizontal range R1 in first, 2, 2, two collisions with the ground, then, 1) a =, , 1, 1, 1, 2) b = 2 3) c =, e, e, e, , 4) 1,2 & 3, , 47. A wall moving with velocity 2cms −1 towards, the ball and ball is moving towards the wall, with a velocity 10cms −1 .It hits the wall, normally and makes elastic collision with, wall. The velocity of ball after collision with, wall in cms −1, 1) 12, 2) 8 3)14 4)16, 48. A body A moves towards a wall with velocity, V. The wall also moves towards the body A, with velocity V0 . After collision the body, moves in opposite direction with velocity V |, 2V0 , which is 1 +, times the velocity V. The, V , , coefficient of restitution is, 1, 1, 1, 2), 3) 1, 4), 4, 3, 2, 49. A sphere A of mass m moving with certain, velocity hits another stationary sphere B of, different mass. If the ratio of velocities of, , 1), , VA 1 − e, the spheres after collision is V = 1 + e ,, B, where e is coefficient of restitution. The, initial velocity of sphere A with which it, strikes is, 2) VA − VB, 1) VA + VB, , 3) VB − VA, , θ1, m1, , 46. A projectile is fixed on a horizontal ground., Coefficient of restitution between the, projectile and the ground is ‘e’. If a, b and c, , 4), , (VB + VA ), , 2, 50. A ball A of mass 3m is placed at a distance d, from the wall on a smooth horizontal surface., Another ball B of mass m moving with velocity, u collides with ball A. The coefficient of, restitution between the balls and the wall and, 205
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, between the balls is e, a) the velocity of ball B after collision is, u ( 3e − 1), 4, b) the velocity of ball B after collision is, u ( 2e + 1), 4, c) after collision, ball A will move away by, , a=, 4., , d ( 2 e + 1), , distance ( 2 e − 1) during the time ball B, returns back to wall., d) after collision, ball A will move away by, d (e + 1), , distance ( 3 e − 1 ) during the time ball B, returns back to wall, 1)a,d, 2) a,c, 3)b,d, 4)c,d, , LEVEL-III - KEY, 01)4, 07)1, 13)2, 19)3, 25)3, 31)4, 37)4, 43)1, 49)1, , 02)2, 08)4, 14)2, 20)3, 26)4, 32)4, 38)3, 44)4, 50)1, , 03)2, 09)3, 15)2, 21)3, 27)2, 33)1, 39)4, 45)4, , 04)4, 10)2, 16)1, 22)1, 28)3, 34)4, 40)1, 46)4, , 05)1, 11)2, 17)1, 23)2, 29)3, 35)2, 41)4, 47)3, , 06)1, 12)3, 18)1, 24)1, 30)2, 36)1, 42)2, 48)3, , LEVEL-III - HINTS, 1., , 2., 3., , W = mg, , L, L, + mg + 2mgL, 2, 2, , u2, ; workdone by gravity = FS cos θ, 2g, here F = mg , S = h,θ = 1800, Maximum frictional force between the slab and, the block, f max = µ N = µ mg ; m=mass of upper block, h=, , Evidently, F < f max, so, the two bodies will move together as a single, unit. If ‘a’ be their combined acceleration, then, a =, , F, m + M, , Therefore, frictional force acting can be obtained, 1 2, as f = Ma ,Using s = at , find S2 and S3 ., 2, W = ( S3 − S2 ) f, 206, , 5., 6., , 7., , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , ( 2m1 − m2 ), , g, 3m1m2 , m2 , T = , , g, m, +, 2, 1 2 , 4m1 + m2 , , , , , 1 2, On 2kg block , W1 = m1 gx = m1 g at ;, 2, , 1, , WT = T × x = T at 2 , 2, , x, On 1kg block, W2 = m2 g, 2, 1, 2, 1, W = ∆ KE = m v 2 − mu, 2, 2, 1, K = mv 2, 2, dK, dv d v , = m, ∴, = mv, v = (m a )v, dt, dt, dt , 1, K.E of man = ( K.E of boy ), 2, 1 2 1 1 m 2 -----------(1), mv = , v1 , 2, 22 2 , , 1, 1m 2, 2, v1 -------(2), Also given, m ( v + 1) =, 2, 2 2, Divide (1) with (2) to get v, dEk , 1, 2, 8. E k = mv = kt ; ⇒ power ‘P’= , =k, 2, dt , In this case, P=Fv, P, K, K', =, Hence, F = =, v, 2 Kt, t, m, where K1 is another constant, 80, 1, × E = mv2, 9., 100, 2, 10. Let ‘x’ be the length of chain inside the tube, Mh , , g, hg, L , , vdv, hg, a=, =, =, ; −, M, h, +, x, dx, h+ x, (h + x), L, v, 0, hg, dx, on integrating both sides, − ∫0 vdv = ∫l −h, h+x, 2.01, (−k ), 1, K, =, K, +, f, i, 11., ∫0.1 x dx = 2 mυi2 − k ln( x) |0.12.01, υ, , f, , =, , 2K, , f, , /m, x, , 2, 12. ∆k =W ⇒ K f = ∫ (4 − x ).dx, 0, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 13. According to the law of conservation of energy,, the total energy of the system remains constant., 95% of P.Ei=K.Ef, 1, 95, 2, × mgl, i.e., m v =, 2, 100, 2g, 14. x =, ; from conservation of mechanical, K, 1 2 1 2, energy ; mgx = Kx + mv , here m = 5kg, 2, 2, 2E, 15. Potential energy of particle at x =, is zero (x, k, 1 2, > 0) ; KE =E or mv = E, 2, 1 2, 16. m g (sin θ − µ cos θ ) x = kx, 2, m1m2, 17. Using reduced mass concept, µ = m + m, 1, 2, 1 2 1 2, µ v = kx, 2, 2, 18. Loss of PE=Gain in KE, 1, 1, mgR cos θ = mv 2 + m ( v1 cos θ − v ) 2, 2, 2, 1, 2, + m (v1 sin θ ) − − − (1), 2, From conservation of linear momentum, m(v1 cos 45o − v ) = mv -------(2), Here v1 is the velocity of ball w.r.t wedge, solve to get speed of wedge (v), 19. ∆KE = Wnet or K f − K i = ∫ Pdt, 20. ac =, , v2, = k 2rt 2 or v = krt, r, , dv, = kr, dt, hence ∴ P = Ft v cos θ = mat v cos θ, , tangential acceleration is, at =, , 21. P = mav = m, , dv 2, v, ds, , s, , v, , 2, P, ds, =, v 2 dv, ; m∫, ∫, 0, v1, , 2 P 1/2, t, m, , ∆ K . E = W = ∫ Pdt, t1, , 1, , 2 Pt 2, 24. Here v = , or, M , NARAYANAGROUP, , 27., 28., , v 22 = 2 gh = 2 g × 4 = 8 g, v 22 − v12 = 4 gr = 4 g × 1 = 4 g, , 1, , ds 2 Pt 2, =, or, dt M , , ∴ v12 = 4 g, , normal reaction at lowest point, R2 =, , mv 22, + mg ,, r, , m v12, − mg, r, 29. mu=(m+M)v; If vθ is Velocity at an angle θ ., At highest point, R1 =, , v2 - vq 2 = 2gl (1- cos q ) ; find vθ, , then Tθ =, , mvθ2, + mg cos θ, r, , mv 2, = mg Þ v2 = gr, 30., r, If θ is angle from highest point, v θ2 − v 2 = 2 gr (1 − cos θ ), , 31. v = g ( 3 h − r ) ; Where h is maximum height at, which the tension in the string is zero., 32. From given condition, v = 5 gL, From energy conservation, , 3m, , t2, , 23., , 25., 26., , A, , 1 2, 2Pt, 22. w = ∆K .E ⇒ Pt = mv ; v =, 2, m, ds, =, dt, , 1, , 2Pt 2, ds = , dt . Integrate to find ‘s’., M , mu = Mv but v = 5gr, If v is velocity at the lowest point, v 2 = 2 gL (1 − sin θ ) but, v 2 = 5gr & r = L - x, where x is minimum distance between 0 and P., T = mg(3cosθ − 2 cosφ ) ; ( f = 600 ; q = 300 ), Let v2 , v1 be the velocities at lowest point, highest, point of vertical circle., , B, , u, , m, d, , m (v / 2), mv 2, −, = − mgL (1 − cos θ ), 2, 2, 7, or cos θ = −, 8, The value of θ given in option (4) can only have a, value of θ as computed above, so (4) is correct, 2, , 207
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JEE-ADV PHYSICS-VOL - II, 2., , 3., , 4., , 5., , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Column I, Column II, a) Force is equal to, p) Work is path, independent, b) For the conservative q) Rate of change, force, of linear momentum, c) Power is equal to, r)Rate of work done, d) Area of P-t curves, s)Product of force to, give, the velocity, t) Work done, u) Negative of the, potential energy, gradient, Match the following, Column I, Column II, a) Work done by, p) Sum of kinetic, a force is, energy and potential, energy is constant, b) Work done by, q) Work done by, a conservative force is conservative and non, conservative forces, c) Change in kinetic, r) Independent of, energy is, path followed, d) Under conservative, s)Independent of time, forces, sum of kinetic, for which force acts, energy and potential, energy is, Match the following, Column I, Column II, a) Mechanical energy is p) Always positive, b) Kinetic energy is, q) Always negative, c) Gravitational potential, r) May be positive or, energy of a body at any, negative, height above surface, of Earth is, d) A body having zero, s) Always zero, momentum has kinetic, energy, Match the following :, Column I, Column II, a) Stable, p) Potential energy is, dU, =0, equilibrium, zero ;, dr, b) Unstable, q) Potential energy is, equilibrium, minimum i.e.,, dU, = zero ,, dr, , c) Neutral equilibrium, , position of body, NARAYANAGROUP, , 7., , dU, = zero ,, dr, , d 2U, is − ve, dr 2, d 2U, =0, s), dr 2, , Match the pairs in two lists given below, List – I, List – II, a) Gravitational force, e) decreases, b) Frictional force, f) conservative force, c) KE of a dropped, g) non-conservative, body, force, d) PE of a dropped, h) increases, body, A body is allowed to fall from a height h above, the ground. Then match the following, List – I, List – II, a) PE = KE, e) at height h/2, b) PE = 2KE, f) constant at any, point, c) KE = 2PE, g) at height 2h/3, d) PE + KE, h) at height h/3, , ASSERTION & REASON TYPE, , 8., , 9., , 10., , 11., , d 2U, is + ve, dr 2, r) Potential energy is, maximum, , d) Most unstable, , 6., , 12., , 13., , In each of the following questions, a, statement is given and a corresponding, statement or reason is given just below it. In, the statements, mark the correct answer as, 1)If both Assertion and Reason are true and, Reason is correct explanation of Assertion., 2) If both Assertion and Reason are true but, Reason is not the correct explanation of, Assertion., 3) If Assertion is true but Reason is false., 4) If both Assertion and Reason are false., Assertion (A) : No work is done by the, centripetal force acting on a body moving along, the circumference of a circle, Reason (R) : At any instant, the motion of, the body is along the tangent to the circle where as, the centripetal force is along the radius vector, towards the centre of the circle, Assertion (A) : No work is done by gravitational, force if a body moves along a horizontal surface., Reason (R) : In horizontal motion, angle, between gravity and displacement is 900, Assertion (A) : Work done by the gravitational, force is zero, in closed path, Reason (R): Gravitational force is a conservative, force, Assertion (A) :The change in kinetic energy of a, particle is equal to work done on it by the resultant, force., Reason (R): Change in kinetic energy is equal to, work done in case of system of one particle., Assertion (A): Graph between potential energy, of a spring versus the extension or compression of, the spring is parabola, Reason (R): Potential energy of a stretched or, compressed spring proportional to square of, extension or compression, Assertion (A): the kinetic energy of a body is, quadrupled, when its velocity is doubled, 209
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , 14., , 15., , 16., , 17., , 18., , 19., , 20., , Reason (R): Kinetic energy is proportional to, square of velocity, Assertion (A) : If a spring is compressed, energy, is stored and when it is elongated, energy is, released., Reason (R): The energy stored in a spring is, proportional to the linear deformation of the, spring., Assertion (A): When a body moving with certain, velocity collides head on with another body of, same mass at rest, the collision being perfectly, elastic head on collision, 100% of its K.E is, transferred to the later, Reason (R) : Both momentum and KE are, conserved in the case of perfectly elastic collision, Assertion (A): If the momentum of a body, increases by 20%, then its KE also increases by, 20%, Reason (R): The KE of a body is directly, proportional to its momentum, Assertion (A): In one dimensional perfectly, elastic collision between two moving bodies of, equal masses, the bodies merely exchange their, velocities after collision., Reason (R) : If a lighter body at rest suffers, perfectly elastic collision with a very heavy body, moving with a certain velocity, then after collision, both travel with same velocities, Assertion (A): Two particles moving in the same, direction do not loose all their energy in a, completely inelastic collision., Reason (R) : Principle of conservation of, momentum holds good for all kinds of collisions, Assertion (A) : n small balls each of mass ‘m’, colliding elastically each second on a surface with, a velocity u. The force experienced by the surface, is 2nmu., Reason (R) : In elastic collision , the ball, rebounds with the same velocity, Assertion (A) : A body of mass " m1 " collides, head on elastically with another body of mass, " m 2 " at rest the ratio of the final energy of the first, body to the final energy of the second body is, , ( m1 − m 2 ), , STATEMENTS TYPE, , 22., , 23., , 24., , 25., , 26., , 27., , 2, , 4 m1 m 2, , Reason (R) : The collision is perfectly elastic and, the coefficient of restitution is 1., 21. Assertion (A) : A body of mass " m1 " collides, another body of mass " m 2 " at rest elastically. The, fraction of energy transferred to the second body is, , 28., , 29, , m1, m1 + m 2, , Reason (R) : In an "inelastic collision" both linear, momentum and KE is conserved, 210, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 30., , 1) Statement I & II are true., 2) Statement I is true, Statement II is false, 3) Statement I is false, Statement II is true, 4) Statement I & II are false, Statement I : An engine A can perform a given, work in 1 hr and engine B can perform the same, work in 1/2 hr. then B has greater power than A., Statement II : Power is the dot product of force, and velocity., This question has statement 1 and statement 2. Of, the four choices given after the statements,choose, the one that best describes the two statements. If, two springs S1 and S 2 of the force constants k1, and k2 , respectively, are stretched by the same, force, it is found that more work is done on spring, S1 than on spring S 2 ., Statement-I : If stretched by the same amount,, work done on S1 , will be more than that of S 2, Statement-II : k1 < k 2, [AIEEE 2012], Statement - I: A quick collision between two, bodies is more violent than slow collision, even, when initial and final velocities are identical., Statement - II: The rate of change of momentum, determines that the force is small or large., Statement - I: If collision occurs between two, elastic bodies their kinetic energy decreases, during the time of collision., Statement- II: During collision intermolecular, space decreases and hence elastic potential, energy increases ., Statement -I: When a body is rotated along a, vertical circle with uniform speed then the sum of, its kinetic energy and potential energy is constant, at all positions, Statement- II: To make a body to move along a, vertical circle, its critical speed at a point is, independent of mass of body, Statement - I: Work done by spring force on a, block connected to the spring may be positive or, negative., Statement - II : spring force is both pushing and, pulling., Statement - I : The normal force on a body by, the floor is not a conservative force, Statement - II : The normal force does no work, at all. Hence it will not store energy in the system., Statement - I : The workdone in pushing a block, is more than the work done in pulling the block on, a rough surface., Statement - II : In the pushing condition normal, reaction in more., Statement - I : A cyclist always bends inwards, while negotiating a curve., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, , Statement - II : By bending he lowers his centre, collision, of gravity., d) mechanical energy of the ball remains the, 31. Statement - I : The work done by all forces on a, same during the collision, system equals to the change in kinetic energy of, 1) a, b, d 2) a, b, c, 3) a, b 4) a, b, c, d, that system. This statement is true even if non- 38. The given plot shows the variation of U, the, conservative forces act on the system., potential energy of interaction between two, Statement II : The total work done by internal, particles with the distance separating them r., forces may be positive., U, OTHER MODEL QUESTIONS, A, 32. Nature of work done by gravitational force, E, a) may be negative, b) may be positive, c) may be zero d) always positive, B, 1) a,b & d are correct, 2) a, b & c are correct, v, D, 3) a,c & d are correct, 4) all correct, 33. When workdone on a particle is positive, C, 1) B and D are equilibrium points, a) K.E remains constant, 2) C is a point of stable equilibrium, b) momentum increases, 3) The force of interaction between the two, c) K.E increases, particles is attractive between points C and, d) K.E decreases, B and repulsive between points D and E on, 1) a, b, 2) b, c 3) b, d 4) a only, the curve, 34. During “elastic collision”, 4) The force of interaction between the, a) there is no loss of kinetic energy, particles is repulsive between points C and A, b) the bodies are perfectly elastic, Which, of the above statement are correct?, c) temporarily some of the kinetic energy is, 1)1, &, 3, 2) 1 & 4, 3) 2 & 4 4) 2 & 3, used to deform the bodies, 39., For, an, isolated, system, in, the absence of any, d) after collision the bodies regain the, dissipative, effect, original shape keeping the total energy, a)KE is conserved, b) PE is conserved, constant, c)energy, is, conserved, d)ME is conserved, 1) only “a” is true 2) a, b, c, d are true, 1), a,, b, 2), a,, b,, d, 3) c, d 4) c only, 3) b, c, d are true 4) a, b, c are true, 40., The, KE, of, a, particle, continuously, increases, 35. Identify the correct statements from the, with, time., Then, following, a) The resultant force on the particles is, a) the collisions between the nuclei and, parallel to the velocity at all instants, fundamental particles are considered as, b) The resultant force on the particle is at an, elastic collisions, angle less than 900 with velocity all the times, b) Emission of an alpha particle by a heavy, c) Its height above the ground level must, nucleus is an “elastic collision”, continuously decrease, c) The collision between two ivory balls is, d) The magnitude of its linear momentum is, considered as “elastic collision”, increasing continuously, d) A running man jumps into a train. It is an, 1) b, d, 2) a, d 3) a, b, c, d, 4) a, b, d, “elastic collision”, 41., Select, the, correct, alternative, (s), 1) only a & b are true, 2) only b & c are true, a) Work done by static friction is always zero, 3) a, b & c are true, 4) b, c & d are true, b) Work done by kinetic friction can be, 36. If force is always perpendicular to motion, positive also, a) KE remains constant, c) Kinetic energy of a system can not be, b) workdone by the force is zero, increased without applying any external, c) momentum remains constant, force on the system, d) speed remains constant, d) Work energy theorem is valid in non1) a,b,d 2) a,b,c3) b,c,d 4) a,b,c,d, inertial frames also, 37. A ball hits a floor and rebounds after an, 1) a, b, 2) a, b, d, 3) b, c, d 4) a, d, inelastic collision. In this case,, 42., Two, Solid, spheres, of, same, material having, a) total energy of the ball and the earth, radii, in, the, ratio, 1, :, 2, are, moving, with same, remains the same, kinetic, energy, on, a, horizontal, path., They, b) total momentum of the ball and the earth is, are, brought, to, rest, by, applying, same, conserved, retarding, force., Then, c) momentum of the ball just after the, collision is same as that just before the, NARAYANAGROUP, , 211
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WORK,ENERGY,POWER & VERTICAL CIRCULAR MOTION, i) their initial momenta are in the ratio 1: 2 2, ii) distance travelled are in the ratio 1 : 1, before coming to rest, iii) the times taken by them coming to rest is, , 1: 2 2, 1. i) and (ii) are correct 2. ii) and (iii) are, correct, 3. All the three are correct 4.(i) and (iii) are correct, , LEVEL-IV - KEY, Matching Type, 1. a – r,t , b – q, c – p,, 2. a – q,u, b – p,u,c – r,s,, 3. a – q, b – s, c – r,, 4. a – r, b – p, c – q,, 5. a – q, b – r, c – s,, 6. a – f, b – g, c – h,, 7. a – e, b – g, c – h,, , d–s, d –t, d–p, d–s, d –p, d–e, d–f, , Assertion & Reason Type, 8) 1, 14)4, 20)1, , 9) 1 10) 1 11) 1 12) 1 13) 1, 15) 1 16) 4 17) 3 18) 1 19) 1, 21) 4, , Statements Type, 22)1, 28)2, 32)2, 38)3, , 23) 3 24) 1 25) 1 26) 3 27) 1, 29) 1 30) 1 31) 1, Other Model Questions, 33)2 34) 2 35) 3 36) 1 37) 3, 39)3 40) 4 41) 2 42)1, , LEVEL-IV - HINTS, 8., 9., 10., 11., 12, , 13., , 212, , As centripetal force and displacement of the, body are perpendicular to each other, work done, is zero., As angle between gravity and displacement is, 900 , work done by the gravitational force is, zero., A force is said to be conservative, if workdone, by the force along any closed path is zero., ∆K .E. = Wnet force . This relation is valid for, particle as well as a system of particles., 1 2, Potential energy stored in a spring U = kx ,, 2, 2 . So, U - x graph is a parabola., U∝x, 1 2, Kinetic energy K = mv , when ‘m’ is, 2, K1 v12, 2, =, constant , K ∝ v and, ., K 2 v2 2, , JEE-ADV, PHYSICS-VOL, - II, JEE MAINS, - VOL - VI, , 1 2, Kx , where ‘x’ may be compression or, 2, elongation and in both cases energy is stored in, the spring., 15. In perfectly elastic collision both momentum and, KE are conserved and transfer of KE is, maximum when m1=m2 =m so both statements, are correct but R is correct explanation to A., P2, 16. K.E =, 2m, m1 − m2 , 2m2 , 17. v1 = m + m u1 + m + m u2, 1, 2 , 1, 2 , , 14. U =, , 2m1 , m2 − m1 , v2 = , u1 + , u2, m, +, m, m, +, m, 1, 2 , 1, 2 , If m1=m2=m, v1=u2 and v2=u1, If m2<<m1,v1=u1 and v2=2u1, so A is true R is false, 18. If it is a perfectly inelastic collision then, m1u1+m2u2=(m1+m2)v, m1u1 + m2u2, P12, P22, v=, +, ; KE =, m1 + m2, 2m1 2m2, As P1 and P2 both simultaneously cannot be, zero, there fore total K.E cannot be lost, 19. In elastic collision K.E remains conserved, therefore the ball rebounds with the same, velocity according to newton’s II law, Fxt=Change in linear momentum, Fx1=mn(u+u) ; F=2nmu, 1, 1, 2, 2, 20. KE1 = m1v1 ; KE2 = m2 v2, 2, 2, 2, , 1 m1 − m2 2, m1 , u1, 2 m1 + m2 , , ( m1 − m2 ), KE1, =, =, KE2 1 2m 2 2, 4m1m2, 1, m2 , u1, 2 m1 + m2 , For perfectly elastic collision, e=1, 21. Fraction of KE transferred to the second body, 4m1m2, =, 2, m + m . In an inelastic collision only, (, , 1, , 2, , 2, , ), , linear momentum is conserved, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, 5., , LEVEL-V, SINGLE ANSWER TYPE QUESTIONS, 1., , 2., , particle moves from point ( 0, 0, 0 ) to point, , A simple pendulum having a bob of mass m is, suspended from the ceiling of a car used in a, stunt film shooting. The car moves up along, an inclined cliff at a speed v and makes a jump, to leave the cliff and lands at some distance., Let R be the maximum height of the car from, the top of the cliff. The tension in the string, when the car is in air is, , ( 2, 4, 0) as, , B), , (2,4,0), y = x2, (0,0,0), , A), , mu 2 cos 2 α, 2, , mu 2 sin 2 α, mu 2 sin 2 α, D) −, 2, 2, The blocks A and B shown in the figure have, masses M A = 5 kg and M B = 4 kg. The system, is released from rest. The speed of B after A, has travelled a distance 1 m along teh incline, is, , 280, units, 5, , B), , 140, units, 5, , 232, 192, units, units, D), 5, 5, A particle is being acted upon by one, dimensional conservative force. In the F–x, curve shown, four points A, B, C, D are, marked on the curve.State which type of, equilibrium is the particle have at position c., , C), 6., , C), 3., , shown in fig. is, , Y, , mv 2, mv 2, (C) mg +, (D) zero, (A) mg (B) mg –, R, R, A particle of mass m is projected at an angle, α to the horizontal with an initial velocity u., the work done by gravity during the time it, reaches its highest point is, A) u 2 sin 2 α, , ur, A force F = ( 3xy − 5 z ) $j + 4 zi$ is applied on a, particle. The work done by the force when the, , F, D, B, , O, , C, , A, , A) stable equilibrium, C) Neutral, , A, 5m, 37°, , 7., , B, , g, g, 3, 3, g (B), g (C), (D), 2 3, 2, 4, 2, A particle is projected along a horizontal, surface whose coefficient of friction varies as, A, µ = 2 , where r is the distance from the origin, r, in metres and A is a positive constant. The, initial distance of the particle is 1m from the, origin and its velocity is radially outwards. The, minimum initial velocity at this point so the, particle never stops is, (A), , 4., , A) ∞, , b) 2 gA c), , NARAYANAGROUP, , 2gA, , d) 4 gA, , A particle of mass, , B) unstable, D) No equilibrium, 10, kg is moving in the, 7, , positive direction of x. Its initial position is x, = 0 & initial velocity is 1 m/s. The velocity at, x = 10 is: (use the graph given), Power (in watts), 4, 2, , 10, , A) 4 m/s, , X, (in m), , B) 2 m/s C) 3 2 m/s D) 100/3 m/s, 213
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WORK POWER ENERGY, 8., , JEE-ADV PHYSICS-VOL - II, , The potential energy of a particle is determined, by the expression U = α ( x + y ) , where α, is a positive constant. The particle begins to, 2, , 2, , move from a point with coordinates ( 3,3 ) ,, only under the action of potential field force., Then its kinetic energy T at the instant when, the particle is at a point with the coordinates, , (1,1), 9., , t(s), (B), K.E, , (A), K.E, , is, , A) 8α, B) 24α, C) 16α, D) Zero, An engine is pulling a train of mass m on a, level track at a uniform speed u. The resistive, force offered per unit mass is f., A) Power expended by the engine is “mfu”., B) The extra power developed by the engine to, mghu, maintain a speed u up a gradient of h in s is, s, , C) The frictional force exerting on the train is mf on, the level track, D) None of above is correct, 10. A body of mass m slides downwards along a, plane inclined at an angle α . The coefficient, of friction is µ . The rate at which kinetic, energy plus gravitational potential energy, dissipates expressed as a function of time i, A) µ mtg2 cos α, B) µ mtg2 cos α (sin α – m cos α ), C) µ mtg2 – sin α, D) µ mtg2 sin α (sin α – m cos α ), 11. In the figure the variation of components of, acceleration of a particle of mass is 1 kg is, shown w.r.t. time. The initial velocity of the, r, particle is u = ( −3i + 4 j) m/s. The total work, done by the resultant force on the praticle in, time from t = 0 to t = 4 seconds is :, ax(in m/s2), , 37°, , t (in sec), , (A) 22.5 J (B) 10 J, (C) 0, (D) None of these, 12. For a particle moving on a straight lint the, variation of acceleration with time is given by, the graph as shown. Initially the particle was, at rest. Then the corresponding kinetic energy, of the particle versus time graph will be, 214, , a(m/s2), , t, , t, (C), K.E, , (D), K.E, , t, , t, , 13. A block of mass m is being pulled up the rough, inclined by an agent delivering constant power P., The coeficient of friction between the block and, the inclined is µ . The maximum speed of the block, during the course of ascent is, P, m, µ, θ, , (A) v =, , P, mg sin ? + µmg cos ?, , (B) v =, , P, mg sin ? − µmg cos ?, , (C) v =, , 2P, mg sin ? − µmg cos ?, , 3P, mg sin ? − µmg cos ?, 14. The spring block system lies on a smooth, horizontal surface. The free end of the spring, is being pulled towards right with constant, speed v0 = 2m/s. At t = 0 sec, the spring of, constant k = 100 N/ cm is unsttretched and, the block has a speed 1 m/s to left. The, maximum extension of the spring is, (D) v =, , 1m/s k = 100N/cm, m, V0 = 2m/s, 4kg, , (A) 2 cm, , (B) 4 cm, , (C) 6 cm, , (D) 8 cm, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , 15. Two equal masses are attached to the two ends, of a spring of spring constant k. The masses, are pulled a part symmetrically to stretch the, spring by a length x over its natural length., The work done by the spring on each mass is, 1, 1, (A) kx2, (B) – kx2, 2, 2, 1, 1, (C) kx2, (D) – kx2, 4, 4, 16. A block of mass m is allowed to slide down a, fixed smooth inclined plane of angle q and, length l . The magnitude of power developed, by the gravitational force when the block, reaches the bottom is, (A), , 2m 2 l(g sin θ)3, , (B) (2 / 3)m3 lg 2 sin θ, , (C) (2 / 3)m 2 l 2 g cos θ (D) (1/ 3)m3lg 2 sin θ, 17. An object of mass (m) is located at the origin, of a vertical plane . The body is projected at, an angle θ with velocity u. The mean power, developed by the gravitational force during the, interval of time till it reaches maximum height, mgu sin θ, (A) mgu sin θ, (B), 2, mgu sin θ, mgu sin θ, (C), (D), 3, 4, 18. The potential energy of a particle varies with, position x according to the relation, U ( x ) = 2x 4 − 27 x the point x =, , 3, is point of, 2, , (A) unstable equilibrium (B) stable equilibrium, (C) neutral equilibrium (D) none of these., 19. A particle, which is constrained to move along, the x–axis, is subjected to a force from the, origin as F (x) = – kx + ax3. Here k and a are, positive constants. For x=0, the functional form, of the potential energy U (x) of particle is, (A), U(x), , (B), U(x), , ˆ , where k is a positive, 20. A force F = − K(yiˆ + xj), constant, acts on a particle moving in the xy, plane. Starting from the origin, the particle is, taken along the positive x-axis to the point (a,, 0), and the parallel to the y-axis to the point, (a, a). The total work done by the force on the, particle is, (A) – 2Ka2 (B) 2Ka2 (C) – Ka2 (D) Ka2, 21. A smooth sphere of radius R is made to, translate in a straight line with a constant, acceleration a = g. A particle kept on the top, of the sphere is released from there at zero, velocity with respect to the sphere. The speed, of the particle with respect to sphere as a, function of θ as it slides down is, (B) Rg (1 + cos θ − sin θ), (A) Rg (sin θ + cos θ), , (C) 4Rg sin θ, (D) 2Rg (1 + sin θ − cos θ), 22. The potential energy of a 1 kg particle free to, move along the x-axis is given by U(x), , x4 x2 , = − 2 J . The total mechanical energy, x, , of the particle is 2J. Then, the maximum speed, in (m/s) is, (C) 3 / 2 (D) 2, (A) 1/ 2 (B) 2, 23. A smooth sphere of radius R is made to translate, in a straight line with a constant acceleration a, = g. A particle kept on the top of the sphere is, released from there at zero velocity with respect, to the sphere. The speed of the particle with, respect to sphere as a function of θ as it slides, down is, (A) Rg (sin θ + cos θ), (B) Rg (1 + cos θ − sin θ), (C) 4Rg sin θ, (D) 2Rg (1 + sin θ − cos θ), 24. A section of fixed smooth circular track of, radius r in vertical plane is shown in the figure., A block is released from position x and leaves, the track at y. The radius of curvature of its, trajectory when it just leaves the track at y is, O, , x, , x, R, , (C), U(x), , (D), U(x), , 53° 37°, , X, , Y, O, , x, , NARAYANAGROUP, , x, , (A) r, , (B), , r, 4, , (C), , r, (D) none of these, 2, 215
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 25. A large slab of mass 5 kg lies on a smooth, horizontal surface, with a block of mass 4 kg, lying on the top of it. The coefficient of friction, between the block and the slab is 0.25. If the, block is pulled horizontally by a force of F = 6, N. The work done by the force of friction on, the slab, between the instants t=2s to t=3s is, , ( g = 10 ms ), −2, , 4 kg, 5 kg, , A) 2.4 J B) 5.55 J C) 4.44J D) 10 J, 26. A chain AB of length l is lying in a smooth, horizontal tube so that a fraction h of its length, l , hangs freely and touches the surface of the, table with its end B. At a certain moment, the, end A of the chain is set free. The velocity of, end A of the chain, when it slips out of tube , is, , 28. The bob of a pendulum is released from a, horizontal position. If the length of the, pendulum is 1.5 m, the speed with which the, bob arrives at the lowermost point, given that, it dissipated 5% of its initial energy against, air resistance?, A) 3.14 m/s, B) 5.28 m/s, C) 1.54 m/s, D) 8.26 m/s, 29. A 1 kg block situated on a rough inclined plane, is connected to spring of a spring constant 100, N m −1 as shown in fig. The block is released, from rest with the spring in the unstretched, position. The block moves 10 cm down the, incline before coming to rest. Find the, coefficient of friction between the block and, the incline. Assume that the spring has a, negligible mass and the pulley is frictionless., K = 100 N/m, , 1 KG, , A, 37°, , h, B, , A) h, , C), , 2g, lh, , B), , l, 2 gl log e , h, , D), , l, 2 gh log e , h, 1, 2g, hl, , 1, kg starts from rest from A, 2, to move in a vertical plane along a smooth, fixed quarter ring of radius 5m, under the, action of a constant horizontal force F = 5 N, as shown. The speed of bead as it reaches, point B is, , 27. A bead of mass, , F, , A, , R = 5m, , velocity, , B, , A) 14.14 m/s, C) 5 m/s, 216, , A) 0.115 B) 1.25, C) 5.2, D) 4.5, 30. A nail is fixed at a point P vertically below the, point of suspension 'O' of a simple pendulum, of length 1m. The bob is released when the, string of pendulum makes an angle 300 with, horizontal. The bob reaches lowest point then, describes vertical circle whose centre coincides, with P. The least distance of P from O is, A) 0.4 m B) 0.5 m C) 0.6, D) 0.8 m, 31. A block is freely sliding down from a vertical, height 4 m on smooth inclined plane. The block, reaches bottom of inclined plane then it, describes vertical circle of radius 1 m along, smooth track. The ratio of normal reactions, on the block while it is crossing lowest point,, highest point of vertical circle is, A) 6 : 1, B) 5 : 1, C) 3 : 1, D) 5 : 2, 32. Mass of the bob of a simple pendulum of, length L is m. If the bob is projected, horizontally from its mean position with, , B) 7.07 m/s, D) 25 m/s, , 4gL , then the tension in the string, becomes zero after a vertical displacement of, A) L/3, B) 3L/4, C) 4L/3, D) 5L/3, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , MULTIPLE ANSWER QUESTIONS, 33. Two blocks, of masses M and 2M, are, connected to a light spring of spring constanat, K that has one end fixed, as shown in figure., The horizontal surface and the pulley are, frictionless. The blocks are released from, when the spring is non deformed. The string, is light., K, , 36. Which of the following is/are conservatve, force(s)?, r 5, r, (A) F = 2r3 rˆ, (B) F = ˆr, r, r 3 ( x i + y j), 3 ( x 2 i + y j), F=, r, 3/2, (C), ( x 2 + y2 ) (D) F = ( x 2 + y 2 )3 / 2, 37. A block of mass 2 kg is hanging over a smooth, and light pulley through a light string. The, other end of the string is pulled by a constant, force F = 40 N. The kinetic energy of the, particle increase 40 J in a given interval of, time. Then : (g = 10 m/s2), , 4Mg, ., K, (B) Maximum kinetic energy of the system is, , (A) Maximum extension in the spring is, , 2M 2g 2, K, (C) Maximum energy stored in the spring is four, times that of maximum kinetic energy of the system., (D) When kinetic energy of the system is maximum,, 2, , 2, , 4M g, ., K, 34. Select the correct alternatives:, (A) Work done by static friction is always zero, (B) Work done by kinetic friction can be positive, also, (C) Kinetic energy of a system can not be increased, without applying any external force on the system, energy stored in the spring is, , (D) Work energy theoram is valid in non-inerial, frames also., 35. Displacement time graph of a particle moving, in a straight line is as shown in figure. select, the correct alternative(s):, C, , S, B, A, O, , t, , (A) Work done by all the forces in region OA and, BC is positive, (B) Work done by all the forces in region AB is, zero, (C) Work done by all the forces in region BC is, negative, (D) Work done by all the forces in region OA is, negative., NARAYANAGROUP, , 2kg, , F =40N, , (A) tension in the string is 40 N, (B) displacement of the block in the given interval, of time is 2 m, (C) work done by gravity is - 20 J, (D) work done by tension is 80 J, 38. In the system shown in the figure the mass m, moves in a circular arc of angular amplitude, 60o. Mass 4m is stationary . Then:, , 4m, 60°, m, B, , m, A, , (A) the minimum value of coefficient of friction, between the same of mass 4m and the surface of, the table is 0.50, (B) the work dine by gravitational force in the block, m is positive when it moves from A to B, (C) the power delovered by the tention when m, moves from A to B is zero, (D) The kinetic energy of m in position B equals, the work done by gravitational force on the block, when its moves from position A to B., 217
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 39. A strip of wood mass M and length l is placed, on a smooth horizontal surface. An insect of, mass m starts from rest at one end of the strip, and walks to the other end in time t, moving, with a constant speed., (A) The speed of the insect as seen from the ground, is <, , l, t, , (B) The speed of the strip as seen from the ground, l M , is , t M + m , , (C) The speed of the strip as seen from the ground, l, , m, , , , is , t M + m , (D) The total kinetic energy of the system is, 1, l , (m + M ) , 2, t , , 2, , (A) positive from 0 to t 1 (B) negative from t 1 to t 2, (C) zero from t 2 to t 3 (D) negative from t 3 to t 4., 43. A smooth track in the form of a quarter circle, of radius 6m lies in the vertical plane. A, particle moves from P1 to P2 under the action, r r, , r, , r, , of forces F1, F2 and F3. Force F1 is always, toward P2 and always 20N in magnitude. Force, r, F2 is always acts horizontally and is always, r, 30N in magnitude. Force F3 always acts tangen-, , tially to the track and is of magnitude 15N., Select the correct alternative(s):, O, , 40. In the figure shown upper block is given a, velocity of 6m/s and lower block 3 m/s. When, relative motion between them is stopped., , 6m, , P2, F2, , F1, , 6m, , F3, , Rough, 1 kg, , 6m/s, , 2 kg, , 3m/s, , Smooth, , (A) Work done by friction on upper block is negative, (B) Work done by friction on both blocks is positive, (C) The magnitude of work done by friction on, upper block is 10J, (D) Net work done by friction is zero., 41. The potential energy U in joule of a particle of, mass 1kg moving in x-y plane obeys the law U, = 3x + 4y, where (x, y) are the co-ordinates of, the particle in metre. If the particle is at rest, at (6, 4) at time t = 0 then:, (A) the particle has constant acceleration, (B) the particle has zero acceleration, (C) the speed of particle when it crosses the y-axis, is 10m/s, (D) co-ordinates of particle at t = 1s are (4.5, 2), 42. Displacement time graph of particle moving, in a straight line is as shown in figure., From the graph we can conclude that work, done on the block is :, 218, , P1, , r, , (A) Work done by F1 is 120J, r, , (B) Work done by F2 is 180J, r, , (C) Work done by F3 is 45 π J, r, , (D) F1 is conservative in nature., 44. A block of mass M is attached with a spring, constant k. The whole arrangement is placed on a, vechile as shown in the figure. If the vehicle starts, moving towards right with an acceleration a (there, is no friction anywhere), then :, k, M, , a, , (A) maximum elongation in the spring is, , Ma, k, , (B) maximum elongation in the spring is, , 2M a, k, , (C) maximum compression in the spring, , 2ma, k, , (D) maximum compression in the spring is zero, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , 45. A small ball of mass m is released from rest at, a height h 1 above ground at time t = 0. At time, t = t0, the ball again comes to rest at a height, h2 above ground. Consider the ground to be, perfectly rigid and neglect air friction. In the, time interval from t = 0 to t = t0, pick up the, correct statements., (A) Work done by gravity on ball is mg (h1 - h2), (B) Work done by ground on ball for duratin of, contact is mg (h1 - h2)., (C) Average acceleration of the ball is zero., (D) Net work done on the ball by all forces, except gravity is mg (h1 - h2)., , COMPREHENSIVE TYPE QUESTIONS, Comprehension-1:, A block of mass m is kept in an elevation which, starts moving downward with an acceleration, ‘a’ as shown in figure. The block is observed, by two observers A and B for a time interval, t0 ., A, , m, , a, , B, , 46. The observer B finds that the work done by, gravity on the block is :, 1, 2 2, A) mg t0, 2, , 1, 2 2, B) − mg t0, 2, , 1, 1, 2, mgat02, D) − mgat0, 2, 2, 47. The observer B finds that the work done by, pseudo-force on the block is, , C), , shown in fig, the particle moves from the origin O, to point A(6m, 6m). the figure shows three paths,, OLA, OMA and OA for the motion of the particle, from O to A., Y, , O, , B), , 1, 2, C) mgat0, 2, , 1, 2, D) − mgat0, 2, , L, , 49. Which of the following is correct?, A) There is equal probability for the force being, conservative or non-conservative., B) Conservative or non-conservative nature of, force cannot be predicted on the basis of given, information., C)The given force is non-conservative., D)The force is conservative., 50. Along which of the three paths is the work done, maximum., A)OA, B) OMA, C) OLA, D) work done has the same value for all the three, paths, 51. Work done for motion along path OA is nearly, A) 383 J, B) 90 J C) 180 J D) 1811 J, , Comprehension - 3 :, One of the forces acting on a certain particle, depends on the particle's position in the xy-plane., r, This force F expressed in newtons, is given by the, r, , expression F = ( xy î + xyĵ) where x and y are in, metres., The particle is moved from O to C, through three different paths :Y(m), , C, , B(0,1), , 1 22, ma t0, 2, O, , Comprehensioni - 2 :, Force acting on a particle moving in the x-y plane, ur, 2, is F = y i$ + x $j N , x and y are in metre. As, , (, , NARAYANAGROUP, , X, , (A), , A) zero, B) −ma 2 t0 C) + ma 2t0 D) − mgat0, 48. According to observer B, the net work done, on the block is :, 1, 2 2, A) − ma t0, 2, , A (6m, 6m), , M, , A(1,0), , X(m), , 52. The work done by this force on path OC is, A), , 1, J, 2, , 1, 2, , B) – J, , C), , 2, J, 3, , D) –, , 2, J, 3, , ), , 219
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 53. The work done by this force on path OAC is, 1, 2, , 2, , 2, , 1, 2, , 2, , 2, , A), , 1, J, 2, , B) – J, , A), , 1, J, 2, , B) – J, , D) – 3 J, C) 3 J, 54. The work done by this force on path OBC is, D) – 3 J, C) 3 J, 55. Which of the following can be negative ?, A) Kinetic energy, B) Potential Energy, C) Chemical Energy, D) All of these, , Comprehension : 5, 56. A smooth vertical rod is released from rest, such that it is constrained to move vertically, , on a smooth wedge (θ = 450 ) . When the, wedge moves through a distance x, the speed, of the rod is :, , 60. The minimum safe value of r2 so that the block, does not fly off the track at C is, (A) 1 m, (B) 2 m, (C) 1.5 m (D) 3 m., 61. The work done by gravitational force from A, to C is, (A) 10 J (B) 20 J, (C) 30 J (D) 40 J, , Comprehension : 6, A chain of length l = πR / 4 is placed. On a smooth, hemispherical surface of radius R with one of its, ends fixed at the top of the sphere. Mass of chain, is 2πkg and R = 1m. (g = 10 m/s2)., 62. The gravitational potential energy of the chain, considering reference level at the base of, hemisphere is, (A) 20J, (B) 20 2 J (C) 40 J (D) 40 2 J., 63. If the chain sliped down the sphere, kinetic, energy of the chain when it has sliped through, π, , an angle θ =, 4, (A) 23.4 J (B) 63.44 J (C) 80 J (D) 97.4J., 64. The tangential acceleration of the chain when, it starts sliding down., , m, m, , 40 , , 1 , , , , 1 , , 2, , (A) π 1 −, , 2, , , θ = 45°, , gx, a) 2gx b), c) gx d) none of these, 2, 57. The work done by the normal reaction on the, rod is :, mgx, 3, a) mgx, b) −, c) − mgx d) −mgx, 2, 2, 58. The work done by the normal reaction on the, wedge is :, mgx, 3, mgx, a) mgx, b) −, c) mgx d), 2, 2, 2, , Comprehension – 5 :, , (C) 10 1 −, , , (D) zero., , a ball and the other end is connected to a fixed, point O. The ball is released from rest at t = 0, with string horizontal and just taut. The ball then, moves in vetical circular path as shown. The time, taken by ball to go from position A to B is t1 and, from B to lowest position C is t2 . Let the velocity, r, r, of ball at B is vB and at C is vC respectively.., O, , A, , θ, 90–θ, , B, , A, , r, rC, 65. If vC = 2 vB then the value of θ as shown is, , C, h2, B, , 59. The force exerted by block on the track at B, is, (A) 10 N (B) 20 N (C) 30 N (D) 40 N, 220, , 1 , , PASSAGE-IV:, One end of a light string of length L is connected to, , A block of mass m = 1kg is released from point A, along a smooth track as shown. Part AB is circular, with radius r1 = 4m and circular at C with radius r 2., Height of point A is h1 = 2m and of c is h2 = 1m., (g = 10 m/s2)., , h1, , 20 , , (B) π 1 −, , 2, , , 1, 2, −1 1, (C) cos, 2, −1, (A) cos, , (B) sin, (D) sin, , −1, , −1, , 1, 4, , , 1, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , r, r, 66. If vC = 2 vB then:, (A) t1 > t2 (B) t1 < t2 (C) t1 = t2, (D) Information is insufficient, r r, r, 67. If vC − vB = vB then the value of θ as shown, in the figure is, 1/3, , 1, (A) cos , 4, −1, , 1/3, , −1 1 , (C) cos , 2, , 1/3, , 1, (B) sin , 4, −1, , 1/3, , −1 1 , (D) sin , 2, , MATRIX MATCHING QUESTIONS, 68. A bob of mass 2 kg is suspended from a vehicle, by a rope of length l = 5m. the vehicle and the, bob are moving at a constant speed v0 . The, vehicle is suddenly stopped by a bumper and, the bob the rope, swings out a maximum, angle of 600 . Match the following., , Column - I, (A) u = 3.5m / s, (B) u = 9.5m / s, (c) T = 15N, (d) T = 35N, Column - II, (p) There will be some point on the trajectory of, object at which speed of the object is zero but, tension in the string is not zero., (q) There will be ome point on the trajectory of, object for which tension in the string is zero but, speed of the object is not zero, (r) There will be some point on the trajectory of, object for which tension in the string is zero but, spee of the object is not zero, (s) The acceleration of the object will be in direction, 70. A particle is suspended from a string of, length R. It is given a velocity u = 3 gR at, the bottom. Match the following., C, , D, , B, u, , A, v0, , COLUMN-I, A) Net force acting on the bob at lowest point just, after the hitting, B) Acceleration of the bob at lowest point, C) Net force acting on the bob at its highest point, D) Acceleration of the bob at its highest point, COLUMN-II (all values in SI units), p) 5 3 vehicle is stopped, q) 10, r) 20, s)10 3, 69. A small object of mass 0.5 kg is attached to an, end of massless 2 meter long inextensible, string with the other end of the string being, fixed. Initially, the string is vertical and the, object is at its lowest position ahving initial, horizontal velocity of magnitude u. The tension, in string is T when the object is at its lowest, position. The object subsequently moves in, vertical plane. The forces acting on object are, tension exerted by string and gravitational pull, by earth. Match the statements in column I, with corresponding results in column II (take, g = 10m/s2), NARAYANAGROUP, , Column - I, A) Velocity at B, B) Velocity at C, , Column - II, p) 7 mg, q) 5gR, , C) Tension in string at B, D) Tension in string at C, , r) 7gR, s) 5 mg, t) None, , INTEGER ANSWER TYPE QUESTIONS, 71. Block A has a weight of 300 N and block B, has weight 50N. Calculate the distance A must, descend from rest before it obtains a speed of, 4 m/s (Neglect the mass of cord and pulleys)., (Take g = 10 m/s2), , B, A, 221
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 72. A particle of mass m moves along a circle of, radius R with a normal acceleration varying, with time as a n = bt 2 , where b is a constant., Find the time dependence of the power, developed by all the forces acting on the, particle, and the mean value of this power, averaged over the first 2 seconds after the, beginning of motion.(m = 1, v = 2, r = 1), 73. Two blocks A and B are connected to each, other by a string and a spring; the string passes, over a frictionless pulley as shown in the, figure. Block B slides over the horizontal top, surface of a stationary block C and the block, A slides along the vertical side of C, both with, the same uniform speed.The coefficient of, friction between the surface and blocks is 0.5., K= 2000N/m. If mass of A is 2 kg calculate, mass of B., , LEVEL-V - KEY, SINGLE ANSWER, 1) D 3) D 3) C 4) C 5) D 6) A 7)C, 8) C 9) B 10) B 11) B 12) D 13) A 14) C, 15) D 16) A 17)B 18) B 19) D 20) C 21) D, 22) C 23) D 24) C 25) B 26) B 27) A, 28) B 29) A 30) D 31) C 32) D, MULTIPLE ANSWER, 33) ABC, 34) B,D, 35) B 36) ABC, 37) ABD, 38) A,B,C,D 39) A,C, 40) A,C, 41) A,C,D 42) A,B,C, 43) B,C,D 44) B,D, 45) ACD, COMPREHENSION, 46) C 47) A 48) C 49) C 50) A 51) B 52) C, 53) A 54) A 55) B 56) C 57) B 58) D 59) B, 60) B 61) A 62) C 63) A 64) A 65) B 66) B, 67) B, MATRIX MATCHING, 68) A-r;B-q;C-s;D-p, 69) A-p,s,B-q,s,C-r,s,D-s 70) A-r,B-q,C-p,B-t, INTEGER ANSWER, 71) 2 72) 2 73) 4 74) 5 75) 3, , B, , C, , LEVEL-V - HINTS, SINGLE ANSWER QUESTIONS, , A, , 1., 74. A small block is given a velocity v from point, A. Given x=3R, R = 20 m and g = 9.8m/s2 .If, the block strikes the point A after it leaves, the smooth circular track in vertical plane ,, the value of v is 7x, find v ?, , 2., , 1, 1 2, mu 2 sin 2 α, 2, Wm = ∆KE = m ( u cos α ) − mu = −, 2, 2, 2, 3. (C) If A moves down the incline by 1 metre, B, 1, metre. If the speed of B is v, 2, then the speed of A will be 2v., From conservation of energy:, Gain in K.E. = loss in P.E., , C, , shall move up by, R, , O, V, , 1, 1, 3, 1, 2, m A ( 2v ) + m B v 2 = m A g × − m Bg ×, 2, 2, 5, 2, , B, , A, X, , 75. A particle is projected along the inner surface, of a smooth vertical circle of radius R, its, velocity at the lowest point being (1/5), , (, , ), , 95 gR . If the particle leaves the circle at, −1, , an angular distance cos (x / 5) from the highest, point, the value of x is, 222, , ur r, Acceleration of car is “g” so g eff = g − a, Velocity at top = u cos α, , 1 g, 2 3, If the particle never stops then it may move till, Solving we get v =, , 4., , x=∞, ∞, 1 2 ∞, V2, 1, mv = ∫ µ mg .dx ⇒, = Ag ∫ 2 .dx, 2, 2, x, 1, 1, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , 5., , WORK POWER ENERGY, , so, V = 2 gA, z and x components are useless points., , 3, 3, , ∴ ∫ dw = ∫ ti − t + 1 j ⋅ x, 4, , 0, 0 4, W, , 2, , 192, 4, so, dw = Fy.dy = 3xy.dy = ∫ 6 x .dx = 5 units, 0, , 7., , 3 2, 3 2, , t − 3 i + − 8 t − t + 4 j dt, , , 8, ∴ W = 10 J, Alternate Solution :, Area of the graph ;, , Fvdx = Pdx, m× v2dv = ( 2 + 0.2x ) dx ; v dv =, 2, , 1, ( 2 + 0.2m) dx, m, , 10, , 7 10, 2, 0.2, v, dv, =, dx, +, xdx , ∫, ∫1, ∫, 10 0, 0, , , ∫a, , v, , 2, , =, , 8., , 4, , x, , dt = 6 = V(x) f − ( −3) ⇒ V(x) f = 3., , and ∫ a y dt = − 10 = V(y) f − ( −4 ) ⇒ V(y) f = −6, Now work done = ∆ KE = 10 J, 12. (D), , 14, 1.4 100 , × 10 +, 10, 10 2 = 14 + 7, , v3 = 63 + 1 = 64 ; v =4m/ s, As the particle moves only under the action of, conservative force, its mechanical energy must be, conserved. So ∆T + ∆U = 0 (T stands for kinetic, energy), or ∆T = ∆U = Ui − U f, = α ( 32 + 32 ) − α (12 + 12 ) = 16α, , 10., , D, , D, , a, , B, , C, , v, t, , 2, , v, , C, , t, , A, , t, , B, , D, A, , C, , A, , The above graphs show v - t graph from a - t, graph & Then v2 - t graph, which are self, explanatory., 13. (A) Let at any time the speed of the block along, the incline upwards be v., P, mdv, − mg sin? − µmgcos? =, v, dt, , the speed is maximum when, θ, , P = F .V, So = mg ( sin α − µ cos α ) µ g cos α t, , = µ mg 2t ( sin α − µ cos α ) cos α, 11. (B) From given graphs :, 3, , 3, 3 2, ax = t and ay = − t + 1 ⇒ v x = t + C, 4, 8, 4, , At t = 0 ; vx = − 3 ⇒ C = − 3, 3, 3, , ∴ v x = t 2 − 3 ⇒ dx = t 2 − 3 dt .... (1), 8, 8, , 3 2, , Similarly ; dy = − t − t + 4 dt, 8, , r uur r, As dw = F ⋅ ds = F ⋅ ( dx i + xy j), NARAYANAGROUP, , dv, =0, dt, , P, ∴ v max = mg sin ? + µmg cos ?, 14. (C) In the frame (inertial w.r.t. earth) of free end of, spring, the initial velocity of block is 3 m/s to left, and the spring unstretched., 3m/s, , 4Kg, , initial state, maximum, extension, state, , 4Kg, A, , .... (2), , Applying conservation of energy between initial, and maximum extension state., 1, 1, mv 2 = kA 2, 2, 2, , or, , A=, 223
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , m, 4, × 3 = 6cm., v=, k, 10, 000, 15. Work done on both block are the same and total, 1, 2, , 2, work done by spring is − kx ., , 1, , reference level mg, , Hence on each block − kx, 4, rr, 16. p = F.v ; = mg sin θ 2gl sin θ, = 2m 2 g 3l sin 3 θ, , 2, , Ans. (A), , 18. U ( x ) = 2 x 4 − 27 x, du, = − ( 8 x3 − 27 ) = 0 for equilibrium, dx, , 3, d 2u, = 24 x 2 > 0 at x =, 2, 2, dx, 3, Hence U is min at x = so stable., 2, x=, , 3, ;, 2, , k, , 19. F ( x ) = – kx + ax 3 ; F( x ) = 0 ; for x = 0 x =, ;, a, So slope is zero at x = 0 x =, , k, a, , maR sin θ + mgR(1 − cos θ) =, , 21. (d), , 1, mv 2, 2, , ma, R θ, , 22. (c) U =, , mg, , d 2U, d 2U, 2, =, 3, x, −, 1, = +ve for x = m1, dx 2, dx 2, , 1, 4, , K max =, 224, , 10 4 50, 3−, = = 5.55 J, 3 3 , 9, 26. Let ‘x’ be the length of chain inside the tube, , VdV, hg, =, Integrate, dx, h+ x, 27. Applying the work - energy theorem, we get, , 9, 4, , 1, mv 2, 2, , v=, , Mh , , g, hg, L , , a=, =, M, (h + x) h + x, L, −, , K max + Ul min = T .E = 2J, K max =, , 2 gr / 5 r, =, g cos 37 2, 25. Maximum frictional force between the slab and the, block, 1, f max = µ N = µ mg = × 4kt10 = 10 N, 4, Evidently, F < f max, so, the two bodies will move together as a single, unit. If a be their combined acceleration, then, F, 6, 2, a=, =, = ms −2, m+M 4+5 3, Therefore, frictional force acting can be obtained, as, 2, f = Ma = × 5 N, 3, 1 2, Using s = at, 2, 1 2 2, 1 2 2 4, s(2) = × (2) = and s(3) = × (3) = 3, 2 3, 3, 2 3, Therefore, work done by friction, =, , x4 x2 dU, − ,, = x3 − x = 0 ⇒ x = 0, x = ±1, 4 2 dx, , U ( ±1) = −, , 2r, mgr 1 2, +0=, + mv, 5, 5, 2, , v12, 2 gr, v=, ; Radius of curvature =, an, 5, , mgh mgu sin θ, 17. avg. Power = t =, 2, 1/2, , F =−, , 23. Applying work energy theorem., 1, maR sin θ + mgR (1 − cos θ ) = mV 2, 2, 24. Apply conservation of energy with O as, , 3, 2, , 1, × mv 2 − 0 = F × R + mg × R, 2, 28. Length of the pendulum , l = 1.5 m, Mass of the, bob = m, Energy dissipated = 5%, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, According to the law of conservation of energy,, the total energy of the system remains constant., At the horizontal position, Potential energy of the bob, EP =0, kinetic energy, 1 2, of the bob , EK = mv total energy, 2, , 1 2, mv ...(ii), 2, As the bob moves from the horizotnal position to, the lowermost point, 5% of its energy gets, dissipated. The total energy at the lowermost point, is equal to 95% of the total energy at the horizontal, point, i.e,, Ex =, , 1, 95, mv2 =, × mgl, 2, 100, , ⇒v=, , 2 × 95 ×1.5 × 9.8, = 5.28m / s, 100, , 1 2, 29. mg (sin 37 − µ cos37 ) x = kx, 2, 30. If V is velocity at the lowest point, o, , o, , V 2 = 2 gL(1 − sin θ ), but, V 2 = 5 gr and r = L - x, where x is minimum distance between 0 and P., find x., 31. Let V2 , V1 be the velocities at lowest point, highest, point of vertical circle., V22 = 2 gh = 2 g × 4 = 8 g, V22 − V12 = 4 gr = 4 g × 1 = 4 g, ∴V12 = 4 g, normal reaction at the lowest point, highest point, are R2 , R1, , mV22, R2 =, + mg ;, r, , mV12, R1 =, − mg, r, , 32. V = g (3h − r ), Where h is maximum height at which the tension in, the string is zero., , MULITIPLE ANSWER, 33., , 4Mg, K, System will have maximum KE when net force on, , NARAYANAGROUP, , WORK POWER ENERGY, the system becomes zero. Therefore 2Mg = T and, 2Mg, K, Hence KE will be maximum when 2M mass has, , ⇒, , T = kx, , x=, , 2Mg, ., K, Applying W/E theorem, , gone down by, , 2Mg 1 4M2g2, 2M2g2, - K., ;, k, =, f, K 2, K2, K2, Maximum energy of spring, kf - 0 = 2Mg., , 2, , 1 4Mg 8M 2 g 2, K. , =, 2 K , K, Therefore Maximum spring energy = 4 × maximum, K.E., , When K.E. is maximum x =, Spring energy =, , 2Mg, ., K, , 1 4M 2g 2 2M 2g 2, .K., =, 2, K, K, , i.e. (D) is wrong., 35. [B,D] In region OA particle is acccelerated, in, region AB particle has uniform velocity while in, region BD particle is deceleration., Therefore, work, done is positive in region OA , zero in region AB, and negative in region BC., r uur, 36.. (A, B, C) Since ; W = ∫ F ⋅ dr, Clearly for forces (A) and (B) the integration do, not require any information of the path taken., For (C) : Wc =, , ∫, , 3 ( xi + yj), , (x, , 2, , +y, , ), , 2 3/2, , ⋅ ( dx i + dy j) = 3∫, , x dx + y dy, , (x, , 2, , + y2 ), , 3/2, , Taking : x2 + y2 = t, 2x dx + 2y dy = dt, ⇒ xdx + ydy =, , dt, dt / 2 3 dt, ⇒ Wc = 3∫ 3/2 = ∫ 3/2, 2, t, 2 t, , which is solvable., Hence (A), (B) and (C) are conservative forces., But (D) requires some more information on path., Hence non-conservative., 37. (A)(B)(D), Free body diagram of block is as shown in figure., , 225
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 73. Normal reaction between blocks A and C will be, zero. Therfore, there will be no friction between, them. Both A and B are moving with uniform speed., Therefore, net force on them should be zero For, equilibrium of A: mAg = kx, m g, , 4., , (2)(10), , A, ∴ x = k = 2000 = 0.01m, For equilibrium of B: ; µmB g = T = kx = mAg, , mA, 2, ∴ mB = µ = 0.5 = 4kg, kx, , T = kx, B, µmBg, , Frictionless, Table, , A, T, T, , 2hK 2, hK1, 3hK 2, hK 2, A) µ K, B) 2µ K C) 2µ K, D) µ K, 1, 2, 1, 1, What is the minimum value of the mass M so, that the block is lifted off the table at the, instant shown in the diagram ? Assume that, the blocks are initially at rest., m, m, A), B), 0, sin 60, tan 600, C) m sin 600, D) none of these, , m, , 60°, , mAg, , LEVEL-VI, , M, , SINGLE ANSWER QUESTIONS, 1., , 2., , An engine is pumping water continuously. The, water passes through a nozzle with a velocity, ν . As water leaves the nozzle, the mass per, unit length of the water jet is m0 . Find the rate, at which kinetic energy is imparted to the, water:, 1, 1, 1, 1, 3, 2, 32, 12, A) m0 v B) m0 v C) m0 v D) m0 v, 2, 2, 2, 2, A hemispherical vessel of radius R moving, with a constant velocity v0 and containing a, ball, is suddenly haulted. Find the height by, which ball will rise in the vessel, provided the, surface is smooth:, , v02, 2v0 2, v02, B), C), D) none of these, 2g, g, g, Two balls of same mass are projected as, shown, by compressing equally (say x) the, springs of different force constants K1 and K 2, by equal magnitude. The first ball is projected, upwards along smooth wall and the other on, the rough horizontal floor with coefficient of, friction µ . If the first ball goes up by height, h , then the distance covered by the second, ball will be:, , 5., , 6., , 7., , 12, , 2E , The speed of the particle at x = , , K , , K1, , 228, , is, , 12, , 8., K2, , mg 2, , mg 2, A) 2mg B), C), D) mg 2 2, 2, 2, A particle moves move on the rough horizontal, 3, ground with some initial velocity V0 . If of, 4, its kinetic energy lost due to friction in time, t0 . The coefficient of friction between the, particle and the ground is, V0, V0, 3V0, V0, A) 2 gt, B) 4 gt, C) 4 gt, D) gt, 0, 0, 0, 0, The total mechanical energy of a particle is E., 2, , A), 3., , A bob of mass m is suspended from a fixed, support with a light string and the system with, bob and support is moving with a uniform, horizontal acceleration. The breaking strength, of the string is mg 2 . Find the workdone by, the tension in the string in the first one second:, , 2E , , . Find the potential energy of the, m , particle at x :, 1 2, 1 2, 2 2, A) zero, B) Kx C) Kx D) Kx, 2, 4, 5, The coefficient of friction between a particle, moving with some velcoity V0 and the rough, , V0 , horizontal surface is 2 gt . Find how much, 0 , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , 9., , WORK POWER ENERGY, , kinetic energy is lost in time t0 due to friction:, A) 1/4, B) 1/2, C) 3/4, D) 2/3, A block A of mass m slides on a smooth slider, in the system as shown. A block c of same mass, hanging from a pulley pulls block A. When the, block A was at position B, the spring was, unstretched. Find the speed of the block A, when AB = OB = L, , 11. A and B are smooth light hinges equidistant, from C, which can slide on ABC. The spring of, force constant K is fixed at its one end C and, connected to light rods AD and BD at point D., A block of mass m is suspended at D. Find the, velocity of the block, when ∠CAD changes, from 30° to 45° . AD = BD = L, A, , C, , B, , (m), A, B, , 90°, Spring constant = K, D, , C, O, , gL KL2 2 , −, A) , , m , 2, , , 1, 2, , , KL2, gL, −, B) , 2m, , , (, , 2, 2 −1 , , , ), , 1, 2, , 1, 2, , 1, 2, , 2, gL KL2 2 , , 2 KL2, gL, 2, 1, −, −, C) , D) 2 − m , m, , , , , 10. A ring ‘A’ of mass ‘m’ is attached to a stretched, spring of force constant K, which is fixed at C, on a smooth vertical circular track of radius, R. Points A and C are diametrically opposite., When the ring slips from rest on the track to, point B, making an angle of 30° with AC., ( ∠ACB = 30° ) spring becomes unstretched., Find the velocity of the ring at B, , (, , ), , A(m), , B, , D, 30°, , C, , KR 2, 2− 3, A) , 2m, , ), , KR 2, 2− 3, B) , m, , ), , (, , (, , 2 −1, , NARAYANAGROUP, , ), , ), , , KL2, B) gL 2 −, 2m, , , C), , , gL, , , (, , , , (, , KL2, 2 −1 −, 4m, , 2, 2 −1 , , , D) g L −, , , , 1, , 2, 2 −1 , , , ), , ), , (, , KL , , 2m , 2, , 2, , ), , 1, , 2, , 1, 2, , 12. Three springs A,B and C each of force constant, K, are connected at O. The other ends of B, and C can slide on smooth sliders. A pan is, hanging from other end of the spring A. When, a block of mass m is placed int he pan, find, the amount of workdone by the gravity on, block system after it stops vibrating. The, spring C does not sag:, , 1, , 1, , 2, + gR , , , 2, , (, , (, , (, , 1, , 2, 2 −1 , , 2, , 2, + gR 3 , , , 2, , 2 KR 2, 2− 3, C) , m, KR 2, D) , 2m, , , KL2, A) gL −, 2m, , , ), , 1, , 2, , 2, + gR 3 , , 1, , 2, , 2, + gR , , , A), , 3m2g2, 2K, , B), , 2 2, mg, K, , C), , 2m2g2, K, , D), , m2g2, 2K, 229
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 13. A rope of length l and mass ‘m’ is connected, to a chain of length l and mass 2m and hung, verticlly as shown. what is the change in, gravitational potential energy if the system is, inverted and hung from same point., , 16. A small body A starts sliding from the height h, down an inclined groove passing into a half circle of radius h/2 (see figure). Assuming the, friction to be negligible, find the velocity of, the body at the highest point of its trajectory, (after breaking off the groove)., , l, , m, X, , l, , 2m, , A) mgl, B) 4mgl, C) 3mgl D) 2mgl, 14. In the figure shown all the surfaces are, frinctionless and mass of block m=1kg, block, and wedge are held initially at rest, now wedge, isgiven a horizontalacceleration of10m /s2 by, applying a force on the wedge so that the block, does not slip on the wedge, the work done by, normal force in ground frame on the block in, 3 sec is, 2, , 10 m/s, m, , A), , 9, gh, 27, , B), , 8, gh, 27, , 27, gh, 8, , D), , 10, gh, 27, , 17. In the figures (a) and (b) AC, DG and GF are, fixed inclined planes, BC = EF = x and AB =, DE = y. A small block of mass M is released, from the point A. It slides down AC and, reaches C with a speed V C. The small block is, released from rest from the point D. It slides, down DGF and reaches the point F with speed, VF. The coefficients of kinetic frictions between, the block and both the surfaces AC and DGF, are m. Calculate V C and V F., A, , M, , C), , D, , θ, , G, , A) 30J B) 60J C) 150J D) 100 3 J, 15. A ring of mass m can slide over a smooth, vertical rod. The rod is connected to a spring, 4mg, of force constant K =, where 2R is the, R, , natural length of the spring. the other end of, the spring is fixed to the ground at a horizontal, distance 2R from the base of the rod. The mass, is released from the height of 1.5R from, ground, then work done by spring, velocity of, the ring as it reaches the ground is, , 3R/2, A, , B, , C, (a), , E, , F, (b), , A) 1.7 m/s B) 2.7 m/s C) 3.7 m/s D) 0.7 m/s, 18. A 0.5 kg block slides from the point A (see, figure) on a horizontal track with an initial, speed of 3m/s towards a weightless horizontal, spring of length 1 m and force constant 2, Newton/m.The part AB of the track is, frictionlessand the part BC has the coefficients, of static and kinetic friction as 0.22 and 0.2, respectively. If the distances AB and BD are, 2m and 2.14 m respectively find the total, distance through which the block moves before, it comes to rest completely (Take g = 10 m/, s2)., , 2R, , mgR, , 2 gR, 2, , B) mgR, 2 gR, , mgR, , 2 gR, C), 2, , mgR, , gR, D), 2, , A), , 230, , A, B, D, C, A) 4.20 m B) 4.14 m C) 4.24 m D) 4.26 m, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, 19. A block of mass 1 kg kept over a smooth, surface is given velocity 2 m/s towards a spring, of spring constant 1 N/m at a distance of 10m., Find after what time block will be passing, through P again, P, m, , WORK POWER ENERGY, at a symmetrical position with respect to the, position of release as shown in the figure. If, Vc and Va be the minimum speeds in clockwise, and anticlock wise directions respectively,, given to the bob in order to hit the peg ‘B’, then ratio Vc : Va is equal to, PegB, , k, V, , PegA, , A) ( 20 + 2π ) sec, , B) 10sec, , C) (10 + 2π ) sec, , D) (10 + π ) sec, , 20. A body is displaced from ( 0, 0 ) to, , (1m,1m ) along, , the path x = y by a force, , ur, F = x 2 $j + yi$ N . The work done by this force, , ), , will be, 4, 5, 3, 7, A) J, B) J, C) J, D) J, 3, 6, 2, 5, 21. Forces acting on a particle moving in a straight, line varies with the velocity of the particle as, α, F = where α is constant. The work done by, υ, this force in time interval ∆t is:, 1, α∆t C) 2α∆t D) α 2 ∆t, 2, 22. A particle of mass m initially at rest starts, moving from point A on the surface of a fixed, smooth hemisphere of radius r as shown. The, particle looses its contact with hemisphere at, point B. C is centre of the hemisphere. The, equation relating θ and θ ' is, , A) α∆t, , 30°, , 30°, , 10 m, , (, , Bob, , B), , A, B, r, , (A) 1:1, , (B) 1: 2 (C) 1: 2, , (D) 1:4, , PREVIOUS IIT QUESTIONS, 24. A wind -powered genrator converts wind, energy into electrical energy. Assume that the, generator converts a fixed fraction of the wind, energy intercepted by its blades into electrical, energy . For wind speed υ , the electrical, powwer output will be proportional to, [IIT-2008], 2, 3, A) υ, B) υ, C) υ, D) υ 4, 25. An ideal spring with spring constant k is hung, from the ceiling and a block of mass M is, attached to its lower end. The mass is released, with the spring initially unstretched Then,the, maxumum extension in the spring is [IIT-2002], 4Mg, 2Mg, Mg, Mg, A), B), C), D), k, k, k, 2k, 26. If W1 , W2 and W3 represent the work done in, moving a particle from A to B along three, different paths 1,2 and 3 respectively (as, shown ) in the gravitational field of a point, mass m, find the correct relation between, W1 ,W2 and W3, [IIT-2003], B, , θ, , m, , θ', C, , (A) 3sin θ = 2cos θ ' (B) 2sin θ = 3cos θ ', (C) 3sin θ ' = 2 cos θ (D) 2sin θ = 3cos θ ', 23. A bob attached to one end of a string, other, end of which is fixed at peg A. The bob is taken, to a position where string makes an angle of, 300 with the horizontal. On the circular path, of the bob in vertical plane there is a peg ‘B’, NARAYANAGROUP, , 1, , 2, , 3, , A) W1 > W2 > W3, , B) W1 = W2 = W3, , C) W1 < W2 < W3, , D) W2 > W1 > W3, 231
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 27. A particle is acted by a force F=kx, where k is, a +ve constant. Its potential energy at x=0 is, zero. which curve correctly represents the, variation of potential energy of the block with, [IIT-2004], respect to x ?, U, , U, , X, , A), , X, , 30. A tennis ball is dropped on a horizontal smooth, surface. It bounces back to its original position, after hitting the surface. The force on the ball, during the collision is proportional to the length, of compression of the ball. Which one of the, following sketches describes the variation of, the kinetic energy K with time t most, appropriately? The figures are only illustrative, and not to the scale., (IIT-2014), , B), K, , A), U, , U, , t, , X, , C), , X, , D), , K, , B), t, , 28. A bob of mass M is suspended by a massless, string of length L. The horizontal velocity υ at, just sufficient ot make it reach the point B.The, angle θ at which the speed of the bob is half, of that A, satisfies [IIT-2008], , K, , C), t, , B, , K, , D), t, L θ, P, , π, π, <θ <, 4, 2, π, 3π, 3π, <θ <π, C) < θ <, D), 2, 4, 4, 29. The work done on a particle of mass m by a, , A) θ =, , force, , π, 4, , B), , , x, y, k, iˆ +, 3/2, 2, ( x 2 + y 2 )3 / 2, ( x + y2 ), , , , ˆj , , , , A, , (k, , being a constant of appropriate dimensions), when the particle is taken from the point ( a, 0 ), along a circular path of radius a bout the origin, in the x-y plane is, [IIT-2013], 2kπ, kπ, kπ, a), b), c), d) zero, a, a, 2a, 232, , 31) A wire, which passes through the hole in a, small bead, is bent in the form of quarter of a, circle. The wire is fixed vertically on ground, as shown in the figure. The bead is released, from near the top of the wire and it slides along, the wire without friction. As the bead moves, from A to B, the force is applies on the wire is, (IIT-2014), , 90°, , B, , a) always radially outwards, b) always radially inwards, c) radially outwards initially and radially inwards, later., d) radially inwards initially and radially outwards, later., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , MULTIPLE ANSWER QUESTIONS, 32. The potential energy of a particle moving along, , x -axis is given by U = 20 + 5sin ( 4π x ) , where, U is in J and x is in metre under the action of, conservative force:, A) if total mechanical energy is 20 J,, then at x = 7/8m, particle is at equilibrium, B) if total mechanical energy is 20 J,, then at x = 7/8m particle is not at equilibrium, C) if total mechanical energy is 20 J,, then at x = 3/8m, particle is at equilibrium, D) if total mechanical energy is 20 J,, then at x = 3/8m, particle is not at equilibrium, 33. A block of mass 1 kg moves towards a spring, of force constant 10 N/m. The spring is, massless and unstretched. The coefficient of, friction between block and surface is 0.30., After compressing the spring, block does not, return back: (g = 10 m/s), A) the maximum value of speed of block for which, it is possible is 3.8 m/s, B) the maximum value of speed of block for which, it is possible is 4.2 m/s, C) if Ei and E f are initial and final mechanical energy,, which is sum of kinetic energy and potential energy,, then work done by friction on a system is (Ei - Ef), D) statement in option (C) is wrong, 34. The spring constant of spring A is twice the, spring constant of spring B. Each of the spring, is cut into two pieces. First piece of spring A is, (4/5) of the total length. Second piece of spring, B is (5/6) of its total length. Both springs are, of equal length initially:, ., A) the ratio of force constant of first piece of spring, B to the first piece of spring A is (12/5), B) the ratio of force constant of first piece of spring, B to the first piece of spring A is 2, C) the ratio of force constant of second piece of, spring A to the first piece of spring B is 5/3, D) the ratio of force constant of second piece of, spring A to the first piece of spring B is 7/5, 35. A particle of mass 1 kg is moving along Xaxis. Its velocity is 6 m/s at x = 0 ., Acceleration-displacement curve and potential, energy-displacement curve of the particle are, shown:, NARAYANAGROUP, , a 4, U(J), , (m/s2), 6, , 160, 0, , 10, , t, , x(m), , 6, , –160J, , A) the work done by all the forces is 704 J, B) the work done by external forces is 350 J, C) the work done by external forces is 384 J, D) the work done by conservation forces is 300J, 36. A particle sides down from rest on an inclined, plane of angle θ with horizontal. The distances, are as shown. The particle slides down to the, position A, where it velocity isν, p, h, , S, A, , H, , θ, , A) (ν 2 − 2gh ) will remain zero, B) (ν 2 − 2gs sin θ ) will remain zero, ν 2 − 2gs ( H − h ) , will remain zero, C) , ( p − s), , , , 2 2gsH , D) ν −, will remain zero, p , , 37. A particle is taken from point A to point B under, the influence of a force field. Now it is taken, back from B to A and it is observed that the, work done in taking the particle from A to B is, not equal to the work done in taking it from B, to A. If W nc and W c is the work done by nonconservative forces and conservative forces, present in the system respectively, ∆U is the, change in potential energy, ∆ k is the change, kinetic energy, then, A) Wnc – ∆U = ∆D B) Wc = – ∆U, C) Wnc + Wc = ∆k, D) Wnc – ∆U = – ∆k, 233
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 38. An engine is pulling a train of mass m on a, level track at a uniform speed u. The resistive, force offered per unit mass is f., A) Power produced by the engine is mfu., B) The extra power developed by the engine engine, to maintain a speed u up inclined plane of h in s is, , C) The frictional force exerting on the train is mf, on the level track, D) None of above is correct, 39. The alternative that gives the conservative, force of the following is, uur, uur, A) F1 = 2 xyi$ + x 2 $j, B) F2 = y 3 $i + xy 2 $j, uur, , C) F3 = yi$ + x $j, D) F4 = xy 2 $i + x2 $j, 40. A man is standing on a plank which is placed, on smooth horizontal surface. There is, sufficienet friction between feet of man and, plank. Now man starts running over plank,, correct statement is/are, A) Work done by friction on man with respect to, ground is nagative, B) Work done by friction on man with respect to, ground is positive, C) Work done by friction on plank with respect to, ground is positive, D) Work done by friction on man with respect to, plank is zero, 41. A small sphere of mass m suspended by a, thread is first taken a side so that the thread, forms the right angle with the vertical and then, released, then, (A) Total acceleration of sphere as a function of θ, measured from the vertical is g 1 + 3cos 2 θ, (B) Thread tension as a function of θ measured, from the vertical is T = 3mg cos θ, (C) The angle θ between the thread and the vertical, at the moment when the total acceleration vector, of the sphere is directed horizontally is cos −1 1/ 3, (D) The thread tension at the moment when the, vertical component of the sphere’s velocity is, maximum will be mg., 42. A particle P is initially at rest on the top pf a, smooth hemispherical surface which is fixed, on a horizontal plane. The particle is given a, velocity u horizontally. Radius of spherical, surface is a., 234, , u, θ, , a, , (A) If the particle leaves the sphere, when it has, , mghν, s, , uur, , P, , ga, a, ,u =, 4, 2, (B) If the particle leaves the sphere at angle θ (fig), , fallen vertically by a distance of, , ag, 3, , then u =, 2, 3, (C) If u = 0 and the particle just slides down the, hemispherical surface, it will leave the surface when, , where cos θ =, , 2, 3, (D) The minimum value of u, for the object to leave, cos θ =, , the sphere without sliding over the surface is ag ., , COMPREHENSION TYPE QUESTIONS, Comprehension-I, The potential energy U (in J) of a particle is given, by ( ax + by ) , where a and b are constants. The, mass of the particle is 1 kg and x and y are the, coordinates of the particle in metre. The particle is, at rest at ( 4a, 2b ) at time t = 0 ., 43. Find the speed of the particle when it crosses, x-axis, A) 2 a 2 + b2, , B), , a 2 + b2, , a 2 + b2 ), 1 2, (, 2, a, +, b, C), D), 2, 2, 44. Find the speed of the particle when it crosses, y-axis, , A) 4 a 2 + b2, C), , 2 ( a 2 + b2 ), , (, , B) 2 2 a 2 + b 2, D), , (a, , 2, , ), , + b2 ), , 45. Find the acceleration of the particle, A) 4, C), , (a, , 2, , + b2 ), , 2 ( a 2 + b2 ), , B) 2 2 ( a 2 + b 2 ), D), , (a, , 2, , + b2 ), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , 46. Find the coordinates of the particle at t = 1, second, A) ( 3.5a,1.5b ), , B) ( 3a, 2b ), , C) ( 3a,3b ), , D) ( 3a, 4b ), , (A), , Comprehension - II, A block of mass m sits at rest on a frictionless table, in a rail car that is moving with speed vc along a, straight horizontal track (fig.) A person riding in the, car pushes on the block with a net horizontal force, F for a time t in the direction of the car’s motion., , Ground, , 51. In terms of F, m, & t, how far did the force, displace the object according to the person in, car?, Ft 2, m, , (B), , F, , Ft 2, + 2v c t, 2m, , (B), , Ft 2, + vct, 2m, , (C), , Ft 2, + vct, m, , (D), , Ft 2, − vc t, 2m, , In the figure the variation of potential energy of a, particle of mass m = 2kg is represented w.r.t. its xcoordinate. The particle moves under the effect of, this conservative force along the x-axis., U(in J), 10, , C), , B) Vc –, , Ft, – Vc, m, , F2t 2, 2m, , 2Ft, m, , D) zero, , (B), , F2t 2, 2F 2 t 2, (C), (D) none of these, m, m, , 50. According to the person on the ground. The, change in KE of block is, 2, , 2, , Ft , , m Vc + , 2, m mvc, (A) , −, 2, 2, , Ft , , m Vc + , 2, m mvc, (B) , +, 2, 2, 2, , 2, (C) mv c, , 2, , Ft , , m Vc + , m (D) None of these, − , 2, , NARAYANAGROUP, , –5, , 5, , –10, , 2, , 10 X(in metre), , –15, , D) zero, , 49. How much did K.E of the block change, according to the person in the car?, (A), , 15, , 12, , Ft, m, , 48. According to a person standing on the ground, outside the train?, Ft, m, , 4Ft 2, m, , 20, , 47. What is the final speed of the block according, to a person in the car?, , A) Vc +, , (D), , (A), , S, , C) −, , 2Ft 2, m, , 52. According to the person on the ground. The, displacement of block is, , S1, , 2Ft, B), m, , (C), , Comprehension-III, , Train, m, , Ft, A), m, , Ft 2, 2m, , 53. If the particle is released at the origin then, (A) it will move towards positive x-axis., (B) it will move towards negative x-axis., (C) it will remain stationary at the origin., (D) its subsequent motion cannot be decided due, to lack of information., 54. x = – 5 m and x = 10 m positions of the particle, are respectively of, (A) neutral and stable equilibrium., (B) neutral and unstable equilibrium., (C) unstable and stable equilibrium., (D) stable and unstable equilibrium., , Passage-IV, Rod AO3 of length L can rotate about A. Initially, rod was at potition AO2 , when spring OB of force, constant K, attached to block B of mass m was at, position OA with unstretched length L. The smooth, block B can slide on rod when pulled by the block, D of mass m through a massless spring and smooth, pulley at O1 ., 235
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 55. Find the velocity of the block B, when the rod, and spring at B make an angle of 30° with, their respective initial positions : (B is the, middle point of the block), A, , C, , O2, , B, O1, D, , (, , 1, , ), , 10mgL − KL2 2 − 3 2 2, 2mgL − KL2, , , , A) , B) , 8m, 4m, , , , , 5mgL − KL2, C) , 4m, , , (, , 1, , ), , 2 −1 2, , , , , 6mgL − KL2, D) , 4m, , , ), , 1, , ), , 1, , (, , 2 −1 2, , , , , (, , 2 −1 2, , , , , 56. Find the work done by the frictional force (if, slider is rough) at the instant when rod and, the spring attached at block B make an anlge, of 30° with their respective initial positions., A), , (, , 1 2, KL 2 − 3, 2, , (, , 2, , − mgL, mgL, 4, , 1 2, KL 2 − 3, 8, , (, , ), , 5, − mgL, 4, , (, , ), , 2, , 2, , < θ ≤ 900 ) if 0 < u ≤ 2 gR, (iv) The magnitude of tension at a height ‘h’ is, calculated, by, using, formula, 0, , M 2, u + [ gR − 3gh], R, 57. If R = 2m, M = 2kg and u = 12m / s . Then, value of tension at lowest position is, (A) 120 N (B) 164 N (C) 264 N (D) zero, 58. Tension at highest point of its trajectory in, above question will be, (A) 100 N (B) 44 N (C) 144 N (D) 264 N, 59. If M = 2kg , R = 2m and u = 10m / s . Then, , velocity of particle when θ = 600 is, (A) 2 5 m / s, (B) 4 5 m / s, (C) 5 2 m / s, (D) 5 m / s, PASSAGE-II:, A bead of mass m is threaded on a smooth circular, wire centre O, radius a, which is fixed in vertical, plane. A light string of natural length ‘a’, elastic, constant =, , −, , 2, B) KL 2 − 3, , C), , ), , ), , (0, , T=, , O3, , θ, , (ii) Particle will complete the circle if u ≥ 5 gR, (iii) Particle will oscillates in lower half, , 3mg, a, , connects the bead to the lowest point A of the wire., The other end of the string is fixed to ring at point, B near point A. The string is slaked initially. The, bead is projected from A with speed u., , 2, 1 2, KL 2 − 1, 2, PASSAGE-I:, A particle of mass M attached to an inextensible, string is moving in a vertical circle of radius. R about, fixed point O. It is imparted a velocity u in horizontal, direction at lowest position as shown in figure., , D), , u, , P, , h, , Following information is being given, (i) Velocity at a height h can be calculated by using, formula v 2 = u 2 − 2 gh, 236, , and breaking strength 3mg, , A, , P, u, , B, 60. The smallest value u0 of u for which the bead, will make complete revolutions of the wire, will, be, (A) u0 = 5 ga, , (B) u0 = 6 ga, , (C) u0 = 7 ga, , (D) u0 = 2 ga, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , 61. If v = 2u0 , the tension T in the elastic string, when the bead is at the highest point B of the, wire is, , 4u02, , 3mu02, (A), (B) 4mg (C) 2mg (D) a − g m, a, , , 62. The elastic energy stored in the string when, the bead is at the highest point B will be, (A), , 3mga, 2mga, (B) 2mga (C) 4mga (D), 2, 2, , MATRIX MATCHING TYPE QUESTIONS, 63. A spherical ball of mass m is kept at the, highest point in the space between two fixed,, concentric spheres A and B as shown. The, sphere A has radius R and sphere B has a, radius ( R + d ) . All surfaces are smooth. The, diameter of ball is slightly less than d. The ball, is given a gentle push so that angle made by, radius vector of the ball with vertical is θ . N A, and N B are the magnitudes of normal reaciton, forces on the ball exerted by spheres A and B, respectively:, h, θ, R, , A, , r, 64. The velocity of a particle is v = ati$ + bt 2 $j,, where t is the time in second. Match the, columns for t = 1 second:, Column-I, A) Acceleration of particle is, B) Tangential acceleration is, C) Radial acceleration is, D) Radius of curvature of path is, Column-II, p) less than ( a 2 + b 2 ), r) less than ( a 2 + b 2 ), , 3/2, , q) less than ab, s) greater than 2b, , 65. A particle of 500 gm mass moves along a, horizontal circle of radius 16m such that, normal acceleration of particle varies with time, as an = 9t 2, Column - I, Column - II, (a) Tangential force, p) 72, on particle at t = 1, second (in newton), (b) Total force on, q) 36, particle at t = 1, second ( in newton), (c) Power delivered by r) 75, total force at t = 1 sec ( in watt), (d) Averge power, s) 6, developed by total force over, first one second (in watt), , INTEGER ANSWER TYPE QUESTIONS, d, , B, , Match the columns:, Column-I, −1 2 , A) θ ≤ cos , 3, , −1 3 , B) θ ≤ cos , 4, , −1 3 , C) θ ≥ cos , 4, Column-II, , −1 2 , D) θ ≥ cos , 3, , p) N B =0 and N A = mg ( 3cos θ − 2 ), q) N B =0 and N A = mg ( 4cosθ − 2 ), r) N A =0 and N B = mg ( 2 − 3cos θ ), s) none of these, NARAYANAGROUP, , 66. A ball leaves the track at B which is at 3m, height from bottom most point of the track., The ball further rises upto 4m height from the, bottom most point before falling down. Find h, ( in m), if the track at B makes an angle 300, with horizontal., , B, h, , 3m, , 67. The displacment x (in m), of a particle of mass, m (in kg) is related to the time t (in second), by t = x + 3 . Find the work done in first six, second. (in mJ), 237
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , 68. Block A of mass 1kg is placed on the rough, surface of block B of mass 3 kg. Block B is, placed on smooth horizontal surfac. Blocks, are given the velocities as shown. Find net, work done by the frictional force. [ in (-) ve J], 4m/sec, , B, , 69. A block of mass 2kg is placed on an inclined, plane of angle 53°, attached with a spring as, shown. Friction coefficient between block and, the incline is 0.25. The block is released from, the rest and when spring is in natural length., Find maximum speed of the block it acquires, after the release in cm/s is found to be nearly, 5n. Find ‘n’(take g = 10 m/s2), 8, k=, , /m, 45N, , 2kg, µ = 0.25, fixed, , A, , h = 3m, , x2, , L, 8, , u, B, , 72. The sphere at P is given a down ward velocity, v0 and swings in a vertical plane at the end of, a rope of l = 1m attached to a support at O., The rope breaks at angle 300 from horizontal,, knowing that it can withstand a maximum, tension equal to four times the weight of the, sphere. Then the value of v0 will be, , 70. Figure shows a light, inextensible string, attached to a cart that can slide along a, frictionless horizontal rail aligned along an x, axis. The left end of the string is pulled over, a pulley, of negligible mass and friction and, fixed at height h = 3m from the ground level., The cartslides from x1 = 3 3 m to x2 = 4 m, and during the move, tension in the string is, kept constant 50 N. Find change in kinetic, energy of the cart in joules. (Use 3 = 1.7) in, form of 10 x n, where n =, , x1, , 71. A particle is suspended vertically from a point, O by an inextensible massless string of length, L. A vertical line AB is is at a distance of L/8, from O as shown. The object is given a, horizontal velocity u. At some point, its motion, ceases to be circular and eventually the object, passed through the line AB. At the instant of, crossing AB, its velocity is horizontal. Find u., [1999], 238, , O, , 8m/sec, , A, , 53°, , A, , ( g = 10m / s ), 2, , A, , l = 1m, θ, , O, , v0, , 73. A block of mass 0.18kg is attached to a spring, of force-constant 2 N/m. The coefficient of, friction between the block and the floor is 0.1, Initially the block is at rest and the spring is, un-stretched. An impulse is given to the block, as shown in the figure.The block slides a, distance of 0.06m and comes to rest for the, first time . The initial velocity of the block in, m/s is V = N /10 . Then N is:, [IIT-2011], , 74. A particle of mass 0.2 kg is moving in one, dimension under a force that delivers constant, power 0.5W to the particle . If the initial speed, (in ms −1 ) after 5s is, [IIT-2013], NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , SUBJECTIVE TYPE QUESTIONS, , COMPREHENSION TYPE, 43) A 44) B 45) D 46) A 47) A 48) A 49) A, 50) B 51) B 52) B 53) D 54) D 55) A 56) C, 57)B 58)B 59) B 60)C 61) D 62) A, MATRIX MATCHING TYPE, 63)A-P, B-S, C-S, D-R 64) A-s; B-r; C-q; D-p, 65) A-s,B-r,C-p,D-q, INTEGER ANSWER TYPE, 66) 7 67) 0 68) 6 69)10 70) 5, , 75. An observer and a vehicle, both start moving, together from rest with accelerations 5 m/s2, and 2 m/s2, respectively. There is a 2 kg block, on the floor of the vehicle, and µ = 0.3, between their surfaces. Find the work done, by frictional force on the 2 kg block as, observed by the running observer, during first, 2 seconds of the motion., 2m/s2, , , , 5m/s, 2kg, , 75) 24 J, 76. Two blocks A and B are placed one over other., Block B is acted upon by a force of 20 N which, displaces it through 5 m. Find work done by, frictional force on block A, A 2kg, 3kg, B, , µ = 0.5, 20N, , m0 =, , dm, dx, , 1 dm dx 2 1, 3, = , v = m0 v, 2 dx dt , 2, , 2., , 1, mv0 2 = mgR (1 − cos θ ), 2, , v0 2, = R − R cos θ = required height, 2g, , 2g, , 3., , 2mgh, 1, 2, K1 x 2 = mgh (or) x =, K1, 2, 2µ mgx0, 1, 2, K 2 x 2 = µ mgx0 (or) x =, K2, 2, , LEVEL-VI - KEY, , NARAYANAGROUP, , 77) 2 tan −1 ( 2 ), , d 1, 1 dm , d, ( KE ) = dt 2 mv2 = 2 dt v 2, dn, , , , , , Frictionless, , SINGLE ANSWER TYPE, 1) A 2) A 3) D 4) D 5) C 6) A, 8) C 9) B 10) B 11) C 12) C 13) A, 15) A 16) B 17) A 18) C 19) D 20) B, 22) C 23) C 24)C 25)B 26)B 27)A, 29)D 30)B 31)D, MULTIPLE ANSWER TYPE, 32) A,C, 33) A,C, 34) A,C, 36) ABCD 37) ABC, 38) ABC, 40) B,C,D 41) ABC, 42)ACD, , 76) 40 J, , LEVEL-VI - KEY, SINGLE ANSWER TYPE QUESTIONS, 1., , 77. A block of mass m is placed inside a smooth, hollow cylinder of radius R kept horizontally ., Initially system was at rest . Now cylinder is, given constant acceleration 2 g in the horizontal, direction by external agent . Find the, maximum angular displacement of the block, with the vertical., , 3 3, , 71) u = gL 2 + 2 , , , 72) 5 73) 4 74) 5, SUBJECTIVE TYPE QUESTIONS, , 2, , 7) A, 14) C, 21) A, 28)D, , h K2, equating x0 = µ . K, 1, , 4., , 5., 35) A,C, 39) A,C, , The accelerations of the blocks along the, string are equal; now apply F = ma for both, the blocks., T sin θ = ma and T cos θ = mg, So, a = g tan θ, Now, T =, , mg, = mg 2 (given), cos θ, 239
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WORK POWER ENERGY, , cos θ =, , JEE-ADV PHYSICS-VOL - II, 1 2, 1, Kx + mg ( AD ) = mv 2, 2, 2, x = AC − CB = 2R − 2R cos30°, , 1, , or θ = 450, 2, , and a = g tan θ = g tan 450 = g, , (, , = R 2− 3, , 1 2, workdone, W = T.S = T at 2, 2 , , (, , ), , 7., , = ( AC sin 30° ) cos 60° =, , So,, , V = V0 − µ gt0, , KR, v=, m, , 12, , 2E , 2E , At x = , , v=, , K , m , 2, or v =, , 12, , 2E , So, at x = , , kinetic energy is equal to total, K , mechanical energy., 12, , is zero, , Finally, CD = L sin 45° =, , 1, of initial K.E, 4, Decrease in PE = Increase in KE + Increase in, elastic PE, , (, , ), , 2 −1, , 2, , 1, , 2, 2 −1 , , , ), , L, 2, , L, 2, , Increase in elastic PE + Increase in KE = Decrease, in PE, 2, , 1 L, L, 1, L, L, K, − + mv2 = mg , − , 2 2, 2, 2, 2 , 2, , , v = gL, , , (, , ), , K L2, 2 −1 −, 4m, , (, , 2 −1, , ), , 1, , 2, , 2, , , , 2mg, 2mg, , W = mg .x = mg., k, K, 16. v 2 (at end of track) = 2gh., K ' x = mg, , =, , (, , , + gR , , , 1, 2, , there is no effect of spring C., , 2, , 1 v0 1 1, 2, K.E left = m = mv0 , 2 2, 42, , , , KL2, v = gL −, 2m, , , R 1 2, = mv, 2 2, , + mg, , K, 1, 1 1, = +, ; K'=, K' K K, 2, , velocity left = v0 / 2, , 1, 1, 1, mgL = × ( 2m ) v 2 + Kx 2 = mv 2 + KL2, 2, 2, 2, , 2, , 2, , 12. The system will adjust in such a way by sliding the, spring C remains unstretched and spring A and B, remains vertical. Thus, effective force constant is, giveny by, , v1 = v0 − ( µ g ) t0, v , = v0 − 0 gt0 = v0 / 2, 2 gt0 , , 9., , (2 − 3), , On solving,, , So, U ( x ) = 0, 8., , 2, , ), , R, 2, , 11. Inititally, CD = L sin 30° =, , 1 2, 2E, or mv = E, m, 2, , 2E , Hence, PE at x = , , K , , (, , 1, KR 2 2 − 3, 2, , 1, 11, K .Ei = mV02 ; K .E f =, mV02, 2, 42, 12, , ( As ∠CBA = 90°), , AD = AB cos 60°, , 2, 1, = mg 2 ( g )(1) 2 = m g 2 / 2, 2, , 6., , ), , 2, , ; x=, , mu 2, = mg cos θ, Let body break at angle θ the h, . (1), 2, u, θ, , 10. Decrease in elastic PE + Decrease in PE = Increase, in KE, 240, , NARAYANAGROUP
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , r r, 29. W= ∫ F .dr, , 34. K A = 2 K , K B = K, , , , xiˆ, yjˆ, , dxiˆ + dyjˆ, = k∫, +, 2, 2 3/2, 2, 2 3/2 , , rA ( x + y ), (x + y ) , , rB, , (, , rB, , = k∫, rA, , (x, , xdx, 2, , +y, , rB, , = k∫, rA, , (x, , 1, 2, , +y, , rB, , = k∫, rA, , (x, , 1, 2, , +y, , ), , 2 3/2, , +, , (x, , ydy, , 2, , + y2 ), , x, d , 2, , 2, , ), , 2 3/2, , ), , 2 3/2, , (x, , 2, , ), , ( K A )2 = 5 ( 2K ) = 10 K, , 3/2, , , y , + d , , 2 , 2, , + y2 ), , r, , r, , B, 1 1, 1, = k − = k − , r rA, rA rB , , r, , But rA = a and rB = a, , ; ∴W = 0, , 30. v ∝ t ; v 2 ∝ t 2 ; K.E ∝ t 2, , MULTIPLE ANSWER CORRECT, 32. U min is at sin ( 4π x ) = −1, U min = 20 − 5 = 15 J, K max = E − U min = 20 − 15 = 5 J, 3 7, 3 7, 4π x = π , π ; x = , ,........., 2 2, 8 8, 33. The spring is compressed by x, Block will not return if µ mg ≥ Kx, , µ mg ( 0.3)(1)(10 ), =, = 0.30m, K, 10, , work done against friction = Ei − E f, 1, 1, mv0 2 − Kx 2, 2, 2, , ( 0.3)(1)(10 )( 0.3 + 2 ) = , , 1, 2, 1, 2, (1) v0 − (10 )( 0.3 ), 2, 2, , on solving , v0 = 3.8 m/sec, 242, , and ( K B ) 2 =, , 6, (K), 5, , 35. Area of ( a − t ) curve = 32 ms −1 = V f − Vi, V f = 32 + Vi = 32 + 6 = 38 ms −1, work done by all forces = ∆KE, , B, B, 1, 2rdr, dr, = k ∫ 3 d (r2 ) = k ∫, =, k, 3, ∫, 2r, 2r, r2, rA, rA, rA, , µ mg ( x + 2 ) =, , ( K B )1 = 6 ( K ), , 1, 1, m (V f 2 − Vi 2 ) = ( 382 − 6 2 ) = 704 J, 2, 2, work done by conservation forces, U i − U f = 320 J, work done by external forces = 704-320 = 384J, 36. h = s sin θ, 1 2, mv + mg ( H − h ) = mgH, 2, 1 2, mv − mgh = zero, 2, =, , rB, , So, xmax =, , 4, L, L, 5L, L , ( LA )2 = ; ( LB )1 = , ( LB ) 2 =, 5, 5, 6, 6, force constant ∝ (1/5 length), 5, 5, ( K A )1 = ( 2 K ) = K ;, 4, 2, , ( LA )1 =, , 2, v 2 − 2 gh =zero ; v − 2 gs ( sin θ ) = zero, , H ( H − h), But sin θ = P = p − s, (, ), , ( H − h), , 2, 2, So, v − 2 gs ( p − s ) = zero and v − 2 gs, , H, = zero, p, , 41. Between A and B, 1, mgl cos θ = mvB2 ; VB = 2 gL cos θ, 2, ar = 2 g cos θ ; at = g sin θ, Now at B ; TB − mg cos θ =, , mvB2, L, , Put VB ⇒ TB = 3mg cos θ, tan ( 90 − θ ) =, ⇒ cos θ =, , at 1, = tan θ ⇒ tan θ = 2, ar 2, , 1, 3, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, 42. Applying conservation of total energy, 1, 1, mu 2 + mga (1 − cos θ ) = mv 2, 2, 2, , mv 2, a, for particle to lose contact N = 0, mg cos θ − N =, , 2, v 2 = ag cos θ ; u + ga ( 2 − 3cos θ ) = 0, , COMPREHENSION QUESTIONS, , r ur, 1 ∂U $ ∂U, a, =F/m= , i+, 43., m ∂X, ∂Y, , (, , ), , $j , , , , = − ai$ + bj since, m = 1 kg, uur, uur, 44. acceleration ax = − a, a y = −b, r, acceleration a = a x 2 + a y 2 = a 2 + b 2, u x = − at , v y = −bt, , 1, a, X = 4a + ax t 2 = 4a − t 2, 2, 2, 1, b, Y = 2 b + a y t 2 = 2b − t 2, 2, 2, particle crosses x-axis, when y = 0, 45. Particle crosses y-axis, when x = 0, , 46. Coordinate at t = 1 sec will be ( 3.5a,1.5b ), 47. V = u + at = 0 +, , Ft, m, , Ft, 48. Vg = V0 +, m, , 1, 1 F 2t 2, 1 F 2t 2, mv 2 =, =, 2, 2 m2, 2 m, 1 F 2, t, 51. S train = 0t +, 2m, 1F 2, .t, 52. S ground = VC t + Strain ; = VC t +, 2m, 55. ∠ABO = 90°, (Since, ∠AOB = 30° and ∠OAB = 60° ), 49., , OB = L cos 30° =, , L, L 3, ; AB =, 2, 2, , L, BC = BA sin 30° =, 4, NARAYANAGROUP, , WORK POWER ENERGY, L, 4, Distance by which D has gone down, = AB = L sin 30° = L, Decrease in PE = Increase in KE + Increase in, elastic PE, , Distance by which B ahs gone down = BC =, , L 1, 1 , 3, mgL + mg = × 2m × v 2 + K L − L, , 4 2, 2 , 2 , , (, , ), , 2, , 1, , 10mgL − KL2 2 − 3 2 2, , , on solving, v = , , 8m, , , , 56. W = ∆KE = 0, W f + WN +W S +W g = 0, WN = Work done by normal reaction = 0 (Acts, perpendicular to displacement), WS = Work done by spring force, 1 , L 3, = 0 − K L −, , 2 , 2 , , 2, , Wg = Work done by force of gravity =, , 5, mgL, 4, , 2, 1, , , 5, 2 2− 3, , W f = − 0 − KL , +, mgL, 2 4, 2, , , , , , , (, , ), , 2, 1, 5, = KL2 2 − 3 − mgL, 8, 4, 57. Put h = 0 T = 164 N, 58. Put h = 2R T= 144 N, , R, 0, 59. At θ = 60 h = R − R cos 60 = 2, R, in v 2 − u 2 = 2 gh, 2, 60. When particle is at highest position, the elastic, force is downwards, , Put h =, , 3mg, ( 2a − a ) = 3mg, a, it v is velocity at height point at B, Fl =, , mu02, = Fl + mg − N, a, 243
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WORK POWER ENERGY, , JEE-ADV PHYSICS-VOL - II, , If V = 0, then KE at lowest point A will be, , 2, 2, v2 (a + b ), R=, =, an, ab, , 3/2, , 1 2, mu0 = [ Elastic energy + gPE ] at B, 2, , =, , 1 3mg 2, , a + mg 2a, 2 a , , 66. h0 = 3 +, , m ( 2v0 ), = 3mg + mg − N, a, 2, , = 3+, , 2, , dx, = 2t − 6, dt, at t = 0, v = −6 ; at t = 6, v = +6, v=, , N = ( 4 − 28 ) mg =-24 mg, negative sign denote it acts downwards and adds, to tension, total tension in string T = 3mg+N, , initial KE =, , 4v 2, , T = 0 − gm, a, , , final KE =, , 62. Elastic PE stored in the string, 68., , 2 3, a = mga, 2, , , d, , 63. h = R + (1 − cos θ ), 2, , , v 2 = 2 gh, , or N = mg ( 3cos θ − 2 ), Ball will lose contact with sphere A, when N = 0, 3cosθ − 2 = 0, , r, 64. v = ati$ + bt 2 $j = ai$ + b $j for t = 1 second, r, a = ai$ + 2bt $j, r, 1/2, a = ( a 2 + 4b 2 ) > 2b, , 244, , ( a 2 + b2 ), , 1/2, , 1, 2, m ( −6 ) = 18m, 2, , 1, 2, m ( 6 ) = 18m, 2, , (1 + 3) v = (1)(8) + ( 3)( 4) = 20 ; v = 5m / sec, for block A, W f =, , MATRIX MATCHING, , =, , 7 h, h ( 3), = ⇒ h= 7m, −, ;, 4 4, 4 4, , 67. x = ( t − 3) = t 2 − 6t + 9, , 4mv02, = 4mg − N, a, , ab, , 2 g ( h − hB ) sin 2 300, uB 2 sin 2 θ, = 3+, 2g, 2g, , 4 ( given ) = 3 + h sin 2 300 − hB sin 2 300, , 61. When v = 2v0 = 2 7 ga , For po int B, , 2, 2 2, , a, 2, b, +, (, ), = ( a 2 + 4b 2 ) −, 2, 2, , ( a + b ) , , , 3/2, , INTEGER ANSWER, , u 02 = 7 ga, , 1 3mg, , 2 a, , < ( a2 + b2 ), , 1, 39, (1) ( 52 − 82 ) = − J, 2, 2, , 1, 27, ( 3 ) ( 52 − 4 2 ) = + J, 2, 2, net work done by friction = -6 J, 69. Conceptual, 70. Change in kinetic energy = Work done by the force;, so W = 50 × 1 (along the string);, so W = 50 Joule, , for block B, W f =, , v, , C, D, , 90-θ, Q, , θ, mg, θ, , 1, 2, , L, .71 L 8, P, , L + Lsinθ, , u, , < ab, , Lcosθ, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - II, , WORK POWER ENERGY, , Now, we have following equations, (1) TQ = 0 Therefore, mg sin θ =, , mv 2, ...... (1), L, , (2) v 2 = u 2 − 2gh = u 2 − 2gL(1 + sin θ) ...... (2), 1, (3) QD = ( range), 2, , ........ (3), , , 3 3, u = gL 2 +, , 2 , , , ∴ f = 4N, , acc. of the block with respect to observes, = 2 − 5 = −3 m / s 2, ∴ displacement of the block w.r.to observes, , 1, = × −3 × 4 = −6 m, 2, ∴ work done by friction w.r. to observers, = −24 Joule, , 76. Limiting friction between the blocks, f L = 10 N, , mv 2, 72. T − mg sin θ =, R, 4mg − mg sin 30 =, , f L = 6 N , Fpseudo = 4 N, , [ amax ] A =, m ( v02 + 2 gl sin 30 ), l, , 10, = 5 m / s2, 2, , i.e., for shipping between A, B, Bmust move with, 5m / s2, 2, , 5m/s, , 5g, v0 =, 2, , 10N, F, , 73. Decrease in mechanical energy = work done, 1, 1 2, 2, against friction mυ − kx = ( µ mg ) x, 2, 2, , v=, , 2 µ gx + k, m, , Putting m = 0.18kg , x = 0.06m, k = 2 Nm −1 ,, µ = 01 we get, , υ = 0.4m / s =, , θ N, , 4, m/s, 10, , ∴N = 4, 74, , ⇒ F = 25 N, , but given F = 20 N, 77. Free body diagram is :, , mg, , 1, 1, mυ 2 − mu 2 = W, 2, 2, , N cosθ = mg, , 1 2, mv = Pt, 2, , ⇒ tan θ = 2, , υ=, , N sin θ = m ( 2 g ), ⇒ θ = tan −1 ( 2 ), , 2 Pt, 2 × 0.5 × 5, =, = 5ms −1, 0.2, m, , 75. FBD of the block,, , Fpseudo, f, NARAYANAGROUP, , ∴ maximum possible angular displacement, , = 2θ, = 2 tan −1 ( 2 ), , *****, , 245
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , SYSTEM OF PARTICLES AND RIGID, BODY DYNAMICS, SYNOPSIS, Centre of mass, Ø, , Ø, Ø, , CM is the point which behaves as if total mass of, the body is supposed to be concentrated at that, point. This point may lie inside (or) outside the, material of body, but always lies within the, boundary of the body. Mass may exist or may, not exist at the location of centre of mass., Centre of gravity: The point through which, weight of the body is supposed to act is called, centre of gravity., , WE-1: If the distance between the centers of the, atoms of potassium and bromine in KBr, (potassium-bromide) molecule is 0.282 × 10 −9 m ,, find the centre of mass of this two particle, system from potassium (mass of bromine =, 80 u, and of potassium = 39 u), Sol. Let position coordinate of potassium, xK = 0, Position co-ordinate of bromine,, xBr = 0.282 × 10−9 m ., ∴ Position co-ordinate of centre of mass., , xCM =, , Coordinates of CM, Coordinates of CM of a system of ‘n’ different, particles, m1 x1 + m2x2 + ......+mnxn, Xcm =, m1 + m2 +......+mn, Ycm =, , m1 y1 + m2 y2 + ......+mnyn, m1 + m2 +......+ mn, , m1 z1 + m2 z2 + ......+mn zn, m1 + m2 + ...... + mn, Case-1: Position of CM of two particle system:, Ø In case of two bodies, the ratio of distance of, centre of mass from the bodies is in the inverse, ratio of their masses. If m1 and m2 are masses, of two bodies separated by a distance ‘d’ then, the sum of moment of weights about centre of mass, is zero., d, m, m1gd1 = m 2 gd 2 ( or ) 1 = 2, d 2 m1, Zcm =, , m1, , m2, , C.M, d2, , d1, d, , In figure, d = d1 + d 2 ., On solving, m1d1 = m2 d 2, , m1d1 = m2 × ( d − d1 ), , m2 d, m1 d, ⇒ d1 = m + m and d2 = m + m, 1, 2, 1, 2, Here, d 1 , d 2 are the distances of CM from m1,m2., Thus CM locates nearer to heavier body., Note: If m1 , m2 are located at x1 , x2 from origin then, , xCM =, , m1 x1 + m2 x2, m1 + m2, , NARAYANAGROUP, , mK xK + mBr xBr, mK + mBr, , 39 × 0 + 80 × 0.282 ×10−9, = 0.189 ×10−9 m, 39 + 80, WE-2:Two blocks of masses 10 kg and 30 kg are, placed on X-axis. The first mass is moved on, the axis by a distance of 2 cm right. By what, distance should the second mass be moved to, keep the position of centre of mass, unchanged., ⇒ xCM =, , Sol., , m1, m2, Mass of the first block, m1 = 10 kg ., Mass of the second block, m2 = 30 kg ., ∆ x CM =, , m1∆ x1 + m 2 ∆ x 2, m1 + m 2, , 2, 10 × 2 + 30 ∆ x2, ∴∆x2 = − ., 40, 3, Therefore, the second block should be moved left, through a distance of 2/3cm to keep the position, of centre of mass unchanged., WE-3: When ‘n’ number of particles of masses, m, 2m,3m,....nm are at distances x1 = 1,, 0=, , x2 = 2, x3 = 3....xn = n units r e s p e c t i v e l y, , Sol., , from origin on the X-axis, then find the, distance of centre of mass of the system from, origin., m (1) + 2 m ( 2 ) + 3 m (3 ) + ... + ( nm ) n, x =, m + 2m + 3m + ... + nm, n ( n + 1)( 2n + 1) , , m (12 + 22 + 32 + .... + n2 ) , 6, = 2n + 1, xcm =, =, m (1 + 2 + 3 + .... + n ), n ( n + 1) , 3, , , 2 , , cm, , 1
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-4: When ‘n’ number of particles of masses, m, 2m,3m,....nm are at distances x1 = 1,, , x2 = 4, x3 = 9,...xn = n units respectively, from origin on the X-axis, then find the, distance of their centre of mass from origin., 2, , Sol. xCM =, , m (1) + 2 m( 4 ) + 3 m ( 9 ) + .... + nm ( n 2 ), m + 2m + 3m + .... + nm, , n ( n + 1) , , , 2, = n ( n + 1), =, ., n ( n + 1), 2, 2, 2, , m (1 + 2 + 3 + ..... + n, 3, , =, , 3, , 3, , m (1 + 2 + 3 + .... + n ), , ), , WE-5: When ‘n’ number of particles each of mass, ‘m’are at distances x1 = a, x2 = ar , x3 = ar 2 ,, ......... xn = ar n , units from origin on the, X-axis, then find the distance of their centre, of mass from origin., , ma + m ( ar ) + m ( ar ) + .... + m ( ar, 2, , Sol. xCM =, , m + m + m + ... + m ( nterms ), , m ( a + ar + ar + .... + ar, 2, , xCM =, , If r > 1 then ⇒ xCM, If r < 1 then ⇒ xCM, , n, , n, , ), , ), , mn, n, n, 1 a ( r − 1) a ( r − 1), = , =, n r − 1 n ( r − 1), n, n, 1 a (1 − r ) a (1 − r ), = , =, n 1 − r n (1 − r ), , WE-6: A 1m long rod having uniform area of cross, section is made of four materials. The first, 0.2 m is made of iron, the next 0.3 m of lead,, the following 0.2 m of aluminium and the, remaining part is made of copper. Find the, centre of mass of the rod. The densities of iron,, lead, aluminium and copper are 7.9 ×103 kg / m3 ,, 11.4 ×103 kg / m3 , 2.7 ×103 kg / m3 and 8.9 ×103 kg / m3, respectively., Sol. mass ( m ) = volume (V ) × density ( d ), m = Area ( A ) × length ( l ) × density ( d ), , m = Ald ., lm, O, , Iron, 0.2m, 0.35m, , Lead, 0.3m, , 0.2m, , Copper, 0.3m, , 0.6m, 0.85m, , Iron part mass, m1 = A×0.2×7.9×103 = 1.58 ×103 A ., , Lead part mass, m2 = A× 0.3×11.4×103 = 3.42 ×103 A ., 2, , Aluminium part mass,, , m3 = A × 0.2 × 2.7 × 103 = 0.54 × 103 A ., Copper part mass, m4 = A× 0.3× 8.9 ×103 = 2.67 × 103 A, Co-ordinate of iron part from end “O” of the rod, x1 = 0.1m ., Co-ordinate of lead part from end “O” of the rod, x2 = 0.35 m ., Co-ordinate of aluminium part from end “O” of, the rod, x3 = 0.6 m ., Co-ordinate of copper part from end “O” of the, rod, x4 = 0.85 m ., ∴ Centre of mass of the rod,, m x + m2 x2 + m3 x3 + m4 x4, xCM = 1 1, m1 + m2 + m3 + m4, , (1.58×10 ×0.1+3.42×10 ×0.35+0.54×10 ×0.6+2.67×10 ×0.85) A, (1.58×10 +3.42×10 +0.54×10 +2.67×10 ) A, 3, , ⇒xCM =, , 3, , 3, , 3, , 3, , 3, , 3, , 3, , ⇒ xCM = 0.481 m from the end “O” of the rod., WE-7:If the centre of mass of three particles of, masses of 1kg, 2kg, 3kg is at (2,2,2), then, where should a fourth particle of mass 4kg, be placed so that the combined centre of, mass may be at (0,0,0)., Sol. Let ( x1 , y1 , z1 ) , ( x2 , y2 , z2 ) and ( x3 , y3 , z3 ), be the positions of masses 1kg, 2kg, 3kg and let, the co-ordinates of centre of mass of the three, particle system is ( xcm , ycm , zcm ) respectively.., xCM =, , m1x1 + m2 x2 + m3x3, 1× x1 + 2 × x2 + 3× x3, ⇒2 =, ,, m1 + m2 + m3, 1+ 2 + 3, , ( or ) x1 + 2 x2 + 3 x3 = 12 ......(1), Suppose the fourth particle of mass 4kg is placed, at (x4,y4,z4) so that centre of mass of new system, shifts to (0,0,0). For x coordinate of new centre, of mass we have, 1× x1 + 2 × x2 + 3 × x3 + 4 × x4, 0=, 1+ 2+ 3+ 4, ⇒ x1 + 2 x2 + 3 x3 + 4 x4 = 0 ..........(2), from equations (1) and (2), 12 + 4 x4 = 0 ⇒ x4 = −3, similarly, y4 = −3 and z4 = −3, Therefore 4kg should be placed at (-3,-3,-3)., Case-2: Center of mass of a system of n particles, in one dimension:, Consider n-particles having masses, m1 , m2 ,...., mn along X-axis. The centre of mass, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-11: Two bodies of 6kg and 4kg masses have, their velocity 5iˆ − 2 ˆj + 10kˆ and 10iˆ − 2 ˆj + 5kˆ, respectively. Then, the velocity of their, centre of mass is, [E-2007 ], Sol. The velocity of centre of mass is, r, r, m1 v1 + m2 v 2, VCM = m + m = 7iˆ − 2 ˆj + 8kˆ, 1, 2, , Effect of external forces on CM, Ø, , Ø, , r, r, 1, We know a cm = ∑ mi a i, M i, r, r, r, Therefore M a cm = ∑ Fext + ∑ Fint, But the internal forces are in the form of, action - reaction pairs. Hence they cancel each, r, r, r, r, other. Thus ∑ Fint = 0 ;∴ M a cm = ∑ Fext, Thus centre of mass is effected by only external, force acting on the system. Internal forces will have, no effect on the motion of centre of mass. When, no external force acts on the system, a) acceleration of centre of mass is zero i.e.,, r, r, r, r, r, r, Fext = M a cm ⇒ M a cm = 0 ⇒ a cm = 0 ., b) Velocity of centre of mass is constant, r, vcm = constant, r, c) Linear momentum of the system is constant p cm =, constant. It is called the law of conservation, of linear momentum., , due to internal forces into many fragments, they, move randomly in different directions. But the centre, of mass follows the same parabolic path as, unexploded bomb. So at any moment the vector, sum of the moments of mass of all the fragments, about centre of mass is zero., (c) When a wheel is rolling on a road, then the paths, of various particles are complicated as they are in, combined motion (translational + rotational). But, the motion of centre of mass is purely translational, and it follows straight line path., Note: Gravitational force between two masses,electric, force between two charges are the examples of, internal forces for the system, Which cannot, produce acceleration in centre of mass of the, system., WE-12: Two particles A and B initially at rest, move, towards each other, under mutual force of, attraction. At an instance when the speed of, A is v and speed of B is 2v,the speedof centre, of mass (CM) is, [E-2008 ], Sol. Internal forces doesn’t change the position of centre, of mass. So velocity of CM is zero., , Mutual forces between two bodies :, Ø, Ø, , Characteristics of centre of mass, Ø, , Centre of mass of a system of particles depends, on mass of particles and their relative positions., Ø For continuous distribution of mass, centre of mass, depends on mass distribution and shape of the, body., Ø Sum of moments of masses about centre of mass, r r, is zero i.e., ∑ mi r i = 0, i, Ø Centre of mass is independent on frame of, reference chosen to locate it., Ø Mass need not be present at centre of mass., Ø The motion of centre of mass is purely translational., Ø The motion of centre of mass is according to, Newton’s 2nd law., Ø Examples for the motion of centre of mass, (a) When a bomb at rest at origin of x,y,z-coordinate, system explodes due to internal forces into many, fragments. These fragments fly off randomly with, different velocities in different directions. But CM, is not effected and remains at rest at the origin., r r, r, mi r i = 0 , where r is position vector of th, ∑, ∴ i, i, i, particle about origin., (b) A bomb is projected on the ground to follow, parabolic path. When it explodes during the motion, NARAYANAGROUP, , When two particles approach each other due to, their mutual interaction, then they always meet at, their centre of mass., To a system of particles m1(x1y1), m2(x2y2), another particle of mass m3 is added so that, centre of mass shifts to the origin then coordinates, of third particle are, x3 = −, , Ø, , (m1 x1 + m 2 x2 ), m3, , ; y3 =, , − ( m1 y1 + m2 y2 ), m3, , In a system of two particles of masses m1 and m2 ,, when m1 is pushed towards m2 through a distance, d then shift in m2 towards m1 without altering CM, −m1, , Ø, , position is m d ., 2, A boy of mass m is at one end of a flat boat of, mass M and length l which floats stationary on, water. If boy moves to the other end,, i) The boat moves in opposite direction through, ml, , a distance d = ( M + m ), ii) The displacement of boy with respect to ground, 1, , Ø, , − Ml, , is d = ( M + m ), A boy of mass m is standing on a flat boat floating, stationary on the surface of water. If the boy starts, moving on the boat with velocity Vr with respect, to boat, then, , 5
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , i) Velocity of the boat w.r.t. ground is V =, , centre of mass of remaining part shifts by a distance, , − mVr, ,, M +m, , −r 2 d, where ‘d’ is the distance of the C.M. of, R2 − r 2, the removed part from the centre of the original, disc. In this case the circular sheet is removed from, the edge of disc, then the shift in centre of mass is, maximum. Here d = R − r ., , ‘ − ’ indicates boat moves in opposite direction to, the velocity of the boy., MVr, M +m, WE-13: A 10kg boy standing in a 40kg boat, floating on water is 20m from the shore of the, river. If he moves 8 metres on the boat towards, the shore, then how far is he from the shore, now?, Sol.Mass of the boy (m)=10kg Mass of the boat, (M)=40kg Distance travelled by boy (l)=8m, Distance travelled by the boat in opposite direction, 1, ii)Velocity of boy w.r.t. ground is V =, , =, , Maximum shift ∆x =, Ø, , From a uniform solid sphere of radius R, if a sphere, of radius ‘r’, is removed, then the centre of mass, , where ‘d’ is the distance of the CM of removed, part from the centre of the original sphere. In this, case spherical cavity is made at the edge of large, sphere, then shift in C.M. is maximum. It is given by, , Distance of the boy from the shore is, (20-8)+(1.6) = 13.6m, , Shift in centre of mass in different cases:, , Ø, , d, , −r 3d, of the remaining part shifts by ∆x = R 3 − r 3 ,, (, ), , ml, 10 × 8, =, = 1.6 m, M + m 10 + 40, , Shift is the distance of final location of centre of, mass of the system from its initial location. Shift in, the centre of mass generally occurs due to, a) Addition of matter b) Removal of matter, c) Change in shape d) Change in mass distribution, a) Addition of mass : Due to addition of mass,, the C.M of a system generally shifts towards or into, , −r 2, R+r, , ∆x =, Ø, , −r 3 ( R − r ), , (R, , 3, , −r, , 3, , ), , d, , ., , To a circular disc of radius R 1 another disc of, radius R2 and of the same material is added then, shift in the CM is, , the region where mass is added. If C1 is the CM, 2, , before addition and C2 is the CM of added mass, , ∆x =, , and C1 C2 = d , then, madded × d , ∆X shift = , , minitial + madded , , Ø, , CM shifts towards the side of added mass, Ø, , R1 + R 2, , d, , 2, , If two spheres of same material and radii r1 and r2, are kept in contact, distance of centre of mass from, the centre of the first sphere is equal to, , where mass is removed. If C1 is the CM of the, , r23, ∆x = r 3 + r 3 ( r1 + r2 ) ., 1, 2, , body before removal and C2 is the CM of the, , Similarly distance of centre of mass from the centre, , removed part and C1C2 = d then, −mremoved × d , ∆X shift = , , M initial − mremoved , , 6, , 2, , b) Removal of mass : Due to removal of mass,, the C.M of a system shifts away from the region, , Ø, , R2 (R 1 + R 2 ), , ‘ − ’ indicates CM shifts opposite to the side of, removed mass, From a uniform circular disc of radius R, if a, circular sheet of radius ‘r’ is removed, then the, , Ø, , d, , r13, of the second sphere is ∆x = 3 3 ( r1 + r2 ) ., r1 + r2, The location of CM of system depends on the, mass distribution within the system. Due to this, the location of CM changes whenever the shape, of system changes and also the relative positions, of particles change., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-14: A circular disc of radius R is removed, from a bigger circular disc of radius 2R such, that the circumferences of the discs touch., The centre of mass of the new disc is at a, distance α R from the centre of the bigger, disc. The value of α is [E-2011], Sol., , imagine a small element of radius ‘r’ and thickness, ‘dx’ at a distance x from ‘O’. Mass of small, element, dm = (π r 2 ) dx ρ . From figure,, , dx, , R, , r, x, , O, , O', , cm, , x, αR, , h, , O1, O, , O2, , r x, Rx, = ⇒ r=, R h, h, , ∫ dm x = ∫ (π r ρ dx ) x, =, ∫ dm ∫ π r dx ρ, h, , R, , xCM, , 2R, , 2, , 0, , h, , 2, , 0, , π (R ) M, 2, , mass of cutoff portion m =, , π (2R ), , 2, , M, =, 4 and, , position of its centre of mass, OO1 = R hence,, for remaining part (new disc), , xCM, , M, M ×0−, 4, =αR =, M, M −, 4, , , × R, , , R, 1, = − ⇒α = −, 3, 3, , Locating the Centre of Mass can be done in four, different ways. They are,, 1) Method of symmetry, 2) Method of Decomposition, 3) Method using theorems of Pappus’s, 4) Method of integration, For continuous distribution of mass, the coordinates of centre of mass are given by, x cm =, , ∫ xdm ; y, ∫ dm, , cm, , =, , ∫ ydm ; z, ∫ dm, , cm, , =, , distance, , the vertex on the line of symmetry is, , 3h, ., 4, , Sol. Consider a cone of height ‘h’ base radius ‘R’ and, density ρ . To find centre of mass of the cone, , 3h, from vertex on its line of symmetry.., 4, , WE-16: If the linear density of a rod of length L, varies as λ = A + Bx , find the position of its, centre of mass., Sol. Let the x-axis be along the length of the rod and, origin at one of its ends. As rod is along x-axis, for, all points on it y and z coordinates are zero., y, x, , ∫ zdm, ., ∫ dm, , WE-15: Distance of centre of mass of a uniform, cone of height ‘h’ and base radius R, from, , NARAYANAGROUP, , 2, , Therefore, centre of mass of cone is at a, , Methods to locate CM:, , Ø, , h, , x4 , h4, h R x , , , h, ∫0 h2 x dx ∫0 x3 dx = 4 0 = 4 = 3h, =, = h, 3 h, 2 2, h3, 4, h R x, , , 2, x, ∫0 h 2 dx ∫0 x dx 3 0 3, 2, , O, , X, dx, L, , z, , Centre of mass will be on the rod. Now consider, an element of rod of length dx at a distance x, from the origin, then dm = λ dx = ( A + Bx ) dx, 7
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , ∫ x dm = ∫ x ( A + Bx ) dx, =, ∫ dm ∫ ( A + Bx ) dx, L, , L, , 0, , xCM, , 0, , L, , L, , 0, , 0, , AL2 BL3, +, 2, 3 = 3 AL + 2 BL = L ( 3 A + 2 BL ), = 2, BL2, 3 ( 2 A + BL ), 3 ( 2 A + BL ), AL +, 2, , xCM, , WE-17: Identical blocks each of mass M and, length L are placed one above the other such, that each extends out by a maximum length, as shown in figure. Find the maximum, extension of the n th block from the top, so that, the blocks will not fall., , JEE-ADV PHYSICS-VOL - III, another block (GH) is placed below the three, blocks in equilibrium, then, c1, B, A, c2, D, C l, F, E l, 2, x2, 4, H, G, The centre of mass (C3) of the upper three block, must lie on the edge of the lower fourth, block i.e. at G. To find x3 consider E as origin., l, 2M (0 ) + M , 2 = l, x3 =, 3M, 6, , ∴ x3 =, , l, 6, , l, l, similarly x4 = , x5 = ,........, 8, 10, , Sol. For a two block system, the centre of mass (C1), of upper block should be at the edge of lower block, i.e. at, , l, distance. But if centre of mass of upper, 2, , block is not resting on the lower block then, the, upper block falls down because of unbalanced, torque created by gravitational force., , B, , A, , c1, , D, , C, , x, , A, , c2, , C l, x, 2, F, E 2, The centre of mass (C2) of (AB) and (CD) block, system must lie on the edge E of third block., To, find x2 consider C as origin. Then, l, M (0) + M , 2 = l, x2 =, 2M, 4, , x2 =, , at, 8, , l , of its bottom block is ., n, , l, x= , n, x, , c1, , B, , l, 2n, , Note-1: When the above blocks are arranged in such a, manner, that each block projects out by same, distance, so that the blocks will not fall then the, distance of projection of each block from the edge, , l, 2, , If a third block (EF) is arranged below the two, blocks then, , D, , for nth block xn =, , Note-2: If the entire system is placed at the edge of a, table, so that the blocks will not fall then the equal, distance of projection of each block from the edge, l , of its bottom block is , , n +1 , , x, c, , x, x, , Table, , l, So, centre of mass of upper two blocks is, 4, , l, distance from edge of lower block. Also, if, 4, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , POSITION OF CENTRE OF, MASS, , S.No, , SHAPE OF THE BODY, , 1, , Circular ring, , At the centre of the ring, , 2, , Circular disc, , At the centre of the disc, , FIGURE, , C, , C, 3, , Thin uniform straight, rod, , 4, , Triangular plate, , At the geometric centre, , A, , C, , B, , At the centroid, C, , Square plate, , 5, , At the point of intersection, of the diagonals, , C, , C, , 6, , Rectangular plate, , At the point of intersection, of the diagonals, , 7, , Hollow or solid, sphere, , At the centre of the sphere, , 8, , Hollow cone, , At a height of h/3 , from the base, , C, , h, C, , Solid cone, or, Pyramid, , 9, , At a height of h/4 from the base, , h, , h, 3, oc =, , h, 4, , OC =, , L, 2, , C, O, , 10, , Solid (or) hollow, cylinder, , At the mid-point of its own, axis, , C, , O, , NARAYANAGROUP, , l, , 9
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , S.No SHAPEOF THEBODY, , 11, , An arc of radius R, subtending an angle, α at its centre, of curvature, , i)A semi-circle of, radius ‘R’, , ii) A quadrant of a, circle of radius ‘R’, , POSITION OF CENTRE OF, MASS, a , 2R, At a distance of, sin , 2, a, , from its centre of curvature, on the axis of symmetry, , Semi-circular disc, , 13, , 14, , Hollow hemi-sphere, (or), Hemi-spherical shell, , 15, , Horse-shoe magnet, , C, α, x, , 2R, from its, p, centre on the axis of, symmetry, , 4R, from, p 2, its centre ‘o’ on the axis of, symmetry, , R, C, , R, from its, 2, centre ‘o’ on the axis of, symmetry, , OC=, , 2R, p, , C, R, π/4, , OC=, , O, , 4R, p 2, , R, C, , centre ‘o’ on the axis of, symmetry, , At a distance of, , π, , O, , 4R, from its, 3p, , 3R, from its, 8, centre ‘o’ on the axis of, symmetry, , x, , O , a, 2R, OC =, sin , 2, a, , , At a distance of, , At a distance of, Solid hemi-sphere, , R, , At a distance of, , At a distance of, 12, , FIGURE, , OC=, , 4R, 3p, , OC =, , 3R, 8, , O, , R, C, o, R, C, o, , At its centre within the, boundary limits, , OC =, , R, 2, , C, , At a distance of, 16, , semi-Circular, annular plate, , OC =, , 4 R12 + R1R2 + R22, 3p (R1 + R2 ), , from its centre of symmetry, 10, , C, R2, , R1, O, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Ø, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Method of decomposition:, In this method C.M of a body or a system can be, determined by decomposing it structurally into small, units of geometrically symmetrical parts. Some, examples are as follows, , =, , a), , b), , =, , WE-19: A thin rod of length L is lying along, X-axis with its ends at x=0 and x=L. Its linear, n, , x, density (mass/length) varies with x as k ,, L, where n can be zero or any positive number. If, the position x of the centre of mass of the rod, is plotted against n, which of the following, graphs best approximates the, dependence of X on n? [AIEEE-2008], XCM, XCM, , +, , +, , 1), , In this method each part is considered as a point, object, hence the system will be converted into the, system of n particles. Now the coordinates of CM, of system w.r.t. some reference point are, , WE-18: A uniform piece of metal sheet is cut in, the form as shown in the fig. Locate the center, of mass of the piece., Y, , Sol., , X CM, , 30, 20, , 20, , n, , XCM, L, 4) L, 2, n, n, o, L K , n, n x xdx L ( n + 1), ∫, xdm, 0, ∫ = L , =, =, L K , n+2, n, dm, ∫, ∫0 Ln x dx, , If n =0, then X CM =, , 10, Y, , L, 2) L, 2, n, o, , XCM, L, 3) L, 2, o, , 1, 1, 1, mi xi ; yCM = ∑ mi yi and zCM = ∑ mi zi, ∑, M i, M i, M i, , xCM =, , L, L, 2, o, , L, 2, , As n increases, the centre of mass shift away from, , 10, , 20 30, , X, , x=, , L, towards x = L which only option (1) is, 2, , satisfying., , 30, 20, , Rigid body: If there is no relative motion between, , 3, , Sol., , 10 1 2, 0 10 20 30, X, Let the mass per unit area be σ .We divide the, structure into parts 1,2 and 3.The mass of part-1 is, 300σ , mass of part-2 is 200σ and that of part-3is, 100σ . The coordinates of centre of mass of part1, are (5, 15); that of part 2 are (20, 5); and that of, part-3 are (15,25)., ( 300σ ) 5 + ( 200σ ) 20 + (100σ )15, X =, ≈ 11.7, 600σ, , cm, , Yc m =, , (300σ )15 + ( 200σ ) 5 + (100σ ) 25 ≈ 13.3, 600σ, , In vector notation rc m = 11.7iˆ + 13.3 ˆj, NARAYANAGROUP, , Ø, , any two particles of the body along the line joining, them by the application of external force,then that, body is called rigid body., No real body is truly rigid, since real bodies deform, under the influence of external forces., , Types of motion of a rigid body, Ø, , Ø, , Translational motion : All particles of the, body move in parallel paths such that displacements, of all the particles are same as that of the body, then its motion is said to be translational., Rotational motion :A body is said to be in, pure rotation if every particle of the body moves in, a circle with same angular velocity and the centers, of all the particles lie on a straight line called the, axis of rotation., 11
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Ø, , Rolling motion : The combination of rotational, , θ, , t, , t, , and translational motion with regard to certain, constraints is called rolling motion., , 0, , 0, , 0, , 4, 3, ∫ dθ = ∫ ωdt = ∫ (ω0 + at − bt ) dt, , Kinematics of rotational motion about, a fixed axis:, Ø, , The kinematic equations for rotational motion with, uniform angular acceleration, 1) ω = ω0 + α t, , 1, 2) θ = ω0t + α t 2, 2, , 3) ω 2 = ω02 + 2αθ, , 1, , 4) θ n = ωo + α n − , 2, , , WE-20: The motor of an engine is rotating about, its axis with angular velocity of 120 rpm.It, comes to rest in 10s, after being switched off., Assuming constant deceleration, calculate the, number of revolutions made by it before, coming to rest., Sol. Here n = 120 rpm = 2rps, , ⇒ θ = ω 0t +, , at 5 bt 4, −, 5, 4, , Moment of force (Torque), Ø, Ø, , Torque is the turning effect of a force about a fixed, point., Magnitude of the torque is given as the product of, magnitude of force and perpendicular distance of, line of action of a force from the fixed point., r, τ = F ( r sin θ ) ⇒ τr = rr × F, S.I. Unit: Nm Dimensional formula: ML2 T − 2 ., z, , τ, , ω0 = 2π n = 4π rad s −1 ; ω = 0 and t = 10s, , ω = ω0 + α t, , F, P, , 0 = 4π + α × 10 or α = −0.4π rads −2, Also, the angle covered by the motor,, , r, O, , 1, θ = ω0t + α t 2, 2, 1, ∴ θ = 4π ×10 + × ( −0.4π ) ×102 = 40π − 20π = 20π rad, 2, , Hence, the number of revolutions completed,, N=, , θ, 20π, =, = 10, 2π, 2π, , WE-21: A wheel rotates with an angular, acceleration given by α = 4at 3 − 3bt 2 ,where t is, the time and a and b are constants. If the wheel, has initial angular speed ω0 , write the, equations for the (a) angular speed (b) angular, displacement, dω, ⇒ dω = α dt, Sol. (a) Since α =, dt, Integrating both sides, we get, ω, , ∫, , ω0, , t, , d ω = ∫ α dt =, 0, , ∫ ( 4 at, , 3, , − 3bt 2 ) dt, , 0, , t, t3, − 3b ⇒ ω = ω0 + at 4 − bt 3, 4, 3, dθ, (b) Since ω =, ⇒ dθ = ω dt, dt, On integrating both sides, we get, ω − ω0 = 4a, , 12, , 4, , t, , θ, , rs, in, θ, , y, , x, , Application: A force of given magnitude applied, at right angles to the door at its outer edge is most, effective in producing rotation., Ø The moment of a force vanishes if either the, magnitude of the force is zero, or if the line of action, of the force passes through the fixed point., Ø If the direction of F is reversed, the direction of, the moment of force is also reversed., Ø If directions of both r and F are reversed, the, direction of the moment of force remains the same., Sign convention : Torque that produces anti, clockwise rotation is taken as positive and, clockwise rotation taken as negative., WE-22: A particle is projected at time t=0 from a, point ‘O’ with a speed ‘u’ at an angle ‘ θ ’ to, horizontal. Find the torque of a gravitational, force on projectile about the origin at time, ‘t’.(x, y plane is vertical plane), r, 1 2$, $ , Sol. r = ( u cos θ ) t i + ( u sin θ ) t − gt j, 2, , , ur, r, r, ur, F = − ( mg ) $j ;, τ = r×F, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , y, u, Ø, , o θ, , x, r , 1, , , τ = ( u cos θ ) t $i + ( u sin θ ) t − gt 2 $j × mg − $j, 2, , , , r, τ = ( u cos θ ) t ( mg ) $i × − $j, , ( ), , ( ), τ = −mg ( u cos θ ) t ( k$ ), , Moment of couple:, A pair of equal and opposite forces with different, lines of action is known as a couple. A couple, produces rotation without translation. If an object, is not on pivot (unconstrained) a couple causes the, object to rotate about its centre of mass., F, , d, F, This couple can produce turning effect (or) torque, on the body. Moment of couple is a measure of, turning effect (τ ) ., ∴τ = moment of couple=magnitude of either force, × perpendicular distance between the forces, ∴τ = Fd, , Mechanical Equilibrium of a rigid body:, Ø, , Ø, , A rigid body is said to be in mechanical equilibrium,, if both its linear momentum and angular momentum, are not changing with time, or equivalently, body, has neither linear acceleration nor angular, acceleration., , Condition for translational equilibrium, The vector sum of the forces, on the rigid body is, n r, r r, r, zero; F1 + F2 + .... + Fn = ∑ Fi = 0, i =1, , If the total force on the body is zero, then the total, linear momentum of the body does not change with, time. P= constant, , Condition for rotational equilibrium :, Ø, , The vector sum of the torques on the rigid body is, n, r r, r, r, τ, +, τ, +, ....., τ, =, zero, 1 2, ∑τi = 0, n, i =1, , NARAYANAGROUP, , If the total torque on the rigid body is zero, the, total angular momentum of the body does not, change with time., The rotational equilibrium condition is independent, of the location of the origin about which the torques, are taken., , Principle of moments:, Ø, , An ideal lever is essentially a light rod pivoted at a, point along its length. This point is called the fulcrum., Two forces F1 and F2 , parallel to each other and, usually perpendicular to the lever act on the lever, at distances d1 and d 2 respectively from the, fulcrum. Ν is directed opposite to the forces F1, and F2 . (N = Reaction at fulcrum) For translational, equilibrium., N – F1 – F2 = 0, N, , d1, , d2, , F2, F1, Ø For rotational equilibrium take the moments about, the fulcrum; the sum of moments must be zero,, d1 F1 − d 2 F2 = 0, N acts at the fulcrum itself and has zero moment, about the fulcrum., Ø In the case of the lever force F1 is usually some, weight to be lifted. It is called the load and its distance, from the fulcrum d1 is called the load arm. Force, F2 is the effort applied to lift the load; distance d 2 of, the effort from the fulcrum is the effort arm., Principle of moments for a lever, Load arm x load = effort arm x effort, Mechanical advantage :, Ø The ratio F1 / F2 is called the Mechanical, Advantage, F d, M . A. = 1 = 2, F2 d1, If the effort arm d 2 is larger than the load arm, the, mechanical advantage is greater than one. It means, that a small effort can be used to lift a large load., WE-23: PQR is a rigid equilateral triangle frame, of a side length ‘L’. Forces F1 , F2 and F3 are, acting along PQ, QR, PR. If the system is in, rotational equilibrium find the relation, between the forces., Sol.Perpendicular distance of any force from centroid, ‘C’ of triangle is L / 2 3 . The forces F1 , F2, produce anti-clockwise turning effect where as F3, 13
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , clockwise turning effect about ‘C’., F2, R, , C, , ∴N1 = f and N2 = W, Taking torque about ‘B’ ; N1 (AO) = W (CB), AB , 0, N1 (AB) cos530 = W, sin 53, 2, , , 2, N1 = W and N2 = W = 10 × 9.8 = 98 N., 3, 2, The frictional force is f = N1 = W= 65 N, 3, , Q, , P, , F1, F3, Since the system is in rotational equilibrium the total, torque acting on the system about the centroid is zero, L, L, L, + F2 ×, − F3 ×, =0, F1 ×, 2 3, 2 3, 2 3, Hence F1 + F2 − F3 = 0 ;∴ F3 = F1 + F2, WE-24: A metre stick is balanced on a knife edge, at its centre. When two coins, each of mass 5g, are put one on top of the other at the 12cm, mark, the stick is found to be balanced at, 45cm.What is the mass of the metre stick?, C, , Toppling:, N, , f, , N, , B, , D, , A, , C E b, , F, , B, f, , A, , W=mg, , A, , D, , F, , E b, , W=mg, , B, f, , C a, , C, , a, , N, D, , F, , E b, , Sol., , 10g, , R, , mg, Since the stick is in rotational equilibrium, the total, torque of all the forces about the resultant ‘R’ is, zero. Taking the turning effects about the point of, action of the resultant R we have, 10 g × 33 = mg × 5; m = 66 g, WE-25:A uniform ladder of mass 10 Kg leans, against a smooth vertical wall making an, angle 530 with it.The other end rests on a, rough horizontal floor.Find the normal force, and the frictional force that the floor exerts, on the ladder., Sol. The ladder is in equilibrium., , W=mg, Suppose a force F is applied at a height b above, the base AE of the block.Further , suppose the, friction ‘f’ is sufficient to prevent sliding.In this case, if the normal reaction N also passes through C then, despite the fact that the block is in translational, equilibrium (F = f and N = mg) an unbalanced, torque(due to the couple of forces F and f ) is, there.This torque has tendency to topple the block, about point E. To cancel the effect of this, unbalanced torque the normal reaction N is shifted, towards right a distance ‘a’ such that, net anti clock, wise torque is equal to the net clock wise torque., Fb, , A, , N1, N2, W, , Fb = mg(a) ⇒ a = mg, Now, as F (or) b (or) both are increased distance, a also increases. But it can not go beyond the right, edge of the block. So in extreme case the normal, reaction passes through E. Now if F or b are further, increased, the block will topple down.This is why, the block having the broader base has less chances, , of toppling in comparison to a block of smaller base., , B, 14, , f, , C, , O, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-26: A uniform cylinder of height h and radius, r is placed with its circular face on a rough, inclined plane and the inclination of the, plane to the horizontal gradually increased., If µ is the coefficient of friction, then under, what conditions the cylinder will, (a) slide before toppling, (b) topple before sliding, N, f=µN, Sol., , Mg sin θ θ, Mg, , Mg cos θ, , (a) The cylinder will slide if, Mg sin θ > µ Mg cosθ ⇒ tanθ > µ .....(1), The cylinder will topple if, h, 2r, ( Mg sin θ ) > ( Mg cos θ ) r ⇒ tan θ > ...(2), 2, h, Thus, the condition of sliding is tanθ > µ and, 2r, . Hence, the, condition of toppling is tan θ >, h, 2r, cylinder will slide before toppling if µ <, h, 2r, (b) The cylinder will topple before sliding if µ >, h, WE-27: A uniform cube of side a and mass m rests, on a rough horizontal table. A horizon tal force, F is applied normal to one of the face at a, point directly above the centre of the face, at, 3a, a height, above the base.What is the, 4, minimum value of F for which the cube begins, to topple about an edge?, Sol. In the limiting case normal reaction will pass through, O. The cube will topple about O if torque of F, exceeds the torque of mg., N, a, 2, , 3a, 4, , F, , O, mg, 3a , a, 2, ⇒ F > mg ; ⇒ F > mg, 4, 2, 3, , , 2, So, the minimum value of F is mg, 3, NARAYANAGROUP, , WE-28: A force F is applied on the top of a cube as, shown in the figure. The coefficient of friction, between the cube and the ground is µ . If F is, gradually increased, find the value of µ for, which the cube will topple before sliding., , F, , a, P, , f, mg, Sol. Let m be the mass of the cube and ‘a’ be the side, of the cube., The cube will slide if F > µmg ---------(1), and it will topple if torque of F about P is greater, than torque of ‘mg’ about P i.e,, 1, a, Fa > mg or F > mg -----------(2), 2, 2, From equations (1) and (2) we see that cube will, 1, topple before sliding if µ > ., 2, , Moment of inertia [Rotational Inertia]:, Ø, Ø, Ø, , A body at rest cannot start rotating itself or a rotating, body cannot stop rotating on its own. Hence, a, body has inertia of rotational motion., The quantity measuring the inertia of rotational, motion is known as moment of inertia., Moment of inertia of a particle of mass m is, , I = mr 2, Where r = perpendicular distance of particle from, axis of rotation., S.I unit: kgm 2 ; Its D.F - ML2, , Dimensional formula : ML , 2, , Ø, , Moment of inertia of a group or system of particles, is I = m1r12 + m2 r2 2 + .........mn rn 2 I = Σmr 2, Where m1, m2 ..........mn are masses of particles, , and r1, r2 ............rn are their perpendicular, distances from axis of rotation., Ø Moment of Inertia in rotational motion is, analogous(similar) to mass in translatory motion., Ø Moment of Inertia of a rigid body depends on the, following three factors., a) mass of the body b) position of axis of rotation, c) Nature of distribution of mass., Note-1: Moment of inertia of a rotating rigid body is, independent of its angular velocity., Note-2: Moment of inertia of a metallic body depends, on its temperature., 15
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-29: Four holes of radius R are cut from a thin, square plate of the side 4R and mass M in XY, plane as shown. Then moment of inertia of, the remaining portion, about z-axis is, Sol. M is the mass of the square plate before cutting, the holes., π, M , π R2 = M, Mass of one hole m = , 2 , 16, 16 R , , Perpendicular axes theorem, Statement: It states that the moment of inertia of, a plane lamina about an axis perpendicular to its, plane is equal to the sum of its moments of inertia, about two mutually perpendicular axes concurrent, with perpendicular axis and lying in the plane of the, body., z, Planar Body, , R, , R, , N, , O, R, , R, , X, , M, P(x,y), , moment of inertia of remaining portion, , I = Isquare − 4Ihole, mR 2, , M, 16R 2 + 16R 2 − 4 , I=, + m ( 2R2 ) , 12, 2, , 8 10π , 2, = −, MR, 3, 16, , , Radius of Gyration(K): Radius of gyration, of a rigid body about an axis of rotation is distance, between the axis of rotation and a point at which, the whole mass of the body can be supposed to be, concentrated so that its moment of inertia would, be the same with the actual distribution of mass., Moment of inertia of a rigid body of mass M is, , =, , Ø, , 8, MR 2 − 10mR 2, 3, , I = MK 2, Where K = radius of gyration, r + r + ............. + r, n, Where n is total number of particles in the body, K=, , 2, 1, , 2, 2, , n, n, , and r1 , r2 .............rn are their perpendicular, distances from axis of rotation., S.I unit: metre, CGS unit: cm, Dimensional formula: M LT , Note: K is not the distance of centre of mass of body, from the axis considered., Ø Radius of gyration of a rigid body depends on, the following two factors, a)Position of axis of rotation., b)Nature of distribution of mass., 0, , 16, , 0, , Iz = Ix + Iy, , Y, , Ø, Ø, , This theorem is applicable to bodies which are, planar., This theorem applies to flat bodies whose thickness, is very small compared to their other dimensions., , Ø K z = K x2 + K y2, WE-30: Two identical rods each of mass M and, length L are joined in cross position as shown, in figure. The moment of inertia of a system, about a bisector would be., B1, B2, , Sol. Moment of inertia of a system about an axis which, is perpendicular to plane of rods and passing, through the common centre of rods, , ML2 ML2 ML2, Iz =, +, =, 12, 12, 6, Again from perpendicular axes theorem, I z = I B1 + I B2 = 2I B1 = 2 I B2 as I B1 = I B2 , ∴ I B1 = I B2 =, , I Z ML2, =, 2, 12, , Parallel axes theorem:, Statement: The moment of inertia of a body about, an axis is equal to the sum of the moment of inertia, of the body about a parallel axis passing through, its centre of gravity and the product of its mass and, the square of the distance between the two parallel, axes., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , IG, , I, , I = IG + Ma, , Ø, , M, a, , 2, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , G, , This theorem is applicable to a body of any shape., , Ø K = KG2 + a 2, WE-31: The moment of inertia of a rod of length, l about an axis passing through its centre of, mass and perpendicular to rod is I. The, moment of inertia of hexagonal shape formed, by six such rods, about an axis pass ing, through its centre of mass and perpen dicular, to its plane will be, Sol. M.I. of rod AB about its centre and perpen, , ml 2, = I ⇒ ml 2 = 12I, 12, A l B, x, , dicular to length =, , WE-33: Find the moment of inertia of a thin, uniform rod about an axis perpendicular to, its length and passing through a point which, l, is at a distance of from one end. Also find, 3, radius of gyration about that axis., l, Sol. i) Let P be the point at a distance from one end., 3, l l l, It is at a distance of − = from the centre, 2 3 6, as shown in the figure., l I l Ic, 3, 6, , P, , C, , l, 2, By parallel axes theorem I = I C + Mr 2, 2, , Ml 2, Ml 2, l, =, +M =, 12, 9, 6, , I, Ml 2 l, =, =, M, 9M 3, WE-34: A uniform cylinder has radius R and, length L. If the moment of inertia of this, cylinder about an axis passing through its, centre and normal to its circular face is mg, equal to the moment of inertia of the same, cylinder about an axis passing through its, centre and normal to its length, then, Sol. Moment of inertia of a cylinder about an axis passing, MR 2, through centre and normal to circular face =, 2, Moment of inertia of a cylinder about an axis, passing through centre and normal to its length, ii) The radius of gyration, K =, , Now moment of inertia of rod about the axis which, is passing through O and perpendicular to the plane, of hexagon, ml 2, Irod =, + mx 2 [ from parallel axes theorem ], 12, 2, , 3 , ml 2, 5ml 2, =, +m, l =, 12, 6, 2 , , Now moment of inertia of system, I system = 6 × I rod = 5ml 2 = 5 ×12 I = 60 I, WE-32: The radius of gyration of a body about an, axis at a distance of 12cm from its centre of, mass is 13cm. Find its radius of gyration about, a parallel axis through its centre of mass., Sol. By parallel axes theorem, M (13 ) = I 0 + M (12 ), 2, , 2, , I 0 = M (132 − 12 2 ) = M ( 25 ), , Its radius of gyration about a parallel axes through, its centre of mass K =, NARAYANAGROUP, , I0, = 25 = 5cm, M, , L2 R 2 , =M +, , 12 4 , L2 R 2 , MR 2, =, M, But 2, +, , 12 4 , , R 2 L2 R 2, =, +, ⇒, 2 12 4, ∴ L = 3R, , R 2 L2, =, ;, 4 12, , 17
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-35: A metal piece of mass 120g is stretched to, form a plane rectangular sheet of area of cross, section 0.54m 2 . If length and breadth of this, sheet are in the ratio 1:6, find its moment of, inertia about an axis passing through its, centre and perpendicular to its plane., Sol.Mass M=120g = 120 × 10−3 kg, b, , Area = lb = 0.54m 2 ⇒ ⇒ .b = 0.54, 6, , , Q l =, , , b, , 6, , WE-38: A rod PQ of mass ‘m’ and length L is rotated, about an axis through ‘P’ as shown in figure., Find the moment of inertia of the rod about, the axis of rotation., Sol. Consider a small element ‘dx’ of the rod which is at, a distance ‘x’ from the end ‘P’. If ‘ θ ’ is the, inclination of rod w.r.t the axis of rotation, the radius, of the circle in which the element rotates is given, r, by sin θ =, ⇒ r = x sin θ, x, , b 2 = 0.54 × 6 ⇒ b = 3.24 = 1.8m, , Q, , M ( l 2 + b2 ), , = 33.3 × 10−3 kgm 2, 12, WE-36: The moment of inertia of HCl molecule, about an axis passing through its centre of, mass and perpendicular to the line joining the, H+ and Cl − ions will be (if the inter atomic, distance is 1A0 )., Sol. r = 1A0 = 10−10 m ; m1 = 1amu ; m2 = 35.5amu, I=, , θ, , M.I. of the element about the axis of rotation is, dI = dm.r 2, , mm, , ≅ 1.624 × 10 −27 kg Q 1 amu = 1.67 × 10 −27 kg , , 1, , Sol., a, , 4, , 3, , 1, , 2, , 1, , =, , 2 2 2 2, 2, 2, ma + ma + mb 2 + ma 2 + mb2 + ma 2, 5, 5, 5, 5, , 8, I = ma 2 + 2mb 2, 5, 18, , where dm is the mass of element dm =, dI =, , m, dx, L, , m, 2, dx ( x sin θ ) . Total M.I. of the rod is given, L, , by I= ∫ dI = ∫0, , L, , m, mL2, sin 2 θ x 2 dx , I =, sin 2 θ, L, 3, , WE-39: Two uniform circular discs, each of mass, 1kg and radius 20cm, are kept in contact about, the tangent passing through the point of, contact. Find the moment of inertia of the, system about the tangent passing through the, point of contact., A, , Sol., , b, , 2, 2, I1 = ma 2 ; I 2 = ma 2 + mb 2, 5, 5, 2, 2, I 3 = ma 2 + mb 2 ;, I 4 = ma 2, 5, 5, Moment of Inertia of the, I = I1 + I 2 + I 3 + I 4, , x, , P, , 1 2, Reduced mass µ = m + m = 0.9726 amu, 1, 2, , Moment of inertia about an axis passing through, centre of mass of two particle system and, perpendicular to the line joining them is, I = µ r 2 = 1.624 × 10−47 kg m 2, WE-37:Four solid spheres each of diameter 2a and, mass m are placed with their centers on the, four corners of a square of side b.Calculate, the moment of inertia of the system about any, side of the square., , dx, , B, Mass M = 1kg, r = 20 × 10−2 m, I1 =, , system, , MR 2, 5MR 2, + MR 2 =, 4, 4, , Similarly I 2 =, , 5MR 2, , I = I1 + I 2, 4, , 10 MR 2 10 × 1× ( 20 × 10, ∴I =, =, 4, 4, , ), , −2 2, , = 0.1kgm 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-40: Two solid sphere ( A and B) are made of, metals of different densities ρ A and ρ B, respectively. If their masses are equal, the ratio, of their moments of inertia (IA / IB) about their, respective diameter is, [E-2007 ], Sol. As two solid spheres are equal in masses, so, 1, 4, 4, 3, 3, RA ρ B 3, ⇒, π, R, ρ, =, π, R, ρ, =, mA = mB, A A, B B ⇒, , 3, 3, RB ρ A , The moment of inertia of sphere about diameter, 2, , R , I, ρ , I, 2, I = mR 2 ⇒ A = A ⇒ A = B , I B RB , IB ρA , 5, , 2, , 3, , WE-41:The moment of inertia of a then circular, disc about an axis passing through its center, and perpendicular to its plane is I. Then, the, moment of inertia of the disc about an axis, parallel to its diameter and touching the edge, of the rim is [E-2008 ], , MR 2, Sol. I =, ⇒ MR 2 = 2 I, 2, M.I. of the disc about tangent in a plane, 5, 5, = MR 2 = I, 4, 2, WE-42: The moment of inertia of a disc, of mass, M and radius R, about an axis which is a, tangent and parallel to its diameter is [E-2010], Sol. About the tangent parallel to the diameter, , I = I g + MR2 =, , 5, MR 2, 2, + MR 2 = MR, 4, 4, , WE-43: Two solid spheres A and B each of radius, R are made of materials of densities ρ A and, , ρ B respectively. Their moments of inertia, about a diameter are I A and I B respectively. The, value of IA/IB is, [E-2012], 4, π R3 ρ A, ρ, IA 3, =, = A, Sol. I B 4, π R3 ρ B ρ B, 3, WE-44: From a complete ring of mass M and radius, R, an arc m aking 300 at centre is removed., What is the moment of inertia of the, incomplete ring about an axis passing, through the centre of the ring and perpen, dicular to the plane of the ring., Sol. Mass of incomplete ring = M − M × π = 11M, 2π 6, 12, NARAYANAGROUP, , 30, , 0, , R, , 11M 2 11, R = MR 2, M.I.of incomplete ring I = , , 12, 12 , Note: If a sector of mass m, rotates about its natural, axis then its M.I. is mR 2, WE-45: A thin wire of length l having density ρ is, bent into a circular loop with C as its, centre, as shown in figure. The moment of, inertia of the loop about the line AB is, [E-2014], 2, , 3, 3, 3ρ l 3, l , 2, =, Sol. I = MR = × l ρ × , , 2, 2, 8π 2, 2π , WE-46:For the given uniform square lamina, ABCD,whose centre is O.Its moment of inertia, about an axis AD is equal to how many times, its moment of inertia about an axis EF ?, [AIEEE-2007], F, D, C, , A, A, 1), , B, , E, , 2) I AD = 3I EF, , 2 I AC = I EF, , 3) I AC = 4 I EF 4) I AC = 2 I EF, Sol. I EF = I GH, , I AC = I BD, , ( dueto symmetry ), ( dueto symmetry ), , I AC + I BD = I 0, ⇒ 2 I AC = I 0 -------(1), and I EF + I GH = I 0, ⇒ 2 I EF = I 0 --------(2), From Eqs (1) and (2), we get, , I AC = I EF, ∴ I AD = I E F +, , I AD =, , md 2, md 2, md 2, =, +, 12, 4, 4, , md 2, = 4 I EF, 3, 19
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , Moment of inertia of some regular rigid bodies, Rigid Body, 1)Circular, ring, of mass M, and radius, R., , Axis of Rotation, 1) ^r to the plane of ring and passing, through its centre, , Radius of, Gyration ( K ), , MR 2, , R, , 2MR 2, , 2R, , 2) ^r to the plane of ring and passing through, its rim, (or) passing through any tangent ^r to, the plane of ring, 3) In the plane of the ring and passing through its, centre, (or) passing through any diameter of ring, , MR 2 / 2, , R/ 2, , 4)In the plane of the ring and passing through its edge, (or) passing through any tangent of ring in its plane., , 3MR 2 / 2, , 3/ 2R, , 1) ^r to the plane of plate and passing through its centre, , MR 2 / 2, , R/ 2, , 3MR 2 / 2, , 3/ 2R, , 2) ^r to the plane of plate and passing through its edge, 2 ) T h i n (or ) passing through any tangent ^r to its plane., circular plate, of mass M 3)In the plane of plate and passing through its centre, and radius R (or) passing through any diameter of plate, 4)In the plane of the plate and passing through its edge, (or) passing through any tangent of plate in its plane., 3 ) T h i n, h o l l o w 1)Passing through its centre or any diameter, sphere of 2) Passing through any tangent, mass M and, radius R, 1)Passing through its centre or any diameter, 4)Solid, sphere, 2) Passing through any tangent, of mass M, and radius R, 5), r, Thin uniform 1) ^ to the length of rod and passing through its centre, rod of mass, 2) ^r to the length of rod and passing through its end, M and, length L, , 20, , Moment of, Inertia ( I ), , MR 2 / 4, , R/2, , 5MR 2 / 4, , 5R / 2, , 2MR 2 / 3, 5MR 2 / 3, , 2R / 3, 5R / 3, , 2MR 2 / 5, 7MR 2 / 5, , 2R / 5, 7R / 5, , ML2 /12, , L/2 3, , ML2 / 3, , L/ 3, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Moment of inertia of some regular rigid bodies, Rigid Body, , Axis of Rotation, , 1) ^ to the plane and passing through its centre, r, , Moment of, Inertia ( I ), , Radius of, Gyration ( K ), , M 2, (L + B2 ), 12, , L2 + B2, 2 3, , 6)Thin, uniform, M 2, (L + B2 ), rectangular 2) ^r to the plane of plate and passing through a corner, 3, plate, of mass M, Length L and 3)In the plane of plate ^r to breadth and passing through, Breadth B. centre of plate ., MB2 /12, , L2 + B2, 3, , B/2 3, , 4)In the plane of plate ^r to breadth and passing through, , MB2 / 3, , B/ 3, , ML2 /12, , L/2 3, , edge of plate, , ML2 / 3, , L/ 3, , 1) ^r to the plane of plate and passing through its centre, , ML2 / 6, , L/ 6, , 2ML2 / 3, , 2L / 3, , edge of plate, 5)In the plane of plate ^r to length and passing through, centre of plate., 6)In the plane of plate ^r to length and passing through, , 7)Thin square, r, plate of mass 2) ^ to the plane of plate and passing through a corner, M and side, 3)In the plane of plate parallel to any side and passing, length L., through centre of plate, , ML2 /12, , L/2 3, , ML2 /12, , L/2 3, , 4)In the plane of plate and passing through any two, opposite corners, , 8)Thin, hollow, cylinder, of mass, M radius, R and, Length L, , 1) About geometrical or natural axis, , MR 2, , 2) Parallel to the length of cylinder and touching its, surface (or) passing through line of contact of cylinder, with floor when it is rolling., , 2MR 2, , æ L2 R 2 ÷ö, ç, M, 3) ^ to the axis of cylinder and passing through its centre ççç12 + 2 ÷÷÷, è, ø, r, , NARAYANAGROUP, , R, , 2R, L2 R 2, +, 12, 2, 21
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Moment of inertia of some regular rigid bodies, Rigid Body, , Axis of Rotation, , Moment of, Inertia ( I ), , Radius of, Gyration ( K ), L2 R 2, +, 3, 2, , æ L2 R 2 ÷ö, çç + ÷, M, r, 4) ^ to the axis of cylinder and passing through one end, çè 3, 2 ÷÷ø, 9)Solid, cylinder of 1)About geometrical or natural axis, Mass M, radius R, and length 2)Parallel to the length of cylinder and touching its, L., surface (or) passing through line of contact of cylinder, , MR 2 / 2, , R/ 2, , 3MR 2 / 2, , with floor when it is rolling., , 3R / 2, , æ L2 R 2 ÷ö, ç, M, 3) ^r to the axis of cylinder and passing through its centre ççç12 + 4 ÷÷÷, è, ø, , 4) ^r to the axis of cylinder and passing through one end, WE-47: Consider a uniform square plate of side a, and mass m. The moment of inertia of this, plate about an axis perpendicular to its plane, and passing through one of its corners is, [AIEEE-2008], 5 2, 1, 7, 2, ma 2 3), ma 2 4) ma 2, 1) ma, 2), 6, 12, 12, 3, Sol. Using parallel axes theorem,, Ml 2 Ml 2 7 Ml 2, I = I G + Mr 2 =, +, =, 12, 2, 12, WE-48:A disc of moment of inertia 4kgm 2 is, spinning freely at 3rads −1 . A second disc of, moment of inertia 2kgm 2 slides down the, spindle and they rotate together. (a) What is, the angular velocity of the combination ?, (b) What is the change in kinetic energy of, the system?, Sol. (a) Since there are no external torques acting, we, may apply the conservation of angular momentum., I f ω f = I iωi ⇒ ( 6kgm2 ) ω f = ( 4kgm2 )( 3rads −1 ), Thus ω f = 2rads −2, (b) The kinetic energies before and after the, collision are, 22, , Ki =, , L2 R 2, +, 12, 4, , L2 R 2 , M +, , 4 , 3, , L2 R 2, +, 3, 4, , 1, 1, I iωi2 = 18 J ; K f = I f ω 2f = 12 J, 2, 2, , The change is ∆K = K f − K i = −6 J, In order for the two discs to spin together at the, same rate, there had to be friction between them., The loss in kinetic energy is converted into thermal, energy., , Angular momentum of a particle, Definition: The moment of linear momentum of a, Ø, , Ø, , Ø, , body w.r.t. an axis of rotation is known as angular, momentum., r, The angular momentum L of the particle with, respect to fixed point O is represented as, r r r, L = r × p = m( r ×v ), , The magnitude of angular momentum vector is, r, L = r p sinθ where p is the magnitude of p and, r, r, θ is the angle between r and p ., It is always directed perpendicular to the plane of, rotation and along the axis of rotation., , Angular momentum of rigid body:, When a rigid body is rotating, then the vector sum, of angular momenta of all the particles of body, about the axis of rotation is called angular, momentum of rigid body. It is equal to the product, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , of moment of inertia and angular velocity., , WE-51: A circular platform is mounted on a vertical, frictionless axle. Its radius is r =2m and its, moment of inertia is I = 200kg − m2 .It is, , ∴ L = ∑ ( ri × mi vi ) = Iω, i, , S.I. Unit: kgm 2 / sec, Dimensional formula : ML2T −1, When a body is rolling its total angular momentum, is the vector sum of its angular momentum about, centre of mass and the angular momentum about a, fixed point on the ground., , Law of conservation of angular momentum:, If there is no external torque acting on the rotating, body (or system of particles), then its angular, momentum is conserved., If τ ext, , r, dL r , = τ ext , Q, dt, , , dL, =0, = 0 then, dt, , ⇒ L = Iω = constant ∴ I1ω1 = I 2ω2, WE-49:A ballet dancer spins about a vertical axis, at 60 rpm with arms outstretched. When her, arms are folded the angular frequency, increases to 90 rpm. Find the change in her, moment of inertia, Sol. By the principle of conservation of angular, momentum I × 60 = I 2 × 90, 2I, Final moment of inertia, I 2 =, 3, , initially at rest. A 70kg man stands on the edge, of the platform and begins to walk along the, edge at speed. V0 = 1.0m / s relative to the, ground. Find the angular velocity of the, platform., Sol. Angular momentum of man = angular momentum, of platform in opposite direction., mv0 r = Iω ⇒ ω = 0.7rad / s, WE-52: A uniform bar of length 6a and mass 8m, lies on a smooth horizontal table. Two point, masses m and 2m moving in the same, horizontal plane with speeds 2v and v, respectively, strike the bar (as shown in fig), and stick to the bar after collision. Calculate, (a) velocity of the centre of mass (b) angular, velocity about centre of mass and (c) total, kinetic energy, just after collision., 2m, , v, C, 8m, 2a, , a, , a, , 2a, , 2I I, =, Change in moment of inertia = I −, 3 3, , 2v, , WE-50: A horizontal disc is freely rotating about, a vertical axis passing through its centre at, the rate of 100 rpm. A bob of wax of mass 20g, falls on the disc and sticks to it a distance of 5, cm from the axis. If the moment of inertia of, the disc about the given axis is 2 × 10 −4 kgm 2 ,, find new frequency of rotation of the disc., −4, 2, Sol. I = Moment of inertia of disc= 2 × 10 kgm, 1, , I 2 = moment of inertia of the disc + moment of, inertia of the bob of wax on the disc, −4, , −4, , = 2 × 10 + mr = 2 × 10 + 20 × 10, 2, , −3, , ( 0.05 ), , 2, , = 2 × 10−4 + 0.5 × 10−4 = 2.5 × 10−4 kgm 2, By the principle of conservation of angular momentum, I1 n1 = I 2 n 2 ⇒ 2 × 10 −4 ×100 = 2.5 × 10−4 n2, n2 =, , 100 × 2, = 80 rpm, 2.5, , NARAYANAGROUP, , m, Sol. (a) As Fext = 0 linear momentum of the system, is conserved, i.e., −2m × v + m × 2v + 0 = ( 2m + m + 8m ) ×V, or V=0 i.e. velocity of centre of mass is zero., (b) As τ ext = 0 angular momentum of the system is, conserved, i.e., m1v1r1 + m2 v2 r2 = ( I1 + I2 + I 3 ) ω, 2, 2, 2, 2mva + m ( 2v )( 2a ) = 2m ( a ) + m ( 2a ) + 8m × ( 6a ) /12 ω, , , , i.e. 6 mva = 30 ma 2ω, , v , ⇒ω = , 5a , , (c) From (a) and (b) it is clear that, the system has, no translatory motion but only rotatory motion., 2, , 1, 1, 3, v , E = I ω 2 = ( 30 ma 2 ) = mv 2, 2, 2, 5, 5a , , 23
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-53: A hoop of radius r and mass m rotating, with an angular velocity ‘ω0 ’ is placed on a, rough horizontal surface. The initial velocity, of the centre of the hoop is zero. What will be, the velocity of the centre of the hoop when it, ceases to slip., (JEE-2013), v, ωr, 2, 2, Sol. mr ω0 = mvr + mr × ⇒ v = 0, r, 2, , WE-55:The pulley of Atwoods machine has a, moment of inertia ‘I’ about its axis and its, radius is ‘R’. Find the magnitude of, acceleration of the two blocks assuming the, string is light and does not slip on the pulley., Sol., ////////////////////, , Rotational dynamics, , Relation between Torque and angular, momentum of a rigid body:, The vector sum of torques acting on various, particles of a rigid body gives the net torque acting, on the body., , τ = ∑τ i, momentum of the body. The time rate of change of, the angular momentum of a particle is equal to the, torque acting on it., and τ =, , dL, , L is total angular, dt, , Relation between torque and angular, acceleration:, dL, But L = Iω, dt, dω, ∴τ = I, ⇒ τ = Iα, dt, This equation is called equation of rotatory motion, and analogous to Newton’s 2nd law in dynamics., WE-54:A uniform rod of mass ‘m’ and length ‘l’ is, suspended by means of two light inextensible, strings at the ends of a rod. Tension in one, string immediately after the other string is cut, is, Sol. mg − T = ma .........(1), τ =, , T2, , T1, , m, M, Suppose the block of mass ‘M’ goes down with, an acceleration ‘a’. The angular acceleration, a, of the pulley is, α =, R, Mg − T1 = Ma ; T2 − mg = ma, a, And T1 R − T2 R = Iα = I, R, M − m ) gR 2, (, Solving the equation, a = I + M + m R 2, (, ), , Rotational kinetic energy:, The sum of the kinetic energies of various particles, of rotating body is called rotational kinetic energy., L2 1 2 1, KE rot =, = Iω = ωL, 2, 2I 2, WE-56: The angular momentum of rotating body, is increased by 20%. What will be the increase, in its rotational kinetic energy?, Sol. Kinetic energy KE=, , T, , T, , ∆E 120 , =, , E 100 , , mg, l, mg, τ, 3g, 2, α= =, =, 2, I, ml, 2l ......(2), 3, l, a = α .......(3), 2, , solving eq (1), (2) and (3) we get,, , 24, , T=, , mg, 4, , 2, , ( or ), , L2, ⇒ E ∝ L2, 2I, , ∆E, = 0.44, E, , ∆E, × 100 = 44%, E, WE-57: A uniform rod of length ‘l’ is held vertically, on a horizontal floor fixing its lower end, the, rod is allowed to fall onto the ground. Find, (i) its angular velocity at that instant of, reaching the ground (ii) The linear velocity, with which the tip of rod hits the floor., Sol. The rod rotates about an axis through one end., From the principle of conservation of mechanical, energy. Loss of P.E of the rod is equal to its gain in, rotational K.E., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Bu=0, , A, a, A, , l, l, 2, , ∴ mg, , Sol., , l 1 2, l 1 ml 2 2, = Iω ⇒ mg = ., ω, 2 2, 2 2 3, , on solving ω =, , 3g, l, , (ii) V = rω or V = lω = l 3 g / l = 3gl, WE-58: A rigid body is made of three identical thin, rods, each of length ‘L’ fastened together in, the form of the letter ‘H’.The body is free to, rotate about horizontal axis that runs along, the length of one of the legs of ‘H’. The body, is allowed to fall from rest from a position in, which the plane of ‘H’ is horizontal. What is, the angular speed of the body when the plane, of ‘H’ is vertical?, Sol. The moment of inertia of the system about one, 2, , 4, mL, + mL2 ; I = mL2, rod as axis I =, 3, 3, Potential energy decreases for B and C, X, A, Y, , B, C, , mgL, 3, + mgL = mgL, 2, 2, By conservation of mechanical energy, the loss in, PE of body is equal to the gain in rotational KE, , 3, 1 4, 3 g, , ∴ mgL = mL2 ω2 on solving ω =, 2, 2 3, , 2 L, , WE-59:A uniform rod AB of mass ‘m’ length ‘2a’, is allowed to fall under gravity with AB in, horizontal. When the speed of the rod is ‘v’, suddenly the end ‘A’ is fixed. Find the, angular velocity with which it begins to, rotate., NARAYANAGROUP, , mv, L1 = L 2 ⇒ mva = Iω, , m (2 a ), 3v, mva =, ω ⇒ ω =, 3, 4a, 2, , WE-60: A particle is projected at time t=0 from a, point ‘O’ with a speed ‘u’ at an angle 450 to, horizontal. Find the angular momentum of, the particle at time t=u/g., r, Sol.Velocity at any time ‘t’ is v = v x $i + v y $j, y, r, , P(x,y), θ, x, O, r, position vector of projectile at time ‘t’ is r = xi$ + y $j, , ur, , since L = m ( r$ × v$ ), , ur, L = m xi$ + y $j × vx $i + v y $j , , , ur, $, $, , , L = m xv y k − yvx k ; L = mk xv y − yvx , , , u, u2, and x =, where vx =, 2, 2g, , (, , )(, , ), , , u u , 0, Q x = ( u cos 45 ) t =, , 2 g , , , v y = u sin 450 − gt =, , 1− 2, v0, 2, , 1, y2, y = u sin 450 t − gt 2 =, 2, 2g, , (, , ), , ur, , 2 −1 ; L =, , $ 3, −kmu, 2 2g, , Work, Power & Angular Impulse, Work: Work done by external torque on, rotating body is W = ∫ τ dθ, If τ is constant , then, W = τ θ, Work energy theorem : Work done by external, torque on rotating body is equal to change in, rotational kinetic energy., 1, 1, τ θ = Iω 2 − Iω02, 2, 2, 25
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Ø, , Work done by retarding torque to stop the rotating, body is equal to initial rotational kinetic energy of, body., 1, τθ = Iω 2 and θ = 2π N ,, 2, where N = no. of rotations made by the body before, coming to rest., WE-61:What is the work done in increasing the, angular frequency of a circular ring of mass, 2kg and radius 25 cm from 10rpm to 20 rpm, about its axis ?, Sol. Work done=increase in rotational kinetic energy, 1, 1, 2, 2, 2, 2, 2, = I ( ω f − ωi ) = MR (ω f − ωi ), 2, 2, 2, 2, 1, 2 2π , π , = × 2 × ( 0.25) , −, = 0.2054 J, 3 3 , 2, , , , Angular Impulse: The large torque acts on a body, Ø, , for relatively very short interval of time is called, impulsive torque., The product of impulsive torque and its time of, action is called angular impulse J. It is a vector. It is, always equal to change in angular momentum., ur, r, r, r, ur, r dL, ∴ J = ∫ τ dt ; As τ =, ; ∫ τ dt = ∆ L, dt, r, ur ur, ∴J = I ω −ω0, , (, , ), , WE-63: A uniform rod of mass ‘m’ and length ‘l’, is on the smooth horizontal surface. When a, constant force ‘F’ is applied at one end of the, rod for a small time ‘t’ as shown in the figure., Find the angular velocity of the rod about its, centre of mass., equilibrium, , Unstretched, a, , θ0, , position, at any time, , θ, , Power: The rate of work done by torque is called, power. Instantaneous power is given by, d, dθ, dW, P=, = (τθ ) = τ, = τω, dt, dt, dt, As the power is a scalar, it is written as P = τ .ω, Average power is, Pave, , 1 2 1 2, Iω − Iω0, Total work done 2, 2, =, =, t, Total time, , WE-62: A motor rotates a pulley of radius 25cm at, 20 rpm. A rope around the pulley lifts a 50kg, block, What is the power output of the motor?, Sol. The tension in the rope is equal to the weight since, there in no acceleration. Thus T=500N Therefore,, τ = TR = ( 500 )( 0.25 ) = 125 Nm, , R, , T, , ur r ur ur, Sol. According to angular impulse J = τ t = L 2 − L1, m, , r, τ t = Iω 2 − I ω 1, , Q ω1 = 0; ω2 = ω ; τ = F, , l, ml 2, 6 Ft, F t=, ω −0 ⇒ω =, 2, 12, ml, WE-64: A fly-wheel of mass 25kg has a radius of 0.2m., It is making 240rpm. What is the torque, necessary to bring to rest in 20s?[E-2010], 2π n 2π × 4, =, Sol. α =, t, 20, MR 2, α = 0.2π Nm, Torque τ = I α =, 2, WE-65: Moment of inertia of a body about an axis, is 4 kg − m 2 . The body is initially at rest and, a torque of 8Nm starts acting on it along the, same axis. Workdone by the torque in 20s, in, joules is, [E-2013], τ, , 2π N 2π ( 20 ) 2π, =, rads −1, 60, 60, 3, The power required is, 2π, , P = τω = (125 Nm −1 ) , rads −1 = 261W, 3, , , Angular velocity , ω =, , 26, , l, 2, , 1, , 2, Sol. τ = Iα ⇒ α = = 2 ; θ = α t = 400 ;, I, 2, ω = τθ = 3200J, WE-66: A wheel which is initially at rest is, subjected to a constant angular accleration, about its axis. It rotates through an angle of, 150 in time t sec. Then how much it rotates in, the next 2t sec [E-2014 ], , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Sol. If angular acceleration is constant, we have, 1, 1, 1, θ = ω0t + α t 2 ⇒ 150 = 0 + α t2 ⇒150 = α t 2........(1), 2, 2, 2, for the second condition (time =3tsec), 1, 2, θ1 = α ( 3t ) .........(2), 2, So, ∆θ = θ 1 − θ = 1200, WE-67: A thin uniform rod of length l and, mass m is swinging freely about a horizontal, axis passing through its end. Its maximum, angular speed is ω . Its centre of mass rises to, a maximum height of, [AIEEE-2009], , Ø, , ωv, O, Ø, , 2, l 2ω2, 1 2 1 ml 2, ω ⇒h=, Sol. mgh = Iω =, 2 3, 6g, 2, WE-68: A pulley of radius 2m is rotated about its, , axis by a force F = ( 20t − 5t ) N (where, t is, measured in seconds) applied tangentially. If, the moment of inertia of the pulley about its, axis of rotation is10kgm 2 , then the number, of rotations made by the pulley before its, direction of motion is reversed, is(AIE 2011), Sol.Given force, F = 20t − 5t 2, 2, , Uniform pure rolling or simply “pure rolling” means, that no relative motion exists at the point of contact, between the body and the surface. Let a disc of, radius R rolls without slipping on a horizontal, stationary surface/ground. For the disc to roll, without slipping, we must have, , ω, t, dω, = 4t − t 2 ⇒ ∫ d ω = ∫ ( 4t − t 2 )dt, 0, 0, dt, 3, t, ⇒ ω = 2t 2 −, 3, When direction is reversed, ω = 0, i.e., t = 0 to 6 s Now,,, , ⇒, , Vcm ω, Rω, , θ, 6, t , dθ = ω dt ⇒ ∫ dθ = ∫ 2t 2 − dt, 0, 0, 3, , , (ii) If v P >v Q ⇒ vc m -R? > 0 ⇒ v c m >R?, ( Forward Slipping ), (iii) If v P <v Q ⇒ vc m -R? < 0 ⇒ v c m <R?, , 6, , 2t 3 t 4 , ⇒θ = , − ⇒ θ = 144 − 108 = 36rad, 3 12 0, , 36, , ∴ Number of rotations, n = 2π = 2π < 6, Rolling Motion, , (Backward Slipping), so, v c m = R ? is the condition for a body to be, in pure rolling on a stationary horizontal surface/, ground. It is sometimes simply called as Rolling., , Pure rolling motion: Motion of a rolling body on, , NARAYANAGROUP, , P, , Vcm, Q, (i) ( v c m )P =vsurface ⇒ vcm − Rω = 0 ⇒ vc m = Rω, , 3, , horizontal surface is the combination of translational, motion and rotational motion where the point of, contact has no relative motion with surface. All, points of rolling rigid body have same angular speed, but different linear velocities. A body rolls on a, surface when the surface has frictional force i.e., torque will act on the rolling body due to frictional, , P, It means that the velocity of point at the top of the, disc ( v1 ) has a magnitude υcm + Rω or 2 υcm and, is directed parallel to the level surface., H rω = v, H, H 2V, V, C, V, 2V, V, +, V=0, C, C, 2V, V -rω=V L V, L, V=0, (C), (B), (A), Rolling, Rotation, Translation, (i) Linear speed at H=2V (max), (ii) Linear speed at L=0 (min), (iii) Linear speed at M= 2V, , Uniform pure rolling:, , 2, FR ( 20t − 5t ) 2, α=, =, = 4t − t 2, I, 10, , θ, , force., The condition that point of contact is instantaneously, at rest requires υcm = Rω, v1, , Non uniform pure rolling:, Ø, Ø, , If a CM = Rα , then no friction arises and the body, is in pure rolling., If a CM > Rα , then static friction arises and acts, opposite to motion of the body to support rotatory, 27
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , motion such that Rα is made equal to a CM to, keep the body still in pure rolling motion., (Instantaneously VCM = Rω ), Ø, , If a CM < Rα , then static friction arises and acts in, the direction of motion of the body to support, translatory motion such that a CM is made equal to, Rα to keep the body still in pure rolling motion., (Instantaneously VCM = Rω ), If the above two conditions fail then the body slips, and the friction present is kinetic., , Total K.E of a rolling body :, Ø, , A rolling body has both translational and rotational, kinetic energies, 1, 2, , 2, (i) KE trans = mVc, , (iii), =, , 1, 2, + K .Erotational, , 2, (ii) KE rot = Iω, , K .Etotal = K .Etranslatory, , 1, K2 , 1, 1, 2, mVc2 + Iω 2 = mVc 1 + 2 , R , 2, 2, 2, , , Vc = Velocity of C.M, K = radius of gyration , R = radius, , Where, , KEtrans. : KErot. : KEtot., , K2, = 1: 2, R, , KE tra, 1, =, 2, motion KE total 1 + K, R2, , Note-2:Fraction of KE associated with rotatory motion, 1, R2, 1+ 2, K, , Note-3: The K.E. of some rolling bodies :, Ø, Ø, Ø, , 3, 4, 7, 2, The K.E. of rolling solid sphere is E = mv, 10, 5 2, The K.E. of a rolling hollow sphere is E= mv, 6, 2, The K.E. of rolling solid cylinder is E = mv, , Ø The K.E. of a hollow cylinder or thin ring is E = mv 2, WE-69: A solid cylinder of mass M has a string, wrapped several times around its circum, ference. The free end of string is attached to, the ceiling and the cylinder is released from, rest. Find the acceleration of the cylinder and, the tension in the string. (MAINS-2014), 28, , Mg, , Sol. Mg − T = Ma ........(1) Also, τ = TR = Iα, I α MRα, ⇒T =, =, .....(2), R, 2, 1, 2, where I = M R, 2, Condition for no slipping, a = α R ....(3), Solving equations (1) ,(2)and (3) we obtain, 2, Mg, a = g and T =, 3, 3, , Rolling of a body on an inclined plane, without slipping :, Ø, , Consider a body of mass M radius R and moment, of inertia I rolling without slipping on an inclined, plane (making an angle θ with the horizontal). For, the body to roll without slipping necessary friction(f), must be present. From the figure we observe that, all forces (other than friction) are acting along the, radius and hence will not produce torque in the, body., N, , K2 , : 1 + 2 , R , , , Note-1: Fraction of KE associated with translatory, , KE rot, =, KE total, , R, , f = µN, , a, Mg sin θ, , θ Mg, , Mg cos θ, , At the point of contact velocity is zero, so the net, torque due to the various forces about the IAOR, is τ = ( Mg sin θ ) R = Iα .... (1), a, ( Mg sin θ ) R = ( MR 2 + Mk 2 ), R, 2, g sin θ R ⇒ a = g sin θ, a= 2, k 2 .... (2), R + k2, 1+ 2, R, , a, ... (3), R, , Torque due to friction fR = I , Ia ( Mk, ⇒ f = 2 =, R, R2, , 2, , ), , g sin θ, Mg sin θ, ⇒ f =, 2, R2 , k , 1, +, 1 + 2 , R2 , k , , , , If µ be the coefficient of friction, then, f ≤ µN ,. where N = Mg cos θ, , tanθ, ... (4), R2 , 1, +, k2 , , , , using(3), we get µmin = , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Condition for a body to roll without slipping:, For a body to roll without slipping, the force of, friction ‘f’ calculated above must be less than or, equal to the maximum value of friction, i.e. µ Mg cos θ, ⇒, , Mg sin θ, tan θ, ≤ µ Mg cos θ ⇒ µ ≥, R2 , R2 , 1 + 2 , 1 + 2 , k , , k , , (i) Velocity of the body when it reaches the bottom is, given by, , v=, , 2 gh, =, k2, 1+ 2, R, , 2 gl sin θ, ( since h = l sin θ ), k2, 1+ 2, R, , (ii) Acceleration of the body is given by a =, , Note-1: If a1 , a2 , a3 and a4 are the accelerations of, centre of masses of rolling solid sphere, solid, cylinder, hollow sphere and hollow cylinder, respectively when they roll down the same inclined, plane then a1 > a2 > a3 > a4, Note-2: If t1 , t 2 , t3 and t4 are the times of travel of rolling, solid sphere, solid cylinder, hollow sphere and, hollow cylinder respectively to reach the bottom, from the top of an inclined plane then, t1 < t 2 < t 3 < t 4, , g sin θ, k2, 1+ 2, R, , 2, tan θ, 7, 2, b) For Hollow sphere ( µ ) = tan θ, 5, , a) For Solid sphere ( µ ) =, , (iii) Time taken by the body to reach the bottom is, given by, Ø, , t=, , 2l (1 + k 2 / R 2 ), g sin θ, , If all these are allowed to roll down from the top of, an inclined plane, they will reach the bottom in the, following order, 1) Solid sphere, 2) Disc (or) Solid cylinder, 3) Hollow sphere 4) Ring (or) Hollow cylinder, , 1, c) For Solid cylinder (or) Disc ( µ ) = tan θ, 3, 1, d) For Ring (or) Hollow Cylinder ( µ ) = tan θ, 2, Note : When a body rolls down without slipping, work, is not done against friction as the point of contact, instantaneously at rest., , Ø Bodies rolling on a horizontal plane :, Body, Square of, Rotational K.E., radius of, Translational K.E., gyration, 1. Solid sphere, (2/5)r2, 2/5, 2, 2. Hollow sphere, (2/3) r, 2/3, 3. Disc (or), Solid Cylinder, r2 / 2, 1/2, 4. Ring (or), Hollow Cylinder, r2, 1, , Rotational K.E Traslational K.E, Total K.E., Total K.E., 2/7, 2/5, , 5/7, 3/5, , 1/3, , 2/3, , 1/2, , 1/2, , Ø Bodies rolling on an inclined plane, Body, Velocity at bottom Acceleration, , Time of descent, , Solid sphere, , 10 gh, 7, , 5, g sin θ, 7, , 14l, 5 g sin θ, , 2., , Hollow sphere, , 6 gh, 5, , 3, g sin θ, 5, , 10l, 3g sin θ, , 3., , Disc (or)Solid cylinder, , 4 gh, 3, , 2, g sin θ, 3, , 3l, g sin θ, , 4., , Ring (or) Hollow cylinder, , gh, , 1, g sin θ, 2, , 4l, g sin θ, , 1., , NARAYANAGROUP, , 29
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , Ø, , A body rolls on a smooth horizontal surface with, speed v and then rolls up a rough inclined plane, of inclination θ ., , θ, (i) The height reached by the body before coming, v2 k 2 , h, =, 1 + , to rest is given by, 2g r 2 , 7v 2, a) For solid sphere, h =, 10 g, , b) For Hollow sphere, h =, , 5v 2, 6g, , c) For Disc (or) Solid cylinder, h =, , 3v 2, 4g, , v2, g, Note : If all these bodies travel with same velocity on, horizontal surface then, i) Solid sphere reaches the minimum height., ii) Ring reaches maximum height., , Instantaneous axis of rotation:, The pure rolling motion is purely rotatory motion, about instantaneous point of contact with ground., The axis passing through instantaneous point of, contact and perpendicular to plane of rotation is, called instantaneous axis of rotation., ∴ The total kinetic energy of rolling body is written, 1, 2, as KE total = I Pω ,, 2, where IP is moment of inertia about instantaneous, axis of rotation., , Rolling bodies over moving platform:, The rolling bodies do not slide on the surface on, which they are moving. If they are rolling on a, moving platform, the point of contact of the body, with the platform should have same velocity as, the platform., Case 1: If point of contact of surface is moving with, velocity u with respect to ground, then, Vcm − ω R = u, , d) For Ring (or) Hollow cylinder, h =, , Angular momentum in case of rotation, about a fixed axis:, Ø, Ø, , When the total external torque is zero, the total, angular momentum of the system is conserved., The general expression for the total angular, N, , momentum of the system is L = ∑ ri × pi, , O, ωR, , ω, , Vcm, , P, , u, , N, ///////////////////////////////////, Case 2: For no sliding on the moving platform,, u = ωR − vcm, , O, , ω, , vcm, , P, , u, , Case 3: For accelerated surface, acm − α R = a, , i=1, , Ø, , For rotation about a fixed axis, the component of, angular momentum perpendicular to the fixed axis, is constant., , Angular Momentum of Rolling Wheel, in combined Rotation & Translation:, Angular momentum of a rolling wheel about an, axis passing through the point of contant P and, perpendicular to the plane of wheel:, , ω, O, , vCM, , P, r r, r r, r, r, L = Ltranslation + Lrotation = m R× vCM + Icmω or, ur, ur, ur, ur, ur, r, 2, L = mωR2 + ICM ω or L = ( ICM + mR ) ω = I Pω, , (, , 30, , ), , α, P, , acm, , a, ///////////////////////////////////, , Direction of friction in case of, translation & rotation combined:, The direction of friction cannot be found by direct, observation in case of rotational motion, as the, body is translating as well as rotating. The direction, of frictional force is determined after deciding the, motion tendency of the point of contact of the body, under consideration with the ground. A rolling, object of mass M and radius R is placed on a, rough horizontal surface. A force F is applied as, shown., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-71: (i) Explain why friction is necessary to, make the disc to roll in the direction indicated., (ii)Give the direction of frictional force at B,, and the sense of frictional torque, before, perfect rolling begins. (iii)What is the force, of friction after perfect rolling begins ?, A, , R, , F, x, O, P, , Ø, Ø, , Acceleration of point P due to translation, , R, 2, , F, at =, ( towards right ), M, , Acceleration of point P due to rotation only,, τ, FR, ar = α R = R =, x ( towards left ), I, I, r, r r, net acceleration of point P is aP = at + ar, aP =, , F FRx, −, ( towards right ), M, I, , (i ), , From the above equation it is clear that motion, tendency at point P depends upon both x and I,, Eq (i) can be written as,, aP =, , F, M, , Rx , , 1 − 2 , K , , ( ii ), , If K 2 > Rx : friction will act in backward direction., If K 2 = Rx : no friction will act., If K 2 < Rx : friction will act in forward direction., WE-70: A disc rotating about its axis with angular, speed ‘ ω0 ’ is placed lightly (without any, translational push) on a perfectly frictionless, table. The radius of the disc is R. What are, the linear velocities of the points A, B and C, on the disc shown. Will the disc roll in the, direction indicated?, A, , ω0, , C, , R, 2, , f, B, Sol: We know that, υ = Rω, For point A: υ A = Rω0 (in the direction of the arrow), For point B: υ B = Rω0 (in the direction opposite, , to arrow) For point C: υC = ω0 (in the, 2, direction of the arrow) The disc will not roll in the, direction indicated. It is because the disc is placed, on a perfectly frictionless table and without friction,, a body cannot roll., R, , NARAYANAGROUP, , C, , ω0, , f, B, Sol: (i)To roll a disc, a torque is required which in turn, requires a tangential force to act on it. As the force of, friction is the only tangential force acting on the disc,, so it is necessarily required for the rolling of a disc., (ii) Frictional force at B opposes the velocity of, B. Therefore, frictional force is in the same direction, as the arrow. The sense of frictional torque is such, as to oppose the angular motion. By right hand, r, r, rule, ω0 act into the plane of paper and τ out of, the paper., (iii) Frictional force decreases the velocity of the, point of constant B. Perfect rolling begins when, this velocity is zero at which the force of friction is, zero., WE-72: A thin hollow sphere of mass ‘m’ is, completely filled with a liquid of mass ‘m’., When the sphere rolls with a velocity ‘v’,, kinetic energy of the system is (neglect, friction), [E-2011], Sol: Total energy = KE + rotational KE, 1, 1 2, , = ( 2m) v2 + mr 2 ω2, 2, 2 3, , 1, 1, 4, = ( 2m ) v 2 + mv 2 = mv 2, 2, 3, 3, WE-73: A round uniform body of radius ‘R’, mass, ‘M’ and moment of inertia ‘I’, rolls down, (without slipping) an inclined plane making, an angle ‘ θ ’ with the horizontal.Then, its, acceleration is, [A-2007], 2 gh, I ;, Sol:, 1+, MR 2, h = L sin θ , v 2 − u 2 = 2as, v2, 2 gL sin θ, g sin θ, a=, =, =, I, I , 2s, , 2 1 +, L 1+, 2 , MR 2, MR , , v=, , 31
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , WE-74:The uniform 50kg pole ABC is balanced, in the vertical position. A 500N horizontal, force is suddenly applied at B. If the coefficient, of kinetic friction between the pole and the, ground is 0.3, determine the initial, acceleration of point A. (Take g = 10ms −2 )., A, , JEE-ADV PHYSICS-VOL - III, , k, M, , m, 4m, B, , 500 N, 2m, , C, , decreases ( ∆U g = − mgx ) , the potential energy, , Sol: N = mg = 500, 500 − µ N, 500 − µ N = ma x ⇒ ax =, m, ( 500 ) − ( 0.3)( 500 ) = 7ms −2, ⇒ ax =, 50, Now, let us calculate torque due to forces about, the centre of mass or centre of gravity G of the, rod, then τ = ( 500 )(1) − ( 0.3)( 500 )( 3), , ⇒ τ = 500 − 450 = 50Nm, A, , α, G, , mg, , B, , 500 N, , N, µN, , C, , τ, 1, 50, ⇒ α = rad s −2, ⇒α =, 1, 2, I, 3, ( 50 )( 6 ), 12, So, acceleration of point A is, , As, α =, , 1, a A = ax − rα = 7 − ( 3) = 6ms −2 ( to the right ), 3, WE-75: A block of mass m=4kg is attached to a, , spring of spring constant ( k = 32 Nm −1 ) by a, rope that hangs over a pulley of mass M=8kg, If the system starts from rest with the spring, unstretched, find the speed of the block after, it falls 1m. Treat the pulley as a disc, so, 1, I = MR 2, 2, , 32, , Sol: Since the rim of the pulley moves at the same speed, as the block, the speed of the block and the angular, velocity of the pulley related by v = ω R When, the block falls by a distance x, its potential energy, 1 2, , of the spring increases ∆U g = + kx , and, 2, , , both the block and the pulley gain KE, 1 2 1 2, , ∆K = mv + Iω , 2, 2, , , From the conservation of mechanical energy,, ∆K + ∆U = 0 ,, 2, , 1 2 1 v 1 2, mv + I + kx − mgx = 0, 2, 2 R 2, , M 2 1 2, 1, m + v + kx − mgx = 0, 2, 2 , 2, Putting m = 4kg, M = 8kg, k = 32Nm −1 , x =1m, 1, 8 2 1, 2, 4 + v + ( 32 )(1) − ( 4 )(10 )(1) = 0, 2, 2, 2, 4v 2 + 16 − 40 = 0 ⇒ v = 2.4 ms −1, WE-76: A uniform rod of length L and mass M is, pivoted freely at one end as shown in the, figure. (a) Find the angular acceleration of, the rod when it is at angle ‘ θ ’ to the vertical., (b)Assuming the rod to start from the vertical, positions, find the angular velocity as the, function of ‘ θ ’. (c) Find the tangential, acceleration of the free end when the rod is, horizontal., , Sol: (a) Figure shows the rod at an angle θ to the, vertical, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , C.U.Q, θ, , CENTRE OF MASS, , α, , 1., , Mg, O, L, sin θ, 2, Net torque about the point O is, L, , τ 0 = Mg sin θ , 2, , Using the second law of motion τ 0 = I 0α, 3 g sin θ, MgL, ML2, sin θ =, α ⇒α =, 2L, 2, 3, (b) From above, we have, 3g sin θ, d ω 3g sin θ, α=, ⇒ω, =, 2L, dθ, 2L, 3 g sin θ, ⇒ ω dω =, dθ, 2L, Integrating within appropriate limits, we get, ω, 3g θ, ∫0 ω dω = 2L ∫0 sin θ dθ, ω2, 3g, 3g, θ, ⇒, = − [ cos θ ]0 =, (1 − cosθ ), 2, 2L, 2L, 3g, ⇒ω =, (1 − cos θ ), L, The above result can also be obtained by using, the Law of Conservation of Mechanical energy,, where we use, Loss inGPE Gainin RKE , of CM of Rod = of Rod, , , , , , L, 1, 1 1, (1 − cos θ ) = I ω 2 = ML2 ω 2, 2, 2, 23, , 3g, ⇒ω =, (1 − cos θ ), L, π, (c) When the rod is horizontal θ = ,, 2, 3g, So, α =, , So, the tangential linear, 2L, 3g, acceleration is at = α L =, 2, This is greater than the acceleration of an object, falling freely., , 2., , 3., , 4., , 5., , LINEAR MOMENTUM OF CENTRE, OF MASS, 6., , ⇒ Mg, , NARAYANAGROUP, , When a force is applied on a body, Newton’s, second law is applicable to, 1) centre of mass, 2) any part of the body, 3) upper most part of body, 4) lower most part of body, Centre of mass of the earth–moon system, lies, 1) on the surface of the earth, 2) on the surface of the moon, 3) with in the earth, 4) at the midpoint of the line joining their centres, A square plate and a circular plate made up, of same material are placed touching each, other on a horizontal table. If the side length, of square plate is equal to diameter of the, circular plate then the centre of mass of the, combination will be, 1) at their point of contact, 2) inside the circular plate, 3) inside the square plate, 4) outside the combination, A uniform straight rod is placed in vertical, position on a smooth horizontal surface and, released. As the rod is in motion, the centre, of mass moves, 1) horizontally, 2) vertically down, 3) in a parabolic path 4) does not move., A disc and a square sheet of same mass are, cut from same metallic sheet. They are kept, side by side with contact at a single point., Then the centre of mass of combination is, 1) at point of contact 2) inside the disc, 3) inside the square, 4) outside the system, , 7., , Two balls are thrown at the same time in, air, while they are in air, the acceleration of, their centre of mass, 1) depends on masses of the balls, 2) depends on the direction of motion of the balls, 3) depends on speeds of the balls, 4) is equal to acceleration due to gravity, Consider a two particle system with the, particles having masses m1 and m2 . If the first, particle is pushed towards the centre of mass, through a distance d, by what distance should, the second particle be moved, so as to keep, the centre of mass at the same position?, [MAINS 2006], m1d, m 1d, m 2d, 1) d 2) m 3) m + m 4) m, 1, 1, 2, 2, 33
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , VECTOR PRODUCT (OR) CROSS, PRODUCT, 8., , →, , →, , →, , →, , →, , →, , →, , →, , →, , If P× Q = R; Q× R = P and R× P = Q then, → →, , →, , 1) P , Q and R are coplanar, →, , →, , 2) angle between P and Q may be less than 900, →, , →, , →, , 3) P + Q + R cannot be equal to zero., → →, , →, , 4) P, Q and R are mutually perpendicular, ROTATIONAL VARIABLES, RELATION, BETWEEN LINEAR AND ANGULAR, VARIABLES,ROTATIONAL, KINEMATICS,TORQUE AND, MECHANICAL EQUILIBRIUM, , 9., , Which of the following equation is wrong, uur ur ur, r r r, 1) t = r ´ F, 2) ar = w ´V, ur ur r, ur r ur, 3) at = a ´ r, 4) V = r ´w, 10. The following pair of physical quantities are, analogous to one another in translatory, motion and rotatory motion., 1) Mass , moment of inertia 2) Force,Torque, 3) Linear momentum , Angular momentum, 4) All, 11. The correct relation of the following is, r r ur, r r ur, 1) τ = r.F, 2) τ = r × F, ur, r F, r r ur, 3) τ = r, 4) τ = r + F, r, 12. Two particles p and q located at distances rp, and ‘ rq ’ respectively from the centre of a, , rotating disc such that rp > rq ., 1. both p and q have the same acceleration, 2. both p and q do not have any acceleration, 3. ‘p’ has greater acceleration than ‘q’, 4. ‘q’ has greater acceleration than ‘p’, 13. When a constant torque is applied on a rigid, body, then, 1) the body moves with linear acceleration, 2) the body rotates with constant angular velocity, 3) the body rotates with constant angular, acceleration, 4) the body undergoes equal angular displacement, in equal intervals of time, 14. Identify the increasing order of the angular, velocities of the following (E-2005), a) earth rotating about its own axis, b) hours hand of a clock, c) seconds hand of a clock, d) fly wheel of radius 2m making 300 rps, 1)a,b,c,d 2)b,c,d,a 3)c,d,a,b 4)d,a,b,c, 34, , JEE-ADV PHYSICS-VOL - III, 15. The direction of following vectors is along the, line of axis of rotation, 1) angular velocity, angular acceleration only, 2) angular velocity, angular momentum only, 3) angular velocity, angular acceleration, angular, momentum only, 4) angular velocity, angular acceleration, angular, momentum and torque, 16. A particle is moving along a fixed circular orbit, with uniform speed. Then true statement from, the following is, 1) angular momentum of particle is constant only, in magnitude but its direction changes from point, to point, 2) angular momentum of particle is constant only, in direction but its magnitude changes from point, to point, 3) angular momentum of particle is constant both, in magnitude and direction, 4) angular momentum of particle is not constant, both in magnitude and direction, 17. Class I lever is that in which, 1) fulcrum is between the load and effort, 2) load is between the fulcrum and effort, 3) effort is between the load and fulcrum, 4) fulcrum, load and effort at one point, 18. If force vector is along X-axis and radius, vector is along Y-axis then the direction of, torque is, 1) along +ve Z-axis 2) along -ve Z-axis, 3) in X-Y plane making an angle 45o with X-axis, 4) in X-Y plane making an angle 135o with X-axis, 19. During rotation of a body, the position vector, is along X–axis and force vector is along, Y–axis, The direction of torque vector is, 1) in the X-Y plane 2) along –ve Z-axis, 3) along +ve Z-axis 4) in the X-Z plane, 20. If the direction of position vector rr is towards, r, south and direction of force vector F is, towards east, then the direction of torque, vector τr is, 1) towards north, 2) towards west, 3) vertically upward, 4) vertically downward, 21. Which of the following is wrong?, 1) Direction of torque is parallel to axis of rotation, 2) Direction of moment of couple is perpendicular, to the plane of rotation of body, 3) Torque vector is perpendicular to both position, vector and force vector, 4) The direction of force vector is always, perpendicular to both the directions of position, vector and torque vector, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 22. A circular disc is rotated along clockwise, direction in horizontal plane. The direction of, torque is, 1) horizontally right side 2) horizontally left side, 3) vertically upwards 4)vertically downwards, 23. Magnitude of torque is maximum in the, following case, 1) radius vector is perpendicular to force vector, 2) radius vector is parallel to force vector, 3) Angle between radius vector and force vector, is 45o, 4) Angle between radius vector and force vector, is 60o, 24. A constant resultant torque rotates a wheel, about its own axis. Then true statement of, the following is, 1) angular velocity of wheel is constant, 2) angular acceleration of wheel is constant, 3)angular acceleration of wheel gradually increases, 4) angular momentum of wheel is constant, 25. A wheel is free to rotate about its own axis, without friction. A rope is wound around the, wheel. If other end of rope is pulled with a, constant force, then true statement from the, following is, 1) constant torque is produced and the wheel is, rotated with constant angular velocity, 2) constant torque is produced and the wheel is, rotated with constant angular acceleration, 3) variable torque is produced and the wheel is, rotated with variable angular velocity, 4) variable torque is produced and the wheel is, rotated with variable angular acceleration, 26. The following pairs of physical quantities are, not analogous to each other in translatory, motion and rotational motion, 1) force, torque, 2) mass, moment of inertia, 3) couple, torque, 4) linear momentum, angular momentum, , ROTATIONAL INERTIA OF SOLID, BODIES, ROTATIONAL DYNAMICS, 27. The moment of inertia of a rigid body depends, on, A) mass of body, B) position of axis of rotation, C) time period of its rotation, D) angular velocity of the body, 1) A and B are true, 2) B and C ar true, 3) C and D are true 4) A and D are true, NARAYANAGROUP, , 28. I1 , I 2 are moments of inertia of two solid, spheres of same mass about axes passing, through their centres If first is made of wood, and the second is made of steel, then, 1) I1 = I 2 2) I1 < I 2 3) I1 > I 2 4) I1 ≤ I 2, 29. A Uniform metal rod is rotated in horizontal, plane about a vertical axis passing through, its end at uniform rate. The tension in the, rod is, 1) same at all points, 2) different at different points and maximum at, centre of rod, 3) different at different points and minimum at axis, of rotation., 4) different at different points and maximum at axis, of rotation, 30. A boiled egg and a raw egg of same mass and, size are made to rotate about their own axis., If I1 and I 2 are moments of inertia of boiled, egg and raw egg, then, 1) I1 = I 2 2) I1 > I 2 3) I1 < I 2 4) I1 = 2 I 2, 31. Raw and boiled eggs are made to spin on a, smooth table by applying the same torque., The egg that spin faster is, 1) Raw egg, 2) Boiled egg, 3) Both will have same spin rate, 4) Difficult to predict, 32. Moment of Inertia of a body depends upon, 1) distribution of mass of the body, 2) position of axis of rotation, 3) temperature of the body, 4) all the above, 33. Of the two eggs which have identical sizes ,, shapes and weights, one is raw and other is, half boiled. The ratio between the moment of, inertia of the raw to the half boiled egg about, central axis is :, 1) = 1 2) > 1 3) < 1 4) not comparable, 34. The radius of gyration of a rotating metallic, disc is independent of the following physical, quantity., 1) Position of axis of rotation 2) Mass of disc, 3) Radius of disc, 4) temperature of disc, 35. A brass disc is rotating about its axis. If, temperature of disc is increased then its, 1) radius of gyration increases, but moment of, inertia remains the same, 2) moment of inertia increases but radius of, gyration remains the same, 3) radius of gyration, moment of inertia both, remain the same, 4) radius of gyration, moment of inertia both, increase, 35
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 36. The radius of gyration of a rotating circular, ring is maximum about following axis of, rotation, 1) natural axis, 2) axis passing through diameter of ring, 3) axis passing through tangent of ring in its plane, 4)axis passing through tangent of ring perpendicular, to plane of ring., 37. Moment of inertia of a thin circular plate is, minimum about the following axis, 1) axis perpendicular to plane of plate passing, through its centre, 2) axis passing through any diameter of plate, 3) axis passing through any tangent of plate in its, plane, 4) axis passing through any tangent perpendicular, to its plane, 38. A ring of mass ‘m’ and radius ‘r’ is melted, and then moulded into a sphere . The moment, of inertia of the sphere will be, 1) more than that of the ring, 2) less than that of the ring, 3) equal to that of the ring, 4) none of the above, 39. Two copper circular discs are of the same, thickness. The diameter of A is twice that of B., The moment of inertia of A as compared to that, of B is, 1) twice as large, 2) four times as large, 3) 8 times as large, 4) 16 times as large, 40. The moment of inertia of a thin square plate, ABCD of uniform thickness about an axis, passing through the centre O and, perpendicular to the plane of the plate is, [IIT1992], I2, I1, I3, A, B, , I4, , O, , C, a) I 1 + I 3, c) 2 I1 + I3, 1) a,b are true, 3) c,d are true, , 36, , D, b) I 2 + I 4, d) I 1 + 2 I 3, 2) b,c are true, 4) b,d are true, , JEE-ADV PHYSICS-VOL - III, 41. Identify the correct order in which the ratio, of radius of gyration to radius increases for, the following bodies., I) Rolling solid sphere II) Rolling solid cylinder, III) Rolling hollow cylinder, IV) Rolling hollow sphere, 1) I, II, IV, III, 2) I, III, II, IV, 3) II, I, IV, III, 4) II, I, III, IV, 42. Identify the increasing order of radius of, gyration of following bodies of same radius, I) About natural axis of circular ring, II) About diameter of circular ring, III) About diameter of circular plate, IV) About diameter of solid sphere, 1) II, III, IV, I, 2) III, II, IV, I, 3) III, IV, II, I, 4) II, IV, III, I, 43. Identify the decreasing order of moments of, inertia of the following bodies of same mass, and same radius., I) About diameter of circular ring, II) About diameter of circular plate, III) About tangent of circular ring ⊥ r to its plane, IV) About tangent of circular plate in its plane, 1) III, IV, II, I, 2) IV, III, I, II, 3) IV, III, II, I, 4) III, IV, I, II, 44. Three dense point size bodies of same mass, are attached at three vertices of a light, equilateral triangular frame. Identify the, increasing order of their moment of inertia, about following axis., I) About an axis ⊥ r to plane and passing through, a corner, II) About an axis ⊥ r to plane and passing through, centre, III) About an axis passing through any side, IV) About ⊥ r bisector of any side, 1) IV,III, II, I, 2) III, II, IV, I, 3) II, IV, III, I, 4) II, III, IV, I, 45. Four point size dense bodies of same mass, are attached at four corners of a light square, frame. Identify the decreasing order of their, moments of inertia about following axes., I) Passing through any side, II) Passing through opposite corners, III) ⊥ r bisector of any side, IV) ⊥ r to the plane and passing through any corner, 1) III, IV, I, II, 2) IV, III, I, II, 3) III, II, IV, I, 4) IV, III, II, I, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 46. A motor car is moving in a circular path with, uniform speed v. Suddenly the car rotates, through an angle θ . Then, the magnitude of, change in its velocity is, θ, 2, θ, 3) 2v tan, 2, , 1) 2v cos, , 2) 2v sin, , θ, 2, , 4) 2v sec, , θ, 2, , 47. An electric motor rotates a wheel at a, constant angular velocity (ω ) while opposing, torque is t . The power of that electric motor, is, tw, t, 1), 2) tw, 3) 2tw, 4), 2, w, 48. A constant power is supplied to a rotating disc., The relationship between the angular velocity, , 52., , 53., , 54., , (ω ) of the disc and number of rotations (n), made by the disc is governed by, 1, , 1) ω ∝ n 3, 3, , 3) ω ∝ n 2, , 2, , 2) ω ∝ n 3, , 55., , 4) ω ∝ n 2, , ANGULAR MOMENTUM &, CONSERVATION OF ANGULAR, MOMENTUM, 49. An ice block is in a trough which is rotating, about vertical axis passing through its centre., When ice melts completely, the angular, velocity of the system, , 56., , 57., , 58., 1) increases, 2) decreases, 3) remains same, 4)becomes double, 50. A circular disc is rotating about its own axis,, the direction of its angular momentum is, 1) radial, 2) along axis of rotation, 3) along tangent, 4) perpendicular to the direction of angular velocity, 51. A ballet dancer is rotating about his own, vertical axis on smooth horizontal floor. I , ω ,, L, E are moment of inertia, angular velocity,, angular momentum, rotational kinetic energy, of ballet dancer respectively. If ballet dancer, stretches himself away from his axis of, rotation, then, NARAYANAGROUP, , 59., , 1) I increases and ω , E decrease but L is constant, 2) I decreases, ω and E increase but L is constant, 3) I increases, ω decreases, L and E are constant, 4) I increases, ω increases but L and E are constant, If polar ice caps melt, then the time duration, of one day, 1) increases, 2) decreases, 3) does not change, 4) zero, A hollow sphere partly filled with water has, moment of inertia I when it is rotating about, its own axis at an angular velocity w . If its, angular velocity is doubled then its moment, of inertia becomes, 2) More than I, 1) Less than I, 3) I, 4) zero, If most of the population on earth is migrated, to poles of the earth then the duration of, a day, 1) increases, 2) decreases, 3) remains same, 4) first increases then decreases, The law of conservation of angular momentum, is obtained from Newton's II law in rotational, motion when, 1) external torque is maximum, 2) external torque is minimum, 3) external torque is zero, 4) external torque is constant, If earth shrinks then the duration of day, 1) increases, 2) decreases, 3) remains same, 4) first increases then decreases to initial value, A circular disc is rotating in horizontal plane, about vertical axis passing through its centre, without friction with a person standing on, the disc at its edge. If the person gently walks, to centre of disc then its angular velocity, 1) increases, 2) decreases, 3) does not change, 4 )becomes zero, A ballet dancer is rotating about his own, vertical axis.Without external torque if his, angular velocity is doubled then his rotational, kinetic energy is, 1) halved, 2) doubled, 3) quadrupled, 4) unchanged, The following motion is based on the law of, conservation of angular momentum, A) rotation of top, B) diving of diver, C) rotation of ballet dancer on smooth, horizontal surface, D) a solid sphere that rolls down on an inclined, plane, 1) A, B and C are true 2) A, B and D are true, 3) B, C and D are true 4) A, C and D are true, 37
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 60. Two bodies with moment of inertia I1 and I 2, , ( I 2 > I1 ), , 61., , 62., , 63., , 64., , 65., , are rotating with same angular, , momentum. If K1 and K 2 are their K.E.s,, then, 1) K 2 > K1 2) K 2 < K1 3) K1 = K 2 4) K 2 ≥ K1, A solid sphere is rotating in free space . If, the radius of the sphere is increased keeping, mass same which one of the following will not, be affected?, 1) Moment of inertia 2) Angular momentum, 3) Angular velocity 4) Rotational kinetic energy, A circular wheel is rotating in horizontal plane, without friction about its axis. If a body is, gently attached to the rim of the wheel then, following is false., 1) Moment of inertia increases but angular, momentum remains same, 2) Angular velocity decreases but angular, momentum remains same, 3) Rotational kinetic energy decreases but angular, momentum remains same, 4) Angular momentum increases but angular, velocity remains same, A uniform metal rod of length 'L' and mass, 'M' is rotating about an axis passing through, one of the ends perpendicular to the rod with, angular speed ' ω ' . If the temperature, increases by "t 0 C" then the change in its, angular velocity is proportional to which of, the following ? (Coefficient of linear expansion, of rod = α ), 1) ω, 2) ω, 3) ω2, 4) 1/ ω, A gymnast standing on a rotating stool with, his arms outstretched, suddenly lowers his, arms, 1) his angular velocity decreases, 2) his angular velocity increases, 3) his moment of inertia remains same, 4) his moment of inertia increases, Angular momentum of the particle rotating, with a central force is constant due to, [AIEEE-2007], 1) constant force, 2) constant linear momentum, 3) zero torque, 4) constant torque, , ROLLING MOTION &ROTATIONAL, KINETIC ENERGY, 66. Solid sphere, hollow sphere, solid cylinder and, hollow cylinder of same mass and same radii, are simultaneously start rolling down from the, top of an inclined plane. The body that takes, longest time to reach the bottom is, 38, , 1) solid sphere, 2) hollow sphere, 3) solid cylinder, 4) hollow cylinder, 67. Solid sphere, solid cylinder, hollow sphere,, hollow cylinder of same mass and same radii, are rolling down freely on an inclined plane., The body with maximum acceleration is, 1) solid sphere, 2) solid cylinder, 3) hollow sphere, 4) hollow cylinder, 68. In the case of following rolling body, translatory and rotational kinetic energies are, equal for, 1) circular ring, 2) circular plate, 3) solid sphere, 4) solid cylinder, 69. A disc is rolling (without slipping) on a, frictionless surface . C is its centre and Q and, P are two points equidistant from C. Let, V p , VQ and Vc be the magnitudes of velocities, of points P,Q and C respectively, then, [IIT-2004], , Q, C, P, , 1) VQ > VC > VP, , 2) VQ < VC < VP, , 1, 3) VQ = VP , VC = VP, 4) VQ < VC > VP, 2, 70. A particle performs uniform circular motion, with an angular momentum L. If the angular, frequency f of the particle is doubled, and, kinetic energy is halved, its angular, momentum becomes :, , 1) 4L, , 2) 2 L, , 3), , L, 2, , 4), , L, 4, , 71. If V is velocity of centre of mass of a rolling, body then velocity of lowest point of that body, is, 1) 2V, 2) V, 3) 2V, 4) Zero, 72. If the velocity of centre of mass of a rolling, body is V then velocity of highest point of that, body is, V, 1) 2V, 2) V, 3) 2V, 4), 2, 73. If x is ratio of rotational kinetic energy and, translational kinetic energy of rolling body, then the following is true, 1, 1) x = 1, 2) x £ 1 3) x ³ 1 4) x =, 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 74. A body is freely rolling down on an inclined, plane whose angle of inclination is q . If ‘a’, is acceleration of its centre of mass then, following is correct, 1) a = gsin q, 2) a < g sin q, 3) a > g sin q, 4) a = 0, 75. A Child is standing with folded hands at the, centre of a platform rotating about its central, axes. The K.E of the system is ‘ K ’. The child, now stretches his hands so that the moment, of, inertia of the system doubles. The K.E, of the system now is, K, K, 1) 2K, 2), 3), 4) 4K, 2, 4, 76. A yo-yo is placed on a rough horizontal, surface and a constant force F , which is less, than its weight, pulls it vertically. Due to this, F, , C, , /////////////////////////, O, 1) frictional force acts towards left, so it will move, towards left, 2) frictional force acts towards right, so it will move, towards right, 3) it will move towards left, so frictional force acts, towards left, 4) it will move towards right so friction force acts, towards right, 77. When the following bodies of same radius, starts rolling down on same inclined plane,, identify the decreasing order of their times, of descent, I) solid cylinderII) hollow cylinder, III) hollow sphere, IV) solid sphere, 1) IV, I, III, II, 2) II, III, I, IV, 3) I, IV, III, II, 4) II, III, IV, I, 78. When the following bodies having same radius, starts rolling down on same inclined plane,, identify the increasing order of their, accelerations, I) hollow cylinder, II) solid cylinder, III)solid sphere, IV) hollow sphere, 1) I, IV, III, II, 2) IV, I, II, III, 3) I, IV, II, III, 4) I, IV, III, II, NARAYANAGROUP, , 79. When a ring is rolling V1, V2, V3 and V4 are, velocities of top most point, lowest point, end, point of horizontal diameter, centre of ring, respectively, the decreasing order of these, velocities is, 1) V2, V1, V4, V3, 2) V2, V1, V3, V4, 3) V1, V2, V3, V4, 4) V1, V3, V4, V2, 80. The increasing order of fraction of total, kinetic energy associated with translatory, motion of the following rolling bodies is, I) circular ring, II) circular plate, III) solid sphere, IV) hollow sphere, 1) I, II, IV, III, 2) IV, I, II, III, 3) I, IV, II, III, 4) IV, I, III, II, 81. A and B are two solid spheres of equal, masses. A rolls down an inclined plane without, slipping from a height H. B falls vertically, from the same height. Then on reaching the, ground., 1) both cannot do work, 2) A can do more work than B, 3) B can do more work than A, 4) both A and B will have different linear speeds, 82. A solid sphere, a hollow sphere and a ring, are released from top of an inclined plane, (frictionless) so that they slide down the, plane. Then maximum acceleration down the, plane is (no rolling):, 1) solid sphere, 2) hollow sphere, 3) ring, 4) same for all, 83. A sphere cannot roll on, 1) a smooth horizontal surface, 2) a smooth inclined surface, 3) a rough horizontal surface, 4) a rough inclined surface., , C.U.Q - KEY, 01) 1, 07) 4, 13) 3, 19) 3, 25) 2, 31) 2, 37) 2, 43) 4, 49) 2, 55) 3, 61) 2, 67) 1, 73) 2, 79) 4, , 02) 3, 08) 1, 14) 1, 20) 3, 26) 3, 32) 4, 38) 2, 44) 1, 50) 2, 56) 2, 62) 4, 68) 1, 74) 2, 80) 3, , 03) 3, 09) 4, 15) 4, 21) 4, 27) 1, 33) 2, 39) 4, 45) 2, 51) 1, 57) 1, 63) 2, 69) 1, 75) 2, 81) 4, , 04) 2, 10) 4, 16) 3, 22) 4, 28) 3, 34) 2, 40) 1, 46) 2, 52) 1, 58) 2, 64) 2, 70) 4, 76) 1, 82) 4, , 05) 2, 11) 2, 17) 1, 23) 1, 29) 4, 35) 4, 41) 1, 47) 2, 53) 2, 59) 1, 65) 3, 71) 4, 77) 2, 83) 2, , 06) 4, 12) 3, 18) 2, 24) 2, 30) 3, 36) 4, 42) 3, 48) 1, 54) 2, 60) 2, 66) 4, 72) 3, 78) 3, , 39
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 8., , LEVEL - I (C.W), CENTRE OF MASS, 1., , A system consists of two masses connected, by a massless rod lies along x–axis. The, distance of centre of mass from O is, m2=0.6kg, , m1=0.4kg, , 9., , O, , x2=7m, , x1=2m, , 1) 2m 2) 3m, 3) 5m, 4) 7m, Four particles, each of mass 1 kg are placed, at the corners of a square OABC of side 1 m., ‘O’ is at the origin of the coordinate system., OA and OC are aligned along positive X-axis, and positive Y-axis respectively. The position, vector of the centre of mass is (in ‘m’), 1, 1 ˆ ˆ, i− j, 1) iˆ + ˆj 2) iˆ + ˆj 3) iˆ − ˆj 4), 2, 2, A thick straight wire of length π m is fixed at, its midpoint and then bent in the form of a, circle. The shift in its centre of mass is, π, 3) 2 m, 4) m, 1) π m 2) 0.5 m, 2, A rigid body consists of a 3kg mass located, ur, at r1 = 2$i + 5 $j m and a 2kg mass located at, r, r 2 = (4iˆ + 2 ˆj) m. The position of centre of mass, is, 19 ˆ , 19 ˆ , 14 ˆ, 14, i m 2) ˆi +, jm, 1) j +, 5 , 5 , 5, 5, 19 ˆ 14 ˆ , j m, 3) i +, 4) 0, 5 , 5, A boat of mass 40kg is at rest. A dog of mass, 4kg moves in the boat with a velocity of, 10m/s. What is the velocity of boat(nearly)?, 1) 4m/s, 2) 2m/s, 3) 8m/s, 4) 1 m/s, Two blocks of masses 10kg and 30 kg are, placed along a vertical line if the first block is, raised through a height of 7cm then the, distance through the second mass should be, moved to raise the centre of mass of the system, by 1cm is, 1)1cm up 2)1cm down 3)2 cm down 4)2 cm up, , 2., , (, , 3., , 4., , (, , 5., , 6., , ), , (, , ), , (, , ), , ), , MOTION OF CENTRE OF MASS,, LINEAR MOMENTUM OF, CENTRE OF MASS, 7., , 40, , Two bodies of different masses 2kg and 4kg, are moving with velocities 2m/s and 10m/s, towards each other due to mutual gravitational, attraction. Then the velocity of the centre of, mass is, , 1) 5ms–1 2) 6ms–1, 3) 8ms–1 4) Zero, If two particles of masses 3kg and 6kg, which are at rest are separated by a distance, of 15m. The two particles are moving towards, each other under a mutual force of attraction., Then the ratio of distances travelled by the, particles before collision is, 1) 2 : 1 2) 1: 2, 3) 1 : 3, 4) 3 :1, Two bodies of 6 kg and 4 kg masses have their, velocity 5iˆ − 2 ˆj + 10kˆ and 10iˆ − 2 ˆj + 5kˆ respectively.Then the velocity of their centre of, mass is, 1) 5iˆ + 2 ˆj − 8kˆ, 2) 7iˆ + 2 ˆj − 8kˆ, , 3) 7iˆ − 2 ˆj + 8kˆ, 4) 5iˆ − 2 ˆj + 8kˆ, 10. A thin uniform rod of length “L” is bent at, its mid point as shown in the figure. The, distance of the centre of mass from the point, “O” is, , θ, O, L, θ, L, θ, sin, 2) cos, 2, 2, 2, 2, L, θ, L, θ, sin, 3), 4) cos, 4, 2, 4, 2, 11. Three identical spheres each of mass ‘m’, and radius ‘R’ are placed touching each, other so that their centres A, B and C lie, on a straight line. The position of their, centre of mass from centre of A is, 2R, 5R, 4R, 1), 2) 2R, 3), 4), 3, 3, 3, 12. A boy of mass 50kg is standing at one end of, a boat of length 9m and mass 400kg. He, runs to the other end. The distance through, which the centre of mass of the boat boy, system moves is, 1) 0, 2) 1m, 3) 2m 4) 3m, 13. A dog weighing 5kg is standing on a flat boat, so that it is 10 metres from the shore. It, walks 4m on the boat towards the shore and, then halts. The boat weighs 20kg and one can, assume that there is no friction between it, and water. The dog from the shore at the end, of this time is, 1) 3.4 m 2) 6.8m 3) 12.6 m 4) 10 m, , 1), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , VECTOR PRODUCT (or) CROSS, PRODUCT, 14. The angular velocity of a rotating body is, ur, ω = 4$i + $j − 2k$ . The linear velocity of the, body whose position vector 2$i + 3 $j − 3k$ is, 1) 5$i +8 $j +14k$, 2) 3$i + 8 $j + 10k$, 3) 8$i − 3 $j + 2k$, 4) -8$i +3 $j +2k$, 15. The area of the triangle whose adjacent sides, are represented by the vector 4$i + 3 $j + 4k$, , (, , and 5i$ in sq. units is, 1) 25, 2) 12.5, , ), , 3) 50, 4) 45, 16. The angle between the vectors $i + $j + k$ and, , (, , ($i − $j − k$ ) is, −1, , 8, 3, , −1, 3) cos, , 8, 3, , 1) sin, , ), , ROTATIONAL VARIABLES, RELATION, BETWEEN LINEAR & ANGULAR, VARIABLES, 17. The linear velocity of a point on the surface, of earth at a latitude of 60° is, 800, m/sec, 3, 5, 3) 800 × m/sec, 18, , 2), , 800π, 3, , 4), , 2000π, m/sec, 27, , m/sec, , 18. A table fan, rotating at a speed of 2400 rpm, is switched off and the resulting variation of, the rpm with time is shown in the figure. The, total number of revolutions of the fan before, it comes to rest is, Rev/min, 2400, , 600, 0, , t(s), 8, , 1) 420, NARAYANAGROUP, , 16, , 2) 280, , 24, , 3) 240, , 1), , π, rads −1, 15, , 2), , π, rads −1, 30, , π, rads −1, 7, 20. The angular displacement of a particle is, given by θ = t3 + t2 + t + 1 then, its angular, , 3), , π, rads −1, 45, , 4), , velocity at t = 2 sec is ......... rads −1, 1) 27, 2) 17, 3) 15, 4) 16, 21. In the above problem, the angular, acceleration of the particle at t = 2 sec is, ......... rads − 2, 1) 14, 2) 16, 3) 18, 4) 24, , ROTATIONAL KINEMATICS,, TORQUE, MECHANICAL, EQUILIBRIUM, , 1 π, 2) sin +, 3 3, 8, −1, 4) cos, 3, −1, , 1), , 19. The average angular velocity of the seconds, hand of a watch if the seconds hand of the, watch completes one revolution in 1 minute, is, , 4) 380, , 22. A stationary wheel starts rotating about its, own axis at uniform angular acceleration, 8rad / s 2 . The time taken by it to complete 77, rotations is, 1) 5.5 sec, 2) 7 sec 3) 11 sec 4) 14 sec, 23. A stationary wheel starts rotating about its own, axis at constant angular acceleration. If the, wheel completes 50 rotations in first 2 seconds,, then the number of rotations made by it in next, two seconds is, 1) 75, 2) 100 3) 125, 4) 150, r, r, 24. If F = 2 ˆi - 3 ˆj N and r = 3iˆ + 2 ˆj m then, torque τr is, $ 3) -12k, $, $, $ 4) -13k, 1) 12k, 2) 13k, 25. A crowbar of length 120 cm has its fulcrum, situated at a distance of 20cm from the load., The mechanical advantage of the crow bar is, 1) 1, 2) 3, 3) 5, 4) 7, ROTATIONAL INERTIA OF SOLID BODIES, 26. Three particles of masses 1gm, 2gm & 3gm, are at 1cm, 2cm, & 3cm from the axis of, rotation respectively then the moment of, inertia of the system & radius of gyration of, the system respectively are .......gm cm2 and, .. cm, 1) 63, 2.449, 2) 60, 4.5, 3) 36, 4.449, 4) 36, 2.449, 41
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 27. A hoop of mass 500gm & radius 10cm is, placed on a nail. then the moment of inertia, of the hoop, when it is rotated about the nail, will be-- kgm2, 1) 0.05 2) 0.02 3) 0.01 4) 0.03, 28. The ratio of moments of inertia of two solid, spheres of same mass but densities in the, ratio 1:8 is, 1) 1 : 4, 2) 4 :1 3) 2 : 1 4) 8 :1, 29. The radius of a solid sphere is R and its density, D. When it is made to rotate about an axis, passing through any diameter of sphere,, expression for its moment of inertia is, 8, 5, 1) pDR, 7, , 2), , 8, pDR 5, 15, , 28, 28, 5, pDR 5, 4) pDR, 15, 5, 30. Four point size bodies each of mass M are fixed, at four corners of a light squre frame of side, length L. The moment of inertia of the four, bodies about an axis perpendicular to the plane, of frame and passing through its centre is, 1) 4ML2 2) 2 2ML2 3) 2ML2 4) 2ML2, 31. Four particles each of mass ‘m’ are placed, at the corners of a square of side length ' l '., The radius of gyration of the system about, an axis perpendicular to the plane of square, and passing through its centre is, , 3), , 1), , l, 2, , 2), , l, 2, , 3) l, , 4), , 2l, , 32. In the above problem the moment of inertia, of four bodies about an axis perpendicular to, the plane of frame and passing through a, corner is, 2) 2ML2, 3) 2 2ML2 4) 4ML2, 1) ML2, 33. In above problem the moment of inertia of, four bodies about an axis passing through, opposite corners of frame is, 1) 2ML2 2) 2ML2 3) ML2 4) 2 2ML2, 34. In the above problem the moment of inertia, of four bodies about an axis passing through, any side of frame is, 1) 4ML2 2) 2 2ML2 3) 2ML2 4) 2ML2, 35. The diameter of a fly wheel is R. Its coefficient, of linear expansion is a . If its temperature is, increased by ∆T the percentage increase in its, 42, , moment of inertia is, 1) 200 × α × ∆T, 2) 100 × α × ∆T, 3) 50 × α × ∆T, 4) 150 × α × ∆T, 36. Three point sized bodies each of mass M are, fixed at three corners of light triangular frame, of side length L. About an axis perpendicular, to the plane of frame and passing through, centre of frame the moment of inertia of three, bodies is, 3ML2, 3) 3ML2 4) 3ML2, 2, 37. In above problem, about an axis perpendicular, to the plane of frame and passing through a, corner of frame the moment of inertia of three, bodies is, , 1) ML2, , 2), , 3ML2, 1) ML2 2) 2ML2, 3) 3ML2 4), 2, 38. In above problem about an axis passing through, any side of frame the moment of inertia of three, bodies is, 3ML2, 3ML2, 2ML2, 1) ML2 2), 3), 4), 2, 4, 3, 39. The radius of gyration of a body is 18 cm when, it is rotating about an axis passing through, centre of mass of body. If radius of gyration of, same body is 30 cm about a parallel axis to first, axis then, perpendicular distance between two, parallel axes is, 1) 12 cm 2) 16 cm 3) 24 cm 4) 36 cm, 40. The position of axis of rotation of a body is, changed so that its moment of inertia, decreases by 36%. The % change in its, radius of gyration is, 1) decreases by 18% 2) increases by 18%, 3) decreases by 20% 4) increases by 20%, 41. A diatomic molecule is formed by two atoms, which may be treated as mass points m1 and, m2 joined by a massless rod of length r. Then, the moment of inertia of molecule about an axis, passing through centre of mass and, perpendicular to the rod is :, 1)zero, 2) ( m1 + m2 ) r 2, m1m2 , , 2, , m1 + m2 , , 3) m + m r, 4) m m r, 1 2 , 1, 2 , 42. I is moment of inertia of a thin square plate, about an axis passing through opposite, corners of plate. The moment of inertia of, same plate about an axis perpendicular to the, plane of plate and passing through its centre, is, 1) I/2, 2) I / 2, 3) 2I 4) 2I, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 43. Mass of thin long metal rod is 2 kg and its, moment of inertia about an axis perpendicular, to the length of rod and passing through its, one end is 0.5kg m 2 . Its radius of gyration is, 1) 20 cm, 2) 40 cm, 3) 50 cm, 4) 1m, , ANGULAR MOMENTUM AND, CONSERVATION OF ANGULAR, MOMENTUM, 44. The diameter of a disc is 1m. It has a mass of, 20kg. It is rotating about its axis with a speed, of 120rotations in one minute. Its angular, momentum in kg m 2/s is, 1)13.4, 2) 31.4, 3) 41.4 4) 43.4, 45. If the earth were to suddenly contract to 1/nth, of its present radius without any change in, its mass, the duration of the new day will be, nearly, 1) 24/n hours, 2) 24n hours, 3) 24/n2 hours, 4) 24n2 hours, 46. A particle performs uniform circular motion, with an angular momentum L. If the angular, frequency f of the particle is doubled, and, kinetic energy is halved, its angular, momentum becomes, 1) 4L, 2) 2L, 3) L/2, 4) L/4, 47. A ballet dancer is rotating about his own, vertical axis at an angular velocity 100 rpm, on smooth horizontal floor. The ballet dancer, folds himself close to his axis of rotation by, which is moment, of inertia decreases to, half of initial moment of inertia then his final, angular velocity is, 1) 50rpm, 2) 100rpm, 3) 150rpm, 4) 200rpm, 48. A circular ring of mass M is rotating about, its own axis in horizontal plane at an angular, velocity ω . If two point size bodies each of, mass m, are gently attached to the rim of ring, at two ends of its diameter, then the angular, velocity, of ring is, Mω, , 2mω, , mω, , 2M ω, , 1), 2), 3), 4), M + 2m, M + 2m, M + 2m, M + 2m, 49. A ballet dancer is rotating at angular velocity, ω on smooth horizontal floor. The ballet, dancer folds his body close to his axis of, rotation by which his radius of gyration, decreases by 1/4th of his initial radius of, gyration, his final angular velocity is, 3ω, 9ω, 9ω, 16ω, 1), 2), 3), 4), 4, 4, 16, 9, 50. A particle of mass m is moving along a circle, of radius r with a time period T. Its angular, momentum is, , NARAYANAGROUP, , 2π mr, 4π mr, 2π mr 2, 4π mr 2, 1), 4), 2), 3), T, T, T, T, 51. If the radius of earth shrinks by 0.2% without, change in its mass, the % change in its, angular velocity is, 1) increase by 0.4% 2) increase by 0.1%, 3) decrease by 0.4% 4) decrease by 0.1%, 52. A metallic circular plate is rotating about its, axis without friction. If the radius of plate, expands by 0.1% then the % change in its, moment of inertia is, 1) increase by 0.1% 2) decrease by 0.1%, 3) increase by 0.2% 4) decrease by 0.2%, 53. A constant torque acting on a uniform circular, wheel changes its angular momentum from A, to 4A in 4sec. The torque acted on it is, 3A, A, 2A, 3A, 1), 2), 3), 4), 4, 4, 4, 2, 54. Density remaining constant, if earth, contracts to half of its present radius,, duration of the day would be (in minutes), 1) 45, 2) 80, 3) 100, 4) 120, 55. A mass is whirled in a circular path with an, angular momentum L. If the length of string, and angular velocity, both are doubled, the, new angular momentum is, 1) L, 2) 4L, 3) 8L, 4) 16L, , ROTATIONAL DYNAMICS, 56. An automobile engine develops 100 KW, when rotating at a speed of 1800 rev/min. The, torque it delivers ( in N-m ), 1) 350, 2) 440, 3) 531 4) 628, 57. An electric motor exerts a constant torque, 5Nm on a fly wheel by which it is rotated at, the rate of 420rpm The power of motor is, 1)110watt, 2)150watt, 3)220watt, 4)300watt, , ROLLING MOTION, 58. A shaft rotating at 3000rpm is transmitting a, power of 3.14KW. The magnitude of the, driving torque is, 1) 6Nm 2) 10Nm, 3) 15Nm 4) 22Nm, 59. A solid sphere rolls down without slipping, from rest on a 30 0 incline. Its linear, acceleration is, 1) 5g/7, 2) 5g/14 3) 2g/3 4) g/3, 60. A hollow sphere rolls down a 30o incline of, length 6m without slipping. The speed of cen, tre of mass at the bottom of plane is, 1) 6ms −1 2) 3ms −1 3) 6 2ms −1 4) 3 2ms −1, , 43
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 3., , A uniform rod of length one meter is bent at, its midpoint to make 900. The distance of, centre of mass from the centre of rod is (in, cm), 1) 20.2, 2) 13.4, 3) 15, 4) 35.36, Particles of masses 1kg and 3kg are at, , 4., , 10. Two objects of masses 200g and 500g have, velocities of 10i m/s and ( 3i + 5 j ) m / s, respectively. The velocity of their centre of mass, is, 5, 2) i − 25 j, 1) 5i − 25 j, 7, 25, 5, j, 4) 25i − j, 3) 5i +, 7, 7, , ( 2i + 5 j + 13k ) m and ( −6i + 4 j − 2k ) m then, instantaneous position of their centre of mass, is, 1, 1) ( −16i + 17 j + 7k ) m, 4, 1, 2) ( −8i + 17 j + 7k ) m, 4, 1, 3) ( −6i + 17 j + 7k ) m, 4, 1, 4) ( −6i + 17 j + 5k ) m, 4, A boat of mass 50kg is at rest. A dog of mass, 5kg moves in the boat with a velocity of 20m/, s. What is the velocity of boat?, 1) 4m/s, 2) 2m/s 3) 8m/s 4) 1 m/s, , 5., , MOTION OF CENTRE OF MASS,, LINEAR MOMENTUM OF, CENTRE OF MASS, 6., , Two bodies of masses 5kg and 3kg are moving, towards each other with 2ms −1 and 4ms −1, respectively. Then velocity of centre of mass is, 1) 0.25ms −1 towards 3kg 2) 0.5ms −1 towards 5kg, 3) 0.25ms −1 towards 5kg 4) 0.5ms −1 towards 3kg, 7. A circular disc of radius 20cm is cut from one, edge of a larger circular disc of radius 50cm., The shift of centre of mass is, 1) 5.7cm 2) -5.7cm 3) 3.2cm 4) -3.2cm, 8. Two particles of masses 4kg and 6kg are, separated by a distance of 20m and are, moving towards each other under mutual, force of attraction, the position of the point, where they meet is, 1) 12m from 4kg body 2) 12m from 6kg body, 3) 8m from 4kg body 4) 10m from 4kg body, 9. A uniform metre rod is bent into L shape with, the bent arms at 900 to each other. The, distance of the center of mass from the bent, point is, L, L, L, L, m 2), m 3), m 4), m, 1), 4 2, 2 2, 2, 8 2, , 46, , VECTOR PRODUCT OR CROSS, PRODUCT, 11. The position of a particle is given by, r, r = $i + 2 $j − k$ and its momentum is, ur, p = 3$i + 4 $j − 2k$ . The angular momentum is, perpendicular to, 1) x-axis, 2) y-axis, 3) z-axis, 4) line at equal angles to all the axes, 12. A uniform sphere has radius R. A sphere of, diameter R is cut from its edge as shown., Then the distance of centre of mass of, remaining portion from the centre of mass of, the original sphere is, , R, 1)R/7, 2) R/14, 3)2R/7 4) R/18, 13. The area of the parallelogram whose adjacent, sides are P = 3i$ + 4 $j; Q = −5$i + 7 $j is, (in sq.units), 1)20.5, 2) 82, 3) 41, 4) 46, r, r, 14. If A = 3i + j + 2k and B = 2i − 2 j + 4k and θ, is the angle between the two vectors, then, sinθ is equal to, 2, 2, 2, 2, 1), 2), 3), 4), 3, 7, 13, 3, , ROTATIONAL VARIABLES,, RELATION BETWEEN LINEAR, AND ANGULAR VARIABLES, 15., , A particle is moving with uniform speed, 0.5m/s along a circle of radius 1m then the, angular velocity of particle is ( in rads-1 ), 1)2, 2)1.5, 3)1, 4) 0.5, 16. The angular velocity of the seconds hand in a, watch is, 1) 0.053 rad/s, 2) 0.210 rad/s, 3) 0.105 rad/s, 4) 0.42 rad/s, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 17. The angular displacement of a particle is given, by θ = t 3 + 2t + 1 , where t is time in seconds., Its angular acceleration at t = 2s is, 1) 14 rad s −2, , 2) 17 rad s −2, , 3) 12 rad s −2, , 4) 9 rad s −2, , ROTATIONAL KINEMATICS,, TORQUE, MECHANICAL, EQUILIBRIUM, 18. A circular disc is rotating about its own axis, at a uniform angular velocity ω . The disc is, subjected to uniform angular retardation by, ω, 2, during 120 rotations. The number of rotations, further made by it before coming to rest is, 1)120, 2) 60, 3) 40, 4) 20, 19. The handle of a door is at a distance 40cm, from axis of rotation. If a force 5N is applied, on the handle in a direction 300 with plane of, door, then the torque is, 1) 0.8 Nm, 2) 1 Nm 3) 1.6 Nm 4) 2 Nm, 20. A door can just be opened with 10N force on, the handle of the door. The handle is at a, distance of 50cm from the hinges. Then, the, torque applied on the door (in Nm) is, 1) 5, 2) 10, 3) 15, 4) 20, 21. A particle of mass m is projected with an initial, velocity u at an angle θ to horizontal.The, torque of gravity on projectile at maximum, height about the point of projection is, , which its angular velocity is decreased to, , mgu 2 sin 2θ, 1), 2, , 2) mgu 2 sin 2θ, , 1, mgu 2 sin θ, 2, 4) mu sin 2θ, 2, 2, 22. A uniform rod is 4m long and weights 10kg., If it is supported on a knife edge at one meter, from the end, what weight placed at that end, keeps the rod horizontal., 1) 8kg, 2) 10kg 3) 12kg 4) 15kg, , 3), , ROTATIONAL INERTIA OF SOLID BODIES, 23. The ratio of moments of inertia of a solid, sphere about axes passing through its centre, and tangent respectively is, 1) 2:5, 2) 2:7, 3) 5:2, 4) 7:2, NARAYANAGROUP, , 24. If I is moment of inertia of a thin circular plate, about an axis passing through tangent of plate, in its plane. The moment of inertia of same, circular plate about an axis perpendicular to its, plane and passing through its centre is, 4I, 2I, 4I, 2I, 2), 3), 4), 5, 5, 3, 3, 25. The moment of inertia of a solid sphere about, , 1), , an axis passing through its centre is 0.8kgm 2 ., The moment of inertia of another solid sphere, whose mass is same as mass of first sphere,, but the density is 8 times density of first, sphere, about an axis passing through its, centre is, 1) 0.1kgm2, , 2) 0.2 kgm 2, , 3) 0.4 kgm 2, 4) 0.5 kgm 2, 26. Moment of inertia of a hoop suspended from, a peg about the peg is, , MR 2, 3MR 2, 2, 1) MR, 2), 3) 2MR, 4), 2, 2, 27. Four particles each of mass 1kg are at the, four corners of square of side 1m. The M.I.of, the system about a normal axis through centre, of square is, 2, , 1) 6 kgm 2 2) 2 kgm 2 3)1.25 kgm 2 4) 2.5 kgm 2, 28. Three identical masses, each of mass 1kg,, are placed at the corners of an equilateral, triangle of side l. Then the moment of inertia, of this system about an axis along one side of, the triangle is, 3 2, 3 2, l, 4) l, 4, 2, 29. A wire of mass m and length l is bent in the, form of circular ring. The moment of inertia, of the ring about its axis is, , 1) 3l 2, , 2) l 2, , 3), , ml 2, ml 2, ml 2, 1) ml, 2), 3), 4), 4π 2, 2π 2, 8π 2, 30. The moment of inertia of a thin uniform rod of, mass M and length L about an axis, perpendicular to the rod, through its centre is, I.The moment of inertia of the rod about an axis, perpendicular to rod through its end point is, 2, , 1), , I, 4, , 2), , I, 2, , 3) 2I, , 4) 4I, 47
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 31. Four point size bodies each of mass m are, fixed at four corners of light square frame of, side length 1m. The radius of gyration of these, four bodies about an axis perpendicular to the, plane of frame passing through its centre is, 1, 1, 2) 2, 3), 4), 1) 2, 2, 2, 32. Uniform square plate of mass 240 gram is, made to rotate about an axis passing through, any diagonal of plate. If its moment of inertia, is 2 × 10−4 kgm 2 then its side length is, 1) 10cm 2) 12cm, 3) 15cm 4) 20cm, 33. Two objects of masses 1kg and 2kg separated, by a distance of 1.2m are rotating about their, centre of mass. Find the moment of inertia of, the system, 1) 0.96kgm 2, 2) 0.48kgm 2, 3) 0.83kgm 2, 4) 0.72kgm 2, 34 The radius of gyration of a body about an axis, at a distance of 4cm from its centre of mass, is 5cm. The radius of gyration about a parallel, axis through centre of mass is, 1) 2cm, 2) 5cm, 3) 4cm, 4) 3cm, 35. The M.I. of a thin rod about a normal axis, through its centre is I. It is bent at the centre, such that the two parts are perpendicular to, each other and perpendicular to the axis. The, M.I. of the system about the same axis will be, 1) 2I, 2) I, 3) I/2, 4) 4I, 36. The moment of inertia of two spheres of equal, masses about their diameters are the same., One is hollow, then ratio of their diameters, 1) 1:5, 2) 1: 5, 3) π :1, 4) 5 : 3, , ANGULAR MOMENTUM AND, CONSERVATION OF ANGULAR, MOMENTUM, 37. A circular disc of mass 4kg and of radius 10cm, is rotating about its natural axis at the rate, of 5 rad/sec. its angular momentum is, 1) 0.25 kgm 2 s −1, 2) 0.1kgm 2 s −1, 3) 2.5kgm 2 s −1, 4) 0.2 kgm 2 s −1, 38. If the mass of earth and radius suddenly, become 2 times and 1/4th of the present value,, the length of the day becomes, 1) 24h, 2) 6h, 3) 3/2h 4) 3h, 39. A child is standing with folded hands at the, centre of a platform rotating about its central, axis. The K.E. of the system is K. The child, 48, , 40., , 41., , 42., , 43., , now stretches his arms so that the M.I. of, the system doubles. The K.E. of the system, now is, 1) 2K, 2) K/2, 3) 4K, 4) K/4, If radius of earth shrinks by 0.1% without, change in its mass, the percentage change in, the duration of one day, 1) decrease by 0.1% 2) increase by 0.1%, 3) decrease by 0.2% 4) increase by 0.2%, A ballet dancer spins about a vertical axis at, 60rpm with his arms closed. Now he stretches, his arms such that M.I. increases by 50%., The new speed of revolution is, 1) 80rpm 2) 40rpm 3) 90rpm 4) 30rpm, A metallic circular wheel is rotating about its, own axis without friction. If the radius of wheel, expands by 0.2%, percentage change in its, angular velocity, 1) increase by 0.1% 2) decrease by 0.1%, 3) increase by 0.4% 4) decrease by 0.4%, A uniform circular disc of radius R is rotating, about its own axis with moment of inertia I at, an angular velocity ω If a denser particle of, mass m is gently attached to the rim of disc, than its angular velocity is, , Iω, I + mR 2, 1) ω 2) I ω ( I + mR ) 3), 4), I + mR 2, Iω, 44. A particle of mass m is rotating along a circular, path of radius r. Its angular momentum is L., The centripetal force acting on the particle is, , L2, L2 m, L2, L2, 1), 2), 3), 4), mr, r, mr 2, mr 3, ur, r, 45. F = ai$ + 3 $j + 6k$ and r = 2$i − 6 $j − 12k$ . The, value of ‘a’ for which the angular momentum, is conserved is, 1) -1, 2) 0, 3) 1, 4) 2, 46. If earth shrinks to 1/64 of its volume with mass, remaining same, duration of the day will be, 1) 1.5h, 2) 3h, 3) 4.5h, 4) 6h, 47. A mass is whirled in a circular path with a, constant angular velocity and its angular, momentum is L. If the length of string is now, halved keeping the angular velocity same, the, new angular momentum is, 1) L/4, 2) L/2, 3) L 4) 2L, 48. A disc rotates with angular velocity ω and, kinetic energy E. Then its angular momentum, E, 2E, ω, 1) Iω, 2) L =, 3) L =, 4) L =, ω, ω, E, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , ROTATIONAL DYNAMICS, 2, kgm 2 . It is, 2, π, rotated by a 60W motor for one minute. The, number of rotations made by the wheel in one, minute is, 1) 90 2) 450 3) 1800, 4) 1200, 50. The shaft of a motor is making 1260rpm. The, torque supplied by the motor is 100Nm. the, power of motor is ( in KW), 1) 100 2) 21, 3) 13.2, 4) 4.8, 51. An electric motor rotates a wheel at a, constant angular velocity 10rps while, opposing torque is 10Nm . The power of that, electric motor is, 1) 120W 2) 628W 3) 314W 4) 3.14W, 52. The work done in increasing the angular, frequency of a circular ring of mass 2kg and, radius 25cm from 10 rpm to 20rpm about is, axis, 1)0.2058J 2)0.2040J 3)0.2085J 4)0.2004J, , LEVEL-I (H.W) - KEY, 01) 4, 07) 2, 13) 3, 19) 2, 25) 2, 31) 3, 37) 2, 43) 4, 49) 3, 55) 2, , 49. A wheel at rest has M.I., , ROLLING MOTION, 53. A ring is allowed to roll down on an incline of, 1 in 10 without slipping. The acceleration of, its center of mass is, 1) 9.8ms −2 2) 4.9ms −2 3) 0.98ms −2 4) 0.49ms −2, 54. A cylinder is released from rest from the top, of an incline of inclination θ and length ‘l’. If, the cylinder roles without slipping, its speed, at the bottom, 1), , 4 gl sin θ, 3, , 2), , 3gl sin θ, 2, , 4 gl, 4 g sin θ, 4), 3sin θ, 3l, 55. For a body rolling along a level surface, without, slipping the translational and rotational kinetic, energies are in the ratio 2:1.The body is, 1) Hollow sphere, 2) solid cylinder, 3) Ring, 4) Solid sphere, 56. A solid sphere and a spherical shell roll down, an incline from rest from same height. The, ratio of times taken by them is, 3), , 21, 25, 21, 25, 2), 3), 4), 25, 21, 25, 21, 57. When a solid sphere is rolling along level, surface the percentage of its total kinetic, energy that is translational is, 1) 29%2) 71% 3) 60%, 4) 40%, 58. A thin ring of mass 1kg and radius 1m is rolling, at a speed of 1ms −1 . Its kinetic energy is, 1) 2J 2) 1J 3) 0.5J 4) zero, , 1), , NARAYANAGROUP, , 1, 2., , 02) 2 03) 4 04) 1 05) 2, 08) 1 09) 1 10) 3 11) 1, 14) 3 15) 4 16) 3 17) 3, 20) 1 21) 4 22) 2 23) 2, 26) 3 27) 2 28) 3 29) 2, 32) 1 33) 1 34) 4 35) 2, 38) 4 39) 2 40) 3 41) 2, 44) 4 45) 1 46) 1 47) 1, 50) 3 51) 2 52) 1 53) 4, 56) 1 57) 2 58) 2, LEVEL-I (H.W) - HINTS, , 06) 3, 12) 2, 18) 3, 24) 2, 30) 4, 36) 4, 42) 4, 48) 3, 54) 1, , m1 x1 + m2 x2, m1 + m2, m x +m x, m y +m y, xcm = 1 1 2 2 ; ycm = 1 1 2 2, m1 + m2, m1 + m2, , . xcm =, , L, θ, Sin , 4, 2, r, r, r, m1 r 1 + m2 r 2, 4. r cm =, m1 + m2, m×v, 5. vb =, m+M, , 3., , d=, , 6. vcm =, , m1v1 + m2 v2, m1 + m2, , r 2d, 7. shift = − 2 2, R −r, 8. m1r1 = m2 r2, L, θ, 2, 2, (or) rcm = cos, rcm = xcm, + ycm, 4, 2, r, r, r, m1 v1 + m2 v 2, 10. v cm =, m1 + m2, r r r, r, 11. r × F = τ ; τ ⊥ x − axis, , 9., , r 3d, 12. shift = 3 3, R −r, , r r, 13. Area of parallelogram = P × Q, ur ur, A× B, 14. sin θ =, AB, v, 15. ω =, r, 2π, 16. ω =, 60, dω, 17. α =, dt, 49
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 6., , LEVEL- II (C.W), CENTRE OF MASS, 1., , A uniform wire is bent into the form of a, rectangle of length L and width W. The, coordinates of its centre of mass from a corner, are, 1) (0, 0), , 2., , 3., , 4., , 5., , L, , 2) , W , 2, , , W, L W , 3) L, , 4) , , 2, 2 2 , A uniform disc of radius R is put over another, uniform disc of radius 2R of same thickness, and density. The peripheries of the two discs, touch each other. The position of their centre, of mass is, 1) at R/3 from the centre of the bigger disc towards, the centre of the smaller disc, 2) at R/5 from the centre of the bigger disc towards, the centre of the smaller disc, 3) at 2R/5from the centre of the bigger disc towards, the centre of the smaller disc, 4) at 2R/5from the centre of the smaller disc, Three particles each 1kg mass are placed at, the corners of a right angled triangle AOB,, O being the origin of the co–ordinate system, OA and OB along +ve x-direction and +ve y –, direction. The position vector of the centre of, mass is (OA = OB = 1m) (in meters), i+ j, i− j, 2( i + j ), 1), 2), 3), 4) (i–j), 3, 3, 3, If three particles of masses 2kg, 1kg and, 3kg are placed at corners of an equilateral, triangle of perimeter 6m then the distance of, centre of mass which is at origin of particles, from 1kg mass is (approximately) ( Assume, 2kg on x-axis, 1, m, 1) 6 m 2) 2m 3), 4) 2m, 2, , Six identical particles each of mass ‘m’ are, arranged at the corners of a regular hexagon, of side length “L”. If the mass of one of the, particle is doubled, the shift in the centre of, mass is, 1) L, , 2) 6L / 7, , NARAYANAGROUP, , 3) L / 7, , L, 4), 3, , 7., , A bomb of mass ‘m’ at rest at the coordinate, origin explodes into three equal pieces. At, a certain instant one piece is on the x–axis, at x=40cm and another is at x=20cm,, y = –60cm. The position of the third piece is, 1) x = 60cm, y=60cm 2) x = –60cm, y= –60cm, 3) x = –60cm, y=60cm 4) x = 60cm, y= –60cm, Particles of masses m,2m, 3m ........... nm gram, are placed on the same line at distances, l,, 2l, 3l, ...... nl cm from a fixed point. The, distance of centre of mass of the particles, from the fixed point in cm in, 1), , (2n + 1)l, 3, , 2), , n(n 2 + l )l, 3), 2, 8., , l, n +1, , 2l, 4) n(n 2 + l )l, , Three particles each of mass 2kg are at the, , corners of an equilateral triangle of side 3 m., If one of the particles is removed, the shift in, the centre of mass is, 1) 0.2m 2) 0.5m, 3) 0.4m, 4) 0.3m, 9. The mass of a uniform ladder of length 5m is, 20 kg. A person of mass 60kg stand on the, ladder at a height of 2m from the bottom. The, position of centre of mass of the ladder and, man from the bottom is, 1)1.256m 2) 2.532m 3) 3.513m 4)2.125m, 10. A uniform thin rod of length 1m and mass 3kg, is attached to a uniform thin circular disc of, radius 30cm and mass 1kg at its centre, perpendicular to its plane. The centre of mass, of the combination from the centre of disc is, 1) 0.375m 2) 0.25m 3) 0.125m 4)0.475m, 11. Four identical particles each of mass “m” are, arranged at the corners of a square of side, length “L”. If one of the masses is doubled,, the shift in the centre of mass of the system., w.r.t. diagonally opposite mass, 1), , L, 2, , 2), , 3 2L, 5, , 3), , L, 4 2, , 4), , L, 5 2, , 51
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 12. A circular hole of radius ‘r’ is made in a, disk of radius ‘R’ and of uniform thickness, at a distance ‘a’ from the centre of the disk., The distance of the new centre of mass from, the original centre of mass is, r, , R, a, , aR 2, ar 2, 1) 2, 2) 2, R − r2, R − r2, a( R 2 − r 2 ), a( R 2 − r 2 ), 3), 4), r2, R2, 13. The centre of mass of the letter F which is cut, from a uniform metal sheet from point A is, 6, A, 2, 4, 2, , 8, 2, , 2, 1) 15/7, 33/7, 2) 15/7, 23/7, 3) 22/7, 33/7, 4) 33/7, 22/7, 14. Two identical thin uniform rods of length L, each are joined to form T shape as shown, in the figure. The distance of centre of mass, from D is, A, , C, , 16. Two particles of equal masses have velocities, r, r, v1 = 4iˆand v 2 = 4jˆ . First particle has an, r, acceleration a1 =(5$i + 5 $j ) ms–2 while the, acceleration of the other particle is zero., The centre of mass of the two particles, moves in a path of, 1) Straight line, 2) Parabola, 3) Circle, 4) Ellipse, 17. Two particles of masses “p” and “q” (p>q), are separated by a distance “d”. The shift, in the centre of mass when the two particles, are interchanged is, 1) d(p+q) / (p–q), 2) d(p–q) / (p+q), 3) d p/(p–q), 4) d q/ (p–q), , VECTOR PRODUCT OR CROSS PRODUCT, 18. The unit vector perpendicular to, ur, ur, A = 2iˆ + 3 ˆj + kˆ and B = iˆ − ˆj + kˆ is, , 2, 2, , MOTION OF CENTRE OF MASS, AND LINEAR MOMENTUM, , B, , 1), , 4 iˆ − ˆj − 5 kˆ, 42, , 2), , 4 iˆ − ˆj + 5 kˆ, 42, , 3), , 4 iˆ + ˆj + 5kˆ, 42, , 4), , 4iˆ + ˆj − 5kˆ, 42, , 19. An electron is moving with speed 2 × 105 m / s, along the positive x-direction in the presence, ur, of magnetic induction B = $i + 4 $j − 3k$ T . The, , (, , ), , magnitude of the force experienced by the, electron in N ( e = 1.6 × 10 − 1 9 C ), D, , 1) 0, 2) L/4, 3) 3L/4 4)L, 15. Figure shows a square plate of uniform, thickness and side length 2 m. One fourth, of the plate is removed as indicated. The, distance of centre of mass of the remaining, portion from the centre of the original square, plate is, O, , 52, , 2) 1/2 m, , 3)1/6 m, , r, , ur, , 1) 18 × 1013, 2) 28 × 10 − 13, 3) 1.6 ×10−13, 4) 73 × 10 −13, 20. A particle of mass 80 units is moving with a, uniform speed v = 4 2 units in XY plane,, along a line y = x + 5 . The magnitude of the, angular momentum of the particle about the, origin is, 1) 1600units, 2) 160 2 units, 3) 152 2 units, , 1) 1/3m, , ur, , ( F = q ( v × B )), , 4) 16 2 units, , 4)1/8m, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL VARIABLES,, RELATION BETWEEN LINEAR AND, ANGULAR VARIABLES, 21. The linear and angular velocities of a body in, rotatory motion are 3 ms–1 and 6 rad/s, respectively. If the linear acceleration is 6, m/s2 then its angular acceleration in rads–2 is, 1) 6, 2) 10, 3) 12, 4) 2, , ROTATIONAL KINEMATICS, TORQUE, AND, MECHANICAL EQUILIBRIUM, 22. A stationary wheel starts rotating about its, own axis at an angular acceleration, , 5.5rad / s 2 . To acquire an angular velocity, 420 revolutions per minute, the number of, rotations made by the wheel is, 1) 14, 2) 21, 3) 28, 4) 35, 23. A circular disc is rotating about its own axis, at constant angular acceleration. If its, angular velocity increases from 210 rpm to, 420 rpm during 21 rotations then the angular, acceleration of disc is, 1) 5.5rad / s 2, , 2) 11rad / s2, , 3) 16.5rad / s 2, 4) 22rad / s2, 24. A circular disc is rotating about its own axis, at uniform angular velocity w . The disc is, subjected to uniform angular retardation by, which its angular velocity is decreased to, w / 2 during 120 rotations. The number of, rotations further made by it before coming to, rest is, 1) 120, 2) 60, 3) 40, 4) 20S, 25. Average torque on a projectile of mass m ,, initial speed u and angle of projection θ, between initial and final positions P and Q ,, about the point of projection is :, , mu 2 sin 2θ, 1), 2, , 2) mu 2 cosθ, , mu 2 cos θ, 3) mu 2 sin θ, 4), 2, 26. A metal rod of uniform thickness and of, length 1 m is suspended at its 25 cm division, with help of a string. The rod remains, horizontally straight when a block of mass 2, NARAYANAGROUP, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , kg is suspended to the rod at its 10 cm division., The mass of rod is, 1) 0.4 kg 2) 0.8 kg, 3) 1.2 kg 4) 1.6 kg, 27. A metallic rod of mass 20 kg and of uniform, thickness rests against a wall while the lower, end of rod is in contact with rough floor. The, rod makes an angle 60° with floor. If the, weight of rod produces a torque 150 N m about, its lower end,the length of rod (g = 10 ms–2), 1) 1.5 m, 2) 2 m, 3) 3 m 4) 4 m, 28. A roller of mass 300 kg and of radius 50 cm, lying on horizontal floor is resting against a, step of height 20 cm. The minimum horizontal, force to be applied on the roller passing, through its centre to turn the roller on to the, step is, 1) 980N, 2)1960N, 3)2940N 4) 3920N, , ROTATIONAL INERTIA OF SOLID, BODIES, 29. A thin rod of mass M and length L is bent into a, circular ring. The expression for moment of, inertia of ring about an axis passing through its, diameter is, , ML2, ML2, ML2, ML2, 2), 3), 4), 2p2, 4p2, 8p2, p2, 30. Two identical circular plates each of mass 0.1, kg and radius 10 cm are joined side by side as, shown in the figure. Their moment of inertia, about an axis passing through their common, tangent is, 1), , 1) 1.25x10-3 kgm 2, , 2) 2.5x10-3 kgm 2, , 3) 1.25x10-2 kgm 2 4) 2.5x10-2 kgm 2, 31. A wheel starting from rest is uniformly, accelerated with α = 4 rad / s2 for 10 seconds., It is then allowed to rotate uniformly for the, next two seconds and is finally brought to rest, in the next 10 seconds. Find the total angle, rotated by the wheel., 1) 200 rad 2) 400 rad 3) 300 rad 4) 480rad, 53
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 32. Two spheres each of mass M and radius R/2, are connected with a massless rod of length, 2R as shown in the figure. The moment of, inertia of the system about an axis passing, through the centre of one of the spheres and, perpendicular to the rod is, Y, , Y, M, , P, , 1, , Q, , M, , 2R, R, 2, , R, 2, , 1, , Y, , Y, , 37. I is moment of inertia of a thin circular plate, about its natural axis. The moment of inertia, of a circular ring whose mass is half of mass, of plate but radius is twice the radius of plate, about an axis passing through any tangent of, ring in its plane is, 2) 4 I, 3) 6 I, 4) 1.5 I, 1) 3 I, 38. The moment of inertia of a uniform rod of, length 2l and mass m about an axis xy passing, through its centre and inclined at an enable, α is, , 1, , 21, 2, 5, 5, MR 2 2) MR 2 3) MR 2 4) MR 2, 5, 5, 2, 21, 33. Moment of inertia of a thin circular plate of, mass M, radius R about an axis passing, through its diameter is I . The moment of, inertia of a circular ring of mass M, radius R, about an axis perpendicular to its plane and, passing through its centre is, I, I, 1) 2I, 2), 3) 4I, 4), 2, 4, 34. The mass of a thin circular plate is M and its, radius is R. About an axis in the plane of, plate at a perpendicular distance R/2 from, centre of plate, its moment of inertia is, , 1), , ml 2, sin 2 α, 1), 3, , 39., , 40., , MR 2, MR 2, 3MR 2, 3MR 2, 2), 3), 4), 4, 2, 4, 2, 35. In a rectangle ABCD (BC = 2 AB). The, moment of inertia is maximum along axis, through, , 1), , E, , A, , 41., , D, , F, , H, , B, , C, , G, , 1) BC, 2) AB, 3) HF, 4) EG, 36. M is mass and R is radius of a circular ring., The moment of inertia of same ring about an, axis in the plane of ring at a perpendicular, 2R, distance, from centre of ring is, 3, 1), , 54, , 2MR 2, 3, , 2), , 4 MR 2, 9, , 3), , 3 MR 2, 8, , 4), , 17 MR 2, 18, , 42., , ml 2, sin 2 α, 2), 12, , ml 2, ml 2, 2, cos α, cos 2 α, 3), 4), 6, 2, The ratio of radii of two solid spheres of same, material is 1 : 2. The ratio of moments of, inertia of smaller and larger spheres about, axes passing through their centres is, 1) 1 : 4, 2) 1 : 8, 3) 1 : 16, 4) 1: 32, is, moment, of, inertia, of, a, thin, circular, ring, I, about an axis perpendicular to the plane of, ring and passing through its centre. The same, ring is folded into 2 turns coil. The moment, of inertia of circular coil about an axis, perpendicular to the plane of coil and passing, through its centre is, I, I, 1) 2I, 2) 4I, 3), 4), 2, 4, A metallic thin wire has uniform thickness., From this wire, two circular loops of radii r,, 2r are made. If moment of inertia of 2 nd loop, about its natural axis is n times moment of, inertia of 1st loop about its natural axis. The, value of n is, 1) 2, 2)4, 3) 2 2, 4) 8, The moment of inertia of a solid cylinder about, an axis parallel to its length and passing, through its centre is equal to its moment of, inertia about an axis perpendicular to the, length of cylinder and passing through its, centre. The ratio of radius of cylinder and its, length is, 3) 1: 3, 4) 1 : 3, 1) 1: 2 2) 1 : 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 43. The moment of inertia of a solid cylinder, about its natural axis is I. If its moment of, inertia about an axis ⊥r to natural axis of, cylinder and passing through one end of, cylinder is 19I/6 then the ratio of radius of, cylinder and its length is, 1) 1 : 2, 2) 1 : 3, 3) 1 : 4, 4) 2 : 3, 44. Two identical circular plates each of mass M, and radius R are attached to each other with, their planes ⊥r to each other .The moment of, inertia of system about an axis passing through, their centres and the point of contact is, 3, MR 2, 5MR 2, 2, 1), 2), 3) MR, 4) MR2, 4, 4, 4, 45. The radius of gyration of rod of length ‘L’ and, mass ‘M’ about an axis perpendicular to its, length and passing through a point at a, distance L/3 from one of its ends is, L, L2, 7, 5, L 2), L, 3), 4), 3, 9, 6, 2, 46. Two point size bodies of masses 2 kg, 3 kg are, fixed at two ends of a light rod of length 1 m., The moment of inertia of two bodies about an, axis perpendicular to the length of rod and, passing through centre of mass of two bodies is, 1)0.6 kgm2 2)0.8 kgm2 3)1 kgm2 4)1.2 kgm2, 47. Three rings each of mass M and radius R, are arranged as shown in the figure. The, moment of inertia of the system about AB is, A, , 1), , B, 3, 7, 2, 2, 1) 3MR 2 2) MR, 3) 5MR 2 4) MR, 2, 2, 48. Three identicalHitachi thin rods each of mass, m and length L are joined together to form, an equilateral triangular frame. The moment, of inertia of frame about an axis perpendicular, to the plane of frame and passing through a, corner is, 2mL2, 1), 3, , 3mL2, 2), 2, , NARAYANAGROUP, , 4mL2, 3), 3, , 3mL2, 4), 4, , ANGULAR MOMENTUM &, CONSERVATION OF ANGULAR, MOMENTUM, 49. A thin uniform circular disc of mass M and, radius R is rotating in a horizontal plane about, an axis perpendicular to the plane at an, angular velocity ω . Another disc of mass M/, 3 but same radius is placed gently on the, first disc coaxially. The angular velocity of, the system now is, 4ω, 3ω, 3ω, 2) ω, 3), 4), 1), 3, 4, 8, 50. A turn table is rotating in horizontal plane, about its own axis at an angular velocity, 90rpm while a person is on the turn table at, its edge. If he gently walks to the centre of, table by which moment of inertia of system, decreases by 25%, then the time period of, rotation of turn table is, 1) 0.5sec, 2) 1sec, 3) 1.5sec 4) 2sec, 51. A uniform cylindrical rod of mass m and length, L is rotating with an angular velocity ω . The, axis of rotation is perpendicular to its axis of, symmetry and passes through one of its edge, faces. If the room temperature increases by, ‘t’ and the coefficient of linear expansion is, α , the change in its angular velocity is, 3, αωt, 1) 2αωt, 2) αω t 3) αωt 4), 2, 2, , ROTATIONAL DYNAMICS, 52. A constant torque of 1000Nm turns a wheel, of M.I. 200kg m 2 about an axis through, centre. The angular velocity after 3s is, 1) 15 rad s −1, 2) 22 rad s −1, 3) 28 rad s −1, 4) 60 rad s −1, 53. If 484J of energy is spent in increasing the, speed of a wheel from 60rpm to 360rpm, the, M.I. of the wheel is, 1) 1.6 kg m 2, 2) 0.3kg m2, 3) 0.7 kg m 2, 4) 1.2 kg m2, 54. The angular frequency of a fan of moment of, inertia 0.1kgm2 is increased from 30rpm to, 60rpm when a torque of 0.03Nm acts on it., The number of revolutions made by the fan, while the angular frequency is increased from, 30rpm to 60rpm, 1) 7.855rev, 2) 6.855rev, 3) 5.855rev, 4) 8.855rev, 55
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 55. A wheel rotating at an angular speed of, 20 rad s −1 is brought to rest by a constant, torque in 4s. If the M.I. is 0.2 kg m 2 the work, done in first 2s is, 1) 50J, 2) 30J, 3) 20J, 4) 10J, , ROLLING MOTION, 56. A sphere of mass m and radius r rolls on a, horizontal plane without slipping with a speed, u. Now it rolls up vertically, then maximum, height it would be attain will be, , 3u 2, 5u 2, 7u 2, u2, 1), 2), 3), 4), 4g, 2g, 10 g, 2g, 57. A circular ring starts rolling down on an, inclined plane from its top. Let v be velocity, of its centre of mass on reaching the bottom, of inclined plane. If a block starts sliding down, on an identical inclined plane but smooth, from, its top, then the velocity of block on reaching, the bottom of inclined plane is, 1), , v, 2, , 2) 2v, , 3), , v, 2, , 4), , 2v, , 58. A thin rod of length L is vertically straight on, horizontal floor. This rod falls freely to one, side without slipping of its bottom. The linear, velocity of centre of rod when its top end, touches floor is, , 3gL, 3gL, 3) 3gL 4), 2, 4, 59. A wheel of radius ‘r’ rolls without slipping with, a speed v on a horizontal road. When it is at a, point A on the road, a small lump of mud, separates from the wheel at its highest point, B and drops at point C on the ground. The, distance AC is, 2V, B, 1), , 2gL, , 2), , A, 1) v, , r, g, , 2) 2v, , C, r, g, , 3) 4v, , r, g, , 4) v, , 3r, g, , angle θ with the horizontal. The frictional, force between the cylinder and the incline is, mg sin θ, 3, 2mg sin θ, 4), 3, , 1) mg sin θ, , 2), , 3) mg cos θ, , 61. A thin metal disc of radius 0.25m and mass, 2kg starts from rest and rolls down an inclined, plane. If its rotational kinetic energy is 4J at, the foot of the inclined plane, then its linear, velocity at the same point is, 1) 1.2ms −1 2) 2.8ms −1 3) 20ms −1 4) 2ms −1, 62. A small sphere of radius R rolls without, slipping inside a large hemispherical bowl of, radius R . The sphere starts from rest at the, top point of the hemisphere. What fraction of, the total energy is rotational when the small, sphere is at the bottom of the hemisphere, 7, 2, 5, 7, 2), 3), 4), 1), 5, 7, 7, 10, 63. A metal disc of radius R and mass M freely, rolls down from the top of an inclined plane, of height h without slipping. The speed of its, centre of mass on reaching the bottom of the, inclined plane is, , 4 gh, 3gh, gh, 2), 3) gh 4), 3, 4, 2, 64. A thin rod of length L is vertically straight on, horizontal floor. This rod falls freely to one, side without slipping at its bottom. The linear, velocity of the top end of the rod with which it, strikes the floor is, 1), , 1), , 2gL, , 2), , 3gL, 2, , 3) 3gL, , 4), , 3gL, 4, , LEVEL-II - (C.W) - KEY, 01) 4, 07) 1, 13) 1, 19) 3, 25) 1, 31) 4, 37) 3, 43) 1, 49) 3, 55) 2, 61) 2, , 02) 2, 08) 2, 14) 3, 20) 1, 26) 3, 32) 1, 38) 1, 44) 3, 50) 1, 56) 3, 62) 2, , 03) 1, 09) 4, 15) 3, 21) 3, 27) 3, 33) 3, 39) 4, 45) 3, 51) 1, 57) 4, 63) 1, , 04) 2, 10) 1, 16) 1, 22) 3, 28) 4, 34) 2, 40) 4, 46) 4, 52) 1, 58) 4, 64) 3, , 05) 3, 11) 4, 17) 2, 23) 1, 29) 3, 35) 2, 41) 4, 47) 4, 53) 3, 59) 3, , 06) 3, 12) 2, 18) 1, 24) 3, 30) 2, 36) 4, 42) 3, 48) 2, 54) 1, 60) 2, , 60. A solid cylinder of mass m rolls without, slipping down an inclined plane making an, 56, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , MR 2, ; I ' = MR 2, 4, 2, MR 2, R, 2, I = I c + Md =, +M , 4, 2, perpendicular distance is maximum when the axis, of rotation passes through AB, hence M.I about, AB is maximum., 2R, 2, and d =, I = I c + Md 2 ; I C = MR, 3, 2, 3, 3, M, 2, MR I = M R 2 = 2 R = 6 I, [ ], I=, ; 1, 1 1, 2, 2 2 , 2, Take small element and use integration, 2, I = MR 2 , M α R 3 ; I α R 5, 5, 1, Mass is same. l = 2π rn ⇒ rα where n is, n, number of turns, , 33. I =, 34., 35., , 36., 37., 38., 39., 40., , 2, , MR 2 ML2 MR 2, =, +, 2, 12, 4, 2, 19 I ML2 MR 2, MR, =, +, 43. I =, ; I1 =, 2, 6, 3, 4, 2, 2, MR, MR, 3, +, = MR 2, 44. I1 =, 2, 4, 4, 45. According to parallel axes theorem, 42., , 2, , ML2, L, 4 ML2 ML2, L, I=, +M =, ∴K =, =, 12, 3, 36, 9, 6, m1m2 2, 46. I = m + m L, 1, 2 , , MR 2, 3, MR 2, 7 MR 2, 2, 2, I, =, +, 2, MR, 47., 2, = 2 + 3MR = 2, 2, mL2, 48. I = 2 I1 + I 2 ; I1 =, ; I 2 = I c + md 2, 3, mL2, 3L, Ic =, and d =, 12, 2, MR 2, M, , , I2 = M + R2, 49. I1ω1 = I 2ω2 ; I1 =, 2, 3 , , 50. I1ω1 = I 2ω2 ; I1 = 100, I 2 = 100 − 25 = 75, , I∝, , 1 2 k2 , Iω, mgh, =, mv 1 + 2 , ; W =τθ 56., 2, t, R , 57. for ring, v1 = 2 gh2 = v for block v 2 = 2gh, k, 1+ 2, R, L, 3g, 58. ω =, ; v = rω and r =, 2, L, 55. τ =, , 59. R = 2v × T, , = 2v, , 2h, g, , 60. f, , k2 , = mg sin θ 2, 2 , k +R , , 1 2 1 MR2 2 mv 2, 61. KErot = 2 Iω = 2 2 ω =, , , 4, , 2, , I1 r1 n2 , I ∝ r 2 and I = r = n , 2, 2 1, 41. I ∝ MR 2 and M ∝ L ∝ R ∴ I ∝ R3 and, 3, I 2 R2 , = , I1 R1 , , 58, , 1, ∆ω, ∆I, ⇒, =−, = 2α∆t, ω, ω, I, 1, 2, 2, 53. W = I 2π ( n2 − n1 ), 52. τ = Iα ; ω = α t, 2, 54. τ = I α , ω 2 − ω02 = 2αθ , θ = 2π N, , 51., , 62., , 1 2, Iω, 2, 1 2 k2 , mv 1 + 2 , 2, R , , 63. v =, , 2 gh, k2, 1+ 2, r, , 1, 2, , 64. mgh = Iω 2, , LEVEL- II (H.W), CENTRE OF MASS, 1., , 2., , 3., , Four particles, each of mass 1kg, are placed, at the comers of square of side one meter in, the XY plane. If the point of intersection of, the diagonals of the square is taken as the, origin,the co-ordinates of the center of mass, are, 1) (1,1)2) (-1, 1), 3) (1,-1), 4) (0,0), Three identical particles each of mass 0.1kg, are arranged at three corners of a square of, side 2m . The distance of the center of mass, from the fourth corner is, 1) 2/3m, 2) 4/3m, 3) 1m 4) 8/3m, A bomb of mass ‘m’ at rest at the coordinate, origin explodes into three equal pieces. At a, certain instant one piece is on the x-axis at, x=60cm and another is at x=40cm, y=60cm., The position of the third piece is, 1) x = −100cm , y = −60cm, 2) x = −60cm , y = −60cm, 3) x = −60cm , y = 60cm, 4) x = 60cm , y = −60cm, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 4., , 5., , 6., , Masses 1kg, 1.5kg , 2kg and M kg are, situated at (2,1,1), (1,2,1), (2,-2,1) and, (-1,4,3).If their center of mass is situated at, (1,1,3/2),the value of M is, 1) 1kg, 2) 2kg, 3) 1.5kg 4) 3kg, Six identical particles each of mass ‘m’ are, arranged at the corners of a regular hexagon, of side length ‘L’. If the masses of any two, adjacent particles are doubled. The shift in the, centre of mass is, L, 3L, 3L, 3L, 2), 3), 4), 1), 8, 16, 4, 8, Three particles each of mass ‘m’ are arranged, at the corners of an equilateral triangle of side, ‘L’. If one of masses is doubled. The shift in, the centre of mass of the system, , 8., , L, , L, , 3L, , L, , 2) 4 3, 3), 4) 2 3, 3, 4, Four identical particles each of mass ‘m’ are, arranged at the corners of a square of side length, ‘l’. If the masses of the particles at the end of a, side are doubled, the shift in the centre of mass, of the system., l, l, l, l, 1), 2), 3), 4), 6 2, 6, 2, 5 2, The co-ordinates of centre of mass of letter E, which is cut from a uniform metal sheet are, (take origin at bottom left corner and width of, letter 2cm every where), 1), , 7., , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 6 cm, 8 cm, , 2 cm, 2cm, 6 cm, , 1) (2cm, 4cm), 3) (3cm, 5cm), , 2) (2.4cm, 5cm), 4) (3.3cm, 5cm), , MOTION OF CENTRE OF MASS,, LINEAR MOMENTUM OF, CENTRE OF MASS, 9., , Two particles of equal mass have velocities, ur, ur, V 1 = 8$i and V 2 = 8 $j . First particle has an, r, −2, acceleration a1 = 5$i + 5 $j ms while the, , (, , ), , acceleration of the other particle is zero.The, centre of mass of the two particles moves is a, path of, 1) straight line, 2) parabola, 3) circle, 4) ellipse, NARAYANAGROUP, , VECTOR PRODUCT OR CROSS PRODUCT, 10. The magnitude of two vectors which can be, represented in the form i + j + ( 2 x ) k is, 18 .Then the unit vector that is, perpendicular to these two vectors is, −i + j, i− j, −i + j, −i + j, 1), 2), 3), 4), 2, 8 2, 2 2, 8, −1, 11. A proton of velocity 3$i + 2 $j ms enters a, , (, , (, , ), , ), , field of magnetic induction 2$i + 3k$ T . The, accel eration produced in the proton in, (specific charge of proton = 0.96 × 108 Ckg −1 ), r ur, F = q v × B , , , 8, 1) 0.96 ×10 6$i + 9 $j + 4k$, , (, , ), , (, ), 2) 0.96 ×10 ( 6$i − 9 $j − 4k$ ), 3) 0.96 ×10 ( $i − $j − k$ ), 4) 0.96 ×10 ( 5$i − 9 $j − 4k$ ), 8, , 8, , 8, , ROTATIONALVARIABLES, RELATION, BETWEEN LINEAR & ANGULAR, VARIABLES, 12. A vehicle starts from rest and moves at uniform, acceleration such that its velocity increases, by 3ms −1 per every second. If diameter of, wheel of that vehicle is 60cm, the angular, acceleration ofw heelis(in rad s-1), 1)5, 2)10, 3)15, 4)20, 13. Starting from rest the fly wheel of a motor, attains an angular velocity of 60 rad/sec in 5, seconds. . The angular acceleration obtained is, 1) 6 rad / s 2, 2) 12 rad / s 2, 3) 300 rad / s 2, 4) 150 rad / s 2, , ROTATIONAL KINEMATICS, TORQUE,, MECHANICAL EQUILIBRIUM, 14. A ceiling fan is rotating about its own axis, with uniform angular velocity ω . The electric, current is switched off then due to constant, opposing torque its angular velocity is, 2ω, reduced to, as it completes 30 rotations., 3, The number of rotations further it makes, before coming to rest is, 1) 18, 2) 12, 3) 9, 4) 24, 59
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 15. A wheel has a speed of 1200 revolutions per, minute and is made to slow down at a rate of, 4 rad/s2. The number of revolutions it makes, before coming to rest is, 1) 143, 2) 272, 3) 314 4) 722, 16. A particle of mass 1kg is projected with an, initial velocity 10ms −1 at an angle of, projection 450 with the horizontal. The, average torque acting on the projectile, between the time at which it is projected and, the time at which it strikes the ground about, the point of projection in newton meter is, 1) 25, 2) 50, 3) 75, 4) 100, 17. A uniform metre scale of mass 1kg is placed, on table such that a part of the scale is beyond, the edge. If a body of mass 0.25kg is hung at, the end of the scale then the minimum length, of scale that should lie on the table so that it, does not tilt is, 1) 30cm, 2) 80cm 3) 70cm 4) 60cm, 18. A heavy wheel of radius 20cm and weight, 10kg is to be dragged over a step of height, 10cm, by a horizontal force F applied at the, centre of the wheel. The minimum value of F, is, 1)20kgwt, 2)1kgwt, 3) 10 3 kgwt, , 4) 10 2 kgwt, , ROTATIONAL INERTIA OF SOLID BODIES, 19. Two discs one of density 7.2 g/cm3 and the, other of density 8.9 g/cm3 are of same mass, and thickness. Their moments of inertia are, in the ratio, 1), , 8.9, 7.2, , 2), , 7.2, 8.9, , 3) ( 8.9 × 7.2 ) :1, 4) 1: ( 8.9 × 7.2), 20. The mass of a circular ring is M and its radius, is R. Its moment of inertia about an axis in, the plane of ring at a perpendicular distance, R/2 from centre of ring is, , MR 2, MR 2, 3MR 2, 3MR 2, 2), 3), 4), 4, 2, 2, 4, 21. Two circular rings each of mass M and radius, R are attached to each other at their rims, and their planes perpendicular to each other, as shown in the figure. The moment of inertia, of the system about a diameter of one of the, 1), , 60, , rings and passing through the point of contact, is, , G, 3, 3, 5, 5, MR 2 2) MR 2 3) MR 2 4) MR 2, 2, 4, 2, 4, 22. The moment of inertia of a thin square plate, , 1), , of mass 1.2 kg is 0.2 kgm 2 when it is made to, rotate about an axis perpendicular to plane, of plate and passing through a corner of plate., The side length of plate is, 1) 0.2m, 2) 0.4m 3) 0.5m 4) 0.8m, 23. Three point masses m1,m2,m3 are placed at, three corners of an equilateral triangle of side, a. The moment of inertia of the system about, an axis coinciding with the altitude of triangle, passing through m 1 is, , m1, , m2 a, 2, , a, 2, , 1) ( m1 + m2 + m3 ) a, , 2, , m3, , 2), , ( m2 + m3 ) a 2, , 6, m2 + m3 ) a, ( m2 + m3 ) a 2, (, 3), 4), 4, 2, 24. From a uniform wire two circular loops are, made (i) P of radius r and (ii) Q of radius nr. If, the moment of inertia of Q about an axis, passing through its centre and perpendicular, to its plane is 8 times that of P about a similar, axis. The value of n is (diameter of the wire, is very much smaller than r or nr), 1) 8, 2) 6, 3) 4, 4) 2, 25. The moment of inertia of a uniform thin rod of, length L and mass M about an axis passing, through a point at a distance of L/3 from one of, its ends and perpendicular to the rod is, 2, , 7 ML2, 1), 48, , 2, , ML2, 2), 6, , ML2, 3), 9, , ML2, 4), 3, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 26. Two small spheres of mass 5kg and 15kg are, joined by a rod of length 0.5m and of, negligible mass. The M.I. of the system about, an axis passing through centre of rod and, normal to it is, 1) 10 kgm 2 2) 1.25 kgm 2 3) 20 kgm 2 4) 8 kgm 2, 27. Ratio of densities of materials of two, circular discs of same mass and thickness is, 5:6.The ratio of their M.I. about natural axes, is, 1) 5:6, 2) 6:5, 3) 25:36 4) 1:1, 28. M.I. of a uniform horizontal solid cylinder of, mass M about an axis passing through its edge, and perpendicular to the axis of cylinder when, its length is 6 times of its radius R is, 39, 39, 49, 49, MR 2 2) MR 2 3) MR 2 4), MR 2, 4, 2, 4, 2, A circular disc of radius R and thickness R/6, has moment of inertia I about an axis passing, through its centre and perpendicular to its, plane. It is melted and recast into a solid, sphere. The M.I. of the sphere about its, diameter as axis of rotation is, 1) I, 2) 2I/3, 3) I/5 4) I/10, The moment of inertia of ring about an axis, passing through its diameter is I. Then, moment of inertia of that ring about an axis, passing through its centre and perpendicular, to its plane is, 1) 2I, 2) I, 3) I/2, 4) I/4, A thin rod of mass 6m and length 6L is bent, into regular hexagon. The M.I. of the, hexagon about a normal axis to its plane and, through centre of system is, 2) 3mL2, 3) 5mL2 4) 11mL2, 1) mL2, If I1 is moment of inertia of a thin rod about, an axis perpendicular to its length and passing, through its centre and I2 is its moment of, inertia when it is bent into a shape of a ring, then (Axis passing through its centre and, perpendicular to its plane), , 1), 29., , 30., , 31., , 32., , 1) I 2 =, , I1, 4π 2, , I2 π 2, =, 3), I1, 3, , NARAYANAGROUP, , 2) I 2 =, , I1, π2, , I2, 3, 4) I = π 2, 1, , 33. The moment of inertia of thin rod of linear density, λ and length l about an axis passing through, one end and perpendicular to its length is, , λl 2, λl 2, λl 3, λl 3, 2), 3), 4), 12, 3, 12, 3, 34. Moment of inertia of a bar magnet of mass M ,, length L and breadth B is I. Then moment of, inertia of another bar magnet with all these values, doubled would be, 1) 8I, 2) 4I, 3) 2I, 4) I, 1), , ANGULAR MOMENTUM & CONSERVATION OF, ANGULARMOMENTUM, 35. A circular disc is rotating without friction, about its natural axis with an angular velocity, ω . Another circular disc of same material and, thickness but half the radius is gently placed, over it coaxially. The angular velocity of, composite disc will be, 4ω, 8ω, 7ω, 16ω, 2), 3), 4), 3, 9, 8, 17, 36. A ballet dancer is rotating about his own, vertical axis on smooth horizontal floor with, a time period 0.5sec. The dancer folds himself, close to his axis of rotation due to which his, radius of gyration decreases by 20%, then, his time period is, 1) 0.1sec 2) 0.25sec 3) 0.32sec 4) 0.4sec, 37. A particle of mass 1kg is moving along the line, y = x + 2 with speed 2m/sec. The magnitude, of angular momentum of the particle about the, origin is, 2) 2 2 kg − m 2 / sec, 1) 4 kg − m 2 / sec, , 1), , 3) 4 2 kg − m 2 / sec 4) 2 kg − m 2 / sec, 38. A uniform metal rod of length L and mass M, is rotating about an axis passing through one, of the ends perpendicular to the rod with, angular speed ω . If the temperature, increases by toC then the change in its, angular velocity is proportional to which of, the following?(Coefficient of linear expansion, of rod= α )., 1), , ω, , 2) ω, , 3) ω 2, , 4), , 1, ω, , 61
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , ROTATIONAL DYNAMICS, 39. A fly wheel of M.I. 6 × 10−2 kgm2 is rotating, with an angular velocity of 20 rad s −1 . The, torque required to bring it to rest in 4s is, 1)1.6Nm 2) 0.6Nm 3) 0.8Nm 4)0.3Nm, 40. When 200J of work is done on a fly wheel its, frequency of rotation increases from 4Hz to, 9Hz. The M.I. of the wheel about the axis of, rotation is (nearly), 1) 0.12kg m 2, 2) 0.2 kg m 2, 3) 0.22 kg m 2, 4) 0.3kg m2, 41. The moment of inertia of a wheel of radius, 20cm is 40 kgm 2 if a tangential force of 80N, applied on the wheel, its rotational K.E. after, 4s is, 1) 16.2J 2) 51.2J, 3) 25.6J 4) 24.8J, , ROLLING MOTION, 42. An initial momentum is imparted to a, homogeneous cylinder, as a result of which it, begins to roll without slipping up an inclined, plane at a speed of v0 = 4m / sec The plane, , 10, 20, 30, 40, 1) 7, 2), 3), 4), 7, 7, 7, 45. A thin metal rod of length 0.5m is vertically, straight on horizontal floor. This rod is falling, freely to a side without slipping. The angular, velocity of rod when its top end touches the, floor is (nearly), , 1) 7rad s −1, , 3) 3.5rad s −1, 4) 2.1rad s −1, 46. What should be the minimum coefficient of, static friction between the plane and the, cylinder, for the cylinder not to slip on an, inclined plane, 1, 1, 2, 2, tan θ 2) sin θ 3) tan θ 4) sin θ, 3, 3, 3, 3, 47. A thin metal disc of radius 25cm and mass, 2kg starts from rest and rolls down on an, inclined plane. If its rotational kinetic energy, is 8J at the foot of this inclined plane, then, linear velocity of centre of mass of disc is, 1) 2 m/s 2) 4m/s 3) 6m/s 4) 8m/s, , 1), , makes an angle θ = 300 with the horizontal., What height h will the cylinder rise to?, , LEVEL-II(H.W) - KEY, 01) 4, 07) 1, 13) 2, 19) 1, 25) 3, 31) 3, 37) 2, 43) 3, , ( g = 10m / s ), 2, , 1) 0.8m, 2) 1.2m, 3) 1.0m, 4) 1.6m, 43. A solid cylinder starts rolling down on an, inclined plane from its top and V is velocity of, its centre of mass on reaching the bottom of, inclined plane. If a block starts sliding down, on an identical inclined plane but smooth, from, its top, then the velocity of block on reaching, the bottom of inclined plane is, , v, 3, 2, v, 2) 2v, 3), v, 4), 2, 2, 3, 44. A wheel of radius 0.2m rolls without slip ping, with a speed 10m/s on a horizontal road. When, it is at a point A on the road, a small lump of, mud separates from the wheel at its highest, point B and drops at point C on the ground., The distance AC is, 1), , B, , 62, , C, , 02) 2, 08) 2, 14) 4, 20) 4, 26) 2, 32) 4, 38) 2, 44) 4, , 03) 1, 09) 1, 15) 3, 21) 3, 27) 2, 33) 4, 39) 4, 45) 1, , 04) 3, 10) 1, 16) 2, 22) 3, 28) 3, 34) 1, 40) 1, 46) 1, , 5) 2, 11) 2, 17) 4, 23) 4, 29) 3, 35) 4, 41) 2, 47) 2, , 06) 2, 12) 2, 18) 3, 24) 4, 30) 1, 36) 3, 42) 2, , LEVEL-II(H.W) - HINTS, m1 x1 + m2 x2, m y + m2 y2, ycm = 1 1, ;, m1 + m2, m1 + m2, , 1., , xcm =, , 2., , 2, 2, rcm = xcm, + ycm, , 3., , ( xcm , ycm ) = ( 0, 0), , ; 0=, , m1 x1 + m2 x2 + m3 x3, m1 + m2 + m3, , m1 y1 + m2 y2 + m3 y3, m1 + m2 + m3, m x + m2 x2 + m3 x3 + m4 x4, xcm = 1 1, m1 + m2 + m3 + m4, md, md, shift =, 6. shift =, M +m, M +m, , 0=, , 4., 5., , A, , 2) 4.2rad s −1, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 45. Mg, , l 1 2, = Iω, 2 2, , , , k 2 / R2 , µ = tan θ 2, , k + 1 , R2, , , 46., , 1 2 1 mR 2 2, KE, Iω = , =, ω, rot, 47., 2, 2 2 , , 5., , 6., , LEVEL- III, MOTION OF CENTRE OF MASS&, LINEAR MOMENTUM, 1., , Seven homogeneous bricks each of length, L,mass M are arranged as shown. Projection, x=, , L, 10, , then x co-ordinate of C.M is, , 7., , -x-x-x-x-x-x0, , 22, 32, 42, 12, L 2), L 3), L, L, 4), 35, 35, 35, 35, 2. The centre of mass of a non uniform rod of, length L whose mass per unit length, , 1), , Kx 2, λ=, ,Where k is a constant and x is the, L, distance from one end is :, 3L, L, K, 3K, 2), 3), 4), 1), 4, 8, L, L, A rope of length 30 cm is on a horizontal table, with maximum length hanging from edge A of, the table. The coefficient of friction between, the rope and table is 0.5. The distance of centre, of mass of the rope from A is, , 3., , 5 15, 5 17, 5 19, 7 17, cm 2), cm 3), cm 4), cm, 3, 3, 3, 3, As shown in figure from a uniform, rectangular sheet a triangular sheet is, removed from one edge. The shift of centre, of mass is, 20 cm, , 60 cm, O, , 30 cm, , 1) 4.2 cm 2) – 4.2cm 3) 6.67 cm 4)– 6.67 cm, , 64, , a, a, a, 1) L/5, 2) L/ 4 3) L/3, 4) L, Two masses ‘m1 ’ and ‘m2’ (m1>m 2 ) are, connected to the ends of a light inextensible, string which passes over the surface of a, smooth fixed pulley. If the system is, released from rest, the acceleration of the, centre of mass of the system will be (g =, acceleration due to gravity), g (m1 - m 2 ), 1), (m1 + m 2 ), , g (m1 - m 2 ) 2, 2), (m1 + m2 ) 2, , g (m1 + m 2 ), g (m1 + m 2 ), 4), (m1 − m2 ), (m1 − m2 ), Two bodies of masses m 1 and m 2 are moving, with velocity v1 and v2 respectively in the, same direction. The total momentum of the, system in the frame of reference attached to, the centre of mass is (v is relative velocity, between the masses), 3), , 8., , 1), , 1), 4., , A circular disc of radius R is removed from a, bigger circular disc of radius 2R such that the, circumference of the discs coincide . The, centre of mass of the new disc is α R from the, centre of the bigger disc. The value of α is, 1) 1/3, 2) 1/2, 3) 1/6, 4) 1/4, Four identical planks each of lengths ‘L’, are arranged one above the other over a, table as shown. Each projects a distance ‘a’, beyond the edge of the one that is below it., What is the maximum possible value of ‘a’, for the system to be in equilibrium without, tripping forward?, , m1 m 2 v, m1 − m 2, , 2 m1 m 2 v, m1 + m2, , 4 m1 m 2 v, m1 + m2, A shell in flight explodes into n equal, fragments k of the fragments reach the, ground earlier than the other fragments., The acceleration of their centre of mass, subsequently will be, 1) g, 2)(n–k)g, (n − k ), (n − k ) g, g, 4), 3), k, n, 3) zero, , 9., , 2), 4), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 10. A body of mass ‘m’ is dropped and another, body of mass M is projected vertically up with, speed ‘u’ simultaneously from the top of a, tower of height H . If the body reaches the, highest point before the dropped body reaches, the ground, then maximum height raised by, the centre of mass of the system from ground, is, 1) H +, , u2, 2g, , uB = 50 m / s , Initially they were 90 m apart., Find the maximum height attained by the, centre of mass of the particles. ( g = 10 m / s 2 ), B, uB, , u2, 2), 2g, 2, , uA, , 0, , 0, 60, 30, ////////////////////////////////////////////////////////////////////, , 3 −1 , 3 −1, g, 1) 4 2 g 2) ( 3 − 1)g 3) 4) 2 g, , , 2, , , 12. A rope thrown over a pulley has a ladder with, a man of mass m on one of its ends and a, counter balancing mass M on its other end., The man climbs with a velocity vr relative to, ladder . Ignoring the masses of the pulley and, the rope as well as the friction on the pulley, axis, the velocity of the centre of mass of this, system is :, m, m, M, 2M, vr 2), vr 3), vr 4), vr, 1), M, 2M, m, m, 13. Two particles of masses 2 kg and 3 kg are, projected horizontally in opposite directions, from the top of a tower of height 39.2 m with, velocities 5 m/s and 10 m/s respectively. The, horizontal range of the centre of mass of two, particles is, 1) 8 2 m in the direction of 2 kg, , 2) 8 2 m in the direction of 3 kg, 3) 8 m in the direction of 2 kg, 8 m in the direction of 3 kg, , A, , 2, , 1 Mu , 1 mu , 3) H +, 4) H +, , , , , 2g m + M , 2g m + M , 11. Two blocks of equal mass are tied with a light, string, which passes over a massless pulley, as shown in figure. The magnitude of, acceleration of centre of mass of both the, blocks is ( neglect friction everywhere ), , 4), , 14. Two particles A and B of mass 1 Kg and 2 Kg, respectively are projected in the directions, shown in figure with speeds u A = 200 m / s and, , 1) 115.55m, 3) 4.55 m, , 2) 145.55 m, 4) 34.55 m, , VECTOR PRODUCT OR CROSS, PRODUCT, 15. At a given instant of time the position vector, of a particle moving in a circle with a velocity, 3iˆ − 4 ˆj + 5kˆ is iˆ + 9 ˆj − 8kˆ . Its angular velocity, at that time is:, 13iˆ − 29 ˆj − 31kˆ, 13iˆ − 29 ˆj − 31kˆ, 1), 2), 146, 146, 13iˆ + 29 ˆj − 31kˆ, 13iˆ + 29 ˆj + 31kˆ, 3), 4), 146, 146, , (, , ), , (, , ), , (, , ), , (, , ), , ROTATIONAL VARIABLES,, RELATION BETWEEN LINEAR, & ANGULAR VARIABLES, 16. Two points P and Q, diametrically opposite on, a disc of radius R have linear velocities v and, 2v as shown in figure. Find the angular speed, of the disc., v, , P, , 2v, Q, , v, 2v, v, v, 2), 3), 4), R, R, 2R, 4R, 17. Point A of rod AB (l =2m) is moved upwards, against a wall with velocity v=2 m/s. Find, angular speed of the rod at an instant when, θ = 600 ., , 1), , v, , A, , θ, B, , 1) 4rad/s 2)1.155rad/s 3) 2rad/s 4) 2.50rad/s, NARAYANAGROUP, , 65
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 18. A uniform circular disc of radius R lies in the, XY plane with its centre coinciding with the, origin of the coordinate system. Its moment, of inertia about an axis, lying in the XY plane,, parallel to the X-axis and passing through a, point on the Y-axis at a distance y =2R is I1 ., Its moment of inertia about an axis lying in a, plane perpendicular to XY plane passing, through a point on the x-axis at a distance x, = d is I 2 . If I1=I2 the value ofd is, 19, 17, 15, 13, R 2), R 3), R 4), R, 2, 2, 2, 2, ROTATIONAL KINEMATICS, TORQUE, MECHANICAL EQUILIBRIUM, 19. A wheel rotating with uniform angular, acceleration covers 50 revolutions in the first, five seconds after the start. Find the angular, acceleration and the angular velocity at the, end of five seconds., 1) 4π rad / s 2 ,80π rad / s, , 1), , 2) 8π rad / s 2 ,40π rad / s, 3) 6π rad / s 2 , 40π rad / s, 4) 6π rad / s 2 ,80π rad / s, 20. A square is made by joining four rods each of, mass M and length L. Its moment of inertia, about an axis PQ, in its plane and passing, through one of its corner is, P, , 45, , 0, , L, , Q, , 4, 8, 10, ML2 3) ML2 4), ML2 +, 3, 3, 3, 21. A shaft is turning at 65rad /sec at time zero., Thereafter, angular acceleration is given by, α = ( −10 − 5t ) rad / s 2 where t is the elapsed, time. Find its angular speed at t =3sec., 1) 25 rad/sec, 2) 12.5rad/sec, 3) 17 rad/sec, 4) 22 rad /sec, , 1) 6ML2, , 66, , 2), , 22. A wheel having radius 10 cm is coupled by a, belt to another wheel of radius 30cm. 1st, wheel increases its angular speed from rest, at a uniform rate of 1.57 rad s–2. The time for, 2nd wheel to reach a rotational speed of 100, rev/min is...(assume that the belt does not slip), 1) 20 sec 2) 10 sec 3) 1.5 sec, 4) 15 sec, 23. An equilateral prism of mass m rests on a, rough horizontal surface with coefficient of, friction µ . A horizontal force F is applied on, the prism as shown. If the coefficient of, friction is sufficiently high so that the prism, does not slide before toppling, the minimum, force required to topple the prism is, , F, a, , a, , a, mg, µmg, mg, µmg, 1), 2), 3), 4), 3, 3, 4, 4, 24. The mass of a metallic beam of uniform, thickness and of length 6 m is 60 kg. The, beam is horizontally and symmetrically lies, on two vertical pillars which are separated, by a distance 3 m. A person of mass 75 kg is, walking on this beam. The closest distance, to which the person can approach one end of, the beam so that the beam does not tilt down, is (neglect thickness of pillars), 1) 30 cm 2) 20 cm 3) 15 cm, 4) 10cm, 25. Two persons P and Q of same height are, carrying a uniform beam of length 3 m. If Q, is at one end, the distance of P from the other, end so that P, Q receive loads in the ratio 5 :, 3 is, 1) 0.5 m 2) 0.6 m 3) 0.75 m, 4) 1 m, 26. A uniform meter scale of mass 1 kg is placed, on table such that a part of the scale is beyond, the edge. If a body of mass 0.25 kg is hung, at the end of the scale then the minimum, length of scale that should lie on the table so, that it does not tilt is, 1) 90 cm 2) 80 cm 3) 70 cm, 4) 60 cm, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 27. A metallic cube of side length 1.5 m and of, mass 3.2 metric ton is on horizontal rough, floor. The minimum horizontal force that, should be applied on the cube at a height 1.2, m from that floor to turn the cube about its, lower edge is, 1) 1.96 × 103 N, 2) 4.9 × 103 N, 3) 1.96 × 104 N, 4) 4.9 × 104 N, 28. A cubical block of mass m and side L rests on, a rough horizontal surface with coefficient of, friction µ . A horizontal force F is applied on, the block as shown. If the coefficient of, friction is sufficiently high so that the block, does not slide before toppling, the minimum, force required to topple the block is, [JEE 2000], F, , ///////////////////, 1) mg/4, 2) infinitesimal, 3) mg/2, 4) mg (1 – u), 29. The center of an equilateral triangle is O., Three forces F1 , F2 and F3 are applied along, AB, BC and AC respectively. The magnitude, of F3 so that the total torque about O should, be zero is, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , angle 15° with the horizontal is, 15, , 0, , ML2, 11ML2, 7 ML2, 10ML2, 2), 3), 4), 12, 24, 12, 24, 31. A thin rod of length L and mass M is bent at, the middle point O at an angle of 600 . The, moment, of inertia of the rod about an, axis passing through O and perpendicular to, the plane of the rod will be, 1), , O, L/2, , 60, , 0, , L/2, , ML2, ML2, ML2, ML2, 2), 3), 4), 6, 12, 24, 3, 32. Four identical solid spheres each of mass M, and radius R are fixed at four corners of a light, square frame of side length 4R such that, centres of spheres coincide with corners of, square. The moment of inertia of 4 spheres, about an axis perpendicular to the plane of, frame and passing through its centre is, 1), , 21MR 2, 42MR 2 84MR 2 168MR 2, 2), 3), 4), 1), 5, 5, 5, 5, 33. In the above problem moment of inertia of 4, spheres about an axis passing through any, side of square is, 1) ( F1 + F2 ), 3), , F1 + F2, 2, , 2) ( F1 − F2 ), 4) 2 ( F1 + F2 ), , ROTATIONAL INERTIA OF SOLID BODIES, 30. A square plate of mass M and edge L is shown, in figure. The moment of inertia of the plate, about the axis in the plane of plate and, passing through one of its vertex making an, NARAYANAGROUP, , 21MR 2, 42MR 2, 84MR 2, 168MR 2, 1), 2), 3), 4), 5, 5, 5, 5, 34. Thickness of a wooden circular plate is same, as the thickness of a metal circular plate but, density of metal plate is 8 times density of, wooden plate. If moment of inertia of wooden, plate is twice the moment of inertia of metal, plate about their natural axes, then the ratio of, radii of wooden plate to metal plate is, 1) 1 : 2, 2) 1 : 4, 3) 4 : 1 4) 2 :1, 67
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 35. A uniform circular disc of radius 'R' lies in, the X-Y plane with the centre coinciding with, the origin. The moment of inertia about an, axis passing through a point on the X-axis at, a distance x = 2R and perpendicular to the, X-Y plane is equal to its moment of inertia, about an axis passing through a point on the, Y-axis at a distance y = d and parallel to the, X-axis in the X-Y plane. The value of 'd' is, 1), , 4R, R, R, R, 2) 17 3) 15 4) 13 , 2, 2, 2, 3, , 36. Two rings of the same radius R and mass M, are placed such that their centres coincide, and their planes are perpendicular to each, other. The moment of inertia of the system, about an axis passing through the common, centre and perpendicular to the plane of one, of the rings is, , MR 2, 3MR 2, 2) MR 2 3), 4) 2MR 2, 2, 2, 37. In the above problem, the moment of inertia, of the system about an axis passing through, the diameters of both rings is, , Their moments of inertia about the axis, passing through the centre and perpendicular, to their planes are in the ratio 1:m. The, relation between m and n is, 1) m=n 2) m = n2, 3) m = n3, 4) m = n4, 40. The moment of inertia of a hollow sphere of, mass M having internal and external radii R, and 2R about an axis passing through its, centre and perpendicular to its plane is, 3, 2, , 2, 1) MR 2), , 62, 13, 31, MR 2 3), MR 2 4), MR 2, 35, 32, 35, , 41. Find moment of inertia of half disc of radius, , R 2 and mass M about its centre. A smaller, half disc of radius R1 is cut from this disc., , 1), , MR 2, MR 2, 3MR 2, 2), 3), 4) MR 2, 4, 2, 2, 38. Four thin metal rods, each of mass M and, length L, are welded to form a square ABCD, as shown in figure. The moment of inertia of, the composite structure about a line which, bisects rods AB and CD is, 1), , ML2, ML2, ML2, 2ML2, 2), 3), 4), 6, 3, 2, 3, 39. Two circular loops A and B are made of the, same wire and their radii are in the ratio 1:n., 1), , 68, , (, , 1), , M, 2, 2, R1 + R 2, 4, , 3), , M 2, R 1 + R 22 ), (, 16, , ), , 2), , M 2, R 1 + R 22 ), (, 8, , 4), , M 2, R 1 + R 22 ), (, 32, , ANGULAR MOMENTUM &, CONSERVATION OF ANGULAR, MOMENTUM, 42. A uniform smooth rod (mass m and length l), placed on a smooth horizontal floor is hit by a, particle (mass m) moving on the floor, at a, distance l / 4 from one end elastically, (e =1).The distance travelled by the centre, of the rod after the collision when it has, completed three revolutions will be, 1) 2π l, 2) cannot be determined, 3) πl, 4) none of these, 43. A bullet of mass m is fired upward in a, direction of angle of projection 60° with an, initial velocity u. The angular momentum of, this bullet when it is crossing highest point, with respect to point of projection is, , 2mu 3, 3mu 3, 1), 2), 5g, 8g, , 2mu 3, 3mu 3, 3), 4), 9g, 16 g, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 44. A particle of mass 5g is moving with a speed, of 3 2cms −1 in X-Y plane along the line, y = x + 4 . The magnitude of its angular, , P, (3/2) R, , momentum about the origin in gcm 2 s −1 is, , 30, 2, 45. A ballot dancer is rotating about his own, vertical axis on smooth horizontal floor with, a time period 0.5 sec. The dancer folds, himself close to his axis of rotation due to, which his radius of gyration decreases by, 20%, then his new time period is, 1) 0.1 sec 2)0.25 sec 3) 0.32 sec 4) 0.4 sec, 46. A smooth uniform rod of length L and mass, M has two identical beads of negligible size,, each of mass m, which can slide freely along, the rod. Initially the two beads are at the, centre of the rod and the system is rotating, 1) zero, , 2) 60, , 3) 30, , 4), , with angular velocity ω0 about its axis, perpendicular to the rod and passing through, its mid point (see figure). There are no, external forces. When the beads reach the, ends of the rod, the angular velocity of the, system is, [JEE - 1988], , Bead, , Bead, , L, 2, , L, 2, , ω0, , C, , v0, , 1) increase continuously as the disc moves away, 2) decrease continuously as the disc moves away, 3) is always equal to 2MRv0, 4) is always equal to MRv0, 48. A disc of mass m and radius R moves in the, X-Y plane as shown in figure. The angular, momentum of the disc about the origin O at, the instant shown is, y, v = ωR, ω, , 3R, x, , O, , 4R, , 5, 7, 2, 2, 1) − mR ω k, 2) mR ω k, 2, 3, 9, 5, 2, 2, 3) − mR ω k, 4) mR ω k, 2, 2, 49. A uniform sphere of mass m, radius r and, moment of inertia I about its centre of mass, axis moves along the x-axis is shown in figure., Its centre of mass moves with velocity v0,, and it rotates about its centre of mass with, , angular velocity ω0 . Let L = ( Iω0 + mv0r)( −k ) ., The angular momentum of the body about the, the origin O is, y, , ω0, 1), , M ω0, M + 3m, , M ω0, M + 6m, , + 6m ) ω0, 4) ω0, M, 47. A uniform circular disc of mass M and radius, R rolls without slipping on a horizontal, surface. If the velocity of its centre is v 0, then, the total angular momentum of the disc about, a fixed point P at a height 3R/2 above the, centre C., 3), , (M, , 2), , NARAYANAGROUP, , v0, r, O, 1) L, only if v0 = ω0 r, 2) greater than L, v0 > ω0 r, 3) less than L, if v0 > ω0 r, 4) L, for all value of v0 and ω0, , 69
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 50. In the figure shown, a ring A is initially rolling, without sliding with a velocity v on the, horizontal surface of the body of the body B, (of same mass as A). All surfaces are smooth., B has no initial velocity. What will be the, maximum height reached by A on B?, , A v, , smooth, , g, 3g, g, 2g, 2), 3), 4), 2, 2, 3, 3, 54. The arrangement shown in figure consists of, two identical uniform solid cylinders each of, mass 5kg on which two light threads are, wound symmetrically. Find the tensions of, each thread in the process of motion. The, friction in the axle of the upper cylinder is, assumed to be absent., , 1), , B, 3v2, 1), 4g, , v2, 2), 4g, , v2, 3), 2g, , v2, 4), 3g, , ROTATIONAL DYNAMICS, 51. Calculate the linear acceleration of the blocks, in the given figure . Mass of block A = 10kg,, mass of block B = 8kg, mass of disc shaped, pulley = 2kg (take g = 10m / s 2 ), , B, A, , 20, 19, 29, 20, m / s 2 2) m / s 2 3) m / s 2 4), m / s2, 19, 20, 20, 29, 52. A block of mass m is attached at the end of, an inextensible string which is wound over a, rough pulley of mass M and radius R as, shown in figure. Assume that string does not, slide over the pulley. Find the acceleration of, the block when released., , 1), , 1) 4.9N, 2) 9.8N 3) 8.8N 4) 5.8N, 55. The top in the figure has a moment of inertia, equal to 4.0 × 10−4 kgm 2 and is initially at rest., It is free to rotate about the stationary axis, AA1. A string wrapped around a peg along the, axis of the top is pulled in such a manner as, to maintain a constant tension of 5.57N . If, the string does not slip while it is unwound, from the peg. what is the angular speed of, the top after 80.0cm of string has been pulled, off the peg., A', F, , A, , M, , R, , m, , 2mg, 2mg, mg, mg, 2), 3), 4), m+M, 2m + M, 2m + M, m+M, 53. A uniform rod of length L and mass M is, pivoted freely at one end (at bottom level ), and placed in vertical position. What is the, tangential linear acceleration of the free end, when the rod is horizontal?, , 1), , 70, , 1)130rad/s 2)142rad/s 3)149rad/s 4)120rad/s, 56. A solid cylinder of mass m=4kg and radius, R=10cm has two ropes wrapped around it, one, near each end. The cylinder is held, horizontally by fixing the two free ends of the, cords to the hooks on the ceiling such that, both the cords are exactly vertical. The, cylinder is released to fall under gravity. Find, the tension along the strings., 1) 6.53N 2) 5.23N, 3) 3.23N 4) 4.43N, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , ROLLING MOTION, 57. Assume the earth’s orbit around the sun as, circular and the distance between their, centres as ‘D’ Mass of the earth is ‘M’ and, its radius is ‘R’ If earth has an angular velocity, ‘ ω0 ’ with respect to its centre and ‘ ω ’ with, respect to the centre of the sun, the total, kinetic energy of the earth is:, 1), , 2, 2, MR2ω02 ω 5 Dω , 1 + , + , , 5 ω 0 2 Rω0 , , , , 2), , 2, MR 2ω02 5 Dω , 1 + , , 5 2 Rω0 , , , , 61. A tangential force F acts at the top of a thin, spherical shell of mass m and radius R. Find, the acceleration of the shell if it rolls without, slipping., F, R, O, , 6F, 6m, 3m, 5m, 2), 3), 4), 5m, 5F, 5F, 6F, 62. A uniform circular ring of radius R is first, rotated about its horizontal axis with an, , 1), , angular velocity ω0 and then carefully placed, on a rough horizontal surface as shown. The, coefficient of friction between the surface and, the rings is µ . Time after which its angular, , 5 Dω 2 , 5, 2 2, 3) 2 MR ω0 1 + 2 Rω , 0 , , ω 2 5 Dω 2 , 2, 2 2, 4) 5 MR ω0 1 + ω + 2 Rω , 0 , 0, , speed is reduced to 0.5ω0 is, ω0, , 58.A solid cylinder of mass 10kg is rolling, perfectly on a plane of inclination 300 . The, force of friction between the cylinder and the, surface of the inclined plane is, 1) 49N 2) 24.5N 3) 49/3N 4) 12.25N, 59 The velocities are in ground frame and the, cylinder is performing pure rolling on the, plank, velocity of point ' A ' would be, A, , ω0 µ R, ω0 g, ω0 R, 2ω0 R, 2), 3), 4), 2g, µg, 2µ R, 2µ g, 63. A uniform circular disc of radius R rolls without, slipping with its center of mass moving along, positive x=axis with a speed v. The velocity, of point P at the instant shown in figure is, 1), , y, C, , VC, , r P, θ, , v, , VP, , 1) 2VC 2) 2VC − VP 3) 2VC + VP 4) 2 (VC − VP ), 60. A carpet of mass M made of inextensible, material is rolled along its length in the form, of a cylinder of radius R and kept along a, rough floor. The carpet starts unrolling, without sliding on the floor, when a negligibly, small push is given to it. The horizontal, velocity of the axis of a cylindrical part of the, carpet, when its radius is reduced to R/2 is, 1), , 14, gR 2), 3, , NARAYANAGROUP, , 7, gR 3), 3, , gR 4), , 2gR, , x, vr sin θ vr cos θ, , j, 1) v p = v +, i +, R , R, , vr sin θ, , 2) v p = v +, R, , , vr cos θ, j, i −, R, , , 3) v p = v +, , vr sin θ, vr cosθ, i+, j, R, R, , 4) v p = v +, , vr sin θ, vr cosθ, i−, j, R, R, , 71
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 64. A uniform solid sphere of radius r is rolling, on a smooth horizontal surface with velocity, v and angular velocity ω = ( v = ω r ) . The, sphere collides with a sharp edge on the wall, as shown in figure. The coefficient of friction, between the sphere and the edge µ = 1 / 5., Just after the collision the angular velocity, of the sphere becomes equal to zero. The, linear velocity of the sphere just after the, collision is equal to, ω, , 1., xcm =, , v, 3v, v, 3), 4), 5, 5, 6, 65. A particle of mass ‘m’ is rigidly attached at, ‘A’ to a ring of mass ‘3m’ and radius ‘r’. The, system is released from rest and rolls without, sliding. The angular acceleration of ring just, after release is, , 2. dm =, , 2), , r, O, , 4., , 2R, 5, , 2), , 5R, 2, , 3), , 7R, 5, , 4), , 9R, 5, , 72, , 02) 1, 08) 3, 14) 1, 20) 3, 26) 4, 32) 4, 38) 4, 44) 2, 50) 2, 56) 1, 62) 4, , 03) 2, 09) 4, 15) 2, 21) 2, 27) 3, 33) 4, 39) 3, 45) 3, 51) 1, 57) 2, 63) 2, , 04) 4, 10) 3, 16) 3, 22) 1, 28) 3, 34) 4, 40) 4, 46) 2, 52) 2, 58) 3, 64) 1, , 05) 1, 11) 1, 17) 3, 23) 1, 29) 1, 35) 2, 41) 1, 47) 4, 53) 2, 59) 2, 65) 2, , 06) 2, 12) 2, 18) 3, 24) 1, 30) 2, 36) 3, 42) 1, 48) 1, 54) 1, 60) 1, 66) 3, , kx 3, ∫0 L dx, , Fractional length hanging,, l, µ, l, 0.5, =, ⇒, =, ⇒ l = 10 cm, L 1+ µ, 30 1 + 0.5, let ‘ ρ ’ be the mass per unit length. The coordinates of 20ρ and 10ρ are (10,0) and (0,5), respectively from ‘A’., , shift =, , − mass of removed part × d, Mass of remaining part Here d=20 cm, , 7., , ( a cm ) y, , =, , Fext, ( m 1 + m 2 ) g − 2T → 1, =, (), M, m1 + m 2, , But T =, , LEVEL-III - KEY, 01) 1, 07) 2, 13) 2, 19) 2, 25) 2, 31) 2, 37) 4, 43) 4, 49) 4, 55) 3, 61) 4, , kx 2, ;x, dx cm, L, , L, , 6., , P, , 1), , L4 , , 4, 3L, = 0L, = L 2, = 3=, 4, L , kx, ∫0 dm ∫0 L dx 3 , , ∫ xdm, , r 2a, Shift of centre of mass x = 2, R − r2, Where r = radius of removed disc, R = radius of original disc, a = distance between the centres, Note:In this question shift must be ∝ R for exact, approach to the solution, CM of bricks, above each brick must not be, Σmi xi, beyond its edge. xcm = Σm ; xcm = L, i, L, L, L, x1 = a + , x2 = 2a + , x3 = 3a +, 2, 2, 2, L, (or) a =, n, , m, , g, g, g, g, 1), 2), 3), 4), 4r, 6r, 8r, 2r, 66. A solid sphere of mass M and radius R is, placed on a rough horizontal surface. It is, stuck by a horizontal cue stick at a height h, above the surface. The value of h so that the, sphere performs pure rolling motion, immediately after it has been stuck is, J, h, R, , ∑m, , 2, 2, Distance of C.M from A, r cm= xcm, + ycm, , 5., , A, , i i, , L, , O, , 1) v, , ∑m x, i, , 3., , edge, , V, , LEVEL - III - HINTS, , 8., , 2m1 m2 g, →( 2), m1 + m2, , 9. acm =, , Theoretical, , ∑m a, ∑m, , i i, i, , 10. hmax = H +, , ( ucm ), , 2, , 2g, 11. Acceleration of system,, mg sin 60 − mg sin 30, a=, 2m, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 3 −1, ma1 + ma 2, a = , g , Now a cm =, 2m, 4 , 3 −1 , here, a1 and a 2 are 4 g at right angles., , , 2a 3 − 1 , g, Hence, acm = 2 = , 4, 2, , , 12. The masses of load, ladder and man are M,, M-m and m respectively. Their velocities are, v(upward), -v and vr -v respectively, , ∴ vcm =, , ∑ mi vi, , t=, , 23. About right base corner ; τ F = τ mg, 24. About a pillar clockwise torque = anticlockwise, torque, 25. Let x is distance of P from other end, , ( xFP ) + ( LFQ ) = W, 26., , ∑m, , i, , M (v) + ( M − m)( − v) + m(v r − v), m, =, =, vr, 2M, 2M, 2h, 13. Range of C.M = vcm, g, m1 v1 + m2 v 2, But v cm = m + m, 1, 2, 14. Maximum height attained by C.M, 2, ucm, = Initial height of C.M +, 2g, r r, r r×v, 15. ω = 2, r, 3iˆ − 4 ˆj + 5kˆ = xiˆ + yjˆ + zkˆ × iˆ + 9 ˆj − 8kˆ, , (, , ) (, , ), , v, 2v, v, =, ; ⇒ x = 2R ; ω =, x x + 2R, 2R, v, ω=, L cos θ, mR 2, mR 2, 2, I1 =, + m ( 2R ) ; I 2 =, + md 2 ; I1 = I 2, 4, 2, 1, θ = 2π N ,θ = ω0t + α t 2, 2, M.I about an axis passing through the diagonal, 2 ML2, Ig =, M.I about the given axis, 3, , 16. ω =, 17., 18., 19., 20., , L , I = I g + 4M , , 2, , 2, , 21. ω = ∫ α dt, 22. At any instant linear acceleration of all points of, coupled belt is same. a = rα, r1α1 = r2α 2 ;find α 2 ; ω2 = ω1 + α 2t, NARAYANAGROUP, , ω2, α 2 Qω1 = 0, , 27., 28., , 29., , 30., , L, 2, , where L is the length of the rod and W is its weight, but W = FP + FQ, If the distance from one end is x then, x, ( 50 − x )1 = x ( 0.25 ) =, 4, 200 − 4x = x ⇒ x = 40cm, ∴ length on the table = 100 - 40 = 60 cm, L, τcw =τAcw ; F×1.2 = mg × ; F = 1.96 ×104 N, 2, (F) (perpendicular distance) = mg(perpendicular, mg, 1 , FL = mg ⇒ F =, distance), 2, 2, If the perpendicular distance of any side of the, triangle from ‘O’ is ‘x’then F1 x + F2 x − F3 x = 0, ⇒ F1 + F2 = F3, From diagram, we get x = AO sin 600, , L, 3, ML2, ML2, ×, I, =, I, =, I, =, ; x, ; z, y, 2 2, 6, 12, 2, ML, I AB =, + Mx 2, 12, 31. Moment of inertia of a uniform rod about one end, mL2, =, ∴ moment of inertia of the system, 3, 2, m ( L / 2), mL2, = 2×, =, 2, 3, 12, 32. I = 4I1 where I1 is M .I of each sphere, =, , I1 = I c + Md 2, 2, L, 2, ; L = 4R, and I c = MR ; d =, 5, 2, 33. I = 2( I c + I1 ) and I1 = I c + Md 2, 2, I c = MR 2 and d = 4R, 5, 34. I1 = 2I 2 ; M1R 12 = 2M 2 R 22, 2, , R1 , M, = 2 2 ;but m ∝ D × R 2, M1, R2 , 73
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 66. Let v be the velocity of the centre of mass of the, sphere and ω be the angular velocity of the body, about an axis passing through the centre of mass., 2, 2, J = Mv ; J ( h − R ) = 5 MR × ω, 2 2, from the above two equations v ( h − R ) = r ω, 5, from the condition of pure rolling, v = Rω, 2R, 7R, h− R =, ⇒h=, 5, 5, , its plane, 5, MR 2, 4, , c) about natural axis, , g), , d) about any tangent, , h) MR 2, , 1, 2, , ⊥ r to its plane, , 5., , Match the following, A disc rolls on ground without slipping., Velocity of centre of mass is v. There is a point, P on circumference of disc at angle θ . Suppose, v p is the speed of this point. Then, match the, following table:, , LEVEL- IV, Matching Type Questions, 1., , 2., , 3., , 4., , Match the following :, List– I, List – II, A. Position of centre of, e. is zero, mass, B. The algebraic sum, f. in non uniform, of moments of all the, gravitational field, masses about centre, of mass, g.is independent, C. Centre of mass and, of frame of, centre of gravity coincide, reference, D. Centre of mass and, h. in uniform, centre of gravity do, gravitational, not coincide, field, Match the following:, List - 1, List - 2, a) torque, e) mass, b) moment of inertia, f) linear momentum, c) angular momentum g) linear acceleration, d) angular acceleration h) force, If R is radius and K is radius of gyration then, in the case of following rolling bodies match, the ratio K 2 : R 2, List - 1, List - 2, a) solid sphere, e) 1 : 1, b) solid cylinder, f) 2 : 3, c) hollow sphere, g) 1 : 2, d) hollow cylinder, h) 2 : 5, Match the following moment of inertia of thin, circular plate about different axes of rotation, List - 1, List -2, 3, 2, , a) about any diameter e) MR 2, b) about any tangent in f), , 76, , 1, MR 2, 4, , Vcm, θ, P, , 6., , Column-I, , Column-II, , a) If θ = 600, , p) v p = 2v, , b) If θ = 900, , q) v p = v, , c) If θ = 1200, , r) v p = 2v, , d) If θ = 1800, Column-I, , s) v p = 3v, Column-II, 2, 2, p) MR, 5, , a) Moment of inertia, of annular disc of inner, radius R1 and outer, radius R2 about, symmetric axis, b) Moment of inertia, , q), , 3, MR 2, 10, , r), , M 2, ( R1 + R22 ), 4, , of elliptical disc of two, radii R1 and R2, about symmetric axis, c) Moment of inertia, , of solid sphere of radius, R about symmetric axis, d) Moment of inertia, , s), , M 2, ( R1 + R22 ), 2, , of solid cone of base, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 7., , radius R about symmetric axis, A solid spherical ball of mass M and radius, R rolls without slipping down a surface, inclined to horizontal at an angle θ ., Considering that ball is uniform solid sphere, and that ball and surface are perfectly rigid., Column-I, Column-II, a) Friction force, p) Zero, involved, b) Minimum value of q) ( 2 / 7 ) Mg sin θ, coefficient of friction, for pure rolling, c) Work done against r) Static friction, frictional force, , 8., , 9., , d) Force of kinetic, s) ( 2 / 7 ) tanθ, friction, A solid sphere, hollow sphere, solid cylinder,, hollow cylinder and ring each of mass M and, radius R are simultaneously released at rest, from top of incline and allowed to roll down, the incline., Column-I, Column-II, a) Time taken to reach p) Solid sphere, bottom is maximum for, b) Angular acceleration q) Hollow cylinder, maximum for, c) Kinetic energy at r) Hollow sphere, bottom is same for, d) Rotational kinetic s) Ring, energy is maximum for, A rigid body of mass M and radius R rolling, without slipping on an inclined plane. The, magnitude of force of friction, Column-I, Column-II, Mg sin θ, a) For ring, p), 2.5, Mg sin θ, b) For solid sphere, q), 3, Mg sin θ, c) For solid cylinder r), 3.5, Mg sin θ, d) For hollow sphere s), 2, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , Assertion & Reason, Read the assertion and reason carefully to mark, the correct option out of the options given below, (a) If both assertion and reason are true and the, reason is the correct explanation of the assertion., (b) If both assertion and reason or true but reason, is not the correct explanation of the assertion., NARAYANAGROUP, , 19., , (c) If assertion is true but reason is false., (d) If the assertion and reason both are false., Assertion : The centre of mass of a two particle, system lies on the line joining the two particles,, being closer to the heavier particle., Reason: Product of mass of one particle and its, distance from centre of mass is numerically equal, to product of mass of other particle and its distance, from centre of mass., Assertion: The centre of mass of system of n, particles is the weighted average of the position, vector of the n particles making up the system., Reason: The position of the centre of mass of a, system is independent of coordinate system., Assertion: The centre of mass of an isolated, system has a constant velocity., Reason: If centre of mass of an isolated system, is already at rest, it remains at rest., Assertion: The centre of mass of a body may lie, where there is no mass., Reason: Centre of mass of a body is a point,, where the whole mass of the body is supposed to, be concentrated., Assertion: A particle is moving on a straight line, with a uniform velocity, its angular momentum is, always zero., Reason: The momentum is zero when particle, moves with a uniform velocity., Assertion: The centre of mass of a proton and, an electron, released from their respective, positions remains at rest., Reason: The centre of mass remain at rest, if no, -external force is applied., Assertion: The position of centre of mass of a, body does not depend upon shape and size of the, body., Reason: Centre of mass of a body lies always at, the centre of the body, Assertion: A shell at rest, explodes. The centre, of mass of fragments moves along a straight path., Reason: In explosion the linear momentum of, the system is not conserved., Assertion: A judo fighter in order to throw his, opponent on to the mattress he initially bend his, opponent and then rotate him around his hip., Reason: As the mass of the opponent is brought, closer to the fighter’s hip, the force required to, throw the opponent is reduced., Assertion: The centre of mass of an electron and, proton, when released moves faster towards, proton., Reason: Proton is lighter than electron., 77
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 20. Assertion: At the centre of earth, a body has, centre of mass, but no centre of gravity., Reason: Acceleration due to gravity is zero at, the centre of earth., 21. Assertion: When a body dropped from a height, explodes in mid air, its centre of mass keeps moving, in vertically downward direction., Reason: Explosion occur under internal forces, only. External force is zero., 22. Assertion: It is harder to open and shut the door, if we apply force near the hinge., Reason: Torque is maximum for the door., 23. Assertion: Moment of inertia of a particle, changes, when the axis of rotation changes., Reason: Moment of inertia depends on mass, and distance of the particles., 24. Assertion: Inertia and moment of inertia are, different quantities., Reason: Inertia represents the capacity of a body, to oppose its state of motion or rest., 25. Assertion: If earth shrink (without change in mass), to half its present size, length of the day would, become 6 hours., Reason: As size of the earth changes its moment, of inertia changes., 26. Assertion: Torque due to force is maximum when, angle between r and F is 900., Reason: The unit of torque is newton-metre., 27. Assertion: Radius of gyration of body is a, variable quantity., Reason: The radius of gyration of a body about, an axis of rotation may be defined as the root, mean square distance of the particles from, the axis of rotation., 28. Assertion: A ladder is more apt to slip, when, you are high up on it than when you just begin to, climb., Reason: At the high up on a ladder, the torque is, large and it is small when one just begins to climb., 29. Assertion: Torque is equal to rate of change of, angular momentum., Reason:Angular momentum depends on moment, of inertia and angular velocity., 30. Assertion:The speed of whirlwind in a tornado, is alarmingly high., Reason: If no external torque acts on a body, its, angular velocity remains conserved., 31. Assertion: The velocity of a body at the bottom, of an inclined plane of given height, is more when, it slides down the plane, compared to, when it, rolling down the same plane., , 78, , Reason: In rolling down, a body acquired both,, kinetic energy of translation and rotation., 32. Assertion: In rolling, all points of a rigid body, have he same linear speed., Reason: The rotational motion does not affect, the linear velocity of rigid body., 33. Assertion: A wheel moving down a perfectly, frictionless inclined plane will undergo slipping (not, rolling motion)., Reason: For perfect rolling motion, work done, against the friction is zero., 34. Assertion: The total kinetic energy of a rolling, solid sphere is the sum of translational and, rotational kinetic energies., Reason: For all solid bodies total kinetic energy, is always twice the translational kinetic energy., , Statement type questions, , 35., , 36., , 37., , 38., , 1) Statement A& B are true, 2) Statement A is true, Statement B is false, 3) Statement A is false, Statement B is true, 4) Statement A & B are false, Consider the following two statements A and B, and identify the correct answer, Statement A : The centre of mass of a system of, particles depends on forces on the particles., Statement B : In the absence of external force,, the centre of mass of system moves with uniform, velocity, Consider the two statements A and B and identify, the correct answer, Statement A : A wooden sphere and a copper, sphere of same radius are kept in contact with, each other the centre of mass will be with in, the wooden sphere., Statement B: Three identical spheres each of, radius R are placed touching each other on, horizontal table. The centre of mass of the system, is located at the point of intersection of the, medians of the triangle formed by the centres of, spheres., Consider the following two statements A and B, and identify the correct choice, Statement A : The rotational kinetic energy of a, rolling body is always greater than its translatory, kinetic energy, Statement B: The maximum value of radius of, gyration of a rolling body can not be greater than, the radius of that body, Consider the following two statements A and B, and identify the correct choice, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Statement A : Spokes are used in a bicycle, wheel to increase the strength of wheel, Statement B: Mass of fly wheel is concentrated, on its rim to increase its moment of inertia, 39. Consider the following two statements A and B, and identify the correct choice, Statement A : Moment of inertia of circular plate, is minimum about its natural axis, Statement B : Inertia of rotation of a rotating, body is proportional to its angular momentum, 40. Consider the following two statements A and B, and identify the correct choice, Statement A : The torques produced by two, forces of couple are opposite to each other., Statement II : The direction of torque is always, perpendicular to plane of rotation of body, 41. Consider the following two statements A and B, and identify the correct choice, Statement A : The torque required to stop a, rotating body in a given time is directly proportional, to its initial angular momentum, Statement B: If radius of earth shrinks then its, rotational kinetic energy increases, 42. Statement A : Mechanical advantage of a lever, is always < 1, Statement B: Mechanical advantage of a lever, can be increased by increasing its effort arm or by, decreasing its load arm., 43. Consider the following two statements A and B, and identify the correct choice, Statement A : When a rigid body is rotating about, its own axis, at a given instant all particles of body, possess same angular velocity., Statement B: When a rigid body is rotating about, its own axis, the linear velocity of a particle is, directly proportional to its perpendicular distance, from axis, 44. Consider the following two statements A and B, and identify the correct choice, Statement A : The moment of inertia of a rigid, body is independent of its angular velocity, Statement B: The radius of gyration of a rotating, metallic disc is dependent on its temperature, 45. Choose correct statement., (A) The position of centre of mass of a system is, dependent on the choice of coordinate system, (B) Newton’s second law of motion is applicable, to the centre of mass of the system., (C) Internal forces cannot change the state of, centre of mass., (D) Internal forces can change the state, NARAYANAGROUP, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 46., , 47., , 48., , 49., , 50., , 51., , 52., , 53., , 54., , of centre of mass, [EAMCET-2012], 1) Both A and B are correct, 2) Both B and C are wrong, 3) Both A and C are wrong, 4) Both A and D are wrong, A shell is projected at some angle with horizontal., When the shell is at its highest point, it explodes, into two pieces., Statement A : : The law of conservation of linear, momentum can be used for the small interval of, explosion, Statement B : The net force on the shell at highest, point is zero, Statement A : Impulsive force on a particle may, change its kinetic energy and its momentum, Statement B : Momentum of a particle changes, only when kinetic energy of the particle changes, A uniform rod is held vertically on a smooth, horizontal surface. Now the rod is released, given, it simultaneously a gentle push, Statement A : Centre of mass of the rod moves, in vertical direction as the rod falls, Statement B : The rod is falling freely, A particle is thrown vertically upward from ground,, while another is thrown simultaneously vertically, downward from some height, Statement A : In the reference frame of centre, of mass of the system, the particles move uniformly, Statement B : Acceleration of the centre of mass, of the system is zero, Statement A : Momentum of a system w.r.t centre, of mass of the system is zero, Statement B : Centre of mass can acceleration, only under the action of external forces., Statement A : Internal force cannot change, kinetic energy of a system of particles, Statement B : Internal forces cannot change, momentum of a system of particles., Statement A : Linear momentum of a system of, particles with respect to centre of mass must be, zero, Statement B : Linear momentum of a system of, particles is the vector-sum of linear momenta of, all particles of the system., Statement A : Work done by a force on a body, whose centre of mass does not move may be nonzero, Statement B : Work done by a force depends, on the displacement of the centre of mass, Statement A : Net work done by internal force, in a system may be zero., Statement B : Net force on the centre of mass of, the system by internal mechanism is zero, 79
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 55. Statement A : In collision between two bodies,, they remain in contact with each other for a very, short interval of time before they separate. During, the period of restitution, the bodies try to regain, their original shape., Statement B : During the period of contact,, bodies exchange their momentum and energy, 56. Statement A : For a system of particles, total, energy of the system can change even if net force, acting on the system is zero, Statement B : If net force on a system of particles, is zero, total momentum can not change., 57. Statement A : In pure rolling motion, net work, done by friction is zero., Statement B : Sum of translational work done, by friction and rotational work done by friction is, zero, 58. Statement A : For a system of particles under, central force field, total angular momentum is, conserved about the centre, Statement B : Torque acting on such a system is, zero about the centre, 59. Statement A : A ball is rolling on a rough, horizontal surface. It gradually slows down and, stops., Statement B :: Force of rolling friction decreases, linear velocity, 60. Statement A : A ring is rolling without slipping on, rough surface as shown in figure. The force of, friction necessary for ring to purely roll is in forward, direction., Statement B : Force of friction is zero when, external force acts at top of ring., 61. Statement A : Velocity acquired by a rolling body, depends on inclination of plane on which it rolls, down without slipping, Statement B : Velocity depends upon height of, descent of body, 62. Statement A : A cylinder rolls up an incline plane,, reaches some height and then rolls down. The, direction of friction force acting on cylinder is up, the incline while ascending as well as descending., Statement B : Direction of force of friction is in, accordance with sense of angular acceleration., 63. Statement A : Angular momentum of a particle, executing uniform circular motion is constant., Statement B : Momentum of a particle executing, uniform circular motion is constant., 64. Two solid spheres (of masses m and 4m and radii, r and 16r) roll down without slipping on an incline., Statement A : Both reach the bottom of incline, with same kinetic energies., 80, , JEE-ADV PHYSICS-VOL - III, Statement B : Both spheres take same time to, reach bottom of the incline., 65. Statement A : A particle in uniform motion may, have non-zero angular momentum about a point, in space., Statement B : A particle may be moving on a, curved path with uniform speed., 66. Statement A : Rolling on a stationary surface, can be treated as pure rotation about the point of, contact, Statement B : Point of contact of the body is, instantaneous centre of rotation, as it is, instantaneously at rest during rolling., , Multi Option Questions, 67. Identify the correct one from the following, statements., A. the position of centre of mass in a co–ordinate, system does not change if a man moves from, the one end to other end on a floating wooden, log in still water., B. When a man moves from one end to other, end on a floating wooden log in still water, it, moves in opposite direction, C. Due to action and reaction the wooden log, floating in still water moves in opposite direction, as the man on it moves from one end to the other, end, 1) B & C are true, 2) A & D are true, 3) A, B & C are true 4) All are correct, 68. If external forces acting on a system have, zero resultant, the centre of mass, A. may move, B. may accelerate, C. must not move D.must not accelerate, 1) A & B are correct 2) B & C are correct, 3) C & D are correct 4) A & D are correct, 69. In which of the following cases, the centre of, mass of a rod is certainly not at its centre?, A. The density increases from left to right upto, the centre and then decrease, B. The density decreases from left to right upto, the centre and then increase, C. The density continuously increases from left, to right, D. The density continuously decreases from left, to right, 1) A & B are correct 2) B & C are correct, 3) C & D are correct 4) A & D are correct, 70. If a circular concentric hole is made in a disc, then about an axis passing through the centre, of the disc and perpendicular to its plane., 1) moment of inertia decreases, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 2) moment of inertia increases, 3) radius of gyration increases, 4) radius of gyration decreases, 71. A rotor of radius r is rotating about its own, vertical axis and a person in contact with inner, wall of rotor remains in equilibrium without, slipping down. If ω is angular velocity of rotor, and µ is minimum coefficient of friction, between person and the wall of rotor then, following is correct, A) µ ∝ ω2 B) µ ∝, , 1, r, , C) µ ∝, , 1, ω2, , D) µ ∝ r, , 1) A and B are true, 2) A and D are true, 3) B and C are true 4) C and D are true, 72. A particle of mass m is executing uniform, circular motion on a path of radius r. If v is, speed and p the magnitude of its linear, momentum, then the radial force acting on, the particle is, pm, vp, mv 2, p2, 2), 3), 4), 1), r, r, r, mr, r, r, 73. In circular motion if v is velocity vector, a is, r, acceleration vector, r is instantaneous, ur, position vector, and p is momentum vector, ur, and ω is angular velocity of particle. Then, r ur, r, 1) v, ω and r are mutually perpendicular, ur r, ur, 2) p, v and ω are mutually perpendicular, r r, r ur, 3) r × v = 0 and r × ω = 0, rr, r ur, 4) r.v = 0 and r.ω = 0, 74. The length of second hand of a watch is 1cm., Then, 1) The linear speed of tip of second hand is, π, cm / s, 30, 2) The linear speed of the tip of second hand is, uncertain, 3) The change in linear velocity vector in 15, π, 2cm / s 2, seconds is, 30, 4) The change in acceleration vector in 15minutes, , π2 2, cm / s 2, 1800, 75. Two particles, each of mass m are attached, to the two ends of a light string of length L, which passes through a hole at the centre of, , is, , NARAYANAGROUP, , a smooth table. One particle describes a, circular path on the table with angular, velocity ω1 , and the other describes a conical, pendulum with angular velocity ω2 below the, table. If l1 and l2 are the lengths of portions, of the string above and below the table, then, l1 ω2, 1) l = ω, 2, 1, , l1 ω22, 2) = 2, l2 ω1, , 1, 1, ml, 1, 1 l cos θ, 3) ω 2 + ω 2 = g, 4) ω 2 + ω 2 = g, 1, 2, 1, 2, 76. A symmetrical body of mass M, radius R and, radius of gyration k is rolling on a horizontal, surface without slipping. If linear velocity of, centre of mass is vc and angular velocity ω ;, then, , 1 2 k2 , 1) the total KE of body is 2 mvc 1 + R 2 , , , 2) the rotational KE is, , 1, MR 2ω 2, 2, , 3) the translational KE is, , 1, Mvc2, 2, , 4) Total energy = 0, 77. A ring type flywheel of mass 100kg and, diameter 2m is rotating at the rate of, 5, rev/sec. Then, 11, 1) the moment of inertia of the wheel is, 100kg − m 2, 2) the kinetic energy of rotation of flywheel is, 5 × 103 J, 3) the angular momentum associated with the, flywheel is 103 joule-sec, 4) the flywheel, if subjected to a retarding torque, 250N − m , will come to rest in 4sec., 78. In which of the following case(s), the angular, momentum is conserved?, 1) The planet Neptune moves in elliptical orbit, around the sun with sun at one focus, 2) A solid sphere rolling on an inclined plane, 3) An electron revolving around the nucleus in, elliptical orbit, 4) An α − particle approaching a heavy nucleus, from sufficient distance., 81
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JEE-ADV PHYSICS-VOL - III, , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , 79. A solid sphere is in pure rolling motion on an, inclined surface having inclination θ, [IIT-2006], 1) frictional force acting on sphere is µmg cos θ, 2) f is dissipative force, 3) friction will increase its angular velocity and, decrease its linear velocity, 4) If θ decreases, friction will decrease., r, 80. The torque τ on a body about a given point is, ur ur, ur, found to be equal to A × L , where A is a, ur, constant vector and L is angular momentum, of the body about that point. From this it, follows that, ur, dL, ur, is perpendicular to L at all instants of time, 1), dt, ur, ur, 2) the component of L in the direction of A does, not change with time, ur, 3) the magnitude of L does not change with time, ur, 4) L does not change with time., 81. A sphere is rolled on a rough horizontal surface., It gradually slows down and stops. The force, of friction tends to, 1) decrease linear velocity, 2) increase linear momentum, 3) decrease angular velocity, 4) increase angular velocity, , LEVEL - IV - KEY, Matching Type, , 1) A → g; B → e; C → h; D → f, 2) a → h; b → e; c → f ; d → g, 3) a - h;, b - g; c - f; d - e, 4) a-f; b-g; c-h; d-e, 5) a-q, b-p, c-s,, d-r, 6) a-s; b-r;, c-p; d-q, 7) a-q,r;, b-s; c-p; d-q, 8) a-q,s;, b-p; c-pqrs; d-q,s, 9) a-s; b-r;, c-q; d-p, , Assertion & Reason Type Questions, 10) 1, 16) 4, 22) 3, 28) 1, 34) 3, , 11)2, 17) 4, 23) 1, 29) 2, , 12)2, 18) 1, 24) 2, 30) 3, , 13) 1, 19) 4, 25) 1, 31) 1, , 14) 4, 20) 1, 26) 2, 32) 4, , 15) 1, 21) 1, 27) 1, 33) 2, , Statement Type Questions, 35) 3 36) 1 37) 3 38) 3 39)3 40) 3, 41) 1 42) 3 43) 1 44) 1 45) 4 46) 2, 47) 2 48) 2 49) 2 50)) 1 51) 3 52) 1, 82, , 53) 2 54) 3 55) 1 56) 1 57) 1 58) 1, 59) 1 60) 3 61) 3 62) 1 63) 2 64) 3, 65) 1 66) 2, , MULTI OPTION QUESTIONS, 67) 3 68) 4, 69) 3 70) 1,3, 71) 3 72) 1,3,4, 73) 1,4 74) 1,3,4, 75) 2,4 76) 1,3, 77) 1,2,3,4, 78) 1,3,4, 79) 3,4 80) 1,2,3, 81) 1,4, , LEVEL-IV-HINTS, Assertion and reasoning type, 10. If centre of mass of system lies at origin, r, then r cm = 0, y, , x, r1, , r2, , ur, ur, m1 r1 + m 2 r2, r cm =, m1 + m 2, ur, ur, ∴ m1 r1 + m 2 r2 = 0 or m1r1 = m 2 r2, , It is clear that if m1 > m 2 then r 2 > r1, 12. External force on the system, d r, Fext = M, v cm If system is isolated i.e., dt, r, Fext = 0 then vcm =constant.Initially if the velocity, of centre of mass is zero then it will remain zero., 13. As the concept of centre of mass is only theoretical,, therefore in practice no mass may, lie at the, centre of mass. For example, centre of mass of a, uniform circular ring is at the centre of the ring where, there is no mass., r, 14. When particle moves with constant velocity v then, its linear momentum has some inite value, ur, r, P = mv . Angular momentum (L) = Linear, momentum (P) x Perpendicular distance of line of, action of linear momentum form the point of, rotation(d). So if d ≠ 0 then L ≠ 0, but if d = 0, then L may be zero. So we can conclude that angular, momentum of a particle moving with constant, velocity is not always zero., 15. Initially the electron and proton were at rest so, then centre of mass will be at rest. When they move, towards each other under mutual attraction then, , ( ), , (, , ), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , 16., , 17., , 18., 19., , 20., , 21., 22., , 23., , 24., , 25., , SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS, , velocity of centre of mass remains unaffected, because external force on the system is zero., The position of centre of mass of a body depends, on shape, size and distribution of mass of the body., The centre of mass does not lie necessarily at the, centre of the body., As the shell is initially at rest and after explosion,, according to law of conservation of linear, momentum, particles move in all direction,such that, total momentum of all parts is equal to zero., Through bending weight of opponent is made to, pass through the hip of judo fighter, to make, its torque zero., The position of centre of mass of electron and, proton remains at rest. As their motion is due to, internal force of electrostatic attraction, which is, conservative force. No external force is acting on, the two particles, therefore centre of mass remain, at rest., At the centre of earth, g = 0. Therefore a body has, no weight at the centre of earth and have no centre, of gravity (centre of gravity of a body is the point, where the resultant force of attraction or the weight, of the body acts). But centre of mass of a body, depends on mass and position of particles and is, independent of weight., Explosion is due to internal forces. As no external, force is involved, the vertical down ward motion, of centre of mass is not affected., Torque = Force x perpendicular distance of line of, action of force from the axis of rotationHence for, a given applied force, torque or true tendency of, rotation will be high for large value of d. If distance, d is smaller, then greater force is required to cause, the same torque, hence it is harder to open or shut, down the door by applying a force near the hinge., The moment of inertia of a particle about an axis of, rotation is given by the product of the mass of the, particle and the square of the perpendicular distance, of the particle from the axis of rotation., For different axis, distance would be different,, therefore moment of inertia of a particle changes, with the change in axis of rotation., There is a difference between inertia and moment, of inertia of a body. The inertia of a body depends, only upon the mass of the body but the moment of, inertia of a body about an axis not only depends, upon the mass of the body but also upon the, distribution of mass about the axis of rotation., When earth shrinks it angular momentum remains, constant. i.e., 2, 2π, L = Iω = mR 2 ×, = cons tan t, 5, T, , NARAYANAGROUP, , ∴ T αIα R 2 . It means if size of the earth changes, then its moment of inertia changes. In the problem, radius becomes half so time period (Length of the, day) will becomes 1/4 of the present value i.e.24/, 4=6 hr., 26. τ = rFsin θ. If θ = 900 then τmax = rF Unit of, torque is N - m., 27. Radius of gyration of body is not a constant, quantity. Its value changes with the change in, location of the axis of rotation. Radius of gyration, of a body about a given axis is given as, , r12 + r22 + ..... + rn2, n, When a person is high up on the ladder, then a, large torque is produced due to his weight about, the point of contact between the ladder and the, floor. Whereas when he starts climbing up, The, torque is small. Due to this reason, the ladder is, more apt to slip, when one is high up on it., uur, r dL, r=, and L = Iω, dt, In a whirlwind in a tornado, the air from nearby, regions gets concentrated in a small space thereby, decreasing the value of its moment of inertia, considerably. Since, Iω = constant, so due to, decrease in moment of inertia of the air, its angular, speed increases to a high value. If no external torque, dL, = 0 or L = constant, acts, then τ = 0 ⇒, dt, ⇒ Iω =constant. As in the rotational motion, the, moment of inertia of the body can change due to, the change in position of the axis of rotation, the, angular speed may not remain conserved., In sliding down, the entire potential energy is, converted into kinetic energy. While in rolling down, some part of potential energy is converted into, K.E. of rotation. Therefore linear velocity acquired, is less., In rolling all points of rigid body have the same, angular speed but different linear speed., Rolling occurs only on account of friction which is, a tangential force capable of, providing torque. When the inclined plane is, perfectly smooth, body will simply slip under the, effect of its own weight., K N = K R + K T This equation is correct for any, body which is rolling without slipping. For the ring, and hollow cylinder only K R = K T i.e., K=, , 28., , 29., 30., , 31., , 32., 33., , 34., , K N = 2K T, 83
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , LEVEL-V, , m, , SINGLE ANSWER QUESTIONS, 1., , A, θ, , A rigid massless beam is balanced by a, particle of mass 4m in left hand side and a, pulley particle system in right hand side. The, , (a), , x, , value of y is:, , 2, tan θ, 3, , (b), , 2, tan θ, 9, , 1, , x, , 5., , Y, m, , 4m, , (c) 6 tan θ, (d) none of these, A disc of mass m is connected with an ideal, spring of stiffness k . If it is released from, rest., it rolls without sliding on an inclined, plane. The maximum elongation of the spring, is :, , m, k, , 2m, 7, , 2., , 5, , (a) 6, (b) 6, (c) 1:1, (d) 11/12, A uniform box is kept on a rough inclined plane., It begins to topple when θ is equal to :, , m, θ, , (a), , 2x, , 6., , x, , θ, , (a) 300, 3., , (b) 600, , (c) tan −1, , 1, 2, , mg sinθ, k, , (b), , a, 60°, , 84, , 3mg sin θ, k, , (d), , 2mg sin θ, k, , A massless thin hollow sphere is completely, filled with water of mass m . If the hollow, sphere rolls with a velocity v . the kinetic, energy of the sphere of water is :(Asune water, is non viscous), , V, , 1, 3, , 1, 2, , (b) mv 2, , (a) mv 2, , 4., , (c), , (d) 45°, , A rod touches a disc kept on a smooth, horizontal plane. If the rod moves with an, accleration a , the disc rolls on the rod without, sliding.Then, the acceleration of the disc w.r.t, the rod is, , a, , 2mg sin θ, 3k, , 7., , 7, mv 2, 10, , 5, 6, , (d) mv 2, , A particle P collides elastically at M with a, speed v . The change in angular momentum of, the particle about the point N during collision, is :, P, , 3a, , (a) 3, (b), (c) zero, (d) a/2, 2, A uniform cylinder of mass m is kept on an, accelerating wedge. If the wedge moves with, an acceleration a = 3g tan θ , the minimum, coefficient of friction between the cylinder and, wedge to avoid relative sliding between them, is, , (c), , θ, M, N, , I, V, , (a) 2mvl cos θ ↓, (c) zero, , (b) 2mvl cos θ ↑, (d) mvl cos θ ↑, NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 8., , 9., , Aball is attached to a string that is attached to 12. A hemispherical shell of mass M and radius, a thick pole. When the ball is hit, the string, R is hinged at point O and palced on a, warps around the pole and the ball spirals, horizontal surface. A ball of mass M moving, inwards sliding on the frictionless surface., with a velocity u inclined at an, Neglecting air resistance, what happens as the, 1, ball swings around the pole?, angle θ = tan −1 strikes the shell at point A, (a) The mechnical energy and angular momentum, 2, are conserved, (as shown in the figure) and stops. What is the, (b) The angular momentum of the ball is conserved, minimum speed u if the given shell is to reach, at the mechanical energy of the ball increases, the horizontal surface OP?, (c) The angular momentum of the ball is conserved, O, P, and the mechanical energy of the ball decreases, (d) The mechanical energy of the ball is conserved, and angular momentum of the ball decreases, A, The free end of a thread wound on a bobbin is, α, u, passed round a nail A hammered into the wall., m, The thread is pulled at a constant velocity’v’, gR, 2 gR, Assuming pure rolling of bobbin, find the, (a) Zero (b), (c), velocity v0 of the centre of the bobbin at the, 5, 3, instant when the thread forms an angle α with, d) it cannot come on the surface for any value of u, the vertical:(R and r are outer and inner radii 13. A hollow sphere of mass 2kg is kept on a rough, off the babbin), horizontal surface. A force of 10 N is applied, at the centre of the sphere as shown in the, figure. Find the minimum value of µ so that, the sphere starts pure rolling., (Take g=10m/s2), V, F = 10N, , vR, R sin α − r, 2vR, (c), R sin α + r, , 30°, , vR, R sin α + r, v, (d), R sin α + r, , (a), , (b), , 10. A thin wire of length L and uniform linear mass, density ρ is bent into a circular loop with centre, at O as shown. The moment of inertia of the, loop about the axis XX ' is :, 1, , x, , x, O, , (a), , ρ L3, 8π 2, , (b), , ρ L3, 16π 2, , (c), , 5ρ L3, 16π 2, , (d), , 3ρ L3, 8π 2, , 11. A partice moves in a circular path with, decreasing speed. Choose the correct, statement, (a) Angular momentum remains constant, r, (b) Acceleration ( a ) is towards the centre, (c) Particle moves in a spiral path wth decreasing, radius, (d) The direction of angular momentum remains, constant, NARAYANAGROUP, , m, , 3 × 0.16, , (a), , (b), , 3 × 0.08, , (c) 3 × 0.1, (d) Data insufficient, 14. A cubical block of side L and mass m is placed, on a horizontal floor. In the arrangement as, shown, a force F is applied at the end of the, plane sheet PQ which is firmly attached with, the block. What should be the minimum value, of PQ so that block may be tipped about an, edge?, 30°, , F, P, , L, , (a) L / 3, , Q, a, , (b) L / 2, L, , (c) L 2 / 3, O, , (d) L 3 / 2, 85
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , 15. A rigid body of moment of inertial I is, projected with velocity V making an angle of, 450 with horizontal. The magnitude of angular, momentum of the projectile about the point of, projection when the body is at its maximum, IV 3, height is given by, where R is the, 2 2 gR 2, radius of the rigid body. The ridid body is:, (a) sphere, (b)spherical shell, (c) disc, (d) none of these, 16. An equilateral triangle ABC formed a uniform, wire has two small identical beads initially, located at A. The triangle is set rotating about, the vertical axis AO.Then the beads are, relased from rest simultanously and allowed, to slide down,one along Ab and other AC as, shown.Meglecting frictional effects,the, quantities that are conserved as beads slides, down are:, A, g, , B, , O, , C, , (a) angular velocity and total energy(kinetic, and potential), (b) total angular momentum and moment of, interia about the axis of rotation, (c) angular velocity and moment of interia, about the aixs of rotation, (d) total angular momentum and total energy, 17. On a smooth horizontal table, a sphere is, pressed by blocks A and B by forces F1 and, , F2 respectively ( F1 > F2 ) exactly normal to, the tangent at the point of contact of blocks, and sphere. A force F is applied on the sphere, along a diameter perpendicular to another, diameter OP, which is the line of action of, forces F1 and F2 . The sphere moves out of, block A and B. FInd minimum value of F. the, coefficient of friction is µ at all contacts:, 86, , F, , Q, , F1, , P, , A, , F2, , B, , µ, (a) ( 7 F1 − 3F2 ), 2, , µ, ( 5 F1 − 3F2 ), 2, µ, (c) µ ( 3F1 − F2 ), (d) ( 3F1 − F2 ), 2, 18. Consider the following statements:, s1 : Zero net torque on a body means always, absence of rotational motion of the body., s2 : A particle may have angular momentum, even though the particle is not moving in a, circle., s3 : A ring of rolling without sliding on a fixed, surface. the centripetal acceleration of each, particle with respect to the centre of the ring, is same., State in order, whether s1 , s2 , s3 are ture or, false., (a) FTT, (b) FFT, (c) TTF, (d) FTF, 19. A uniform rod of length L (in between the, supports) and mass m is placed on two suports, A and B. The rod breaks suddenly at length, L/10 from the support B. Find the reaction at, support A immediately after the rod breaks., , (b), , L/10, , A, , D, , B, , L, , 9, 19, mg, 9, mg B), mg C), mg, D), 40, 40, 2, 20, 20. A uniform disc of mass m and radius R is, rolling up a rough inclined plane which makes, an angle of 300 with the horizontal. If the, coefficients of static and kinetic friction are, each equal to µ and the only forces acting, are gravitatinal and frictional, then value of, µ for maximum magnitude of the frictional, force acting on the disc is, 1, 1, 1, 1, A), B), C), D), 3, 2 3, 3 3, 3, , A), , NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 21. On a particle moving on a circular path with a, constant speed v , light is thrown from a, projectors placed at the centre ofthe circular, path. The shadow of the particle is formed on, the wall. The velocity of shadow up the wall is, Wall, v, , φ, , (A), , ? 2 ( r2 − r1 ) + ? 1r2, ? 2 ( r2 − r1 ) + ? 1r1, (B), r2, r2, , (C) ? 1, (D) ? 2, 24 . Two vertical walls are separated by a distance, of 2 metres. Wall ‘A’ is smooth while wall B is, rough with a coefficient of friction µ = 0.5 A, uniform rod is probed between them. The, length of the longest rod that can be probed, between the walls is equal to, 2m, , (A) v sec 2 f, (B) v cos 2 f, (C) v cos f, (D) None of these, 22. A rod of length l is travelling with velocity, v and rotating with angular velocity ? such, , ?l, = v . The distance travelled by end, that, 2, B of the rod when rod rotates by an angle,, p, is, 2, A, , ω, , V, , B, , (a), , 2l, , (b), , P, Q, , Wall, A, , Wall, B, , (a) 2 metres, , (b) 2 2 metres, , 17, metres, 2, 25. A disc is rotating at an angular velocity ω0 .A, constant retardation torque is applied on it to, stop the disc. After a certain time at which, some number of rotation of the disc have been, performed so that total angle rotated is θ1 and, , (c) 2 metres, , (d), , 2, th of these rotatios will further stop, 3, the disc. Find the retarding force., , that only, , 5, l, 2, , (c) 3l (d) 4l, , 23. A large rectangular box has been rotated with, a constant angular velocity ? 1 about its axis, as shown in the figure. Another small box kept, inside the bigger box is rotated in the same, sense with angular velocity ? 2 about its axis, (which is fixed to floor of bigger box). A particle, P has been identified, its angular velocity about, AB would be, , 11ω02, 9ω02, 5ω02, 3ω02, a), 2), 3), 4), 14θ1, 14θ1, 14θ1, 14θ1, 26. A square plate hinged at A, of side a and mas, M is placed in ( x − z ) plane. The plate is, allowed to fall upto ( x − y ) plane. Find its, angular velocity., Z, , A, ω1, y, , C, ω2, , B, , NARAYANAGROUP, , r1, r2, , D, , x, , A, , P, , 3g , a) , , a 2 , , 1/2, , 1/2, , 24 g , b) , 7 a , , 12 g , c) , 7a , , 1/2, , 1/2, , 12 g , d) , , 7 2a , , 87
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , 27. A disc of mass m and radius r is placed on a, rough horizotal surface. A cue of mass m hits, the disc at a height h from the axis passing, through centre and parallel to the surface. The, disc will start pure rolling for., a) h <, , r, 3, , b) h =, , r, 2, , c) h >, , r, 2, , d) h ≥, , 31. A disc is given an initial angular velocity ω0, and placed on rough horizontal surface as, shown. The quantities which will not depend, on the coefficient of friction is/are, , r, 2, ω0, , MULTIPLE ANSWER QUESTIONS, 28. A solid cylinder is rolling down the inclined, plane without slipping. Which of the following, is/are correct, (a) The friction force is dissipative, (b) The friction force is necessarily changing, (c) The friction force will aid rotation but opposes, translation, (d) The friction force is reduced if θ is reduced, 29. An impulse I is applied at the end of a uniform, rod if mass m. then :, (a) KE of translaton of the rod is, , I2, 2m, , (b) KE of rotation of the rod is, , I2, 6m, , (c) KE of rotation of the rod is, , 3I 2, 2m, , (a) The time until rolling begins, (b) The displacement of the disc until rolling begins, (c) The velocity when rolling begins, (d) The work done by the force of friction, 32. A thin rod AB of mass M and length L is, rotating with angular speed ω0 about vertical, axis passing through its end B on a horizontal, smooth table as shown. If at some instant the, hinge at end B of rod is opened then which of, the following statement is/are correct about, motion of rod ?, A M, L, B, , I, , 2I 2, (d) KE of the rod is, m, , 30. A uniform rod of mass m and length l is in, equilibrium under the constraints of horizontal, and vertical rough surfaces. Then :, y, , 1, θ, O, , x, , (a) the net torque of normal reaction about O is, equal to, , mgl, cos θ, 2, , (b) the net rorque due to friction about O is zero, (c) the net torque due to normal reactions is, numerically equl to the net torque due to the, frictional force abou the CM of the rod, (d) all of the above, 88, , ω0, , (a) The angular speed of rod after opening the, hinge will remain ω0, (b) The angular speed of rod after opening the, hinge will be less than ω0, (c) In the process of opening the hinge the kinetic, energy of rod will remain conserved., (d) Angular momentum of rod will remain, conserved about centre of mass of rod in the, process of opening the hinge, 33. A cylinder rolls without slipping on a rough, floor, moving with a speed v. It makes an elastic, collision with smooth vertical wall. After impact, (a) it will move with a speed v initially, (b) its motion will be rolling without slipping, (c) its motion will be rolling with slipping initially, and its rotational motion will stop momentarily at, some instant, (d) its motion will be rolling without slipping only, after some time, NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 34. A sphere of radius 0.10m and mass 10kg rests, in the corner formed by a 30° inclined plane, and a smooth vertical wall. Choose the correct, options, , f, , mg, 30°, , N2, , (a) N1 = 56.5 N, (b) N 2 = 113 N, (c) f = 0, (d) f ≠ 0, 35. If the resultant of all the external forces acting, on a system of particles is zero, then from an, inertial frame, one can surely say that, (a) linear momentum of the system does not change, in time, (b) kinetic energy of the system changes in time, (c) angular momentum of the system does not, change in time, (d) potential energy of the system does not change, in time, 36. A rod of length ‘l’ is pivoted smoothly at O is, resting on a block of height h. If the block, moves with a constant velocity V, pick the, current alternatives., , θ, , V, , h, , O, , (a) angular velocity of rod is, , V cos ?, h, , 2V 2 cos3 ? sin ?, h2, V cos 2 ?, h, , (d) tangential velocity of free end of rod is, , lV cos 2 ?, h, NARAYANAGROUP, , COMPREHENSION TYPE QUESTIONS, Passage - I :, A string is wrapped several times on a cylinder of, mass M and radius R. the cylinder is pivoted about, its axis of symmetry. A block of mass m tied to the, string rests on a support so that the string is slack., the block is lifted upto a height h and the support is, removed. (shown in figure), , R, m, m, , h, , h, , 38. What is the angular velocity of cylinder just, before the string becomes taut, (a) zero, , (b) angular acceleration of rod is, , (c) angular velocity of rod is, , 37. A wheel is under pure rotational motion about, an axis passing through its centre. It moves, with constant angular velocity., a) if angular velocity is increasing then acceleration, of particles on a spoke if moved from centre to, periphery remains constant, b) acceleration of particles on a spoke if moved, from centre to periphery continuously increases, c) acceleration of particles on a spoke if moved, from centre to periphery continuously increases and, on peripherial points,it remains same, d) accelerations of particles in both the cases remain, same, , (b), , 2gh, R, , (c), , gh, R, , (d), , 2 gh, R, , 39. When the string experience a jerk, a large, impulsive force is generated for a short, duration, so that contribution of weight mg can, be neglected during this duration. Then what, will be speed of block m, just after string has, become taut, 2gh, (a) 1 + M , m , , (b), , 2gh, M, , 1 + 2m , , gh, (c) 1 + M (d), m , , gh, M, , 1 + 2m , , 40. If M = m, what fraction of KE is lost due to, the jerk developed in the string, (a) 1/2, (b) 2/3, (c) 1/3, (d) 1/4, 89
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , Passage-II :, A man of mass 100 kg stands at the rim of a turn, table of radius 2m, moment of inertia 4000 kg. The, table is mounted on a vertical smooth axis, through, its center. The whole system is initially at rest. The, man now walks on table with a velocity 1m/s relative, to earth, 41. With what angular velocity will the turn table, rotate, (a) 0.5 rad/sec, (b) 0.1 rad/sec, (c) 0.05 rad/sec, (d) 0.2 rad/sec, 42. Through what angle will the turn table have, rotated when the man reaches his initial, position on it, π, rad/sec, 11, 2π, (c), rad/sec, 11, , 3π, rad/sec, 11, 4π, (c), rad/sec, 11, , Passage - IV :, ., Three particles each of mass m can slide on fixed, frictinless circular tracks in the same horizontal plane, as shown . Particle m1 (= m)moves with veocity, v0 and hits patricle m2 (= m), the cofficient of, restitution being e = 0.5. Assume that m2 and, m3 (=m)are at rest initially and lie along a radial line, before impact, and the spring is initially unstretched., V0, , k, m3, 2R, R, , (b), , (a), , 43. Through what angle will it have rotated when, the man reaches his initial position relative to, earth, π, (a) rad/sec, 5, 2π, (c), rad/sec, 11, , 2π, (b) rad/sec, 5, π, (d) rad/sec, 11, , 46. Velocity of m2 immediately after impact is, v0, 2, The maximum veloctiy of m3 is, , (a), 47., , v0, 4, , (b), , 3, 5, , (c), , (d), , 3, , 3v0, 2, 3, , (b) 10 v0, (c) v0, (d) v0, 4, 2, 48. The maximum stretch of the spring is, (a), , 3, m, v0, 4, 5R, , (b), , 3, 5, , (d) 3 v m, 0, , (c) v0, , m, 5R, , 3, m, v0, 2, 5R, , 10, , 5R, , Passage - V :, A plank of length 20 m and mass 1 kg is kept on a, horizontal smooth surface. A cylinder of mass 1kg, is kept near one end of the plank. The coefficient, of friction between the two surfaces is 0.5. The, plank is suddenly given a velocity 20m/s towards, left., m, , m, , B, O, , C, , 44.Calculate angular velocity in terms of V and L, 12V, V, 7V, 3V, (a), (b), (c), (d), 7L, L, 12 L, 2L, 45. If insect reaches the end B when the rod has, turned through an angle of 900 calculate V, interms of L, 3, 7, 1, 2, 2gL (b), 2gL (c), gL (d), 2gL, (a), 7, 12, 12, 7, 90, , 3v0, 4, 3, , (a) v0, , Passage-III :, A homogeneous rod AB of length L and mass M is, hinged at the centre O in such a way that it can, rotate freely in the vertical plane. The rod is initially, in horizontal position. An insect S of the same mass, M falls vertically with speed V on point C, midway, between the points O and B. Immediately after, falling, the insect starts to move towards B such, that the rod rotates with a constant angular, velocityω ., , A, , m2, , m1, , l = 20m, , 49. Which of the following statement is correct?, (a) Intitial acceleration of cylinder is 5m / s 2 towards, left, (b) Initial acceleration of cylinder is 5m / s 2 towards, right, (c) Initial acceleration of cylinder is 10 m / s 2 towards, right, (d) Initial acceleration of cylinder is 10 m / s2 towards, left, NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 50. Which of the following statement is correct?, (a) Pure rolling of cylinder takes place immediately, (b) Intitally cylinder slips and then pure rolling begins, (c) Pure rolling never begins, (d) There is no lose in kinetic energy during its entire, motion., 51. Velocity of plank when pure rolling begins is, (a) 10m/s (b) 1.5sec (c) 20m/s (d) 25m/s, 52. Time in which plank and cylinder separate, (a) 1 sec (b) 1.5 sec (c) 2.5 sec (d) 2 sec, Passage - VI :, A ring of radius R is rolling purely on the outer, surface of a pipe of radius 4R. At some instant, the, centre of the ring has a constnat speed of v., 53. The acceleration of the point on the ring which, is in contact with the surface of the pipe is, (a) 4v 2 / 5R (b) 3v 2 / 5R (c) v2 / 4R (d) zero, 54. The acceleration of the point on the ring which, is farthest from the centre of the pipe at the, given moment is :, (a) 4v 2 / 5R (b) 3v 2 / 5R (c) 3v 2 / 4R (d) 6v 2 / 5R, Passage - VII :, A uniform rod of mass ‘m’ and length L is released, from rest, with its lower end touching a frictionless, horizontal floor. At the initial moment, the rod is, inclined at an angle of 300 with the vertical., 55. Then,the value of normal reaction from the, floor just after release,will be:, a) 4mg/7 b) 5mg/9 c) 2mg/5 d) mg/5, 56. In the above problem, the initial acceleration, of the lower end of the rod will be:, a) g 3 / 4 b) g 3 / 5 c) 3 g 3 / 7 d) g 3 / 7, Passage-VIII:, One end of an ideal spring of unstretched length, lO = 1m , is fixed on a frictionless horizontal table., The other end has a small disc of mas 0.1 kg attahed, to it. The disc is projected with a velocity, ?0 = 11 m / s perpendicular to the spring:, , v1, 11l0, 10, , v0, O, l0, NARAYANAGROUP, , 57. Choose the correct statement, (a) Linear momentum of disc is conserved as the, spring force is always perpendicular to velocity of, disc., (b) Angular momentum of disc about fixed end of, spring is conserved., (c) Kinetic energy of disc is conserved, (d) Angular velocity of disc remains constant, 58. In the subsequent motion of disc, maximum, elongation of spring is l0 /10 . The velocity of, disc at this instant is:, (a) 11 m / s, (b) 10 m / s, (c) 5 m / s, (d) 7 m / s, 59. What is the force constant of spring?, (a) 210 N / m, (b) 100 N / m, (c) 110 N / m, (d) 200 m / s, Passage - IX:, A thin uniform rod of mass m and length L is hinged, at one end and from other end a light string is, attached. The string is wound over a frictionless, pullely (having mass 2m) and a block of mass 2m, is connected to string on other side of pulley as, shown. The system is released from rest when the, rod is making an angle of 370 with horizontal.Based, on above information answer the following, questions:, H2, , mL, 37°, , 2m, , H, , 60. Just after release of the system from rest,, acceleration of block is, (a), , 72 g, 48 g, , downwards (b), , downwards, 121, 119, , (c), , 90 g, 90 g, , downwards (d), , upwards, 121, 121, , 61. Just after release of the system, the resultant, force exerted by hinge on rod is, (a) 0.7mg (b) 0.92mg (c) 0.53mg (d) mg, 91
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ROTATIONAL DYNAMICS, 62. Just after release of the system from rest, the, resultant force exerted by hinge H 2 on pulley, is, 46, mg in upward direction, (a), 121, 46, mg in downward direction, (b), 121, 438, mg in upward direction, (c), 121, 438, mg in downward direction, (d), 121, , MATRIX MATCHING QUESTIONS, 63. For the following statements, except gravity, and contact force between the contact, surfaces, no other force is acting on the body., Column I, Column II, (a) When a sphere is (p) Upward direction, in pure–rolling on a, fixed horizontal surface., (b) When a cylinder (q) vcm > R ω, is in pure rolling on, a fixed inclined plane in upward, direction then friction force acts in, (c) When a cylinder is (r) vcm < R ω, in pure rolling, down a fixed incline plane,, friction force acts in, (d) When a sphere of (s) No frictional force, radius R is rolling acts, with slipping on a fixed horizontal, surface, the relation between vcm and ω is, (t) Work done by the, frictional force is zero, 64. A uniform disc is acted upon by some forces, and it rolls on a horizontal plank without, slipping from north to south. The plank, in turn, lies on a smooth horizontal surface. Match the, following regarding this situation :, Column I, Column II, (a) Frictional force on (p) May be, the disc by the surface, directed towards, north, (b) Velocity of the, (q) May be directed, lowermost point of the disc towards south, (c) Acceleration of, (r) May be zero, centre of mass of the disc, (d) Vertical component (s) Must be zero, of the acceleration of, centre of mass, 92, , JEE-ADV PHYSICS-VOL - III, 65. A rigid body of mass M and Radius R rolls, without slipping on an inclined plane of, inclination θ under gravity Match the type of, body with magnitude of the force of friction., C olum n I, C olum n II, a) For ring, , p), , Mg sin θ, 2.5, , b) For solid sphere, , q), , Mg sin θ, 3, , c) For solid cylinder, , r), , Mg sin θ, 3.5, , d) For hollow, , s), , Mg sin θ, 2, , spherical shell, 66. A rigid body is rolling without slipping on, horizontal surface.At given instant BD is, perfectly horizontal and CD is perfectly, vertical., C, , D, , B, , ω = v/R, R, , Column I, a) Velocity at pointA, vA, , Column II, p) v 2, , b) Velocity at point B, vB, , q) Zero, , c) Velocity at point C, vC, , r) v, , d) Velocity at point D, vD, s) 2v, 67. A horizontal table can rotate about its axis. A, block is placed at a certain distance from, center as shown in figure. The table rotates, such that block does not slide. Select possible, direction of net acceleration of block at the, instant shown in figure. Then match the, columns., 4, , 3, , 2, 1, , NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, Column I, Column II, a) When rotation is, p) 1, clockwise with constant ω, b) When rotation is, q) 2, clockwise with decreasing ω, c) When rotation is, r) 3, clockwise with increasing ω, d) Just after clockwise, s) 4, rotation begins form rest, 68. An uniform disc rolls without slipping on a, rough horizontal surface with uniform angular, velocity. Point O is the centre of disc and P is, a point on disc as shown. In each situation of, column I a statement is given and the, corresponding result are given in column –II., Match the statements in coloumn-I with the, results in column-II, P, , O, , Column I, a) The velocity of, disc, b) The acceleration, point P on disc, , Column II, p) Change at point P in, magnitude with time, q) Is always directed of, from that point not, towards centre of disc., r) is always zero, , c) The tangential, acceleration of point, P on disc, d) The acceleration, s) is non-zero and, of point on, remains constant in, disc which is in contact magnitude, with rough horizontal surface, 69. A light string is wrapped on a pulley and two, blocks of masses m1 and m2 are attached to, free end of string as shown in figure. T1 and, , T2 are the tension in string on two sides of, pulley. In column I, some information is, mentioned about friction between of inertia of, pulley, while in colunm II the effect of the, information mention in column I on the motion, of system is given. Match the entries of, column I with the entries of column II., NARAYANAGROUP, , Column I, , Column II, , 1. No friction between, p. Angular acceleration, pulley and string, and, of pulley is 0, moment of inertia of, pulley is not, negligible., 2. Friction is there, between pulley and, string, and pulley is, light., , q. T1 = T2, , 3. Friction is not there, and pulley is light., , r. T1 ≠ T2, , D. Friction is there and, pulley is having, some moment of, inertia., , s. Angular, acceleration of, pulley ≠ 0, , 70. A smooth ball of mass m moving with a uniform, velocity v0 strikes a smooth uniform rod AB, of equal mass m, lying on a frictionless, horizontal table. The ball strikes the rod at, one end A, perpendicular to the rod, as shown, in the figure. The collision is perfectly elastic., Some physicaL quantities pertaining to this, situation are given in COLUMN-1 while their, values are given in COLUMN-2 in a different, order . Match the values in COLUMN-II and, the quantities in COLUMN-I, m, , v0, , A, , m, , B, , Column-I, A) Final kinetic energy of ball, , Column-II, 2, p), 5, , Initial kinetic energy of ball, B) Impulse delivered to the rod, Initial momentum of ball, , q), , 3, 5, , C) Angular momentum of rod about its, centre of mass, Initial angular momentum of the ball about, the centre of mass of the rod, , r), , Final kinetic energy of rotation of the rod, Final kinetic energy of translation of the rod, D) Final kinetic energy of rotation of the rod, Final kinetic energy of translation of the rod, , 9, 25, , s) 3, , 93
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , LEVEL-V - KEY, , INTEGER TYPE QUESTIONS, 71. A rod of mass m and length l is released, from rest from vertical position as shown in, the figure. The normal force as a function of θ ,, which is exerted on the rod by the ground as it, falls downward, assuming that it does not slip, , SINGLE ANSWER QUESTIONS, 1.D 2.C 3. A 4.D 5. D 6. A 7. C, 8. D 9. A 10.D 11. D 12. D 13. B 14. A, 15. C 16. D 17. A 18. D 19. A 20. C 21. A, 22. A 23. B 24. D 25. C 26. A 27. B, MULTIPLE ANSWER QUESTION, 28. C,D, 29. A,C,D 30. A,B,C,D, 31. C,D, 32. A,C,D 33. A,C,D, 34. A,B,C 35. A,B, 36. B,C 37.B,C, COMPREHENSION QUESTIONS, 38. A 39. B 40. C 41. C 42.A 43. A 44. A, 45. B 46. B 47.B 48. A 49. B 50. B 51. B, 52.B 53. A 54. D 55. A 56. C 57. B 58. B, 59. A 60. A 61. C 62. C, MATRIX MATCHING TYPE, 63. A → s, t; B → p; C → p; D → q, r, 64. A → p, q, r; B → p, q, r; C → p, q, r; D → s, 65. A → s; B → r; C → q; D → p, 66. A → q; B → p; C → s; D → r, 67. A → r; B → s; C → q; D → p, 68. A → p; B → q, s; C → p; D → q, 69. A → p, q; B → q, s; C → p, q; D → p, q, r, s, 70. A → r ; B → P ; C → P ; D → S, INTEGER TYPE QUESTIONS, 71. 2 72. 3 73. 4 74. 5, , 3cos θ − 1 , then n =, n, , , , , is mg , , 2, , θ, , 72. One end of a uniform rod of mass M and, length L is supported by a frictionless hinge, which can with stand a tension of 1.75 Mg. The, rod is free to rotate in a vertical plane. The, maximum angle should the rod be rotated from, the vertical position so that when left, the hinge, does not break is, , π, n, , 73. A thin uniform bar of mass m and length 2L is, held at an angle 300 with the horizontal by, means of two vertical inextensible strings, at, each end as shown in figure. If the string at, the right end breaks, leaving the bar to swing, the tension in the string at the left end of the, bar immediately after string breaks is, T=, , n, mg, 13, , LEVEL-V - HINTS, DETIAL SOLUTINS, SINGLE ANSWER QUESTIONS, 1., , a=, , ( 2m − m ) g = g, 3m, , 3, , T = 2mg − 2m×, 30°, , R, starts rolling, 16, down without slipping from the top of another, sphere of radius R = 1 m. The angular velocity, of the sphere in rad s −1 , after it leaves the, surface of the larger sphere is 8 x n. Where n, = --., , 74. A uniform sphere of radius, , 94, , &T 1 = 2T =, , for T in lower string ,, , g, = 4mg / 3, 3, , 8mg, 3, , 8mg, 11mg, + mg =, 3, 3, Let has take movent about hinge, T 1 + mg =, , 4mg × x =, , 11mg, x 11, ×y⇒ =, 3, y 12, NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 2., , 3., , x, 2, (taking moment aobut right bottom corner), 1, ⇒ tan θ =, 2, ma cos θ − f = mA → (1), mg sin θ × x = mg cos θ ×, , f × r = Iα → ( 2 ) ; A = rα → ( 3), From about equation A =, A=, , a, 3, , 1 2, a, 0, where θ = 60 & I = mr , 3, 2, , , 4., ma sinθ, N, f, ma, , O, p, mg cosθ, mg sinθ, mg, , θ, , ma sinθ, , mg sin θ − ma cos θ − f = ma .....(1), N = ma sin θ + mg cos θ .....(2), f = µ N ..........(3), , τ ( p ) = mg sin θ =, , 3, mR 2 × α ......(4), 2, , a = Rα ........(5), Solving above equitions we get m µ, 5., , 6., , 7., 8., , We use work energy principal, 1, 2mg sin θ, mg sin θ × x − kx 2 = 0 ⇒ x =, 2, k, Liquid gets only tracanslatory motion in side any, 1 2, rollny mosum. This kE = mv, 2, taking component , ∆L = mv sin θ × l , , durring strike, , − mv sin θ l = 0, The force is conservative. so ME is conserved, (friction less) & Angular momentum ( L = Iω ), decreases due to reverse swing compare to that of, initial swing before strike., , NARAYANAGROUP, , 9., , When the thread is pulled, the bobbin rolls to the, right. Resultant velocity of point B along the thread, is υ = υ0 sin α − ω r , where υ0 sin α is the, component of translation velocity along the thread, and ω r linear velocity due to rotation. As the, bobbin rolls without slipping, υ0 = ω R. Solving the, obtained equations, we get υ0 =, , υR, ., R sin α − r, , 1, 3, 3 l 3ρl 3, 2, 2, 2, 10. I = mR + mR = mR = ρ l = 2, 2, 2, 2 2π , 8π, 2, , 11. L = mVR, hear V dicreases so L never be, r, r, constant, Also ar & at both are acting & thind point, r, contradicts the given question. But direction L is, always constant., 12. There is neither torque nor angular momentam about, the O. (because line of action passes through O), So u = 0, 13. Form given figure we get, N = F sin θ + mg → (1) ; F cos θ − f = ma → ( 2), , fR = Iα → ( 3), Slove above equation we can easily get, µ = 0.08 × 3, 14. Block will be tipped about point O., , M O = ( F cos 300 ) a = F sin 300 L + mgL / 2, , F=, , (a, , mgL, 3−L, , ), , For F > 0, a 3 > L ; a ≥ L / 3., 15. At highest point, velocity = V cos 450 = V / 2, , V, V 2 sin 2 450 V 2, L = m .r ⊥ ; Here r⊥ = h =, =, 2g, 4g, 2, L=, , mV 3, IV 3, =, ( given ) ; So, m = I2, 2, 4 2 g 2 2 gR, 2 R, , I=, , 1, mR 2 . It is a disc., 2, , V cos 45°, h, , 95
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 40. For M = m,K1 =, ω, , K 0 − K1 1, K, = 3, 0, , V, θ, , h, , Vcosθ, , 41 to 43, 41. By conservation of angular momentum on the mantable system,, r, r, Li = L f or 0 + 0 = I m ωm + I t ωt, , d? 2V 2 cos 3 ?.sin ?, =, (B) a =, dt, h2, 37. Acceleration is given by ω 2 r and it increases if, moved from centre to periphery. Accelerations of, all end points will be same. If ω is varying, 2, 2, acceleration will be ( rα ) + ( rω ) . On, moving from centre to periphery acceleration, (When ω is varying) will increses, 1/2, , COMPHRENSION TYPE QUESTIONS, 38 to40, 38. Just before the string becomes taut, the block falls, freely, so v 0 = 2gh. There is no tension in the, string, so nothing causes the cylinder to spin, so, ω 0 = 0., , 39. When the string experiences a jerk, the large, impulse developed is of very short duration so that, the contribution of weight mg can be neglected, during this time interval., The angular momentum of the system is conserved,, as the tension is internal force for the system. Thus, r, r, we have Li = Lf, 1, mv1R + MR 2 ω1 = mv0 R = m 2ghR, 2, The string is inextensible, so v1 = Rω1 . On solving, 2gh, , for ω1 we get ω1 = R 1 + ( M / 2m) , , , v1 = Rω1 =, , , , 2gh, , 1 + ( M / 2m) , , The final kinetic energy K1 is given by, 2, 1, 11, 1, 1, 2, 2 v1 , K1 = mv12 + Iω12 = mv1 + MR 2 , 2, 2 2, R , 2, 2, , , mv02, 1, M 2 1 , = m + v1 = , , 2 1 + ( M / 2m) , 2, 2, K0, 1, , , =, Q K o = mvo2 , , ,, 1 + ( M / 2m) , 2, , NARAYANAGROUP, , 2K 0, , so the fraction lost is, 3, , ωt = −, , Imω m, v 1, ω = = rad / s, I1 where m r 2, , ω t = −100 ( 2) ×, 2, , 1/ 2, 1, = − rad / s, 4000, 20, , Thus the table rotates clockwise (opposite to man), with angular velocity 0.05 rad/s., 42. If the man completes one revolution relative to the, table then θ mt = 2π;2π = θm − θ t, 2π = ω m t − ω t t (where t is the time taken), t = 2π / (ω m − ω t ) = 2π ( 0.5 + 0.05), , Angular displacement of table is, θ t = ω t t = −5.05 × ( 2π / 0.55) = −, , 2π, radian, 11, , 43. If the man completes one revolution relative to the, earth, then θ m = 2π, 2π, , 2π, , time = ω = 0.5, m, During this time, angular displacement on the table,, θ t = ω t ( time ) = −0.05 ×, , 2π, ,, 0.5, , π, θ t = − radian , θ t = 36° in clockwise direction, 5, , 44 to 45, A angular momentum of system of rod and insect,, just after collision = intial angular momentum of, insect about o, 2, ML2, L, L , ∴, + M ω = Mυ, 4, 4 , 12, , ω, θ, , 0, x, , B, , ω=, , 12V, 7L, , Mg, 99
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , Let a be the angular acceleration of rod about hinge, and a be the acceleration of block just after the, system is released from rest., The free body diagram of rod, pulley and block, are shown in figure. In free body diagram of rod,, the force exerted by hinge on rod is not shown., , ω=0, L, α, 2, , α, , 30°, , a acceleration of cm, , For rod, mg ×, c, , , mL2 , = − Ia I =, , 3 , , For block, 2mg − T = 2ma, , From eqns (1), )2) and (3), 12 g, , 3g, , 4 mg, , we get α = 7l ; a = 7 ; N = 7, Acceleration of lower end of rod is, →, , ax =, , ( ), , ( ), , ( ), , L, L 12 g, 3 ˆ 3 3g ˆ, ×, −i =, −i, α cos 300 −iˆ = ×, 2, 2 7l, 2, 7, , 57-59, , 0, From constraint, a = La cos37 =, , 4 La, 5, , Solving above equations, we get, , v1, , 72 g, 90 g, 98mg, ,a =, ,T =, 121, 121L, 121, Reaction force exerted by H 2 on pulley, a=, , 11l0, 10, , v0, , L, cos 37 0 − T × L cos 37 0, 2, , O, , is, N1 = 2T + 2mg =, , l0, , Angular momentum of disc about fixed end is, conserved, as the spring force passes through O, , Now draw the complete free body diagram of rod, as follows, , 11l , 10?, m?0l0 = m?1 0 ; ?1 = 10 m / s = 0, 11, 10 , From conservation of energy we get, , 1, 1, 1 l , mV02 = mV12 + K 2 0 , 2, 2, 2 10 , ⇒ K 2 = 210 N / m, , 2, , 438, mg, 121, , Lα, 2, , T, T cos37°, , T sin37°, , mg sin37°, R1, R2 mg, , mg cos37°, , 0, 0, ⇒ R1 + T cos37 − mg cos 37 =, , 59- 62, N1, , mL, a, 2, , ⇒ R2 + T sin 37 0 = mg sin 37 0, Net reaction force on rod due to hinge is,, , T, , F = R12 + R22, T, , T, , α, , 2mg, mg, , a, 2mg, , NARAYANAGROUP, , MATRIX MATCHING TYPE, 63. At the time of pure rolling,friction = 0and work by, friction is zero., Due to downward reletive motion the direction, friction is upwards., Due to downward reletive motion, friction is, upwards., There are two cases possible,q and r, 101
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, , 64. There are possibilities of p,q,r that dependence on, relative velocity of contact point. The velocity of, COM is always parallel to the motion of rolling, body., f =, , 65., , mg sin θ, mr 2 , 1 +, ;, I , , , For =, , mg sin θ, 2, , For hollowsphere f =, , mg sin θ, 2.5, , For solid cylider f =, , mg sin θ, 3, , For solid sphere f =, , 66. V p = 2V sin, , ⇒ v0 = v +, , mg sin θ, 3.5, , 67. Angular velocity is constant,tangential acceleration, is zero.But radial acceleration does not zero., 68. (a) at point P ,tangential velocity changes with time., At the point contact of pure rolling frictional is, zero.So acceleration is towards centre., 69. A → As there is no friction between pulley and, rope and pulley so T1 = T2 and torque on pulley, so angular acceleration is 0., B → Since there is friction so acceleration of pulley, is non - zero. So, by torque equation, T1 − T2 = Iα, But as I = 0 ⇒ T1 = T2, C → Since pulley is light, no friction between rope, , and pulley so T1 = T2 and torque on pulley is 0, and α = 0 ., D → Friction is there between pulley and rope, and moment of inertia of pulley is not negligible., So, pulley has angular acceleration and by equation, T1 − T2 = Iα ⇒ T1 ≠ T2, This condition is valid only when M 1 ≠ M 2, If M 1 = M 2 then T1 = T2 = mg and α = 0, , 102, , ⇒ v + v0 = V +, , lw, 2, , .......(4), , 2, l, l ml , ω, Also mv0 = mv + , 2, 2 12 , , θ, 2, , 70. mv − mv0 = − ∫ Fdt = − J, , , lω , V + 2 − v , , lω, a =1 = − , 0 − v0, , ⇒ V + 2 − v = v0, , , , .......(2), , ⇒ v0 = v + V, , ........(3), , .......(5), , From (3) and (5) we get v =, , lω, 6, , So, from (4), we get v + v0 =, , 2, lω ......... (6), 3, , Solving (5) and (6), we get ω =, v0, , F, , 12v0, 5l, , V, , F, ω, , V, , just before, impact, , During, impact, , So, we get v =, , Just after, impact, , 3v0, 2v, ,V = 0, 5, 5, , 1 2, mv, Final KE of ball 2, 9, ⇒, =, =, Initial KE of ball 1 mv 2 25, 0, 2, ⇒, , Im pulse Delivered to rod, 2, =, Initial Momentum of ball (mv0 ) 5, , ( L1 )rod, ( L1 )ball about CM, , ......(1), , mV = J, mv0 = mv + mV, , lω, 6, , ⇒, , ml 2 , , ω, Iω, ( L1 )rod 12 , 2, =, =, =, l ⇒ L, ( 1 )ball mv l 5, mv0 , 0, , 2, 2, , ( K rod )rotation, ( K rod )translation, , ml2 , , ω, 12 , , =, =3, mv 2, NARAYANAGROUP
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ROTATIONAL DYNAMICS, , JEE-ADV PHYSICS-VOL - III, 71. At angle θ, 1 2, l, I ω = mg (1 − cos θ), 2, 2, , 3g, ω2 =, (1 − cos θ), l, , or,, , …(i), Differentiating w.r.t. q,, α=, , l, sin θ, 2, m l2, 3, , mg, , an =, , =, , l 2 3g, ω =, (1 − cos θ), 2, 2, , 3 g sin θ, 2 l, , … (ii), , l, 2, , 3, 4, , and at = α = g sin θ, , 3 g sin θ, 3, , = m g sin θ cos θ −, (1 − cos θ) , 2, 4, , , mv 2, = mg cos θ, R+r, where v is the speed of the centre of the sphere at, that moment and θ is the coresponding angle. The, speed v can be found by using the Law of, conservation of energy,, , 3, 3, , mg sin θ cos θ − 1, 2, 2, , , , x, , C, an, , θ, , 4, 6 3g, mg and α =, 13, 13 L, 74. The equation of motion for the centre of the sphere, at the moment of breaking off, N = 0 is, T=, , f = ma x = m (at cos θ − a n sin θ), , =, , τ TL cos 300 3 3T, α= =, =, 2, I, 2ml, m ( 2L ), 12, now just after the string breaks acceleration of point, A in vertical directon should be zero solving above, equations we get, , y, , at, , r, , Further, m g − N = ma y or,,, , h, , N = m ( g − at ), , N = m [ g − (at sin θ + an cos θ)], 3, 3 g cos θ, , , = m g − g sin 2 −, (1 − cos θ) , 4, 2, , , , =, , mg, [4 − 3 sin 2 θ − 6 cos θ + 6 cos2 θ], 4, , =, , mg, (1 − 3cos θ) 2 ., 4, , The rod does not slip until, , R, θ, , mg, , N =0, , 1, θ = cos−1 ., 3, , i.e.,, , 2M Lω , 72. 1.75Mg = Mg +, , , L 2 , , According to which mgh =, 2, , L, 1 ML2 2, ω, (1 − cos θ ) =, 2, 2 3, Solving (i) and (ii) we get, , (i), , 73., , ∑F, , x, , =0, , (ii), θ = 60º, , ; a = ∑ Fy = mg − T, y, m, m, , NARAYANAGROUP, , mv 2 Iω 2, +, 2, 2, , 2 2, where I = mr , v = rω and, 5, , h = ( R + r )(1 − cos θ ), , Mg, , ax = 0, , θ, , From the equations we get, ω=, , 10 g ( R + r ), 17r 2, , ⇒ ω = 40rads −1 ., , 103
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , LEVEL-VI, , k, c, m, , SINGLE ANSWER QUESTIONS, 1., , The point P of a string is pulled up with an, acceleration g. then the acceleration of the, hanging disc (w.r.t ground) over which the, string is wrapped, is, , Q, , (a) τ C ≠ 0 for ct < t0, (c) x =, , P, , m, , 2., , (a), , 2g, ↓, 3, , (c), , 4g, ↓, 3, , (b), , g, ↑, 3, , (d), , g, ↓, 3, , Q, , Q, , (b) f = 0 for t < t0, , ma, , where x = deformation of the spring, k, , 1, 2, , (d) ( KE )max = ma2t02 , where ( KE )max is the maximum, KE of the rolling body, , 5., , A sphere of mass m1 is placed on a plank of, mass m2 . The coeffcient of friction between, the plank and sphere is µ . If the inclined plane, is smooth, the frictional force between the, plank and sphere :, , A linear impulse ∫ Fdt acts at a point C of the, smooth rod AB . The value of x is so that the, end A remains stationary just after the impact, is :, A, , (a), , l, 4, , (b), , l, 3, , (c), , l, 6, , (d), , l, 5, , O, , m1 µ, m2, , x, C, , µ=0, , Fdt, , θ, , 3., , B, , (a) depends on m1, (b) depends on m2, (c) 0, (d) = µ m1 g cos θ, Four beads each of mass m are glued at the, top, bottom and the ends of the horizontal, diameter of a ring of mass m . If the ring rolls, without sliding with the velocity v of its , the, kinetic energy of the system (beads +ring) is:, , 6., , Two light vertical springs with equal natural, lengths and spring constants k1 and k2 are, sparated by a distance l . Their upper ends, are fixed to the ceiling and their lower ends to, the ends A and B of a light horizontal rod AB., A vertical downwards force F is applied at point, C on the rod. AB will remain horizontal in, equilibrium if the distance AC is :, , m, , m, , m, , k1, , 4., , 104, , (a) 5mv 2 (b) 4mv 2 (c) 2mv 2 (d) mv 2, A rolling body is connected with a trolley car, by a spring of stiffness k . It does not slide, and remains in equilibrium relative to the, accelerating trolley car. If the trolly car is, stopped after a time t = t0 :( the rolling body, touches the trolly), , k2, , x, C, , A, , (a), , l, 2, , lk1, , B, , lk 2, , (b) k + k (c) k, 1, 2, 1, , lk2, , (d) k + k, 1, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 7., , 8., , 9., , ROTATIONAL DYNAMICS, , Let I be the moment of inertia of a uniform, square plate about an axis AB that passes, through its centre and is parallel to two of its, sides. CD is a line in the plane of the plate, that passes through the centre of the plate and, makes an angle θ with AB. Then the moment, of inertia of the plate about the axis CD is equal, to :, (a) I (b) I sin 2 θ (c) I cos 2 θ (d) I cos2 (θ / 2 ), Two point masses A of mass M and B of mass, 4M are fixed at the ends of a rod of length l, and of negligible mass. The rod is set rotation, about an axis perpendicular to its length with, a uniform anguular speed. The work required, for rotating the rod will be minimum when the, distance of axis of rotation from the mass A is, at, 2, 8, 4, l, (b) l, (c) l, (d), (a) l, 5, 5, 5, 5, A spool of mass M and internal and external, radii R and 2R hanging from a rope touches a, curved surface, as shown. A block of mass m, plaed on a rough surface inclined at an angle, α with horizontal is attached with other end of, the rope. The pulley is massless and system, is in equilibrium. Find the coefficient of friction, , 11. A ring of mass M and radius R lies in x-y, plane with its centre at origin as shown. The, mass distribution of rings is non-uniform such, that at any point P on the ring, the mass per, unit length is given by λ = λ0 cos 2 θ ( where, λ0 is a positive constant). Then the moment, of inertia of the ring about z- axis is, y, M, , P, , R, θ, , 1, MR 2, 2, , x, , MR, , MR, , (d) 5λ, , 12. As shown in figure, the hinges A and B hold a, uniform 400 N door in place. the upper hinge, supports the entire weight of the door. find the, resultant force exerted on the door at the, h, , where h is, 2, the distance between the hinges., , hinges . the width of the door is, , y, , M, A, , 3mg sin α + 2 Mg cos α, , α, , (b), , (c) 2λ, , 3mg + 2 Mg, (a) 3mg − 2 Mg, , m, , (a) MR 2, , 400N, , (b) 3mg cos α − 2Mg sin α, , B, , 3mg cos α + 2 Mg sin α, , (c) 3mg sin α − 2Mg cos α, 3mg + 2 Mg tan α, , (d) 3mg − 2Mg tan α, 10. A ring of mass m and radius R is rolling down, on a rough inclined plane of angle θ with, horizontal. Plot the angular momentum of the, ring about the point of contact of ring and the, plane as a function of time., 1) L, , 2) L, , a), , (a) 312 N (b) 280 N (c) 412 N (d) 480 N, 13. A thin wire of lenght L is is bent into a circular, wire of uniform linear density ρ . When, circular wire is in a vertical plane find the, moment of inertia of loop about an axis, BC,pasing through centre of the loop and, which makes an angle θ with the tangent at, the topmost point of the loop, A, , b), t, , 3) L, , θ, , B, θ, , t, , 4) L, , C, , c), , d), t, , NARAYANAGROUP, , t, , ρ L3, a), 8π 2, , b), , ρ L3, 2 2π 2, , ρ L3, c), 4π 2, , ρ L3, d), 3π 2, 105
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, 14. A box of mass 1 kg is mounted with two, cylinders each of mass 1kg, moment of, inertia0.5kg m 2 and radius 1m as shown in, figure, Cylinders are mounted on their control, axis of rotation and this system is placed on a, rough horizontal surface, the rear cylinder is, connected to battery operated motor which, provides a torque of 100n-m to this vcylinder, via a belt as shown. if sufficient friction is, present between cylinder and horizontal, surface for pure rolling, find acceleration of, m, . ( Neglect mass of motor,,, s2, belt and other accessories of vehicle)., , the vehicle in, , Electric meter, m, , 17. A uniform rod AB of length three times the, radius of a hemisphered bowl remains in, equilibrium in the bowl as shown. Neglecting, friction find the inclination of the rod with the, horizontal., r, θ, 3r, , A, , (a) sin −1 (0.92), (b) cos−1 (0.92), (c) cos −1 (0.49), (d) tan −1 (0.92), 18. A particle of mass m is released from rest at, point A in the figure falling freely under gravity, parallel to the vertical Y-axis. the magnitude, of angular momentum of particle about point, O when it reaches B is, ( whereOA=b and AB=h), O, , m, m, m, m, (b) 10 2 (c) 25 2 (d) 30 2, 2, s, s, s, s, 15. Two identical rings Aand Bare acted upon by, , 1, times the radius from the centre, 2, of the ring. if the angular acceleration of the, rings is the same, then, (a) τ A = τ B (b) τ A > τ B (c) τ A < τ B, distance, , (d) Nothing can be said about τ A and τ B as data, are insufficient, 16. A uniform plank of weight W and total length, 2L is placed as shown in figure with its ends in, contact with the inclined planes. the angle.of, friction is 150 . determine the maximum value, of the angle a at which slipping impends., L, α, , (a) 18.10, 106, , A, , h, , Y, , B, , mh, (b) mb 2 gh (c) mb 3 gh (d) 2mb gh, bg, 19. The end B of the rod AB which makes an angle, θ with the floor is being pulled with a constant, (a), , velocity V0 as shown in the figure. The length, of the rod is l . At the instant when θ = 370, , Y, A, , l, V0, , θ, O, , X, , B, , 2, (a) Velocity of end A is V0 downwards, 3, , W, L, , 60°, , b, θ, , (a) 20, , torques τ A and τ B respectively.A is rotating, about an axis passing through the centre of, mass and perpendicular to the plane of the, ring. B is rotating about a chord at a, , B, , C, , 5 V0, 3 l, (c) angular velocity of rod is constant, (d) velocity of end A is constant, , (b) angular velocity of rod is, 45°, , (b) 48.40 (c) 36.2 0, , (d) 88.80, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , 20. A block having equilateral triangular crosssection of side a and mass m is placed on a, rough inclined surface, so that it remains in, equilibrium as shown in figure. The torque of, normal force acting on the block about its, centre of mass is, , 23. A uniform rod oflength l is released from the, position shown in the figure. The acceleration, due to gravity is g . There is no friction at any, surfae. Find the intial angular acceleration of, the rod., , l, 60°, , θ, (a), , (a), , 3, mga sin θ, 2, , 1, (b) 2 3 mga sin θ, , 1, (c) 2 3 mga cos θ (d) Zero, 21. A thin horizontal uniform rod AB of mass m, and and length l can rotate freely about a, vertical axis passing thorough its end A. At a, certian moment the end B starts experiencing, a constant force F which is always, perpendicular to the original position of the, stationary rod and directed in a horizontal, plane. The angular velocity counted relative, to the intial position is, , 30°, , 3 3g, 5 3g, 3 3g, 5 3g, (b), (c), (d), 10l, 7l, 11l, 19l, , 24. Consider an arrangement shown in the figure., The pulley P is frictionless and the threads are, massless. The mass of the spools is m and, moment of inertia of the spool is, , 1, mR 2 . The, 2, , mass of the disc of radius R is also m. The, surface below the spool is rough to ensure, pure rolling of spool. The mass of the block is, m and the surface below the block is smooth., Find the initial acceleration of the block when, the system is released from rest., P, , R, R/2, , Spool, , B, m, , (a), , 6F, sin φ, ml, , 6F, cos φ, ml, , (b), , 8F, 8F, sin φ, cos φ, (d), ml, ml, 22. Ablock of mass m moves on a horizontal circle, against the wall of a cylindrical room of radius, R. The floor of the room, on which the block, moves, is smooth but the friction coefficient, between the wall and the block is µ . The block, , R, Disc, , Rough, , Smooth, , (c), , is given an initial speed V0 . The power, developed by the resultant force acting on the, block as a function of distance travelled s is, , (a), , 4, g, 37, , (b), , 2, g, 37, , (c), , 8, 10, g (d), g, 37, 37, , 25. Find the moment of inertia of a hemisphere of, mass M and radius R shown in the figure,, about an axis AA' tangential to the, hemisphere., A, , 9 , 2, mR, 20 , , (a) I = , , R, M, , (a), , µ m 03, e, R, , (c), , µ m V 03, R, , −3 s, µ, , NARAYANAGROUP, , (b) −, (d), , µmV03 −3Rµs, e, R, , µ mV03 −3Rµs, e, R, , A', , 7 , 2, mR, 20 , , (c) I = , , 13 , 2, mR, 20 , , (b) I = , , 3 , 2, mR, 20 , , (d) I = , , 107
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , MULTIPLE ANSWER QUESTIONS, 26. A wheel rolls purely between a rough horizontal, surface below it and a horizontal plank above, it under the action of a horizontal force F, uur, uur, applied on the plank. If at any time v p and vc, represent velocity of plank and velocity of, centre of mass of, wheel and, uur, uur, a p and ac represent acceleration of plank, and acceleration of centre of mass of wheel, repectively then which of the following is/are, correct., , (a) The rod is in translational and rotational, equilibrium., (b) The rod is in rotational equilibrium only., (c) The magnitude of the froce exerted by the rod, on the pivot is 503N, (d) The rod is in rotational equilibrium about P only, 29. A light rod of length 4 m can be maintained in, equilibrium position as shown in the figure if, we apply single force on it., Y, 4m, X, , Vp, , F, , 37°, 2kg, , Vc, , uur, uur, (a) v p = 2 vc, uur, uur, (c) v p = vb, , uur, uur, (b) a p = 2 ac, uur, uur, (d) a p = ac, , 27. A small block of mass m is released from rest, from position A inside a smooth hemispherical, bowl of radius R as shown in figure. Choose, the wrong option(s), , 5kg, , The required force, (a) would have magnitude of 77, (b) Would have a line of action making an angle of, tan −1 (17 / 9 ) with negative x- axis, (c) would be appiled at a distance of, , A, , R, , B, , (a) Acceleration to block is constant throughout, (b) Acceleration of block is g at A, (c) Acceleration of block is 3g at B, (d) Acceleration of block is 2g at A, 28. Consider a uniform rod of mass 40 kg and, length 8m, pivoted about a point P 3m from, one end as shown in the figure. Few external, forces are acting on the rod as shown in figure., 3m, , 50 N, ∆, P, , 108, , m1, , m2, , F, F, , (b) a1 = m, , (a) a2 = 0, , 1, , 20 N, , mark out the correct statement (s)., Take g = m / s 2 , , the right end, (d) the rod cannot be maintained in equilibrium, under the action of a single force., 30. Two particles of masses m1 and m2 aree, connected with a rigid rod of length l . If a force, F acts perpendicular to the rod then (a1 & a2, are instantaneous acceleration of m 1 & m2), , 5m, , 200 N, , 100 N, , 48, m from, 17, , F, , (c) aCM = m + m, 1, 2, , (d) α =, , F ( m1 + m2 ), m1 m2 l, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , 31. A uniform solid sphere of mass m is placed on, a sheet of paper on a horizontal surface. The, coefficient of friction between paper and, sphere is µ . If the paper is pulled horizontally, with an acceleration, , D, , P, , B, S, , C, P, , a, , (a) the tension in the string is equal to mg sin θ, (b) force acting on the cylinder is, , mg sin θ, 2, , (c) tension in the string is equal to, , mg sin θ, 2, , (d) frictional force acting on the cylinder is zero, 32. A rigid body is in pure rotation, that is,, undergoing fixed axis rotation. Then which of, the following statement(s) are true, (a) You can find two points in the body in a plane, perpendicular to the axis of rotation, having same velocity, (b) You can find two points in the body in a plane, perpendicular to the axis of rotation having same, acceleration, (c) Speed of all the particles lying on the curved, surface of a cylinder whose axis coincides with the, axis of rotation is same, (d) Angular speed of the body is same as seen, from any point in the body, 33. A rough disc of mas m rotates freely with an, angular velocity ω . If another rough disc of, m, and same radius but spinning in, 2, opposite sense with angular speed ω is kept, , mass, , on the first disc. Then:, (a) the final angular speed of the dise is ω3, (b) the net work done by friction is zero, (c) the friction does a positive work on the lighter, disc, −mR ω, 3, 2, , (d) the net work done by friction is, , 2, , 34. In the figure, the disc D does not slip on the, surface S, the pulley P has mass and the string, does not slip on it. The string is wound around, the disc., NARAYANAGROUP, , (a) The acceleration of the block B is double the, acceleration of the centre of D, (b) The force of friction exerted by D on S acts to, the left, (c) The horizontal and the vertical sections of the, string has the same tension, (d) The sum of the kinetic energies of D and B is, less than the loss in the potential energy of B as it, moves down, 35. A triangular block ABC of mass m and side 2a, lies on a smooth horizontal plane is shown., There point masses of mass m each strikes, the block at A, B and C with speed as shown., After the collision the particle come to rest., Then:, , m, , A, , Y, , v, 60°, , X, m, , 60°, B, , 60°, , v, C, , v, m, , (a) the centre of mass of ∆ABC remains, stationary after collision, (b) the centre of mass of ∆ABC moves with a, velocity v along x- axis after collision, (c) the triangular block rotates with an angular, velocity ω =, , 2 3mva, about its centriod axis, I, , perpendicular to its plane, (d) a point lying at a distance of, , , 1, , , from, 2 3ma , , centroid G on perpendicular bisector of BC is at, rest just after collision, 109
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, 36. A rod leans against a stationary cylindrical, body as shwon in figure, and its right end slides, to the right on the floor with a constant speed, v. Choose the correct option(s), α, , ω, , R, θ, , R, X, , (a) the angular speed ω is, , − Rv 2 ( 2 x 2 − R 2 ), x2 ( x2 − R2 ), Rv, , (b) the angular acceleration α is, (c) the angular speed ω is, , 3/2, , x x2 − R2, Rv, , x x2 − R2, , (d) the angular acceleration α is, , − Rv 2 ( 2 x 2 − R 2 ), x2 ( x2 − R2 ), , 3/2, , 7, 11ρ0 Al 3, mgl, I, =, (d), 39, 36, 39. The torque τ on a body about a given point is, found to be equal to A x L where A is a constant, vector, and L is the angular momentum of the, body about that point. From this it follows that, dL, is perpendicular to L at all instants of time, (a), dt, (b) the component of L in the direction of A does, not change with time, (c) the magnitude of L does not change with time, (d) L does not change with time, 40. Consider a sphere of mass ‘m’ radius ‘R’ doing, pure rolling motion on a rough surface having, r, velocity v 0 as shown in the Figure. It makes, an elastic impact with the smooth wall and, moves back and starts pure rolling after some, time again., , (c) KE =, , 37. The uniform 120 N board shown in figure is, supported by two ropes. A 400 N weight is, suspended one-fourth of the way from the left, end. Choose the correct options, T2, , θ, , 30° T, 3, 0.25L, , 0.75L, , O, , 400N, , (b) T2 = 371N, (a) T1 = 185 N, (c) T2 = 185 N, (d) tan θ = 0.257, 38. The KE and moment of inertia about the given, end point of a rod of mass m and length l and, cross sectional area A which is rotating with, , g, as shown in the Fig. will be [ density, l, x, of the rod varies as ρ = ρ 0 1 + , x is the, l, , distance measured from O) ], ω=, , ω=, , g, l, , O, , (a) KE =, 110, , 7, mgl, 36, , V0, , (b) I =, , 7 ρ0 Al 3, 36, , (a) Change in angular momentum about ‘O’ in the, entirem otion equals2m v0R in magnitude., (b) Moment of impulse provided by the wall during, impact about O equals 2mv0R in magnitude., 3r, (c) Final velocity of ball will be v 0, 7, 3r, (d) Final velocity of ball will be – v 0, 7, 41. If a cylinder is rolling down a rough inclined, with initial sliding., (a) after some time it may start pure rolling, (b) after sometime it must start pure rolling, (c) it may be possible that it will never start pure, rolling, (d) cannot conclude anything, 42. Which of the following statements are correct., (a) friction acting on a cylinder without sliding on, an inclined surface is always upward along the, incline irrespective of any external force acting on, it., (b) friction acting on a cylinder without sliding on, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , an inclined surface is may be upward may be, downwards depending on the external force acting, on it., (c) friction acting on a cylinder rolling without sliding, may be zero depending on the external force acting, on it., (d) nothing can be said exactly about it as it depends, on the friction coefficient on inclined plane., , COMPREHENSION TYPE QUESTIONS, Passage - I : (43-45), A small particle of mass m is given an initial velocity, v0 tangent to the horizontal rim of a smooth cone at, a radius r0 from the vertical centerline as shown at, point A. As the particle slides to point B, a vertical, distance h below A and a distance r from the vertical, centerline, its velocity v makes an angle θ with the, horizontal tangent to the cone through B., 43. The value of θ is, r0, D, h, , A, α, , r, , 2gr0, gr0, gr0, 4gr0, (b), (c), (d), ., tan α, 2 tan α, tan α, tan α, , (a), , Passage - II : (46-48), A rod AB of mass 3m and length 4a is falling freely, in a horizontal position and c is a point distant a, from A. When the speed of the rod is u, the point c, collides with a particle of mass m which is moving, vertically upwards with speed u. If the impact, between the particle and the rod is perfectly elastic, find, C, , G, , A, , B, u, , l, 2, , 3mg, , 46. The velocity of the particle immediately after, the impact, (a), , 29, u down, 19, , (b), , 19, u down, 29, , 29, 27, u,up, u down, (d), 19, 19, 47. The angular velocity of the rod immediately, after the impact, , (c), , B, , α, , 19u, 12u, 29u, 19u, (b), (c), (d), 12a, 19a, 19a, 29a, 48. The speed of B immediately after the impact, is, , (a), , (a), , cos −1, , v0 r0, (r0 − h tan α ) v20 + 2gh, , (a), , vr, , −1, 0 0, (b) cos ( r + h tan α ) v 2 + 2 gh, 0, 0, , (c), , (d), , cos −1, , v0 r0, , ( r0 − h tan α ), , cos −1, , (b), , 19, u up, 27, , 27, 27, u down, u up, (d), 19, 19, Passage - III : (49-50), An uniform rod of mass m=30kg and length, l=0.80m is free to rotate about a horizontal axis O, passing through its centre. A particle P of mass, M=11.2kg falls vertically through a height, , (c), , v02 − 2 gh, , v0 r0, r0 v02 + 2gh, , 44. The speed of particle at point B, (a) v 20 + 2gh, , (b), , v 20 − 2gh, , (c) v 20 + gh, , (d), , 2v02 + 2gh, , 45. The minimum value of v0 for which particle will, be moving in a horizontal circle of radius r0., NARAYANAGROUP, , 19, u down, 27, , h=, , 36, m and collides elastically with the rod at a, 245, , distance, , l, from O. At the instant of collision the, 4, 111
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, rod was stationary and was at an angle α = 37 0, with horizontal as shown in figure, , 37°, , (a) µ g, , 0, l, 4, , K, M, , 2K, M, , (b), , ( g = 10ms −2 ), 7 −1, 9 −1, ms, (b) 7ms −1 (c) ms (d) 1ms −1, 9, 7, Passage - IV : (51-53), A uniform thin cylinder M and radius R is attached, to two identical massless springs of spring constant, K, which are fixed to the wall, as shown. The spring, are attached to the axle of the disc symmetrically, on either side at distance d from its centre. The, axle is massless and both the springs and the axle, are in horizontal plane. The unstretched length of, each spring is L. The disc is initially at its equilibrium, position with its centre of mass (CM) at a distance, L from the wall. The disc rolls without slipping with, ur, velocity V 0 = V0iˆ . The coefficient of friction is µ, , (a), , M, K, , 3M, 5M, (d) µ g, K, 2K, Passage - V : (54-56), A wheel of radius R, mass m with an axle of radius, r is placed on a horizontal surface. Its moment of, inertia is I = mR 2 .Unwinding a rope from its axel, a force F is applied to pull it along a horizontal, surface. The friction is sufficient enough for its pure, rolling ( ∠θ = 00 ), wheel, , F, θ, R, r, , Axle, , 54. Find the linear acceleration of the wheel, , F ( I / m ) − Rr , , (c), d, , V0, , F ( 2 I / m ) − 2 Rr , ( I / m ) + r , 2, , (b), , 2 F ( I / m ) − Rr , ( I / m ) + r 2 , , (, , ), , F I 2 / m − Rr , , , (d), 2, ( I / m ) + r , , 55. Find the condition for which frictional force acts, in backward direction, , R, X, L, , 51. The net external force acting on the disc when, its CM is at displacement x with respect to, its equilibrium position is, 2 Kx, 4 Kx, (a) − Kx (b) −2Kx (c) −, (d) −, 3, 3, 52. The centre of mass of the disc undergoes SHM, with angular velocity ω , equal to, 112, , M, 2K, , (b) µ g, , (a) I / m + r 2 , ) , (, , 2d, , 4K, 3M, , (d), , (c) µ g, , l, 4, , 49. Calculate angular velocity of the rod just after, collision is, (a) 1 rad/s (b) 3 rad/s (c) 2 rad/s (d) 4 rad/s, 50. Velocity of particle P after collision is, , Y, , 2K, 3M, , (c), , 53. The maximum value of V0 for which the disc, will roll without slipping., , h, , l, 2, , (a), , (a) ( I / m ) > Rr, , (b) ( 2 I / m ) > Rr, , I 2, I , (c) m > Rr, (d) , > Rr, m 2, , , 56. Find the condition for which frictional force acts, in forward direction, , (a) ( I / m ) < Rr, , (b) ( 2 I / m ) < Rr, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, I 2, (c) m < Rr, , , , ROTATIONAL DYNAMICS, , I , (d) , < Rr, m 2, , Passage - VI : (57-59), Consider a cylinder of mass M = 1kg and radius, R=1 m lying on a rough horizontal plane. It has a, plank lying on its stop as shown in the figure., m = 1kg, , 60°, A, , M, B, , A force F = 55 N is applied on the plank such that, the plank moves and causes the cylinder to roll., The plank always remains horizontal. There is no, slipping at any point of contact., 57. The acceleration of cylinder is, (a) 20 m/s2 (b) 10 m/s2 (c) 5 m/s2 (d) 12 m/s², 58. The value of frictional force at A is, (a) 7.5 N (b) 5.0 N (c) 2.5 N (d) 1.5 N, 59. The value of frictional force at B is, (a) 7.5 N (b) 5.0 N (c) 2.5 N (d) 1.5 N, Passage - VII :(60-62), A cabin is falling freely and inside thecabin a disc, of mass M and radius R is made to undergo, uniform pure rolling motion with the help of some, external agent. Inside the cabin wind is blowing in, horizontal direction which imparts an acceleration, a to all the objects present in cabin in horizontal, direction. [Disc still performs uniform pure rolling, motion]. A very small particle gets separated from, disc from point P and after some time it passes, through the centre of disc O. Based on above, information, answer the following questions:, , P, , g, , 37°, , 60. The time taken by particle to reah from, P to O is, , NARAYANAGROUP, , 4 15 R, 3 8a, , (c) 3, , (b) 4, , 6R, 7a, , (d), , 6R, 4a, , 3 15 R, 4 8a, , 61. The angular velocity of disc is, 1 7a, 8a, 4 7a, 16, 8a, (a) ×, (b), (c) ×, (d) ×, 3 6R, 15R, 9 6R, 9, 15R, 62. The revolution made by disc in time interval, computed in Q.No. (i) is, , (a) 8, , R, , 0, , (a), , (b), , 6, 5p, , (c), , 5p, 2, (d), 6, 3p, , Passage - VIII :(63-65), A disc of a mass M and radius R can rotate freely, in vertical plane about a horizontal axis at O. distant, r from the centre of disc as shown in the figure., The disc is relased from rest in the shown position., , M,R, 0, , C, , 63. The angular acceleration of disc when OC, rotates by an angle of 370 , is, , 8rg, (a) 5 R 2 + 2r 2 , , , , 5rg, (b) 4 R 2 + 2r 2 , , , , 10rg, (c) 3 R 2 + 2r 2 , , , , (d), , 8rg, 5R 2, , 64. The angular velocity of disc in above described, case is, (a), , 8 gr, 5 R 2 + 2r 2 , , (b), , 6 gr, 5 R 2 + 2r 2 , , (c), , 12 gr, 5 R 2 + 2r 2 , , (d), , 12 gr, 5R 2, , 65. Reaction force exerted by hinge on disc at this, instant is, Mg, 2, 2, (a) 5 ( R + 2r 2 ) × g ( R 2 + 6r 2 ) + ( 4 R 2 ), 2, , 113
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, Mg, 2, 2, 2, (b) 5 ( R + 2 r 2 ) × 3 ( R + 6r ), , 4Mg, Mg, 2, ×, R, 4R2, 2, 2, 2, (c) 5 ( R + 2r ), (d) 5 ( R + 2r 2 ) ×, , MATRIX MATCHING TYPE QUESTIONS, , applied on cylinder at different positions with, respect to its centre O in each of four, situations of column-1, due to which magnitude, of acceleration of centre of mass of cylinder, is ‘a’ Match the appropriate results in columnII for conditions of columnI, Column-I, , L, as shown in the figure. The right support is, 2, , F, , F, , 66. A rod of length L and weight w is kept in, equilibrium on the two support separated by, , R/2, , R, O, , (a), , taken out at time t = 0., Match the following questions based on the, above information, , F, , O, , (b), , O, , (c), , O, , (d), , w,L, , R/2, F, , Column-II, (p) Friction force on cylinder will not zero, L, 2, , F, F, r) a ≠, m, m, s) friction force acting on cylinder is zero, 68. Column I, Column II, (Object), (Moment of inertia), , q) a =, , Column I, Column II, (a) The moment, (p) 3g/7, of inertia of the rod, about the support point at t = 0 is, (b) The angular, , (q), , 12 g, 7L, , acceleration of rod about, the support point at t = 0 is, (c) The linear, , (r), , (a) Uniform rod, M, , 30°, l=R, , 4ω, 7, , (b) Uniform semicircular ring., , acceleration of centre, of mass of rod at t = 0 is, (d) The normal, , 8MR 2, p), 11, , 7 ω L2, (s) 48 g, , q), , MR 2, 12, , Axis is perpendicular, to plane of ring, M, , reaction on the rod, by the support at t = 0 is, , ωL2, (t) 3 g, , 67. A uniform solid cylinder of mass m and radius, R is placed on a rough horizontal surface, where friction is sufficient to provide pure, rolling. A horizontal force of magnitude F is, 114, , [π = 22/7], , (c) Uniform triangular, , 13MR 2, r), 8, , plate of mass M, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , R, 60°, , ω0, , 60°, , v0, , R, , R, , (V0 > Rω0), , (C), A, , MR, 8, M from which circular Portion of radius, R is then removed M.I of remaining mass, about axis which is perpendicular to plane, of plate and passing through its centre, (d) Uniform disk of initial mass, , 2, , s), , R, 2R, , 69. In each situation of column-I, a uniform disc, of mass m and radius R rolls on a rough fixed, horizontal surface as shown. At t=0(initially), the angular velocity of disc is ωo and velocity, , m2, , (D), , α, , m1, , F, , Column-II, p) The angular momentum of disc about point A, (as shown in figure) remains conserved., q) The kinetic energy of disc after it starts rolling, without slipping is less than its initial kinetic energy., r) In the duration disc rolls with slipping, the, friction acts on disc towardsleft, s) Before rolling starts acceleration of the disc, remain constant in magnitude and direction., t) Final angular velocity is independent of friction, coefficient between disc and the surface., , of centre of mass of disc is V0 (in horizontal, , INTEGER ANSWER TYPE QUESTIONS, , direction). The relation between V0 and ω0 for, each situation and also initial sense of, rotation is given for each situation in columnI. Then match the statements in column-I with, the corresponding results in column-II, Column-I, , 70. A plank of mass m1 with a uniform solid sphere, of mass m2 placed on it rests and a force F is, applied to the plank. The acceleration of the, plank provided there is no sliding between, the plank and the sphere is, , F, n, m2, 7, , m1 +, , then the, , value of n is, , ω0, v0, , (V0 > Rω0), , (A), A, , ω0, v0, (B), , A, , NARAYANAGROUP, , (V0 > Rω0), , 71. A uniform cylinder of radius r is rotating about, its axis at the angular velocity ω0 . It is now, placed into a corner as shown in figure. The, coefficient of friction between the wall and the, cylinder as well as the ground and the cylinder, is µ . The number of turns, the cylinder, completes before it stops, are given by, ω20 r 1 + µ 2 , , , the value of n is, nπg µ(1 + µ) , , 115
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, WALL, , 72. The pulley shown in figure, has a radius 10 cm, and moment of inertia 0.5 kg-m2 about its axis., Assuming the inclined planes to be frictionless,, the acceleration of the 4.0 kg block is, , 1, n, , that, , value of n is, , 74. A ball of radius R=20cm has mass m=0.75kg, and moment of inertia (about its diameter), I = 0.0125 kgm 2 . The ball rolls without sliding, over a rough horizontal floor with velocity, V0 = 10 ms −1 towards a smooth vertical wall. If, coefficient of restitution between the wall and, the ball is e=0.7, velocity V of the ball long, after the collision is ( g = 10 ms −2 ), 75. A uniform square plate of mass ‘m’ is, supported as shown. If the cable suddenly, breaks, assuming centre of mass is on, horizontal ine passing through A determine ;, The reaction at A is, , 2.0kg, , mg, that n is, n, , 4.0kg, , B, 45°, , 45°, , 73. In the arrangement shown in figure, ABC is a, straight, light and rigid rod of length 90cm. End, A is pivoted so that the rod can rotate freely, about it, in vertical plane. A pulley, having, internal and external radii R=7.5cm and r=5cm, is fixed to a shaft of radius 5cm. The pulley shaft system can rotate about a fixed, horizontal axis O. B is point of contact of the, pulley and the rod. From free end C of the rod, a mass m2 = 2kg is suspended by a thread., Another thread is wound over the shaft and a, block of mass m1 = 4kg is suspended from it., If coefficient of friction between the rod and, the pulley surface is µ = 0.4 and moment of, inertia of pulley-shaft system about axis O is, I = 0.045 kg − m2 , the acceleration of block, , A, C, , b, , b, , 76. In the figure shown there is a fixed wedge ‘W’, of inclination θ . A is a block, B is a disc and, ‘C’ is a solid cylinder. A, B and C each has, mass ‘m’. Assuming there is no sliding, anywhere and string to be of negligible mass, find :, The friction force acting on the cylinder due, to the wedge is, , mg, (1 + n sin θ ) that n is, 15, B, , m1 , when the system is released, ( g = 10 ms −2 ) is, A, , B 30cm C, , 60cm, , C, A, , R, , W, fixed, , O, r, , θ, , m1, , 116, , m2, , 77. In the figure shown a uniform ringh of mass, m is placed on arough horizontal fixed surface., The coefficient of friction between left half of, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , ring and table is µ1 whereas between right half, and table is µ2 at the moment shown. The ring, has angular velcoity in clockwise sense in the, figure shown. At this moment find the, magnitude of acceleration, , ( in, , m3, , 48kg , then m is, 2, , m / s2 ) of, , m4, m3, , centre C of ring. [Given g = 10 m / s 2 ], µ1, , m2, , 80. An isosceles right triangular plate ABC of, mass m is free to rotate in vertical plane about, a fixed horizontal axis through A. It is, supported by a string such that the side AB is, horizontal. The reaction at the support A is, , µ2, , A, , m1, , C, , p ( mg ), , thus p + q = − ., , q, 78. In the given diagram a sphere of mass m and, radius R is rolling without slipping on a rough, , l, , A, , inclined surface of inclination ( p / 6 ) . Centre, of mass of sphere is at C which is, , R, 3, , l, , distiance from centre in a direction parallel to, inclined plane. Moment of a intertia of the, sphere about point of contact is I 0 (given). At, the given instant sphere is rotating with, constant velcity ? 0 . Calculate the angular, accel eration of sphere at this instant to near, est integer?, ω0, , m, R, R 3, R, , No slipping, , C, Centre of, mass, 30°, , [Given that m = 2kg , R = 0.5m, , g = 10m / s 2 ,, , ? 02 = 3 in, , SI unit, , and, , I 0 = 10kg − m 2 ], 79. Figure shows an arrangement of masses, hanging from a ceilling. In equilibrium each, rod is horizontal, has negligible mass and, extends three times as far to the right of the, wire supporting is as to the left. If mass m4 is, NARAYANAGROUP, , C, , 81. The densities of two solid spheres A and B of, the same radii R vary with radial distance r as, r, r, ρA (r ) = k , ρB ( r ) = k ,, and, R, , R, respectively, where k is a constant. The, moments of inertial of the individual spheres, about axes passing through their centres are, 5, , IA n, I A and I B , respectively. If I = 10 , the value, B, of the n is, 82. A horizontal circular platform of radius 0.5 m, and mass 0.45 kg is free to rotate about its, axis. Two massless spring toy-guns, each, carrying a steel ball of mass 0.05 kg are, attached to the platform at a distance 0.25 m, from the centre on its either sides along its, diameter (see figure). Each gun, simultaneously fires the balls horizontally and, perpendicular to the diameter in opposite, directions. After leaving the platform, the balls, 117
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, have horizontal speed of 9 ms −1 with respect, to the ground. The rotational speed of the, platform in rads-1 after the balls leave the, platform is, , MULTIPLE ANSWER QUESTIONS, 26. A,B, 27. A,C, 28. A,C, 29. A,B,C 30. A,B,C 31. B,D, 32. C,D, 33. A,D, 34. A,B,D, 35. B,C, 36. C,D, 37. A,B,D, 38. A,B, 39. A,B,C 40. A,B,D, 41. A,C, 42. B,C, COMPREHENSION QUESTIONS, 43. A 44. A 45. C 46. A 47. B 48. C 49. D, 50. C 51. D 52. D 53. C 54. A 55. A 56. A, 57. B 58. A 59. C 60. A 61. B 62. D, 63. A 64. C 65. A, MATRIX MATCHING TYPE, 66. A → s ; B → q ; C → p ; D → r, 67. A → p ; B → q , s ; C → p , r ; D → p , r, 68. A → q; B → p; C → s; D → r, 69. A → p, q, r; B → p, q, r; C → p, q; D → p, q, r, INTEGER TYPE QUESTIONS, 70. 2 71. 8 72. 4 73. 1 74. 2 75. 4 76. 7, 77. 4 78. 1 79. 4 80. 5 81. 6 82. 4 83. D, , 83. A ring of mass M and radius R is rotating with, angular speed ω about a fixed vertical axis, passing through its centre O with two point, M, and rest at O. These, 8, masses can move radially outwards along two, maseless rods fixed on the ring as shown in, the figure. At some instant the angular speed, , masses each of mass, , of the system is, , 8, ω and one of the masses is, 9, , 3, R from O. At this instant, 5, the distance of the other mass from O is, (JEE_ADV-15), , LEVEL-VI - HINTS, SINGLE ANSWER TYPE, , at a distance of, , 1., , 2mg-T=ma ; TR = I α, α=, , 2., , a, g, ; solving α =, R, 3, , m1 g sin θ = Fpseudo ........(1), , m1 g sin θ − f = m1 g sin θ ⇒ f = 0, 3., O, , 1 2, m 4v +, 2 , , ( 2v ) + ( 2v ), 2, , 2, , + 02 +, , , 1 2 1 2 1, mv + I ω = m 4v 2 + 4v 2 + mv 2 = 5mv 2, 2, 2, 2, V, 2V, 2v, , V, , V, , 2v, , LEVEL-VI - KEY, SINGLE ANSWER QUESTIONS, 1. D 2. C 3. A 4.B 5. C 6. D 7.A, 8.C 9.B 10.D 11.A 12.C 13.A 14.A, 15.A 16.C 17.B 18.B 19.B 20.B 21.A, 22.B 23.A 24.A 25.B, 118, , V, , 4., , 5., , V, , V, , f = 0 for t < t0 until it can stop no friction acts, because it neither slides nor rotates due to action, of the rolling, Let J be the impluse acting on the rod, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, J = mvcm ; Jx =, , 6., , ROTATIONAL DYNAMICS, , 1, ml 2ω . Since the end A is, 12, , l, J 12 Jx l, stationary VA = Vcm − ω = − 2 = 0, 2, m ml 2, Let x be the distance from the end of spring constant, K1. If y is the elongation of the rod, kasking, moments about the point C,, , ⇒ K1 yx = K 2 y ( L − x ), 7., 8., , I AB, , 1, = ml 2 = ICB ( irrespective an angle), 6, , 1 2, Iω, 2, Let x is the distance of CM from A, W=, , I = Mx 2 + 4 M × 2 ( l − x ) × ( −1) = 2Mx − 8Ml, dI, 4, = 10 − 8Ml = 0; x = l, dx, 5, , 9., , Torque about point of contact of the spool will be, zero, 2, Mg, 3, Equating the forces acting on the block along and, perpendicular to the incline, N + T sin α = mg cos α, T .3R = Mg .2 R or T =, , 11. Divide the ring into infinitely small lengths of mass, dm. Even though mass distribution is non- uniform,, each mass dm1 is at same distance R from origin, Q MI of ring about z − axis is, , = dm1 R 2 + dm2 R 2 + ......dmn R 2 = MR 2, 12. Only a horizontal force acts at hinge B, because, hinge A is assumed to support the door,s weight.Let, us take torques about A as axis., , ∑F, ∑F, , x, , = 0 or V − 400 N = 0, We find from these that, H = 100 N and V = 400 N, →, To find the resultant force R, on the hinge at A,, , we have R =, , + (100 ), , 2, , ) = 412 N, , Y, θ, , 90°, , X, , X', , C, Y', , Now I BC + I B ' C ' = I 0 = IZ = MR 2, , MR 2, 1, L2, ρ L3, =, = ( ρ L) 2 = 2, 2, 2, 4π, 8π, , 14. For whole sytem f1 − f 2 = 3 (1) a.... (1), , T α, , For rear cylinder 100 − f1 = 0.5 ( a ) .... ( 2 ), , mg, , For rear cylinder f 2 = 0.5 ( a ) ....( 3 ), , 3mg sin α + 2Mg cos α, 3mg cos α − 2Mg sin α, , 10. L = ( mg sin θ ) Rt ; Since, τ =, , 100 = 4a; a = 20m / s 2, ∆L, so, L = τ ( ∆t ), ∆t, , The curve between L and time t will be a straight, line., NARAYANAGROUP, , 2, , L = 2π R so R = L / 2π, 1, 1, I ' XX = MR 2 ; I 'YY = MR 2, 2, 2, I 0 = I Z = I ' XX + I 'YY = MR 2, , so I BC, , µ=, , ( 400, , 13. M = ρ L ;, , 2, 2, , , mg sin α + Mg cos α = µ mg cos α − Mg sin α , 3, 3, , , f, , or F2 − H = 0, , y, , 2, or N = mg cos α − Mg sin α, 3, Also mg sin α + T cos α = f = µ N, , N, , =0, , 15. I B ' ( in new given condition ), 2, , =, , 1, R , MR 2 + M , = MR 2 = I A, , 2, 2, 119
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JEE-ADV PHYSICS-VOL - III, dV − µ, ∫V V = R ∫0 dS, 0, V, , ⇒, , ROTATIONAL DYNAMICS, , S, , T2, p, a1, , V, −µ, ⇒ ln, =, (S), V0, R, ⇒ V = V0 e, , α2, , − − − − − (1), , Now power co n s umed by friction,, P = − f .V, = − µ NV, mV 2, = −µ, V, R, mV 3, = −µ, − − − − − − − ( 2), R, substitutethe value of V from eq (1) , we get, −3 µ S, , −µ m 3 R, P=, V0 e, R, 23. Suppose C is the point through which, theinstantaneous axis of rotation passes and G is, the centre of mass of the rod. From the geometry, of the figure, C, 30°, 30°, , l/2, , 60°, , G, mg, , co, s, , 30, , °, , NB, , B, , R/2, , a, B, , T1, , a2, , − µS, R, , R, , NA, , m, , α1, , T2, , f, , mg, , , , , Now a = a 1 R −, , R, , 2, , a1R, .....(i) ; a1 = a1R, 2, a2 = a1 ( 2 R ) + a 2 ( R ) .....(iii), , ⇒ a=, , ......(ii), , For the block ma = T1 ......(iv), For the spool ma1 = T2 − T1 − f .......(v), , mR 2 , , a1 = T2 R + T1R + fR, 2 , , .....(vi), , For the disc ma2 = mg − T2, , ........(vii), , 1, 2 2, mR a = T2 R, 2, , , ......(viii), , 4, g, 37, , For these equations, we get a =, , 25. Moment of inertia about an axis through, the flat face of hemisphere, , l/2, , I0, , 30°, , A, , IG, , I, , A, , CG = l cos30, The moment of inertia about C, 0, , I=, , 5, R, 8, , 2, ml, 5, + m ( l cos300 ) = ml 2, 12, 6, 2, , G, 3, R, 8, , If a is the angular acceleration, then, 5 2, l, 3 3g, 0, ml a = mg cos 30 ⇒ a =, 6, 2, , , , , 10l, 24. Suppose the acceleration of the block B is a,, acceleration of disc is a2 and the acceleration fo, centre of mass of spool is a1 , also suppose the, angular accelerations of spool and disc are a1 and, a 2 respectively.., , A', , I0 =, , 2, mR 2, 5, , From the parallel axis theorem, the moment of, inertial through c.m. of the hemisphere, 2, , NARAYANAGROUP, , 3 83 , 2, I0 = I0 − m R = , mR, 8 320 , , 121
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , vertical components of force exetred by pivot on, rod, respectively., , Using parallel axis theorem, Moment of inertia about, the axis AA' ,, , 13 , 5 , I = I G + m R ⇒ I = mR 2, 20 , 8 , , 200 N, , 50 N, Rx, , MUTLIPLE ANSWER QUESTIONS, 26. For pure rolling of wheel wrt horizontal surface, below it, For pure rolling of wheel wrt plank, , 100 N, , Rω, , Vc, , VP = Vc + Rω = 2Vc, , dv p, , ∴, , =, , 2dv c, , (105) + ( 480 ), , R =, , 2, , 2, , ; 503 N, , The rod will exert equal and opposite force on the, pivot., 29. For equilibrium of the rod, let us say force R is, appilied whose X and Y components are, RX and RY as shown in figure., , ω, , Vc = Rω, , 20 N, , RY = 400 + 100 − 20 = 480 N, , Then ,, , Vc +Rω, , 400 N, , RX = 200 − 50 = 150 N, , Vp, , Vc, , Ry, , R, , 2, , RX = 36 N , RY = 48 + 20 = 68 N, Ry, , ; a p = 2ac, , R, , X, , dt, dt, 27. Acceleration of block is not constantthrought., , 4-X, 36 N, , 2, , Acceleration of block at B is V / R where, , Rx, , V = 2 gR ., , 60 N, , 2, , 20 N, , 28. The rod is in translation equilibrium in any case as, it is privoted, now let us check for its rotational, equilibrium., For rotational equilibrium, the net torque acting, about any point must be zero, 3m 1m, 4m, , 200 N, , 50 N, , 48N, , For rotational equilibrium, x × 20 = 48 ×× ( 4 − x ), ⇒ x=, 30., , acm =, , 48, m ; So, R = 362 + 682 ; 77 N ., 17, , F, l, ; F × = Iα, m1 + m2, 2, , l, a1 = acm + α, 2, , 31. ma − f = ma0, , ∆, p, , fR = I α ; I =, , 100 N, 400 N, Let us take the torque about P., , ∑τ, , 20 N, , = 400 ×1 − 20 × 5 −100 × 3 = 0, , [ Taking clockwise as + ve, and anticlock wise as − ve ], So rod is in rotational equilibrium, also., If a body is in rotational equilibrium then, , ∑τ, , = 0 about any point., The force exterted by a pivot on rod maintains the, translational equilibrium in horizontal and vertucal, directions. Let Rx and Ry be the horizontal and, 122, , ext, , l, 2, , ; a2 = acm − α, , 2mR 2, ; a0 = Rα, 5, , 32. All points in the body, in plane perepndicualr to the, axis of rotation, revolve in concentric circles. All, points lying on the circle of same radius have same, speed (and also same magnitude of acceleration), but different directions of velocity (also different, direction of acceleration), Hence there cannot be two points in teh given plane, with same velocity or with same acceleration. As, mentioned above, points lying on circle of same, radius have same speed., Angular speed of body at any instant w.r.t any point, on the body is same by definition., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, , From energy-conservation,, , Y, , I? 2, = Mg × r sin 370, 2, O, X, , P, , Let us solve this situation wrt cabin, frame of reference. Acceleration of detached, particle wrt cabin in horizontal direction is a, (towards right) and is zero in vertical direction. Let, angular velcoty of disc be ? and velocity of its, centre of mass be v , then form pure rolling motion, condition, v − R? = 0 i.e., v = R? Initial, velocity of particle is,, →, , u p = ( v − R? cos37 ) i + ( R? sin 37, 0, , ^, , 0, , MR 2, ?2, 3, ⇒, + Mr 2 , = Mgr ×, 5, 2, 2, , ⇒, , ? =, , 12 gr, 5 R 2 + 2r 2 , , From FBD of disc,, , Rx − Mg sin 37 0 = Mar = M? 2 r, RY, RX, , )j, ^, , 37°, , O, , ω2r, , R? ^ 3R? ^, =, i+, j, 5, 5, , ar, , For required situation, position vector of final, location of particle is,, →, , r = ( R sin 37 + vt ) i + ( R cos37, ^, , 0, , 0, , )j, ^, , where t is the time taken by particle to, reach from P to O., , R cos37 0 =, , at, , Mg cos37 0 − Ry = Mat = Mra, ⇒ Rx =, , 3R?, 4, t⇒t =, 5, 3?, R?, 5, , 0, and R sin 37 + vt = , , 1 2, t + at, 2, , ?t, 2, =, 2p 3p, , R = Rx2 + Ry2 =, , Mg, 5 R 2 + 2r 2 , , , 2, 2 2, 2 2, g, R, +, 6, r, +, 4, R, (, ), (, ) , , , 66. (A) I P = I CM + Ma 2, O, C, , 37°, , C′, , From t = Ia, MR 2, , ⇒ Mg × r cos37 0 = , + Mr 2 a, 2, , , 8rg, ⇒a =, 5 R 2 + 2r 2 , 126, , Mg 4 R 2 , , , 5 R 2 + 2r 2 , , MATRIX MATCHING QUESTIONS, , 63-65, , O, , 3Mg R 2 + 6r 2 , , ,, 5 R 2 + 2r 2 , Ry =, , 4, 15 R, 8a, where v = R? ⇒ ? =, ,t= ×, 15 R, 3, 8a, Number of revolutions made in time t is,, , n=, , Mg, , =, , 7WL2, ML2 ML2, =, +, (S), 48 g, 12, 16, , L, 12 g, = I Pα ⇒ α =, (Q), 4, 7L, L, 3g, (C) a = α =, (P), 4, 7, 3 Mg 4W, =, (D) N = Mg −, (R), 7, 7, , (B) W ×, , 67. Assume friction to be absent and horizontal Force, is applied at a distance x above centre, a=, , F, mR 2, and Fx =, α, m, 2, , or Rα =, , 2Fx, mR, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , ROTATIONAL DYNAMICS, Now we substitute the value of a p from equation, (4) in equation (5), hence, F − m 2 ac = m 1 ( ac + αr ) … (6), , R, if a = Rα then x =, 2, , The fricton force will be zero and a =, , F, m, , R, ,Friction force is, 2, F, towards left anf a ≠, m, R, if a < Rα or x > Fricton force towards right, 2, F, and a ≠, m, 1 2, 1, 2, 2, 68. (a) I = ml × sin θ = ml, 3, 12, 2, b) I = I COM + mr ( centre of mass of the reaging, , 5, 2, , From equations (2) and (3) or , αr = ac, , If a > Rα or x <, , Substituting this value in equation (6), we have, or,, , ac =, , 71. The free body diagram is shown in the figure., 1, 2, , The initial energy = I ω20, where I is the moment of inertia of the cylinder, 1, 2, , and is given by I = Mr 2, (M = Mass of the cylinder and r = Radius), ∴ Initial Kinetic energy of cylinder, =, , … (1), R, , N, f2, Mg, , Here, there is no motion of the centre of gravity of, the cylinder, hence,, R + µN = Mg … (2) ; N = µR … (3), Solving for R and N,, R=, , Mg, 1, ( + µ2 ) … (4) ;, , N =, , µMg, , (1 + µ2 ), , … (5), , The total initial energy is dissipated against frictional, forces., , … (4), , m2, ac, , r ωo, 1, 2 2, ∴ Mr ω0 = ( µN + µR ) .2πn ; n = 8πg, 4, 2, , m1, f, , Substituting the value of f from equation (2) in, equation (1), we get, F − m 2ac = m 1a p, … (5), , 2, , where n is the number of turns made by the cylinder, before it stops., Putting the values of N and R, and solving for n, gives the final result.., , F, ap, , NARAYANAGROUP, , 1, Mr 2 ω02, 4, f1, , … (3), , a p = a c + αr, , F, , 7, , 70. The situation is as shown in the figure., Here we have,, F − f = m1 ap, … (1), and, f = m 2 ac, … (2), , and,, , or,, , F, 7, m 2 + m1, 2, , 7, , INTEGER TYPE, , Further,, , 7, 7, , , m 1ac or, F = m 2 + m 1 ac, 2, , , 2, , Now a p = ac + αr = 2 ac =, ., 2, m1 + m 2, , 2R, will be art r =, from the centre of the ring, π, 69. Angular momentum each conserved about the point, of contact with the ground.Angular mo mentum also, is conserved in all cases about any point on the line, passing through point of contact and parallel to the, velocity of the centre of mass. Then kinetic energy, decreases in all cases due to work done by the, friction. We have to calculate the relative velocity, of contact point and the direction of friction in A,B, and D towards the left. and in case of C, the friction, direction towards right. Those direction never be, changed in any given cases., , 2, f × r = I α = m 2 r 2α, 5, , F − m 2 ac =, , 72., , ( T2 − T1 ) R = Iα, , 127
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , GRAVITATION, SYNOPSIS, Kepler’s Laws :, Ø, , Kepler’s first law or law of orbits: Every planet, revolves around the sun in elliptical orbit with the, sun at one of its focii., , F1, , Ø, Ø, Ø, , F2, , rp, , Ø, Ø, Ø, , 2b, , 2a, As shown in the fig., sun may be at F 1 or F2 . Here, a and b denote the lengths of semi major and semi, minor axes., The nearest position of the planet from the sun is, called perihelion., The farthest position of the planet from the sun is, called aphelion., A planet of mass m is moving in an elliptical orbit, around the sun(S) of mass ‘M’, at one of its focii., , P, , GM 1 + e , 1− e , , and V p =, , , a 1− e , 1+ e , If e>1 and total energy (K.E +P.E)>0, the path of, the satellite is hyperbolic and it escapes from its, orbit., If e<1 and total energy is negative, it moves in an, elliptical path., If e=0 and total energy is negative, it moves in, circular path., If e = 1 and total energy is zero, it will take parabolic, path., The path of the projectile thrown to lower heights, is parabolic and thrown to greater heights is, elliptical., Kepler’s second law or Law of Areas: The, radius vector joining the planet to the sun sweeps, out equal areas in equal intervals of time., VA =, , Ø, Ø, Ø, , Ø, , rA, , M, S, , m, A, , O, , GM, a, , dA , Areal Velocity of radius vector joining the, dt , planet to sun remains constant. Mathematically, dA, = constant, dt, , 2b, , dl, dθ, , sun, , C, , Ø, , Ø, , SO, OA, , But A =, , c, ⇒ c = ea, a, , 1, 1, 1, ( dl ) r = ( rdθ ) r = r 2 dθ, 2, 2, 2, , So,, , d 1 2 , r dθ = constant, dt 2, , , Similarly ra = a + c = a + ea = a (1 + e ), , ⇒, , 1 2 dθ 1 2, r, = r ω = constant, 2 dt 2, , From conservation of angular momentum at A and, , 1 mr 2ω I ω, L, =, =, = constant, 2 m, 2 m 2m, L = constant, As the gravitational force on planet by sun is central,, torque is zero and hence angular momentum of the, planet is constant., , From fig, rp = a − c = a − ea = a (1 − e ), , ⇒, , P, we have mV p rp = mVA rA, Vp, VA, , Ø, , p, , a, , Eccentricity of the elliptical path e =, e=, , Ø, , r, , =, , rA 1 + e, =, rp 1 − e, , From conservation of energy, we have, , NARAYANAGROUP, , Ø, , 129
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, Ø, Ø, Ø, Ø, , This law is a consequence of law of conservation, of angular momentum., dA L mVr Vr, =, =, =, dt 2 m 2 m, 2, Areal velocity of radius vector of the planet is, independent of mass of the satellite., As angular momentum is conserved,, , m(Vmax )( rmin ) = m(Vmin )( rmax ), ⇒, , Ø, Ø, , V max, 1+ e, =, V min, 1− e, , Ø, , rmax + rmin (1 + e ) a + (1 − e ) a, =, =a, 2, 2, , Hence, T 2 ∝ a 3, where ‘ a ’ is length of semi major axis of ellipse, The gravitational force between the planet and the, Sun provides the necessary centripetal force for, the planet to go round the Sun., If M = mass of Sun, m = mass of planet and, r = average distance of the planet from the Sun,, then, F =, , GMm, = mrω 2, 2, r, , GM, 4π 2, =, r3, T2, , 2π , , as ω =, , T , , , r3, ⇒ T 2 ∝ r3, GM, W.E - 1: An artificial satellite is in an elliptical, orbit around the earth with aphelion of 6R, and perihelion of 2R where R is Radius of the, earth = 6400Km. Calculate the, eccentricity of the elliptical orbit., Sol: We know that, T 2 = 4π 2, , perigee ( rp ) = a (1 − e) = 2R............(1), apogee ( ra ) = a(1 + e) = 6R ............(2), Solving (1) & (2), eccentricity (e) = 0.5, 130, , 1, Sol : Given rP = rE and TE = 1yr, 4, From Kepler’s third law, T 2α r 3, 2, , Here, V perihelion = Vmax and Vaphelion = Vmin, Kepler’s laws can be applied to natural and artificial, satellites as well., Kepler’s third law or Law of periods : The square, of period of revolution of a planet around the sun is, proportional to cube of the average distance of, planet (i.e., semi major axis of elliptical orbit) from, the sun., rmean =, , W.E - 2: The mean distance of a planet from the, sun is approximately 1/4 times that of earth, from the sun. Find the number of years, required for planet to make one revolution, about the sun., , 3, , 3, , TP , rP , rP 2, = ⇒ TP = TE , TE , rE , rE , 3, , 3, , r 2 1 2, TP = (1) E = = 0.125 Yrs, 4rE 4 , , W.E- 3:The speed of the planet at the perihelion, P be Vp and the Sun - planet distance SP, be rp as shown in Fig. Relate { rp,Vp }to the, corresponding quantities at the aphelion, { rA ,VA }. Will the planet take equal times to, traverse BAC and CPB ?, Sol: The magnitude of the angular momentum at P is, Lp = mpVprp,, The magnitude of the angular momentum at A, is LA = mpVArA, rp, , rA, , s, sun, , 2b, , According to law of conservation of angular, momentum, m prpVp = mprAVA or, , Vp, VA, , =, , rA, rP, , Here rA >rp hence Vp >VA., The area SBAC bounded by the ellipse and, the radius vectors SB and SC is larger than, SBPC in Fig. From Kepler’s second law,, equal areas are swept in equal times., Hence, the planet will take a longer time to, traverse BAC than CPB., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , W.E- 4: Let us consider that our galaxy consists, of 2.5 × 1011 stars each of one solar mass., How long will this star at a distance of 50,000, light years from the galastic entre take to complete one revolution ? Take the diameter of, the Milky way to be 105 ly.G = 6.67 × 10–, Nm 2Kg-2. (1 ly = 9.46 × 1015 m ), Sol: Here M = 2.5 × 1011 solar mass, , Ø, Ø, Ø, , = 2.5 × 1011 × (2 × 1030 )kg = 5.0 × 10 41 kg, r = 50,000 ly = 50,000 × 9.46 × 1015 m, = 4.73 × 1020 m, , 1, 4π2 r3 2, , , T=, , GM , , Ø, , 4π 2 r 3, GT, , Ø, Ø, Ø, , 11, , We know that, M =, , Ø, , 2, 1, , 4 × (22 / 7) 2 × (4.73 × 10 20 )3 2, =, −11, 41 , (6.67 × 10 ) × (5.0 × 10 ) , , = 3.53 × 1014 s., , Ø, Ø, , Newton's Law of Gravitation:, , Basic Forces in Nature:, Basic forces are classified into four categories, a) Gravitational Force, b) Electromagnetic Force, c) Strong nuclear Force, d) Weak nuclear Force, Relative strengths of basic forces between, protons :, Basic force, Gravitaional, , Ø, Ø, , Long range, (upto infinity), , Relative, strength, 1, , Short range, (<<1 fm), , 10, , Electromagnetic, , Long range, (upto infinity), Short range, (1fm), , 10, , m1m2, r2, Where G is universal gravitational constant, , by F = G, , Ø, , 2, −2, G = 6.67 ×10−11 Nm kg, , Ø, , (or) 6.67 × 10−8 dyne cm 2 gm−2, G is a scalar, Dimensional formula M, , Ø, , In Vector form F =, , 10, , 38, , →, , −1, , L3T − 2 , , − Gm1m2 ∧ − Gm1m2 →, r=, r, r2, r3, , r, , 36, , Gravitational Force:, This force is between any two massive particles., It is always attractive force., It is a conservative force., It is independent of medium present between, the masses., It can provide radial acceleration., It is communicated through a particle called as, Graviton., , NARAYANAGROUP, , The magnitude of gravitational force of, attraction between two point masses is given, , 31, , Weak nuclear, , Strong nuclear, , Ø, Ø, Ø, Ø, , Range, , Electromagnetic Force :, This force exists between any two charged, particles., This force is either attractive or repulsive., It is communicated through Photons., Strong Nuclear Force :, This force may act between any pair of nucleons in, the nucleus., It is charge independent., It is spin dependent ., It is communicated through π mesons., Weak Nuclear Force :, They are responsible for radioactive decay like, β - decay.., They act between leptons, positrons,, µ -mesons, neutrinos and Hadrons etc., It is communicated through weak bosons., , Ø, Ø, Ø, , Here r∧ is the unit vector in the direction of r and, ‘-’ sign indicates that the force is attractive., Properties of gravitational force:Gravitational force acts along the line joining the, two interacting particles i.e., gravitational force is, a central force., Gravitational force is independent of the presence, of other particles., Gravitational force form an action - reaction pair., So gravitational force obeys Newton IIIrd Law, →, →, F12, F21, →, B, A, r, uur, uuur, ⇒ F12 = − F21 ⇒ F12 + F21 = 0, 131
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, Ø, , F - d2 graph is a rectangular hyperbola as, shown., Force, , W.E- 5: If two particles each of mass 'm' are placed, at the two vertices of an equilateral triangle, of side 'a', then the resultant gravitational, force on mass m placed at the third vertex is, , FA, , 2, , (distance), , Ø, , Ø, , Sol:, , Gravitational force is a conservative force i.e., the, amount of work done by gravitational force in, displacing a body from one place to other place, is independent of the path traversed., The net gravitational force on any particle is the, vector sum of all individual gravitational forces on, it, by all other particles in the system., F1,net = F12 + F13 + F14 + ......... + F1n, , we can express this equation more compactly as, , r, vector sum F1, n e t =, Ø, , Ø, , n, , ∑, , i= 2, , r, F1 , i, , Sun exerts gravitational force on earth , but earth, does not move towards sun because the, gravitational pull of the sun on the earth provides, the necessary centripetal force to earth, so the orbit, is stable., If gravitational force is proportional to r–n then, time period of a planet Ta r, , n +1, 2, , v∝r, , 1-n, 2, , UNIVERSAL GRAVITATIONAL CONSTANT:, //////////, , M, , l/2, , A m, r, , θ, , B, , M, , FB, , FR, A, , 600, , 0, , 60 B, , FR = FA2 + FB2 + 2 FA FB cos 600, = 3F, , [Q FA = FB = F ], , Gm2 , FR = 3 2 , a , W .E- 6: If four identical particles each of mass, m, are kept at the four vertices of a square of, side length a, the gravitational force of, attraction on any one of the particles is, m, , FR, a, , F, , Sol:, a, , a, , FR =, , F m, 2, , 2Gm Gm2, +, a2, 2a 2, , FR = 2 F + F ' =, 1, Gm 2 , 2+ , 2, 2, , a, , along the diagonal, , towards the opposite corner., W.E- 7: Four particles, each of mass M and, equidistant from each other, move along a, circle of radius R under the action of their, mutual gravitational attraction. The speed of, each particle is, (2014A), Sol: Let a be the distance between two particles., , a, , a, , M, , R, a, , universal gravitational constant (G) can be, , FR, M, , 132, , 1, , F, m, , From Cavendish experiment the value of, , kθ r 2, calculated by G =, Mml, M - Mass of heavier sphere, m - Mass of lighter sphere, k - Torsion constant ; θ - Angle of twist, , m, , a, , M, , Ø, , 30 0, 30 0, , C, , a, , F, , F, M, MV 2, R, , The resultant gravitational force on any one of the, particle is given by FR =, , GM 2 2 + 1 , ., , 2 , a2, , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, W.E - 11: Mass M is split into two parts m and, (M-m), which are then separated by a certain, distance. What is the ratio of (m/M) which, maximises the gravitational force between the, parts ?, Sol: If r is the distance between m and (M-m), the, gravitational force between them will be, F =G, , m( M − m) G, = 2 ( mM − m 2 ), r2, r, , For F to be maximum dF/dm = 0 as M & r are, constants., ⇒, , d G, , ( mM − m 2 ) = 0, dm r 2, , , ⇒, , m 1, = . So, the force will be maximum when, M 2, , ;, , mV 2, K, K, = 52 ⇒V2 =, r, r, mr 3 2, , 2π r, mr, = 2π r, But T =, V, K, , 32, , ;, , ∴T 2 α r 7 2, , W.E - 13:Three spherical balls of masses 1kg, 2kg, and 3kg are placed at the corners of an equilateral triangle of side 1m. Find the magnitude of the gravitational force exerted by 2, kg and 3 kg masses on 1 kg mass., Sol: If F 1 is the force of attraction between 1kg, 2kg, 1× 2, , masses, then, F1 = G × (1) 2 ⇒ F1 = 2G, A1 kg, F1 600 F, 2, B, , 2 kg, , C, , 3 kg, , If F2 is the force of attraction between 1kg, 3kg, 1× 3, , masses, then, F2 = G × (1) 2 ⇒ F2 = 3G, The angle between the forces F 1 and F 2 is 60o. If, ‘ FR ’ is the resultant of these two forces then, 134, , ⇒ FR = (2G) 2 + (3G) 2 + 2 × 2G × 3G × cos 60o, , ⇒ FR = 19 G, , W.E -14:Two particles of masses 1Kg and 2Kg are, placed at a distance of 50cm. Find the initial, acceleration of the first particle due to, gravitational force., Sol: Gravitational force between two particles is, Gmm, 6.67 ×10−11 ×1× 2, 1 2, =, = 5.3×10−10 N, r2, (0.5)2, The acceleration of 1Kg particle is, , F=, , ⇒M −2m = 0, , the parts are equal., W.E - 12:Imagine a light planet revolving around, a very massive star in a circular orbit of radius, r with a period of revolution T. On what, power of ‘r’ will the square of time period, depend on the gravitational force of, attraction between the planet and the star is, proportional to r −5 2 ?, Sol: The gravitational force provides necessary, centripetal force, , FR = F12 + F22 + 2 F1F2 cos θ, , F 5.3 ×10−10, =, = 5.3 × 10−10 ms −2 towards, m1, 1, the 2Kg mass, W.E -15:An infinite number of particles each of, mass m are placed on the positive X-axis at, 1m,2m,4m,8m,.... from the origin. Find the, magnitude of the resultant gravitational, force on mass ‘m’ kept at the origin., y, a1 =, , Sol, , m m m m m, O 1 2 4 8 x, The resultant gravitational force, Gm 2 Gm 2 Gm 2, +, +, + .........., 1, 4, 16, 1 1, , = Gm 2 1 + + + ......... , 4 16, , , F=, , , , , , 1, a , 4, = Gm 2 , 2 , 1 = Gm Q S∞ =, , 1− , 1− r , 3, , 4, W.E -16:In a double star system, two stars of, masses m 1 and m 2 separated by a distance, 'x' rotates about their centre of mass. Find, the common angular velocity and Time, period of revolution., , x, r1 c.m r2 m2, 1, The gravitational force between the masses, provides the necessary centripetal force., , Sol: m, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Gm m, , GRAVITATION, , 1 2, = m1 r1ω, i.e., ––––(1), x2, The distance of centre of mass from m1 is, , r1 =, , 2, , m2 x, m1 + m 2, , -------- (2), , From (1) and (2), , G ( m1 + m 2 ), ⇒ω=, x3, , ω2 =, , T =, , Gm1m 2, mm x, = 1 2 ω2, x2, m1 + m 2, , 2π, ; T = 2π, ω, , G ( m1 + m 2 ), x3, , x3, G ( m1 + m 2 ), , W.E- 17:In Cavendish’s experiment, let each small, mass be 20g and each large mass be 5 kg. The, rod connecting the small masses is 50 cm long,, while the small and the large spheres are, separated by 10.0 cm. The torsion constant is, 4.8 ×10−8 kgm2 s −2 and the resulting angular, deflection is 0.40 . Calculate the value of, universal gravitational constant G from this, data., Sol: Here, m = 20 g = 0.02kg , M = 5 kg, r = 10cm = 0.1m, l = 50cm = 0.5m, , θ = 0.40 = ( 0.40 )( 2π / 3600 ) = 0.007 rad,, , k = 4.8 × 10 −8 kgm 2 s −2, Thus, from G =, , kθ r 2, Mml, , ( 4.8 × 10 ) ( 0.007 )( 0.1), G=, −8, , W.E -19:A particle of mass m is situated at a, distance d from one end of a rod of mass M, and length L as shown in fig. Find the, magnitude of the gravitational force, between them., x, dm, m, d, L, Sol : Consider an element of mass 'dm' and length, 'dx' at a distance 'x' from the point mass., M, dx ., Mass of the element dm =, L, Gravitational force on ‘m’ due to this element is, M , Gm dx , L, , dF =, x2, , ⇒F=, , GmM, L, , ⇒F=, F=, , (d + L ), , ∫, d, , ;, , −1 , x −2 dx = GmM x , L, −1, , , ( d + L), , GmM −1 , L x d, , r2, , T2, , Ms =, , (or), , 4 × (22 / 7)2 × (1.5 × 1011 )3, (6.67 × 10, , −11, , NARAYANAGROUP, , d, , 1, 1 , −, , d ( d + L) , , W.E -20:The gravitational force acting on a, particle, due to a solid sphere of uniform, density and radius R, at a distance of 3R, from the centre of the sphere is F1 . A spherical, hole of radius (R/2) is now made in, the, sphere as shown in the figure. The sphere with, hole now exerts a force F2 on, the same, particle. Ratio F1 to F2 is (2013E), , R, 2, , 4π 2 r 3, GT2, , given, r = 1.5 × 108 km = 1.5 × 1011m;, T = 365 days = 365 × 24 × 60 × 60 s, ∴ Ms =, , GmM, L, , ( d + L), , 2, , W.E -18:The mean orbital radius of the Earth, around the Sun is 1.5 ×108 km. Estimate the, mass of the Sun., Sol: As the centripetal force is provided by the, gravitational pull of the Sun on the Earth, 4π 2, , =, , , dx, , , GmM d + L − d , GmM, , =, L (d + L)d d (d + L), , = 6.72 × 10 −11 Nm 2 kg −2 ., , = M e r ω 2 =M e r, , ∫, d, , 5 × 0.02 × 0.5, , GM s M e, , F=, , M, L, x2, , ( d + L ) Gm , , ) × (365 × 24 × 60 × 60), , 2, , ≈ 2 × 1030 kg, , m, , R, , 3R, , Sol: Let mass of the removed sphere = M., Then mass of the original sphere = 8M (since, mass ∝ R 3 ), F1 =, , 8GMm, GMm, −, and F2 = 8GMm, 2, 9R2, 9R 2, 5R , , , 2 , , F, , 50, , 1, Therefore, F = 41, 2, , ( on simplifying), 135
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, Ø, , Ø, , Ø, Ø, , Acceleration due to gravity (g):, It is the acceleration gained by the body due to, gravitational force of attraction . The value of ‘g’ is, constant at a given place but differs from place to, place., The force with which a body is attracted towards, the centre of a planet is called weight of that body, on the planet., GM, Relation between G and g is, g = 2, R, −2, , , LT, ‘g’ is a vector quantity with , as its, −, 2, dimensional formula. Its SI unit is m s ., , ∆g d, =, g, R, Ø Decrease in g at small heights is more than decrease, in g at small depths., Ø Decrease in g at large heights is less than decrease, in g at large depths., Ø ‘g’ becomes zero at the centre of the earth., Graphical representation of variation of ‘g’, with height and depth: The variation of g with, the distance r from the centre of the earth is shown, below, , Ø, , i) Above the earth : g h =, , Variation of 'g' :, Variation of g with altitude : If g and g h are, acceleration due to gravities on the surface of the, Earth and at height ‘h’ above the surface of the, Earth of mass M and radius R then, GM, GM, 2, g = 2 and gh =, ( R + h), R, R2, gh = g , ( R + h )2, , , Ø, Ø, , Ø, , ∆g, 2h, ≈, Fractional change,, g, R, , Ø, Ø, , Variation of g with depth: If g is acceleration, due to gravity at the surface of the earth and gd is, acceleration due to gravity at a depth ‘d’ below, the surface of the earth, then, GM 4, on surface, g = 2 = π GR ρ, R, 3, 4, at a depth, g d = π G ( R − d ) ρ, 3, d, g d = g 1 − , R, , Ø, , Ø, , gR 2, (R + h)2, , 1, gR 2, (Q R + h = r) ⇒ g h ∝ 2, 2, r, r, ⇒ gh versus r graph is a curve as shown., gh =, , -2, , g=9.8ms, , g∝, , g r, , , , , , , 2h , , For small values of h, g h ≈ g 1 −, , R , , Thus, as the height increases, the value of g, decreases., The decrease in value of g at height h, 2h , ( h << R) is ∆g = g − gh ≈ R g, , Ø, , Fractional change,, , inside, the earth, , 1, r2, , distance from centre of earth (r), , i) Inside the earth: g d =, , x, , g, (R − d), R, , g, (r) (Q R − d = r) ⇒ g d ∝ r, R, ⇒ gd versus r graph is a straight line passing, through the origin as shown in fig., Variation of ‘g’ with latitude :, Consider an object of mass m at latitude λ of the, earth due to rotation of a earth, the value of, acceleration due to gravity gλ at a given place is, gd =, , Ø, , given by g λ = g − rω 2 cos λ, where rω 2 cos λ is the component of centrifugal, acceleration along the radius of the Earth., pole, r, , P, R a, , Equator, , O, , FC COS λ, FC=mrω, , 2, , λ, , Thus, as depth increases, the acceleration due to, gravity decreases., The decrease in value of g at depth ‘d’ is, d , ∆g = g, R, , 136, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , where r is the radius of the circle in which the, object is revolving. Here r = Rcos λ, ∴ g λ = g − ω R cos λ, where ω is the angular velocity. R is radius of the, earth and λ is latitude of the place, Special cases :, At the poles λ = 900, 2, , Ø, , 2, , ∴ gλ =90o = g − ω R ( 0), , (, , ∴ gλ, , Ø, , 2, , ), , (= 90 ), o, , ∴ gλ, , Ø, , (Qcos90, , o, , = 0), , At the equator λ = 00, , ∴gλ =0o = g −ω2 R(1), , Ø, Ø, , 2, , = g -- (maximum), , ( ), , ( =0 ), o, , (Qcos0 =1), , 2, , o, , = g − ω 2 R .... (minimum), , Here, Rω 2 = 0.034ms −2 for the Earth., The value of ‘g’ at poles does not depend on the, speed of rotation of the earth, but at the equator, ‘g’ decreases with the increase of speed of rotation, of earth., If earth suddenly stops its rotation, then the, acceleration due to gravity at poles remains, constant, and acceleration due to gravity at equator, increases by ω 2 R ., Variation of ‘g’ due to shape of the earth:, The earth is elliptical in shape. It is flattened at the, poles and bulged at the equator. So radius at, equator is greater than the radius at poles., P, Rp, , Re, , E, , 1, , the value of g at the equator is, R2, minimum and at the poles is maximum., Ø Lines joining the places on the earth having same, values of g are called isograms., W.E -21: A star 2.5 times the mass of the sun is, reduced to a size of 12 km and rotates with, a speed of 1.5 rps. Will an object placed on, its equator remain stuck to its surface due, to gravity ? (Mass of the sun = 2 × 1030 kg)., , As gα, , NARAYANAGROUP, , Sol: Acceleration due to gravity, g =, =, , 6.67 × 10−11 × 2.5 × 2 × 1030, (12000)2, , GM, R2, , = 2.3 × 1012 ms−2, , Centrifugal acceleration, = r ω2 = r (2 πf ) 2 = 12000(2π × 1.5) = 1.1 × 10 ms, 2, , 6, , −2, , Since, g > rω2 , the body will remain stuck with, the surface of star., W.E- 22:What is the time period of rotation of the, earth around its axis so that the objects at, the equator becomes weightless?, (g=9.8m/s2, Radius of earth = 6400km), Sol: When earth is rotating the apparent weight of a, body at the equator is given by, , Wapp = mg − mRω 2, If bodies are weightless at the equator, 0 = mg − mRω 2 ⇒ g = Rω 2, ⇒ω=, , g, R, , Time period, T =, , R, 2π, = 2π, g, ω, , 6.4 ×106, = 5078s = 84 minute 38s, 9.8, W.E- 23: The height at which the acceleration due, to gravity becomes g/9 ( where g is the, acceleration due to gravity on the surface of, the earth) in terms of the radius of the earth, (R) is, (2009A), T = 2π, , 2, , g, R, 1, R , =, Sol: Given = g , ⇒, 9, R+h 3, R+h, 3R = R + h ⇒ 2R = h, W.E -24: How much above the surface of earth does, the acceleration due to gravity reduce by 36%, of its value on the surface of earth., Sol: Since g reduces by 36%, the value of g there is, 64, g., 100, If h is the height of location above the surface of, earth, then,, , 100-36=64%. It means, g ' =, , 137
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, g'= g, , R2, , ( R + h), , 2, , experienced by a unit mass placed at that point., ur, uur F, ∴ Gravitational field strength, Eg = m, 0, , 64, R2, ⇒, g=g, 2, 100, ( R + h), , 8, R, R 6.4 × 106, 6, =, ⇒h= =, = 1.6 ×10 m, 10 R + h, 4, 4, W.E -25:Find the percentage decrease in the weight, of the body when taken to a depth of 32Km, below the surface of earth., Sol: Weight of the body at depth d is, , Units of gravitational field strength are Nkg −1 or, ms −2 and dimensional formula is LT −2, It is a vector quantity. It is always directed radially, towards the centre of mass of the body producing, the field., Note: In the earth's gravitational field,, , ⇒, , ur, ur, uur F m g ur, 0, Eg =, =, =g, m0, m0, , d, mg ′ = mg 1 − , R, , % decrease in weight =, d, R, , = ×100 =, , mg − mg ′, × 100, mg, , Ø, , 32, × 100 = 0.5%, 6400, , W.E 26:A man can jump 1.5m on the Earth., Calculate the approximate height he might, be able to jump on a planet whose density is, one-quarter that of the Earth and whose radius, is one-third that of the Earth., Sol: We know that, in case of Earth,, g=, , Ø, , P, Rp, , ( 4π / 3 ) R 3 ρ = 4 π G R ρ, GM, =, G, ×, , , R2, R2, 3, , , Similarly, for the other planet whose radius, density, g′ =, , hmax =, , 1, , g, Rρ =, 12, , , ⇒, , r, , r, Theoretically gravitational field due to a particle, extends upto infinite distance around it, The value of Eg is zero at r = ∞ ., If the system has a number of masses, then resultant, gravitational field intensity can be found out by using, the principle of superposition., r, r, r, r, i.e. Eg = Eg1 + Eg 2 + Eg3 + ........, , g, = 12 ., g′, , u2, 1, ⇒ hmax ∝ (here u is constant), 2g, g, , It is the region or space around a massive particle, in which its gravitational influence is felt., , Ø, , Gravitational field strength (or) Intensity, of Gravitationl Field:, , Ø, , Gravitational field strength at any point in a, gravitational field is defined as the gravitational force, , 138, , − GM r, r, r3, , E, , Gravitational Field:, , Ø, , E, , R, and, 3, , h′ g, = = 12 ⇒ h′ = 12h = 12 × 1.5 = 18m, h g′, , Ø, , Re, , In vector form the above formula is E g =, , ρ, 4π G R ρ , is , g ′ = 3 3 4 , , , 4, , 1 4π G, , 12 3, , Hence, in the earth’s gravitational field, the intensity, of gravitational field is nothing but acceleration, due to gravity 'g'., The intensity of gravitational field at a distance r, GM, from a point mass ‘M’ is given by E g = 2, r, The direction of the force F and hence of E is from, P to O as shown in fig., , Null Point :, It is the point in a gravitational field at which resultant, field intensity is zero., If two particles of masses m1 & m2 are separated, by a distance r, the distance of null point from m1, is given by, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , Gm1, Gm2, =, 2, x2, (r − x), , E1 −E2 = 0 ⇒, , GRAVITATION, x=, , ;, , r, m2, +1, m1, , Ø, , Field due to Circular Disc:, Gravitational field intensity due to a circular disc of, mass M at any point on the axial line, Q, , m2, , m1, E1, , E2, , R2 + x 2, , R, , x, , Eg θ, x, , O, , p, , r, , W.E-27:Two bodies of masses 100Kg and 10,000Kg, are at a distance of 1m apart. At what distance, from 100kg on the line joining them will the, resultant gravitational field intensity be zero?, Sol:, , or Eg =, , 2GM, (1 − Cosθ ) (in terms of ‘ θ ’), R2, , Field due to Hollow Sphere (or), Spherical Shell (E or I):, , G × 100 G × 10, 000, =, 2, x2, (1 − x ), , Ø, , 2, ⇒ 100 x 2 = (1 − x ) ⇒ x =, , 1, 11, , Gravitational field intensity due to a uniform, spherical shell, E, , Field due to Circular Ring:, Ø, , Gravitational field intensity due to a uniform circular, ring of mass M at any point at a distance ‘x’(from, the centre of the ring) on its axis is, Eg =, , (x, , GMx, 2, , +R, , ), , 2 3/2, , uuur, along PO, , GM, R2, , E∝, , 1, r2, , r, R, At a point inside the spherical shell,, , (E ), , g inside, , =0,, , (E ), g, , centre, , = zero, , At a point outside the spherical shell,, dE sin θ, , dE, , (E ), g, , surface, , =, , GM, (here r = R), R2, , =, , GM, (here r > R), r2, , R, dE cos θ, 0, , θ, θ, , dE cos θ, , dE, , (E ), , p, , g, , dE sin θ, , x, , At the centre of the circular ring, E g = 0, , E max, , 2GM, =, 3 3R 2, , NARAYANAGROUP, , Field due to Solid Sphere(uniform mass, density):, Gravitational field intensity due to a solid sphere, , Gravitational field intensity is directed towards the, centre of the circular ring., , Eg is maximum , at x =, , outside, , R, and, 2, , E, GM, R 2 E∝ r, , E∝, , 1, r2, , R, , r, , Eg = 0 ( at the centre of solid sphere ), ( Eg )inside =, , GMr, R3, , ( for r < R ), 139
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, At a point on the surface of the solid sphere,, , GM, ( for r = R ), R2, At a point outside the solid sphere,, ( Eg )outside = GM, ( for r > R ), r2, E g = 0 (at infinite distance), Field due to Straight Rod: A rod of length, 2l,density ρ ,placed along y- axis,such that mid, point of rod coincides with origin.The gravitational, field intensity at a point P(a,0) is, ( Eg ) surface =, , Eg =, , 2G ρ, 2Gρ, l, =, a, a l 2 + a2, , Ø, , a2, l2, , Ø, , a, , P, , l, , Ø, , 2Gρ, if l is ∞ ,then, Eg =, a, , r, r, Since Fg = mEg, A thin rod of mass M and length L is bent into a, semi circle, gravitational force on a particle of mass, , m placed at the centre of curvature is F=, Ø, Ø, , GMm, 2π, L2, , The amount of work done in bringing a unit mass, from infinity to a certain point in the gravitational, field of another massive object is called as, gravitational potential at that point due to massive, object., , Ø, , Let W is the work done and m0 is the test mass, then, , Ø, , W, m0, , As this work done is negative, the gravitational, potential is negative., S.I unit : J/Kg, 0 2 −2, Dimensional formula : M L T , , Ø, , The gravitational potential at a point P which is at a, distance r from a point mass M is given by, V =−, , Ø, , GM, r, , o, , r, , M, , P, , If the system has a number of masses m1 , m2 ,, m3 .......... mn at distances r1 , r2 , r3 .......... rn from, the point p, the resultant gravitational potential at a, point p can be written as, V = V1 +V2 + V3 + .........Vn, m m m, m , ⇒ V = −G 1 + 2 + 3 + ........... + n , r, r, r, rn , 1, 2, 3, n, , ⇒ V = −G ∑, , i =1, , mi, ri, , Potential due to Circular Ring:, , R 2 + x2, , Ø, , R, m, , If x >>>R ⇒ F =, , GMm, , then for a distant point,, x2, , ring behaves as point mass., 140, , V =, , Potential due to a Point Mass:, , dm, , Ø, , 2G M m, 3 3R 2, , Gravitational Potential:, , A thin rod of mass M and length L is bent in to a, complete circle,then resultant force on a particle, placed at its centre is zero., A point mass m is at a distance x from the center, of the ring of mass M and radius R on its, axis.gravitational force between the two is, GMmx, F=, 32, ( R2 + x2 ), , x, , R, , maximum force, 2, , Ø, , Ø, θ, , Force is maximum, at x = ±, Fmax =, , , 2G ρ , a2, a2, Eg =, 1 − 2 + higher powers of 2 + − − − , a 2l, l, , l, , GMm, x ,then force varies linearly, R3, , as distance ‘x’, , 1, 1+, , If x <<<R ⇒ F =, , Gravitational potential due to a circular ring, at a, distance r from the centre and on the axis of a ring, of mass M and radius R is given by, V =, , − GM, R +r, 2, , O, 2, , R, , r, , P, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, At r = 0,, , V = −, , GM, ,, R, , GRAVITATION, , i.e., at the centre of the, , ring gravitational potential is −, , GM, R, , Gravitational Potential due to a Spherical, , Ø, , Vsurface =, , Shell:, Let M be the mass of spherical shell and R is its, radius, V =, , Ø, , 3, x2 , Vinside = −GM , − 3 (if x<R), 2R 2R , At a point on the surface of the solid sphere,, , −GM, r, , Ø, , Ø, Vinside, , Ø, , R, O, , At a point outside the solid sphere, Voutside =, , 3 GM 3, = Vsurface ., 2 R, 2, The variation of V with x is as shown:, , At the centre, x=0 ⇒ Vc = −, , r, , V, , Ø, , −G M, R, , Vcentre = −, , GM, (r = 0 at centre), R, , Ø, , r, , Ø, Ø, , −GM, (If r > R), r, , Ø, , M, R, , At infinity, V∞ = 0, The variation of magnitude of V with r is as, shown ( For a spherical shell ), , x, , In case of solid sphere potential is maximum at, centre., Newton’s Shell Theorem : Gravitational potential, at a point outside of a solid (or) hollow sphere of, mass M is same as potential at that point due to a, point mass of M separated by same distance., Hence, the sphere can be replaced by a point mass., Gravitational potential difference:The amount, of work done in bringing a unit mass between two, points in the gravitational field is called as the gravitational potential difference between the two points., Wb − Wa , , m0 , , 1 1, Wab = − mo (Vb − Va ) = −Gmmo − , rb ra , , GM, −, R, R, , r, , Relation between gravitational field and, potential:, Ø, , Gravitational Potential due to a Solid, Sphere:, At a point inside the solid sphere,, Vinside =, , Outside, , ∆ V = Vb − Va = − , , V, , Ø, , m surface, Inside, , GM, ,, R, At a point outside the spherical shell,, , Voutside =, , Centre, , (If r = R), , Vinside = Vsurface = Vcentre = −, , Ø, , −3 GM, 2 R, −GM, R, , At a point on the surface of the spherical shell,, V surface =, , −GM, x, , (If x>R), , At a point inside the spherical shell, (If r < R), , −GM, =, R, , −GM, (If x=R), R, , −GM, (3R 2 − x 2 ), 2R3, , NARAYANAGROUP, , Gravitational field and the gravitational potential are, →, , related by E = −gradientV = −gradV, →, ∂V ∧ ∂V ∧ ∂V ∧ , E = −, i+, j+, k, ∂y, ∂ z , ∂x, , 141
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, ∂V, = Partial derivative of potential function, ∂x, V with respect to x, i.e., differentiate V wrt x, assuming y and z to be constant., The above equation can be written in the following, forms., , Ø, , Here,, , Ø, , − dV, , If gravitational field is along x-direction, dx, , E=, , Ø, , only., , ∧, ∧, ∧, uur, uur, dV = − E . dr , (where dr = dx i + dy j + dz k, →, , Ø, , →, , ∧, , ∧, , Ø, , Ø, Ø, , V =−, , ∧, , and E = Ex i + E y j + Ez k ), , ∝, , Ø, , ∝, , b) The negative slope of V - r curve gives E, The amount of work done by the gravitational force, in bringing a body from infinity to any point in the, gravitational field is defined as the gravitational, potential energy at that point., , Consider a system consists of three particles, of masses m 1, m2 and m 3 located at A, B and C, respectively.Total potential energy ‘U’ of the, , dU, dr, , ém m, ù, ê 1 2 m2 m3 m1m3 ú, system is U = -G ê rr + rr + rr ú, 23, 13 úû, êë 12, , uur, ⇒ dU = − F . dr, u, r ur uur, r ur uu, r, ⇒ ∫ dU = − ∫ F . dr ⇒ U − U0 = −∫ F. dr, r0, , Z, C, r13, , r0, , m1, A, , We generally choose the reference point at infinity, and assume potential energy to be zero there. If, ur uur, r ur uur, , ⇒ U = − ∫ F . dr = −W as ∫ F . dr = W , ∞, ∞, , , Ø, , Reference, Level, , h1, , Y, , X, , If a body is moving only under the influence of, gravitational force, from law of conservation of, , Gravitational Potential Energy of a, System of Particles:, Ø, , The gravitational potential energy for a system of, n particles is given by, , h2, P.E= 0, , m2, B, , mechanical energy U1 + K1 = U 2 + K2, , Potential energy of a body or a system is the, negative work done by the conservative forces, in bringing it from infinity to present position., P.E=+mgh2, , m3, r23, , r12, , O, , we take r0 = ∞ and U 0 = 0 then,, r, , Gm1m2, r, , Gravitational Potential Energy of Three, Particle System:, , →, , u0, , The gravitational potential energy of two particles, of masses m1 and m2 separated by a distance r is, given by U = −, , Gravitational potential energy:, , For a conservative field, F = −, , W U, = ⇒ U = mV, m m, , Gravitational Potential Energy of Two, Particle System:, , ur, , Note:a) If E is given V can be calculated by the, r, r ur uur, V, =, dV, =, −, formula, ∫, ∫ E. dr, , If a particle moves opposite to the field direction, then work done by the field will be negative. So, potential energy will increase and change in, potential energy will be positive., If a particle moves in the direction of the field, work done is positive, so potential energy, decreases and change in potential energy is, negative., potential energy exists for only conservative forces, and it does not exist for non conservative forces., By the definition of gravitational potential,, , U =, , ∑U, , i, , Gm1 m 2 Gm 2 m 3, , =−, +, + ............, r23, r12, , , P.E=-mgh1, 142, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Ø, , For n particle system there are, , GRAVITATION, n ( n − 1), , For a sphere of radius ‘x’, mass of the sphere, 4, = π x 3 ρ , where ρ = density of sphere, 3, −4, 2, Gravitational potential on the surface = π Gρ x, 3, Gm, (since gravitational potential = −, x, −G 4, −4, =, × π x3ρ =, π Gx 2 ρ ), x, 3, 3, Work done by the agent in increasing the surface, from x to x + dx is, , pairs and, , 2, the potential energy is calculated for each pair and, added to get the total potential energy of the, system., , Gravitational Potential Energy of a body, in Earth’s Gravitational Field:, Ø, , If a point mass ‘m’ is at a distance r from the centre, of the earth, then, U = −, , GMm, r, , Ø, , On the surface of earth,, , Ø, , GMm, GM , , = − mgR Q g = 2 , R, R , , At a height ‘h’ above the surface of earth,, , −Gm ( dm ), , U surface = −, , 16π 2, −4, , =, π G x 2 ρ ( 4π x 2 dx ρ ) =, G ρ 2 x 4 dx, 3, 3, , , , GMm, R+h, The difference in potential energy of the body of, mass m at a height h and on the surface of earth is, Uh = −, , Ø, , Therefore, total work done, =, , ∆U = U h − U surface, , Ø, , GMmh, =, ( R + h) R, , GMmh, h, , R 2 1 + , R, , ∆U =, , ⇒, , 1, R, , If, , 1 mgh, =, R+h, h, , 1+, , R, Gravitational potential energy at the centre of the, earth is given by, U c = mVc = −, , Ø, , 2, , h << R, ∆U ≈ mgh, Work done in lifting a body of mass m from earth, surface to a height h above the earth’s surface is, , W = U h − U surface ; W = GMm −, , Ø, , mgh, h, 1+ ., R, , 3 GMm, 2 R, , 3, −3GM, Here, Vc = Vs =, (It is minimum but not, 2, 2R, zero. However ‘g’ at centre of earth is zero), Self potential energy of a uniform sphere of, mass ‘M’ and radius ‘R’:, It is the amount of work done to bring identical, massive particles to construct a sphere of mass M, radius R and density ρ, , NARAYANAGROUP, , R, −16π 2, − 16π 2 G ρ 2 R 5, G ρ 2 ∫ x 4 dx =, 0, 3, 15, , , −16π GR , =, , 15, , , , 1 , GMm GMm , 1, =−, −−, = GMm −, , R+h , R , R R+h, , =, , = Gravitational potential × dm, , x, , Ø, , 5, , 2, , , M , −3 GM 2, =, , 4, 5 R, π R 3 , 3, , , = Gravitational self potential energy of a sphere., Self potential energy of a thin uniform shell of, , Gm 2, 2R, Ø Change in the gravitational potential energy in lifting, a body from the surface of the earth to a height, equal to ‘nR’ from the surface of the earth, , mass ‘m’ and radius ‘R’ is −, , ∆U =, , GMmh GMm(nR) GMmn mgRn, =, =, =, R( R + h) R( R + nR) R(n + 1) n + 1, , W.E -28: The gravitational field due to a mass, distribution is given by E = - K x 3 in xdirection. Taking the gravitational potential, to be zero at infinity, find its value at a distance, x., Sol: The potential at a distance x is, x, , K, −K, −K , V = − ∫ Edx = ∫ 3 dx = 2 = 2, x, 2x ∞ 2x, ∞, W.E-29: A particle of mass m is placed at the centre, of a uniform spherical shell of equal mass and, radius a. Find the gravitational potential at, a point P at a distance a/2 from the centre., x, , 143
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, Sol: The gravitational potential at P due to particle at, centre is V1 =, , − Gm −2Gm, =, a/2, a, , The potential at P due to shell is V2 =, , −Gm, a, , W.E - 33: Find the gravitational potential, energy of a system of four particles, each, of mass m placed at the vertices of a, square of side l.Also obtain the gravitational potential at centre of the square., m, m, , −3Gm, a, W.E -30:The gravitational field in a region is given, r, r r, by E = − ( 20 Nkg −1 ) ( i + j ) . Find the, , The net potential at P is V1 + V2 =, , gravitational potential at the origin (0,0), r uur, Sol: V = −∫ E. dr = − ∫ Ex .dx + ∫ Ey .dy =20x+20y, , ⇒ V=0 at the origin (0,0)., W.E -31: Calculate the gravitational potential at, the centre of base of a solid hemisphere of mass, M, radius R., Sol: Consider a hemispherical shell of radius r and, thickness dr. Its mass is given by, , O, , diagonal pairs with distance 2 l ., U = −4, , V=−, , dr, r, R, , Since all points of this hemispherical shell are at, the same distance r from centre O, potential at O, due to it is, dV =, , −Gdm −3GMrdr, =, r, R3, , x=, , −3GM, ., 2R, W.E -32: The gravitational field in a region is, given by the equation E = (5 i + 12 j) N / kg ., If a particle of mass 2kg is moved from the, origin to the point (12m, 5m) in this region,, the change in the gravitational potential, energy is, ( 2012 E), r uur, Sol: dV = − E. dr, ∴ V = ∫ dV =, 0, , Change in gravitational potential energy dU= mdV, = 2(-120) = -240 J, 144, , 4Gm, 4 2Gm, =−, r, l, , , 2l , Q r = 2 , , , , W.E - 34: Two bodies of masses m and 4m are placed, at a distance r. The gravitational poten, tial at a point on the line joining them, where thegravitational field is zero is, (2011A), Sol: Position of null point from mass m is, , R, , = −(5i + 12 j). (12i + 5 j) = −(60 + 60) = −120, , Gm 2, Gm 2, 2Gm 2 , 1 , −2, =−, 2+, , , l, l , 2l, 2, , The gravitational potential at the centre of the, square is, V = Algebraic sum of potential due to each, particle, , 2, , dm, M, , l, , m, m, Sol :The system has four pairs with distance l and two, , 3Mr 2 dr, dm =, 2, π, r, dr, =, (, ) R3, 2, π R3, 3, M, , l, 2, , r, r, =, 3, 4m, +1, m, , r, 3, m, , Ø, , Null point, , 4m, , 3 12 −9Gm, ∴ potentialV = −Gm + =, r, r 2r , Escape Velocity:, The minimum velocity required for a body in order, to escape from the gravitational field of a planet is, defined as escape velocity., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Ø, , GRAVITATION, , For a body to just escape to infinity , it is projected, with a minimum velocity such that its total energy, must be zero (PE+KE= 0)., −GMm 1, + mVe2 = 0, R, 2, , If Ve is the escape velocity from the surface of the, planet then Ve =, , 2GM, R, , ⇒ h=, , than the escape velocity ( v > v e ) , total energy is, , 4, , positive, the body escapes from gravitational, influence of the planet and enter into interstellar, space with certain velocity., By law of conservation of energy, 1 2 GMm 1 2, mv −, = mv∞, 2, R, 2, , − GMm, R+h, For object to escape TE=PE+KE=0, PEbody =, , ⇒ ve =, , energy. TE surface = TEmax.height, GMm 1, GMm, + mv 2 = −, R, 2, R+h, , i.e., the body will move in interplanetary or, interstellar space with a velocity v 2 − ve2, Salient features regarding escape velocity:, Ø Escape velocity depends on the mass, density and, radius of the planet from which the body is, projected., Ø Escape velocity does not depend on the mass of, the projected body and its direction of projection, and the angle of projection., Ø Escape velocity from the surface of earth, =11.2Km/s, Ø Escape velocity from the surface of moon, =2.31 Km/s, Ø There is no atmosphere on moon, because r.m.s., velocity of molecules is greater than the escape, velocity on the moon (i.e., vrms > ve ), Ø, , 1, GMmh, R + h 2GM, mv 2 =, ⇒, =, 2, R ( R + h), h, Rv2, , R ve2 , R, v2, + 1 = e2 ⇒ = 2 −1, h v , h, v, NARAYANAGROUP, , 2GM, 2GM , , = v 2 − v e2 Q v 2e =, R, R , , , ⇒ v∞ = v 2 − ve2, , 2GM, = 2gh ( R + h ), R+h, , Case I: If the velocity of projection v < ve then,, Total energy is negative. The body goes to certain, maximum height and then falls back.To find this, maximum height ,we use law of conservation of, , −, , ⇒ v 2∞ = v 2 −, , GMm 1 2 GMm, ⇒ mve =, R+h, 2, R+h, , Here, gh is acceleration due to gravity at height h., Behaviour of a Body Projected Vertically Up, with Different Velocities from the Surface of a, Planet:, Ø Consider a body of mass ‘m’ projected with a, velocity ‘v’ from the surface of a planet of mass, ‘M’ and radius ‘R’, , v, , − 1, , v, , , , Case II:-If the velocity of projection v = v e then, total, energy of the body just becomes zero, so that the, body just escapes from the planet and goes to, infinity and the body possess zero velocity at infinity., Case III:- If a body is projected with a velocity greater, , , , Also Ve = 2 gR and Ve = 2 π R ρ G R, 3, , (Where ρ is the mean density of the planet), Escape Velocity of a body From certain height, above the surface of a planet:, Ø At a height ‘h’ above the surface of a planet, , KEbody = − PEbody =, , R, 2, e, 2, , Ø, , Escape velocity on sun is maximum. As v rms < v e ,, hence, even the lightest molecules can not escape, from there. Sufficient amount of hydrogen is present, in the atmosphere of the sun since the escape, velocity on the sun is very high., If a body falls freely from infinity, then, it reaches, the earth with a velocity of 11.2 Km/s., 145
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, W.E -35: If Earth has mass nine times and radius, twice that of the planet mars, calculate the, velocity required by a rocket to pull out of the, gravitational force of Mars. Take escape speed, on surface of Earth to be 11.2 km/s, , W.E - 37: A planet in a distant solar system is 10, times more massive than the earth and its, radius is 10 times smaller. Given that escape, velocity from the earth is 11km/s, the escape, velocity from the surface of the planet is, , Sol: Here, M e = 9M m , and Re = 2 Rm, , Sol: Given M P = 10 M e ; RP =, , ve (escape speed on surface of Earth )=11.2 km/s, , We know that v e =, , Let Vm be the speed required to pull out of the, gravitational force of mars., We know that, ve =, , 2GM e, and v m =, Re, , M m Re, ×, =, M e Rm, , ⇒ vm =, , Earth Satellites, Satellites: The bodies revolving round a planet in, its gravitational field are defined as satellites., Orbital speed of Satellites: The velocity of a, satellite revolving around the earth of mass M and, radius R in a circular orbit of radius 'r' at a height, 'h' from the surface of earth is called orbital velocity., , 1, 2, ×2 =, 9, 3, , 2, (11.2 km / s ) = 5.3km / s, 3, , vo =, , W.E- 36: A rocket is fired with a speed v = 2 gR, near the earth's surface and directed upwards., (a) Show that it will escape from the earth., (b) Show that in interstellar space its speed, , 146, , (, , ), , GM, , ( R + h), , 3, , ( R + h )3, ( R + h) 3, = 2π, GM, gR 2, , For a satellite orbiting very close to earth., , h << R then, v o =, , GM , , Q g = R 2 , , 1, mv2 = 2mgR, 2, is greater than the energy required for escaping, (= mg R), the rocket will escape., (b) If u is the velocity of the rocket in interstellar, space (free from gravitational effects) then by, conservation of energy,, 2, 2, 1, 1, 1, m 2 gR − m 2gR = mv2, 2, 2, 2, 2, v = 4gR - 2gR or v = 2gR, , ), , Ø, , gR 2, ( R + h), , GM, =, ( R + h), , Time period T = 2π, , And as initial KE of the rocket, , (, , GM, =, r, , Angular velocity ω =, , is v = 2gR ., , GMm , = 0−, = mgR, R , , 2 GM, R, , =10 × 11 = 110km/s, , 2GM m, Rm, , Sol : (a) As PE of the rocket at the surface of the, earth is (–GMm/R) and at infinity is zero, energy, required for escaping from earth, , (2008A), , 2GM P, 100 × 2GM e, =, = 10v e, RP, Re, , ∴ vP =, , vm, 2GM m, Re, ×, Dividing, we get v =, Rm, 2GM e, e, =, , Re, 10, , GM, =, R, , gR, , 1, 2, 3, R3 ⇒ T ∝ R, For two satellites revolving around the earth in, ω2 ∝, , Ø, , different circular orbits of radii r1 and r2 at vertical, v1, r2, R + h2, heights h1 and h2 , v = r = R + h, 2, 1, 1, , Ø, , Orbital velocity for a satellite close to the surface, , Ø, , of earth v0 = 7.92 kms −1 ≈ 8kms −1, Orbital velocity is independent of mass of the, satellite., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , Ø, Ø, , It is always along the tangent to the orbit., Relation between escape and orbital velocities is, , Ø, , ve = 2 v0, If the speed of the orbiting body 'v' is such that, , Ø, , vo < v < ve then its orbit changes from circle to, ellipse., If the satellite revolves close to the earth surface,, (h<<R),then, Time period of revolution,, , T =, , 1, =, 2π, , Ø, , GM, 1, =, 3, r, 2π, , 1, 2π, , GM, r3, , =, , 1, 2π, , Ø, Ø, , G M, r, , =, , The kinetic energy of the satellite is,, 1, 1 GM , m v 20 = m , , 2, 2 r , , or K =, , GMm, 2r, , GMm, 2r, When the satellite revolves in an orbit of radius ‘r’, (here r = R+h) potential energy is negative, means, that the satellite is moving in the gravitational field, of the planet. The planet and the satellite form a, bound system. If it is to be escaped out of the, , The total energy is E = K + U = −, , GMm, , g, R, , Ø, , will have to be given to it., Trajectories of a body projected with different, velocities:, An object revolves around a planet only when it is, projected with sufficient velocity in a direction, perpendicular to the gravitational force of attraction, of the planet on the object., , G M m 2r, , For satellites τ =0 and L = constant in a given, orbit because gravitational force is central force., Angular momentum of the satellite depends on mass, of the satellite, mass of the planet and radius of the, orbit., A satellite behaves like a freely falling body., When a satellite revolves round a planet in an, elliptical orbit, then its orbital speed is not uniform., The mass of a planet can be determined with the, help of its satellite., If the satellite is travelling in the same direction as, the rotation of earth i.e. west to east , the time, interval between two successive times at which it, will appear vertically overhead to an observer at a, T sTe, fixed point on the equator is T = T − T, e, s, , NARAYANAGROUP, , −GMm, r, , gravitational field, some additional energy 2( R + h ), , If the satellite revolves close to the earth surface,, , L = m v0r = m r, , Ø, Ø, , Ø, , GM, ( R + h)3, , Angular Momentum: The angular momentum of, the satellite is given by, , Ø, , Ø, , GM , , Q g = R 2 , , gR 2, ( R + h )3, , (h<<R)then n =, , Ø, , U=, , Ø, , 3π, Gρ, , 1, 1, =, T, 2π, , The potential energy of the system is, , K =, , Frequency of Revolution(n): The number of, revolutions made by the satellite in one second is, called the frequency of revolution(n)., n=, , Energy of Orbiting Satellite:, Ø, , R, = 84.6min =1.41 hr, g, , T = 2π, , (or), , , 2π 2π 2π , =, −, Since ωrel = ω sat − ωearth ⇒, , T, Ts Te , , , point of projection, part of ellipse, , speed v, , v<, , earth, hyperbola, v > (2gr ), parabola, v=, , (gr ), , circle, v=, , (gr ) ≈ 8 kms −1, , (2 gr ) ≈ 11 km s −1, , ellipse, (2 gr ) > v >, , (gr ), , If v < gr object falls on the surface of earth, If v = gr object revolve in a circular orbit., If, , gr < v < 2 gr object revolves in an, , elliptical orbit., 147
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, If v = 2gr object escapes from the field and, follows parabolic path., If v > 2gr object escapes from the field and, follows hyperbolic path., Special Cases :, v2, , 4, , 3, , v1, , 2, v 1, , h1, R, M, , v=0, Escape, , Case I:Work done to lift an object at rest from the, surface of a planet to a height h is, TEi = TEsurface = -, , GMm, -G M m, +0=, R, R, , TEf = TEheight = -, , GMm, -GMm, +0=, R+h, R+h, , ÞW=, , GMm, GMm, R, R+h, , GMmh, mgh, =, R (R + h ) 1 + h, R, , Case II: Work done to shift an object at rest from the, surface of planet in to an orbit in which object, revolves around the planet is, TEi = TEsurface =, , -GMm, -GMm, +0 =, R, R, , -G M m, , 1, , -G M m, , 2, TEf = TEorbit = R + h + 2 m v 0 = 2 (R + h ), , Work done W= TEf –TEi =, , GMm, GMm, R, 2 (R + h ), , é R + 2h ù, ú, W = G M m êê, ú, êë 2 R ( R + h ) úû, , Case III: Work done to shift an object revolving around, the planet from one orbit in to another orbit is, GMm, , 1, , -GMm, , 2, TEi = (TE )h1 = - R + h + 2 mv1 = 2 R + h, (, 1, 1), , -GMm, , 2, TEf = (TE )h2 = - R + h + 2 mv 2 = 2 (R + h ), 2, 2, Work done, , GMm, , 1, , GMm, GMm, W = TEf –TEi = 2 ( R + h ) - 2 (R + h ), 1, 2, , Case IV: Work done (or) additional energy to be, imparted for an object to just escape an object, which is initially revolving around the planet close, to the surface is, , -GMm GMm, GMm, +, =R, 2R, 2R, TEf = O(object escapes only when its TE, becomes zero (or) positive), Work done (or) additional energy imparted to the, object is, GMm, W = DE = TE f -TEi =, = KE of the object, 2R, Hence, an object (satellite) revolving around the, planet escapes when, 1)It’s KE is doubled (increases by 100%), 2)It’s velocity is increased to 2 times of present, value (increases by 41.4%), Additional velocity imparted to the body, , (, , = ve –v0 = 2 v0 - v0, , ), , 2 -1 v 0 = 3.2km / s (nearly), Note: In the above case if theobject initially revolves, around the planet at a height h from the surface, -GMm, then it’s TE = 2 ( R + h ), Additional energy required to escape the, GMm, object is 2 (R + h), W.E -38: A satellite orbits the earth at a height of, 400 km above the surface. How much energy, must be expended to rocket the satellite out, of the gravitational influence of earth? Mass, of the satellite is 200 kg , mass of earth =, 6.0 × 1024 kg, radius of earth = 6.4 × 106 m,, G = 6.67 × 10–11 Nm 2 kg–2., Sol : Total energy of orbiting satellite at a height, -GMm, , h = 2(R + h), Energy expended to rocket the satellite out of, the earth’s gravitational field. = − (Total energy, GMm, , of thesatellite) = 2(R + h ), =, , 148, , ù, GMm éê, h 2 - h1, ú, ê, 2 ëê (R + h1 )(R + h 2 ) úûú, , TEi =, , h2, , Work done W = TEf –TEi =, , ÞW=, , (6.67 × 10−11 ) × (6 × 10 24 ) × 200, 2(6.4 × 106 + 4 × 105 ), , = 58.85 ×108 J, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, W.E - 39: A body is projected vertically upwards, from the surface of the earth with a velocity, equal to half of escape velocity of the earth., If R is radius of the earth, maximum height, attained by the body from the surface of the, earth is, (2009E)(2011M), Sol: From law of conservation of energy,, TEsurface = TEMax Height, 2, , −, , GMm 1 ve , GMm 1, 2, + m = −, + m (0), R, 2 2, R+h 2, , −, , GMm 1 2GM, + m, R, 2 4R, , GMm 2 2GM , , =−, Q v e =, , R+h , R , , , On simplifying, we get h = R/3, W.E -40: A particle is fired vertically upwards from, the surface of earth reaches a height 6400Km., Find the initial velocity of the particle ., Sol: TE. on the surface of the earth, = TE. at the highest point, 1, GMm, GMm, mV 2 −, = 0−, 2, R, R+h, 1, mgh, ⇒ mV 2 =, 2, h, 1+ R, , given, h = R =6400Km, , So, V 2 = gh ⇒ V = gh, , ⇒ V = 10 × 6400 ×103 = 8Km/s, W.E -41: If a satellite is revolving around a planet, of mass M in an elliptic orbit of semi major, axis a, then show that the orbital speed of the, satellite when it is at a distance r from the, 2 1, focus will be given by V = GM − , r a, , GMm, 2a, GMm, which is conserved. So, KE + PE = −, 2a, At position ‘r’,orbital speed of the satellite is V., 1, −GMm, 2, Then, KE = mV and PE =, 2, r, 1, GMm, GMm, 2, =−, So, mV −, 2, r, 2a, , GRAVITATION, W.E- 42: A rocket is fired ‘vertically’ from the, surface of mars with a speed of 2 kms-1. If 20%, of its initial energy is lost due to martian atmospheric resistance, how far will the rocket, go from the surface of mars before returning, to it?, Mass of mars = 6.4 × 1023 kg;, radius of mars = 3395 km;, Sol: From the law of conservation of energy, −GMm 80 1, −GMm, 2, +, +0, mV =, R, 100, 2, R+h, , , , 1 , 1, 3 2, GMm −, = 0.4 m ( 2 × 10 ), R R+h, ⇒, , = 1−, , R, 3,395, =, = 3,888.9km, 0.873 0.873, , ∴ The required height up to which the, rocket will go = 3,888.9 − 3, 395 = 493.9km, W.E -43: Two heavy spheres each of mass 100Kg, and radius 0.1m are placed 1m apart on a, horizontal table. What is the gravitational, field and potential at the mid point of the, line joining their centres., Sol: Gravitational field at the mid point of the line, joining their centres is given by, r, GM, GM r, r, E=, r =0, −r ) +, 2 (, 2 ( ), ( r 2), ( r 2), , Gravitational potential at the mid point of the line, joining their centres is given by, V =, , − GM −GM −4GM, +, =, r, ( r 2) ( r 2 ), , =−, , 4 × 6.67 × 10 −11 × 100, = − 2.7 × 10 − 8 J / kg, 1, , W.E- 44:The gravitational potential difference, between the surface of a planet and a point, 20m above it is 16J/kg. Calculate the, workdone in moving a 2kg mass by 8m on a, slope of 600 from the horizontal., Sol: The vertical height through which the body, has to be raised = 8sin 600 = 4 3 m. The P.D, .D, for a distance of 20m is 16J/kg. Hence, the, P.D for a distance of 4 3 m for a mass of 2kg, is, , NARAYANAGROUP, , 3.395 ×106 × 1.6 ×106, 6.67 × 10−11 × 6.4 × 1023, , R+h =, , Sol: Total energy of the system is E = −, , 2 1, ⇒ V = GM − , r a, , GM , R , 1−, = 1.6 ×106, R R + h , , 4 3, × 16 × 2 ; 11 J, 20, 149
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, W.E - 45: The mass of a spaceship is 1000kg. It is, to be launched from the earth surface out into, free space. The value of ‘g’ and ‘R’ are 10ms-2, and 6400km respectively.The required energy, for this work will be (2012A), Sol: To launch the spaceship out into free space,, from energy conservation, −, E=, , GMm GM, = 2, R, R, , GMm, +E =0, R, , , 10, mR = mgR = 6.4 × 10 J, , , W.E - 46: The minimum energy required to launch, a satellite of mass m from the surface of a, planet of mass M and radius R in a circular, orbit at an altitude of 2R is, (2013A), Sol: From the law of conservation of energy, −GMm, −GMm, + KEimparted =, R, 2 × 3R, KEimparted =, , R, 4M, C, , 6R, Sol: The neutral point N is at a distance ‘ x ’ from, mass M , given by, x=, , 6R, = 2R, 4M, +1, M, , The mechanical energy at the surface of M is, Et =, , 1 2 GMm 4GMm, mv −, −, 2, R, 5R, , At the neutral point N, the speed approaches, zero. The mechanical energy at N is purely, potential., 150, , 1 2 GM 4GM, GM GM, v, −, =−, −, 2, R, 5R, 2R, R, 1/2, , 2GM 4 1 , 3GM , − ⇒v =, or v =, , , R 5 2, 5R , 2, , W.E-48:A 400kg satellite is in a circular orbit of, radius 2RE about the earth. How much energy, is required to transfer it to a circular orbit of, radius 4RE? What are the changes in the kinetic and potential energies?, Sol: Mass of satellite m = 400 kg, Initial radius of circular orbit = 2RE, Final radius of circular orbit = 4RE, , GM m, , x, , d, =, m2, +1, m1, , From the principle of conservation of, mechanical energy, , GM m, , W.E -47:Two uniform solid spheres of equal, radii R, but mass M and 4M have a centre, to centre separation 6R, as shown in Fig., The two spheres are held fixed.A projectile of mass ‘m’ is projected from the surface, of the sphere of mass M directly towards the, centre of the second sphere. Obtain an, expression for the minimum speed ‘ v ’ of the, projectile so that it reaches the surface of the, second sphere., , N, , GMm 4GMm, −, 2R, 4R, , E, Initial total energy Ei = − 4R, E, , GMm 1 5 GMm , 1 − = , , R 6 6 R , , M R mv, 0, , EN = −, , E, Final total energy E f = − 8 R, E, , The change in the total energy is ∆E = E f − Ei, ∆E = −, ∆E = −, , Q, , GM E m GM E m, +, 8RE, 4 RE, , GM E m GM E mRE, = 2 , 8RE, RE 8, , GM E, = g = 9.81 m / s 2, RE2, , ∆E =, , gm R E, 8, , ∴∆E =, , ( RE = 6.37 × 106 m ), , 9.81 × 400 × 6.37 × 106, = 3.13 × 10 9 J, 8, , Kinetic Energy is reduced, ∆K = K f − Ki = −3.13 × 109 J, , The change in potential energy is twice the, change in total energy, ∴ ∆U = U f − U i = −6.25 × 109 J ., , Binding Energy:, The energy required to remove the satellite from, its orbit to infinity is called binding energy of the, system., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 1, GMm, 2, mV0 −, + B .E . = 0, 2, r, , Binding energy (BE) =, Ø, Ø, , For a satellite, , Ø, , 1, U, dU, F = −, K .E =, = E, and, 2 ,, dr, 2, r, Here, U → potential energy, E → total energy, gy, The same is true for electron-nucleus system, , GMm, 2r, , F∝, , KE, PE, PE, = 2,, = −1, = −2 ,, TE, KE, TE, , PE : KE :TE =−2:1: −1, , because there also, the electrostatic force, Fe ∝, , Energy graph for a satellite is, , Ø, , KE, , 1, r, , L = mvr = constant ⇒ v ∝ ., , V1, , P2, , PE, , r2, , V2, , Change in Orbit of a Satellite:, Energy required to shift a satellite from an orbit of, , Ø, , =, , r1, , O, , P1, , For satellites in different orbits v ∝, , radius r1 into an orbit of radius r2 is, E = E2 − E1, , 1, r2, , For a given satellite in an orbit ,, , r, , TE, , Ø, , Important Features Regarding Satellite:, If the law of force obeys the inverse square law, , 1, r, , V2, , GMm 1, 1 , − , 2 r1 r2 , , r2, , mgR 1, 1 , E =, −, , 2 r1 r2 , , P2, , 2, , ( or ), Ø, , O, , When the satellite is transferred to a higher orbit, , ( r2 > r1 ) then variation in different quantities are, as shown in the following table., Quantity, Orbital Velocity, Time period, , Variation, , Relation with r, , 1, r, , Decreases, , v∝, , Increases, , T ∝r, , 3/2, , 1, , Linear momentum, , Decreases, , p∝, , Angular momentum, , Increases, , L∝ r, , Kinetic energy, , Decreases, , 1, K∝, r, , Potential energy, , Increases, , U ∝−, , Total energy, , Increases, , E∝−, , Binding energy, , Decreases, , r, , P1, , W.E -49:A satellite of mass ‘m’ and radius ‘R’ is, orbiting the Earth in a circular orbit of radius, ‘r’. It starts losing energy due to air resistance, at a rate of C Js −1 . The time taken by the, satellite to reach the Earth is --Sol : E =, , NARAYANAGROUP, , 1, r, 1, BE ∞, r, , dE GMm 1 dr, −GMm, ⇒, =, =C, 2r, dt, 2 r 2 dt, , ⇒ dt =, t, , ∫ dt =, 0, , 1, r, , V1, , r1, , GMm dr, 2C r 2, , GMm, 2C, , ∫, , R, r, , 1, GMm 1, 1 , dr ⇒ t =, − , r2, 2 C r, R, , W.E - 50:Two satellites of same mass are launched, in the same orbit round the earth so as to, rotate opposite to each other. They soon, collide inelastically and stick together as, wreckage. Obtain the total energy of the, system before and just after the collision., Describe the subsequent motion of the, wreckage., , 151
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , Geostationary and Polar Satellites, , Sol : In case of satellite motion, energy of a satellite, GMm, an orbit is given by E = 2r, , in, , So the total energy of the system before collision, E i = E1 + E 2 = 2E = -, , GMm, r, , As the satellites of equal mass are moving in, opposite direction and collide inelastically, the, velocity of wreckage just after collision, by, conservation of linear momentum will be, i.e., v = 0, mv - mv = 2mv,, i.e., just after collision wreckage comes to rest in, the orbit. So energy of the wreckage just after, collision will be totally potential and will be, EP = -, , GM (2 m ), 2 GMm, =r, r, , And as after collision the wreckage comes to rest, in the orbit, it will move along the radius towards, the earth under its gravity., W.E -51: A launching vehicle carrying an artificial, satellite of mass m is set for launch on the, surface of the earth of mass M and radius R., If the satellite intended to move in a circular, orbit of radius 7R, the minimum energy, required to be spent by the launching vehicle, on the satellite is, (2010 E), Sol: Here r = R + h = 7R, Orbital velocity V0 =, , GM, r, , Ø, Ø, Ø, Ø, Ø, Ø, , Ø, Ø, Ø, Ø, Ø, , TE of satellite in its orbit = PE + KE, , Ø, Ø, , GMm 1, + mV02, r, 2, , Ø, , = −, =−, , GMm 1 GM , + m, , r, 2 r , , GMm 1 G M , GMm, = − 7 R + 2 m 7 R = − 14 R, , , , GMm, R, If the minimum energy to launch the satellite in to, , TE of satellite on the earth = −, , its orbit is Emin , then,, −, 152, , GMm, GMm, 13GMm, + Emin = −, ⇒ Emin =, R, 14R, 14 R, , Ø, Ø, , Communication satellites:, The satellites which remain stationary with respect, to the Earth are known as communication satellites., For example INSAT 1A, 1B, 2A etc, If anything is gently released from a, satellite, then it starts moving with the velocity of, the satellite and itself becomes a satellite., The satellite is projected from west towards east, so that maximum benefit of the motion of the earth, may be obtained., A satellite moving in a stable orbit does not need, any energy from an external source., Conditions for geo-stationary satellite:, The plane of orbit of the satellite should coincide, with geo-equatorial plane, The velocity of the satellite must be in the same, direction as that of earth i.e., from west towards, east., The period of revolution of satellite must be equal, to the period of rotation of earth about its own axis, i.e., 24 hrs., Time period of revolution of geo-stationary satellite, with respect to earth is infinity., The height of the geo-stationary satellite from the, surface of the earth is nearly 36000 KM., The relative velocity of geo-stationary satellite with, respect to earth is zero., The orbit of the geo-stationary satellite is called, the ‘Parking Orbit”., Polar Satellites:, These are low altitude (500 km to 800 km) satellites, They go round the poles of earth in north-south, direction, Polar satellites have a time period of 100 minutes, nearly, These satellites can view polar and equatorial, regions at close distances with good resolution., These satellites are useful for remote sensing,, meteorology and environmental studies of earth., , Weightlessness:, Ø, , Weightlessness is a phenomenon in which the object, is in a state of free fall, , Ø, Ø, , Wapp = m ( g − a ), Here a = g ⇒ W app = 0, A pendulum will not vibrate in an artificial satellite, since g = 0 inside the satellite., , Therefore, T, , = 2π, , l, =∞, g, , ⇒ Frequency = 0, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, Ø, , GRAVITATION, , Condition for Weightlessness in a Satellite:, The force acting on the astronaut of mass ‘m’, GMm, m v 20, is 2 − FR =, r, r, , Ø, , here FR is the reactional force, The reactional force on the floor of the satellite is, zero. Hence, there is the state of weightlessness, in a satellite i.e.,, , Ø, , GMm, m v 20, =, r2, r, , As the frame of reference attached to the satellite, is an accelerated frame, whose acceleration, towards the centre of the earth is a =, , v 20 GM, = 2 =g, r, r, , C.U.Q, KEPLER’S LAWS, 1., , 2., , 3., , 4., , 5., , The time period of an earth’s satellite in, circular orbit is independent of, 1) the mass of the satellite, 2) radius of its orbit, 3) both the mass and radius of the orbit, 4) neither the mass of the satellite nor the radius of, its orbit, If the earth is at one-fourth of its present, distance from the sun, the duration of the year, would be, 1) half the present year, 2) one-eighth the present year, 3) one-fourth the present year, 4) one -sixteenth the present year, The radius vector drawn from the sun to a, planet sweeps out ___ areas in equal time, 1) equal, 2) unequal 3) greater 4) less, If the area swept by the line joining the sun, and the earth from Feb 1 to Feb 7 is ‘A’, then, the area swept by the radius vector from Feb, 8 to Feb 28 is, 1) A, 2) 2A, 3) 3A, 4) 4A, The motion of a planet around sun in an, elliptical orbit is shown in the following figure., Sun is situated at one focus. The shaded areas, are equal. If the planet takes time ‘ t1 ’ and, ‘ t2 ’ in moving from A to B and from C to D, respectively, then, , NARAYANAGROUP, , D C, S, 1) t1 > t2, , A B, 2) t1 < t2, , 3) t1 = t2 4) Incomplete information, 6. Two satellites are revolving around the earth, in circular orbits of same radii. Mass of one, satellite is 100 times that of the other. Then, their periods of revolution are in the ratio, 1) 100:1, 2) 1:100 3) 1:1, 4) 10:1, 7. According to Kepler’s second law, line joining, the planet to the sun sweeps out equal areas, in equal time intervals. This suggests that for, the planet, 1) radial acceleration is zero, 2) tangential acceleration is zero, 3) transverse acceleration is zero, 4) All, NEWTON’S LAW OF GRAVITATION, 8. If F g and Fe are gravitational and electrostatic, forces between two electrons at a distance 0.1, m then F g / Fe is in the order of, 1) 1043, 2) 10–43, 3) 1035, 4) 10–35, Gm1m2, is valid, 9. F =, r2, 1) Between bodies with any shape, 2) Between particles, 3) Between any bodies with uniform density, 4) Between any bodies with same shape, 10. Fg, Fe and Fn represent the gravitational,, electro-magnetic and nuclear forces, respectively, then arrange the increasing order, of their strengths, 1)Fn,Fe,Fg 2)Fg,Fe,Fn 3)Fe,Fg,Fn 4)Fg,Fn, Fe, 11. Find the false statement, 1) Gravitational force acts along the line joining the, two interacting particles, 2) Gravitational force is independent of medium, 3)Gravitational force forms an action-reaction pair, 4) Gravitational force does not obey the principle, of superposition., 12. Law of gravitation is not applicable if, A)Velocity of moving objects are comparable, to velocity of light, B) Gravitational field between objects whose, masses are greater than the mass of sun., 1)A is true, B is false 2) A is false, B is true, 3)Both A & B are true 4) Both A&B are false, 153
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 13. Among the following the wrong statement is, 1) Law of gravitation is framed using Newton’s, third law of motion, 2) Law of gravitation cannot explain why gravity, exists, 3) Law of gravitation does not explain the presence, of force even when the particles are not in physical, contact, 4) When the range is long, gravitational force, becomes repulsive., 14. Out of the following interactions, weakest is, 1) gravitational, 2) electromagnetic, 3) nuclear, 4) electrostatic, 15. Neutron changing into Proton by emitting, electron and anti neutrino. This is due to, 1) Gravitational Force 2) Electromagnetic Force, 3) Weak Nuclear Force 4) Strong Nuclear Force, 16. Attractive Force exists between two protons, inside the Nucleus. This is due to, 1) Gravitational Forces 2) Electromagnetic Forces, 3)Weak Nuclear Forces 4)Strong Nuclear Forces, 17. Repulsive force exist between two protons out, side the nucleus. This is due to, 1) Gravitational Forces 2) Electromagnetic Forces, 3)Weak Nuclear Forces 4)Strong Nuclear Forces, 18. Radioactive decay exist due to, 1)Gravitational Forces 2)Electromagnetic Forces, 3)Weak-Nuclear Forces, 4)Strong-Nuclear Forces, 19. Two equal masses separated by a distance, d attract each other with a force (F). If one, unit of mass is transferred from one of them to, the other, the force, 1) does not change 2) decreases by (G/d2), 3) becomes d 2 times 4) increases by (2G/d2), 20. Which of the following is the evidence to show, that there must be a force acting on earth and, directed towards Sun?, 1) Apparent motion of sun around the earth, 2) Phenomenon of day and night, 3) Revolution of earth round the Sun, 4) Deviation of the falling body towards earth, 21. Six particles each of mass ‘m’ are placed at, the corners of a regular hexagon of edge, length ‘a’. If a point mass ‘ m0 ’ is placed at, the centre of the hexagon,then the net, gravitational force on the point mass is, 1), , 154, , 6Gm 2, a2, , 2), , 6Gmm0, 3) Zero, a2, , 4), , 6Gm, a4, , 22. If suddenly the gravitational force of attraction, between earth and satellite revolving around, it becomes zero, then the satellite will (2002A), 1) Continue to move in its orbit with same velocity, 2) Move tangential to the original orbit with the, same velocity, 3) Becomes stationary in its orbit, 4) Move towards the earth, RELATION BETWEEN g AND G,, VARIATION OF g, 23. If the speed of rotation of earth about its axis, increases, then the weight of the body at the, equator will, 1) increase 2) decrease 3) remains unchanged, 4) some times decrease and sometimes increase, 24. The ratio of acceleration due to gravity at a, depth ‘h’ below the surface of earth and at a, height ‘h’ above the surface for h<<R, 1) constant only when h<<R, 2) increases linearly with h, 3) increases parabolically with h 4) decreases, 25. If the gravitational force of earth suddenly, disappears, then,, 1) weight of the body is zero, 2) mass of the body is zero, 3) both mass and weight become zero, 4) neither the weight nor the mass is zero, 26. Which of the following quantities remain, constant in a planetary motion, when seen from, the surface of the sun., 1) K.E, 2) angular speed, 3) speed, 4) angular momentum, 27. Average density of the earth, (2005A), 1) does not depend on ‘g’, 2) is a complex function of ‘g’, 3) is directly proportional to ‘g’, 4) is inversely proportional to ‘g’, 28. A person will get more quantity of matter in, kg-wt at, 1) poles 2) a latitude of 600 3) equator 4) satellite, 29. A pendulum clock which keeps correct time, at the surface of the earth is taken into a mine,, then, 1) it keeps correct time 2) it gains time, 3) it loses time, 4) none of these, 30. Two identical trains A and B move with equal, speeds on parallel tracks along the equator., A moves from east to west and B moves from, west to east. Which train will exert greater, force on the track?, 1) A, 2) B 3) they will exert equal force, 4) The mass and the speed of each train must be, known to reach a conclusion., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 31. Assuming the earth to be a sphere of uniform, density, the acceleration due to gravity, 1) at a point outside the earth is inversely, proportional to the square of its distance from, the centre, 2) at a point outside the earth is inversely, proportional to its distance from the centre, 3) at a point inside is zero, 4) at a point inside is inversely proportional to its, distance from the centre., 32. If earth were to rotate faster than its present, speed, the weight of an object, 1) increase at the equator but remain unchanged at, poles, 2) decrease at the equator but remain unchanged, at the poles, 3) remain unchanged at the equator but decrease, at the poles, 4) remain unchanged at the equator but increase at, the poles, 33 The time period of a simple pendulum at the, centre of the earth is, 1) Zero 2) infinite 3) less than zero 4) two second, 34. A body of mass 5 kg is taken into space. Its, mass becomes, 1) 5 kg, 2) 10 kg 3) 2 kg, 4) 30 kg, 35. If the mean radius of earth is R, its angular, velocity is ω and the acceleration due to, gravity at the surface of the earth is ‘g’ then, the cube of the radius of the orbit of a satellite, will be, 1), , Rg, ω2, , 2), , R2g, ω, , 3), , R 2ω, R2g, 4), g, ω2, , 36. If R=radius of the earth and g =acceleration, due to gravity on the surface of the earth,, the acceleration due to gravity at a distance, (r<R) from the centre of the earth is, proportional to, 1) r, 2) r2, 3) r–2, 4) r–1, 37. If R = radius of the earth and g = acceleration, due to gravity on the surface of the earth, the, acceleration due to gravity at a distance (r>R), from the centre of the earth is proportional to, 1) r, 2) r2, 3) r–2, 4) r–1, 38. Earth is flattened at poles and bulged at, equator. This is due to, 1) revolution of earth around the sun in an elliptical, orbit, 2) angular velocity of spinning motion about its axis, is more at equator, 3) centrifugal force is more at equator than poles, 4) more centrifugal force at poles than equator, NARAYANAGROUP, , GRAVITATION, 39. Tidal waves in the sea are primarily due to, 1) the gravitational effect of the moon on the earth, 2) the gravitational effect of the sun on the earth, 3) the gravitational effect of the Venus on the earth, 4) the atmospheric effect of the earth itself, 40. Consider earth to be a homogeneous sphere., Scientist A goes deep down in a mine and, Scientist B goes high up in a balloon. The, gravitational field measured by, 1) A goes on decreasing and that of B goes on, increasing, 2) B goes on decreasing and that of A goes on, increasing, 3) Each decreases at the same rate, 4) Each decreases at different rates., GRAVITATIONAL FIELD INTENSITY,, GRAVITATIONAL POTENTIAL,, POTENTIAL ENERGY AND, WORKDONE, 41. Intensity of gravitational field inside the hollow, spherical shell is, 1) Variable 2) minimum 3) maximum 4) zero, 42. The work done by an external agent to shift, a point mass from infinity to the centre of the, earth is ‘W’. Then choose the correct relation., 1) W=0, 2) W>0, 3) W<0, 4) W ≤ 0, 43. The intensity of the gravitational field of the, earth is maximum at, 1) centre of earth, 2) equator, 3) poles, 4) same everywhere, 44 Let VG and EG denote gravitational potential, and field respectively, then choose the wrong, statement., 1) VG = 0, EG = 0, 2) VG ≠ 0, EG = 0, 3) VG = 0, EG ≠ 0, 4) VG ≠ 0, EG ≠ 0, 45. Two identical spherical masses are kept at, some distance. Potential energy when a mass, ‘m’ is taken from the surface of one sphere to, the other, 1)increases continuously 2)decreases continuously, 3) first increases, then decreases, 4) first decreases, then increases, 46. A thin spherical shell of mass ‘M’ and radius, ‘R’ has a small hole. A particle of mass ‘m’ is, released at its mouth. Then, 1) the particle will execute S.H.M inside the shell, 2) the particle will oscillate inside the shell, but the, oscillations are not simple harmonic, 3) the particle will not oscillate, but the speed of, the particle will go on increasing 4) none of these, 155
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 47. The gravitational field is a conservative field., The work done in this field by moving an object, from one point to another, 1) depends on the end-points only, 2) depends on the path along which the object is, moved, 3) depends on the end-points as well as the path, between the points., 4) is not zero when the object is brought back to, its initial position., 48. A hole is drilled through the earth along a, diameter and a stone is dropped into it. When, the stone is at the centre of the earth, it has, finite a) weight b) acceleration, c) P.E., d) mass, 1) a & b, 2) b & c 3) a, b & c 4) c & d, 49. A body has weight (W) on the ground. The, work which must be done to lift it to a height, equal to the radius of the earth is, 1) equal to WR 2) greater than WR, 3) less than WR 4) we can’t say, 50. A gravitational field is present in a region. A, point mass is shifted from A to B, along, different paths shown in the figure. If W1 , W2, and W3 represent the work done by, gravitational force for respective paths, then, 1) W1 = W2 = W3 2) W1 > W2 > W3, Path 1, Path 2, A, , 3) W1 > W3 > W2 4) none of these, , 1), , 51. The energy required to remove an earth, satellite of mass ‘m’ from its orbit of radius ‘r’, to infinity is, GMm, r, , 2), , − GMm, 2r, , 3), , GMm, 2r, , 4), , Mm, 2r, , 52. A hollow spherical shell is compressed to half, its radius. The gravitational potential at the, centre, 1) increases 2) decreases 3) remains same, 4) during the compression increases then returns, to the previous value., 53. For a satellite projected from the earth’s, surface with a velocity greater than orbital, velocity the nature of the path it takes when, its energy is negative, zero and positive, respectively is, 156, , ESCAPE SPEED, 56. The earth retains its atmosphere, due to, 1) the special shape of the earth, 2) the escape velocity being greater than the mean, speed of the molecules of the atmospheric gases., 3) the escape velocity being smaller than the mean, speed of the molecules of the atmospheric gases., 4) the sun’s gravitational effect., 57. Ratio of the radius of a planet A to that of, planet B is ‘r’. The ratio of accelerations due, to gravity for the two planets is x. The ratio, of the escape velocities from the two planets, is, 2) r / x 3) r, 4) x / r, 1) rx, 58. The ratio of the escape velocity and the orbital, velocity is, , B, , Path 3, , 1), , 1) Elliptical, parabolic and hyperbolic, 2) Hyperbolic, parabolic and elliptical, 3) Elliptical, circular and parabolic, 4) Parabolic, circular and Elliptical, 54. If a satellite is moved from one stable circular, orbit to a farther stable circular orbit, then, the following quantity increases, 1) Gravitational force2) Gravitational P.E., 3) linear orbital speed 4) Centripetal acceleration, 55. If the universal gravitational constant, decreases uniformly with time, then a satellite, in orbit will still maintain its, 1) weight, 2) tangential speed, 3) period of revolution 4) angular momentum, , 2, , 2), , 1, 2, , 3) 2, , 4) 1/2, , 59. The escape velocity from the earth for a rocket, is 11.2 km/sec. Ignoring the air resistance, the, escape velocity of 10 mg grain of sand from, the earth will be (in km/sec), 1) 0.112, 2) 11.2, 3) 1.12, 4) None, 60. The escape velocity for a body projected, vertically upwards from the surface of earth, is 11 km/s. If the body is projected at an angle, of 450 with the vertical, the escape velocity, will be, 1) 11 2 km / s, 2) 22 km / s, 3) 11km / s, 4 ) 22 2 km / s, 61. A missile is launched with a velocity less than, the escape velocity. The sum of its kinetic, and potential energy is, 1) Positive 2) Negative, 3) Zero, 4) May be positive or negative depending upon its, initial velocity, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 62. The escape velocity of a body depends upon, its mass as, 1) m0, 2) m1, 3) m3, 4) m 2, 63. The magnitude of potential energy per unit, mass of the object at the surface of earth is, ‘E’. Then escape velocity of the object is, 1) 2E, 2) 4E 2, 3) E 4) E / 2, 64. A space station is set up in space at a distance, equal to earth’s radius from earth’s surface., Suppose a satellite can be launched from space, station. Let V1 and V2 be the escape velocities, of the satellite on earth’s surface and space, station respectively. Then, 1) V2 = V1, 2) V2 < V1, 3) V2 > V1, , GRAVITATION, , 71., , 72., , 4) No relation, , EARTH SATELLITES, 65. The minimum number of geo-stationary, satellites required to televise a programme all, over the earth is, 1) 2, 2) 6, 3) 4, 4) 3, 66. When a satellite going around the earth in a, circular orbit of radius r and speed v loses, some of its energy , then, 1)r and v both increase 2)r and v both decrease, 3) r will increase and v will decrease, 4) r will decrease and v will increase, 67. The satellite is orbiting a planet at a certain, height in a circular orbit. If the mass of the, planet is reduced to half, the satellite would, 1) fall on the planet, 2) go to orbit of smaller radius, 3) go to orbit of higher radius, 4) escape from the planet, 68. A satellite is revolving round the earth in an, elliptical orbit. Its speed will be, 1) same at all points of the orbit, 2) different at different points of the orbit, 3) maximum at the farthest point, 4) minimum at the nearest point, 69. An artificial satellite of the earth releases a, packet. If air resistance is neglected, the point, where the packet will hit, will be, 1) ahead, 2) exactly below 3) behind, 4) it will never reach the earth, 70. A satellite is moving in a circular orbit round, the earth. If any other planet comes in between, them, it will, NARAYANAGROUP, , 73., , 74., , 75., , 76., , 77., , 1) Continue to move with the same speed along, the same path, 2) Move with the same velocity tangential to original, orbit., 3) Fall down with increasing velocity., 4) Come to rest after moving certain distance along, original path., A space-ship entering the earth’s atmosphere, is likely to catch fire. This is due to, 1) The surface tension of air 2) The viscosity of air, 3) The high temperature of upper atmosphere, 4) The greater portion of oxygen in the, atmosphere at greater height., An astronaut orbiting the earth in a circular, orbit 120 km above the surface of earth, gently, drops a ball from the space-ship. The ball will, 1) Move randomly in space, 2) Move along with the space-ship, 3) Fall vertically down to earth, 4) Move away from the earth, Following physical quantity is constant when, a planet that revolves around Sun in an, elliptical orbit ., 1) Kinetic energy, 2) Potential energy, 3) Angular momentum, 4) Linear velocity, A satellite launching station should be, 1) Near the equatorial region, 2) Near the polar region, 3) On the polar axis, 4) At any place, When a satellite in a circular orbit around the, earth enters the atmospheric region, it, encounters small air resistance to its motion., Then, 1) its angular momentum about the earth decreases, 2) its kinetic energy decreases, 3) its kinetic energy remains constant, 4) its period of revolution around the earth increases, The period of a satellite moving in circular orbit, near the surface of a planet is independent of, 1) mass of the planet, 2) radius of the planet, 3) mass of the satellite 4) density of planet, Out of the following statements, the one, which correctly describes a satellite orbiting, about the earth is, 1) There is no force acting on the satellite, 2) The acceleration and velocity of the satellite, are roughly in the same direction, 3) The satellite is always accelerating about the earth, 4) The satellite must fall, back to earth when its, fuel is exhausted., 157
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 78. When an astronaut goes out of his space-ship, into the space he will, 1) Fall freely on the earth 2) Go upwards, 3) Continue to move along with the satellite in the, same orbit., 4) Go spiral to the earth, 79. When the height of a satellite increases from, the surface of the earth., 1) PE decreases, KE increases, 2) PE decreases, KE decreases, 3) PE increases, KE decreases, 4) PE increases, KE increases, 80. A satellite S is moving in an elliptical orbit, around the earth. The mass of the satellite is, very small compared to the mass of the earth, 1) the acceleration of S is always directed towards, the centre of the earth, 2) the angular momentum of S about the centre of, the earth changes in direction, but its magnitude, remains constant, 3) the total mechanical energy of S varies, periodically with time, 4) the linear momentum of S remains constant in, magnitude, 81. If S1 is surface satellite and S 2 is geostationary satellite, with time periods T1 and, T2 , orbital velocities V1 and V2 ,, 1) T1 > T2 ; V1 > V 2, 2) T1 > T2 ;V1 < V2, 3) T1 < T2 ; V1 < V2, 4) T1 < T2 ; V1 > V2, 82. A relay satellite transmits the television, programme from one part of the world to, another part continuously because its period, 1) is greater than period of the earth about its axis, 2) is less than period of rotation of the earth about, its axis., 3) has no relation with the period of rotation of the, earth about its axis., 4) is equal to the period of rotation of the earth, about its axis., 83. The following statement is correct about the, motion of earth satellite., 1) It is always accelerating towards the earth, 2) There is no force acting on the satellite, 3) Move away from the earth normally to the orbit, 4) Fall down on to the earth, 84. An artificial satellite of mass m is revolving, round the earth in a circle of radius R. Then, work done in one revolution is, mgR, 1) mgR, 2), 3) 2π R × mg 4) Zero, 2, , 158, , 85. A satellite is revolving round the earth. Its, kinetic energy is Ek . How much energy is, required by the satellite such that it escapes, out of the gravitational field of earth, E, 1) 2 Ek, 2) 3 Ek, 3) k, 4) infinity, 2, 86. If the universal gravitational constant, increases uniformly with time, then a satellite, in orbit will still maintain its, 1) weight, 2) tangential speed, 3) period of revolution 4) angular momentum, 87. Two satellites of masses m 1 and m 2, (m1 > m2) are revolving around earth in circular, orbits of radii r1 and r2 (r1 > r2) respectively., Which of the following statements is true, regarding their velocities V 1 and V 2., V1, , 88., , 89., , 90., , 91., , 92., , 93., , 94., , V2, , 1) V1 = V2 2) V1 < V2 3) V1 > V2 4) r = r, 1, 2, An earth satellite is moved from one stable, circular orbit to another larger and stable, circular orbit. The following quantities increase, for the satellite as a result of this change, 1) gravitational potential energy, 2) angular velocity 3) linear orbital velocity, 4) centripetal acceleration, A satellite is revolving in an elliptical orbit in, free space; then the false statement is, 1) its mechanical energy is constant, 2) its linear momentum is constant, 3) its angular momentum is constant, 4) its areal velocity is constant, When a satellite falls into an orbit of smaller, radius its speed, 1) decreases, 2) increases, 3) does not change, 4) zero, Two artificial satellites are revolving in the, same circular orbit. Then they must have the, same, 1) Mass, 2) Angular momentum, 3) Kinetic energy, 4) Period of revolution, If satellite is orbiting in space having air and, no energy being supplied, then path of that, satellite would be, 1) circular, 2) elliptical, 3) spiral of increasing radius, 4) spiral of decreasing radius, A satellite in vacuum, 1) is kept in orbit by solar energy, 2) previous energy from gravitational field, 3) by remote control, 4) No energy is required for revolving, Two heavenly bodies s1 & s2 not far off from, each other, revolve in orbit, 1) around their common centre of mass, 2) s1 is fixed and s2 revolves around s1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 3) s2 is fixed and s1 revolves around s2, 4) cannot say, 95. If V , T , L , K and r denote speed, time period,, angular momentum, kinetic energy and radius, of satellite in circular orbit, b) Lα r1/2, a) V α r −1, 3/2, c) T α r, d) Kα r −2, 1) a,b are true, 2) b,c are true, 3) a,b,d are true, 4) a,b,c are true, 96. Two similar satellites s1 and s2 of same mass, ‘m’ possess around completely inelastic, collision while orbiting earth in the same, circular orbit in opposite direction then, 1) total energy of satellites and earth system become, zero, 2) the satellites stick together and fly into space, 3) the combined mass falls vertically down, 4) the satellites move in opposite direction, , ENERGY OF ORBITING SATELLITES, 97. For a planet revolving round the sun, when it, is nearest to the sun, 1) K.E. is min and P.E. is max., 2) Both K.E. and P.E. are min, 3) K.E. is max. and P.E. is min, 4) K.E. and P.E. are equal, 98. A body is dropped from a height equal to, radius of the earth. The velocity acquired by, it before touching the ground is, 1) V= 2 gR 2) V=3gR 3) V= gR 4) V=2gR, 99. When projectile attains escape velocity, then, on the surface of planet , its, 1) KE > PE, 2) PE > KE, 3) KE = PE, 4) KE = 2 PE, 100. A satellite is moving with constant speed ‘V’, in a circular orbit around earth. The kinetic, energy of the satellite is, 1, 3, 2, 2, 1) mV, 2) mV 2 3) mV, 4) 2mV 2, 2, 2, , GEOSTATIONARY AND POLAR, SATELLITES, , 101. The orbit of geo-stationary satellite is circular,, the time period of satellite depends on, (2008 E), 1) mass of the Earth, 2) radius of the orbit, 3) height of the satellite from the surface of Earth, 4) all the above, 102. Polar satellites go round the poles of earth in, 1) South-east direction 2) north-west direction, 3) east-west direction 4) north-south direction, 103. A geo-stationary satellite has an orbital period, of, 1) 2 hours 2) 6 hours 3) 24 hours 4) 12 hours, 104. The time period of revolution of geo-stationary, satellite with respect to earth is, NARAYANAGROUP, , GRAVITATION, 1) 24 hrs 2) 1 year 3) Infinity, 4) Zero, 105. A synchronous satellite should be at a proper, height moving, 1) From West to East in equatorial plane, 2) From South to North in polar plane, 3) From East to West in equatorial plane, 4) From North to South in polar plane, 106. The orbital angular velocity vector of a geostationary satellite and the spin angular, velocity vector of the earth are, 1) always in the same direction, 2) always in opposite direction, 3) always mutually perpendicular, 4) inclined at 23 1/2° to each other, 107. It is not possible to keep a geo-stationary, satellite over Delhi. Since Delhi, 1) is not present in A.P 2) is capital of India, 3) is not in the equatorial plane of the earth, 4) is near Agra., 108. The angle between the equatorial plane and, the orbital plane of a geo-stationary satellite is, 1) 450, 2) 00, 3) 900, 4) 600, 109. The angle between the equatorial plane and, the orbital plane of a polar satellite is, 1) 450, 2) 00, 3) 900, 4) 600, , WEIGHTLESSNESS, 110. Pseudo force also called fictitious force such, as centrifugal force arises only in, 1) Inertial frames, 2) Non-inertial frames, 3) Both inertial and non-inertial frames, 4) Rigid frames, 111. Feeling of weightlessness in a satellite is due to, 1) absence of inertia, 2) absence of gravity, 3) absence of accelerating force, 4) free fall of satellite, , C.U.Q - KEY, 1) 1, 7) 3, 13) 4, 19) 2, 25) 1, 31) 1, 37) 3, 43) 3, 49) 3, 55) 4, 61) 2, 67) 4, 73) 3, 79) 3, 85) 1, 91) 4, 97) 3, 103) 3, 109) 3, , 2) 2, 8) 2, 14)1, 20) 3, 26)4, 32) 2, 38) 3, 44) 3, 50) 1, 56) 2, 62) 1, 68) 2, 74) 1, 80) 1, 86) 4, 92) 4, 98) 3, 104) 3, 110) 2, , 3) 1, 9) 2, 15)3, 21) 3, 27) 3, 33) 2, 39) 1, 45) 3, 51) 3, 57) 1, 63) 1, 69) 4, 75) 1, 81) 4, 87) 2, 93) 4, 99) 3, 105)1, 111) 4, , 4) 3, 10)2, 16) 4, 22) 2, 28) 3, 34) 1, 40) 4, 46) 4, 52) 2, 58) 1, 64) 2, 70) 2, 76) 3, 82) 4, 88) 1, 94) 1, 100)1, 106) 1, , 5) 3, 11) 4, 17) 2, 23) 2, 29) 3, 35) 3, 41) 4, 47) 1, 53) 1, 59) 2, 65)4, 71) 2, 77) 3, 83) 1, 89) 2, 95) 2, 101)4, 107) 3, , 6) 3, 12)3, 18) 3, 24) 2, 30) 1, 36) 1, 42) 3, 48) 4, 54) 2, 60) 3, 66) 4, 72) 2, 78) 3, 84) 4, 90) 2, 96) 3, 102) 4, 108) 2, , 159
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 8., , LEVEL - I (C.W), KEPLER’S LAWS, 1., , 2., , If ‘A’ is areal velocity of a planet of mass M,, its angular momentum is, 2, 1) M/A, 2) 2MA, 3) A2 M 4) AM, A planet revolves round the sun in an elliptical, orbit of semi minor and semi major axes x and, y respectively. Then the time, period of revolution is proportional to, , 3, , 3., , 4., , 5., , 6., , 3, , 3, , 1) ( x + y ) 2 2) ( y − x ) 2 3) x 2, 4) y 2, Let ‘A’ be the area swept by the line joining, the earth and the sun during Feb 2012. The, area swept by the same line during the first, week of that month is, 1) A/4, 2) 7A/29 3) A, 4) 7A/30, A satellite moving in a circular path of radius, ‘r’ around earth has a time period T. If its, radius slightly increases by 4%, then, percentage change in its time period is, 1) 1%, 2) 6%, 3) 3%, 4) 9%, The time of revolution of planet A round the, sun is 8 times that of another planet B. The, distance of planet A from the sun is how many, times greater than that of the planet, B from the sun, (2002A), 1) 2, 2) 3, 3) 4, 4) 5, The distance of Neptune and Saturn from the, Sun are respectively 1013 and 1012 meters and, 3, , their periodic times are respectively Tn and, Ts . If their orbits are circular, then the value, of Tn / Ts is, 1) 100, 7., , 2) 10 10, , 3), , 1, 4) 10, 10 10, , The Earth moves around the Sun in an, elliptical orbit as shown in the figure. The ratio, OA, = x . Then, ratio of the speed of the Earth, OB, at B and at A is nearly, B, O, , Sun, , 160, , Earth, A 1), , x 2) x 3) x x, , 4) x 2, , 9., , The period of moon’s rotation around the earth, is nearly 29 days. If moon’s mass were 2 fold, its present value and all other things remain, unchanged, the period of moon’s rotation, would be nearly (in days), 2) 29 / 2 3) 29 3 4) 29, 1) 29 2, If the mass of earth were 2 times the present, mass, the mass of the moon were half the, present mass and the moon were revolving, round the earth at the same present distance,, the time period of revolution of the moon would, be (in days), 1) 56, 2) 28, 3) 14 2 4) 7, , LAW OF GRAVITATION, 10. Two spheres of masses m and M are situated, in air and the gravitational force between them, is F. The space between the masses is now, filled with a liquid of specific gravity 3. The, gravitational force will now be, 1), , F, 9, , 2) 3F, , 3) F, , 4), , F, 3, , 11. The gravitational force between two bodies is, 6.67 ×10−7 N when the distance between their, centres is 10m. If the mass of first body is 800, kg, then the mass of second body is, 1) 1000 kg 2) 1250 kg 3) 1500 kg 4) 2000 kg, 12. Two identical spheres each of radius R are, placed with their centres at a distance nR,, where n is integer greater than 2. The, gravitational force between them will be, proportional to, 1) 1 R 4, 2) 1 R 2, 3) R2, 4) R4, 13. A satellite is orbiting round the earth. If both, gravitational force and centripetal force on the, satellite is F, then, net force acting on the, satellite to revolve round the earth is, 1) F/2, 2)F, 3)2F, 4) Zero, 14. Mass M=1 unit is divided into two parts X and, (1–X). For a given separation the value, of X for which the gravitational force between, them becomes maximum is, 1) 1/2, 2) 3/5, 3) 1, 4) 2, ACCELERATION DUE TO GRAVITY AND, ITS VARIATION, , 15. If g on the surface of the earth is 9.8 m / s 2 ,, its value at a height of 6400 km is (Radius, of the earth = 6400km)., 1)4.9ms–2 2)9.8ms–2 3)2.45ms–2 4)19.6ms–2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 16. If g on the surface of the earth is 9.8ms −2 , its, value at a depth of 3200km (Radius of the, earth = 6400km) is, 1) 9.8ms −2 2) zero 3) 4.9ms −2 4) 2.45ms −2, 17. If mass of the planet is 10% less than that of, earth and radius of the planet is 20% greater, than that of earth then the weight of 40kg, person on that planet is, 1)10 kg wt 2)25 kg wt 3)40 kg wt 4)60 kg wt, 18. The angular velocity of the earth with which it, has to rotate so that the acceleration due to, gravity on 600 latitude becomes zero is, 1) 2.5 × 10−3 rad s −1, , 2) 1.5 × 10−3 rad s −1, , 3) 4.5 × 10−3 rad s −1, 4) 0.5 × 10−3 rad s −1, 19. Assume that the acceleration due to gravity, on the surface of the moon is 0.2 times the, acceleration due to gravity on the surface, of the earth. If Re is the maximum range of, a projectile on the earth's surface, what is, the maximum range on the surface of the, moon for the same velocity of projection, 1) 0.2 Re, 2) 2 Re, 3) 0.5 R e 4) 5Re, 20. The value of acceleration due to gravity on, the surface of earth is x. At an altitude of ‘h’, from the surface of earth, its value is y. If R is, the radius of earth, then the value of h is, x , y , 1) y −1 R 2) x −1 R 3), , , , , , y, R 4), x, , x, R, y, , GRAVITATIONAL FIELD, INTENSITY, 21. The point at which the gravitational force, acting on any mass is zero due to the earth, and the moon system is (The mass of the earth, is approximately 81 times the mass of the, moon and the distance between the earth and, the moon is 3,85,000km.), 1) 36,000km from the moon, 2) 38,500km from the moon, 3) 34500km from the moon, 4) 30,000km from the moon, 22. Masses 2 kg and 8 kg are 18 cm apart. The, point where the gravitational field due to them, is zero is, 1)6 cm from 8 kg mass 2) 6 cm from 2 kg mass, 3)1.8 cm from 8 kg mass 4) 9 cm from each mass, NARAYANAGROUP, , 23. Particles of masses m1 and m2 are at a fixed, distance apart. If the gravitational, field, r, r, strength at m 1 and m 2 are I 1 and I 2, respectively. Then,, , ur, , uur, , 1) m1 I1 + m2 I 2 = 0, , ur, , uur, , uur, , ur, , uur, , ur, , 2) m1 I 2 + m2 I1 = 0, , 3) m1 I1 − m2 I 2 = 0, 4) m1 I 2 − m2 I1 = 0, GRAVITATIONAL POTENTIAL,, POTENTIAL ENERGY, 24. The PE of three objects of masses 1kg, 2kg, and 3kg placed at the three vertices of an, equilateral triangle of side 20cm is, 1) 25G, 2) 35G, 3) 45G, 4) 55G, 25. A small body is initially at a distance ‘r’ from, the centre of earth. ‘r’ is greater than the, radius of the earth. If it takes W joule of work, to move the body from this position to, another position at a distance 2r measured, from the centre of earth, how many joules, would be required to move it from this position, to a new position at a distance of 3r from the, centre of the earth., 1) W/5, 2) W/3, 3) W/2, 4) W/6, 26. A body of mass ‘m’ is raised from the surface, of the earth to a height ‘nR’ (R -radius of, earth). Magnitude of the change in the, gravitational potential energy of the body is, (g - acceleration due to gravity on the surface, of earth), (2007M), mgR, n , n−1, mgR, 4) ( n−1), 1) , mgR 2) , mgR 3), n, n+1, n , 27. A person brings a mass 2 kg from A to B. The, increase in kinetic energy of mass is 4J and, work done by the person on the mass is −10J ., The potential difference between B and A is, ....... J / kg, 1) 4, 2) 7, 3) -3, 4) -7, 28. The work done in lifting a particle of mass ‘m’, from the centre of earth to the surface of the, earth is, 1) –mgR, , 2), , 1, mgR 3) Zero, 2, , 4) mgR, , 29. The figure shows two shells of masses m1 and, m2 . The shells are concentric. At which point,, a particle of mass m shall experience zero, force?, 161
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, m1, m2, D, , C B, , 1) A, , 2) B, , 3) C, , 4) D, , A, , 30. Energy required to shift a body of mass ‘m’, from an orbit of radius 2R to 3R is (2002A), 1), , GMm, 12 R, , 2), , GMm, 3R 2, , 3), , GMm, 8R, , 4), , GMm, 6R, , 1, 1) V e =, , Ve, 4, , 3) Ve = 2Ve1, 4) Ve1 = 2Ve, 37. A satellite revolves in a circular orbit with, speed V =, , 1, Ve . If satellite is suddenly stopped, 3, , and allowed to fall freely on to earth, the speed, with which it hits earth’s surface is, , ESCAPE & ORBITAL VELOCITIES, 31. The ratio of escape velocities of two planets if, g value on the two planets are 9.9m / s 2, and 3.3m / s 2 and their radii are 6400km and, 3200km respectively is, 1) 2.36 : 1 2) 1.36 : 1 3) 3.36 : 1 4) 4.36 : 1, 32. The escape velocity from the surface of the, earth of radius R and density ρ, 1) 2 R, , 2π ρ G, 3, , 2) 2, , 2π ρ G, 3, , R, 2π Gρ, 4), g, R2, 33. A body is projected vertically up from surface, of the earth with a velocity half of escape, velocity. The ratio of its maximum height of, ascent and radius of earth is, 1) 1 : 1, 2) 1 : 2, 3) 1 : 3, 4) 1 : 4, 34. A spaceship is launched in to a circular orbit, of radius ‘R’ close to the surface of earth. The, additional velocity to be imparted to the, spaceship in the orbit to overcome the earth’s, gravitational pull is (g = acceleration due to, gravity), , 3) 2π, , 1) 1.414Rg 2) 1.414 Rg, 3) 0.414Rg 4) 0.414 gR, 35. The escape velocity from the earth is 11 km/, s. The escape velocity from a planet having, twice the radius and same density as that of, the earth is (in km/sec), 1) 22, 2) 15.5, 3) 11, 4) 5.5, 36. An object of mass ‘m’ is at rest on earth’s, surface. Escape speed of this object is Ve ., Same object is orbiting earth with h = R , then, escape speed is Ve1 . Then, , 162, , 2) Ve = 2Ve1, , 1), , gR 2), , gR, 3, , 3), , 2gR 4), , 2, gR, 3, , 38. A space station is set up in space at a distance, equal to earth’s radius from the surface of, earth. Suppose a satellite can be launched, from the space station also. Let, v1 and v2 be the escape velocities of the, satellite on the earth’s surface and space, station respectively, then, 1) v2 = v1, , 2) v2 < v1, , 3) v2 > v1, , 4) 1, 2 and 3 are valid depending on the mass of, satellite., , EARTH SATELLITE, 39. The orbital speed for an earth satellite near, the surface of the earth is 7 km/sec. If the, radius of the orbit is 4 times the radius of the, earth, the orbital speed would be, 1) 3.5 km/sec, 2) 7 km/sec, 3) 7 2 km/sec, 4) 14 km/sec, 40. Two satellites are revolving round the earth, at different heights. The ratio of their, orbital speeds is 2 : 1. If one of them is at a, height of 100km, the height of the other, satellite is ( in km), 1) 19600, 2)24600 3) 29600 4) 14600, 41. A satellite of mass m revolves around the, earth of radius R at a height x from its surface., If g is the acceleration due to gravity on the, surface of the earth, the orbital speed of the, satellite is, 1/2, , 1) gx, , gR 2 , 2) , , R+x, , 3), , gR, gR 2, 4), R−x, R+x, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 42., , Two satellites M and N go around the earth, in circular orbits at heights of RM and RN, respectively from the surface of the earth., Assuming the earth to be a uniform sphere, of radius RE , the ratio of velocities of the, VM, satellites V is, N, 2, , 43., , 44., , 45., , 46., , RN + RE, RM , RN + RE, RN, 1) 2), 3) R + R 4), RM + RE, RM, M, E, RN , A satellite of mass ‘m’ revolves round the earth, of mass ‘M’ in a circular orbit of radius‘r’ with, an angular velocity ‘ ω ’. If the angular, velocity is ω /8 then the radius, of the orbit will be, 1) 4r, 2) 2r, 3) 8r, 4) r, The moon revolves round the earth 13 times, in one year. If the ratio of sun-earth distance, to earth-moon distance is 392, then the ratio, of masses of sun and earth will be, 1) 365 2) 356 × 10−12 3) 3.56 × 105 4) 1, A satellite is launched into a circular orbit of, radius R around the earth. A second satellite, is launched into an orbit of radius 1.01 R. The, time period of the second satellite is larger, than that of the first one by approximately, 1) 0.5%, 2) 1.5%, 3) 1%, 4) 3%, An astronaut orbiting in a spaceship round the, earth has a centripetal acceleration of, 2.45m / s 2 . The height of spaceship from, earth’s surface is (R= radius of earth), 1) 3R, 2) 2R, 3) R, 4) R / 2, , GRAVITATION, 49. Two satellites of masses 400 kg, 500 kg are, revolving around earth in different circular, orbits of radii r1, r2 such that their kinetic, energies are equal. The ratio of r 1 to r2 is, 1) 4 : 5 2) 16 : 25 3) 5 : 4, 4) 25 : 16, 50. The kinetic energy needed to project a body, of mass m from earth’s surface (radius R ) to, infinity is, mgR, mgR, 2) 2mgR 3) mgR, 4), 2, 4, GEOSTATIONARY AND POLAR SATELLITES, 51. Orbital speed of geo-stationary satellite is, 1) 8km/sec from west to east, 2) 11.2km/sec from east to west, 3) 3.1km/sec from west to east, 4) Zero, , 1), , LEVEL-1 (C.W) - KEY, 1)2, 7)2, 13)2, 19)4, 25)2, 31)1, 37)4, 43)1, 49)1, , 2)4, 8)4, 14)1, 20)1, 26)1, 32)1, 38)2, 44)3, 50)3, , 3) 2, 9)3, 15)3, 21)2, 27)4, 33)3, 39)1, 45)2, 51)3, , 4)2, 10)3, 16)3, 22)2, 28)2, 34)4, 40)1, 46)3, , 5)3, 11)2, 17)2, 23)1, 29)4, 35)1, 41)2, 47)3, , 6)2, 12)4, 18)1, 24)4, 30)1, 36)3, 42)2, 48)1, , LEVEL-1(C.W) - HINTS, 1., , dA, L, =, dt 2 M, , 2., 3., , From Kepler’s 3rd law, T 2α r 3, For 29 days - A, For 1 day - A/29 ,, For 1 week - 7A/29,, , 4., , T 2α r 3 , T ×100 = 2 R ×100, , ∆T, , 3 ∆R, , 3, , ENERGY OF SATELLITES, 47. A satellite moves around the earth in a circular, orbit with speed ‘v’. If ‘m’ is mass of the, satellite then its total energy is, , 5., , T r 2, From Kepler’s 3rd law, T 2α r 3 , 1 = 2 , T2 r1 , , 1, 1, 3, mv2, 2) mv2 3) – mv2 4) mv2, 2, 2, 2, 48. The K.E. of a satellite in an orbit close to the, surface of the earth is E. Its max K.E. so as, to escape from the gravitational field of the, earth is., 1) 2E 2) 4E 3) 2 2 E 4) 2 E, , 6., , T r 2, From Kepler’s 3rd law, T 2α r 3 , 1 = 2 , T2 r1 , , 7., , From conservation of angular momentum, , 1), , NARAYANAGROUP, , 3, , mv r = Constant, v1r1 =v 2 r2, 8. Time period does not depend upon the mass of the, satellite, , 163
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 8., , LEVEL - I (H.W), KEPLER’S LAWS, 1., , In planetary motion, the areal velocity of, position vector of a planet depends on angular, velocity (ω ) and the distance of the planet, from sun (r). If so, the correct relation for areal, velocity is, (2003E), 1), , dA, ∝ ωr, dt, , 2), , dA, ∝ ω2 r, dt, , dA, dA, ∝ ωr 2, ∝ ωr, 4), dt, dt, If a and b are the nearest and farthest, distances of a planet from the sun and the, planet is revolving in elliptical orbit, then, square of the time period of revolution of that, planet is proportional to, , 3), 2., , 1) a 3, 2) b3, 3) ( a + b ), 4) ( a − b ), Let ‘A’ be the area swept by the line joining, the earth and the sun during Feb 2007. The, area swept by the same line during the first, week of that month is, 1) A/4, 2) 7A/29 3) A, 4) 7A/30, The period of a satellite in an orbit of radius, R is T. Its period of revolution in an orbit of, radius 4R will be, 1) 2T, 2) 2 2T 3) 4T, 4) 8T, The period of revolution of an earth's satellite, close to the surface of earth is 60 minutes., The period of another earth's satellite in an, orbit at a distance of three times earth's radius, from its surface will be (in minutes), 1) 90, 2) 90 × 8 3) 270, 4) 480, If a planet of mass m is revolving around the, sun in a circular orbit of radius r with time, period T, then mass of the sun is, 1) 4π 2r 3 / GT, 2) 4π 2r3 / GT 2, 3, , 3., , 4., , 5., , 6., , 7., , 3, , 3) 4π 2 r / GT, 4) 4π 2r3 / G2T 2, The period of revolution of a planet around the, sun in a circular orbit is same as that, of period of similar planet revolving around a, star of twice the radius of first orbit and ‘M’ is, the mass of the sun and mass of star is, 1) 2M, 2) 4M, 3) 8M, 4) 16M, , NARAYANAGROUP, , A planet moves around the sun in elliptical, orbit. When earth is closest from the sun, it is, at a distance r having a speed v. When it is at, a distance 4r from the sun its speed is, v, v, 3) 2v, 4), 4, 2, A planet of mass ‘m’ is in an elliptical orbit, , 1) 4v, , 9., , 2), , about the sun ( m << M ) with an orbital time, period ‘T’. If ‘A’ be the area of the orbit then, its angular momentum is, 1), , 2mA, T, , 2) mAT, , 3), , mA, 2T, , 4) 2mAT, , LAW OF GRAVITATION, 10. The gravitational force between two particles, of masses m1 and m2 seperated by certain, distance in air medium is F. If they, are taken to vacuum and separated by the, same distance, then the gravitational force, between them will be, 1) greater than F, 2) less than F, 3) F, 4) Zero, 11. The mass of a ball is four times the mass of, another ball. When these balls are separated, by a distance of 10cm, the gravitational force, between them is 6.67 ×10−7 N . The masses of, the two balls are ( in kg), 1) 10 , 20 2) 5 , 20 3) 20 , 30 4) 20 , 40, 12. Gravitational force between two point masses, m and M separated by a distance r is F. Now, if a point mass 3m is placed next to m, the, force on M due to m becomes, 1) F, 2) 2F, 3) 3F, 4) 4F, 13. Three uniform spheres each of mass m and, diameter D are kept in such a way that each, touches the other two, then magnitude of the, gravitational force on any one sphere due to, the other two is, 1), , 3G m 2, D2, , 2), , 2 3G m 2, 3), D2, , 3G m 2, 4), 4D 2, , 3G m 2, D2, , 14. A 3 kg mass and a 4 kg mass are placed on x, and y axes at a distance of 1 metre from, the origin and a 1 kg mass is placed at the, origin. Then the resultant gravitational force, on 1 kg mass is, 1) 7G, 2) G, 3) 5G, 4) 3G, 165
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , GRAVITATIONAL FIELD INTENSITY, , ACCELERATION DUE TO, GRAVITY AND ITS VARIATION, 15. The height at which the value of g is half that, on the surface of earth of radius R is, 1) R, 2) 2R, 3) 0.414R 4) 0.75R, 16. The depth at which the value of g becomes, 25% of that at the surface of earth is (in Km), 1)4800, 2)2400, 3) 3600, 4) 1200, 17. If the radius of earth decreases by 10%, the, mass remaining unchanged, then the, acceleration due to gravity, 1) decreases by 19%, 2) increases by 19%, 3) decreases by more than 19%, 4) increases by more than 19%, 18. The acceleration due to gravity at the poles, is 10ms −2 and equatorial radius is 6400km for, the earth. Then the angular velocity of rotation, of the earth about its axis so that the weight, of a body at the equator reduces to 75% is, 1, 1, rads -1 4), rads-1, 200, 400, 19. The maximum horizontal range of a projectile, on the earth is R. Then for the same velocity, of projection, its maximum range on another, 1, , 1, , 1) 1600 rads -1 2) 800 rads-1 3), , planet is, , 5R, 4, , . Then, ratio of acceleration due, , to gravity on that planet and on the earth is, 1) 5:4, 2) 4:5, 3) 25:16 4) 16:25, 20. A particle hanging from a massless spring, stretches it by 2 cm at earth’s surface. How, much will the same particle stretch the spring, at a height of 2624 Km from the surface of, the earth? (Radius of earth = 6400 Km), 1) 1cm, 2) 2cm, 3) 3cm, 4) 4cm, 21. The value of acceleration due to gravity ‘g’, on the surface of a planet with radius double, that of earth and same mean density as that, of the earth is ( g e = acceleration due to, gravity on the surface of earth), 1) g p = 2ge 2) g p = ge / 2 3) g p = ge / 4 4) g p = 4ge, 22. If g is acceleration due to gravity on the, surface of the earth, having radius R, the, height at which the acceleration due to gravity, reduces to g/2 is, 1) R/2, 166, , 23. There are two bodies of masses 100Kg and, 1000Kg separated by a distance 1m. The, intensity of gravitational field at the mid point, of the line joining them will be, 2) 2.4 ×10−7 N / kg, 1) 2.4 ×10−6 N / kg, 3) 2.4 ×10−8 N / kg, 4) 2.4×10−9 N / kg, 24. Masses 4 kg and 36 kg are 16 cm apart. The, point where the gravitational field due to them, is zero is, 1)6 cm from 4 kg mass 2)4 cm from 4 kg mass, 3)1.8 cm from 36 kg mass 4)9 cm from each mass, 25. Two particles of masses 4Kg and 8Kg are, kept at x = - 2m and x = 4m respectively. Then,, the gravitational field intensity at the origin is, 1) G, 2) 2G, 3) G/2, 4) G/4, 26. Three particles each of mass m are kept at, the vertices of an equilateral triangle of side, L. The gravitational field at the centre due to, these particles is, , 2), , 2R, , 3), , R, 2, , 4), , (, , ), , 2 −1 R, , 1) Zero 2), , 3GM, L2, , 3), , 9GM, L2, , 4), , 2GM, L2, , GRAVITATIONAL POTENTIAL,, POTENTIAL ENERGY, 27. Three particles each of mass m are placed at, the corners of an equilateral triangle of side, b. The gravitational potential energy of the, system of particles is, −Gm2, −3Gm2, −3Gm2, −Gm2, 2), 3), 4), 2b, b, 2b, b, 28. If W is the weight of a satellite on the surface, of the earth, then the energy required to launch, that satellite from the surface of earth into a, circular orbit of radius 3R is (here R is the, radius of earth), 1) 5WR/6 2) 6WR/5 3) 2WR/3 4) 3WR/2, 29. A body of mass m is lifted from the surface of, earth to a height equal to R/3 where R is the, radius of earth, potential energy of the body, increases by, 1) mgR/3 2) mgR/4, 3) 2mgR/3 4) mgR/9, 30. An object of mass 2 Kg is moved from infinity, to a point P. Initially that object was at rest, but on reaching P its speed is 2m/s. The, workdone in moving that object is -4J., Then potential at P is ...........J/Kg., 1) 8, 2) - 8, 3) 4, 4) - 4, , 1), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 31. If mass of earth is M, radius is R, and, gravitational constant is G, then workdone to, take 1Kg mass from earth surface to infinity, will be, GM, 2R, 32. A body of mass m is placed on the earth surface, is taken to a height of h = 3R, then, change in, gravitational potential energy is, GM, 2R, , 1), , 2), , GM, R, , 3), , 2GM, R, , 4), , mgR, 2mgR, 3mgR, mgR, 2), 3), 4), 4, 3, 4, 6, 33. A body is released from height 5R where R is, the radius of the earth. Then that body reaches, the ground with a velocity equal to, , 1), , 5 gR, 3, , 1), , 2), , 3 gR, 5, , 3) 5gR, , 4) 3gR, , 34. The difference in PE of an object of mass 10kg, when it is taken from a height of 6400km to, 12800 km from the surface of the earth is, , (M, , E, , = 6 ×1024 kg ), , 2) 1.565 × 108 J, , 3) 2.65 × 108 J, 4) 4.5 × 108 J, 35. If the gravitational potential energy of a body, at a distance r from the centre of the earth is, U, then its weight at that point is, 2), , U2, r, , 3) U 2r, , 4), , U, r, , ESCAPE & ORBITAL SPEEDS, 36. The escape velocity of an object on a planet, whose radius is 4times that of the earth and ‘g’, value 9 times that on the earth, in kms-1, is, 1) 33.6, 2) 67.2, 3) 16.8, 4) 25.2, 37. The escape velocity of a sphere of mass ‘m’ is, given by, 1), , 2GMm, Re, , 2), , 2GM, Re2, , 3), , 2GMm, 4), Re2, , 2GM, Re, , 38. A body is projected up with a velocity equal to, 3/4th of the escape velocity from the surface, of the earth. The height it reaches is (Radius, of the earth is R), 1) 10R/9, 2) 9R/7, 3) 9R/8 4) 10R/3, 39. A space craft is launched in a circular orbit, very close to earth. What additional velocity, NARAYANAGROUP, , surface is Ve . The escape velocity of the, same body from a height equal to 7R from, earth surface will be, 1), , Ve, 2, , 2), , Ve, 2, , 3), , Ve, 2 2, , 4), , Ve, 4, , 42. Escape velocity of a body from the surface of, the earth is V1 and from an altitude equal to, twice the radius of the earth , escape velocity, is V2 . Then,, 1) V1 =V2 2) V1 = 7 V2 3) V1 = 3V2 4) V1 = 2 V2, , EARTH SATELLITES, , 1) 1.045 × 108 J, , 1) U, , should be given to the space craft so that it, might escape the earth’s gravitational pull, 1) 20.2Kms −1 2) 3.25Kms−1 3) 8Kms −1 4) 11.2Kms−1, 40. If the escape velocity on earth is 11.2 km/s,, its value for a planet having double the radius, and 8 times the mass of earth is..(in km/sec), 1) 11.2, 2) 22.4, 3) 5.6, 4) 8, 41. The escape velocity of a body from earth, , 43. The ratio of the orbital speeds of two satellites, of the earth if the satellites are at heights, 6400km and 19200km (Radius of the earth =, 6400 km), 1) 2 :1, 2) 3 :1 3) 2:1, 4) 3:1, 44. An artificial satellite is revolving in a circular, orbit at height of 1200 km above the surface, of the earth. If the radius of the earth is, 6400km and mass is 6 ×1024 kg, the orbital, velocity is ( G = 6.67 ×10−11 Nm 2 / kg 2 ), , 1) 7.26kms −1, 2) 4.26kms −1, 3) 9.26kms −1, 4) 2.26kms −1, 45. The mean radius of the orbit of a satellite is 4, times as great as that of a parking orbit of, the earth. Then its period of revolution around, the earth is, 1) 4 days 2) 8 days, 3) 16 days 4) 96 days, 46. If the mass of earth were 4 times the present, mass, the mass of the moon were half the, present mass and the moon were revolving, round the earth at twice the present distance,, the time period of revolution of the moon would, be (in days), 1) 56 2, 2) 28 2 3) 14 2 4) 7 2, 167
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 47. A satellite of mass ‘m’ revolves around earth of, mass M in a circular orbit of radius,’r’ with angular, velocity ω . If radius of the orbit becomes 9r, then, angular velocity of this orbit is, ω, ω, 1) 9ω, 2), 3) 27ω, 4), 27, 9, 48. An artificial satellite of mass ‘m’ is revolving, around in a circular orbit of radius ‘r’. If the, mass of earth is M, angular momentum of the, satellite with respect to the centre of earth is, (2012M), 1), , GM m2 r, , 55. The height of geo-stationary satellite above, the centre of earth is (in km), 1) 6400, 2) 12800 3) 36000 4) 42000, , WEIGHTLESSNESS, 56. How much faster than its normal rate should, the earth rotate about its axis so that the, weight of the body at the equator becomes, zero (R = 6.4 ×106 m, g = 9.8m / s 2 ) (in times), 1)nearly17 2)nearly12 3)nearly 10 4)nearly 14, , LEVEL-1(H.W) - KEY, 1) 3, 7) 3, 13) 4, 19) 2, 25) 3, 31) 2, 37) 4, 43) 1, 49) 1, 55) 4, , 2) 2m GMr, GM, , 3) 2M Gmr, 4), r, 49. Two satellites of masses 400kg, 500 kg are, revolving around earth in different circular, orbits of radii r1 , r2 such that their kinetic, energies are equal. The ratio of r1 , r2 is, 1) 5 : 4 2) 16:25 3) 45 : 4 4) 25:16, 50. Angular momentum of a satellite revolving, round the earth in a circular orbit at a height, R above the surface is L. Here R is radius of, the earth. The magnitude of angular, momentum of another satellite of the same, mass revolving very close to the surface of, the earth is, 1)L/2, 2)L/ 2, 3) 2 L, 4)2L, , ENERGY OF SATELLITES, 51. The K.E. of a satellite is 104 J . Its P.E. is, 1) −104 J 2) 2 ×104 J 3) −2×104 J 4) −4×104 J, 52. Energy required to move a body of mass ‘m’, from an orbit of radius 3R to 4R is, GMm, 1) 2R, , GMm, , GMm, , 3) 1, 9) 1, 15) 3, 21) 1, 27) 3, 33) 1, 39) 2, 45) 2, 51) 3, , 4) 4, 10) 3, 16) 1, 22) 4, 28) 1, 34) 1, 40) 2, 46) 2, 52) 4, , GEO-STATIONARY AND POLAR, SATELLITES, 54. Imagine a geo-stationary satellite of earth, which is used as an inter continental telecast, station. At what height will it have to be, established, 1) 10 3 m 2) 6.4×103m 3) 35.94 ×106 m 4)infinity, , 5) 4, 11) 2, 17) 4, 23) 2, 29) 2, 35) 4, 41) 3, 47) 4, 53) 1, , 6) 2, 12) 1, 18) 1, 24) 2, 30) 2, 36) 2, 42) 3, 48) 1, 54) 3, , LEVEL-1(H.W) - HINTS, 1., , 2., 3., 4., 5., , dA, L, dA, =, ⇒, ∝ vr ∝ ω r 2, dt 2m, dt, From Kepler’s 3rd law, T 2 ∝ r 3, , dA, A, A′, A, = constant ⇒, =, ⇒ A′ =, dt, 28 7, 4, , From Kepler’s 3rd law, T 2 ∝ r3, From Kepler’s 3rd law, T 2 ∝ r3, , 2, 6. From Kepler’s 3rd law, mrw =, , GMm, r3, T = 2π, 2, r, GM, , 2, 7. From Kepler’s 3rd law, mrw =, , GMm, r3, T, =, 2, π, r2, GM, , GMm, , 2) 6R, 3) 12 R, 4) 24 R, 53. K.E. of an orbiting satellite is K. The minimum, additional K.E. required so that it goes to, infinity is, 1) K, 2) 2K, 3) 3K, 4) K/2, , 168, , 2) 3, 8) 2, 14) 3, 20) 1, 26) 1, 32) 3, 38) 2, 44) 1, 50) 2, 56) 1, , 8., , From Kepler’s 2nd law,, , 1, rv = Constant, 2, , dA L, =, dt 2m, 10. Gravitational force does not depend upon the, medium between the masses., Gm1m2, 11. Fg =, Given, m2 = 4m1, R2, G m 1m 2, 12. F =, ; Gravitational force between, r2, two point masses is independent of the presence, of other masses., , 9., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , midpoint of a side and (ii) at the centre of the, triangle are, respectively., , LEVEL - II (C.W), , 1) 0,, , KEPLER’S LAWS, 1., , 2., , 3., , If the Earth shrinks such that its density, becomes 8 times to the present value,, then new duration of the day in hours, will be, (2008M), 1) 24, 2) 12, 3) 6, 4) 3, A planet moves around the sun. At a given, point P, it is closest from the sun at a distance, d1 ,and has a speed V1 . At another point Q,, when it is farthest from the sun at a distance, d 2 , its speed will be, d2V1, d1V1, d12V1, d22V1, 1), 2) d, 3) d, 4) 2, d2, d1, 1, 2, If a graph is plotted between T 2 and r 3 for, a planet then, its slope will be, 2, GM, T, 4π 2, 1), 2), 2, GM, , 4π, , 3) 4π GM 4) Zero, 3, r, Two different artificial satellites orbiting with, same time period around the earth having, angular momenta 2:1. The ratio of masses of, the satellites is, 1) 2:1, 2) 1:2, 3) 1:1, 4)1:3, The ratio of Earth’s orbital angular momentum, (about sun) to its mass is 4 .4 × 1015 m2s–1., The area enclosed by the earth’s, orbit is approximately, 1) 1 × 1022 m2, 2) 3 × 1022 m2, 22, 2, 3) 5 × 10 m, 4) 7 × 1022 m2, , θ, , 4., , 5., , 3), 8., , 7., , Gravitational force between two point masses, m and M separated by a distance r is F. Now, if a point mass 3m is placed next to m, the, total force on M will be, 1) F, 2) 2F, 3) 3F, 4) 4F, If three particles,each of mass M are placed, at the three corners of an equilateral triangle, of side a, the force exerted by this system on, another particle of mass M placed (i) at the, , 170, , 3 GM, a2, , 2, , ,, , GM, a2, , 2, , 4) 0, 0, , Two masses ‘M’ and ‘4M’ are at a distance, ‘r’ apart on the line joining them, ‘P’ is point, where the resultant gravitational force is zero, (such a point is called as null point). The, distance of ‘P’ from the mass ‘M’ is, , r, r, 4r, 2r, 2), 3), 4), 1) 5, 3, 5, 3, 9. If the mass of one particle is increased by 50, % and the mass of another particle is, decreased by 50 %, the force between them, 1)decreases by 25%, 2)decreases by75 %, 3) increases by 25%, 4)does not change, 10. If the distance between the sun and the earth, is increased by three times , then the, gravitational force between the two will, 1) remain constant, 2) decrease by 63%, 3) increase by 63%, 4) decrease by 89%, 11. Two lead balls of masses m and 5m having, radii R and 2R are separated by 12R. If, they attract each other by gravitational, force, the distance covered by small sphere, before they touch each other is, 1) 10 R 2) 7.5 R, 3) 9 R, 4) 2.5 R, 12. Three identical particles each of mass “m”, are arranged at the corners of an equilateral, triangle of side “L”. If they are to be in, equilibrium, the speed with which they must, revolve under the influence of one another’s, gravity in a circular orbit circumscribing the, triangle is, , LAW OF GRAVITATION, 6., , 4GM 2, ,0, 2), 3a 2, , 4GM 2, 3a 2, , 1), 13., , 3Gm, L, , 2), , Gm, L, , 3), , Gm, 3L, , 4), , 3Gm, L2, , Two particles each of mass ‘m’ are placed, at A and C are such AB=BC=L. The, gravitational force on the third particle placed, at D at a distance L on the, perpendicular bisector of the line AC is, Gm2, Gm2, 1) 2 along BD 2), along DB, 2L2, L, Gm2, Gm2, 3) 2 along AC 4) 2 along BD, L, L, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , ACCELERATION DUE TO, GRAVITY AND ITS VARIATION, 14. The height at which the value of acceleration, due to gravity becomes 50% of that at the, surface of the earth. (Radius of the earth =, 6400 km) is, 1) 2650, 2) 2430, 3) 2250, 4) 2350, 15. The radius and density of two artificial, , 2) 2gR 3) gR 4) 4gR, 1) 4 gR, 21. A projectile is fired vertically upwards from, the surface of the earth with a velocity Kve ,, where ve is the escape velocity and K<1.If R, is the radius of the earth, the maximum height, to which it will rise measured from the centre, of the earth will be(neglect air resistance), , satellites are R1 , R2 and ρ1, ρ 2 respectively.., The ratio of acceleration due to gravities on, them will be, R1ρ2, 2) R ρ, 2 1, , R2 ρ2, 1) R ρ, 1 1, , R1ρ1, 3) R ρ, 2 2, , R2ρ1, 4) R ρ, 1 2, , 16. A man weigh ‘W’ on the surface of earth and, his weight at a height ‘R’ from surface of earth, is (R is Radius of earth), 1), , W, 4, , 2), , W, 2, , 3) W, , 4) 4W, , 17. The acceleration due to gravity at the latitude, 450 on the earth becomes zero if the, angular velocity of rotation of earth is, 1), 18., , 2, gR, , 2) 2gR, , 2g, R, , 3), , 4), , 5R, 2, , Acceleration due to gravity on moon is 1/6 of, the acceleration due to gravity on earth. If the, ratio of densities of earth and moon is 5/3, then, radius of moon in terms of radius of earth will, be, 1), , 5, Re, 18, , 2), , 1, Re, 6, , 3), , 1, 3, Re, Re 4), 2 3, 18, , ESCAPE SPEED, 19. The mass of a planet is half that of the earth, and the radius of the planet is one fourth that, of earth. If we plan to send an artificial satellite, from the planet, the escape velocity, will be, ( Ve = 11kms −1 ), 1)11kms −1, , 2) 5.5 kms −1, , 3) 15.55 kms −1, , 4) 7.78 kms −1, , 20. If a rocket is fired with a velocity, V = 2 gR, near the earth’s surface and goes upwards, its, speed in the inter-stellar space is, NARAYANAGROUP, , 1), , 1− K 2, R, , 2), , R, 1− K 2, , R, 1+ K 2, If the radius of earth shrinks by 0.2%, without any change in its mass , escape, velocity from the surface of the earth, 1) increases by 0.2% 2) decreases by 0.2%, 3) increases by 0.1% 4) increases by 0.4%, , 3) R(1 − K 2 ), 22., , 4), , 23. If d is the distance between the centres of, the earth of mass M1 and moon of mass M2,, then the velocity with which a body should, be projected from the mid point of the line, joining the earth and the moon, so that it just, escapes is, G(M1 + M2 ), d, , 1), , 2G (M 1 + M 2 ), d, , 3), , 2), 4), , G(M 1 + M 2 ), 2d, , 4G (M 1 + M 2 ), d, , 24. The escape velocity of a planet having mass 6, times and radius 2 times as that of earth is, 1), , 3ve 2) 3ve, , 3), , 2ve, , 4) 2ve, , 25. If ‘ ve ’ is the escape velocity of a body from a, planet of mass ‘M’ and radius ‘R’. Then the, velocity of the satellite revolving at height ‘h’, from the surface of the planet will be, 1) ve, , R, R, 2R, R+h, 2) ve, 3) ve, 4) ve 2 ( R + h ), R, R+h, R+h, , 26. A particle falls towards earth from infinity. The, velocity with which it reaches earth’s surface is, 1) v = 2gR, 2) v = 2gR, 3) v = gR, 4) v = R/g, 171
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , EARTH SATELLITES, 27. Two satellites P, Q are revolving around earth, in different circular orbits. The velocity of P, is twice the velocity of Q. If the height of P, from earth’s surface is 1600 km. The radius, of orbit of Q is ( R =6400 km)., 1)1600km 2)20000km 3)32000km 4)40000km, 28. A planet is revolving around the sun. Its, distance from the sun at apogee is rA and that, at perigee is rP . The masses of planet and sun, are ‘m’ and M respectively, VA is the velocity, of planet at apogee and VP is at perigee, respectively and T is the time period of, revolution of planet round the sun, then identify, the wrong answer., 2, 1) T =, , π2, 3, ( rA + rP ), 2Gm, , 2, 2) T =, , π2, 3, ( rA + rP ), 2GM, , 3) v ArA = v P rP, 4) v A < v P ; rA > rP, 29. Suppose the gravitational force varies, inversely as the nth power of distance, then, the time period of a planet in circular orbit of, radius ‘R’ around the sun will be proportional, to, (2004A), n +1 , , , 2 , , n −2 , , , 2 , , n −1 , , 3) R n 4) 2 , R, R, R, 30. An artificial satellite is revolving around the, earth in a circular orbit. Its velocity is onethird of the escape velocity. Its height from, the earth’s surface is ( in Km), 1) 22400, 2) 12800 3) 3200, 4) 1600, 1), , 2), , GRAVITATIONAL POTENTIAL ENERGY, 31. The work done to increase the radius of orbit, of a satellite of mass ‘m’ revolving around a, planet of mass M from orbit of radius R in to, another orbit of radius 3R is, 2GMm, GMm, GMm, GMm, 2), 3), 4), 1), 3R, 3R, 6R, 24R, 32. A stone is dropped from a height equal to nR,, where R is the radius of the earth, from the, surface of the earth. The velocity of the stone, on reaching the surface of the earth is, , 172, , 2g(n +1)R, 2 gR, 2gnR, 2), 3), 4) 2 gnR, n, n +1, n +1, 33. Three particles of equal mass ‘m’ are situated, at the vertices of an equilateral triangle of side, ‘L’. The work done in increasing the side of, the triangle to 2L is, 1), , 2G2m, Gm2, 3Gm2, 3Gm2, 2), 3), 4), 2L, 2L, 2L, L, 34. A small body is at a distance ‘r’ from the, centre of mercury, where ‘r’ is greater than, the radius of Mercury. The energy required, to shift the body from r to 2r measured from, the centre is E. The energy required to shift it, from 2r to 3r will be, , 1), , E, E, E, 3), 4), 2, 3, 4, 35. Escape velocity of a body of 1kg mass on a, planet is 100m/s. Gravitational potential, energy of the body at the planet is, 1) -5000J 2) -1000J 3) -2400J 4) 5000J, 36. By what percent the energy of the satellite, has to be increased to shift it from an orbit of, , 1) E, , 2), , 3r, 2, 1) 66.7% 2) 33.3%, 3) 15% 4) 20.3%, 37. At what height from the surface of earth, the, total energy of satellite is equal to its potential, energy at a height 2R from the surface of earth, (R = radius of earth ), 1) 2R, 2) R/2, 3) R/4, 4) 4R, GEO-STATIONARY SATELLITES, 38. A geo-stationary satellite is orbiting the earth, at a height 6R above the surface of the earth,, where R is the radius of earth. The time period, of another satellite revolving around earth at a, height 2.5 R from earth’s surface is, 1) 12 2hr 2) 12 hr, 3) 6 2hr 4) 6 hr, , radius ‘r’ to, , LEVEL-II (C.W) KEY, 1)3, 7)2, 13)2, 19)3, 25)4, 31)2, 37)2, , 2)3, 8)2, 14)1, 20)2, 26)2, 32)3, 38)3, , 3)1, 9)1, 15)3, 21)2, 27)3, 33)3, , 4)1, 10)4, 16)1, 22)3, 28)1, 34)3, , 5)4, 11)2, 17)3, 23)4, 29)1, 35)1, , 6)4, 12)2, 18)1, 24)1, 30)1, 36)2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 27. Given, vP = 2vQ, , LEVEL - II (H.W), , GM, GM, =2, → rQ = 4rP, rP, rQ, 2, 28. T =, , 4π rA + rP , GM 2 , , 1., , rA + rP , , Q r =, , 2 , , , 3, , 2, , KEPLER’S LAWS, , By the law of conservation of angular, momentum v ArA = v P rP, 1 2 mgh, mv =, k, 2, h, 29. F = n , F = m r ω ; 30. 2, 1+, r, R, , 1), , 31. Workdone = change in TE, GMm GMm, −, Workdone =, 2R, 6R, , 2., , −3Gm2, L, , Final potential energy U f = −, W= U f − Ui = −, , 3Gm2, 2L, , 3Gm2, 2L, , 3., , 3Gm2 3Gm2, −−, =, L , 2L, , , GMm GMm GMm, +, =, 2R, R, 2R, , E1 = −, , ∴ PE (U ) = −, , 36. E =, , 174, , 5., , GMm, = −5000J, R, , −GMm, −GMm, −GMm, W = E2 − E1 =, , E2 =, 2r, 2r1, 2r2, , −GMm −GMm , =, 37., r = R + h, 2r, 3R , , 38., , 4., , GMm GMm GMm, +, =, 3R, 2R, 6R, , 2GM, GM, = 100 ⇒, = 5000, R, R, , 35. ve =, , T1, =, T2, , 3, 1, 3, 2, , R, R, , =, , (7 R ), , 3, , 7R , , , 2 , , 3, , 2) π ×104 3) 2π ×104 4) 4π ×104, , Two metal spheres each of radius ‘r’ are kept, in contact with each other. If d is the density, of the material of the sphere, then the, gravitational force between those spheres is, proportional to, 2, r2, d, 1) d r, 2) d r 3), 4) 2, d, r4, Two lead spheres of same radius are in contact, with each other. The gravitational force of, attraction between them is F. If two lead, spheres of double the previous radius are in, contact with each other, the gravitational force, of attraction between them will be, 1) 2F, 2) 32F, 3) 8F, 4) 16F, The gravitational force between two bodies is, decreased by 36% when the distance between, them is increased by 3m. The initial distance, between them is, 1) 6 m, 2) 9 m, 3) 12 m, 4) 15 m, Two particles of masses ‘m’ and ‘2m’ are at a, distance ‘3r’ apart at the ends of a straight, line AB. C is the centre of mass of the system., The magnitude of the gravitational force on a, unit mass placed at C due to the masses is, 2 6, , 34. E = U 2 − U1, ⇒−, , π, × 104, 2, , LAW OF GRAVITATION, , 1 2 mgh, mv =, h, 32. 2, 1+, R, , 33. Initial potential energy U i =, , Two satellites S1 and S2 are revolving round a, planet in coplanar and concentric circular, orbits of radii R1 and R2 in the same direction, respectively. Their respective periods of, revolution are 1hr and 8 hr. The radius of the, orbit of satellite S1 is equal to 104 km. Their, relative speed when they are closest, in kmph, is :, , 7Gm, 9Gm, 3Gm, 3), 4), 2, 2, 4r, 4r, 2r 2, If the distance between two bodies is, increased by 25%, then the % change in the, gravitational force is, 1) Decreases by 36%, 2) Increases by 36%, 3) Increases by 64%, 4) Decreases by 64%, , 1) Zero, 6., , 2 4, , 2), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 7., , Three point masses each of mass ‘m’ rotate, in a circle of radius r with constant angular, velocity ω due to their mutual gravitational, attraction. If at any instant, the masses are, on the vertex of an equilateral triangle of side, ‘a’, then the value of ω is, Gm, Gm, 3Gm, 2), 3), 4) Zero, 3, 3, a, a, 3a3, The angular momentum (L) of earth revolving, round the sun is proportional to r n , where r is, the orbital radius of the earth. The value of, ‘n’ is :(assume the orbit to be circular), 1, 1, 2) 1, 3) −, 4) 2, 1), 2, 2, Four particles of masses m, 2m,3m and 4m are, placed at the corners of a square of side length, a. The gravitational force on a particle of mass, m placed at the centre of the square is, 1), , 8., , 9., , 1) 4 2, , Gm2 3 2Gm2, 2Gm2, 2 2Gm2, 2), 3), 4), a2, a2, a2, a2, , ACCELERATION DUE TO, GRAVITY AND ITS VARIATION, 10. If the radius of the earth is made three times,, keeping the mass constant, then the weight of, a body on the earth’s surface will be as, compared to its previous value is, 1)one third 2)one ninth 3)three times 4)nine times, 11. The difference in the value of ‘g’ at poles and, at a place of latitude 450 is, Rω 2, Rω 2, Rω 2, 2), 3), 4), 1) Rω, 2, 4, 3, 12. The angular velocity of earth’s rotation about, its axis is ‘ ω ’. An object weighed by a spring, balance gives the same reading at the equator, as at height ‘h’ above the poles ., The value of ‘h’ will be, 2, , ω 2 R2, ω 2 R2, 2ω 2 R 2, 2ω 2 R 2, 1), 2), 3), 4), g, 2g, g, 3g, 13. The radius and acceleration due to gravity of, 1, 1, moon are and that of earth, the ratio of, 4, 5, the mass of earth to mass of moon is, 1) 1:80, 2) 80:1, 3) 1:20, 4) 20:1, NARAYANAGROUP, , GRAVITATION, 14. The difference in the value of ‘g’ at poles and, at a latitude is, , 3, Rω 2, 4, , then latitude angle is, , 1) 600, 2) 300, 3) 450, 4) 950, 15. A particle hanging from a spring stretches it, by 1 cm at earth’s surface. Radius of earth is, 6400 km. At a place 800 km above the earth’s, surface, the same particle will stretch the, spring by, 1)0.79 cm 2)1.2 cm 3) 4 cm 4)17 cm, 16. A tunnel is dug along a diameter of the earth., The force on a particle of mass ‘m’ placed in, the tunnel at a distance x from the centre is, 1), , GM e m, GM e m, GMem, GM e mR 3, x, x, 2), 3), 4), x, R3, R2, R3 x, , ESCAPE SPEED, 17. The mass of the Earth is 9 times that of Mars., The radius of the Earth is twice that of Mars., The escape velocity of the Earth is 12 km/sec., The escape velocity on Mars, is .... km/sec, 1) 4 2km / sec, 2) 2 2km / sec, 4) 8 2km / sec, 3) 6 2km / sec, 18. The escape velocity of a body from earth is, 11.2 km/s. If a body is projected with a velocity, twice its escape velocity, then the velocity of, the body at infinity is (in km/s), 1) 19.4, 2) 194, 3) 1.94, 4) 0.194, 19. A particle is kept at rest at a distance R, (Earth’s radius) above the earth’s surface. The, minimum speed with which it should be, projected so that it does not return is, GM, GM, GM, GM, 2), 3), 4), R, 2R, 3R, 4R, 20. 16 kg and 9 kg are separated by 25m. The, velocity with which a body should be projected, from the mid point of the line joining the two, masses so that it just escape is, 1), , 1) g, 2) 2gR 3) G 4) 2 G, 21. The escape velocity from earth is 11 km/, sec.The escape velocity from a planet having, nine times the radius and one third of density, as earth is ___ Km/sec, 1) 11, 2) 22 3 3) 33 3 4) 44 3, 175
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , EARTH SATELLITES, 22. The speed of a satellite that revolves around, earth at a height 3R from earth’s surface is, ( g=10 m / s 2 at the surface of earth, radius of, earth R = 6400 km) (in kms-1), 1) 2 2, 2) 4, 3) 4 2, 4) 8, 23. If an artificial satellite is moving in a circular, orbit around earth with speed equal to one, fourth of Ve from earth, then height of the, satellite above the surface of the earth is, 1)7R, 2) 4R, 3) 3R, 4) R, 24. Two satellites A and B go round a planet P in, circular orbits having radii 4R and R, respectively. If the speed of the satellite A is, 3V, the speed of the satellite B is, 1) 12V, 2) 6V, 3) 4V, 4)3V, 25. A satellite moving in a circular path of radius, ‘r’ around earth have a time period T. If its, radius slightly increases by ∆r , the change in, its time period is, 1), , 3T, , 2 r, , , ∆r, , , 3T2 , 3) 2 r 2 ∆r, , , T, 2) , r, , 27. A stone is dropped from a height equal to 3R,, above the surface of earth. The velocity of, stone on reaching the earth’s surface is, R, 3, gR 3) 2gR 4) gR, 2), 2, 2, 28. If ‘g’ is acceleration due to gravity on the, earth’s surface, the gain in the potential, energy of an object of mass ‘m raised from, the surface of the earth to a height equal to, the radius ‘R’ of the earth is, 1) 2 mgR, 2) mgR 3) mgR/4 4) mgR/2, 29. The work done in bringing three particles each, of mass 10 gm from large distance to the, 176, , 2), , LEVEL-II(H.W) - KEY, 1)2, 7)2, 13)2, 19)1, 25)1, 31)1, , 3T2 , 4) ∆r, 2 r , , ENERGY OF SATELLITES,, GRAVITATIONAL POTENTIAL ENERGY, , g, , E, E, E, 3), 4), 2, 3, 4, 31. By what percent, the energy of the satellite, has to be increased to shift it from an orbit of, radius ‘r’ to ‘3r’, 1) 66.7%, 2) 33.3% 3) 15%, 4) 20.3%, 32. At what height from the surface of earth, the, total energy of satellite is equal to its potential, energy at a height 3R from the, surface of earth ( R = radius of earth ), 1) 4R, 2) 3R, 3) 2R, 4) R, , 1) E, , , ∆r, , , 26. A satellite is orbiting earth in an orbit with a, velocity 4 km/sec, then acceleration due to, gravity at that height is (in ms-2 ), 1) 0.4, 2) 0.62, 3) 0.87, 4) 1.21, , 1), , vertices of an equilateral triangle of side 10, cm is(approximately), 1) 10 −13 J 2) 2 ×10−13 J 3) 4 ×10 −13 J 4) 10 −11 J, 30. The energy required to shift the body, revolving around a planet from r to 2r is E, (measured from centre of planet). The energy, required to shift it from 2r to 4r is, , 2)2, 8)1, 14)2, 20)4, 26)2, 32)4, , 3)4, 9) 1, 15)1, 21)3, 27)2, , 4)3, 10)2, 16)1, 22)2, 28)4, , 5)2, 11)2, 17)1, 23)1, 29)2, , 6)1, 12)2, 18)1, 24)2, 30)2, , LEVEL-II(H.W) - HINTS, 1., , 2π R, T 2α R 3 , V0 = T , Rel. velocity = V01 − V02, , 3., , Gm1m2, , Here, m = (Vol)(den), R2, For solid spheres in contact Fα R 4, , 4., , F=, , Gm1m2, 1, ⇒F∝ 2, 2, R, R, , 5., , F=, , Gm1m2, , Fnet = F2 − F1, R2, , 6., , F1 d22 ∆F, F −F, = 2;, ×100 = 1 2 ×100, F2 d1 F, F1, , 7., , F = F12 + F22 + 2 F1 F2 cos θ, , 2., , F=, , 3GMm, = mrω 2, 2, a, , Here, r =, , a, 3, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , 8., 9., , GRAVITATION, , mv 2 GMm, = 2, r, r, Find individual forces and calculate resultant, L = mvr ;, , GM , ve, 23. v = 4 , v0 = R + h , , , , 12. F = mg ′, g ′ = g − Rω 2 cos 2 φ, , ∆ T 3 ∆r , = , T, 2 r , , GM, M, ⇒ g α 2, 2, R, R, , 2, 2, 14. Rω cos λ =, , vh, =, 26., vs, , 3, Rω 2, 4, , 16., , e∝g, , gh, R, =, 2, g, ( R + h), , Here,, , 3Gm1m2 3Gm1m2, 29. W = −U = − −, =, r, r, , , , vm, M m Re, =, 17., ve, M e Rm, , 30. TE =, , 18. velocity at infinity, v∞ = v 2 − v 2e = 2 (11.2 ) − (11.2 ), 2, , 2, , 31. TE =, , GMm, , 19. F = ( R + h )2, , 20. v =, , ( R + h), , 2, , =, , −GMm, ;, 2r, , −GMm ∆E, × 100%, ;, 2r, E, , GMm, , 32. T.E = P.E ;, , Mv 20, and centripetal force F = ( R + h ), GMm, , ;, , 21. vα R ρ, g, 22. ‘g’ at a height 4 R =, 16, g, ( 4R ), and v 0 = g ( 4 R ) →, 16, , NARAYANAGROUP, , ( R + h), , 2, , =, , 1, mv 02, 2, , LEVEL - III, 1., , Mv 20, ( R + h), , 4G ( M 1 + M 2 ), d, , 2, , 1 1 , 28. P.E. = −GMm − , R1 R2 , , GM ' m, M' M, F=, but 3 = 3, 2, x, x, R, , ⇒, , R, R , , gh = g , , R+h, R+h, , 1 2 ( mg ) Rh ( mg ) R ( nR ), 27. 2 mv = R + h = R + nR, (, ), , 2, , 15., , RB, RA, , 25. According to Kepler’s II law, T 2α R 3, differentiating on both sides w.r.t ‘r’, , 2, 2, 11. gφ = g − Rω cos φ , g poles = g, , 13. g =, , GM, v, ⇒ A=, R, vB, , 24. v =, , GMm, 10. W = mg =, r2, , A point mass is orbiting a significant mass M, lying at the focus of the elliptical orbit having, major and minor axes given by 2a and 2b, respectively. Let r be the distance between, the mass M and the end point of major axis., Velocity of the particle can be given as, 1), , ab GM, r, a3, , ab GM, 3), 2r r 3, , 2), , ab GM, r, b3, , 2ab, GM, 3, 4) r a + b , 2 , 177
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 2., , A planet of mass m revolves in elliptical orbit, around the sun so that its maximum and, minimum distances from the sun are equal to, ra and rp respectively. The angular momentum, of this planet relative to the sun is, 1) L = m, , GMrp ra, , (r, , p, , 3) L = M, , + ra ), , Gmrp ra, , (r, , 4) L = M, , + ra ), , P, , 8., , 2GMrp ra, , (r, (r, , p, , + ra ), , p, , + ra ), , Gmrp ra, A satellite moving in elliptical orbit around earth, as shown. The minimum and maximum distance, of the satellite from earth are 3 units and 5, units respectively. The distance of satellite from, earth when it is at ‘P’ is —— ( units ), p, , 3., , 2) L = m, , T, T, 2T, 2, T 3), 2), 4), 3, 6, 3π, 8, A homogeneous spherical heavenly body has, a uniform and very narrow frictionless duct, along its diameter. Let mass of the body be M, and diameter be D. A point mass m moves, smoothly inside the duct. Force, exerted on this mass when it is at a distance, s from the centre of the body is, 3, , 1), , 1), 9., , GMm, GMm πGMm s 8GMm, ., 3 3), 2, s, 2), 4), 2, 3, R, −, s, D, /, 2, (, ), (, ), D, s, , Two concentric shells of different masses m1, and m2 are having a sliding particle of mass, m . The forces on the particle at position, I , II and III are, , 1) 4, 2) 3, , S, , 4., , E, , O, , Q, , 6., , 7., , r2, , 3) 3.75, m1, , 4) 6, The longest and the shortest distance of a planet, from sun is R1 and R2 . Distance from sun when, it is normal to major axis of orbit is, R1R2, 2R1R2, R +R, R +R 2, 2), 1) 1 2, 3) R + R 4) R + R, 2, 2, 1, 2, 1, 2, A satellite is orbiting just above the surface of, a planet of average density D with period T. If, G is the universal gravitational constant, the, 3π, is equal to, quantity, G, 1) T 2 D, 2) 3π T 2 D 3) 3π D 2T 4) D2T, A planet revolves around sun in an elliptical, orbit of eccentricity ‘e7’. If ‘T’ is the time, period of the planet then the time spent by the, planet between the end of the minor axis and, close to sun is, 2, 1, , 5., , II, , 178, , 1) 0,, , r1, , I, , 2, , 1 e , Te, e, , πT, 1) T −, 3) − 1 4), 2), π, e, 4 2π , π, , An artificial satellite revolves around earth in, circular orbit of radius r with time period T.., The satellite is made to stop in the orbit which, makes it fall onto earth. Time of fall of the, satellite on to earth is given by, , Gm1 G( m1 + m2 ) m, ,, 2, 2, r2, r1, Gm2, Gm1, 2) r 2 , 0, r 2, 2, 1, , III, r3, , m2, , 3), , G(m1 + m2 )m Gm2, G(m1 + m2 )m G(m2 )m, , 2 ,0 4), ,, ,0, 2, 2, 2, r1, r2, r1, r2, , 10. Suppose a vertical tunnel is dug along the, diameter of earth assumed to be a sphere of, uniform mass having density ρ . If a body of, mass m is thrown in this tunnel, its acceleration, at a distance y from the centre is given by, m, y, , 4π, G ρ ym, 3, 4, 3) πρ y, 3, , 1), , 3, π Gρ y, 4, 4, 4) π G ρ y, 3, , 2), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 11. A point mass M is at a distance S from an, infinitely long and thin rod of linear density D., If G is the gravitational constant then, gravitational force between the point mass and, the rod is, 1) 2, S, M, , 3), , MGD, S, , MGD, 2S, , 2), 4), , 2 MGD, 3 S, , R, from the centre of, 2, sphere. The gravitational force on a particle, , is located at a distance, , R, from the centre of, 2, , the sphere on the line joining both the centres, of sphere and cavity is ( opposite to the centre, of cavity ) [ here g =, , GM, where M is the, R2, , mass of the sphere ]., , mg, mg, 3mg, mg, 1), 2), 3), 4), 2, 8, 16, 4, 13. Four masses ‘m’ each are orbiting in a circle of, radius ‘r’ in the same direction under, gravitational force. Velocity of each particle is, m, m, , (, , r, , m, , 1), 2), , m, , Gm 1+ 2 2, r, 2, , (, , Gm, 1+2 2, r, , ), , Gm, r, , 4), , 8 R apart as shown in fig. The force of, attraction between the ring and the sphere is, , NARAYANAGROUP, , 2 2 GmM, 27 R 2, GmM, 2), 8R 2, , 1), , M, , 3), , GmM, 9R 2, , 2 GmM, 9 9R 2, 15. A mass m extends a vertical helical spring of, spring constant k by x m at the surface of, earth. Extension in spring by the same mass, at height h metre above the surface of earth, is, , 4), , 2, R2, GMm, GMm, R + h), (, x, 2, 1) k R + h 2), 3), x 4), ( ), ( R + h), kR2, R2, 16. A straight rod of length L extends from x = a, to x = L + a . Find the gravitational force it, exertson a pointm assm at x = 0 is, ( if the linear density of rod µ = A + Bx 2 ), , A, , , , , , 1) Gm a + BL , , 1, , , 1 , , , , , 2) Gm A a − a + L + BL, , , A , A, , , 3) Gm BL +, 4) Gm BL − , , a + L, a, , , 17. A mass m is placed in the cavity inside a, hollow sphere of mass M as shown in the, figure. The gravitational force on m is, , ), , Gm1+ 2 2 , , , 2r 2 , 14. The centres of a ring of mass m and a sphere, of mass M of equal radius R, are at a distance, , 3), , R, , 8R, , MGD, S, , R, 12. A cavity of radius, is made inside a solid, 2, sphere of radius R . The centre of the cavity, , of mass m at a distance, , m, , r, m, R, , 1), , GMm, R2, , 2), , GMm, r2, , GMm, , 3) ( R − r ) 2, 4) Zero, 18. A spherical shell is cut into two pieces along, a chord as shown in the figure. P is a, point on the plane of the chord.The, gravitational field at P due to the upper part, is I1 and that due to the lower part is I 2 ., What is the relation between them?, 179
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , P, , 1) I1 > I 2, 2) I1 < I 2, 3) I1 = I 2, , 4) no definite relation, 19. The magnitudes of the gravitational field at, distance r1 and r2 from the centre of a uniform, sphere of radius R and mass M are E1 and, , E2 respectively. Then:, E1 r1, 1) E = r if r1 < R and r2 < R, 2, 2, , E1 r22, =, if r1 < R and r2 < R, 2), E2 r12, , 3), , E1 r12, =, if r1 < R and r2 < R, E2 r22, 20. Two masses 90 kg and 160 kg are 5 m apart., The gravitational field intensity at a point 3m, from 90 kg and 4 m from 160 kg is, 1)10 G, 2)5 G, 3) 5 2G 4) 10 2G, 21. Gravitational field intensity at the centre of, the semi circle formed by a thin wire AB of, mass ‘m’ and length ‘L’ is, Y, , (), (), , Gm 2 $, i, L2, Gm 2 $, X, j, 2), A, B, π L2, 2π Gm $, 2π Gm $, i, j, 3), 4), 2, L, L2, 22. Two equal masses each ‘m’ are hung from a, balance whose scale pans differ in vertical, height by ‘h’. the error in weighing is, 1, 1) π G ρ mh, 2) G ρ mh, 3, 8, 4, 3) π G ρ mh, 4) π G ρ mh, 3, 3, 180, , GM, L, , 2) − 64, , GM, L2, , GM, L, 25. The gravitational potential of two, homogeneous spherical shells A and B of same, surface density at their respective centres are, in the ratio 3 : 4 . If the two shells, coalesce into single one such that surface, density remains same, then the ratio of, potential at an internal point of the new shell, to shell A is equal to :, 1) 3 : 2, 2) 4 : 3, 3) 5 : 3, 4) 5 : 6, 26. A thin uniform annular disc ( see figure ) of, mass M has outer radius 4R and inner radius, 3R. The work required to take a unit mass, from point P on its axis to infinity is, , 4) − 16, , 3) Zero, , 4), , (), , 2R, R, R, R, 2) 2π, 3) 2π, 4) 2π, g, 2g, 3g, g, 24. Four particles each of mass M are located at, the vertices of a square with side L. The, gravitational potential due to this at centre of, square is, , 1) 2π, , 1) − 32, , E1 r13, =, if < R and r2 < R, E2 r23 r1, , 1), , 23. If earth were to rotate on its own axis such, that the weight of a person at the equator, becomes half the weight at the poles, then its, time period of rotation is (g=acceleration due, to gravity near the poles and R is the radius, of earth) (Ignore equatorial bulge), (2013M), , P, , (), , 4R, , 1), , (, , 2GM, 4 2 −5, 7R, , 2) −, , 3), , GM, 2R, , 4), , ), , (, , 2GM, 4 2 −5, 7R, , 2GM, 5R, , (, , ), , 2 −1, , NARAYANAGROUP, , )
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 27. The gravitational force in a region is given by,, ur, F = ayi$ + ax $j . The work done by gravitational, force to shift a point mass m from ( 0,0,0 ) to, , ( x0 , y0 , z0 ), , is, , 32. A ring has non-uniform distribution of mass, having mass ‘M’ and radius ‘R’. A point mass, m0 is moved from A to B along the axis of the, ring. The work done by external agent against, gravitational force of ring is, M, , 1) max0 y 0 z0 2) max0 y0 3) − max0 y0 4)0, 28. Two identical thin rings each of radius ‘R’ are, co-axially placed at a distance ‘R’. If the rings, have a uniform mass distribution and each has, mass m1 and m2 respectively, then the work, done in moving a mass ‘m’ from the centre of, one ring to that of the other is:, Gm(m1 − m2 )( 2 − 1), 2), 2R, , 1) Zero, , Gm1m( 2 + 1), Gm 2(m1 + m2 ), 4), m2 R, R, 29. The gravitational field in a region due to a, certain mass distribution is given by, , 3), , R, , A, R, 1), , ), , 2), , GMm0 1, 1 , −, , , R 5, 2, , 4), , 2, , , , 1) −, , 25, N, 3, , 2) −, , 50, N, 3, , 3), , 25, N, 3, , 4) Zero, , 30. A particle of mass 1kg is placed at a distance, of 4m from the centre and on the axis of a, uniform ring of mass 5kg and radius 3m. The, work done to increase the distance of the, particle from 4m to 3 3m is, G, 2) 4 J, , G, 1) J, 3, , G, 4) J, 6, , G, 3) J, 5, , 31. Consider two configurations in fig (i) and fig(ii), 3m, 2m, , a, m, , a, , a, a, fig (i), , 2m m, , a, fig (ii), , 3m, , The work done by external agent in changing, the configuration from fig(i) to fig(ii) is, , 6Gm2 , 1 , 1+, 2) −, , , a , 2, , 1) Zero, , 3) −, , 6Gm 2, a, , 1 , , 1 −, , 2, , , NARAYANAGROUP, , 4), , 6Gm2 1 , −, 2− , a , 2, , GMm0 1, 1 , −, , R 2, 5, , GMm0, 5R, 33. Two concentric spherical shells A and B of, radii R and 2R and masses 4M and M, respectively are as shown. The gravitational, potential at point ‘p’ at distance ‘r’ (R <r<2R), from centre of shell is (r = 1.5R), , 3), , B, , . The work done by the field, in moving a particle of mass 2 kg from (2m,1m), to 3 m, 2m along the line 3x+4y=10 is, , , , R, , GMm 0, 2R, , ur, E = 4$i − 3 $j N / kg, , (, , B, , 2R, r, P, , 1) −, , 4GM, R, , A, R, , 9G M, , 4GM, , 19GM, , 3) −, 4) − 6 R, 2) − 2 R, 3R, 34. The potential energy of a body of mass ‘m’ is, given by U=px+qy+rz. The magnitude of the, acceleration of the body will be, 1), , p+q+r, m, , 2), , p 2 + q2 + r 2, m, , p3 + q3 + r 3, p 4 + q4 + r 4, 4), m, m, 35. A particle is placed in a field characterized by, a value of gravitational potential given by, uur, V = -kxy, where ‘k’ is a constant. If Eg is the, gravitational field then, , 3), , ur, 1) E g = k xi$ + y $j and is conservative in nature, , (, (, (, , ur, 2) E g = k yi$ +, ur, 3) E g = k xi$ +, , nature, , ur, , (, , ), x $j ) and is conservative in nature, y $j ) and is non conservative in, , ), , $ $, 4) Eg = k yi + x j and is non conservative in, nature, , 181
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 36. A thin rod of length ‘L’ is bent to form a semi, circle. The mass of the rod is ‘M’. What will, be the gravitational potential at the centre of, the circle?, −GM, −GM, −π GM, −π GM, 1), 2), 3), 4), L, 2π L, 2L, L, 37. If the gravitational field intensity at a point is, GM, , then potential at a distance ‘r’ is, r 2.5, 2GM, 2GM, GM, − GM, 1), 2) 3.5, 3) - 1.5 4) 3.5, 1.5, r, 3r, 3r, r, 38. In a certain region of space, the gravitational, field is given by −k / r , where r is the distance, and k is a constant. If the gravitational, potential at r = r0 be V0 , then what is the, expression for the gravitational potential V?, 1) k log ( r / r0 ), 2) k log ( r0 / r ), 3) V0 + k log ( r / r0 ), 4) V0 + k log ( r0 / r ), 39. Distance between the centre of two stars is, 10a. the masses of these stars are M and, 16M and their radii a and 2a respectively. A, body of mass m is fired straight from the, surface of the larger star towards the surface, of the smaller star. What should be its, minimum initial speed to reach the surface of, the smaller star ?, , GM 1 5GM 3 GM 3 5 GM, 2), 3), 4), 2 a, 2 a, 2, a, a, 40. There is a crater of depth R/100 on the surface, of the moon (radius R). A Projectile is fired, vertically upward from the crater with velocity,, which is equal to the escape velocity v from, the surface of the moon. Find the maximum, height attained by the projectile. (2003A), 1) 90R, 2) 95 R, 3) 99.5 R 4) 50 R, 41. Gravitational acceleration on the surface of a, , 42. Two spherical planets P and Q have the same, uniform density ρ , masses M P and M Q and, surface areas A and 4A respectively. A, spherical planet R also has uniform density, ρ and its mass is ( M p + M Q ) . The escape, velocities from the planets P, Q and R are, VP ,VQ and V R , respectively. Then (2012 I), 1) VQ >VR >VP, , 1, 2, 43. A spherically symmetric gravitational system of, 3) VR /VP =3, , 6, g , where g is the gravitational, 11, , acceleration on the surface of the earth. The, average mass density of the planet is 2/3 times, that of the earth. If the escape speed on the, surface of the earth is taken to be 11kms −1 ,, the escape speed on the surface of the planet, in kms−1 will be, (2010 I), 1) 3, 2) 6, 3) 9, 4) 12, 182, , 4) VP /VQ =, , particles has a mass density, , ρ for r ≤ R, ρ = 0, 0 for r > R, , Where ρ0 is a constant. A test mass can, undergo circular motion under the influence of, the gravitational field of particles. Its speed v, as function of distance r from the centre of the, system is represented by, ( 2008 I), v, , v, , 1), , 2), r, , R, v, , r, , R, , v, 4), , 3), , 1), , planet is, , 2) VR >VQ >VP, , r, , R, , r, , R, , 44. A satellite is moving with a constant speed ‘v’, in a circular orbit about the earth. An object, of mass ‘m’ is ejected from the satellite such, that it just escapes from the gravitational pull, of the earth. At the time of its ejection, the, kinetic energy of the object is, 1), , 1 2, mv, 2, , 2) mv2, , 3), , 3 2, mv, 2, , 4) 2mv 2, , LEVEL-III-KEY, 1) 1, 7) 2, 13)4, 19)1, 25)3, 31)3, 37)1, 43)3, , 2) 2, 8) 3, 14)1, 20)4, 26)1, 32)2, 38)3, 44)2, , 3) 1, 9) 4, 15)4, 21) 4, 27)2, 33)4, 39)4, , 4) 4, 10) 4, 16) 2, 22)3, 28)2, 34)2, 40) 3, , 5) 1, 11) 1, 17) 4, 23)1, 29)2, 35)2, 41)1, , 6) 1, 12)2, 18)3, 24)1, 30)4, 36)4, 42)2,4, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 39. Let there are two stars 1 and 2 as shown below., M, a, , C1, , 16M, r1 P, , r2, , C2, , 2a, 1, , 2, , Let P is a point between C1 and C2 , where, gravitational field strength is zero. Hence, GM G (16M ) r2, =, ; r = 4,, r12, r22, 1, , r1 + r2 = 10a, , 4 , ∴ r2 = 4 + 1 (10a ) = 8a, , , , r1 = 2a, Now, the body of mass m is projected from the, surface of large star towards the smaller one., Between C2 and P it is attracted towards 2 and, between C1 and P it will be attracted towards 1., Therefore, the body should be projected to just, cross point P because beyond that the particle is, attracted towards the smaller star itself., 1 2, From conservation of mechanical energy mv, 2, = potential energy of the body at P, − potential energy at the surface of larger star.., GMm 16GMm , 1 2, , ∴ 2 mvmin = r − r, 1, , 2, GMm 16GMm , − −, −, 2a , 10a − 2a, 1 2, 45 GMm, mvmin = , 2, 8 a, , 3 5 GM , , , 2 a , 40. Speed of particle at A,, vA = escape velocity on the surface of, vmin =, , 2GM, earth =, R, At highest point B, vB = 0, Applying conservation of mechanical energy,, decrease in kinetic energy, = increase in gravitational potential energy, 1, = mv 2A = U B − U A = m (VB − VA ), 2, NARAYANAGROUP, , v2A, = VB − VA, 2, 2, GM, GM −GM , R , , =−, − 3 (1.5R 2 ) − 0.5 R −, , 100 , R, R + h R , , , 2, , 1, 1, 3 1 99 1, =−, +, − , ., R, R + h 2 R 2 100 R, Solving this equation, we get h = 99.5R, , 4, , G π R3 ρ, , 41. g = GM = 3, 2, 2, R, R, , R∞, , g∞ρ R ;, , g, ρ, , Now escape velocity, ve = 2 gR, ve ∞ g ×, , ve ∞ gR ;, , ( ve ) planet = (11kms −1 ), 42. V P =, VR =, , 2G M, R, , , VQ =, , g, ∞, ρ, , g2, ρ, , 6 3, −1, ×, 121 2 = 3kms, , 2G8M, = 2VP, 2R, , 2G 9 M, = 91 / 3V P, 91 / 3 R, , 43. For r ≤ R ;, , mv2 GmM, → (1), = 2, r, r, , , 3 , here, M = 3 π r ρ 0, , , substituting in Eq(1) we get v∞r, 4, , i.e., v − r graph is a straight line passing through, origin. for r > R, mv 2, =, r, , 3, , Gm π R 3 ρ 0, 4, , r2, , or v ∞, , 1, r, , The corresponding v − r graph will be as shown, in option (3), 44. In circular orbit of a satellite, potential energy, = −2 × ( kinetic energy ) = −2 × 1 mv 2 = − mv 2, 2, Just to escape from the gravitational pull, its total, mechanical energy should be zero. Therefore, its, kinetic energy should be + mv 2, 185
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 5., , LEVEL - IV, Matching Type Questions, 1., , Match the Columns, Column-I, A) Concept of, elliptical path, B) Gravitational, attraction force, , Column - II, P) at the poles on the, surface of earth, Q) Decreases as we go, upwards from the, surface of earth, R) Kepler’s 1st law, , C) Acceleration due, to gravity, D) Acceleration due to S) Kepler’s 2nd law, gravity is maximum, T) Newton’s Law, A satellite of mass m is moving in a circular, orbit of radius r = (Re+h) around earth of, radius Re and mass M e, and density of earth, ρ . Match the following, Column-I, Column - II, , 2, , A) Orbital velocity of the satellite, , P) T, , = 2π, , r3, GM e, , GM e m, 2r, GMem, C) Potential energy of the satellite R) −, 2r, GMem, D) Total energy of the satellite S) −, r, , B) Kinetic energy of the satellite., , Q), , GM, , 3., , e, E) Time period of the satellite. T), r, Column-I, Column-II, The gravitational potential, energy at, A)Surface of earth, P) 0, , B) At height h =, C) At infinity, , 4., , R, 3, , Q) -mgR, 3, 2, 3, S) - mgR, 4, , R) - mgR, , D) At centre of earth, Column-I, Column-II, A) Acceleration due to gravity P) g, at north pole of earth when earth, rotates with angular speed w, B) Acceleration due to gravity, , æ, , xö, , Q) g çççè1- R ÷÷÷ø, , at height x from surface of earth x<<R æ 2x ö, C) Acceleration due to gravity R) g çççè1- R ÷÷÷ø, at depth x, D) Acceleration due to gravity, S) g - Rw 2, at equator due to rotation of earth, with angular speed w, 186, , Bodies can be projected from surface of earth, to make them to orbit as a satellite or escape, from gravitational pull of planet. Column-I, gives the position of body after it is projected, and Column-II gives values of velocity of, projection., Column-I, Column-II, A) Velocity of projection of, P) 11.2 km/s, body from surface of earth, for body to orbit around surface, of earth close to it., B) Velocity of projection of, Q) 7.8 km/s, body to orbit around earth in, geostationary orbit, C) Velocity of body to orbit, R) <7.8 km/s, around earth in any orbit., D) Velocity of body to escape, S) » 3km/s, from surface of earth, 6. Column-I, Column-II, (position), (Gravitational intensity), A) At centre of earth, P) GM/R2, B) At r<R (inside earth) Q) 0, C) At r=R, R) GMr/R3, D) At r>R (outside earth) S) GM/r2, 7. Column-I, Column-II, A) Gravitational force P) Sun-synchronous orbit, B) Escape speed, Q)Geo-synchronous orbit, C) Geo-stationary, R) Conservative force, satellite, D) Remote sensing S) Independent of, satellite, mass of body, 8. If our planet suddenly shrinks in size, still, remaining perfectly spherical with mass, remaining unchanged., Column-I, Column-II, A) Duration of the day, P) increase, B) Kinetic energy of rotation Q) unchanged, C) Duration of the year, R) decrease, 9. When a planet moves around the sun, Column-I, Column-II, A) Its angular momentum, P) increases, B) When it is near the sun its speed Q) constant, C) When it is near the sun its, R) decreases, potential energy, 10. A satellite is revolving round the earth in an, elliptical orbit., Column-I, Column-II, (A) Gravitational force exerted by earth and, P) Zero, centripetal force at some points only can be, (B) Work done by gravitational force in, Q) Equal, some small parts of orbit can be, (C) In comparison of centripetal force at some, R) Greater, point magnitude of gravitational force can be, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 11. Two satellites S1 and S2 revolve round a planet, in coplanar circular orbits in opposite, sense.Their periods of revolution are 1hr. and, 8hrs. respectively. The radius of orbit of S1 is, 104 km., Column-I, Column-II, A) Speed of Ist satellite, P) π ×104 km / h, B) Speed of IInd satellite, Q) 3π ×104 km / h, C) Minimum magnitude of, R) 2π ×104 km / h, relative velocity between, the two satellites, , ASSERTION AND REASONING TYPE, QUESTIONS, In the following questions, each question, contains Assertion and Reason. Each question, has four choices (1), (2),(3) and (4) out of which, only one is correct., (1) Assertion is True, Reason is True;, Reason is a correct explanation for Assertion, ., (2)Assertion is True, Reason is True; Reason, is NOT a correct explanation for Assertion ., (3)Assertion is True, Reason is False., (4) Both Assertion ,Reason are false, 12. Assertion: Particle of mass m dropped into a hole, made along the diameter of the earth from one end, to the other end possess simple harmonic motion., Reason:Gravitational force between any two, particles is inversely proportional to the square of, the distance between them., 13. Assertion: Escape velocity is independent of the, angle of projection., Reason: Escape velocity from the surface of earth, is 2gR where R is radius of the earth., 14. Assertion: Work done by gravitational field in, cyclic process is zero, Reason: Work done by conservative field does, not depend upon path., 15. Assertion: Gravitational potential is zero inside a, shell., Reason: Gravitational potential is equal to double, the work done in bringing a unit mass from infinity, to a point inside the gravitational field., 16. Assertion: In elliptical orbit around the sun,, the earth is closer to the sun during summer than, during winter in northern hemisphere., Reason: The angular momentum of the earth about, the sun is not conserved., NARAYANAGROUP, , GRAVITATION, 17. Assertion: A spherical shell produces no, gravitational field anywhere., Reason: The field due to various mass elements, cancels out, everywhere inside and outside the, shell., 18. Assertion: For the planets orbiting around the sun,, angular speed, linear speed, KE change with time,, but angular momentum remains constant., Reason: No torque is acting on the rotating planet., So its angular momentum is constant., 19. Assertion: The change of weight with height h near, the earth’s surface is proportional to h 0, Reason: Since gravitational potential is given by, V = −GM / r ., 20. Assertion: A particle is at a height R from the, surface of earth. (Here R is radius of earth). If, mass of particle is m then its gravitational potential, energy is mgR., Reason: If a particle is slowly lifted above the, surface of earth then work is done by external, agent . Work done by external agent is wasted in, the form of heat energy., 21. Assertion: Two particles are to be projected from, the surface of earth so that particles just leave the, gravitational field of earth. One particle is projected, vertically upward and another is at an angle of 450, with vertical. Speed given to both particles is same., Reason: Escape speed does not depend upon, angle of projection., 22. Assertion: For a satellite revolving very near to, earth’s surface the time period of revolution is given, by 1 h 24 min., Reason: The period of revolution of a satellite, depends only upon its height above the earth’s, surface., 23. Assertion: Kepler’s second law can be, understood by conservation of angular momentum, principle, Reason: Kepler’s second law is related with areal, velocity which can further be proved to be based, on conservation of angular momentum as, , ( dA / dt ) = ( r 2ω ) / 2, , 24. Assertion: If earth suddenly stops rotating about, its axis then the acceleration due to gravity will, become same at all the places., Reason: The value of acceleration due to gravity, is independent of rotation of earth., 25. Assertion: Orbital velocity of a satellite is greater, than its escape velocity., 187
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GRAVITATION, , 26., , 27., , 28., , 29., , 30., , 31., , 32., , 33., , 34., , Reason: Orbit of a satellite is beyond the gravita, tional field of earth, Assertion: When a planet moves in elliptical, orbit around sun, its angular momentum about sun, remains conserved., Reason: Total energy of the planet remains, conserved, Assertion: Satellite is put in an orbit at a height, where air resistance is present, then orbital velocity, of the satellite will decrease., Reason: Due to air resistance a lot of heat will be, produced which may burn satellite, Assertion: The magnitude of gravitational potential, at the surface of solid sphere is less than, that of the centre of sphere., Reason: Due to solid sphere, gravitational potential, is same within the sphere., Assertion: smaller the orbit of a planet around the, sun, shorter is the time it takes to complete., Reason: According to Kepler’s third law of, planetary motion, square of time period is, proportional to cube of mean distance from the sun., Assertion: The value of acceleration due to gravity, does not depend upon mass of the body, Reason: Acceleration due to gravity is a constant, quantity, Assertion: Earth does not retain hydrogen, molecules and helium atoms in its atmosphere,, but does retain much heavier molecules, such as, oxygen and nitrogen., Reason: Lighter molecules in the atmosphere have, translational speed that is greater or closer to escape, speed of earth., Assertion: If a particle projected horizontally just, above the surface of earth with a speed greater, than escape speed, then it will escape from, gravitational influence of earth., Reason: Escape velocity is independent of its, direction., Assertion: If time period of a satellite revolving in, circular orbit in equatorial plane is 24h, then it, must be a geo-stationary satellite., Reason: Time period of a geo-stationary satellite, is 12 h, Assertion: Two satellites are following one another, in the same circular orbit. If one satellite tries to, catch another ( leading one ) satellite, then it can, be done by increasing its speed without changing, the orbit., Reason: The energy of earth - satellite system in, circular orbit is given by E = + ( GMm) / ( 2r ) , where, r is the radius of the circular orbit., , 188, , JEE-ADV PHYSICS-VOL - III, , STATEMENT TYPE QUESTIONS, 1) Statement-1 is true, Statement-2 is true, 2) Statement-1 is true, Statement-2 is false, 3) Statement-1 is false, Statement-2 is true, 4) Statement-1 is false, Statement-2 is false, 35. Statement-1: When a body is projected with velocity, v = v0 (where v 0 is orbital velocity) then, path of the projectile is circular., Statement - 2 : Gravitational force between body, and the earth provides the centripetal force., 36. Statement - 1 : For a mass M kept at the centre of, a cube of side ‘a’, the flux of gravitational field, passing through its sides is 4π GM. (2008A), Statement - 2 : If the direction of a field due to a, point source is radial and its dependence on the, distance ‘r’ from the source is given as r 1/2. Its flux, through a closed surface depends only on the, strength of the source enclosed by the surface, and not on the size or shape of the surface, 37. Statement-1: Orbiting satellite or body has K.E., of always less than that of Potential energy., Statement - 2 : For any bound state, the magnitude, of potential energy is always twice that of kinetic, energy (K.E.), 38. Statement-1 : There is almost no effect of rotation, of earth at poles., Statement - 2 : Because rotation of earth is about, polar axis., 39. Statement-1 : The force of attraction due to a, hollow spherical shell of uniform density, on a point, mass situated inside it is zero., Statement - 2 : The gravitation field due to the, shell inside the shell will be zero., 40. Statement-1 : The angular momentum uder a, central force is constant., Statement - 2 : A force directed towards a fixed, point following inverse square law is conservative, 41. Statement-1 : A satellite moves round the earth in, a circular orbit under the action of gravity.A person, in the satellite experience a zero gravity field in the, satellite, Statement - 2 : The contact force by the surface, on the person is zero, 42. Statement-1 :A body becomes weightless at the, centre of earth., Statement - 2 : As the distance from centre of, earth decreases acceleration due to gravity, increases., 43. Statement-1 : The speed of satellite always remains, constant in an orbit, Statement - 2 : The speed of a satellite depends, on its path., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, 44. Statement-1 : If earth suddenly stops rotating about, its axis, the acceleration due to gravity at poles, remains unchanged., Statement - 2 : Rotation of earth is about polar, axis so the centripetal acceleration of any pole is, zero., 45. Statement-1 : Total energy of an orbiting satellite, depends only on the semi major axis and not on its, eccentricity., Statement - 2 : For zero eccentricity the path of, the satellite will be circular., 46. Statement-1 : A planet is moving around the sun in, an elliptical orbit. Then from aphelion point to, perihelion point the speed of the planet continuously, increases., Statement - 2 : According to Kepler’s second, law, the areal velocity of the line joining planet to, the sun is constant., 47. Statement-1 : Acceleration due to moon’s gravity, on moon’s surface is ge/6.Acceleration due to, earth’s gravity on moon’s surface, is g e/(60)2, Statement - 2 : The distance of moon from earth’s, centre is approximately equal to 60 times the radius, of earth., 48. Statement-1 : Space rockets are usually launched, in the equatorial line from west to east., Statement - 2 : The acceleration due to gravity is, minimum at the equator., , MORE THAN ONE CORRECT, CHOICE, 49. Which of the following is correct, A) An astronaut in going from Earth to Moon will, experience weightlessness once., B) When a thin uniform spherical shell gradually, shrinks maintaining its shape, the gravitational, potential at its centre decreases., C) In the case of spherical shell, the plot of V, versus r is continuous., D) In the case of spherical shell, the plot of, gravitational field intensity I versus r is continuous, 50. An object is weighed at the North pole by a beam, balance and a spring balance, giving readings of, WB and W S respectively. It is again weighed in the, same manner at the equator, giving reading of, WB′ and WS ′ respectively. Assume that the, acceleration due to gravity is the same every where, and that the balances are quite sensitive., , A) WB = WS, , B) WB′ = WS ′, , C) WB = WB′, , D) WS ′ < WS, , NARAYANAGROUP, , GRAVITATION, 51. For a planet moving around the sun in an elliptical, orbit, which of the following quantities remain, constant ?, A) The total energy of the ‘sun planet’ system, B) The angular momentum of the planet about the, sun., C) The force of attraction between the two, D) The linear momentum of the planet, 52. If a satellite orbits as close to the earth’s surface, as possible, A) its speed is maximum, B) time period of its rotation is minimum, C) the total energy of the earth plus satellite system, is minimum, D) the total energy of the earth plus satellite system, is maximum, 53. A satellite to be geo-stationary, which of the, following are essential conditions?, A) it must always be stationed above the equator, B) it must be rotate from west to east, C) it must be about 36,000km above the earth, surface, D) its orbit must be circular, and not elliptical, , LEVEL-IV- KEY, MATCHING TYPE QUESTIONS, 1)A-R, B-T, C-Q, D-P, 2)A-T,, B-Q,, C-S, D-R, E-P, 3)A-Q, B-S, C-P, D-R ; 4)A-P, B-R, C-Q,D-S, 5)A-Q, B-S, C-R, D-P; 6) A-Q, B-R,C-P,D-S, 7)A-R, B-S, C-Q, D-P; 8) A-R, B-P, C-Q, 9) A-Q, B-P, C-R, ; 10)A-Q, B-P, C-R, 11)A-R,B-P, C-Q, ASSERTION - REASONING TYPE, 12.2, 13.1, 14.1, 16.4, 17.4, 18.1, 20.4, 21.1, 22.1, 24.3, 25.4, 26.2, 28.3, 29.1, 30.3, 32.1, 33.4, 34.4, STATEMENTS, 35.1, 36. 1, 37.1, 39.1, 40.1, 41.3, 43.3, 44.1, 45.1, 47.1, 48.1, , 15.4, 19.2, 23.1, 27.2, 31.1, , 38.1, 42.2, 46.1, , MORE THAN ONE CHOICE, 49. A,B,C 50. A,C,D 51. A,B 52. A,B,C, 53. A,B,C,D, 189
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, LEVEL-IV-HINTS, 12. As the particle moves along the hole in to and fro, motion it executes SHM, Gm1m2, 1, F=, ⇒F∝ 2, 2, r, r, 13., Using only energy, 1, m v e2 =, 2, , (G, , M m, , )/, , conservation,, , R, , 2h , 2h , 19. g ′ = g 1 − ; ∆g = g − g ′ = g , R, , R, dW d, d, d 2 gh, =, ( m∆g ) = m ∆g = m , dh dh, dh, dh R , 2mg, = constant, R, 20. Potential energy of particle at a height R is, GMm, mgR, GPE = −, =−, 2R, 2, =, , 2GM, = 2 gR = 11.2 km s −1, R, Which does not depend upon angle of projection, and mass of the projected body., 22. The time period of satellite which is very near to, , 21. Ve =, , 23., , R, = 84min = 1 h 24min, g, , dA 1, dA 1 2, = rv ⇒, = r ω, dt, 2, dt, 2, , ⇒, , 190, , g ′ = g − ω 2 Re cos 2 λ, , If ω = 0 then g ′ = g . The value of g will be same, at all places., , 14. Work done by conservative field in cyclic, process is zero, 15. Work will be done only in bringing the unit, mass from infinity upto the surface of shell because, inside the shell there is no gravitational field and in, moving inside the shell no work will be done., 16. Assertion is true otherwise the orbit would be, unstable., 17. Although no gravitational field is produced inside a, symmetric shell, it produces a field at points outside, the shell., r, 18. The torque on a body is given by τ = dL / dt . ., In case of planet orbiting around the sun no torque, is acting on it., ⇒ L (angular momentum) = constant., , earth is given by T = 2π, , 24. The value of g at any place is given by, , mr 2ω L, =, = constant ⇒ L = constant., 2m, 2m, , 25. Ve = 2 v0, 26. Angular momentum is conserved because net, torque is zero, 27. Decrease in speed and production of heat, both, take place., 28. Vin =, , GM, 3R 2 − r 2 , 2 R3 , , At surface Vs =, Vin =, , GM, R, , [ at r = R ], , 3, Vs ⇒ Vin > Vs, 2, , V is not same everywhere as indicated by Vin, 29. According to Kepler’s third law of motion, T 2 ∝ a3 ,, When a is smaller, shorter is the time, period., 30. Acceleration due to gravity is given by g = GM / r 2 ., Thus it does not depend on mass of body ., 3KT, 3RT, , v=, m, M, So, for lighter gas molecules v is greater which is, enough to take these molecules away from, earth’s atmosphere., , 31. From kinetic theory of gas, v=, , 2GM, = 2 gR = 11.2 k m s−1, R, Which does not depend upon angle of projection, and mass of the projected body. Escape velocity, is independent of direction of projection., 33. For a satellite to be geo-stationary, the necessary, requirements are:, 1. Its orbit must be in equatorial plane and, circular., 2. Its time period must be 24 h;, 3. Its sense of rotation must be same as that of, earth about its own axis., 34. Here, Assertion is wrong because as speed of one, satellite increases, its kinetic energy and hence, total energy increases,, i.e., total energy becomes less negative and hence, r increases, i.e., orbit changes., , 32. Ve =, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 5., , LEVEL - V, SINGLE ANSWER TYPE QUESTIONS, 1., , A spherical hollow is made in a lead sphere of, radius R such that its surface touches the, outside surface of the lead sphere and passes, through the centre. The mass of the lead, sphere before hollowing was M . The force of, attraction that this sphere would exert on a, particle of mass m which lies at a distance d, from the centre of the lead sphere on the, straight line joining the centre of the sphere 6., and the hollow is, GMm, A), d2, , B), , , , , , GMm , 1, , 1, 2, 2, , d, R , , 8 1 , , 2d , , , , 3., , 1 1, 1, 1 , C) GM R r D) GM R R , , 0 , 0, Two rings having masses M and 2M ,, respectively, having same radius are placed, coaxially as shown in figure., 3R, P, , , , , , GMm , 1, , GMm, 1, , 2, C) d 2 , R, , D) 8d 2, 8 1 , , 2d , , , , 2., , A body starts from rest from a point at a, distance r0 from the centre of the earth. It, reaches the surface of the earth whose radius, is R . The velocity acquired by the body is, 1 1, 1 1, A) 2GM R r B) 2GM R r , , 0 , 0, , R, , A thin rod of length L is bent to form a circle., Its mass is M . What force will act on, the, mass m placed at the centre of the circle?, GMm, 2 GMm, 4 2GMm, A), B), D) zero 7., 2 2 C), 2, 4 L, L2, L, A solid sphere of uniform density and mass, M has radius 4 m . Its centre is at the origin, of the coordinate system. Two spheres of radii, 1 m are taken out so that their centres at, P 0, 2,0 and Q 0, 2,0 respectively. This, leaves two spherical cavities. What is the, gravitational field at the origin of the, coordinate axes? Z, , If the mass distribution on both the rings is, non-uniform, then gravitational potential at, point P is, GM 1, 2 , GM 2 , , 1, A) , B) , C) zero, , , R 2, R 2 , 5, D) cannot be determined from given information, A point P lies on the axis of a fixed ring of, mass M and radius R , at a distance 2R from, its centre O . A small particle starts from P, and reaches O under gravitational attraction, only. Its speed at O will be:, A), , P, , O, , Q, , Y, , X, , 4., , 31GM, Gm, A), B), C) 31GM D) zero, 1024, 1024, The gravitational potential due to earth at 9., infinite distance from it is zero. Let the, gravitational potential at a point P be, 5 J kg 1 . Suppose, we arbitrarily assume the, gravitational potential at infinity to be, 10 J kg 1 , then the gravitational potential at, P will be, A) 5 J kg 1, B) 5 J kg 1, C) 15 J kg 1, D) 15 J kg 1, , B), , 2GM, R, , 2GM, 5 1, D) zero, R, The gravitational field due to a mass, distribution is given by, E K / x 3 in x direction., Taking the gravitational potential to, be zero at infinity, its value at distance x is:, 2K, K, K, 3K, A) 2, B) 2, C) 2, D) 2, x, 2x, x, 2x, An artificial satellite of earth is launched in, circular orbit in equatorial plane of the earth, and satellite is moving from west to east., With respect to a person on the equator, the, satellite is completing one round trip in 24 h ., Mass of earth is, M 6 1024 kg . For this, situation, orbital radius of the satellite is:, B) 6400 km, A) 2.66 104 km, C) 36, 000 km, D) 29, 600 km, C), , 8., , 2GM , 1 , 1, , , R , 5, , , , , , 191, 191
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 10. A satellite is orbiting around earth in a circular, orbit of radius r . A particle of mass m is, projected from satellite in forward direction, 2, with velocity v , times orbital velocity, 3, (this velocity is given with respect to earth)., During subsequent motion of the particle, its, minimum distance from the centre of earth is, r, 2r, 4r, A), B) r, C), D), 2, 3, 5, 11. The satellite is moving in an elliptical orbit, about the earth as shown in figure:, P, , 15. A satellite revolving around the planet in a, circular orbit is to be raised to a bigger circular, orbit. The required energy can be supplied to, the satellite for achieving the bigger orbit:, A) in one stage, B) in minimum two stages, C) in minimum four stages, D) in minimum three stages, 16. A spherical uniform planet is rotating about, its axis. The velocity of a point on its equator, is v . Due to rotation of planet about its axis, the acceleration due to gravity g at equator, 1, is, of g at poles. The escape velocity of a, 2, particle on the pole of planet in terms of ve :, A) ve 2v B) ve 3v C) ve v D) ve v / 2, , S, , E, , MULTI ANSWER TYPE QUESTIONS, , Q, , The minimum and maximum distance of, satellite from earth are 3 units and 5 units,, respectively. The distance of satellite from the, earth when it is at P is equal to, A) 4 units, B) 3 units, C) 3.75 units, D) none of these, 12. An exploratory rocket of mass m is in orbit, about the sun at a radius of RES /10 (one tenth, of the radius of the earth’s orbit about the sun)., To exit this orbit, it fires its engine over a short, period of time. This quickly doubles the, velocity of the rocket while halving its mass, (due to fuel consumption). Immediately after, the burn, what is the kinetic energy of the, rocket? Take mass of sun as M S, GM S m, , 10GM S m, , 20GM S m, , 5GM S m, , C) R, D) R, A) 2 R B) R, ES, ES, ES, ES, 13. A shell is fired vertically from the earth with, speed vesc / N , where N is some number, greater than one and vesc is escape speed on, the earth. Neglecting the rotation of the earth, and air resistance, the maximum altitude, attained by the shell will be ( RE is radius of, the, earth):, RE, R, NR, N 2R, B) E2 C) 2 E D) 2 E, A) 2, N 1, N, N 1, N 1, 14. A planet of small mass m moves around the, sun of mass M along an elliptical orbit such, that its minimum and maximum distance from, sun are r and R respectivley. Its period of, revolution will be:, 3, , A) 2, , r R, 6GM, , C) , 192, , 2GM, , potential energy EP curves for a twoparticle system. Name the point at which the, system is bound., Ek, , Energy, , A, , B) 2, , r R, , 3, , D) 2, , r R, GM, , C, , D Position, , A) A, B) B, C) C, D) D, 18. A tunnel is dug along a chord of the earth at a, perpendicular distance R / 2 from the earth’ss, centre. The wall of the tunnel may be assumed, to be frictionless. A particle is released from, one end of the tunnel. The pressing force by, the particle on the wall, and the acceleration, of the particle vary with x (distance of the, particle from the centre of earth ) according, to, Pressing force, , Pressing force, , A), , B), x=R/2, x=R, , x, , x=R/2, x=R, , x, , acceleration, , acceleration, , C), , D), x=R/2, x=R, , 3GM, , B, , Ep, , 3, , 3, , r R, , 17. Figure shows the kinetic energy Ek and, , x, , x=R/2, x=R, , x, , 19. A solid sphere of uniform density and radius 4, units is located with its centre at the origin O, of coordinates. Two spheres of equal radii 1, unit, with their centres at A 2,0, 0 and
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , COMPERHENSION TYPE QUESTIONS, , B 2, 0, 0 respectively, are taken out of the Comprehension-22-24:, solid leaving behind spherical carvities as, A solid sphere of mass m radius r is placed inside, shown in the figure. Then, a hollow thin spherical shell of mass M and radius, R as shown in figure. A particle of mass m is, placed on the line joining the two centres at a, distance x from the point of contact of the sphere, and the shell. Find the magnitude of the resultant, gavitational force on this particle due to the sphere, and the shell if, A) the gravitational force due to this object at the, origin is zero, B) the gravitational force at the point B 2, 0, 0 is, zero, C) the gravitational potential is the same at all points, 22. r < x < 2r, of the circle z 2 y 2 36, Gmm 2r x , Gmm x r , D) the gravitational potential is same at all points of, A), B), 3, 2r, 2r 3, the circle y 2 z 2 4, Gmm x r , Gmm 2 x r , 20. A double star consists of two stars having, C), D), 3, masses M and 2M . The distance between, r, r3, their centres is equal to r . They revolve under 23. 2r < x < 2R, their mutual gravitational interaction. Then,, Gmm, Gmm, Gmm, 2Gmm, which of the following statements are not, 2 B), 2 C), 3 D), 2, A), correct?, 4 x x, x r, x r, x r, A) Heavier star revolves in orbit of radius 2r / 3, 24. x > 2R, B) Both the stars revolve with same speed, period, 2GMm Gmm, GMm, 2Gmm, of which is equal to 2 / r 3 2GM 2 / 3, A) x r 2 x r 2 B) 2 x R 2 x r 2, , , , , , , C) Kinetic energy of the heavier star is twice that, , , , , GMm, Gmm, GMm, Gmm, of the other star, D) Havier star revolves in orbit of radius r/3, C) x R 2 x r 2 D) x R 2 x r 2, , , , , , , 21. Two satellites S1 and S 2 are revolving around Comprehension-25-27:, the earth in coplanar concentric orbits in the, In the graph shown, the PE of earth-satellite system, is shown by solid line as a function of distance r, opposite sense. At t 0 , the positions of, (the separation between earth’s centre and satellite)., satellites are shown in the diagram. The period, The total energy of the two objects which may or, of S1 and S 2 are 4 h and 24 h, respectively.., may not be bounded to earth are shown in figure, The radius of orbit of S1 is 1.28 10 4 km. For, by dotted lines., this situation mark the correct statement(s)., Y, , X, , O, , A, , B, , Z, , m', , M, , x, , m, , E2, Energy, S1, , S2, , 0, , E2 total, r0, , r, , E1, , A) The angular velocity of S 2 are observed by S1, at t 12 h is 0.486 rad s 1 ., B) The two satellites are closest to each other for, the first time at t 12 h and then after every 24 h, they are closest to each other., C) The orbital velocity of S1 is 0.64 104 km ., D) The velocity of S1 relative to S 2 is continuously, changing in magnetic and direction both, , E1 total, V(r), , Based on the above information answer the, following questions:, 25. Mark the correct statement(s):, A) The object having total energy E1 is bounded one, B) The object having total energy E2 is bounded one, C) Both the objects are bounded, D) Both the objects are unbounded, , 193, 193
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GRAVITATION, , JEE-ADV PHYSICS-VOL - III, , 26. If object having total energy E1 having same 31. What is the possible orbital speed of the, PE curve as shown in figure, then, satellite in moving around the planet in circular, orbit in a plane which is perpendicular to the, A) r0 is the maximum distance of object from, axis of planet?, earth’s centre, A) R G , B) 2R G , B) this object and earth system is bounded one, G, C) the KE of the object is zero when r r0, C) R 2 G , D) R, D) all the above, 2, 27. If both the objects have same PE curve as, shown in figure, then, 32. If an object is projected radially outwards from, the surface such that it reaches upto a, A) for objects having total energy E2 all values of, maximum distance of 3R from the axis then, r are possible, what, should be the speed of projection?, B) for object having total energy E2 values of r r0, 2, are only possible, A) R G, B) 2 R G ln 3, C) for object having total energy E1 all values of r, 3, are possible, 4, 2, D) none of the above, C) R G, D) R G ln 3, 3, 3, Comprehension-28-30:, A triple star system consists of two stars, each of 33. Assume that the planet is rotating about its, mass m , in the same circular orbit about central, axis with time period T . How far from the axis, 30, of, the planet do the synchronous, star with mass M 2 10 kg . The two outer, telecommunications, satellites orbit?, stars always lie at opposite ends of a diameter of, A) RT G , B) 2RT G , their common circular orbit. The radius of the, circular orbit is r 1011 m and the orbital period of, G, each star is 1.6, C) RT 2 G , D) RT, 107 s ., v, 2, Comprehension-34-36:, Two planets of equal mass orbit a much more, m, m, massive star. Planet m1 moves in a circular orbit of, M, radius 1108 km with a period of 2 years Planet, v, m2 moves in an elliptical orbit with closest distance, 20, 11, 2, 2, and farthest distance, r1 1 108 km, [Take 2 10 and G 10 Nm kg ], 3, r2 1.8 108 km , as shown:, 28. The mass m of the outer star is, 16, 11, 30, 1030 kg, m, A), B) 10 kg, 15, 8, Star r, r, 15, 8, A, P, 30, 30, 10 kg, C), D) 10 kg, 16, 11, m, 29. The orbital velocity of each star is, 5, 5, 34. Using the fact that the mean radius of an, 10 103 m / s, 10 105 m / s, A), B), 4, 4, elliptical orbit is the length of the semimajor, 5, 5, axis, find the period of m2 ’s orbit., 10 102 m / s, 10 104 m / s, C), D), 4, 4, A) 3.31 years, B) 2.21 years, 30. The total mechanical energy of the system is, C) 4.25 years, D) 1.52 years, 1375, 1375, 35, 38, , , 10, J, , , 10, J, 35. What is the mass of the star?, A), B), 64, 64, A) 5.29 1020 Kg, B) 1.49 1025 Kg, 1375, 1375, 34, 37, 10 J, 10 J, C) , D) , C) 1.49 1029 Kg, D) 1.49 1030 Kg, 64, 64, Comprehension-31-33:, Consider a hypothetical planet which is very 36. Compare the speed of planet m2 at P with, that at A., long and cylindrical. The density of the planet, is , its radius is R ., A) VP 2.4VA, B) VP 3.6VA, R, C) VP 4.2VA, D) VP 1.8VA, 1, , 2, , 1, , 2, , 194
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , LEVEL -V - HINTS, , INTEGER TYPE QUESTIONS, 37. A planet revolves about the sun in elliptical, orbit of semi-major axis 2 1012 m . The areal, velocity of the planet when it is nearest to the, sun is 4.4 1016 m / s . The least distance, between planet and the sun is 1.8 1012 m ., The minimum speed of the planet in km/s is k/, 10. Determine the value of k ., 38. The gravitational potential energy of a satellite, revolving around the earth in circular orbit is 4 MJ. Find the additional energy (in MJ) that, should be given to the satellite so that it, escapes from the gravitational field of the, earth., 39. A particle is projected from the earth’s surface, with an initial speed of 4 km/s. What will be, themaximum height attained by the particle ?, 40. Earth is a sphere of uniform mass density. If, the weight of the body is 10n N half way down, the centre of earth find the value of n . The, body weighed 100 N on the surface., 41. An infinite collection of equal masses of 2 kg, are kept on a horizontal line x axis at, Find, the, positions x 1, 2, 4,8,....... ., gravitational potential at x 0 in GJ units., 42. Three uniform spheres, each having a, mass M 5kg and radius a 2.5m are kept, in such a way that each touches the other two., Find the magnitude of the gravitational force, in GN on any of the spheres due to the other, two., 43. A chord of length 64m is used to connect a, 100kg astronaut to a spaceship whose mass is, much lager than that of the astronaut. Estimate, the value of the tension in 102 N in the chord., Assume that the spaceship and the astronanut, fall on a straight line from the earth’s centre., The radius of the, earth is 6400 km., 44. Two satellites of mass ratio 1: 2 are revolving, around the earth in circular orbits such that, the distance of the second satellite is four times, as compared to the distance of the first, satellite. Find the ratio of their centripetal, forces., LEVEL - V -KEY, SINGLE ANSWER QUESTIONS, 1) B, 2) D 3) D 4) B 5) B 6) A, 7) A, 8) B 9) A 10) A 11) A 12) B, 13) A, 14) C 15) B 16) A, MULTI - ANSWER QUESTIONS, 17) A, B, C, D 18) B, C 19) A, C, D, 20) A, C, 21) A, B, C, D, COMPERHENSION QUESTIONS, 22) C, 23) B 24) D 25) A 26) D, 27) A, 28) B 29) D 30) B 31) C, 32) B, 33) D 34) A 35) C 36) D, INTEGER TYPE QUESTIONS, 37) 4, 38) 2 39) 1 40) 5 41) 4 42) 4, 43) 3, 44) 8, , SINGLE ANSWER TYPE QUESTIONS, 1., , F1 gravitational force between sphere of mass, 2, M and the particle of mass m GMm / d ., F2 gravitational force between the removed part, and the particle of mass m GM 1m / d R / 2 2, where M 1 mass of the removed part, GM 1m, 3, , F, , 3, 2, 2, 4 R, 4 R M, R, , , , 8 d , 3 2, 3 8, 8, 2, , where M 1 mass of the removed part, GMm, 2, F, , 3, 2, 2, 4 R, r R M, R, , , , So,, 8 d , 3 2, 3 8, 8, 2, , Hence, required force, F F1 F2 , , 2., , GMm, GMm, , 2, 2, d, R, , 8 d , 2, , , , , , , GMm , 1, , 1, , 2, 2, , d, R , , 8 1 , , , 2d , , , , ., , Field produced by circular rod at its centre is zero., Hence no force is acting on the particle placed, at the centre., 3. Let = density of sphere, R radius of sphere,, r radius of spherical cavity.., 4, 3, Mass of complete sphere R M, 3, 3, 3, M 1, Mr, M, , 3, Here, m R3 , 64 ., 4, , , , , , Now, I R I I P I Q Here, I R 0 , also I P I Q ., 4. According to the problem, as the potential at , increases by 10 J kg 1 , hence potential w i l l, increase by the same amount everywhere (potential, gradient will remain constant)., Hence, potential at point P 10 5 5 J kg 1 ., GMm 1 2 GMm, 5. r 2 mv R, 0, 1 2 GMm GMm v 2 GM GM, mv , , , , 2, R, r0, 2, R, r0, 6., , 1, 1, v2 2G M v , R, r, , 0 , , 1, 1 , 2G M , , , R, r, , 0 , , ., As all the points on the periphery of either ring are, at same distance from point P , the potential at point, P due to whole ring can be calculated as, 195, 195
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 11. From given information, semi-major axis is equal, to 4 units. Let e be the eccentricity of ellipse, then, where x is axial, distance from the centre of the ring. This expression, b, 1, is independent of the fact whether the distribution, ae 1 e , of mass is uniform or non-uniform., 1, 4, GM G 2M, , So, V at P is V , 3, 2R, 5R, 5, GM 1, 2 , Semi-minor axis,, , , ., R 2, 1, 5 , b a 1 e 4 1 units 15 units ., GMm, GMm 1 2, 16, , mv, 7. , 2, 2, 2, 2, So, required distance b 2 12 4 units ., R 4R, R, 2GM , 1 , GM s m 1 2 1 m, 2, v2 , 1, 12. , mv , 2v , , , R , 5, R, Es 2, 2 2, , , 8. V Edx, 10 , GMm, GMm 1 2, 9. Here time period of satellite w.r.t observer on, , mv, 13. , equator is 24h and the satellite is moving from, R, R, , h, 2, west to east, so angular velocity of satellite wrt, 1 2, 1, 1, 2GM h, earth’s axis of rotation (considered as fixed) is,, v GM , v2 , 2 2, 2, R Rh, Rh R, , , ,, where, T, and, T, are, time, periods, S, e, v, h, Ts Te, e, v 2 ve2, Given v , of satellite and earth, respectively., Rh, N, , R, 1, 4, hr 1.45 10 rad / s ., Substitute h , 6, 1 N 2, 1/, 2, GM, GM, a3 , , r , From, v , T, , 2, , 14., , (Kepler’s Law), r, r, Gm , GM, 2, , r 3/ 2 , rR, r R, , and a , ., Hence, T , , 2, 2, GM, 7, 4, , r 2.66 10 m 2.66 10 km ., 15. Energy is needed to raise to higher orbits.Two stages, 10. As the velocity of particle is less than orbital velocity, are needed. One for incresing the orbit and the, of satellite, the particle goes in elliptical orbit of semisecond for the correction., major axis less than r ., GM, GMm mv2 mg m GM, = . 2 v, =, 16., 2, R R, 2R, R, R, 2, 2GM, As v e , v e 2v, R, r, V GM /, , , , R2 x 2, , , , 1, , MULTI ANSWER TYPE QUESTIONS, , v1, , 17. For all points, EP EK (numerically). So, totally, energy is negative. Thus, the system is a bound, system corresponding to all the points., 18. Net force towards centre of earth, 2, m0 , v0 r m0 v1r1 , where m0 is the mass of, 3, = mg mgx / R ;, particle and v0 is orbital speed equal to, Normal force N mg sin ., , Let r1 be the minimum distance and v1 be velocity, of part icle at this position, t hen, , GM, r, , , , v1r1 , , From energy conservation,, , 2, v0 r, 3, , N, mg, , , x, , m0 2 / 3 v02 GMm0 m0v12 GMm0, , , , r, r1, 2, 2, , Solving above equations, we get r1 , 196, , r, ., 2, , Thus pressing force N , , R/2, , mgx R, R 2x
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , mg, is constant and independent of x ., 2, Hence b ., Tangential force F ma mg cos , N, , Q g cos , , gx, R, , x2 , x, , F, , R2, gx, R2 4x2 ., 4 a, R, , Gmm, , x r, , 2, , When x 2 R, m', , M, x, m, , R, Curve is parabolic and at x , a 0 . Hence C ., 2, the sphere and shell both will contribute to the, 19. The mass of the sphere with cavities is symmetrically, GMm, GMm, situated with respect to the origin. The circle, Fsphere , Fshell , 2, 2, 2, 2, y z 36 has a radius of 6 units and all points, x r, x R, on it are at a distance of 6 units from the centre of, GMm, GMm, the sphere where the whole mass of the sphere is, F, , 2, 2, supposed to be concentrated. The circle is outside, x R x r ., the sphere. Similar is the case with circle, Comprehension - 25 - 27:, y2 z2 4 ., 25. For the system to be bounded one, total energy of, r, the system must be negative. So, objective having, 20. M(r-x)=2Mx Mr-Mx =2Mx x=, total energy E1 is bounded and other is unbounded., 3, 3, 26., Total, energy = KE + PE = negative, so, system is, T1 R1 2, bounded., , 21. From Kepler’s law T 2 R3 . So,, T2 R2 , At t r0 , total energy = PE and KE = 0, 2, 2, For r r0 , total energy = PE and KE = negative,, T2 3, 24 3, 4, R2 R1 1.28 10 km, which is not possible, so r0 is the maximum distance, 6 , T1 , of object from earth’s centre., 3.22 10 4 km Orbital velocity of S1 is, 27. The object having total energy E2 is bounded as, 2 R1 2 1.28 104, its total energy is positive and it is never intersection, 4, v1 , , , 0.64, , , 10, km, PE cure, so all values of r are possible for it., T1, 4, Comprehension - 30-32 :, Orbital velocity of S 2 is, 28. Fmm = Gravitational force between two outer stars, 4, v2 , , 2 R 2, 2 1.28 10, , T2, 4, , 0.64 10 4 km ., At t 12 h the two satellites are closest to each, other and after every 24 h they come at same, position relative to each other. It is clear that, direction of v2 w.r.t. v1 is changing continuously, in both magnitude and direction., Angular velocity of S 2 w.r.t S1 at t 12 h is, v1 v2 / R2 R1 0.468 rad s 1 ., This can be easily proved by writing the basic, equations., , COMPREHENSION QUESTIONS, Comprehension-22 - 24:, m', , M, , When 2r x 2 R, , x, m, , then the force will be due to sphere only., , Gm 2, ., 4r, FmM Gravitational force between central star and, GmM, outer star , ., r2, For circular motion of outer star,, 2, G m 4m , mv, Fmm FmM , v2 , r, 4r, 2 r, T period of orbital motion v, 11, 16 2 r 2, 150, , 8 10 30 1030 kg ., m, 4M , 2, 16, , 8, GT, 11, 2 r, 2 r 2 10 10 , 29. T , v, , v, T, 1.6 10 7, 5, , 10 104 ms 1 ., 4, 30. Total mechanical energy = K.E + P.E., , , , , , , 197, 197
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , X distance between planet and sun and , angular speed of planet about sun. From Kepler’s, second law areal velocity of planet is constant. At, farthest, G 4 M m 2G M, Gm , . position, m, , , , dS, 1, 2, 4r, r, r , , A, 2 R r 1 2 R r 2 R r , 2, dt 2, Gm , m, 1, 20, 11 11, 30 , , M 10 10 11, 1, 2A, 2 R r vB or vB , (least speed), (Using, r , 4, 3, 8, 10, 2, 2R r, 11, , 1375, 30, 30 , values) vB 40 km / s thus, k 4 ., 1038 J ., 2 10 10 , 32, 64, , , 38. PE 4MJ TE 2MJ, Comprehension - 31 - 33 :, The additional energy required to make the satellite, 31. Consider a closed cylindrical Gaussian surface of, escape 2MJ ., radius r and length l around the planet., mghRe 1 2, v2, h, 1, Gauss’s law gives, 2, 39. R h 2 mv, v, 2, e, 2g , 2 G R, R, 2 rlE 4 G l R2 E , r, d, To rotate a satellite around the planet in the orbit 40. mg mg 1 where d is the depth, R, radius r, centripetal force is provided by the, Gravitational force., 1 mg, mg 1 , 50 N, 2, 2, 2 G R, mv, 2, 2, , , m, , i.e.,, v 2 G R, 41. Gravitational potential at a point at a distance of r, r, r, i.e., independent of radius r., GM, from any mass m , 32. Work done by gravitational force,, r, 3R 1, 2, Therefore, potential, at, x, 0 is, W 2 G R m , dr, R r, GM GM GM GM, V , , , , ...., W 2 G m R 2 ln 3 ., 1, 2, 4, 8, Applying work energy theorem, we get, 1 1 1, , GM, V GM 1 2 3 .... V , ,, 1, 2, 2, mv 2 G mR ln 3 v 2 R G ln 3, 1 1/ 2, 2 2 2, , 2, 1, as the common ratio of successive terms is ., 2 , vT, G, , RT ., 2, 33. v r , r r , T , 2, 2, In GJ units, V 2 m 4 ., Comprehension - 34 -36:, 42. The net gravitational force acting due to any two, 34. Mean radius of planet,, sphere on the third is the vector sum of their, r r, individual gravitational force. Therfore,, 3/ 2, m2 1 2 1.4 108 km Now,,, T r, , 2, F F F F 2 F 2 2FF cos60, 1, 2GMm Gm, 2 mv 2 , , r, 2r, 2, , , time period of m 2, , or T2 2 1.4 , 35. T , , 3/ 2, , 1 .4 1 0 8 , T1 , , 8, 10, , , 3/2, , 3.31 year, , 3, 2 r 3/ 2, 4 2 r , M , GM, GT 2, 2, , , , 2, , 3, , 4 3.14 10 11 , 29, 6.67 10 11 2 365 25 3600 2 1.49 10 kg, , 36., , vr cons tan t, , INTEGER TYPE QUESTIONS, , 37. Area covered by line joining planet and sun in time, 1 2, dt is ds 2 X d ; Area velocity, dS, 1, , X, dt, 2, 198, , 2, , d, 1, X 2 where, dt, 2, , 1, , 2, , 1, , 2, , 1 2, , Putting F1 F2 F 3 F1, G m m Gm 2, 2, Where F , a2, a, 2, 5, In G Newton, F 2.5 2 4, , , , , , 43. The tension in the string T is given as, GMm, m R l 2, 2, T Fgr mar T , Rl, , , , , , GMm R , T m R l , , , R2 R l , , 2, , 2, , GMm R , T m R l , , , R2 R l , 2, , 2
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 4., 3mgl 3 100 10 6, , 3 102 N, 3, R, 6400 10, mv 2 GMm, 2, 44. Centripetal force f , r, r, Where m is the mass of the satellite and m is the, mass of the satellite of earth (mass M), 2, f1 1 4r1 16, , , 8, f 2 2 r12, 2, T , , LEVEL - VI, SINGLE ANSWER QUESTIONS, 1., , 5., , Find the potential energy of the gravitational, interaction of a point mass m and a rod of, mass m and lengt l if they are along a straight, line. Point mass is at a distance of a from the, end of the rod., m, , a, 2, , Gm 2a l , Gm, 2a l , ln , B) U l 2 ln 2a l , l, , , 2a l , 2, 2, 2, Gm, 2a l , Gm, 2a l , U, , ln, ln, U, , C), , D), , , l2, l2, 2a l , 2a l , , A) U , , 2., , P, , 6., a, , A), C), 3., , GMm, L2 a 2, GMm, , L, M, , GMm, B) a L2 a 2 , GMm, , D) a L2 a 2 2, A planet of mass m moves along an ellipse, around the sun so that its maximum and, minimum distances from the sun are equal to, r1 and r2 respectively. Find the aqnglular, momentum of this planet relative to the centre 7., of the sun. Mass of the sun is M., 2GMr1r2, 2GMr1r2, 2, A) m, B) m r r, 1 2, r1 r2 , a L2 a 2, , C) m, , 2GMr12 r22, r r2 , , 2, D) m, , 2GMr1r2, r r2 , , 2, G PR 2, 3, 2, B, GP 2 R 2, B), 3, R/2, A, 2, R, GPR 2, C), 5, 2 2 2 2 2, GP R, D), 3, A ring of radius R 4m is made of a highly, dense meterial. Mass of the ring is, m1 5.4 109 kg distributed uniformly over uts, circumference. a highly dense partice of mass, m2 6 108 kg is placed on the axis of he ring, A), , Mass M is distributed uniformly along a line, of length 2L. A particle of mas m is at a point, that is a distance a above the centre of the, line on its perpendiculr bisector (Point P in, figure). The gravitational force that the line, exerts on the particle is, , L, , Inside a uniform shere of density there is a, spherical cavity whose centre is at a distance, l from the centre of the sphere. Find the, strengh of the gravitational field inside the, cavity., 2, 4, A) E G l, B) E G l, 3, 3, 4 2, 4, 2 2, D) E G l, C) E G l, 3, 3, Inside a fixed sphere of radius R and uniform, density , there is spherical cavity of radius, R, such that surface of the cavity passes, 2, through the centre of the sphere as shown in, figure. A particle of mass m is released from, rest at centre B of the cavity. Calculate velocity, with which particle strikes the centre A of the, sphere. Neglect earth’s gravity. Initially, sphere and particle are at rest., , at a distance x0 3 m from the centre., Neglecting all other forces, except mutual, gravitational interaction of the two, calculate, (i) displacement of the ring when particle is at, the centre of ring and, (ii) speed of the paricle at that instant., A) i 0.4m ii 16cms 1 B) i 0.3m ii 18cm / s, C) i 0.2m ii 12cm / s D) i 0.6m ii 24cm / s, A cosmic body A moves to the sun with velocity, v0 (when far from the sun) and aiming, parameter l the arm of the vector v0 relative, to the centre of the sun . Find the minimum, distance by which this body will get to the sun., Mass of the sum is M., , 199, 199
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 2, , , lv02 , GM , GM, , 1, , , 1, 1, , , A) v 2 , B), 2, , GM , v, , 0, 0, , , 2, , , lv02 , GM , , 2, 1, , , 1, , , C) v 2 , D) GMlv0 1, GM, , , 0, , , , 8., , Two satellites S1 and S 2 revolve around a, planet in coplanar circular orbits in the opposit, sense. The periods of revolutions are T and, T rspectively. Find the angular speed of S 2, as observed by an astronaut in S1 , when they, are closest to each other., , R, 2R, , P1 3, A) P 7, 2, , P1 4, B) P 9, 2, , 1, , P1 7, P1 3, C) P 3 D) P 8, 2, 2, 12. A projectile of mass m is fired from the surface, of the earth at an angle 600 from the, 2, , 1 , 1 , 2 n 3 1, 2 n 3 1, , , , , W, W, 2, A), 13 B), , , T n 1, T 2 n 3 1, , , , , 13 , 23 , 2 n 1, 2 n 1, , , , , W, W, C), 23 D), 13 , T n 1, T n 1, , , , , , 9., , 11. The density of the core of a planet is 1 and, that of the outer shell is 2 . The radii of the, core and that the planet are R and 2R, respectively. Gravitational acceleration at the, surface of the planet is same as at a depth R., 1, Find the ratio ., 2, , GM e, , vertical. The initial speed v0 is equal to R ., e, How high does the projectile rise? Neglect air, resistance and the earth’s rotation., V0, , , Re, Re, A), B), A particle of mass m is placed on centre of, 2, 5, curvature of fixed, uniform semi - circular ring, of radius R and mass M as shown in figure., Re, Re, Calculate :, C), D), (a) interaction force between the ring and the, 4, 8, particle and, 13. Find the velocity of a satellite travelling in an, (b) work required to displace the particle from, elliptical orbit, when it reaches point C on the, centre of curvature to infinity., end of the semiminor axis., 2GM, GM, V, c, a F , b, , , A), 2, M, a, R, R, g, 2, B) Vc R, A) Vc R, b, g, a, GMm, 2GMm, b 2, B) a F , R, R, a, m, Re, , rmax, , c, , GMm, 2GMm 2, g, g, b, , 2, C) Vc R, D) Vc R 2, R, R, R2, a, a, 2GMm, GMm, b, D) a F , MULTIPLE ANSWER QUESTIONS, R2, R, 14. A cannon shell is fired to hit a target at a, 10. Given a thin homogeneous disc of radius a and, horizontal distance R. However it breaks into, two equal parts at its highest point. One partd, mass m1 . A particle of mass m2 is placed at a, A returns to the cannon. The other part, distance l from the disk on its axis of, A) Will fall at a distance R beyond the target, symmetry. Initially both are motionless in free, B) Will fall at a distance 3R beyond the target, space but they ultimately collide because of, C) Will hit the target, graviational attraction. Find the relative, D) Have nine times the kinetic energy of A, velocity at the time of collision. Assume a<<1. 15. A particle mooving with kinetic energy 3 J, 1, makes an elastic collison (head - on) with a, , 2 1 , , 2 1 2, stationary particle which has twice its mass,, A) 2G m1 m2 B) 2G m1 m2 a l , During impact :, , , a, l, , , , , , , A) The minimum kinetic energy potential of system, 2, , 2 1 , , 1 , 21, is 1 J, 2G, m, , m, , C) 1 2 a l D) 2G m1 m2 a l , C) a F , , 200, , , , , , , , , ,
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , B) The minimum elastic potential energy of the, system is 2 J, C) Momentum and total energy are conserved at, energy instant, D) The ratio of kinetic energy to potential energy, of the system first decreases and then increases., 16. Consider a thin spherical shell of uniformly, density of mass M and radius R :, A) The gravitational field inside the shell will be zero, GM, , 2, , B) The gravitational self energy of shell is, 2R, C) Attractive force experience by unit area of the, GM 2, , A) The total energy of the two satellites plus earth, GMm, , system just before collision is , r, B) The total energy of the two satellites plus earth, 2GMm, , system just before collision is , r, C) The total energy of two satellites plus earth, GMm, , system just after collision is , 2r, D) The combined mass (two satellites) will fall, towards the earth just after collision., , COMPREHENSION QUESTIONS, , shell pull the other half is, Comprehension - 20 -22, 2R2, D) Net gravitational force with which one, GM 2, , hemisphere of the shell arrracts other, is, 8R 2, 17. A satellite moves in an elliptical orbit about, the earth. The minimum and maximum distance, of the satellite from the centre of earth are, 7000 km and 8750 km respectively. For this, situation mark the correct statement(s)., [Take M e 6 10 24 kg ], A) The maximum speed of the satellite during its, motion is 5.64 km/s, B) The minimum speed of the satellite during its, motion is 4.51 km/s, C) The length of major axis of orbit is 15750 km, D) None of the above, 18. The gravitational potential changes uniformly, from -20J/kg to -40J/kg as one moves along, x -axis from x 1m to x 1m . Mark the, correct statement about gravitational field, intensity of origin., A) The gravitational field intensity at x 0 must be, equal to 10N/kg., B) The gravitational field intensity at x 0 may be, equal to 10N/kg., C) The gravitational field intensity at x 0 may be, greater than 10N/kg., D) The gravitational field intensity at x 0 must not, be less than 10N/kg., 19. Consider two satellites A and B of equal mass, m, moving in same circular orbit about earth,, but in opposite sense as shown in figure. The, orbital radius is r. The satellites undrgoes a, collision which is perfectly inelastic. Which is, perfectly inelastic. For this situation, mark out, the correct statement(s)., [Take mass of earth as M]., A, , B, r, , Earth, , V2, , P, , A, S, V1, , A planet of mass m is moving in an elliptical orbit, around the sun of mass M. The semi major axis of, its orbit is a, eccentricity is e., 20. Find speed of planet V1 at perihelion P, A), , GM 1 e , a 1 e , , C), , GM 1 e , a 3 1 e , , B), , 1 e GM, a, 1 e, 2, GM 1 e , a 1 e 2 , , D), , 21. Find speed of planet V2 at aphelion A., A), , GM 1 e , a 1 e , , GM 1 e , a 1 e , , B), , 2, 2, GM 1 e , GM 1 e , C), D), a 1 e 2 , a 1 e 2 , 22. Find total energy of planet interms of given, parameters., GMm, GMm, GMm, GMm 2, A) , B) , C) , D) , 4a, 8a, 2a, 2a, Comprehension - 23 - 25, m, h, A, B, , R, 2R, , Sphere of mass M and radius R is surrounded by a, spherical shell mass M and radius 2R as shown. A, small particle of mass m is released from rest from, a height h R abovethe shell. There is a hole, in the shell., 201, 201
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , C) Relative to earth’s centre, angular momentum, 23. In what time will it enter the hole at A ?, of A when it is in elliptical path would be less than, hR 2, 2hR 2, angular momentum of B, B), A) 2, D) During the whole process angular momentum, GM, GM, of B would be more than angular momentum of A, 2, hR, 28. If r2 3r1 and time period of revolution ford, C), D) None of these, GM, B be T than time taken by A in moving from, 24. What time will it take to move from A to B?, position 1 to position 2 is, R2, R2, 3, T 2, 3, T 2, B) , A) , B) T, A) T, C), D), GMh, GMh, 2, 3 3, 2, 3, R2, INTEGER, TYPE, QUESTIONS, C) , D) None of these, GMh, 24, 25. With what approximate speed will it collide at 29. A mass of 6 10 kg is to be compressed in a, sphere in such a way that the escape velocity, B?, from its surface is 3 108 m / sec. Find the, 2GM, GM, 3GM, GM, radius of the sphere (in mm)., A), B), C), D), R, 2R, 2R, R, Comprehension - 26 - 28, 30. Two equal masses are held at a distance of 3.0, cm in a line and released simultaneously. What, r, will be the separation between them after 2, sec ?, Earth, B, 2, 1, r, 31. Two satellites S1 and S 2 are to be set in the, A, orbits of R / 4 and R / 6above the earth’s, sufrace. They revolve around the earth in a, coplanar circular orbit in the opposite sense., What will be the ratio of speed of projection, Two satellites A and B are revolving around the, from the earth’s surface ?, earth in circular orbits of radius r1 and r2, 32. Distance between the centre of two stars is, respectively with r1 r2 . Plane of motion of the two, 10a. The masses of these stars are M and 16M, and their radii are a and 2a, respectively. A, are same. At position 1, A is given an impulse in the, body of mass m is fired straight from the, direction of velocity by firing a rocket so that it, surface of the larger star towards the smaller, follows an elliptical path to meet B at position 2 as, star. What should be its minimum speed to, shown. Focal lengths of the elliptical path are r1, reach the surface of the smaller star (round, and r2 respectively. At position 2, A is given another, GM, off to the nearest integer in the unit of, ), implulse so that velocities of A and B at 2 become, a, equal and the two move together., ?, For any elliptical path of the satellite time period of, revolution is given by Kepler’s planetary law as 33. Two particles A and B of masses 1kg and 2kg,, respectively, are kept at a very large, T 2 r 3 where a is semi-major axis of the ellipse, separation. When they are released, they, move under their gravitational attraction. Find, r1 r2, in this case. Also angular momentum, which is, the speed (in 10 5 m / sec) of A when that of B, 2, is 3.6 cm/hr., of any satellite revolving around the earth will remain, a constant about earth’s centre as force of, 34. An artificial satellite is moving in a circular orbit, gravity on the satellite which keeps it in elliptical, around the earth with a speed equal to half the, path is along its position vector relative to the earth, magnitude of the escape velocity from the, centre., earth. If the satellite is stopped suddenly in its, 26. When A is given its first impulse at that, orbit and allowed to fall freely on the earth,, moment, Find the speed (in km/s) with it hits the surface, A) A, B and earth centre are in same straight line, of earth, B) B is a head of A angularly, g 10m / sec2 and R 6400km ., C) B is behinds of A angularly D) None of the above, 27. If the two have same mass, 35. A Large spherical mass M is fixed at one, A) A would have more potential energy than B while, position and two identical point masses m are, on their initial circular paths, kept on a line passing through the centre of M, B) A would have more kinetic energy than B while, (see figure). The point masses are coneceted, on their initial circular paths, by a rigid massless rod of length l and this, 2, , 1, , 202
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , assembly is free to move along the line, connecting them. All three masses interact only, through their mutual gravitational interaction., When the point mass nearer to M is at a, distance r 3l from M, the tension in the rod, M , is zero for m k , . The value of k is, 288 , M, , m, r, , Fnet , , x0, , , , 6) B, 13) C, , 25) D, 34) 8, , LEVEL - VI - HINTS, SINGLE ANSWER QUESTIONS, , 4., , m, dm .dx, l, Gm. dm Gm 2 dx, dU , , a x l a x, l/2, , l / 2, , dU , , 0, , GMma.dx, La x, 2, , L, , 2 3/ 2, , , , r2, , , , 2GmM, adx, 1/ 2, 2, 2, 2, 2L a x a x2 , , GMm, a L2 a 2, , r1, , 90 S, , V1, P, , Applying conservation of angular momentum and, mechanical energy at P and P ' ., r1 , mv1r1 sin 900 mv2 r2 sin 900 or v2 r v1, 2, , LEVEL -VI-KEY, , U , , P', , 2 F sin , , V2, , l, , SINGLE ANSWER QUESTIONS, 1) A, 2) C 3) B 4) B 5) A, 7) A 8) C 9) D 10) A 11) C 12) A, MULTI - ANSWER QUESTIONS, 14) A, D 15) A, B, C, D 16) A, B, D, 17) A, B, C 18) B, C, D 19) A, B, D, COMPREHENSION QUESTIONS, 20) A, 21) B 22) D 23) A 24) C, 26) B, 27) B,C 28) C, INTEGER TYPE QUESTIONS, 29) 9, 30) 3 31) 1 32) 3 33) 2, 35) 7, , 1., , L, , 0, , 3., , m, , x L, , 1 1, 1 r12 2, m 2 1 v1 GMm , 2 r2, , r2 r1 , 2GMr2, 2GM, 2 r1 r2 , v1 , or v1 r r, r1 r1 r2 , 2 , 1, angular momentum of planet, L mv1r1, 2GMr1r2, or L m r r, 1 2, Let, E gravitational field at x due to the complete, sphere E1 field due to hole, and E2 field due to the remaining portion, E E1 E2 , , E2 E E1, , .... i , , GM, Gm, 2 , 2, x, R, , x, , , , 2, , , Gm 2 l / 2 dx, l l / 2 a x, , 3, , Gm 2 a l / 2 , Gm 2 2a l , , 1n , , , 1n , , , l, l, a l /2 , 2a l , x, , dx, , x=0, , Q, , P, x, , ax, a, , m, , , , 5., , 4, 4 R, 3, Here, M R 0 and m . 0, 3, 3 2, Applying conservation of mechanical energy., ncrease in kinetic energy = decrease in gravitational, potentiial energy, 1 2, mv U B U A m VB VA , or, 2, v 2 VB VA , ..... (i), , Potential at A : VA potential due to complete, sphere - potential due to cavity, , F, , 2., , 1.5GM Gm 2GM 1.5GM, , R R, R, R, /, 2, , , Gm, dM, , , M , 3, dM , dx F a 2 x 2, 4 R, R 3, 4 3, , , 2L , Here, m , and M R , 3 2, 6, 3, Components F cos (i.e., parallel to line) cancel, Substituting, the, values,, we, get, each other. Net force will be perpendiculat to the, 2, rod., GM , R 1.5Gm, VB 3 1.5R 2 0.5 , R , 2 R / 2, , , x, , dx, , , , 203, 203
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , , 1 1 1 2 1/ 3, 2 , ., T T 2/ 3 T 1, , , , 1, 2 / 3 1, 1 2/3, , 11 GM 3GM, , 8 R, R, , , G R3 11, 4, , . R3 G R 2, R 2, 6, 3, , , 6., , 1, VB VA G R 2 So, form eq. (i), 3, 2, v, G R2, 3, (a) Let x be the displacement of ring. Then, displacement of the particle is x0 x, or, , 3.0 x m., , 5.4 10 x 6 10, , 8, , Sun, , l, , 7., v0, , Let r be the minimum distance and m the mass of, cosmic body A. applying conservation of angular, momentum (about sun) and conservation of, mechaical energy, we have,, ...... (i), , 1 2 1 2 GMm, mv0 mv , ...... (ii), 2, 2, r, Solving these two equation, we have, 2, , , lv02 , GM , 1 , r 2, 1, v0 , GM, , , , , , 0, , s1, , 0, , 2Gm dM , sin , R2, , 2GMm, 2GMm, GMm, sin .d , (b) U , 2, 2, R, R, R, , V , , a, , 0, , a, G dm , , , 0, r, , m , G 12 2 r.dr , 2Gm1, , a , a, r, , Now applyuing conservation of mechanical, energy, decrease in gravitational P.E. = increase in, 1 2, vr U i U f ; Here = reduced, K.E. or, 2, m1m2, mass m m and vr relative speed., 1, 2, 1 mm, , , , Gm1m2 2Gm1m2 , , , l, a, , , , 2, 1 2, 2 m m vr , , 1, 2 , , s2, , v2, , T r 3/ 2 or rT 2/ 3, , r2, , 2/3, , /2, , , , /2, , 11. Let m1 be the mass of the core and m2 the mass, of outer shell., , 8., , r1 T1 , r T , 2, 2, , 2 F sin (towards C), , 2 1, vr 2G m1 m2 a l , , , , v1, , T , , , T , , 2/3, , , , 1, , 2/3, , 1 1 r , 2 r2 2 r1, 2 . 1 , , v v, T, T1, T2 T1 r2 , 21 2 1 2, , r, r2 r1, r2 r1, 1 1, r2, 204, , , , /2, , GMm, R, i.e, this much energy is required to dispalce the, particle from centre of curvature to infinity., 10. Potential at centre of disc., , v, , r1, , B, , Binding energy U , , 90, , mv0l mvr ., , , , O m, , M, M , dM , Rd .d, , R , , 0, , Solving, we get x 0.3m, , A, , F, A, , , , 3 x, , dm, , d, , 9., , Centre of mass will not move. Hence,, , 9, , r, , C, , Gm1 G m1 m2 , 2, g A g B (given) Then R 2 , 2R, 4m1 m1 m2 , , (or), , 4, 3, 4, 4, 4, , 4 R 3 1 R 3 .1 2 R R 3 2, 3, 3, 3, 3, , , 1 7, 4 1 1 7 2 3, 2
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , 2, 12. Let v be the speed of the projetile athighest point, 1, u, 3, m, , , Minimum, kinetic, energy, of, system, 1J, and rmax its distance from the centre of the earth., 2, 3, Applying conservation of angular momentum and, Maximum potential energy of system = 2J, mechanical energy,, 16. Say the shell has acquired a mass m and further a, mv0 sin mvrmax, ................. (i), mass dM is to be added, 1 2 GM e m 1 2 GM e m, M GMdM, GMdM, mv0 , mv , ........., (ii), dW, , V, dm, , W, , , or,, 0 R, Re, rmax, 2, 2, R, , Solving these two equation with given data we get,, rmax , , GM 2, , self energy U, 2R, Say F is now the attractive force per unit area. If, the shell expands from R to R+dR then work done, by attractive force is F 4R2dR since this is the, work done by gravitational field, this may be equal, to reduction in gravitational potential energy., , 3Re, 2, , hmax rmax Re , , or the maximum height, vc, , Re, 2, , m, r, , 1 dU GM 2, , or, R 2 FdR dU or, F , 4 R 2 dR 2 8 R 4, 13., Now total force experienced by one hemisphere, 1r, , F R 2 (vector addition adds only, , From the property of an ellipse we can show that, F ds, ra, components perpendicular to the equation plane), Total energy of a satellite in an elliptical orbit is,, GM 2, F, , 1 2 GMm GMm, GMm, Net, gravitational, force, , G, E, , mvc , , 8R 2, 2a, 2, r, 2a, V, b, , a, , M, , A, , or, , 1 2 GMm, GMm, mvc , , 2, a, 2a, , GM, g, or vc , or vc R, ;, a, a, , as, , 17., GM gR, , 2, , MULTI ANSWER QUESTIONS, , , , E, Vp, , From angular momentum conservation, , vp r, 8750, 14. By the principle of conservation of momentum the, A , 1.25, mv, , r, , mv, r, horizontal velocity of the other part will be the, p, p, A A, v A rp 7000, horizontal velocity of the other part will be greater, than u cos . Hence it will strike the point beyond, From energy conservation,, the target. The kinetic energy of their part will, mv 2p GM e m mv A2 GM e m, be nine times that of the part A., , , , 15. In a head on elastic collision between two particles,, 2, rp, 2, rA, the kinetic energy becomes minimum and potential, energy becomes maximum and potential energy, 1 1, becomes maximum at the instant when they move, v A2 v 2p GM e , r r , with a common velocity. The momentum and energy, p , A, are conserved at energy instant., v A 4.51 km / s and v p 5.64 km / s, Let m and u be the mass and initial velocity of the, first particle, 2m be the mass of second particle, and v be the common velocity. Then,, 18. From given information, Ex 10 N / kg, 1, u, mu 2 3 J mu m 2m v or v , 2, 3, , But, no information is provided about E y and Ez, So, E Ex2 E y2 Ez2 10 N / kg if E y E z 0, 205, 205
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, 10 N / kg if E y and/or Ez are non-zero, , 19. Just before collision, the total energy of two, satellites is,, E, , GMm GMm GMm, , , 2r, 2r, r, , Comprehension - 23 - 25, 23. Acceleration due to gravity near the surface of shell, can be assumed tod be uniform h R ., g, , Let orbital velocity is v1 then from momentum, conservation,, mv mv 2m v1, , G 2M , , 2R , t, , v1 0, , 2, , , , GM, 2R2, , h, , From, , 1 2, gt, 2, , 2h, hR 2, 2, g, GM, , As velocity of combined mass just after collision is, Gm, Gmh, zero, the combined mass will fall rowards earth. At 24. U A 2 gh 2 2 h , 2R, R, this instant, the total energy of the system only consist, of the gravitational potential energy given by, From A to B, field due to shell is zero, but field due, to sphere is non-zero., GM 2m, U, 2r, R, R2, t, , , Hence, AB, COMPREHENSION TYPE, UA, GMh, , Comprehension - 20 - 22, , 25., , V2, , P, , r1, , ae, , r2, , A, , S, V1, , r1 a ae a 1 e , , GM GM , GM, 2 , , , , R , R, 2R, , ........ i , , Comprehension - 26 - 28, , 20. mV1r1 sin 90 mV2 r2 sin 90, , V1r1 V2 r2 ........ ii , , 27. U , , According to conservation of energy at P and A, 1, GMm 1, GMm, mV12 , mV22 , ........ iii , 2, r1, 2, r2, , From i , ii and iii V1 , 21. V2 , , K A U A K B U B K A U A U B mv v A vB , 1 2, mvB m VA VB vB 2 v A vB , 2, , 2a, , r2 a ae a 1 e , , K A 0 : Potential between A and B due to shell is, constant. From energy conservation we can write,, , GM, a, , 1 e , 1 e , , , , V1r1, GM 1 e , V2, r2, a 1 e , , 1, GMm, 2, 22. Total Energy E 2 mV1 r, 1, GMm, Substitute V1 and r1 and simplify ; E , 2a, , GMm, GMm, and KE , r, 2r, , for same m, if r1 r2 , KE1 KE2 and U1 U 2, Also, angular momentum of A before reaching the, position 2(during elliptical path) wasd less than that, of B. Since, some impulse is needed (at position 2), for A in direction of its motion to equal its speed, and angular momentum equal to that to B., 28. In the continuation with same comprehension, , r, , r2 r2 / 3 2r2, 2, , 3, , Let time period of A in elliptical path is T0, 3, , , , 2, T 2 2, T0 r , ; T0 , T r2 , 3 3, , , , Time taken by A to move from 1 to 2 ; t , 206, , T0 T 2, , 2 3 3
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JEE-ADV PHYSICS-VOL - III, , GRAVITATION, , INTEGER TYPE QUESTIONS, 29. As ve , , 2GM, R, , The linear momentum of the system is comserved, 11, , R, , 33. The net force acting on the system is, , m1 m2 F 1 F2 0, , 2GM 2 6.67 10 6 10, , 2, ve2, 3 108 , , 24, , R 9mm, 30. Since force and acceleration are same mass and, they have zero initial velocity, the sepration will be, the same, which is 3.0 cm, , mA VA mBVB, 2, m, v A B vB 3.6 7.2 cm / hr, mA, 1, , , , 7.2 10 7.2 102 7.2 10, , , 2 105, 3, 3, 3.6 10, 3600, 3.6 10, , In 105 m / sec, ; v A 2, , GMm1, GMm1, 1, 2, , 31. 2 m1v1 , R, R, 2 R , 4, , , 34. Given v0 ve / 2, 1/ 2, , 1, GMm 2, m 2 v 22 , R, 2, , 2 R , 6, , , v1, , v2, , GM , , , Rh, , GMm 2, , R, , , , , On solving, h R ., From the law of conservation of energy,, , 21, 1, 20, , 32. At the point P between the stars, let the gravitational, field intensity be zero. So, G 16 m , x2, , , , GM, , 10 a x , , 1/2, , 1 2GM , , , 2 R , , , , 1, GM, GMm, mv 2 , R, R h 2, , or,, 2, , 1 2 GMm GMm GMm, mv , , , 2, R, 2R, 2R, , GM, gR, x, or, v , , 4 x 4 x 40a x 8a, R, 10a x, 1/2, If means that if the body crosses the point P, it is, 10 6.4 106 8 km / sec., attracted by the other star. Thus the critical, velocity is the velocity of the body just to reach the 35. For m closer to M, point P, which can be given as, GMm Gm 2, 1 2, 2 ma .......... 1, 2, V mv ,where V potential difference, 9, l, l, 2, between A and P., and for the other m, , , , V Potential at A Potential at P, G 16 M, , 2a, , , , , , , G M G 16 M, , 8a , 8a, , , , , , GM , , 2a , , , , Gm 2 GMm, , ma .......... 2 , l2, 16l 2, Solving (1) and (2), we get K = 7, , 45GM, 90GM, 45GM, , ; V 2 V , 8a, 8a, 4a, , , , 3 GM, 2.414, GM, 3, , 2, a, 2, a, , Rounded of to the nearest integer in, , GM, is 3., a, 207, 207
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 5., , LEVEL-V, SINGLE ANSWER QUESTIONS, 1., , Two simple pendulums of length 1 m and 16 m, respectively are given small displacements at, the same time in the same direction. The, number of oscillations ‘N’ of the smaller, pendulum for them to be in phase again is, (A), , 2., , 3., , 4, 3, , 3, 4, , (B), , (C) 4, , (D), , 1, 16, , K, , π, 3, seconds for a pendulum of 1m hung from the, roof. The acceleration of the car is (g = 10ms −2 ), , M, , The driver of a car records a period of, , (B) 15ms −2, (A) 10ms −2, (D) 34.5 ms–2., (C) 17.2ms −2, A bob of mass M is hung using a string of, length l . A mass m moving with a velocity u, pierces through the bob and emerges out with, , 3mu, 2MA, , 1, (B), 2π, , 1 2mu , , m, L, (A), , v 1, +, 2L π, , 6., , l, l, l, l, (B), (C), (D), 4, 2, 3, 12, A particle of mass m oscillating as given by, 3, , U(y) = K y with force constant K has an, amplitude A. The maximum velocity during the, oscillation is proportional to, (C) A3/2 (D) A1/2, (A) A, (B) A3, 8., , K, m, , A particle at the end of a spring executes, simple harmonic motion with a period t 1, while, the corresponding period for another spring is, t2. If the period of oscillation with the two, springs in series is T, then, (A) T = t1 + t 2, , v 1, (B) 2 2 L + π, , , 2, , v 1, , , +, (C) 2 L + π m (D), 2L π, v, K, , , 58, , 7., , 1 3mu , , K, m, , M, π M, π 2M, 5π M, (B), (C), (D), K, 2 K, 2 K, 6 K, A uniform rod of length l is mounted so as to, rotate about a horizontal axis perpendicular, to the rod and at a distance x from the centre, of mass. The time period will be the least when, x is, , (A) 2π, , (A), , 2m, 3MA, , (C) 2 π 3MA , (D) 2π 2MA , , , , , A block of mass m moves with a speed v, towards the right block which is in equilibrium, with a spring attached to rigid wall. If the, surface is frictionless and collisions are elastic,, the frequency of collisions between the masses, will be :, , v, , m, K, , A/2, mean, position, , u, horizontally. The frequency of small, 3, oscillations of the bob, considering A as, amplitude is,, , 1, (A), 2π, , M, A/2, , velocity, , 4., , A block of mass M is connected to a spring of, force constant k and is placed on a smooth, horizontal surface. The block is displaced and, compressed the spring by “A”. The block is, left free to move from this position, when the, block is at a distance A/2 from mean position, it collides elastically with the identical block., Time to be taken by block to move from, extreme position to mean position is...., , K, , m, , 9., , (B) T 2 = t12 + t 22, , (C) T −1 = t −21 + t −21, (D) T −2 = t1−2 + t 2−2, A body of mass m, is attatched to a vertical, rod of mass M and length L, hung from a, pivoted support. A spring of constant K fixed, to a support on the left as shown and is, attached to the rod at a distance from the pivot., The frequency of the oscillation is :, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, 11. If the mass of the the pulleys shown in figure, are small and the cord is inextensible, the, angular frequency of oscillation of the system, is, , M, , x, k, , L, , m, , 1, 2π, , A), , (C) 2π, , ka, , K, 1, (B), ( M + 2m), 2π, , K, M, , + 2m , 3, , , A, , K, M, , + 2m , 3, , , B, m, , M + 2m, (D) 2π, K, , 10. A smooth semicircular wire track of radius R, is fixed in a vertical plane. One end of massless, 3R, is attached to the, 4, lowest point O of the wire track . A small ring, of mass m which can slide on the track is, attached to the other end of the spring. The, ring is held stationary at point P such that the, spring makes an angle of 60º with the vertical., , kb, x, , (A), , K a + Kb, m, , (B), , Ka Kb, ( K a + Kb ) m, , (C), , K a Kb, 4 m ( K a + Kb ), , (D), , 4K a Kb, ( K a + Kb ) .m, , spring of natural length, , mg, . Considering, R, the instance when the ring is released , the, free body diagram of the ring, when aT is, tangential acceleration, F is restoring force and, N is normal reaction is, , The spring constant is K=, , 12. A ball is suspended by a thread of length l at, the point O on an inclined wall as shown. The, inclination of the wall with the vertical is α ., The thread is displaced through a small angle, β, , ( > α ) away from the vertical and the ball is, , released. The period of oscillation of, pendulum( the collision between the wall and, the ball is elastic) is, P, α β, , R, , C, , l, , P, 60°, , O, , Q, aT, N, , F, P, , F, , (A), , 60°, , mg, , P, , (B), , aT, , O, , 60°, , N, P, , F, 60°, , O, , (B) 2π, , l sin α, g, , (D) 2π, , α , l, l, −2, .cos −1 , g, g, β , , F, , 60°, , (C), , α , l, l, +2, .cos −1 , g, g, β , , mg, , O, , N, , (A) 2π, , mg, aT, , NARAYANAGROUP, , 60°, , (D), , aT, , P, mg, , (C) 2π, , l sin α, g cos β, , O, , 59
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 13. Two masses m1 and m2 are suspended together, by a massless spring of spring constant k (Fig)., When the masses are in equilibrium, m1 is, removed. Frequency and amplitude of, oscillation of m2 are, K, , m g, , K, , mg, , 17. The block of mass m1 shown in figure is, fastened to the spring and the block of mass, m2 is placed against it. The blocks are pushed, a further distance ( 2 / k )( m1 + m2 ) g sin θ from, mean position against the spring and released., The common speed of blocks at the time of, separation is, , 2, (A) ω = m ; A = K, 1, , 1, (B) ω = m ; A = K, 2, , K, m2, m1, , m1, , K, m1 g, (C) ω = m ; A = K, 1, , K, m2 g, (D) ω = m ; A = K, 2, , 14. Two light springs of force constant k1 and k2, and a block of mass m are in one line AB on a, smooth horizontal table such that one end of, each spring is fixed on rigid supports and the, other end is free as shown in the figure. The, distance CD between the free ends of the, spring is 60 cm. If the block moves along AB, with a velocity 120 cm/s in between the springs,, calculate the period of oscillation of the block., (k1=1.8N/m, k2 = 3.2N/m, m=200gm), 120 cm/sec, A, B, K1 C, D K2, , m2, , k, θ, , A), , 3, ( m1 + m2 ).g sin θ, k, , 2, ( m1 + m 2 ).g sin θ B), k, , m1 + m2 , 5, ( m1 + m 2 ).g sin θ, , .g sin θ D), k, k, , , 18. In figure a sharp blow by some external agent, imparts a speed of 2 m/s to the block towards, left when it is at equilibrium position The, potential energy of the spring when the block, is at the right extreme is, ( k = 100 N / m , M = 1 kg and F = 10 N ), , C), , M, , m, , k, F, , 60 cm, (A) 1.41 S (B) 2.82 S (C) 5.64 S(D) 1.92 S, 15. A solid sphere of radius R is floating in a liquid, of density ρ with half of its volume, submerged. If the sphere is slightly pushed and, released, it starts performing simple harmonic, motion. The frequency of these oscillations is, 1 2R, 1 R, 1 3g, 1 5g, B), C), D), 2π 3g, 2π g, 2π 2 R, 2π 3R, 16. A mass m is undergoing SHM in the vertical, direction about the mean position y0 with, amplitude A and angular frequency ω . At a, distance y from the mean position, the mass, detaches from the spring. Assume that the, spring contracts and does not obstruct the, motion of m. The distance y (measured from, the mean position) such that the height h, attained by the block is maximum is (Aw2 > g), g, 2g, 3g, 4g, A) 2, B) 2, C) 2, D) 2, ω, ω, ω, ω, , A), , 60, , A) 4.5 J, B) 4 J, C) 0.5 J, D) 2.5 J, 19. A rectangular plate of sides a and b is, suspended from a ceiling by two parallel, strings of length L each (figure). The, separation between the strings is d . The plate, is displaced slightly in its plane keeping the, strings tight. The time period of oscillation is, , L, a, L, , A) 2π g B) 2π, , d, , L, , b, L+a, La, La, C) 2π g .b D) 2π g .d, g, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , 20. A small block oscillates back and forth on a, smooth concave surface of radius R (Figure)., 24. In the fig the pulley has a mass M, radius r, The time period of small oscillations is, and the string does not slip over it. The time, period of oscillation is, A) 2π, R, R, R, R, B) 2, D), C) π, g, g, g, g, 21. Assume that a tunnel is dug across the earth., A ball is dropped into it. The time where it, moves with a speed of gR is, , A) 2π, , R, R, π R, π R, B) π, C), D), g, g, 2 g, 2 g, 22. A uniform rod of mass m and length l is, suspended through a light wire of length l and, torsional constant k as shown in figure. The, time periods of the system for small oscillations, in the vertical plane about the suspension point, and angular oscillations in the horizontal plane, about the centre of the rod are, , A) 2π, , 13l, ml 2, ; 2π, 12 g, 12 K, , A) 2π, , l, ml 2, ; 2π, g, 3K, , B) 2π, , k l, m, , C) 2π, , k, , B), m, , m, L, , NARAYANAGROUP, , m, , 2π, R, , C) 2π, , 2m + 3MR 2, 2k, , m, k, , m+M, k, 25. The average kinetic energy of a particle of, mass m undergoing S.H.M with angular, frequency ω and amplitude A, over half of one, time period is, D) 2π, , (A), , 1, mω 2 A2, 2, , (B), , 1, mω 2 A2, 4, , (C) mω 2 A2, (D) 2mω 2 A2, 26. The time periods of systems depicted below, under identical conditions are respectively, , 13l, ml 2, ; 2π, 12 g, 3K, , l, ml 2, l, 2, π, ;, 2, π, D), g, 12 K, 23. Two small balls, each of mass m are connected, by a light rigid rod of length L . The system is, suspended from its centre by a thin wire of, torsional constant k . The rod is rotated about, the wire through an angle θ0 and released. The, tension in the rod as the system passes through, the mean position is, , A), , B), , 2m + M, 2k, , C), , θ 04, , 2, , + 4m g L, 2L, 2, , k, k, , k, k, , m, , m, , m, m, from left to right, , 2 2, , k 2θ 04 + m 2 g 2 L2, L, , kθ02, kθ 2 + mgL, D) 0, 2L, L, , m, 4m, m, ; (ii) 2π, ; (iii) 2π, k, k, 4k, (A) ii,i,ii,iii (B) i,i,iii,ii (C) i,ii,ii,iii (D) i,i,ii,iii, 27. Two identical simple pendulums each of length, ‘ l ’ are connected by a weightless spring as, shown., (i) 2π, , 61
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, In equilibrium, the pendulums are vertical and, the spring is horizontal and undeformed. The, time period of small oscillations of the linked, pendulums, when they are deflected from their, equilibrium positions through equal, displacements in the same vertical plane in, the opposite directions and released, is, , (A) 2π, , l, , l, (B), g, , 2π, , l, 2 kl , , g +, , m , , , l, , 30. A coin of mass m is placed on a horizontal, platform which is undergoing S.H.M. about a, mean position O in horizontal plane. The force, of friction on the coin is f. While the coin does, not slip on the platform f is, (A) directed towards O always., (B) directed away from O always, (C) directed towards O when the coin moves in, wards, (D) maximum when coin and platform are at rest, 31. x = A sin 2 ωt + B cos2 ωt + C sin ωt, cos ω t, represents S.H.M. for, (A)any value of A, B and C (except C = 0), (B) A = -B, C = 2B, amplitude = B 2, , 2π, , (C), , m, , K, , m, , l, kl , , 2π, g + (D), m, , , 2l m, +, g k, , 28. A mass of 0.98 kg suspended using a spring, of constant k= 300 N m −1 is hit by a bullet of, 20gm moving with a velocity of 3m/s vertically., The bullet gets embedded and oscillates with, the mass in a vertical plane. The amplitude of, oscillation will be :, (A) 0.15cm (B) 0.35 cm (C) 1.2cm (D) 12 m, , MULTIPLE ANSWER QUESTIONS, 29. Two blocks A and B each of mass m are, connected by a massless spring of natural, length l and spring constant K. The blocks, are initially resting on a smooth horizontal floor, with the spring at its natural length as shown., A third identical block C also of mass m, moves, on the floor with a speed v along line joining, B and A and collides with B elastically. Then, C, B, A, m, m, m, (A) The frequency of oscillation of the system AB, , 1 2K, 2π m, B) The K.E. of the system at maximum compression, of the spring is mv 2 / 4, is, , (C) The maximum compression of the spring is v, , m, K, , (D) The maximum compression of the spring is v, , m, 2K, , 62, , (C) A= B, C=0, (D) A=B, C=2B,amplitude= B, 32. A simple pendulum of length L and mass m is, vibrating with an amplitude a. Then the, maximum tension in the string is not, (A) mg, , a 2 , (B) mg 1 + L , , , 2, a 2 , a , mg, 1 + , , (C) mg 1 + (D), 2, L, 2 L , , , , , 33. Three simple harmonic motions in the same, direction having the same amplitude and same, period are superposed. If each differ in phase, from the next by 45°, then, (A) the resultant amplitude is (1 + 2 )a, (B) the phase of the resultant motion relative to the, first is 90°, (C) the energy associated with the resulting motion, is (3 + 2 2 ) times the energy associated with any, single motion, (D) the resulting motion is not simple harmonic, , COMPREHENSION TYPE QUESTIONS, Passage-I, Two identical blocks P and Q have mass m each., They are attached to two identical springs initially, unstretched. Now the left spring (along with P) is, compressed by A/2 and the right spring (along with, Q) is compressed by A. Both the blocks are, released simultaneously. They collide perfectly, inelastically. Initially time period of both the blocks, was T., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, A/2, , OSCILLATIONS, 40. The angle θ as shown is, , A, P Q, , 34. The time period of oscillation of combined mass, is, T, T, (A), (B) 2T (C) T (D), 2, 2, 35. The amplitude of combined mass is, A, A, 2A, 3A, (A), (B), (C), (D), 4, 2, 3, 4, 36. The energy of oscillation of the combined mass, is, 1 2, 1 2, 1 2, 1 2, kA, (A) kA (B) kA (C) kA (D), 2, 4, 8, 16, Passage-II, For SHM to take place force acting on the body, should be proportional to - x or F = -kx. If A be, the amplitude then energy of oscillation is 1/2 KA2., 37. Force acting on a block is F = (-4x + 8). Here F, is in newton and x is the position of block on, x-axis in meters, (A) Motion of the block is periodic but not simple, harmonic, (B) Motion of the block is not periodic, (C) Motion of the block is simple harmonic about the, origin, x=0, (D) Motion of the block is simple harmonic about x=2m, 38. If energy of oscillation is 18 J, between what, points does the block will oscillate?, (A) between x = 0 and x = 4 m, (B) between x = -1m and x = 5 m, (C) between x = -2m and x = 6m, (D) between x = 1m and x = 3 m, 39. The amplitude of oscillation is, (A) 4 cm (B) 2 cm (C) 1 cm (D) 3 cm, Passage-III Two non-viscous, incompressible and, immiscible liquids of densities ρ and 1.5 ρ are, poured into the two limbs of a circular tube of radius, R and of small cross-section kept fixed in a vertical, plane as shown in fig. Each liquid occupies onefourth the circumference of the tube., , θ, h1, 1.5p, , θ, θ, , h2, p, , 1, , NARAYANAGROUP, , 2, , −1 1 , −1 3 , −1 2 , A) tan B) tan C) tan D) zero, 5, 2, 3, 41. If the whole liquid column is given a small, displacement from its equilibrium position, the, time period of these oscillations is, , R, 2R, 3R, R, B) 2π, C) 2π, D) 2π, 6.11, 3, 2, 9.8, Passage-IV, Two identical balls A and B, each of mass 0.1 kg,, are attached to two identical massless springs. The, spring-mass system is constrained to move inside, a rigid smooth pipe bent in the form of a circle as, shown in figure. The pipe is fixed in a horizontal, plane. The centers of the balls can move in a circle, of radius 0.06 m. Each spring has a natural length, of 0.06 π meter and spring constant 0.1 N/m., Initially, both the balls are displaced by an angle θ, A) 2π, , =, , π, radian with respect to the diameter PQ of the, 6, , circle (as shown in fig.) and released from rest., , A, , x = Rθ, P, , θ, , O θ, , θ = π/6, , B, x = Rθ, Q, , 42. The frequency of oscillation of ball B is, 1, 1, 6, 2, Hz (B), Hz (C) Hz (D) Hz, π, 2π, π, π, 43. The speed of ball A when A and B are at the, two ends of the diameter PQ is (in m/s), (A) 0.0314 (B) 0.0628 (C) 0.1256 (D) zero, 44. The total energy of the system is, (A) 3.9x10-4 J (B) 10-4 J (C) 6x10-4 J (D) 2x10-4 J, Passage-V, You are riding a four wheeler automobile of mass,, 3000kg. Assume that you are examining the, oscillation characteristics of its suspension system., The suspension sags by 15 cm when the automobile, is placed on it. Also, the amplitude of oscillations, decreases by 50% during one complete oscillation., Shock absorber supports 750 kg (g=10 m/s2), , (A), , 63
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OSCILLATIONS, 45. The spring constant of the spring is (in N/m), (A) 20x104 (B) 10x104 (C) 5x104 (D) 40x104, 46. The time period of one oscillation is, (A) 0.77 s (B) 1 s, (C) 2 s, (D) 1.414 s, , MATRIX MATCHING QUESTIONS, , JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL50. Match the following:, Column-I, Column-II, (A) A constant force acting (p) the time period, along the line of SHM affects, (B) A constant torque acting (q) the frequency, along the arc of angular SHM, affects, (C) A particle falling on the (r) the mean, block executing SHM when, position it crosses the mean, position affects, (D) A particle executing SHM (s) the amplitude, when kept on a uniformly accelerated car affects, 51. For a particle executing S.H.M. along a, straight line, match the statements in columnI with statements in column-II. (Note that, displacement given in column-I is to be, measured from mean position.), Column-I, Column-II, (A)Velocity-time graph will be (p) Straight line, (B) Acceleration-velocity, graph may be, (q) Circle, (C) Acceleration-displacement (r) Ellipse, graph will be, (D) Acceleration-time graph (s) Sinusoidal curve, will be, 52. In column-I equations describing the motion, of a particle are given and in column-II possible, nature of the motions. Match the entries of, column-I with the entries of column-II., Column-I, Column-II, , 47. In case of seconds pendulum, match the, following. Consider shape of earth also, Column - I, Column - II, (A) At pole, (p) T > 2s, (B) On a satellite, (q) T < 2s, (C) At mountain, (r) T = 2s, (D) At centre of earth (s) T = 0 (t) T = ∞, 48. Match the following:, A spring block system executes SHM in such, a way that the block is having velocity v when, it crosses the mean position. Now the changes, have been made in such a way that the velocity, while crossing the mean position gets doubled, without changing mass of the block. In, Column-I some statements (incomplete) are, given and corresponding completions are given, in Column-II. Match the entries of Column-I, with the entries of Column-II. Assume that, the system is horizontal., Column-I, Column-II, (A) The frequency of oscillation, (p) 2, will change by a factor of, (B) The amplitude of oscillation, (q) 2, will change by a factor of, (p) Oscillatory, (A) y = Ae(ωt +φ ), (C) The magnitude of maximum, (r) 1, (B) y = B sin ωt + C cos ωt (q) Periodic, acceleration will change by a, factor of, (C) y = A sin (ωt + kx ), (r) S.H.M, (D) Maximum PE increases by a, (s) 4, (s) Rectilinear, (D) y = kx, factor of, 53., A, particle, moves, according, to the law given, 49. Match the following:, x, v and a are, below, in, the, column-I, where, Column - I, displacement, velocity and acceleration, (A) Linear combination of two SHMs, respectively and ω , A are +ve constants. Then, (B) y = A sin ω1t + A sin (ω2t + φ ), match the following., (C) Time period of a pendulum of infinite length, Column-I, (D) Maximum value of time period of an, (A) a = −ω 2 x3, (B) a = −ω 2 x 2, oscillating pendulum, Column-II, R, (R is radius of the earth, g, (q) SHM for equal frequencies and amplitude, (r) Superposition may not always be an SHM, (s) Amplitude will be 2A for ω1 = ω2 and, if phase difference of π / 2, , (p) T = 2π, , 64, , 2, (C) a = −ω A sin, , πx, where ( − A ≤ x ≤ A), 2A, , (D) v = a = −ω 2 A2 − x 2, Column-II (Nature of motion of particle), (p) Motion is periodic, (q)Motion is oscillatory, (r) Motion is not simple harmonic, (s) Mechanical energy is conserved, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , 54. For a particle under going linear S.H.M. about, x = 0 , choose the correct possible, combination. Symbols have their usual, meanings., Column-I, Column-II, r r, (A) v . a > 0, (p) Extreme position, (B) Velocity may be negative (q) Mean position, (C) Acceleration is negative (r) 0 < x < A, r, (D) vr × ar = 0, (s) − A < x < 0, , ASSERTION & REASON QUESTIONS, (A) Statement-I is true, Statement-II is true;, Statement-II is a correct explanation for, Statement-I., (B) Statement-I is true, Statement-II is true;, Statement-II is NOT a correct explanation for, Statement-I., (C) Statement-I is true; Statement-II is false, (D) Statement-I is false ; Statement-II is true, 55. Statement-I: When a girl sitting on a swing stands, up, the periodic time of the swing will increase., Statement-II: In standing position of a girl, the, length of the swing will decrease., 56. Statement-I: Two simple harmonic motions are, π, , given by y1 = 10 sin 3π t + and, 4, , , (, , ), , y2 = 5 sin 3π t + 3 cos 3π t ., , These have, , 59. Statement-I: In a simple pendulum performing, S.H.M. the net acceleration is always between, tangential and radial accelerations except at lowest, point., Statement-II: At lowest point tangential, acceleration is zero., 60. Statement-I: Time period of spring block system, is the same whether in an accelerated or in an, inertial frame of reference., Statement-II: Mass of the block of spring block, system and spring constant of the spring are, independent of the acceleration of the frame of, reference., 61. Statement-I: If the amplitude of a simple, harmonic oscillator is doubled, its total energy, becomes four times., Statement-II: The total energy is directly, proportional to the square of the amplitude of, vibration of the harmonic oscillator., , INTEGER ANSWER QUESTIONS, 62. In the given spring block system if k = 25π 2, Nm-1, find time period of oscillation., K, 5kg, , 5kg, , K, , K, , K, , 63. Consider a liquid which fills a uniform U-tube,, as shown in the figure upto a height h = 10 m., The angular frequency of small oscillations of, the liquid in the U-tube is (g = 10 ms-2), , amplitudes in the ratio 1 : 1., Statement-II: y1 & y2 represents two waves of, amplitudes 5& 5 3 .So the resultant amplitude is 10., 57. Statement-I: During the oscillations of simple, h, pendulum, the direction of its acceleration at the, mean position is directed towards the point of, suspension and at extreme position it is directed 64. A uniform plank of mass m = 1 kg, free to move, towards the mean position., in the horizontal direction only, is placed at, the top of a solid cylinder of mass 2m and, Statement-II: The direction of acceleration of a, radius R. The plank is attached to a fixed, simple pendulum at the mean position or at the, wall by means of a light spring of spring, extreme position is decided by the tangential and, constant k = 7 N/m2. There is no slipping, radial components of force by gravity., between the cylinder and the plank, and, 58. Statement-I: A particle is moving along x − axis., between the cylinder and the ground. The, The resultant force F acting on it is given by, angular frequency of small oscillations of the, F = − ax − b, where a and b are both positive, system is, m, k, constants. The motion of this particle is not S.H.M., Statement-II: In S.H.M. resultant force must be, proportional to the displacement from mean, R, position., 2m, NARAYANAGROUP, , 65
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 65. Two simple pendulums A and B having lengths, l and l / 4 respectively are released from the, position as shown in fig. Calculate the time (in, seconds) after which the two strings become, parallel for the first time., (Take l =, , 90, m and g = 10 m / s 2 ), 2, π, θ2 θ1, , 49), 50), 51), 52), 53), 54), , A-q,r;B-q,r,s;C-p;D-p, A-r,s ; B-r,s; C-p,q,s ; D-r,s, A - s; B - q,r ; C -p; D - s, A - p,q,r; B - p,q,r ; C - p,q,r; D - s, A-p,q,r,s; B-r, s ; C-p,q,r,s ; D-p,q,r,s, A - r, s; B - q,r,s ; C - p,r ; D - p,q,r,s, ASSERTION AND REASON TYPE, 55) D 56) B 57) A 58) D 59) D 60) A 61) A, INTEGER TYPE QUESTIONS, 62) 1 63) 1 64) 2 65) 1 66) 2, , l/4, , l, , LEVEL -V - HINTS, B, , SINGLE ANSWER TYPE, 1., , A, , 66. In the arrangement shown in fig. pulleys are, light, springs are ideal and, , K1 = 25π 2 N / m, K 2 = 2 K1 , K3 = 3K1 and K 4 = 4 K1, are the force constants of the springs., Calculate the period of small vertical, oscillations of block of mass m = 3kg ., , k2, , N 1 = ( N − 1) 16 ; N = ( N − 1)4 = 4 N − 4, , 4, 3, i.e., when 4 oscillations of the shorter and three of, the longer pendulum are over they will be in phase., So, choice (a) is correct and rest of the choices, are wrong., π, Given: T =, second, l = 1m As the car, 3, accelerates it will given a pseudoforce on bob and, the bob will tilt itself from the vertical by an angle, 3N = 4 ; N =, , k4, , 2., m, , k1, , We know that time period is directly proportional, to the square root of its length. Two different, pendulums can be in phase at the earliest with a, difference in the number of oscillations being a, maximum of one. NTshorter = ( N − 1)TLonger, , a, θ = tan −1 and the time period of oscillation is, g, , k3, , l, , T = 2π, , g 2 + a2, , LEVEL-V - K E Y, SINGLE ANSWER TYPE, 01) A 02) D 03) C 04) C 05) D 06) D, 08) B 09) C 10) C 11) C 12) D 13) B, 15) A 16) A 17) B 18) C 19) A 20) A, 22) A 23) B 24) A 25) B 26) D 27) B, , 2, , MULTIPLE ANSWER TYPE, 29) A,B,D 30) A,C,D 31) A,B,D 32) A,C,D 33) A,C, , COMPREHENSION QUESTIONS, 34) C 35) A 36) D 37) D 38) B 39) A 40) A, 41) A 42) A 43) B 44) A 45) C 46) A, , MATRIX MATCH TYPE, 47) A-q ; B-t ; C-p ; D-t, 48) A - r; B - p; C-p; D-s, 66, , π , 2, andL 3 = 4π, , , 07) C, 14) B, 21) C, 28) B, 3., , squaring both sides with T, , 1, g + a2, 2, , ⇒, , g 2 + a 2 = 36, , a 2 = 36 2 − 10 2 ; a = 36 2 − 10 2 = 34.5ms −2, so, choice (d) is correct., Using momentum conservation we get,, u, 2 mu, mu = Mv + m., ∴ v=, 3, 3 M, K.E. at mean position of the bob, 1, 1 4 m2u 2, 2m 2u 2, 2, Mv, =, M, =, =, 2, 2 9 M2, 9M, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, Comparing with K .E. =, , OSCILLATIONS, , 1, M ω 2 A2 at mean, 2, , 4 m 2u 2, 2, position, We get ω =, 9M 2 A 2, , 2mu, Or ω =, 3MA, , 1 2mu , , , 2π 3MA , ∴ Frequency (C) is correct., Time taken to collide on left wall and get back to, 2L, . Time, the mass attached with spring is t1 =, v, to get the spring compressed once and to come, , 7., , Since total energy can be the maximum P.E.,, 1 2, 1 2, 3, 3, 3, 3, we get, mv + Ky = KA ⇒ mv = K A − y ;, 2, 2, , (, , 2 K ( A3 − y 3 ), m, maximum velocity at mean position (y=0) is, , V=, , = 2π v ⇒ v =, , 4., , back is, t 2 =, , T 2π, =, 2, 2, , 2 KA 3, 3/ 2, ⇒ Vmax ∝ A, m, so choice (C) is correct., Vmax =, , 8., , m, , For first spring. t1 = 2π k , 1, m, , For second spring t 2 = 2π k , ..... (2), 2, The effective force constant in their series, combination is k =, , mean, position, , ⇒, , T2 =, , 4π 2 m(k1 + k 2 ), ....., k1k 2, , (3), , From equations (1) and (2), we obtain, , T, M, t PQ = + tQO ......(1); T = 2π, ............(2), 6, K, 6. Time period of the rod oscillating about a horizontal, axis with center of mass at a distance x is, , m m, t12 + t 22 = 4π 2 + ∴ t 2 + t 2 = T 2 [from eq(3)], 1, 2, k1 k 2 , x, , dT, =0, For least T,, dx, , 1, −, , 1, 2, 2 , 2, Mgx. (1 + Mx ) .2 Mx − ( I + Mx ) , dT, 2, = 2π , =0, dx, Mgx, , , , , , 1, Mgx, 2Mx = I + Mx 2, 2, 2 I + Mx, 3, , Mgx.Mx = ( I + Mx 2 ) 2 Solve to get x and, substitute back in T to get xleast =, , NARAYANAGROUP, , k1k 2, Therefore, time period, k1 + k 2, m(k1 + k 2 ), , k1k 2 , , PQ = QO; t PO = t PQ + tQO ;, , ∴, , ..... (1), , of combination T = 2π , , Extreme, position, , I + Mx 2, T = 2π, Mgx, , m, , k, , Time period of spring T = 2π , , k being the force constant of spring., , m, m, =π, ∴ Average, K, K, , t +t, time between two successive collisions , t = 1 2, 2, 1, collision frequency =, t, P, Q, O, , 5., , ), , l, 12, , θ, , 9., , y, , m, , As the rod is displaced by y towords the spring,, the spring will get compressed by y and the angular, y 1 2 1 2 1 2, shift, tan θ = ; Iω + Ky + mv = Constant, x 2, 2, 2, , , 1 ML2, 1, 1, + mL2 ω 2 + Ky 2 + mv 2 = Constant, , 2 3, 2, 2, , 1 M, 1, 1, , + m v 2 + Ky 2 + mv 2 = Constant, , 2 3, 2, 2, , 67
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, , 2nd Method, , Differentiating w.r.t. time , we get,, , 1 M, 1, 1, , + m 2va + K 2 yv + m 2va = 0, , 2 3, 2, 2, , Ky, , K, ∴ ω = M, , + 2m , 3, , So, choice (c) is correct and choices (a) and (b), are incorrect. Choice (d) is not possible since, it is, incorrect dimensionally., , ka/4, , a = - M + m + m, 3, , , =, m, , m, N, 60°, P, , F, , 10. free body diagram is 60°, , mg, aT, , O, , 11. Let T be the tension in the cord and xa and xb the, displacements of pulleys A and B respectively. Now, assume that pulley B is fixed ; then extension of, x, spring x b = or x = 2 x b . Similarly if we imagine, 2, that pulley A is fixed, x = 2 x a . But neither pulley B, nor pulley A is fixed. ∴ x = 2x a + 2 x b ....(1), From free body diagram of pulleys,, 2T = k b x b .... (2) or 2T = ka xa .... (3), , 12. If α < β , the ball collides with the wall and rebounds, with same speed. The motion of ball from A to Q is, one part of a simple pendulum. Time period of ball, = 2(t AQ ) . Consider A as the starting point (t = 0) ., Equation of motion is x(t ) = A cos ωt, x(t ) = l β cos ωt , because amplitude = A = lβ, time from A to Q is the time t when x becomes, −l α = lβ cos ωt, −lα . ⇒, ⇒, , −α , t = t AQ = 1/ ω cos −1 , , β , , The return path from Q to A will involve the same, time interval. Hence time period of ball = 2 t AQ, 2, α, l, −α , l, l, −1 α , cos −1 − = 2, cos−1 , = 2 π g − 2 g cos β ., ω, g, , β, β , , If k eq denotes equivalent spring constant,, , T, = x = 2 x a + 2x b From equations (2) and (3), k eq, xa =, , 2T, 2T, 1, and x b =, ka, k b i.e., k eq = , 1, 1, 4 + , ka kb , , Hence, , k eq, =, m, , ω =, , T, , k ak b, 4 m (k a + k b ), , T, , kbxb, , α β, , l, , Q, , A, lα, , lβ, , 13. (i) When m1 is removed only m2 is left. Therefore,, k, , angular frequency : ω = m, 2, (ii) Let x1 be the extension when only m2 is left., Then, kx1 = m2g or x1 =, , m2 g, k, , …(1), , Similarly, let x2 be extension in equilibrium when, both m1 and m2 are suspended. Then,, A, , B, , (m1 + m2)g = kx2 ; ∴ x2 =, , ( m1 + m2 ) g, k, , …(2), , From Eqs. (1) and (2) amplitude of oscillation :, kbxb, T, 68, , T, , A = x2 – x1 =, , m1 g, k, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , y = A sin ωt + A cos ωt , ∴ amplitude = 2A, , then, the total mechanical energy of the liquid, , column is : E = { A(h + x )ρ + A(h − x)ρ} ., , h+x, , a2, ω2 y2, =, (B) 2, (C) a α − x (D) a = Aω 2 sin ωt, v, A2 − y 2, d y, = −ω 2 y we can match, 2, dt, 2 3, 53.(A) a = −ω x , if x = −ve, a = +ve, x = +ve, a = −ve, 52. by using the equation, , a, , a, , 4, , 1, dx 1, (2 Ahρ) + ( 2 Aρg ) (h2 + x 2 ), 2, dt 2, , After differentiating the total energy and equating, it to zero, one finds acceleration = −ω2 x ., The angular frequency of small oscillations, ω , is:, ω=, , 2 Aρg, =, 2 Ahρ, , 1, , 1 dx 1 2 1, d x , m + kx + 2 m , 2 dt 2, 2, dt 2 , , k, 2 = 2k, =, 62. Equivalent spring constant = keq, k, 5, 2k +, 2, 5×5 5, =, Reduced mass of system = µ =, 5+5 2, , 1, 1, = 5π, =1s, 5, k, 25π 2, 63. Suppose that the liquid is displaced slightly from, equilibrium so that its level rises in one arm of the, tube, while it is depressed in the second arm by the, same amount, x .If the density of the liquid is ρ ,, NARAYANAGROUP, , 2, , 2, , After differentiating the total energy and equating, it to zero, one finds acceleration = −ω 2 x, The angular frequency,, , 2k ×, , 2k, , 2, , 2, , 61. E ∝ A2, INTEGER ANSWER QUESTIONS, , 5, , 10, = 1 rad/s, 10, , 11, 1 7 dx 1 2, 1 d x , + 2m.R 2 , = ( m) + kx, 22, 2 4 dt 2, R dt 2 , , 3, 2, ASSERTION & REASON QUESTIONS, , µ, = 2π, keq, , =, , 2, , E=, , velocity in S.H.M, , ∴T = 2π, , g, h, , 64. Suppose that the plank is displaced from its, equilibrium position by x at time t , the centre of, the cylinder is, therefore, displaced by x/2., ∴ the mechanical energy of the system is given by,,, E = K .E. (Plank) + P.E. (spring) + K.E. (cylinder), , acceleration in S.H.M, 54., , h–x, , 2, , =, , 2, , (D) v = a = −ω 2 ( A2 − x 2 ) not S.H.M, , 2, , h+x, h − x, , + A(h + x)ρ⋅ g ⋅, + A(h − x ) ⋅ ρ ⋅ g ⋅, 2, 2 , , , π, , 51. ( A)v = Aω cos ωt = Aω sin ωt + , 2, , , (B) a = −ω 2 x 2 , x = −ve, a = − ve , x = +ve, a = −ve, x = −ve, a = +ve, πx, 2, a, =, −, ω, A, sin, ,, (C), if x = +ve, a = −ve, 2A, , dx , , dt , , 1, 2, , R, R, C) T = 2π, D) T = 2π, , is maximum value., g, g, 50. A constant force and a constant torque affect only, the mean position. In third case, as the block falls, on mean position, the mean position is not affected., In a car, a constant pseudo force will act which will, affect only the mean position., , θ2 θ1, , 4k, 7m, , = 2 rad/sec, , l/4, , l, , 65., , ω=, , 1, , Let T = 2π, , l, ; θ1 = θ 2, g, , 2, , 4π, θ cos , T, , , 2π, t = −θ cos , , T, , 2π, , 4π, t = 2nπ ± , t +π , t ;, T, T, , , , , for n = 0, t = T / 2 for n = 1, t = T / 6, 3T / 2, , = 5π, , the first time meeting, t =, 66., , T π, =, 6 3, , l, = 1s, g, , 1, 1, 1, 1 , T = 2π 4m , +, +, +, , K1 K 2 K 3 K 4 , 73
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 4., , LEVEL-VI, SINGLE ANSWER QUESTIONS, 1., , Two light springs of force constants k1 , k2 and, a block of mass m are in the line AB on a, smooth horizontal table such that one end of, each spring is fixed on rigid supports and the, other end is free as shown: The given values, are k1 = 1.8 N / m ; k2 = 3.2 N / m ;, m = 200 gm ; L = 60 cm ; If the block moves, along AB with a velocity 120 cm/sec in between, the springs, then period of oscillation of the, block is[Q assume all collisions are elastic]:, (A) 1.04 sec, , k2, , k1, , L, 5., , (B) 0.785 sec, , m, , (C)1 sec, , A, 2., , A L - shaped bar of mass M is pivoted at one, of its end so that it can freely rotate in a, vertical plane as shown. If it is slightly, displaced from its equilibrium position then the, frequency of oscillation is...., A, 1 g 3 10 , (A) 2π L 4 , , , L θ0, 1 g 10 , (B) 2π L 4 , B, , , , B (D)2.82 sec, C L D, The following figure shows a particle of mass,, m , attached with four identical springs, each, of length l . Initial tension in each spring is, F0 . The period of small oscillations of the, particle along a line perpendicular to the plane, of the figure is (neglect gravity), , (C), , 1, 2π, , g, L, , ( 2 ) (D) 2π, 1, , g, 2L, , C, A uniform cylinder of mass m and radius R, is in equilibrium on an inclined plane by the, action of a light spring of stiffness k , gravity, and reaction force acting on it. If the angle of, inclination of the plane is φ , then angular, frequency of small oscillations of the cylinder, is..............., , k, , (A) 2, , k, 2k, (B) 2, m, m, , m, , ml, ml, 2π, (B), F0, F0, , (A) π, , Particle, , R, , 2k, (D), 3m, , (C) 2, , φ, , 2k, m, , m, , ml, ml, (C) 2π 2 F (D) π 4 F, 0, 0, 3., , A V-shaped glass tube of uniform cross-section, is kept in a vertical plane as shown. The, angular frequency of small oscillations of, liquid in a tube is, , 6., , A body A of mass m1 and body B of mass m2, are interconnected by a massless spring as, shown. The body A performs free vertical, harmonic oscillations with the amplitude A and, frequency f . The maximum value of f such, that body B does not leave the surface is, , A, , h, , (A), (C), 74, , {, , g sin α .sin β, h, , g, h, , α, , (B), , 1, , m1, , 1, (B) 2π, , β, g sin α .sin β, h, , (D), , B, , g ( sin α + sin β ), h, , m1 + m2, m2, , g, , A, m1 + m2 g, , , m1 A, , (A) 2π , , m2, , 1, 2π, 1, (D) 2π, (C), , g, A, m2 g, ., m1 A, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, 7., , OSCILLATIONS, , In the arrangement shown, the sleeve of mass, M is fixed between two identical springs, whose combined force constant is k . The, sleeve can slide without friction over a, horizontal bar AB. The arrangement rotates, with a constant angular velocity ω about a, vertical axis passing through the middle of the, bar. The period of small oscillations of the, sleeve is.............., , ω, , (C) π, , m, k, , (D) π, , (A) π, , ρ, ρ1, , (B ) π, , }, ρ2, , l1, (C) π, , (D) π, 9., , ρl, g, , l, ρg, , l, , m, k − mω 2, , ρl, ( ρ1 + ρ2 ) g, 1, 1 , +, , , ρ 2 , ρ1, , NARAYANAGROUP, , (A), , 3k g, +, 2m l, , (C), , 3k 3g, +, m 2l, , (B), , 3k 3g, +, 2m l, , (D), , 3k 2 g, +, m 3l, , 11. A thin-walled tube of mass m and radius R, has a rod of mass m and very small cross, section soldered on its inner surface. The sideview of the arrangement is as shown, , m, R, , ( ρ1 + ρ2 ) l, ρg, , (, , ρ1 + ρ 2, , ), , A uniform rod of mass, m , and length l, remains in equilibrium inside a smooth, hemisphere of radius R as shown. The period, of small oscillations of the rod, is.... Q r 2 = R 2 − l 2 / 4 , , R, g, , O, , A vertical pole of length l , density ρ , area of, cross section A ,floats in two immiscible liquids, of densities ρ1 and ρ2 .In equilibrium position, the bottom end is at the interface of the liquids., When the cylinder is displaced vertically, the, time period of oscillation is..............., , 8., , (C) 2π, , l, , m, k − mω 2, , (B) 2π, , m, k, , m, , l+R, g, 10. A thin uniform vertical rod of mass m and, length l pivoted at point O is shown in Fig., The combined stiffness of the springs is equal, to k . The mass of the spring is negligible., The angular frequency of small oscillation is, , B, , M, , (A) 2π, , R, , (D) 2π, , sleeve, A, , l2, + r2, (A) 2π 12, gr, l2, + r2, (B) 2π 4, gr, , The entire arrangement is placed on a rough, horizontal surface. The system is given a, small angular displacement from its, equilibrium position, as a result, the system, performs oscillations. The time period of, resulting oscillations if the tube rolls without, slipping is, (A) 2π, , 4R, g, , (B) 2π, , 2R, g, , (C), , 2π, , R, g, , (D) None of these, 75
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , l/2, l/2, , 2mg, given an extension of, from the, k, equilibrium position and released. The time, period of oscillations of the block is, , }, , l, , }, , 12. An elastic string of constant k is attached to, a block of mass m as shown. The block is, , }, , OSCILLATIONS, , l/4, }, , K, 4ml, , 2l, , 3ml, , l, , 2 ml, , 2l, 3, , (A) π 3( kl + 2mg ) + π 3g (B) π kl + 2 mg + π, , String of stiffness 'k', , m, m π, , (B) 4, + 3, k 3, , , m, (A) 2π, k, , m, m, m, m, +2 3, + 3, (D) π, k, k, k, k, 13. A ring of mass m can freely slide on a smooth, vertical rod. The ring is symmetrically attached, with two springs, as shown, each of stiffness, k . Ring is displaced such that each spring, makes an angle θ with the horizontal. If the, ring is slightly displaced vertically, then time, period is............, (C) 2π, , 4 ml, , l, , (C) π kl + 2mg + π 3 (D) π kl + 2 mg + π 3, 15. Consider a swing whose one end is fixed above, the other rope by “b”. The distance between, the poles of the swing is a . The lengths l1 and, , l2 are such that l12 + l2 2 = a 2 + b 2 as shown., The period of the small oscillations of the, swing, by neglecting the height of the swinging, person, is...., , b, l1, l2, , a, θ, , θ, m, , (A) 2π, , m, k, , (B) 2π, , 76, , l1 + l2, g, , l1l2, (C) 2π g a + b, (, ), , m sin θ, 2k, , 2π, m, π, m, (D), sin θ 2k, sin θ 2k, 14. A light inextensible string carrying a spring is, passed over two smooth fixed pulleys as, shown. If the rod is slightly displaced about, the hinge from is equilibrium position, then, time period is., (C), , (A) 2π, , (B) 2π, , l1l2, ga, , l1 + l2, (D) 2π g a + b, (, ), 2, , 2, , 16. A thin uniform rod of mass m and length l is, hinged at one end making an angle α from, the horizontal on the ceiling of a room. The, other end is supported by a vertical massless, string. The angular frequency of small, oscillations of this system is, (A), , g, l sin α, , (B), , 3g, l sin α, , (C), , 3g, 5l sin α, , (D), , 3g, 2l sin α, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , 17. Four identical bars of mass m , length l are 20. Two particles move parallel to x − axis about, the origin with same amplitude ‘ a ’ and, connected by pins at A,B,C and D. The bars, frequency ω . At a certain instant they aree, are attached to four springs of same stiffness, found at a distance a / 3 from the origin on, as shown. The entire system can move in, opposite sides but their velocities are in the, horizontal plane. In the equilibrium position, same direction. What is the phase difference, shown, θ = 450 . If the corners A and C aree, between the two ?, given small displacements toward each other, −1 7, −1 5, −1 4, −1 1, and released, then time period of vibration is, (A) cos, (B) cos, (C) cos, (D) cos, 9, 9, 9, 9, 2M, 21., From, what, minimum, height, must, the, system, h, (A) 2π, be released when spring is unstretched so that, 3K, B, A, , 45°, , C, , D, , (B) 2π, , 4m, K, , (C) 2π, , M, 4K, , M, K, 18. A rod of mass m and length l is pivoted at a, point O in a car whose acceleration towards, , after perfectly inelastic collision ( e = 0 ) with, ground, B may be lifted off the ground: (Spring, constant = k ), (A) mg / ( 4k ), , m A, , (B) 4mg / k, , (D) 2π, , h, , (C) mg / ( 2k ), , 2m B, , (D) None of the above, left is a0 . The rod is free to oscillate in a, vertical plane. In the equilibrium state the rod 22. The mean velocity of a particle performing, S.H.M. with time period of 0.6s and amplitude, remains horizontal then other end is suspended, of 10 cm averaged over a time interval during, by a spring of stiffness k . The time period of, which it travels a distance of 5 cm, starting, oscillations is...., from extreme position is, a0, 1)0.5 m/s 2)0.7 m/s 3)0.3m/s 4)1.04 m/s, lm, 2, π, 23. The time period of small oscillations in a, (A), ma0 + kl, vertical plane performed by a ball of 40gm, fixed at the middle of horizontally stretched, K, 2lm, 2, π, string of 1m length when the constant tension, (B), ma, +, kl, 0, O, 10N in wire is:, (A) 0.1 s, (B) 0.2 s (C) 0.3 s (D) 0.4 s, lm, 24., The, maximum, velocity of a point undergoing, 2, π, (C), 3ma0 + 4kl, simultaneous two oscillations given by, x1 = a cos wt and x2 = a cos 2wt is:, 2lm, (D) 2π 3ma 6kl, (A) 2.74aw (B) 2aw (C) 1.414aw (D) aw, 0 +, 25., A, physical pendulum is positioned such that is, 19. A body A of mass moving with velocity v while, centre of gravity is vertically above, passing through its mean position collides in, suspension point. From that position the, perfect inelastically with a body B of same mass, pendulum started moving toward the stable, which is connected to a vertical wall through a, equilibrium and passed it with an angular, spring whose spring constant is k . After, velocity ω . By neglecting the friction at point, collision it sticks to B and executes S.H.M., of suspension, the period of small oscillations, of the pendulum is, Find the amplitude of resulting motion:, 2π, 4π, π, 5π, m, m, m, m, (A), (B), (C), (D), v, v, v, v, (A), (B), (C), (D), ω, ω, ω, ω, k, 2k, k, 2k, NARAYANAGROUP, , 77
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 26. A fly wheel of mass 35kg swings as pendulum, about a knife-edge as shown. The period of, oscillation is 1.22s. , the moment of inertia of, the fly wheel about its centre is, (A) 1.15 kg .m 2, (B) 2.4 kg.m 2, , MULTIPLE ANSWER QUESTIONS, 30. A simple pendulum consists of a bob of mass, m and a light string of length l as shown., Another identical ball moving with the small, velocity v0 collides with the pendulum’s bob, and sticks to it. For this new pendulum of mass, 2m , mark out the correct statement(s)., , 0.3m, , (C) 3.2 kg .m 2, 0.4m, , (D) 0.75 kg .m 2, 27. A thin uniform plate shaped as an equilateral, triangle with a height ‘h’ performs small, oscillations about the horizontal axis coinciding, with one of its sides. The time period of, oscillation is....., h, h, 2h, 2h, (B) π, (C) π, (D) 2π, g, g, g, g, 28. A smooth horizontal disc rotates about the, vertical axis ‘O’ with constant angular velocity, ω as shown. A thin uniform rod AB of length, ‘ l ’ performs small oscillations about the, vertical axis A fixed to the disc at a distance, ‘ a ’ from axis of rotation of the disc. The, angular frequency of the oscillations of rod is, , (A) 2π, , 3a, ω, (A), 2l, O, A, B, a, ω, (B), a, I, l, 3a, 3a, ω (D), ω, (C), l, l, rod fixed at A, 29. Two balls with masses m1 = 1kg and m2 = 2kg, are slipped on a thin smooth horizontal rod as, shown. The balls are interconnected by a light, spring of stiffness K = 24 N / m . The left, hand ball ( m1 ) is imparted the initial velocity, 12 cm/s towards second ball. The amplitude, of oscillations made by the arrangement, is..................., , m1, (A) 3 cm, 78, , K, , (B) 2 cm, , m2, (C) 1 cm, , (D) 4 cm, , l, v0, , m, , m, l, , (A) Time period of the pendulum is 2π g, (B) The equation of motion for this pendulum is, θ=, , g , v0, t, sin , 2 gl, l , , (C) The equation of motion for this pendulum is, θ=, , g , v0, cos , t, 2 gl, l , , (D) Time period of the pendulum is 2π, , 2l, g, , 31. Figure (a) shows a spring of force constant k, fixed at one end and carrying a mass m at, the other end placed on a horizontal frictionless, surface. The spring is stretched by a force F., Figure (b) shows the same spring with both, ends free and a mass m fixed at each free, end. Each of the spring is stretched by the, same force F. The mass in case (a) and the, masses in case (b) are then released. Which, of the following statements are true ?, , A, , k, , B, m, , F, , F, , O, m, , m, , (a), , (A) While oscillating, the maximum extension of, the spring is more in case (a) than in case (b), (B) The maximum extension of the spring is same, in both cases., (C) The time period of oscillation is the same in, both cases., (D) The time period of oscillation in case (a) is 2, times that in case (b)., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , COMPREHENSION QUESTIONS, , Passage-II, , Passage-I, In the arrangement given below, both the springs, are in their natural lengths. The coefficient of friction, between m2 and m1 is µ . There is no friction, , Two identical blocks A and B, each of mass, m = 3kg , are connected with the help of an ideal, spring and placed on a smooth horizontal surface, as shown. Another identical block C moving with, , between m1 and the surface. If the blocks are, displaced slightly, they together perform simple, harmonic motion., , velocity v0 = 0.6 m / s collides with A and sticks, to it, as a result, the motion of system takes place, in some way., k = 2000, m→ m, m, –N/m, C, A, B, , m2, k1, , k2, , m1, , 32. The frequency of oscillations is, , 1, (A) 2π, , ( k1 + k2 ), ( m1 + m2 ), , 1, (B) 2π, , k1k2, ( m1 + m2 )( k1 + k 2 ), , (C), , 1, 2π, , (D), , 1, 2π, , ( k1 + k2 )(m1 + m2 ), m1m2, , ( m1 + m2 ), ( k1 + k2 ), , 33. If the frictional force on m2 is acting in the, same direction of its displacement from mean, position, then, m1 m2, (A) k < k, 1, 2, , m1 k1, (B) m > k, 2, 2, , (C) m1k2 = m2 k1, (D) frictional force is never in the direction of, displacement, 34. If friction on m2 acts in the direction of, displacement, then maximum possible, amplitude of SHM is, (A), , (C), , µ m2 g ( m1 + m2 ), m1k 2 − m2 k1, , (B), , µ m1 g ( m1 + m2 ), m1k1 − m2k 2, , (D), , NARAYANAGROUP, , µ m1 g ( m1 + m2 ), m2 k1 − m1k 2, µ m2 g ( m1 + m2 ), m1k1 − m2 k2, , Based on this information, answer the following, questions:, 35. After the collision of C and A, the combined, body and block B would, (A) oscillate about centre of mass of system and, centre of mass is a rest., (B) oscillate about centre of mass of system and, centre of mass is moving, (C) oscillate but about different locations other than, the centre of mass., (D) not oscillate, 36. Oscillation energy of the system, i.e., part of, the energy which is oscillating (changing), between potential and kinetic forms, is, (A) 0.27 J (B) 0.09 J, (C) 0.18 J (D) 0.45 J, 37. The maximum compression of the spring is, (A) 3 30 mm, , (B) 3 20 mm, , (C) 3 10 mm, , (D) 3 50 mm, , Passage-III, A small block of mass m is fixed at upper end of a, massive vertical spring of spring constant, k = 4mg / L and natural length '10L ' . The lower, end of spring is free and is at a height L from fixed, horizontal floor as shown. The spring is initially, unstreched and the spring-block system is released, from rest in the shown position., , m, , 10L, , L, 79
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 38. At the instant the speed of block is maximum,, the magnitude of force exerted by the spring, on the block is, mg, (A), (B) mg (C) zero (D) None of these, 2, 39. As the block is coming down, the maximum, speed attained by the block is, 3, 3, gL (D), gL, 2, 2, 40. Till the block reaches its lowest position for, the first time, the time duration for which the, spring remains compressed is, , (A) gL, , L, , (B) 3gL (C), , 1, , π, , L, 2, sin − 1, 4g, 3, , π, (D) 2, , L, , −1, (A) π 2 g + 4 g sin 3 (B) 4, , (C) π, , L, +, 2g, , L, +, g, , L, 1, sin −1, 4g, 3, , L, +, 2g, , L, 2, sin − 1, 4g, 3, , Passage-IV, A block of mass m is suspended from one end of, a light spring as shown. The origin O is considered, at distance equal to natural length of spring from, ceiling and vertical downward direction as +ve yaxis. When the system is in equilibrium a bullet of, mass m / 3 moving in vertical upward direction with, velocity v0 strikes the block and embeds into it., As a result, the block (with bullet embedded into, it) moves up and starts oscillating., l, k, , O, m, , m/3, , Y, , v0, , Based on above information, answer the following, questions:, 41. Select the correct statement(s):, (A) The block-bullet system performs S.H.M., , 42. The amplitude of oscillation would be:, 2, , (A), , 4mg mv0, (B), , +, 12k, 3k , 2, , 2, , mv0 mg , +, , 12k 3k , 2, , 2, , 2, , mv0 mg , mv0 4mg , +, +, (C), (D), , 6k k , 6k 3k , 43. The time taken by block-bullet system to move, 2, , 2, , mg, initial equilibrium position ) to, k, y = 0 (natural length of spring) is: [A, represents the amplitude of motion], , from y =, , (A), , 4m, 3k, , −1 mg , −1 4 mg , cos 3kA − cos 3kA , , , , , , , (B), , 3k, 4m, , −1 mg , −1 4 mg , cos 3kA − cos 3kA , , , , , , , 4m −1 4mg , −1 mg , sin , − sin , , , 6k , 3kA , 3kA , (D) None of the above, , (C), , Passage-V, A block of mass m is connected to a spring of, spring constant k and is at rest in equilibrium as, shown in (a). Now, the block is displaced by h, below its equilibrium position and imparted a speed, v0 towards down as shown in figure (b). As a, result of the jerk, block executes simple harmonic, motion about its equilibrium position. Based on, above information answer the following questions:, [Where A is the amplitude of oscillation,, k, h, δ = sin −1 , ω =, ], m, A, , mg, , about y =, k, (B) The block-bullet system performs oscillatory, mg, , motion but not S.H.M. about y = k, (C) The block-bullet system performs S.H.M., 4 mg, y=, 3k, , about, (D) The block-bullet system performs oscillatory, motion but not S.H.M. about, 80, , y=, , 4mg, 3k, , Equilibrium, position, , m, h, , (a), v0 m, (b), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , Passage-VII, , 44. The amplitude of oscillation is:, 2, , (B), , (A) h, , mv0, + h2, k, , m, v0 + k, (D) None of these, k, 45. The equation for the simple harmonic motion, is:, (C), , Two blocks of masses 3kg block is attached to a, spring with a force constant, k k = 900 N / ma, which is compressed 2m initially from its equilibrium, position. When 3kg mass is released, it strikes the, 6kg mass and the two stick together in an inelastic, collision., , 2m2 = 6kg, , (A) y = − A sin [ωt + δ ] (B) y = − A cos [ωt + δ ], π, , (C) y = A sin ωt + δ + , 2, , , k, , π, , (D) y = A sin ωt + δ + , 4, , 46. Find the time taken by block to cross the mean, position for the first time:, δ, (A), ω, , π, −δ, π −δ, π −δ, (B) 2, (C), (D), ω, 2ω, ω, , Passage-VI, A plank of mass M is placed on a smooth, horizontal surface. Two light identical springs each, of stiffness k are rigidly connected to struts at the, ends of the plank as shown.When the springs are, in their unextended position the distance between, their free ends is 3l . A block of mass m is placed, on the plank and pressed against one of the springs, so that it is compressed by l . To keep the blocks, at rest it is connected to the strut by means of a, light string, initially the system is at rest. Now the, string is burnt., , m1 = 3kg, , 2m 1m, 49. The common velocity of the blocks after, collision is, (A) 10 m/s (B) 30 m/s (C) 15 m/s (D) 2 m/s, 50. The amplitude of resulting oscillation after the, collision is:, 1, 1, m (B), m (C) 2m (D) 3m, 2, 3, 51. The velocities of a particle executing S.H.M., are 30 cm/s and 16 cm/s when its displacements, are 8 cm and 15 cm from the equilibrium, position. Then its amplitude of oscillation in, cm is:, (A) 25, (B) 21, (C) 17, (D) 13, (A), , Passage-VIII, A cube made of wood having specific gravity 0.4, and side length a is floated in a large tank full of, water., , string, struts, , 3l, , l, , m, , struts, , M, , 47. Maximum displacement of plank is:, ml, 5ml, 3ml, 2ml, (B), (C), (D), m+M, m+M, m+M, m+M, 48. Time period of oscillation of block:, , (A), , 2Mm, 2 Mm, (A) ( 2π + 3) k M + m (B) ( π + 6 ) k M + m, (, ), (, ), Mm, Mm, (C) ( π + 3) k M + m (D) ( 2π + 6 ) k M + m, (, ), (, ), NARAYANAGROUP, , 52. Which action would change the depth to which, block is submerged ? ( γ wood < γ water ), (A)more water is added in the tank, (B) atmospheric pressure increases, (C) the tank is accelerated upwards, (D) A small coin is placed over the cube, 53. If the cube is depressed slightly, it executes, SHM from it’s position. What is it’s time, period?, A) 2π, , a, g, , B) 2π, , 5a, 2a, 4π, C) 2π, D), 2g, 5g, 5, , a, g, 81
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 54. What can be maximum amplitude of it’s, vertical simple harmonic motion ?, (A), , a, 2, , (B) 0.4a, , (C) 0.6a, , (D) 0.2a, , Passage-IX, A block is attached to a spring and is placed on a, horizontal smooth surface as shown in which spring, is unstretched. Now the spring is given an initial, compression 2x0 and block is released from rest., Collisions with the wall PQ are elastic., , Passage-X, A block is tied within two springs, each having, spring constant equal to k . Initially the springs, are in their natural length and horizontal as shown., The block is released from rest., The springs are ideal, acceleration due to gravity, is g downwards. Air resistance is to be neglect., The natural length of spring is l0 ., , k, , k, , x>0, , 58. If the decrease in height of the block till it, , k, , reaches equilibrium is 3l0 then the mass of, the block is:, , x0, 55. Find the time period of motion of the block:, (A), , 2π, 3, , m, k, , (B), , 4π, 3, , m, k, , 3π, 2, , (C), , m, π m, (D), 2 k, k, , 56. Write its equation of motion indicating position, as a function of time:, (A) x = −2 x0 cos ω t 0 < t <, , T, 2, , (A), , 2π T, , <t <T, (B) x = −2 x0 cos ωt +, , 3 2, , , (C) x = − x0 cos ωt 0 < t <, , (C), , T, 2, , t (B), , x, (C), , 82, , (D), , 2, , 4 gl 0, 3, , (B), , 2 gl0, 3, , (D) None of these, , speed becomes zero is 8l0 then the mass of, the block is:, (A), (C), , 2kl0, g, 3kl0, g, , (B), , 2kl0, g, , (D) None of these, , Passage-XI, , t, , x, t, , 2 gl0, , 60. If the decrease in height of the block till its, , πT, , (D) x = −2 x0 cos ωt + < t < T, 3 2, , 57. Draw x − t (position-time) graph for one, period. Treating position of block in, unstretched position of spring as origin, x, x, , (A), , 2kl0, 2kl0, 3kl0, (B), (C), (D) None of these, g, g, g, 59. If the block is under equilibrium and the angle, made by the spring with horizontal is 600 then, the mass of the block is:, (A), , t, , For a particle oscillating along x-axis according to, equation x = A sin ωt, 61. The mean value of its velocity averaged over, first 3/8 of the period is, (A) 0.3 Aw (B) 0.1 Aw (C) Aw (D)0.5 Aw, 62. The magnitude of its mean velocity vector, averaged over first 3/8 of the period is:, (A) 0.3Aw (B)0.55Aw (C) Aw (D) 0.1Aw, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , 63. The mean value of its speed averaged over, first 3/8 of the period is:, (A) 0.3Aw (B) 0.55Aw (C) Aw, (D) 0.1Aw, , Passage-XII, The distance travelled in a time, t by a particle, moving along x-axis according to an equation, , x = A cos wt is given by S = nA + S 0 , when time, T , t can be written as t = n + t0 such that, 4, , t0 < T / 4 and distance travelled in that t0 is S 0 ., 64. The distance S is, when ‘n’ is even number:, (A), , , nπ , , S = A ( n + 1) − cos ωt +, , 2 , , , , (B), , , nπ , , S = A ( n + 1) − cos ω t −, 2 , , , , (C), , , nπ , , S = A n + sin ωt −, 2 , , , , (D), , , nπ, , S = A ( n + 1) − sin ωt −, 2, , , , (B), , , nπ, , S = A ( n + 1) − cos ω t −, 2, , , , (C), , , nπ, , S = A n + sin ωt −, 2, , , , (D), , , nπ , , S = A ( n + 1) − sin ωt −, 2 , , , , (C) v = ω, , (A, , (D) v = ω, , ( A + B) − ( x + y), , 2, , + B 2 ) − ( x2 + y 2 ), 2, , 2, , 69. The acceleration of particle is given by [Q r, is position vector], (B) a = ω 2 r, (A) a = −ω 2 r, (C) a = ω r, (D) a = −ω r, 70. Two particles ‘A’ and ‘B’ start SHM at t = 0 ., Their positions as function of time are given, , 65. The distance S is, when ‘n’ is odd number., , nπ , , S = A ( n + 1) − cos ωt +, 2 , , , , (B) v = A2 cos2 ωt + B 2 sin 2 ωt .ω, , MATRIX MATCHING QUESTIONS, , , , , , (A), , 68. The velocity of particle is given by, (A) v = A2 + B 2 ω, , , , , , , , , , 66. The resultant amplitude of oscillations, resulting from super position of, , by X A = A sin ωt and X B = A sin ( ωt + π / 3), Column-I, Column-II, 5π, (A) Minimum time when, (p), 6ω, x is same, π, (B) Minimum time when velocity, (q), 3ω, is same, (C) Minimum time, π, after which vA < 0 and vB < 0, (r), ω, (D) Minimum time, π, (s), after which xA < 0 and xB < 0, 2ω, 71. A particle of mass 2 kg is moving on a straight, , x1 = 3cos ωt , x2 = 5cos (ω t + π / 4 ) and, , line under the action of force F = ( 8 − 2 x ) N ., , x3 = 6sin ωt is, (A) 14, (B) 7, , The particle is released from rest at x = 6m ., For the subsequent motion, match the, following (all the values in the Column-II are, in their SI units):, Column-I, Column-II, (A) Equilibrium position is at x, (p) π / 4, (B) Amplitude of SHMs (, q) π / 2, (C) Time taken to go, directly from x = 2 to x = 4, (r) 4, (D) Energy of SHM is, (s) 6, (E) Phase constant of SHM assuming (t) 2, , (C) 4, , (D) 2, , Passage-XIII, A point moves in the plane xy according to the, law x = A sin ωt ; y = B cos ω t ; where A,B & ω, are positive constant., 67. The equation for trajectory for path taken by, particle is, x2 y2, +, =1, A2 B 2, , (A) x 2 + y 2 = A2, , (B), , (C) y = Bx, , (D) y = Ax + Bx 2, , NARAYANAGROUP, , ., , equation of the form A sin ( ωt + φ ), 83
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 72. Match the following, Column-I lists the various modes of oscillations, of masses connected to springs. Column-II, lists the corresponding frequencies of, oscillations when executing S.H.M., Match them properly., Column-I, k, k, , 74. Column I shows spring block system with a, constant force permanently acting on block, match entries of column-I with column-II., Column-I, x=0, k, , (A), , m, , F = 2mg, , x=0, , (A), , m, , m, , k, , (B), , m, , F = 2mg, , k, , (B) m, , 2m, , g, , g, , 2k, (C), , k, , k, , k, , m, (C), , (D), , m, , m, , k, , F = 2mg, , F = 2mg, , k/2, m, , (D), , Column-II, (p) Time period of oscillation T = 2π, , Column - II, , 1, (p), 2π, , 3k, 2m, , initially relaxed when force is applied, , 1, (q), 2π, , 2k, m, , 1, k, 1 3k, (s), 2π 3m, 2π m, 73. A mass m is subjected to a force, F = ( at − bx ) iˆ initially the mass lies at the, origin at rest. Here x refers to the x, coordinate of the mass, t refers to the time, elapsed. All the values are in S.I. unit (i.e., F , m , t , x , a and b are constants). Now match, the column-I with column-II. (All values in, column-II are in S.I.Units.), Column-I, Column-II, (A) Maximum velocity attained, by the mass, (p) 1, (B) Average velocity of the particle, during the subsequent motion, (q) 2, (C) Average acceleration of the particle, during subsequent motion, (r) 0, (D) Position of particle at t =, , π, 2, , 2mg, spring is, k, initially relaxed when force is applied, (r) Maximum velocity attained by block is before, , (q) Amplitude of oscillation is A =, , (r), , 84, , m, spring is, k, , (s), , π, −1, 2, , m, , force is applied block is in equilibrium 2 g k , , , position, (s) Maximum magnitude of acceleration of block, When force is applied block is in equilibrium is 2g., (t) Velocity of block when spring is in natural length, is zero. If block acquire natural length., 75. In column-I, the projection of a particle, moving along a circle (uniformly) in x − y, plane with its centre of origin along x and y, axes, while in Column-II, the description of, particle’s motion is given. It is given that, particle’s angular velocity is constant and, equal to ω and the radius of circle is, π, A, δ ≠ 0, or π . For this situation match the, 2, column-I with column-II., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , Column-I, (A) x ( t ) = A sin ( ωt + δ − π / 2 ) ;, (B) x ( t ) = A sin ( ωt + δ ) ;, (C) x ( t ) = A sin (ωt ) ; x ( t ) = A cos ( ωt ), (D) x ( t ) = A cos ( ωt + δ ) ;, Column-II, (p) Uniform circular motion (clockwise), (q) Uniform circular motion (anti-clockwise), (r) At t = 0 particle is neither on x − axis nor on, y -axis, (s) At t = 0 particle is either on x -axis or on, y − axis., 76. A simple harmonic oscillator consists of a, block attached to a spring with k = 200 N / m ., The block slides on a frictionless horizontal, surface, with equilibrium point x = 0 . A graph, of the block’s velocity v as a function of time, t is shown. Correctly match the required, information in Column-I with the values given, in Column-II (use π 2 = 10 ), V[m/5], 2π, 0.10, , 0.20, , O, , t(s), , -2π, , Column-I, (A) The block’s mass in kg, (B) The block’s displacement at, t = 0 in maters, (C) The block’s acceleration, , Column-II, (p) −0.20, , at t = 0.10 s in m / s 2, (D) The block’s maximum kinetic, energy in joules, , (r) 0.20, , 77. Statement-I: During the oscillations of simple, pendulum, the direction of its acceleration at the, mean position is directed towards the point of, suspension and at extreme position it is directed, towards the mean position., Statement-II: The direction of acceleration of a, simple pendulum at the mean position or at the, extreme position is decided by the tangential and, radial components of force by gravity., 78. Statement-I: For a particle of mass 1 kg, executing, S.H.M., if slope of restoring force vs displacement, graph is = -1, then the time period of oscillation, will be 6.28 s., Statement-II: If 1 kg mass is replaced by 2 kg, mass and rest of the information remains same as, in statement-1, then the time period of oscillation, will remain 6.28 s., 79. Statement-I: In simple harmonic motion, the graph, between velocity and the displacement is an ellipse., Statement-II: In simple harmonic motion the phase, difference between velocity and displacement is, π /2., 80. Statement-I: Two cubical blocks of same material, and of sides a and 2a , respectively are attached, rigidly and symmetrically to each other as shown., The system of two blocks is floating in water in, such a way that upper surface of bigger block is, just submerged in the water. If the system of blocks, is displaced slightly in vertical directions, then the, amplitude of oscillation on either side of equilibrium, position would be different., , a, , (q) -200, , (s) 4.0, , STATEMENT MODEL, (A) If Statement-1 is true, Statement-2 is true;, Statement-2 is the correct explanation for, Statement-1., (B) If Statement-1 is true, Statement-2 is true,, Statement-2 is not a correct explanation for, Statement-1., (C) If Statement-1 is true; Statement-2 is false., (D) If Statement-1 is false; Statement-2 is true., NARAYANAGROUP, , 2a, Statement-II: The force constant on two sides of, equilibrium position in the above-described situation, is different., 85
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 81. Statement-I: Three pendulums are suspended, from ceiling as shown., , 84. A rod of mass m and length l hinged at one, end is connected by two springs of spring, constants k1 and k2 so that it is horizontal at, equilibrium. What is the angular frequency, of the system ? (in rad/s), , ( Takel=lm,b=1/4m,K1 = 16N / m, K2 = 63N / m, m = 3kg ), These three pendulums are set to oscillate as shown, by arrows, and it is found that all three have same, time period. Now, all three are taken to a place, where acceleration due to gravity changes to 4/9th, of its value at the first place. If spring pendulum, makes 60 cycles in a given time at this place, then, torsion pendulum and simple pendulum will also, make 60 oscillations in same (given) time interval., Statement-II: Time period of torsional pendulum, is independent of acceleration due to gravity., 82. Statement-I: A circular metal hoop is suspended, on the edge by a hook. The hoop can oscillate, side to side in the plane of the hoop, or it can, oscillate back and forth in a direction perpendicular, to the plane of the hoop. The time period of, oscillation would be more when oscillations are, carried out in the plane of the hoop., Statement-II: Time period of physical pendulum, is more if moment of inertia of the rigid body about, corresponding axis passing through the pivoted, point is more., , K1, O, b, l, , K2, , 85. A uniform disc of mass m and radius, 80, R=, m is pivoted smoothly at P. If a, 23π 2, uniform ring of mass m and radius R is welded, at the lowest point of the disc, find the period, of SHM of the system (disc + ring). (in, seconds), P, , INTEGER ANSWER QUESTIONS, 83. A weightless rigid rod with a small iron bob at, the end is hinged at point A to the wall so that, it can rotate in all directions. The rod is kept, in the horizontal position by a vertical, inextensible string of length 20 cm, fixed at, its midpoint. The bob is displaced slightly, perpendicular to the plane of the rod and, string. Find period of small oscillations of the, system in the form, , πX, s , the value of X is, 10, B, l = 20cm, , A, Bob, l, , 86, , l, , R, R, , m, , 2R = r, , A, , R, R, , m, , B, , 86. In the figure shown a plate of mass 60g is at, rest and in equilibrium. A particle of mass, m = 30 g is released from height 4.5 mg/k from, the plate. The particle sticks to the plate., Neglecting the duration of collision find the, time from the collision of the particle and plate, to the moment when the spring has maximum, compression. Spring has force constant 1 N/, m. Calculate the value of time in the form, π / x and find the value of x ., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- IV, m, , OSCILLATIONS, , l, 2 cos θ, , , LEVEL - VI - HINTS, 1., , t = t1 + t2 + t3 , let C ', D ' be the extreme positions, of the springs k2 , k1 respectively.., ∴, , 87. A small body of mass m is connected to two, horizontal springs of elastic constant k , natural, length 3d / 4 . In the equilibrium position both, springs are stretched to length d , as shown., What will be the ratio of period of the motion, , t1 = t DD ' + tD ' D =, , 2 L t = t + t = T2 = π m, ; 3 CC ' C 'C 2, k2, v, Let x is the small displacement from mean position, perpendicular to plane of figure then:, , t2 = t DC + tCD =, , 2., , A, , θ, , (Tb / Ta ) if the body is displaced horizontally, , k, , k, , LEVEL-VI - KEY, SINGLE ANSWER QUESTIONS, 2) A 3) B 4)A 5) C 6) B 7) B, 9) A 10) C 11) A 12) B 13) C 14) A, 16) D 17) A 18) D 19) B 20) A 21) B, 23) B 24) A 25) B 26) A 27) C 28) A 29) B, MULTIPLE ANSWER QUESTIONS, 30) A,B, 31) B,D, COMPREHENSION QUESTIONS, 32) A 33) B 34) A 35) B 36) A 37) A 38) B, 39) C 40) B 41) C 42) B 43) A 44) B 45) A, 46) C 47) B 48) D 49) A 50) C 51) C 52) D, 53) C 54) C 55) B 56) B,A 57) B 58) C 59) C, 60) B 61) A 62) C 63) B 64) B 65) C 66) B, 67) B 68) B 69) A, MATRIX MATCHING QUESTIONS, 70) A-q; B-p; C-s ; D-r 71) A-r; B-t; C-q; D-r; E-q, 72) A-q; B-p; C-s; D-r 73) A-q; B-p ; C-r ; D-s, 74) A-p,q,r,s,t ; B-p,q,r,s,t ; C-p,q,r,s;D-p,q,r,s, 75) A-p,q,r ; B - p,q,r ; C - p,q,s ; D -p,q,r, 76) A - r ; B - p; C - q; D - s, 1) D, 8) B, 15) B, 22) A, , x, , F0+dF0, , m, restoring force on particle due to only two springs, , d, m, , C, , θ, , F0+dF0, , by a small distance where Ta is the time period, when the particle oscillates along the line of, springs and Tb is time period when the particle, oscillates perpendicular to the plane of the, figure ? Neglect effects of gravity., d, , T1, m, =π, 2, k1, , 3., , = 2 ( F0 + dF0 ) sin θ , From symmetry, total, restoring force due to all four springs =, x, ≈ 4 ( F0 + dF0 ), 4 ( F0 + dF0 ) sin θ :, l, 4 F0, 4F0, ≈, x = −ω 2 x, x (Q dF0 .x → 0 ) ⇒ a = −, l, ml, 2π, ∴T =, =?, ω, The diagram in disturbed condition is:, , x, ∆l, , y x', , x, , Excess length of, liquid which, provide restoring, force, , h, α, , β, , ∴ ∆l = x + x ' ......................(1) From diagram:, , α, x, , β, y, , y, , x', , ASSERTION & REASON QUESTIONS, , 77) A 78) C 79) B 80) A 81) D 82) A, INTEGER ANSWER QUESTIONS, 83) 4 84) 8 85) 2 86) 5 87) 2, NARAYANAGROUP, , sin α , y = x sin α = x 'sin β ⇒ x ' = , x : ....(2), sin β , 87
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JEE- ADV PHYSICS-VOL- IV, , OSCILLATIONS, , 1, ( 3m ) vcm2, 2, The remaining energy is oscillating between kinetic, and potential energy during the motion of blocks:, Oscillating energy = maximum elastic potential, energy, Passage-III, 38. When speed of block is maximum, net force on, block is zero. Hence at that instant spring exerts a, force of magnitude ‘ mg ’ on the block., 39. kx = mg .................(1), 1, 1, 2, mg ( L + x ) = kx 2 + mvmax ................(2), 2, 2, 3, gL, From (1) & (2) : we get vmax =, 2, 40. x = A sin ωt ; where x = L / 4, , of the system =, , moving with velocity v , then by using, v = ω A2 − x2 , we can find the amplitude of, motion., 2, 2, 3k 2 mg , 3k 2, 2, v0 , A −, A − ( y − y0 ) =, , =, 4m , 3k , 4m , 4, 2, , ⇒ t0 = ? ( Q from compression to mean position), and t2 =, , 2π π, =, 4ω 4, , y=0, y, m, m, 3, , v, , A, , v0, , 4m, 3, , y, , A 4m, 3, , As the bullet strikes the block with velocity v0 and, gets embedded into it, the velocity of the combined, mass can be computed by using the principle of, momentum conservation., m, 4m, v, V0 =, V ⇒v= 0, 3, 3, 4, Let new mean position is at distance y from origin,, 4m, 4mg, g ⇒y=, then ky =, . Now, the block, 3, 3k, executes S.H.M. about mean position defined by, 4mg, y=, . At t = 0 , the combined mass is at a, 3k, displacement of y − y0 from mean position and is, NARAYANAGROUP, , β θ, α, , L, ; ∴ t = t1 + t2 = ?, g, , Passage-IV, 43. Initially in equilibrium let the elongation in spring by, mg, y0 , then ⇒ y0 =, k, , 2, , 2, , mv02 mg , mv, mg , ⇒ 0 = A2 − , ⇒, A, =, +, , , 12k, 12k 3k , 3k , To compute the time taken by the combined mass, mg, from y =, to y = 0 , we can either go for, k, equation method or circular motion projection, method., , y=0, y = y0, y – y0, 4mg, y=, 3k, α = θ+β, , θ α −β, =, ω, ω, y − y0 mg, y 4mg, cos α =, =, ; cos β = =, A, 3kA, A 3kA, mg , −1 4mg , cos −1 , − cos , , 3kA , 3kA , so, t =, ω, , Required time, t =, , =, , 4m, 3k, , −1 mg , −1 4mg , cos 3kA − cos 3kA , , , , , , , Passage-V, 44. The angular frequency of simple harmonic motion, , k, m, The velocity of block, when it is at a displacement, of y from mean position is given by,,, is given by, ω =, , v = ω A2 − y 2 , where A is the amplitude of, oscillation. For given initial condition,, , v0 =, , 2, k, mv, A2 − h 2 ⇒ A2 = 0 + h 2, m, k, 2, , ⇒ A=, , mv0, + h2, k, 93
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JMAINS, - C.W - VOLIV, - II, JEE- ADV, PHYSICS-VOL-, , OSCILLATIONS, 45. To have the equilibrium of simple harmonic motion,, it is best to represent simple harmonic motion as, uniform circular motion., Y, , P, δ, , h, , y, , , , , k ( M + m ) Mm , l, Mm, , 3l, , = ( 2π + 6 ), , Mm, k ( M + m), , Passage-VII, 49. Angular frequency of oscillation is given by, , t=0, , At t = 0 , let particle is making an angle δ with, h, h, −ve, x − axisasshown,then sin δ = ⇒ δ = sin −1 , A, A, At time, y = − A sin (ωt + δ ), So, the equation of simple harmonic motion is,, 2, k, mv0, h , y=−, t + sin −1 , + h 2 sin , k, A , m, 46. To compute the time taken by the block to cross, mean position for the first time we can make use of, circular motion representation:, , h, π − sin −1 , π −δ, A, t=, =, ω, k, m, , k, = 10 3 . velocity of 3 kg block at the, m, instant it hits the 6kg block is given by, ω=, , v = ω A2 − x 2 = 10 3 22 − 12 ⇒ 30 m / s, During collision moment is conserved thus, , m1v1 = ( m1 + m2 ) vC ⇒ 3 × 30 = 9 × vC, or, hence, common velocity of combined mass =, , ( vC ) = 10 m / s, 50. After collision angular frequency of oscillation has, changed however the instantaneous position of, block does not change. Also equilibrium position, is still at relaxed length of spring., Q, , Passage-VI, 48. (1) CM remains at rest ; block moves 5t on plank, when system comes to rest, 5ml, −m [ 5l − ∆x ] + M ∆x = 0 ; ∆x =, m+M, 1 m.M 2, 1 2, (2) In CM frame , v m / M = kl, 2 m+M , 2, , M +m, vv / M = , k l, Mm , (3) Consider motion of block w.r.t.plank, kx, aM =, right ward ; F = − ( kx + maM ) or, M, d 2x k, , 1 1 ω 2 = kmM, =, −, k, + x, =, x, +, a, , , M, , (M + m) k, dt 2 m, M m, , Mm, ( M + m) k, , Time taken by block to travel 3l between springs, , 94, , 3l, , =, , k ( M + m), l, Mm, Time period of oscillation of block, = vm / M, , , , = 2 2t1 +, , , , , X, , π, Time to get the spring relaxed is t1 = 2, , 3l, , k, 9kg, P, , Hence x = 1m, , 1m, , Now angular velocity ω ' =, , ω'=, , k, ( m1 + m2 ) ;, , 900, = 10, 9, , Velocity = ω ' A '2 − x2 or 10 = 10 A' 2 − 12, Hence A ' = 2m, , 51. v 2 = ω 2 ( A2 − x 2 ), , For x = 8cm ; 302 = ω 2 ( A2 − 82 ) ..............(1), , For x = 15cm;16 2 = ω 2 ( A2 − 152 ) ..............(2), Solving (1) and (2),, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , THERMAL PROPERTIES OF MATTER-I, SYNOPSIS, , Ø, , Heat and Temperature, Heat:, Ø, Ø, Ø, , Different Thermometric Scales:, , Heat is a form of energy which produces the, sensation of hotness., Heat is the thermal energy of the body., Heat flows from bodies at high temperature to, bodies at low temperature., , Concept of Temperature:, Ø, Ø, Ø, Ø, Ø, Ø, , Temperature is a physical quantity which measures, the degree of hotness or coldness of a body., Temperature determines the direction of flow of, heat between two bodies in thermal contact with, each other until both acquire same temperature., When two bodies are at same temperature, then they are said to be in thermal equilibrium, with each other., In thermal equilibrium the heat in the two bodies, may or may not be equal., Temperature is the macroscopic property of a body, or a system., Temperature is a scalar quantity., , Ø, , Ø, , Ø, Ø, , Measurement of temperature requires:, (a) the construction of an instrument (i.e.,, thermometer), (b) the calibration of the thermometer., Construction of thermometer depends on some, physical property of matter (such as pressure,, volume, emf, resistance etc) that changes with, temperature., Calibration of the thermometer depends on fixing, certain points on the thermometer., The fixed points are ( ice point) the temperature of, the melting point of ice and (steam point) the, temperature of the boiling point of water under, normal atmospheric pressure., , Ø, , Ø, , Ø, , Ø, , NARAYANAGROUP, , 80°R, , 80, , C−0, F − 32, K − 273, R −0, =, =, =, (or), 100 − 0 212 − 32 373 − 273 80 − 0, C, F − 32 K − 273 R, =, =, =, (or), 100, 180, 100, 80, , Temperature difference on different scales is, ∆C ∆F ∆K ∆R, =, =, =, 5, 9, 5, 4, Common reading on Celsius and Fahrenheit scales, is −40° i.e. −40°C = −40°F ., C F − 32, X X − 32, =, ⇒ =, 5, 9, 5, 9, ., X = −40°, Temperature of the core of the sun is 107 K while, of its surface 6000 K. Normal temperature of, , Since,, , Ø, , human body is 310.15K ( = 37 0 C = 98.6 o F ), , The distance between the LFP and UFP of a, thermometer is called the fundamental interval., Fundamental interval = ( UFP ) − ( LFP ) ., The fundamental interval is divided into equal parts., These parts are given arbitrary numerical values of, temperature known as thermometric scale., , UFP Total No.of, divisions, 100°C 100, 180, 212°F, 373.15K 100, , C F − 32 K − 273 R, =, =, =, 5, 9, 5, 4, , Thermometric Scales:, Ø, , Thermometric LFP, scale, Celsius scale, 0°C, Fahrenheit scale 32°F, Kelvin scale (or) 273.15 K, Absolute scale, Reaumur scale, 0°R, On any thermometric scale, Reading - LFP, = constant, UFP - LFP, X−L, = constant., (or), U−L, , Relation Between Temperatures of Different, Scales:, , Measurement of Temperature, Ø, , The Celsius ( °C ) , Fahrenheit ( °F ) , Kelvin (K),, Reaumur (R), Rankine (Ra) are commonly used, thermometric scales., , while NTP implies 273.15K ( = 00 C = 32o F ) ., , Faulty Thermometer:, Ø, Ø, , If the reading on a faulty thermometer is ‘X’ and its, lower and upper fixed points are L and U, respectively then, correct reading on Celsius scale is, 1
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , Ø, , C−0 X−L, =, 100, U−L, , Correct reading on Fahrenheit scale is, F − 32 X − L, =, 180, U−L, , Ø, , Correct reading on Kelvin scale is, , Ø, , K − 273 X − L, =, ., 100, U−L, , Error in measurement by faulty thermometer, = measured value - true value, Correction = - Error, WE-1:The graph between two temperature scales, A and B is shown in Fig. Between upper fixed, point and lower fixed point there are 150, equal divisions on scale A and 100 on scale B., The relation between the temperatures in two, scales is given by ____, , 1, 9, tC = tC + 32, 2, 5, 320, = −24.60 C, or tC = −, 13, WE-4: An accurate Celsius thermometer and a, faulty Fahrenheit thermometer register 60°, and 141° respectively when placed in the same, constant temperature enclosure. What is the, error in the Fahrenheit thermometer?, C F − 32, 60 F − 32, =, ⇒F =1400 F, ⇒, Sol : From =, 5, 9, 5, 9, Error = 141- 140 = 10 F ; Correction = −10 F, , Types of Thermometers:, Primary standard thermometers, (gas thermometers), , 0, , Temperature ( A), , Constant volume Constant pressure, 0, , 180, , Gravimetric thermometers, (Compensated Air Thermometer), , Secondary Thermometers, , VtA = 1500, 300, O, , Expansion, (liquid), Thermometers, , VtB = 1000, , Radiation, Pyrometers, , Principle of Thermometry :, If X is a property that varies linearly with, temperature T as X = a T + b, where, ‘a’ and ‘b’ are constants then, , 0, , 0, , 100 Temperature ( B), , tA − 30 tB − 0, =, 150, 100, WE-2: What is the temperature for which the, reading on Kelvin and Fahrenheit scales, are same ?, Sol : On the Kelvin and Fahrenheit scales, , Sol : When t B = 0, t A = 300 ∴, , K - 273.15 F - 32, =, 100, 180, , X −X , t = t 0 ×100°C is general equation, X100 − X0 , used to measure temperature t., or X t = X 0 (1 + α t ), , Measurement of Temperature Based, on Triple point, , (if X=K=F), , X - 273.15 X - 32, =, 100, 180, 9, X = (255.38) = 574.6, 4, \ 574.6 K = 574.6°F.., WE-3:At what temperature is the Fahrenheit, scale reading equal to half that on the, Celsius scale ?, 9, 1, Sol : As t F = tC + 32 and t F = tC ,, 5, 2, 2, , Resistance, Thermometers, , Ø If the value of thermometric property at 0K,, 273.16K and T KK is 0, XTr and X respectively, Ø, Ø, , TK, X, X, =, i.e., TK = ( 273.16 ), K, ,, 273.16 XTr, XTr, When a constant volume gas thermometer is used, to measure temperature of a body then, TK = ( 273.16 ), , P, K, PTr, , Where PTr is pressure of a given amount of gas at, triple point of water and P is the pressure at a, temperature which is to be determined, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, WE-5: Two absolute scales X and Y have triple, points of water defined to be 300 X and 450Y., How are TX and TY related to each other ?, Sol : Here, temperature 300 on absolute scale, X=273.16K (Triple point of water), ∴ Value of temperature T X on absolute, scale X =, , 273.16, TX, 300, , Similarly, value of temperature T Y on absolute scale, Y =, , 273.16, TY, 450, , Since both these values are equal,, 273.16, 273.16, 2, TX =, TY, ∴TX = TY, 300, 450, 3, WE-6:The readings corresponding to the ice point, and steam point for a constant pressure gas, thermometer are 500 cc, and 545 cc. If the, reading corresponding to room temperature, be 510 cc, find the room temperature?, Sol. Given : V0=500cc; V100=545cc. and Vt=510 cc., , THERMAL PROPERTIES OF MATTER–I, WE-9:The resistance of a platinum resistance, thermometer is found to be 11.0 ohm when, dipped in a triple point cell. When it is dipped, in a bath,resistance is found to be 28.887 ohm., Find the temperature of the bath in 0C on, platinum scale., Sol. In terms of triple point of water,, é, R ùú, TK = ê 273.16, K, ê, RTr úû, ë, 28.887, = 717.32 K, 11.0, Now as TC = TK - 273.15, , so TK = 273.16´, , TC = 717.32 - 273.15 = 444.17 0 C, , WE-10:Graph shows the relation between, Centigrade and Fahrenheit scales of, temperature . Find slope in each graph?, Sol., Case (i) :, F, , Vt − V0 , 510 − 500 , × 100 = , 100, 545 − 500 , 100 − V0 , , Using, t = V, , = 22.22o C, WE-7: A constant volume gas thermometer shows, pressure readings of 50 cm and 90 cm of, mercury at 00C and 1000C respectively.What, is the temperature on gas scale when the, pressure reading is 60 cm of mercury ?, Sol. Given that P 0=50cm of Hg,P100=90 cm of Hg, Pt= 60 cm of Hg, t=, , Pt - P0, ´100 = 60 - 50 ´100 = 250 C, P100 - P0, 90 - 50, , WE-8:The resistance of a platinum wire is 15Ω, at 200C. This wire is put in a hot furnace and, the resistance of the wire is found to be 40Ω., Find the temperature of the hot furnace if, temperature coefficient of resistance of, platinum is 3.6 × 10−3 0C −1, , 2120 F, , Slope =, 320 F, 1000C, , -17.780 C O, , 25, + 40 × 20 ≈ 7745, 3.6 × 10−3, 7745, ⇒ t2 =, ≈ 5160 C, 15, , 15t 2 =, , NARAYANAGROUP, , C, , A plot of Fahrenheit temperature (F) versus, Celsius temperature (C), , 9, F = C + 32, 5, Slope of the graph, , Case (ii):, , (Q y = mx + c ), m =9/5, , C F − 32, 5, 160, =, ;C = F −, (Q y = mx + c ), 5, 9, 9, 9, , Slope of the graph , m = 5/9, C, , R2 (1 + α t2 ), Sol. Rt = R0 (1 + α t ) ⇒ R = (1 + α t ), 1, 1, 40 (1 + α t2 ), =, ⇒ 40 − 15 = α (15t2 − 40t1 ), 15 (1 + α t1 ), , 9, 5, , Slope =, , O, , 320F, , 5, 9, , F, , -17.780C, , A plot of Celsius temperature (C) versus, Fahrenheit temperature (F), 3
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , Thermal Expansion of Solids:, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, , Thermal expansion of solids is different for isotropic, and anisotropic solids., Thermal expansion is same in all directions of, isotropic solids., Metals, amorphous solids like glass and crystals, like rock salt are the examples of isotropic solids., Thermal expansion is different in different directions, of anisotropic solids., Many of the crystals such as crystalline CaCO3 ,, Galena are the examples of anisotropic solids., Solid substances such as cast iron, silver iodine,, silica glass, rubber, leather, ice, lead etc., contract, on heating., The interatomic force of attraction depends on the, distance between atoms., On heating solids expand due to increase in, interatomic spacing, which is a result of, asymmetrical lattice vibrations., The molecules possess both KE and PE. So the, KE and PE of molecules increase when the body, is heated., The increase in KE may be in the form of, a) translational KE, b) vibrational KE, c) rotational KE, In solids the increase in KE is in vibratory KE and, rotatory KE., The increase of KE results in rise in temperature., , Coefficient of Linear Expansion:, Ø, , α =, , Ø, Ø, Ø, , Due to the presence of intermolecular attraction,, the molecules possess PE., At a particular distance of separation the force of, attraction is maximum, potential energy is minimum, and stability is maximum., The atoms are in a specific state of vibration, at a, particular distance of separation (ro) and makes, solid to have a definite size., The graph between the interatomic distance and, potential energy is a curve called potential energy, curve., , E, , 1 dl , , r, , E1, E0, , Ø, Ø, , Ø, Ø, , 4, , The coefficient of linear expansion of a solid depends, on the nature of the material and scale of, temperature. (it is independent on dimension of, material ), The linear expansion of a solid l2-l1=e= l1 α (t2- t 1), It depends on three factors., a) Its original length (l1), b) The nature of the material ( α ), c) Change in temperature (t2 - t 1), Increase in length ∆l = lα∆t, ∆l, Fractional change in length, = α∆t, l, Percentage change in length, , Ø, , For anisotropic solids, if αx , αy and α z are, coefficients of linear expansions along x, y and z, directions respectively then the average coefficient, of linear expansion is α =, , Ø, , ∆l, × 100 = α∆t × 100, l, , αx + α y + αz, 3, , Numerical value of coefficient of linear expansion, of a solid is αC when the temperature is measured, in Celsius scale and its value is α F when the, temperature is measured in Fahrenheit scale then, a) α F, , 5, 9, = α C (or) α C = α F, 9, 5, , b) α F < α C ., A composite rod is made by joining two rods of, different materials and of same cross section. If, l1, l2 are their initial lengths at t1 °C , then, (a) the increase in length of composite rod at t2 °C, , r2, T1, , r1, r0, , / 0C, , l 2 = l 1 1 + α ( t 2 − t1 ) , , Ø, , 0, E2, , l 2 − l1, / 0C, l1 × ( t 2 − t1 ), , α in differential form α = l0 dt , length of the solid after heating, , Potential Energy Curve:, Ø, , The ratio of increase in length of a solid per degree, rise in temperature to its original length is called, coefficient of linear expansion ( ∝ ), , T0, , is given by ∆l = ( l1α1 + l2α 2 )( t2 − t1 ), b) The effective coefficient of linear expansion of, the composite rod is given by α =, , l1α 1 + l2 α 2, ., l1 + l2, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , Ø, , Variation of density of substance with, temperature, When a solid is heated its volume increases and, hence its density decreases, as mass remains, constant., If ρ 1 and ρ 2 are densities of a solid at t 10C and, t20C, and as m1=m2 ; ρ 1 V1 = ρ 2 V2, ρ 1V1= ρ 2V1[1+ γ (t2-t1)], ρ 1= ρ 2 [ 1+ γ (t2 - t 1)], If ρ t and ρ 0 are densities at t 0 C and 00 C., ρ0, ρt =, (1+?t), , Ø, , Ø, Ø, Ø, , Ø, , (or) ρ t =ρ o (1+?t)-1 ., , ρt ≈ ρo (1-?t), For anisotropic materials γ is the sum of linear, coefficients in three mutually perpendicular, directions., γ = αx + α y + αz ., , Ø, Ø, Ø, , For isotropic solids γ = 3α, Ø, , Ø, , Ø, Ø, Ø, Ø, Ø, , Ø, , 6, , Same Expansion In Different Rods:, , Applications:, Between the rails a gap is left to allow for their, expansion in summer. If l is the length of the rail, and ∆t is the change in temperature then the gap, is given by ∆ l = l α ∆ t, A wire of length l is bent in the form of a ring with, a small gap of length x1 at t 10C. On heating the, ring to t 20C the gap increases to x2 in length. The, coefficient of linear expansion of wire, x −x, α= 2 1, x1 (t 2 − t1 ), Gap behaves like the material for all thermal, expansions., Telephone wires are loosely connected between, the poles in summer, to allow for their contraction, in winter., Concrete roads are laid in sections and gaps are, provided between them to allow for expansion., Pipes used to convey steam from boiler must have, loops to prevent cracking of pipes due to thermal, expansion., Huge iron griders used in the construction of, bridges and buildings are allowed to rest on rollers, on either side providing scope for expansion., Hence the damage to the structure can be avoided., When a drop of water falls on a hot glass chimney,, the portion of the spot where the water falls,, contracts and the remaining portion expands.So,, the glass chimney breaks (brittle nature of the glass, also)., , Pyrex glass is used to prepare test tubes for heating, purpose because its linear expansion coefficient is, small. ( α = 3x10-6 0C-1), Silica glass (quartz) is used for making bulbs of, thermometer because of low linear expansion, coefficient. ( α = 0.5 x10-6 0C-1), Invar is an alloy of Iron, Nickel and Carbon . Invar, has very low linear expansion coefficient, so used, in wrist watches, pendulum clocks and standard, scales., A hole is drilled at the centre of a metallic plate., When plate is heated, the diameter of hole, increases., When two holes are drilled on a metal plate and, heated the distance between the holes increases., When a solid and hollow spheres with same outer, radius made up of same metal are heated to same, temperature then both expand equally., Platinum is used to seal glass because their, coefficients of expansion are almost same., If two rods of different materials have the same, difference between their lengths at all temperatures, only when their linear expansions are equal., ∆l1 = ∆l2 ; l1α1∆t = l2α 2 ∆t, l1 α 2, Then l1α1 = l2α 2 , l = α, 2, 1, if the constant difference in their lengths is x then, xα 2, xα1, l1 =, l2 =, ,, α1 ~ α 2, α1 ~ α 2 , x = l2 ~ l1, , Bimetallic Strip:, t, , t, , l, t, t, , r1, l, , Ø, Ø, , R, , θ, , r2, , where t is thickness of each strip, Bimetallic strip works on the principle that different, metals expand differently for the same rise in, temperature., If a bimetallic strip made of brass and iron is heated, brass bends on convex side (α b > α i ), NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, Ø, Ø, , If it is cooled brass bends on concave side., Radius of curvature of a bimetallic strip., θ=, , Ø, Ø, , Pendulum Clocks:, Variation of Time Period of Pendulum Clocks :, , l, dl l − l, ( or ) θ = = 2 1 ; l = l (α 2 − α1 ) ∆T, R, dr r2 − r1 R, 2t, , ∴ R=, , Ø, , THERMAL PROPERTIES OF MATTER–I, , 2t, (α 2 − α1 ) ∆T, , l0, g, If temperature is increased by ∆ t,, T0 = 2π, , (Qα 2 > α1 ), , l0 (1 + α ∆t ), g, (by using Binomial expansion), T = 2π, , t = thickness of each metal strip used., Bimetallic strip can be used as temperature sensor, in thermometers and fire alarms., As an automatic switch or circuit breaker in electric, iron, refrigerators, incubators, thermostats, flash, lights etc., As a balance wheel in wrist watches., , T = T0 (1 +, , α, ∆t, 2, ∆T = increase in time period ., , ⇒ ∆T = T − T0 = T0, , Measuring Tapes:, Ø, Ø, Ø, Ø, , Measuring tapes are made of invar steel because, of its least coefficient of linear expansion., Measuring tapes made of metals show correct, reading only at a temperature at which they are, constructed or calibrated., When a metal scale expand it shows less value than, true value and vice versa, If lm is the measured reading of the length of a, , Ø, , body at t20C and lc is the correct length of the, , Ø, Ø, Ø, , Ø, , K=, , P, P, =, ∆V γ∆t, V, , ⇒ P = K γ∆t, , ⇒ P = 3 K α ( t 2 − t1 ), where K is bulk modulus, NARAYANAGROUP, , The total expansion of brass rods should be equal, to that of steel rods . ∆l1 = ∆l2, , t2 > t1 , , where αS , α b are the coefficients of linear, expansions of the scale and the body respectively., A metal scale calibrated at particular temperature, does not give the correct measurement at any other, temperature., When scale expands correction to be made, ∆l = lα∆t Correct reading = l + ∆l (t2> t 1), When scale contract correction to be made, ∆l = lα∆t Correct reading = l − ∆l (t2< t 1), l= measured value, % error in the measurement = α∆t ×100, If a cube of coefficient of cubical expansion γ is, heated, then the pressure to be applied on it to, prevent its expansion is P then ∆V = V γ∆t, , Pendulum clocks looses time in summer and gains, time in winter, 1, The loss or gain per day = a? t × 86400 Sec., 2, Compensated pendulum length is always constant, at all temperatures, so it shows correct time at all, temperatures., , Grid Iron Pendulum:, , body at calibration temperature t1°C then, , lc = lm 1+ ( αS −αb ) ( t 2 − t1 ), , α, ∆t ), 2, , n1l1α1 = n2l2α 2, , Thermal Stress:, Ø, Ø, , It is developed due to prevention of expansion of, a solid when it is heated., A rod of length l0 clamped between two fixed walls., For ∆t Change in temperature, Young’s modulus, F / A Fl0, F, Y=, =, =, ∆l / l0 A∆l Aα∆t (Q ∆ l = l 0α ∆ t ), F, = Y α ∆t, A, Thermal force F = YAα∆t ., Thermal force is independent of length of rod., Thermal stress = Yα∆t, For same thermal stress in two different rods, heated through the same rise in temperature,, or, , Ø, Ø, Ø, , Ylα1 = Y2α 2, , Ø, , Two rods of different metals having the same area, of cross section A are placed between the two, 7
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , WE-14: An iron rod of length 50cm is joined at an, end to copper rod of length 100cm at 200C., Find the length of the system at 1000C and, average coefficient of linear expansion of the, system., , inertia of the wheel and C is the torsional, rigidity of its spring. The wrist watch keeps, accurate time at 250C. How many seconds, would it gain a day at –250C if the balance, wheel made of Aluminium ?, , ( airon = 12 ´10-6 / 0 C and acopper = 17´10-6 / 0 C. ), Sol. Increase in length of composite rod is, ∆l = ∆l1 + ∆l2 = ( α1l1 + α 2l2 )∆t, , ( Given, α Al =25.5 × 10–6/ 0C ), , = (12 × 10 −6 × 50 + 17 × 10−6 ×100) × (100 − 20), , = 0.192cm, Length of the composite rod at 1000 C is l + ∆l, = 150.192 cm, Average linear expansion co-efficient, ∆l, 0.192, =, = 16 × 10 − 6 / 0 C, l ∆ t 150 × 80, WE-15: Density of gold is 19.30 g/cm3 at 200C., Compute the density of gold at 90 0 C by, α avg =, , adding steam to it. (α = 14.2 ×10 −6 / 0 C ), , ρ2 V1, V1, −1, Sol : ρ = V = V (1 + 3α∆T ) = (1 + 3α∆T ), 1, 2, 1, , or, , ρ2, = (1 − 3α ∆T ) or, ρ1, , ρ 2 = ρ1 (1 − 3α∆T ), WE-16: Uniform pressure P is exerted on all sides, of a solid cube of bulk modulus, B and volume, coefficient of expansion γ , at temperature t 0C., By what amount should the temperature of, cube be raised in order to bring its volume, back to the value it had before the pressure, was applied ?, Sol : As, , ....(i), , If ∆T is the required increase in, temperature, ∆V = γ V ∆T ....(ii), From eqns. (i) and (ii),, γ V ∆T =, , VP, B, , P, , or ∆T = γ B, , WE-17 :The balance wheel of a mechanical wrist, watch has a frequency of oscillation given by, f =, , 1, 2π, , C / I , where I is the moment of, , NARAYANAGROUP, , C, 1, =, I 2π k, , 2π, , f ∝, , As, , C, M, , 1, 1, ; f∝, k, T, , Q I = MK 2 , df, , −dT, , −dk, , ⇒ f = T = k, , dk, dT, = α dt ⇒, = +α dt, k, T, , Number of seconds gained/day, dT = ( 8.64 × 104 ) (α dT ) =110.2 s/day, , WE-18: A clock with a metallic pendulum is, 5seconds fast each day at a temperature of, 15 o C and 10 seconds slow each day at a, temperature of 30 o C. Find coefficient of, linear expansion for the metal., Sol. The time lost or gained per day is, 1, , ∆t = 2 α∆T × 86400 [as 1 day = 86400 s.], If graduation temperature of clock is T 0 then gain, in time at 150C is, 1, (α ) (T0 -15) × 86400 ........ ( i ), 2, , 5 =, , = (19.30 ) 1 − 3 (14.2 ×10−6 ) ( 70 ) = 19.24 g / cm3, , P, VP, B=, , ∆V =, ∆V / V, B, , 1, Sol : f =, , At 30oC clock is loosing time thus, 10 =, , 1, 2, , α (30 – T0) 86400 ...... (ii), , Dividing equation (ii) by (i), we get, 2 (T0 – 15) = (30 – T0) or T0 = 20oC, Thus from equation (i), 5=, , 1, 2, , α [20 – 15] 86400, , α = 2.31 × 10–5 / o C, WE-19:A steel bar of cross sectional area 1 cm 2 and, 50 cm long at 300C fits into the space between, two fixed supports. If the bar s now heated to, 2800 C, what force will it exert against the, supports ?, ( a for steel = 11 × 10–6/0C and, Young's modulus for steel = 2 × 10111 N/m2 ), Sol :Force exerted on the supports, =Stress × Area of cross section = Y a A (t2 - t1 )., = 2 × 10111 × 11 × 10–6 × 10–4 × 250 = 55000N., 9
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, WE-20: A mass of 2kg is suspended from a fixed, point by a wire of length 3m and diameter, 0.5 mm. Initially the wire is just unstretched,, the mass resting on a fixed support. By how, much must the temperature fall if the mass is, to be entirely supported by the wire (Given Y, for wire = 206 G Pa, α =11 × 10–6/0C), Sol :Contraction due to cooling is equal to the, stretching produced by the weight ‘mg’., ∴∆L =, , mgL, 2 × 9.8 × 3, =, AY π ( 0.25 )2 ×10−6 × 206 ×109, , Now the contraction due to cooling, = Lα∆t =3 × 11, 1 × 10–6 × ∆t, solving ∆t = 440 C, WE-21:A metallic rod of length l cm and crosssectional area A cm2 is heated through t°C., After expansion if a mechanical force is, applied normal to its length on both sides of, the rod and restore its original length, what, is the value of force? The young's modulus, of elasticity of the metal is E and mean, coefficient of linear expansion is α per, degree Celsius., Sol :Change in the length = ∆l = lα t, Length of rod at t°C is l + l αt, Decrease in length due to stress = ∆l, But length of rod remains constant ∴∆lt − ∆l = 0, stress F l + ∆ lt, ∴ ∆ l = ∆ lt = l α t ⇒ E =, = ×, strain A −∆ lt, , ∴F =, , EA∆lt − EAl αt, EAαt, =, =−, l + ∆lt, l + l αt, (1 + αt ), , Here, negative sign indicates that forces are, compressive in nature., WE-22:Two metal rods are fixed end to end between, two rigid supports as shown in figure. Each, rod is of length l and area of cross-section is, A.When the system is heated up,determine the, condition when the junction between rods does, not shift ?(Y1 and Y2 are Young's modulus of, materials of rods, α1 and α 2 are coefficients, of linear expansion), 10, , L, , L, , α1Y1, , α2Y2, , Sol :Since, each rod is prevented from expansion, so,they are under compression and mechanical, strain.The strain in each rod is zero., e1 e2, e1, F, e2, F, = =0, = α1∆T −, = α 2 ∆T −, l, l, AY1 ; l, AY2 ; l, , α1∆T −, , F, F, = 0 and α 2 ∆T −, =0, AY1, AY2, , α1∆T =, , F, F, .........(1) and α2∆T =, ......(2), AY1, AY2, , Dividing (1) by (2), we get α1Y1 = α 2Y2, WE-23: A bimetallic strip of thickness 2 cm, consists of zinc and silver rivetted together., The approximate radius of curvature of the, strip when heated through 500 C will be :, (linear expansivity of zinc and silver are, 32 × 10–6/0C and 19 × 10–6 /0C respectively), 2t, Sol : Radius of curvature R = (α − α )∆T, 2, 1, , R=, , 2 ×1, = 30.77m, ( 32 − 19) ×10−6 × 50, , WE-24: A steel rail 30m long is firmly attached, to the road bed only at its ends. The sun raises, the temperature of the rail by 500C, causing, the rail to buckle. Assuming that the buckled, rail consists of two straight parts meeting in, the centre, calculate how much the centre of, the rail rise? Given, α steel = 12 × 10–6/0C., x+Vx, , y, , x+Vx, , 2x, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, Sol :Let the initial length be 2 x and the final total length, be 2 ( x + ∆ x ) as shown ., Let y be the height of the centre of the buckled rail., Clearly, ∆x = α x ∆T and, , THERMAL PROPERTIES OF MATTER–I, Sol :Given d=20cm, V = V0 (1 + γ t ) = V0 (1 + 3α t ) (since γ =3α ), , change in volume = V − V0 = 3V0α t, 3, , 4 d, −6, = 3 × π × 23 × 10 × 100, 3 2, , y = ( x + ∆x )2 − x 2 = 2 x ( ∆x ) = 2 x 2α∆t, y = x 2α∆T, , [neglecting ( ∆x )2], , Thus, y = [15 2 (12 ×10−6 ) 50]cm = 0.52m, WE-25:When composite rod is free, composite, length increases to 2.002m from temperature, 200C to 1200C. When composite rod is fixed, between the support, there is no change in, component length. Find Y and α of steel if, Ycu = 1.5 × 1013 N / m 2 α cu = 1.6 ×10−5 /0C, , Steel, , Copper, 0.5m, 2m, , Sol. ∆l = lsα s ∆T + lcα c ∆T, , 3, , 4 0.2 , −6, = 3 × π × 23 × 10 × 100, 3 2 , , = 28.9cc (1cc = 10−6 m3 ), W.E.28:A wooden wheel of radius R is made of two, semi circular parts (see figure).The two parts, are held together by a ring made of a metal, strip of cross-sectional area S and length L.L, is slightly less than 2π R .To fit the ring on, the wheel,it is heated so that its temperature, rises by ∆T and it just steps over the wheel.As, it cools down to surrounding temperature. It, presses the semi-circular parts together.If the, coefficient of linear expansion of the metal is, α and its Young’s modulus is Y, then the force, that one part of the wheel applies on the other, part is (AIEEE 2012 ), , 0.002 = 1.5α s + 0.5 × 1.6 × 10 − 5 ´100, , 1.2 ×10 −5, = 8 ×10−6 0C, 1.5, If there is no change in composite length, thermal, force of steel and copper rod should be equal, Fst=Fcu ; Y s A α s ∆t = Y cu A α Cu ∆t, αs =, , R, , Ys αc, α c 1.5 × 1013 × 1.6 × 10−5, = ; Y = Yc ×, =, Yc αs s, αs, 8 × 10 −6, Ys = 3 × 1013 N / m 2, WE-26: A metal rod of Young’s modulus F and, coefficient of thermal expansion α is held at, its two ends such that its length remains, invariant.If its temperature is raised by t o c ,then, the linear stress developed in it is(AIE-2011 ), F, FL, Sol : ∆L = α L∆T = AY ⇒ Stress = = Y α∆T = Y α t, A, WE-27: An aluminium sphere of 20cm diameter is, heated from 0o c to 100o c . Its volume changes, by (given that the coefficient of linear, , expansion for aluminium (α A1 = 23×10−6 / o C ), (AIEEE 2011 ), , NARAYANAGROUP, , F, , Sol :, T, , T, , Increase in length, ∆L = α L ∆ T ⇒, , ∆L, = α∆T, L, , the thermal stress developed is, T, ∆L, =Y, = Y α∆T ; T = SY α∆T, S, L, , From the FBD of one part of the, wheel, F = 2T, Where F is the force applied by one part of, the wheel on other part, F = 2 SY α∆T, 11
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 9., , C.U.Q, MEASUREMENT OF TEMPERATURE, 1., , 2., , 3., , Temperature of gas is a measure of, 1) the average translational kinetic energy of the, gas molecules, 2) the average potential energy of the gas molecules, 3) the average distance of the gas molecules, 4) the size of the molecules of the gas, Celsius is the unit of, 1) Temperature, 2) Heat, 3) Specific heat, 4) Latent heat, On the Celsius scale the absolute zero of, temperature is at, 1) 00 C 2) −320 C, , 3) 1000 C, , 4) −273.150 C, , 4. The correct value of 00 C on the Kelvin scale is, 1) 273.150 C, 5., , 6., , 7., , 2) 273.160 C, , 0, 3) 2730 C, 4) 273.2 C, The standard scale of temperature is, 1) the mercury scale, 2) the gas scale, 3) the platinum resistance scale 4) liquid scale, Melting and Boiling point of water on, Fahrenheit scale of temperature respectively, 1) 212 0 F , 320 F, 2) 320 F , 2120 F, , 3) 00 F ,1000 F, 4) 320 F ,1320 F, For measurements of very high temperature, say around 50000 C (of sun) , one can use:, 1) Gas thermometer, 2) Platinum resistance thermometer, 3) Vapour pressure thermometer, 4) Pyrometer( Radiation thermometer), , 8., , Mercury boils at 3560 C . However, mercuryy, thermometers are made such that they can, measure temperatures upto 5000 C . This is, done by, 1) maintaining vacuum above the mercury column, in the stem of the thermometer, 2) filling Nitrogen gas at high pressure above the, mercury column, 3) filling Nitrogen gas at low pressure above the, mercury column, 4) filling oxygen gas at high pressure above the, mercury column, , 12, , 10., , 11., 12., , 13., , For measuring temperature near absolute, zero,the thermometer used is, 1) thermoelectric thermometer, 2) radiation thermometer, 3) magnetic thermometer, 4) resistance thermometer, Which of the following scales of temperature, has only positive degrees of temperature?, 1) Centigrade, 2) Fahrenheit scale, 3) Reaumur scale, 4)Kelvin scale, Which of the following is the smallest rise in, temperature?, 1) 1oF, 2) 1oR, 3) 1K, 4) 1oC, The temperature at which two bodies appear, equally hot or cold when touched by a person, is, 1) 0oC, 2)37oC, 3) 25oC, 4) 4oC, The range of clinical thermometer is, 1) 37oC to 42oC, 2) 95oF to 110oF, , 4) 950 C to1040 C, 3) 900 F to1120 F, 14. Which of the following is the largest rise in, temperature?, 1) 1oF, 2) 1oR 3) 1K, 4) 1oC, , THERMAL EXPANSION OF SOLIDS, 15. Solids expand on heating because, 1) the K.E. of the atoms increases., 2) the P.E. of the atoms increases, 3) total energy of the atoms increases., 4) the K.E. of the atoms decreases., 16. Expansion during heating, 1) occurs only in solids., 2) decreases the density of the material, 3) occurs at same rate for all liquids and gases., 4) increases the weight of the material., 17. When a metal bar is cooled, then which one of, these statements is correct., 1)Length, density and mass remain same., 2) Length decreases, density increases but mass, remains same, 3)Length and mass decrease but density remains, the same., 4)Length and density decrease but mass, remains the same., 18. When a metal bar is heated, the increase in, length is greater, if, 1) the bar has large diameter 2) The bar is long., 3) the temperature rise small 4) Small diameter, 19. A ring shaped piece of a metal is heated, If, the material expands, the hole will, 1) contract 2) expand 3) remain same, 4) expand or contract depending on the width, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 20. A solid ball of metal has a spherical cavity, inside it. The ball is cooled.The Volume of, the cavity will, 1) decrease, 2) increase, 3) remain same, 4) have its shape changed, 21. The substance which has negative coefficient, of linear expansion is, 1) lead, 2) aluminum, 3) iron, 4) invar steel, 22. Two spheres of same size are made of same, material but one is hollow and the other is, solid. They are heated to same temperature,, then, 1) both spheres will expand equally., 2) hollow sphere will expand more than solid one, 3) solid sphere will expand more than hollow one, 4) hollow sphere will expand double that of solid, one, 23. If temperature of two spheres of same size, but made of different materials changes by, ∆T then, 1) both expands equally, 2) sphere with greater α expands or contracts, more than other., 3) sphere with greater α expands or contracts less, than other., 4) both contracts equally., 24. The linear expansion of a solid depends on, 1) its original mass, 2) nature of the material and temperature, difference., 3) the nature of the material only, 4) pressures, 25. The coefficient of linear expansion of a solid, depends upon, 1) the unit of pressure, 2) the nature of the material only, 3) the nature of the material and temperature, 4) unit of mass, 26. If α c and α k denote the numerical values of, coefficient of linear expansions of the solid,, expressed per 0C and per Kelvin respectively,, then., 1) α c > α k, 2) α c < α k, 3) α c = α k, 4) α c = 2 α k, 27. If α c and α f denote the numerical values of, coefficient of linear expansion of a solid,, expressed per 0C and per 0F respectively, then, 1) α c > α f, 2) α f > α c, 3) α f = α c, 4) α f + α c = 0, NARAYANAGROUP, , THERMAL PROPERTIES OF MATTER–I, 28. The coefficient of linear expansion of a metal, , 29., , 30., , 31., , 32., , 33., , 34., 35., , 36., , 37., , rod is 12x10-6 / 0C, its value in per 0F, 20, 15, × 10 − 6 / 0 F, × 10 −6 / 0 F, 2), 1), 3, 4, 3) 21.6 × 10 −6 / 0 F 4) 12 × 10 −6 / 0 F, The coefficient of volume expansion is, 1) equal to the coefficient of linear expansion., 2) twice the coefficient of linear expansion, 3) equal to the sum of coefficients of linear and, superficial expansions., 4) Twice the coefficient of areal expansion., Always platinum is fused into glass, because, 1) platinum is good conductor of heat, 2) melting point of platinum is very high, 3) they have equal specific heats, 4) their coefficients of linear expansion are equal, Two metal strips that constitute a bimetallic, strip must necessarily differ in their., 1) length, 2) mass, 3) coefficient of linear expansion 4) resistivity, Thermostat is based on the principle of, 1) equal expansion of two rods of different lengths., 2) different expansion of two rods of different, lengths., 3) different expansion of two rods of same length, 4) equal expansion of two rods of same length., A pendulum clock shows correct time at 0 0 C., At a higher temperature the clock., 1) looses time, 2) gains time, 3) neither looses nor gains time 4)will not operate, To keep the correct time modern day watches, are fitted with balance wheel made of, 1) steel 2) platinum 3) invar 4) tungsten, A brass disc fits into a hole in an iron plate., To remove the disc., 1) the system must be cooled, 2) the system must be heated, 3) the plate may be heated (or) cooled, 4) the disc must be heated, When hot water is poured on a glass plate, it, breaks because of, 1) unequal expansion of glass, 2) equal contraction of glass, 3)unequal contraction of glass 4)glass is delicate, When the temperature of a body increases, 1) density and moment of inertia increase, 2) density and moment of inertia decrease, 3) density decreases and moment of inertia, increases., 4) density increases and moment of inertia, decreases., 13
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 38. In balance wheel of watch, the factors that, make its oscillations uniform are, 1) tension in string, 2) moment of inertia of balance wheel, 3) temperature, 4) pressure, 39. When a metal ring is heated, 1) the inner radius decreases and outer radius, increases, 2) the outer radius decreases and inner radius, increases, 3) both inner and outer radii increases, 4) both inner and outer radii decreases, 40. A cube of ice is placed on a bimetallic strip at, room temperature as shown in the figure. What, will happen if the upper strip of iron and the, lower strip is of copper?, , 46., , 47., , Ice, , Fe, ...................Cu....................., , 41., , 42., , 43., , 44., , 1) Ice moves downward 2) Ice moves upward, 3) Ice remains in rest, 4) None of the above, To withstand the shapes of concave mirrors, against temperature variations used in high, resolution telescope, they are made of, 1) quartz, 2) flint glass, 3) crown glass 4)combination of flint and silica, The holes through which the fish plates are, fitted to join the rails are oval in shape because, 1) bolts are in oval shape, 2) to allow the movement of rails in the direction, of length due to change in temperature., 3) to make the fitting easy and tight, 4) only oval shape holes are possible, A semicircular metal ring subtends an angle, of 1800 at the centre of the circle. When it is, heated, this angle, 1) remains constant, 2) increases slightly, 3) decreases slightly 4) becomes 3600, The diameter of a metal ring is D and the, coefficient of linear expansion is α . If the, temperature of the ring is increased by 10 C,, the circumference and the area of the ring will, increases by, 1) π Dα , 2π Dα, 2) 2π Dα , π D 2α, , π Dα, π D 2α, 4) π Dα ,, 2, 2, 45. The moment of inertia of a uniform thin rod, about its perpendicular bisector is I. If the, temperature of the rod is increased by ∆t ,, the moment of inertia about perpendicular, bisector increases by (coefficient of linear, , 3) π Dα ,, , 14, , 48., , 49., , 50., , 51., , expansion of material of the rod is α )., 1)Zero 2)I α ∆t 3)2 I α ∆t, 4)3 I α ∆t, A bimetal made of copper and iron strips, welded together is straight at room, temperature. It is held vertically so that the, iron strip is towards the left hand and copper, strip is towards right hand. The bimetal strip, is then heated. The bimetal strip will, 1) remain straight, 2) bend towards right, 3) bend towards left 4) have no change, If L1 and L2 are the lengths of two rods of, coefficients of linear expansion α 1 and α 2, respectively the condition for the difference in, lengths to be constant at all temperatures is, 1) L1 α 1 = L2 α 2, 2) L1 α 2 = L2 α 1, 4) L1 α 22 = L2 α 12, 3) L1 α 12 = L2 α 22, When a copper ball is cooled the largest, percentage increase will occur in its, 1) diameter 2) area 3) volume 4) density, The coefficients of linear expansion of P and, Q are α1 and α 2 respectively. If the coefficient, of cubical expansion of ‘Q’ is three times the, coefficient of superficial expansion of P, then, which of the following is true ?, 1) α 2 =2 α1 2) α1 =2 α 2 3) α 2 =3 α1 4) α1 =3 α 2, The substance which contracts on heating is, 1) silica glass, 2) iron, 3) invar steel, 4) aluminum, PQR is a right angled triangle made of brass, rod bent as shown. If it is heated to a high, temperature the angle PQR., , P, , 1) increases, 2) decreases, 3) remains same, Q, R, 4) becomes 1350, 52. A brass scale gives correct length at 00C. If, the temperature be 250C and the length read, by the scale is 10 cm. Then the actual length, will be, 1) more than 10 cm 2) less than 10 cm, 3) equal to 10 cm, 4) we can not say, 53. The coefficient of volume expansion is, 1) twice the coefficient of linear expansion., 2) twice the coefficient of real expansion., 3) thrice the coefficient of real expansion., 4) thrice the coefficient of linear expansion, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , 54. When a metal sphere is heated maximum, percentage increase occurs in its, 1) density, 2) surface area, 3) radius, 4) volume, 55. A solid sphere and a hollow sphere of same, material have same mass. When they are, heated by 50°C, increase in volume of solid, sphere is 5 c.c. The expansion of hollow, sphere is, , 1) 5 c.c., 3) Less than 5 c.c., , 2) more than 5 c.c., 4) None, , C.U.Q - KEY, 1) 1, 7) 4, 13)2, 19)2, 25)2, 31)3, 37)3, 43)1, 49)1, 55)2, , 2) 1, 8)2, 14)2, 20)1, 26)3, 32)3, 38)2, 44)4, 50)1, , 3)4, 9)3, 15)1, 21)1, 27)1, 33)1, 39)3, 45)3, 51)3, , 4)1, 10)4, 16)2, 22)1, 28)1, 34)3, 40)1, 46)3, 52)1, , 5)2, 11)1, 17)2, 23)2, 29)3, 35)1, 41)1, 47)1, 53)4, , 6)2, 12)2, 18)2, 24)2, 30)4, 36)1, 42)2, 48)4, 54)4, , LEVEL - I (C.W), MEASUREMENT OF TEMPERATURE, 1., , 2., , 3., , 4., , 5., , If the temperature of a patient is 40o C his, temperature in the Fahrenheit scale will be, 1) 720F, 2) 960F 3) 1000F 4) 1040F, The freezing point on a thermometer is marked, as 20 0 and the boiling point as 150 0. A, temperature of 600C on this thermometer will, be read as, 1) 400, 2) 650, 3) 980, 4) 1100, A Celsius thermometer and a Fahrenheit, thermometer are put in a hot bath. The reading, on Fahrenheit thermometer is just 3 times the, reading on Celsius thermometer. The, temperature of the hot bath is, 1)26.670C 2)36.670C 3)46.67 0C 4)56.67 0C, Oxygen boils at −183o C . This temperature is, approximately, 1) 215o F, 2) −297 o F 3) 329o F 4) 361o F, A mercury thermometer is transferred from, melting ice to a hot liquid. The mercury rises, to 9/10 of the distance between the two fixed, points. Find the temperature of the liquid in, Fahrenheit scale, 1)194o F, 2) 162o F 3) 112o F 4) 113o F, , NARAYANAGROUP, , 6., , 7., , A Centigrade and a Fehrenheit thermometer, are dipped in boiling water. The water, temperature is lowered until the Fahrenheit, thermometer registers 140o . What is the fall, in temperature as registered by the, Centigrade thermometer, 1) 300, 2) 400, 3) 600, 4) 800, Two absolute scales A and B have triple points, of water defined to be 200 A and 300 B (given, triple point of water is = 276.16 K). The, relation between TA and TB is, 1) TA = TB, , 2), , TB =, , 3, TA, 2, , 2, 3, 3) TB = TA, 4) TB = TA, 3, 4, 8. The temperature coefficient of resistance of wire, is 12.5 × 10−4 / Co . At 300 K the resistance of, the wire is 1 ohm. The temperature at which, resistance will be 2 ohm is, 1) 1154 K 2)1100 K 3)1400 K 4) 1127 K, 9. The reading of Centigrade thermometer, coincides with that of Fahrenheit thermometer, in a liquid. The temperature of the liquid is, 1) −40o C 2) 0o C 3) 100o C 4) 300o C, 10. The pressure of a gas filled in the bulb of a, constant volume gas thermometer at 0 0C and, 1000C are 28.6 cm and 36.6 cm of mercury, respectively. The temperature of bulb at which, pressure will be 35.0 cm of mercury will be, 1) 800C, 2) 700C 3)550C, 4) 400C, , THERMAL EXPANSION OF SOLIDS, 11. The coefficient of linear expansion of a metal, is 1 × 10-5/0C. The percentage increase in area, of a square plate of that metal when it is, heated through 1000C is, 1) 0.02% 2) 0.1% 3) 0.001% 4) 0.2%, 12. The length of each steel rail is 10m in winter., The coefficient of linear expansion of steel is, 0.000012/0C and the temperature increases by, 150C in summer. The gap to be left between, the rails, 1) 0.0018m 2) 0.0012m 3) 0.0022m 4) 0.05m, 13. A clock while keeps correct time at 300 C has, a pendulum rod made of brass. The number, of seconds it gains (or) looses per second when, the temperature falls to 100 C is [ α of brass, = 18 × 10-6 /0 C ], 1) 18 × 10-6 sec, 2) 18 × 10-5 sec, 3) 0.0018 sec, 4) 0.018 sec, 15
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 14. A metal plate of area 1.2 m2 increases its area, by 2.4 × 10-4 m2 when it is heated from 00 C to, 1000 C. The coefficient of cubical expansion of, the metal expressed in per 0C is, 1)2 × 10-6 2)4 × 10-6 3)6 × 10-6 4) 3 × 10-6, 15. The length of a metal rod at 00C is 0.5m.When, it is heated, its length increases by 2.7mm. The, final temperature of rod is (coeff. of linear, expansion of metal = 90 × 10-6/0C), 1) 200C, 2) 300C, 3) 400C 4) 600C, 16. The density of a substance at 00C is 10 g/c.c., and at 100 0C its density is 9.7g/c.c. The, coefficient of linear expansion of the substance, is., 1)10-4/0C, 2)3 × 10-4/0C, 3)6 × 10-4/0C, 4)9 × 10-4/0C, 17. What force should be applied to the ends of, steel rod of a cross sectional area 10 cm2 to, prevent it from elongation when heated form, 273 K to 303 k? ( α of steel 10-5 0C-1, Y = 2 ×, 1011 Nm −2 ), 1)2 × 104N, 2)3 × 104N, 4, 3)6 × 10 N, 4)12 × 104 N, 18. The inner diameter of a brass ring at 273 K is, 5 cm. To what temperature should it be heated, for it to accommodate a ball 5.01 cm in, diameter. ( α = 2 × 10-5 /0 C), 1) 273 K 2)372 K 3) 437 K, 4) 173K, 19. A metal sheet having size of 0.6 × 0.5 m2 is, heated from 293 K to 5200 C. The final area, of the hot sheet is { α of metal=2 × 10-5 /0 C], 1) 0.306 m2, 2) 0.0306 m2, 2, 3) 3.06 m, 4) 1.02m2, 20. A crystal has linear coefficients 0.00004/0C,, 0.00005/0C, 0.00006/0C. Coefficient of cubical, expansion of the crystal is, 1)0.000015/0C, 2) 0.00015/0C, 0, 3) 0.00012/ C, 4) 0.00018/0C, 21. A wire of length 60 cm is bent into a circle with, a gap of 1 cm. At its ends, on heating it by, 1000C, the length of the gap increases to 1.02, cm. α of material of wire is, 1) 2 × 10-4/0C, 2) 4 × 10-4/0C, 3) 6 × 10-4/0C, 4) 1 × 10-4/0C, , LEVEL - I (C.W) - HINTS, 1., , F − 32 C − 0 , C −0 X − L, =, , 2. , =, 100 , 180, 100 U − L, , 3., , F − 32 C − 0 , =, , and F=2C, 100 , 180, , 4., , F − 32 C − 0 , =, , , 100 , 180, , 6., 7., , 8., , 16, , 2) 3, 8) 3, 14) 4, 20) 2, , 3) 1 4) 2 5) 1 6) 2, 9) 1 10) 1 11) 4 12) 1, 15) 4 16) 1 17) 3 18) 2, 21) 1, , F − 32 C − 0, =, 180, 100, , Size of the degree on absolute scale A= size of the, degree on absolute Scale B, (276.16)TA (276.16)TB, =, 200, 300, R 2 − R1, C − 32 C − 0 , α =, =, , R 1 t 2 − R 2 t 1 9. 180, 100 , , Pt − P0, 10. t = P − P × 100, 100, o, , 11. β = 2α ,, , 12. l2 − l1 = l1α ( t2 − t1 ), , ∆A, 100 = β ∆t100, A, , ∆T 1, = α∆t, T, 2, l2 − l1, 15. t2 − t1 = l α, 1, , 13., , A2 − A1, 3, 14. β = A ( t − t ) , γ = 2 β, 1 2, 1, , d0 − dt, γ, 16. γ = d ∆t , α = 3, t, , 17. ∆l =, , Fl, ---- (1), AY, , Increase in length, ∆l = l α ∆ t ---- (2), from (1) and (2);, , r2 − r1, 18. ∆t = rα, 1, , Fl, = l α ∆ t ⇒ F = YAα∆T, AY, , 19. A2 = A1 (1 + β∆t ), , 20. γ = α x + α y + α z, , l2 − l1, 21. α = l ∆t (gap can be taken as l1 ), 1, , LEVEL - I (C.W) - KEY, 1) 4, 7) 2, 13) 2, 19) 1, , F − 32 9 , = , 5. , 10 , 180, , LEVEL - I (H.W), 1., , What is the temperature on Fahrenheit scale, corresponding to 30o C, 1) 86o F, , 2) 52o F, , 3) 62o F 4) 72o F, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 2., , 3., , 4., , 5., , 6., , 7., , A faulty thermometer has its fixed points, marked at 6o and 96o . What is the correct, temperature on the Centigrade scale when this, thermometer reads 87 o, 1) 83o C, 2) 93o C, 3) 90o C 4) 85o C, The temperature at which Celsius reading is, half the Fahrenheit reading, 2) 20o C 3) 160o C 4) 80o C, 1) 40o C, The normal boiling point of liquid hydrogen is, −253o C . What is the corresponding, temperature on absolute scale, 1) 22 K, 2) 20 K, 3) 274 K 4) -20 K, A faulty thermometer has 90.5o C and 0.5o C, as upper and lower fixed points respectively., What is the correct temperature if this faulty, thermometer reads 15.5 o C, 1) 16.67o C 2) 16o C 3) 15o C 4) 15.5o C, The temperature of a substance increases by, 27o C . On the Kelvin scale this increase is, equal to, 1) 300 K 2) 2.46 K 3) 27 K, 4) 7 K, A Fahrenheit thermometer registers, 107 o while a faulty Celsius thermometer, , registers 42o . Find the error in the later.., 1) 0.6o C 2) 0.72o C 3) 1.2o C 4) 7.2o C, 8. A platinum wire has a resistance of 2. 62 Ω at, 15 0C and 3.29 Ω at 80 o C. Find the, temperature coefficient of the resistance of, platinum wire., 1) 4.18 x 10 -3 oC-1, 2)9.34 x 10 -3 oC-1, 3) 1. 934 x 10 -3 oC-1 4)934 x 10 -3 oC-1, 9. The Fahrenheit and Kelvin scales of, temperature will give the same reading at, 1) –40, 2) 313, 3) 574.25 4) 732.75, 10. The pressure of hydrogen gas in a constant, volume gas thermometer is 80.0cm at 00C,, 110cm at 1000C and 95.0 cm at unknown, temperature t. Then t is equal to, 1) 500C 2) 750C, 3) 950C, 4) 1500C, , THERMAL EXPANSION OF SOLIDS, 11. A brass sheet is 25 cm long and 8 cm breadth, at 00 C. Its area at 1000C is ( α = 18 ×10−6 / 0 C ), 1) 207.2 cm2, 2) 200.72 cm2, 2, 3) 272 cm, 4) 2000.72 cm2, NARAYANAGROUP, , THERMAL PROPERTIES OF MATTER–I, 12. A metal rod having a linear coefficient of, expansion 2 × 10-5 /0 C has a length 1m at, 250C, the temperature at which it is shortened, by 1 mm is (1983 E), 1) 500C, 2) -500C, 3) -250C 4) -12.50C, 13. A clock with an iron pendulum keeps correct, time at 150C. If the room temperature rises, to 200C, the error in seconds per day will be, (coefficient of linear expansion for iron is, 0.000012/0C), 1) 2.5sec, 2) 2.6sec 3) 2.4sec 4) 2.2sec, 14. A steel rod of length 0.5km is used in the, construction of a bridge. It has to withstand, a temperature change of 400C. The gap that, is allowed for its expansion is [ α = 10-6/0C], 1) 0.02cm, 2) 0.02mm 3) 2m 4) 20 mm, 15. A wire of length 100cm increases in length by, 10-2 m when it is heated through 1000 C. The, coefficient of linear expansion of the material, of the wire expressed in /K units is, 1) −1×10−6 2) 1×104 3) 1×10−4 4) 10−2, 16. The variation of density of a solid with, temperature is given by the formula, d1, d1, 1) d 2 =, 2) d 2 =, 1 − γ (t2 − t1 ), 1 + γ (t2 − t1 ), d1, d1, 4) d 2 = 1 + 2γ (t − t ), 1 − 2γ (t2 − t1 ), 2, 1, 17. An iron bar whose cross sectional area is 4cm2, is heated from 00C and 1000C. The force, required to prevent the expansion of the rod, is [Y of Iron = 2 × 1012 dyne / cm2, α of Iron = 12 × 10-6 /0 C], 1) 0.96 × 108 N, 2) 0.96 × 107 N, 3) 9.6 × 107 N, 4) 96 × 103 N, 18. A hole is drilled in a copper sheet. The, diameter of the hole is 4.24 cm at 27.0 0 C ., What is the change in the diameter of the hole, when the sheet is heated to 2270 C ? α for, copper = 1.70 × 10 −5 K −1, , 3) d 2 =, , 1) 1.44 ×10−2 cm, , 2) 14.4 ×10−2 cm, , 3) 144 × 10 −2 cm, 4) 0.144 × 10−2 cm, 19. Distance between two places is 200km. α of, metal is 2.5 × 10-5 /0 C. Total space that must, be left between steel rails to allow a change, of temperature from 360F to 1170F is, 1)2.25km 2)0.225km 3)22.5km 4)0.0225km, 17
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 20. A crystal has a coefficient of linear expansion, 12 ×10 −6 / 0 C in one direction and, 244 × 10−6 / 0 C in every direction at right, angles to it . Then the coefficient of cubical, expansion of crystal is, 2) 500 × 10−6 / 0 C, 1) 450 × 10−6 / 0 C, 3) 244 × 10−6 / 0 C, 4) 36 × 10−6 / 0 C, 21. When a thin rod of length ‘ l ’ is heated from, t01C to t 02 C length increases by 1%. If plate, of length 2 l and breadth ‘ l ’ made of same, material is heated form t 0 1 C to t 02 C,, percentage increase in area is, 1) 1, 2) 2, 3) 3, 4) 4, , LEVEL - I (H.W) - KEY, 1)1, 7)1, 13)2, 19)2, , 2)3, 8)1, 14)4, 20)2, , 3)3, 4)2, 9)3, 10)1, 15)3 16)1, 21) 2, , 5)1, 11)2, 17)4, , F − 32 C − 0 , =, , , 100 , 180, , 3., , F − 32 C − 0 , =, , ; F=2C, 100 , 180, , 4., , K=C+273, , 6., , ∆0C = ∆ 0 K, , 8., , α =, , 2., , 17. ∆l =, , Fl, = l α ∆ t ⇒ F = YAα∆T, AY, 18. L2 − L1 = L1α∆T, 19. l2 − l1 = l1a ∆t, from (1) and (2), , 20. γ = α x + α y + α y, ∆A , 21. , 100 = β × ∆t × 100, A , , β = 2α, , LEVEL - II (C.W), , 1., , C−0 X −L, =, 100 U − L, , C−0 X −L, =, 100 U − L, F − 32 X − L, =, 7., 180, U −L, , 5., , 2., , 3., , F − 32 K − 273 , =, , But F = K = x, 100 , 180, Pt − P0, 10. t = P − P × 100, 100, o, , 9., , 11. A2 = A1 (1 + β∆t ) , β = 2α, , 4., , l2 − l1, 12. t2 − t1 = l ∆t, 1, 1, 2, , 13. loss or gain in t ime per day = α∆t 86400, , 18, , ∆l = l α ∆ t ---- (2), , MEASUREMENT OF TEMPERATURE, , R 2 − R1, R 1 t 2 − R 2 t1, , 14. l2 − l1 = l1α∆t, , Fl, ---- (1), AY, , Increase in length, , 6)3, 12)3, 18)1, , LEVEL - I (H.W) - HINTS, 1., , d0, 16. dt = (1 + γ∆t ), , l2 − l1, 15. α = l ( t − t ), 1 2, 1, , 5., , The resistance of a certain platinum resistance, thermometer is found to be 2.56 Ω at 0oC and, 3.56 Ω at 1000 C . When the thermometer is, immersed in a given liquid, its resistance is, observed to be 5.06 Ω . The temperature of, the liquid, 1) 45 0C, 2) 250 0C 3)225 0C 4)120 0C, A constant volume gas thermometer shows, pressure readings of 50 cm and 90 cm of, mercury at 0o C,100o C respectively, The, temperature of the bath when pressure reading, is 60 cm of mercury., 1) 45 0C 2) 30 0C 3) 25 0C 4) 20 0C, On a hypothetical scale A the ice point is 420, and the steam points is 1820 For another scale, B. The ice point is –100 and steam point in 900., If B reads 600. The reading of A is., 1) 160 0, 2) 140 0, 3) 120 0, 4) 110 0, The upper and lower fixed points of a faulty, mercury thermometer are 2100F and 340 F, respectively. The correct temperature read by, this thermometer is, 1) 220F, 2) 800F 3) 1000F 4) 1220F, A Fahrenheit thermometer registers 1100 F, while a faulty Celsius thermometer registers, 440 C . Find the error in the later, 1) 0.37 0, , 2) 0.87 0, , 3) 0.67 0, , 4) 0.48, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , THERMAL EXPANSION OF SOLIDS, When a rod is heated from 250C to 750C, it, expands by 1 mm. When a rod of same, material but with 4 times the length is heated, from 250C to 500C. The increase in length is, 1) 1mm, 2) 1.5mm 3)1.6mm 4)2 mm, 7. An iron metal rod is to maintain an accuracy, of one part per million. The coefficient of, linear expansion of iron is 1 × 10-5 /0 C. The, minimum variations in temperature of the rod, could be, 1) ± 10C 2) ± 50C, 3) ±0.10C 4) ± 0.010C, 8. Two metal rods have coefficients of linear, expansion 1.1 × 10-5 /0 C and 1.65 × 10-5 /0 C, respectively. The difference in lengths is 10cm, at all temperatures. Their initial lengths must, be respectively., 1) 40 cm and 50 cm 2) 40 cm and 30 cm, 3) 50 cm and 60 cm 4) 30 cm and 20 cm, 9. Two rods of same length and same diameter, are drawn from equal masses and same, quantity of heat is supplied to the two rods., Find the ratio of expansions if specific heats, of the material is 2/3 and that of coefficient of, linear expansion is 1/2, 1) 4/3, 2) 1/2, 3) 3/4, 4)1/3, 10. Two rods of different materials having, coefficients of thermal expansion α1 , α 2 and, young’s modulus Y 1, Y 2 respectively are fixed, between two rigid walls. The rods are heated, such that they undergo the same increase in, temperature. There is no bending of rods. If, α1 : α 2 =2:3, thermal stress developed in the, rods are equal provided Y 1:Y2 is equal to, 1) 2:3, 2) 1:1, 3) 3:2, 4) 4:9, 11. Two uniform metal rods one of aluminium of, length l1 and another made of steel of length, 6., , l2 and linear coefficients of expansion αa and, αs respectively are connected to form a single, rod of length l1+l 2 . When the temperature of, the combined rod is raised by t o C , the length, of each rod increases by the same amount., l1, Then l +l is, 1 2, αs, 1) α + α, a, s, , αa, 2) α + α, a, s, , NARAYANAGROUP, , αa, 3) α, s, , αs, 4) α, a, , 12. When the temperature of a body increases, from t to t+ ∆ t, its moment of inertia increases, from I to I + ∆ l. The coefficient of linear, expansion of the body is α . The ratio ∆ I/I is, (2012 E), 3) α ∆ t 4) 2 α ∆ t, 1) ∆ t/t 2) 2 ∆ t/t, 13. There is some change in length when a 33000, N tensile force is applied on a steel rod of area, of cross-section 10−3 m 2 . The change of, temperature required to produce the same, elongation of the steel rod when heated is, (Y= 3 × 1011 N / m 2 , α = 1.1×10−5 / 0 C ), 1) 200 C, 2) 150 C 3) 100 C, 4) 00 C, 14. Brass scale of a Barometer gives correct, reading at 00C. coefficient of linear expansion, of brass is 18 × 10-6 /0C. If the barometer, reads 76cm at 200C, the correct reading is, ( γ Hg=18 × 10-5 /0 C), 1 76.426 cm, 2) 75.7cm, 3) 76.642 cm, 4) 76.264 cm, 15. A thin brass sheet at 10°C and a thin steel, sheet at 20°C have the same surface area., The common temperature at which both would, have the same area is (Coefficient of linear, expansion for brass and steel are respectively,, 19 × 10–6/°C are 111 × 10–6/°C), 1) –3.75°C 2) –2.75°C 3) 2.75°C 4)3.75°C, 16. A pendulum clock gives correct time at 200C, at a place where g= 10m/s2. The pendulum, consists of a light steel rod connected to a, heavy ball. If it is taken to a different place, where g = 10.01m/s2 at what temperature the, pendulum gives correct time ( α of steel is, 10–5/0C), [2007 E], 1) 300C, 2) 600C, 3) 1000C 4) 1200C, 17. Two rods of lengths L1 and L2 are welded, together to make a composite rod of length, (L1+L2). If the coefficient of linear expansion, of the materials of the rods are α1 and α 2, respectively, the effective coefficient of, linear expansion of the composite rod is, [2012 E], L1α1 − L2α 2, L1α1 + L2α 2, 1) L + L, 2) L + L, 1, 2, 1, 2, 3) α1α 2, , 4), , α1 + α 2, 2, , 19
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 18. A clock pendulum made of invar has a period, of 0.5sec at 200C. If the clock is used in a, climate where the temperature averages to, 300C, how much time does the clock loose in, each oscillation. For invar α = 9 × 10−7 0C −1, 1) 2.25 × 10−6 sec, 2) 2.5 × 10−7 sec, 4) 1.125 × 10−6 sec, 3) 5 ×10−7 sec, 19. A steel scale is correct at 0°C. The length of a, brass tube measured by it at 40°C is 4.5m. The, correct length of the tube at 0°C is, (Coefficients of linear expansion of steel and, brass are 11 × 10-6/°C and 19 × 10-6/°C, respectively)., 1) 4.001m 2) 5.001 m 3)4.999m 4)4.501m, 20. The ratio of lengths of two rods is 1 : 2 and the, ratio of coefficient of expansions is 2 : 3. The, first rod is heated through 600C. Find the, temperature through which the second rod is, to be heated so that its expansion is twice that, of first is, 1) 600C, 2) 400C 3) 300C, 4) 100C, , LEVEL - II (C.W)-KEY, 1) 2, 7) 3, 13) 3, 19) 4, , 2) 3 3) 2 4) 4 5) 3 6) 4, 8) 4 9) 3 10) 3 11)1 12)4, 14) 2 15) 1 16) 4 17) 2 18) 1, 20) 2, , 8., , l1α1 = l2α 2 , l1 − l2 = 10, , 9., , Q1 = Q2 ;, , ms1 ( ∆t )1 = ms2 ( ∆t )2, , 10. Thermal stress Y1a 1∆t = Y2 a 2 ∆t, 11. l1α a ∆t1 = l2α s ∆t2, 12. from I = MR2,, 13. ∆t =, , ∆I, ∆R ∆I, = 2×, = 2α∆t, ,, I, R I, , F, YAα, , 14. True value =scale reading l − ( γ − α ) ∆t , β1t1 − β 2t2, ∆l ∆g, =, = α∆t, 15. t = β − β, 16., l, g, 1, 2, ∆L1, ∆L2, = L1α1 ;, = L2α 2, ∆t, ∆t, ∆L ∆L1 ∆L2, =, +, ; ( L1 + L2 ) α = L1α1 + L2α 2, ∴, ∆t, ∆t, ∆t, 1, 18. ∆T = α∆t ; 19. lc = lm 1 + ( αS ~ αb ) ( t 2 − t1 ) , 2, ∆t1 α 2 l2, 20. ∆l = α l ∆t ⇒ ∆t = α × l, 2, 1, 1, , 17., , LEVEL - II (H.W), , LEVEL - II (C.W)- HINTS, 1., 3., , 4., , 5., , t=, , Rt − R0, P − P0, × 100 2. t = t, R100 − R0, P100 − P0, , MEASUREMENT OF TEMPERATURE, 1., , Re ading − LFP, = constant, UFP − LFP, X A − LA X B − LB, =, U A − L A U B − LB, , 2., , Re ading − LFP, = constant, UFP − LFP, F − 32 X − L, =, 180, U −L, C − 0 F − 32, =, 100, 180, , 3., , 6., , As α1 = α 2, , 7., , ∆l, = 1/106, l, , 20, , , 100, , , e1, e2, ⇒ l ∆t = l ∆t, 1, 1, 2, 2, ∆l, ⇒, = α ∆t, l, , The resistance of a resistance thermometer, has values 2.70 Ω and 3.70 Ω at 00C and 1000C, respectively. The temperature at which the, resistance is 3.10 Ω is, 1) 300C, 2) 400C, 3) 600C, 4) 700C, A gas thermometer measures the temperature, from the variation of pressure of a sample of, gas. If the pressure measured at the melting, point of lead is 2.20 times the pressure, measured at the triple point of water find the, melting point of lead., 1) 600K, 2) 420K, 3) 790 K 4) 510 K, On a hypothetical scale X, the ice point is 400, and the steam point is 1200. For another scale, Y the ice point and steam points are –300 and, 1300 respectively. If X-reads 500 The reading, of Y is, 2) –80, 3) –100, 4) –120, 1) –50, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , THERMAL EXPANSION OF SOLIDS, 4., , 5., , 6., , 7., , 8., , 9., , 0, , A steel tape is calibrated at 20 C. when the, temperature of the day is -10 0 C, the, percentage error in the measurement with, the tape is ( α =12 × 10-6 /0 C), 1) 3.6% 2) 0.36% 3) 0.18% 4) 0.036%, The temperature coefficient of resistance of, wire is 12.5 × 10−4 . At 300K the resistance of, wire is 1Ω . The temperature at which, resistance will be 2Ω is, 1)827K, 2)854K 3)1527K 4) 1127K, The diameter of iron wheel is 1cm. If its, temperature is increased by 7000C What is, the increase in circumference of the wheel?, ( α =12 × 10-6 /0 C), 1)0.0264cm 2)0.264 cm 3)2.64cm 4)26.4 cm, If a cylinder of diameter 1.0cm at 300C is to, be slid into a hole of diameter 0.9997 cm in a, steel plate at the same temperature, the, minimum required rise in the temperature of, the plate is: (Coefficient of linear expansion, of steel=12 × 10-6/0C), 1) 250C, 2) 350C, 3) 450C 4) 550C, The initial lengths of two rods A and B are in, the ratio 3:5 and coefficients of linear, expansion are in the ratio 5:3. If the rods are, heated from 340C to 650C, the ratio of their, expansion will be, 1)1:1, 2) 3:5, 3) 1:2, 4) 2:3, A thin copper wire of length L increases in, , flask made of the same glass as that rod, measures a volume of 1000 c.c at 00C. The, volume it measures at 1000C in c.c. is, 1) 1018 cc, 2) 918 cc 3) 818 cc 4) 718 cc, 12. A pendulum clock runs fast by 5 seconds per, day at 200c and goes slow by 10 seconds per, day at 35 0C. It shows correct time at a, temperature of, 1) 27.50C 2) 25.0C, 3) 30.0C 4) 33.0C, 13. A second's pendulum clock having steel wire, is calibrated at 20°C . When temperature is, increased to 30°C , then how much time does, the clock loose or gain in one week ?, [ α steel = 1.2 × 10−5 ( °C ) ], −1, , 14., , 15., , length by one percent when heated from t10 C, and t2 0 C . The percentage change in area, when a thin copper plate having dimension, 2L × L is heated from t10C to t20 C is, 1)1 %, 2) 3 %, 3) 2 %, 4) 4%, 10. The brass scale of a barometer gives correct, reading at 10oC. The barometer reads 75 cm, at 30oC. What is the atmospheric pressure at, 0oC (in cm Hg), , (α, , Brass, , = 20 × 10−6 / 0 C ; γ Hg = 175 ×10 −6 / 0 C ), , 1) 74.8, 2) 75.03, 3) 70, 4) 60, 11. Two marks on a glass rod 10cm apart are, found to increase their distance by 0.06mm, when the rod is heated from 00C to 100C. A, , NARAYANAGROUP, , 16., , 17., , 1) 0.3628s, 2) 3.626s 3) 362.8 s 4) 36.28s, A metre scale made of steel is calibrated at, 200C to give correct reading. Find the distance, between 50 cm mark and 51 cm mark if the, scale is used at 100C. Coefficient of linear, expansion of steel is 1.1 × 10–5/0C, 1) 1.00011 cm, 2) 1.0011 cm, 3) 1.011 cm, 4) 1.000011 cm, A thin brass sheet at 20°C and a thin steel, sheet at 30°C have the same surface area., The common temperature at which both would, have the same area is (Coefficient of linear, expansion for brass and steel are respectively,, 19 × 10–6/°C are 111 × 10–6/°C), 1) –6.250C, 2) +6.250C, 0, 3) –3.25 C, 4) +3.250C, Distance between two places is 200 km. α of, steel is 12 × 10–6/°C. Total space that must, be left between steel rails to allow for a change, of temperature from 36°F to 117°F is (in km), 1)1.08, 2)0.108, 3)0.8, 4) 0.0108, Two thin metal strips, one of brass and the, other of iron are fastened together parallel to, each other. Thickness of each strip is 2 mm. If, the strips are of equal length at 0°C. The radius, of the arc formed by the bimetallic strip when, heated to 80°C is (Coefficient of linear, expansion of brass = 19 × 10-6/°C &, of iron = 12 × 10-6/°C)., 1) 3.57m 2) 2.67m 3) 3.12m, 4) 4.56m, , 21
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 18. A brass wire 1.8 m long at 270 C is held taut, with little tension between the two rigid, supports. If the wire is cooled to a temperature, of −390 C , the tension developed in the wire,, if its diameter is 2.0 mm, Coefficient of linear, expansion of brass = 2.0 × 10−5 K −1 ; Young’ss, modulus of brass = 0.91× 1011 Pa, 1) 3.8 × 102 N, 2) 5.8 × 102 N, 3) 7.8 × 102 N, 4) 6.8 × 102 N, , LEVEL-II ( H.W ) - KEY, 1) 2 2)1 3) 3, 7) 1 8) 1 9) 3, 13)4 14)1 15)2, , 4) 4, 10)1, 16)2, , 5) 4 6)1, 11)1 12)2, 17)1 18)1, , 1, 13. ∆T = α∆t × 7 × 86400, 2, , 14. lc = lm (1 + ( αS − α b ) ( t 2 − t1 ) ), 15., , ( A1 )br 1 + β Br ( t − 20 ) = ( A1 )st 1 + β St ( t − 30 ), Given ( A1 )br = ( A1 ) st, 2t, ∆C ∆F, =, 17. r = (α − α ) ∆ T, 5, 9, 1, 2, 18. Decrease in length due to tention, , 16. ∆l = α l ∆t ,, ∆l =, , Rt − R0, × 100, R100 − R0, , t=, , 2., , T, P, P, =, TK = ( 273.16 ), K, ⇒, TTr PTr, PTr, , 7., , Re ading − LFP, 120 − x 130 − y, = constant,, =, UFP − LFP, 80, 160, ∆l, × 100 = a ∆ t × 100, ∆l = l α ∆ t ⇒, l, R − R1, 1, α= 2, −4 =, ;, 12.5, ×, 10, R1t2 − R2t1, t2 − 54, Increase is circumference = π Dα∆t where D =, diameter of the wheel, r2 = r1 (1 + α∆t ), ⇒ r2 − r1 = r1 α ∆ t, , 8., , ∆l = l1α∆t ⇒, , 9., , ∆A = Aβ ∆ t ⇒, , 3., 4., 5., 6., , from (1) and (2), , THERMAL EXPANSION, 1., , 2., , ∆A, × 100 = ß∆t × 100, A, , 11. ∆l = α l ∆t ; V2 = V1 (1 + 3α∆t ), 1, a ( 35 − t ) × 86400 = 10 ---- (1), 12., 2, 1, a ( t − 20 ) × 86400 = 5 ----- (2), 2, Solving 1 and 2, , 3., , Two rods of the same length, have radii in the, ratio 3:4. Their densities are respectively 8000, and 9000 kg/m3. Their specific heats are in the, ratio of 2:3. When the same amount of heat is, supplied to both, the changes in their lengths, are in the ratio. (If their linear coefficients are, in the ratio 5:6), 1) 1:1, 2) 5:2, 3) 5:12, 4) 12:5, A solid sphere of radius r and mass m is, spinning about a diameter as axis with a speed, ω 0. The temperature of the sphere increases, by 1000C without any other disturbance. If the, coefficient of linear expansion of material of, sphere is 2 × 10-4 /0 C, the ratio of angular, speed at 1000C and ω 0 is, 1) 1:1 2) 1:1.04, 3) 1.04:1, 4) 1:1.02, Two rods of different materials and identical, cross sectional area, are joined face to face at, one end and their free ends are fixed to the, rigid walls. If the temperature of the, surroundings is increased by 30°C, the, magnitude of the displacement of the joint of, the rod is (length of rods l1 =l2 =1unit, ratio of, their young's moduli, Y1 /Y2 =2, coefficients of, linear expansion are α 1 and α 2 ), 1) 5(α 2 − α1 ), 2) 10(α1 − α 2 ), 3) 10(α 2 − 2α1 ), , 22, , Fl, = l α ∆ t ⇒ F = YAα∆T, AY, , LEVEL - III, , ∆l1 l1 α1, = ×, ∆l2 l2 α 2, , 10. H1 = H 2 1 − ( γ Hg − α ) ∆t , , Fl, ---- (1), AY, , Increase in length due to rise in temperature, ∆l = l α ∆ t ---- (2), , LEVEL-II (H.W) - HINTS, 1., , ( A2 )brass = ( A2 )steel, , 4) 5(2α1 − α 2 ), NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 4., , 5., , 6., , A wire of length L0 is supplied heat to raise, its temperature by T. if γ is the coefficient of, volume expansion of the wire and Y is Young’s, modulus of the wire then the energy density, stored in the wire is, 1 2 2, 1 2 2 3, 2) γ T Y, 1) γ T Y, 2, 3, 1 2 2, 1 γ 2T 2, 3), 4) γ T Y, 18, 18 Y, A uniform solid brass cylinder of mass M=0.5, Kg and radius R=0.03m is placed in, frictionless bearings and set to rotate about, its geometrical axis with an angular velocity, of 60 rad/s. After the cylinder has reached, the specified state of rotation, it is heated, (without any mechanical contact) from room, temperature 200C to 1000C. The fractional, change in angular velocity of the cylinder is, ( α =2 × 10-5/0C), 1) -3.2 × 10-3, 2) 3.2 × 10-3, -3, 4) -2.3 × 10-3, 3) 2.3 × 10, Calculate the compressional force required to, prevent the metallic rod of length l cm and, cross-sectional area A cm2 when heated, through t °C , from expanding along length, wise. The Young's modulus of elasticity of the, metal is E and mean coefficient of linear, expansion is α per degree Celsius, EAα t, , 7., , 8., , THERMAL PROPERTIES OF MATTER–I, , E Aα t, , 1) EAα t, 2), 3), 4) Elα t, 1 + αt, 1−αt, An iron rod of length 50 cm is joined to an, aluminium rod of length 100cm. All, measurements refer to 20oC. The coefficient, of linear expansion of iron and aluminium are, 12 × 10−6 / °C and 24 × 10−6 / °C respectively.., The average linear expansion coefficient of, composite system is :, 1) 36 ×10−6 / °C, 2) 12 × 10−6 / °C, 3) 20 × 10−6 / °C, 4) 48 × 10−6 / °C, A rod of length 20 cm is made of metal. It, expands by 0.075 cm when its temperature is, raised from 0°C to 100°C . Another rod of a, different metal B having the same length, expands by 0.045 cm for the same change in, temperature. A third rod of the same length is, composed of two parts, one of metal A and the, other of metal B. This rod expands by 0.060, cm for the same change in temperature. The, portion made of metal A has the length :, 1) 20 cm 2) 10 cm, 3) 15 cm 4) 18 cm, , NARAYANAGROUP, , 9., , A thin circular metal disc of radius 500.0 mm, is set rotating about a central axis normal to, its plane. Upon raising its temperature, gradually, the radius increases to 507.5 mm., The percentage change in the rotational kinetic, energy will be, 1) 1.5%, 2) -1.5%, 3) 3% 4) -3%, , 10. A steel wire AB of length 100 cm is fixed rigidly, at points A and B in an aluminium frame as, shown in the figure. If the temperature of the, system increases through 100 0 C, then the, excess stress produced in the steel wire, relative to the aluminium?, , α Al = 22 × 10 −6 / 0 C and α steel = 11× 10−6 / 0 C, young's Modulus of steel is 2 × 1011 Nm–2 ., Aluminium frame, , A, , steel wire, , 1) 2.2 × 108 Pa, 3) 0.2 × 108 Pa, , B, , 2) 22 × 108 Pa, 4) 220 × 108 Pa, , 11. An equilateral triangle ABC is formed by, joining three rods of equal length and D is the, mid-point of AB. The coefficient of linear, expansion for AB is α1 and for AC and BC is, , α 2 . The relation between α1 and α 2 , if, distance DC remains constant for small, changes in temperature is (2010 E), A, D α1 B, , α2 α2, , C, 1) α1 = α 2, , 2) α1 = 4α 2, , 1, 4) α1 = α 2, 2, 12. A cube of edge (L) and coefficient of linear, expansion ( α ) is heated by 10C. Its surface, area increases by, 1) 6 α L2, 2) 8 α L2 3) 12 α L2 4) 2 α L2, , 3) α 2 = 4α1, , 23
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 13. An iron ball of diameter 6cm and is 0.01 mm, too large to pass through a hole in a brass plate, when the ball and plate are at a temperature of, 200C. The temperature at which (both for ball, and plate) the ball just passes through the hole, , (, , is αiron = 12 × 10, , 680C, , −6 0, , / C ; αbrass = 18 × 10, , 480C, , 280C, , −6 0, , / C, , ), , 400C, , 1), 2), 3), 4), 14. A rod of length 2 m is at a temperature of, 200 C . The free expansion of the rod is 0.9mm., If the temperature is increased to 500 C , the, stress produced when the rod is fully, prevented to expand, Y = 2 × 1011 N / m 2 , α = 15 × 10−6 / 0 C, 1) 9 × 107 N / m 2, 2) 4.5 × 107 N / m 2, 4) 3 × 107 N / m 2, 3) 5 × 107 N / m2, 15. The coefficient of linear expansion for a certain, metal varies with temperature as α (T ) . If L0, is the initial length of the metal and the, temperature of metal changed from T0 to, , T ( T0 > T ) then,, T, 2) L = L0 1+ ∫T α (T ) dT , , , , 1) L = L0 ∫T α (T ) dT, T, , 0, , 0, , T, , L, =, L, 0 1 − ∫ α (T ) dT , 3), 4) L > L0, T0, , 16. A steel tape is placed around the earth at the, equator. When the temperature is 0 0 C, neglecting the expansion of the earth, the, clearance between the tape and the ground if, the temperature of the tape rises to 300C, is, , nearly (α steel = 11×10−6 / K ), 1) 1.1 km, 2) 0.5 km 3) 6400 km 4) 2.1 km, 17. The variation of lengths of two metal rods A, and B with change in temperature are shown, in figure. The coefficients of linear expansion, α A for the metal A and the temperature T will, be :, Length(mm), , 1) α A = 3 × 10−6 / °C , 500°C, 2) α A = 3 × 10−6 / °C , 222.22°C, 3) α A = 27 × 10−6 / °C ,500°C, 4) α A = 27 × 10−6 / °C , 222.22°C, 18. The coefficient of linear expansion of an in, homogeneous rod changes linearly from α1 to, α 2 from one end to the other end of the rod., The effective coefficient of linear expansion, of rod is, α1 + α2, 1) α1 + α 2, 2), 3) α1α2 4) α1 − α2, 2, 19. A rod of steel is 5m long and 3cm diameter at, a temperature of 200C. Find the free expansion, of the rod when the temperature is raised to, 650C. Find the respective pulls exerted if (i), the ends do not yield and (ii) the ends yield by, 0.12 cm. Y = 2 × 105 MN / m2 and α =12×10−6, per 0C, 1) 0.27 cm, 42.41 KN, 76.34 KN, 2) 0.27 cm, 76.30 KN, 42.39 KN, 3) 0.27 cm, 38.63 KN, 78.23 KN, 4) 0.27 cm, 78.23 KN, 38.63 KN, 20. Two bars are unstressed and have lengths of, 25cm and 30cm at 200C as shown in Figure., Bar (1) is of aluminium and bar (2) is of steel., The cross-sectional area of bars are 20cm2 for, aluminium and 10cm2 for steel. Assuming that, the top and bottom supports are rigid, stress, N, in Al steel bars in, when the temperaturee, mm 2, is 700C. (Nearly ), (Ya = 0.70 ×105 N / mm2 .Ys = 2.1× 105 N / mm2 ., α a = 24 × 10−6 / 0C and α s = 12. × 10−6 / 0C ), , Al, , 25cm, , St, , 30cm, , A, B, , 1006, 1004, 1002, 1000, 998, 996, 994, O, 24, , (Given α = 9 × 10 −6 / °C ), B, , O, , Temperature( C), , T, , 1) 75, 150, , 2) 25, 50 3) 50, 100 4)100, 200, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , Assertion & Reason Type, , 20. Contraction of the two bars due to compressive, stress = Elongation of the two bars due to rise of, temperature, , Sl , Sl , Y + Y = (α L∆t ) Al + (α L∆t ) St, Al St, Force in steel = force in aluminium, , 5., , S Al × Α Al = S St × Α St, , LEVEL - IV, 6., , Matching Type Questions, 1. List - I, List - II, a)Isotropic solids, e) Expands on melting, b) Ice, f) Equal expansion in all, directions, c)Anisotropic solids g) Contracts on heating, d) Copper, h) unequal expansion in, different directions, 2. List - I, List - II, a) Thermal expansion e) Pendulum clock, b) α , β , γ, f) Depends on, Dimension, Material,, Temperature, c) Bimetallic strip, g) Depends on nature of, the material only, d)Invar steel, h) Balance wheel, of a watch, 3. List - I, List - II, a) Bimetal thermostat e) Pendulum clock, b) Compensated, f) Invar steel, pendulum, c) Metal tape, g) Differential, expansion of metals, d) Elinvar, h )Hair spring, 4. List - I, List - II, a. Thermal stress, e. 3α∆t100, , 8., , 9., , 10., , 11., , b. Loss in time of a, , f. (α − α ) ∆ t, 2, 1, , d, , 12., , pendulum clock per sec, c. percentage increase, in volume of a solid, , g. Yα∆t, , 13., , d. Radius of circular arc, , h. (1/ 2 ) α∆t, , of bimetallic strip, 26, , 7., , 1)A and R are correct and R is correct explanation, for A, 1) A and R are correct and R is not correct, explanation for A, 3) A is true and R is false, 4) Both A and R are false, Assertion (A): A thick and thin metallic rods of, same material heated through same rise of, temperature then thermal stress is same., Reason (R): Thermal stress is independent of area, of cross section., Assertion (A): An iron ball strucked in a brass, plate is removed by heating the system., Reason (R): The coefficient of linear expansion, of brass is more than that of iron., Assertion (A): Invar steel is used to prepare clock, pendulum., Reason (R): The coefficient of linear expansion, of invar steel is Infinity., Assertion (A): When hot water is poured in a thick, glass tumbler then the tumbler breaks., Reason (R): Glass is a bad conductor of heat, Assertion (A): To have same difference between, the lengths of two metallic rods their initial lengths, of 0ºC should be in the inverse ratio of their, coefficient of linear expansion., Reason (R): If the lengths of two metallic rods at, 0º C are in the inverse ratio of their, coefficient of linear expansion then the change in, the lengths due to same rise of temperature is same., Assertion (A): A solid on heating undergoes, expansion only because of increasing the amplitude, of the simple oscillators., Reason (R): A solid on heating undergoes, expansion only because of increasing the inter, atomic distance., Assertion (A): A metallic plate containing circular, hole is heated then the size of the hole increases., Reason (R): The expansion of the solid always, takes place radially outwards., Assertion (A): Platinum is used to fuse into glass, tube., Reason (R): Both platinum and glass have almost, same values of coefficient of linear expansion., Assertion (A): A thin rod and a thick rod made of, same material having same length are heated through, same range of temperature. Then both the rods, expand equally., Reason (R): The linear expansion e = lα∆t, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, , 14. Assertion (A): A thin rod and a thick rod made of, same material having same length are given same, amount of heat θ . Then the thin rod expands more., Reason (R): The linear expansion depends upon, initial length of the rod only ., 15. Assertion (A): A platinum wire can be sealed, through glass. But a brass one cannot be sealed, through glass., Reason (R): Coefficient of linear expansion of, platinum and that of brass have different values., 16. Assertion (A): Two rods of the same material have, the same lengths but diameter are in the ratio of, 1:2. If 1000cal of heat is supplied to the two rods, separately the ratio of their linear expansion is 4:1., Reason (R): The linear expansion e = lα∆t, 17. Assertion (A): The linear coefficients of expansion, of a crystal along three perpendicular axes are, −α α, 7α, , . Its volume coefficient is, 2 5, 10, Reason (R): for anisotropic, α,, , solids, , γ = α x + α y + αz ., , Statement type questions, , 18., , 19., , 20., , 21., , Options :, 1. Statement 1 is true and statement 2 is true, 2. Statement 1 is true and statement 2 is false, 3. Statement 1 is false and statement 2 is true, 4. Statement 1 is false and statement 2 is false, Statement 1: Fahrenheit is the smallest unit, measuring temperature., Statement 2: Fahrenheit was the first temperature, scale used for measuring temperature., Statement 1: A brass disc is just fitted in a hole in, a steel plate. The system must be cooled to loosen, the disc from the hole., Statement 2: The coefficient of liner expansion, for brass is greater than the coefficient of linear, expansion for steel., Statement 1: When a bimetallic strip made of iron, and brass is heated then it bends in the form of, concave towards Brass., Statement 2: The coefficient of linear expansion, of iron is less than brass., Statement 1: The linear expansion does not, depend on nature of the material, initial length, and, rise in temperature., Statement 2: The coefficient of linear expansion, depends on nature of the material and system of, temperature., , NARAYANAGROUP, , 22. Statement 1: Gas thermometers are more sensitive, than liquid thermometers., Statement 2: Expansion in gases is more, prominent than liquids., , More than one option Type Questions, 23. When a rod is heated, its linear expansion, depends on, a) initial length, b) area of cross section, c) mass, d) temperature rise, 1) only a is correct, 2) a & d are correct, 3) b & c are correct 4) a & c are correct, 24. The numerical value of coefficient of linear, expansion is independent of units of, a) length b) temperature c) area d) mass, 1) only (a) is correct, 2) (a) & (b) are correct, 3) (a) ,(b) & (c) are correct, 4) (a) ,(c) & (d) are correct, 25. Expansion during heating, (a) occurs in solids only, (b) causes decrease in weight, (c) is due to increase of interatomic spacing, 1)only (a) is wrong, 2)(a),(b) & (c) are wrong, 3) (a) & (b) are wrong, 4) (a) ,(b) & (c) are correct, 26. When a copper solid sphere is heated, its, (a) moment of inertia increases, (b) Elasticity decreases, (c) density decreases, (d) mass increases, 1) only (b) is true, 2) (a) & (b) are true, 3) (a),(b) & (c) are true, 4) all are true, 27. Due to thermal expansion with rise in, temperature, (a)Metallic scale reading becomes lesser than true, value, (b) Pendulum clock goes fast, (c) A floating body sinks a little more, (d) The weight of a body in a liquid increases, 1) only (a) is correct, 2) (a) & (b) are correct, 3) (a),(b) & (d) are correct, 4) (a),(c) & (d) are correct, 27
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER–I, 28. Which of the following statements are true, (a) Rubber contracts on heating, (b) Water expands on freezing, (c) Water contracts on heating from 0°C to 4°C, (d) Water expands on heating from 4°C to 40°C, 1) (a) is correct, 2) (b) and (c) are correct, 3) (c) & (d) are correct, 4) all are correct, 29. When a metal ring having some gap is heated, a) length of gap increases, b) radius of the ring decreases, c) the angle subtended by the gap at the centre, remains same, d) length of gap decreases, 1) only d is correct, 2) a and b are correct, 3) a & c are correct, 4) all are correct, , LEVEL - IV - KEY, Matching Type Questions, 1) a-f, b-g, c-h, d-e, 3) a-g, b-e, c-f, d-h, , 2) a-f, b-g, c-h, d-e, 4) a- g, b-h, c-e, d-f, , Assertion & Reason Type, 5) 1 6) 1 7) 3 8) 1 9) 1 10) 1, 11) 1 12) 1 13) 1 14) 3 15) 3 16) 1, 17) 1, , Statement type questions, 18)2, , 19)1, , 20) 3 21) 3 22)1, , More than one option type questions, 23) 2 24) 4 25) 3 26) 3 27) 4 28) 4, 29) 3, , LEVEL - IV - HINTS, Assertion & Reason Type, 5., , 6., , Thermal stress = yα∆t, Thermal stress is independent of area of cross, section., Brass expands more on heating than Iron, , (α Brass, 7., 8., , 9., 28, , > α Iron ), , Linear expansion of Invar steel is very very less., Glass is bad conductor of heat and due to uneven, expansion and contraction glass tumbler breaks., , 10. In case of en-harmonic oscillators, increase in, amplitude is not equal on both sides. To have equal, amplitude on both sides atoms change their, position, with this, interatomic distance increases, and substance expand., 11. To have same angular separation, among the, atoms, expansion of solids takes place radially, outwards., 12. Platinum and glass expand equally on heating and, contracts equally on cooling α pt = α glass ., 14. Linear expansion also depends on rise in, temperature. The rise in temperature is more in thin, rod, than thick rod when both are given same, 1, m, 15. Brass and glass are not having equal expansions, on heating and equal contractions on cooling, , amount of heat. dQ = ms ∆t ; ∆t α, , [α Brass, , ≠ α Glass ], , But [α Platinum = α Glass ] so platinum wire can be, sealed through glass but brass can not be sealed., 16. e ∝ ∆t ∝, , 1, 1, ∝ 2 ⇒ e ∝ 12 ⇒ ∴ e1 = 4, m π r lρ, r, e2 1, , Statement Type Questions, 20. As brass expands more than Iron on heating,, bimetallic strip bends with brass on convex side, [α Brass > α Iron ], 21. Linear expansion depends on nature of material,, initial length, rise in temperature and scale of, temperature. Where as coefficient of linear expansion, depends on nature of material and scale of, temperature., , More than one option questions, 24. α is independent on length, area and mass. It, depends on temperature., 25. Expansion occurs due to increase of interatomic, spaces., 26. When copper solid sphere is heated, radius, increases, moment of inertia increases volume, increases, mass remains constant so density, decreases. Elasticity also decreases., , l1 α 2, ∆l1 = ∆l2 = l1α1∆t = l2α 2 ∆t ; l = α, 2, 1, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , THERMAL PROPERTIES OF MATTER - II, γA =, , SYNOPSIS, Thermal Expansion of Liquids, , m1 = mass of empty bottle, m2 = mass of the bottle with liquid, m3 = mass of the bottle with remaining liquid, mexp = m2 - m3 ; mremain = m3 - m1, ∆t = rise in temperature, , Introduction:, Ø, Ø, Ø, , Liquids do not have any shape of their own. volume, of liquid is equal to volume of the container, Linear and superficial expansions have no meaning, for liquids., As liquids possess definite volume, they experience, volume expansion only., , Coefficient of Real Expansion( γ R ):, Ø, Ø, Ø, , γA =, , Percentage change in volume of a liquid is given, ∆V , by V × 100 = γ R (t2 − t1 ) ×100 ., , , , Coefficeint of Apparent Expansion( γ A ):, Ø, Ø, Ø, , Ø, , When a liquid is heated both liquid and container, expand., Expansion which depends upon expansion of vessel, is called apparent expansion., The apparent expansion of liquids depends on, a) Initial volume of liquid, b) Rise in temperature. Ø, c) Nature of liquid., d) Nature of material of container., The apparent increase in volume per unit original, volume per 10 C rise in temperature is called, coefficient of apparent expansion of liquid., γA =, , Ø, , V2 − V1, / 0C ⇒ V = V [1 + γ (t − t )], 2, 1, A 2, 1, V1 (t2 − t1 ), , Ø, , γ A can be determine by using specific gravity bottle Ø, method, , NARAYANAGROUP, , m2 − m3, /0 C, ( m3 − m1 ) ∆t, , Relation between ( γ R ) and ( γ A ):, , Expansion which does not depend upon the, Ø, expansion of vessel is called Real expansion., The real expansion of liquids depends on, a) Initial volume of liquid ,b) Rise in temperature, c) Nature of liquid, The real increase in volume per unit original volume, per 1 0C rise in temperature is called coefficient of, real expansion, V −V, γ R = 2 1 / 0 C ⇒ V = V [1 + γ (t − t )], 2, 1, R 2, 1, V1 (t2 − t1 ), , Ø, , mass of the liquid expelled, mass remained × change in temperature, , The coefficient of real expansion of a liquid is equal, to the sum of coefficient of apparent expansion of, the liquid and coefficient of volume expansion of, the vessel., γR = γA + γg, , ∆V, , On heating Level of liquid, , γR > γg= > γA>0, , Rises, , γR < γg= > γA<0, , Falls, , γR = γg= > γA=0, , Remains same, , γg = 0 = > γA =γR, , Rises, , γR=-ve => γA = γR Rises, , If the same liquid is heated in two vessels X and Y,, real expansion of liquid is independent on nature of, the vessel, then, γ R = γ AX + 3α X and γ R = γ AY + 3αY, γ AX + 3α X = γ AY + 3αY ;, γ AX − γ AY = 3(αY − α X ), Here γ AX , γ AY denote coefficients of apparent, expansion of liquid in vessels X and Y respectively., αX and αY are coefficients of linear expansion of, vessels X and Y respectively., 29
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , WE-1.Volume of the bulb of a mercury thermometer Ø The temperature at which the density of liquid is, at 00 C is V0 and area of cross section of the, x −1 0, 1, times of its density at 0 0 C is t 3 = γ C, capillary tube is A 0 , coefficient of linear, x, R , expansion of glass is α g , and the cubical, 0, expansion of mercury is γ m . If the mercury fills WE-2. A block floats in water at 4 C so that 0.984, of its height is under water. At what, the bulb at 00C, find the length of mercury, 0, temperature of water will the block just sink, column in thermometer at T C, in water? Neglect expansion of block., Sol: Expansion of mercury = V0γ mT, ( γ R for water = 2.1× 10−4 / 0 C ), Expansion in glass bulb = V0 3α gT, Apparent expansion in mercury, Sol: Let d1 - density of water at 4 0C, d2 - density of water at t 0C, = V0γ mT – V0 3α gT ; i.e., At l = V0T ( γ m − 3α g ), d3 - density of block at t 0C, l=, , V0T ( γ m − 3α g ), At, , =, , V0T ( γ m − 3α g ), , A0 (1 + 2α gT ) (Q A0= At), , Variation of Density of Liquid With, Temperature:, Ø, , Ø, , Ø, , Ø, , Ø, , For a liquid if the temperature increases volume, m, , increases and hence density decreases. Q d = , V, , For calculating the change in density , the coefficient, of real expansion of the liquid is to be considered., , water are equal d 2 =, 0.984 =, , d1, = d3, 1 + γ∆t, , 1, 1 + 2.1 × 10 − 4 ( t − 4 ), , 1 + 2.1× 10−4 ( t − 4 ) =, , 1, 0.984, , d0 = dt(1+γRt) or dt = d 0 (1 − γ Rt ), 8 ×104, where, d0 = density of liquid at 00C, ⇒t −4=, = 77.430 C, 492 × 2.1, dt = density of liquid at t 0C, It must be heated to ' t ' = 77.43 + 4 = 81.430 C, γR = Coefficient of real expansion of liquid, If d1 and d2 are the densities of a liquid at WE-3:A sphere of diameter 7cm and mass 266.5g, temperatures t 1 and t2 respectively, floats in a bath of liquid at 0 0C. As the, d1 = d2 [1+ γR (t2 - t1)], temperature is raised, the sphere just sinks at, a temperature of 350C. If the density of the, If d 0 and dt are densities of liquid at 00 C and, liquid at 0 0C is 1.527 g/cm3 find the coefficient, d0 − dt 0, of cubical expansion of the liquid., γ =, / C, t 0 C , then R d × t, Sol: The sphere will sink in the liquid at 350C, when its, 0, density becomes equal to the density of liquid at, If d1 and d2 are densities of liquid at t1 0 C and, 350C. The density of sphere,, d1 − d 2 0, m, 4, 0, ρt = ; V = π r 3, t 2 C ,then γ R = d t − d t / C, 1 2, 2 1, V, 3, , Ø, , The temperature at which the density of a liquid is, , Ø, , , 0, x, x% less than that at 0 0 C is t1 = (100 − x)γ C, , R , The temperature at which the density becomes x%, 100 − x 0, of the density at 00 C is t2 = , C, xγ R , , 30, , d1, ; d1 = 1g / c.c ; d = 0.984 g / c.c., 3, 1 + γ∆t, The block sinks when the density of the block and, d2 =, , ρt =, , 266.5, 4 22 7 , ×, ×, 3 7 2 , , 3, , = 1.483 g / c m 3, , Now ρ 0 = ρ t [1 + γ∆T ], 1.527 = 1.483[1 + γ × 35] ;1.029 = 1 + γ × 35, γ =, , 1.029 − 1, = 0.00083 / 0 C, 35, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , If water is taken above Hg and the system is heated, then observed (apparent) expansion of water will, be equal to its real expansion. This is the principle, involved in dilatometer., The unoccupied volume or volume of air present in, WE-4: A long cylindrical metal vessel, having a, the vessel will be constant only when both container, linear coefficient (α), is filled with a liquid, and liquid have same thermal expansion. (or) same, upto a certain level. On heating it, it is found, increase in volume, that the level of liquid in the cylinder, remains the same. What is the volume, coefficient of expansion of the liquid ?, [EAMCET 2013(M)], Un ocupied, volume, Sol: Volume of liquid V=Al. Increase in volume, , Condition for constant volume of, unoccupied space in the container at all, temperatures:, Ø, , ∆V = ( ∆A) l (Ql is constant ), , Liquid, , Let Vc and Vl are volumes of container and liquid., Also ? c and ? l are coefficients of expansion of, container and liquid respectively., Unoccupied volume remain constant, DVc = DVl ; i.e., VC ? C Dt = Vl ? l Dt, , For container: ( ∆V )C = ( ∆A) l = Aβ∆tl = V (2α )∆t, , VC ? C = Vl ? l, , Ø, , To compensate expansion of container using, mercury (Dilatometer), A small amount of Hg is taken in a container such, that its expansion is equal to expansion of container, made of glass, Let Vc and Vl are volumes of container and liquid., , Also γc and γ l are coefficients of container and, liquid respectively., , Water, Hg, , \ DVc = DVl ; Vcγ cDt = Vlγ l Dt, , For liquid: ( ∆V )l = V γ l ∆t, If the level of the liquid remains same, , ( ∆V )C = ( ∆V )l ; V (2α )∆t = V γ l ∆t, ⇒ γ l = 2α, , Volume of Liquid Expelled:, Ø, , A container of volume VC at temperature t 10C is, completely filled with a liquid. If the container is, heated to t 20C, then volume of liquid over flown is, V1 = initial volume of the liquid = initial volume of, the container, , V2(liquid ) = V1 [1 + γ l (t2 − t1 ) ], V2(container ) = V1 [1 + γ c (t2 − t1 )], Volume of liquid over flow is, , (V2 )l − (V2 )c = V1 (γ l − γ c )(t2 − t1 ), , WE-5. A 250cm3 glass bottle is completely filled, VHg, γg, 1, with water at 500C. The bottle and water are, =, =, V c γ c = Vl γ l ;, heated to 600C. How much water runs over, Vc, γ Hg, 7, If:a) the expansion of bottle is neglected, 1, b) the expansion of bottle is included?, So if glass container is filled with Hg upto, th of, 7, ( βglass = 1.2 ×10−5 / K & γ water = 60 ×10−5 / 0C ), its volume then expansion of container is, compensated. Due to expansion of container level Sol. Water overflown = (final volume of water) –, decreases and due to expansion of Hg level, (final volume of bottle ), increases. The total expansion would be zero., (a) If the expansion of bottle is neglected:, NARAYANAGROUP, , 31
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JEE- ADV PHYSICS-VOL- V, ρ1 = density of liquid at 300C, , THERMAL PROPERTIES OF MATTER - II, Ø, , ρ 2 = density of liquid at 400C, Mass expelled ∆m1 = M1 − M 2 = V1 ρ1 at 300C, 20 V1 ρ1, ∆m2 = M 1 − M 3 = V2 ρ 2 at 400C; 18 = V ρ, 2 2, , 10, V1 × 1.5 × 103, 9 × 1.2, =, 3 ; 1 + γ∆t =, 9 V1 (1 + γ∆t ) × 1.25 × 10, 10, 10.8, =, =1.08; γ∆t = 1.08 − 1 ; 3α = 0.08, 10, 0.08, −3 0, α=, 3 ∴α = 2.6 ×10 / C, , Anomalous Expansion of Water:, Ø, Ø, Ø, Ø, , Liquids, in general expand in volume with rise of, temperature., Pure water when heated contracts from 00 C to, 3.980 C (40 C) and expands from 40C onwards. It, is called anomalous expansion of water., Water has negative expansion coefficient in the range, at 0 0 C to 4 0 C and positive expansion coefficient, above 40 C., At 40 C water occupies minimum volume and hence, density becomes maximum (1gm /cc)., V, , 1.00013, 1.00000, 10, , 4, t0C, , Ø, , The density of water increases from 00 C to 40C, and decreases with rise in temperature from 4 0C, , 1.0 gm/cc, , d, , Ø, , Correction For Barometric Reading:, The brass scale of barometer is usually calibrated, at 0 0C . If observation is taken at different, temperature, then a correction is needed for brass, scale. Suppose the height of mercury at 00 C is, H 0 and true scale reading is H. If α is the, coefficient of linear expansion of brass, then true, height of brass scale at temperature t is, H t = H (1 + α t ) ........(1), As atmospheric pressure is constant at all, temperatures, so we have, Pressure at 00 C =pressure at t 0 C, H 0 ρ 0 g = Ht ρt g, .......(2), Here ρ 0 and ρt are the densities of mercury at, 00 C and t 0 C respectively. Also we have, , ρ0, , γ being volume coefficient of mercury.., 1+ γ t, Thus from equation (1) and (2), we, ρo, g, have H o ρo g = H (1 + α t ), 1+ γ t, ρt =, , H o = H (1 + α t )(1 + γ t )−1 ; H o = H (1 + α t )(1 − γ t ), H o = H 1 + α t − γ t − αγ t 2 , , H o = H [1 − (γ − α )t ] ; [Q γα is neglected ], WE-11. A Barometer with brass scale, which is, correct at 00C, reads 75cm on a day when the, air temperature is 200C. Calculate correct, reading at 00C. (Coefficient of real expansion, of mercury =0.00018/0C and coefficient of, linear expansion of brass =0.0000189/0C.), , Sol: We know H0 ≅ H (1 − ( γ − α ) t ) ., In this problem, H = 75cm, t = 200C,, tC4C, γ = 0.00018/0C, α = 0.0000189/0C, H0 = 75.000[1–(0.00018–0.0000189)20], The experiment with Hope’s apparatus establishes, =74.758 cm, that water has maximum density at 40C., Dilatometer is a convenient apparatus to study the, anomalous expansion of water., 0, , Ø, , Ø, , Water in a lake freezes when temperature of water, falls below 0o C . If temperature of surroundings of, lake is −T 0C then temperature just below surface, of lake is 0o C and at bottom of lake is, 40 C or 277 K, The coefficient of volume expansion of water at 40C, is zero., , NARAYANAGROUP, , o, , 33
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, 8., , C.U.Q, 1., , 2., , 3., , 4., , 5., , 6., , 7., , Expansion of liquids on heating is different, from that solids, since the expansion of liquids, is, 1) much more than solids because molecular, spacing in them is less, 2) much more than solids because molecular, spacing in them is more, 3) much less than solids because molecular spacing, in them is more., 4) much less than solids because molecular spacing, in them is less, A liquid with coefficient of real volume, expansion ( γ ) is filled in container of material, having coefficient of linear expansion ( α ). If, liquid over flows on heating., 1) γ = 3α 2) γ >3α, 3) γ < 3α 4) γ = α, On heating a liquid of coefficient of real, expansion γ in a container having coefficient, of linear expansion α / 3 . The level of liquid, in the container will, 1) rise 2) fall 3)remains same 4) over flows, A long cylindrical vessel of volume V and, coefficient of linear expansion α contains a, liquid. The level of liquid has not changed on, heating. The coefficient of real expansion of, the liquid is., V −α, V +α, V, 2), 3), 4) 3α, 1), V, V, V −α, The liquid whose coefficient of real expansion, is equal to 1.5 times the coefficient of areal, expansion of container and heated then the, level of the liquid taken in the container, 1) rises, 2) falls, 3) remains same, 4) first rises and then falls, A metal ball suspended from the hook of a, spring balance is kept immersed in a liquid, other than water. On increasing the, temperature of this liquid, the reading in the, spring balance., 1) Increases, 2) Decreases, 3) Remains same 4) May increases or decreases., A metal ball immersed in alcohol weights W 1, at 00C and W2 at 500 C Assuming that the, density of the metal is large compared to that, of alcohol then., 1) W1 = W2, 2) W1>W2, 3) W1< W2, , 34, , 4) W1 > W2, , 9., , 10., , 11., , 12., 13., , A block of wood is floating on water at 200C, with certain volume above the water level. The, temperature of water is slowly increased the volume, 1) increases, 2) decreases, 3) remains same, 4) first decreases and then increases., A glass is full of water at 40 C when it is, (a)cooled (b) heated then, which one of the, following is correct, 1) water level decreases, increases, 2)water level increases, decreases, 3) water level decreases, decreases, 4)water over flow in both the cases, The top of a lake is frozen when the air in, contact with the lake surface is at − 50 C the, temperature of water in contact with the bottom, of the lake will be, 1) − 50 C 2) 40 C 3) 0 0 C 4) − 40 C, A metal sphere is suspended in water at, 0 0 C by a thread when water is heated to 40 C, the tension in the thread, 1) decreases 2) increases 3) remains same, 4) first increases and then decreases, Water has maximum density at, 1) 00C, 2) 40C 3) 250C, 4) 370C, A Sealed glass jar is full of water. When it is, kept in a freezing mixture, it is broken because, 1) water expands from 4 0C to 0 0C, 2) ice expands while melting, 3) water expands due to freezing, 4) ice expands since its temperature falls below 00C, C.U.Q -KEY, 1) 2, 2) 2 3) 3 4) 4 5) 3 6) 1, 7) 3, 8) 2 9) 4 10) 2 11) 1 12) 2, 13) 1, , LEVEL-I (C.W.), EXPANSION OF LIQUID, 1., , 2., , The coefficient of real expansion of liquid is, γRand the coefficient of apparent expansion of, the liquid is γA. The coefficient of cubical, expansion of the vessel is γ. If γR : γA =4:1, then γA: γ is, 1) 3:1, 2) 1:3, 3) 4:1, 4) 1:4, γ A of liquid is 7/8 of γ R of liquid. α g of vessel, is, γ, γ, γ, γ, 1) R, 2) R 3) R 4) R, 8, 12, 24, 36, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 3., , 4., , 5., , 6., , The apparent coefficient of expansion of liquid,, 1) 12.6 ×10−4 / 0C, 2) 0.8 × 10 −4 / 0C, when heated in a copper vessel is C and when, 3) 1.25 × 10−5 / 0C, 4) 1.25 × 10−4 / 0C, heated in a silver vessel is S. If A is the linear, coefficient of expansion of Copper, linear 10. A weight thermometer contains 52g of a liquid, at 100C. When it is heated to 1100C,2 g of the, expansion coefficient of silver is, liquid is expelled. The coefficient of real, C + S − 3A, C + 3A − S, 1), 2), expansion of the liquid is[αglis 9 × 10-6/0C], 3, 3, 1)27 × 10-6/0C, 2) 427 × 10-6/0C, S + 3A − C, C + S + 3A, 4) 473 × 10-6/0C, 3) 373 × 10-6/0C, 3), 4), 3, 3, LEVEL - I (C.W) - KEY, The density of a liquid at 1000C is 8.0 g/cm3, 1), 2, 2) 3 3) 2 4) 2 5) 1 6) 1, and at 00C is 8.4 g/cm3, the coefficient of cubical, 7) 3, 8) 1 9) 4 10) 2, expansion of the liquid is, 1) 10-4 / 0C, 2) 5 × 10-4 / 0C, LEVEL - I (C.W) - HINTS, -4 0, 4) 4 × 10-4 / 0C, 2) 8 × 10 / C, γR =γA +γg, If γ is the coefficient of a real expansion of a 1., 7, liquid then the temperature at which density, 2. γ A = γ R and γ A = γ R − 3α, 0, 8, of a liquid is 1% of its density at 0 C is, 3., 99, 1, 100, 1, 2), 3), 4), 1), γ, 99γ, γ, 100γ, A one litre flask contains some mercury. It is, found that at different temperatures the volume 4., of air inside the flask remain same. The volume, of mercury taken in the flask is (coefficient of, 5., linear expansion of glass is 9 ×10−6 0 C and, coefficient of volume expansion of Hg is 6., −4, , 7., , 8., , 9., , THERMAL PROPERTIES OF MATTER - II, , [EAMCET 2008(M)], 1.8 ×10, C )., 1) 150ml 2) 750ml 3)1000ml 4)700ml, A liquid occupies half of a vessel at a particular, temperature. The volume of the unoccupied, part remains constant at all temperatures. If, α and γ are the coefficients of linear and real, expansions of a vessel and liquid, then γ is, 1) 3α, 2) 3α /2, 3) 6 α, 4) 9 α, A glass bulb of volume 250cc is filled with, mercury at 200 C and the temperature is raised, to 1000 C .If the coefficient of linear expansion, of glass is 9 × 10−6 / 0 C Coefficient of absolute, expansion of mercury is 18 × 10−5 / 0 C .The, volume of mercury overflows, 1) 3.06cc, 2)2.94cc 3)6.12cc 4)7.73cc, If on heating a liquid through 800C, the mass, , For copper vessel γ R = ( γ A )c + 3α c, For silver vessel γ R = ( γ A ) s + 3α s, , γ =, , d o − dt 0, / C, dt ( t2 − t1 ), , given dt =, , d 0 − dt, d0, , t1 = 00 C , t2 = t ; γ =, dt × t, 100, , Vl γ l = Vg γ g, , 0, , 1, th of mass still remaining, the, 100, coefficient of apparent expansion of the liquid, is, , expelled is, , NARAYANAGROUP, , 7., , Vl γ l = Vg γ g and γ g = 3α ⇒ Vl =, , 8., , ∆V = V ( γ l − γ g ) ∆t, , 9., , γA =, , Vg, 2, , mass of the liquid exp elled, mass of theliquid remaining × ∆t, , mass of the liquid exp elled, 10. γ A = mass of theliquid remaining × ∆t ,, , γ R = γ A + 3α, , LEVEL-I (H.W.), EXPANSION OF LIQUID, 1., , The coefficient of real expansion of a liquid is, 7 × 10-4/0 C.The co efficient of linear expansion, of the vessel is1×10−5 / 0 C . The coefficient, of apparent expansion of the liquid is, 1) 7 × 10-4/ 0C., 2) 6 × 10-5/ 0C., 3) 67 × 10-5/ 0C., 4) 73 × 10-5/ 0C., 35
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, 2., , 3., , 4., , 5., , 6., , 7., , The coefficient of real expansion γR of a liquid, is 5 times the coefficient of linear expansion of, the material of the container in which the liquid, is present. The ratio of the coefficient of, apparent expansion and real expansion of the, liquid is, 1) 5:2, 2) 1:5, 3) 2:5, 4) 5:1, When a liquid in a glass vessel is heated, its, apparent expansion is 10.30 × 10-4/0C. Same, liquid when heated in a metalic vessel, its, apparent expansion is 10.06 × 10-4 /0 C. The, coefficient of linear expansion of metal is, ( αglass = 9 × 10-6/0C), [EAMCET 2013, 2010(M) 2012(E)], 1) 51 × 10-6/0C, 2) 43 × 10-6/0C, -6 0, 3) 25×10 / C, 4) 17 × 10-6/0C, Coefficient of real expansion of mercury is, 0.18 × 10-3/0C. If the density of mercury at 00C, is 13.6 g/cc its density at 473K will be, 1) 13.12 g/c.c., 2) 13.65g/c.c., 3) 13.51 g/c.c., 4) 13.22 g/c.c., If coefficient of real expansion of a liquid is, 1 0, / C. The temperature at which its, 5500, density is 1% less than density at 00C is, 1) 55.50C, 2)1000C, 3) 990C, 4) 10C, The coefficient of cubical expansion of liquid, and glass are in the ratio of 8:1. The volume, of the liquid to be taken into 800cc container, so that the unoccupied portion remains, constant is, 1) 10cc 2) 100cc, 3) 80cc, 4) 8cc, The fraction of the volume of a glass flask must, be filled with mercury so that the volume of, the empty space may be the same at all, temperatures is, , (α, 8., , 36, , glass, , = 9 × 10−6 / 0C , γ Hg = 18.9 × 10 −5 / 0C ), , 9., , A glass vessel just holds 50gm of a liquid at, 00C. If the coefficient of linear expansion is, 8 × 10 −6 / 0C . The mass of the liquid it holds at, 800C is [ coefficient of absolute expansion of, liquid = 5 × 10 −4 / 0C (nearly), 1) 46 g 2) 48 g, 3) 51 g, 4) 42 g, 10. A weight thermometer contains 51 g of, mercury at 200C and 50 g of mercury at 1000C., The coefficient of apparent expansion of, mercury in glass vessel is, 2) 2.5 × 10-3 / 0C, 1) 25 × 10-5 / 0C, 4) 4 × 10-4 / 0C, 3) 2 × 10-5/ 0C, , LEVEL - I (H.W) - KEY, 1) 3, 7) 2, , 2) 3, 8) 3, , 3) 4, 9) 2, , 4) 1 5) 1, 10) 1, , 6) 2, , LEVEL - I (H.W) - HINTS, 1., , γR =γA +γg, , 3., , For copper vessel γ R = ( γ A )c + 3α c, , 2. γ R = 5α , γ R = γ A + 3α, , For silver vessel γ R = ( γ A ) s + 3α s, 4., , dt = d o / (1 + γ∆t ), , 5., , given γ R =, , 1 0, d −d, / C ; d0=100; d t=99, γ = 0 t, dt × t, 5500, , 6., , γl 8, =, γg 1 ;, , Vl γ l = Vg γ g, , 7., , Vl γ l = Vg γ g ⇒, , 8., , ∆V = V ( γ l − γ g ) ∆t, , 9., , If x is mass of the liquid expelled. γ A =, , Vg, Vl, , =, , Vg, γl, γ, ⇒, −1 = l −1, γg, Vl, γg, , x, ( m − x) ∆ t, , m2 − m3, 10. γ a = ( m − m ) ∆ t, 3, 1, , LEVEL-II (C.W.), , 1, 1, 1, 1, 1), 2), 3), 4), 2, 7, 4, 5, EXPANSION OF LIQUID, 3, A glass flask of volume 200cm is completely, 1. Coefficient of apparent expansions of a liquid, filled with mercury at 200C. The amount of, in two different vessels are a and b. then the, mercury that overflow when the flask is heated, real coefficient of expansions of liquid, if the, to 800C (Coefficient of volume expansion of, ratio of volume expansion of vessels is x : y, glass is 27 × 10-6/ 0 C, γ of mercury 0.18 × 10-3/, 0, bx − ay, ay + bx, ay − bx, ay + bx, C), 1), 2), 3), 4), 1) 2.16cm3 2) 0.032cm3 3) 1.84cm3 4) 2.40cm3, x− y, x+ y, x−y, x− y, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 2., , 3., , 5., , 6., , 7., , the rod changes by, 1) 00C, 2) 100C, 3) 200C 4) 1700C, A barometer with a brass scale correct at 0 0C, reads 70 cm of mercury on a day when the air, temperature is 400C. The correct reading at, 0 0C is (Coefficient of real expansion of, mercury is 0.00018/0C and coefficient of linear, expansion of brass is 0.000018/0C), 1) 60.5cm 2) 69.5cm 3) 20.5cm 4) 50.00cm, A solid floats in a liquid at 200C with 75% of it, immersed. When the liquid is heated to 1000C,, the same solid floats with 80% of it immersed, in the liquid. Calculate the coefficient of, expansion of the liquid. Assume the volume of, the solid to be constant., 1) 8.33×10–4/0C, 2) 83.3×10–4/0C, 3) 833×10–4/0C, 4) 0.833×10–4/0C, 10. The volume of mercury in the bulb of a, thermometer is 10-6 m3. The area of crosssection of the capillary tube is 2 × 10-7 m2. If, the temperature is raised by 100 0C, the, increase in the length of the mercury column, , A flask contains 100c.c of a liquid at 100C., When it is heated to 1100C increase in volume, of the liquid appears to be 2 c.c. Find the 8., coefficient of real expansion of the liquid., ( α of flask is 11× 10−6 / 0 C ), 1) 2.33 × 10-4 /0C, 2) 3.33 × 10-4 /0C, 3) 23.3 × 10-4 /0C, 4) 33.3 × 10-4 /0C, 0, At 0 C the densities of a cork and a liquid in, which the cork floats are d 1 and d2 respectively., 9., The coefficient of expansion for the material, of the cork and the liquid are γ and 100 γ, respectively. If the cork sinks when the, temperature of the liquid is ‘t0C’ then the ratio, d2, d1 is, , 4., , THERMAL PROPERTIES OF MATTER - II, , 1), , 1 + 100γt, 1 + γt, , 2), , 1 + γt, 1 + 100γt, , 3), , 100 + γt, 1 + γt, , 4), , 1 + γt, 100 + γt, , A wooden block of density 860 kg/ m3 at 00C, −5 0, is ( γ Hg = 18 × 10 / C ) [EAMCET 2009(M)], is floating on benzene liquid of density, 1) 18 cm 2) 0.9 cm, 3) 9 cm 4) 1.8 cm, 900 kg/m3 at 00C. The temperature at which, 11., A, non-conducting, body, floats, in a liquid at 200C, the block just submerge in benzene is, with 2/3 of its volume immersed in the liquid., [ γ wood = 8 × 10 −5 / 0 c, γ benzene = 12 × 10−4 / 0 C ], When liquid temperature is increased to 1000C,, 1) 240 c 2) 420 c, 3) 160 c, 4) 320 c, 3/4 of body volume is immersed in the liquid., Then the coefficient of real expansion of the, A sphere of mass 180g and diameter 6 cm floats, liquid is.... (neglecting the expansion of, on the surface of a liquid. When the liquid is, 0, container of the liquid) [EAMCET 2011(E)], heated to 35 C, the sphere sinks in the liquid., 0, -3, If the density of liquid at 0 C is 2 gcm . The, 2) 15.6 ×10−4 / 0 C, 1) 1.56 ×10−4 / 0 C, coefficient of real expansion of liquid is, 3) 1.56 ×10−5 / 0 C, 4) 15.6 ×10−5 / 0 C, 2) 81.4 × 10-4/0C, 1) 71.4 × 10-4/0C, 12. A glass flask of volume one litre is filled, 3) 91.4 × 10-4/0C, 4) 61.4 × 10-4/0C, completely with mercury at 00C. The flask is, A vessel contains a liquid filled with 1/10th of, now heated to 1000C. Coefficient of volume, its volume. Another vessel contains same liquid, expansion of mercury is 1.82 × 10-4/0C and, upto 1/8th of its volume. In both cases the, coefficient of linear expansion of glass is 0.1, volume of empty space remains constant at all, × 10-4/0C. During this process, amount of, temperatures. Then the ratio of coefficient of, mercury which overflows is, linear expansions of the two vessels is, [EAMCET 2013(E)], 1) 2:5, 2) 5:2, 3) 4:5, 4) 5:4, 1) 21.2 cc 2) 15.2 cc 3) 2.12 cc 4) 18.2 cc, The co–efficient of linear expansion of iron is, 11/180 of volume coefficient of expansion of, mercury which is 18 × 10−5 / 0C .An iron rod is, 10m long at 270C. The length of the rod will be, decreased by 1.1mm then the temperature of, , NARAYANAGROUP, , LEVEL - II (C.W) - KEY, 1) 1, 7) 2, , 2) 1, 8) 2, , 3) 1, 9) 1, , 4) 2 5) 1, 10) 3 11) 2, , 6) 3, 12) 2, , 37
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , LEVEL - II (C.W) - HINTS, 1., , γa x, = ;, γb y, , 2., , Coefficient of apparent expansion, , γA =, , 2., , x a −γR, =, y b −γR, , 4., , If the boat sinks in benzene, VBdBg = Vldlg, , 3., , d 2 di (1 + γ 2 ∆t ), =, d1 d i (1 + γ 1∆t ), , d f = d i (1 + γ∆t ) ;, , ; VBdB = Vldl, , d (wood) at t0C = d (benzene) at t0C, , ( d0 )wood, , ( d0 )benzene, (1 + γ w∆t ) (1 + γ B ∆t ), 5., 6., , V1 =, , 7., , α iron =, , 8., , H 0 ρ0 g = Ht ρt g ; H 0 ρ0 = H (1 + α t ), , V, V, , V2 =, 10, 8, 11, γ Hg, 180, , γR =, , 4., , 1 + 60γ Hg, , 4) 1 + 60γ, Fe, 0, A boat is floating in water at 0 C such that, 97% of the volume of the boat is submerged, in water. The temperature at which the boat, will just completely sink in water is, 0, , ; ∆l = lα∆t, , 5., , ρ0, (1 + γ t ), , X 2 − X1, / 0C, X 1 ( t 2 − t1 ), , V2 − V1, 11. V1 = 2V/3, V2= 3V/4, γ = V ∆t, 1, , 6., , 12. ∆V = V (γ l − γ s )∆t, , LEVEL-II(H.W.), EXPANSION OF LIQUID, , 38, , 1 + 60γ Fe, , 1 − 60γ Fe, , ( γ R = 3×10−4 / C) (nearly), , ; V1γ l = Vg γ g, , 10. ∆V = A× ∆l ⇒ V γ∆t = A × ∆l, , 1., , 2) 1 + 60γ, Hg, , 3) 1 − 60 γ, Hg, , ∴ H 0 = H 1 − ( γ − α ) t , , 9., , 1 + 60γ Fe, 1) 1 + 60γ, , Hg, , =, , d − dt, m, γ = 0, ; dt = ; V = 4 π r 3, dt × t, V, 3, , γR, 100γ R, 101γ R, 101γ R, 2), 3), 4), 303, 101, 300, 100, When a block of iron floats in mercury at 00 C,, a fraction k1 of its volume is submerged, while, at the temperature 600C, a fraction k2 is seen, to be submerged. If the coefficient of volume, expansion of iron is γFe and that of mercury is, γHg, then the ratio k1/k2 can be expressed as, , 1), , V2 − V1, V1 ( t2 − t1 ) and γ R = γ A + 3α, , 3., , If the coefficient of real expansion γ R is 1%, more than coefficient of apparent expansion, ,linear expansion coefficient of the material is, , The ratio of coefficients of apparent, 7., expansions of the same liquid in two different, vessels is 1:2. If α1 and α2 are the coefficient, of linear expansions then coefficient of real, expansion of the liquid is, 1) 2α1 - α22) 3α1 - 4α2 3) α1 - 2α2 4) 6α1 - 3α2, , 1) 100C 2) 1030C, 3) 600C, 4) 500C, A sphere of diameter 8cm and mass 275 g floats, in a bath of liquid. As the temperature is raised,, the sphere begins to sink at a temperature of, 3, 400 C. If the density of the liquid is 1.5g/cm at, 00C, find the coefficient of cubical expansion, of the liquid. Neglect the expansion of the, sphere, 2) 25×10−5 /0 C, 1) 125×10−4 /0 C, 6, 0, −, 3) 15 ×10 / C, 4) 115×10−3 /0 C, The coefficient of volume expansion of mercury, is 20times the linear expansion of glass. Find, the volume of mercury that must be poured in, to a glass vessel of volume V so that the, volume above the mercury remain constant at, all temperatures, 3V, V, 3V, V, 2), 3), 4), 40, 20, 20, 30, If γ (apparent) of a liquid in a vessel is 76% of, γ (real) of that liquid, the coefficient of linear, expansion of the vessel is, 1) 8% of γ (real), 2) 16%of γ (real), 3) 24%of γ (real), 4) 25.3%of γ (real), , 1), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 8., , THERMAL PROPERTIES OF MATTER - II, , The height of the mercury column in a 2., barometer provided with a brass scale, corrected at 0 0C is observed to be 74.9 cm at, 150C.Find the true height of the column at 00C., , (a, , = 20×10-6 / 0C and ? Hg = 175×10-6 / 0 C ), , b, , 1) 74.72cm, 3) 74.12cm, , 2) 79.92cm, 4)72.64 cm, , LEVEL - II (H.W) - KEY, 1) 4, 7) 1, , 2) 1, 8) 1, , 3) 1, , 4) 2, , 5) 1, , 6) 3, , 3., , LEVEL - II (H.W) - HINTS, 1., 2., 3., , ( γ A )1, (γ A )2, , =, , 1, 2, , 1, , γ R − 3α1, , ; 2 = γ − 3α, R, 2, , γ , , γ R = γ A + A ; γ R = (γ A + 3α ), 100 , , d1, v d, Vdg = v ρ g ; V = ρ ; k1 = ρ at 00 C, , 1, k2 =, , 1, , 1) (γ − γ ) ∆t, 2, 1, 4., , d2, k1 d1 ρ 2, 0, =, at, t, C;, ρ2, k 2 d 2 ρ1, , mg =, , 5., , γ =, , 97V, ρ 0 g ; mg = V ρ t g ; 97V, 100, 100, , d0 − dt, m, , dt =, dt × t, V, , 6., , γ Hg = 20α g, , 7., , γA =, , 8., , 1., , 76, γR, 100, , ρ 0 g = V ρt g, , 1, , 2) (γ − γ ) ∆t, 1, 2, , 3) (γ 1 − γ 2 )∆ t, 4) (γ 2 − γ 1 )∆t, The loss in weight of a solid when immersed in, a liquid at 00C is W0 and at t 0 C is ‘W’. If cubical, coefficient of expansion of the solid and the, liquid are γ s and γ l then W =, , ρ0, 1 + γ Fet, k1 d 0 (1 + γ Fet ), =, ×, =, k2, d0, (1 + γ Hg t ) ρ0 1 + γ Hg t, , 4., , A piece of metal weighs 46g in air. When it is, immersed in a liquid of specific gravity 1.24 at, 270C, it weighs 30g. When the temperature of, the liquid is raised to 420C, the metal piece, weighs 30.5g.Specific gravity of liquid at 420C, is 1.2. Calculate the coefficient of linear, expansion of the metal., 1) 2.4 × 10−5 0C, 2) 3.4 × 10−5 0C, 5, 0, −, 3) 2.9 × 10 C, 4) 24 × 10−5 0C, A piece of metal floats on mercury. The, coefficients of volume expansion of the metal, and mercury are γ 1 and γ 2 respectively. If the, temperature of both mercury and metal are, increased by an amount ∆t ,the fraction of the, volume of the metal submerged in mercury, changes by the factor., , 1) W0 [1 + (γ s − γ l )t ], 5., , 2) W0 [1 − (γ s − γ l )t ], , 3) W0 [1 + (γ l − γ s )t ], 4) W0 [1 − (γ l − γ s )t ], The density of a liquid of coefficient of cubical, expansion γ is ρ at 00C when the liquid is heated, to a temp T, the change in density will be, , ; VHg γ Hg = Vg γ g, , 1), , ; γ g = γ R −γ A, , 3), , − ργ T, 1+ γ T, , − (1 + γ T ), , 2), 4), , ργ T, 1+ γ T, , γ (1 + γ T ), , γT, γT, ρ, 6. A uniform pressure P is exerted on all sides of, H 0 ρ 0 g = Ht ρt g ; H 0 ρ 0 = H (1 + α t ) (1 + 0γ t ), a solid cube at temperature t0 C. By what, amount should the temperature of the cube be, ∴ H 0 = H 1 − ( γ − α ) t , raised in order to bring its volume back to the, original value before the pressure was applied,, LEVEL-III, if the bulk modulus is B and volume coefficient, is γ?, A mercury thermometer contains 2c.c. of Hg., 0, 0, 0, at 0 C. Distance between 0 C and 100 C, P, B, 1, γP, 3), 4), 1), 2), marks on the stem is 35cm and diameter of, γB, γP, γBP, B, the bore is 0.02cm then γA of liquid is, LEVEL - III - KEY, 1) 0.000055/0C, 2) 0.000066/0C, 1) 1, 2) 1 3) 4 4) 1,4 5) 1 6) 2, 3) 0.00055/0C, 4) 0.000058/0C, , NARAYANAGROUP, , 39
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , LEVEL - III - HINTS, 1., 2., , ASSERTION & REASON, , ∆V = V γ A ∆t ... (1); ∆V = Al = π r 2l .... (2), from (1) and (2) V γ A ∆t = π r 2 l, m1 − m2 = V1 ρ1 ; m1 − m3 = V2 ρ 2, , V2 = V1 (1 + γ∆t ), , 3., , f1 =, , d metel, d metel, (1 + γ 2 ∆t ), f2 =, ×, ;, d mercury, (1 + γ 1∆t ) d mercury, , f 2 (1 + γ 2 ∆t ) f 2 − f1, =, = ( γ 2 − γ 1 ) ∆t, f1 (1 + γ 1∆t ) ;, f1, , 4., , W0 = V0 d o g ,, W0 V0 d0, =, Wt Vt dt, , 5., , 6., , =, , V0 dt [1 + γ l t ], , V0 [1 + γ s t ] dt, , ρ − ρ t = γρ t T, , 40, , 5., , γT ρ, ; ρ − ρt = (1 + γ T ), , When temperature increases density decreases, PV, PV, ⇒ ∆V =, ∆V = V γ∆t ...... (1); B =, B, ∆V, PV, = V γ∆t ...... (2), K, P, PV, = V γ∆t ; ∆t =, from (1) and (2), Kγ, K, , MATCHING TYPE QUESTIONS, , 2., , 4., , Wt = Vt d t g, , LEVEL-IV, 1., , 3., , 6., , 7., , 1) Both A and R are true and R is the correct, explanation of A, 2) Both A & R are true but R is not the correct, explanation of A., 3) A is true but R is false, 4) Both A and B are false, Assertion(A): It is observed that when a liquid is, heated in a vessel its level does not change., Reason (R): coefficient of real expansion of the, liquid = coefficient of volume expansion of the vessel, Assertion(A): Real expansion of liquid does not, depend upon material of container., Reason (R): Liquids have no definite shape. They, acquire the shape of the containers in which they, are taken., Assertion(A): A wooden block is floating on a, liquid.When the temperature of the liquid is, increased the volume of the block immersed in, the liquid increases., Reason (R):As temperature increases,the density, of liquid decreases., Assertion(A): when a liquid in a container is, heated first the level of the liquid falls down and, then rises ., Reason (R): when the liquid in a container is, heated first the container undergoes expansion, and generally the expansion of the liquid is, greater than that of solid., Assertion(A): when a beaker containing liquid, is heated the centre of mass of the system first falls, down then rises up above the initial position., Reason (R): The liquid in the beaker undergoes, expansion on heating and the expansion of liquid is, more than that of beaker., , LIST -I, LIST -II, a) App. expansion e) Nature of vessel, and liquid, b) Real expansion f) Nature of liquid, MORE THAN ONE ANSWER QUESTIONS, c) γ A, g) Nature of Vessel &, 8. Certain volume of a liquid is taken in a long, liquid & temperature, glass tube and its temperature is increased at, d) γR, h) Nature of, a uniform rate, the rate of increase in the length, liquid and temp., of the liquid depends on, LIST -I, LIST-II, a) length of the liquid, a) γ g is +ve & < γR e) liquid level does not, b) area of cross section of the glass tube, change, c) coefficient of expansion of glass, 1) only (a) is correct, 2) (a) & (b) are correct, b) γ g is - ve, f) liquid level increases, 3) (b)&(c) are correct 4)(a),(b)&(c)are correct, continuously, c) γ g = γR, g) liquid level decreases. 9. A metal ball suspended from a spring balance, is immersed in water at 4°C . If the, d) γ g > γR, h) liquid level first, temperature of water is changed the reading, decreases and then, in the balance, increases., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , (a) may decreases, (b) increases, (c) may remains same, 1) only (c) is true, 2) (b) is true, 3) (a) & (c) are true, 4) (b) & (c) are true, 10. Identify the correct statements from the, following:, a) The apparent expansion of liquid depends on, the expansion of material of the container, b)The real expansion of the liquids depends on the, density of the liquid., c)The expansion of liquid with respect to the, container is called the apparent expansion., 1) Only a & b are true, 2) Only b & c are true, 3) a,b & c are true, 4) Only a & c are true, 11. A liquid of coefficient of real expansion γ is, partly filled in a vessel of coefficient of linear, expansion γ /3. When the system is heated,, then., a) The volume of space above liquid remains same., b) The level of liquid relative to vessel remains same., c) The fraction of volume of liquid in vessel remains, same., 1) Only (a) is correct, 2) Only (b) & (c) are correct, 3) Only ( c) is true 4) All are true, , and then heat is absorbed by the liquid, liquid, expands more than solids., , THERMAL EXPANSION OF GASES, SYNOPSIS, Introduction:, Ø, Ø, Ø, , Boyle’s Law:, Ø, , LEVEL-IV - KEY, 1)3, 7)1, , 2)2, 8)4, , 3)1, 9)2, , 4)2, 10)3, , 5)1, 11)4, , 6)1, , LEVEL-IV - HINTS, 03. For a given temperature, γ R = γ g , volume, expansion of vessel = volume expansion of the liquid, 04. Liquids acquire the shape of the container but their, real expansion does not depends on nature of the, container, 05. As temperature increases, Buoyancy decreases,, apparent weight increases and volume of immersed, part increases., Density of liquid is inversely proportional to, ρ 0 − ρt, temperature γ = ρ ∆t, t, , 6., , 7., , The liquid in the container heated first the level of Ø, liquid falls because heat is absorbed by the container, and then heat is absorbed by the liquid, liquid Ø, expands more than solids., The liquid in the container heated first the level of Ø, liquid falls because heat is absorbed by the container, , NARAYANAGROUP, , Gases have no definite shape and volume. The, Gases completely occupies the entire volume of the, vessel in which it is filled., The state of given mass of gas can be described in, terms of three parameters called pressure, volume, and temperature., Keeping one parameter constant, the relation, between other two can be established. Hence there, are three gas laws., Statement : At constant temperature, the, volume of given mass of a gas is inversely, proportional to its pressure., Explanation : i) Let P and V be the pressure and, volume of given mass of a gas at constant, temperature. According to Boyle's law, V α 1 ( at, P, 1, const. temp.) ⇒ V = K or PV = K (constant), P, K depends on mass of the gas & the constant, temperature at which the gas is kept., For a given mass of a gas and at a given temperature., PV, 1 1 = P2V 2, The shape of the graph plotted between pressure, (P) of given mass of a gas and its volume (V) at, constant temperature is a rectangular hyperbola. It, is also called as 'isotherm', Temperature is constant, T=constant, P, The curve is isothermal curve, , V, , Boyle's law generally holds good only at low, pressure and high temperatures., A gas which obey Boyle's law under all conditions, of temperature and pressure is called ideal gas., Real gases obey gas laws only at low, pressure and high temperatures., 41
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, Ø, Ø, Ø, Ø, Ø, Ø, , All Gases are real gases only., Attraction between the molecules of perfect gas is, zero., Ideal or perfect gas obeys gas laws at all, temperatures and pressures without any, limitations., Hydrogen or Helium behaves closely as, perfect gas. Hence they are preferred in constant, volume gas thermometers., The graphs drawn between P & V at constant, temperature of a gas are called isotherms, PV-P graph is a horizontal straight line parallel to, pressure axis., a), PV, , b), , P, , Ø, , P−, , 1, graph is a straight line passing through origin, V, P, , Ø, , i), , 1, V, , Two vessels of volumes V1 and V2 containing a gas, under pressures P 1 and P 2 respectively are joined, at the same temperature. Then the common pressure, , PV + PV, 2 2, P= 1 1, PV = PV, 1 1 + PV, 2 2 ⇒, V1 + V2, Ø, , If second vessel is an evacuated then, P(V1 + V2 ) = PV, 1 1 ⇒, , Ø, , PV, P= 1 1, V1 + V2, , Q P2 = 0, , Boyle's law in terms of density :, , \, , P1 r1, =, P2 r2, P, , θ, , P0A + mg = P1A ; i.e., H r gA + Ah r g = P1A, (or) P 1 = (H + h) r g \ P1 = (H + h) cm of Hg, , p0, , Consider a gas of mass m, pressure P, volume V, m, and density r then V = .But From Boyle's law, m r, PV = constant or P = constant, r, , Since mass of the gas is constant, P a r ., ∴ At constant temperature, for a given mass of, gas, pressure is directly proportional to its density, , 42, , ρ, , A graph plotted between pressure and density at, constant temperature is a straight line passing, through origin., Quill tube is used to verify Boyle's law. A quill tube, is a capillary tube with narrow uniform bore and, one end is closed. A mercury pellet is introduced, into the capillary tube such that an air column is, trapped between the mercury pellet and closed end., In quill tube, Pl = constant. (P is pressure and l is, length of air column)., In quill tube experiment, PV = constant (or), PA l = constant \ P l = constant., (Q A , area of cross section of the bore is constant), Where P is pressure of enclosed air and l is length, of enclosed air column. If H is atmospheric pressure, (mercury barometer) and h is length of mercury, column, pressure P in different cases is as shown in, the figure., When the open end of the tube is vertically, upwards, then, Volume of the enclosed air = A l 1, and Pressure of the air = P 1, From free body diagram of mercury pellet, , h, p1, , l1, , ii) When the open end of the, tube is vertically downwards,, then, Volume of the enclosed air,, V2 = Al2 and Pressure of the, air = P2, , For equilibrium of mercury pellet,P0A=mg+P2A, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, mg, (or), A, hAr g, P2 = H r g A, (or) P 2 = (H – h) r g, , i.e., P2 = Po -, , l2, p2, , h, p0, , \ P2 = (H – h) cm of Hg, iii) When the tube is horizontal, volume of the, enclosed air, V 3 = Al3 Pressure of the air = P 3, i.e., P 3A = P 0A Þ P3 = P 0 \ P3 = H cm of Hg, , p0, , p3, , h, , l3, , iv) When the tube is making an angle q with vertical,, with open end upwards, then, Volume of enclosed air = A l4, and pressure of the air = P 4, For equilibrium of mercury pellet, we have, Ahrg cos q, P4A = P0A + mg cos q (or) P4 = P 0 +, A, , p0, , N, , θ, , mg sinθ, , l4, , p, , 4, , mg cosθ, , P4 = H r g + h r g cos q = (H+hcos q )cm of Hg, c) According to Boyle's law, P l = constant, (H + h) l 1=(H–h) l 2 =H l 3 =(H+h cos θ ) l 4, d), , If l1 is length of the enclosed air column in the quill, tube with its open end vertically upwards and l2, with its open end downwards, then atmospheric, , THERMAL PROPERTIES OF MATTER - II, d) h – 1/ l Graph : From the graph between h and, 1, , H can be calculated., l, 1/l, , h, , H, , The graph drawn between ‘h’ (excess pressure), and 1/ l (where l is the length of the air column) is, a straight line making a negative intercept on ‘h’, axis whose magnitude gives atmospheric pressure., , Motion of An Air Bubble In A Liquid, When an air bubble rises from bottom to surface of, a lake, its volume increases. If V1 and V 2 are the, volumes of air bubble at the bottom and at the top, of the lake and temperature is assumed to be, constant, then according to Boyle's law, P1V1 = P2V2, , ; (H + h)V1 = HV2, , Where h is depth of the lake and H is atmospheric, pressure on water barometer. (i.e., nearly 10m of, water column height), a) If V2 = nV1 then (H + h) V1 = H(nV1), æ, hö, H+h = Hn, \ h = (n – 1) H and n = çççè1 + ø÷÷÷, H, , b) if r1 and r2 are the radii of the bubble at the, bottom and at the top and if r 2 = nr1,, then (H + h)V1 = HV2, i.e., (H+h), , 4 3, 4, 3, p r1 = H p (nr1 ), 3, 3, , (or) (H + h) r13 = Hn3 r13 \ H + h = Hn3, 1, , Þ h=, , (n3, , æ, h ö3, – 1) H and n = ççç1 + ÷÷÷, è Hø, , pressure H can be calculated as P1l1 = P2l 2 ; i.e.,, , ( H + h) l 1 = ( H - h) l 2, , i.e., H = h ( l1 + l2 ), ( l2 − l1 ), NARAYANAGROUP, , V2, h, V1, , 43
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, Faulty barometer :, , air, , 2, 1, P2 V1 = PV, 2 3, 3, 2, , A, , l, , WE-2. The volume of an air bubble increases by, x% as it raises from the bottom of a water lake, to its surface. If the water barometer reads H,, the depth of the lake is, Sol: P1V1 = P2V2; (H + h) dgV1 = HdgV2, HV, ( H + h )V1 = HV2 ; H + h = 2, V1, , L, , H, , 50, 1, P2 = P2, 100, 2, 4, ∴V3 = V1, 3, , Case 2 : P2V2 = P3V3 ; P3 =, , B, , V −V , ∆V , h = H 2 1 ; h = H , , V , V1 , Let H t be the true atmospheric pressure in terms, of mercury height. Let H f be the faulty reading, given by the barometer. If l is length of air column, trapped, l = ( L − H f ) ., Pressure of air trapped = ( H t − H f ) = P, As temperature is constant. We can use Boyle’s, law. i.e. Pl =constant., ⇒ ( Ht − H f, , )( L − H ) =constant, f, , Then the above equation can be represented as, , ( H − H )( L − H ), = ( H − H )( L − H ), 1, 1, , 1, , 2, , 1, 1, , 1, 2, , 1, 2, , Where H1 , H 2 = true values, H11 , H 21 = faulty values, WE-1. A given mass of ideal gas has volume V at, pressure P and room temperature T. Its pressure, is first increased by 50% and then decreased, by 50% (both at constant temperature). The, volume becomes, Sol: PV = constant ;, 150 × P1 3, = P1, Case 1: P1V1 = P2V2 ; P2 =, 100, 2, 3, 2, PV, PV, ∴V2 = V1, 1 1 =, 1 2, 2, 3, 44, , Here, , ∆V, x, ×100 =, V, 100, , ∴h =, , Hx, 100, , WE-3. The density of an air bubble decreases by, x% as it raises from the bottom of a lake to its, surface. The water barometer reads H. The, depth of the lake is, Sol: P1V1=P2V2, V, , d, , d −d , h = H 2 − 1 = h 1 − 1 = H 1 2 , V1 , d2 , d2 , , Q d1 − d 2 = x ;, ∆d, × 100 = x ;, d, , d 2 = d1 − x = 100 − x, x , h=H, , 100 − x , , WE-4. An ideal gas is trapped between a mercury, column and the closed lower end of a narrow, vertical tube of uniform bore. The upper end, of the tube is open to the atmosphere., (Atmospheric pressure is 76cm of mercury)., The length of the mercury and the trapped gas, columns are 20cm and 43cm, respectively., What will be the length of the gas column, when the tube is tilted slowly in a vertical plane, through an angle of 60 0. Assume the, temperature to be constant., Sol. Boyle’s law holds good because the temperature is, constant so, P 1V1=P2V2, P1Al1 = P 2Al2 or P 1l1=P2l2, (A is constant), since the bore is uniform, P1 = 76 + 20 = 96 cm of Hg; l1= 43cm;, P2 = 76 + hcos θ = 76+20 cos 600, = 76+10=86 cm of Hg ; l2= ?, i.e 96 × 43=86l2 → l2 =48cm of Hg, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , WE-5. A column of Hg of 10cm length is contained, in the middle of a narrow horizontal 1m long, tube which is closed at both ends. Both the, halves of the tube contain air at a pressure 76, cm of Hg. By what distance will the column, of Hg be displaced, if the tube is held vertical, ? (Assume temperature to be constant), , B, , Sol., , L+Y, , A, L, , L-Y, , L, , Patm (76 cm Of Hg), , Hg, , x, , 100 cm, 100-x, , Sol : When mercury is poured on the top of the piston,, due to increase in pressure, the volume of air will, decrease according to Boyle's law. If final mercury, column of height x is poured on the piston then gas, pressure in equilibrium can be given as, Pf = ( 76 + x ) cm, , 10cm, , P atm(76 cm Of Hg), , of Hg, , If initially the length of air column on each, side is L, according to the given problem,, 2L+10=100,i.e., L=45 cm......(1), Now if the tube is held vertical, the Hg, column will be displaced downward by y, such that, PB+10=PA..........(2), applying Boyle’s law to air enclosed inside A,, , As atmospheric pressure is equivalent to the, pressure due to a mercury column of height 76cm., If A be the area of cross section of cylinder then, according to Boyle's law P1V1 = P2V2 o r, (76)(100 A) = (76 + x )(100 - x ) A, , LP0, P0LA=PA(L–y)A, i.e., PA = ( L − y ) .......(3), , WE-7. A gas is enclosed in a vessel of volume V at, a pressure P. It is being pumped out of the, vessel by means of a piston-pump with a stroke, volume ν . What is the final pressure in the, vessel after 'n' strokes of the pump ? Assume, temperature remains constant., , While for air enclosed inside B,, , P0 LA = PB ( L + y ) A, i.e., PB =, , LP0, ( L + y ) .......(4), , Substituting the values of P A and PB from, equation (3) and (4) in (2), with L = 45 and, P0 = 76 cm of Hg, we get, 45 × 76, 45 × 76, 2, −, = 10; y 2 + 684 y − ( 45 ) = 0, 45, −, y, 45, +, y, (, ) (, ), , or y =, , −684 ±, , , (684 ), , 2, , 2, + 4 (45 ) , , , 2, , or y = −342 + 345 ; 3cm, WE-6. A vertical cylinder of height 100cm contains, air at a constant temperature and its top is, closed by a frictionless piston at atmospheric, pressure (76cm of Hg) as shown figure (a). If, mercury is slowly poured on the piston, due to, its weight air is compressed. Find the, maximum height of the mercury column, which can be put on the piston., NARAYANAGROUP, , or 7600 = 7600 + 24x − x 2 or x = 24 cm., , Sol. According to ideal gas equation PV=nRT,at, constant temperature for a given mass, PV= P 1V1, Now as stroke volume is ν during 1st stroke for, constant mass (say m) volume changes from V to, ( V + ν ) and so if pressure changes from P to P 1,, the above equation yields, V , , PV = P 1(V+ ν ), i.e., P1 = P V + ν ......... (1), , , , , After the first stroke, the gas left in the vessel has, again volume V but at pressure P1 (with mass, m1<m). Now the second stroke will take place from, these initial conditions and if P 2 is the pressure of, the gas in the cylinder at the end of 2nd stroke,, , , P1V = P2(V + ν ), i.e., P2 = P1 V + ν , V, , , , , , 45
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , V0, Vt, =, 273.15 273.15 + t, , pressure.Then, , æ, ö, 1, \ Vt = V0 çç1 +, t÷, çè 273.15 ø÷÷, , V, , V t = V 0 (1 + a t ) Also , α =, , Substituting the value of P 1 from equation (1) in the, V , P2 = P , , V + ν, , above, , If V1 & V2 are the volumes of the given mass of the, gas at temperatures t 1 & t 2 respectively, , 2, , Repeating the same for n strokes, the pressure of Ø, the gas in the vessel after nth stroke will be,, , , 1, V , Pn = P , = P, , , V + ν, 1 + ( ν / V) , n, , æ m1 m 2 ö÷, çç +, ÷, çè P, P ÷÷ø, 1, , =, , Graphs between volume and temperature, of a gas in different cases:, At constant pressure, the density of a given mass, of a gas is inversely proportional to its temperature, on Kelvin scale. i.e., d ∝, , of a given mass of gas changes from d1 to d 2 so, that d1T1 = d 2T2 ., D, v, , a), A, 0, -273, , P1P2 ( m1 + m 2 ), (P2 m1 + m 2 P1 ), , a gas at 00 C and t 0C respectively at constant, 46, , tC, , OB CD, =, 273 BC, , Slope =, , ∆v, ∆T, , T, v, , The volume of a given mass of gas at, constant pressure is directly proportional to the, absolute temperature. V αT (at constant pressure), , Let V0 and Vt be the volumes of the given mass of, , 0, , O, , θ, , Charles constant pressure law, , Ø, , tan θ =, , θ, , b), , 2, , V V, ⇒ 1= 2, T1 T2, , θ C, v0, , B, , v, , Charles Law:, Ø, , 1, (or) dT = constant., T, , If temperature changes from T1 to T2 , the density, , 2, , (m1 + m 2 ), , 1, / 0 C =0.00366/0C, 273.15, , Here a is called volume coefficient of the gas., , ., , constant, the common pressure reached will be, P, = const., Sol. According to Boyle's law, r, km 2, \ V1 = m1 = km1 and V2 =, P2, r1, P1 æ, ö, m, m, ÷, 2, ç 1, Total volume = k çç P + P ÷÷÷, è 1, 2 ø, Let mixture has common pressure P and common, density ρ ., ( m1 + m 2 ), r=, æm, m ö, k çç 1 + 2 ÷÷÷, çè P, P ÷ø, Þ P = kr =, , V −V, , 0, 2, 1, The volume coefficient of gas is α = V t − V t / C, 1 2, 2 1, , α=, , n, , WE-8.Two chambers, one containing m 1g of a gas, at P1pressure and other containing m2g of a, gas at P2 pressure are put in communication Ø, with each other. If temperature remains, , 1, , Vt − V0, V0 t, , p2, p1, , c), , p1 > p2, , θ1 θ2, , T, , Charles law of constant volume:, Ø, , The pressure of a given mass of a gasses is directly, proportional to the absolute temperature at constant, volume. This law is also known as, Gay - Lussac's law. PαT (at constant volume), NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, ⇒, , Ø, , THERMAL PROPERTIES OF MATTER - II, Ø, , P1 P2, =, T1 T2, , Let P 0 and P t be the pressures of given mass of a Ø, gas at 00C and t 0C respectively at constant volume., Ø, P0, Pt, =, Then, 273.15 273.15 + t, 1, , , ∴ Pt = P0 1 +, t, 273.15 , , , Real Gas, , Pt − P0, Pt = P0 (1 + β t ) Also β =, P0t, , Ø, , If P1 & P2 are the pressures of the given mass of a, gas at temperatures t 1 & t2 respectively then the Ø, Ø, pressure coefficient of gas is, β=, , P2 − P1, 1, =, / 0 C =0.00366/0C, Pt, 273.15, 1 2 − p2 t1, , Ø, , Here β is called pressure coefficient of the gas, , Graphs between pressure and, temperature of a gas in different cases:, Ø, , If two vessels of same volume containing a gas at, T1K and T2 K temperatures and at pressures, , P1 and P2 are connected by a narrow tube then, the common pressure, , P=, , P1T2 + P2T1, T1 + T2 ., , D, P, , a), A, 0, -273, , 0, , tC, , tan θ =, , OB CD, =, 273 BC, , Slope =, O, , θ, , ∆P, ∆T, , PV, T, , per unit mass of given gas is called, , PV, = r Þ PV = rT, T, For a mass m of the gas PV = mrT .... (1), The value of 'r' depends on the nature of the gas, i.e., its value is different for different gases . S.I., unit of 'r' is J kg–1 K–1., , Dimensional formula - M 0 L2T −2θ −1 , , T, , P, , When one mole of the gas is considered, the ratio, , c), , V1, θ1, , V1>V 2, , θ2, , T, , Ideal Gas, , Ø, , The ratio, , i.e.,, Ø, , P, b), , The gas which obeys Boyle's law at low, pressure and high temperature is called a real, gas., Real gas has finite size molecules., Long before reaching the absolute zero, the gas, converts into liquid., Total internal energy (u) = ∑ P.E + ∑ K.E of all the, molecules., Since intermolecular forces are not absent, P.E of, the molecules of a real gas is not zero., Ideal Gas (Or Perfect Gas) Equation :, An ideal gas of mass ‘m’ having pressure ‘p’ volume, PV, = constant, ‘V’, at temperature ‘T’ ∴, T, PV, The ratio, depends on both nature and mass, T, of the gas., , 'specific gas constant' and is represented by 'r'., , θ, O, , Ø, , Ø, , θ C, P0, , B, , Permanent gases such as hydrogen, air, helium,, oxygen and nitrogen obey the laws with sufficient, accuracy at low pressures and high temperatures., At absolute zero, an ideal gas remains in gaseous, state., Total internal energy (u) of an ideal gas, = ∑ K.E of all the gas molecules., Since intermolecular forces are absent among the, ideal gas molecules, P.E of the molecules becomes, zero., , A gas which obeys Boyle's law and Charles', laws strictly at all temperatures and pressures, is called a perfect or an ideal gas., The ideal gas has point size molecules, , NARAYANAGROUP, , PV, T, , is the same for all gases and is called, , Universal gas constant 'R'., i.e.,, , PV, T, , = R Þ PV = RT, , For 'n' moles of a gas, the ideal gas equation is, PV = nRT, 47
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, Ø, Ø, , S.I. unit of R is J mol–1 K–1 and dimensional, Ø, 2, –2, –1, –1, formula of R is ML T K mol ., When a gas of mass 'm' and molecular weight 'M' Ø, is considered, the number of moles of the gas is, given by n = m/M ; substituting in PV=nRT,, we get PV =, , Ø, , Ø, , m, RT .... (2), M, , From (1) and (2) r = R/M, The ideal gas equation in terms of mass of the gas Ø, and specific gas constant is PV = m r T, , When pressure and volume are constant for a given, PV =, , m, RT, M, , m µ, , R, Ø, i.e., K = N where NA is Avogadro's number of, , If a gas with physical parameters (P1,V1,T1) is mixed, with another gas with parameters (P2,V2,T2) and if, resultant mixture is with parameters (P,V,T) then, , Total number of molecules ( N ), Avagadro number of molecules ( N A ), , \ PV = N, , N, , RT, , A, , (or) PV = NKT, , Ø, , Calculation of Universal gas constant :, Universal gas constant is the gas constant for one, mole of a gas. It is same for all gases, since at the, same temperature and pressure, one mole of any, gas occupies the same volume., Ø, Experimentally, it was found that, 1 mole of any, gas occupies a volume of 22.4 litres at N.T.P., At N.T.P, Normal pressure,, P = hdg = 76 × 13.6 × 980 dyne cm-2, Normal temperature, T = 273K, Volume of 1 gram mole of gas = 22400 c.c, Universal gas constant, R = PV, T, , 76 × 13.6 × 980 × 22400, =, 273, , T, 1, K m, , m= , 1 = 2, T, T m 2 T1, , PV, PV, PV, 1 1, + 2 2 =, T1, T2, T, , Where N = Number of molecules present in the, gas, K = Boltzmann's constant,, where n = no.of moles of gas,, S.I. unit of R is J mole -1 K-1, , 48, , PV, PV, 1 1, 2 2, ; T = T, 1, 2, , Boltzmann's constant (K) is defined as universal, gas constant per molecule., , n=, , Ø, , P1, P, = 2, d1T1 d 2T2, , ideal gas, , A, , Ø, , ⇒, , Ideal gas equation in terms of Boltzmann's, constant :, , molecules . We know, PV = n R T and, , Ø, , R=N 0K Where K=Boltzmann's constant, N0 = Avogadro's number, The gas equation in terms of density ‘d’, for a given, P, mass of a gas is, = constant., dT, , An air bubble rises from bottom of a lake to the, top. If V1 and V2 are the volumes of air bubble,, , T1 and T2 are the temperatures at bottom and top, PV, PV, 1 1, 2 2, of the lake then T = T, 1, 2, , ( H + h ) V1, T1, , =, , HV2, ( H + h ) r13 = Hr23, (or), ., T2, T1, T2, , At the top of mountain a thermometer reads T1K, temperature and barometer reads H1 cm of Hg, pressure. At the bottom of mountain they read T2 K, temperature and H2 cm of Hg pressure, respectively. If d1 and d 2 are densities of air at the, top and bottom of a mountain, then, P1, P, H, H, d, HT, = 2 ; 1 = 2 (or) 1 = 1 2 ., d1T1 d 2T2 d1T1 d 2T2, d 2 H 2 T1, , R = 8.314 × 107 erg mole -1 K-1, WE-9. P–V diagrams of same mass of a gas are, -1, -1, -1, -1, -1 -1, =8.314 J g mole K = 8314 J kg mole K, drawn at two different temperatures T1 and, =1.987cal mole-1K-1 =0.0821 lt atm mole-1K-1, T2. Explain whether T1 > T2 or T2 > T1, Significance of R:, Sol: Keeping the pressure of the gas remains constant., The value of "R" does not depend on the mass of, According to Charles’s law V ∝ T, gas or its chemical formula., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , P, , T2, T1, V1 V2, , PV, PV, 1 1, = 2 2, T1, T2, , Now, , P = constant, , ⇒, , ( 76×13.6 + 250) 980 × , , 4, 3, π ( 0.18), 3, , T1, , V, , In the graph is pressure kept constant and volumes, are compared, =, , From Charles’s law at constant pressureV ∝ T, From the graph V2 > V1 so T 2 > T 1, , ( 76 ×13.6 ) × 980 , , 4, 3, π ( 0.2), 3, , (or), , 313, , T1 = 283.37 K ; T1 = 283.37 − 273 = 10.370 C, , WE-10. 4g of hydrogen is mixed with 11.2 litre of WE-12. A faulty barometer tube is 90cm long and, He at STP in a container of volume 20 litre. If, it contains some air above mercury. The, the final temperature is 300 K. Find the, reading is 74.5cm when the true atmospheric, pressure 76cm. What will be the ture, pressure., atmospheric pressure if the reading on this, Sol: 4g hydrogen = 2 moles hydrogen, barometer is 74cm? (H = 10cm of water, 1, column), 11.2 litre He at STP = mole of He, 2, P = PH + PHe = ( nH + nHe ), , RT, V, , 1 8.31 × ( 300 ), , = 2+ , 5, 2, 2 ( 20 × 10 − 3 ) = 3.12 × 10 N / m, , , WE-11. An air bubble starts rising from the bottom, of a lake. Its diameter is 3.6 mm at the bottom, and 4mm at the surface. The depth of the lake, is 250 cm and the temperature at the surface, is 400C. What is the temperature at the, bottom of the lake? Given atmospheric, pressure = 76cm of Hg and g = 980cm/s2, Sol: At the bottom of the lake, volume of the bubble, V1 =, , 4, 3, π ( 0.18 ) cm 3, 3, , Pressure on the bubble P 1 = Atmospheric, pressure + Pressure due to a column of 250cm of, water = ( 76 ×13.6 + 250 ) 980 dyne/cm2, At the surface of the lake, volume of the, 4, 3, V2 = π ( 0.2 ) cm3, bubble, 3, Pressure on the bubble ; P 2 = atm. pressure, , = ( 76 ×13.6 × 980 ) dyne/cm2, , T2 = 273 + 40 0 C = 313 K, NARAYANAGROUP, , Sol., , 90cm, , 15.5cm, , 16.0cm, , 74.5cm, , 74.0cm, , Let the area of cross-section of the tube be A cm 2, and true pressure be H cm of mercury. Since the, temperature is constant., Boyle’s law can be applied, to the air enclosed in the upper part of the, barometer tube, thus, , P1 = ( 76.0 − 74.5 ) = 1.5cm of mercury, V1 = A × ( 90 − 74.5 ) = A ×15.5cm3, P2 = ( H − 74.0 ) cm of mercury, V2 = A × ( 90 − 74.0 ) = A ×16cm3, Applying Boyle’s law PV, 1 1 = PV, 2 2, , 1.5 × ( A ×15.5 ) = ( H − 74 ) × ( A ×16 ), H − 74 =, , 1.5 ×15.5, 16, , H = 75.45cm, 49
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THERMAL PROPERTIES OF MATTER - II, , JEE- ADV PHYSICS-VOL- V, , 4) increase or decrease depending on the chemical, composition of gas, m, 4. At constant pressure density of a gas is, ρ = . From the ideal gas equation ,, V, 1) directly proportional to absolute temperature, 2) inversely proportional to absolute temperature, PM, ∴ρ=, 3) independent of temperature, RT, 4) directly proportional to square root of absolute, Ø From this equation we can see that ρ – P graph, temperature, is straight line passing through origin at, 5. The slope of T-P graph for a given mass of a, constant temperature, gas increases, the volume of the gas, 1) increases, 2) decreases, Ø ( ρ ∝ P ) for a given gas and ρ − T graph is, 3) does not change 4) may increase or decrease, rectangular hyperbola at constant pressure 6. Which of the following methods will enable the, volume of an ideal gas to be made four times, 1, , ρ ∝ ., 1) double the absolute temperature and pressure, T, , 2) halve the absolute temperature and double the, Ø Similarly for a given mass of a gas ρ − V graph, pressure., 3) quadruple the absolute temperature at constant, 1, , pressure, is a rectangular hyperbola ρ ∝ , V, 4) quarter the absolute temperature at constant, , , pressure, ρ, ρ, ρ, 7. An ideal gas is that which, 1) cannot be liquefied 2) can be easily liquefied, P= constant, 3) has strong inter molecular forces, m= constant, m=constant, 4) has a large size of molecules., T= constant, 8., In a gas equation, PV = RT, V refers to the, m=constant, volume of, V, P, T, 1) any amount of a gas, 2)1gram mass of a gas, 3) 1 gram mole of a gas 4) 1litre of a gas, C.U.Q, 9. For a constant volume gas thermometer one, should fill the gas at, 1. When the volume of a gas is decreased at, 1) high temperature and high pressure, constant temperature the pressure increases, 2) high temperature and low pressure, because the molecules, 3) low temperature and low pressure, 1) strike unit area of the walls of the container more, 4) low temperature and high pressure, often., 10. The molar gas constant is the same for all, gases because at the same temperature and, 2) strike the unit area of the walls of the container, pressure, equal volumes of gases have the, with higher speed, same, 3) strike the unit area of the wall of the container, 1) number of molecules, with lesser speed., 2) average potential energy, 4) move with more kinetic energy, 3) ratio of specific heats 4) density, 2. Boyle's law is represented by the equation, PV=K (K is not constant), K depends on 11. A box contains x molecules of a gas. How, will the pressure of the gas be effected if the, 1) pressure of the gas, 2) volume of the gas, number of molecules is made 2x?, 3) mass of the gas, 4) all the above, 1) Pressure will decrease., 3. A closed vessel contains some gas at, 2) Pressure will remain unchanged., atmospheric pressure and room temperature., It is then given a high speed by placing it in a, 3) Pressure will be doubled., fast moving train. The temperature of the gas, 4) Pressure will become three times, 1) will increase, 2) will decrease, 3) will remain unchanged., , Expression of density for different cases :, , 50, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 12. According to Charles’s law,, 1) at constant pressure, volume of gas is, proportional to its absolute temperature., 2) at constant pressure, the volume of a gas is not, proportional to its absolute temperature., 3) at constant gauge pressure, the molecular, volume of a gas is proportional to its absolute, temperature., 4) at constant volume, the absolute pressure is, proportional to absolute temperature., 13. The density of an ideal gas, 1) is directly proportional to its pressure and, absolute temperature, 2) is directly proportional to its pressure and, inversely proportional to its absolute temperature, 3) is inversely proportional to its pressure and, directly proportional to its absolute temperature, 4) is inversely proportional to both its pressure, and absolute temperature of the gas, 14. The relation between volume V, pressure P, and absolute temperature T of an ideal gas, is PV = xT, where x is a constant. The value, of x depend upon, 1) the mass of the gas molecule, 2) the average kinetic energy of the gas molecules, 3) P, V and T, 4) the number of gas molecules in volume V., 15. The air of the atmosphere becomes cool at, higher altitudes due to, 1) decrease in density 2) variation in pressure, 3) expansion of the air, 4) height above the surface of the earth, 16. If pressure and temperature of an ideal gas, are doubled and volume is halved, the number, of molecules of the gas, 1) becomes half, 2) becomes two times, 3) becomes 4 times, 4) remains constant, 17. If gas molecules undergo, inelastic collision, with the walls of the container, 1) temperature of the gas will increase, 2) temperature of the gas will fall, 3) pressure of the gas will increase, 4) neither temperature nor the pressure change, 18. A gas in an airtight container is heated from, 25oC to 90oC. The density of gas will, 1) increase slightly 2) increase considerably, 3) remain the same 4) decrease slightly, NARAYANAGROUP, , THERMAL PROPERTIES OF MATTER - II, 19. A volume V and temperature T was obtained,, as shown in diagram, when a given mass of, gas was heated. During the heating process, the pressure is, B, V, , 1) increased, 2) decreased, , A, , 3) remains constant, O, , T, , 4) changed erratically, 20. A P-V diagram is obtained by changing the, temperature of the gas as shown. During this, process the gas is, A, , P, , 1) heated continuously, B, , 2)cooled continuously, , V, , O, , 3) heated in the beginning but cooled towards the, end, 4) cooled in the beginning but heated towards the, end, 21. The critical temperature of the gas is the, temperature, 1) at which Charles’s law is obeyed, 2) at which Boyle’s law is obeyed, 3) above which the gas cannot be liquefied, 4) at which all molecular motion ceases, 22. The P-T graph for the given mass of an ideal, gas is shown in figure. Then the volume, B, , 1) increases, , P, , 2) decreases, , A, , 3) remains constant, O, , T, , 4) data insufficient, , C.U.Q. - KEY, 1) 1, 7) 1, 13) 2, 19) 1, , 2) 3, 8) 3, 14) 4, 20) 3, , 3) 3, 9) 2, 15) 3, 21) 1, , 4) 2 5) 2 6) 3, 10) 1 11) 3 12) 1, 16) 2 17) 4 18) 3, 22) 1, 51
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , A gas is kept at 130C in a vessel, If the volume, LEVEL-I (C.W.), of the gas is kept constant and is heated, the, pressure will be doubled to its initial pressure, GAS LAWS, at a temperature, 1) 576 K, 2) 286 K, 3) 143 K 4) 73 K, A vessel containing 10 litre of air under a, 10., State, the, equation, corresponding, to 8g of O 2, pressure of 1MPa is connected to a 4 litre, is, empty vessel. The final air pressure in the, vessel assuming that the process is isothermal., 1) PV=8RT, 2)PV=RT/4, 1) 7/5 MPa 2) 5/7MPa 3) 1MPa 4) 10MPa, 3)PV=RT, 4)PV=RT/2, Two vessels of volume 10 and 5 litres contain, IDEAL GAS EQUATION, air at 5 atmospheres and x (Unknown) 11. A given amount of gas is heated until both its, atmospheres. When they are connected, pressure and volume are doubled. If initial, together with a small tube the resultant, temperature is 270 C, its final temperature is, pressure is '6' atmospheres find the value of, 1) 300 K 2) 600 K 3) 1200 K 4) 900K, 'x', 12. At. N.T.P. 28 g of Nitrogen occupies 22.4 litres., 1)8 atm 2) 16 atm 3) 4 atm, 4) 2 atm, What is the mass of 5.6 litres of, An air bubble rises from the bottom of a lake, Nitrogen at 38cm of Hg pressure and 2730C, and its radius is doubled on reaching the, temperature, surface. If the temperature is constant the, 1) 7 g, 2) 48 g 3) 1.75 g, 4) 1.5 g, depth of the lake is. (1 atmospheric pressure 13. A vessel of volume 4 litres contains a mixture, = 10m height of water column), of 8g of O2, 14 g of N2 and 22 g of CO2 at, 1) 7m, 2) 70m 3) 10m, 4) 0.7m, 270C .The pressure exerted by the mixture is, If an air bubbles rises from the bottom of a, 1) 10 atmosphere, 2) 5 × 106 N/m2, 1, 4) 6 × 105 N/m2, 3) 7.69 × 105 N/m2, mercury tank to the top its volume become 1, 9., , 1., , 2., , 3., , 4., , 2, , 5., , 6., , 7., , 8., , 52, , times. When normal pressure is 76 cm of Hg, then the depth of the Hg tank is, 1) 38 cm 2) 132 cm, 3) 76 cm 4) 49 cm, A quill tube contains a mercury column of, length 19cm. The length of air column is 24cm, when it is held vertically. On inverting it with, its open end downwards the length of air, column will be, (atmospheric pressure = 76cm of Hg), 1) 20cm 2) 30cm 3) 40cm 4) 35cm, At what temperature will the volume of a gas, be twice the volume at 270 C at a given, pressure., 1) 3270 C, 2) 540 C 3) 1270 C 4) 1000C, If the temperature of a gas is increased by 1K, at constant pressure its volume increases by, 0.0035 of the initial volume. The temperature, of the gas is, 1) 100K 2) 150K 3) 300K 4) 285.7K, A cylinder contains a gas at temperture of 270C, and a pressure 1MPa. If the temperature of, the gas is lowered to -230C, the change in, pressure is, 1) 1MPa 2) 5/6MPa 3) 1/6MPa 4) 5MPa, , LEVEL - I (C.W) - KEY, 1) 2 2) 1, 7) 4 8) 3, 13) 3, , 3) 2, 9) 1, , 4) 1 5) 3 6) 1, 10) 2 11) 3 12) 3, , LEVEL - I (C.W) -HINTS, 1., 2., , PV = P1V1 + P2V2 ; P2 = 0, PV = P1V1 + P2V2, , 3., , P1V1 = P2V2, , ; ( H + h) r13 = Hr23, , 4., , P1V1 = P2V2, , ; ( H + h)V1 = HV2, , 5., , ( H + h)l1 = ( H − h)l2, , 7., , V ∝T ;, , 8., , P1 T1, ∆T ∆P, =, =, P2 T2 ; T, P, , 9., , P1 T1, =, ;, P∝T, P2 T2, , V1 T1, 6. V ∝ T ; V = T, 2, 2, , ∆V ∆T, =, V, T, , 10. PV =, , m, RT, M, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, PV, PV, 1 1, 2 2, 11. T = T, 1, 2, , m, 12. PV = RT , PV ∝ mT, M, , m, 13. PV = RT ;, M, , RT, P=, V, , 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , m1 m2 m3 , +, +, , , M1 M2 M 3 , , THERMAL PROPERTIES OF MATTER - II, 9., , State the equation corresponding to 4g of N 2is, 1) PV = 8RT, 2) PV = RT/7, 3) PV = RT, 4)PV = RT/2, , IDEAL GAS EQUATION, , 10. A gas at temperature 27 0 C and pressure 30, atmosphere is allowed to expand to one, atmospheric pressure. If the volume becomes, LEVEL - I (H.W), 10 times its initial volume, the final, temperature becomes, GAS LAWS, 1) 1000 C 2) 3730 K 3) 3730 C 4) − 1730 C, If a given mass of a gas occupies a volume, 100cc at one atmospheric pressure and a 11. 16 g of O gas and x g of H gas occupy the, 2, 2, temperature of 1000C. What will be its volume, same volume at the same temperatrue and, at 4 atmospheric pressure, the temperature, pressure. Then x =, being the same?, 1)1/2g, 2)1g, 3) 8g, 4) 16 g, 1) 100cm3 2) 400cm3 3) 25cm3 4) 200cm3 12. An enclosure of volume 3 litre contains 16 g of, A vessel containing 9 litres of an ideal gas at, oxygen, 7 g of nitrogen and 11 g of carbon 760 mm pressure is connected to an evacuated, di-oxide at 27°C . The pressure exerted by the, 9 litre vessel. The resultant pressure is, mixture is approximately, 1) 380mm 2) 760mm 3) 190mm 4) 1140mm, [R = 0.0821 lit atm mole-1 K-1], A bubble rises from the bottom of a lake 90m, 1) 1 atmosphere, 2) 3 atmosphere, deep on reaching the surface, its volume, 3) 9 atmosphere, 4) 8.3 atmosphere, becomes (take atmospheric pressure equals, LEVEL - I (H.W) - KEY, to 10 m of water ), 1), 3, 2) 1 3) 3 4) 2 5) 2 6) 1, 1)4 times 2)8 times 3)10 times 4) 3 times, 7) 2 8) 3 9) 2 10) 4 11) 2 12) 4, An air bubble rises from the bottom to the, surface of lake and it is found that its diameter, LEVEL - I (H.W) -HINTS, is doubled. If the height of water barometer is, 11m, the depth of the lake in meters is, 2.PV = P1V1 + P2V2, 1. P ∝ 1, 1) 70m 2) 77m, 3) 7.7m 4) 78m, V, The temperature of a gas contain in a closed 3. (H + h)V = HV, 4. ( H + h) r13 = Hr23, 0, vessel increased by 2 C when the pressure is, increased by 2% the initial temperature of the, V1 T1, ∆P ∆T, =, 5., ,, 6., ,, V ∝T V = T, P∝T, gas is, P, T, 2, 2, 1)200K, 2)100K 3) 2000 C 4) 1100 C, P T, The volume that a gas occupies at 343K if its 7. P ∝ T , P1 = T1, P ∝T, 1 = 1, 8., P2 T2, P2 T2, volume at -250 C is 7.5 litre is (The process is, isobaric), PV, PV, m, 1 1, 2 2, 1) 10.29 lit 2) 102.9 lit 3) 1.029 lit 4) 1029 lit, RT, 9. PV =, 10. T = T, M, 1, 2, A car tyre has air at 1.5 atm at 300 K.If P, increases to 1.75 atm with volume same, the, m, m, 11., of oxygen =, of hydrogen, temperature will be ____, M, M, 1) 350 0 C 2) 350K 3) 300 0 C 4) 300K, RT m1 m2 m3 , m, A gas at 6270 C is cooled that its pressure, +, +, PV =, RT ; P =, , , 12., becomes 1/3 of its initial value at constant, V M1 M2 M 3 , M, volume. Its final temperature is, 1) 900 K 2) 600 K, 3) 300 K 4)100K, , NARAYANAGROUP, , 53
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , GAS LAWS, 1., , 2., , 3., , 4., , 5., , A vessel contains a gas under a pressure of, 5 × 105 pa. If 3/5 of the mass of the gas is flown, out,What will be the gas pressure if the, temperature being maintained constant,, 1) 50 MPa 2) 2MPa 3) 0.2MPa 4) 0.5MPa, When an air bubble of radius ‘r’ rises from the, bottom to the surface of a lake, its radius, becomes 5r/4 (the pressure of the atmosphere, is equal to the 10m height of water column). If, the temperature is constant and the surface, tension is neglected, the depth of the lake is, 1) 3.53 m 2) 6.53 m 3) 9.53 m 4) 12.53m, How much should the pressure of the gas be, increased to decrease the volume by 10% at, constant temperature ?, 1)10%, 2)9.5%, 3)11.11%, 4)5.11%, 1 litre of oxygen at a pressure of 1 atmosphere, and 2 litres of nitrogen at a pressure of 0.5, atmosphere are introduced in a vessel of 1 litre, capacity without any change in temperature., The total pressure in atmosphere is, 1) 1, 2) 2, 3) 3, 4) 4, Two closed vessels of equal volume contain, air at 105kPa at 300K and are connected, through a narrow tube. If one of the vessels is, now maintained at 300K and the other at 400K, then the pressure becomes., 1) 120kPa 2) 105kPa 3) 150kPa 4)300kPa, , IDEAL GAS EQUATION, 6., , 7., , 8., , 54, , A vessel is filled with an ideal gas at a pressure, of 10 atmospheres and temp 27 0 C . Half of, the mass of the gas is removed from the vessel, the temperature of the remaining gas is, increased to 87 0 C . Then the pressure of the, gas in the vessel will be, 1) 5 atm 2)6 atm, 3) 7 atm, 4)8 atm, Two identical containers connected by a fine, capillary tube contain air at N.T.P. if one of, those containers is immersed in pure water,, boiling under normal pressure then new, pressure is, 1) 76 cm of Hg, 2)152 cm of Hg, 3) 57 cm of Hg, 4) 87.76 cm of Hg, At the top of a mountain a thermometer read, 70 C and barometer reads 70 cm of Hg. At the, bottom of the mountain the barometer reads, 76cm of Hg and thermometer reads 270 C. The, density of air at the top of mountain is ______, times the density at the bottom., 1) 0.99, 2) 0.9 3) 0.89, 4) 0.95, , 9., , 10., , 11., , 12., , 13., , During an experiment an ideal gas is found to, obey an additional gas law VT = constant. The, gas is initially at temperature T and pressure, P. When it is heated to the temperature2T, the, resulting pressure is, 1) 2P, 2) P/2, 3) 4P, 4) P/4, During an experiment an ideal gas is found to, obey an additional law VP2 = constant. The, gas is initially at a temperature 'T' and volume, 'V'. When it expands to a volume 2V, the, temperature becomes, T, 1)T, 2) 2T, 3) 2 T 4) 2, At the bottom of a lake where temperature is, 7 0 C the pressure is 2.8 atmosphere. An air, bubble of radius1 cm at the bottom rises to the, surface. Where the temperature is 27 0 C ., Radius of air bubble at the surface is, 2) 41 3, 3) 51 3, 4) 61 3, 1) 31 3, The gas in vessel is subjected to a pressure of, 20 atm at a temperature 270C. The pressure, of the gas in a vessel after one-half of the gas, is released from the vessel and the, temperature of the remainder is raised by 500C, is, [EAMCET 2011(M)], 1) 8.5 atm 2) 11.7 atm 3) 17 atm 4) 10.8 atm, An ideal gas is initially at temperature T and, volume V. Its volume is increased by ∆V due, to an increase in temperature ∆T , pressure, remaining constant. The physical quantity, δ=, , ∆V, V ∆T, , varies with temperature as, [EAMCET 2010(M)], T+∆T, , T+∆T, , T, 1), , 2), , T, , T, , T, , T, 3), , T+∆T, T+∆T, , 4), , T, , T, T, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, 2., , An air bubble of volume V0 is released by a fish, at a depth h in a lake. The bubble rises to the, surface. Assume constant temperature and, standard atmospheric pressure P above the lake., The volume of the bubble just before reaching, 9., the surface is, (d is the density of water)., 1) V0 +, , hgd, P, , 2), , V0 ( P + hgd ), P, , temperature T1 and T2 respectively were, connected with narrow capillary tube. The gas, reaches a common pressure P and a common, temperature T. The ratio P/T is equal to, , V0, + hgd, 4) (V0 + V0 dg ), P, If the pressure of a gas contained in a closed, vessel increases by x% when heated by 10 C ,, its initial temperature is, 1) (100/x) Kelvin, 2) (100/x) Celsius, , 3), 3., , x + 100 , Kelvin, x, , , , 3) , 4., , 5., , 6., , 100 − x , Celsius, x, , , , 4) , , A closed vessel contains 8 g of oxygen and 7g, of Nitrogen. Total pressure at a certain, temperature is 10 atm. When all the oxygen is, removed from the system without change in, temperature then the pressure will be, 1) 10 × 7/15atm, 2)10 × 8/15atm, 3) 10 × 8/16 atm, 4) 10 × 8/32 atm, A cylinder contains gas at a pressure of 2.5, atm. Due to leakage, the pressure falls to 2, atm, after sometime. The percentage of the, gas which is leaked out is, 1) 40, 2) 15, 3) 20, 4) 25, 0, The volume of a gas at 0 C is 546cc.at constant, pressure it is heated from 300C to 500C the, change in volume is, 1) 20cc 2) 40cc, 3)10cc, 4) 273cc, , IDEAL GAS EQUATION, , 7., , 8., , 56, , temperature of one litre sphere constant at 270C,, if temperature of two litre sphere is increased to, 1270C, then the final pressure is, 1) 110 cm of Hg, 2) 120 cm of Hg, 3) 150 cm of Hg, 4) 200 cm of Hg, Two containers of equal volume containing the, same gas at pressure P1 and P2 and absolute, , P, , 1 P1, , P, , 3), , P2 , , 2) 2 T + T , 2 , 1, , 1, 2, 1) T + T, 1, 2, , P1T2 + P2T1, T1 + T2, , 4), , P1T2 − P2T1, T1 − T2, , 10. During an experiment an ideal gas is found to, obey an additional law V2P= constant. The gas, is initially at a temperature T and volume V., When it expand to a volume 2V, the, temperature becomes., 1) T, 2) 2T, 3) T 2, 4) T/2, 11. The density of a gas at N.T.P. is 1.5 g/lit. its, density at a pressure of 152cm of Hg and, temperature 270 C, , 1., A flask is filled with 13 g of an ideal gas at, 270C its temperature is raised to 520C. The, mass of the gas that has to be released to 2., maintain the temperature of the gas in the flask, at 520C and the pressure remaining the same 3., is, 1)2.5 g 2)2.0 g, 3)1.5 g, 4)1.0 g, 4., A one litre sphere and a two litre sphere are, connected with a capillary tube of negligible, volume. They contain an ideal gas at 270C at a, pressure of 100cm of Hg. Keeping the, , 1), , 273, g / lit, 100, , 2), , 150, g / lit, 273, , 3), , 1, g / lit, 273, , 4) 1.5 g/ lit, , LEVEL - II (H.W) - KEY, 1) 2, 7) 4, , 2) 2, 8) 2, , 3) 1, 9) 2, , 4) 3 5) 3 6) 2, 10) 4 11) 1, , LEVEL - II (H.W) -HINTS, P ∝ ρ , ρ ∝ m When temperature and volume are, constant P ∝ m, , PV, 1 1 = PV, 2 2, , ; ( P + hdg )V0 = PV, , P2 T2, =, P1 T1 and, , P2 − P1 T2 − T1, =, P1, T1, , m, m, RT , P ∝, M, M, P1 m1 M 2, =, ×, , ( P1 + P11 ) = 10, P1 = 5, P2 m2 M 1, , PV =, , NARAYANAGROUP
Page 732 : JEE- ADV PHYSICS-VOL- V, 5., 6., 7., , ∆ P ∆m, P ∝ m when V&T are constant ⇒ P = m, 1, ∆V = V α∆T , here α = 273 ; ∆V = 546, m, 1, PV =, RT , m ∝ , mass released = m1 - m2, M, T, 1, , 8., , THERMAL PROPERTIES OF MATTER - II, , P, T1, , 4., , 1, , PV1 PV2 P V1 P V2, +, =, +, T1, T1, T1, T2, , PV, PV PV PV, 1, + 2 =, +, T1, T2, T, T, 2, 10. Given that PV = constant. From Gas equation, T T, P ∝ ; V 2 = constant ⇒ TV = constant, V V, P1, P2, 11. d T = d T, 5., 1 1, 2 2, , 9., , LEVEL - III, GAS LAWS, 1., , 1) T1 > T2, 2) T2 > T1, 3) T1 = T2, 3, d 4) T1 = T2, For an ideal gas V-T curves at constant, , T2, , pressures P1 & P2 are shown in figure, from the, figure, V, P1, 1) P1>P2, P2, 2) P1<P2, 3) P1=P2, , T, , 4) P1 < P2, A Volume V absolute temperature T, diagram was obtained when a given mass of, gas was heated. During the heating process, from state 1 to 2, the pressure, , V, , The graph drawn between pressure and volume, in Boyles law experiment is shown in figure, for different molecular weights then, , 1) Remains constant, , 2, 1, , 1) M 2 < M 1, , M2, M1, , P, , 2., , 2) M 1 < M 2, , 6., , 3) M 1 = M 2, , V 4) M 3 = M, 1, 2, The graph drawn between pressure and volume, in Boyle’s law experiment is shown in figure, for different masses of same gas at same, temperature then, 1) m2 > m1, , P, , 3., , m2, m1, , 2) m1 > m2, 3) m1 = m2, , 7., V, 4) m13 = m2, In Boyles experiment for a given gas at, different temperatures the graph drawn, between pressure and density are straight, lines as shown then, , NARAYANAGROUP, , 2) Decreased, 3) Changed erratically, , 4) Increased, T, Two identical containers each of volume V0 are, joined by a small pipe. The containers contain, identical gases at temperature T0 and pressure, P0. One container is heated to temperature 2T0, while maintaining the other at the same, temperature. The common pressure of the gas, is P and n is the number of moles of gas in, container at temperature 2T0., 4, P0, 3, , 1) P = 2 P0, , 2) P =, , 2 PV, 0 0, 3) n = 3RT, 0, , 3PV, 0 0, 4) n = 2 RT, 0, , A cycle tube has volume 2000 cm3. Initially, th, , 3, the tube is filled to of its volume by air, 4, at pressure of 105 N/m2. It is to be inflated to, a pressure of 6 × 105 N/m2 under isothermal, 57
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , 8., , 9., , conditions. The number of strokes of pump, 12. A closed container of volume 0.02m3, contains a mixture of neon and argon gases,, which gives 500 cm3 air in each stroke, to, at a temperature of 27°C and pressure of, inflate the tube is, 1× 105 Nm-2 . The total mass of the mixture is, 1) 21, 2) 12, 3) 42, 4) 11, 28g. If the gram molecular weights of neon and, A horizontal uniform glass tube of 100cm, argon are 20 and 40 respectively. Find the, length is sealed at both ends contains 10 cm, masses of the individual gases in the container,, assuming them to be ideal. (Universal gas, mercury column in the middle, the temperature, constant R = 8.314 J/mol.k), and pressure of air on either side of mercury, 1) m1 = 4 g, m2 = 24 g 2) m1 = 8 g, m2 = 20 g, o, column are respectively 31 C and 76cm of, 3) m1 = 16 g, m2 = 12 g 4) m1 = 12 g, m2 = 16 g, mercury , if the air column at one end is kept, LEVEL - III - KEY, at 0oC and the other end at 273oC then, o, 1) 1 2) 1 3) 1 4) 2 5) 1 6) 3, pressure of air which is at 0 C is, 7) 1 8) 3 9) 4 10) 3 11) 3 12) 1, (in cm of Hg ), LEVEL - III - HINTS, 1)76, 2)88.2, 3)102.4, 4)122, A closed hollow insulated cylinder is filled with 1., m, PV =, RT ;, gas at 00C and also contains an insulated piston, M, of negligible weight and negligible thickness at, P = constant, At constant pressure, the middle point. The gas on one side of the piston, 1, M2, V∝, from graph, 0, P, is heated to 100 C. If the piston moves 5cm, the, M1, M, volume V 2 > V1 then, length of the hollow cylinder is, [2011 E], v1 v2, V, M2 < M1, 1)13.65cm 2)27.3cm 3)38.6cm 4)64.6cm, , 10. Two thermally insulated vessels 1 and 2 are 2., filled with air at temperature (T1, T 2), volume, (V1, V2) and pressure (P1, P2) respectively. If, the valve joining the two vessels is opened,, the temperature inside the vessel at, equilibrium will be, 1) T1 + T2, 2) T1T2 (P1V1 + P2V2) / (P1V1T1 + P2V2T2), 3., 3) T1T2 (P1V1 + P2V2) / (P1V1T2 + P2V2T1), 4) (T1+T2) / 2, 11. Two identical vessels A and B with frictionless, pistons contain the same ideal gas at the same, temperature and the same volume V. The masses, of gas in A and B are m A and m B respectively.., The gases are allowed to expand isothermally to 4., the same final volume 3 V. The change in pressure, of the gas in A and B are found to be ∆P and 1.5, ∆P respectively. Then, , 58, , 1) 9mA = 4mB, , 2) 3mA = 2mB, , 3) 2mA = 3mB, , 4) 4m A = 9mB, , m, RT, M, same gas is used at, PV =, , P = constant, m2, m1, , P, v1 v2, , P, , V, , V ∝ m from graph, V2 > V1 then m2 > m1, , V, , P, = constant ; At, dT, constant pressure, T2, 1, d ∝ from graph, T, P = constant, d2> d1 then T2 < T1, , T1, , d1, , constant pressure, , d2, , d, , P1, P2, V = constant, T1, , T2, , T, , PV, = constant at, T, constant volume, P ∝ T from the graph, temperature T 2 > T1, then P2 > P1, NARAYANAGROUP
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THERMAL PROPERTIES OF MATTER - II, 3., , 4., , Match List I and List II, List-I, List-II, A) Barometer, E) Charles’s law, B) specific gas constant F) J mole -1 K-1, C) gas thermometer, G) Boyle’s law, D) universal gas constant H) J Kg-1 K-1, Match List I and List II, List-I, List-II, A) Boyle’s Law, E) PV = NkBT, B) Charles’s law, C) Gay-Lussac’s law, , P, =constant,at, T, constant volume, G) V αT at, constant pressure, , F), , D) Equation of state of H) PV, constant, 1 1 =, an ideal gas, , ASSERTION & REASON TYPE QUESTIONS, 1) Both assertion (A) and reason (R) are, correct and R gives the correct explanation, 2) Both assertion (A) and reason (R) are, correct but R does not give the correct, explanation, 3) A is true but R is false, 4) Both A and R are false, , 5., , 6., , 7., , 8., , 9., , 60, , Assertion (A): Real gases do not obey the ideal, gas equation., Reason (R): In the ideal gas equation, the volume, occupied by the molecules as well as the inter, molecular forces are ignored., Assertion (A): Gases are characterised with two, coefficients of expansion, Reason (R): When heated both volume and, pressure increase with the rise in temperature., Assertion (A): PV/T=constant for 1 gram of gas., This constant varies from gas to gas., Reason (R):1 gram of different gases at NTP, occupy different volumes., Assertion (A):PV/T=constant for 1 mole of gas., This constant is same for all gases., Reason (R): 1 mole of different gases at NTP, occupy same volume of 22.4 litres., Assertion (A): At constant pressure when a gas is, heated from 40 to 410C, the increase in volumes is, 1/273 of its initial volume at 273 K, , JEE- ADV PHYSICS-VOL- V, Reason (R):Volume coefficient of gas is, , 1 0, /C, 273, , 10. Assertion (A): Volume of gas at 500C is ‘V’., Keeping the pressure constant, the temperature is, doubled. Volume becomes 2V., Reason (R): At constant pressure, the volume of, gas is directly proportional to its absolute, temperature., 11. Assertion (A): Pressure of gas is same every where, inside a closed container, Reason (R):The gas molecules under go elastic, collisions among themselves and with walls of the, container, 12. Assertion (A): Gases obey Boyle's law at high, temperature and low pressure only., Reason (R): At low pressure and high, temperature, gases would behave like ideal gases., 13. Assertion (A):The air pressure in a car tyre, increases during driving, Reason (R): Temperature of air in the tyre, increases due to friction of tyre with road. Increase, in temperature results in an increase in pressure, according to Charles’s law [EAMCET 2012 M], , STATEMENT TYPE QUESTIONS, Options :, 1. Statement 1 is true and statement 2 is true, 2. Statement 1 is true and statement 2 is false, 3. Statement 1 is false and statement 2 is true, 4. Statement 1 is false and statement 2 is false, 14. Statement-1 : The pressure of a given mass of, gas varies linearly with its absolute temperature. The, volume of the gas shall remain constant in the, process., Statement-2 :When pressure is directly, proportional to temperature, P-T graph is a straight, line., 15. Statement-1 : Real gas approaches ideal gas, behaviour at low pressures and high temperatures., Statement-2 :At low pressure , density of gas is, very low., 16. Statement-1 : All molecules in a gas move with, the same speed., Statement-2 : Average velocity of molecules of a, gas sample is zero., , MORE THAN ONE OPTION QUESTIONS, 17. According to Boyle's Law PV=C the value of, C depends on., A)Mass of the gas B)Type of gas C) Temperature, 1) A,B, 2) B,C 3) A,C 4) A,B,C, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 18. Select the correct graphs, A) the P-1/V graph at constant temperature is a, rectangular hyperbola., B) the PV-V graph is a straight line parallel to the, Y-axis., C) P-V graph at constant temperature is a straight, line passing through the origin, D) V-T graph at constant pressure is a straight line, passing through the origin., 1) A,B, 2) B,D, 3) C,D, 4) A,D, 19. Which of the following processes will, quadruple the pressure, A) Reduce V to half and double T, B) Reduce V to 1/8th and reduce T to half, C) Double V and half T, D) Increase both V and T to double the values., 1) B,C 2) A,B, 3) C,D, 4) A,D., 20. Following operation are carried out on a sample, of ideal gas initially at pressure P volume V, and Kelvin temperature T., A) At constant volume, the pressure is increased, fourfold., B) At constant pressure, the volume is doubled, C) The volume is doubled and pressure halved., D) If heated in a vessel open to atmosphere, onefourth of the gas escapes from the vessel. Arrange, the above operations in the increasing order of final, temperature, 1) A, B, C, D, 2) C, B, A, D, 3) B, A, D, C, 4) D, C, B, A, 21. Real gases approaches ideal gas at high, temperature and low pressure because, A.interatomic separation is large, B. size of the molecule is negligible when compared, to inter atomic separation, 1) A & B are true, 2) only A is true, 3) only B is true, 4) A & B are false, 25. The parameter that determine the physical, state of gas are :, A) Pressure, B) Volume, C) Number of moles, D) Temperature, 1) A & B 2) A,B & C 3) A,B & D 4)A,C &D, 22. The parameter that determine the physical, state of gas are :, A) Pressure, B) Volume, C) Number of moles, D) Temperature, 1) A & B 2) A,B & C 3) A,B & D 4) A,C &D, 23. In the equation PV=constant, the numerical, value of constant depends upon, A) temperature, B) mass of the gas, C) system of units used D) nature of the gas, 1) A & B 2) B & C, 3) C & D, 4) All, NARAYANAGROUP, , THERMAL PROPERTIES OF MATTER - II, 24. PV = n RT holds good for, A) Isobaric process, B) Isochoric process, C) Isothermal process, D) Adiabatic process, 1) A & B 2) A,B & C, 3) A,B & D, 4) All, , LEVEL-IV - KEY, MATCHING TYPE QUESTIONS, 1) A-F, B-H, C-E,D-G. 2) A-F, B-E,C-H, D-G, 3)A-G,B-H,C-E,D-F. 4)A-H.,B-G,C-F,D-E, ASSERTION & REASON TYPE QUESTIONS, , 5)1, 11)1, , 6)1, 12)1, , 7)1, 8)1, 13) 1, , 9)1, , 10)1, , STATEMENT TYPE QUESTIONS, 14)3, , 15)1, , 16)3, , More than one option Type questions, 17)3, 23)4, , 18)2, 24)4, , 19)2, , 20)4, , 21)1, , 22)3, , LEVEL-IV - HINTS, 5., 6., 7., , 8., 9., , 10., 11., 12., 13., , The real gas obeys all gas laws at low pressure and, high temperature. Ideal gas obeys all gas laws at all, temperatures and pressures., under constant pressure and volume gases have, same coefficient of expansions i.e. α = β, 1 mole of any gas will occupy same volume (at, PV, = R = constant, N.T.P), T, but for one gram of any different gases will occupy, different volumes (at N.T.P), PV, = r = constant(for a given gas), T, PV = nRT, R is universal gas constant, it is constant for all gases, ∆V, α=, , ∆t = 10 C, V ∆t, 1, α=, 273.150 C, From Charles’s law at constant pressure, V ∝ T, pressure developed in a gas is due to elastic collision, of gas molecules with walls and is same through, out the container., Ideal (or) perfect gas obeys gas laws at all, temperature and pressures without any limitations., When the vehicle is in motion temperature of tyre, and hence the temperature of gas increases due to, friction. Hence P ∝ T at constant volume., , 61
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JEE-JEE, ADV, PHYSICS-VOLV, ADV, -PHY- VOL - V, , THERMAL PROPERTIES OF MATTER - II, , 3 . A cube of coefficient of linear expension α s is, floating in a bath containing a liquid of, , LEVEL - V, , coefficient of volume expertion γ l . When the, , THERMAL EXPANSION OF SOLIDS, AND LIQUIDS, , temperature is raised by ∆T , the depth upto, which the cube is submerged in the liquid, remains the same. Then the relation between, , SINGLE ANSWER QUESTIONS, 1., , Three identical rods of equal length L are, joined to form an equilateral triangle ABC as, shown in figure. D is the midpoint of AB. The, coefficient of linear expansion is a1 for AB,, and a 2 for AC and BC. If α1 = 4α 2 , the change, in time period of the systam is, A, , D, , α2, , α1, , α s and γ l is, A) γ l = 3α s, 4., , B, , C) γ l = 2α s, D) γ l = α s / 2, A heavy brass bar has projections at its ends, as shown in the figure. Two fine steel wires,, fastened between the projections, are just taut, (zero tension) when the whole system is at, , 00 C . What is the tensile stress in the steel, , α2, , wires when the temperature of the sytem is, raised to 3000 C ?, Given that, , C, , (A) (α1 + α 2 )L∆t, , B) γ l = 3α s / 2, , asteel = 12 × 10−6 0C −1 Ysteel = 2 × 1011 N m−2, , 2α + α, (B) 1 2 L∆t, 2, , (α1 + 2α 2 )L∆t, (D) Zero, 2, In a vertical U-tube containing a liquid, the two, arms are maintained at different temperatures,, t1 and t 2. The liquid columns in the two arms, have heights l1 and l2 respectively. The, coefficient of volume expansion of the liquid, is equal to, , abrass = 20 × 10−6 0C −1, , (C), , 2., , 5., , (A) 48 × 107 Nm−2 (B) 84 ×107 Nm−2, (C) 32 × 104 Nm −2 (D) 24 ×104 Nm −2, The variation of lengths of two metal rods, A and B with change in temperature are shown, in Fig. The coefficients of linear expansion, , a A for the metal A will be nearly:, (Given a B = 9 × 10−6 / 0C ), , t2, , 62, , L1, , t1, , Length(mm), , L1, , l1 − l 2, (A) l t − l t, 2 1, 1 2, , l1 − l2, (B) l t − l t, 1 2, 2 2, , l1 + l2, (C) l t + l t, 2 1, 1 2, , l1 + l2, (D) l t + l t, 1 1, 2 2, , 1006, 1004, 1002, 1000, 998, 996, 994, , A, , B, , Temperature (°C), , (A) 13 × 10 −6 / 0C (B) 27 × 10 −6 / 0C, (C) 18 × 10 −6 / 0C (D) 43 × 10 −6 / 0C
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JEE- ADV PHYSICS-VOL- V, , MULTIPLE ANSWER QUE STIONS, 6., , THERMAL PROPERTIES OF MATTER - II, C. Incandescent lamp, D. Electric fuse, Column II, p. Radiation from a hot body, q. Energy conversion, r. Melting, s. Thermal expansion of solids, , A metallic circular disc having a circular hole, at its centre rotates about an axis passing, through its centre and perpendicular to its, plane. When the disc is heated:, (A) its angular speed will decrease, (B) its diameter will decrease, (C) its moment of inertia will increase, INTEGER TYPE QUESTIONS, (D) its angular speed will increase, 7. A bimetallic strip is formed out of two identical 10. A steel rod of length 5 m is fixed between two, support. The coefficient of linear expansion of, strips one of copper and the other of brass., steel is 12.5 × 10–6/°C. Calculate the stress, The co-efficients of linear expansion of the two, (in 108 N/m2) in the rod for an increase in, metals are α C and α B . On heating, the, temperature of 40°C. Young’s modulus for steel, temperature of the strip goes up by ∆T and, is 2 × 1011 Nm−2, the strip bends to form an arc of radius of, curvature R. Then R is, Passage for Q no 11,12, (A) proportional to ∆T, Two rods of different metals having the same area, (B) inversely proportional to ∆T, of cross section A are placed between two rigid, (C) proportional to | α B − α C |, walls. For the first rod l1,α1, Y1 , and for the second, (D) Inversely proportional to | α B − α C |, rod l2 , α 2 , Y2 are the physical quantitities with usual, MATRIX MATCHING TYPE QUESTIONS, meanings. Now the temp of the system is increased, 8. Whenever a liquid is heated in a container,, by t 0C ., expansion in liquid as well as container takes, −1, place. If γ is the volume expansion coefficient, Given [ l1 = 20 cm , α 1 = 18 × 10 − 6 ( 0 C ), of liquid and a is coefficient of linear expansion, of the container match the entries of Column I, Y1 = 10 × 1011 N / m 2 , l2 = 18cm, and Column II, −1, Column I, Column II, α 2 = 20 × 10−6 oC, Y2 = 9 × 1011 Nm −2 ], (i) Liquid level rises with, (A) γ = 2α, 11. The force with which the rods act on each other, respect to container, is 9F KN then F =, (ii) Liquid level remains, (B) 2α < γ < 3α, 12. The length of the smallar rod at 1000 C is 3.6, same with respect to container, L cm then L =, (iii) Liquid level drops with (C) γ = 3α, respect to container, LEVEL - V-KEY, (iv) Liquid level remains, (D) γ > 3α, SINGLE ANSWER QUESTIONS, same with respect to ground, 1) D, 2)A, 3)C, 4) A, 5) B, 9. Column I gives some devices and Column II, MULTIPE ANSWER QUESTIONS, gives some processes on which the functioning, 6) A,C 7) B. D, of these devices depend. Match the devices, MATRIX MATCHING TYPE, in Column I with the processes in Column II., 8) (i) → (D), (ii) → (C), (iii) → (B), (iv) → (A), (IIT 2007), Column I, 9) (A → s, B → q, C → p, q, D → q, r), A. Bimetallic strip, INTEGER TYPE QUESTIONS, B. Steam engine, 10) 1, 11) 4, 12)6, , ( ), , 63
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JEE- ADV PHYSICS-VOL- V, , THERMAL PROPERTIES OF MATTER - II, , MATRIX MATCHING TYPE QUESTIONS, 8., , Change in volume of a liquid w.r.t. container, , l1, , l2, C, , A, , ∆V = V0 [ γ − 3 ∝]∆T, α = co-effecient of linear expansion of material of, container and γ is volume expansion co-effecient, , 9., , of liquid, (a) If γ > 3α level of liquid rises w.r.t. container as 2., well as w.r.t. ground, (b) γ = 3α level of liquid remains same w.r.t., container but rises w.r.t. ground, (c) 2α < γ < 3α level of liquid falls w.r.t. container, but rises w.r.t. ground, (d) γ = 2α level of liquid falls w.r.t. container but, remains same w.r.t. ground, Conceptulal, , INTEGER TYPE QUESTIONS, 10. Thermal stress = Yα ∆T, = 2 × 1011 × 12.5 × 10–6 × 40 = 1×108 N / m 2, Answer is 1., 11. The increase in length due to expansion = decrease, in length due to elastic force, 3., , ( l1α1 + l2α 2 ) =, , F l1 l2 , + , A y1 y2 , , Fl2, |, 12. l2 = l2 + l2α t − Ay, 2, , LEVEL - VI, SINGLE ANSWER QUESTIONS, 1., , Two rods AB and BC of equal cross-sectional, area are joined together and clamped between, two fixed supports as shown in the figure. For, 4., the rod AB and road BC lengths are l1 and l2, coefficient of linear expansion are α1 and α 2 ,, young’s modulus are Y1 and Y2 , densities are, ρ1 and ρ2 respectively. Now the temperaturee, of the compound rod is increased by θ ., Assume that there is no significant change in, the lengths of rod due to heating. then the time, taken by a transverse wave pulse to travel from, end A to other end C of the compound rod is, directly proportional to, , m2, , B, , (A), , l2 Y1 + l1Y2, , (C), , l1Y2 + l2 Y1, , (B) 2 l2 Y1 + l1Y2, (D), , l2 Y1 − l1Y2, , Two wires A and B of the same corss sectional, area, young’s modulli Y1 , Y2 and coefficients, of linear expansion α1, α 2 respectively are, joined together and fixed between rigid, supports at either ends. The tension in the, compound wire when the wire A is heated and, wire B is cooled at different temperature is, same when wire A alone in cooled at same, temperature as wire B earlier. The correct, option is, α, , Y, , α1, , 2Y2, , 1, 2, A) α > 2Y, 2, 1, , α, , Y, , α1, , Y2, , 1, 2, B) α < 2Y, 2, 1, , C) α > Y, D) α > Y, 2, 1, 2, 1, A thermostated chamber at a height h above, earth’s surface maintained at 300 C has a clock, fitted with uncompensated pendulum. The, maker of the clock for chamber mistakenly, designed it to maintain correct time at 200 C ., It is found that when the chamber is borught, to earth’s surface the clock in it clicked correct, time. Re is the radius of Earth. The linear, coefficient of the material pf pendulum is, h, , h, , 5R, , R, , A) R, B) 5R, C) h e, D) he, e, e, INTEGER TYPE QUESTIONS, The system shown in figure consists of 3, springs and two rods. If the temperature of the, rod is increased by ∆T , then the total energy, 99 2 2, kL α (∆T ) 2 ., stored in three springs is β ×, 484, Determine the value of β . The spring are, initially relaxed and there is no friction, anywher. For rods the coefficient of linear, expansion is α, , 65
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , TRANSMISSION OF HEAT, Temperature gradient:, , SYNOPSIS, Ø, , These are three distinct modes of heat transfer: Ø, conduction, convection and radiation, , Conduction, Ø, Ø, Ø, Ø, , It is the transmission of heat without the actual Ø, movement of the particles of the medium., It takes place mainly in solids., It takes place in metals due to free electrons., Consider a good conductor in the shape of uniform, rod of length l, whose opposite parallel faces are, maintained at different steady temperatures θ1 and Ø, , θ2 such that θ1 > θ2 ., , Ø, , Ø, , A (θ1 − θ 2 ) t, KA (θ1 − θ 2 ) t, or Q =, l, l, , Here K is the constant of proportionality called, coefficient of thermal conductivity or thermal, conductivity of the material of the block, Co-efficient of thermal conductivity is defined, as the rate of flow of heat per unit area per unit, temperature gradient in steady state, Q/t, K=, θ −θ , A 1 2 , l , , Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, 66, , θ2, , Under steady state conditions the amount of heat Q, flowing from hot face to the cold face of the block is, Q∝, , Ø, , Q, , is called the temperature gradient. It tells, l, how the temperature changes per unit distance, moved in the direction of heat flow., If dθ is small change in temperature in the direction, of heat flow, across small element of length dx then,, dθ, . Here the negative, temperature gradient = −, dx, sign indicates the decreses in temperature as the, distance increases in the direction of heat transfer, In steady state, temperature gradient is same along, the length of the conductor., , Thermal resistance : (R), , l, θ1 A, , (θ1 − θ2 ), , Ø, , Thermal resistance R of a conductor of length l ,, cross-section A and conductivity K is given by, l, R=, KA, Ø The S.I unit of R is K / W, Ø Dimensional formula of R is M −1 L−2T 3θ, Note:If one end of a metal rod is kept in a steam jacket, and other end is kept in an ice block, then the amount, KA(θ1 − θ 2 )t, of ice that melts is m =, lLice, Here l =Length of metal rod, Ø Diffusivity (D) is the ratio of thermal conductivity, , Ø, , The S.I unit of K is W/m-K, The C.G.S unit of K is cal/s-cm-oC, Dimensional formula of K is MLT −3θ −1, K depends on the nature of the metal., K is independent of length, area of cross section Ø, and temperature difference., For a perfect conductor K =∝ ., For a perfect insulator K= 0, If K value is more, it is a good conductor of heat, If K value is less, it is a bad conductor of heat., , (K) to thermal capacity per unit volume ( ms / V ), of a material., The rate of flow of heat across the material of a, block between the parallel faces is given by, dQ, dθ , = − KA , ; Here the negative sign indicates, dt, dx , the decrease in temperature as the distance increases, in the direction of heat transfer., Q /t is the rate of heat flow (or) rate of energy, , transfer, which is equal to power P. Then, , l , ∆θ = P , = PR . This equation is useful for, KA , , solving problems when heat flows through layers of, materials placed in series or parallel., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, W.E-5: A metal rod AB of length 10x has its one, end in ice at 0º C and the other end B in water, at 100º C . If a point P on the rod is maintained, at 400º C , then it is found that equal amounts, of water and ice evaporate and melt per unit, time. The latent heat of evaporation of water, is 540 cal/g and latent heat of melting of ice is, 80 cal/g. If the point P is at a distance of λ x, from the ice end A, find the value of λ . [neglect, any heat loss to the surrounding.], (JEE2009), Sol., , λx, , Thermal resistance of the element is dR =, , =, , 0, , l, 1 1 , − = l, , K π ( a − b ) b a K π ab, , = λ =9, , θ2, , θ1, K1, l1, , Ø, , b, , K2, θ, , A, , l2, , Let two rods of same cross-sectional area having, lengths l1 and l2 and co-efficient of thermal, conductivities K1 and K2 are connected in series., For the 1st rod θ1 − θ = P R1 ......... (i), For the 2nd rod θ − θ 2 = P R2 ........ (ii), , l, , For combined rod θ1 − θ 2 = P ( Reff ) .... (iii), , Sol., , Eq. (i) + (ii) gives (θ1 −θ2 ) = P( R1 + R2 ) ..... (iv), , dx, , x, , Hence , the effective thermal resistance for rods in, series is the sum of thermal resistances of each rod., Ø, , l, , Consider a small element of width dx at a distance, x from one end as shown in figure., From similar triangles, , Reff = R1 + R 2, , from eq. (iii) and (iv), , b, , a, , 68, , l, , Conduction of heat through a Composite slab, a) When different rods of same cross, sections are connected in series., , W.E-6: Find the value of the thermal resistance of, the non-uniform cylindrical rod of thermal, conductivity K and length l as shown in figure., , a, , l, , −1, , b−a , a +, x , 1 , l , =, , b−a , Kπ , −, , , , l , , , , 400 KS, 300 KS, ; λ xL = (10 − λ ) xL, ice, vapour, , 400, 300, =, λ × 80 (10 − λ ) 540, , −2, , dx, 1 b−a , =, R=∫, a+, x dx, 2, Kπ r, K π ∫0 l , 0, l, , 100°C, (steam), , 0°C(ice) 400°C, , dx, Kπ r 2, , ∴ Thermal resistance of the cylinder is, , (10 − λ)x, , P, , dmice dmvapour, =, dt, dt, , r −a b−a, b−a, =, ⇒ r = a +, x, x, l, l , , In series, Reff = R1 + R 2, l1 + l2, l, l, = 1 + 2, Keff A K1 A K2 A, , ⇒, , l1 + l2, l, l, = 1 + 2, K eff, K1 K 2, , l , , Q R =, , KA, , , , ⇒ K eff =, , ( l1 + l 2 ) K1K 2, l1 K 2 + l2 K1, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , W.E-7: A room has a window fixed with a pane of, area 1.2m2. The glass has thickness 2.2mm. If, the temperature outside the room is 36ºC and, Ø, the temperature inside is 26ºC, (a) calculate the heat flowing into the room, every hour., l1 + l2 + ... + ln l1, l, l, =, + 2 + ... + n, K eff, K1 K 2, Kn, (b) If the same single pane window is replaced, by double paned window with an air gap of, b) When different slabs of same thickness are, 0.50 cm between the two panes calculate the, connected in parallel, heat flowing into the room every hour., Ø Let two rods of same length having cross-sectional, Kg = 0.80Wm−1K −1; Kair = 0.0234Wm−1K −1., areas A1,A2 and thermal conductivities K1,K2 are, arranged in parallel., Sol:(a) We assume that one side of the pane is at 36ºC, and the other side (inside the room) is at 26ºC., t, θ1, θ2, Given Thickness of the window pane, K1, A1, d = 2.2mm = 2.2 × 10−3 m, Q, Area of the window pane A = 1.2m 2, Ø, , 2K K, , 1 2, If l1 = l2 then K eff = K + K, 1, 2, If n rods of different materials and same area of, cross sections are connected in series then, effective, thermal conductivity is, , A2, , K2, , Ø, , The quantity of heat supplied is distributed between, , Therefore heat flown into the room per hour is, , Q Q Q, the two rods ; i.e = 1 + 2 (or) P = P1 + P2, t, t, t, , Q = 4364 × 3600 = 1.57 × 107 J, , ⇒, ∴, , Ø, , θ1 − θ2 θ1 − θ2 θ1 − θ 2, =, +, Reff, R1, R2, , 1, 1 1, RR, = +, Reff = 1 2, (or), Reff R1 R2, R1 + R2, , In parallel,, ⇒, , 1, 1, 1, = +, Reff R1 R2, , K ( A1 + A2 ) K1 A1 K 2 A2, =, +, l, l, l, , ∴ K eff =, , Ø, , A(θ2 −θ1 ) 0.8×1.2×10, Q, =K, =, = 4364 J / s., t, d, 2.2×10−3, , K1 A1 + K 2 A2, A1 + A2, K1 + K 2, 2, , If n rods of same length and different area of cross, sections of different materials are connected in, parallel then, effective thermal conductivity is, l, , K eff, , =, , K1 A1 K 2 A2, K A, +, + ..... + n n, l, l, l, , K A + K 2 A2 + ... + K n An, = 1 1, A1 + A2 + .... + An, , NARAYANAGROUP, , Rg =, , dg, K g Ag, , =, , 2.2 ×10 −3, = 2.29 ×10−3 K/ W, 0.8 ×1.2, , Thermal resistance for the air gap, Ra =, , da, 0.5×10−2, =, =178×10−3 K / W, Ka Aa 0.0234×1.2, , Net thermal resistance RT = R g + Ra + Rg, , If A1 = A2 then K eff =, , K eff ( A1 + A2 + ..... + An ), , (b) When single pane window is replaced by a, double paned window we have two layers of glass, and one layer of air between them., Thermal resistance for glass, , (, , = 2.29 × 10−3 + 178 ×10 −3 + 2.29 × 10−3, , ), , −3, , = 182.6 × 10 K / W, , P=, , ∆θ, 10, =, = 55 J / s, RT 182.6 × 10 −3, , Therefore heat flown into the room per hour is, Q = P × t = 1.98 ×105 J, 69
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, W.E-8: Three rods AB,BC and BD made of the, same material and having the same area of, cross section have been joined as shown in the, figure. The ends A,C and D are held at, temperatures 200 C , 800 C and 800 C, respectively. If each rod is of same length, then, find the temperature at the junction B of the, three rods ( 2010 E ), , A, , B, , 00C, , 00C, , Brass, , Steel, θ, Cu, , 0, , 100 C, , Let θ is the junction temperature. Given A = 4cm2, lcu = 46cm, lBr = 13cm, & lsteel = 12cm, , C, , Kcu = 0.92, KBr = 0.26&Ksteel = 0.12CGS, . . .units, , Q, Q, Q, + , In steady state t = t , cu brass t steel, , D, , KCu A(100 −θ ) KBr A(θ − 0) Ksteel A(θ − 0), =, +, lcu, lBr, lsteel, , Sol., , A, , R, , R, , B, , 0, , 20 C, , C, 0, 80 C, , 0.92A(100 −θ ) 0.26A(θ −0) 0.12A(θ −0), =, +, 46, 13, 12, ⇒ θ = 40º C, , R, D, 80 C, , Q, ∴ , t cu, , 0, , Let θ B is the temperature of the junction, , =, , KCu A (100 − θ ), = 4.8 cal / sec, lCu, , W.E-10: A cylinder of radius R made of a material, of thermal conductivity K1 is surrounded by, Q, Q, Q, In steady state, t + t = t , cylindrical shell of inner radius R and outer, DB CB BA, radius 2R made of a material of thermal conKA(80 −θB ) KA (80 − θB ) KA(θB − 20), +, =, ductivity K 2 . The two ends of the combined, l, l, l, system are maintained at two different tem80 − θ B + 80 −θ B =θ B − 20 ⇒θ B = 60º C, peratures. There is no loss of heat across the, W.E-9:Three rods of Copper, Brass and Steel are, cylindrical surface and system is in steady state., welded together to form a Y-shaped structure., What is the effective thermal conductivity of, 2, the system?, (EAM-2013M), Area of cross - section of each rod is 4 c m ., K, End of copper rod is maintained at 1000 C ,, 2, , R, , Heat flow, , 2R, , where as ends of brass, steel are kept at 0 C ., K, Lengths of the copper, brass and steel rods are, θ2, θ1, 46,13 and 12 cms respectively. The rods are, thermally insulated from surroundings except, 2, 2, 2, at ends . Thermal conductivities of copper, Sol. A1 = π R , A2 = π 4 R − R, brass and steel are 0.92,0.26 and 0.12 CGS, Two cylinders are in parallel,, units respectively . Rate of heat flow through, ∴ Keff ( A1 + A2 ) = K1 A1 + K2 A2, copper rod is :, (JEE-2014), Sol., K π R 2 + K 2 3π R 2, ( K1 + 3K 2 ), K eff = 1 2, ;i.e., K eff =, 2, π R + 3π R, 4, 0, , 1, , (, , 70, , ), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, W.E-14: A cylinder of radius R and length l is made, up of a substance, whose thermal conductivity, K varies with the distance x from the axis as, K=K 1x+K2. Determine the effective thermal, conductivity between the flat faces of the, cylinder., , surface of ice will be at 00c. If A is the area of the, lake, heat escaping through ice in time dt,, 0 − ( −θ ) , dQ = KA , dt, y, , -00C, , dx, , Ice, y, axis, , dy, , 00C, Sol.Let us subdivide the entire cylinder into a number of, coaxial cylindrical shells of infinitesimally small, thickness dx ., Cross sectional area of the shell is 2π x ( dx ) ., Ø, Using the expression for the effective thermal, conductivity., , K eff =, , ∑AK, ∑A, i, , R, , i, , ; K eff =, , i, , 1, K ( 2π xdx ), ∑ Ai ∫0, , R, , =, , 1, ( K1x + K2 ) 2π x( dx), π R2 ∫0, R, , (, , ), , Ø, , 2, 1, Ø, K1 x 2 + K 2 x dx = ( 2 K1 R + 3K 2 ), 2 ∫, R 0, 3, W.E-15. Two identical rods are joined at their, middle points. The ends are maintained at Ø, constant temperatures as indicated. Find the, temperature of the junction, =, , 0, , 75 C, , 0, , 180 C, , 250C, , Sol: Under study state condition, Let θ be the temperature of the junction., 180 −θ θ − 75 θ − 50 θ − 25, =, +, +, ⇒θ = 82.5ºC, R, R, R, R, , Growth of ice layer on ponds, Ø, , 72, , Now due to escaping of this heat if dy thickness, of water in contact with lower surface of ice freezes,, dQ = mL = ρ ( dyA ) L [ as m=ρ dV=ρ A dy ], , dy K θ 1, y, =, × ⇒ t = ρ L ∫ ydy = 1 ρ L y 2, 0, dt, ρL y, Kθ, 2 Kθ, It is clear that time taken to double and triple the, thickness will be in the ratio, t1 : t 2 : t3 = 12 : 22 : 32 , i.e., t1 : t2 : t3 =1 : 4 : 9 ....., The time taken to change the thickness of ice layer, , from y1 to y2 is t ∝ ( y22 − y12 ), The time intervals to change thickness from 0 to y,, from y to 2y and so on will be in the ratio, ∆t1 : ∆t 2 : ∆t3 = (12 − 02 ) : ( 22 − 12 ) : ( 32 − 22 ), , i.e., ∆t1 : ∆t2 : ∆t3 = 1 : 3 : 5 ..., 0, , 50 C, , Water, , When atmospheric temperature falls below 00C the, water in the lake will start freezing. Let at any time, t, the thickness of ice layer in the lake be y and, temperature of water in contact with the lower, , W.E 16. The thickness of ice in a lake is 5cm and, the atmospheric temperature is - 100 C ., Calculate the time required for the thickness, of ice to grow to 7cm. Thermal conductivity of, ice = 4 × 10−3 cal / cm − s − 0 C ; density of ice, = 0.92 g / cm3 and latent heat of fusion of ice, = 80 cal/g, ρL 2, ( x2 − x12 ), 2 KT, 92 ×10−2, 80 2 2, =, 2 × 4 ×10−3 × 100 ( 7 − 5 ), = 22080s = 6.13Hr, , Sol: ∆t =, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , Electrical, Thermal & Fluid Analogy, Electric Current, , Thermal current, , Heat flows from higher tem, Electric charge flows, perature to lower temperature, from higher potential to, lower potential, The rate of flow of charge The rate of flow of heat may, is called the electric, be called heat current, i.e., H = dQ, dq, current, i.e, I =, dt, dt, The relation between, Similarly, the heat current may, the electric current, be related with the, and the potential, temperature difference as, difference is given by, θ −θ, H= 1 2, Ohm's law,i.e,, R, V1 − V2, I=, where R is the thermal, R, where R is the electrical resistance of the conductor, resistance of the, conductor, The electrical resistance The thermal resistance may, is defined as, be defined as, l, R=, ρl l, R= =, KA, A σA, where K=thermal, where resistivity and, conductivity of conductor, electrical conductivity, dq, V −V, =i = 1 2, dt, R, σA, = (V1 − V2 ), l, , NARAYANAGROUP, , dQ, θ −θ, =H = 1 2, dt, R, KA, =, (θ1 − θ2 ), l, , Fluid current, Fluid flows from higher, pressure to lower pressure, The rate of flow of fluid is, called Fluid current, V=, , volume, =Area×Velocity, Time, , V=, , P, R, , Where R is fluid, Resistance, , R=, , P 8ηl, =, V πr 4, , co-efficient of viscosity, , V=, , P, R, , 73
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , Ingen hausz experiment, , Radial flow of heat, Ø, , Consider two thin spherical shells of radii r1 and Ø If a number of identical rods of different metals are, coated with wax and one of their ends is put in, r2 . A medium of thermal conductivity ‘K’ is, boiling water, then in steady state the square of, contained between these shells. A heater is placed, length of the bar over which wax melts is directly, at the centre of the shells. Heat is conducted through, proportional to the thermal conductivity of the, the medium radially from inner to the outer shell., Let the temperatures of the inner and the outer shells, L2, 2, L, ∝, K, ⇒, =Constant, metal.i.e, be θ1 and θ2 at steady state., K, W.E-18. In the Ingen hausz method to compare the, thermal conductivities of different substances,, dθ, the length upto which wax melted in copper, and zinc rods are 9.3 cm and 5cm respectively., r P, Compare their thermal conductivities., dr, r, Sol.Given lc = 9.3cm and lz = 5cm, 1, , Ø, , 4π Kr1r2 (θ1 − θ2 ), dr, 4π K, ∫r r 2 = − H θ∫ dθ ⇒ H = ( r2 − r1 ), , r2, , θ2, , 1, , 1, , Thickness of the shell , ( r2 − r1 ) =, , 4π Kr1r2 (θ1 − θ 2 ), H, , W.E-17. A hollow sphere of glass whose external, and internal radii are 11 cm and 9 cm respectively is completely filled with ice at 0ºC and, placed in a bath of boiling water. How long, will it take for the ice to melt completely?, given thatdensity ofice = 0.9 g/cm 3, latent heat, of fusion of ice = 80 cal/g and thermal, conductivity of glass = 0.002 cal / cm-sºC., 4π Kr1r2 ∆θ, Sol: In steady state, rate of heat flow H = r − r, 2, 1, Substituting the values,, H=, , ( 4)(π )( 0.002)(11)( 9)(100 − 0), (11− 9), , dQ, = 124.4 cal / s, dt, , ;, , dQ, dm, = L, dt, dt, , Total mass of ice, m = ρice ( 4π r ) = ( 0.9 )( 4 ) π ( 9 ), = 916 g, ∴ Time taken for the ice to melt completely, 2, 1, , t=, , m, 916, =, = 589s, ( dm / dt ) 1.555, , ( θ1 ) and outer ( θ0 ) surfaces are fixed (θ1 > θ0 ) ,, find the heat flow through the jacket. (Apply, the heat conduction equation to steady state, radial heat flow corresponding to cylindrical, symmetry)., Sol: Consider a cylindrical shell of thickness ‘dr’ and, radius ‘r’., Cross sectional view of steam pipe is as shown in, figure., , R0 r, , dr, , R1, , (or), , dm dQ / dt 124.4, ∴, =, = 1.555 g / s, =, L, 80, dt , , 74, , 2, , K c lc2 9.3 , = 2 =, = 3.46, K, l, 5, , , z, z, Choose an element of radial thickness dr at a radial, Thus, the thermal conductivity of copper is 3.46, distance ‘r’ from the centre of shells. Let dθ be, times that of zinc., the temperature difference across it. The rate of, flow of heat through the element, W.E-19. A steam pipe with a radius R1 is sur dθ , rounded by an insulating jacket with an outer, 2 dθ , H = KA , = K π r dr (or), , , dr , radius of R0 . If the temperature of the inner, Q, , r2, , Let L be the length of the cylinder., The heat transferred per second,, R dr, dQ, dθ, dθ, q=, = − KA, = − K ( 2π rL ), ⇒∫, R r, dt, dr, dr, 0, , 2, , 1, , 2π LK, =−, q, , ∫, , θ0, , θ1, , dθ, , ∴q =, , 2π KL (θ1 − θ0 ), R, ln 0 , R1 , , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , W.E-20. A steam pipe of radius 5 cm carries steam, at 100ºC. The pipe is covered by a jacket of, insulating material 2cm thick having a thermal conductivity 0.07 W/m-K. If the tempera- Ø, ture at the outer wall of the pipe jacket is 20ºC,, how much heat is lost through the jacket per, meter length in an hour?, Ø, Sol.Thermal resistance per meter length of an element, at a distance r and thickness dr is, , k = 0.07 W / m-K, , (b)Forced convection, Ø, , 7cm r, , Ex: Hot air rises by natural convection. sea breeze,, land breeze, trade winds, monsoons etc. are also, due to free convection., In natural convection, gravity plays an important, role and it always takes place vertically carrying, the heat upwards., Natural convection can not take place in a gravity, free space such as orbiting satellite or freely falling, lift. Ex: (i) Ventilators placed below the roof allow, hot air to escape (ii) working of chimney., , dr, , 5cm, , If the fluid particles are forced to move by an, external agent (like fan or by a blower or by a pump, etc.,) it is called forced convection., Ex: heat transfer to parts of human body (blood, circulation)., , Convection co-efficient, Ø, , In forced convection, the rate of heat flow is, proportional to the surface area and the, 1, , QR = , temperature difference between the surface and the, KA, , fluid. P = Q / t = hA ∆θ ., r, 1 7×10 dr, Ø Where h is the coefficient of convection. A is surface, Total resistance R = ∫ r d R =, 2π K ∫5×10 r, area over which fluid moves, ∆θ is the difference, 1, 1, 7, ln (1.4 ) = 0.765 K/W, =, ln =, of temperature between the surface and the fluid., 2π K 5 ( 2π )( 0.07 ), W.E-21. A runner moves along the road at 2 ms-1 in, Temperature difference, still air that is at a temperature of 29oC. His, Heat current H = Thermal resis tan ce, surface area is 1.4m2, of which approximately, 85% is exposed to the air. Find the rate of, convective heat loss from his skin at a, (100 − 20) = 104.6W, =, temperature 35 o C to the outside air?, 0.765, Coefficient of convection for dry air and bare, Heat lost in one hour = Heat current × time, skin at wind speed 2ms-1 is 22W/m2-oC., 5, = (104.6 )( 3600 ) J = 3.76 × 10 J, Q, = h A ∆θ ; h=22 W/m2-oC, Sol:, Convection, t, The process in which heat is transferred from one, ∆θ =35-29=6oC, place to other by the actual movement of particles, A = 85% of the surface area of the runner, of medium due to difference in density ., 1.4 × 85, 2, i.e., A =, Convection takes place in fluids. It is more, 100 ≅ 1.2m, predominant in liquids., Q, = 22 ×1.2 × 6 = 160W, Heat transfer in mercury takes place by, t, conduction not by convection., Radiation:, Convection is of two types. (a) Natural or Free, Ø It is the process of transmission of heat from one, convection (b) Forced convection., place to another without any material medium., (a) Natural or Free convection, If the fluid particles move only due to the Ø It is the fastest process of heat transmission., Thermal radiation:, temperature difference (density difference), it is, Ø Heat energy transferred by means of, called natural convection., electromagnetic waves is Thermal radiation., , dr, dR =, K ( 2π r ), , 2, , −2, , −2, , 1, , Ø, Ø, Ø, Ø, Ø, , NARAYANAGROUP, , 75
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , Nature and properties of radiant energy:, Ø, Ø, Ø, Ø, Ø, , Reflecting power (r):, , It consists of long wavelength electromagnetic, radiation., The wave length of these waves is nearly 800nm to, 4,00,000 nm. (or 7800Ao to above 1 mm), It occupies the infrared region of the, electromagnetic spectrum., velocity is 3 x 108 ms-1 like light waves., The intensity of radiant energy obeys inverse square, 1, ; where I = Intensity of radiation, d2, d = distance from the source, , law. I ∝, , r=, , Transmitting power (t):, t=, , Ø, Ø, Ø, Ø, , Black body radiation:, Ø, , Every object emits and absorbs radiant energy at, all temperatures except at absolute zero., The energy emitted by a body does not depend on Ø, the temperature of the surroundings., The rate of emission increases with the increase in, the temperature of the body., If two bodies continuously emit and absorb same, amount of energy, then they are in thermo dynamic, equilibrium., The radiant energy emitted by a body depends on, the nature of the surface of the body, surface area, of the body and temperature of the body., , Ø, , At a given temperature, for a given wavelength, range, the ratio of energy absorbed to the energy Ø, incident on the body is absorptive power., , Ø, 76, , Amount of radiant energy absorbed, Amount of radiant energy incident, , aλ = 1 .( for black body), , 1646K, 1449K, , Ελ, , 1259K, , (λ), , The ratio of radiant energy emitted by a surface to, radiant energy emitted by a black body under same, Ø, conditions is called emissivity., For a perfect black body emissivity e= 1 and e = 0, Ø, (for perfect reflector);e=0.97(for human skin), , ∴ aλ =, , Y, , 904K, , Absorptive power (aλ ) :, Ø, , Distribution of energy in black body, spectrum, , 908K, , The amount of energy emitted per second per unit, surface area of a body at a given temperature for a, given wavelength range (λ and λ + dλ ) is called, emissive power., , Emissivity (e):, Ø, , A body which completely absorbs all the heat, radiation incident on it is called a perfect black body, (or), A body which emits the radiation of all wavelengths, when it is at high temperature is called perfect black, body., Ferry’s black body and Wien’s black body are, examples of artificial black bodies., , 1095K, , Emissive power (eλ ) :, Ø, , Amount of radiant energy transmitted, Amount of radiant energy incident, , aλ + r + t = 1 , ‘ aλ ’ is absorptive power, ‘r’ is, reflecting power & ‘t’ is the transmitting power., , Prevost’s theory of heat exchange:, Ø, , Amount of radiant energy reflected, Amount of radiant energy incident, , X, , Observations from graph, Intensity of radiation increases with increase of, wavelength., For a particular wavelength ( λ m ) the intensity of, radiation emitted is maximum. Beyond λ m intensity, of radiation emitted decreases., Area under the curve ( E λ versus λ ) represents, total energy emitted per sec per unit area by a black, body corresponding to all the wavelengths., , Wien’s displacement law:, Ø, , Wavelength corresponding to the maximum, intensity ( λ m ) shift towards left (or smaller, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , wavelength side) along the axis (i.e decreases ) as Ø, the temperature of the body is increased. So the, , Radiant energy emitted by a hot body per second, , wavelength ( λ m ) varies inversely as the absolute, , Stefan - Boltzmann’s law:, , Ø, , P = eAσT 4, , 1, temperature of the body. λ m ∝ or λ m T = b, T, where ‘b’ is known as Wien’s constant, , If a black body at absolute temperature T is, surrounded by an enclosure at absolute, temperature To , then the rate of loss of heat energy, per unit area by radiation is given by, , b = 2.9×10−3 m− K, , E = σ (T 4 − T0 4 ), , Y, , Ø, , λm, , For any hot body, E = σ e(T 4 − T04 ), Where ‘e’is the emissivity of the body., Radiant power of any body, P = σ Ae (T 4 − T04 ), , Temperature of the Sun & Solar constant :, Ø, (T), , X, , The temperature of the Sun can be determined by, assuming it to be a black body., , Y, λm, , R, , 1, T, , λ m1, λm2, , Ø, Ø, Ø, , X, , υm2, T, = 2 =, T1, υ m1, , On increasing temperature of a body, its colour, changes gradually from red → orange → yellow, → green → blue → violet., Thus the temp of violet star is maximum and temp, of red star is minimum., Sun is a medium category star with λ m = 4753A 0, (yellow colour) and temp about 6000K., , Stefan’s law:, Ø, , The amount of heat radiated by a black body per, second per unit area is directly proportional to the, fourth power of its absolute temperature., E ∝ T 4 ⇒ E = σ T 4 ( σ = Stefan’s constant ), , σ = 5.67 ×10−8Wm−2 K −4, Ø, , Dimensional formula of σ is M L T K, , NARAYANAGROUP, , Earth, , 1 0, , −3, , −4, , , , Ø, , Total amount of radiant energy emitted by the sun, , Ø, , 2, 4, per second = ( 4π R ) σ T ;R=Radius of the Sun, let ‘r’ be the mean distance between the sun and, , the earth and S 0 be the solar constant. Then the, energy received per second by the sphere of radius, , ( r ) is 4π r 2S0, Solar constant: Solar constant is defined as “the, rate at which the radiant energy of the sun received, by perfectly black surface, normal to unit area in, the absence of atmosphere, when kept at distance, equal to the mean distance of earth from the Sun”, 4π R 2σ T 4 = 4π r 2S0, , r 2 S , r 2 S , T = 0 ⇒ T = 0 , R σ , R σ , Hence, the surface temperature of the Sun can be, determined., 1/4, , 4, , 77
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, Ø, , At a give temperature, for a given wavelength range,, the ratio of emissive power to absorptive power of, a body is constant and that constant is equal to the, emissive power of a perfect black body at the given, temperature and wavelength., , eλ, = constant = Eλ, i.e., aλ, Ø, , The warming of earth atmosphere and surface due, , θ0, θ1, , θ2, t1, , ⇒, , dQ, dθ, = K (θ − θ 0 ) ⇒ − ms, = K (θ − θ 0 ), dt, dt, , ⇒−, , Ø, , a) The excess of temperature (θ − θ 0 ) must, , Ø, , not be large i.e. (θ − θ 0 ) ≤ 30 − 350, b) Newton’s law of cooling is a special case of, Stefan’s law, because this law can be derived from, Stefan’s law., The curve between rate of cooling and temperature, difference is a straight line passing through origin., Q R ∝ (θ − θ0 ) , , dθ, = Rate of fall of temperature or rate, dt, , Y, , R, , of cooling., Rate of loss of heat of a hot body due to cooling, dQ, dθ, = ms, ., dt, dt, , Ø, , Here ‘m’, ‘s’ are mass and specific heat of the body. Ø, The rate of cooling of a hot body is directly, proportional to the mean excess temperature of, the body above that of the surroundings, provided, the difference in temperature of the body and that, of surroundings is small., dθ, dθ, θ +θ, , = K 1 2 − θ s here, = Rate of cooling., dt, 2, , dt, , body respectively. θ s is temperature of surroundings, and K is the cooling constant., , X, , (θ−θ0), , Curve between the rate of cooling and body, temperature., , R = K (θ − θ 0 ) = Kθ − Kθ 0, Y, , R, , θ0, (θ), , θ1 ,θ 2 are the initial and final temperatures of the, , 78, , X, , Limitations of Newton’s law of cooling:, , dθ, K, =, (θ − θ 0 ) (K= Cooling constant), dt ms, , where −, , t2, , surroundings, Ø, , Newton’s law of cooling, Rate of loss of heat ∝ (θ − θ 0 ), , Time, , θ 2 = θ 0 + (θ1 − θ 0 ) e − Kt + C , θ 0 =temperature of the, , to greenhouse gases( CO2, N2O,C.F.C, O3 ) is called, green house effect., Ø, , Cooling curve, Y, , i.e. good absorbers are good emitters., Fraunhoffer lines in solar spectrum can be explained, on the basis of Kirchhoff’s law., , Green house effect, Ø, , Ø, , Temperature, , Kirchhoff’s law :, , X, , −Κθ0, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, Ø, , TRANSMISSION OF HEAT, , Because log e (θ − θ0 ) = −Kt + log e A .This is the Ø, equation of a straight line, hence the curve between, , log (θ − θ0 ) and ‘t’ will be a straight line ., , Ø, , Y, , In Newton’s law of cooling, if air is dry then the, law of cooling remains valid even at high, temperatures., To determine specific heat of a liquid: In an, experiment of Newton’s law of cooling m1 is the, mass of calorimeter, m2 is the mass of water in it, , loge (θ−θ0), , or m3 is the mass of liquid in it. If t1 and t2 are the, times taken by water and liquid to cool from, temperatures θ2 to θ1 in the surroundings of, temperature θ s , then specific heat of liquid can be, calculated from, , X, t, , Ø, , Curve between log e R and log e (θ − θ 0 ), , t1 m1 s1 + m 2 s 2, =, t 2 m1 s1 + m 3 s 3, , Y, , Where s1 , s2 and s3 are the specific heats of, calorimeter, water and liquid respectively., loge R, , Additional formulae:, Ø, loge (θ−θ0), , Ø, , Ø, , Ø, , X, , If the rates of cooling of two bodies are same then, the rate of fall of temperature of the body with highest Ø, dθ, 1, ∝, heat capacity will be the least. i.e., dt ms, if R is constant, If two liquids are cooled under identical conditions, (i.e. surface area, temperature difference and time, difference are same) then their rates of cooling will, be the same. R1 = R2 ⇒ (θ −θ0 )1 = (θ −θ0 )2, If two liquids are cooled under identical, circumstances then their rates of fall of temperature, will not be same. The rate of fall of temperature, , Ø, , rate of fall of temperature, NARAYANAGROUP, , dθ, 1, ∝, dt r3 ρ S, , ), , Initial rate of fall of temperature of a spherical body, of radius ‘r’ can be found by substituting V =, , i.e.,, , Ø, , 1, ms, When a solid sphere of radius r, density ρ and, Ø, specific heat S is heated to temperature θ and then, , cooled in an enclosure to temperature θ0 , then its, , (, , dθ eA 4 4, =, T − T0, dt Vρs, , 4 3, πr, 3, , and A = 4πr 2 in the above formula., , ( Rθ ) of that liquid will be minimum whose specific, heat is maximum and vice versa. i.e.Rθ ∝, , Rate of fall of temperature of a body at temperature, T kept in surroundings of temperature T0 is, , (, , ), , dθ e4πr 2, 3e, =, T 4 − T04 =, dt 4 πr 3ρs, rρs, 3, , (T4 − T04 ), , Initial rate of fall of temperature of cubical body of, side length L is, , (, , dθ e6L2 4, =, T − T04, dt L3ρs, , =, , (, , 6e, T 4 − T04, Lρs, , ) (Q A = 6L2 , V = L3 ), , ), , Ratio of initial rates of fall of temperature of a sphere, and a cube of same material in the same surroundings, , ( dθ / dt )1, is dθ / dt, (, )2, , e L T4 − T4 , = 1 14 04 , 2e 2 r T2 − T0 , 79
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , W.E-22. The plots of intensity versus wavelength W.E-25. If a black body is radiating at T = 1650 K,, for three black bodies at temperatures T1,T2, at what wavelength is the intensity maximum?, and T3 respectively are shown in fig. Their, −3, temperatures are shown in fig. How their Sol: According to Wien’s law, λmT = 2.9×10 mK, temperatures are related?, 2.9 × 10 −3, λmax =, , Y, , T3, , (I), , T2, , T1, , X, , (λ), , 1, T, from graph λ2 > λ3 > λ1 ∴T1 > T3 > T2, W.E-23. Variation of radiant energy emitted by Sun,, filament of Tungsten Lamp and welding arc, as a function of its wavelength shown in fig., Identify the temperatures of Sun, filament, lamp and welding arc., , Sol: From Wien’s displacement law λm ∝, , Y, , W.E-26. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer, surface area of the two bodies are the same., The two bodies emit total radiant power at the, same rate. The wavelengths λA and λB corresponding to maximum spectral radiancy in the, radiation for A and B respectively differ by 1.00, µ m. If the temperature of A is 5802 K. Find, (a) the temperature of B , (b) λB, Sol: Given eA = 0.01, eB = 0.81 and TA = 5820 K, Power radiated P = eσ T 4 and e ATA4 = eBTB4, 1, , 1, , e 4, 0.01 4, ∴TB = A × TA = , × 5802 = 1934 K, 0.81 , eB , , T1, X, , Sol: From Wien’s displacement law, Sun-T3, tungsten filament T 2, welding arc-T1, , W.E-24. The frequency (ν m ) corresponding to, which energy emitted by a black body is, maximum, may vary with temperature T of the, body as shown in figure. Which of the curves, represents correct variation?, 1, Sol: From Wien’s displacement law λm ∝, T, ∴υ ∝ T Hence graph A is correct, Y, DA, C, , From Eqs. (i) and (ii) λB = 1.5 × 10−6 m = 1.5 µ m, W.E-27. Two spherical bodies A (radius 6cm) and, B(radius 18cm) are at temperature T1 and T2 ,, respectively. The maximum intensity in the, emission spectrum of A is at 500 nm and in, that of B is at 1500 nm. Considering them to, be black bodies, what will be the ratio of the, rate of total energy radiated by A to that of B., (JEE-2010), TA λB 1500, Sol. λmT = constant ; T = λ = 500 = 3, B, A, , Rate of total energy radiated P ∝ AT 4 ∝ R 2T 4, 2, , B, , 80, , ......(i), , λB TA 5802, =, =, = 3 ∴ λ = 3λ ......(ii), B, A, λ A TB 1934, , T2, , (λ), , ∴ λB − λA = 1µ m (given), , ∴ λB − λ A = 1× 10−6 m, From Wien’s displacement law,, , T3, , O, , = 1.8 µ m, , as TB < TA , λB > λA, , E, , µm, , 1650, , T, , X, , 4, , 2, PA RA TA 6 , 4, ∴ =, × = × ( 3) = 9 :1, PB RB TB 18 , , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , W.E-28. Find the temperature of an oven if it radiates 8.28 cal per second through an opening, whose area is 6.1 cm2. Assume that the, radiation is close to that of a black body., Sol. The emittance of the oven, E =, , r2 , 2.0 , ∴ I2 = I1 12 = 102 = 2.55 ×103 W / m 2, 0.4 , r2 , The power at the surface of the sphere, 2, , (, , 2, , P = e σ AT 4 (or) 2.55 × 103 × 4π ( 0.4 ) =, 2, , From Stefan’s law, E = σ T 4 ,, Where σ = 5.67 × 10−8W / m2 − K 4, T4 =, , ), , P = I 2 4π r22 = 2.55 ×103 × 4π ( 0.4) and, , 8.28 × 4.2, = 5.7 × 10 4 Watt / m 2, 6.1 × 10 − 4, , e × 5.67 × 10−8 × 4π ( 0.4 ) × ( 523), 2, , E, 5.7 ×104, = 1×1012 K 4, ,=, −8, σ, 5.67 ×10, , ∴e =, , 2.55 × 103, 5.67 × 10 −8 × ( 523), , 4, , 4, , = 0.61, , ∴T = 10 K = 1000 K, W.E-29.Three very large plates of same area are W.E-31: One end of a rod of length 20cm is inkept parallel and close to each other. They are, serted in a furnace at 800K. The sides of the, considered as ideal black surface and have very, rod are covered with an insulating material, high thermal conductivity . The first and third, and the other end emits radiation like a black, plates are maintained at temperatures 2T and, body. The temperature of this end is 750 K in, 3T respectively. Find the temperature of the, the steady state. The temperature of the surmiddle ( i.e. second ) plate under steady state, rounding air is 300K. Assuming radiation to, ( 2012 JEE ), be the only important mode of energy transfer, Sol., between the surrounding and the open end of, T, 2T, 3T, the rod, find the thermal conductivity of the, rod. Stefan’s constant σ = 6.0 ×10−8W / m2 − K 4, 3, , 1, , Sol., , Furnace, 800 K, Let T1 is the temperature of the middle plate, Under steady state, rate of emission = rate of, absorption, σ ( 2 A )(T1 ) = σ A ( 2T ) + σ A ( 3T ), 4, , 4, , 1, , 4, , 97 , 2T14 = 16T 4 + 81T 4 ⇒ T1 = T, 2 , W.E-30 : A sphere with diameter of 80cm is held at, a temperature of 250ºC and is radiating energy. If the intensity of the radiation detected, at a distance of 2.0m from the sphere’s centre, is 102 W/m 2, What is the emissivity of the, sphere?, Sol. From inverse square law I ∝, , 4, , 1, r2, , I2, r2, Intensity on the surface of the sphere, =, = 12, I1 Intensity at a distance 2.0m from the centre r2, NARAYANAGROUP, , 20cm, , 750 K, Airtemp, 300 K, , Quantity of heat flowing through the rod per second in steady state., dQ K . A.dθ, =, .... (i), dt, x, Quantity of heat radiated from the end of the rod, per second in steady state:, , (, , dQ, = Aσ T 4 − T04, dt, , From Eqs. (i) and (ii), , ), , --- (ii), K .dθ, = σ T 4 − T04, x, , (, , ), , K × 50, 4, 4, = 6.0 ×10−8 ( 7.5 ) − ( 3) × 108, , , 0.2, K=74 W/mk, 81
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, W.E-32. A body cools from 80ºC to 50ºC in 5 min- 3., utes. Calculate the time it takes to cool from, 60ºC to 30ºC. The temperature of the surroundings is 20ºC., Sol: According to Newton’s law of cooling, dθ, θ +θ, , = K 1 2 − θ0 , dt, 2, , , For decrease of temperature from 800 C to 500 C, 80 − 50, 80 + 50, , =K, − 20 ......(1), 5, 2, , , 4., , For decrease of temperature from 600 C to 300C, 60 − 30, 60 + 30, , =K, − 20 ..........(2), t, 2, , From eq’s (1) and (2) , t = 9 min, W.E-33. Two spheres made of same material have, their radii in the ratio 1 : 3. They are heated, to the same temperature and kept in the same, surroundings at a moderate temperature. 5., Show that the ratio of their initial rates of fall, of temperature is 3:1 if the bodies are cooled, by natural convection and radiation., Sol. The rate of fall of temperature of a hot body is given, , by, , dθ eAσ 4, =, T − T04, dt, ms, , (, , (, , ), , 6., , ), , 2, 4, 4, dθ e 4π R σ T − T0, dθ, 1, ∴, =, ⇒, ∝, 4, dt, dt, R, π R3 ρ s, 3, , ( dθ / dt )1, ( dθ / dt )2, , R1, , Q = 1:3 , R2, , Their ratio of fall of temperature is 3 : 1., , C.U.Q, , 1., , 2., , 82, , =, , R2 3, =, R1 1, , 7., , 8., 9., , Two rods of different materials having different, lengths and same cross sectional areas are, joined end to end in a straight line. The free, ends of this compound rod are maintained at, different temperatures. The temperature, gradient in each rod will be, 1) same, 2) zero, 3) directly proportional to thermal conductivity of rod, 4) inversely proportional to thermal conductivity of, the rod, A piece of paper wrapped tightly on a wooden, rod is found to get charged quickly when held, over a flame compared to similar piece when, wrapped on a brass rod. This is because, 1) brass is good conductor and wood is a bad, conductor of heat, 2) brass is a bad conductor of heat, 3)wood contains large number of free electrons, 4) wood is a good conductor of heat, When heat flows through a wire of uniform, cross section under steady state, then, 1) temperature gradient is same every where, 2) temperature at a particular point remains same, 3) rate of heat flow is same at all cross sections, 4) all the above, Temperature is analogous to, 1) charge, 2) potential difference, 3) electric field strength 4) force, On heating one end of a rod the temperature, of the whole rod will be uniform when, 1) k = 1 2) k = 0 3) k = 100 4) k = ∞, For an ideal conductor thermal resistance is, 1) unity 2) infinity, 3) zero 4) 1000, A metal rod of area of cross section A has, length L and coefficient of thermal conductivity, K.The thermal resistance of the rod is, , L, KL, KA, A, 2), 3), 4), Metals are good conductors of heat because, KA, A, L, KL, 10. Thermal conductivity of a metal rod depends, 1) they contain large number of free electrons, on, 2) their atoms are relatively apart, 1) area of cross section 2) temperature gradient, 3) their atoms collide frequently, 3) time of flow of heat, 4) all the above, 4) they have reflecting surfaces, 11. Coefficient of thermal conductivity, In steady state, 1) depends upon nature of the material of the body, 1) heat received is partly conducted and partly, 2) is independent of dimensions of the body, radiated, 2) heat is not absorbed, 3) both 1 and 2, 3) both 1 and 2 4) all the heat is conducted, 4) depends on temperature difference, , 1), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 12. If the end of metal rod is heated, then the rate, of flow of heat does not depend on, 1) area of the end of the rod 2) mass of the rod, 3) time, 4) temperature gradient, 13. In the following solids thermal conductivity is, maximum for, 1) copper 2) aluminium 3) gold 4) silver, 14. For a perfect insulator coefficient of thermal, conductivity is, 1) zero, 2) infinite 3) one, 4) two, 15. It is hotter at some distance over the flames, than in front of it because, 1) air conducts heat upwards only, 2) heat is radiated upwards only, 3) convection of heat occurs upwards only, 4) heat is radiated downwards only, 16. The process in which rate of transfer of heat, maximum is, 1) conduction 2) convection 3) radiation, 4) in all these heat is transferred with the same speed, 17. By which of the following methods could a cup, of hot tea loss heat when placed on metallic, table in a class room, a) conduction, b) convection, c) radiation, d) evaporation of liquid, 1) a,b, 2) b,c 3)a,b,c 4) a,b,c,d, 18. The thermal radiations are similar to, 1) X-rays, 2) cathode rays, 3) α − rays, 4) γ -rays, 19. The temperature at which a black body ceases, to radiate energy is, 1) 0 K, 2) 273 K, 3) -273 K, 4) at all temperatures, 20. The intensity of energy radiated by a hot body, at a distance r from it varies as, 1, 1, 4) 3, 4, r, r, 21. When a body has the same temperature as that, of its surroundings, 1) it does not radiate heat, 2) it radiates same quantity of heat as it receives, from the surroundings, 3) it radiates less quantity of heat as it receives, from the surroundings, 4) it radiates more quantity of heat as it receives, from the surroundings., , 1) r2, , 2), , NARAYANAGROUP, , 1, r2, , 3), , TRANSMISSION OF HEAT, 22. One half of a slab of ice is covered with black, cloth and the other half with white cloth. This, is then placed in sunlight. After some time the, pieces of cloth are removed. Then, 1) ice has melted equally under both the pieces, 2) more ice has melted under white cloth, 3) more ice has melted under black cloth, 4) it will depend on the medium in which ice is, placed, 23. Compared to a person with white skin another, person with dark skin, will experience :, 1) less heat and more cold, 2) more heat and more cold, 3) more heat and less cold, 4) less heat and less cold, 24. Which of the following statements is wrong?, 1) rough surfaces are better radiators than smooth, surfaces, 2) highly polished mirror like surfaces are very good, radiators, 3) black surfaces are better absorbers than white, ones, 4) black surfaces are better radiators than white, ones, 25. The physical factor distinguishes thermal, radiation from visible light is, 1) wavelength, 2) pressure, 3) temperature, 4) amplitude, 26. If we place our hand below a lighted electric, bulb. We feel warmer because of, 1) convection, 2) radiation, 3) conduction, 4) both 1 and 2, 27. Heating effect of the incoming solar radiation, is maximum at local noon because, 1) atmospheric absorption is zero, 2) sun’s rays travel through minimum air, thickness, 3) solar rays are vertical to the ground, 4) outgoing radiation is minimum, 28 The absorptivity of Lamp black and platinum, black is, 1) 0.91 2) 0.98 3) 1.00 4) 0.99, 29. Absorptive power of a white body and of a, perfectly black body respectively are, 1) 1,0, 2) 0,1, 3)-1,-1, 4) ∞ ,0, 0, 30. Three bodies A,B,C are at -27 C, 00C, 1000C, respectively. The body which does not radiate, heat is, 1) A 2) B 3) All the bodies radiate heat 4) C, 83
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, 31. At high temperature black body spectrum is, 1) continuous absorption 2) line absorption, 3) continuous emission 4) line emission, 32. A black body emits, 1) radiations of all wave lengths 2)no radiation, 3) radiation of single wave length, 4) radiation of selected wave length, 33. The best laboratory approximation to an ideal, black body is, 1) a lump of charcoal heated to high temperature, 2) a glass surface coated with coal tar, 3) a metal coated with black dye, 4) a hollow enclosure blackened inside with lamp, black and have a small hole, 34. A black body does not, 1) emit radiation, 2) reflect radiation, 3) absorb radiation, 4) emit and absorb radiation, 35. If the amount of heat energy received per unit, area from the Sun is measured on Earth, Mars, and Jupiter, it will be, 1) the same for all, 2) in decreasing order of Jupiter, Mars, Earth, 3) in increasing order of Jupiter, Mars, Earth, 4) in decreasing order of Mars, Earth, Jupiter, 36. The colour of a star is a measure of its, 1) age, 2) temperature, 3) size, 4) distance from the earth, 37. A polished metal plate with a rough black spot, on it is heated to about 1400K and quickly, taken into a dark room. Then, 1) the spot will appear brighter than the plate, 2) the spot will appear darker than the plate, 3) heat conduction is easier downward, 4) it is easier and more convenient to do so, 38. If ‘p’ calorie of heat energy is incident on a, body and absorbs ‘q’ calories, its coefficient, of absorption is, 1) p/q, 2) p - q 3) q/p 4) q + p, 39. The velocity of heat radiation in vacuum is, 1) Equal to that of light 2)Less than that of light, 3) Greater than that of light, 4) Equal to that of sound, 40. Distribution of energy in the spectrum of a, black body can be correctly represented by, 1) Wien’s law, 2) Stefan’s las, 3) Planck’s law, 4) Kirchhoff’s law, 84, , 41. Four pieces of iron are heated to different, temperatures. The colours exhibited by them, are respectively red,yellow, orange and white, respectively. The one that is heated to the, highest temperature will exhibit the colour, 1) White 2) Yellow 3) Red 4)Orange, 42. A star which appears blue will be, 1) much hotter than the sun, 2) colder than the Sun, 3) as hot as the Sun 4) at -2730 C, 43. If a star is colder than the Sun it appears, 1) Yellow 2) Red 3) Blue 4) Violet, 44. The amount of radiation emitted by a perfectly, black body is proportional to, 1) temperature on ideal gas scale, 2) fourth root of temperature on ideal gas scale, 3) fourth power of temperature on ideal gas, scale, 4) source of temperature on ideal gas scale, 45. At a given temperature, the ratio between, emissive power and absorptive power is same, for all bodies and is equal to the emissive power, of black body.This statement is called, 1) Newton’s Law, 2) Planck’s law 3, ), Kirchoff’s law, 4) Wien’s, law, 46. If the sun become twice hotter, it will radiate, 1) energy sixteen times larger, 2) predominantly in the infrared, 3) predominantly in the ultra violet, 4) energy sixteen times smaller, 47. Three identical spheres of different materials, iron, gold and silver are at the same, temperature. The one that radiates more, energy is, 1) Gold, 2) Silver, 3) Iron, 4) All radiate equally, 48. Cooling graphs are drawn for three liquids a,b, and c. The specific heat is maximum for liquid, , Y, T, e, m, p, , a, b, c, X, , Time, 1) a, 2) b, 3) c, 4) for all the three a,b and c, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 49. A black body of temperature T is inside a, chamber of temperature T0 . Now the closed, chamber is slightly opened to Sun such that, temperature of black body (T ) and chamber, (T0 ) remain constant, 1) Black body will absorb more radiation from the, Sun, 2) Black body will absorb less radiation from the, Sun, 3) Black body emits more thermal energy, 4) Black body emits thermal energy equal to the, thermal energy absorbed by it., 50. The rate of cooling of a body is, 1) independent of the nature of the surface of, the body, 2) independent of the area of the body, 3) dependent on the excess of temperature of, the body above that of the surroundings., 4) independent of the temperature of the, surroundings., 51. A cube, a sphere and a circular plate made of, same material and having same mass are, heated to same high temperature. The body, that cools at the least rate when left in air at, room temperature is, 1) Sphere, 2) Cube, 3) Circular plate, 4) All at the same rate, 52. Newton’s law of cooling is applied in laboratory, for the determination of the, 1) Specific heat of gases 2) Latent heat of gases, 3) Specific heat of liquids 4)Latent heat of liquids, 53. Newton’s law of cooling is a special case of, 1) Kirchoff’s law, 2) Wien’s law, 3) Stefan-Boltzmann’s law 4) Planck’s law, 54. The amount of heat energy radiated per second, by a surface depends upon:, 1) Area of the surface, 2) Difference of temperature between the surface, and its surroundings, 3) Nature of the surface 4) All the above, 55. Four identical copper cylinders are painted. If, they are all heated to the same temperature, and left in vacuum which will cool most rapidly., 1)Painted shiny white 2)Painted rough black, 3)Painted shiny black 4)Painted rough white, 56. If T B and T s are the temperatures of the body, and the surroundings and TB − TS is of very, high value, then the rate of cooling in natural, convection is proportional to, NARAYANAGROUP, , TRANSMISSION OF HEAT, 1) TB, , 4, , 2) TS 3) (TB − TS ), 4, , 5, 4, , 4), , 5, 54, , 4, TB − TS , , , , 57. Newton’s law of cooling is a law connected with, 1) Conduction 2) Convection, 3) Radiation, 4) Convection and Radiation, 58. Newton’s law of cooling holds good provided, the temperature difference between the body, and the surroundings is, 1) large, 2) small, 3) very large, 4) any value, 59. A block of steel heated to 1000C is left in a, room to cool. Which of the curves shown in, the figure, represents the correct behaviour, , Y, T, e, m, p, , 60., , 61., , 62., , 63., , C, B, A, , X, Time, 1) A, 2) B, 3) C, 4)A and C, Let there be four articles having colours blue,, red, black and white. When they are heated, together and allowed to cool, the article that, cool earlier is, 1) Blue 2) Red, 3) Black 4) White, Which of the following qualities are best suited, for cooking utensils?, 1) High specific heat and low thermal conductivity, 2) High specific heat and high thermal conductivity, 3) Low specific heat and low thermal conductivity, 4) Low specific heat and high thermal conductivity, The bulb of a thermometer is spherical and that, of another is cylindrical. Equal quantity of, mercury is filled in them. Then, 1) thermometer with spherical bulb will respond, quickly, 2) thermometer with cylindrical bulb will respond, slowly, 3) thermometer with spherical bulb will respond, slowly, 4) thermometer with cylindrical bulb will respond, quickly, Which of the following methods of flow of heat, is (are) based on gravitational attraction?, 1) Conduction, 2) Convection, 3) Radiation, 4) All of these, 85
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TRANSMISSION OF HEAT, , JEE- ADV PHYSICS-VOL- V, , 64. Two rods A and B of same metal and of same 73. A surface at temperature T0 K receives power, cross-section have length in the ratio 1:2. One, P by radiation from a small sphere at, end of each rod is at O 0C and temperature of, temperature T < T0 and at a distance d. If both, 0, 0, other ends are 30 C and 40 C respectively., T and d are doubled, the power received by, Which of the rod will have higher flow of heat?, the surface will become, 1) Rod A 2) Rod B 3) Both will have same, 1) P, 2) 2P, 3) 4P, 4) 16P, 4) Depends upon the shape, 74. Two circular disc A and B with equal radii are, 65. While measuring the thermal conductivity of, blackened. They are heated to same, liquids, the upper part is kept hot and lower, temperature and are cooled under identical, cooled so that, conditions. What inference do you draw from, 1) convectional flow is stopped, their cooling curves (R is rate of cooling)?, 2) radiation is stopped 3) conduction is easier, Y, 4) it is easier to perform the experiment, 66. The amplitude of radiations from a cylindrical, A, R, heat source is related to the distance is, 1, d, 2, 3) A ∝ d, 4) A ∝ d, Kirchoff’s law states that, 1) a body absorbs radiation of shorter wavelengths, and emits radiation of higher wavelength, 2) a body absorbs radiation of any wavelength but, emits radiation of specific wavelengths, 3) a body absorbs and emits radiation of same, wavelengths, 4) none of these, If pressure on a gas is increased from P to 2P,, then its heat conductivity, 1) increases, 3) decreases, 3) becomes zero, 4) remains unchanged, Two layers of cloth of equal thickness provide, warmer covering than a single layer of cloth, of double the thickness,because they, 1) behave like a thermos 2) have lesser thickness, 3) allow heat from atmosphere to the body, 4) enclose between them a layer of air, In a room containing air, heat can go from one, place to another by, 1) conduction, 2) convection, 3) radiation, 4) all of these, The reflectance and emittance of a perfectly, black body are respectively, 1) 0,1, 2) 1,0, 3) 0.5,0.5 4) 0,0, Wien’s displacement law fails at, 1) low temperature, 2) high temperature, 3) short wavelength 4) long wavelength, , 1) A ∝ 1 / d 2, 67., , 68., , 69., , 70., , 71., , 72., , 86, , 2) A ∝, , B, , (θ−θ0), , X, , 1) A and B have same specific heats, 2) Specific heat of A is less, 3) Specific heat of B is less 4)Nothing can be said, 75. A solid at temperature T1 is kept in an, evacuated chamber at temperature, T2 ( T2 > T1 ). The rate of increase of, temperature of the body is proportional to, 1) T2 − T1 2) T22 − T12 3) T23 − T13 4) T24 − T14, 76. The thermal radiation emitted by a body is, proportional to Tn where T is its absolute, temperature. The value of n is exactly 4 for, 1) a blackbody, 2) all bodies, 3) bodies painted black only, 4) polished bodies only, 77. A blackbody does not, 1) emit radiation, 2) absorb radiation, 3) reflect and refract radiation 4) All the above, 78. In summer, a mild wind is often found on the, shore of a calm river. This is used due to, 1)difference in thermal conductivity of water and soil, 2) convection currents, 3) conduction between air and the soil, 4) radiation from the soil, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , 79. A heated body emits radiation which has, maximum intensity near the frequency ν 0 . The, emissivity of the material is 0.5. If the absolute, temperature of the body is doubled,, 1) the maximum intensity of radiation will be near, , C.U.Q - KEY, , the frequency 2ν 0, 2) the maximum intensity of radiation will be near, ν0, 2, 3) the total energy emitted will increase by a factor, of 32, 4) the total energy emitted will increase by a factor, of 8, 80. Radiation is passing through a transparent, medium, then, 1) the temperature of medium increases, 2) the temperature of medium decreases, 3) the temperature of medium does not alter, 4) the temperature of medium first increases and 1., then becomes steady., 81. The graph shown in the adjacent diagram,, represents the variation of temperature T of, two bodies x and y having same surface area,, with time (t) due to emission of radiation. Find, the correct relation between emissive power(E) 2., and absorptive power(a) of the two bodies, , the frequency, , Y, T, , 3., , y, X, , 4., , t, , X, , 1) E x > E y and a x < a y, 2) E x < E y and a x > a y, 3) E x > E y and a x > a y, 5., 4) E x < E y and a x < a y, 82. In which of the following process, convection, does not takes place primarily?, 1) Sea and Land breeze, 2) Boiling of water, 3) Warming of glass of bulb due to filament, 4) Heating of air around a furnace, NARAYANAGROUP, , 01) 1, 07) 4, 13) 4, 19) 1, 25) 1, 31) 3, 37) 1, 43) 2, 49) 4, 55) 2, 61) 4, 67) 3, 73) 3, 79) 1, , 02) 3, 08) 3, 14) 1, 20) 2, 26) 2, 32) 1, 38) 3, 44) 3, 50) 3, 56) 1, 62) 4, 68) 4, 74) 2, 80) 3, , 03) 4, 09) 1, 15) 3, 21) 2, 27) 2, 33) 4, 39) 1, 45) 3, 51) 1, 57) 4, 63) 2, 69) 4, 75) 4, 81) 3, , 04) 1, 10) 4, 16) 3, 22) 3, 28) 2, 34) 2, 40) 3, 46) 1, 52) 3, 58) 2, 64) 1, 70) 2, 76) 2, 82) 3, , 05) 4, 11) 3, 17) 4, 23) 2, 29) 2, 35) 3, 41) 1, 47) 4, 53) 3, 59) 1, 65) 1, 71) 1, 77) 3, , 06) 2, 12) 2, 18) 1, 24) 2, 30) 3, 36) 2, 42) 1, 48) 1, 54) 4, 60) 3, 66) 2, 72) 4, 78) 2, , LEVEL - I (C.W), CONDUCTION, In a steady state of heat conduction the, temperature of the ends A and B of a rod 100cm, long are 0º C and 100º C . The temperature, of the rod at a point 60cm distant from the end, A is, 1) 0º C 2) 40º C 3) 60ºC 4) 100º C, An aluminium meter rod of area of cross section, 0, 4cm 2 with K = 0.5cal /gm- C is observed that at, steady state 360 cal of heat flows per minute., The temperature gradient along the rod is, 1) 30 C / cm 2) 60C/ cm 3) 120C/ cm 4) 200C/ cm, One end of metal bar of area of cross section, 5cm2 and 25cm in length is in steam other in, contact with ice, the amount of ice melts in one, minute is (Lice=80cal/gm,K=0.8cgs units ), 1) 16 gm 2) 12 gm 3) 24 gm 4)36 gm, Which of the following rods made of same, material will conduct more heat in a given time, when their ends are maintained at the same, temperature difference., 1) l = 1m, r = 1cm 2) l = 2m, r = 2cm, 3) l = 3m, r = 1cm 4) l = 100c m, r = 2cm, A cylindrical rod with one end in a steam, chamber and the other end is in ice. It is found, that 1gm of ice melts per second. If the rod is, replaced by another one of same material, double the length and double area of cross, section, The mass of ice that melts per second, is, 1) 2 gm 2) 4 gm, 3) 1 gm 4) 0.5 gm, 87
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , WIEN’S DISPLACEMENT LAW AND, KIRCHHOFF’S LAW, 6., , In an atomic bomb, the temperature of, 10million degrees is developed at the moment, of explosion. In what region of the spectrum, do the wavelength corresponding to maximum, energy density lie?, , ( b = 0.28 ×10, , 7., , S.I.unit ), 1) Ultra-violet, 2) Visible, 3) Infra-red rays, 4) x-rays, Solar radiation emitted by sun resembles that, emitted by a black body at a temperature of, 6000 K. Maximum intensity is emitted at a, wavelength of about 4800A0. If the sun was, cooled down from 6000 K to 3000 K. then the, peak intensity would occur at a wavelength of, 1) 4800 A0 2) 9600 A 0 3) 2400 A 0 4) 19200A0, −2, , NEWTON’S LAW OF COOLING, 13. A body at 500 C cools in a surroundings, maintained at 300 C . The temperature at, which the rate of cooling is half that of the, beginning is, 1) 16.320 C 2) 26.30 C 3) 400 C 4) 46.30 C, 14. A body cools from 700C to 600C in 8 minute., The same body cools from 600C to 500C in, 1) 8 minutes, 2) less than 8 minute, 3) More than 8 minute, 4) 1 or 2 or 3 depending on the specific heat of the, body, , LEVEL - I (C.W) - KEY, 01) 3 02) 1 03)2 04) 4 05) 3 06) 4, 07) 2 08) 3 09) 1 10) 4 11) 1 12) 3, 13) 3 14) 3, , STEFAN’S LAW, Two spheres of the same material have radii, 1m and 4m and temperature 4000K and 2000K, respectively. The energy radiated per second, by the first sphere is :, 1) greater than that of the second, 2) less than that of the second, 3) equal in both cases, 4) the information is incomplete to draw any, conclusion, 9. Two objects A & B have exactly the same, shape and are radiating the same power. If, their temperatures are in the ratio 3 :1 then, the ratio of their emissivities is., 1) 1: 9 2) 9 : 1, 3) 3 : 1, 4) 1 : 3, 10. A black body at 1270C emits the energy at the, rate of 106 J/m2 s. The temperature of a black, body at which the rate of energy emission is, 16 ×106 J/m2 s is, 1) 5080C 2) 2730C 3) 4000 C 4) 5270C, 11. An incandescent light bulb has a tungsten filament that is heated to a temperature 3×103 K, when an electric current passes through it. If, the surface area of the filament is approximately 10-4 m 2 and it has an emissivity of 0.3,, the power radiated by the bulb is nearly, ( σ = 5.67 ×10−8W / m2 − K 4 ), 1) 138 w 2) 175 w 3) 200 w 4) 225 w, 12. Two black bodies at 3270C and 6270C are, suspended in an environment at 270C. The, ratio of their emissive powers is, 1) 15 : 8 2) 16 : 3 3) 3 : 16 4) 5 : 8, , LEVEL - I (C.W) - HINTS, , 8., , 88, , 1., , Q=, , 2., , Q=, , KA ( ∆θ ) t, θ1 − θ 2 θ − θ 2, =, ⇒, L, l, l, KA (θ1 − θ 2 ) t, KA (θ 1 − θ 2 ) t, 3. Q = mL =, , (, , l, , K π r2, , ) ( ∆θ ) t ⇒ Q ∝ r, , l, , 2, , 4., , Q=, , 5., , Q = mL =, , 6., , λm T = b, , 7., , λm × T = constant, λ1T1 = λ2T2, , 8., , P1 r1 T1 , P = σ AT 4 = σ 4π r 2T 4 ⇒ P = r × T , 2, 2 2, , 9., , e1 T2 , P = eσ AT 4 ⇒ e = T , 2, 1, , l, , l, , KA (θ1 − θ 2 ) t ⇒ m2 = A2 × l1, m1 A1 l2, l, , 2, , E1 T1 , = , 10. E = σ T 4 ⇒, E2 T2 , , 12. E = σ (T − T, 4, , 4, 0, , ), , 4, , 4, , 4, , 11., 1. P = eAσ T 4, , E1 TB1 − T0, =, ⇒, E2 TB2 4 − T0 4, 4, , 4, , dθ, = − K (θ − θ 0 ) ⇒ R ∝ θ − θ 0, dt, 14. Rate of cooling decreases with fall of temperature., Hence, time increases.., , 13., , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , LEVEL - I (H.W), CONDUCTION, 1., , 2., , 3., , 4., , 5., , 6., , 7., , In the steady state the two ends of a meter 8., rod are at 300 C and 200 C , the temperature, at the 40 th cm from the end at higher, temperature is, 1) 220 C 2) 260 C 3) 250 C 4) 240 C, 9., A rod of length 1 m having cross-sectional area, −1, 0.75m2 conducts heat at 6000 Js . Then the, temperature difference across the rod is, if, K = 200 Wm −1K −1, 10., 1) 200 C 2) 400 C 3) 800 C 4) 1000 C, A 3cm cube of iron one face at 1000 C and the, , TRANSMISSION OF HEAT, emission radiations rates are 140A0 and 4200, A0 respectively :, 1) 1 :30 2) 30 :1 3) 42 : 14 4) 14 : 42, , STEFAN’S LAW, The rate of radiation from a black body at 00C, is E. The rate of radiation from this black body, at 2730C is, 1. 2E, 2. E/2, 3. 16 E, 4. E/16, Two bodies of same shape, same size and same, radiating power have emissivities 0.2 and 0.8., The ratio of their temperature is, 1), , 2) 2 :1 3) 1: 5 4) 1: 3, 3 :1, Two spheres have radii 1m, 2m are at same, temperatures, have emissivities e, 2e then ratio of radiant energy emitted per second is, other in a block of ice at 00 C . If K of, 1) 1:2, 2) 1:4, 3) 1:8, 4) 1:1, iron = 0.2 CGS units and L for ice is 80 cal/gm, 11. The radiant power of a furnace of surface area, then the amount of ice that melts in 10 minof 0.6 m2 is 34.2 KW. The temperature of the, utes is (assume steady state heat transfer), furnace is [ σ = 5.7 x 10-8 Wm-2K-4], 1) 450 g 2) 900 g 3) 350 g 4) 500 g, 1) 3400 K 2) 1012 K 3) 1000 K 4) 5700 K, Heat is flowing through two cylindrical rods of 12. How many watt of energy is required to keep, same material. The diameters of the rods are, a black body in the form of a cube of side 1cm, in the ratio 1 : 2 and their lengths are in the, at 2000K? (Temperature of surrounding is, ratio 2 : 1. If the temperature difference be270C and σ = 5.67 × 10−5 Wm −2 K −4 ), tween their ends is same, then the ratio of, 1) 444 KW, 2) 544 KW, amounts of heat conducted through them per, 3) 644 KW, 4) 64 KW, unit time will be, NEWTON’S LAW OF COOLING, 1) 1 :1, 2) 2 : 1, 3) 1 : 4 4) 1 : 8, 13., The rates of cooling of a body at temperatures, One end of a cylindrical rod is kept in steam, chamber and the other end in melting Ice. Now, 1000 C and 800 C are x1 and x2 respectively ,, 0.5 gm of ice melts in 1 sec. If the rod is rewhen placed in a room of temperature 400 C, placed by another rod of same length, half the, x1, diameter and double the conductivity of the first, then, x2 is, rod, then rate of melting of ice will be (in gm/, sec) ( 2008 E ), 1) 4/5, 2) 5/4, 3) 3/2, 4) 2/3, 1) 0.25 2) 0.5, 3) 1 4) 2, 14. A vessel full of hot water is kept in a room and, it cools from 800C to 750C in T 1 minutes, from, WIEN’S DISPLACEMENT LAW AND, 750C to 700C in T2 minutes and from 700C to, KIRCHHOFF’S LAW, 650C in T3 minutes. Then, The wavelength of maximum energy released, 1. T1=T2 =T3, 2. T1>T2 >T3, during an atomic explosion was 2.93 × 10−10 m ., −3, 3. T1<T2 =T3, 4. T1<T2 <T3, Given that Wien’s constant is 2.93 ×10 mK ,, the maximum temperature attained must be, LEVEL - I (H.W) - KEY, of the order of, 01) 2 02) 2 03) 1 04) 4 05) 1 06) 2, 1) 10−7 K, 2) 107 K, 07) 2 08) 3 09) 2 10) 3 11) 3 12) 2, 3) 10−13 K, 4) 5.86 × 107 K, 13) 3 14) 4, What will be the ratio of temperatures of sun, and moon if the wavelengths of their maximum, , NARAYANAGROUP, , 89
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , LEVEL - I (H.W) - HINTS, , 1., 2., 4., , KA ( ∆θ ) t, θ −θ, θ − θ1, ⇒ 1 2 =, L, l, l, KA ( ∆θ ) t, Q KA ( ∆θ ), =, 3. mL f =, t, l, l, 2, KA (θ1 −θ2 ) t, Q K π r ( ∆θ ), Q=, ⇒ =, l, t, l, Q=, , Q, t, ∴ , Q, , t, , 5., , 2., , , 2, , 1 = K 1 × r1 , , , K 2 r2 , , 2, , mL f =, , 3., , KA ( ∆θ ) t, l, , m, 2, , K2 d2 , t 2, , ⇒, =, , , m, K 1 d 1 , d = diameter, , t 1, 6., 7., , λ mT = b, λ m T = constant ⇒ λ1T1 = λ2T2, , Three rods A,B and C have the same, dimensions. Their conductivities are K A , K B, and K C respectively. A and B are placed end, to end, with their free ends kept at certain, temperature difference. C is placed separately, with its ends kept at same temperature, difference. The two arrangements conduct, heat at the same rate K c must be equal to, KA + KB, KAKB, 1, 1) KA+KB 2) K K 3) ( KA +KB ) 4) K + K, 2, A B, A, B, Two rods (one semi-circular and other straight), of same material and of same cross-sectional, area are joined as shown in the figure. The, points A and B are maintained at different, temperatures. The ratio of heat transferred, through a cross-section of a semicircular rod, to the heat transferred through a cross section, of the straight rod in a given time is, Semi circular rod, , Straight rod, , 4, , 8., , P1 T1 , P = σ AT 4 ⇒ P = T (given that P1 = E), 2, 2, , 9., , T e , P = eAσ T ⇒ 2 = 1 , T1 e2 , , 1, , 4., , 4, , 4, , 2, , P1 r1 T1 , = × , 10. P = eσ 4π r T ⇒, P2 r2 T2 , , (, , 11., , 2, , ), , P = eσ A T, , 2, , 4, , 4, , (, , 4, 12. P = σ A T − Ts, , 4, , ), , 5., , dθ, ∝ (θ − θ 0 ) ⇒ R ∝ θ − θ 0, 13., dt, 14. Rate of cooling decreases with fall of temperature., Hence, time increases.., , LEVEL -II (C.W), CONDUCTION, 1., , 90, , The co-efficient of thermal conductivity of, copper, mercury and glass respectively Kc ,Km, and K g such that Kc > K m > K g if the same, quantity of heat is flow per sec per unit area of, each and corresponding temperature gradient, are X c , X m and X g :, 1) X c = X m = X g, 2) X c > X m > X g, 3) X c < X m < X g, 4) X m < X c < X g, , 6., , A, B, 1) 2 :π, 2) 1:2, 3) π : 2, 4) 3:2, Two identical slabs are welded end to end and, 20cal of heat flows through it for 4min. If the, two slabs are now welded by placing them one, above the other, and the same heat is flowing, through two ends under the same difference of, temperatures, the time taken is, 1) 1min 2) 2 min 3) 4 min 4) 16 min, A slab consists of two parallel layers of copper, and brass of the same thickness and having, thermal conductivities in the ratio 1 : 4. If the, free face of brass is at 100 ºC and that of, copper at 0º C , the temperature of interface, is, 1) 80º C 2) 20º C 3) 60º C 4) 40º C, Two metal plates of same area and thickness, l 1 and l 2 are arranged in series. If the thermal, conductivities of the materials of the two plates, are K1 and K 2 . The thermal conductivity of, the combination is, 2K1K2, K +K, 1) K + K, 2) 1 2, 2, 1, 2, 3), , K1K2 ( l 1 + l 2 ), K1l 2 + K2l1, , 4) K1 + K2, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , Two hollow spheres of same material one with 11 A cube of side 10cm is filled with ice of density, 0.9gm/c.c. Thickness of the walls of the cube is, double the radius of the other and double the, 1mm and thermal conductivity of the material of, thickness of the other filled with ice, The ratio, of times in which ice gets melted in the two, the cube is 0.01 C.G.S. units. If the cube is placed, spheres is, in steam bath maintained at a temperature of, 1) 2 :1, 2) 1 : 2, 3) 4 : 1, 4) 1 : 4, 1000 C , the time in which ice completely melts, is ( Lice = 80 cal / gm ), 8. A wall has two layers A and B, each made of a, different material. Both the layers have the, 1) 6 sec 2) 12 sec 3) 24 sec 4) 48 sec, same thickness. The thermal conductivity of, WIEN’S DISPLACEMENT LAW, the material of A is twice that of B. Under, AND KIRCHHOFF’S LAW, thermal equilibrium, the temperature, 12. A black body is at a temperature of 2800 K., difference across the wall is 36º C . The, The energy of radiation emitted by this object, temperature difference across the layer A is, with wavelength between 499 nm and 500 nm, 1) 6º C 2) 12º C 3) 18º C 4) 24º C, is U1 , between 999 nm and 1000 nm is U 2 and, between 1499 nm and 1500 nm is U 3 . The, 9. Two rods of length l and 2l thermal, Wien’s constant b = 2.80 × 106 nm K. Then, conductivities 2Kand Kare connected end to, 1) U1 = 0 2) U 3 = 0 3) U1 > U 2 4) U 2 > U1, end. If cross sectional areas of two rods are, equal, then equivalent thermal conductivity of 13. When the temperature of a black body, increases, it is observed that the wavelength, the system is, corresponding to maximum energy changes, 5, 8, from 0.26 µ m to 0.13 µ m . The ratio of the, 1) 6 K 2) 1.5K 3) 1.2 K 4) 9 K, emissive powers of the body at the respective, 10. Three rods of identical cross-sectional area, temperatures is :, and made from the same metal form the sides, 16, 4, 1, 1, of an isosceles triangle ABC right angled at, 1), 2), 3), 4), 1, 1, 4, 16, B . The points A and B are maintained at, temperatures T and 2T , respectively, in the 14. For an enclosure maintained at 1000K the, maximum radiation occurs at wavelength λ ., steady state. Assuming that only heat, m, If the temperature is raised to 2000K, the peak, conduction takes place, temperature of point, will shift to, (CBSE, PMT 1998), C is, 1) 0.5 λm 2) λm, 3) 4 λm, 4)8 λm, 7., , ( ), , ( ), , STEFAN’S LAW, , A, , C, , B, , 3T, 2 +1, , 1), 3), 3, , (, , T, , ), , 2 −1, , NARAYANAGROUP, , 2), , T, 2 +1, , 4), , T, 2 −1, , 15. The power radiated by a black body is P and it, radiates maximum energy around the, wavelength λ0 . If the temperature of the black, body is now changed so that it radiates, maximum energy around a wavelength 3λ0 / 4 ,, the power radiated by it will increase by a factor, of, 1) 4/3, 2) 16/9, 3) 64/27, 4) 256/81, 16. The rates of heat radiation from two patches, of skin each of area A, on a patient’s chest, differ by 2%. If the patch of the lower temp is, at 300K and emissivity of both the patches is, assumed to be unity, the temp. of other patch, would be., 1) 306K 2) 312K 3) 308.5K 4) 301.5K, 91
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TRANSMISSION OF HEAT, , JEE- ADV PHYSICS-VOL- V, , 17. A spherical black body with a radius of 12cm, NEWTON’S LAW OF COOLING, radiates 450W power at 500K. If the radius 25. A body cools from 500C to 450C in 5 min and to, were halved and the temperature doubled, the, 400C in another 8 min. The temperature of the, power radiated in watts would be, surrounding is, 1) 225, 2) 450, 3) 900, 4) 1800, 1) 340C 2) 300C 3) 430C 4) 370C, 18. If the temperature of the sun were to increase 26. A hot body is placed in cold surroundings. It’s, from T to 2T and its radius from R to 2R, then, rate of cooling is 30C per minute when its, the ratio of the radiant energy received on earth, temperature is 700C and 1.50C per minute, to what it was previously, will be, when its temperature is 500C it’s rate of cooling, 1) 4, 2) 16, 3) 32, 4) 64, when its temperature is 400C., 19. The radiation emitted by a star A is 10,000 times, 1. 0.250C / min, 2. 0.50C / min, that of the Sun. If the surface temperature of, 3. 0.750C / min, 4. 10C / min, the sun and the star A are 6000K and 2000k 27. A calorimeter of water equivalent ‘5g’ has, respectively, the ratio of the radii of the star A, water of mass 55 g upto a certain level. Another, and the Sun is, identical calorimeter has a liquid of mass ‘38g’, 1) 300 : 1 2) 600: 1 3) 900 : 1 4) 1200:1, upto same level. As both of them cool in the, 20. Two electric bulbs have filaments of lengths L, same surroundings from 500C to 460C, water, and 2 L, diameters 2d and d and emissivities, takes 80 s where as the liquid takes 32 s to, 3e and 4e. If their temperatures are in the ratio, cool. If the specific heat of water is 1 cal/g-0C,, 2 : 3, their powers will be in the ratio of, the specific heat of the liquid in cal/g-0C is, 1) 8 : 27 2) 4 : 27 3) 8 : 3, 4) 4 : 9, 1) 0.8, 2) 0.4, 3) 0.5, 4) 0.2, 21. If the absolute temperature of a black body is, LEVEL - II (C.W) - KEY, doubled the percentage increase in the rate of, 01) 3 02) 4 03) 1 04) 1 05) 1 06) 3, loss of heat by radiation is, 07) 3 08) 2 09) 3 10) 1 11) 2 12) 4, 1) 15% 2) 16% 3)1600% 4) 1500%, 13) 4 14) 1 15) 4 16) 4 17) 4 18) 4, 22. A sphere and cube of same material and same, 19) 3 20) 2 21) 4 22) 3 23) 1 24) 4, volume are heated upto the same temperature, 25) 1 26) 3 27) 3, and allowed to cool in the same surroundings., LEVEL - II (C.W)- HINTS, The ratio of the amounts of radiations emitted, KA ( ∆θ ) t, will be, ∆θ , ⇒ K, 1. Q =, = const, 1/3, 2/3, l, l , , 4π, 1 4π , π , :1 3) :1 4) , 1) 1:1 2), :1, 1, ∆θ 1, 3, 2 3 , 6, , ∝ ⇒ X ∝ ;Since, K c > K m > K g, K, l K, 23. A black metal foil is warmed by radiation from, For, same, quantity, of, heat flow per sec per unit, a small sphere at temperature T and at a, ∴, distance d it is found that the power received, area of each X c < X m < X g, by the foil is P. If both the temperature and 2. When A and B are in series, the distance are doubled, the power received, l1 + l2 l1, l, 2K A KB, by the foil will be, =, + 2 ⇒ Keff =, Keff, K1 K2, K A + KB, 1) 64P, 2) 16P, 3) 4P, 4) 8P, 24. A very small hole in an electric furnace is used, 2K AK B , for heating metals. The hole nearly acts as a, 2, , A ( ∆θ ), black body. The area of the hole is 200 mm .To, Q KA + KB , ........ (i), =, keep a metal at 727º C , heat energy flowing, t, 2l, through this hole per sec, in joules, Q K C A ( ∆θ ), is (σ = 5.67 × 10−8Wm −2 K −4 ) (EAM-2014E), For rod C =, ........ (ii), t, l, 1) 22.68 2) 2.268 3) 1.134 4) 11.34, From (i) and (ii) we get value of KC, 92, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, 16. P = σ AT 4, , LEVEL-II (H.W), , ∆P, ∆T, , P∝T ⇒, × 100 = 4 , × 100 , P, T, , 4, , CONDUCTION, , ∴ temp of other patch = T ± ∆T, , (, , ), , 17. P = σ AT ⇒ P = σ 4π R T, 4, , 2, , P1 R1 T1 , =, × , P2 R 2 T2 , , 2, , 1., , 4, , 4, , (, , 2., , ), , 4, 2, 4, 18. P = σ AT ⇒ P = σ 4π R T, 2, , P1 R1 T1 , = × , P2 R2 T2 , , 19. P = σ ( 4π R, , 2, , ), , K A , K B and K C respectively they are kept at, same temperature difference. The rate of heat, flow through C is equal to rate of combined, , 4, , R, T ⇒ A =, R sun, 4, , 20. P = eσ ( 4π R 2 ) T 4 ⇒, , P1 e1 r1 l1 T1 , = × × × , P2 e2 r2 l 2 T2 , , Three metal rods of coefficient of thermal, conductivities K, 2K,3K conducts heats of 3Q,, 2Q, Q per seconds through unit area then the, ratio of temperature gradients, 1) 9:3:1 2) 9:1:1 3) 3:1:1, 4) 1:1:1, Three rods A,B and C have the same, dimensions. Their conductivities are, , PA Tsun , ×, , PSun T A , , 2, , 4, , 21. P = σ AT 4 ⇒, , T 4 , P2 − P1, ×100 = 2 − 1 × 100, P1, T1 , , 22. Given, (Volume ) sphere = (Volume )cube, , heat flow through A and B then K C must be, equal to, KAKB, KAKB , 1, 1) KA+KB 2) K + K 3) ( KA +KB ) 4) 2 K +K , A B, A, B, 2, 3. Two rods one is semi circular of thermal, conductivity K1 and other is straight of thermal, conductivity K 2 and of same cross-sectional, area are joined as shown in the figure. The, points A and B are maintained at same, temperature difference. If rate of flow of heat, is same in two rods then, K1 / K 2 is, Semi circular rod, , 1, , 4, R 3 3, π R3 = a3 ⇒ = , , 3, a 4π , Here, R, a are radius of this sphere and side of the, cube, , 24., 26., 27., , 94, , A, , B, , 1) 2 :π, 2) 1:2, 3) π : 2, 4) 3:2, 4. Two identical rods of same metal are first, P, A, 4π R 2, 4, P = eσ A T 4 − TS ⇒ 1 = 1 =, welded in series and then in parallel are, P2, A2, 6a 2, maintained at same temperature difference, 4, 2, then the ratio of heats conducted in same time, eσ AT 4, I 2 T2 d1 , I=, ⇒ = × hereI =P, 2, is, 1, d, I1 T1 d 2 , 1) 1:1, 2) 1:2, 3) 1:4, 4) 1:3, dθ, θ1 + θ 2, , 5. Two slabs A and B of equal surface area are, =K, − θ0 , 25., P = σ AT 4, placed one over the other such that their surfaces, dt, 2, , , are completely in contact. The thickness and, dθ, ∝ (θ − θ 0 ) ; R ∝ θ − θ0, coefficient of thermal conductivities of slab A is, dt, twice that of B. The first surface of slab A is, From Newton’s law of, cooling, maintained at 1000 C , while the second surface, dθ, 1, ∝, ⇒ t ∝ ms, of slab B is maintained at 250 C . The, dt, ms, temperature at the contact of their surfaces is, W + m1 s1 t1, =, (2008 E), ∴ W +m s, t 2 ; W= thermal capacity of, 2 2, 1) 150 C 2) 62.50 C 3) 550 C, 4) 850 C, calorimeter, , (, , 23., , Straight rod, , ), , (, , ), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , 6. Three metal rods of same lengths and same, WIEN’S DISPLACEMENT LAW, area of cross - sections having conductivities, AND KIRCHHOFF’S LAW, 1,2,3 units are connected in series. Then their, 12. The temperature of a furnace is 22270C and, effective conductivity will be, the intensity is maximum in its spectrum nearly, 1) 2 units 2) 1.6 units 3) 2.4 units 4) 2.8 units, at 12000A0. If the intensity in the spectrum of, 7. Two hollow spheres of same thickness are filled, with ice. The ratio of their diameter is 1 : 2, star is maximum nearly at 4800A0, then the, and the ratio of thermal conductivities of the, surface temperature of the star is, materials is 2 : 3. The ratio of times in which, 1) 84000C 2)62500C 3) 72000C 4)59770C, the ice gets melted in the two spheres is, 13. Black body at a temperature of 1640 K has, 1) 3 : 4 2) 4 : 3, 3) 3 : 8, 4) 8 : 3, the wavelength corresponding to maximum, 8. Three rods of same dimensions have, emission equal to 1.75 µ m. Assuming the, thermal conductivity 3K,2K and K. They are, moon to be a perfectly black body, the temarranged as shown in figure. Then the, perature of the moon, if the wavelength corretemperature of the junction in steady state is, ( 2009 E ), sponding to maximum emission is 14.35 µ m,, 50 C, is, 1) 100 K 2) 150 K 3) 200 K 4) 250 K, 2K, 14., A particular star (assuming it as a black body), 3K, 0, , 0, , 100 C, K, 00C, , 200 0, 100 0, 50 0, C 2), C 3) 75 0 C 4), C, 3, 3, 3, 9. Three rods of lengths L,2L and 3L having thermal conductivities 3K,2K and K are connected, end to end. If cross sectional areas of three, rods are equal then equivalent thermal conductivity of the system is., 1) 18K / 13 2) 36K / 13 3) 9K / 13 4)12K / 13, 10. Three rods of identical cross sectional area and, made from the same metal form the sides of, an equilateral triangle ABC. The points A and, B are maintained at temperatures 3 T and, T respectively . In the steady state, the temperature of the point C is TC . Assuming that, only heat conduction takes place, the value of, TC / T is equal to, , 1), , 1+ 3, 1− 3, 1+ 2, 1− 2, 2), 3), 4), 2, 2, 2, 2, 11. A hollow metal cube, with side 0.5m and wall, thickness 5x10-3m is filled with ice. It is, immersed in water tank maintained at 1000C., Calculate the amount of ice melted in 335sec., ( Conductivity of metal, = 0.5Wm-1 K -1, Latent heat of fusion of, ice=335x103Jkg-1), 1) 15kg 2) 15g, 3) 1.5kg, 4) 1.5g, , 1), , NARAYANAGROUP, , has a surface temperature of about 5 ×104 K ., The wave length in nano-meters at which its, radiation becomes maximum is ( b = 0.0029mk), (2003M), 1) 48, 2) 58, 3) 60, 4) 70, , STEFAN’S LAW, 15. The power radiated by a black body is ‘P’ and, it radiates maximum energy around the, wavelength λ0 . If the temperature of the black, body is now changed so that it radiates, maximum energy around a wavelength, λ 0 / 2 ,the power radiated becomes(2012 E ), 1) 4P, 2) 16P, 3) 64P, 4) 256P, 16. There is a temperature difference of 1K, between two black patches of skin on patient’s, chest and each patch having area A. The, radiant heat emitted from them is differ by 2, % , then temperatures of two patches may be., 1) 100 K, 101 K, 2) 300 K, 301 K, 3) 200 K, 201 K, 4) 400 K, 401 K, 17. A black body radiates energy at the rate of E, watt/ m 2 at a high temperature TK when the, T , temperature is reduced to K ., 2, radiant energy is (2007E), , 1), , E, 2, , 2) 2E, , 3), , E, 4, , 4), , Then, , E, 16, 95
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , 18. The radiant energy from the Sun, incident 26. Two bodies have thermal capacities in the ratio, normally at the surface of earth, is 20kcal/m2 3:4 and the rates of loss of heat in the ratio, min. What would have been the radiant energy, 3:5. Their rates of cooling will be in the ratio, incident normally on the earth, if the sun had a, of, temperature twice of the present one?, 1) 9 : 20 2) 4 : 5 3) 5 : 4, 4) 1 : 1, 1. 40 kcal/(m2 - min) 2. 80kcal/(m2-min), 27. A calorimeter of water equivalent 6 g has wa3. 160 kCal/(m2-min) 4. 320 Kcal/(m2-min), ter of mass 64 g up to a certain volume. An19. A star behaves like perfect Black body, other identical calorimeter has liquid of mass, emitting radiant energy. The ratio of radiant, 50 g and specific heat 0.6 cal/g-0C upto same, energy emitted per sec by this star to that, level. If both of them cool in the same surroundings through same range of temperature, emitted by another star having 8 times the, and the time taken for the water to cool is 140, radius of the former but having Kelvin, s, the time taken for the liquid to cool is, temperature one fourth of the former is, 1) 72 s 2) 140 s 3) 36 s, 4) 120 s, 1) 1:4, 2) 1:8, 3) 4:1, 4) 1:16, LEVEL - II (H.W) -KEY, 20. Two spherical bodies have radii R,2R and, 01), 1, 02) 1 03) 3 04) 3 05) 2 06) 2, emissivities e,2e. If the temperature ratio is, 07), 1, 08) 1 09) 1 10) 1 11) 1 12) 4, 2:1 then the powers will be in the ratio, 13) 3 14) 2 15) 2 16) 3 17) 4 18) 4, 1) 1:1, 2) 2:1, 3) 3:1, 4) 4:1, 19) 3 20) 2 21) 4 22) 3 23) 1 24) 3, 21. The temperature of a black body is increased, 25) 2 26) 2 27) 1, by 50%. The amount of radiation emitted by, the body increases by, LEVEL - II (H.W) - HINTS, 1)50%, 2)100% 3)225% 4)406.25%, KA ( ∆θ ) t, 22. A solid sphere is at a temperature T K. The, sphere is cut into two halves. the fraction of 1. Q =, l, energy emitted per second by the half sphere, ∆θ Q, ∆θ, Q, to that by complete sphere is, =X ⇒ X ∝, , let, , ∝, 1, , 1, , 3, , 256, , radiated becomes 81 P0 . The shift in, wavelength corres-ponding to the maximum, energy will be, λ0, λ, λ, λ, 1) + 0 2) + 0 3) −, 4) − 0, 4, 2, 2, 4, , NEWTON’S LAW OF COOLING, 25. A body cools from 700 C to 500 C in 5 minutes., Temperature of surroundings is 200 C . Its, temperature after next 10 minutes is (2008 M), 1) 250 C 2) 300 C 3) 350 C 4) 450 C, 96, , l , , 1, , 1), 2) 4, 3) 4, 4) 1 6, 2, 23. A black metal foil is warmed by radiation from, a small sphere at temperature ‘T’ and at a, distance d. It is found that the power received, by the foil is P. If both the temperature and, the distance are halved, the power received, by the foil will be in the ratio, 1) 1:4, 2) 1:8, 3) 1:16, 4) 1:64, 24. Power radiated by a black body is P0 and the, wavelength corresponding to maximum energy, is around λ0 . On changing the temperature of, the black body, it was observed that the power, , K, , l, , X1 : X2 : X3 =, , 2., , K, , Q1 Q2 Q3, :, :, K1 K2 K 3, , Q, Q , Q, = + , t C t A t B, , K c A ( ∆θ ) K A A (∆ θ ) K B A ( ∆θ ), =, +, l, l, l, 3., , Q, Q, = , , t semicircular rod t straightrod, , K1 A ( ∆ θ ) K 2 A ( ∆ θ ), =, πr, 2r, r = radius of semi circle., KA ( ∆θ ) t, 4. Q =, l, Q1 K s A ( ∆θ ) t, l, =, ×, ( here K s = K p ), Q2, 2l, K p 2 A ( ∆θ ) t, 2 K (100 − θ ) K (θ − 25 ), Q, Q, =, 5. t = t ⇒, 2l, l, , , 6., , A, , , , B, , Reff = R1 + R2 + R3, l1 + l 2 + l 3, l, l, l, = 1 + 2 + 3, K eff, K1, K2, K3, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, 2., , Three rods of material ‘x’ and three rods of, material y are connected as shown in figure. All, the rods are of identical length and crosssection. If the end A is maintained at 600C and, the junction E at 100C, find effective Thermal, Resistance. Given length of each rod=l, area, 6., of cross-section=A, conductivity of x=K and, conductivity of y=2K., , C, X, , X, , B, , A, , X, , E, , Y, Y, , between the outer and inner surfaces of the, shell is not to exceed ‘T’, the thickness of the, shell should not be less than, 1), , 4π KR 2T 4π KR 2, 2), TP, P, , 4), , 4π R 2T, KT, , A ( T2 − T1 ) K , , f , with f equals to, x, , , x, T2, , 4l, 7l, 4 KA, 7 KA, 2), 3), 4), 3KA, 6 KA, 3l, 3l, A cylinder of radius R made of a material of 7., thermal conductivity K1is surrounded by a, cylindrical shell of inner radius R and outer, radius 3R made of material of thermal, conductivity K2 . The ends of the combined, system are maintained at two different, temperatures. There is no loss of heat across, the cylindrical surface and the system is in 8., steady state. The effective thermal, conductivity is, , 4π R 2 T, KP, , The temperature of the two outer surfaces of, a composite slab, consists of two materials, having coefficients of thermal conductivity K, and 2K and thickness x and 4x respectively, are T2 and T1(T2>T1). The rate of heat transfer, through slab, in a steady state is, , Y, D, , 3), , 4x, , K, , 2K, , T1, , 1), , 3., , K1K2, , 4., , 5., , 98, , K + 3K, , 1) 1, 2) 1/2, 3) 2/3, 4) 1/3, A and B are two points on a uniform metal ring, whose centre is O. The angle AOB = θ . A and, B are maintained at two different constant, temperatures. When θ = 180 0 , the rate of total, heat flow from A to B is 1.2W. When θ = 900 ,, this rate will be, 1) 0.6 watt2) 0.9 watt 3) 1.6 watt 4) 1.8 watt, Two ends of a conducting rod of varying cross, section are maintained at 2000 C and 00 C, respectively, ., In, steady, state, , K + 8K, , 1, 2, 4) 1 9 2, 1) K1 + K2 2) K + K 3), 4, 1, 2, Water is being boiled in a flat bottomed kettle, placed on a stove. The area of the bottom is, 300cm2 and the thickness is 2mm. If the amount, of steam produced is 1gm min-1, then the, difference of the temperature between the, inner and the outer surface of the bottom is, (thermal conductivity of the material of the, kettle 0.5cal cm-1s-1C-1, latent heat of the steam, is equal to 540calg-1), 1) 120C 2)20C 3)0.120C 4) 0.0120C, A point source of heat of power ‘P’ is placed at, the centre of a spherical shell of mean radius, ‘R’. The material of the shell has thermal, conductivity ‘K’. If the temperature difference, , A, , B, , C, , D, 00C, , 2000C, , X, , X, , 1) temperature difference across AB and CD are, equal, 2) temperature difference across AB is greater, than that of across CD, 3) temperature difference across AB is less than, that of across CD, 4) temperature difference may be equal or different, depending on the thermal conductivity of the rod, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, 9., , TRANSMISSION OF HEAT, , Three rods of same dimensions are arranged 13. The temperature of an isolated black body falls, as shown in the figure. They have thermal, from T1 to T2 in time ‘t’. Let ‘c’ be a constant,, conductivities K1 , K 2 and K 3 . The points P and, then......, Q are maintained at different temperatures for, 1, 1, 1 , 1 , the heat to flow at the same rate along PRQ, 1) t = c T − T 2) t = c T 2 − T 2 , 2, 1 , 2, 1 , and PQ. Which of the following options is, correct?, 1, 1, 1 , 1 , 3) t = c T 3 − T 3 4) t = c T 4 − T 4 , R, , , K1, , P, , 1) K3 =, , K2, , Q, , K3, , 1, ( K1 + K2 ), 2, , 2) K 3 = K1 + K 2, , K1 K 2, 3) K 3 = K + K, 4) K3 = 2 ( K1 + K 2 ), 1, 2, 10. A boiler is made of a copper plate 2.4 mm thick, with an inside coating of a 0.2 mm thick layer, of tin. The surface area exposed to gases at, 2, 7000 C is 400 cm . The maximum amount of, steam that could be generated per hour at, atmospheric pressure is, Kcu = 0.9 cal / cm − s −0 & ktin = 0.15 cal / cm − s −0 C , , , and Lsteam = 540 cal / g, , , 1)5000Kg 2)1000Kg 3)4000Kg 4)200Kg, 11. Water in a lake is changing into ice at 00C., when the atmospheric temperature is 100C. If, the time taken for 1cm thick ice layer to be, formed is 7hour, the time required for the, thickness of ice to increase from 1cm to 2cm is, 1) 7hour 2) 14 hour 3) <7hour 4) >14 hour, , RADIATION, 12. Two metallic spheres S1 and S2 are made of, the same material and have identical surface, finish. The mass of S 1 is three times that of S 2., Both the spheres are heated to the same high, temperature and placed in the same room, having lower temperature but are thermally, insulated from each other. The ratio of the, initial rate of cooling of S 1 to that S 2 is :, 1) 1/3, , 2) (1 / 3)3 3) 1/ 3 4), , NARAYANAGROUP, , 1, , 3 /1, , 2, , 1, , , , , , 2, , 1, , , , 14. A star behaves like a perfectly black body, emitting radiant energy. The ratio of radiant, energy per second by this star to that emitted, by another star having 8 times the radius of, former, but having temperature, one-fourth, that of the former in Kelvin is, 1) 1 : 4, 2) 1 : 16 3) 4 : 1, 4) 16 : 1, 15. A sphere of density ‘d’, specific heat capacity, ‘c’ and radius ‘r’ is hung by a thermally insulating, thread in an enclosure which is kept at lower, temperature than the sphere. The temperature, of the sphere starts to drop at a rate which is, proportional to, 1) c / r3d 2) 1 / r3dc 3) 3r3dc 4) 1 / rdc, 16. Two bodies A and B have thermal emissivities, of 0.01 and 0.16 respectively. The outer, surface areas of the two bodies are the same., The two bodies radiate energy at the same, rate. The wavelength λB corresponding to the, maximum spectral radiancy in the radiation, from ‘B’ is shifted from the wavelength, corresponding to the maximum spectral, radiancy in the radiation from ‘A’ by 1.00 µm ., If the temperature of ‘A’ is 5802 K,, 1) the temperature of ‘B’ is 1934K, 2) λB = 1.6µ m, 3) the temperature of B is 11604K, 4) the temperature of B is 2901K, 17. Assuming the sun to be a spherical body of, radius R at a temperature ‘T’ K, evaluate the, total radiant power incident on earth, at a, distance r from the sun. (Take r0 is radius of, earth ‘ σ ’ Stefan’ss constant), 4π r0 2 R 2σ T 4, 1), r2, , π r0 2 R 2σ T 4, 2), r2, , π r02 R 2σ T 4, 3), 4π r 2, , 4), , R 2σ T 4, r2, 99
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , NEWTON’S LAW OF COOLING, 18. Two solid spheres are heated to the same, temperature and allowed to cool under, identical conditions. Assume that all the, surfaces have the same emissivity and ratios, of their radii, specific heats and densities are, respectively 1: α ,1: β ,1: γ . Then the ratio of, initial rates of loss of heat and initial rates of, fall of temperature is, 1) αβγ :1,1: α 2, 2) 1: α 2 , αβγ :1, 2, 3) α :1, αβγ :1, 4) α 2 :1,1: αβγ, 19. A heated object (at time t = 0 and temperature, T = T0 ) is taken out of the oven to cool and, placed on a table near an open window. Write, an expression for its temperature as function, of time T, where Ts is the surrounding, temperature., , Y, log e (?-? 0 ), , 2), X, , 0, t, Y, log e (?-? 0 ), , 3), 0, , X, t, , Y, log e (?-? 0 ), , 1) T = Ts − ( T0 + Ts ) e − kt 2) T = Ts + (T0 + Ts ) e− kt, , 4), 3) T = Ts + ( T0 − Ts ) e − kt 4) T = Ts − ( T0 − Ts ) e − kt, 0, X, t, 20. A system ‘S’ receives heat continuously from, an electrical heater of power 10W. The 23. If a piece of metal is heated to temperature θ, temperature of S becomes constant at 500C, and then allowed to cool in room which is at, when the surrounding temperature is 200C., temperature θ0 , the graph between the, After the heater is switched off, S cools from, temperature T of the metal and time t will be, 35.10C to 34.90C in 1 minute. The heat capacity, closest to ( 2013 JEE ), of S is, 1) 750J(0C)-1, 2) 1500J(0C)-1, Y, Y, 3) 3000J(0C)-1, 4) 6000J(0C)-1, 21. According to Newton’s Law of cooling, the rate, T, of cooling of a body is proportional to ( ∆θ ) ,, where ∆θ is the difference of temperature of, the body and the surroundings, then n is equal, to, 1) 2, 2) 3, 3) 4, 4) 1, n, , 22. A liquid in a beaker has temperature θ ( t ) at, time ‘t’ and ‘ θ0 ’ is temperature of, surroundings, then according to Newton’s law, of cooling the correct graph between loge, (θ − θ0 ) and t is (2012 JEE), , θ0, , T, , 1), , 2), X, , 0, , X, , O, , t, , t, , Y, , Y, T, , T, , .3) θ0, , 4), X, , O, , t, , θ0, X, , O, , t, , Y, log e (?-? 0 ), , LEVEL - III - KEY, , 1), X, , 0, t, , 100, , 01) 2, 07) 3, 13) 3, 19) 3, , 02) 2, 08) 3, 14) 3, 20) 2, , 03) 4, 09) 3, 15) 4, 21) 4, , 04) 4, 10) 3, 16) 4, 22) 1, , 05) 1 06) 4, 11) 4 12) 2, 17) 2 18) 2, 23) 3, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , ASSERTION & REASON TYPE, , 4., , 5., , 6., , 7., , ORDER ARRANGING QUESTIONS, , 1) Both A and R are true, R is correct 13. I) A sphere II) A cube III)A thin circular plate., explanation of A, All made of the same material having the same, 2) Both A and R are true, R is not correct, mass are initially heated to 2000C. Identify the, explanation of A, order in which the objects cool faster when left, 3) Both A and R are true, in air at room temperature, 4) Both A and R are false, 1) III,II and I, 2) II, I and III, Assertion (A): Woolen clothes keep the body warm, 3) I, II and III, 4) II, III and I, in winter, 14. A beaker full of hot water is kept in a room and, Reason (R): Air is a bad conductor of heat, it cools from, Assertion (A): The radiation from the sun surface, I) 900C to 800C in t1 sec., II) 800C to 700C in t2 sec., varies as the fourth power of its absolute temperature., III) 700C to 600C in t3 sec. IV) 600C to 500C in t4 sec., Reason (R): Sun is not a black body., If the room temperature is 100C, identify the, Assertion (A): Thermal radiations are, order in which the times of cooling increases, electromagnetic radiation with wave lengths greater, 1) IV,III,II & 1, 2) I,III,IV & II, than visible light., Reason (R): Thermal radiations can propagate, 3) I, II,III & IV, 4) III, I, II & IV, through vacuum., 15. In the following, which statement is correct, Assertion (A): Two metallic spheres of same size,, A) A hot body emits hot radiations only., one of copper and other of aluminium heated to the, B) A cold body absorbs the radiations only., same temperatures, will cool at the same rate when, C) A cold body emits cold radiations only., they are suspended in the same enclosure., D) All the bodies emits and absorbs radiations, Reason (R): The rate of cooling of a body is directly, simultaneously, proportional to the excess of temperature of the, 1) A, 2) B, 3) C, 4) D, body over the surroundings., , LEVEL -IV - KEY, , STATEMENT TYPE QUESTIONS, Options :, 1. Statement A is true and statement B is true, 2. Statement A is true and statement B is false, 3.Statement A is false and statement B is true, 4.Statement A is false and statement B is false, 8. (A) Heat transfer by conduction and convection, require a material medium., (B) Heat transfer by radiation doesn’t effect the, medium through which it passes., 9. (A) A body of low thermal capacity gets heated or, cooled quickly., (B) Good emitters are bad reflectors., 10. (A) Greater the mass of radiating body, slower will, be cooling, (B)Greater the temperature of the surroundings,, lower will be cooling, 11. (A) Water can be boiled inside the artificial satellite, by convection, (B) Heavy liquid can be boiled in artificial satellite by, convection, 12. (A) Black body radiation is white, (B) Emissive power of a body is proportional to its, absorptive power, NARAYANAGROUP, , MATCHING TYPE QUESTIONS, 1) A-F, B-H, C-E, D-G 2) A-G, B-E, C-H, D-F, 3) A-F, B-G, C-E, D-H, ASSERTION & REASON TYPE QUESTION, 4) 1 5) 2 6) 2 7) 1, STATEMENT TYPE QUESTION, 8) 1 9) 1 10) 1 11) 4 12) 1, ORDER ARRANGING QUESTIONS, 13) 1 14) 3 15) 4, , LEVEL -IV - HINTS, 4., 5., 6., , 7., , Some air is trapped in pores of woolen clothes will, act as bad conductor of heat., Radiation emitted from the sun is measured with the, help of Stefan’s law by considering the sun as, blackbody, but sun is not a perfect black body., Thermal radiations are electromagnetic radiations, which belongs to infrared region. Electromagnetic, energy does not require material medium for their, propagation., Rate of cooling depends on surface area and, temperature difference between the body and the, surroundings., 103
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JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , TRANSMISSION OF HEAT, , LEVEL-V, , T–T0, , SINGLE ANSWER QUESTIONS, 1., , 0.5 mole of an ideal gas at constant, temperature 27°C kept inside a cylinder of, length L and cross-section area A closed by, a massless piston. The cylinder is attached, with a conducting rod of length L , cross-section, area (1/9) m2 and thermal conductivity k,, whose other end is maintained at 0°C. If piston, is moved such that rate of heat flow through, the conducing rod is constant then find velocity, of piston when it is at height L/2 from the, bottom of cylinder. [Neglect any kind of heat, loss from system], , (B) 2 min, (∆T)0, , (C) 3 min, 0.37 (∆T)0, (0.37)2 (∆T)0, O, , 5., , 6., L, , 27°C, 0°C, L, , (A), , k, 50R, , (B), , k, 100R, , (C), , k, 110R, , (D), , k, 90R, , Two thin walled spheres of different materials,, one with double the radius and one-fourth wall, thickness of the other, are filled with ice. If, the time taken for complete melting of ice in, the sphere of larger radius is 25 minutes and, that for smaller one is 16 minutes, the ratio of, thermal conductivities of the materials of, larger sphere to the smaller sphere is :, (A) 4 : 5 (B) 25 : 1 (C) 1 : 25 (D) 8 : 25, 3. The power radiated by a black body is P, and, it radiates maximum energy around the, wavelength λ0. If the temperature of the black, body is now changed so that it radiates, maximum energy around a wavelength 3λ0/4,, the power radiated by it will increase by a factor of, (A) 4/3, (B) 16/9 (C) 64/27 (D) 256/81, 4. A calorimeter of negligible heat capacity, contains 100 cc of water at 400C. The water, cools to 350C in 5 min. The water is now, replaced by k-oil A equal volume at 400C, Find, the time taken for the temp to become 350C, (Given densities of water and K-oil are, respectively 1000 and 800 kg.m-3; and their, specific heats are respectively: 420 and 2100, J/kg-0C), , 2., , 104, , (A) 1 min, , T, , 8., , 9., , 2τ, , Time (t), , (D) 4 min, , A and B are two points on a uniform metal ring, whose centre is C. The angle ABC = θ . A and, B are maintained at two different constant, temperatures. When θ = 180°, the rate of total, heat flow from A to B is 1.2 W. When θ = 90°,, this rate will be, (A) 0.6 W (B) 0.9 W (C) 1.6 W (D) 1.8 W, Two metallic spheres S1 and S2 are made of, the same material and have got identical, surface finish. The mass of S1 is thrice that of, S2. Both the spheres are heated to the same, high temperature and placed in the same room, having lower temperature but are thermally, insulated from each other. The ratio of the, initial rate of cooling of S1 to S2 is (IIT-95), 1, 3, , 1, , 3, , æ1ö, , 13, , (B) 3, (C), (D) ççè ÷÷÷ø, 3, 1, A black body is at a temperature of 2880 K., The energy of radiation emitted by this object, with wavelength between 499 nm and 500 nm, is U1, between 999 nm and 1000 nm is U2 and, between 1499 nm and 1500 nm is U3. The, Wein constant, b = 2.88 × 106 nm-K. Then, (IIT - 1998), (A) U1 = 0 (B) U3 = 0 (C) U1 > U2 (D) U2 > U1, Three discs, A, B and C having radii 2 m, 4 m, and 6 m respectively are coated with carbon, black on their outer surfaces. The wavelengths, corresponding to maximum intensity are 300, nm, 400 nm and 500 nm respectively. The, power radiated by them are QA, QB and QC, respectively (IIT - 2004), (A) QA is maximum, (B) QB is maximum, (C) QC is maximum, (D) QA = QB = QC, In which of the following process, convection, does not take place primarily ? (IIT- 2005), (A) Sea and land breeze (B) Boiling of water, (C) Warming of glass bulb due to filament, (D) heating air around a furnace, (A), , 7., , τ, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , 10. Variation of radiant energy emitted by sun,, filament of tungsten lamp and welding arc as, a function of its wavelength is shown in figure., Which of the following option is the correct, match?, Eα, T3, , 14. In the figure shown, AB is a rod of length 30, cm and area of cross-section 1.0 cm2 and, thermal conductivity 336 S.I. units. The ends, A & B are maintained at temperatures 200 C, and 400 C respectively. A point C of this rod is, connected to a box D, containing ice at 00 C,, through a highly conducting wire of negligible, heat capacity. The rate at which ice melts in, the box is, , T2, , 1, 4, , 1, 4, , 1, 4, , 1, 65 , 97 , 97 , (a) T (b) T (c) T (d) ( 97 ) 4 T, 2 , 4 , 2 , 13. Three rods AB, BC and BD of same length l, and cross-sectionsl area A are arranged as, shown. The end D is immersed in ice whose, mass is 440 gm. Heat is being supplied at, constant rate of 200 cal/sec from the end . Time, in which whole ice will melt (Latent heat of, fusion of ice is 80 cal/gm), , K,l, 200 cal/sec, A, , 2K,l, B, , 10cm, , Ice, , (A) 40/3 min, (C) 20/3 min, , (a) 84 mg/s (b) 84 g/s (c) 20 mg/s (d) 40 mg/s, 15. Which of the following graphs shows the, correct variation in intensity of heat radiations, by black body and frequency at a fixed, temperature?, (a) EV, , (c) Eλ, , NARAYANAGROUP, , (b) Eλ, , UV Visible Infra-red, , UV Visible Infra-red, , 3500K, 2500K, 1500K, , 1500K, 2500K, 3500K, , V, , V, , (d) Eλ, , Infra-red Visible UV, , Infra-red Visible UV, , 3500K, , 1500K, , 2500K, , 2500K, , 1500K, V, , 3500K, V, , 16. Three different arrangemnets of materials 1,, 2 and 3 to from a wall. Thermal conductivities, , a., 1, , (B) 700 sec, (D)indefiniely long time, , D, 0°C, , K/2l, , D, , 20cm, highly conducting wire, , 100°C, C, , Ice, , B, 40°C, , 20°C, , λ, , (A) Sun-T1,tungsten filament-T2, welding arc-T3, (B) Sun-T2,tungsten filament-T1, welding arc-T2, (C) Sun-T3,tungsten filament-T2, welding arc-T1, (D) Sun-T1,tungsten filament-T3, welding arc-T2, 11. A solid copper sphere ( density ρ and specific, heat C) of radius r at a n initial temeprature, 200K is suspended inside a chamber whose, walls are at almost 0K. The time required for, the temeprature of the sphere to drop to 100K, is, A) 1.7 ρ rc B) 2.7 ρ rc C) 3.3 ρ rc D) 4.2 ρ rc, 12. Three very large plates of same area kept, parallel and close to each other. They are, considered as ideal black surfaces and have, very high thermal conductivity. The first and, third plates are maintained at temperature 2T, and 3T respectively. The temperature of the, middle (i.e. second) plate under steady state, condition is (IIT JEE-2012), , C, , A, , T1, , are k1 > k 2 > k3 . The left side of the wall is, 200C higher than the right side. Temperature, difference ∆T across the material 1 has, following relation in three cases:, (a) ∆Ta > ∆Tb > ∆Tc, c., b., (b) ∆Ta = ∆Tb = ∆Tc, 2 3, 1 3 2, 3 1 2, (c) ∆Ta = ∆Tb > ∆Tc, (d) ∆Ta = ∆Tb < ∆Tc, 105
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TRANSMISSION OF HEAT, MULTIPLE ANSWER TYPE, 17. Two identical objects A and B are at, temperatures TA and TB, respectively. Both, objects are placed in a room with perfectly, absorbing walls maintained at a temperature, T (TA > T > TB). The objects A and B attain the, temperature T eventually. Select the correct, statements from the following., (A) A only emits radiation, while B only absorbs it, until both attain the temperature T, (B) A loses more heat by radiation than it absorbs,, while B absorbs more radiation than it emits, until, they attain the temperature T, (C) Both A and B only absorb radiation, but do, not emit it,until they attain the temperature T, (D) Each object continuous to emit and absorb, radiation even after attaiing the temperature T, 18. Two solid spheres are heated to the same, tempearature and allowed to cool under, identical conditions. Compare; (i) initial rates, of fall of temperature, and (ii) initial rates of, loss of heat. Assume that all the surfaces, have the same emissivity and ratios of their, radii, specific heats and densities are, respectively 1: a,1: ß ,1: ? ., (A) aß? :1 (B) 1: a 2 (C) ß = a? (D) 1: a 3, 19. Two bodies A and B have thermal emissivities, of 0.01 and 0.81 respectively. The outer, surface areas of the two bodies are the same., The two bodies emit total radiant power at the, same rate. The wavelength λB corresponding, to maximum spectral radiancy in the radiation, from B shifted from the wavelength, corresponding to maximum spectral radiancy, in the radiation from A, by 1.00 µ m . If the, temperature of A is 5802 K : (IIT - 1994), (A) the temperature of B is 1934 K, (B) λB = 1.5µ m, (C) the temperature of B is 11604 K, (D) the temperature of B is 2901 K, 20. A 100 cm long cylindrical flask with inner and, outer diameter 2 cm and 4 cm respectively is, completely filled with ice as shown in the, figure. The constant temperature outside the, flask is 400C., (Thermal conductivity of the flask is, 0.693 W / m 0 C , Lice = 80cal / gm & ln 2 = 0.693 )., 106, , JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , (a) Rate of heat flow from outside to the flask is 80 π J / s, π, kg / s, (b) The rate at which ice melts is, 4200, (c) The rate at which ice melts is 100 π kg / s, (d) Rate of heat flow from outside to flask is 40 π J / s, 21. A metal cylinder of mass 0.5 kg is heated, electrically by a 12 W heater in a room at, 15 0 C. The cylinder temperature rises, nuniformly to 250C in 5 min and finally, becomes constant at 450C. Asuming that the, rate of heat loss is proportional to the excess, temperature over the surroundings, (a) The rate of loss of heat of the cylinder to, surrounding ar 200C is 2 W, (b) The rate of loss of heat of the cylinder to, surrounding at 450C is 2 W, 240, J / kg 0C, (c) Specific heat capacity of metal is, ln(3/ 2), (d) None fof these, 22. When we consider convection with radiation, in Newton’s law of cooling while temperature, of the object in consideration is sightly higher, than the environment temperature. Choose, correct statements about rate of heat loss., (a) directly proportional to emissivity, (b) directly proportional to Stefan’s constant, (c) directly proportional to surface area, (d) directly proportional to temperature difference, of body and room., , MATRIX MATCHING TYPE QUESTIONS, 23. Match the following Column I and II, Column I, (A) Wien’s displacement explains, (B)Planck’s law explains, (C) Kirchhoff’s law explains, (D) Newton’s law of cooling explains, Column II, (p) Why days are hot and nights cold in deserts, (q) Why a blackened platinum wire, when gradually, heated, appears first dull red and then blue, (r) The distribution of energy in black body, spectrum at shorter as well as longer wavelengths, (s) Why some stars are hot ter than other, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , 24., , Column I, A) A perfect reflecting body, B) A perfect black body, C) An ordinary smooth body, D) An ordinary rough body, Column II, P) absorbs radiation, Q) reflects radiation, R) emits radiations, S) transfer heat, 25. Entries in cloumn I consists of diagrams of, thermal conductors. The tupe of conductor &, direction of heat flows are listed below., Entries column II consists of the magnitude, of rate of heat flow belonging to any of the, entries in column I. If temperature difference, in all the cases is (T1-T2), Column I, , (B), , B is thin spherical shell A is a solid sphere r A<rB, (C), , B is thin spherical shell, A is a solid sphere r A<rB, Body A is being heated by a heater of constant, power ‘P’, (D), , (B), , (A), , 3R, , 3R, 2R, T1, , k0, , k0, Thick cylindrical, shell, flow along axis, , Thick spherical, shell, radical flow, , (D), (C), , B is thin spherical shell, A is a solid sphere, rA ≈ rB, Body A is being heated by a heater of constant, power ‘P’, Column II, (P) TA = TB, , T1 R, , T2, , R, , T2, , 3R, , 3R, , T1, , 2R, , R, , T2, T1, , R, T1, k0, T2, , X solid cylinder,, flow along axis,, variable k as, k = k0(1+x/(3R)), , Thick cylindrical, shell, radical flow, , COMPREHENSIVE TYPE QUESTIONS, , Column II, (P) 6π k0 R ( T1 − T2 ), , π k0 R, (Q) 3 ln 2 (T1 − T2 ), , 4π k0 R, (T1 − T2 ), ln 2, 26. A & B are two black bodies of radii rA and rB, respectively, placed in surrounding of, temperature T 0 . At steady state the, temperature of A & B is T A & TB respctively., Column I, , (R) π k0 R ( T1 − T2 ), , (Q) TA < TB, (R) Heat received by A is more than heat radiated, by it at steady state., (S) Radiation spectrum of A & B is distinguishable, (T) Steady state can’t be achieved, , (S), , (A), , Passage : 1, A body is kept inside a container the temeprature, of the body is T1 and the temeprature of the, container is T2 . the rate at which body absorbs, the energy is α . The emissivity of the body is e., The radiation striking the body is either absorbed, or reflected., 27. After a long time, the temperature of the body, will be, A) T1, B) T2, C) T1 +, , A & B are solid sphere r A=rB, Body ‘B’ is being heated by a heater of constant, power ‘P’, NARAYANAGROUP, , (T1 − T2 ), , D) none of these, 2, 28. At what rate, the body will emit the radiant, energy, A) If t is the time, rate is ( T1 − T2 ) t B) e, C) both of the above, D) none of the above, 107
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JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , TRANSMISSION OF HEAT, 29. At what rate the body will basorb the radiant, energy, A) α , but α ≠ e, B) ( T1 − T2 ) / t , where t is the time, C) e, but e = α, D) None of the above, 30. A good absorber is, A) good reflector, B) poor reflector, C) average reflector D) assessment not possible, , ASSERTION & REASON TYPE QUESTIONS, Note : Each question contains STATEMENT1 (Assertion) and STATEMENT-2 (Reason)., Each question has 5 choices (A), (B), (C), (D), and (E) out of which ONLY ONE is correct., (A) Statement-1 is True, Statement-2 is True;, Statement-2 is a correct explanation for Statement-1., (B) Statement-1 is True, Statement-2 is True;, Statement-2 is NOT a correct explanation for, Statement-1., (C) Statement -1 is True, Statement-2 is False., (D) Statement -1 is False, Statement-2 is True., (E) Statement -1 is False, Statement-2 is False, 31. Statement-1 : As the temperature of the blackbody, increases, the wavelength at which the spectral, intensity (Eλ) is maximum decreases., Statement-2 :The wavelength at which the spectral, intensity will be maximum for a black body is, proportional to the fourth power of its absolute, temperature., 32. Take a test tube nearly filled with water and, put in the bottom a piece of paraffin wax, round, which a piece of thick copper wire has been, wound so that wax remains at the bottom., Temperature required for melting wax is 520C., Statement -1: Heat the water at top where it will, be seen to boil vigorously without melting wax., Statement-2 : Convection currents are not set-up, when water columm is heated from top amd water, is poor conductor of heat., , INTEGER TYPE QUESTIONS, 33. A rod of length l with thermally insulated, lateral surface is made of a material whose, thermal conductivity varies as K = C/T. where, C is a constant. The ends are kept at, temperatures T1 and T2. The temperature at, a distance x from the first end where the, T2 , , T1 , , temperature is T1 , T = T1 , value of n?, 108, , n x / 2l, , . Find the, , 34. A solid copper sphere of density ρ, specific, heat c and radius r is at temperature T1. It is, suspended inside a chamber whose walls are, at temperature 0 K. The time required for the, temperature of sphere to drop to T 2 is, rρc 1, 1 , 3 − 3 . Find the value of x? Take the, xeσ T2 T1 , , emmissivity of teh sphere to be equal to e., 35. Two identical conducting rods are first,, connected independently to two vessels, one, containing water at 100°C and the other, containing ice at 0°C. In the second case, rods, are joined end to end and are connected to, the same vessels. If q1 and q2 (in g/s) are the, rates of melting of ice in two cases, then find, the ratio of q 1/q2., 36. Two spherical bodies A (radius 6 cm) and B, (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the, emission spectrum of A is at 500 nm and in, that of B is at 1500 nm. Considering them to, be black bodies, what will be the ratio of total, energy radiated by A to that of B?(IIT-2010), , LEVEL - V - KEY, SINGLE ANSWER QUESTIONS, 1) B 2) D 3)D, 8)B, 9)C 10)D, 15) C 16) B, , 4)B, 11)A, , 5)C 6)D 7)D, 12) C 13) A 14) D, , MULTIPE ANSWER QUESTIONS, 17) B, D, 20) A,B, , 18) A,B, 21) A,C, , 19) A,B, 22) C,D, , MATRIX MATCHING TYPE, 23), A → Q; B → S; C → R; D → P, 24) A → Q,S; B → P,R,S; C → P,Q,R,S; D → P,Q,R,S), 25), A → R; B → P; C → S; D → Q, 26), A → Q,S; B → P; C → S; D → Q,S, , COMPREHENSION TYPE QUESTIONS, 27) B 28) B 29) C 30) B, , ASSERTION REASON TYPE QUESTIONS, 31) C 32) A, , INTEGER TYPE QUESTIONS, 33) 2 34) 9 35) 4 36) 9, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , LEVEL - V - HINTS, , 5., , SINGLE ANSWER QUESTIONS, 1., , ∆Q ∆w, kaθ, =, = work done per unit time =, ∆t, ∆t, L, , Rate of flow of heat I1 = 1.2 =, , nRT, AV ′, Power = F × Velocity = PAV' =, V, , ⇒, , 2., , 0.5R(300), kaθ, AV ′ =, V, L, 0.5R(300), kaθ, AV ′ =, L, L, A., 2, , ⇒ V′ =, , ka 27 , k, , =, R 300, 100R, , Rate of flow of heat I 2 =, 6., , or, , [R = thermal resistance], •, , 4πK∆θ, 4πK∆θ, =, r2 − r1 , t, 2 , r r , r, 12, mL 4πK∆θ , 4, , =, m = ρ × πr3 , 2, , time, (t / r ) , 3, r, , tr, ρL K , time, ∝, constant, = ×, r, K, time tr , , R, , 3., , R, , ∴Area of sphere,, 3m , A = 4πr 2 or A = 4π , , 4πρ , , 4, ∆T eσT , 3 , , =, 4π × , , ∆t s , 4πρ , , ∴, , S1 m1 , =, S2 m 2 , , 3λ 0, T = b = λ 0 T0 or, ∴, 4, , ∴, , S1 1 , = , S2 3 , , 1/3, , 7., , From the law of cooling, we have, , or λ m =, , kt , −, , , As the properties of the exponential functions, demand, in equal intervals of time the fall in the, ‘temperature difference’ has the same ratio., NARAYANAGROUP, , 2/3, , S 3m , or 1 = , , S2 m , , −1 / 3, , 1/3, , 1, = , 3, , b 2.88 × 106 nm K, =, or λ m = 1000nm, T, 2880 K, , ∴Energy of radiation is maximum, , Since the intial and the final temperature differences, 800 × 2100 × 5, = 2min, 1000 × 4200, , m, m, , According to Wien’s displacement law,, λ m T = Wien’s constant (b), , ( 35 − T0 ) = ( 40 − T0 ) e ?sV , , ∴ t2 =, , 2/3, , 1/3, , 4T, T= 0, 3, , are the same, we must have ?1.s1 .t2 = ?2 .s2 .t1, , ……….. (ii), , S = (Constant) × m–1/3, , (D). Let T 0 = initial temperature of the black body., ∴ λ0T0 = b (constant), Power radiated = P 0 = C.T04. (c = constant), Let T = new temperature of black body., , 4, , 2/3, , ∴Rate of cooling, , t, 2rKs, KL, 8, 25 4, 1 KS, ; K = 25, =, =, S, 16, trK L, 2 KL, , 4, 256 , Power radiated = c.T4 = (cT04 ) = P0 , , 3, 81 , , 4., , ∆T eσAT 4, ∆T eσT 4 A, =, or, =, …… (i), ∆t, ms, ∆t s m, 1/3, , 2, , t/4, , ∴ ms∆T = eσAT 4 ∆t, , 4 3, 3m , For a sphere, m = πr ρ or r = , , 3, 4πρ , , 1, , t, , ∆T, 16, = × 0.3 = 1.6W, 3R /16 3, , According to Stefan’s law, ∆Q = eσAT 4 ∆t, Also, ∆Q = ms∆T, , dr, 1 r2 − r1 , (D). ∫ dR = ∫ 4πr 2 K = 4πK r r , 21 , , Q=, , ∆T, ∆T, ∴, = 0.3, R/4, R, , For θ = 90°, equivalent resistance between A &, B is 3R/16 (R/4 and 3R/4 in parallel), , where V → volume, V' → velocity, ⇒, , R = Total thermal resistance of the ring, ∆T = difference in temperature between A and B, for θ = 180° equivalent resistance between A and, B is R/ 4 (R/2 & R/2 in parallel), , U2, Eλ, , U1, , at, U3, , λµ, , λ m = 1000 nm, , ∴Thus U2 > U1, U2 > U3, λ, , U1 ≠ 0, U 2 ≠ 0, , ∴Option (d) is correct, 109
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JEE- ADV PHYSICS-VOL- V, 6., , TRANSMISSION OF HEAT, , The three rods of same material and crosssectional area from the sides of a triangle ABC., The points A, B and C are maintained at, temepratures T, T 2, , and, , (, , 3T, , 9., , The emissive power of a black body at T=300K, is 100 W / m 2 . Consider a body B of area, A = 10m 2 , coefficient of reflectivity r = 0.3,, and absorptivity a = 0.2. If its temperature is, 300K, then markout the correct statement., , ), , 2 +1, , respectively. Assuming that only heat, conducting takes place and the system is in, steady state, find the angle at B. The, temeprature difference per unit length along, CB and CA is equal., A(T), , a) The emissive power of B is 20W / m 2, b) The emissive power of B is 200W / m 2, c) The power emitted by B is 20W, d) The power emitted by B is 180W, 10. A spherical shell of inner radius R1 and outer, radius R2 is having variable thermal, , B T 2, , 7., , conductivity given by K = a0Tr . Where ‘r’ is, the distance from the centre. Two surfaces of, the shell are maintained at temperature T1, , 3T , C, , 2 +1 , , A) 300, B) 450, C) 600, D) 900, Temperature of a body ? is slightly more than, , (inner surface) and T2 (outer surface),, respectively ( T1 > T2 ) . The heat current, flowing through the shell would be, , the temperature of the surrounding ?0 , its rate, of cooling ( R ) versus temperature of body, , ( ? ) is plotted, its shape would be:, , a), , 4π a0 ( T12 − T22 ), R2 − R1, , (b) R, , (a) R, , c), θ, , θ, , θ, , 8., , d), , (d) R, , (c) R, , θ, , A body cools from 800 C to 700 C in 10, minutes. Find the time required further for it, to cool from 700 C to 600 C . Assume the, temerature of the surrounding to be 300 C ., 4, , , 5, , , (a) 10log e , 3, , (b) 10log e , 4, , 4, log e , 3, 10 ×, (c), 5, log e , 4, , 5, log e , 4, 10 ×, (d), 4, log e , 3, , NARAYANAGROUP, , × R1 R2 b), , 4π a0 R12 R22 (T12 − T22 ), R22 − R12, , 4π a0 (T1 − T2 ) R1 R2, R2 − R1, , 4π a0 (T12 − T22 )( R1 + R2 ), , 2, , R2 − R1, 11. A radiator whose temperature is T 0C, is used, to heat the room in the cold weather. The, radiator is able to maintain a room temperature, of 300C when outside temperature is -100C and, 150C when outside temperature is -300 C., Determine the temperature of the radiator, [Assume Newton’s law of cooling to be valid], a) 850C, b) 150C, c) 98.60C d) 1500C, 12. Two thin walled spheres of different materials,, one with double the radius and one-fourth wall, thickness of the other, are filled with ice. If, the time taken for complete melting of ice in, the sphere of larger radius is 25 min and that, for smaller one is 16 min, the ratio of, thermal conductivities of the materials of, larger sphere to the smaller sphere is, (a) 4 : 5, (b) 25 : 1 (c) 1 : 25 (d) 8 : 25, 113
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JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , TRANSMISSION OF HEAT, 13. An object is being heated by a heater supplying, 60 W heat. Temperature of surrounding is, 20°C and the temperature of object becomes, constant at 50°C. Now the heater is switched, off. What is the rate at which the object will, lose heat when its temperature has dropped, to 30°C, (a) 20 W (b) 30 W (c) 40 W (d) 60 W, 14. The power radiated by a black body is Po and, the wavelength corresponding to the maximum, , The ratio of the surface temperature of star A, to that of star B, TA :TB , is approximately:, A) 2 : 1, B) 4 : 1, C) 1 : 2, D) 1 : 1, 17. An irregular rod of same uniform material as, shown in figure is conducting heat at a steady, rate. The temperature gradient at varoius, sections versus area of cross section graph, will be:, , energy is around λ0 On changing the, temperature of the black body, it was observed, 256, P0, 81, The change in the wavelength corresponding, to maximum intensity, that pow er radiated is increased to, , (a) increases by, , λ0, 4, , (b) decreases by, , A) dT/dx, , B) dT/dx, , λ0, 4, , λ0, λ0, (d) decreases by, 2, 2, 15. The container A is constantly maintained at, 1000 C and insulated container B in the figure, contains ice at 00 C . Different rods are used, to connect them. For a rod made of copper, it, takes 30 minutes for the ice to melt and for a, rod of steel of same cross-section taken in, different experiment it takes 60 minutes for, ice to melt. When these rods are, simultaneously connected in parallel, the ice, melts in:, , A, , (c) increases by, , A, , Star B, , 114, , 18. A solid copper sphere of dimater 10mm, is, cooled to a temperature of 150K and is then, placed in an enclosure at 290 K. Assuming that, all interchange of heat is by radiation, calculate, the initial rate of rise of temperature of the, sphere. The sphere may be treated as a black, body ρcopper = 8.93 ×103 kg / m3 ,, , 500 1000, 2000, Wavelength (nm), , s = 3.7 × 102 JKg −2 K −1 ; σ = 5.7 × 10−8 Wm −2 K −4, A) 0.68 K/s B) 0.068 K/sC) 0.34 K/s D) 0.034 K/s, 19. The temperatures across two different slabs, A and B are in the steady state (as shown in, Fig.) The ratio of thermal conductivities of A, and B is, Temperature (°C), , Intensity, , Star A, , 3000, , D) dT/dx, , A, , B, , A) 15 minutes, B) 20 minutes, C) 45 minutes, D) 90 minutes, 16. The spectra of radiation emitted by two distant, stars are shown below., , 0, , C) dT/dx, , A, , A) 2:3, , 60, 50, 40, 30, 20, 10, 0, , B) 3:2, A, , B, , C) 1:1, 0, , 1 2 3 4 5 6, (cm), , D) 5:3, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , MULTIPLE ANSWER QUESTIONS, 20. The temperature drop through a two-layer, furnace wall is 900°C. Each layer is of equal, area of cross section. Which of the following, actions will result in lowering the temperature, θ of the interface?, Other 1000°C, , Inner Outer, layer layer, , (a) heat flow through A and E slabs are same, (b) heat flow through slab E is maximum, (c) temperature difference across slab E is smallest, (d) heat flow through C = heat flow through B +, heat flow through D., dθ, = − k (θ − θ 0 ),, dt, the constant k is proportional to:, (A) A; surface area of the body, (B) S is the specific heat of the body, , 23. In Newton’s law of cooling, , θ, , (A) By increasing the thermal conductivity of outer, layer, (B) By increasing the thermal conductivity of inner, layer, (C) By increasing thickness of outer layer, (D) By increasing thickness of inner layer, 21. Seven identical rods of material of thermal, conductivity k are connected as shown in Fig., All the rod are of identical length l and crosssectional area A. If the one end B is kept at, 100°C and the other end is kept at 0°C. The, temperatures of the junctions C, D and, E(θC , θD and θE ) be in the steady state?, C, k, , k, , D, , k, , A, , k, , k, E, , k, , 1, being the mass of the body, m, , (D) e is the emissivity of the body, , COMPREHENSION TYPE QUESTIONS, Passage-I, Consider a spherical body A of radius R which is, placed concentrically in a hollow enclosure H, of, radius 4R as shown in the figure. The temperature, of the body A and H are TA and TH respectively.., Emissivity, transmittivity and reflectivity of two, bodies A and H are (eA , eH ), ( t A , tH ) and ( rA , rH ), respectively. (Assume no absorption of the thermal, energy by the space in between the body and, enclosure as well as outside the enclosure and all, radiations to be emitted and absorbed nomral to, the surface.)[Take σ × 4π R 2 × 300 4 = β J / s ], , B, , k, , (C), , H, , (A) θC > θE > θD, , A, , (B) θE = 50o Cand θD = 37.5o C, , 6R, P, , (C) θE = 50 C, θC = 62.5 Cand θD = 37.5 C, 0, , o, , o, , (D) θE = 50o C, θC = 60o Cand θD = 40o C, 22. A composite block is made of slabs A, B, C, D, and E of different thermal conductivities, (given in terms of a constant K) and sizes, (given in terms of length, L) as shown in the, figure. All slabs are of same width. Heat Q, flows only from left to right through the blocks., Then in steady state, 0, , heat, , 1L, A, , 5L, B, , 3K, , 6L, E, , 1L, , 3R, Q, , 24. The temperature of A (a perfect black body), is TA = 300 K and temperature of H is, , TH = 0 K . For H take eH = 0.5 and t H = 0.5,, For this situation mark out the correct, statement(s)., a) The rate at which A loses the energy is β J / s ., b) The rate at which spherical surface containing P, β, J /s., 2, c) The rate at which spherical surface containing Q, receives the energy is β J / s ., d) All of the above, , receives the energy is, 2K, , C, , 4K, , D, , 5K, , 3L, 4L, , NARAYANAGROUP, , 6K, , 115
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JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , TRANSMISSION OF HEAT, 25. In above question, if body A has, eA = 0.5, rA = 0.5 and for H, eH = 0.5, rH = 0.5,, then mark out the correct statement(s)., β, a) The rate at which A loses energy is, 2, b) The rate at which the spherical surface containing, P receives the energy is zero., c) The rate at which the spherical surface containing, Q receives the energy is β d) All of the above, 26. Consider two cases, first one in which A is a, perfect black body and the second in which A, is a non-black body. In both the cases,, temperature of body A is same equal to 300K, and H is at temperature 600K. For H,, t = 0 and a ≠ 1 . For this situation, mark out, the correct statement(s)., a) The bodies lose their distinctiveness inside the, enclosure and both of them emit the same radiation, as that of the black body., b) The rate of heat loss by A in both cases is the, same and is equal to β J / s ., c) The rates of heat loss by A in both the cases are, different., d) From this information we can calculate exact, rate of heat loss by A in different cases., 27. In the previous question if the enclosure is, considered as perfect black body and is, maintained at same temperature as that of, temperature of body A, then in the two cases, a) the body A emits radiation at the same rate., b) the body A emits radiation at different rates, c) the temperature of body A remains constant., d) None of the above, Passage-II, A highly conducting solid sphere of radius R, density, ρ and specific heat s is kept in an evacuated, chamber. A parallel beam of thermal radiation of, intensity I is incident on its surface. Consider the, sphere to be a perfectly black body and its, temperature at certain instant considered as t = 0, is T0 . [Take Stefan’s constnat as σ ]. Answer the, following questions based on above information., 28. The equation which gives the temperature T, of the sphere as a function of time, is, T, t 3dt, T dT, t 3dt, dT, =∫, =, a) ∫T0, b), 4, 4, ∫T0 4σ T ∫0 4 Rρ s, 0 4 Rρ s, I − 4σ T, T, , c) ∫T0, 116, , dT, 3t, =, d), 4, I − 4σ T, 8Rρ s, , ∫, , T, , T0, , 3dT, 5t, =, 4, I − 4σ T, 4Rρ s, , 29. The maximum attainable temperature of the, sphere is, 1, , I 2 I , a) , b) , , 4σ , 2σ , , 1, , 3, , I , c) , , 4σ , , 1, , 4, , d) Never occurs, , MATRIX MATCHING TYPE QUESTIONS, 30. Suppose that both ends of the rod are kept at, a temperature of T0C, and that the initial, temperature distribution along the rod is given, by T = (1000C) sin πx / L , where x is measured, from the left end of the rod. Let the rod be of, copper, with length L and cross-section area, A. Column I represents graph of certain, physical quantities as we move from left to, right end of rod. Column II represents those, physical quantities., Cross-section Area A, Heat, T °C, , T°C, L, , (A) Y, , O, , X, x=L, , (B), , Y, , O, , X, , x = L/2, x=L, , (C), , Y, , Graph coincides, with x-axis, O, , X, x=L, , (p) Initial temperature gradient, (q) Initial temperature, (r) Finl temperature distribution along rod., (s) Final rate of heat transfer along rod., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , TRANSMISSION OF HEAT, , INTEGER TYPE QUESTIONS, 31. A few rods of materials X and Y are connected, as shown in Fig. The cross-sectional areas of, all the rods are same. If the end A is maintained, at 80°C and the end F is maintained at 10°C., If the temperature of junctions B and E in, steady state are, , 39.48 o C, 60.52 o C, and, . Find n 1, n1, n2, , and n2 . Given that thermal conductivity of, material X is double that of Y., , 35. A metal rod AB of length 10x has its one end, A in ice at 0 0C and the other end B in water at, 1000C. if a point P on the rod is maintained at, 4000C, then it is found that equal amounts of, water and ice evaporate and melt per unit time., The latent heat of evaporation of water is 540, cal/g and latent heat of melting of ice is 80 cal/, g. If the point P is at a distance of λ x from the, ice end A, find the value of λ . (Neglect any, heat loss to the surrounding.), , LEVEL VI - KEY, , C, L, 80°C, A X B, , SINGLE ANSWER QUESTIONS, , Y, L, , E Y F, , X, , L, , D, 32. A hot body placed in air is cooled down, according to Newton’s law of cooling, the rate, of decrease of temperature being k times the, temperature difference from the, surrounnding. Starting from t = 0, The time in, which the body will lose half of the maximum, heat is, , x ln 2, . Find the value of x., 2k, , 33. One end of a uniform rod of length 1 m is, placed in boiling water while its other end is, placed in m elting ice.A point P on the rod is, maintained at a constant temperature of, 800°C. The mass of steam produced per, second is equal to the mass of ice melted per, second. If specific latent heat of steam is 7, times the specific latent heat of ice, then the, distance of P from the steam chamber is n/18, m. Find the value of n?, 34. Two indentical conducting rods are first,, connected independently to two vessels, one, containing water at 1000 C and the other, containing ice at 0 0C. In the second case, rods, are joined end to end and are connected to, the same vessels. If q1 and q2 ( in g/s) are the, rates of melting of ice in two cases, then the, ratio of q1 / q2 is, NARAYANAGROUP, , 1) B 2) C, 3) B 4) D 5) B 6) D 7) B, 8) C 9) A 10) B 11) D 12) D 13) A 14) B, 15) B 16) A 17) B 18) B 19) B, MULTIPLE ANSWER QUESTIONS, 20) A,B 21) A,C 22) A,B,C,D 23) A,C, , COMPREHENSION TYPE QUESTIONS, 24) D 25) D 26) C 27) B 28) A 29) C, , MATRIX MATCHING TYPE QUESTIONS, 30), , A-Q, B-P, C-RS, , INTEGER ANSWER TYPE QUESTIONS, 31) 2 32) 2 33) 2 34) 4 35) 9, , LEVEL VI - HINTS, SINGLE ANSWER QUESTIONS, 1., , dQ, , (B). For θ-t plot, rate of cooling =, dt, = slope of the curve., dθ, dt, , AT P,, , = |tan(180-φ2)| = tanφ2 = k(θ2 - θ0), , where k = constant., At Q,, tan ϕ, , dθ, =| tan(180 − ϕ1 ) |= tan ϕ1 = k(θ1 − θ0 ), dt, θ −θ, , 2, 0, 2, ∴ tan ϕ = θ − θ, 1, 1, 0, , 2., , dθ, = −k(θ − θ0 ), where k = constant, dt, ∴, , ∫, , θ, , θi, , t, dθ, = − ∫ k ⋅ dt, 0, θ − θ0, , or [ln (θ − θ0 )]θθi = −kt or, ln(θ − θ0 ) − ln(θi − θ0 ) = − kt, θ − θ0, − kt, or θ − θ = e, or, i, 0, , θ = θ0 + (θ i − θ0 )e− kt ., 117
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JEE- ADV PHYSICS-VOL- V, , dQ, = k1 (T − T0 ), dt, Where T is the temperature of object and T0 is the, , From Newton’s law of cooling,, , temperture of the surroundings. From the law of, conduction,, , dQ, = k2 (T2 − T1 ) Where T2 and, dt, , T1 are temeratures of the bodies from which heat is, , time ∝, , tr, K, , t, 25 4 2rK s 1 K s K L = 8, ;, =, =, 16, trK L, 2 K L K s 25, 14. Let initial temperature is T0 and final (new), temperature is T, while corresponding wavelength, , transferred and to which it is transferred, respectively. ⇒ For equilibrium, , corresponding to maximum intensities are ?0 and, , For first case,, , p0 T0 , = , From Stefan’s law,, p T , , k1 (T − T0 ) = k 2 (T2 − T1 ), , k1 (T − 30 ) = k2 30 − ( −10 ) = k 2 × 40, For second case,, , k1 (T − 15 ) = k2 15 − ( −30 ) = k2 × 45, , ⇒, , 12., , TRANSMISSION OF HEAT, , ∫, , T − 30 40, =, ⇒ T = 1500 C, T − 15 45, , 1, dr, =, 2, 4pr K 4pK, , dR = ∫, , r2 − r1 , , , r2 r1 , , [R = thermal resistance], , Q=, , ? respectively.., , 4, , T , T 4, 256 4 , ⇒ T = 81 = 3 ⇒ T = 3, , 0, 0, 4, , From Wien’s law, ?0T0 = ?T, , ?0 T 4, 3, = = ⇒ ? = ?0, ? T0 3, 4, So wavelength corresponding to maximum energy, at new temperature decreases by, , 4pK? ? 4pK? ?, =, r2 − r1 , t , 2, , , r , r1r2 , , 15. Q = it where i = heat flow rate =, , ∆T 100, =, R, R, , 100 , , 100 , , Also, ∴ Q = R × 30 ⇒ R2 = Q × 60, , , 2 , , r2, , 1, , mL 4pK? ? , 4, , =, m = ? × pr 3 , , t , time, 3, , 2, r , ?L K , = × constant, time tr , , 1 , , Now, Q = R + R treqd, 1, 2 , ∴ t reqd =, , Q, = 20, Q 1 Q 1, min., , × , ×, 100 30 100 60, , T, , λ, , 1000, , A, MB, 16. λMATA = λMBTB (Weins law); T = λ = 500 = 2 :1, B, MA, , t/4, , t, , NARAYANAGROUP, , ?0, ., 4, , 100 , 100 , ∴Q = , ( 30 ) ⇒ R1 = , × 30, R, Q , 1 , , r1, , R, , 4, , R, , 17. H = KA, ∴A, , dT, dx, , is same in stready state condition,, , dT, = constant, dx, , ∴ Hyperbolic graph, , 119
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JMAINS, - C.W - VOL V, - II, JEE- ADV, PHYSICS-VOL-, , TRANSMISSION OF HEAT, , If the body lose half of the maximum loss that it, , INTEGER ANSWER TYPE QUESTIONS, 31. (n1 = 1 and n2 = 2) We first find the thermal, resistances of the different rods shown in the figure., , can, then decrease in temp, , 1 L, , If body loses this heat in time t, then its temperature, , These are given as R AB = k . A, x, 1 ( πL / 2), 1 (πL / 2), R BDE =, ., ., ;, kY, A, kX, A, , T1 − T0 T1 + T0, = 2, 2 , , at tiem t’ will be T1 − , , 1 L, , ; R EF = k . A, Y, Also, k X = 2k Y , Now in steady state the amount, of heat flow from end E to F remains constant as, there is no absorption of heat. Then we must have, that the amount of heat coming at junction B is equal, to the amount of heat having B and same statement, can be given for junction E. If temperatures of, junctions B and E are taken as TB and TE,, respectively, then we have for junction β ., , R BCE =, , Putting these values in Eq. (i), we have, T1 − T0, = T0 + (T1 − T0 )e − kt', 2, T −T, 1, ln 2, or 1 0 = (T1 − T0 )e − kt' or e− kt' = or t ' =, 2, 2, k, , 33. Let, , TB − TE TB − TE , +, ....(i), π 2π , , or 80 − TB = , , Similarly for junction E, we can write, , TB − TE, TB − TE, T − 10, +, + E, 1/ k y .πL / 2A 1/ k x .πL / 2A 1/ k y .L / A, 2 ( TB − TE ) 4 ( TB − TE ), +, = TE − 10, π, π, , ⇒x=, , 34., , Solving equation (i) and (ii), we get, TE = 19.74o C and TB = 60.52o C, , dT, = k(T − T0 ), dt, , 35, , where T0 is the temperature of the surrounding. If, T1 is the initial temperature and T is the temperature, , Melting, Ice, 1–X, , (L), , ⇒ Qsteam = 7 mL and Qice = mL, 800 − 100, 800 − 0, = 7ℜA, i.e., Qsteam = 7Qice ⇒ ℜA, x, 1− x, , 3, ( TB − TE ) = TE − 10 ..........(ii), 4π, , T, , Q, , X, , (7L), , TB − TE TB − TE TE − 10, +, =, R BCE, R BDE, R EF or, , 32. (2) We have −, , P, , Boiling, water, , TB − TC, TB − TE, 80 − TB, =, +, 1/ k x . L / A 1/ k Y .πL / 2A 1/ k x .πL / 2A, , or, , dmsteam d, = mice = m, dt, dt, Q, , TA − TB TB − TE TB − TE, =, +, R AB, RBCE, R BDE or, , or, , Q, T −T , = mc 1 0 , 2, 2 , , 1, m, 9, , Temperature difference, dm , = L, , Thermal resis tan ce, dt , dm, 1, 1, ∝, ;q ∝, dt Thermal resis tan ce, R, The rods are in parallel in the first case and they, q1, 2R, are in series in the second case q = ( R / 2) = 4, 2, dmice dmvapour, =, dt, dt, Q1 λx, , P, , (10 –λ)x Q2, , t, , dT, at any time t, then ∫ (T − T ) = −k ∫ dt, 0, T, 0, 1, , T−T , , 0, or ln(T − T0 ) TT = −kt or ln , = −kt, T, −, T, 1 0, , or T = T0 + (T1 − T0 )e − kt, (i), The body continues to lose heat till its temperature, becomes equal to that of the surrounding. The loss, of heat Q = mc(T1 − T0 ), 122, , Q1 =, , Q1 =, , KA400, = m × 80, t, KA ( 400 − 100 ) t, , (10 − λ ) x, , t = m × 540, , Dividing both, λ = 9, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , CALORIMETRY, SYNOPSIS, , Ø, , θ2, , INTRODUCTION, Ø, , Ø, , Heat is a form of energy in transit, that flows from, a body at high temperature to a body at low, temperature., The S.I. Unit of heat is Joule (J), , Ø, , The C.G..S. Unit of heat is calorie (Cal), , Ø, , 1 Cal = 4.186 J ≅ 4.2J, The quantity of heat required to warm a given, substance depends on its mass ( m), the change, in temperature ( ∆θ ) and nature of the substance., i.e ∆ Q = mS ∆ θ, Where ∆Q = quantity of heat gained, m = mass of substance, ∆θ = rise in temperature, S= specific heat (depends on nature of the, material), , Specific Heat, Ø, , Ø, Ø, Ø, Ø, , Ø, Ø, Ø, Ø, Ø, , The amount of heat required to rise the, temperature of unit mass of a substance through, 10C is called specific heat of the material of the, body., 1 ∆Q, S=, m ∆θ, The S.I unit of S is J/kg- K, The C.G.S unit of S is Cal./g- 0C, Dimensional formula of S is L2T −2θ −1, Molar specific heat capacity, 1 ∆Q, C=, (n = number of moles), n ∆θ, The SI unit of C is J/mole - K, The C.G.S. unit of C is Cal / mole - ºC, Dimensional formula of C is ML2T −2θ −1mole −1, Gases will have two specific heats, (i) Specific heat at constant volume (Cv), (ii) Specific heat at constant pressure (Cp)., Specific heat depends only on the nature of, material and unit of temperature. Usually, temperature dependence of specific heat is, neglected., , NARAYANAGROUP, , If specific heat varies with temperature then heat, energy given to a substance is, Q = m ∫ Sdθ, θ1, , Ø, , Water has largest specific heat among solids and, liquids. So it is used as coolant in automobile, radiators., Ø Among solids, liquids, and gases specific heat is, maximum for Hydrogen. (3.5 Cal/g- ºC ) and, minimum for radon and actinium. ( ≈ 0.022 Cal/g- ºC), Ø Specific heat slightly increases with increase of, temperature., Ø Among liquids specific heat is minimum for, mercury., Ø The value of specific heat may lie between 0 and, α., Ø In isothermal process, the value of specific heat is, infinity but in adiabatic process its value is zero., Ø Specific heat of all substances is zero at 0K., Ø Substances with highest specific heat are bad, conductors of heat and with low specific heat are, good thermal and electrical conductors., Ø The substance with large specific heat warms up, slowly and cools down slowly., S.No. Substance, Specific heat, CGS Value S.I Value, a), Water, 1cal g-1 0C-1, 4186 JKg-1 K-1, -1 0 -1, b), Ice, 0.5 cal g C, 2100 Jkg-1 K-1, c), Steam, 0.47 cal g-1 0C-1 1970 Jkg-1 K-1, , Thermal capacity or Heat capacity, Ø, , It is the amount of heat required to rise the, temperature of the body by 10C, ∆Q, ∆θ, The S.I.unit of H is JK-1, The C.G.S unit of H is Cal oC-1, Dimensional formula of H is ML2T −2θ −1, Thermal capacity depends on mass and nature of, the substance., H=, , Ø, Ø, Ø, Ø, , 123
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , Water equivalent, Ø, Ø, Ø, Ø, Ø, Ø, , Mass of water which has same thermal capacity, as that of the substance is called water equivalent, It is numerically equal to heat capacity (ms) in CGS, units, The S.I unit of water equivalent is kg, The C.G.S unit of water equivalent is g., Dimensional formula of water equivalent is ML0T 0, If mW , ms are masses of water and substance and, , W.E. 4: Two spheres of radii in the ratio 1:2, have, specific heats in the ratio 2:3. The densities, are in the ratio 3:4. Find the ratio of their, thermal capacities., Sol: Thermal capacity of a body = mS., The ratio of thermal capacities, 4 3, π r1 ρ1 S1 3 , m1S1, V1 ρ1 S1, r, ρ S , =, = 3, = 1 1 1 , ., 4, m 2 S 2 V2 ρ 2 S 2, π r23 ρ 2 S 2 r2 ρ 2 S 2 , 3, , r1 1 S1 2 ρ1 3, Here, r = 2 ; S = 3 ; ρ = 4, 2, 2, 2, The ratio of thermal capacities, , SW , S S are their specific heats respectively then,, mW × SW × ∆θ = mS × S S × ∆θ, ⇒ mW ×1 = mS S S , ∴ mW = mS S S, W.E-1:A lead piece of mass 25g gives out 1200, calories of heat when it is cooled from 900 C, to 100 C . What is its (i) specific heat (ii) thermal capacity (iii) water equivalent., Sol: Mass of lead piece (m) = 25 g = 0.025 kg, Heat energy given out ( dQ ) = 1200 × 4.2 J, 1 dQ, (i) specific heat S =, m dθ, 1, 1200 × 4.2, =, ×, = 2520JKg −1K −1, 0.025, 80, (ii) Thermal capacity = mS = 0.025 × 2520, = 63 J/K, 63, Kg = 0.015 Kg, (iii) Water equivalent, 4200, W.E-2 : The specific heat of a substance varies, , as ( 3θ 2 + θ ) ×10−3 cal /g − º C. What is the, amount of heat required to rise the, temperature of 1kg of substance from, 10ºC to 20ºC?, Sol.:For small change in temperature d θ , heat, required, dQ = mSdθ ., θ2, , ∴ Q = ∫ mSdθ, θ1, , 20, , θ2, ∴ Q = ∫ 1000 3θ + θ ×10 dθ = θ +, 2 10, 10, 20, , (, , 2, , ), , −3, , 3, , 3 202 3 102 , = 20 +, − 10 +, = 8200− 1050 = 7150cal, 2 , 2 , , W.E-3: Find the water equivalent of copper block, of mass 200g. The specific heat of copper is, 0.09 cal / g 0C ., Sol: Water equivalent w= mS = 200 × 0.09 = 18g, 124, , 3, , 1, 1 3 2 , = =, 2 4 3 16, , Ø, Ø, , Ø, , CALORIMETRY, Calorimetry means measurement of heat., A device in which heat measurement can be made, is called ‘calorimeter’., , Law of method of mixtures (or), Principle of calorimetry, , If two liquids at different temperatures are mixed, together, the heat lost by hot body is equal to the, heat gained by the cold body. This is called law, of method of mixtures., Ø When three substances of different masses m1,, m2 and m3 ,specific heats S1,S2,S3 and at different, temperatures θ1 , θ2 , and θ3 respectively are, mixed, then the resultant temperature is, m S θ + m2 S2θ2 + m3 S3θ3, θ= 1 1 1, (when state of, m1S1 + m2 S2 + m3S3, contents does not change), Ø Two liquids of masses m 1 and m 2 and specific, heats S1 and S 2 respectively are mixed. Then, the specific heat of the mixture is, m S + m2 S 2, ∴ S mix = 1 1, ( m1 + m2 ), W.E 5: 10 litres of hot water at 70ºC is mixed, with an equal volume of cold water at 20º C ., Find the resultant temperature of the water., (Specific heat of water = 4200 J/kg -K), m1S1θ1 + m2S2θ2, Sol. Resultant temperature, θ = m S + m S, 1 1, 2 2, Here, m1 = m2 = 10kg ,, (since mass of 1 litre of water is 1 kg)., θ1 = 70º C;θ2 = 20º C, and S1 = S 2 = 4200 J / kg − K, 10 × 4200 × 70 + 10 × 4200 × 20, θ=, = 45º C, 10 × 4200 + 10 × 4200, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, W.E 6. A sphere of aluminium of 0.047 kg is, placed for sufficient time in a vessel, containing boiling water, so that the sphere, is at 1000C. It is then immediately transferred, to 0.14 kg copper calorimeter containing 0.25, kg of water at 200C. The temperature of water, rises and attains a steady state at 23 0 C, Calculate the specific heat capacity of, aluminium., ( Scu = 386 J / Kg − K ; S w = 4180 J / Kg − K ), Sol: Heat lost by aluminium sphere =, (heat gained by water) + (heat gained by, calorimeter), 0.047 × SAl × (1000 − 230 ) = 0.25× 4180( 230 − 200 ), +0.14 × 386 ( 230 − 200 ), ∴ S Al = 911J / Kg − K, , W.E 7: The temperature of equal masses of three, different liquids A, B and C are 12ºC, 19ºC, and 28ºC respectively. The common, temperature when A and B are mixed is 16ºC, and when B and C are mixed is 23ºC. What, should be the common temperature when A, and C are mixed?, Sol. Given θ A = 12º C ,θ B = 19º C and θC = 28º C ., Let S A , S B and S C are the specific heats of, respective liquids., When liquid A and B are mixed, Heat gain = Heat lost, , mS A (16 − 12 ) = mS B (19 − 16 ), 4, or S B = S A ........(i ), 3, When liquid B and C are mixed, Heat gain = Heat lost, , mS B ( 23 − 19 ) = mSC ( 28 − 23) or, SB =, , 5, S C .....(ii ), 4, , 15, SC, 16, When A and C are mixed, let equilibrium temperature of mixture is θ , then, Heat gain = Heat lost, , CALORIMETRY, , CHANGE OF STATE, Ø, Ø, , Melting:, Ø, Ø, Ø, Ø, Ø, , NARAYANAGROUP, , The change of state from solid to liquid is called, melting (and the reverse process is called fusion)., Every solid melts at a definite temperature called, melting point., The melting point remains constant until the entire, amount of solid substance melts., During melting, solid and liquid states are in, equilibrium., Melting point of a solid depends on its nature,, pressure and presence of impurities., , Vaporisation (Boiling) :, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, Ø, , From (i) and (ii), we get S A =, , mSA (θ −12) = mSC ( 28 −θ ) ⇒θ = 20.26º C, , Matter exists in three states or phases such as, solid, liquid and gas., A transition from one of these states to, another is called change of state., Melting of ice or vaporisation of water is an, example for change of state. During change of, state, temperature remains constant., , Ø, Ø, , The change of state from liquid to vapour at a, particular temperature is called vaporisation.(and, the reverse process is called condensation), Every liquid vaporises at a definite temperature, called boiling point., The boiling point remains constant until the entire, amount of liquid vaporises., During vaporisation liquid and gaseous states are, in equilibrium., Boiling point of a liquid depends on nature of the, liquid, applied pressure and presence of impurities., , Evaporation:, The escape of molecules from the free surface of, a liquid is called evaporation, Refrigerators, air coolers etc., work on the, principle of cooling produced by evaporation., Body temperature is also controlled by, evaporation of sweat., Evaporation is a slow process where as boiling, is a quick process., Evaporation takes place at all temperatures, whereas boiling takes place at a particular, temperature., Evaporation takes place only at the surface of a, liquid where as boiling occurs through out the, liquid., The rate of evaporation depends on the nature of, the liquid., The rate of evaporation is more when the difference, in temperatures of the liquid and the surrounding, air is high., 125
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , Effect of pressure on melting point and, boiling point, Ø, , Ø, , Ø, , Effect of pressure on M.P and B.P can be, explained with Clausius-Clapeyron relation which, can be derived on the basis of thermodynamics, , Ø, , dP, L, i.e. dT = T (V − V ), f, i, , Ø, , L = Latent heat of fusion, Vi = Initial volume, Vf = Final volume, , Ø, , dP, is, dT, positive. i.e boiling point of every liquid rises with, increase in pressure, Ø In case of melting for substances like wax and, sulphur which expands on melting, V f > Vi and hence melting point rises with increase, in pressure., Ø In case of substances like ice and rubber which, contracts on melting V f < Vi and hence V f − Vi, negative i.e. melting point is lowered with increase, in pressure., Ø On mountains, it is difficult to cook food because, with increase in height, pressure decreases and, hence the boiling point of water decreases., W.E-8: The melting point of ice is 0ºC at 1 atm., At what pressure will it be -1ºC?, , 1 , , −3 3, (Given, V2 − V1 = 1 −, × 10 m ), 0.9 , , Sol. Here ∆T = ( −1 − 0 ) = −1, T = 273 + 0 = 273K, 1 , , −3 3, and V2 − V1 = 1 −, × 10 m (given), 0.9, , , L = 80cal / g, , Ø, Ø, , 80 × 4.2 ×103, 1 , , −3, 273 1 −, ×10, 0.9, , , , ∴∆P = 110.8 ×105 N / m 2 = 110.8 atm, , P2 − P1 = 110.8 atm ⇒ P2 = 110.8 + P1 = 111.8 atm, 126, , The temperature and pressure at which solid, liquid, and vapour states co-exist is called triple point., The triple point of water is, 273.16K (0.010C) and pressure 0.006 atm., (0.459 cm of Hg), Negative slope of ice line showing that the melting, point decreases with increasing of the pressure, Sublimation is the change from a solid to the vapour, state without the intermediate liquid state and the, reverse process of direct condensation of vapour, to solid is called hoar frost., Phase diagram of water, , pressure, , c ice line, , steamline A, , 0, , 0.006 atm, , hoar frost line, B, , Ø, , Under normal conditions of pressure, solids like, camphor, iodine, arsenic, etc., do not melt when, heated, but they under go “sublimation”., , Latent Heat :, Ø, , ∆P, L, we have, ∆T = T (V − V ), 2, 1, ∆P, =, ( −1), , The melting of ice when pressure is applied and, resolidification on removal of pressure is called, regelation. Snow-ball preparation is due to, regelation., Skating is possible on snow due to the formation, of water below the skates., , Triple point :, , In case of boiling as volume of vapour V f is always, greater than volume of liquid Vi hence, , Regelation of ice:, , It is the amount of heat required to change unit, mass of a substance from one state to another, state without any change of temperature., L=, , Ø, Ø, , Q, ( J / Kg or Cal / g ), m, , ∴ Q = mL ; L = Latent Heat, Latent heat of fusion of ice is 80Cal/g or, 0.335 × 106 J/kg., Latent heat of vapoursiation of water is 540 Cal/, g or 2.26 × 106 J/kg, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, Ø, , CALORIMETRY, , The amount of heat required to convert m gram, of ice at –100C to m gram of steam at 1000C is, 0, , m gm, , -10 C, , Q1 = msice ∆q, 0, , m gm, , 0C, , Q2 = mLice, 0, , of ice = 3.4 × 105 J Kg −1 ), Sol: The heat released as the water cools down from, 250 C to 00 C is, , m gm, , 0C, , W.E-9 : A piece of ice of mass 100 g and at, temperature 00 C is put in 200 g of water at, 250 C . How much ice will melt as the, temperature of the water reaches 00 C ?, (specific heat capacity of water, = 4200J kg −1 K −1 and latent heat of fusion, , Q3 = msw ∆q, , Q = mS ∆θ = ( 0.2 )( 4200 )( 25 ) = 21000 J, , m gm, , 0, , 100 C, , Q4 = mLsteam, , The amount of ice melted by this heat is, , m gm, , 0, , 100 C, , m=, , Q, 21000, =, = 62 g, L 3.4 × 10 5, , W.E-10 : The following graph represents change, of state of 1 gram of ice at −200 C . Find the, net heat required to convert ice into steam at, 1000 C, Sice = 0.53cal / g − 0 C, , Y, , Boiling, Point, , Melting, Point, , 0, , T( C), X, , Ø, , Total heat required Q = Q1 + Q2 + Q3 + Q4, During melting or boiling the heat absorbed by, the substance is used in increasing the molecular, distances, Temperature, F, , 0CB, ice, , D, , 0, , C, , A, , Heat, , Length of BC ∝ Latent Heat of fusion of ice, Length of DE ∝ Latent Heat of Vapourisation, Length of DE > Length of BC ( i.e. always Lv > L f ), 1, Slope of AB ∝ mS, solid, , 1, slope of CD ∝ mS, Liquid, Note:, (i) Latent heat of a substance becomes zero at, critical temperature., (ii) Latent heat depends on the nature of a substance, and pressure., NARAYANAGROUP, , 0, a, -20 Q1, , b, , e, , c, Q2, , Q3, , Q4, , Q(cal), , Sol: In the figure :, a to b: Temperature of ice increases until it reaches, its melting point 00 C ., , 0, , 100 C, , d, , 100, , Q1 = mSice 0 − ( −20 ) = (1)( 0.53)( 20 ) = 10.6cal, b to c: Temperature remains constant until all the, ice has melted, , Q2 = mL f = (1)( 80 ) = 80cal, c to d : Temperature of water again rises until it, reaches its boiling point 1000 C, Q3 = mS water [100 − 0] = (1)(1.0 )(100 ) = 100cal, d to e : Temperature is again constant until all the, water is transformed into the vapour phase, Q4 == mLv = (1)( 539 ) = 539cal, Thus, the net heat required to convert 1g of ice at, −200 C into steam at 1000 C is, Q = Q1 + Q2 + Q3 + Q4 = 729.6cal, 127
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, W.E-11 : A calorimeter of water equivalent 83.72, Kg contains 0.48 Kg of water at 35ºC. How, much mass of ice at 0ºC should be added to, decrease the temperature of the calorimeter, to 20ºC., (SW= 4186J / Kg-K and Lice = 335000 J / Kg ), Sol. Heat capacity of the calorimeter = 83.72J K-1, From law of method of mixtures,, Heat lost by calorimeter , , +, = Heat gained by the ice, , Heat lost by water, , , 83.72×15+ 0.48×4186×15 = m×( 335000 +83720), ∴ m = 0.07498 Kg, W.E-12: A steam at 100ºC is passed into 1 kg of, water contained in a calorimeter of water, equivalent 0.2 kg at 9ºC till the temperature, of the calorimeter and water in it is increased, to 90ºC. Find the mass of steam condensed, in kg ( SW=1 cal/g ºC, & Lsteam = 540 cal/, g)(EAM-14E), Sol.Let, m be the mass of the steam condensed., mass of the steam passed into calorimeter,, m2 = 1kg = 1000 g., Water equivalent of calorimeter,, m1Sl = 0.2 kg = 200g, θ1 = temperature of the steam = 100ºC, , θ2 = temperature of the water = 9ºC, θ3 = resultant temperature = 90ºC, From law of method of mixtures,, Heat lost = heat gained (calorimeter + water), m Lsteam + SW (θ1 − θ3 ) = [ m1S1 + m2 SW ] (θ3 − θ2 ), m 540 + 1(100 − 90 ) = [ 200 + 1000 × 1] ( 90 − 9 ), , ⇒ m = 176 g = 0.176kg ≈ 0.18kg, W.E.13: 1g steam at 100ºC is passed in an insulating vessel having 1g ice at 0ºC. Find the equilibrium composition of the mixture. (Neglecting heat, capacity of the vessel)., Sol. Available heat from steam, mL = 1× 540 = 540 cal, Heat required for melting of ice and to rise its temperature to 100º C = mice Lice + mwater S water ∆θ, = (1× 80 ) + 1×1× (100 − 0) = 180 cal, 128, , Let m be the mass of steam condensed, then, 180 1, m× 540 = 180 ⇒ m = 540 = 3 g, Final contents :, , Water = 1 +, steam = 1 −, , 1, 4, =, g, 3, 3, , ,, , 1, 2, = g, 3, 3, , W.E-14: 20g of steam at 100ºC is passed into 100g, of ice at 0ºC. Find the resultant temperature, if latent heat of steam is 540 cal/g, latent heat, of ice is 80 cal/ g and specific heat of water is, 1 cal/gºC., Sol. For steam, Heat lost by the steam in condensation, Q1 = ms Ls = 20 × 540 = 10800cal .........(1), For ice, Heat gained by the ice in melting and to rise its, temperature from 0o C to 1000 C is, Q2 = mice Lice + mice S w ∆t, , = 100 × 80 + 100 ×1 ×100 = 18000cal ......(2), From eq. (1) and (2) ;, Q2 > Q1, Let θ = resultant temperature of the mixture, According to law of method of mixtures, Heat lost by steam = Heat gained by ice, ms Ls + ms Swater (100 −θ ) = mice Lice + mice Swater (θ − 0), , ( 20×540) +20×1(100−θ) = (100×80) +(100×1×θ), ⇒ θ = 40º C, , Note: The temperature of mixture can never be lesser, than lower temperature and can never be greater, than higher temperature, θ L ≤ θ mix ≤ θ H, If ‘m’ g of steam at 1000 C is mixed with ‘m’ g of, ice at 00 C then, a) Resultant temperature of mixture is 1000 C, b) Mass of steam condensed =, , m, g, 3, , c) Mass of steam left uncondensed =, d) The final mixture contains, , 2m, g, 3, , 4m, g of water and, 3, , 2m, g of steam both at 1000 C, 3, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , W.E-15 : 6 gm of steam at 1000 C is mixed with 6, gm of ice at 00 C . Find the mass of steam left, uncondensed, ( L f = 80cal / g , Lv = 540cal / g ,, SWater = 1cal / g − 0 C ), , Saturated and Unsaturated Vapours :, , Sol: For steam, Heat lost by the steam in condensation, Q1 = ms Ls = 6 × 540 = 3240cal .........(1), For ice, Heat gained by the ice in melting and to rise, , its temperature from 0o C to 1000 C is, Q2 = mice Lice + mice S w ∆t, , = 6 × 80 + 6 ×1 ×100 = 1080cal ......(2), From eq (1) and (2) Q1 > Q2, i.e , the total steam did not condensed into water., Let ‘m’ gm of steam is condensed into water by, giving 1080cal. of heat ., 1080, = 2 gm, mLs = 1080 ; m =, 540, ∴ mass of the steam left uncondensed = 6 − 2 = 4g, W.E-16:A piece of ice(heat capacity =2100J/Kg 0C, and latent heat = 3.36 × 105 J / Kg ) of mass m, grams is at −5º C at atmospheric pressure. It, is given 420 J of heat so that the ice starts, melting. Finally when the ice-water mixture, is in equilibrium, it is found that 1gm of ice, has melted. Assuming there is no other heat, exchange in the process. Find the value of, m., (JEE-2010), Sol. Here, heat given is used to increase the, temperature of the ice to 0º C and to melt 1gm of, ice., Given m is mass of ice in gm., , (, , W.E.17.When a small ice crystal is placed into, super cooled water, it begins to freeze, instantaneously. What amount of ice is, formed from 1kg of water super cooled to, −8º C ., Sol. mL = m1S∆θ ; m× 80 = 1000 ×1× 8 ; m = 100g, , ), , (a)When the pressure exerted by a vapour is, maximum it is called saturated vapour, when, pressure exerted is not maximum, it is called, unsaturated vapour., (b)Saturated vapours do not obey the gas laws and, saturated vapour pressure of liquid is independent, of volume occupied. But unsaturated vapour obey, the gas laws., (c)At boiling point of a liquid saturated vapour, pressure is equal to atmospheric pressure at that, place., NOTE:, , Heat, Specific Heat, Molar specific, Heat, Thermal, capacity, Water, Equivalent, , Ø, , 1., , 2., , Ø, , 3., , Super cooling :, Most liquids, if cooled in a pure state in a perfectly, clean vessel, with least disturbance, can be, lowered to a temperature much below the normal, freezing point, without solidifying. This is known, as super cooling or super fusion., In super cooling, water can be cooled upto, −10ºC without becoming solid., , NARAYANAGROUP, , SI, Joule, Joule/Kg -K, Joule/m ol K, , CGS, ( Practical ), Calories, Cal/ g-0 C, Cal/mol -0 C, , Joule/Kg, , Cal/ 0C, , Kg, , g, , C.U.Q, , ∴ 420 = m × 2100 × 5 +1 × 3.36 × 105 × 10−3, , ⇒ m = 8 gm ., , Units, , Physical, Quantity, , 4., , Heat capacity of a substance is infinite. It, means, 1) heat is given out, 2) heat is taken in, 3) no change in temperature whether heat is taken, in (or) given out, 4) all of the above, The heat capacity of a material depends upon, 1) the structure of a matter, 2) temperature of matter, 3) density of matter, 4) specific heat of, matter, Heat required to raise the temperature of one, gram of water through 1 0C is, 1) 0.001 Kcal 2) 0.01 Kcal, 3) 0.1 Kcal, 4) 1.0 Kcal, In defining the specific heat, temperature is, represented in 0F instead of 0C. Then the, value of specific heat will, 1) decrease, 2) increase, 3) remain constant, 4) be converted to heat capacity, 129
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CALORIMETRY, 5., , 6., , 7., , 8., , 9., , 10., , 11., , 130, , Which of the following states of matter have, two specific heats ?, 1) Solid 2) Gas 3) Liquid 4) Vapour, The specific heat of a gas in an isothermal, process is, 1) infinity, 2) zero, 3) negative, 4) remains constant, Why the specific heat at a constant pressure, is more than that at constant volume, 1) There is greater inter molecular attraction at, constant pressure, 2) At constant pressure molecular oscillation are, more violent, 3) External work need to be done for allowing, expansion of gas at constant pressure, 4) Due to more reasons other than those, mentioned in the above, The ratio [Cp / Cv] of the specific heats at a, constant pressure and at a constant volume of, any perfect gas, 1) can’t be greater than 5/4, 2) can’t be greater than 3/2, 3) can’t be greater than 5/3, 4) can have any value, During melting process, the heat given to a, solid is used in (generally), 1) increasing the temperature, 2) increasing the density of material, 3) increasing the average distance between the, molecules, 4) increasing the average K.E. of the molecules, When two blocks of ice are pressed against, each other then they stick together (coalesce), because, 1) cooling is produced, 2) heat is produced, 3) increase in pressure, increase in melting point, 4) increase in pressure, decrease in melting point, Ice is found to be slippery when a man walks, on it This is so because, 1) increase in pressure causes ice to melt faster, 2) increase in pressure causes ice to melt slower, 3) its surface is smooth and cold, 4) ice is colder, , JEE- ADV PHYSICS-VOL- V, 12. Cooking is difficult on mountains because, 1) water boils at low temperature, 2) water boils at high temperature, 3) water does not boil, 4) it is cool there, 13. Paraffin wax expands on melting. The melting, point of wax with increasing pressure, 1) increases, 2) decreases, 3) remains same 4)we can’t say, 14. In a pressure cooker cooking is done quickly, because, 1) the cooker does not absorb any heat, 2) it has a safety valve, 3) boiling point of water rises due to increased, pressure, 4) it is a prestige to cook in a cooker, 15. A large block of ice is placed on a table when, the surroundings are at 00C, 1) ice melts at the sides 2) ice melts at the top, 3) ice melts at the bottom, 4) ice does not melt at all, 16. Which of the following at 1000C produces, most severe burns ?, 1) Hot air 2) Water, 3) Steam 4) Oil, 17. The latent heat of vaporisation of water is, more than latent heat of fusion of ice, why, 1) On vaporisation much larger increase in volume, takes place, 2) Increase in kinetic energy is much larger on, boiling, 3) Kinetic energy decreases on boiling, 4) Volume decreases when the ice melts, 18. The latent heat of vaporisation of a substance, is always, 1) greater than its latent of fusion, 2) greater than its latent heat of sublimation, 3) equal to its latent heat of sublimation, 4) less than its latent heat of fusion, 19. A piece of ice at 0 0C is dropped into water at, 00C. Then ice will, 1) melt, 2) be converted to water, 3) not melt, 4) partially melt, , C.U.Q-KEY, 01) 3 02) 4 03) 1 04) 1 05) 2 06) 1, 07) 3 08) 3 09) 3 10) 4 11) 1 12) 1, 13) 1 14) 3 15) 3 16) 3 17) 1 18) 1, 19) 3, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , LEVEL - I (C.W), HEAT CAPACITY (OR), THERMAL CAPACITY, 1., , 2., , The ratio of densities of two substances is, 2:3 and that of specific heats is 1 : 2. The, ratio of thermal capacities per unit volume is, 1) 1 : 2 2) 2 : 1, 3) 1 : 3 4) 3 : 1, Two spheres of copper of diameters 10cm and, 20 cm will have thermal capacities in the ratio, 1, 1, 1, 1, 2), 3), 4), 8, 2, 4, 6, Two liquids A and B of equal volumes have, their specific heats in the ratio 2 : 3. If they, have same thermal capacity, then the ratio, of their densities is, 1) 1 : 1 2) 2 : 3 3) 3 : 2, 4) 5 : 6, Specific heat of aluminium is 0.25 cal/g-0c., The water equivalent of an aluminium vessel, of mass one kilogram is, 1) 40 cal/ 0C, 2) 250 g, 0, 3) 250 cal/ C, 4) 40 g, , 1), 3., , 4., , LAW OF MIXTURES (OR), CALORIMETRY AND CHANGE OF, STATE PRINCIPLE, 5., , 6., , 7., , The quantity of heat which can rise the, temperature of x gm of a substance through, t1°C can rise the temperature of y gm of, water through t2°C is same. The ratio of, specific heats of the substances is, 1) yt1 / xt2, , 2) xt2 / yt1, , 3) yt2 / xt1, , 4) xt1 / yt2, , Two liquids A and B are at 300C and 200C, respectively. When they are mixed in equal, masses the temperature of the mixture is, found to be 260C. The ratio of specific heats, is, 1) 4 : 3 2) 3 : 4 3) 2 : 3, 4) 3 : 2, 0, M g of ice at 0 C is mixed with M g of water, at 100c. The final temperature is, 1) 80C 2) 60C 3) 40C, 4) 00C, , NARAYANAGROUP, , CALORIMETRY, 8., , A beaker contains 200g of water. The heat, capacity of the beaker is equal to that of 20g, water. The initial temperature of water in the, beaker is 200C. If 440g of hot water at 920C, is poured in it, the final temperature, (neglecting radiation loss) will be nearly, 1) 580C 2) 680C 3) 730C 4) 780C, 9. If 10g of the ice at 00C is mixed with 10g of, water at 1000C, then the final temperature of, the mixture will be, 1) 50C 2) 100C 3) 100 K 4) 00C, 10. 10 grams of steam at 1000 C is mixed with 50, gm of ice at 00 C then final temperature is, , 1) 200 C 2) 500 C 3) 400 C 4) 1000 C, 11. The heat energy required to vapourise 5kg, of water at 373 K is, 1) 2700 K.cal, 2) 1000 K.cal, 3) 27 K.cal, 4) 270 K.cal, 12. Two liquids A and B are at temperatures of, 750C and 1500C respectively. Their masses, are in the ratio of 2 : 3 and specific heats are, in the ratio 3 : 4. The resultant temperature, of the mixture, when the above liquids, are, mixed (Neglect the water equivalent of, container ) is, 1) 1250C 2) 1000C 3) 500C 4) 1500C, 13. 1g of ice at 00 C is mixed 1g of steam at, 1000 C . The mass of water formed is, 1) 1.33g 2) 1 g, 3) 0.133 g 4) 13.3g, 14. A piece of metal of mass 112g is heated to, 1000 C and dropped into a copper calorimeter, of mass 40g containing 200g of water at, 160 C . Neglecting heat loss, the specific heat, of the metal is nearly, if the equilibrium, temperature reached is 240 C, , ( Scu = 0.1cal / g −º C ), 1) 0.292 cal / gm −0 C, 2) 0.392 cal / gm −0 C, 3) 0.192 cal / gm −0 C, 4) 0.492 cal / gm −0 C, 131
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JEE- ADV PHYSICS-VOL- V, 4., , CALORIMETRY, , A copper block of mass 500 gm and specific, heat 0.1 cal/gm 0 C heated from 300 C to, , 11. Quantity of heat lost in condensation of 10, gm of steam at 1000 C is, , 2900 C , the thermal capacity of the block is, , 1) 2.26 × 105 J, 3) 22.6 J, , 1) 50cal / 0 C 2) 50gm 3) 5cal / 0 C 4) 5gm, , LAW OF MIXTURES (OR), CALORIMETRY & CHANGE OF, STATE PRINCIPLE, 5., , 75 gm of copper is heated to increase its, temperature by 100 C . If the same quantity, of heat is given to ‘m’ gm of water, to have, same rise in temperature is ( specific heat of, copper = 420 J / Kg −0 C ), 1) 7.5gm, , 6., , 7., , 9., , 3) 10gm, , 4) 2.5gm, , Two liquids are at 400 C and 300 C . When they, are mixed in equal masses, the temperature, of the mixture is 360 C . Ratio of their specific, heats is, 1) 3:2, 2) 2:3, 3) 4:3, 4) 3:4, If 10g of the ice at 00C is mixed with 10g of, water at 100C, then the final temperature of, the mixture will be, 1) 50C, , 8., , 2) 5gm, , 2) 00C, , 3) 1000 C 4) 400C, , 5 gm of steam at 1000 C is passed into, calorimeter containing liquid. Temperature of, liquid rises from 320 C to 400 C . Then water, equivalent of calorimeter and contents is, 1) 40 g 2) 375 g 3) 300 g 4) 160 g, , 12. Two liquids at temperatures 600 C and 200 C, respectively have masses in the ratio 3:4, their specific heats in the ratio 4:5 . If the, two liquids are mixed, the resultant, temperature is (2000 E), 1) 700 C 2) 500 C 3) 400 C 4) 350 C, 13. Steam at 1000 C is passed into 22 grams of, water at 200 C . When resultant temperature, is 900 C , then weight of the water present is, 1) 27.33 g 2) 24.8 g 3) 2.8 g 4) 30 g, 14. A calorimeter of water equivalent 100 grams, contains 200 grams of water at 100 C . A solid, of mass 500 grams at 450 C is added to the, calorimeter. If equilibrium temperature is, 250 C then, the specific heat of the solid is, (in cal / g − 0 C ), 1) 0.45 2) 0.1, 3) 4.5, 4) 0.01, 15. Two liquids of masses m and 5 m at, temperatures 3θ , 4θ are mixed. If their, specific heats are 2S,3S respectively, the, resultant temperature of mixture is, 1), , 1) 400 C 2) 300 C 3) 200 C, NARAYANAGROUP, , 4) 100 C, , 55, θ, 17, , 3), , 44, θ, 17, , 4), , 33, θ, 17, , LEVEL - I ( H.W )-HINTS, , 4) 600 C, , of ice at 00 C . Find the resultant temperature, (if latent heat of steam is 540 cal/g , latent, heat of ice is 80 cal/g and specific heat of, water is 1 cal/g-0C), , 2), , 01) 2 02) 3 03) 4 04) 1 05) 1 06) 1, 07) 2 08) 2 09) 2 10) 1 11)2 12) 4, 13) 2 14) 1 15) 1, , of water at 800 C then the final temperature, is, , 10. 50 g of steam at 1000 C is passed into 250 g, , 66, θ, 17, , LEVEL - I ( H.W )-KEY, , M gram of ice at 00 C is mixed with 3 M gram, , 1) 300 C 2) 400 C 3) 500 C, , 2) 2.26 × 104 J, 4) 44.52 × 104 J, , 1., , H1 ρ1 S1 , H = mS = ρVS ⇒ H = ρ S , 2, 2 2 , , 2., , H, ρ S r , 4, H = mS = ρ πr3S ⇒ 1 = 1 × 1 × 1 , H 2 ρ 2 S2 r2 , 3, , 3., , mASA = mBSB ⇒ ρ AVAS A = ρ BVBS B, , 3, , 133
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, 4., , Thermal capacity , H = mS, , 5., , Q1 = Q2 ⇒ m1S1∆θ1 = m2 S 2 ∆θ 2, , 6., , Heat lost by 1st liquid = Heat gained by 2nd liquid., , LEVEL - II (C.W), SPECIFIC HEAT, 1., , mS1 ( 400 − 360 ) = mS 2 ( 360 − 300 ), , ( m × Lice ) > mSw (10) ∴ final temp is 0º C, , 7., 8., , 2., , Heat lost by steam = Heat gained by the, calorimeter and contents., msteam × Lv + ms × 1(1000 − 400 ) = mS ( 400 − 320 ), , 9., , 80M + M ×1× (θ 0 − 0 ) = 3M ×1× ( 800 − θ 0 ), , 10. Heat lost by steam = Heat gained by ice., , = mw L f + mw S w (θ 0 − 0 ), , 3., , 11. Q = mLsteam, 12. Heat lost = Heat gained, m1S1θ1 + m2 S2θ2, m1S1 + m2 S2, , 13. msteam Lv + msteam S w (1000 − 900 ), , 4., , msteam = mass of steam converted into water, , 5., , ∴ mass of water = 22 g + msteam, , 14. Heat lost by solid = Heat gained by calorimeter, and water, ms Ss ( 45 − 25 ) = ( mc Sc + mw S w ) ( 25 − 10, 0, , 15. From principle of calorimetry, θ=, , 134, , m1S1θ1 + m2 S2θ2, m1S1 + m2 S2, , Three liquids A,B and C of masses 400gm,, 600 gm and 800 gm are at 300c, 400c and 500c, respectively. When A and B are mixed, resultant temperature is 360C when B and C, are mixed resultant temperature is 440C Then, ratio of their specific heats are, 1) 2:1:1 2) 3:2:1 3) 2:2:1 4) 1:4:9, 1gm of ice at 00C is converted to steam at, 1000C the amount of heat required will be, ( LSteam = 536 cal / g ), , = mw S w ( 900 − 200 ), , 0, , 1) 90 gm 2) 90 cal / 0 C 3) 9 gm 4) 9 cal / 0 C, Two beakers A and B contain liquids of, masses 300 g and 420 g respectively and, specific heats 0.8 cal / g −0 C and 0.6 cal / g −0 C ., The amount of heat on them is equal. If they, are joined by a metal rod, 1) heat flows from the beaker B to A, 2) heat flows from A to B, 3) no heat flows, 4) heat flows neither from A to B nor B to A, , LAW OF MIXTURES (OR), CALORIMETRY AND CHANGE OF, STATE PRINCIPLE, , msteam Lv + ms × S w (1000 − θ 0 ), , θ=, , A metal block absorbs 4500 cal of heat when, heated from 300 C to 800 C . Its thermal, capacity is, , 0, , 0, , ), , 1) 756 cal 2)12000 cal 3)716 cal 4)450 cal, 50g of copper is heated to increase its, temperature by 100C. If the same quantity, of heat is given to 10 g of water, the rise in its, temperature is, ( Scu = 420J/kg/0C and S w = 4200J/kg/0C ), , 6., , 1) 50C, 2) 60C, 3) 70C, 4) 80C, A liquid of mass ‘m’ and specific heat ‘S’ is at, a temperature ‘2t’. If another liquid of thermal, capacity 1.5 times, at a temperature of t/3 is, added to it, the resultant temperature will be, 1), , 4, t, 3, , 2) t, , 3), , t, 2, , 4), , 2, t, 3, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , Boiling water at 1000C and cold water at t 0C, are mixed in the ratio 1:3 and the resultant, maximum temperature was 370C. Assuming, no heat losses, the value of ‘t’ is, 1) 40C, 2) 90C, 3) 120C, 4) 160C, The fraction of ice that melts by mixing equal, masses of ice at -10°C and water at 60°C is, , 7., , 8., , 6, 11, 5, 11, 2), 3), 4), 11, 16, 16, 15, 9. Power of a man who can chew 0.3 kg ice in, one minute is ( in cal/s), 1) 400, 2) 4, 3) 24, 4) 240, 10. The final temperature, when 10 g of steam at, 1000 C is passed into an ice block of mass 100g, , 7., , , m1 1 , = , m1S1 × ∆θ1 = m2 S 2 ∆θ2 Given,, m2 3 , , 8., , 1), , (L, , steam, , = 540 cal / g , Lice = 80 cal / g ; S water = 1 cal / g 0C ), , is, 1) 00C, , 2)15.70C 3) 16.90C, , 9., , Quantity of heat on A = Quantity of heat on B, , 3., , m A × S A × θ 1 = m B × S B × θ 2 ⇒ θ1 > θ 2, When A, B are mixed, , SPECIFIC HEAT, 1., , 2., , mA S A ( ∆θ ) A = mB SB ( ∆θ ) B ...... (i), When B, C are mixed, m B S B ( ∆ θ ) B = mC S C ( ∆ θ )C ...... (ii), , )A = m, , C, , SC (∆θ, , )C, , 4., , Q = mLsteam+mSw(100-0)+mLs, , 5., , Q1 = Q2 ⇒ mSc ×∆θ1 = mSw ×∆θ2, From principle of calorimetry, , 6., , θ=, , m1S1θ1 + m2 S2θ2, m1S1 + m2 S2, , (G i v e n , m 2 S 2, NARAYANAGROUP, , = 1.5 × m 1 S 1 ), , A calorimeter takes 200 cal of heat to rise its, temperature through 100 C . Its water, equivalent in gm is, 1) 2, 2) 10 3) 20 4) 40, Three different substances have the specific, heats in the ratio 1:2:3 and the temperature, increases in the ratio 3:2:1 when the same, heat is supplied to the three substances. The, ratio of their masses is, 1) 1:1:1 2) 1:2:3, 3) 3:2:1 4) 4:3:4, , LAW OF MIXTURES (OR), CALORIMETRY & CHANGE OF, STATE PRINCIPLE, , From (i) and (ii) we get relation between SA and SC., When A and C are mixed, , m AS A (∆θ, , t, , LEVEL - II (H.W), , ∆Q, ∆θ, , 2., , mL f, , msteam × Lv + msteamSw (1000 −θ 0 ) = miceLf + miceSw (θ0 −0), , 01) 2 02) 2 03) 3 04) 3 05) 1 06) 2, 07) 4 08) 2 09) 1 10) 1, , H =, , P=, , 10. Heat lost by steam = Heat gained by ice, , LEVEL - II (C.W) - KEY, , 1., , Here a part of ice is melted because heat given by, water when it comes to 0ºC is less than the heat, required for ice to melt completely., Let m| is the mass of the ice melted., , m′Sice (10 ) + m′Lice = mwater Sw ( 60 ), , 4) 20.40C, , LEVEL - II (C.W) - HINTS, , Heat lost by hot water = Heat gained by cold, water., , 3., , Equal masses of 3 liquids A, B and C have, temperatures 10º C , 25º C and 40º C, respectively. If A and B are mixed, the mixture, has a temperature of 15º C . If B and C are, mixed ,the mixture has a temperature of, 30º C . If A & C are mixed the temperaturee, of the mixture is (EAM-2012M), 1) 16º C 2) 35º C, , 3) 20º C 4) 25º C, 135
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, 4., , 1 gram of ice at −100 C is converted to steam, , 10. The amount of steam at 1000 C that should, , at 1000 C the amount of heat required is, , be passed into 600 g of water at 100 C to make, , ( Sice = 0.5 cal / g −º C ), , the final temperature as 400 C will be, 1) 40 g, 2) 30 g 3) 20 g, 4) 45 g, , ( Lv = 536 cal / g & L f = 80 cal / g , ), , 5., , 1) 861 cal, 2)12005 cal, 3)721 cal, 4)455 cal, 30 gram of copper is heated to increase its, , LEVEL-II ( H.W ) - KEY, 01) 3 02) 4, 07) 2 08) 3, , temperature by 200 C if the same quantity of, heat is given to 20 gram of water the rise in, its temperature., ( S w = 4200 J / kg − K & S cu = 420 J / kg − K ), 6., , 7., , 1) 50 C 2) 60 C 3) 30 C, 4) 80 C, A liquid of mass ‘m’ and specific heat ‘c’ is, heated to a temperature 2T. Another liquid, of mass ‘m/2’ and specific heat ‘2c’ is heated, to a temperature T. If these two liquids are, mixed, the resulting temperature of the, mixture is, 1) (2/3)T 2) (8/5)T 3) (3/5)T 4) (3/2)T, A tap supplies water at 100 C and another tap, , LEVEL-II ( H.W ) - HINTS, , 8., , (ρ, 9., , ice, , = 0.9 ρ water , Lice = 80cal / gm ), , H=, , 2., , Q = mS ( ∆θ ) = const ⇒, , 3., , When A & B are mixed, mS A ( 5 ) = mS B (10 ), , m1 S2 ∆θ 2, = ×, m2 S1 ∆θ1, , ∴ S A = 2S B, When B & C are mixed, mS B ( 5 ) = mSC (10 ), ∴ S B = 2SC So, S A == 4SC ; When A & C are mixed, , mS A (θ − 10 ) = mSC ( 40 − θ ), 4., , Q = m× Sice ×10 + mLice+mSw(100-0)+mLs, , 5., , Q1 = Q2 ⇒ mcu Scu ×∆θ1 = mwSw ×∆θ2, , 6.From principle of calorimetry; θ =, 7., , θ =, , m1 S1θ 1 + m 2 S2 θ 2, m1 S1 + m 2 S 2 ( Given, m1+m2=20), , 8., , Q = m × Lice = ρice (V ) Lice, , kg of water at 30 C into ice at 0 C in one, minute ( L ice = 336000 J/Kg; and, , 9., , P=, , 0, , S water = 4200 J / kg / K ) will be, 1) 77 kW, 3) 38.5 kW, , m1 S1θ1 + m 2 S2 θ 2, m1 S1 + m2 S 2, , From principle of calorimetry, , 1) 360 cal., 2) 500 cal., 3) 72 cal., 4) 720 cal, The power of a system which can convert 10, 0, , 136, , ∆Q, ∆θ, , 1., , at 1000 C . How much hot water must be taken, so that we get 20kg of water at 350 C, 1) 40/9 kg, 2) 50/9 kg, 3) 20/9 kg, 4) 130/9 kg, Find the amount of heat supplied to decrease, the volume of an ice-water mixture by 1 cm3, without any change in temperature., , 03) 1 04) 3 05) 3 06) 4, 09) 1 10) 2, , 2) 55 kW, 4) 40 kW, , Q, ; Q = m× Sw × 30 + m× Lice, t, , 10. Heat lost by steam = Heat gained by water, msteam × Lv + msteamSw (1000 − 400 ) = mwater Sw ( 400 −100 ), NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, 6., , LEVEL - III, LAW OF MIXTURES (OR) CALORIMETRY, & CHANGE OF STATE PRINCIPLE, 1. 30g oficeat00C and 20 g of steam at 1000C are, mixed. The composition of the resultant, mixture is, 1) 40g of water and 10g steam at 1000C, 2) 10g of ice and 40g of water at 0 0C, 3) 50g of water at 1000C, 4) 35g of water and 15g of steam at 1000C, 2. 30 gms of water at 30ºC is in a beaker. Which, of the following, when added to water, will, have greatest cooling effect? (Specific heat, of copper = 0.1 cal/gmºC), 1) 100gm of water at10ºC, 3) 3gm of ice at 0ºC 4) 18gm of copper at 0ºC, 3., , ‘n’ number of liquids of masses m,2m,3m,4m,, ....... having specific heats S, 2S,3S, 4S, ...., are at temperatures t, 2t, 3t, 4t . . . . are mixed., The resultant temperature of mixture is, 1), , 2n ( n + 1), 2) 3 ( 2n + 1) t, , 3n, t, 2n + 1, 3n ( n + 1), , 3) 2 ( 2n + 1) t, 4., , 3n ( n + 1), , 4) ( 2n + 1) t, , Steam is passed into a calorimeter with water, having total thermal capacity 110 cal/gm and, initial temperature 30ºC. If the resultant, temperature is 90ºC, the increase in the mass, of the water is, 1) 12 gm 2) 1.2 gm 3) 5 gm, , 5., , 4) 12.4 gm, , 2 kg of ice at −20ºC is mixed with 5 kg of, water at 20ºC in an insulating vessel having, a negligible heat capacity. The final mass of, water in the vessel. ( The specific heat of, water and ice are 1k cal/kg0C and 0.5 k cal/, kg0C respectively and the latent heat of, fusion of ice is 80 k cal/kg) is, 1) 7 kg, , 2) 6 kg, , NARAYANAGROUP, , 3) 4 kg, , ( Lsteam = 21×105 J / kg and Lice = 3.36×105 J / kg )., 7., , 8., , 2) 15gm of water at 0ºC, , 4) 2 kg, , A thermally insulated vessel contains some, water at 00 C . The vessel is connected to a, vacuum pump to pump out water vapour. This, results in some water getting frozen. The, maximum percentage amount of water that, will be solidified in this manner will be, , 1) 86.2% 2) 33.6% 3) 21%, 4) 24.36%, The specific heat of a substance varies with, temperature as s=0.20+0.14 θ +0.023 θ 2 (cal/, gmºC) .Heat required to raise the temperature, of 2 gm of the substance from 50 C to 150 C is, ( θ is in ºC ), 1) 24 cal 2) 56 cal 3) 82 cal 4) 100 cal, In an industrial process 10 kg of water per, hour is to be heated from 200 C to 800 C . To, , do this steam at 1500 C is passed from a boiler, into a copper coil immersed in water. The, steam condenses in the coil and is returned, to the boiler as water at 900 C . How many, kilograms of steam is required per hour, (specific heat of steam = 1cal/gm, latent heat, of vapourisation = 540 cal/gm)?, 1) 1gm 2) 1 kg 3) 10 gm 4) 10 kg, 9. A heater melts 0ºC ice in a bucket completely, into water in 6 minutes and then evaporates, all that water into steam in 47 minutes 30 sec., If latent heat of fusion of ice is 80 cal/gram,, latent heat of steam will be (specific heat of, water is 1 cal /gam-ºC), 1) 536 Cal/gram, 2) 533.3 Cal/gram, 3) 540 Cal/gram, 4) 2.268 × 106 J/Kg, 10. Ice at 0ºC is added to 200gm of water initially, at 70ºC in a vacuum flask. When 50gm of ice, has been added and has all melted, the, temperature of flask and contents is 40ºC., When a further 80gm of ice is added and has, all melted, the temperature of whole become, 10ºC. Neglecting heat lost to surroundings, the latent heat of fusion of ice is, 1) 80 cal/gm, 2) 90 cal/gm, 3) 70 cal/gm, 4) 540 cal/gm, 137
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , Now the system contains (200+50)gm of water, at 40ºC, so when further 80gm of ice is added, , d) when A,B and C all are mixed, , 8 L f = 670 + 3w.....( ii ), , 3., from (i) & (ii) we get L f ., , LEVEL - IV, Matching Type Questions, 1., , One gram of ice at 00 C is heated to change, to steam at 1000 C having volume 1673cc at, , 4., , Column-II, , (event), , Heat required, , energy, q) 497cal / g, , expanding the gas, , q) j/kg-oC, , c) Heat current, , r) j/sec, , d) Latent heat, , s) j/kg, , In a container of negligible mass m grams of, , Column-I, , Column-II, , a) If m=20gm, mass of steam, in the mixture (in mg), , p)114.8, , b) If m=20gm, mass of water, , q)76.4, , r)5.2, , s)100, , of the mixture (in 0 C ), , Assertion & Reason Type Questions, , The temperature of mixture, , b) when A and C are mixed, , b) Heat capacity, , d) If m=10gm, final temperature, , temperatures are θ , 2θ and 3θ respectively.., , a) when A and B are mixed, , p) watt, , of the mixture (in 0 C ), , Three liquids A,B and C having same specific, heats have masses m,2m and 3m.Their, , Column-I, , a) Specific heat, , c) If m=20gm, final temperature, , s) 577cal / g, , and boiling process, 2., , Column-II, , in the mixture (in mg), , c) Change in kinetic energy r) 100cal / g, d)Change in potential energy, , Column-I, , that has temperature 200 C . If no heat is lost, to the surroundings at equilibrium, match the, following., , a) Change in total potential p) 40cal / g, , b) External work done in, , Match the following., , steam at 1000 C is added to 100gm of water, , normal pressure, Column-I, , 13 , s) θ, 5, , Read the following Questions and Choose if, , Column-II, , 1) both, Assertion and Reason are true and the, Reason is correct explanation of the Assertion., , 5, p) θ, 2, , 2) both, Assertion and Reason are true, but Reason, is not correct explanation of the Assertion., , 5, q) θ, 3, , 3) Assertion is true, but the Reason is false., 4) Both, Assertion and Reason are false., , c) when B and C are mixed, , NARAYANAGROUP, , 7, r) θ, 3, , 5., , Assertion: The specific heat capacity of a body, depends on the material of the body., , 139
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JEE- ADV PHYSICS-VOL- V, , CALORIMETRY, , 6., , Reason: The specific heat capacity of a body, depends on heat given, , 12. Statement-1:At room temperature ice does not, sublimate from ice to steam., , Assertion: Latent heat of fusion of ice is 336000, , Statement-2: The critical point of water is much, above the room temperature., , J kg −1, , 13. Statement-1 : When two liquid samples with, Reason: Latent heat refers to change of state, without any change in temperature., 7., , 8., , temperature T1 and T2 but same specific heat, capacities are mixed the equilibrium temperature, , Assertion: Change of state is an example of, isothermal process., , of the mixture is, , T1 + T2, 2, , Reason: Change of state from solid to liquid, occurs only at melting point of solid and change, of state from liquid to gas occurs only at boiling, point of liquid. Thus, there is no change of, temperature during change of state., , Statement-2:The amount of heat lost by the hotter, liquid is equal to the amount of heat absorbed by, the cooler liquid., , Assertion: Specific heat of a substance during, change of state is infinite., , Matching Type Questions, , LEVEL - IV- KEY, , Reason: During change of state ∆Q = mL ,, specific heat does not come in., , 1) a-s, b-p, c-r, d-q, , 2) a-q, b-p, c-s, d-r, , 3) a-q, b-s, c-p,r, d-s, , 4) a-r, b-p, c-s,d-p, , Assertion & Reason Type Questions, , Statement Type Questions, , 5) 3, , Options :, , Statement Type Questions, , 1. Statement 1 is true and statement 2 is true, , 9) 1, , 6) 2, , 7) 1, , 10) 1 11) 3, , 8) 2, 12) 1 13) 3, , 2. Statement 1 is true and statement 2 is false, , LEVEL - IV - HINTS, , 3. Statement 1 is false and statement 2 is true, 9., , 4. Statement 1 is false and statement 2 is false, , 1., , Statement-1: Specific heat capacity is the cause, of formation of land and sea breeze., , Specific heat depends on nature of the material but, not depend on heat given., , 2., , From definition we know that temperature remains, constant during change of state. From experiments, we get the value of Lice., , 3., , The process in which temperature remain constant, is called isothermal process., , Statement-2: The specific heat of water is more, than that of land., 10. Statement-1:When a solid melts or a liquid boils,, the temperature does not increase when heat is, supplied., , During the change of state temperature remain, constant until total amount of substance can, undergoes change of state., , Statement-2:The heat supplied is used to increase, internal kinetic energy., 11. Statement-1:Melting of solid causes no change, in internal energy., Statement-2: Latent heat is the heat required to, melt a unit mass of solid., 140, , 4., , During the change of state ∆θ = 0, ∴S =, , 1 ∆Q, Q, = ∞ ; From definition L =, m ∆θ, m, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , THERMODYNAMICS, SYNOPSIS, Ø, , Ø, , Ø, Ø, , Ø, , Ø, , Ø, , Ø, Ø, , Ø, , Ø, , Ø, Ø, , Thermodynamics: Thermodynamics is a, branch of science which deals with exchange of, heat energy between bodies and conversion of, the heat energy into mechanical energy and vice, versa., The state of a gas in thermodynamics is specified, by macroscopic variables such as pressure,, volume, temperature, etc.,, A particular portion of matter or a restricted region, of space under investigation is called system., Thermodynamic system: If the state of a, system is represented by pressure (P), volume, (V),temperature (T) and Entropy(s) then it is called, a thermodynamic system., Thermo dynamical process : If the state of, a system changes in such a way that any of P, V,, T, etc., changes, then the process is called thermo, dynamical process., Mechanical equilibrium : There is no, unbalanced force between the system and its, surroundings., Thermal equilibrium : There is a uniform, temperature in all parts of the system and is same, as that of surroundings., When bodies are in thermal equilibrium, no, exchange of heat takes place., Chemical equilibrium : There is a uniform, chemical composition throughout the system and, surroundings., Thermodynamic equilibrium:A system is, said to be in thermodynamic equilibrium when it is, in a state of thermal, mechanical and chemical, equilibrium., Zeroth law of thermodynamics: If two, isolated bodies A and B are in thermal equilibrium, independently with a third body C, then the bodies, A and B will also be in thermal equilibrium with, each other., Zeroth law of thermodynamics leads to the concept, of temperature(T)., Temperature is the measure of degree of hotness, , NARAYANAGROUP, , Ø, Ø, , Ø, , or coldness of a body., Temperature determines the direction of flow of, heat when two bodies are placed in thermal contact., Heat always flows from the body at higher, temperature(hot body) to the body at lower, temperature(cold body)., Temperature is a scalar quantity., Its S.I unit is Kelvin(K) and, C.G.S unit is degree Celsius( o C )., Its dimensional formula is M 0 L0T 0 K 1 or, M 0 L0T 0θ 1 ., , Ø, , Ø, Ø, Ø, , Ø, , Ø, , Heat, It is the thermal energy that exchanges between, two systems due to the temperature difference, between them., Its SI unit is joule and C.G.S unit is calorie., Its dimensional formula is [ML2 T-2], Calorie: It is the quantity of heat required to raise, the temperature of 1g of water by 1o C ., Standard(Mean) Calorie: It is the quantity of, heat required to raise the temperature of 1g of, water from 14.5o C to 15.5o C ., Its value is, 1calorie = 4.186J ≈ 4.2J ., Heat is measured by using calorimeter., Heat is a path dependent quantity., Internal work : It is the work done by one, part of a system on its another part., Ex: The work done by a gaseous system against, intermolecular forces., External work: When the work is done by, (or) on the system on (or ) by the surroundings,, then it is called external work., , P-V graph or indicator diagram :, Ø, , The graph drawn between pressure of a gas on yaxis and its volume on x-axis is called indicator, diagram. This graph is useful to calculate external, work done by the system during thermodynamic, change., 29
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , H∝W ⇒, , Workdone by a gas during change in, volume:, Ø, , When a gas expands, then for a small, change(Quasi-statically) in volume dV, small, workdone is dW=PdV = Area of narrow strip, (almost rectangle) as shown in figure. Here pressure, P is almost constant., , W, , J= H where J= Mechanical equivalent of heat., , Mechanical equivalent of heat (J):, Ø, Ø, , It is the amount of work necessary to produce unit, amount of heat energy., J is not a physical quantity. It is simply a conversion, factor between mechanical work and its equivalent, heat energy., , Ø, , Values of ‘J’:, , Pressure, , The value of J depends on the units of work W, and heat H., When W is in Joules, H is in Cal, then, , P, , J = 4.186J / cal ≈ 4.2J / cal, V1, , When W and H both are expressed in joules, J=1., , volume, , V2, , If the volume changes from V1 to V2, then the total, external work done by the system is, , Applications of Joule’s law, Ø, , W = ∫ PdV =Area under P-V curve., V1, , Ø, , Area under P-V graph is equal to the external work, done during the process., If the volume changes from V1 to V2 at constant, pressure(isobarically), , The height from which ice is to be dropped to melt, it completely is, JL, , V2, , Ø, , Ø, , Ø, , W = P∫ dV = P (V2 − V1 ), V2, , V1, , Pressure, , h= g ; where L= Latent heat of ice., The raise in temperature of water when it falls from, a height h to the ground is,, gh, ∆θ =, ; where ‘S’ is specific heat of water, JS, When a bullet of mass m moving with a velocity v, is stopped abruptly by a target and all of its heat, energy liberated is retained by bullet , then the, increase in temperature is., ∆θ =, , V1, , V2, , Sign convention:, If the work is done by the system, then work done, is (+) ve. (QdV > 0 ⇒W > o), If the work is done on the system, then it is, (-) ve. (QdV < 0 ⇒W < 0), , Relation between work and heat, (Joule’s law):, Ø, , 30, , x v2 , , , rise in its temperature is ∆θ =, 100 2 JS , , Volume, , Work is a path dependent quantity, , Ø, , v2, 2 JS, , If the bullet absorbs x% of heat liberated, then, , P, , Ø, , W=JH., , The amount of heat produced is directly, proportional to the amount of mechanical work, done., , Ø, , When a block of ice of mass M is dragged with, constant velocity 'v' on a rough horizontal surface, of coefficient of friction µ , through a distance, 'd', then the mass of ice melted is,, µMgd, m=, ; where m=mass of ice melted., JL, In order to melt all the ice completely, the block, JL, , should be dragged through a distance d = µ g, Now, the time taken to melt completely is given, by t =, , JL, d, =, v µ gv, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, Ø, , Ø, , When a block is dragged on a rough horizontal, surface of coefficient of friction µ , then the rise in, µ gd, temperature of block is, ∆θ =, JS, If a bullet at a temperature less than its melting point, just melts when abruptly stopped by an obstacle, and if all the heat produced is absorbed by the bullet,, then, J ( mS ∆ θ + mL ) =, , Ø, , 1, mv 2, 2, , Where L= Latent heat of fusion of the material of, the bullet , S= Specific heat of the bullet, ∆θ = rise in temperature before it melts., A metal ball falls freely on the ground from a height, ‘h1’ and bounces to height ’ h2’. If the ball absorbs, all the heat energy generated, then raise in, temperature of the ball is ∆θ =, , Ø, , g ( h1 − h2 ), JS, , When a body rotating with angular speed ω is, suddenly stopped, if it absorbs all the heat, generated, then rise in temperature of body is, ∆θ =, , Ø, , THERMODYNAMICS, , Iω 2, 2 JmS, , (QW = τθ = τωt ), , Where I=Moment of inertia of the given body, A drilling machine drills a hole to a metal plate in a, time 't'. The machine is operated by a torque ' τ ', with constant angular speed ω . If the heat, generated is completely absorbed by the plate, then, the raise in temperature of the plate is ∆θ =, , τωt, JmS, , (QW = τθ = τωt ), Where, m= mass of the plate, S= specific heat of the material of the given plate, WE. 1: A 10kw drilling machine is used for 5, minutes to bore a hole in an aluminium block, of mass 10 × 103 kg. If 40% of the work done, is utilised to raise the temperature of the block,, then find the raise in temperature of the, aluminium block ?, (Specific heat of Aluminium = 0.9 Jkg-1 k-1), Sol. Work done by the drilling machine in 5 min, W= power x time =10 × 103 × 5 × 60 = 3 × 106J, The energy utilised to rise the temperature of the, 40, 6, = 12 ´105 J, block = 40% of W = 3 × 10 ´, 100, Heat gained by aluminium block =, mass × specific heat × increase in temperature., NARAYANAGROUP, , 12´10 5 = (10´103 )´ 0.9´Dt, , \ Dt =, , 12´10 5, = 133.3°C, 0.9´10 4, , WE.2 : Hailstones fall from a certain height. If they, melt completely on reaching the ground, find, the height from which they fall. (g=10 ms-2,, L = 80 calorie/g and J = 4.2 J/calorie.), Sol. On reaching the ground, a hailstone of mass M, losses potential energy which is converted into heat, energy required to melt it. In SI, potential energy, lost = heat energy required for melting the hailstone, Mgh = ML Þ gh = L, , 80´ 4.2 ´1000, 10, = 33.6 ´1000m = 33.6 km., WE.3 : A girl weighing 42 kg eats bananas whose, energy is 980 calories. If this energy is used, to go to height h find the value of h.(J=4.2J/, calorie), Sol. Energy gained by the girl in eating bananas, = 980 calories = 980 ´ 4.2 J., W=H (in S.I.), 980 × 4.2 = mgh ⇒ 980×4.2 = 42×9.8×h, , h=, , L, ,, g, , h=, , 980´4.2, =10m, 42´9.8, WE.4 : A lead bullet of mass 21 g travelling at a, speed of 100 ms −1 comes to rest in a wooden, block. If no heat is taken away by the wood,, then find the raise in temperature of the wood., (Specific heat of lead = 0.03 calorie/g °C.), Sol. kinetic energy of the bullet = heat gained by the, Þh =, , bullet,, , 1 2, mv = mSDt, 2, , v2, (100)2, =, = 39.680C, 2S 2´0.03´ 4.2´1000, WE.5 : The height of the Niagara falls is 50m. If, J = 4.2 ×107 erg / cal . Then the difference of, temperature of water at the top and bottom, of the falls is, Sol. P.E. is converted into heat, mgh = JmS ∆t, Dt =, , ∆t =, , gh 980 × 5000, =, = 0.1170 C, 7, JS, 4.2 ×10, , 31
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, WE.6 : A piece of ice at 00 C falls from rest into, a lake of water which is also at 00 C and 0.5%, of ice melts. Find the minimum height from, which the ice falls., Sol. Let a mass 'm' of ice falls from height h., Loss in potential energy = mgh,, Heat produced, H =, , 0.5, mLice, 100, , First law of thermodynamics:, Ø, , 0.5 Lice, h=, 100 × g, 0.5 × ( 4.2 J / cal ) ( 80 × 103 cal / kg ), 100 × 9.8 m / s 2, , Ø, = 171.43m, , Internal energy:, Ø, , Internal energy of a system is the energy possessed, by the system due to molecular motion and, molecular configuration. The energy due to, molecular motion is called internal kinetic energy, U k and that due to molecular configuration is, U = UK +U p, Regarding internal energy it is worth nothing that, 1) Change in internal energy is path independent, and depends only on the initial and final states of, the system i.e, , ∫, , For bulk changes ∆Q = ∆U + PdV, Ø, , f, , 3, , ∆U1 =∆U2 =∆U3 ; ∆ U = U f − U i, , 2) Change in internal energy in a cyclic process is, always zero as for cyclic process U f = U i , so, , Ø, Ø, Ø, , 32, , Ø, Ø, Ø, , When heat is added (flows into) to the system dQ, is + ve(+dQ), When heat is taken (flows out) from the system, dQ is -ve (-dQ), When gas expands work is done by the gas, dw is, positive (+ dW), When gas compresses work is done on the gas,, then work done by the gas dW is negative (-dW), When internal energy of system increases, dU is +ve (+dU), When internal energy of system decreases, dU is -ve(-dU), , Significance and limitations of first law:, , that ∆ U = U f − U i = 0, 3)In case of ideal gas as there is no molecular, attraction U p = 0 ,i.e, internal energy of an ideal, gas is totally kinetic and is given by, , Ø, , 3, 3, nRT with ∆U = nR∆T, 2, 2, 4) In case of gases, whatever be the process, , Ø, , U = UK =, , dU is state dependent but path independent and, dQ,dW are path dependent., , Sign convention :, , 2, 1, , All the heat added to a system is partially utilised, to do the external work and remaining to increase, its internal energy., The differential form of first law of thermodynamics, is, dQ=dU+dW, where dQ = heat added,, dU = Increase in internal energy, dW= work done = PdV, , ∴ dQ = dU + PdV, , internal potential energy U p ,i.e, , i, , P f V f − PV, nR ∆ T, i i, =, γ −1, γ −1, , 5)The internal energy of ideal gas depends only on, its temperature T. When T increases U also, increases and vice versa., 6)Internal energy of real gases depends upon, temperature, pressure and volume., 7)Real gases consists of both kinetic energy and, potential energy due to intermolecular forces., , 0.5, mLice, 100, , since,W = JH; mgh = J ×, , =, , ∆ U = nC V ∆ T =, , Ø, Ø, , It is a consequence of law of conservation of, energy., This law is applicable to any process in nature., This law is applicable to all the three phases of, matter., First law of thermodynamics does not indicate the, direction of heat transfer. It does not tell any thing, about the conditions under which heat can be, transformed into work., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, WE.7: When heat energy of 1500J is supplied to a, gas the external workdone by the gas is 525J, what is the increase in its internal energy, Sol. Heat energy supplied ∆Q=1500J, External workdone ∆W=525J, By 1st law of thermodynamics ∆Q = ∆U+∆W, ∴∆U = ∆Q − ∆W =1500 – 525 = 975J., WE.8:When 1g of water at 100°C is converted into, steam at 100°C, it occupies a volume of 1671cc, at normal atmospheric pressure. Find the, increase in internal energy of the molecules of, steam., Sol. 1 atmosphere = 1.013 x 105 Nm-2 ;, volume of 1gm of water, V1 = 1cc = 10-6m3;, , THERMODYNAMICS, WE.11:Consider the vaporization of 1g of water at, 1000C to steam at 1000C at one atmospheric, pressure. Compute the work done by the water, system in the vaporization and change in, internal energy of the system., Sol. To change a system of mass m of liquid to vapour,,, heat required is Q = mLvapour, The process takes place at constant pressure, and, so the work done by the system is the work in an, isobaric process. W = P∆V, Where ∆V = (Vvapour − Vliquid ), From first law of thermodynamics, ∆U = Q − W = mLv − P (Vvapour − Vliquid ), , Latent heat of vaporization of water, , m, , , Q ρ = & ρ = 1gm / cc , V, , , Volume of steam = 1671 cc = 1671 x 10-6 m3, External work done dW = P(V2 - V1), = 1.013 x 105 (1671 x 10-6 - 1 x 10-6), = 1.013 x 105 x 1670 x 10-6, = 1.013 x 167 = 169.2 J., Latent heat of vaporisation of steam= 540 cal/g, So, heat supplied to convert 1g of water into steam,, , Lvapour = 22.57 ×105 J / kg, Q = (1.00 × 10−3 )( 22.57 × 105 ) = 2.26 × 103 J, , Q No. of moles = weight /gram molecular weight, 1, ∴ Moles of water in 1g = = 0.0556 mole, 18, Vvapour =, , nRT ( 0.0556 )( 8.315)( 373), =, = 1.70 × 10−3 m3, 5, P, 1.013 × 10, , DQ = 540 x 4.2J = 2268J, , The density of water is, , By first law of thermodynamics, , 1.00 × 103 kg / m 3 = 1.00 g / cm3, , DU = DQ -DW = 2268 -169.2 = 2098.8J, , WE. 9: Calculate the external workdone by the, system in KCal, when 40 KCal of heat is, supplied to the system and internal energy, rises by 8400 J., Sol. dQ=dU+dW, 8400, dU = 8400 J =, KCal = 2 KCal., 4200, ∴ 40 KCal = 2 KCal+external work done, The external work done = 40 - 2 = 38 KCal, WE.10:In a thermodynamic process pressure of a, fixed mass of a gas is changed in such a, manner that the gas releases 20J of heat and, 8J of work is done on the gas. If initial internal, energy of the gas was 30J, what will be the, final internal energy?, Sol. We know that, dQ = dU+dW, Since heat is released by the system, dQ = -20J., and work is done on the gas, dW = -8J, dU = -20 - (-8) = -20 + 8 = -12J, ⇒ U f − U i = −12 J, U f = U i − 12 = 30 − 12 = 18 J, NARAYANAGROUP, , Vliquid = 1.00 × 10−6 m3, Thus the work done by the water system in, vaporization is W = P (Vvapour − Vliquid ), , = (1.013 × 10 5 )(1.70 × 10 −3 − 1.00 × 10 −6 ) = 172 J, , The work done by the system is positive since the, volume of the system has increased., From first law,, ∆U = Q − W ⇒ ∆U = 2.26 ×103 − 172 = 2.09 ×103 J, , Specific heats of a gas:, Ø, Ø, , Gases have two types of specific heats., a) Specific heat at constant volume (cv), b) Specific heat at constant pressure( c p), Specific heat of all substances is zero at 0K., , Specific heat of a gas at constant pressure ( c p ), Ø, , It is the heat required to rise the temperature of 1g, of a gas by 1o C at constant pressure., , ∴cp =, , 1 dQ , , dQp = mc p dT, m dT p ;, 33
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, Ø, , ∴∆Q p = mc p ∆T , if c p is constant., , f , f , , C v = R, C p = 1 + R, 2, 2, , , ∆Q p = m ∫ c p dT , if c p depends on temperature., , Specific heat of a gas at constant volume ( cv ), Ø, , γ = 1+, , It is the heat required to rise the temperature of 1g, , Cp , , ∴ γ =, , CV , , , 2, f, , CP, CV and γ values of different gases:, , of a gas by 1o C at constant volume, , ∴ cv =, , For a gas having 'f' degrees of freedom,, , 1 dQ , , , m dT v, , S.No Atomicity of gas CP, , dQv = mcv dT, , 1., , Monoatomic, , 5, R, 2, , ∆Qv = m ∫ cv dT , if cv depends on temperature., , 2., , Diatomic, , 7, R, 2, , 3, R, 2, 5, R, 2, , SI unit of both c p , cv is J/Kg-K, , 3., , Tri non-linear&, , 4R, , 3R, , ∴ ∆Qv = mcv ∆T , if cv is constant., , Ø, , CGS unit is cal/g- o C, , Molar specific heats, Ø, , (, , C ,C, P V, , ), , When the above specific heats c p , cv are defined, per 1mole of gas, then they are said to be molar, , 4., , specific heats and represented by CP , CV ., , Ø, , 1 dQ 1 dU, Cv = , = , n dT v n dT, , Ø, , , , (Q ∆W = 0 ), , , ( but, , C = M c), , ⇒ M ( c p − cv ) = R, , Where r is specific gas constant and c p , cv are, expressed in J/Kg-K., 34, , 5, = 1.67, 3, 7, = 1.4, 5, 4, = 1.33, 3, , 7, R, 2, , 9, = 1.29, 7, , When n1moles of a gas is mixed with n2 moles of, another gas., , ? mixture =, , Ø, , γR, R, and Cv =, γ −1, γ −1, R, M, , 9, R, 2, , n 1C v1 +n 2 C v 2, n 1 +n 2, , (C p ) mixture = (C v ) mixture + R =, , ( Cp ,CV are molar specific heats ), , c p − cv = r =, , Tri linear, , then, (C v ) mixture =, , Cp - CV= R, where R is universal gas constant, R= 8.314 J/ mol-K ≈ 2cal / mol − K, , CP =, , CP, CV, , ? Of mixture of gases:, , SI unit of both molar specific heats is J/ mol-K, Cp is greater than CV and,, , CP, =γ, CV, , γ=, , and poly atomic, , of a gas, , 1 dQ , C, =, p, , , These are,, n dT p, , Ø, , CV, , n1C p1 +n 2C p 2, n1 +n 2, , C p (mixture), C v (mixture), , n1 + n2, n, n, = 1 + 2, Also γ, γ1 −1 γ 2 −1, mixture −1, At constant pressure, fraction of heat absorbed that, , dU C v 1, is converted into internal energy is dQ = C = ?, p, Ø, , At constant pressure, fraction of heat absorbed that, is converted into external workdone is, , dW nRdT, R, 1, =, =, =1−, dQ nC p dT C p, ?, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, WE.12:Calculate the difference between the two, specific heats of nitrogen, given that the, density of nitrogen at N.T.P is 1.25 g/litre, and J= 4200 J/KCal., ( in KCal/kg-K), Sol. PV = mrT, P, ., The difference in specific heats, r =, ρT, P = 1.013 × 105 N/m2; T = 273 K;, 10-3 kg 1.25´10-3 kg, r = 1.25´ 3 3 = 3, = 1.25kg / m3, -6 3, 10 cm, 10 ´10 m, , 1.013´105, = 296.8J/kg K, 1.25´273, ∴ The difference of specific heats, = 0.0768 KCal/kg-K., WE. 13:Four moles of a perfect gas is heated to, increase its temperature by 20C absorbs heat, of 40 cal at constant volume. If the same gas, is heated at constant pressure find the amount, of heat supplied., Sol. At constant volume dQ = nCvdT = dU = 40, At constant pressure, \r =, , dQ = dU + nRdT = 40 + (4´ 2´ 2) = 56 cal, , WE.14: When an ideal diatomic gas is heated at, constant pressure fraction of the heat energy, supplied which increases the internal energy, of the gas is, Sol. Heat used in increasing the internal energy is, Q1 = Cv dT ; Heat absorbed at constant pressure, to increase the temperature by dT is Q2 = C p dT, , ∴, , Q1 Cv, 1, 1, =, =, =, Q2 C p C p / Cv γ, , Q1 5, for diatomic gas, γ = 7 / 5; ∴ Q = 7, 2, WE.15:A quantity of heat Q is supplied to a, monoatomic ideal gas which expands at, constant pressure. The fraction of heat that, goes into work done by the gas is, , Sol. C p dT = Cv dT + dW ; ∴ dW = ( C p − Cv ) dT, Fraction of heat converted into work, , ( C p − C v ) dT = 1 − C v = 1 − 1, dW, =, dQ, C p dT, Cp, γ, For monoatomic gas, γ = 5 / 3, dW, 1, 3 2, ∴, = 1− = 1− =, dQ, γ, 5 5, , NARAYANAGROUP, , ., , THERMODYNAMICS, WE. 16: The specific heat capacity of a metal at, low temperature(T) is given as, 3, , T , C p ( kJK kg ) = 32 , . A 100 g vessel, 400 , of this metal is to be cooled from 20K to 4 K, by a special refrigerator operating at room, −1, , −1, , (, , 0, , ), , temperature 27 C .The amount of work, required to cool the vessel is, Sol. Heat required to change the temperature of vessel, by a small amount dT, dQ = mC p dT . Total heat required, T , , 3, , Q = m ∫ 32 , dT = 0.001996kJ, 20, 400 , 4, , work done required to maintain the temperature, of sink to T2, Q −Q , , 1, 2, W = Q1 − Q2 = Q Q2 For T2 = 20K ;, , 2, , , 300 − 20 , W1 = , 0.001996 = 0.028 kJ, 20 , For T2 = 4K, 300 − 4 , 0.001996 = 0.148 kJ, 4 , ∴ The work required to cool the vessel from, 20K to 4K is W2-W1=0.148-0.028=0.12KJ, As temperature is changing from 20 K to 4 K, work done required will be more than W1 but, , W2 = , , less than W2 ., WE 17 : The P-V diagram represents the, thermodynamic cycle of an engine, operating, with an ideal monoatomic gas. The amount, of heat, extracted from the source in a single, cycle is, [JEE MAIN 2013], , 2P0, P0, , B, , A, V0, , C, D, , 2V0, , Sol. PV, 0 0 = nRTA ; 2 PV, 0 0 = nRTB ; 2 P0 2V0 = nRTC, Heat supplied H = nC V ∆TAB + nCP ∆TBC, 35
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, PV , 2P V , 2P V, 4P V, = nCV 0 0 − 0 0 + nC P 0 0 − 0 0 , nR, nR, nR, nR , , , , , Ø, , , 0 0 , , 0 0 , = n 2 R nR + n 2 R nR = PV, 0 0, , , , , , 2, , Ø, Ø, , 3, , PV, , 5, , 13, , 2PV, , Thermodynamic processes:, Ø, , Ø, , Ø, Ø, Ø, Ø, Ø, Ø, , Quasi-static process: A quasi-static process, can be defined as an infinitesimally slow process, in which the system remains in thermal and, mechanical equilibrium with the surroundings at, each and every intermediate stage., i.e., temperature, pressure are almost constant, during infinitesimal small change in the state of gas., It is an ideal process. In practice it does not occur., Isothermal process:In this process, the, pressure and volume of gas change, but, temperature remains constant. Hence internal, energy is also constant., i.e., dT = 0; dU = 0, The system is in thermal equilibrium with the, surroundings., It takes place in a thermally conducting vessel. Hence, heat exchanges between system and surroundings., In this process dQ = dW, It is a slow process., It obeys the Boyle’s law i.e. PV=Constant, Specific heat is infinity., , Ø, , Ø, , P, , Ø, , Ø, Ø, Ø, Ø, , Ø, , Indicator diagram, , Ø, , dP, P, =−, Slope of isothermal curve, tan θ =, dV, V, dP, Isothermal bulk modulus, −, =P, dV V, , work, , Ø, , V, , Ø, , Ø, , 36, , Ø, Ø, , The workdone during the isothermal change at, temperature T for n moles of gas is, V , W = 2.303nRT log10 2 , V1 , , P, , 1, =2.303 nRT log10 P , , , , 2, , , , Isothermal process is ideal. In nature, no process, is perfectly isothermal. But we can say melting of, ice, boiling of water are approximately isothermal., In these two processes internal energy increases, even temperature is constant., Adiabatic process: The pressure, volume and, temperature of a gas change but total heat remains, constant i.e., dQ=0 (Q=constant)., , work, , dP, P, = −γ, dV, V, The slope of adiabatic curve is γ times to that of, the isothermal curve., The adiabatic bulk modulus of gas is γ p, i.e. γ times isothermal bulk modulus., The workdone by the system during the adiabatic, expansion is, nR, W=, (T -T ) = nCv(T1-T2)= −nCV∆T, γ −1 1 2, , Slope of adiabatic curve, tan θ =, , PV, 1 1 − P2V2, γ, γ −1, It takes place in a non conducting vessel. Hence, no exchange of heat takes place between system, and surroundings., Adiabatic expansion causes cooling and, compression causes heating., Eg:- Sudden bursting of tube of bicycle tyre,, Propagation of sound in gases, In this process dU = −dW, =n, , Ø, , Ø, , Indicator diagram is, , V, , P, , Ø, , There should not be any exchange of heat between, the system and surroundings. All the walls of the, container and the piston must be perfectly insulating., It is a quick process., The internal energy changes certainly as, temperature changes., In the adiabatic process P, V & T are related as, (i) PV γ = constant(ii) TV γ −1 =constant, (iii) P1−γ T γ =constant, In this process specific heat is zero., , Cp, , (T1 − T2 ), , =, , Comparison between isothermal, and adiabatic curves :, Ø, , When expanded to the same volume from the, same initial state., P, 1-isothermal, 2-adiabatic, , 1, 2, V1, , V2, , V, , Work done : W1>W2, Final pressure: P 1>P2, , Final temperature: T 1>T2, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, For monoatomic gas = 5 : 3 : 2, For diatomic gas = 7: 5 : 2, For non-linear poly atomic gas = 4 : 3 : 1, , (B) If the change is adiabatic. For the gas in the left, chamber,, , Ø, Ø, , γ, , If a system after undergoing through a series of, changes comes back to its initial state, the process, is called cyclic., In a cyclic process (the system finally reaches the, same initial state), workdone is equal to the area, enclosed by the cycle., It is +ve if the cycle is clockwise., It is -ve if the cycle is anticlockwise., P, P, , W=+ve, , ( ), , γ, , P ( 5V ) = P′ V ′ ............(iii), , Cyclic process :, , W=-ve, , V, V, In the cyclic process as U f = U i ,, ∆ U = U f − U i = 0 and so the first law implies, ∆Q = 0 + ∆W , i.e ∆Q = ∆W , heat supplied is, equal to the work done (area of the cycle), WE.18:A piston divides a closed gas cylinder into, two parts. Initially the piston is kept pressed, such that one part has a pressure P and volume, 5V and the other part has pressure 8P and, volume V, the piston is now left free. Find the, new pressure and volume for the isothermal, , and for the gas in the right chamber, , (, , ), , γ, , 8 P (V ) = P′ 6V − V ′ ..........(iv), γ, , dividing (iv) by (iii), 3/2, , 6V − V ′ , 6V, 4, 8, 10, = 1+, , = 3 / 2 or, V′ = V, i.e, V′ , 5, 5, 3, V′, , , Substituting it in Equation (iii), 5V × 3 , P′ = P , , 10V , , 3/2, , =, , 3 3, P = 1.84 P, 2 2, , 10, 8, So P′ = 1.84 P ; V ′ = V and 6V − V ′ = V, 3, 3, WE.19:In a cyclic process shown in the figure an, ideal gas is adiabatically taken from B to A,, the work done on the gas during the process, B → A is 30 J, when the gas is taken from, A → B the heat absorbed by the gas is 20J., What is the change in internal energy of the, gas in the process A → B ., , and adiabatic process. ( γ = 1.5), PA, , Sol., P, 5V, , 8P, V, , L, , P′, V′, , 30J, , P′, , L, , (, , 6V − V ′, , P ′ , with volume V ′ , in the left side and, in the right side, Case - (A) if the change is isothermal., For the gas enclosed in the left chamber,, , ), , P × 5V = P ′V ′ .........( i ), , While for the gas in the right chamber, , ), , 8 P × V = P′ 6V − V ′ .........(ii ), , Solving these for V ′ and P ′ , We get, V′ =, 38, , B, v, , Sol: WBA = −30J,QBA = 0 ; ∆U BA = −WBA = 30 J, Now, DUAB =-DUBA =-30J, , (6V-V′ ), , Final pressure will be same on both sides. Let it be, , (, , 20 J, , 30, 13, 48, V and P′ = P and 6V − V ′ = V, 13, 6, 13, , WE. 20:A gas undergoes a change of state during, which 100 J of heat is supplied to it and it, does 20 J of work. The system is brought back, to its original state through a process during, which 20 J of heat is released by the gas. What, is the work done by the gas in the second, process ?, Sol. dQ1 = dU1 + dW1, , 100 = dU1 + 20 ⇒ dU1 = 80 J, dQ2 = dU 2 + dW2, , (∴ dU1 = −dU 2 ), , −20 = −80 + dW2 ⇒ dW2 = 60 J, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , WE. 21:An ideal gas is taken through the cycle, A → B → C → A , as shown in the figure. If the, net heat supplied to the gas in the cycle is 5 J,, what is the work done by the gas in the process, , PV, , γ, , = constant for an adiabatic process,, , we get γ = 3 / 2, WE. 24:When 5 moles of an ideal gas is compressed, isothermally, its volume decreases from 5 litre, to 1 litre. If the gas is at 27°C, find the work, , C→A, , 2, , Comparing it with the equation, , C, , , , V(m ), 1, , A, 10, , , , , done on the gas log10 5 = -0.6990 ., , , , Sol. In the case of 'n' moles, work done on the gas, , B, , 3, , 2, , P(N/m ), , Sol : ∆ WAB = P∆V = (10 )( 2 − 1) = 10J, ∆ WBC = 0 (as V = constant), , From first law of thermodynamics, ∆Q = ∆W + ∆U ; ∆U = 0 (process ABCA is cyclic), ∴ ∆Q = ∆WAB + ∆ WBC + ∆WCA, ∴ ∆WCA = ∆Q − ∆WAB − ∆WBC = 5 − 10 − 0 = –5J, , WE.22:An ideal monoatomic gas is taken round, the cycle ABCDA as shown in the P-V, diagram. Compute the work done in this, process., , 1, , V , W = nRT log e 2 , V1 , V , W = nRT × 2.3026 × log10 2 , V1 , , 1, ∴ W = 5 × 8.314 × 300 × 2.3026 × log10 , 5, = 5 × 8.314 × 300 × 2.3026 × (-0.6990), = -2.007 × 104J, WE. 25:A gas is expanded to double its volume by, two different processes. One is isobaric and, the other is isothermal. Let W1 and W2 be the, respective work done, then find W1 and W2, Vf, , 5P0, , A, , B, , , − 1, Vi, , , Sol: W1 = Pi (V f − Vi ) = PV, i i, , = nRT ( 2 − 1) = nRT, , Pressure (P), C, D, 2V0 3V0, 5V0 6V0, Volume(V), Sol. Total work done, = Area under P-V curve (parallelogram), = Base x Height = (6V0 – 3V0) (5P0 – 3P0), = (3V0)(2P0) = 6P0V0 units, WE. 23:During an adiabatic process, if the pressure, of an ideal gas is proportional to the cube of, its temperature, find γ ., Sol. For an ideal gas of one mole PV = RT, 3P0, , During an adiabatic process P ∝ T3, P = kT3 ; where k is a constant, 3, , PV , k 3 3, Þ P=, P V, R , R3 , , P = k, , P 2V3 = cons tan t ; PV3 / 2 = constant, NARAYANAGROUP, , æV f, W2 = nRT loge ççç, çè Vi, , ÷÷ö = nRT log 2 = W log 2, 1, e( ), e( ), ÷÷, ø, , WE. 26:Temperature of 1 mole of an ideal gas is, increased from 300 K to 310 K under isochoric, process. Heat supplied to the gas in this process, is Q = 25 R, where R = universal gas constant., What amount of work has to be done by the, gas if temperature of the gas decreases from, 310 K to 300 K adiabatically?, Sol. ∆Q = nCV ∆T, ∴25R = (1) (CV) (310–300) or CV =, , 5, R, 2, , As the gas is diatomic γ = 1.4, Now work done in adiabatic process, W=, , nR (T1 - T2 ), g -1, , =, , (1)( R)(310 - 300), 1.4 -1, , = 25 R, , 39
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, WE. 27:A tyre pumped to a pressure of 6, atmosphere suddenly bursts. Room, temperature is 25 o C. Calculate the, temperature of escaping air. ( γ = 1.4.), Sol. From P11−γ T1γ = P21−γ T2γ Here, P 1 = 6 atm ;, P2 = 1 atm ; T1 = 273+25 = 298K; γ = 1.4, 1.4, (6)(1-1.4)(298)1.4 = (1)(1–1.4) T2, T21.4, , 1.4 −0.4, , = (298) 6, , =, , 1.4, , 1, , WE. 28:Three samples of the same gas A, B and, C ( γ = 3/2) have initially equal volume. Now, the volume of each sample is doubled. The, process is adiabatic for A, isobaric for B and, isothermal for C. If the final pressures are, equal for all three samples, find the ratio of, their initial pressures, Sol: Let the initial pressure of the three samples be P A,, 3, , 3, , PB and P C then PA (V ) 2 = ( 2V ) 2 P, PB = P ; PC (V) = P (2V), 3/2, ∴ PA : PB : PC = (2) : 1 : 2 = 2 2 :1: 2, WE.29: An ideal gas mixture filled inside a, balloon expands according to the relation, PV2/3=constant. What will be the temperature, inside the balloon, Sol. PV2/3 = constant ⇒ , , nRT 2/3, (V ) = constant, V , , Ø, , In this process the gas obeys an additional law in, the form of PV x = constant,, (where x is +ve or − ve constant) along with, ideal gas equation PV = nRT, , Ø, , In this process external Work done is W=, , WE. 30: Work done to increase the temperature, of one mole of an ideal gas by 300C, if it is, expanding under the condition V α T, , ( R = 8.314 J / mol / K ), , (EAM-2012), , Sol: We have, V α T 2 / 3 ; but PV = RT, T, , PV α T ; PV αV 2 ⇒ Pα V, , − nR ∆ T, x −1, , R, , R, , In this process molar specific heat, C= γ − 1 − x − 1, WE. 31:P - V diagram of a diatomic gas is a straight, line passing through origin. What is the molar, heat capacity of the gas in the process, Sol. P - V diagram of the gas is a straight line passing, through origin. Hence,, –1, P ∝ V or PV = constant ⇒ x = −1, Molar heat capacity in the process, C=, , R, R, +, Here, γ = 1.4 (for diatomic gas), g -1 1 - x, R, , R, , ∴ C = 1.4 − 1 + 1 + 1 or C = 3 R, , WE. 32: Find the molar heat capacity in a process, of a diatomic gas if it does a work of Q/4 when, a heat of Q is supplied to it, 2 ( dU ), 5 , Sol. dU = CvdT = 2 R dT (or) dT =, , , , , 5R, , From first law of thermodynamics, dU = dQ – dW = Q –, , Q 3Q, =, 4, 4, , Now molar heat capacity, dQ, Q ´ 5 R = 5 RQ, C=, =, 3Q , dT, 2 ( dU ), 2, , , =, , 10, R, 3, , 4 , , T .V −1/3 = constant ⇒ V ∝ T 3, Temperature increases with increase in volume., , 40, , Polytrophic process:, , Ø, , 2, (or) log T2 = 2.4742 - (0.7782), 7, = 2.4742 - 0.2209 = 2.2533., Anti log of 2.2533 = 178.7, ∴ T2 = 178.7K Þ t2 = 178.7 - 273 = - 94.3OC., , 2k 3/2, V, 3, , work done W = ∫ PdV = k ∫ V 1/2 dV =, , (298)1.4, 60.4, , (298)1.4 1.4 (298)1.4 298, T2 = 0.4 =, =, 0.4, 2, 6, , 1.4, (6), (6) 7, , 3, , ∴ P = k V ⇒ P = kV 1/2, , 2, 3, , is, , WE. 33 :A monoatomic gas undergoes a process, given by 2dU + 3dW = 0, then what is the, process, Sol: dQ = dU + dW ⇒ dQ = dU −, , 2 dU, dU, =, 3, 3, , 1, 1 3, nRdT, = nCv dT = n. RdT =, 3 2, 2, 3, C=, , 1 dQ R, = ;It is not isobaric as C is not, n dT, 2, , 5R, , equal to, ; It is not adiabatic as C ≠ 0, 2, It is not isothermal as C ≠ ∞ so it is a polytrophic, process., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , WE.34 :The relation between U, P and V for an, ideal gas is U = 2 + 3PV. What is the, atomicity of the gas., Sol : For an adiabatic process dQ = 0 = dU + dW, or 0 = dU + PdV –––– (1), From the given equation dU = 3 (PdV + VdP), Substituting dU from (1), -PdV = 3 (PdV + VdP), dV , dP , = −3 , , V , P , , or 4P (dV) + 3V (dP) = 0 or 4 , , On integrating, we get, In (V4) + In (P3) = constant,In (V4P3) = constant, ⇒ V4P3= constant or PV4/3 = constant, i.e., γ =, , 4, i.e., gas is polyatomic., 3, , WE.35:One mole of a monoatomic ideal gas, undergoes the process A → B in the given, P - V diagram. What is the specific heat for, this process ?, P, 6P0, , B, , 3P0, , A, V0, , 5V0, , V, , Sol. Specific heat, ∆Q, 1, W, C=, =, ( ∆U + W ) = Cv +, ∆T ∆T, ∆T, For the given process, 9P, W = 4V0 0 = 18PV, 0 0 (Q W= area of P – V graph), 2, Also, ∆T = T2 − T1, =, , (6 P0 )(5V0 ) (3P0 )V0, R, , -, , R, , =, , 27 PV, 0 0, R, , 3, and Cv = R, 2, \ C = Cv +, , W 3R 18PV, 3R 2 R 13R, 0 0, = +, =, +, =, ö, DT 2 æç 27PV, 0 0÷, 2, 3, 6, ÷÷, ççè, R ø, , WE. 36: If c P and c v denote the specific heats of, nitrogen per unit mass at constant pressure, and constant volume respectively, then, R, R, 1) c P − cV =, 2) c P − cV =, 28, 14, 3) c P − cV = R 4) c P − cV = 28 R, (AIEEE-2007), Sol. Mayer formula c P, NARAYANAGROUP, , − cV =, , R, R, =, M, 28, , WE. 37:When a system is taken from state i to state, f along the path iaf, it is found that Q = 50, cal and W = 20 cal. Along the path' ibf Q = 36, cal. W along the path ibf is, (AIEEE-2007), a, , f, , i, , b, , Sol. From first law of thermodynamics,, dQ = dU + dW, For path iaf, 50 = ∆U + 20 ⇒ ∆U = 30 cal, For path ibf, dW = dQ − dU = 36 − 30 = 6 cal, WE. 38:A thermally insulated vessel contains an, ideal gas of molecular mass M and ratio of, specific heat γ . It is moving with speed v and, is suddenly brought to rest. Assuming no heat is, lost to the surroundings. its temperature, increases by (in Kelvin ) (JEE MAIN-2011), 1 2, 1, m R, 2, Sol. mv = du = nCV dT ; 2 mv = M γ − 1 ∆ T, 2, WE. 39: 100 g of water is heated from 300C to 500C, ignoring slight expansion of the water, the, change in its internal energy is (specific heat, of water is 4180J/Kg/K) (JEE MAIN-11), Sol. dQ = dU + dW ;dW= 0 (Q dV is neglected ), ∴ dQ = dU = mS ∆θ, = (100 ) (10 − 3 ) ( 4180 )( 20 ) = 8360 J ≈ 8.4 K J, , WE. 40 : Five moles of hydrogen initially at STP, is compressed adiabatically so that its, temperature becomes 673 K. The increase in, internal energy of the gas, in kilo joule is, (R = 8.3 J/mol-K; γ = 1.4 for diatomic gas), (EAM-2014), Sol. Work done by an ideal gas in adiabatic expansion, dU = n, , R, 8.3 , dT = 5 , ( 300 ) = 41500 J, γ −1, 1.4 − 1 , , WE. 41 : The volume of one mole of the gas is, changed from V to 2V at constant pressure p., If γ is the ratio of specific heats of the gas,, change in internal energy of the gas is, (EAM-2014), R, , PdV, , Sol. dU = nCv dT = n γ − 1 dT = n, γ −1, =n, , P ( 2V − V ), PV, =, γ −1, γ −1, 41
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, WE. 42 :Three moles of an ideal monoatomic gas, undergoes a cyclic process as shown in the, figure. The temperature of the gas in different, states marked as 1, 2, 3 and 4 are 400K, 700K,, 2500K and 1100K respectively. The work done, by the gas during the process 1-2-3-4-1 is, (universal gas constant is R) (EAM-2013), 2, P, , 1, , 3, , 4, , Sol. Process 1 → 2 and 3 → 4 are polytrophic and, process 2 → 3 and 4 → 1 are isobaric., From the graph, P, = K ⇒ PV −1 = K ⇒ x = −1, V, , Work done W = W1→2 + W2→3 + W3 →4 + W4→1, =, , nR, nR, [T1 − T2 ] + P2 (V3 − V2 ) +, [T3 − T4 ] + P1 [V1 − V4 ], x −1, x −1, , =, , nR, nR, [T1 − T2 ] + nR (T3 − T2 ) +, [T3 − T4 ] + nR [T1 − T4 ], x −1, x −1, , =, , nR, [T1 − T2 + T3 − T4 ] + nR (T3 − T2 + T1 − T4 ), x −1, =, , nR, T ( n = 3) i.e PαT, V, So, volume remains constant for the graphs AB and, CD., So, no work is done during processes for A to B, and C to D., , and P =, , WAB = WCD = 0 and WBC = P2 (VC − VB ), , = nR (TC − TB ) = 3R ( 2400 − 800 ) = 4800 R, , v, , P ∝V ⇒, , Sol. Processes A to B and C to D are parts of straight, line graphs of form y = mx, , WDA = P1 (VA − VD ) = nR (TA − TD ), = 3R ( 400 − 1200 ) = −2400 R, Work done in the complete cycle, W = WAB + WBC + WCD + WDA, , = 0 + 4800 R + 0 + ( −2400 ) R, = 2400 R = 19953.6 J ≈ 20 kJ, , WE. 44 : An ideal gas is subjected to a cyclic process, ABCD as depicted in the p-V diagram given, below, , A, , 3R, [ 400 − 700 + 2500 − 1100], −1 − 1, , B, , P, , C, , + 3R ( 2500 − 700 + 400 − 1100 ), =, , 3R, [1100] + 3R (1100 ), −1 − 1, , 1, = 3R (1100 ) 1 − = 1650 R, 2, , WE. 43 : 3 moles of an ideal monoatomic gas, performs ABCDA cyclic process as shown in, figure below. The gas temperatures are, TA = 400K , TB = 800K , TC = 2400K and TD = 1200K ., The work done by the gas is (approximately), , D, V, Which of the following curves represents the, equivalent cyclic process?(EAM-2009), , A, , p, , C, , D, , D, , C, T, , B, , A, , A, , B, , D, , C, , 4) V, , 3) P, , C, , D, T, , D, , B, , T, , (EAMCET-2010), C, , A, 2) V, , 1) P, , ( R = 8.314 J / mol K ), B, , B, , T, , T, 42, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, Sol. A → B → C → D → A is clockwise process., During A → B , pressure is constant and B → C ,, 1, , it means T is constant., V, During process C → D, both P and V changes, , process follows Pα, , THERMODYNAMICS, Ø, , Reversible process:, Ø, , 1, and process D → A follows Pα which means, V, T is constant., , Comparison of P-V curves of various, processes :, , isochoric, , isobaric, , K=P, , K =∞, , isothermal, K=γP, adiabatic, , Ø, , V, , Free expansion :, Gas, , Vacuum, , Figure shows an insulated cylinder divided into, two parts by a thin massless fixed piston. Volume, of left compartment is filled with an ideal gas and, the right compartment is vacuum. If we release, the piston, gas fills the whole space of the cylinder, rapidly. In this expansion no heat is supplied to, the gas as walls are insulated. ∴ ∆Q = 0 . As the, piston is fixed no work is done by the gas, ∆W = 0, and hence internal energy remains constant., ∴∆U = 0 , T is constant, Such an expansion is called "free expansion"., , A process which can be retraced back in such a, way that the system passes through the states as in, direct process and finally the system acquires the, initial conditions, leaving no change anywhere else,, is called reversible process. Any quasi-static, process can be reversible., , Conditions for a process to be reversible:, Ø, , P, K=0, , These two statements of the second law are, equivalent to each other. Because, if one is violated,, the other is also automatically violated., , (a) There should be no loss of energy due to, conduction, convection or dissipation of energy, against any resistance, like friction, viscosity etc., (b) No heat should be converted into magnetic or, electric energy., (c) The system must always be in thermal,, mechanical and chemical equilibrium with the, surroundings.(i.e the process must be quasi-static), Examples : In practice, there is no reversible, process. But approximately we can give the, following examples., i) The process of change of state from ice into water, is a reversible process., ii) The process of change of state from water to, steam., iii) The gradual extension and compression of an, elastic spring is approximately reversible., iV) The electrolysis process is reversible if internal, resistance is negligibly small., v) Slow compression and expansion of an ideal, gas at constant temperature., , Irreversible process:, Ø, , Second law of thermodynamics :, , In this process the system does not pass through, the same intermediate states as in the direct process., Most of the processes occurring in nature are, irreversible., Examples :1) Diffusion of gas, 2) Dissolving of salt in water, 3) Sudden expansion or compression of gas, , Ø, , Clausius statement: It is impossible for a self, , Heat engine :, , Ø, , acting machine unaided by any external agency to, transfer heat from a cold reservoir to a hot reservoir., In other words heat can’t flow by itself from a, colder to a hotter body., Kelvin-Planck Statement: It is impossible, for any heat engine to convert all the heat absorbed, from a reservoir completely into useful work. In, other words 100% conversion of heat into work, is impossible., , NARAYANAGROUP, , Ø, , Ø, Ø, , The device, used to convert heat energy into, mechanical energy is called a heat engine., For conversion of heat into work with the help of a, heat engine the following conditions required., i) There should be a reservoir at constant higher, temperature ‘ T1 ’ from which heat is extracted. It is, called the source., ii) Working substance which undergoes, thermodynamic cyclic changes(ex: ideal gas)., iii) There should be a reservoir at constant lower, 43
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , Ø, , temperature ‘ T2 ’ to which heat can be rejected., This is called the sink., The source and sink should have very high thermal, capacity., , P, A, , T1, , B, Q1, , Working of heat engine :, , D, , T2, , a) Schematic diagram of heat engine, , C, Q2, v, , Engine, , Source, , Sink, , T1, Q1, , Q2, , The P-V diagram of the cycle is shown in the figure., In process AB heat Q1 is taken by the working, substance at constant temperature T 1 and in process, CD heat Q2 is liberated by the working substance, at constant temperature T2. The net work done is, the area enclosed by the cycle ABCDA. After doing, the calculations for different processes we can show, Q2, T2, that : Q = T, 1, 1, Therefore, efficiency of the Carnot engine is, Q, T, η = 1− 2 = 1− 2, Q1, T1, , T2, , W=Q1 - Q2, b) Engine derives an amount ‘Q1’ of heat from the, source., c) A part of this heat is converted into work ‘W’., d) Remaining heat ‘Q2’ is rejected to the sink., Thus Q1 = W + Q2 or the work done by the engine, is given by W = Q1 − Q2, e)The energy Q2 is unavailable in the universe,, which causes increase in entropy of universe., , Efficiency of heat engine :, Ø, , Efficiency of heat engine (η ) is defined as the, fraction of total heat supplied to the engine which, is converted into work., Mathematically, , Ø, , η=, , W, Q1, , =, , Q1 − Q2, Q1, , = 1−, , Q2, Q1, , According to this , efficiency is 100% if Q2 = 0,, that is no heat is rejected to the cold reservoir or, sink. That is the entire heat absorbed must be, converted to mechanical work , which is impossible, according to Second law of Thermodynamics., , Carnot or Reversible or Ideal heat engine:, Ø, , 44, , When the working substance is an ideal gas and it, is subjected to cyclic process consisting of, isothermal expansion, adiabatic expansion,, isothermal compression and adiabatic, compression, then such heat engine is called, Carnot engine. The cyclic process is called Carnot, cycle., Carnot Cycle :Carnot cycle consists of the, following four stages (i) Isothermal expansion, (process AB), (ii) Adiabatic expansion (process, BC), (iii) Isothermal compression (process CD),, and (iv) Adiabatic compression (process DA)., , Ø, Ø, , As T2 is always less than T1 , η < 1 . i.e., the value, of η can never be equal or greater than 1. When, the temperature of sink T2 =0 K, then η can be 1, or 100% . But it is impossible., For Carnot engine η is independent of the nature, of working substance. It depends on only the, temperatures of source and sink., The efficiency of an irreversible engine is always, less than or equal to that of reversible engine when, operated between the same temperature limits., , ∴ always ηir ≤ η r, , Refrigerator:, Engine, T1, Q1, , Q2, , T2, , W=Q1 - Q2, The refrigerator is just the reverse of heat engine., In refrigerator the working substance extracts an, amount of heat Q2 from the cold reservoir (Sink), at a lower temperature T2 . An amount of external, work W is done on the working substance and, finally an amount of heat Q1 is rejected to the hot, reservoir at a higher temperature T1 ., , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 3, 15, P0 V0 + P0 V0 = 9P0 V0, 2, 2, Obviously, the processes B → C and C → A, involve the abstraction of heat from the gas., Work done per cycle, Efficiency =, Total heat supplied per cycle, ∴∆Q1 =, , Sol. Efficiency =, =, , WE. 48:Efficiency of a heat engine whose sink is, at temperature of 300 K is 40%. To increase, the efficiency to 60%, keeping the sink, temperature constant, the source temperature, must be increased by, T2, 40 3, Sol. T = 1 − η = 1 − 100 = 5, 1, , 5, 5, ⇒ T1 = T2 ⇒ T1 = × 300 = 500 K, 3, 3, New efficiency η ′ = 60%, T2, 60 2, = 1 −η ′ = 1 −, =, ', T1, 100 5, 5, × 300 = 750 K ; ∆T = 750− 500 = 250K, 2, WE.49:A Carnot engine, having an efficiency of, η = 1/10 as heat engine, is used as a, refrigerator. If the work done on the system, is 10 J, the amount of energy absorbed from, the reservoir at lower temperature is, (JEE MAIN-2007), Sol. Coefficient of performance of refrigerator, T1 ′ =, , 1 − η 1 − 1/10, Heat extracted, =, =9 =, η, 1/10, workdone, ∴ Heat extracted = 9 × 10 = 90J, , 2P0, P0, , B, , A, V0, , 46, , C, D, , PV, 0 0, nCV ∆ T1 + nC p ∆T2, , =, , PV, 0 0, 3, 5, nR (TB − TA ) + nR (TC − TB ), 2, 2, , =, , PV, 0 0, 3, 5, ( 2 PV, ( 4PV, 0 0 − PV, 0 0)+, 0 0 − 2 PV, 0 0), 2, 2, , =, , PV, 1, 0 0, =, = 15.4%, 3, 5, 6.5, PV + 2 PV, 2 0 0 2 0 0, , WE. 51: A Carnot engine, whose efficiency is 40%, takes heat from a source maintained at a, temperature of 500 K. It is desired to have an, engine of efficiency 60%. Then , the intake, temperature for the same exhaust (sink), temperature must be(JEE MAIN-12), T2, T, Sol. η = 1 − T ; 0.4 = 1 − 2 ⇒ T2 = 300 K, 1, , 500, , 300, 300, T2, = 1 − 1 ⇒ T11 =, = 750 K, 1, T1, T1, 0.4, WE.52:A diatomic ideal gas is used in a car engine, as the working substance. Volume of the gas, increases from V to 32V during the adiabatic, expansion part of the cycle. The efficiency of, the engine is, (JEE MAIN-2010), 0.6 = 1 −, , 1.4 −1, γ −1, Sol. TV, = T2 ( 32V1 ), = T2V2γ −1 ⇒ TV, 1 1, 1 1, , 1.4−1, , β=, , WE. 50: Helium gas undergoes through a cycle, ABCD (consisting of two isochoric and, isobaric line) as shown in figure. Efficiency, of this cycle is nearly : (Assume the gas to be, close to ideal gas), (JEE MAIN-2012), , Area under P − V diagram, ∆QAB + ∆QBC, , ∴η =, , 1, P0 V0, 1, i.e., η= 2, =, 9P0 V0 18, , work done in cycle, × 100, heat absorbed, , ⇒, , T, 1 3, T2 1, = ⇒ η = 1− 2 = 1− =, T1 4, T1, 4 4, , Entropy(s):, Ø, , The thermodynamic coordinate or parameter that, gives the measure of disorder is called entropy., We cannot measure entropy, but we can measure, change in entropy during thermodynamic change., If ‘ds’ is the small change in entropy at temperature, dQ, T, Where dQ is exchange of heat between system, , T, then ds =, , 2V0, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, WE. 57:Pressure versus temperature graph of an, ideal gas is as shown in figure. Density of, the gas at point A is ρ 0 . Density at B will be, , C, , P, , P, , B,C, , B, , 1), , 2), , P, , D, , 3P0, , B, , P0, , P, , 2T0, , o, , C, , T, D, , 3), , B, , C, , A, , D, , 4), , B, , PM, P, or ρ ∝, Sol. ρ =, RT, T, , A, , V, , V, , P0, P, P 3 P0, T = T and T = 2 T, A, B 0, 0, , Sol. Along process AB, CD temperature is constant, , 3 P, 3, 3, P, T = 2 T ∴ ρ B = 2 ρ A = 2 ρ0, B, , WE. 58 :In the P-V diagram shown in figure ABC, is a semicircle. The work done in the process, ABC is :, , 1, 1, , ρ ∝, V, V, ρ − V graph will be a rectangular hyperbola., Along BC and DA, V is constant ⇒ ρ is constant, , ( isothermal process) i.e., P ∝, , WE. 60 :A thermodynamic system undergoes cyclic, process ABCDA as shown in figure. The work, done by the system is :, , P(atm), , P, C, , 3, , V, , V, , A, T0, , A,D, , A, , 3P0, , B, , C, , B, , A, , 1, 1, , 2, , 2P0, , V(L), , Sol. WAB is negative (volume is decreasing) and WBC, is positive (volume is increasing) and since, , P0, , O, D, , A, , 2v0, , v0, , WBC > WAB, , ∴ Net work done is positive, Q Workdone=area between semicircle, , = π ( pressure radius )( volume radius ), , 1, π, = π (1atm ) litre = atm ( litre ), 2, 2, WE. 59 : Pressure versus temperature graph of an, ideal gas is as shown in figure corresponding, , density ( ρ ) versus volume(V) graph will be:, , V, , Sol. WBCOB = − Area of triangle BCO = −, , PV, 0 0, 2, , WAODA = + Area of triangle AOD = +, , PV, 0 0, 2, , ∴Wnet = 0, WE. 61 :P-V plots for two gases during adiabatic, processes are shown in the figure. Plots 1 and, 2 should correspond respectively to:, , P, , C, , P, , 1, , B, , 2, , D, A, , V, , T, 48, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 3., , Sol. In adiabatic process, Slope of P-V graph,, , dP, P, = −γ, dV, V, , Slope ∝ γ (with negative sign), From the given graph, (slope)2 >(slope)1, , ∴γ 2 > γ 1, , 4., , Therefore, '1' should correspond to O2 ( γ = 1.4 ), and '2' should correspond to He ( γ = 1.67 ) ., WE. 62 : One mole of diatomic ideal gas undergoes, a cyclic process ABC as shown in figure., The process BC is adiabatic. The temperature, at A,B and C are 400 k, 800k and 600 k,, respectively. Choose the correct statement., [JEE MAIN 2014], , 5., , 6., , P, B, , 800 K, , 7., , A, 400 K, , 600 K, C, V, , 1) The change in internal energy in whole, cyclic process is 250 R, 2) The change in internal energy in the, process CA is 700 R, 3) The change in internal energy in the, process AB is -350 R, 4) The change in internal energy in the, process BC is -500R, Sol: Process A → B, ∆U = nCV ∆T = 1×, , C→, ∆U, , 5R, ( 800 − 400 ) = 1000R, 2, , 5R, A ; ∆U = nCV ∆T = 1× 2 ( 400 − 600 ), , cycle, , = −500R, , = 0 , and ∆U AB + ∆U BC + ∆U CA = 0, , 1000 R + ∆U BC − 500R = 0 ; ∴ ∆UBC = −500R, , C.U.Q, 1., , 2., , Water is used in car radiators as coolant, because, 1) its density is more, 2) high specific heat, 3) high thermal conductivity 4) free availability, Of the following specific heat is maximum for, 1) Mercury 2) Copper 3) Water 4) Silver, , NARAYANAGROUP, , 8., , Heat is, 1) kinetic energy of molecules, 2) potential and kinetic energy of molecules, 3) energy in transits, 4) work done on the system, Which of the following does not characterise, the thermodynamic state of matter, 1) Volume, 2) Temperature, 3) Pressure, 4) Work, The thermal motion means, 1) motion due to heat engine, 2) disorderly motion of the body as a whole, 3) motion of the body that generates heat, 4) random motion of molecules, Heat required to rise the temperature of one, gram of water through 10 c is, 1) 0.001 K cal, 2) 0.01 K cal, 3) 0.1 K cal, 4) 1.0 K cal, Heat capacity of a substance is infinite. It, means, 1) heat is given out, 2) heat is taken in, 3) no change in temperature whether heat is, taken in (or) given out, 4) all of the above, For a certain mass of a gas Isothermal relation, between ‘P’ and ‘V’ are shown by the graphs, at two different temperatures T1 and T2 then, P, , T1, , T2, , v, , 1)T1 = T 2, 2)T1>T2, 3)T1 < T2 4)T1 ≥ T2, Certain amount of heat supplied to an ideal, gas under isothermal conditions will result in, 1) rise in temperature, 2) doing external work and a change in, temperature, 3) doing external work, 4) an increase in the internal energy of the gas, 10. The temperature range in the definition of, standard calorie is, 9., , 1) 14.5 0 C to 15.5 0 C 2) 15.5 0 C to 16.5 0 C, 3) 10 C to 2 0 C, , 4) 13.5 0 C to 14.5 0 C, 49
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 11. The pressure p of a gas is plotted against its, absolute temperature T for two different, constant volumes V1 and V2, where V1 > V2 ., If p is taken on y–axis and T on x–axis, 1) the curve for V1 has greater slope than the curve, for V2, 2) the curve for V2 has greater slope than the curve, for V1, 3) the curves must intersect at some point other, than T = 0, 4) the curves have the same slope and do not, intersect, 12. dU + dW = 0 is valid for, 1) adiabatic process, 2) isothermal process, 3) isobaric process, 4) isochoric process, 13. In a given process dW=0, dQ<0 then for a gas, 1)temperature–increases 2)volume–decreases, 3)pressure–decreases 4)pressure–increases, 14. A piece of ice at 0 0 C is dropped into, water at 0 0 C . Then ice will, 1) melt, 2) be converted into water, 3) not melt, 4) partially melt, 15. The temperature determines the direction of, net change of, 1) gross kinetic energy, 2) intermolecular kinetic energy, 3) gross potential energy, 4) intermolecular potential energy, 16. The direction of flow of heat between two, gases is determined by, 1) average kinetic energy 2) total energy, 3) internal energy, 4) potential energy, 17. Heat is absorbed by a body . But its temperature, does not raised. Which of the following statement, explains the phenomena ?, 1) Only K.E. of vibration increases, 2) Only P.E. of inter molecular force changes, 3) No increase in internal energy takes place, 4) Increase in K.E. is balanced by decrease in, P.E., 18. Zeroth law of thermodynamics gives the, concept of, 1) pressure 2) volume 3) temperature 4) heat, 19. We need mechanical equivalent of heat, because, 1) it converts work into heat, 2) in C.G.S system, heat is not measured in the, units of work, 50, , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 3) in SI system, heat is measured in the units of, work, 4) of some reason other than those mentioned above, When we switch on the fan in a closed room., The temperature of the air molecules, 1) increases, 2) decreases, 3) remains unchanged, 4) may increase or decrease depending on the, speed of rotation of the fan., Which type of molecular motion does, contribute towards internal energy for an ideal, monoatomic gas, 1) translational, 2) rotational, 3) vibrational, 4) all the above, In which of the following processes the, internal energy of the system remains, constant ?, 1) Adiabatic, 2) Isochoric, 3) Isobaric, 4) Isothermal, The internal energy of a perfect monoatomic, gas is, 1) complete kinetic, 2) complete potential, 3) sum of potential and kinetic energy of the, molecules, 4) difference of kinetic and potential energies of, the molecules, Which of the following is constant in an, isochoric process ?, 1) Pressure, 2) Volume, 3) Temperature, 4) Mass, How does the internal energy change when, the ice and wax melt at their normal melting, points?, 1) Increases for ice and decreases for wax, 2) Decreases for ice and increases for wax, 3) Decreases both for ice and wax, 4) Increases both for ice and wax, In the free expansion of a gas, its internal, energy, 1) remains constant, 2) increases, 3) decreases, 4) sometimes increases , sometimes decreases, The internal energy of an ideal gas depends, upon, 1) only its pressure, 2) only its volume, 3) only its temperature, 4) its pressure and volume, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, 28. On compressing a gas suddenly, its, temperature, 1) increases, 2) decreases, 3) remains constant, 4) all the above, 29. When heat is added to a system at constant, temperature, which of the following is possible., 1) Internal energy of system increases, 2) Work is done by the system, 3) Neither internal energy increases nor work done, by the system, 4) Internal energy increases and work is done by, the system, 30. The first law of thermodynamics is based on, the law of conservation of, 1) energy 2) mass 3) momentum 4) pressure, 31. A given mass of a gas expands from the state, A to the state B by three paths 1,2 and 3 as, shown in the figure. If W1, W2 and W3, respectively be the work done by the gas along, the three paths then, A, P, , 3, 2, , 1, B, V, , 1) W1 > W2 > W3, 2) W1 < W2 < W3, 3) W1 = W2 = W3, 4) W1 < W2 = W3, 32. A given system undergoes a change in which, the work done by the system equals to the, decrease in its internal energy. The system, must have undergone an, 1) isothermal change, 2) adiabatic change, 3) isobaric change, 4) isochoric change, 33. A closed vessel contains some gas at a given, temperature and pressure. If the vessel is, given a very high velocity, then the, temperature of the gas, 1) increases, 2) decreases, 3) may increase or decrease depending upon the, nature of the gas, 4) does not change, 34. Unit mass of liquid of volume V1 completely, turns into a gas of volume V2 at constant, atmospheric pressure P and temperature T. The, latent heat of vaporization is “L”. Then the, change in internal energy of the gas is, 1) L, 2) L+P(V2 - V1), 3) L - P(V2-V1), 4) Zero, NARAYANAGROUP, , THERMODYNAMICS, 35. In an isobaric (constant pressure) process. the, correct ratio is, 1) ∆Q : ∆U = 1 : 1, 2) ∆Q : ∆U = 1 : γ−1, 3) ∆Q : ∆U = γ−1: 1, 4) ∆Q : ∆U = γ : 1, 36. In an isobaric process, the correct ratio is, 1) ∆Q : ∆W = 1 : 1, 2) ∆Q : ∆W = γ : γ−1, 3) ∆Q : ∆W = γ−1: γ, 4) ∆Q : ∆W = γ : 1, 37. Air in a thermally conducting cylinder is, suddenly compressed by a piston, which is then, maintained at the same position with the, passage of time ?, 1) The pressure decreases, 2) The pressure increases, 3) The pressure remains the same, 4) The pressure may increase or decrease, depending upon the nature of the gas, 38. Which of the following states of matter have, two specific heats ?, 1) Solid 2) Gas 3) Liquid, 4) Plasma, 39. The specific heat of a gas in an isothermal, process is, 1) infinity, 2) zero, 3) negative, 4) remains constant, 40. Why the specific heat at a constant pressure, is more than that at constant volume ?, 1) There is greater inter molecular attraction at, constant pressure, 2) At constant pressure molecular oscillations are, more violent, 3) External work need to be done for allowing, expansion of gas at constant pressure, 4) Due to more reasons other than those mentioned, in the above, 41. The ratio Cp / Cv of the specific heats at a, constant pressure and at a constant volume of, any perfect gas, 1) can’t be greater than 5/4, 2) can’t be greater than 3/2, 3) can’t be greater than 5/3, 4) can have any value, 42. Which of the following formula is wrong ?, Cp, , R, 1) Cv = γ − 1, , 2), , γR, 3) Cp = γ − 1, , 4) Cp - Cv = 2R, , CV, , =γ, , 51
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 43. Two identical samples of gases are allowed to, expand to the same final volume (i), isothermally (ii) adiabatically. Work done is, 1. more in the isothermal process, 2. more in the adiabatic process, 3. equivalent in both processes, 4. equal in all processes, 44. Which of the following is true in the case of a, reversible process ?, 1) There will be energy loss due to friction, 2) System and surroundings will not be in thermo, dynamic equilibrium, 3) Both system and surroundings retains their, initial states, 4) 1 and 3, 45. The ratio of the relative rise in pressure for, adiabatic compression to that for isothermal, compression is, , 1, 1, 3) 1 − γ, 4), γ, 1− γ, 46. Ratio of isothermal elasticity of gas to the, adiabatic elasticity is, 1) γ, , 1, 1, 3) 1 − γ, 4), γ, 1− γ, The conversion of water into ice is an, 1) isothermal process, 2) isochoric process, 3) isobaric process, 4) entropy process, For the Boyle’s law to hold good, the, necessary condition is, 1) Isobaric, 2) Isothermal, 3) Isochoric, 4) Adiabatic, An isothermal process is a, 1)slow process, 2)quick process, 3) very quick process, 4) both 1 & 2, Two samples of gas A and B, initially at same, temperature and pressure, are compressed to, half of their initial volume, 'A' isothermally and, 'B' adiabatically. The final pressure in, 1) A and B will be same, 2) A will be more than in B, 3) A will be less than in B, 4)A will be double that in B, In which of the following processes all three, thermodynamic variables, that is pressure, volume and temperature can change, 1) Isobaric, 2) Isothermal, 3) Isochoric, 4) Adiabatic, 1) γ, , 47., , 48., , 49., , 50., , 51., , 52, , 2), , 2), , 52. During adiabatic expansion the increase in, volume is associated with, 1) increase in pressure and temperature, 2) decrease in pressure and temperature, 3) increase in pressure and decrease in, temperature, 4)Decrese in pressure and increase in temperature, 53. A gas is being compressed adiabatically. The, specific heat of the gas during compression is, 1) zero, 2) infinite, 3) finite but non zero, 4) undefined, PV , 54. The gas law , = constant is true for, T , 1) isothermal change only, 2) adiabatic change only, 3) Both isothermal & adiabatic changes, 4) neither isothermal nor adiabatic change, 55. During adiabatic compression of a gas, its, temperature, 1) falls, 2) raises, 3) remains constant, 4) becomes zero, 56. The work done on the system in an adiabatic, compression depends on, 1) the increase in internal energy of the system, 2) the decrease in internal energy, 3) the change in volume of the system, 4) all the above, 57. The ratio of slopes of adiabatic and isothermal, curves is, , 1, 3) γ 2, 4) γ 3, γ, 58. Two steam engines ‘A’ and ‘B’, have their, sources respectively at 700 K and 650 K and, their sinks at 350 K and 300K. Then, 1) ‘A’ is more efficient than ‘B’, 2) ‘B’ is more efficient than ‘A’, 3) both A and B are equally efficient, 4) depends on fuels used in A and B, 59. If the temperature of the sink is decreased,, then the efficiency of heat engine, 1) first increases then decreases, 2) increases, 3) decreases, 4) remains unchanged, 60. An ideal heat engine can be 100% efficient if, its sink is at, 1) 0K, 2) 273K, 3) 00C, 4) 00F, 1) γ, , 2), , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, 61. If the temperature of a source increases, then, the efficiency of a heat engine, 1) increases, 2) decreases, 3) remains unchanged, 4) none of these, 62. When heat is added to a system then the, following is not possible?, 1) Internal energy of the system increases, 2) Work is done by the system, 3) Neither internal energy increases nor work is, done by the system, 4) Internal energy increases and also work is done, by the system, 63. A sink, that is the system where heat is, rejected, is essential for the conversion of, heat into work. From which law the above, inference follows?, 1) Zeroth 2) First 3) Second 4)Both 1 & 2, 64. The efficiency of a heat engine, 1) is independent of the temperature of the source, and the sink, 2) is independent of the working substance, 3) can be 100%, 4) is not affected by the thermal capacity of the, source or the sink, 65. An ideal heat engine working between, temperatures TH and TL has efficiency η . If, both the temperatures are rised by 100K, each, then the new efficiency of the heat, engine will be, 1) equal to η, 2) greater than η, 3) less than η, 4) greater or less than η depending upon the nature, of the working substance, 66. The efficiency of the reversible heat engine, , THERMODYNAMICS, 68. The adiabatic and isothermal elasticities Bφ, and Bθ are related as, , P, I, , 1) ηr > η I, , 2) ηr < ηI, , 3) η r ≥ η I, 4)ηr > 1 and η I < 1, 67. In a heat engine, the temperature of the, working substance at the end of the cycle is, 1) equal to that at the beginning, 2) more than that at the beginning, 3) less than that at the beginning, 4) determined by the amount of heat rejected to, the sink, , NARAYANAGROUP, , II, V, , 70., , 71., , 72., , 73., , is η r , and that of irreversible heat engine is, , ηI . Which of the following relation is correct?, , B, , B, , φ, θ, 1) B = γ 2) B = γ 3) Bφ − Bθ = γ 4) Bθ − Bφ = γ, φ, θ, 69. For the indicator diagram given below, select, wrong statement ?, , 74., , 75., , 1) Cycle - II is heat engine cycle, 2) Net work is done on the gas in cycle - I, 3) Workdone is positive for cycle - I, 4) Workdone is positive for cycle - II, By opening the door of a refrigerator inside, a closed room, 1) you can cool the room to a certain degree, 2) you can cool it to the temperature inside the, refrigerator, 3) you can ultimately warm the room slightly, 4) you can neither cool nor warm the room, Which of the following will extinguish the fire, quickly ?, 1) Water at 1000C, 2) Steam at 1000C, 0, 3) Water at 0 C, 4) Ice at 00C, Which of the following is true in the case of, molecules, when ice melts ?, 1) K.E is gained, 2) K.E. is lost, 3) P.E is gained, 4) P.E. is lost, When two blocks of ice are pressed against, each other then they stick together because, 1) cooling is produced, 2) heat is produced, 3) increase in pressure will increase in melting point, 4) increase in pressure will decrease in melting, point, A cubical box containing a gas with internal, energy U is given velocity V, then the new, internal energy of the gas, 1) less than U 2)more than U 3) U 4) zero, A cubical box containing a gas is moving with, some velocity. If it is suddenly stopped, then, the internal energy of the gas, 1) decreases, 2) Increases, 3) remains constant, 4) may increases or decreases depending on the, time interval during which box comes to rest., 53
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THERMODYNAMICS, 76. Which one of the following is wrong, statement., 1) During free expansion, temperature of ideal gas, does not change., 2) During free expansion, temperature of real gas, decreases., 3) During free expansion of real gas temperature, does not change., 4) Free expansion is conducted in adiabatic, manner., 77. A common salt is first dissolved in water and, extracted again from the water. In this, process,, 1) entropy decreases, 2) entropy increases, 3) entropy becomes zero, 4) entropy remains constant., 78. A large block of ice is placed on a table where, the surroundings are at 0 0C, then, 1) ice melts at the sides, 2) ice melts at the top, 3) ice melts at the bottom, 4) ice does not melt at all, 79. Which of the following substance at 1000C, produces most severe burns ?, 1) Hot air 2) Water, 3) Steam 4) Oil, 80. What energy transformation takes place, when ice is converted into water, 1) Heat energy to kinetic energy, 2) Kinetic energy to heat energy, 3) Heat energy to latent heat, 4) Heat energy to potential energy, 81. Which of the following laws of, thermodynamics leads to the interference, that it is difficult to convert whole of heat, into work ?, 1) Zeroth 2) Second 3) First 4) Both 1 & 2, 82. Starting with the same initial conditions, an, ideal gas expands from volume V 1 to V2. The, amount of work done by the gas is greatest, when the expansion is, 1) isothermal, 2) isobaric, 3) adiabatic, 4) equal in all cases, 83. The second law of thermodynamics implies, 1) whole of heat can be converted into mechanical, energy, 2) no heat engine can be 100% efficient, 3) every heat engine has an efficiency of 100%, 54, , JEE- ADV PHYSICS-VOL- VI, 4) a refrigerator can reduce the temperature to, absolute zero, 84. In the adiabatic compression the decrease in, volume is associated with, 1)increase in temperature and decrease in, pressure, 2) decrease in temperature and increase in, pressure, 3) decrease in temperature and decrease in, pressure, 4) increase in temperature and increase in pressure, 85. Which of the following is true in the case of, an adiabatic process where γ = C P / CV ?, 1) P1−γ T γ = constant, 2) Pγ T 1−γ = constant, 3) PT γ = constant, 4) Pγ T = constant, 86. If an ideal gas is isothermally expanded its, internal energy will, 1) increase, 2) decrease, 3) remains same, 4) decrease or increase depending on nature of, the gas, 87. For an adiabatic process the relation between, V and T is given by, 1) TV γ = constant, 2) T γ V = constant, 3) TV 1−γ = constant, 4) TV γ −1 = constant, 88. The temperature of the system decreases in, the process of, 1) free expansion, 2) adiabatic expansion, 3) isothermal expansion, 4) isothermal compression, 89. Heat engine rejects some heat to the sink., This heat, 1)converts into electrical energy., 2)converts into light energy., 3)converts into electromagnetic energy, 4)is unavailable in the universe., 90. For an adiabatic change in a gas, if P, V,T, denotes pressure, volume and absolute, temperature of a gas at any time and γ is the, ratio of specific heats of the gas, then which, of the following equation is true?, 1) T ? P1− ? = const., 2) T 1-? P ? = const., 3) T ?-1 V ? = const., , 4) T ? V ? = const., NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, 91. PV versus T graph of equal masses of H2,, He and CO2 is shown in figure. Choose the, correct alternative ?, 3, , 2, 1, , PV, , T, , 92., , 93., , 94., , 95., , 96., , 97., , 98., , (1) 3 corresponds to H2, 2 to He and 1 to CO2, (2) 1 corresponds to He, 2 to H2 and 3 to CO2, (3) 1 corresponds to He, 3 to H2 and 2 to CO2, (4) 1 corresponds to CO2,2 to H2 and 3 to He, If the ratio of specific heats of a gas at, constant pressure to that at constant volume, is γ , then the change in internal energy of, the mass of gas, when the volume changes, from V to 2V at constant pressure P, is, 2) PV, 1) R /(γ − 1), 3) PV / (γ − 1), 4) γ PV/ (γ − 1), Heat is added to an ideal gas and the gas, expands. In such a process the temperature, 1) must always increase, 2) will remain the same if the work done is equal, to the heat added, 3) must always decrease, 4) will remain the same if change in internal energy, is equal to the heat added, First law of thermodynamics states that, 1) system can do work, 2) system has temperature, 3) system has pressure, 4) heat is form of energy, The material that has largest specific heat is, 1) mercury, 2) water, 3) hydrogen, 4) diamond, The law obeyed by isothermal process is, 1) Gay-Lussac’s law, 2) Charles law, 3) Boyle’s law, 4) Dalton’s law, Which law defines entropy in thermodynamics, 1) zeroth law, 2) First law, 3) second law, 4) Stefan’s law, For the conversion of liquid into a solid, 1) orderliness decreases and entropy decreases, 2) orderliness increases and entropy increases, 3) both are not related, 4) orderliness increases and entropy decreases, , NARAYANAGROUP, , THERMODYNAMICS, 99. Among the following the irreversible process, is, 1) free expansion of a gas, 2)extension or compression of a spring very slowly, 3)motion of an object on a perfectly frictionless, surface, 4) all of them, 100. Which of the following processes are nearly, reversible ?, a. Heat conduction, b. Electrolysis, c. Diffusion, d. Change of state, 1) Only a, 2)Both b and d, 3) Only c, 4) All of the above, 101. Gas is taken through a cyclic process, completely once. Change in the internal, energy of the gas is, 1) infinity 2) zero 3) small, 4) large, 102. What will be the nature of change in internal, energy in case of processes shown below ?, , P, , P, v, , P, , v, , P, v, , v, , 1) + ve in all cases, 2) – ve in all cases, 3) – ve in 1 and 3 and + ve in 2 and 4, 4) zero in all cases, 103. Which of the following is incorrect regarding, the first law of thermodynamics ?, 1) It introduces the concept of internal energy, 2) It introduces the concept of entropy, 3) It is applicable to any process, 4) It is a restatement of principle of conservation, of energy., 104. The temperature of the system decreases in, the process of, 1) free expansion, 2) isothermal expansion, 3) adiabatic expansion, 4) isothermal compression, 55
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 105. In a process the pressure P and volume V of, an ideal gas both increase, 1) It is not possible to have such a process, 2) The workdone by the system is positive, 3) The temperature of the system increases, 4) 2 and 3, 106. The heat capacity of material depends upon, 1) the structure of a matter, 2) temperature of matter, 3) density of matter, 4) specific heat of matter, 107. Heat cannot flow by itself from a body at lower, temperature to a body at higher temperature is, a statement or consequence of, 1) I st law of thermodynamics, 2) IInd law of thermodynamics, 3) conservation of momentum, 4) conservation of mass, 108. For an isothermal process, 1) dQ = dW, 2) dQ = dU, 3) dW = dU, 4) dQ = dU + dW, 109. When thermodynamic system returns to its, original state, which of the following is NOT, possible?, 1) The work done is Zero, 2) The work done is positive, 3) The work done is negative, 4) The work done is independent of the path, followed, 110. A liquid in a thermos flask is vigorously, shaken. Then the temperature of the liquid, 1) is not altered, 2) increases, 3) decreases, 4) none, 111. The PV diagram shows four different possible, paths of a reversible processes performed on a, monoatomic ideal gas. Path A is isobaric, path B, is isothermal , path C is adiabatic and path D is, isochoric . For which process does the, temperature of the gas decrease?, P, A, , P0, 1, P, 2 0, , D, C, V0, , 56, , B, , 2V0, , (1) Process A only (2) Process C only, (3)Processes C & D (4)Processes B, C & D, 112. Two completely identical samples of the same, ideal gas are in equal volume containers with, the same pressure and temperature in, containers labeled A and B. The gas in container, A performs non-zero positive work W on the, surroundings during an isobaric process before, the pressure is reduced isochorically to 1/2 of, its initial amount. The gas in container B has, its pressure reduced isochorically to 1/2 of its, initial value and then the gas performs same, non-zero positive work W on the surroundings, during an isobaric process. After the processes, are performed on the gases in containers A and, B, which is at the higher temperature?, 1)The gas in container A, 2)The gas in container B, 3) The gases have equal temperature., 4) The value of the work W is necessary to, answer this question., 113. Which of the following conditions of the Carnot, ideal heat engine can be realised in practice?, 1) Infinite thermal capacity of the source, 2) Infinite thermal capacity of the sink, 3) Perfectly non conducting stand, 4) Less than 100% efficiency, 114. A heat engine works between a source and a, sink maintained at constant temperatures T1, and T2 . For the efficiency to be greatest, 1) T1 and T2 should be high, 2) T1 and T2 should be low, 3) T1 should be high and T2 should be low, 4) T1 should be low and T2 should be high, 115. The heat engine would operate by taking heat, at a particular temperature and, 1) converting it all into work, 2) converting some of it into work and rejecting, the rest at lower temperature, 3) converting some of it into work and rejecting, the rest at same temperature, 4) converting some of it into work and rejecting, the rest at a higher temperature ., , V, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, 1) 2, 7) 3, 13)3, 19)2, 25)1, 31)2, 37)1, 43)1, 49)1, 55)2, 61)1, 67)1, 73)4, 79)3, 85)1, 91)1, 97)3, 103)2, 109)2, 115)2, , 2) 3, 8) 3, 14) 3, 20)1, 26)1, 32)2, 38)2, 44)3, 50)3, 56)1, 62)3, 68)1, 74)3, 80)4, 86)3, 92)3, 98)1, 104)3, 110)2, , C.U.Q - KEY, 3) 3 4) 4, 9) 3 10)1, 15)2 16)1, 21)1 22)4, 27)3 28)1, 33)4 34)3, 39)1 40)3, 45)1 46)2, 51)4 52)2, 57)1 58)2, 63)3 64)2, 69)3 70)3, 75)2 76)3, 81)2 82)2, 87)4 88)2, 93)2 94)1, 99)1 100)2, 105)2 106) 4, 111)3 112)2, , THERMODYNAMICS, 5., 5) 4, 11)2, 17)2, 23)1, 29)2, 35)4, 41)3, 47)1, 53)1, 59)2, 65)3, 71)1, 77)2, 83)2, 89)4, 95)3, 101)2, 107)2, 113)4, , 6) 1, 12)1, 18)3, 24)2, 30)1, 36)2, 42)4, 48)2, 54)3, 60)1, 66)3, 72)3, 78)3, 84)4, 90)1, 96)3, 102)4, 108)1, 114)3, , From what height a block of ice must fall into, a well so that, , melted? (g = 10 m/s2), 1) 300 m 2) 336 m 3) 660 m, 4) none, Two identical balls ‘A’ and ‘B’ are moving with, same velocity. If velocity of ‘A’ is reduced to, half and of ‘B’ to zero, then the rise in, temperatures of ‘A’ to that of ‘B’ is, 1) 3 : 4, 2) 4 : 1, 3) 2 : 1, 4) 1 : 1, A 50kg man is running at a speed of 18kmh −1 ., If all the kinetic energy of the man can be, used to increase the temperature of water, from 200 C to 300 C , how much water can be, heated with this energy?, 1) 15 g 2) 20 g, 3) 30 g, 4) 40 g, A man of 60 kg gains 1000 cal of heat by, eating 5 mangoes. His efficiency is 56%. To, what height he can jump by using this energy?, 1) 4m 2) 20 m, 3) 28 m 4) 0.2 m, , 6., , 7., , 8., , FIRST LAW OF THERMODYNAMICS, 9., , LEVEL - I (C.W), JOULE’S LAW, 1., , 2., , 3., , A piece of lead falls from a height of 100m on, a fixed non-conducting slab which brings it to, rest. If the specific heat of lead is 30.6 cal/kg, °C, the increase in temperature of the slab, immediately after collision is, 1) 6.72°C 2) 7.62°C 3) 5.62°C 4) 8.72°C, Hailstones fall from a certain height. If only, 1% of the hailstones melt on reaching the, ground, find the height from which they fall., (g=10ms-2,L=80calorie/g & J = 4.2J/calorie), 1) 336 m 2) 236 m 3) 436 m 4) 536 m, From what minimum height a block of ice, has to be dropped in order that it may melt, completely on hitting the ground ?, , JL, J, mgh, 3), 4), g, Lg, J, Two spheres A and B with masses in the ratio, 2 : 3 and specific heat 2 : 3 fall freely from, rest. If the rise in their temperatures on, reaching the ground are in the ratio 1 : 2 the, ratio of their heights of fall is, 1) 3 : 1 2) 1 : 3, 3) 4 : 3, 4) 3 : 4, , 10., , 11, , 12., 13., , 1) mgh 2), , 4., , 1, th of its mass may be, 100, , NARAYANAGROUP, , 14., , How much work to be done in decreasing the, volume of an ideal gas by an amount of, 2.4 x 10–4 m3 at constant normal pressure of, 1 x 105 N/m2 ?, 1) 28 joule 2) 27 joule 3) 24 joule 4)25joule, Find the external work done by the system in, kcal, when 20 kcal of heat is supplied to the, system and the increase in the internal energy, is 8400 J (J=4200J/kcal) ?, 1) 16 kcal 2) 18 kcal 3) 20 kcal 4)19 kcal, Heat of 30 kcal is supplied to a system and, 4200 J of external work is done on the system, so that its volume decreases at constant, pressure. What is the change in its internal, energy ? (J = 4200 J/kcal), 1) 1.302 x 105 J, 2) 2.302 x 105 J, 5, 3) 3.302 x 10 J, 4) 4.302 x 105 J, Air expands from 5 litres to 10 litres at 2 atm, pressure. External workdone is, 1) 10J, 2) 1000J 3) 3000 J 4) 300 J, Heat given to a system is 35 joules and work, done by the system is 15 joules. The change, in the internal energy of the system will be, 1) – 50 J 2) 20 J, 3) 30 J 4) 50 J, A gas is compressed at a constant pressure of, 50 N/m2 from a volume 10m3 to a volume of, 4m3. 100J of heat is added to the gas then its, internal energy is, 1) Increases by 400J, 2) Increases by 200J, 3) Decreases by 400J, 4) Decreases by 200J, 57
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , CP, CV AND THEIR RELATIONS, , 15. Find the change in internal energy in joule, when 10g of air is heated from 30°C to 40°C, , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 58, , (c = 0.172 kcal/kg/KJ = 4200 J/kcal), v, 1) 62.24 J 2)72.24 J 3)52.24 J 4)82.24 J, The temperature of 5 moles of a gas at constant, volume is changed from 1000C to 1200C. The, change in internal energy is 80J. The total heat, capacity of the gas at constant volume will be, in joule/kelvin is, 1) 8, 2) 4, 3) 0.8, 4) 0.4, When an ideal diatomic gas is heated at, constant pressure, the fraction of heat energy, supplied which is used in doing work to, maintain pressure constant is, 1) 5/7 2) 7/2, 3) 2/7, 4) 2/5, When a monoatomic gas expands at constant, pressure, the percentage of heat supplied that, increases temperature of the gas and in doing, external work in expansion at constant, pressure is, 1) 100%, 0, 2) 60%, 40%, 3) 40%, 60%, 4) 75%, 25%, For a gas, the difference between the two, specific heats is 4150J Kg-1 K -1 and the ratio, of specific heats is 1.4. What is the specific, heat of the gas at constant volume in, J Kg-1 K -1?, 1)8475, 2) 5186, 3)1660 4) 10375, The specific heat of air at constant pressure, is 1.005 kJ/kg K and the specific heat of air at, constant volume is 0.718 kJ/kgK. Find the, specific gas constant., 1) 0.287 kJ/kg K, 2) 0.21 kJ/kg K, 3) 0.34 kJ/kg K, 4) 0.19 kJ/kg K, The specific heat of Argon at constant volume, is 0.3122kJ/kg/K. Find the specific heat of, Argon at constant pressure if R = 8.314 kJ/k, mole K. (Molecular weight of argon = 39.95), 1) 5203 2) 5302, 3) 2305, 4) 3025, If the ratio of the specific heats of steam is, 1.33 and R = 8312J/k mole K find the molar, heat capacities of steam at constant pressure, and constant volume., 1) 33.5 kJ/k mole,, 25.19 kJ /kg K, 2) 25.19 kJ/k mole,, 33.5 kJ/kg K, 3) 18.82 kJ/k mole,, 10.82 kJ/k mole, 4) 24.12 kJ /k mole,, 16.12 kJ/k mole, , DIFFERENT THERMODYNAMIC, PROCESSES, 23. One mole of an ideal gas undergoes an, isothermal change at temperature 'T' so that, its volume V is doubled. R is the molar gas, constant. Work done by the gas during this, change is, (2008 M), 1) RT log4 2) RT log2 3) RT log1 4) RT log3, 24. One mole of O 2 gas having a volume equal to, 22.4 litres at 0 0C and 1 atmospheric pressure, is compressed isothermally so that its volume, reduces to 11.2 litres. The work done in this, process is, 1)1672.5J 2)1728J 3) –1728J 4) –1572.5J, 25. The isothermal Bulk modulus of an ideal gas, at pressure 'P' is, 1) P, 2) γP, 3) P/2, 4) P / γ, 26. Diatomic gas at pressure ‘P’ and volume ‘V’, is compressed adiabatically to 1/32 times the, original volume. Then the final pressure is, 1) P/32 2) 32 P 3) 128 P 4) P / 128, 27. The pressure and density of a diatomic gas, ( γ = 7 / 5) change adiabatically from (P, d) to, (P1, d 1). If, , d1, P1, = 32 , then, should be, d, P, , 1) 1/128, 2) 32, 3) 128, 4) none of the above, 28. An ideal gas at a pressure of 1 atmosphere, and temperature of 270 C is compressed, adiabatically until its pressure becomes 8, times the initial pressure, then the final, temperature is ( γ = 3 / 2), 1) 6270 C 2) 5270C 3) 4270C 4) 3270C, 29. The volume of a gas is reduced adiabatically, 1, of its volume at 270C, if the value of, 4, γ = 1.4, then the new temperature will be, , to, , 1) 350 × 40.4 K, 2) 300 × 40.4 K, 3) 150 × 40.4K, 4) None of these, 30. Two moles of an ideal monoatomic gas at 270C, occupies a volume of V. If the gas is expanded, adiabatically to the volume 2V, then the work, done by the gas will be (γ = 5 / 3), 1) –2767.23J, 2) 2767.23J, 3) 2500J, 4) –2500J, NARAYANAGROUP
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THERMODYNAMICS, , LEVEL - I (H.W), JOULE’S LAW, 1., , A piece of aluminium falls from a height of, 200m on a fixed non conducting slab which, brings it to rest. If the specific heat of, aluminium is 210 Cal/kg0C. the increase in, temperature of the slab immediately after, collision (assume that there is no loss of heat) is, 1) 2.2oC, 2) 3.3oC, 3) 4.4oC 4) 5.5oC, 2. Hail stones fall from certain height. If only, 2% of the mass of the hail stone melt on, reaching the ground,, the height from which, they fall is ( g = 10 ms-2, L = 80 cal/gm and, J = 4.2J/cal), 1) 33.6 km 2) 67.2 km 3) 672 m 4) 336 m, 3. From what minimum height a block of ice has, to be dropped in order that 0.5% of ice melts, on hitting the ground ?, 1) 171.43m 2) 17.14m 3) 161.43m 4) 1.714km, 4. Two spheres ‘A’ and ‘B’ of masses in the ratio, 1 : 2. Specific heats in the ratio 2 : 3 falls, from heights ‘h1’ and ‘h2’. On reaching the, ground rise in temperatures are equal, then, h1/h2 =, 1) 3 : 2 2) 2 : 9, 3) 2 : 3 4) 2 : 1, 5. 2kg ice block should be dropped from ‘x km’, height to melt completely. The 8 kg ice should, be dropped from a height of, 1) 4x Km, 2) x Km 3) 2x Km 4) x/2 Km, 6. Two metal balls having masses 50g and 100g, collides with a target with same velocity. Then, the ratio of their rise in temperatures is, 1) 1 : 2 2) 4 : 1, 3) 2 : 1, 4) 1 : 1, 7. A brick weighing 4 Kg is dropped into a 1m, deep river from a height of 2m. Assuming that, 80% of the gravitational potential energy is, finally converted into thermal energy find, this thermal energy in calories is, 1) 15, 2) 17, 3) 23, 4) 27, 8. A man of 60 kg gains 1000 cal of heat by, eating 5 mangoes. His efficiency is 28%. To, what height he can jump by using this energy?, 1) 2m 2) 20 m, 3) 28 m, 4) 0.2 m, , FIRST LAW OF THERMODYNAMICS, 9., , 60, , 2kg of water is converted into steam by, boiling at atmospheric pressure. The volume, changes from 2 x 10–3 m3 to 3.34 m3. The work, done by the system is about, 1) – 340 kJ 2) – 170 kJ 3) 170 kJ 4) 340 kJ, , JEE- ADV PHYSICS-VOL- VI, 10. Find the external workdone by the system in, K cal, when 12.5 k cal of heat is supplied, to the system and the corresponding, increasing in internal energy is 10500 J, (J = 4200 J/kcal), 1)15 K cal 2)12.5 kcal 3)10.0 kcal 4)7.5 kcal, 11. Heat of 20 K cal is supplied to the system and, 8400 J of external work is done on the system, so that its volume decreases at constant, pressure. The change in internal energy is, (J = 4200 J /kcal), 1) 9.24 × 104 J, 2) 7.56 × 104 J, 3) 8.4 × 104 J, 4) 10.5 × 104J., 12. A gas expands from 40 litres to 90 litres at a, constant pressure of 8 atmospheres. Work, done by the gas during the expansion is, 1) 4×10-4J 2) 4×104J 3) 4×103J 4) 4×102J, 13. To a system 300 joules of heat is given and it, does 60 joules of work. How much does the, internal energy of the system change in this, process? (in joule), 1) 240 2) 156. 5, 3) –300 4) –528. 2, 14. A gas under constant pressure of 4.5x105Pa, when subjected to 800KJ of heat, changes the, volume from 0.5m3 to 2 m3 . The change in, the internal energy of the gas is, 2) 5.25 × 105 J, 1) 6.75 × 105 J, 3) 3.25 × 105 J, 4) 1.25 × 105 J, , CP, CV AND THEIR RELATIONS, , 15. Find the change in internal energy in joule when, 20gm of a gas is heated from 20oC to 30oC, (cV = 0.18 kcal/kg K; J = 4200J/kcal), 1) 72.8 J 2) 151.2 J, 3) 302 J 4) 450 J, 16. When two moles of a gas is heated from O 0 to, 100C at constant volume, its internal energy, changes by 420J. The molar specific heat of, the gas at constant volume, 1) 5.75 J K-1 mole-1, 2) 10.55 J K-1 mole-1, -1, -1, 3) 21 J K mole, 4) 42 J K-1 mole-1, 17. A cylinder of fixed capacity 67.2 litres contains, helium gas at STP. Calculate the amount of, heat required to raise the temperature of the, gas by 150C ? ( R = 8.314 J mol −1 k −1 ), 1) 561.19 J 2)570.9 J 3)580.9 J 4)590.9 J, 18. When a diatomic gas expands at constant, pressure, the percentage of heat supplied that, increases temperature of the gas and in doing, external work in expansion at constant, pressure is, 1) 60%, 40%, 2) 40%, 60%, 3) 28.57%, 71.42%, 4) 71.42%, 28.57%, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 19. The molar specific heat of a gas at constant, volume is 20 Joule mol-1 K-1. The value of γ, for it will be, 1), , 11, 10, , 2), , 7, 5, , 3), , 5, 3, , 4), , 3, 2, , 20. The specific heat of air at constant pressure, is 1.005 kJ/kg/K and the specific heat of air, at constant volume is 0.718 kJ/kg/K. If the, universal gas constant is 8.314 kJ/k mole K, find the molecular weight of air ?, 1) 28.97 2) 24.6, 3) 22.8 4) 19.6, 21. Calculate the specific heat of a gas at, constant volume from the following data., Density of the gas at N.T.P.=19×10-2 kg/m3,, (Cp/Cv) = 1.4, J = 4.2 × 103 J/kcal; atmospheric, pressure = 1.013 × 105 N/m2. ( in kcal / kg k ), 1) 2.162 2) 1.612, 3) 1.192, 4) 2.612, 22. If the ratio of specific heats of neon is 1.667, and R = 8312 J/k mole K find the specific heats, of neon at constant pressure and constant, volume. (Molecular weight of neon =20.183), 1) 1.029, 0.6174, 2) 1.831, 0.921, 3) 1.621, 0.421, 4) 0.862, 0.246, , DIFFERENT THERMO, DYNAMICAL PROCESS, 23. One mole of an ideal gas expands, isothermally to double its volume at 27°C., Then the work done by the gas is nearly, 1) 2760 cal 2) 414 cal 3) 1380 cal 4) 600 cal, 24. One mole of an ideal gas expands at a, constant temperature of 300 K from an initial, volume of 10 litres to a final volume of 20, litres. The work done in expanding the gas is, (R = 8.31 J/mole –K) ( in joules), 1) 750, 2) 1728, 3) 1500, 4) 3456, 25. The isothermal bulk modulus of a perfect gas, at normal pressure is, 1) 1.013 ×105 N / m2, 2) 1.013 ×106 N / m2, 3) 1.013 × 10−11 N / m2, 4) 1.013 × 1011 N / m2, 26. A gas for which γ =1.5 is suddenly compressed, to 1/4 th of the initial volume. Then the ratio, of the final to initial pressure is, 1) 1:16 2) 1:8, 3) 1:4, 4) 8:1, 27. The pressure and density of a monoatomic, gas ( γ = 5/3) change adiabatically from, d, , P2, 2, (P1, d1) to (P2, d2). If d = 8 then P should be, 1, 1, , NARAYANAGROUP, , 1), , 1, 32, , 2) 32, , 3) 128, , 4), , 1, 8, , 28. Air is filled in a motor tube at 270C and at a, pressure of 8 atmospheres. The tube suddenly, bursts, then temperature of air is [Given γ, of air = 1.5], 1) 27.50C, 2) 750C, 3) 150 K 4) 1500C, 29. A mono atomic gas initially at 27 0 C is, compressed adiabatically to one eighth of its, original volume. The temperature after, compression will be, 1) 100C, 2) 8870C, 3) 9270C 4) 1440C, 30. One gm mol of a diatomic gas ( γ = 1.4) is, compressed adiabatically so that its, temperature rises from 270C to 1270C. The, work done will be, 1) 2077.5 joules, 2) 207.5 joules, 3) 207.5 ergs, 4) 205.5 joules, 31. A container of volume 2m3 is divided into two, equal compartments, one of which contains, an ideal gas at 400 K. The other compartment, is vacuum. The whole system is thermally, isolated from its surroundings. The partition, is removed and the gas expands to occupy, the whole volume of the container. Its, temperature now would be, 1) 400 K 2) 250 K, 3) 200 K 4) 100 K, 32. At 27oC and pressure of 76 cm of Hg the, volume of a diatomic gas is 2000 cm3. If it is, compressed adiabatically to a volume of 1000, cm3, what are its pressure and temperature?, ( γ =1.4), 1) 200.5 cm of Hg, 122.9oC, 2) 180.4 cm of Hg, 84.2oC, 3) 120 cm g kg 80oC, 4) 162.4 cm of Hg 92oC, 33. The work done on a gas when it is compressed, isothermally at 27oC to half of the initial, volume is (nearly), 1) 3436 J, 2) -1718 J, 3) +1718 J, 4) -3436J, , HEAT ENGINE, 34. A Carnot engine has the same efficiency, between 800 K to 500K and x K to 600 K., The value of ‘x’ is, 1) 1000 K 2) 960 K 3) 846 K 4) 754 K, 35. A Carnot’s engine working between 270 C and, 1270 C takes up 800 J of heat from the, reservoir in one cycle. What is the work done, by the engine, 1) 100J 2) 200J, 3) 300 J 4) 400 J, 61
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JEE- ADV PHYSICS-VOL- VI, 2., , A steel ball of mass 0.1kg falls freely from a, height of 10m and bounces to a height of 5.4m, from the ground. If the dissipated energy in, this process is absorbed by the ball, the rise, in its temperature is (specific heat of steel, 460 Jkg −1K −1 , g = 10ms −2 ), 1) 0.01 0C, , 3., , 4., , 5., , THERMODYNAMICS, , 2) 0.1 0C, , 3) 1 0C, , 4) 1.1 0C, , A lead bullet (specific heat= 0.032cal / gm 0C ), is completely stopped when it strikes a target, with a velocity of 300m/s. The heat generated, is equally shared by the bullet and the target., The rise in temperature of bullet will be, 1) 16.7 0 C 2) 1.670 C 3) 167.4 0 C 4) 267.4 0 C, A block of ice falls from certain height and, completely melts. If only 3/4th of the energy, is absorbed by the block, the height of the fall, , should be ( L = 363SI unitsand g =10ms−2 ), 1) 48.4m 2) 84.4m, 3) 88.4m 4) 44.8m, A lead bullet of mass 21g travelling at a speed, of 100 ms −1 comes to rest in a wooden block. If, no heat is taken away by the wood, the rise in, temperature of the bullet in the wood nearly is, (Sp. heat of lead 80cal/kg 0 C ), 1) 250 C 2) 280 C 3) 330 C, 4) 150 C, , FIRST LAW OF THERMODYNAMICS, 6., , 7., , 8., , 9., , When 20J of work was done on a gas, 40J of, heat energy was released. If the initial internal, energy of the gas was 70J, what is the final, internal energy?, 1) 50J, 2) 60J, 3) 90J, 4) 110J, , CP,CV AND THEIR RELATIONS, , 1), , 4H, 3, , 2), , 5H, 3, , 3) 2H, , 4), , 7H, 3, , 1, mole of Helium gas is contained in a, 2, container at S.T.P. The heat energy needed to, double the pressure of the gas, keeping the, volume constant (heat capacity of the, gas= 3 Jg − 1 K − 1 ) is, 1) 3276J, 2) 1638J, 3) 819J 4) 409.5J, 11. How much heat energy in joules must be, supplied to 14gms of nitrogen at room, temperature to rise its temperature by 400 C, , 10., , at constant pressure? (Mol.wt.of N 2 =28gm,, R=constant), 1) 50R 2) 60R, 3) 70R, 4) 80R, 12. The volume of 1kg of hydrogen gas at N.T.P., is 11.2 m3 . Specific heat of hydrogen at, constant volume is 100.46J kg −1K −1 . Find the, specific heat at constant pressure in J kg −1K −1 ?, 1) 120.2, 2) 142.2, 3) 163.4, 4)182.3, 13. 3 moles of a monoatomic gas requires 60cal, heat for 50 C rise of temperature at constant, volume, then heat required for 5 moles of, same gas under constant pressure for 100 C, rise of temperature is (R=2 cal/mole-k), 1) 200cal 2) 400cal 3) 100cal 4) 300cal, 14. One mole of a monoatomic gas is mixed with, one mole of a diatomic gas. What will be the, value of γ ., 1) 1.5, 2) 1.54, 3) 1.4, 4) 1.45, , A quantity of heat ‘Q’ is supplied to a, monoatomic ideal gas which expands at, constant pressure. The fraction of heat that, goes into workdone by the gas is, 1) 2/5, 2) 3/5, 3) 2/3, 4) 1, For hydrogen gas C p − Cv = a and for Oxygen, , 15. The triatomic gas is heated isothermally. What, percentage of the heat energy is used to, increase the internal energy ?, 1)0%, 2) 14%, 3)60%, 4)100%, , gas C p − Cv = b , where C p and Cv are molar, specific heats. Then the relation between ‘a’, and ‘b’ is, 1) a=16b 2) b=16a 3) a=4b, 4) a=b, The H calories of heat is required to increase, temperature of one mole of monoatomic gas, from 200 C to 300 C at constant volume. The, quantity of heat required to increase the, temperature of 2 moles of a diatomic gas from, 200 C to 250 C at constant volume is, , 16. One mole of an ideal gas ( γ = 7 / 5 ) is, adiabatically compressed so that its, temperature rises from 270 C to 350 C . The, work done by the gas is (R=8.47J/mol-K), 1)-160J 2)-168J, 3)150J, 4)120J, 17. The tyre of a motor car contains air at 150 C if, the temperature increases to 350C, the, approximate percentage increase in pressure, is (ignore the expansion of tyre), 1) 7, 2) 9, 3) 11, 4) 13, , NARAYANAGROUP, , DIFFERENT THERMODYNAMIC, PROCESSES, , 63
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 18. A given mass of a gas is compressed, isothermally until its pressure is doubled. It is, then allowed to expand adiabatically until its, original volume is restored and its pressure is, then found to be 0.75 of its initial pressure., The ratio of the specific heats of the gas is, approximately, 1) 1.20 2) 1.41, 3) 1.67, 4) 1.83, 19. One mole of oxygen is heated at constant, pressure starting at 00 C . The heat energy, that must be supplied to the gas to double its, volume (R is the molar gas constant) is, 1) 2.5 × 273 × R, 2) 3.5 × 273 × R, 3) 2.5 × 546 × R, 4) 3.5 × 546 × R, 20. The equation of a certain gas can be written, as, , 3, 1) R, 2, , 5, 2) R, 2, , 7, 3) R, 2, , HEAT ENGINE, , 24., , 25., , 64, , condenser has a temperature of 270 C . The, theoretical coefficient of performance is, 1) 5, 2) 8, 3) 6, 4)6.5, A Carnot’s engine whose sink is at a, temperature of 300K has an efficiency of 40%., By how much should the temperature of the, source be increased so as to increase the, efficiency to 60%?, 1)250K 2)275K, 3)300K 4)325K, A refrigerator placed in a room at 300K has, inside temperature 200K. How many calories, of heat shall be delivered to the room for each, 2KiloCal of energy consumed by the, refrigerator ideally ?, 1) 4K.cal, 2)2K.cal 3)8K.cal 4)6Kcal, An ideal Carnot’s engine whose efficiency is, 40% receives heat at 500K. If the efficiency, is to be 50% then the temperature of sink will, be, 1)600K, 2)800K, 3)1000K, 4)250K, Two Carnot engines A and B are operated in, succession. The first one, A receives heat from, a source at T1=800Kand rejects to a sink at, , 2) 2, 8) 4, 14) 1, 20) 2, , 3) 3, 9) 2, 15) 1, 21) 1, , 4) 1, 10) 2, 16) 2, 22) 1, , 5) 4, 11) 3, 17) 1, 23) 4, , 6) 1, 12) 2, 18) 2, 24) 4, , LEVEL - II (C.W)-HINTS, 1., , 1 2, 1, mv = mS ∆T ;But mv 2 = mgh, 2, 2, 1, ⇒ v 2 = 2 gl sin θ ; × 2 gl sin θ = S ∆T, 2, , 2., , ∆h = h1 − h2 ;, , 3., , 1 1 2, × mv = JmS ∆T, 2 2, , 5., , 1 2, mv = JmS ∆T, 2, , 7., , dW, 1, = 1−, dQ, γ, , 4) 2R, , 21. In a mechanical refrigerator, the low, temperature coils are at a temperature of, −230 C and the compressed gas in the, , 23., , LEVEL - II (C.W) - KEY, 1) 3, 7) 1, 13) 4, 19) 2, 25) 1, , T7 / 5, = cons tan t . Its specific heat at, P2 / 5, , constant volume will be, , 22., , T2K. The second engine B receives heat, rejected by the first engine and rejects to, another sink at T3=300K. If the efficiencies, of two engines are equal, then the value of T 2, is, 1)489.4K, 2)469.4K 3)449.4K 4)429.4K, , 8., , mg ∆h = mS ∆T, 4., , 3, mgh = JmL, 4, , 6. dQ = U f −U i + dW, , Both are diatomic gases and CP − CV = R for all, gases, , 9., , dQV = nCV dT ⇒, , 10., , ( dQ )V, , dQ1 n1C1 dT1, =, dQ2 n2C2 dT2, , = nCV dT, , 12. cP = cV + r where r =, , 11. ( dQ ) P = nCP dT, PV, mT, , 13. ( dQ )V = nCV dT, CP − CV = R ⇒ CP = CV + R, , ( dQ ) P = nCP dT, n1γ 1 + n2γ 2, n1 + n2, 15. Isothermal process, dU = 0, , 14. γ =, , 16. dW =, , nR (T1 − T2 ), γ −1, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , ∆T, ∆P, × 100 =, × 100, 17. PαT ⇒, T, P, ∆P, ⇒, ×100 = 7 ( approximately ), P, 18. For isothermal process , PV, 1 1 = PV, 2 2, γ, γ, For adiabatic process, PV, 1 1 = PV, 2 2, , 3., , 4., , γ, , V 3 , 2P = P V γ, 2 4 , , 5., , T2 V2, 19. T = V ;Heat to be supplied = nC p (T2 − T1 ), 1, 1, , R, 20. P1−γ T γ = constant ; CV =, γ −1, Q2, T2, 21. β = W = T − T, 1, 2, , T2, T2, 22. η1 = 1 − ; η2 = 1 − 1, T1, T1, , T2, T3, 25. 1 − T = 1 − T, , 1, , 1, , 2, , LEVEL-II (H.W), A copper block of mass 5kg slides down along, a rough inclined plane of inclination 300 with a, constant speed. The increase in the, temperature of the block as it slides down, through 100cm assuming that the loss of, mechanical energy goes into copper block as, thermal energy. (specific heat of copper, 420 J kg −1K −1 , g = 10ms −2 ), 1) 1.19 × 10−3 0 C, 2., , 2) 1.0 × 10−3 0C, , 3) 1.5 × 10−3 0C, , 4) 2.5 × 10−3 0C, , 7., , The density of a substance is 400kgm −3 and, , 8., , that of another substance is 600kgm−3 . The, heat capacity of 40cc of first substance is, equal to that of 30cc of second substance. The, ratio of their specific heats is, 1) 1: 6, 2) 6:1, 3) 9:8, 4)8:9, For hydrogen cP − cV = m and for nitrogen, , 2) 2.38 × 10−3 0C, , 3) 1.19 × 10−2 0 C, 4) 2.38 × 10 −2 0 C, A brass sphere of mass 0.2kg falls freely from, a height of 20m and bounces to a height of 8m, from the ground. If the dissipated energy in, this process is absorbed by the sphere the rise, in its temperature is (specific heat of brass =, 360 J kg −1K −1 , g = 10ms −2 ), , In a thermodynamic process the pressure of a, fixed mass of gas is changed. In this process, gas releases 20J heat and 8J work is done on, the gas. If initial internal energy of the gas is, 30J, then final internal energy is, 1) 2J, 2) 42J, 3)18J, 4) 58J, , CP,CV AND THEIR RELATIONS, , JOULE’S LAW, 1., , 1) 0.5 × 10 −3 0C, , FIRST LAW OF THERMODYNAMICS, 6., , Q2, T2, 23. β = W = T − T, 1, 2, T2, 24. η = 1 − T, , A lead bullet of 10g travelling at 300m/s, strikes against a block of wood and comes to, rest. Assuming 50% of heat is absorbed by, the bullet, the increase in its temperature is, (sp-heat of lead is 150J/Kg-K), 1) 1000C, 2) 1250C 3) 1500C 4) 2000C, Water falls from a height 500m, what is the, rise in temperature of water at bottom if whole, energy remains in the water ? (J=4.2), 1) 0.96 0C 2) 1.02 0 C 3) 1.16 0 C 4) 0.3 0 C, A ball is dropped on a floor from a height of, 2m. After the collision it rises up to a height, of 1.5m. Assume that 40% of the mechanical, energy lost goes as thermal energy into the, ball. Calculate the rise in temperature of the, ball in the collision. (Heat capacity of the ball, is 800JK −1 ), , 9., , cP − cV = n , where cP and cV refer to specific, heats per unit mass respectively at constant, pressure and constant volume. The relation, between 'm' and 'n' is, 1) n=14m, 2) n=7m 3) m=7n 4) m=14n, 294 joules of heat is required to rise the, temperature of 2 mole of an ideal gas at, constant pressure from 300C to 350C. The, amount of heat required to rise the, temperature of the same gas through the, same range of temperature at constant, volume (R=8.3 Joules/mole-K) is, 1)12.6J, 2)211J, 3)29.4K, 4)37.8J, , 1) 0.33 0 C 2) 0.66 0C 3) 0.77 0C 4) 0.88 0 C, NARAYANAGROUP, , 65
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THERMODYNAMICS, 10. 1672cal of heat is given to one mole of oxygen, at 00 C keeping the volume constant. Raise in, temperature is (cP=0.2cal/gm0k and R=2cal/, mole/K), 1) 33.60 C 2) 36.30 C 3) 63.30 C 4) 334.40 C, 11. 0.5 mole of diatomic gas at 270 C is heated at, constant pressure so that its volume is tripled., If R=8.3J mole-1k-1then work done is, 1) 4980J, 2) 2490J 3) 630J, 4) 1245J, 12. The volume of 1kg of oxygen gas at NTP is, 0.7m3 . Specific heat of oxygen at constant, volume is 653 Jkg −1k −1 . The specific heat of, oxygen at constant pressure in Jkg −1k −1 is, (atmospheric pressure 105Pa), 1) 713, 2) 813, 3)913, 4)1013, 13. When an ideal diatomic gas is heated at, constant pressure, its internal energy is, increased by 50cal then the work done by the, gas is, 1) 30cal 2)50cal, 3)70cal, 4)20cal, 14. A gaseous mixture consists of 16g of helium, and 16g of oxygen. The ratio C p / CV of the, mixture is, 1) 1.4, 2) 1.54, 3) 1.59, 4)1.62, , DIFFERENT THERMODYNAMIC, PROCESS, 15. 0.1 moles of diatomic gas at 270 C is heated at, constant pressure, so that the volume is, doubled.If R=2 cal.mol −1 k −1 , the work done is, 1) 150cal, 2) 60cal 3) 40cal, 4)30cal, 16. If a triatomic gas is heated at constant pressure,, percentage of the heat energy which is used to, increase the internal energy is, 1) 75%, 2) 14%, 3) 60%, 4)100%, 17. Two moles of an ideal monoatomic gas at, 270 C occupies a volume of V. If the gas is, expanded adiabatically to the volume 2V, then, the work done by the gas will be, 1) −2767.23J, 2) 2767.23J, 3) 2500J, 4) −2500J, 18. One mole of an ideal gas with γ = 1.4 is, adiabatically compressed so that its, temperature rises from 270C to 350C. The, change in the internal energy of the gas is, , ( R = 8.3 J mol, 66, , −1, , k −1 ), , JEE- ADV PHYSICS-VOL- VI, 1)-166J 2)166J, 3)168J, 4)-168J, 19. The volume of air increases by 5% in its, adiabatic expansion. The percentage, decrease in its pressure will be, 1)5%, 2)6%, 3)7%, 4)8%, 20. Certain perfect gas is found to obey the law, PV3/2 = constant, during adiabatic process. If, such a gas at initial temperature T is, adiabatically compressed to half of the initial, volume, its final temperature will be, 1) 2T, 2) 2T, 3) 2 2T, 4) 4T, , HEAT ENGINE, 21. The coefficient of performance of a Carnot, refrigerator working between 300 C & 00 C is, 1) 10, 2) 1, 3) 9, 4) 0, 22. A Carnot engine has efficiency of 40% when, sink temperature is 300K. In order to, increase the efficiency by 50%, the source, temperature will have to be increased by, 1) 2750K 2) 3250K, 3)3800K 4)2500K, 23. A refrigerator works between 30C and 400C., To keep the temperature of the refrigerator, constant, 600 calories of heat is to be removed, every second. The power required is, 1) 33.78Watt, 2) 337.8 Watt, 3) 7.77Watt, 4) 10.77Watt, 24. Two Carnot engines ‘A’ and ‘B’ are operated, in succession. The first one, A receives heat, from a source at T1 = 800 K and rejects to a, sink at T2 K. The second engine B receives, heat rejected by the first engine and rejects, to another sink at T3 = 300 K . If the work, outputs of two engines are equal, then the, value of T2 is, 1)100K, 2)300K, 3)550K, 4)700K, 25. One of the most efficient engines ever, developed operates between 2100K and, 700K. Its actual efficiency is 40%. What, percentage of its maximum possible efficiency, is this?, 1) 40%, 2) 60%, 3)66.67% 4)33.37%, LEVEL - II (H.W ) - KEY, 1) 3 2) 1 3) 3 4) 3 5) 4 6) 3, 7) 3 8) 4 9) 2 10) 4 11) 2 12) 3, 13) 4 14) 4 15) 2 16) 1 17) 2 18) 2, 19) 3 20) 1 21) 3 22) 4 23) 2 24) 3, 25) 3, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 7., , 8., , 9., , One cubic meter of an ideal gas is at a pressure, of 105 N / m 2 and temperature 300K. The gas, is allowed to expand at constant pressure to, twice its volume by supplying heat. If the, change in internal energy in this process is, 104 J, then the heat supplied is, 1) 105 J 2) 10 4 J 3) 11× 10 4 J 4) 2.2 ×105 J, 1kg of water and 1kg of steel are heated, through 1K. The change in their internal, energies are, (Specific heat of steel = 460 J kg-1K-1 ;, Specific heat of water = 4200Jkg-1K-1), 1) 460 J, 4200 J, 2) 4200 J, 460 J, 3) 1000 J, 4200 J, 4) 460 J, 1000 J, Consider the melting of 1g of ice at 00C to water, at 00C at atmospheric pressure. Then the, change in internal energy of the system, (density of ice is 920kg/ m3) ?, 1) 334 J 2)420 J, 3) 540 J, 4)680 J, , CP, CV AND THEIR RELATIONS, 10. A cylinder of fixed capacity 67.2 litres, contains helium gas at S.T.P. The amount of, heat required to rise the temperature of the, gas by 150C is (R =8.31 J mol-1K-1), 1) 520 J 2) 560.9 J, 3) 620 J 4) 621.2 J, 11. 14 g of N 2 gas is heated in a closed rigid, container to increase its temperature from, 230 C to 430 C . The amount of heat supplied, to the gas is, 1) 25 cal 2) 50 cal 3) 100 cal 4) 30 cal, 12. 70 cal of heat is required to rise the, temperature of 2 moles of an ideal gas at, constant pressure from 30ºC to 35ºC. What is, the amount of heat required to rise the, temperature of same gas through the same, range at constant volume?(R = 2cal mole-1K-1), 1) 28 J 2) 50 Cal, 3) 75 J, 4) Zero, 13. The relation between internal energy U,, pressure P and volume V of a gas in an, adiabatic process is : U = a + bPV Where ‘a’, and ‘b’ are constants. What is the value of the, ratio of the specific heats?, a, 1), b, , 68, , b +1, 2), b, , a +1, 3), a, , b, 4), a, , 14. The ratio of specific heats of a gas is γ . The, change in internal energy of one mole of the, gas when the volume changes from V to 2V, at constant pressure “P” is, PV, , 1) γ − 1, , PV, , 3) γ − 1, , 2) PV, , 4) γ, , DIFFERENT THERMODYNAMIC, PROCESSES, , 15. γ for a gas is 5/3. An ideal gas at 270C is, compressed adiabatically to 8/27 of its, original volume. The rise in temperature of, the gas is, 1) 4500C, 2) 3750C 3) 2250C 4) 4020C, 16. One mole of a gas expands with temperature, T such that its volume, V = kT2, where k is a, constant. If the temperature of the gas, changes by 600C then the work done by the, gas is, 1) 120 R 2) R ln 60 3) kR ln 60 4) 60 kR, 17. A monoatomic ideal gas, initially at, temperature T1, is enclosed in a cylinder fitted, with a frictionless piston. The gas is allowed, to expand adiabatically to a temperature T 2, by releasing the piston suddenly. If L 1 and L 2, are the lengths of the gas column before and, after expansion respectively, then T1/T2 is, given by, 2, , 1), , L1 3, , , L2 , , L1, , 2) L, 2, , L2, , 3) L, 1, , 2, , 4), , L2 3, , L1 , , 18. Three samples of the same gas 'x' ,' y' and, 'z', for which the ratio of specific heats is γ =3/, 2, have initially the same volume. The, volumes of each sample is doubled , by, isobaric process in the case of 'y' and by, isothermal process in the case of 'z'. If the, initial pressures of the samples 'x', 'y' and 'z', are in the ratio 2 2 : 1 : 2 , then the ratio of, their final pressures is, 1) 2 : 1 : 1 2) 1 : 1 : 1 3) 1 : 2 : 1 4) 1 : 1 : 2, 19. n moles of an ideal gas undergo a process in, which the temperature changes with volume, as T = kV2. The work done by the gas as the, temperature changes from T0 to 4T0 is, 1) 3nRT0, , 5, , , 2) nRT0, 2, , 3, , 3) nRT0 4) Zero, 2, NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 20. 'm' grams of a gas of a molecular weight M is, flowing in an isolated tube with velocity 2v. If, the gas flow is suddenly stopped the rise in its, temperature is ( γ = ratio of specific heats;, R = universal gas constant; J = Mechanical, equivalent of heat), 1), 3), , 2 Mv 2 ( γ − 1), RJ, , mv 2 γ, 2 RJ, , 2), 4), , mv 2 ( γ − 1), M 2 RJ, , Mv2 γ, 2 RJ, , 21. Heat is supplied to a diatomic gas at constant, pressure. The ratio of ∆ Q: ∆ U: ∆ W is, 1)5 : 3 : 2 2) 5 : 2 : 3 3)7 : 5 : 2 4)7 : 2 : 5, 22. A given quantity of an ideal gas at pressure P, and absolute temperature T obeys P α T 3, during adiabatic process. The adiabatic bulk, modulus of the gas is, 2, 3, 1) P 2) P, 3) P, 4) 2P, 3, 2, 23. An ideal gas is taken through a cyclic thermo, dynamical process through four steps. The, amounts of heat involved in these steps are, Q1 = 5960 J , Q2 = −5585 J , Q3 = −2980 J , Q4 = 3645 J ;, respectively, The corresponding works, involved are W1 = 2200 J , W2 =−825 J , W3 =−1100 J, and W4 respectively. Find The value of W4, and efficiency of the cycle, 1) 1315 J, 10%, 2) 275 J, 11%, 3) 765 J, 10.82%, 4) 675 J , 10.82%, , HEAT ENGINE, 24. A Carnot's engine is made to work between, 2000C and 00C first and then between 00C and, –2000C. The ratio of efficiencies of the engine, in the two cases is, 1) 1.73:1, 2) 1:1.73, 3) 1:1, 4) 1 : 2, 25. A scientist says that the efficiency of his heat, engine which operates at source temperature, 1270C and sink temperature 270C is 26%, then, 1) it is impossible, 2) it is possible but less probable, 3) it is quite probable, 4) data is incomplete, 26. Efficiency of a Carnot engine is 50% when, temperature of outlet is 500 K. In order to, increase efficiency upto 60% keeping, temperature of intake the same, what is the, temperature of outlet, 1) 200 K, 2) 400 K 3) 600 K 4) 800 K, NARAYANAGROUP, , 27. An ideal refrigerator has a freezer at a, temperature of –130C. The coefficient of, performance of the engine is 5. The, temperature of the air (to which heat is, rejected) will be, 1) 3250C, 2) 325 K 3) 390C, 4) 3200C, 28. The heat reservoir of an ideal Carnot engine, is at 800 K and its sink is at 400 K. The amount, of heat taken in it in one second to produce, useful mechanical work at the rate of 750 J is, 1)2250 J, 2)1125 J, 3)1500 J 4) 750 J, 29. A Carnot engine works between 2000 C and, 00C. Another Carnot engine works between, 00C and - 2000 C . In both cases the working, substance absorbs 4 kilocalories of heat from, the source. The efficiency of first engine will, be, 100, 200, 173, 273, 2), 3), 4), 173, 473, 273, 373, 30. In the above problem, the output of second, engine is, 1) 29.3 × 103 Cal, 2) 12.3 × 103 Cal, , 1), , 3) 12.3 × 103 joule, 4) 2.93 × 103 joule, 31. In the above problem, the ratio of outputs of, two engines is, 1)0.577, 2) 0.377, 3)0.777, 4)0.177, , GRAPHS, 32. An ideal monoatomic gas is taken round the, cycle ABCDA as shown in the diagram. The, work done during the cycle is, C(2P,2V), , B(2P,V), , 1) PV, P, 2) 2PV, A(P,V), D(P,2V), 3) 3PV, 4) 4PV, V, 33. The figure shows P-V graph of an ideal one, mole gas undergone to cyclic process ABCA,, then the process B → C is, , 2P0, , 1) Isobaric, , B, , 2) Adiabatic, P, , P0, , A, V0 V, , C, 2V0, , 3) Isochoric, 4) Isothermal, 69
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 34. On a T-P diagram, two moles of ideal gas, perform process AB and CD. If the work done, by the gas in the process AB is two times the, work done in the process CD then what is the, value of T1/T2?, T, A, , T1, T2, , volume, (litres) 10, 10, 30, Pressure (KPa), , D, P, , B, , 1)103 π J, 2) 10 2 π J 3) 10 4 π J 4) 107 π J, 39. A thermodynamic system is taken through the, cycle PQRSP process. The net work done by, the system is, , (4P,3V), , P, , 1) Zero, 2) 3PV, 3) 6PV, P A (P,V) C (P,3V), 4) 9PV, V, In the given elliptical P - V diagram, , 300kPa, 100kPa, , S, , R, , Q, , P, 100cc, , 300cc, , V, , 1) 20 J 2) – 40 J, 3) 400 J, 4) – 374 J, 40. A cyclic process performed on one mole of an, ideal gas. A total 1000 J of heat is withdrawn, from the gas in a complete cycle. Find the work, done by the gas during the process B → C., , P2, P1, , P, V1, , V2, , V, , T, , V, , 1) The work done is positive, 2) The change in internal energy is non-zero, π , 3) The work done = − 4 ( P2 − P1 )(V2 − V1 ), , , 4) The work done = ( π )(V2 − V1 ) 2 = π ( P2 − P1 )2, 37. A system changes from the state (p1,v1) to, (p2,v2) as shown in the diagram. The workdone, by the system is, , 400 K, 300 K, , 1) -1531 J, 2) -1631 J, 3) -1731 J, 4) -1831 J, , B, A, V, , 41. An ideal gas is taken through A → B →C → A,, as shown in figure. If the net heat supplied to, the gas in the cycle is 5J, the work done by, the gas in the process C → A is, , (P2 ,V2 ), , 2, , C, , B, , 3, , (P1,V1 ), 1, , 1)12x104J, , C, , V(m ), , 5, P 4, 3, 5, −2, 2, ×10 Nm, 1, , 70, , 30, , B, , C, , 1) 1/2, 2) 1, 3) 2, 4) 4, 35. A Sample of an ideal monoatomic gas is taken, round the cycle ABCA as shown in the figure., The work done during the cycle is, , 36., , 38. The heat energy absorbed by a system in going, through a cyclic process shown in figure is, , 1, , 2 3 4 5 3, V(m ), , 2)12x108J, , 3)12x105J 4) 6x104 J, , A, , -2, , P(Nm ), , 1) -5J, , 2) -10J, , 10, , 3) -15J, , 4) -20J, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, 3., , THERMODYNAMICS, , Read the following and write the correct pairs., Pressure temperature graph of n moles of an, ideal gas is shown ( ρ - density)., , 6., , C, , 4P0, , 7., , 2P0, P0, , B, A, T0, , Column-I, , D, 8., , 2T0, Column-II, , A) ρ-V graph p), , 3, , 4, , 2, , 1, , 4, , B) P-V graph q), , 1, , 3, , 2, , C) ρ-P graph, , r), , 9., , 1,3, 2,4, , 10., , D) V-T graph s), , 3, , 4, , 2, , 1, , 11., ASSERTION & REASON TYPE QUESTIONS, 1) Both assertion (A) and reason (R) are correct, and R gives the correct explanation, 2) Both assertion (A) and reason (R) are correct, but R does not give the correct explanation, 3) A is true but R is false, 4) Both A and R are false, 4. Assertion (A) : Two systems which are in thermal, equilibrium with a third system are in thermal, equilibrium with each other., Reason (R) : The heat flows always from a system, at higher temperature to a system at a lower, temperature, 5. Assertion (A) : According to the principle of, conservation of energy total heat can be converted, into mechanical work, NARAYANAGROUP, , 12., , 13., , Reason (R) : Due to various losses, it is, impossible to convert total heat into mechanical, work, Assertion (A) : According to Joule, heat and work, are related, Reason (R) : For every 1 cal. of heat we can get, 4.186 J of mechanical work., Assertion (A) : Reversible systems are difficult, to find in real world, Reason (R) :Most processes are dissipative in, nature, Assertion (A): Thermodynamic process in nature, are irreversible, Reason (R) :Dissipative effects can not be, eliminated, STATEMENT TYPE QUESTIONS, Options :, 1. Statement 1 is true and statement 2 is true, 2. Statement 1 is true and statement 2 is false, 3. Statement 1 is false and statement 2 is true, 4. Statement 1 is false and statement 2 is false, Statement-1: First law of thermodynamics, specifies the conditions under which a body can, use its heat energy to produce the work., Statement-2:Second law of thermodynamics, states that heat always flows from hot body to, cold body by itself, Statement-1: Zeroth law of thermodynamics, gives us the concept of energy, Statement-2: Internal energy is dependent on, temperature, Statement-1: Heat given to an ideal gas under, isothermal conditions is used completely to do, external work., Statement-2: The change in internal energy in a, thermodynamic process is independent of the path., Statement-1:It is impossible to derive continuous, supply of work by cooling a body to a temperature, lower than that of the coldest of its surroundings, Statement-2: Heat engine can convert whole of, the heat energy supplied to it into useful work, Statement-1:: Monoatomic, diatomic and, polyatomic gases are adiabatically compressed, P2 , , such that compression ratio is P . Then, 1, monoatomic gas will have maximum final volume, out of these three gases., Statement-2:Monoatomic gas has least degree, of freedom., 73
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 14. Statement-1: Two vessels A and B are, connected to each other by a stopcock .Vessel A, contains a gas at 300K and 1 atmosphere pressure, and vessel B is evacuated. The two vessels are, thermally insulated from the surroundings. If the, stopcock is suddenly opened, the expanding gas, does no work., Statement-2:Since ∆ Q = 0 and as the gas, expands freely so ∆W = 0 and from the first law, of thermodynamics it follows that ∆U is also zero, for the above process., , OTHER MODEL QUESTIONS, 15. According to second law of thermodynamics, Statement - I:All heat can be converted into work, Statement - II: The efficiency of a heat, engine is always lesser than unity, Statement - III: It is not possible to transfer, heat from lower to higher temperature of it self, 1)both I and II are true, 2)both II and III are true, 3)both I and III are true 4) I, II, III are true, 16. The second law of thermodynamics is the, generalisation of the fact that, Statement - I:Heat always flows from hot body, to cold body by itself, Statement - II: Heat can flow from cold body to, hot body itself, Statement - III: It is impossible for a self acting, machine unaided by any external agency to transfer, heat from cold body to hotter body, 1) I & II 2) II & III, 3) I & III 4) I, II & III, 17. The V–T diagram of an ideal gas for the process, A → B → C (straight lines) is as shown in the, figure. In the process A → B → C, A, , PARAGRAPH TYPE QUESTIONS, Read the question to answer Q. 18 to 21 :, A monoatomic ideal gas sample is given heat, Q. One fourth of this heat is used as work, done by the gas and rest is used for, increasing its internal energy., 18. The molar specific heat for the gas in this, process is, 3, R, 3) 2R, 4) 3R, 1) R 2), 2, 2, 19. The equation of process in terms of, volume and temperature is, V, V, 1), = constant, 2), = constant, T, T, 3) VT = constant, 4) V T = constant, 20. The P V diagram for the process is, P, 1), , V, , V, P, , P, 2), , 4), V, , V, , 21. An ideal gas under goes a thermodynamic, cycle as shown in fig. Which of the following, graphs represents the same cycle?, V, , C, , B, , A, , O, , T, A, , C, , v, , P, 3), , 1), , B, , B, , P, , P, , 3), , C, , A, , C, , B, T, , A) pressure is always increasing, B) for some interval pressure decreases but, finally pressure is more than initial pressure, C) pressure first increases then remains, constant, D) graph AB is unpredictable about pressure, 1) A, B, C are correct, 2) B, D are correct, 3) C only correct, 4) B only correct, 74, , O, , P, 2), , T, , A, , O, , P, , B, , B, , A, , 4), C, V, , 1) A, C, , T, , 2) B,D, , C, V, , 3) A,D, , 4) B,C, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, Questions No. 22 to 23, The figure shows P–V diagram of a, thermodynamic cycle, P, 2P0, , B, , C, , P0, A, O, V0, , 26. If work done by the gas in cylinder B is WB &, work done by the gas in cylinder A is W A then, 1) WA = –WB, 2) |WA| > |WB|, 3) |WA| < |WB|, 4) we can’t say anything, 27. What will be the compressive force in, connecting rod at equilibrium, 1) PS, 2) 2 PS, 3) 23/2 PS, 4) zero, Questions No. 28 and 29, The dot in figure represents the initial state, of a gas. An adiabat divides the p-V, diagram into regions 1 and 2 as shown., , D, , 3V0, , 22. The work done by the cycle is, 1) 2P0V0, 2) 3P0V0 3) P0V0 4) 6P0V0, 23. If TA, TB , TC and T D are the respective, temperature at A, B, C and D. Then, choose, the correct statement if T A = T0, 1) The maximum temperature during the cycle, occurs at C., 2) T D = 3T0 3) T B = 2T0 4) all the above, Questions No. 24 to 27, Two cylinder A and B having piston connected, by massless rod (as shown in figure). The, cross-sectional area of two cylinders are same, & equal to ‘S’. The cylinder A contains 'm' gm, of an ideal gas at Pressure P & temperature, T0. The cylinder B contain identical gas at, same temperature T 0 but has different mass., The piston is held at the state in the position, so that volume of gas in cylinder A & cylinder, B are same & is equal to V0. The walls &, piston of cylinder A are thermally insulated,, where as cylinder B is maintained at, temperature T0 . The whole system is in, vacuum. Now the piston is slowly released and, it moves towards left & mechanical equilibrium, is reached at the state when the volume of gas, in cylinder A becomes, , V0, . Then (here γ for, 2, , gas = 1.5), , B, , A, 24. The mass of gas in cylinder B, 1) 2 2 m 2) 3 2 m 3) 2 m 4) m, 25. The change in internal energy of gas in, cylinder A, 1) ( 2 – 1) PV0, 3), , THERMODYNAMICS, , PV0, ( 2 − 1), , NARAYANAGROUP, , 2) 2( 2 – 1) PV0, 4) zero, , p, 2, 1, V, 28. For which of the following processes, the, corresponding heat supplied to the system, Q is positive, 1) the gas moves up along the adiabat,, 2) it moves down along the adiabat,, 3) it moves to anywhere in region 1,, 4) it moves to anywhere in region 2., 29. As the gas moves down along the adiabatic,, the temperature, 1) increases, 2) decreases, 3) remains constant, 4) variation depends on type of gas, LEVEL - IV -KEY, Matching Type Questions, 1)a-q, b-r, c-s, d-p, 2)a-s, b-p, c-r, d-q, 3)A-r, B-q, C-p, D-s, Assertion & Reason Type Questions, 4) 1, 5) 2 6) 1 7) 1 8) 1, Statement Type Questions, 9) 3, 10) 3 11) 1 12) 2 13) 1 14) 1, More than one option questions, 15) 2, 16) 3 17) 3, Paragraph Type Questions, 18) 3, 19) 2 20) 2 21) 1 22) 1 23) 4, 24) 2, 25) 2 26) 3 27) 3 28) 4 29) 2, , LEVEL - IV-HINTS, Matching Type Questions, 3., , Process A - B is an isothermal process i.e, T= constant, Hence Pα, , 1, or P - V graph is rectangular, V, 75
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, hyperbola with increasing 'P' and decreasing V., 1, . Hence ρ - V graph is also a rectangular, V, hyperbola with decreasing V and increasing ' ρ '., ρα, , PM, ., RT, Hence ρ - P graph will be a straight line passing, through origin with increasing ρ and P.., Process B - C is an isochoric process, because, P-T graph is straight line passing through origin, V = constant, hence P - V graph will be a straight line parallel to, P - axis with increasing P., Since V = constant hence ρ will also be constant., Hence ρ -V graph will be a dot. ρ -P graph will, be a straight line parallel to P-axis with increasing, P, because ρ =constant., Process C - D inverse of A - B;D - A inverse of B - C, ρ, P, V, T, ρα P ⇒ P =, , A P0, , V0, , T0, , ρ0, , B 2P0, , V0, 2, , T0, , 2ρ0, , C 4P0, , V0, 2, , 2T0, , 2ρ0, , D 2P0, , V0, , 2T0, , ρ0, , T0 , 0, hence V0 = nR P and ρ 0 = RT, 0, 0, From the above the corresponding graphs will be, obtained, Assertion & Reason Type Questions, (A) According to the statement of zeroth law of, thermodynamics (A is true). It leads to the concept, of temperature., (R) Heat transfers from hot body to cold body on, its own (R is true).Temperature decides the, direction of flow of heat, (A) According to law of conservation of energy, (R) According to second law of thermodynamics, Both are true and different, (A) According to the statement of Joule's law, (R) 1 calorie = 4.186 Joule, Process is reversible only if it is in quasi-state, (system in equilibrium with the surroundings at, every stage) and there are no dissipative effects, which is difficult to find in real world., Dissipative forces cannot be eliminated., , PM, , 4., , 5., , 6., 7., , 76, , 8., , Thermodynamic process in nature are irreversible due, to the dissipative forces like friction, viscosity etc., In nature dissipative forces are present every, where hence they can not be eliminated, Statement Type Questions, 9. 1)It does not tell any thing about the conditions, under which heat can be transformed into work.(I, is false), 2) According to to the statement of second law of, thermodynamics (II is true), 10. 1) Zeroth law of thermodynamics leads to the, concept of temperature (I is false), 2) Except in some exceptional cases internal, energy is a measure of temperature (II is true), 11. 1)In isothermal expansion gas absorbs heat and, does work as there is no change in internal energy, (dU = 0) dQ = dW, 2)Internal energy is a measure of temperature and, is independent of the path., 12. 1) It is true according to second law of, thermodynamics taking the example of refrigerator., Hence it is true., 2)All heat can not be converted in to work, according to Kelvin-plank statement. Hence B is, wrong., V1, , 13. Statement:1- P V γ = P V γ ⇒ V2 = (P / P ) 1 γ, 1 1, 2 2, 2, 1, γ is least for monoatomic gas. Hence its final, volume is maximum, Statement:2- monoatomic gas has 3 degrees of, freedom. The diatomic gas has 5 degrees of, freedom Polyatomic gas has 6 degrees of freedom., Hence statement 2 is correct., 14. Statement:1-Due to free expansion work done is, zero, Statement:2-According to 1st law of, thermodynamics, More than one option questions, 15. Statement:1-All heat can not be converted into, work according to Kelvin - Planck statement., Q2, , Statement:2- η = 1 − Q for Q2 = 0 , η = 1, 1, , An ideal engine with η = 1 is never possible i.e, heat released to the cold reservoir can never be, made zero. Hence η < 1, Statement:3- According to Clausius statement, which is true, , NARAYANAGROUP
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 16. Statement:1- Heat always flows from hot body, to cold body, hence it is true., Statement:2- Heat cannot transfer from cold body, to hot body on its own, hence it is wrong., Statement:3- According to Clausius statement, which is true., 17. According to the ideal gas equation PV = nRT as, temperature increases pressure first increases as, volume decreases in the case A → B, B → C , V αT hence P = constant., , Paragraph Type Questions, 18. From 1st law of thermodynamics dQ = dU + dW, 1 3, ∴dU = dQ 1 − = dQ, 4 4, , 1, 4, , But dW = dQ ;, nCV dT =, , 3, nC P dT ⇒ CP = 2 R, 4, , 1, 4, , 1, 4, , RT dV 1 dT, =, ⇒ Vα T, ;, V V, 2 T, , (∴TαV ), , 20. V α T ; PV = nRT, , )( 3V, , NARAYANAGROUP, , 0, , P0 ×3V0, = 3T0 ; TC has the maximum temperature., nR, m, RT in cylinder A ---------(1), 24. PV, 0 0 =, M, TD =, , PBVB =, , mB, RT in cylinder B ----------(2), M, γ, , V0 , From adiabatic relation PV = Pf , 2, γ, 0 0, , Pf = 2 2 P0 = PB ;, , ), , dividing (1) by (2), , 3 , , here VB = V0 , 2 , , PV, m, 0 0, ⇒, =, ⇒ mB = 3 2 m, 3, , mB, 2 2 P0 V0 , 2 , , 25. Change in internal energy of a gas in cylinder A., ∆U = −W =, , Hence it represents a straight line, 21. Process AB in the given figure is an isobaric process., During this process V αT, But PV = nRT (or ) PV ∝ T, ∴ during this process pressure P remains constant., Process BC is an isochoric process . During this, process, volume decreases and temperature, remains constant . Hence, pressure increases during, this process. Hence, on P-V diagram, process AB, will be a straight line parallel to V-axis and BC will, be a straight line parallel to P-axis and CA will be, rectangular hyperbola. Hence C is correct while D, is wrong., On P – T diagram, process AB will be straight line, parallel to T-axis, during which temperature, increases, process BC will be a straight line passing, through origin, during which temperature and, pressure both decrease and process CA will be, straight line parallel to P-axis during which pressure, increase., Hence (A) is correct while (B) is wrong., 22. Net work done by the gas during the cyclic process, is given by area enclosed, , (, , 2 P × 3V0, PV, 2 PV, 0 0, = 2T0 ; TC = 0, = 6 0 0 = 6T0, nR, nR, nR, , 2, , ⇒ PV = nRV 2 ⇒ PαV, , W = 2 P0 − P0, , TB =, , PV, 0 0, = T0, nR, , PV, m, 0 0, =, PBVB mB, , 19. dW = dQ ; PdV = nC P dT, But P =, , TA =, , 23. From ideal gas equation, , − V 0 = P0 ( 2V0 ) = 2 P0V 0, , =, , (P V, f, , f, , − PV, i i), , γ −1, , V0, − PV, 0 0, 2, =2, 3, −1, 2, , 2 2 P0, , (, , ), , 2 − 1 PV, 0 0, , 26. Work done in A, PV − Pf V f PV, 0 0 − 2 2 P0 (V0 / 2 ), WA = i i, =, 3, γ −1 , −1, 2, , (, , ), , = 2 PV, 0 0 1 − 2 = 0.828 PV, 0 0, , Vf , , Work done in B; WA = nRT0 log V , , , i, , , , 3V , 3V , 3, WA = Pf V f log 0 = 2 2 P0 0 log , 2, 2 , 2V0 , = 1.719 PV, ∴ WA < WB, 0 0, , 27. Compressive, force= Pf (area) = 2 2 P ( S ) = 2 2PS, 28. If the gas goes into the region '2' the workdone by, the gas increases as the area under the curve, increases so heat is needed to be supplied., 29. The temperature of the gas decreases as the gas, does work at the expense of internal energy., 77
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 5., , LEVEL - V, THERMODYNAMICS, SINGLE ANSWER QUESTIONS, 1., , 2., , An amount Q of heat is added to a mono atomic, ideal gas in a process in which the gas performs, a work Q/2 on its surrounding. Find equation, of the process., (A) PV1/3 = constant (B) PV −1/4 = constant, 6., (C) PV1/4 = constant (D) PV −1/3 = constant, Figure shows a cycle ABCDA undergone by 2, moles of an ideal diatomic gas. The curve AB, is a rectangular hyperbola and T1 = 300 K and, T2 = 500 K. Determine the work done by the, gas in the process A → B., , Monoatomic , diatomic and triatomic gases, whose initial volume and pressure are same,, each is compressed till their pressure becomes, twice the initial pressure. Then :, (A) if the compression is isothermal, then their final, volumes will be same, (B) if the compression is adiabatic, then their final, volumes will be different, (C) if the compression is adiabatic , then, monoatomic gas will have maximum final volume, (D) All of these, A cyclic process ABCD is shown in the p − V, diagram. Which of the following curves, represent the same process ?, P, B, , A, , V, D, , C, , C, , D, BC and DA are Hyperbolas, , A, , V, , D, , B, T1, , 3., , T2, , T, , *, R . The number in, 12, , the numerator is not readable. What may be, this number ?, (A) 25, (B) 21, (C) 41, (D) 42, 4. An ideal gas can be expanded from an initial, state to a certain volume through two different, processes,, (I) PV2 = K and (II) P = KV2 , where K is a, positive constant. Then, choose the correct, option from the following., (A) Final temperature in (I) will be greater than in (II), (B) Final temperature in (II) will be greater than in (I), (C) Work done by the gas in both the processes would, be equal, (D) Total heat given to the gas in (I) is greater than in (II), , A cyclic process is shown in the P-T diagram., Which of the curves show the same process, on aP-V diagram ?, C, , B, , p, A, O, , A), , C, , B, , p, , T, B), , C, , B, , p, A, , A, , V, C), , V, , C, , D), , p, , B, , p, B, , A, , V, 78, , B, , A, , (A) -3.326 kJ, (B) 4.326 kJ, (C) 2.326 kJ, (D) 3.326 kJ, One mole of an ideal monoatomic gas, undergoes a process defined by U = a V, where U is internal energy and V is its volume., The molar specific heat of the gas for this 7., process is found to be, , C, , C, , A, , V
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JEE- ADV PHYSICS-VOL- VI, 8., , THERMODYNAMICS, , An ideal monoatomic gas is taken round the 12. A thermodynamic system undergoes cyclic, process ABCDA as shown in figure. The work, cycle ABCDA as shown in following P-V, done by the system is, diagram. The work done during the cycle is, (3P, V), C, , (3P, 3V), D, , P, , ↑, P, , B, , 2P0, B, (P, V), , O, , 9., , C, , 3P0, A, (P, 3V), , O, , P0, , V→, , (A) PV, (B) 2PV, (C) 4PV (D) 3PV, Ideal gas is taken through the process shown, in the figure :, , D, , A, V0, , V, , 2V0, , P0 V0, (D) zero., 2, P, 13. In a cyclic process shown in the figure an ideal, C, gas is adiabatically taken from B to A, the work, B, done on the gas during the process B to A is, 30 J, when the gas is taken from A to B the, heat absorbed by the gas is 20 J. The change, A, in internal energy of the gas in the process A, T, to B is :, (A) in process AB, work done by system is positive, P, (B) in process AB, heat is rejected, A, (C) in process AB, internal energy increases, 20J, (D) in process AB internal energy decreases and in, 30J, process BC, internal energy increases., B, 10. The specific heat of solids at low temperatures, varies with absolute temperature T according, V, to the relation S = AT 3, where A is a constant., (A) 20 J (B) –30 J, (C) 50 J, D) –10 J, The heat energy required to raise the 14. The figure shows two paths for the change of, temperature of a mass m of such a solid from, state of a gas from A to B. The ratio of molar, T = 0 to T = 20 K is :, heat capacities in path 1 and 2 is :, (A) 4 × 104 mA, (B) 2 × 103 mA, P, (C) 8 × 106 mA, (D) 2 × 106 mA., 11. In the P–V diagram shown in figure ABC is a, 2, semicircle. The work done in the process ABC, A, B, is, , (A) P0 V0 (B) 2 P0 V0 (C), , 1, , P(atm), , V, , C, , 3, B, 1, , A, 1, , (A)zero, π, (C) − atm − L, 2, , 2, , V(L), , π, (B) atm − L, 2, , (D) 4 atm–L., , (A) > 1 (B) < 1 (C) 1 (D) data insufficient, 15. 3 moles of an ideal mono atomic gas performs, a cycle as shown in fig. If gas temperature, TA = 400 K TB = 800K, TC = 2400 K, and, TD = 1200K. Then total work done by gas is, P, B, A, , C, D, →T, , (A) 2400 R (B) 1200 R (C) 2000 R (D) Zero, 79
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 16. A metal block of density 5000 kg/m3 and mass, 2 kg is suspended by a spring of force constant, 200 N/m. The spring block system is, submerged in water vessel. Total mass of, water in vessel is 300 gm and in equilibrium, the block is at a height 40 cm above the bottom, of vessel. The specific heat of material of block, is 250J/kg/k and that of water is 4200 J/kg/k., Neglect the heat capacities of vessel and the, spring. If the support is broken the rise in, temperature of water, when block reaches, bottom of vessel is, (A) 0.0012°C, (B) 0.0049°C, (C) 0.0028°C, (D) 0.0°C, 17. One mole of Argon undergoes a process given, by PV3 / 2 = const. If heat obtained by gas is Q, and molar specific heat of gas in the process, is C then which of the following is correct if, temperature of gas changes by -26 K (assume, Argon as an ideal gas), (A) C = 0.5 R, Q = 13 R (B) C = -0.5 R, Q = 1.3 R, (C) C = -0.5 R, Q = 13 R (D) C = 0, Q = 13 R, 18. 2 kg of ice at –20°C is mixed with 5 kg of water, at 20°C in an insulating vessel having a, negligible heat capacity. Calculate the final, mass of water remaining in the container. It is, given that the specific heats of water and ice, are 1 kcal/ kg°C and 0.5 kcal/kg°C while the, latent heat of fusion of ice is 80 kcal/kg, [IIT - 2003], (A) 7kg (B) 6kg, (C) 4 kg (D) 2 kg, 19. Steam at 100°C is passed into 1.1 kg of water, contained in a calorimeter of water equivalent, 0.02 kg at 15°C till the temperature of the, calorimeter and its contents rise to 80°C. The, mass of the steam condensed in kilogram is, (A) 0.130 (B) 0.065 (C) 0.260 (D) 0.135, 20. A block of ice at –10°C is slowly heated and, converted to steam at 100°C. which of the, following curves represents the phenomenon, qualitatively ? [IIT - 2000], A), , 21. The mass of Hydrogen molecule is, 3.32 × 1027 kg . If 10 23 hydrogen molecules, strikes a fixed wall of area 2cm2 at angle of, 4 5 0 to the normal persecond and rebound, elastically with a speed of 103 ms −1 then the, pressure on the wall, A) 2.45 ×103 Nm−2, B) 2.347 ×103 Nm−2, C) 3.264 ×103 Nm−2, D) 1.864 ×103 Nm−2, 22. A closed container of volume 0.02m2 contains, a mixture of Neon and Argon gases, at a, temperature of 270 C and pressure of, , 1×105 Nm−2 . The total mass of the mixture is, 28g. If the gram molecular weights of Neon, and argon are 20 and 40 respectively, masses, of the individual gass are [ IIT- 1994], A) 4g,24g, B) 8g,20g, C) 12g,16g, D) 6g, 22g, 2, 23. In an adiabetic process, R = Cv . The, 3, pressure of the gas will be proportional to, A) T 5 / 3 B) T 5 / 2, C) T 5 / 4, D) T 5 / 6, 24. The heat supplied to one mole of an ideal, monoatomic gas in increasing temeprature, from T0 to 2T0 is 2RT0 . Find the process to, which the gas follows, A) PV = constant, B) P/V = constant, C) V /P = constant, D) PV 2 = constant, 25. One mole of monoatomic ideal gas follows a, proces AB, as shown. The specific heat of the, 13R, process is, . find the value of x on P - axis., 6, P, , 3P0, , B), , B, , X, A, V0, , T, , T, Heat supplied →, , C), , Heat supplied →, D), , T, , T, Heat supplied →, , 80, , Heat supplied →, , 5V0, , V, , A) 4P0, B) 5P0, C) 6P0, D) 8P0, 26. 5.6 Litre of helium gas at STP is adiabatically, compressed to 0.7 litre. Taking the initial, temperature to be T1 , the work done in the, process is, (IIT JEE-2011), 9, 3, 15, 9, RT1 (d) RT1, (a) RT1 (b) RT1 (c), 8, 2, 8, 2
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , (A) | ∆E A |<| ∆E B | < | ∆E C | if temperature in every, process decreases, (B) | ∆E A |>| ∆E B | > | ∆E C | if temperature in every, process decreases, (C) | ∆E A |>| ∆E B |>| ∆E C | if temperature in every, process increases, (D) | ∆E B |<| ∆E A |<| ∆E C | if temperature in every, process increases, 31. Select the correct alternatives for an ideal gas:, MULTIPLE ANSWER QUESTIONS, (A) The change in internal energy in a constant, 28. Molar heat capacity of an ideal gas, pressure process from temperature T 1 to T 2 is equal, to nCv (T2 – T1), where Cv is the molar specific, varies as C = C ? + aT , C = C? + ßV, heat at constant volume and n the number of moles, and C = C? + ap , where a, ß and a are, of the gas., constants. Find the equations of the, (B) The change in internal energy of the gas and the, process for an ideal gas in terms of the, work done by the gas are equal in magnitude in an, variables T and V., adiabatic process., (C) The internal energy does not change in an, −, aT, /, R, R / ßV, ), (A) Ve (, = const (B) T .e( ) = const, isothermal process., (D) Va = nT, (C) V = anT, (D) No heat is added or removed in an adiabatic, 29. A gas undergoes change in its state from, process., position A to position B via three different paths 32. Diagram of a cyclic process ABCA is as shown, as shown in Fig. Select the correct alternatives:, in fig. Choose the correct alternative., 27. Two moles of ideal helium gas are in a rubber, balloon at 300 C . The balloon is fully, expandable and can be assumed to require no, energy in its expansion. The temperature of, the gas in the balloon is slowly changed to, 350 C . The amount of heat required in raising, the temperature is nearly (take R = 8.31 J/, mol.K) (IIT JEE-2012), (A) 62 J (B) 104 J (C) 124 J (D) 208 J, , P, , P, , 1, 2, , A, , A, , B, , 3, , B, , V, , C, , (A) Change in internal energy in all the three paths, is equal., V, (B) In all the three paths heat is absorbed by the, gas., (A) ∆QA →B is negative (B) ∆U B →C is positive, (C) Heat absorbed/released by the gas is maximum, (C) ∆U C →A is negative (D) ∆WCAB is negative, in path (1)., (D) Temperature of the gas first increases and then 33. Figure shows the P-V diagram of a cyclic, process. If dQ is the heat energy supplied to, decreases continously in path (1)., the system, dU is change in the internal energy, 30. An ideal gas undergoes an expansion from a, of the system and dW is the work done by the, state with temperature T1 and volume V1, system, then which of the following relations, through three different polytropic processes, is/are correct, A, B and C as shown in the P–V diagram. If, | ∆E A |,| ∆E B | and | ∆E C | be the magnitude of, C, changes in internal energy along the three, B, paths respectively, then:, P, D, , P, , A, A, B, C, V1, , V2, , V, V, , (A) dQ = dU - dW, (C) dQ = dW, , (B) dU = 0, (D) dQ = - dW, 81
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THERMODYNAMICS, , JEE- ADV PHYSICS-VOL- VI, , 34. Two gases have the same initial pressure,, P, volume and temperature. They expand to the, A, 3, same final volume, one adiabatically and the, other isothermally, B, 2, D, (A) the final temperature is greater for the isothermal, 1, process, C, (B) the final pressure is greater for the isothermal, process, 0, 1, 2, 3, V, (C) the work done by the gas is greater for the, (A) The process during the path A → B is isothermal, isothermal process, (B) Heat flows out of the gas during then path, (D) all the above options are incorrect, B→C→D, 35. During the process A–B of an ideal gas :, (C) Work done during the path A → B → C is zero, (A) work done on the gas is zero, (D) Positive work is done by the gas in the cycle, (B) density of the gas is constant, ABCDA, (C) slope of line AB from the T–axis is inversely 40. C and C denote the molar specific heat, V, P, proportional to the number of moles of the gas, capacities of a gas at constant volume and, (D) slope of line AB from the T–axis is directly, constant pressure, respectively, Then, proportional to the number of moles of the gas., [IIT-2009], (A) CP – CV is larger for a diatomic ideal gas then, P, for a monoatomic ideal gas, B, (B) CP + CV is larger for a diatomic ideal gas then, for a monoatomic ideal gas, (C) C P / CV is larger for a diatomic ideal gas then, A, for a monoatomic ideal gas, (D) C P . CV is larger for a diatomic ideal gas then, T, for a monoatomic ideal gas, 36. 1 kg of ice at 0ºC is mixed with 1.5 kg of water 41. An ideal gas is taken from state A ( pressure,, at 45ºC [latent heat of fusion = 80 cal/g]. Then, P volume V ) to the state B ( preesure P / 2 ),, (A) the temperature of the mixture is 0ºC, volume2V) along a straight line path in P-V, (B) mixture contains 156.25 g of ice, diagram. Select the correct statement from the, (C) mixture contains 843.75 g of ice, following, (D) the temperature of the mixture is 15ºC., a) The work done by the gas in the process A to B, 37. In a thermodynamic process helium gas obeys, exceeds the work that would be done by it if the, the law TP–2/5 = constant. If temperature of 2, system were taken from A to B along an isotherm, moles of the gas is raised from T to 3T, then, b) In T-V diagram, the path AB becomes a part of, (A) heat given to the gas is 9RT, a parabola, (B) heat given to the gas is zero, c) In P - T diagram, path AB becomes a part of a, (C) increase in internal energy is 6RT, hyperbola, (D) work done by the gas is –6RT., d) In going from A to B, the temeprature T of the, 38. A gas is found to obey the law P 2V = constant., first increases to a maximum value and then decrases, The initial temperature and volume are T0 and 42. During melting of a slab of ice at 273 K at, V0. If the gas expands to a volume 3V0, then, atmospheric pressure., (A)final temperature become 3 T0, a) positive work is done by ice-water system on, the atmosphere, (B) internal energy of the gas will increase, b) positive work is done on ice-water system by, T0, the atmoshpere, (C) final temperature becomes, 3, c) the internal energy of the ice-water system, (D) internal energy of the gas decreases., increases, 39. The figure shows the P-V plot of an ideal gas, d) the internal energy of the ice-water system, taken through a cycle ABCDA. The part ABC, decreases, is a semi-circle and CDA is half of an ellipse. 43. One mole of an ideal gas in inital state A, Then, [IIT-JEE-2009], 82
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , undergoes a cyclic process ABCA, as shown, in figure. Its pressure at A is P0 . Choose the, correct option(s) from the following :, (IIT JEE-2010), , (p) positive during isothermal expansion., (q), , V, , V0, , (r), , B, , 4V0, , C, , P1 V1 V2 , , (1 - γ ) V1 , , γ −1, , γ −1, , , − 1, , , , , − 1, , , , compression, , during adiabatic process, , (s) Negative from most initial state to the final state., 46. The figure shows a cyclic process ABCDA., , A, , P, T0, , T, , (a) internal energies at A and B are the same, (b) work done by the gas in process AB is, PV, 0 0l n 4, P, (c) pressure at C is 0, 4, , P1 V1 V2 , , ( γ - 1) V1 , , 30, , A, , 20, 10, , D, B, , C, V, , 10, , 20, , Column I (Process), Column II, A) AB, (p) W < 0, T0, B) BC, (q) Q > 0, (d) temperature at C is, 4, C) CD, (r) W > 0, MATRIX TYPE QUESTIONS, D) DA, (s) Q < 0, 44. In Column I some statements or expressions 47. Column I contains a list of processes involving, related to first law of thermodynamics are, expansion of an ideal gas. Match this with, given and in column II some processes are, Column II describing the thermodynamic, mentioned. Match the entries of column I, with, change during this process. (IIIT 2008), the entries of column II., Column I, Column I, A) An insultated container has two chambers, (A) dU = nCVdT is valid for, separated by a valve chamber I contains an ideal, (B) Temperature of the system can change in, gas and the chamber II has vacuum., (C) Q = dU + W is valid for, (D) The process in which heat exchange, Column II, I, (p) Adiabatic process, II, (q) Isothermal process, (r) Polytropic process, (s) Free expansion between the system and, Ideal gas, Vacuum, surroundings is zero, B) An ideal monoatomic gas expands to twice its, 45. There isan idealgassam ple.The ratio ofC P, original volume such that or remains constant its, and CV for gas sample is γ . In its initial state, its pressure is P1 and volume is V1. Now it is, 1, pressure Pα 2 , where V is the volume of the gas, expanded isothermally from volume V 1 to V 2 ., V, Then it is compressed adiabatically from, C) An ideal moniatomic gas expands to twice its, volume V2 to V1 again, original volume such that its pressure, Regarding the above situation, match the, 1, following, Pα 4/3 Where V is its volume, Column I, V, (A) Heat given to system (i.e. ideal gas sample), D) An ideal monoatomic gas expands such that its, (B) Work done by gas during adiabatic, pressure P and volume V followsthe behaviour, (C) Change in internal energy of gas sample, shown in the graph, (D) Change in internal energy of gas sample, Column II, 83
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, P, , V1, , V, , 2V1, , Column II, p) The temperature of the gas decreases, q) The temperature of the gas increases, r) The gas loses heat, s) The gas gains heat, 48. The figure shows a cylic process ABCDA., P, 30, , A, D, , 20, 10, , B, , C, V, , 10, , 20, , 50. One mole of a monatomic ideal gas is taken, through a cycle ABCDA as shown in the P - V, diagram. Column-II gives the characteristics, involved in the cycle. Match them with each, of the processes given Column-I., (IIT -2011), P, 3P, , 1P, , B, , A, , C, , D, , 0 1V 3V, Column-I, (A) Process A - B, (B) Process B - C, (C) Process C - D, (D) Process D - A, Column-II, (p) Internal energy decreases, (q) Internal energy increases, (r) Heat is lost, (s) Heat is gained, (t) Work is done on the gas., , 9V, , Column I (Process), Column II, A) AB, (p) W < 0, B) BC, (q) Q > 0, C) CD, (r) W > 0, D) DA, (s) Q < 0, 49. Work done and heat supplied are represented, COMPREHENSIVETYPE QUESTIONS, on y and x-axes respectively for two gases, showing an isotherm and two isobars. The Passage : 1, A closed and isolated cylinder contains ideal gas., scales of two axes are same. The initial state, An adiabatic separator of mass m, cross sectional, of, two, gases, aresame., area A divides the cylinder into two equal parts each, tan θ1 = 1, tan θ 2 = 2 / 5 and tan θ3 = 2 / 7, ., with volume V0 and pressure P0 in equilibrium, match the options of the two columns., Assume the separator to move without friction., y, 51. If the piston is slightly displaced by x, the net, 1, force acting on the piston is, 2, , W, , P0 ?A 2 x, (A), V0, , 3, θ3, , θ2, , θ1, Q, , COLUMN - I, a) straight line 1 corresponds to, b) straight line 2 corresponds to, c) straight line 3 corresponds to, d) y - axis corresponds to, COLUMN - II, P) isothermal process, Q) monoatomic gas, R) diatomic gas, S) adiabatic process, 84, , x, , 2P0 ?A 2 x, (B), V0, , P0 ?A 2 x, 3P0 ?A 2 x, (C), (D), V0, (?-1)V0, 52. Identify the correct statement, (A) The process is adiabatic only when the piston, is displaced suddenly, (B) The process is isothermal when the piston is, moved slowly, (C) The motion is periodic for any displacement of, the piston, (D) The motion is SHM for any displacement of, the piston
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 53. The time period of oscillation for small, displacements of the piston is, , mV0, (A) 2π 2P ?A 2, 0, , mV0, (B) 2π 4P ?A 2, 0, , P0, k, B, x, , A, , (C) 2π, , mV0 (?-1), 2P0 ?A 2, , (D) none of these, , Gas, , Passage : 2, A calorimeter of mass m contains an equal mass of 57. The force constant of spring is, water in it. The temperature of the water and, (A) 189.6 N m-1, (B) 18.96 N m-1, calorimeter is t2. A block of ice of mass m and, (C) 1896 N m-1, (D) 2896 N m-1, temperature t 3 < 0o C is gently dropped into the 58. Change in internal energy of the gas is, (A) 1246.5 J, (B) 124.65 J, calorimeter. Let C1 , C2 and C3 be the specific, (C) 200 J, (D) 12.46 J, heats of calorimeter, water and ice respectively and 59. Heat supplied by heater during this process is, L be the latent heat of ice., (A) 129.65 J, (B) 1296.5 J, 54. The whole maxture in the calorimeter becomes, (C) 12.96 J, (D) 250 J, ice if, Passage : 4, (A) C1t 2 + C2 t 2 + L + C3 t 3 > 0, A mercury barometer is defective. When an, accurate barometer reads 770 mm, the defective, (B) C1t 2 + C 2 t 2 + L + C3 t 3 < 0, one reads 760 mm. When the accurate one reads, 750 mm, the defective one reads 742 mm. then, (C) C1t 2 + C2 t 2 − L − C3 t 3 > 0, 60. The length of air coloumn when accurate, (D) C1t 2 + C 2 t 2 − L − C3 t 3 < 0, barometer reads 770mm is, 55. The whole mixture in the calorimeter becomes, A) 76 mm B) 74 mm C) 72mm D) 70mm, water if, 61. The reading of the accurate barometer, when, defective are reads 752mm is, (A) ( C1 + C2 ) t 2 − C3 t 3 + L > 0, A) 760mm B) 764mm C) 758mm D) 761mm, (B) ( C1 + C2 ) t 2 + C3 t 3 + L > 0, (C) ( C1 + C2 ) t 2 − C3 t 3 − L > 0, (D) ( C1 + C2 ) t 2 + C3 t 3 − L > 0, 56. Water equivalent of calorimeter is, mC1, , mC2, , (A) mC1 (B) C (C) C (D) None of these, 2, 1, Passage : 3, Two mole of an ideal monatomic gas are, confined within a cylinder by a mass less spring, loaded with a frictionless piston of negligible, mass and crossectional area 4 × 10-3m2. The, spring is initially in ill relaxed state. Now the, gas is heated by a heater for some time. During, this time the gas expands and does 50J of work, in moving the piston through a distance of, 0.01m. The temperature of gas increases by, 50k., , ASSERTION & REASON TYPE QUESTIONS, , Note : Each question contains STATEMENT-1, (Assertion) and STATEMENT-2 (Reason)., Each question has 5 choices (A), (B), (C), (D), and (E) out of which ONLY ONE is correct., (A) Statement-1 is True, Statement-2 is True;, Statement-2 is a correct explanation for Statement1., (B) Statement-1 is True, Statement-2 is True;, Statement-2 is NOT a correct explanation for, Statement-1., (C) Statement -1 is True, Statement-2 is False., (D) Statement -1 is False, Statement-2 is True., (E) Statement -1 is False, Statement-2 is False, 62. Statement -1 : If an ideal gas expands in vacuum, in an insulated chamber, ∆Q, ∆U and ∆W all are, zero., Statement -2 : Temperature of the gas remains, constant., 85
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 63. Statement -1 : At a given temperature the specific 71. Figure shows the variation of internal energy, heat of a gas at constant volume is always greater, (U) with the pressure (P) of 2.0 mole gas in, than its specific heat at constant pressure., cyclic process abcda. The temperatures of gas, Statement -2 : When a gas is heated at constant, at c and d are 300 and 500 K, respectively., volume some extra heat is needed compared to that, The heat absorbed by the gas during the, at constant pressure for doing work in expansion., process is x × 100R ln 2 . Find the value of x., 64. Statement -1 : Internal energy change is zero if, the temp is constant, irrespective of the process, being cyclic or non-cyclic., U, a, d, Statement -2 : dU = n C vdT for all process and is, independent of path., 65. Statement -1 : As the temperature of the, b, c, blackbody increases, the wavelength at which the, spectral intensity (Eλ) is maximum decreases., P, Statement -2 :The wavelength at which the spectral, P, 2P, intensity will be maximum for a black body is, proportional to the fourth power of its absolute 72. The heat absorbed by a system in going, through the cyclic process shown in Fig is, temperature., x × 5πJ . Find the value of x., 66. Statement -1 : The specific heat of a gas in an, adiabatic process is zero but it is infinite in an, V (in cm3), isothermal process., Statement -2 : Specific heat of a gas directly, proportional to heat exchanged with the system and, Ellipse, inversely proportional to change in temperature., 300, 100, INTEGER TYPE QUESTIONS, 67. A 50 gm lead bullet (sp. heat 0.020 cal/g) is, initially at 300 C . It is fired vertically upward, –2, P(Nm ), 5, 5, with a speed 84 m/s. On returning to the, 1 × 10 3 × 10, starting level, it strikes a slab of ice at 00 C . 73. 1/R (R is universal gas constant) moles of an, ( A×100 ) mg of ice is melted. Find the value of, ideal gas (γ = 1.5) undergoes a cyclic process, ‘A’., (ABCDA) as shown in figure. Assuming the, 68. An ideal gas (CP/CV = γ ) is taken through a, gas to be ideal. If the net heat exchange is 10x, process in which the pressure and the volume, Joules, find the value of x ? [In 2 = 0.7], are related as P = a Vb. The value of b for which, P, the specific heat capacity in the process is zero, 0, , is b = −, , xγ, . Find the value of x., 2, , 3, , 2 × 10 atm, , A, , 0, , B, , 3, 69. A vessel contains helium, which expands at, 1 × 10 atm, C, D, constant pressure when 15 kJ of heat is, supplied to it. Wat will be the variation of the, T, 300K, 400K, internal energy of the gas? (in kJ), 70. When a quantity of liquid bismuth at its melting 74. A metal rod AB of length 10x has its one end A, in ice at 0°C and the other end B in water at, point is transferred to a calorimeter containing, 100°C . If a point P on the rod is maintained at, oil, then the temperature of oil rises from, 400°C, then it is found that equal amounts of, 12.5°C to 27.6°C. The experiment is repeated, water and ice evaporate and melt per unit time., under identical condition except that bismuth, The latent heat of evaporation of water is 540, is solid, the temperature of the oil rises to, cal/g and latent heat of melting of ice is 80 cal/, 18.1°C. The specific heat of bismuth is 0.032, g. If the point P is at a distance of λ x from the, cal/g’C, The latent heat of fusion of bismuth, ice end A, find the value of λ . [Neglect any, is 6.7 K cal/g. Then determine the value of K1, heat loss to the surrounding], Melting point of bismuth is 271°C., , 86
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JEE- ADV PHYSICS-VOL- VI, 75. An ideal diatomic gas under goes a process in, which its internal energy changes with volume, as given U = cV2/5 where c is constant. Find, the ratio of molar heat capacity to universal, gas constant R ?, 76. An ideal gas is taken through a cyclic, thermodynamic process through four steps., The amount of heat involved in four steps are, Q1 = 6000J,Q2 =−5000J,Q3 =−3000J andQ4 = 4000J, respectively. If efficiency of cycle is 10x% then, find value of x?, 77. A piece of ice (heat capacity, and latent heat 81., = 2100 J kg −1 0C −1, , THERMODYNAMICS, (a) net change in the heat energy, (b) net work done, (c) net change in internal energy., (d) Take R = 8.32 Jmol-1., p, 2atm, , 1atm, , A, , D, 300K, , B, , C, 400K, , 0.05 kg steam at 373 K and 0.45 kg of ice at, 253 K are mixed in an insulated vessel. Find, = 3.36 × 105 J kg −1 ) of mass m grams is at, the equilibrium temperature of the mixture., 0, Given, Lfusion = 80 cal/g = 336 J/g L vaporisation, 5 C at atmospheric pressure. It is given 420, = 540 cal/g = 2268 J/g,, J of heat so that the ice starts melting. Finally, when the ice water mixture is in equilibrium, it, Sice=2100 J/kg K = 0.5 cal g/g K, is found that 1 gm of ice has melted. Assuming, and swater =4200 J/gk K = 1 cal/gK [IIT-2006], there is no other heat exchange in the process, 82. An ice cube of mass 0.1 kg at 0ºC is placed in, the value of m is, (IIT JEE-2010), an isolated container which is at 227ºC. The, 78. A diatomic ideal gas is compressed, specific heat capacity c of the container varies, with temperature T according to the empirical, 1, adiabatically to, of its initial volume. In the, relation c = A + BT, where A = 100 cal/kg-K, 32, and B = 2 × 10−2 cal/kg-K 2. If the final, initial temperature of the gas is Ti (in Kelvin), temperature of the container is 27ºC,, and the final temperature is a Ti , the value of, determine the mass of the container. (Latent, a is, (IIT JEE-2010), heat of fusion for water = 8 × 104 cal/kg, specific, SUBJECTIVE TYPE QUESTIONS, heat capacity of water = 10+3 cal/kg-K)., 79. When a system is taken from state i to state f 83. A diatomic gas is enclosed in a container by a, along the path iaf, it is found that Q = 50 cal, movable piston of cross-sectional area A = 1, and W = 20 cal. Along the path ibf, Q = 36 cal, m2 at 300 K, as shown in the figure. The length, (figure) (a) What is W along the path ibf ? (b), of the gas column is 1 m. The gas is now heated, If W = - 13 cal for the curved return path f i,, to 400 K isobarically. (i) Find the new height, what is Q for this path ? (c) Take U i = 10 cal., of the piston. (ii) Now the gas is compressed, to its initial volume adiabatically. Find the final, What is U f ?, (d) If U b = 22 cal, what, temperature of the gas., is Q for the process ib and for the process bf ?, 84. Two moles of an ideal monatomic gas is taken, P, through a cycle ABCA as shown in the P-Ta, f, diagram (figure). During the process AB,, pressure and temperature of the gas vary such, that PT = constant. If T1 = 300 K, calculate., (a) The work done on the gas in the process, b, AB and, V, O, (b) The heat absorbed or released by gas in, 80. Two moles of helium gas undergo a cyclic, each of the process., process as shown in the figure. Assume the gas, Give answers in terms of the gas constant R., to be ideal calculate net, 87
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , (a) Now, work done by the gas in a process is, given by, , LEVEL - V - KEY, SINGLE ANSWER QUESTIONS, 1)A 2)A 3) D 4)B 5)D 6)B 7)B, 8)C 9)B 10)A 11)B 12)D 13)B 14)B, 15)A 16)B 17)C 18)B 19)A 20) A 21)A, 22)A 23)B 24)B 25)C 26) A 27) D, MULTIPE ANSWER QUESTIONS, 28) A,B,C 29) (A, B, C) 30) (A, C), 31) (A, B, C, D) 32) (A, B, D) 33) (B, C), 34) (A, B, C) 35) (A, B, D) 36) (A, B), 37) (B, C, D)38) (A, B) 39) (B, D) 40) (b) (d), 41) A,B,D 42) B,C 43) (a), (b), (c), (d), MATRIX MATCHING TYPE, 44)(A → p, q, r, s), (B → p, r),, (C → p, q, r, s), (D → p, s), 45)(A → p),(B → q,r, s),(C → p, q),(D → p, q), 46) (A) → (s) ; (B) → (q,r), (C) → (q) ; (D) → (p,s), 47) A → q, ( b ) → p,q ( c ) → p,s ( d ) → q,s, 48) (A → s, B → r, C → q, D → p, s), 49) (A → P, B → Q, C → R, D → S), 50) A → p,r,t ; B → p,r ; C → q,s ; D → r,t, ,t, COMPREHENSION TYPE QUESTIONS, P-I 51) B 52) C 53) A, P-II 54) D 55) D 56) B, P-III 57) C 58) A 59) B, P -IV 60) C 61) D, ASSERTION REASON TYPE QUESTIONS, 62) B 63) D 64) A65) C 66) A, INTEGER TYPE QUESTIONS, 67) 9 68) 2 69) 9 70) 6 71) 4 72) 2 73) 7, 74) 9 75) 5 76) 2 77) 8 78) 4, SUBJECTIVE TYPE QUESTIONS, 79) (a) 6 cal (b) –43 cal (c) 40 cal (d) 18 cal, 80) A) 1153.4 J (B) 1153.4 J (C) 0, 81) 273 K, 82) 0.49 kg, 83) (A) 4/3 m (B) 448.4 k, 84) (A) 1200 R (B) 1200 R ln 2, , LEVEL - V - HINTS, SINGLE ANSWER QUESTIONS, 1., , 2., , ∆U = ∆W (polytropic process), nR∆T, nR∆T, 1, =−, ⇒r=, γ −1, r −1, 3, Evidently, for the process A → B, V∝, , 1, T, , or VT = constant (say, k), , or TdV + VdT = 0, , 88, , or dV = −, , VdT, T, , W = ΣPdV, , or WAB = Σ, , nRTdV, = Σ (− nRdT), V, , 500, , WAB =, , ∫, , −nRdT = – (2 mol) (8.314 J/mol-K), , 300, , 3., , [500 – 300]K = –3.326 kJ, nRT, = a. V, ncvT = a. V, γ −1, PV = a ( γ − 1) V P.V 2 = constant, R, R, 1, C=, −, Here r =, γ −1 r −1, 2, 1, , nRT 2, V = K ; TV = constant ;, v, T, 1, nRT, = KV 2 ; 3 = constant ; T ∝ v 3, T ∝ Process (B), v, v, v, , 4., , (B). Process (A)., , 5., , (D). Pi = P, Pf = 2 P, Vi = V, Vf = ?, For adiabatic process PVγ = (2 P) Vfγ, 1, ⇒ Vf = , 2, , 1/ γ, , V ;, , γ ↑ Vf ↑, , for monoatomic gas γ is max. For isothermal process, PV = (2 P) Vf ⇒ Vf = V/2., Independent on nature of gas., 6. From given P-V diagram A → B is isobaric and, C → D is ischoric so option b is correct., 7. For A → B, PαT ⇒ V = constant, For B → C, P = constant and T = decreases ⇒ V, decreases, for C → A, T = constant and p decreases ⇒ V, increases, 8. In cycle process W = area of cycle on P – V diagram, 9. A → B temperature is constant and pressure is, increasing ∴ volume is decreasing. So A → B, work done is negative., B → C pressure is constant and temperature is, decreasing. So volume decreases. Hence work, done is negative, 20, , 20, , 10. Q = ∫ mS dT = ∫ (AT 3 )dT = 4 × 104 mA ., 0, 0, 11. WAB is negative (volume is decreasing) and WBC is, positive (volume is increasing) and since,, |WBC| > |WAB|, ∴ Net work done is positive and area between, semicircle which is equal to, , π, atm − L ., 2, , 12. WBCOB = – Area of triangle BCO = +, , P0 V0, 2
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , WAODA = + Area of triangle AOD = +, , H = msice ∆T = (2kg) × 0.5× 20 = 20 kcal, ∴ After this, heat available = (100 – 20) = 80 kcal, , P0 V0, 2, , ∴ Wnet = 0 ., 13. WBA = – 30 J, QBA = 0,, ∴ ∆U BA = − WBA = 30 J, Now, ∆U AB = −U BA = −30 J ., 14. Molar heat capacity C =, , ∆Q, ∆T, , ∆U + ∆W, dT, ∆U is same in both the paths but ∆W2 > ∆W1, C1, ∴ C 2 > C1 or C < 1 ., 2, , or C =, , 15. WAB = WCD = 0 (V = const), , This heat will now be gained by ice at 0°C to melt, into water at 0°C. Let m kg of ice melt., ∴ m × 80 = 80 ∴m = 1 kg, Out of 2 kg of ice, 1 kg of ice melts into water and, 1 kg of ice remains unmelted in container., ∴ Amount of water in container = 5 + 1 = 6kg, 19. Heat lost by steam = Heat gained by water +, calorimeter, ∴mL + ms (100 – 80) = 1.12 × s × (80 – 15), or m[540 + (1×20)] = 1.12 × 1× 65, or m =, , 1.12 × 1× 65 65, =, kg or m = 0.13 kg, 560, 500, Water, , 100°C, III, , 20., , WBC = 3R (TC - TB ) ü, ý P = const., WDA = 3R (TA - TD ) þ, ∴ W = 3R(TA + TC − TB − TD ) = 2400R, , IV, Steam, , 0°C, –10°C, , Ice, I, , II, , Water, , The change of ice at 10°C into steam at 100°C, occurs in four stages ; it is represented by curve, 16. When the block is in equilibrium in the water the, (a)., spring stretches by x, I stage – The temperature of ice changes from –, Kx + weight of liquid displaced = weight of block, 10°C to 0°C., 2, II Stage – Ice at 0°C changes in water at 0°C., × 1000 × 10 = 2 × 10 ∴ x = 0.08 m, 200x +, The state changes as heat is supplied., 5000, III, Stage, –, Water at 0°C changes into water at, 1 2, U, =, kx, 100°C., In equilibrium energy stored in spring, 2, IV Stage – Water at 100°C change into steam at, 1, 2, 100°C., U = × 200 × (0.08) = 0.64J, 21. Momentum of the molecule, 2, when support in booken & block reaches bottom, p = mv = 3.32 × 10−24 kgms −1, of the vessel, ∆ p = 2 p cos 450 = 2 p, U + Mgh = MW SW ∆T + M × SB ∆ T, Where M is mass of block, p = ∆p × f = 1.414 × 3.32 ×10−24 ×1023 = 0.469 N, ∴ 0.64 + 2 × 10×0.4=[0.3×4200+2×250] ∆T, 0.4695, ∴ ∆T = 0.0049°C, p=, = 2.447 ×103 Nm−2, −4, 2 ×10, 17. If a gas under goes a thermodynamic process PVx, 22. Let m2 m2 be masses of Neon and Argon, R, R, −, = const, then molar specific heat C =, respectively m + m = 28, γ −1, , x −1, , for Argon r = 5 / 3, \ C=, , R, R, = - 0.5 R, 5, 3, -1, -1, 3, 2, , Q = nCDT = 1 ´ - 0.5R ´ - 26 = 13R, , 18. Heat lost by water at 20°C = ms W ∆T, H = 5 × 1 × 20 = 100 kcal, At first, ice at (–20°C) will take heat to change into, ice at 0°C., , 1, , PV = nRT, , 2, , m, m , PV = 1 + 2 RT (or), m1 m 2 , , m m , PV = 1 + 2 RT, 20 40 , , m1 m2 1×105 0.02, +, =, ⇒ 2m1 + m2 = 32, 20 40 8.314 × 300, m1 + m2 = 28, , 89
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , R, clculate γ, 23. R = 2CV / 3 use CV =, γ −1, then use T γ P1−γ, 24. ∆Q = nC ( ∆T ) Molar heat capacity, C=, , 5 − 3x , R, R, +, = (1) RT0 , = 2 RT0, ( γ − 1) (1 − x ), 2 (1 − x ) , , 13R, 3, ∆W, 2R, Cv = R, = C − Cv =, 25. C =, 6, 2, ∆T, 3, PV, ∆T = 0 0 ( 5n − 3) ∆W = 2 P0V0 ( 5n + 3) ∴ n = 6, R, nR ( T1 − T2 ), 2/3, 2/3, 26. W =, T1 ( 5.6 ) = T2 ( 0.7 ), r −1, nR × 3T1 9, 2/3, = nRT1, = T1 × 4 W =, ⇒ T2 = T1 ( 8 ), 2/3, 2, 1, 9, But n =, ⇒ W = RT1 ., 4, 8, 27. For an isobaric process, Q = nC P ∆T, , 5 , 5, = 2 R ( 5 ) = 2 × × 8.31× 5 = 208 J ., 2, 2, , , , MULTIPE ANSWER QUESTIONS, 28. A, B, C., (i) Form first law of termodynamics,, , dQ = dU + dW dQ = C dT = C? dT + PdV, dQ, dV, RT dV, C=, = C? + P, = C? +, ...... (1), dT, dT, V dT, since it varies as, C = C? + aT .... (2), On comparing expression (1) and (2),, , dV, RT dV a , ....(3), , dT =, V, V dT R , aT, On integrating eqn. (3), we have log V −, =, R, constant or V = (another constant) eaT / R, , we have aT = , , or V = Ve −( aT / R ) = constant .....(4), Eqn. (4) is the equation of process., (ii) We have C = C? +, , RT dV, V dT, , Comparing it with given molar heat, capacity, C = C? + ßV Hence, we have, RT dV, dV ß dT, = ßV 2 =, ....... (5), V dT, V, R T, 90, , Above equation integration yields, 1 ß, − = log T + constant, V R, , log T +, , R, = another constat ..... (6), ßV, , or Te( R / ßV ) = another cosntant ... (7), (iii) On comparing C = C? +, , RT dV, V dT, , and C = C? + aP we have, , RT dV, = aP, V dT, , RT dV, = aP, V dT, , Since PV = RT for one mole, hence, , dV, = a or V = aT or V = an T for n moles, dT, 29. Internal energy (U) depends only on the initial and, final states. Hence, ∆U will be same in all the three, paths. In all the three paths. In all the three paths,, work done by the gas is positive and the product, PV or temperature T is increasing. Therefore,, internal energy is also increasing. So, from the first, law of thermodynamics, heat will be absorbed by, the gas. Further, area under P – V graph is maximum, in path 1 while ∆U is same for all the three paths., Therefore, heat absorbed by the gas is maximum in, path1. For temperature of the gas, we can see that, product PV first increases in path 1 but whether it, is decreasing or increasing later on we cannot say, anythng about it unless the exact values are known, to us., 30. Initial state is same for all the three processes (say, initial internal energy = E0)., In the final state, V A = VB = VC and P A > P B > PC, PAVA > P BVB > P CVC, EA > E B > E C, If T 1 < T 2, then E 0 < E f for all the three processes, and hence, (E0 – EA) < (E0 – EB) > (EC – E0), , ∆E A < ∆E B < ∆E C, If T1 < T2, then E 0 < Ef for all the three processes, and hence (EA – E 0) > (EB – E 0) > (EC – E 0), ∆E A > ∆E B > ∆E C, 31. A, B, C, D., a. ∆U = Q − W = nC P ∆T − P∆V, = nCP ∆T − nR∆T = n(C P − R)∆T, = nC V ∆T = nCV (T2 − T1 ), b. ∆Q = ∆U + ∆W, But ∆Q = 0 for adiabatic process; hence
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , This is less than 1kg so the resulting temperature is 0o c ., ∆U = −∆W or, ∆U = ∆W, 37. In adiabatic process ∆Q = 0, c. ∆U = nC V ∆T = 0 (∴∆T = 0), 3 , d. ∆Q = 0 (in adiabatic change), ∆U = ncv ∆T = 2 R ( 3T − T ) = 6 RT, 2 , 32. A, B, D., During process A and B, pressure and volume both, ∆W = −∆U = −6RT ∆W = −∆U = −6RT, are decreasing. Therefore, temperature and hence, 2 2 2, internal energy of the gas will decrease 38. P 2V = constant n R T .V = constant, V2, (T ∝ PV) or ∆VA →B = negative. Further,,, nRT, T2, ∆WA →B is negative., P=, ⇒, = constant, V, V, In process B to C, pressure of the gas is constant, while volume is increasing. Hence, temperature To2 T 2, =, ⇒ T = 3 To T α V here V increases, should increase or ∆U B →C = positive. During C Vo 3Vo, to A volume is constant while pressure is increasing., ∴ T increases hence internal energy increases, Therefore, temperature and hence internal energy 39. Temperature at B > temperature at D, ∴ ∆U is negative (for B → C → D), of the gas should increase of ∆U C →A = positive., Also, W is negative (for B → C → D), During process CAB, volume of the gas is, Tracing is clockwise on PV diagram., decreasing. Hence, work done by the gas is, ∴ Total W is positive. ∴ (B) and (D) are Correct, negative., 40., C, and CV for diatomic is greater than monoatomic., 33. In a cylic process, the system returns to its initial, P, So,, CP + CV, CP. CV is greater for diatomic ideal, state. Hence the change in the internal energy dU =, gas, (B) and (D) are Correct, 0. Therefore, choice (b) is correct. From the first, law of thermodynamics,, dQ = dU + dW = dW (Q dU = 0), P, nRT, |, 41., Hence choice (c) is also correct., P = − cV + c| V = −cV + c, V, , Initial, Final, p, , 34., V, , T = −kV 2 + k | V So T - V graph in parabola, nRT, cnRT, V=, P=−, + c|, P = − cV + c|, P, P, So P - T graph in also parabola, 42. Since on melting of ice volume decreases. So, atmosphere is doing +ve work on ice water system., , 35. P–T graph is a straight line passing through origin., Therefore,, ∆Q = ∆U + ( −∆W ) ⇒ ∆U > 0 Q ∆Q > 0, V = constant, 43., Internal, energy of an ideal gas depends on, ∴ work done on the gas is zero., temperature, m 1, V, PV, 4V, Further density of the gas ρ = ∝, WBC = nRT ln 2 = (1)( R ) 0 0 ln 0, V V, V1, R, V0, Volume of the gas is constant. Therefore, density, of gas is also constant., P, P, = constant P at C = 0, = PV, 0 0 ln 4 For CA, T, 4, nR , PV = nRT or P = , T, T0, V , Hence all option are correct.., T at C =, i.e., slope of P–T line ∝ n., 4, 36. On bringing of 1.5kg of water 45o c to 0o c heat, MATRIX MATCHING TYPE QUESTIONS, released = 1500 × 1× 45 cal., 44 & 45. See the conditions of different process, mass of ice that melts with this heat = 46. (A) Volume is constant , pressure is decreasing, therefore temperature decreases., 1500 × 45, = 843.75 gm, ⇒ ∆U < 0 ⇒ ∆Q < 0, 80, (B) Pressure constant and volume increases, 91
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , ⇒ Temperature increases, 2P γA 2 x, (C) Volume is constant and pressure increases., ⇒F= 0, V0, Hence temperature will increase, (D) Volume decreases therefore work done is 53. From force fine restoring force constant. and then, negative., use expression for time period, 47. ∆Q ≠ 0, ∆T ≠ 0 . So process is neither isothermal Paragraph-II, nor adiabatic., 54. Heat lost by water and container, 48. (A) Volume is constant , pressure is decreasing, mC1(t2 – 0) + mC2(t2 – 0) + mL < mC3(0 – t 3), therefore temperature decreases., ∴C1t2 + C2t2 + C3t3 + L < 0, 55. Heat lost = mC 1(t2 – 0) + mC2(t2 – 0), ⇒ ∆U < 0 ⇒ ∆Q < 0, Heat gained = mC3(0 – t 3) + mL, (B) Pressure constant and volume increases, ∴ C1t2 + C2t2 > C3t2 > C3t3 + L, Temperature, increases, ⇒, ∴ C1t2 + C2t2 + C3t3 – L > 0, (C) Volume is constant and pressure increases., Hence temperature will increase, mC1, (D) Volume decreases therefore work done is 56. mC1 = MC2 ∴ M = C, 2, negative., Paragraph-III, 49. For isothermal process W = Q ( ∆U = 0 ), 57-59 When the gas is heated it expands & pushes the, piston by x. If k is force constant of spring and A is, For isothermal process, slope of straight line on, area of crossection of the piston. It P 0 is, W = Q curve is 1 or tan θ = 1. Thus straight line 1, atmospheric pressure then at equilibrium of piston, represents isothermal process, (A) → P. For, the pressure of the gas on the piston, isobatic process., P = P0 +, , f +2, Q=, R∆T and W = R∆T ( one mole gas ), f , , The increase in the volume of the gas by small, movement of x of piston is, , W 2 , =, , Q f +2, , For monoatomic gas. f = 3 and slope of straight, line on W - Q curve = 2/5, So, straight line 2 corresponds to monoatomic, gas and straight line 3 corresponds to diatomic gas, ( f = 5 ). Heat is not absorbed in adiabatic process., The straight line coinciding vertical axis corresponds, to adiabatic process., 50. A-p,r,t ; B-p,r ; C-q,s ; D-r,t, (A) Volume decreases, ∴ Temperature decreases, (B) Pressure decreases, ∴ Temperature decreases, (C) Volume increases, ∴ Temperature decreases, PV ( P )( 9V ) 9 PV, =, =, (D) TD =, mR, nR, nR, ( 3P )( 3V ) = 9 PV, TA =, ., nR, nR, , COMPREHENSION TYPE QUESTIONS, Paragraph-I, 51, 52. ∆Q = 0 for system P0 V0γ = P1V1γ = P2 V2γ, , γAx , γAx , ⇒ P1 = P0 1 −, ⇒ P2 = P0 1 +, , V0 , V0 , , , 92, , kx, A, , dv = Adx, x, , W = ∫ Pdv = P0 Ax +, 0, , 1 2, kx, 2, , Putting A = 4 × 10-3m2, x = 0.1m, W = 50 J., P0 = 1.013 × 105 Nm–2 in above equation, ∴ k = 1896 Nm −1, , Paragraph-IV, 60. Let A be the cross section and L be the length, p1 = 770 − 760 = 10mm p2 = 750 − 742 = 8, , p1v1 = p2v2 ⇒ 10 L = 8 ( L + 18 ) ⇒ 2 = 72mm, , 61., , ( P0 − 752)(80) = 10 × 72, , ⇒ P0 = 761mm, , ASSERTION-RESON TYPE QUESTIONS, 62. (B) Work done in a free expansion = 0, 63. (D): Definition based, 64. For the ideal gas internal energy depends on, temperature, 65. From Wein’s law l m T=constant i.e. peak emission, wavelength l m µ, , 1, . Hence as T increases l, T, , decreases., 66. (A), ∆Q = nC∆T for adiabatic ∆Q = 0 ∴ C = 0, For isothermal ∆T = 0 ∴ C = ∞, , m
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JEE- ADV PHYSICS-VOL- VI, , INTEGER TYPE QUESTIONS, 67., , 1 2, mv + mC ∆T = ∆mice Lice, 2, 42cal + 30cal = ∆mice .80 cal / gm, ∆mice =, , 72, ; 0.9 gm ; 900 mg, 80, , 68. (2), Given that P = aVb or PV–b = a, Comparing with PVr = constant, we have r = –b, R, r −1, R, R, R, Here, C = 0 = C v − γ − 1 ∴O = γ − 1 − − b − 1 or b = −γ, , we know that C = Cv −, , 69. (9), Here ∆Q = 15000J (given) In an isobaric process, ∆Q = nC p ∆T, ∆U = nC V ∆T ( always ), , ∆U nCV ∆T 1, ∆Q, 3, =, = ⇒ ∆U =, ∆Q nCp ∆T γ, γ or ∆U = 5 × 15 = 9 kJ, , THERMODYNAMICS, Q = work done by the gas, W, = area of ellipse = −π ab, Where a and b are semi-major and semi-minor axes, of the ellipse, respectively, which are given from, the figure as, a = 1.0 × 105 N/m and b = 100 × 10–6 m3, Thus area of ellipse is, Q = W = π × 1.0 × 105 × 100 × 10−6, = 3.14 × 10 = 3.14 J, 73. For cyclic process ∑ Q = ∑W, Net change in heat energy, = WAB + WBC + WCD + WDA, = nR(TB −TA ) + nRTB ln, , PB, P, + nR(TD −TC ) + nRTD ln D, PC, PA, , = 70 J Answer is x = 7., 300, ∆Q , = 10 x − λ x = mLv ….. (1), ∆t l, KA, , 74. , , 10 x, , 70. (6), 100° C, Energy with 5 kg of H 2O at 20°C to become ice at 0°C, λx, E1 = 5000 × 1 × 20 = 100000 cal, 400° C, Energy to raise the temperature of 2 kg ice from –, 400, ∆Q , 20°C to 0°C, , = λ x = mL f, E1 = 5000 × 0.5 × 20 = 20000 cal, and ∆t w, ….. (2), (E1 – E 2) = 80000 cal is available to melt ice at 0°C., KA, So only 1000 g or 1 kg of ice would have melt., 300, λ x Lv, ×, =, So, the amount of water available 1 + 5 = 6 kg, dividing, 10 x − λ x 400 L f, 71. (4), Change in internal energy for cyclic process ( ∆U) = 0 ⇒ 3 λ = 540, 4 (10 − λ ) 80 λ = 9(10 − λ ) 10λ = 90 ⇒ λ = 9, For process a → b, (P − constant), 75. From given relation we can write T ∝ V 2/5, Wa →b = P∆V = nR∆T = −400R, ........(1), For process b → c (T − constant), ideal gas equation ⇒ PV = nRT ................ (2), Wb→c = −2R(300)1n 2, from (1), (2) PV ∝ V 2/5 (or), For process c → d(P − constant) Wc→ d = +400R, PV 3/5 = const ............... (3), For process d → a(T − constant), Comparing equation (3) with polytropic equation, 3, Wd →a = +2R(500)1n 2 Net work,, PV x = const , we have x = 5, (∆W) = Wa →b + Wb→ c + Wc →d + Wd →a, R, R, ∆W = 400R 1n 2, ∴C =, +, = 5R ∴ C = 5, γ −1 1 − x, R, dQ = dU + dW, first law of thermodynamics., 76. (2)The amount of heat supplied = 10,000 J, dQ = 400 R 1n 2, The work done = 2000 j % efficiency, 72. (2), The cycle is clockwise; thus net work is done by, W, 2000, the gas. As in a cyclic process no change in internal, × 100 =, × 100 = 20, =, Q, 10000, energy takes place, heat supplied is equal to the, work done by the gas in one complete cycle. So in, 5, −3, 77. m ( 2100 )( 5 ) + 1( 3.36 × 10 ) × 10 = 420, this case heat supplied to the gas is given as, 93
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 11m + 336 = 420 11m = 420 − 336 = 84 m = 8 gm ., 1 , γ −1, 78. Ti v = αTi v , 32 , , Q1 = 0.05 × 2268 × 1000 = 113400J, Heat lost by water of steam to cool to 273 K, , γ−1, , Q 2 = 0.05 ´ 4200 ´ 100 = 21,000 J, ∴ Total heat available from steam Q=134400 J, , Heat gained by ice to raise temp from, 253 k to 273 k = Q11 = 0.45 × 2100 × (273 - 253), = 18900 J, Heat gained by ice during metting at 273 k = Q12, , 2, , α = ( 32 ) 5 = ( 32 ) 5 = ( 25 ) 5 = 4 ., 7, , t, , 2, , SUBJECTIVE TYPE QUESTIONS, 79. (a) ∆U = Q − W = same for both paths, Wibf = 36 − (50 − 20) = 6 cal, , Q12 = 0.45 ´ 336 ´ 9000 = 151200 J, , (b) Q curved - (-13) = -(50 - 20), Q curved = - 43 cal, (c) U f = Ui + 30 = 40 cal, (d) Qi b = Wi b + ∆Ui b = 6 + (22 − 10) = 18 cal, , Total heat required Q11 to Q12 = Q1 = 170100 J Q1 > Q, ∴ Whole of ice will not melt., Equilibrium temp of mixture = 273 k = 0°C, 82. Heat lost by container, , Qb f = Q i f − Q i b = 36 − 18 = 18 cal, , Q=, , 80. Number of moles, n = 2, Helium is a mono-atomic gas, therefore,, CV =, , Heat, , = n × CP × (TB − TA ), , 2, = 2 × 8.32 × 400 ln = 4613.6 J, 1, (∆QCD ) = nC p (TD − TC ), , Total heat gained = 10700 cal, 21600 m = 10700 ∴ m = 0.49kg, T, , T, , 400, , 1, 2, (a) P = constant V = V ⇒ V2 = 300 V1, 1, 2, , 4, 4, hA h 2 = m, 3, 3, , 4, T1 = , (b) T1 V = T2 V, 3, 84. (a) PT = const. = 2 P1T1 = 600P1, γ −1, 1, , γ −1, 2, , × 400 = 448.4k, , T, , V=, , B, 1, 600P1 2nRT, nRT nRT 2, W = ∫ Pdv ∫, dT, ×, =, T, 600P1, P, 600P1, A, 2T1, , ∆U =, , 2R, (T1 − 2T1 ) = − 900R, γ −1, , ∆Q = ∆U + W = − 2100R For process B , →C, , 5 3, ∆Q = nCP ∆T = 2 × × RT1 = 1500 R, 3 2, = 1153.4J, , 81. Heat lost by steam during condensation = MSLV, , 94, , Q 2 = 0.1 × 1000 × 27 = 2700 cal, , (b) For process A , →B, , P , = nRT ln D , PA , , 1, = 2 × 8.32 × 300ln = −3460.2J, 2, (∆W )Net = 4160 + 4613.6 − 4160 − 3460.2, , melting, , ∴ W = − 1200 R -1200R, , 5, = 2 × × 8.32 × (300 − 400) = −4160 J, 2, DA, , in, , γ −1, , VC , PB , = (∆W )BC = nRT ln V = nRT ln P , B, C, , (∆Q), , ice, , 0°C to 27°C, , h2 A =, , Since Process BC is isothermal, therefore ∆U = 0, BC, , by, , Heat gained by water when its temp rises from, , 5, = 2 × × 8.32(400 − 300 ) = 4160 J, 2, , (∆Q), , gained, , Q1 = 0.1 ´ 80000 = 8000 cal, , The gas undergoes cyclic process., Since, internal energy is property of the sytem, the, net change in internal energy during the cyclic, process is zero., The net change in the heat energy is equal to the net, work done., 83., ( ∆Q) Net = ( ∆Q)AB + ( ∆QBC ) + ( ∆QCD ) + ( ∆QDA ), AB, , ∫ (A + BT)dt = − 21600 m cal., , 500, , 3, 5, R ; CP = R, 2, 2, , (∆Q), , 300, , For process C , →A, 2P , ∆Q = ∆W = nRTl n 1 = nR2T1l n 2 = 1200 R l n 2, P1
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 4., , LEVEL - VI, THERMODYNAMICS & KTG, SINGLE ANSWER QUESTIONS, 1., , Molar specific heat at constant volume for an, ideal gas is given by C v = a + bT (a and b are, constants), T is temperature in Kelvin, then, equation for adiabatic process is (R is universal, gas constant), (A) T a e bT V R = constant (B) T R e bT V a = constant, , 2., , (A) m = m e−, 0, −, , (C) m = m e−, 0, , 1, T, , 5., , 2 µg , 1, 2, v 0 t + µgt , L , 2, , , 6., , 3µg , 1, 2, v 0 t − µgt , L , 2, , 2, , ∈α L, , (A) ms + w °C / S, , V, E, , C, B, , (A) 4.67 kJ, (C) 2.67 kJ, , (C), 7., , D, , A, PA, , PB, , 5, 5, 5, 5, (B), (C), (D) ., 2 ln 2, 3, 4 ln 2, 6, A vessel of water equivalent W kg contains m, kg of water of specific heat S. When water, evaporates at the rate of α kgS -1 , the, temperature of the vessel and water in it falls, from T1 °C to T2 °C in t s. If m>> α t and a, fraction Eof the heat needed for evaporation, is taken from the vessel and the water then, average rate of fall of temprature is, (L = average latent heat of vapourisation in J, kg–1), , (A), , 1, µg , 2, v0 t − µgt , L, 2, , , 2 µg, , VA, , (A) 3, (B) 5/2, (C) 5/3, (D) 7/2., An ideal monoatomic gas undergoes a cyclic, process ABCA as shown in the figure. The ratio, of heat absorbed during AB to the work done, on the gas during BC is, , 2V0, , (D) m = m e− L ( v0 t −µgt ), 0, Figure shows a cyclic process ABCDBEA, performed on an ideal cycle. If P A = 2 atm, P B, = 5 atm and PC = 6 atm, V E – VA = 20 litre, find, the work done by the gas in the complete, process. (1 atm. pressure = 1 × 105 Pa), VE, , 2, , v0, , M0, , 3., , V, , (C) T b e aT V R = constant (D) T a e R V bT = constant, An ice cube of mass M0 is given a velocity v0, on a rough horizontal surface with coefficient, of friction µ . The block is at its melting point, and latent heat of fusion of ice is L. The block, receives heat only due to the friction forces, and all work is converted into heat. Find the, mass of the remaining ice block after time t., , (B) m = m e, 0, , Volume versus temperature graph of two moles, of helium gas is as shown in figure. The ratio, of heat absorbed and the work done by the gas, in process 1–2 is, , PC, , (B) 3.67 kJ, (D) 1.67 kJ, , P, , ∈α L, °C / s, w, , ∈α, , (B) L(ms + w) °C / s, (D), , ∈α L, °C / s, ms, , An ideal gas expands isothermally from a, volume V1 and V2 and then compressed to, original volume V 1 adiabatically. Initial, pressure is P1 and final pressure is P3. The, total work done is W. Then :, (IIT - 2004), (A) P3 > P1, W > 0, (B) P3 < P1, W < 0, (C) P3 > P1, W < 0, (D) P3 = P1, W = 0, 95
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 8., , The work done by a certain material when 11. An ideal gas goes through a cycle consisting, of isochoric adiabatic and isothermal lines., temperature changes from T0 to 2T0 m while, The isothermal process is perform at minimum, 2, pressure remains constant is 3β T0 where β, temperature. If the absolute temperature, varies K times with in the cycle then find out, is a constant. Draw curve between volume (V), its efficiency., and temeprature (T) of the material., A), , B), , T, D) V, , C) V, , 9., , T, , T, , T, , An ideal gas is made to undergo a, thermodynamic process given by V ∝ T 2 ; find, the molar heat capacity of the gas for the above, process., , R, (a) ?− 1, ( ), , ?R, (b) ?− 1, ( ), , (a) 1 −, (c), , ln K, l, , (b) 1 +, , ln K, l, , (d), , ln K, l, , l, ln K, , 12. A cycle is made of three process iso baric,, adiabatic and isothermal. Isothermal process, 2? − 1 , ?−1 , R, R, has minimum temperature. Absolute, (c) , (d) , , , ?−1 , ?+1 , temperature charges by K times withing the, 10. Determine the average molar heat capacity of, cycle. Find out the efficiency., an ideal gas under going a process shown in, P, fig., , Pressure (P), , 1, , 2, , (KT), , A, 2P0, Qr, , P0, , 3, , B, V1, , V0, , ? −1 , R, ?, −, 1, , , , (a) , , ?− 2 , (c) , R, ?− 4 , , V2, , V, , V3, , 2V0 3V0, , Volume (V), , 96, , Qg, , 3? − 2 , R, ?, −, 1, , , , (b) , , ?+1 , (d) , R, ?−1 , , ln K, , (a) 1 − K − 1, (, ), , ln K, , (b) 1 + K − 1, (, ), , ln K, , (c) 1 + K + 1, (, ), , (d) None
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 13. A calorimeter contains 4.00 kg of water at 17. 450 g water at 40°C was kept in a calorimeter, of water equivalent 50 g, 25 g ice at 0°C is, 20.00 C . What amount of ice at − 10 0 C must, added and simultaneously 5 g steam at 100°C, be added to cause the resulting mixture to rach, was passed to the calorimeter. The final, thermal equilibrium at 5.00 C . Assume that, temperature of the calorimeter will be, heat transfer occurs only between the water, ( Lice = 80 cal / g , Lsteam = 540 cal / g ), and ice., (a) 0°C (b) 100°C (c) 40°C (d) None of these, (a) 0.55kg, (b) 0.66 kg, 18. 5 moles of nitrogen gas are enclosed in an, (c) 2.51kg, (d) 0.25kg, adiabatic cylindrical vessel. The piston itself, is a rigid light cylindrical container containing, 14. A 0.50 kg ice cube at −100 C is placed in 3.0, 3 moles of helium gas. There is a heater which, kg of coffee at 200 C . What will be the final, gives out a power 100cal to the nitrogen gas., A power of 30 cal is transferred to helium, temperature of mixture? Assume that specific, through the bottom surface of the piston. The, heat of tea is same as that of water., rate of increment of temperature of the, (a) 5.10 C (b) 80 C (c) 100 C, (d) 60 C, nitrogen gas assuming that the piston, 15. A thermally insulated vessel contains some, moves slowly, is, water at 0o C . The vessel is connected to a, vacuum pump to pump out water vapour, as a, result of this intense evaporation, some of the, He, water gets freezed. If latent heat of, vaporization at 0o C are L V = 580 cal / g and, , N2, , L f = 80 cal / g the maximum percentage, , amount of water that can be solidified in this, (a) 2 K/s (b) 4 K/s (c) 6 K/s (d) 8 K/s, manner, is (approximately), 19. When water is boiled at 2 atm pressure the, a) 1 2 % b) 1 8 %, c) 8 8 % d) 1 0 0 %, latent heat of vapourization is 2.2×106 J/kg and, 16. If specific heat capacity of a substance in solid, the boiling point is, 120°C. At 2 atm, and liquid state is proportional to the, pressure 1 kg of water has a volume of 10−3 m3, temperature of substance, then temperature, and 1 kg of steam has volume of 0.824 m3 . The, Vs time plot for the substance is best present, increase in internal energy of 1 kg of water, by [Assume heat is supplied to substance at, when it is converted into steam at 2 atm, constant rate and inital temperature is less than, pressure and 120°C is, melting point of substance.], T, , T, , (a), , t, , T, , (b), , t, , T, , (c), , t, , (d), , t, , [1 atm pressure, =1.013×105 N/ m 2 ], (a) 2.033 J, (b) 2.033 × 106 J, 6, (c) 0.167 × 10 J, (d) 2.267 × 106 J, 20. A vertical cylinder of cross-section area A, contains one mole of an ideal mono-atomic gas, under a piston of mass M. At a certain instant, a heater which supplies heat at the rate q J/s, is switched ON under the piston. The velocity, with which the piston moves upward under the, condition that pressure of gas remains constant, is [Assume no heat transfer through walls, of cylinder], 97
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, ρ0, , monoatomic gas is present in both the cylinders, at normal atmospheric pressure p0 . Both the, , M, , gases occupy same volume V0 , initially.., Now the piston of the left cylinder is, compressed in adiabatic manner so that volume, 2, q, (a) constant and equal to 5 × p A + Mg, 0, , V0, and then the, 2, left piston is clamped. Again the adiabatic slider, B is removed so that the two cylinders come, in thermal equilibrium. Assume all other, surfaces except A to be adiabatic. For this, situation, mark out the correct statement(s)., , of the left portion becomes, , 3, q, (b) constant and equal to 5 × p A + Mg, 0, , (c) varying, (d) constant but can’t be determined from given, information, 21. If CV for an ideal gas is given by Cv = 3 + 2T, where T is the absolute temperature of gas,, then the equation of adiabatic process for this, gas is, (a) V RT 2 = constant, (b) V RT 2e2T = constant, (c) V RT 2e2T = constant, , (d) V RT 3e2T = constant, , MULTIPLE ANSWER QUESTIONS, 22. One mole of an ideal monoatomic gas is taken, from A to C along the path ABC. The, tem perature of the gas at A is T0. For the, process ABC :, (A) work done by the gas is RT 0, (B) change in internal energy of the gas is, (C) heat absorbed by the gas is, , 11, RT0, 2, , 11, RT0, 2, , P, 2P0, , P0, , P0, n , p0,V0, , n , p0 ,V0, , A, B, , a) Just after the removal of adiabatic separator B,, the pressure in the left and right chambers are 2γ p0, and p0 , respectively.., b) After the removal of adiabatic separator B, the, gas in right chamber expands under constant, pressure process., c) Workdone by the gas of the right chamber on, surroundings during its expansion is 0.22 p0V0 ., d) During the expansion of gas in right chamber,, the energy transferred from the left chamber to right, chamber is 0.55 p0V0 where γ =, , C, , 5, ., 3, , COMPREHENSION TYPE QUESTIONS, P0, , A, , V0, , Comprehension - I, The molar specific heat of a gas is defined as, , B, , 2V0, , V, , (D) heat absorbed by the gas is, , 13, R T0, 2, , ., , (R = universal gas constant), 23. Two cylinders are connected by a fixed, diathermic partition A, and a removable, adiabatic partition B is placed adjacent to A as, shown in the figure. Inititally n moles of an ideal, 98, , dQ, Where dQ = heat absorbed, ndT, n = mole number dT = change in temperature, 24. A gas with adiabatic exponent ‘ γ ’ is expanded, C=, , according to the law p = α V . The initial, volume is V0 . The final volume isη V0 .(η > 1 )., The molar heat capacity of the gas in the, process is
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JEE- ADV PHYSICS-VOL- VI, R, (B), 2(γ − 1), , R, (A) (γ + 1), 2, , (C), , THERMODYNAMICS, , R (γ + 1), 2 (γ − 1), , (D), , R (γ − 1), 2 (γ + 1), , 25. An ideal gas whose adiabatic exponent is γ ., is expanded so that the heat transferred to the, gas is equal to decrease in its internal energy., The molar heat capacity in this process is, R, R, R, (B), (C), (D) R(γ − 1), γ −1, 1− γ, γ +1, 26. The equation of the above process in the, variables T, V is, , (A), , (A) TV, (C), , (γ −1), , γ , , , TV γ −1 , , = constnat, , γ −1 , , , (B) TV 2 , , = constnat (D), , = constnat, , γ −1 , , , TV γ , , = constnat, , 27. The speed of piston is, (A) 0.05 m/s, (B) 0.5 m/s, (C) 5 m/s, (D) None of these, 28. The final pressure of the gas is, (A) 1.8 atm, (B) 0.18 atm, (C) 1.5 atm, (D) None of these, 29. The average power of the heater is, (A) 49.36 KW, (B) 48.36 W, (C) 4.936 KW, (D) None of these, Comprehension - III, The rectangular box shown in the figure has a, partition which can slide without friction along the, length of the box. Initially each of two chambers of, the box have one mole of a mono-atomic ideal gas, ( γ = 5/3) at a pressure P 0, volume V 0 and, temperature T 0. The chamber on the left is slowly, heated by an electric heater. The walls of box and, partition are thermally insulated. Heat loss through, lead wire of heater is negligible. The gas in left, chamber expands, pushing the partition until the final, pressure in both chambers becomes 243p0/32., , Comprehension - II, The system shown in the figure is in equilibrium., The piston is massless, frictionless and insulated., All walls of the chamber are also insulated. When, heat is generated inside the lower chamber the, piston slowly moves upwards by 2 m and the liquid, comes out through an orifice so that it can rise to a, maximum height of 5m above the orifice level. The 30. Final temperature of the gas in right chamber, is, lower chamber contains 2 moles of an ideal, (A) 2.25 T 0 (B) 4.5 T 0 (C) 8.75 T 0 (D)12.93 T 0, monoatomic gas at 500K., 2, 31. Final temperature of the gas in left chamber is, Area of orifice a = 0.05 m, 2, (A) 2.25 T 0 (B) 4.5 T 0 (C) 8.75 T 0 (D) 12.93 T 0, Area of the piston A = 1 m, 3, 3, 2, 32. The work done by the gas in the right chamber, Density of the liquid, ρ = 10 kg/m , g =10 m/s, 5, 2, is, Atmospheric pressure, P atm = 10 N/m, (A) 5.5 T 0 J, (B) 10.5 T 0 J, Orifice, (C) 25.5 T0 J, (D) None of these, PASSAGE-IV:, An ideal monoatomic gas undergoes a pressure, Liquid, , H1 = 5m, , pV n = constant. The adiabatic constant for gas is, y. During the process, volume of gas increases from, V0 to rV0 and pressure decreases from p0 to, , Gas T=500K, H2 = 2m, , p0, Based on above information, answer the, 2r, , following questions:, 33. The value of n is, (a), , 2log r, log 2r, log 2r, log 2r, (b), (c), (d), log 2r, log r, 3log r, 3, 99
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , 34. The molar heat capacity of the gas for the MATRIX MATCHING TYPE QUESTIONS, process is, 38. Column I shows certain thermodynamic system, and column II represents thermodynamic, R(n − γ ), R(n − 1), (a), (b), properties., (n − 1)(γ − 1), (n − γ )(γ − 1), Column -I, A), R, R, R, +, (c), (d), Hg, γ −1, n −1 γ, PASSAGE-V:, 2000 mole of an ideal diatomic gas is enclosed in a, gas, h, vertical cylinder fitted with a piston and spring as, shown in the figure. Initially, the spring is, compressed by 5 cm and then the electric heater, An ideal gas is filled in a cylinderical vessel of height, starts supplying energy to the gas at constant rate, h which is enclosed by a massless thermally, of 100 J/s and due to conduction through walls of, insulating piston. Mercury is filled above the piston, cylinder and radiation, 20 J/s has been lost to, as shown. Now gas is slowly supplied heat., surroundings., Mercury does not spill., [ k = 1000 N / m, g = 10 m / s 2 , Atmospheric, , B), , pressure, p0 = 10 N / m , Cross-section area, 5, , 2, , of piston A0 = 50cm2 Mass of piston, m = 1 kg, R = 8.3 kJ / mol-K], Based on above information, answer the following, questions:, gas, , h, , k, , m, , 35. The initial pressure of the gas is, (a) 1 N / m 2, , (b) 1.02 N / m 2, , (c) 1.10 N / m 2, (d) 1.12 N / m 2, 36. Work done by the gas in t = 5 s is, (a) 300 J (b) 400 J (c) 114.3 J (d) 153.6 J, 37. Increase in temperature of gas in 5 s is, (a) 6.9 × 10−3 K, (b) 6.9 × K, (c) 83 × 10−4 K, , 100, , (d) 96 × 10−4 K, , A cylindrical vessel is enclosed by a light piston., The piston is connected to ceiling by an ideal spring, as shown in figure., Spring is initially relaxed and then heat is supplied, slowly to the ideal gas in the vessel. The system is, kept in open atmosphere., C), , Level of sand, , gas, , h, , A thermally insulated cylindrical vessel is enclosed, by a light thermally insulated piston., Some sand is kept on top of piston as shown in, figure. The system is kept in open atmosphere. Now, sand grains are removed slowly one by one.
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JEE- ADV PHYSICS-VOL- VI, , D), , THERMODYNAMICS, 40. Match the entries of column I and II, Column I, Column II, , Level of sand, , V, , P, B, , gas, , i) A, , h, , A, , C, , A), C, , 1, , B, V, , A good conducting cylindrical vessel is enclosed, by a light thermally insulated piston., Some sand iskept on top of piston as shown in, figure. Now sand grains are, added slowly one by one., Column -II, (p)Internal energy of the gas is increasing, (q) Pressure of the gas is increasing., (r) Temperature e of the gas is decreasing, (s) Work done by as is positive, (t) Molar heat capacity of the gas is positive, 39. Match the entries of column I and II., Column I, Column II, , (i), , B, , 4, , 4, , 1, , ii) A, , 1, , (ii), , B), 4, , P, A, , D, , B, , C, , 4, , 2, , iii), , C), , 4, , P, C, , B, , D), , A, , C, V, , 41 An ideal gas undergoes two processes A and, B. One of these is isothermal and the other is, adiabatic., COLUMN-I, COLUMN-II, A, , p) Heat supplied, , B, 4, , 3, , V, , (iii), , C), , 2, , during curve A is, positive, , 1, , V, , T, , (b) P, , V, , P, 1, , 3, , A, , 2, , 2, , (iv), , q) Work done by gas, , D), 3, , V, , P, , V, , 1, , C, , D, , T, , P, 3, , A, , V, , (a) P, , 1, , V, , 2, , B, , V, , 3, , 3, , C, T, , P, , 2, , 2, , B, , V, , B, , V, , 1, , B) D, , C, , D, , iv), , 3, , T, P, , A, , A, , 3, , A), , P, , P, , V, , 2, , T, , B, , 1, V, , T, , V, , in both processes, positive, 101
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, (c) P, , (d) V, A, , B, , A, , r) Internal energy, , s) W is negative, , B, V, , increases in adiabatic, process, , P, , whereas ? U is, positive, , (d) P, , INTEGER TYPE QUESTIONS, A, , s) Temperature of gas, , B, V, , 43. One mole of an ideal monatomic gas undergoes, the process P = aT, where a is a constant. The, work done by the gas if its temperature, , 50R, in process B is, increases by 50 K is, . Find the value of x., constant, x, 42. A sample of gas goes from state A to state B 44. A hot body placed in air is cooled down, in four different manners, as shown by the, according to Newton’s law of cooling, the rate, graphs. Let W be the work done by the gas, of decrease of temperature being k times the, and ? U be change in internal energy along, temperature difference from the surrounnding., the path AB., Starting from t = 0, The time in which the body, Correctly match graphs with the statements, will lose half the maximum heat it can lose is, provided., x ln 2, . Find the value of x., COLUMN-I, COLUMN-II, 2k, , (a) V, , 45. A gaseous mixture enclosed in a vessel, consisits of 1 g mole of gas A with ( γ1 = 5 / 3) and, A, , B, , p) Both W and ? U, P, , are positive, (b) P, B, , q) Both W and ? U, , A, T, , are negative, (c) T, A, , r) W is positive, , B, V, , whereas ? U is, nagative, 102, , another gas B with ( γ 2 = 7 / 5) at a temperature, T. The gases A and B do not react with each, other and assume to be ideal. Find the number, of gram moles of the gas B, if γ for the gaseous, mixture is (19/13)., 46. A lead ball at 30°C is dropped from a height h., The ball is heated due to the air resistance, and it completely melts just before reaching, the ground. The molten substance falls slowly, on the ground. Latent heat of fusion of lead is, 22200 J/kg. Specific heat capacity of lead =, 126 J/kg–°C and melting point of lead = 330°C., Assume that all mechanical energy lost is used, to heat the ball., Find the value of h in km ? (Use g=10 m/s2), 47. One end of a uniform rod of length 1 m is placed, in boiling water while its other end is placed in, melting ice. A point P on the rod is maintained, at a constant temperature of 800°C. The mass, of steam produced per second is equal to the
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , mass of ice melted persecond. If specific, latent heat of steam is 7 times the specific, latent heat of ice, then the distance of P from, the steam chamber is n/18 m. Find the value, of n?, 48. One mole of an ideal Po monoatomic gas is, taken 4po through a thermo dynamic process, shown in the p-V diagram. The heat supplied, to 2po the system in this process is, , K × ( π + 10 ) p0V0 .Determine the value of K., ρ0, 4ρ0, , 2ρ0, , B, , A, , VD, , the same pressure P0 and at the same, temperature. What work has to be performed, in order to increase isothermally the volume, of one part of gas η times compared to that of, the other by slowly moving the piston ?, x, , P1A, F, agent, , P2A, , 51. A monatomic ideal gas of two moles is taken, through a cyclic process starting from A as, shown in the figure. The volume ratios are, VB, V, = 2 and D = 4 . If the temperature T at, A, VA, AA, , C, , V, , VB, VA, O, , B, A, TA, , TB, T, , (A) The temperature of the gas at point B, (B) Heat absorbed or released by the gas in, each process, (C) The total work done by the gas during the, complete cycle, (Express your answer in terms of the gas, constant R), [IIT-2001], , LEVEL - VI - KEY, , V, , 49. One mole of an ideal gas whose pressure, changes with volume as P = αV , where α is a, constant, is expanded so that its volume, increase η times., Find the change in internal energy and heat, capacity of the gas., 50. A piston can freely move inside a horizontal, cylinder closed from both ends. Initially, the, piston separates the inside space of the, cylinder into two equal parts each of volume, V0 , in which an ideal gas is contained under, , D, , SINGLE ANSWER QUESTIONS, 1) A 2) B 3) C 4) B 5) C 6) A 7) C, 8) C 9) C 10) B 11) A 12) A 13) B 14) A, 15) C 16) D 17) C 18) A 19) B 20) A 21) D, MULTIPLE ANSWER QUESTIONS, 22) A,C, 23) A,B,C,D, COMPREHENSION TYPE QUESTIONS, PI: 24) C 25) B, 26) B, P:II 27) B 28) A, 29) A, PIII: 30) A 31) D 32) D, PIV: 33) C 34) A, PV: 35) D 36) C 37) A, MATRIX MATCHING TYPE QUESTIONS, 38) A-PST, B-PQST, C-RS, D-Q, 39) I-A, II-C, III-A, IV-D, 40) I-A; II-C; III-B; IV-D, 41) A-P,Q;B-R,S;C-P,Q; D-R,S, 42) A-S; B-Q; C-R; D-Q, INTEGER ANSWER TYPE QUESTIONS, 43) 2 44) 2 45) 2 46) 6 47) 2 48) 1, SUBJECTIVE TYPE QUESTIONS, 49) =, , 50), 51), , nR γ + 1 , 2 γ − 1 , , 2, , , η − 1 2 , 2, 2, , = − PV, ln, V, −, 0 , 0 0, V0 − ln V0 , , , η +1, , , A) 600 K B) 1500 R C) 600 R, , A is 27ºC, Calculate, 103
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, Heat of rise Heat to change , , , , m kg of ice + m kg of ice , from − 100 C to 00 , towater , , , , Heat to raise, Heat lost by 4 kg , , , , + m kg of water, = of water cooling , from 00 C to 5.00 C from 200 C to 50 C , , , , , miceCice ? T + mice L f + miceCw? T, m ( 2100 ) 00 C − ( −100 C ) + m ( 333), , + m ( 4186 ) ( 5.0 C − 0.0 C ), 0, , 0, , m ( 2100 )(10 ) + 333m + m ( 210 ) = 2510kJ, , 14. In this situation there are three possibilities regarding, the final state of the mixture:, 1. All ice 2. A mixture of ice water at 00 C, 3. All waer Energy release required to bring the, 3.0 kg of water at 200 C down to 00 C ,, , Q1 = mwCw ( 20 C − 0 C ), 0, , 0, , = ( 3)( 4186 )( 20 ) = 251kJ, Energy required to change the ice, from −100 C to 00 C ,, , Q2 = miceCice 00 C − ( −100 C ) , , Heat to raise Heat to change , , , , 0.50 kg, 0.50 kg of ice + , +, 0, from − 10 C to , , of ice, , , , 0, 0 C, , , to water, Heat to raise Heat lost by, , , , , 0.50 kg, = 3.0 kg of, , of water water cooling , , , , 0, 0, from 0 C toT from 20 C toT , , Hence,, , 10.5 + 167 + 0.50 ( 4186 ) T =, , ( 3)( 4186 ) ( 200 C − T ), , Which on solving yields T = 5.10 C, 15. Let mi be the amount of water that gets freezed, out of Mg of water 00 C due to evaporation., Now, mi × 80 = ( M − m1 ) 580, , mi 580, =, = 0.878, M 660, So, maximum % = 87.8% ; 88%, , ⇒, , 16. S = k1T, , dQ, = k2 dQ = msdl ⇒ dQ = k 2 dt, dt, , msdl = k2 dt, ⇒ ∫ mk1Tdt = ∫ k2 dt mk1, , T2, = k 2t, 2, , T 2 ∝ t ⇒ y 2 = 4ax, M.P., , = ( 0.50 )( 2100 )(10 ) = 10.5kJ, , M.P., time, , Energy required to change the ice to water at 00 C, , Q3 = mice L f = ( 0.50 )( 333) = 167 kJ, A total of 10.5kJ + 167 kJ = 177kJ, energy is required to bring ice at −100 C, to water at 200 C down to 00 C is 250kJ., Thus the mixture must end up all water, somewhere between 00 C and 200 C ., Let the final temperature be T. Now, applying energy conservation,, 106, , After that the nature of existance of matter change, so proportionally constant cahnges., S = k3t dθ = msdt Similarly, mk3, B.P., M.P., , 1, t, , T2, = k2 t T 2 ∝ t, 2
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, T0V0, , ?−1, , = T1 (V0 / 2), , ?−1, , pV, , 0 0, ⇒ T1 = 2?−1T0 Where T0 = nR, , When adiabatic separator is removed, temperature, on both sides are different, temperature on left is, , T1 while temperature on right is T0 , so heat transfer, takes place from left chamber to right chamber and, the gas in right chamber starts expanding, as piston, (rigth side) is free to move, the pressure remains, , C=, pdV =, , −2RdT , R, R, +, =, γ − 1 ( γ − 1)dT 1 − γ, , −2, −2RdT RT, dT, ( γ − 1) dV, dV =, RdT, =−, γ −1, V, γ −1, T, 2 V, , log T = −, , − ( γ−1), ( γ − 1), log V = log V 2 + log K, 2, , γ−1, logTV 2, , or, = constant, constant as p0 . So, expansion in right chamber is, taking place at constant temperature. During this Comprehension-II (27-29), Assume datum at the position of the orifice. Initial, expansion of gas in right chamber, the volume of, energy of the liquid, gas in left chamber remains constant but temperature, and pressure change. For the instant just after the, removal of adiabatic separator we can take the, H = 5m, pressure of gas in left chamber as the same just, Datum, 0, , before the removal i.e., p0 × 2 ? ., Let the final temperature on both the side be, T, then the energy transferred from left to, right is used to increase the temperature of, the right part and to expand the gas., i.e., nCV (T1 − T ) = nC p ( T − T0 ), , = nCV ( T − T0 ) + W, ?−1, 5T0 + 3T1 = 5 + 3 × 2 T ; 1.22T, , 0, 0, ⇒T =, 8, 8, , , , So, the energy transferred is,, , Q = nCV (T1 − T ) =, , 3nR ?−1, ( 2 − 1.22 ) T0, 2, , (, , ), , Work done, W = p0 V f − Vi = nR (T − T0 ), , = 0.22 nRT0 ; 0.22 p0V0, COMPREHENSION TYPE QUESTIONS, Comprehension I: (24-26), , =, , R, dV, +p, γ −1, dT, , R, R R( γ + 1), R, R, =, + =, + αV., γ −1, 2αV γ − 1 2 2( γ − 1), , Q = −∆U = ∆U + W or W = −2∆U, , 108, , H1 = 5m, H = 2m, , Gas, H2 = 2m, , V=, , a, 0.05, 2gh =, 2 × 10 × 5 = 0.5 m / s, A, 1, , 1, 1, Ei = − ρgAH12 = (103 )(10)(52 ) = −1.25 × 105 J, 2, 2, 1, 2, , 2, and final energy E r = − ρA(H1 − h) + ρgAhh 0, , 1, = − (103 )(10)(1)(5 − 1) 2 + (103 )(10)(1)(2)(5), 2, −0.45 × 105 + 105, , = 0.55 p0V0, , C=, , Liquid, , Work done by the expanding gas is, W = Er – Ei = 1.25 × 105 – 0.45 × 105 + 105, = 1.8 × 105 J, Initial pressure of the gas, p1 = Patm + ρgH1, = 105 + (103)(10)(5) = 1.5 × 105 Pa, Final, pressure, of, the, gas,, p2 = Patm + ρg(H1 − h + h 0 ), , = 105 + (103)(10)(5 – 2 + 5) = 1.8 × 105 Pa, Initial volume, V 1 = AH2 = (1)(2) = 2m3, Final volume, V2 = A(H2 + h) = (1)(2 + 2) = 4m3, pV, , p V, , p V, , (1.8)(4), , 1 1, 2 2, 2 2, Since T = T or T2 = p V T1 = (1.5)(2), 1, 2, 1 1
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, , MATRIX MATCHING TYPE QUESTIONS, 3)Similar to 1, 38., 2(PV= Constant), 4) 1, A) dQ = du +dw, T ( Constant ), As heat is supplied to the gas, the internal energy, 2, 3 V ( Constant ), of the gas increases, It’s temperature also increases, and some part of heat supplied is converted into, 3, 1 ( Adiabatic), work which rises the level of position. Hence, change in internal energy Positive., TV γ −1 = Constant, V, Temperature ⇒ increases Work done ⇒ positive, Downward force acting on piston = weight of, 3, 2, mercury = constant.Hence pressure inside the gas, is also constant, b) Downward force on position is due to spring As, 1, the level of position rises, the force due to spring is, T, also increases., P, B, = Constant, Hence neutralize the downward force, pressure 40) i) Process A, V, of gas gets increased., ( PV n = constant; then P1−nT n = Constant), c) As the position rises work done is positive.., P, d) change in internal energy is zero - temperature, C, =constant. Pressure increases as downward force, on piston increases, 39) 1) 1, , 2 P ( Constant), , A, , V, , VαT, , 2, , V, , Therefore,, 4, , 2, , 3 V ( Constant ), , 3, , 1, , 2, , 4 ( V constnat ), , VαT, 4, 2) 1, , P, = Constant, T, ii) Process A, , 1 V ( Constant ), 2, , P ( Constant), ⇒, V, 4, , 3, , 2, , 3 ( PV = K ), , T ( Constant ), , P, = Constant, T, , Process C, P = Constant, , D, , Proces B, , C, , T, , B, , 4 ( V constnat ), , 4, 1 PV = K, T ( Constant ), 110, , B( V =Constnat ), , ⇒ T = Constant, , 1, , 3, , Proces B, C, ( V =Constnat ), ⇒, , VαT, , 2, , P, = Constant, T, , Process C, A, ( P = Constant), T, , 3, , B, , A, , D, , C, T
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JEE- ADV PHYSICS-VOL- VI, B, , iii) Process A, , THERMODYNAMICS, , Proces B, , V = k1, , ∴W = + ve And U = -ve, b) process A is adiabatic and B is isothermal for, A:- It is a adiabatic compression So temp increases . So, internal energy increases., B is isothermal so temperature is constant, c) Process A is isothermal and B is adiabatic, for A:- Q=W [w is +ve], Q is +ve For B:- w = -u, Same as is first case., d) (R,S) process B is isothermal and A is adiabatic, same as in second case., , A, , P = Constant, , P, = k1, T, , Process C, , V = k2, , B, , Proces D, , C, , P = Constant, , P, = k2, T, Hence, , k2 > k1, , 42. a) Apply, , P, , A, D, , has more P than A, ∴ TB > TA Hence ∆U is, +ve. The area enclosed in Pv graph is -ve, ( are, a along y –axis), , B, , b) From the graph TB < TA ; so, ∆U = − ve ., , C, , The graph is P = - M T + K, , T, , − mpv, PV, =T, P =, +k, Applying PV = RT , ⇒, R, R, , iv) Process C, A, ( P = Constant), Process A, B, (PV= Constant), , ⇒ PT, , k, PV, = k ⇒ p = −cpv + k ⇒ 1 = −v +, p, T, From the graph W would be -ve, , Also apply, , n1, 1− x1, , Proces B, , ⇒ 1+ v =, , C, , (PV= Constant), , PV, = constant A & B have same V but B, T, , ⇒ PT, , n2, 1− x2, , c), , n1 > n 2, , k, p, , VT = K, , TA > TB , ∆U = -ve, , PV = RT, , PV PV 2 = cons tan t, , VB > VA, P, , ∴W + ve, , d) TA > TB , ∆U =-ve, , B, , V, , V, , +ve area, A, O, , C, V, , –ve area, , 41 a ): process A isothermal process and ‘B’, adiabatic ., For A:- Q = u + w[u = 0] Q = w, Here w is +ve [V2 > V1 ] So Q is +ve, For B:- Q = u + w[Q = 0] w =, , nR, [T2 − T1 ], 1− γ, , is, , –ve area, , P, , P, , INTEGER ANSWER TYPE QUESTIONS, 43. (2), P = αT1/ 2, where α is constant, , (, , .........(i), n = 1 mol, monatomic, , ), , W = ∫ P dV = ∫ αT1 / 2 dV, , 111
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 4., , LEVEL - V, , Which of the following graphs correctly, represent the variation of ß = –, , THERMODYNAMICS & KTG, 1., , dV dP, V, , with P, , for an ideal gas at constant temperature?, (IIT - 2002), (A) β, (B) β, , SINGLE ANSWER QUESTIONS, During an experiment, an ideal gas is found, P2, , to obey a condition ρ = constant [ρ=density of, , 2., , the gas]. The gas is initially at temperature T,, pressure P and density ρ. The gas expands, such that density changes to ρ /2 –, (A) The pressure of the gas changes to 2P, (B) The temperature of the gas changes to 2T, (C) The graph of the above process on the P-T, diagram is parabola, (D) The graph of the above process on the P-T, diagram is straight line, Pressure versus temperature graph of an ideal, gas of equal number of moles of different, volumes are plotted as shown in figure., Choose the correct alternative :, (A)V1 = V2, V3 = V4 and V2 > V3, (B) V1 = V2, V3 = V4 and V2 < V3, (C) V1 = V2 = V3 = V4, (D) V4 > V3 > V2 > V1., , P, (D) β, , (C) β, , 5., , P, P, Pressure versus temperature graph of an ideal, gas is shown in figure. Density of the gas at, point A is ρ0 . Density at B will be, P, , P0, , A, , T0, , 4, , T, , 2T0, , 3, 3, 4, ρ0 (B) ρ0, (C) ρ0 (D) 2ρ0, 4, 2, 3, An ideal gas is initially at temperature T and, volume V. It volume is increased by ∆V due, to an increase in temperature ∆T , pressure, remaining constant. The quantity, δ = ∆V / V ∆T varies with temperature as, (IIT - 2000), , (A), , 6., , 2, 3, 1, T, , One mole of an ideal gas undergoes a process, P=, , B, , 3P0, , P, , 3., , P, , (A), , (B) δ, , δ, , P0, 2, , V , 1 + 0 . Here, P 0 and V 0 are constants., V, , Change in temperature of the gas when, volume is changed from V = V 0 to V = 2V0 is :, (A) −, , 2P0 V0, 5R, , (B), , (C) −, , 5P0 V0, 4R, , (D) P0V0 ., , (C), , δ, , T, , T+∆T, , T, , (D) δ, , T, , T+∆T, , T, , T+∆T, , T, , 11P0 V0, 10R, T, , T+∆T, , T, , T, , 23
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 7., , The air tight and smooth piston of a, cylinderical vessel are connected with a string,, as shown. Initially pressure and temperature, , LEVEL -V-KEY, SINGLE ANSWER QUESTIONS, 1) B, 2) A 3)B, 4)A 5)B, 6)C, 7)C 8) D 9) D 10) A, , of the gas are P0 and T0 . The atmospheric, pressure is also P0 . At a later time, tension in, 3, the string is P0 A where A is the cross8, sectional area of the cylinder. At this time, the, temperature of the gas has become., , 8., , 9., , 3, 3, 11, 13, B) T0, C) T0 D) T0, A) T0, 8, 4, 8, 8, A real gas behaves like an ideal gas if its, (IIT JEE-2010), (a) pressure and temperature are both high, (b) pressure and temperature are both low, (c) pressure is high and temperature is low, (d) pressure is low and temperature is high, Two cylinders fitted with pistons and placed, as shown, connected with string through a, small tube of negligible volume, are filled with, , a gas at pressure P0 and temperature T0 . The, radius of smaller cylinder is half of the other., , LEVEL - V - HINTS, SINGLE ANSWER QUESTIONS, 1., , P2 P′ 2, P, =, ⇒ P′ =, Hence from (1), T ′ = T 2 ., ρ ρ/ 2, 2, , 2., , (slope)1 2 < (slope)3 4 ∴ V2 > V3, Also, V1 = V2 and V3 = V4., 3., , b), , 3, P0, 5, , c), , 2, P0, 5, , P0, 2, , and at V = 2V0 , P =, , R/2, , (n = 1), , 4P0, 5, , 4P , (2V0 ) 0 , PV, 5 = 8P0 V0, ∴ Tf =, =, nR, R, 5R, , 1m, , 4, P0, 5, , At V = V0, P =, , P0 , (V ), PV, PV 2 0, ∴ Ti =, =, = 0 0, nR, R, 2R, , 1m, , a), , PT = constant hence P-T curve is a hyparbola., In isochoric process V = constant and P ∝ T, Therefore, P–T graph is a straight line passing, through origin. But since, , nR , 1, P=, T slope of the straight line ∝, V, V , , If the temperature is increased to 2T0 , find, the pressure, if the piston of bigger cylinder, moves towards left by 1 metre?, R, , P2, P 2 RT, kM , =k⇒, = k ⇒ PT = , R ........... (1), ρ, PM, , d), , 8 1 P V 11P0 V0, ∴ ∆ T = Tf − Ti = − 0 0 =, ., 10R, 5 2 R, , 5, P0, 4, , 10. One mole of an ideal gas undergoes a process, 4., , −1, , 2V0 2 , P = P0 1 + , , where, V , , P0 ,V0 are, , constants. Change in temperature of the gas, when volume is changed from V = V0 to, , V = 2V0 is:, 4 PV, 3 PV, 2 PV, 7 PV, 0 0, 0 0, 0 0, 0 0, a), b), c), d), 5 nR, 4 nR, 3 nR, 9 nR, 24, , At a constant temperature, for a given mass of gas,, PV = constant, according to Boyle’s law, ∴ PV = constant ∴ PdV + VdP = 0, or, , dV, V, =−, dP, P, , or β =, , or −, , dV, 1, =+, VdP, P, , 1, or β P = 1, P, , The equation between β and P is of the form xy =, constant which represents a rectangular hyperbola.
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JEE- ADV PHYSICS-VOL- VI, , THERMODYNAMICS, 2., , Assume that the temperature remains, essentially constant in the upper parts of the, atmosphere. The atmospheric pressure varies, with height as. (the mean molecular weight of, air is M, where P0 = atmospheric pressure at, ground reference), A) P e, 0, , −3Mgh, 2 RT, , − Mgh, , B) P e 2 RT, 0, , −3Mgh, , 3., , 4., , − Mgh, , D) P e RT, C) P e RT, 0, 0, Assume a sample of a gas in a vessel. The, speeds of molecules are between 2 m/s to 5, m/s, The number of molecules for speed v (m/, s) is given by n = 7v − v 2 − 10 . The most, probable speed in the sample is, (a) 3.5 m/s (b) 5 m/s (c) 10 m/s (d) 4 m/s, Tyre of a bicycle has volume 2 × 10−3 m3 ., Initially the tube is filled to 75% of its volume, by air at atmospheric pressure of, p0 = 105 N / m 2 . When a rider rides the bicycle, the area of contact of tyre with road, is A = 24 × 10−5 m 2 . The mass of rider with, bicycle is 120 kg. The number of strokes which, delivers, V = 500 cm3 volume of air in each, stroke required to inflate the tyres is, [Take g = m / s 2 ], (a) 10, (b) 11, (c) 20, (d) 21, , Comprehension - I, A small spherical monoatomic ideal gas bubble, 5, , γ = is trapped inside a liquid of density, 3, , , ρl (see figure). Assume that the bubble does not, , exchange any heat with the liquid. The bubble, contains n moles of gas. The temperature of the, gas when the bubble is at the battom is T0 the height, of the liquid is H and the atmospheric pressure is, P0 (Neglect surface tension)., (IIT - 2008), ρ0, , Liquid, , H, y, , 26, , 5., , 6., , As the air bubble moves upward, besides, buoyancy force, the follwing forces are acting, on it, A) Only the force of gravity, B) The force due to gravity and the force due to, pressure of the liquid, C) The force due to gravity, the force due to, pressure of the liquid and the force due to viscosity, of the liquid, D) The force due to gravity and the force due to, viscosity of the liquid, When the gas bubble is at height y from the, bottom, its temperature is, P0 + ρl gh , , A) T0 , P0 + ρl gy , , 2/5, , P0 + ρl g ( H − y ) , , B) T0 , P0 + ρl gH , P0 + ρl gH , , C) T0 , P0 + ρl gy , , 2/5, , 3/5, , P0 + ρl g ( H − y ) , , D) T0 , P0 + ρl gH , , 7., , 3/5, , The buoyancy force acting on the gas bubble, is, ( P + ρ gH ) 2/5, , 0, l, A) ρl nRgT0 ( P + ρ gy ) 7/5, 0, l, , ρl nRgT0, , B) ( P + ρ gH )2/5 {( P + ρ g ( H − y )}3/5, 0, l, 0, l, C) ρl nRgT0, , ( P0 + ρl gH )3/5, ( P0 + ρl gy )8/5, , ρl nRgT0, , D) ( P + ρ gH )3/5 {( P + ρ g ( H − y)}2/5, 0, l, 0, l, Comprehension - II, A very tall vertical cylinder is filled with a gas of, molar mass M under isothermal conditions, temperature T. The density and pressure of the gas, at the base of the container is ρ0 and p0, respectively
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JEE- ADV PHYSICS-VOL- VI, 8., , THERMODYNAMICS, , Select the incorrect statement, (A) Pressure decreases with height, (B) The rate of decrease of pressure with height is, a constant, , ⇒, , V, P, , ∫, , ⇒, , dP, = −ρg where ρ is density of the gas at a, dh, height h, , P0, , (C), , RT, M, Select the incorrect statement if gravity is, assumed to be constant throughout the, container, (A) Both pressure and density decreases, exponentially with height, , (B) The variation of pressure is P = P0, (C) The variation of density ρ = ρ e, 0, , 2., , ⇒, , P = P0 e− rt / V0, , Since pressure decrased with height, , Consider a small volume ∆V of air of mass ∆m, , P∆V = nRT, ∆m , P∆V = , RT, M , , Mgh, RT, , (D) The molecular density decreases as one moves, upwards., 10. Select the correct statement, (A) The density of gas cannot be uniform, throughout the cylinder, (B) The density of gas cannot be uniform throughout, the cylinder under isothermal conditions, , P=, , ∆m RT, ρ RT, PM, ⇒P=, ⇒ρ=, ∆v M, M, RT, , ∴dp =, p, , dρ ρMg, =, (C) The rate of change of density, dh, RT, , ∫, , P0, , (D) All of the above, , PM, g dh, RT, h, , dp, Mg, =−, dh, p, RT ∫0, , p, − Mgh, ln = −, P, RT, 0, , LEVEL VI - KEY, , P = P0, 3., , LEVEL VI - HINTS, , − Mgh, e RT, , n = 7? − ?2 − 10, For most probable velocity, , SINGLE ANSWER QUESTIONS, 1., , P − rt, =, P0 V0, , ln, , dp = pgdh, , Mgh, −, e RT, , SINGLE ANSWER QUESTIONS, 1) D, 2) D, 3) A, 4) D, COMPREHENSION TYPE QUESTIONS, P:I, 5) D 6) B 7) B, PII:, 8) B 9) C 10) D, , t, , dP, 1, = − ∫ r dt, P, V0 0, , ⇒, , (D) P = ρ, 9., , dP, = −rP, dt, , dn, =0, d?, , The temperature of the gas is constant, therefore, , (Q n is maximum at this velocity), , PV = constant, , 7 − 2? = 0 ⇒ ? =, , ⇒, , P, , 7, = 3.5 m / s, 2, , dV, dP, +V, =0, dt, dt, , 4., , The pressure of air in the tube is, 27
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JEE-ADV PHYSICS-VOL- I, , WAVES, , WAVES, SYNOPSIS, Propagation, wave, , Introduction, There are essentially two ways of transporting, energy from the place where it is produced to, the place where it is desired to be utilized., The first involves the actual transport of matter., For example, a bullet fired from a gun carries, its kinetic energy with it which can be used at, another location. The second method by which, energy can be transported is much more useful, and important, it involves what we call a wave, process., A wave is a disturbance that propagates in, space, transports energy and momentum from, one point to another without the transport of, matter. Waves are every where whether we, recognize or not, we encounter waves on a, daily basis. Sound waves, visible light waves,, radio waves, ripples on water surface,, earthquake waves and waves on a string are, just a few examples of waves., Waves can be one, two or three dimensional, according to the number of dimensions in which, they propagate energy. Waves moving along, strings are one dimensional, ripples on liquid, surface are two dimensional, while sound and, light waves are three dimensional., , Types of Waves, Waves can be classified in a number of ways, based on the following characteristics, , On the basis of necessity of medium, i)Mechanical waves: Require medium for, their propagation e.g., Waves on string and, spring, waves on water surface, sound waves,, seismic waves., ii) Non-mechanical waves: Do not require, medium for their propagation are called e.g.,, Electromagnetic waves like, light, heat, (Infrared), radio waves, -rays, x-rays etc., , On the basis of vibration of particle:, , 1), , On the basis of vibration of particle of medium, waves can be classified as transverse waves, and longitudinal waves., Transverse waves: i) Particles of the, medium vibrates in a direction perpendicular, to the direction propagation of wave, , NARAYANAGROUP, , Vibration of particle, , ii) It travels in the form of crests(C) and, troughts(T), C, , C, , T, , 2), , T, , iii) Transverse waves can be transmitted, through solids, they can be setup on the surface, of, liquids. But they cannot be trnasmitted, into liquids and gases., iv) Medium should posses the property of, rigidity, v) Transverse waves can be polarised., vi) Movement of string of a sitar or violin,, movement of the membrane of a Tabla or, Dholak, movement of kink on a rope waves, setup on the surface of water., Longitudinal waves : i) Particles of a medium, vibrate in the direction of wave motion., Propagation of wave, Vibration of Particle, , ii) It travels in the form of compression (C), rarefaction (R)., Maximum Pressure, and density, C, , R, , C, , R, , Minimum Pressure, and density, , iii) These waves can be transmitted through, solids, liquids and gases because for, propagation, volume elasticiy is necessary., iv) Medium should posses the property of, elasticity., v) Longitudinal waves can not be polarized., vi) Sound waves travel through air, vibration, of air column in organ pipes vibration of air, column above the surface of water in the tube, of resonance apparatus., 1
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , On the basis of energy propagation, i) Progressive wave: These waves advances, in a medium with definite velocity. These waves, propagate energy in the medium. Eg: Sound, wave and light waves., ii) Stationary wave: These waves remains, stationary between two boundaries in medium., Energy is not propagated by these waves but it, is confined in segments (or loops) e.g., Wave in, a string, waves in organ pipes., , 5., , particle to complete one vibration about its mean, position, Or it is the time taken by the wave to travel a, distance equal to one wavelength., Time period = 1/Frequency T 1/ n, Wave pulse: It is a short wave produced in a, medium when the disturbance is created for a, short time., , Simple Harmonic wave, When a wave passes through a medium, if the, particles of the medium execute simple harmonic, vibrations, then the wave is called a simple, harmonic wave. A graph is drawn (fig.) with the, displacement of the particles from their mean, positions, at any given instant of time, on the, y-axis and their location from origin on x-axis., y, , x, T, , , , Characteristics of wave:, 1. Amplitude (A): Maximum displacement of a, vibrating particle of medium from its’ mean, position is called amplitude., Wavelength : It is equal to the distance, travelled by the wave during the time in which, any one particle of the medium completes one, vibration about its mean position., Or, Distance travelled by the wave in one time, period is known as wavelength., Or, It is the distance between the two successive, points with same phase., C, , , , wave train., , 7. Wave function: It is a mathematical, , A, , 2., , 6. Wave train: A series of wave pulse is called, , C, , , C, , T, , T, Transverse wave, , R, , C, , R, , C, , R, , description of the disturbance created by a wave., For a string, the wave function is a displacement., For sound waves it is a pressure or density, fluctuation where as for light waves it is electric, or magnetic field., Now let us consider a one dimensional wave, travelling along x-axis. During wave motion, a, particle with equilibrium position x is displaced, some distance y in the direction perpendicular, to the x-axis. In this case y is a function of, position (x) and time (t)., i.e., y = f(x, t). This is called wave function., Let the wave pulse be travelling with a speed v., After a time t, the pulse reaches a distance vt, along the +x-axis as shown. Thus the motion of, the particle P1 at distance ' x ' at time ‘ t ’ is, same as the motion of the particle P at time, t 0 at position x0 x vt . Hence the wave, functioon now can be represented as, y f ( x vt ) ., , C, , v, , , , Longitudinal wave, , 3. Frequency (n): Frequency of vibration of a, , 4., , 2, , particle is defined as the number of vibrations, completed by particle in one second., (Or)it is the number of complete wavelengths, traversed by the wave in one second., Unit of frequency is hertz (Hz) or per second., Time period (T): Time period of vibration of, particle is defined as the time taken by the, , =, , In general, then we can represent the transverse, position y for all positions and times, measured, in stationary frame with the origin at O, as, y(x, t) = f(x – vt), ............. (i), NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , Similarly, if the pulse travels to the left, the 12. Phase: Phase gives the state of the vibrating, particle at any instant of time as regards to its, transverse position of elements of the string is, position and direction of motion., described by, y(x, t) = f(x + vt), ............. (ii) Phase is the angular displacement from its mean, position. = ( t kx), The function y, sometimes called the wave, function, depends on the two variables x and t. If phase is constant then the shape of wave, remains constant., For this reason, it is often written y(x, t), which, is read “y as a function of x and t”., Equation of Progressive Wave :, Note-1: The equation y = f(vt – x) represents the 1. If during the propagation of a progressive wave,, displacement of the particle at x = 0 as time, the particles of the medium perform SHM about, passes, their mean position, then the wave is known as a, y, y, harmonic progressive wave., A, A, t, o, o, t, 2., Suppose a plane simple harmonic wave travels, T/2, /2, T, , -A, -A, from the origin along the positive direction of, Representation, Representation, x-axis from left to right as shown in the figure, x, , of y f t , v, , , y, , of y f ( x vt ), , Note-2: If order of a wave function to represent a, wave, the three quantities x, v, t must appear in, combinations ( x vt ) or ( x vt ) ., B x vt , , 2, , y x vt , x vt , Ae, represents, travelling, waves, Thus, , y x 2 v 2t 2 ,, , 8., , 9., , , , 2, , etc.,, while, , , , x vt , A sin 4 x 2 9t 2 , etc. do not represent a wave., Harmonic wave: If a travelling wave is a sin, or cos function of x vt the wave is said to, be harmonic or plane progressive wave., The differential form of wave equation:, All the travelling waves satisfy a differential, equation which is called the wave equation. It, 2, , 2 y, 2 y, v, , v, 2, 2 ; where, k, t, x, It is satisfied by any equation of the form, , is given by, , y f x vt , , 10. Angular wave number (or) propagation, constant (k): Number of wavelengths in the, distance 2 is called the wave number or, propagation constant i.e., k , , 2, , , It is unit is rad/m., , 11. Wave velocity (v): It is the distance travelled, by the disturbance in one second. It only depends, on the properties of the medium and is, independent of time and position., , v n , , T 2 k, NARAYANAGROUP, , v, y, , o, , x, , x, , The displacement y of a particle at O from its, mean position at any time t is given by, y A sin t . --- (1), The wave reaches the particle P after time, x, t ., v, So that the motion of the particle ‘ P ’ which is, at a distance ‘ x ’ at a time ‘ t ’ is same as motion, of the particle at x 0 , at the earliear time, x, t ., v, Hence the displacement ‘ y ’ of the particle ‘ P ’, x, at ‘ x ’ at a time ‘ t ’ in equation (1) by t ., v, , x, , y A sin t A sin t kx k , v, v, , In general along x-axis, y = A Sin ( t kx), + sign for a wave t ravelling along -ve, X direction, - sign for a wave travelling along +ve, X direction, where y is displacement of the particle after a, time t from mean position, x is displacement of, the wave, A is Amplitude., is angular frequency or angular velocity, 2 / T = 2 n, k is propagation constant & k = 2 / , For a given time ‘t’, y x graph gives the shape, of pulse on string., 3
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , Various forms of progressive wave, function:, , y, , (i) y A sin t kx (or) y A sin( kx t ), , o, , kx, , (ii) y A cos(t kx) (or) y A cos(kx t ), , , 2 , , x, (iii) y A sin t , , , , Change in Phase with time for a constant x,, i.e., at a fixed point in the medium, , t x, (iv) y A sin 2 , T , , x, t x, t, 2 1 ; t 2 2 , 2, T , T , , t, , 1, , y, , 2 , T, (v) y A sin t x , T , , , t, , 2, (vi) y A sin vt x , , t, , x, (vii) y A sin t , v, t x, (viii) y A sin 2 , T , , General Expression for a Sinusoidal Wave, Y A sin kx t (or), Y A sin(t kx ), where is the phase constant, just as we learned, in our study of periodic motion. This constant, can be determined from the initial conditions., , Positive and Negative Initial Phase, Constants., In general, the equation of a harmonic wave, travelling along the positive x-axis is expressed, as y A sin kx t . Where is called, the initial phase constant. It determines the initial, displacement of the particle at x = 0 when t = 0., i)Posit ive, init ial, phase, constant, y Asin(kx t ) .The sine curve starts from, the left of the origin., , (For the wave travelling in positive x-direction), 2, 2, t t , t2 t1 , t, 2, 1, T, T, 2 t, , T, 2, Time difference, Phase difference , T, , Variation of Phase with Distance, At a given instant of time t = t, phase at x = x1,, t x , 2 1 , 1, T , (For the wave travelling in positive x-direction, and phase at x = x2,, , x, , x, , 2, , t x , 2 2 , T , , y, , t, , y, , t, , , , o, , kx, , ii) Negative initial phase constant, y Asin(kx t ) . The sine curve starts from, the right of the origin., 4, , x x , 2, , , , 1, , 2, 2, x2 x1 x, , , , 2, x, , , i.e., Phase difference , , 2, Path difference, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , Particle Velocity: The rate of change of, displacement y w.r.t time t is known as particle, velocity., Hence from y A sin t kx , Particle velocity, v p , , y, A cos t kx , t, , Maximum particle velocity v p max A, y, y, , t, k x, Particle, velocity, at, a given position and time is, , equal to negative of the product of wave velocity, with slope of wave at that point i.e., , Also, , v, , particle, , y , , x , , vWave , , Particle velocity = –(wave velocity) × slope of, wave curve, v, v, , WAVES, Intensity is defined as power per unit area., E P 1, I, v 2 A2 2 2 f 2 A 2 v, S t S 2, If frequency f is constant then I A2, , Reflection and Refraction of Waves :, When waves are incident on a boundary between, two media a part of incident waves returns back, into the initial medium (reflection) while the, remaining is partly absorbed and part ly, transmitted into the second medium (refraction), Boundary conditions: Reflection of a wave, pulse from some boundary depends on the nature, of the boundary., Rigid end: When the incident wave reaches a, fixed end, it exerts an upward pull on the end,, according to Newton’s third law the fixed end, exerts an equal and opposite downward force, on the string. It result as inverted pulse or phase, change of ., Crest (C) reflects as trough (T) and vice-versa., T, , Time changes by and Path changes by, 2, 2, C, , ENERGY, POWER AND INTENSITY, OF A WAVE:, If a wave given by y A sin(t kx) is propagating through a medium, the particle velocity, y, A cos(t kx), will be v p , t, If is the density of the medium, kinetic energy of the wave per unit volume will be, , C, , F, , T, F, , Free end: When a wave or pulse is reflected, from a free end, then there is no change of phase, (as there is no reaction force)., Crest (C) reflects as crest (C) and trough (T), reflects as trough (T), Time changes by zero and, Path changes by zero., C, , C, , C, , 2, , Note: Exception: Longitudinal pressure waves, suffer no change in phase from rigid end. i.e.,, compression pulse reflects as compression, and its maximum value will be equal to energy, pulse. On the other hand if longitudinal pressure, per unit volume i.e., energy density U., wave reflects from free end, it suffer a phase, 1, U A2 2, change of , i.e., compression reflects as, 2, rarefaction and vice-versa., The energy associated with a volume V S x, Wave in a combination of string, will be (where ‘S’ is the area of cross section)., (i) Wave goes from thin to thick string, 1, 2 2, Rigid boundary-x, Incident, +x, E U V A S x, wave, 2, Rarer, Denser, The power (rate of transmission of energy) will, , , 1 y , 1, , 2 A2 cos 2 (t kx ), 2 t , 2, , be, , P, , E 1, v 2 A2 S, t 2, , x, , as t v, (Speed of wave) , , NARAYANAGROUP, , Reflected, wave, , Transmitted wave, , 5
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , Speed of transverse wave in a string, , Incident wave yi ai sin t k1 x , Reflected wave, , i) Let a transverse pulse is travelling on a, stretched string as shown in fig(a)., , yr ar sin t k1 x , , v, , a), , ar sin t k1 x , Transmitted wave, yt at sin t k2 x , , T cos, , (ii) Wave goes from thick to thin string, Incident, wave, , Free boundary, -x, , b), , +x, , T cos, , dl, , , T sin, , , T, , T sin, C, , Denser, , ii) Now consider a small element of length dl, on this pulse as shown fig (b). Let this element, is forming an arc of radius R and subtending an, angle 2 at center of curvature C., iii)We can see that two tensions T are acting on, the edges of dl along tangential directions as, shown., iv)The horizontal components of these tensions, cancel each other, but the vertical components, add to form a radial restoring force in downward, direction, which is given as, , Rarer, , Transmitted wave, , Reflected, wave, , Incident wave yi at sin t k1 x , Reflected wave yr ar sin t k1 x 0 , , ar sin t k1 x , Transmitted wave yt at sin t k2 x , Note: Ratio of amplitudes: It is given as follows, ar k1 k2 v2 v1, at, 2k1, 2v2, , , , , and, ai k1 k2 v2 v1, ai k1 k 2 v1 v2, , FR 2T sin 2T , , ( as sin ), T, , The Speed of A Travelling Wave, , v, , ii) Let a crest shown by a dot () moves a, distance x in time t . The speed of the wave, is v x t ., iii) We can put the dot () on a point with any, other phase. It will move with the same speed, v (otherwise the wave pattern will not remain, fixed)., iv) The motion of a fixed phase point on the, wave is given by, y sin( kx ωt ) ., v) For the same particle displacement ‘y’ at, two different positions, kx ωt constant ----(1), k x ωt 0, x ω, x ω, t k v t k, 2 n, v 2 / n, ( 2 n and k 2 / ), , dl, R, , .....(1), , dl , , 2 R , , v) If ‘ ’ be the mass per unit length (Linear, density) of the string, the mass of this element is, given as dm dl . In the reference frame moving, with wave speed, wave will appear as stationary, but dl appears to be moving toward left with speed, v then we can say that the acceleration of this, element in that reference frame is, , i) Let a wave moves along the +ve x-axis with, velocity ‘v’ as shown in fig., , 6, , T, , , , a, , v2, R, , .....(2), , Now from equations (1) and (2) we have, 2, dl dl v, dmv 2, FR , or T , R, R, R, T, or v , .....(3), , , Spacial cases:, 1., , If A is the area of cross-section of the wire then, linear density M / L AL / L A, T, S, T, , ; where S Stress , A, , A, If string is stretched by some weight then, v, , 2., , T = Mg v , , Mg, , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, v, , WAVES, Velocity at the bottom vB 0, ( tension TB 0 ), T, , T, Velocity at the top vT , , 3., , If suspended weight is immersed in a liquid of, density and = density of material of the, suspended load then, v, , ( tension TB mg lg ), The average velocity of wave, , gl, vT vB, , 2, 2, The time taken by the transverse pulse, generated at bottom to reach the top is given by, vavg , , t, , 4., , 5., , 6., , 7., , mg, gl, , , l, , 2, , l, g, , vavg, , Mg 1 / , T Mg 1 v , , , Note: Velocity at a distance x from bottom v gx, If v1 , v2 are the velocities of transverse waves, The time taken to reach the point P from bottom, while the load is in air medium and in water, x, x, medium respectively, the relative density of mais vx v 2 g, 2, v1, avg, terial of load is d 2 2, W.E-1:, A, longitudinal, progressive wave is given, v1 v2, by the equation y = 5 × 10–2 sin (400 t + x)m., If v1 , v2 and v3 are the velocities of transverse, Find (i) amplitude (ii) frequency (iii) wave, waves while the load is in air, in water and in a, length and (iv) velocity of the wave. (v) velocity, liquid mediums respectively, the relative den1, v12 v32, x m at, and, acceleration, of, particle, at, d, , sity of material of load is, ., 6, v12 v22, t = 0.01 s (vi) maximum particle velocity and, If the temperature a string varies through , acceleration., then the thermal force(tension) developed due, Sol. Comparing with the general equation of the, to elasticity of string is T YA, progressive wave y = Asin( t + kx) we find,, v, = 400 and k = , We find, (i) A = 5 × 10–2 m., YA, Y, 400, v , , n, , , 200 Hz, (ii), , , 2, 2, where Y = Young’s modulus of elasticity of string,, 2 2, A = Area of cross section of string,, , , 2m, (iii), Temperature coefficient of thermal expansion,, k, , , 400, = Density of wire , v, , , 400ms 1, (iv), A, k, , Velocity of wave in vertical strings. If a thick, string is suspended vertically then, (v) v p A cos t t kx 10 3ms 1, Top, , a p A2 sin t kx 4 104 ms 2, (vi) vmax A 20 ms 1, , l, , x, , amax A2 8 104 ms 2, , Bottom, , NARAYANAGROUP, , 7
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , W.E-2:The wave function of a pulse is given by W.E-4: A copper wire is held at the two ends by, y=, , 3, , 2x + 3t , , 2, , where x and y are in metre and, , rigid supports. At 30oC, the wire is just taut,, with negligible tension. Find the speed of, transverse waves in this wire at 10°C if Y =, 1.3×1011N/m2, 1.7 10 5 / C and, , t is in second., (i) Identify the direction of propagation., 9 103 kg / m3, (ii) Determine the wave velocity of the pulse., Sol. (i) Since the given wave function is of the form, Y , y = f(x + vt), therefore, the pulse travels along Sol. v , , the negative x-axis., (ii) Since 2x + 3t = constant for the same particle, 1.3 1011 1.7 105 30 10 , , 70m / s, displacement ‘y’. Therefore, by differentiating, 9 103, with respect, to, time,, we, get, W.E-5: A 4 kg block is suspended from the ceiling, dx 3, dx, of an elevator through a string having a linear, , 1.5m / s, 2 3 0 v , dt, 2, dt, mass density of 19.2 × 10–3 kg m–1. Find the, W.E-3:Figure shows a snapshot of a sinusoidal, speed with which a wave pulse can travel on, the string if the elevator accelerates up at, travelling wave taken at t = 0.3s. The, 2 ms–2? (g = 10 ms–2), wavelength is 7.5 cm and the amplitude is 2, cm. If the crest P was at x = 0 at t = 0, write, M g a, T, the equation of travelling wave., , Sol. v , , , y, 4 10 2 , –1, 3 =50ms ., t = 0.3s, 2cm, 19.2 10, x, W.E-6: A uniform rope of length 12 m and mass, 1.2cm, 6 kg hangs vertically from a rigid support. A, block of mass 2 kg is attached to the free end, of the rope. A transverse pulse of wavelength, Sol. The wave has travelled a distance of 1.2 cm in, 0.06 m is produced at the lower end of the, 0.3s., Hence, speed of the wave,, rope. What is the wavelength of the pulse, v 1.2 / 0.3 4cm / s and 7.5cm, when it reaches the top of the rope ?, 2 2, Sol. Now as v T / , k, , 0.84cm 1, 7.5, Angular frequency vk 4 0.84, 3.36rad / s, v, Since the wave is travelling along positive xdirection and crest (maximum displacement) is, at x = 0 at t = 0, we can write the wave equation, P, , , , , , as, y A sin kx t , 2, , , (or) y x, t A cos kx t , , 8, , 6 2 g, , Therefore, the desired equation is,, , vT, T, T , vB, TB, , y x, t 2 cos 0.84 x 3.36 t cm, , So, T 2 B 2 0.06 0.12m, , 2g, , 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , W.E-7: A uniform rope of mass 0.1 kg and length W.E-9:Two blocks each having a mass of 3.2kg, 2.45m hangs from a ceiling. (a) Find the speed, of transverse wave in the rope at a point 0.5m, distant from the lower end, b) Calculate the, time taken by a transverse wave to travel the, full length of the rope (g = 9.8 m/s2), Sol. a) If M is the mass of string of length L, the mass, of length x of the string will be (M/L)x., , TAB, 64, , AB, 10 103 6400 80m / s, Tension in string CD is T 3.2kg 32 N, Thus speed of transverse waves in string CD is, v AB , , v, , T , , vCD , , Mx, g, L, , So, v , , T, , , , Mgx, gx ......... 1, M , L , L, , Hence x = 0.5m, So, v 0.5 9.8 2.21m / s, b) v , t, , dx, dx dt dx, gx , gx, dt, dt, L, , 1 1 2, x dx t 2, g, , dt , 0, , are connected by wire CD and the system is, suspended from the ceiling by another wire, AB. The linear mass density of the wire AB, is 10 g/m and that of CD is 80 g/m. Find the, speed of a transverse wave pulse produced in, AB and CD and ratio of speeds of transverse, pulse in AB to that in CD., Sol. Tension in string AB is TAB 6.4kg 64 N, Thus speed of transverse wave in string AB is, , 0, , L g, , Here, L = 2.45 m , t 2 2.45 / 9.8 1s, W.E-8:The strings, shown in figure, are made, ofsame material and have same cross-section., The pulleys are light. The wave speed of a, transverse wave in the string AB is v1 and in, CD it is v2 . Find v1 / v2 ., Sol: If T1 and T2 are the tensions in strings AB and, CD respectively then T2 = 2T1., A, , v AB 80, , 4 :1, vCD 20, WE-10 A progressive wave travels in a medium, M 1 and enters into another medium M 2 in, which its speed decreases to 75% . What is, the ratio of the amplitude and intensity of, the, a. Reflected and the incident waves, and, b. Transmitted and the incident waves?, Sol. let Ai , Ar and At be the amplitudes of the incidents, reflected, and transmitted waves., Given that, velocity in the medium refracted is, 75% of that in the initial medium., 3, v2 v1, 4, , 400 20m / s , , v2, 1 3 1, 1, Ar v2 v1 v1, , , 4, , Ai v2 v1 v2 1 3 1, 7, v1, 4, a., , A, , r, i.e., the required ration is A 1: 7 and, i, , I A2 , , T1, , T1, , T2, D, , v1 1, v1, T1, As v T v T v , 2, 2, 2, 2, NARAYANAGROUP, , TCD, 32, , DC, 80 103, , Ir, 1, , I i 49, , 3, 2 , At, 2v2, 2v / v, 6, 4, , 2 1 , v, 3, 2, 1, 1 7, b. Ai v2 v1, v1, 4, , i.e., t he required ratio is, I A2 , , At, 6 : 7 and, Ai, , It 36, , I i 49, 9
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , W.E-11: A long wire PQR is made by joining two, , L, , 1, , L, , wires PQ and QR of equal radii as shown., M o ax 2 dx T dt, , PQ has length 4.8m and mass 0.06kg. QR has, 0, 0, length 2.56 m and mass 0.2kg. The wire PQR, L, 2 M ax 3 2 , is under a tensioon of 80N. A sinusoidal wave, t, o, , T t 0, pulse of amplitude 3.5cm is sent along the, , 3, , wire PQ from the end P. No power is dissipated, , 0, during the propagation of the wave pulse., 3, 3, 2 , a. Find the time taken by the wave pulse to reach, t, M o L 2 M o 2 , , , the other end R of the wire ., 3 T , b. The amplitudes of reflected and transmitted W.E-13: A stretched string is forced to transmit, wave pulse after incident on the joint Q., transverse waves by means of an oscillator, Q, coupled to one end. The string has a diameter, of 4 mm. The amplitude of the oscillation is, P, R, 48m, 2.56m, 104 m and the frequency is 10 Hz. Tension in, R, P, the string is 100N and mass density of wire, l1, l2, Sol. a., 4.2 103 kgm 3 . Find, , , (a) the equation of the waves along the string, 0.06 1, M, (b) the energy per unit volume of the wave, 1 1 , kg / m, l1, 4.8 80, (c) the average energy flow per unit time across, any section of the string, M, 0.2, 20, 2 2 , , kg / m, Sol.(a) Speed of transverse wave on the string is, l2, 2.56 256, T, v, A , T, 80, A, v1 , , 80m / s, 1, 1, 100, 80, v, T, , v2 , 2, , 80, , , 256 4 32m / s , 20 / 256, A, , , l, l, t t1 t2 1 2 4.8 2.56, V1 V2 80, 32, = 0.06 + 0.08 = 0.14sec, , , (4.2 103 ) (4.0 103 )2, 4, , 43.53ms 1, 2 n 20 rad / s 62.83 rad/s, , k 1.44m 1, v, Equation of the waves along the string, y ( x, t ) A sin( kx t ), , v v , 32 80, Ar 2 1 Ai , 3.5 1.5cm, 32 80, v2 v1 , (104 m ) sin (1.44m 1 ) x (62.83rads 1 )t , thus Ar 1.5cm and -ve sign represents that the, reflected pulse suffers a phase difference of (b) Energy per unit volume of the string,, radian., 1, 2 2, u energy density A, 2, 2v2 , 2 32, At , 3.5 2cm ., Ai , 1, 80 32, u (4.2 103 )(62.83) 2 (10 4 ) 2, v1 v2 , 2, W.E-12: A wave pulse starts propagating in +ve, 8.29 102 Jm 3, X-direction along a non-uniform wire of, length ‘L’, with mass per unit length given by (c) Average energy flow per unit time P = power, 1, , M o x and under a tension of TN. Find, 2 A2 ( Sv) (u )(Sv ), 2, , the time taken by the pulse to travel from the, lighter end (x = 0) to the heavier end., , P (8.29 102 ) (4.0 103 ) 2 (43.53), 4, dx, T, T, 5, 1, Sol. v dt M x, 4.53 10 Js, o, b., , 10, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , Principle of Superposition:, 1., , 2., , The displacement at any time due to a number, of waves meeting simultaneously at a point in a, medium is the vector sum of the individual, displacements due to each one of the waves at, that point at the same time., , If y1 , y2 , y3........... are the displacements at a, particular time at a particular position, due to, individual waves, then the resultant, displacement., , y y1 y2 y3 ........., , WAVES, , y1 a1 sin t , y2 a2 sin t then by the, , principle of superposition y y1 y2, y a1 sin t a2 sin t A sin t , , where A a12 a22 2a1a2 cos and, tan , , a2 sin , a1 a2 cos , , Since Intensity (I), , Amplitude A , , 2, , 2, , I a , 1 1 , I 2 a2 , Therefore, the resultant intensity is given by, , I I1 I 2 2 I1I 2 cos , 3., , Important applications of superposit ion, principle., i) Interference of waves: Adding waves that, differ in phase., ii) Formation of stationary waves: Adding, waves that differ in direction., iii) Formation of beats: Adding waves that, differ in frequency., , Interference of Sound waves, 1., , 2., 3., 4., 5., 6., , When two waves of same frequency, same, wavelength, same velocity (nearly equal, amplitude) moves in the same direction. Their, superimposition results in the interference., Due to interference the resultant intensity of, sound at a point is different from the sum of, intensities due to each wave separately., Interference is of two type (i) Constructive, interference (ii) Destructive interference, In interference energy is neither created nor, destroyed but is redistributed., For observable interference, the sources, (producing interfering waves) must be coherent., Let at a given point two waves arrives with phase, difference and the equation of these waves is, given by, , A, a2, , Table: Constructive, and destructive interference, When the waves meet, a point with same phase,, constructive interference, is obtained at that point, (i.e., maximum sound), , Phase difference between Phase difference, the waves at the point of = 180° (or) (2n1);, observation =0°(or) 2n n = 1,2,......, Phase difference between Phase difference, , the waves at the point of, 2n 1, observation = n, 2, i.e., even multiple of /2) (i.e., odd multiple of /2), Resultant amplitude at, the point of observation, will be maximum, Amax = a1 + a2, If a1 = a2 = a0, Amax = 2a0, , Resultant amplitude at, the point of observation, will be minimum, Amax = a1a2, If a1 = a2 Amax = 0, , Resultant intensity at the Resultant intensity at the, point of observation will point of observation will, be maximum, be minimum, Imax I1 I 2 2 I1 I 2, I min I1 I 2 2 I1 I 2, , , , , 2, , I1 I 2, , , , If I1 = I2 = I0 Imin = 0, , , , When the waves meet a, point with opposite phase,, destructive interference, is obtained at that point, (i.e., minimum sound), , , , , , 2, , I1 I 2, , , , If I1 = I2 = I0 Imin = 4I0, , , a1, , NARAYANAGROUP, , 11
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , 7., , I max I1 I 2 , , , I min I1 I 2 , , 2, , a1, , 1, a1 a2 a2, , , , a1 a2 a1 1 , a, , 2, , , x, 2, x 2 cos ------(1), For maximum intensity path difference, x N --------(2), From equations (1) and (2) we get, 2 cos N 2cos N, at least p is Ist maxima N 1, 1, cos 60, 2, x, tan x D tan 60 x 3D, D, , 2, , From figure, we get cos , , 2, , W.E-14: Two loud speakers L1 and L2 , driven by, a common oscillator and amplifier, are, arranged as shown. The frequency of the, oscillator is gradually increased from zero, and the detector at D records a series of, maxima and minima., D, , 40m, , L1, 9m, , Standing Waves or Stationary Waves:, , L2, , If the speed of sound is 330 m/s then the frequency at which the first maximum is observed is, Sol., , D, , 40m, , L1, 9m, , 41m, , L2, , It is clear from figure that the path difference, between L1D and L2 D is x 41 40 1m, For maximum x N where N = 1, 2, 3...., v, for Ist maximum N = 1, , n, v, 330, x 1 1 1, n 330 Hz, n, n, W.E-15: Two coherent narrow slits emitting of, wavelength in the same phase are placed, parallel to each other at a small separation, of 2 . The sound is detected by moving a, detector on the screen S at a distance, D from the slit S1 as shown in figure., P, x, 2, S1, , O, S2, D, S, , Find the distance x such that the intensity at, P is equal to the intensity at O., Sol., , 12, , 1., , When two sets of progressive wave trains of, same type (both longitudinal or both transverse), having the same amplitude and same time period/, frequency/wavelength travelling with same, speed along the same straight line in opposite, directions superimpose, a new set of waves are, formed. These are called stationary waves or, standing waves., These waves are formed only in a bounded, medium., In practice, a stationary wave is formed when a, wave train is reflected at a boundary. The, incident and reflected waves then interface to, produce a stationary wave., Suppose that two super imposing waves are, incident wave y1 a sin t kx and reflected, wave y2 a sin t kx , (As y2 is the displacement due to reflected wave, from a free boundary), Then by principle of superposition, y = y1 + y2 = a sin t kx sin t kx , ( sin C + sin D 2sin, , CD, CD, cos, ), 2, 2, , y 2a cos kx sin t, (If reflection takes place from rigid end, then, equation of st ationary wave will be, y 2a sin kx cos t ), 2. As this equation satisfies the wave equation., 2, 2 y, 2 y, , v, . It represents a wave, t 2, x 2, 3. As it is not of the form f ax bt , the wave is, not progressive., 4. Amplitude of the wave ASW = 2a cos kx., 5. Nodes (N): The points where amplitude is, minimum are called nodes., i)Distance between two successive nodes is / 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, ii) Nodes are at permanent rest., iii) At nodes air pressure and density both are, high., , WAVES, 2., , 3., 6., , Antinodes (A): The points of maximum, amplitudes are called antinodes., (i) The distance between two successive, antinodes is / 2, (ii) At antinodes air pressure and density both, are low., (iii) The distance between a node (N) and, adjoining antinode (A) is / 4, , 7., , 4., , Amplitude of standing waves in two, different cases:, , Reflection at open end, or free boundary, , Reflection at closed end, or rigid boundary, , ASW = 2a cos kx, , ASW = 2a sin kx, , Amplitude is maximum, when cos kx = 1, kx 0, 2,........n, , Amplitude is maximum, when sin kx = 1, 3 2n 1, kx , ...., 2 2, 2, , , n, x 0, , ......., 2, 2, 2, Where k , , and n = 0, 1, 2, 3....., , 3, , ..............., 4 4, 2, Where k , , and n = 1, 2, 3,......., , (fundamental note) which are actually produced, by the instrument are called overtones. e.g. the, tone with frequency immediately higher than the, fundamental is defined as first overtone., Octave: The tone whose frequency is doubled, the fundamental frequency is defined as Octave., i) If n2 = 2n1 it means n2 is an octave higher than, n1 or n1 is an octave lower than n2., ii) If n2 = 23 n1, it means n2 is 3-octave higher or, n1 is 3-octave lower., iii) Similarly, if n2 = 2nn1 it means n2 is n-octave, higher or n1 is n octave lower., Unison: If time period is same i.e., two, frequencies are equal then vibrating bodies are, said to be in unison., , Standing Waves on a String, , 5., 6., , Consider a string of length l, stretched under, tension T between two fixed points., If the string is plucked and then released, a, transverse harmonic wave propagates along it’s, length and is reflected at the end., The incident and reflected waves will, superimpose to produce transverse stationary, waves in a string., Nodes (N) are formed at rigid end and antinodes, (A) are formed in between them., Number of antinodes = Number of nodes –1, Velocity of wave (incident or reflected wave), , 7., , T, ., , Frequency of vibration (n) = Frequency of wave, , 1., , Table: : Amplitude in two, different cases, , Overtone: The harmonics other than the first, , 2., 3., 4., , x, , is given by v , , v 1 T, , , For obtaining p loops (p-segments) in string, it, l, has to be plucked at a distance, from one, 2p, fixed end., , , Amplitude is minimum, when cos kx = 0, , Amplitude is minimum, when sin kx = 0, , kx , , 3 2n 1, 3 2n 1, kx , ...., , ...., 2 2, 2, 2 2, 2, , x, , 3, , ........., 4 4, , , n, x 0, , ......., 2, 2, , 8., , 9., , Fundamental mode of vibration, , Terms related to the Application of Stationary, wave, 1. Harmonics: The frequency which are the, integral multiple of the fundamental frequency, are known as harmonics e.g. if n be the, fundamental frequency, then the frequencies n,, 2n, 3n ...... are termed as first, second, third ...., harmonics., NARAYANAGROUP, , l, , i) Number of loops p = 1, l, ii) Plucking at (from one fixed end), 2, 13
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , vi) Positions of nodes:, l 2l 3l, xN 0, , , .........l, P P P, vii) Positions of antinodes:, , 1, 1 2l, 2, iv) Fundamental frequency or first harmonic, , iii) l , , n1 , , 1 T, 1 T, , 1 2l , , x AN , , 10. Second mode of vibration:, , Sonometer, 1., l, , l, , i) Number of loops p = 2, l, l, (from one fixed end), ii) Plucking at, 2 2 4, iii) l 2, iv) Second harmonic or first over tone., n2 , , 1, 2, , 1, 3, , 2., , It is an apparatus, used to produce resonance, (matching frequency) of tuning fork (or any, source of sound) with stretched vibrating string., It consists of a hollow rectangular box of light, wood. The experimental set up fitted on the box, is shown below., Experimental, wire, , T, 3 T, , 3n1, 2l , , 12. More about string vibration, i) In general, if the string is plucked at length, l, , then it vibrates in p segments (loops) and, 2p, p T, 2l , ii) All even and odd harmonics are present. Ratio, of harmonic = 1 : 2 : 3......, iii) Ratio of over tones = 2 : 3 : 4 ......, 2l, iv) General formula for wavelength ;, P, where P = 1, 2, 3,..... correspond to 1st, 2nd, 3rd, modes of vibratio of the string., v, v) General formula for frequency n P , 2l, , we have the pth harmonic n n p , , Rider Bridge, Tension, T = mg, , Resonance Box, , T 1 T, , 2n1, l , , 11. Third mode of vibration:, i) Number of loops p = 3, l, 1, (from one fixed one), ii) Plucking at, 23 6, 3 3, 2l, 3 , iii) l , 2, 3, iv) Third harmonic or second over tone., n3 , , l 3l 5l, 2P 1 l, ,, ,, ....., 2 P 2P 2 P, 2P, , 3., 4., , The box serves the purpose of increasing the, loudness of the sound produced by the vibrating, wire., If the length of the wire between the two bridges, is l, then the frequency of vibrat ion is, 1 T, T, , 2l , r 2 , (r = Radius of the wire, = Density of material, of wire) = mass per unit length of the wire, Resonance: When a vibrating tuning fork is, placed on the box, and if the length between the, bridges is property adjusted then if, (n)Fork = (n)string rider is thrown off the wire., n, , 5., , Laws of string, , i) Law of length: If T and are constant then, 1, nl constant n1l1 n2l2, l, n, l, , If % change is less than 5% then, n, l, n, l, 100% 100%, or, n, l, ii) Law of mass: If T and l are constant then, n, , n, , 1, n, 2, 1 , , n2, 1, , If % change is less than 5% then, 14, , n, 1 , , n, 2 , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, or, , WAVES, , W.E-16: The vibrations of a string of length 60, , n, , 100% , 100%, n, , , cm fixed at both ends are represented by the, , iii) Law of density: If T, l and r are constant, 1, n1, 2, then n n const n , 2, 1, If % change is less than 5% then, n, 1 n, , , 100% , 100%, or, n, 2 , n, , iv) Law of tension: If l and are constant, then n T, , , n, n, T, M1, const 1 1 , n2, T2, M2, T, , If % change is less than 5% then, n 1 T, n, T, , 100% , 100%, or, n 2 T, n, T, , Tuning fork, i) It is a U shaped metal bar made of steel or an, alloy with a handle attached at the bend., ii) When it is struck against a hard rubber pad,, its prongs begin to vibrate as shown in figure(a)., A, , A, N, , A, N, , x , equation. y 4 sin cos 96t Where x, 15 , and y are in cm and t in sec., a)What is the maximum displacement at, x = 5 cm ?, b)What are the nodes located along the string ?, c) What is the velocity of the particle at, x = 7.5 cm and t = 0.25 s ?, d) Write down the equations of component waves, whose superposition gives the above wave., , Sol. a) For x = 5cm, y 4sin 5 /15 cos 96t , (or) y 2 3 cos 96t , So y will be maximum when cos (96 t) = 1 i.e.,, (ymax)x = 5 2 3cm, b)At nodes amplitude of wave is zero., x, x , 4sin 0 or , 0, , 2,3......, 15, 15 , So x = 0, 15, 30, 45, 60 cm [as length of, string = 60cm], , c) As y 4sin x /15 cos 96t , dy, x , 4 sin sin 96t 96 , iii)A tuning fork emits a single frequency note,, dx, 15 , i.e., a fundamental with no overtones., iv) A tuning fork may be considered as a, So the velocity of the particle at x = 7.5cm and, vibrating free bar as shown figure(b) that has, t = 0.25s,, been bent into U-shape., vpa = -384 sin(7.5/15) sin (96 × 0.25), v) Two antinodes are formed one at each free, end of the bar which are in phase., vpa = -384 1 0=0, vi) The frequency of a tuning fork of arm length, ‘l’ and thickness ‘d’ in the direction of vibration, x , , d) y = y1 + y2 with y1 2 sin 96t , is given by, 15 , , Y, d, d Y , n v, x , , , v , y2 2 sin 96t , l2, l2 , , 15 , , where Y is the Young’s modulus and is the, W.E-17: A guitar string is 90 cm long and has a, density of the material of the tuning fork., fundamental frequency of 124 Hz. Where, vii) Using the tuning fork we can produce, transverse waves in solids and longitudinal, should it be pressed to produce a fundamental, waves in solids, liquids and gases., frequency of 186 Hz?, viii) Transverse vibrations are present in the, prongs. Longitudinal vibrations are present in Sol. Since T is constant we have n 1, the shank., l, ix) Loading or waxing a tuning fork increases its, n, 124, inertia and so decreases its frequency, while, l2 1 l1 , 90 60cm, n2, 186, filing a tuning fork decreases its inertia and so, increases its frequency., Thus, the string should be pressed at 60cm from, x) When tuning fork is heated its frequency, an end., decreases due to decrease in elasticity., NARAYANAGROUP, , 15
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WAVES, , JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , W.E-18: A wire having a linear mass density W.E-21 A sonometer wire has a length of 114 cm, , 5.0×10–3 kg/m is streched between two rigid, supports with a tension of 450 N. The wire, resonates at a frequency of 420 Hz. The next, higher frequency at which the same wire, resonates is 490 Hz. Find the length of the, wire., Sol. Suppose the wire vibrates at 420 Hz in its nth, harmonic and at 490 Hz in its (p + 1)th harmonic., 490 p 1, , or p 6, 420, p, , between two fixed ends. Where should two, bridges be placed to divide the wire into three, segments whose fundamental frequencies are, in the ratio 1 : 3 : 4 ?, Sol. In case of a given wire under constant tension,, fundamental frequency of vibration n 1/ l , , 1 1 1, l1 : l2 : l3 : : 12 : 4 : 3, 1 3 4, l1 72cm; l2 24cm; l3 18cm, 6, 450, 900, l , 2.1m, 420 , First bridge is to be placed at 72 cm from, 2l 5.0 103, 420, one end., W.E-19: The equation of a standing wave, Second bridge is to be placed at 72 + 24 = 96, produced on a string fixed at both ends is, cm from one end, where ‘y’ is measured in cm. What could be, W.E-22: An aluminium wire of cross-sectional, the smallest length of string?, area 10–6 m2 is joined to a copper wire of the, Sol. Comparing with y 2 A sin kx cos wt, same cross-section. This compound wire is, , stretched on a sonometer, pulled by a load of, 20cm, We have k , 10, 10 kg. The total length of the compound wire, If the string vibrates in ‘p’ loops then length of, between two bridges is 1.5 m of which the, p, p, aluminium wire is 0.6 m and the rest is the, , l, string ‘l’ is, ., copper wire. Transverse vibrations are set, 2, 2, up in the wire in the lowest frequency of, , , l, , , 10, cm, is, minimum, if, p, =, 1, excitation for which standing waves are, ' ', 2, formed such that the joint in the wire is a, W.E-20: The equation for the vibration of a string, node. What is the total number of nodes, fixed at both ends, vibrating in its third, observed at this frequency excluding the two, x, at the ends of the wire ? The density of, harmonic is given by y 0.4 sin cos 600t, aluminium is 2.6 104 kg/m3., 10, where x and y are in cm, Sol. As the total length of the wire is 1.5 m and, 1) What is the frequency of vibration?, out of which LA = 0.6 m, so the length of, 2) What are the position of nodes?, copper wire, 3) What is the length of string?, Lc = 1.5 - 0.6 = 0.9 m. The tension in the whole, 4) What is the wavelength and speed of, wire is same (=Mg = 10g N) and as fundamental, transverse waves that can interfer to give this, frequency of vibration of string is given by, vibration?, Sol. Comparing with, 1 T, 1 T, n, , [ A ], y 2 A sin kx cos t we have, 2 L 2 L A , 1) 600 gives n = 300 Hz, 1, T, x, 1, T, so nA , and nc , ------- (1), 0, 2) To get the position of nodes sin, 2 LA A A, 10, 2 Lc c A, x, Now as in case of composite wire, the whole, N where N = 0, 1, 2...., i.e.,, wire will vibrate with fundamental frequency, 10, Hence nodes occur at x = 0, 10, 20 cm..., n p A nA pC nC -------- (2), 3) Since the string is in 3rd harmonic, Substituting the values of f A and f c from, , 2, 2, , , 20 cm , Eqn.(1)in(2), l 3 gives l = 30cm; , k, /10, , , 2, pA, T, 4)Speed of wave v = n = 300 × 20=60ms–1., 2 0.6 A 2.6 103, 16, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , pc, , 2 0.9, , WAVES, , WE-24: A wire of density 9×103 kg/m3 is stretched, , T, A 1.0401104, , p A 2 2.6 2 1 1, , , pc 3 10.4 3 2 3, So that for fundamental frequency of composite, string, pA = 1 and pc = 3, i.e., aluminium string, will vibrate in first harmonic and copper wire, at second, overtone as shown in figure., , i.e.,, , l, , n nA 3nC, This in turn implies that total number of nodes, in the string will be 5 and so number of nodes, excluding the nodes at the ends = 5 - 2 = 3, W.E-23: A wire of uniform cross-section is, stretched between two points 1 m apart. The, wire is fixed at one end and a weight of 9 kg, is hung over a pulley at the other end produces, fundamental frequency of 750 Hz., (a) What is the velocity of transverse waves, propagating in the wire ?, (b) If now the suspended weight is submerged, in a liquid of density (5/9) that of the weight,, what will be the velocity and frequency of the, waves propagating along the wire ?, Sol. a) In case of fundamental vibrations of string, , / 2 L , i.e.,, , 2 1 2m, L, , Now as v n and n = 750 Hz,, vT 2 750=1500m/s, b) Now as in case of a wire under tension, , v, T, T, v T B B, , vT, TT, , T, , B, vB 1500 T 1500, A, , From v n nB , , NARAYANAGROUP, , mg 1 l / b , mg, , n, , 1 Y L, 2L L, , 1 9 1010 4.9 104, , 35Hz, 2 1, 9 103 1, W.E-25: A string 120 cm in length sustains a, standing wave, with the points of string at, which the displacement amplitude is equal, to 2 mm being separated by 15.0 cm, Find, the maximum displacement amplitude., Sol. From figure. points A, B, C, D and E are having, equal displacement amplitude., Further, xE xA 4 15 60cm, A, , E, , B, , C, , D, , 2l 2 120, , 60, n, n, 2 120, n, 4, 60, So, it corresponds to 4th harmonic., Also, distance of node from A is 7.5 cm and, no node is between them. Taking node at origin,, the amplitude of stationary wave can be written, as, Asw Amax sin kx, , As , , Asw 2mm; k , , M, , v, , between two clamps 1 m apart and is subjected, to an extension of 4.9 × 10–4 m. What will be, the lowest frequency of transverse vibrations, in the wire ? (Y = 9 × 1010 N/m2), Sol. In case of fundamental vibrations of a string, , 2 2, , and x = 7.5 cm, , 60, , , 2, , 2 Amax sin , 7.5 Amax sin, 4, 60, , Hence, Amax 2mm, , , , Sound Waves :, 1000m / s, , vB 1000, , 500 Hz, lB, 2, , Sound is a form of energy propagated in the form, of longitudinal waves. This energy causes the, sensation of hearing on reaching the ear. Any, vibrating body could be a source of sound., Longit udinal mechanical waves can be, transmitted in all the three states of matter namely,, solids, liquids and gases. According to their, 17
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, range of frequencies longitudinal mechanical waves, are divided into the three categories., 1) Longitudinal waves having frequencies below, 20Hz are called infrasonic waves. These are, created by earthquakes, elephants and whales., Infrasonic waves can be heard by snakes., 2) Longitudinal waves having range, of, frequencies lying between 20Hz and 20kHz are, called audible sound waves. The audible, wavelength is 16.5 mm to 16.5m at S.T.P when, velocity of sound is 330 m/s. These are, generated by tuning forks, streched stings and, vocal cords., The human ear can detect these waves., 3) Longitudinal waves having frequencies, greater than 20 kHz are called ultrasonics. The, human ear can’t detect these waves. These, waves can be produced by high frequency, vibrations of a quartz crystal under an alternating, electric field. These waves can be detected by, mosquito, fish and dog etc., , where density of medium, v velocity of, wave, A - Amplitude, n - Frequency, Human ear responds to sound intensities over a, wide range from 10-12 W/m2 to 1 W/m2., In a spherical wave front ( i.e. wave starting, from a point source) , the amplitude varies, inversely with distance from position of source, 1, 1, i .e, A I 2, r, r, I, r, S, r, , 2, , In a cylindrical wave front (i.e. wave starting, from a linear source ), the amplitude varies, inversely as the square root of distance from, 1, 1, I, the axis of source i.e. , A , r, r, , Application of ultrasonic waves :, i) The fine internal cracks in a metal can be, detected by ultrasonic waves., ii) They are used for determining the depth of, the sea and used to detect submarine., iii) They can be used to clean clothes and fine, machinery parts, iv) They can be used to kill animals like rats,, fish and frogs etc., , r, , Sound level in decibles is given by, , I , 10log , I0 , If 1 and 2 be the sound levels corresponding, to sound intensities I1 and I2 respectively. Then,, I, I, 1 10 log 1 and 2 10 log 2, I0, I0, , Characteristics of Sound, Hearing of sound is characterised by following, three parameters., 1) Loudness (Refers to Intensity) :, It is the sensation received by ear due to intensity, of sound, Greater the amplitude of vibration, greater will, be intensity ( I A2 ) and so louder will be sound., The loudness being the sensation, depends on, the sensitivity of listener’s ear. Loudness of a, sound of a given intensity may be different for, different listeners., The average energy transmitted by a wave per, unit normal area per second is called intensity, E, of a wave. I , . ItsSI Unit : W/m2, At, It is the average power transmitted by a wave, through the given area., P, I avg. ; I 2 2 n 2 A2 v, area, 18, , , I, I , 2 1 10 log 2 log 1 , I0, I0 , , , I , (or) 2 1 10log 2 , I1 , 2), , Pitch (Refers to Frequency):, The shrillness or harshness of sound is known, as pitch. Pitch depends on frequency. Higher the, frequency, higher will be the pitch and shriller, will be the sound., , 3), , Quality or Timber (Refers to Harmonics):, It is the sensation received by ear due to, waveform. Quality of a sound depends on, number of overtones. i.e, harmonic present., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , Velocity of Sound, The equation for velocity of sound through a, , vt v0 , , v0t, v0 0.61t C, 546, , E, , , Note:, 1) When temperature rises by 1o C then velocity of, sound increases by 0.61 m/s, where E modulus of elasticity; density, 2) The velocity of sound increases with increase, As modulus of elasticity is more for solids and, in humidity. Sound travels faster in moist air, less for gases, so, than in dry air at the same temperature, because, vsolids vliquids vgases, density of humidity air is less than that of dry, Y, air., In case of solids v ,, moist air dry air vmoist air vdry air, where Y is Young's modulus,, 3) The velocity of sound at constant temperature in, B, a gas does not depend upon the pressure of air., In case of fluids (liquids and gases) v , 4) Amplitude, frequency, phase, loudness, pitch,, quality donot effect velocity of sound., where B is the Bulk modulus, W.E-26: Find the speed of sound in a mixture of, Velocity of sound in Gases :, 1 mol of helium and 2 mol of oxygen at 27°C., Newton’s formula :, C Pmix 19 R / 6 19, Newton assumed that the propagation of sound, , , in a gas takes place under isothermal conditions. Sol. mix C, 13R / 6 13, Vmix, Isothermal Bulk modulus , B P, nM n M, 1 4 2 32, M mix 1 1 2 2 , P, vs , n1 n2, 1 2, , 68, 103 kg / mol ;, 1.013 x105, 3, 280ms 1, At S.T.P. v , 1.29, RT, 19 8.314 300, Which is less than the experimental value, v mix, , , M, 13, 68 10 3 / 3 401m / s, mix, ( 332 m / s ), 3, Laplace’s correction: Laplace assumed that W.E-27: A window whose area is 2m opens on a, the propagation of sound in a gas takes place, street where the street noise result in an, under adiabatic conditions., intensity level at the window of 60dB . How, much ‘acoustic power’ enters the window via, Adiabatic, Bulk, modulus,, B, , , P, , sound waves. Now if an acoustic absorber is, P, PV, nRT, RT, fitted at the window, how much energy from, v , , , , street will it collect in five hours ?, , m, m, M, where V = volume, m is mass, M = molecular, I , weight. T is absolute temperature, Sol. Sound level 10log I , o, For air = 1.4. Therefore, I , I, At STP v0 280 1.4 330ms 1 , which agrees, 60 10 log 106 I 106 I, o, with the experimentally calculated value., Io, Io , Velocity of sound in a gas is directly proportional, I 106 10 12 106 W / m 2, to the square root of the absolute temperature, E, 1, E IAt, but intensity I , vt, T t 273 2, At, , , , v, , T, , vo, To 273 , E 10 6 2 5 3600 36 103 J, Various forms of longitudinal wave:, t , , vt vo 1 , , As we know, during a longitudinal wave, 546 , propagation the particles of the medium oscillate, medium is given by v , , , , NARAYANAGROUP, , , , 19
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, to produce pressure and density variation along the, direction of the wave. These variations result in, series of high and low pressure (and density), regions called compression and rarefactions, respectively. Hence the longitudinal wave can, be in terms of displacement of particles called, displacement wave y(x, t) or in terms of change, in pressure called pressure wave P( x, t ) or, change in density called density wave d ( x, t ) ., , 1) Pressure Wave:, i)A longitudinal sound wave can be expressed, either in terms of the longitudinal displacement, of the particles of the medium or in terms of, excess pressure produced due to compression, or rarefaction. (at compression, the pressure is, more than the normal pressure of the medium, and at rarefaction the pressure is lesser than the, normal)., ii) If the displacement wave is represented by, y A sin(t kx) then the corresponding, pressure wave will be represented by, dy, P B, (B = Bulk modulus of elasticity of, dx, medium), P BAk cos(t kx ) P0 cos(t kx), where P0 pressure amplitude BAk, iii) Pressure wave is / 2 out of phase(lags), with displacement wave. i.e. pressure is, maximum when displacement is minimum and, vice-versa., Note1:At the centre of compression and, rarefaction particle velocity is maximum and, at the boundary of compression and rarefaction, particles are momentarly of rest. This is, explained as in a harmonic progressive wave, vp, dy, , vP = -(slope of y-x ) v , v, dx, Since the change in pressure of the medium, , vp , P B v , , i.e., for a given medium, B and v are constants., Where v p is maximum, p is also maximum,, which is true at y = 0, Note 2: As sound sensors (e.g ear or mike ) detect, pressure changes, description of sound as, pressure wave is preferred over displacement, wave., dy , P B , dx , , 20, , 2) Density wave form : Let o be the normal, density of the medium and be the change in, density of t he medium during t he wave, propagation., Then fraction of change in volume of the element, v, , , m, , v , , v, 0, , , According to definition of Bulk’s modulus, , , v , B P P 0 , v , , , , 0, .p, B, , 0, (p )max Cos (kx t ), B, 0 Ak Cos ( kx t ), , , p max BAk , max Cos ( kx t ),, where ( ) max 0 Ak is called density, amplitude. Thus the density wave is in phase, with the pressure wave and this is 900 out of, phase (lags ) with the displacement wave as, shown in the figure., , +A, O, -A, , P0+BAK, P0, P0-BAK, d0+d0AK, d0, d0-d0AK, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , Note 1: The relation between density amplitude and WE-30: A firework charge is detonated many, metres above the ground. At a distance of 400, , m from the explosion, the acoustic pressure, pressure amplitude is () max (p ) max , B, reaches a maximum of 10.0 N / m 2 . Assume, that the speed of sound is constant at 343 m/s, P 1, Note 2: Average Intensity I 2 A2v, throughout the atmosphere over the region, S 2, considered, the ground absorbs all the sound, In terms of pressure amplitude, sound intensity, falling on it, and the air absorbs sound energy at the rate of 7.00 dB/km. What is the, 2, 2, , p, , , 1, , p, 1, sound level (in decibels) at 4.00 km from the, , , max, I 2 max v , explosion?, 2, 2, v, Bk , Sol. r 400m, r 1 4000m,, , [ (P)max BAk , k , and B v 2 ], 1.2kg / m3 , v 343m / s, v, 2, 10, Pmax, Thus intensity of wave is proportional to square, I, , 1.21102W / m 2, of pressure amplitude or displacement amplitude, 2 v 2(1.2)(343), or density amplitude and is independent of, 2, 1, I1 r , frequency., as I 2 1 , r, I r , W.E-28: What is the maximum possible sound, level in dB of sound waves in air? Given that, I (400), 1.21 103 W / m 2, I1 , 3, density of air 1.3kg / m , v 332m / s and, 4000, 5, 2, 1.21103 , I , atmospheric pressure P 1.01 10 N / m ., 90.8 dB, 10 log 1 log , 12 , Sol. For maximum possible sound intensity, pressure, I , 110, , amplitude of wave will be equal to atmospheric, At a distance of 4 km from the explosion, absorption from the air will decrease the sound, 5, 2, pressure, i.e., p0 P 1.01 10 Nm, level by an additional amount,, 2, 5 2, (7)(3.60) 25.2 dB, (1.0110 ), p, I 0 , 1.18 107 W / m 2, At 4 km, the sound level will be, 2 v 2 1.3 332, f 90.8 25.2 65.6 dB, I, 107, Organ pipes, SL 10log 10log 12 190dB, I0, 10, Organ pipe: An organ pipe is a cylindrical, tube of uniform cross section in which a gas is, WE-29: The faintest sounds the human ear can, trapped as a column., detect at a frequency of 1000 Hz correspond, Open pipe : If both ends of a pipe are open, to an intensity of about 1.00 1012 W / m 2 ,, and a system of air is directed against an edge,, which is called threshold of hearing. The, standing longitudinal waves can be set up in the, loudest sounds the ear can tolerate at this, tube. The open end is a displacement antinode., frequency correspond to an intensity of about Due to finite momentum, air molecules undergo, certain displacement in the upward direction, 1.00W / m 2 , the threshold of pain. Determine, hence antinode takes place just above the open, the pressure amplitude and displacement, end but not exactly at the end of the pipe., amplitude associated with these two limits., Due to pressure variations, reflection of, longitudinal wave takes place at open end and, Sol. Pmax 2 vI, hence longitudinal stationary waves are formed, in open tube., 2(1.20)(343)(1.00 10 12 ), 2.87 105 N / m2, , Pmax, 2.87 10 5, , v (1.2)(343)(2 1000), ( 2 n) ; 1.11 1011 m, , A, , , (a), , NARAYANAGROUP, , (b), , (c), , 21
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, fig: a) For fundamental mode of vibrations or, I harmonic, , L 1 ; 1 2L, 2, V, n1 , ------ (1), 2L, , V 1n1 ; V 2 Ln1, fig:b) For the second harmonic or first, overtone,, L 2, 2V, , V 2 n2 V Ln2 n 2 2L ------ (2), fig:c) For the third harmonic or second, overtone,, , 2, L 3 3 3 L, 2, 3, 2, 3V, ------- (3), V 3 n3 V Ln3 n 3 , 2L, 3, From (1), (2) and (3) we get,, n1 : n2 : n3 ..... 1: 2 : 3 : ......, i.e. for a cylindrical tube, open at both ends, the, harmonics excitable in the tube are all integral, multiples of its fundamental., 2L, , , where, In the general case,, p, p 1, 2,....., V pV, p harmonic frequency 2l , where, p 1, 2,....., Closed pipe: If one end of a pipe is closed,, then reflected wave is 180 out of phase with, the wave. Thus the displacement of the small, volume elements at the closed end must always, be zero. Hence the closed end must be a, displacement node., th, , If n1 is the fundamental frequency, then the velocity, of sound waves is given as, V, , V 1n1 V 4 Ln1 n1 4L ----- (1), , figure b) for third harmonic or first overtone., , 4, L 3 2 , 2 L, 4, 3, 4, 3V, ---- (2), V 2 n2 , V Ln2 n2 , 3, 4L, figure c) for fifth harmonic or second, overtone., , 4, L 5 3 , 3 L, 4, 5, 4, 5V, V 3 n3 , V Ln3 n3 , ---- (3), 5, 4L, From (1), (2) and (3) we get,, n1 : n2 : n3 ..... 1: 3 : 5 : ......, 4L, , In, the, general, case,, , 2 p 1 , where, p 0,1, 2,....., th, p harmonic frequency , , (b), , ,, , End Correction, Due to finite momentum of air molecules in organ, pipe reflection takes place not exactly at open, end but some what above it. Hence antinode is, not formed exactly at the open end rather it is, formed at a little distance away from open end, outside it., The distance of antinode form the open end is, known as end correct (e)., It is given by e = 0.6 r , where r = radius of, pipe., , l l', , (c), , figure a) for the fundamental mode of, vibration or I harmonic :, , L 1 1 4L, 4, 22, , 4L, , where p = 1, 2...., , l' l, , (a), , 2 p 1V, , Effective length in open organ pipe l ' l 2e , Effective length in closed organ pipe l ' l e , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , Note: When the end correction is considered, then, i)the fundamental frequency of open pipe, V, V, n, n, 2 l 1.2r , 2 l 2e , ii) The fundamental frequency of closed pipe, V, V, n, n, , , 4, l, 0.6r , 4 l e, , Velocity of sound, (Resonance column apparatus) :, , W.E-32: A tuning fork of frequency 340 Hz is, vibrated just above a cylindrical tube of, length 120 cm. Water is slowly poured in the, tube. If the speed of sound in air is 340 m/s., Find the minimum height of water required, for resonance. (v = 340m/s), v, with p = 1, 3, 5,............., Sol: n p, 4L, So length of air column in the pipe, , L, , l2, , If l1 , l2 and l3 are the first, second and third, , resonating lengths then l1 e ....(1), 4, 3, l2 e , ...(2), 4, 5, l3 e , --- (3), 4, From equations (1) and (2), 1) 2 l2 l1 2) V n 2n l2 l1 , , 1, l 3l1, 3) 2 e 2, 4) l3 l2 l2 l1 l3 2l2 l1, 2, W.E-31: A tube of certain diameter and length, 48 cm is open at both ends. Its fundamental, frequency of resonance is found to be 320 Hz., The velocity of sound in air is 320 m/s., Estimate the diameter of the tube. One end, of the tube is now closed. Calculate the, frequency of resonance for the tube., v, v, Sol. n0 2 L 2e 2 L 2 0.6r as e 0.6r , So substituting the given data,, 320 100, 10, 320 , or r cm, 2 48 1.2 r , 6, , So, D = 2r = 2 × (10/6) = 3.33cm., Now when one end is closed,, v, nc , 4 L 0.6r , , , 320 100, 163.3Hz, 4 48 0.6 10 / 6 , , NARAYANAGROUP, , pv, 25 p cm with p = 1,3,5,....., 4n, , i.e., L = 25cm, 75cm, 125cm, Now as the tube is 120 cm , so length of air, column must be lesser than 120 cm, i.e., it can, be only 25 cm or 75 cm. Further if h is the height, of water filled in the tube,, L + h = 120 cm or h = 120 – L, So h will be minimum when Lmax = 75cm, (h)min = 120 – 75 = 45cm., , BEATS, , It is the phenomenon of periodic change in the, intensity of sound when two waves of slightly, different frequencies travelling in same, direction superpose with each other., Maximum Intensity of sound(Waxing) is, produced in the beats when constructive, Interference takes place., Minimum Intensity of sound(Waning) is, produced in the beats when destructive, Interference takes place., , Analytical treatment of Beats:, Equations of waves producing beats are given, as y1 a sin 1t and y2 a sin 2t let 1 2, Resultant wave equation is, 1 2 , 2 , t sin 1, , t, 2 , 2 , , y y1 y2 2a cos , , 1 2 , t, 2 , , y A t cos , , 23
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, Here A t , , 2 t, 2a sin 1, , Uses of Beats:, To determine unknown frequency of a tuning fork, with the help of a standard tuning fork., ii) To tune the stretched string of a musical, instrument to a particular frequency., iii) To detect the presence of dangerous gases in, mines., Note:, 1. When wax is added to the arms of one of the, tuning forks then its frequency decreases., i.e. n1 n, 24, , When arms of one of the tuning forks are filed then, its frequency increases., i.e., n1 n, The following table gives the relation for beats, produced when sounded together under different, conditions., , 2, , Amplitude is function of time. Frequency of, n n, variation of amplitude 1 2, 2, n1 n2, Frequency of resultant wave = 2, The variation in the intensity of sound between, successive maxima or minima is called one beat., The number of beats per second is called beat, frequency. I f n1 and n 2 are the frequencies of, the two sound waves that interfere to produce, beats then, Beat frequency = n1 ~ n2, The time period of one beat (or) the time interval, between two successive maxima or minima is, 1, n1 ~ n2, The time interval between a minima and the, 1, immediate maxima is, 2n1 ~ n2 , As the persistence of human hearing is about, 0.1 sec, beats will be detected by the ear only, if beat period is t 0.1sec or beat frequency, n n1 n2 10 Hz, Maximum number of beats that can be heard by, a human being is 10 per second., If more than 10 beats are produced then no. of, beats produced are same but no. of beats heard, are zero, If a1 , a2 are amplitudes of two sound waves that, interfere to produce beats then the ratio of, maximum and minimum intensity of sound is,, 2, I max a1 a2 , , , I min a1 a2 , i), , 2., , 3., , Fork, , Frequency, , Relation n when, 1, 1, n > n n < n, , Wax is added, to 1st fork, , n11 n1, , n = n2n1 n = n1n2, , Wax is added, nd, to 2 fork, , n22 n2, , n = n1n2 n = n2n1, , n11 n1, , n = n1n2 n = n2n1, , n12 n1, , n = n2n1 n = n1n2, , st, , 1 fork is, filled, nd, , 2 fork is, filled, , W.E-33: The frequency of tunning fork ‘A’ is, 250 Hz. It produces 6 beats/sec, when sounded, together with another tunning fork B. If its, arms are loaded with wax then it produces 4, beats/sec. Find the frequency of tuning fork, B., Sol. n nA nB = 6 beats / sec, If wax is added to the tunning fork A then its, frequency decreases. i.e., n1A nA and, given n1 4 beats / sec n, This is possible when nA nB n, 250 nB 6 nB 244 Hz, , W.E-34: A tunning fork of frequency of 512 Hz, when sounded with unknown tunning fork, produces 5 beats/sec. If arms of the unknown, fork are filed then it produces only 3 beats /, sec. Find the frequency of unknown tunning, fork., Sol. n n1 n2 , n1 512 Hz , n 5 beats/sec, If arms of unknown fork are filed then its frequency increases., i.e., n12 n2 and given n1 3 beats / sec., This is possible when n n2 n1, 5 n2 512 n2 517 Hz, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , W.E-35: The lengths of two open organ pipes are, l and l l l l . If v is the speed of, sound, find the frequency of beats between, them., v, Sol. Beat frequency n1 n2 v , 2l 2 l l , 1, v l v , l vl, 1 1 1 1 2, 2l , l 2l , l 2l, W.E-36: If two sound waves,, y1 =0.3sin596 [t-x/330] and, , y 2 =0.5sin604 [t -x/330] are superposed,, what will be the (a) frequency of resultant, wave (b) frequency at which the amplitude of, resultant waves varies (c) Frequency at which, beats are produced. Find also the ratio of, maximum and minimum intensities of beats., Sol. Comparing the given wave equation with, y A sin t x / v as k / 1/ v we find, that here,, A1=0.3, 1 2n1 596 n1=298Hz, , and A2= 0.5, 2 2n2 604 n2 = 302 Hz, a) The frequency of the resultant, , n1 n2 298 302 , , 300 Hz, 2, 2, b) The frequency at which amplitude of resultant, wave varies:, navg , , n1 n2 298 302 , , 2 Hz, 2, 2, c) The frequency at which beats are produced, nb = 2nA = n1 – n2 = 4Hz, d) The ratio of maximum to minimum intensities, of beat, nA , , 2, , 2, , I max A1 A2 , 0.3 0.5 64, , , 16, 2, 2 , Imin A1 A2 , 0.3 0.5 4, W.E-37:The frequency of a tuning fork ‘x’ is 5%, greater than that of a standard fork of, frequency ‘K’. The frequency of another fork, ‘y’ is 3% less than that of ‘K’. When ‘x’ and, ‘y’ are vibrated together 4 beats are heard per, second. Find the frequencies of x and y., Sol. Let the frequency of standard fork be K, nx K , , 5 K 105 K, , 100 100, , NARAYANAGROUP, , ny K , , 3K 97 K, , 100 100, , 105, 97, K, K, 100, 100, On solving, K = 50 Hz, 105, 50 52.5 Hz, The frequency of x , 100, 97, 50 48.5Hz, Similarly frequency of y , 100, W.E-38: A string under a tension of 129.6 N, produces 10 beats per sec when it is vibrated, along with a tuning fork. When the tension, in the string is increased to 160 N, it sounds, in unison with the same tuning fork. Calculate, the fundamental frequency of the tuning fork., Sol. Let ‘n’ be the frequency of fork., n nx n y 4 , , The wire frequency would be n 10 , In case of a wire under tension n T, , n 10, 129.6, , n = 100 Hz, 160, n, W.E-39 Two open organ pipes 80 cm and 81 cm, long found to give 26 beats in 10 sec, when, each is sounding its fundamental note. Find, the velocity of sound in air., Sol. Number of beats per second, v, v, n , , 2l1 2l2, , , , , 26, v, v, 2v, , ~, 2.6 , 10 160 162, 160 162, , v, , 2.6 160 162, 33696cms 1 337 ms 1 ., 2, , Doppler’s effect: Whenever there is a relative, motion between a source of sound and the, observer (listener), the frequency of sound heard, by the observer is different from the actual, frequency of sound emitted by the force., The frequency observed by the observer is, called the apparent frequency. It may be less than, or greater than the actual frequency emitted by, the sound source. The difference depends on the, relative motion between the source and observer., , 1. When observer and source are stationary, O, S, , , , , 25
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , 2., , i) Sound waves propagate in the form of spherical, v vm vo , wavefronts (shown as circles), (from observer to source) n ' v v v n, ii) The distance between two successive circles, m, S , , is equal to wavelength , iii) Number of waves crossing the observer =, If medium is stationary i.e., vm = 0 then, Number of waves emitted by the source, v v0 , n' , n, iv) Thus apparent frequency n ' = actual, v vS , frequency (n)., When source is moving but observer is Sign convention for different situation, i) The direction of v is always taken from source, at rest, to observer., ii) If the velocities vo , vs in the direction of v, S, S, O, O, S, then positive +ve is taken., , , y, x, iii) If the velocities vo , vs in the opposite, i) S1, S2, S3 are the positions of the source at, direction of v then positive -ve is taken., three different positions., ii) Waves are represented by non-concentric Note:- i) Doppler effect in sound is asymmetric., ii) Doppler effect in light is symmetric., circles, they appear compressed in the forward, iii), Doppler’s effect in vector form is written, direction and spread out in backward direction., as, iii) For observer (X), 1, , 3, , 2, , Apparent wavelength ' Actual, wavelength ( ), , r, S, , Apparent frequency n ' Actual frequency (n), For observer (Y) : ' n ' n, , 3. When source is stationary but observer, is moving, O, , O, S, , y, , , , x, , i) Waves are again represented by concentric, circles., ii) No change in wavelength received by either, observer X or Y., iii) Observer X (moving towards) receives, wave fronts at shorter interval thus n ' n ., iv) Observer Y receives wavelengths at longer, interval thus n ' n, , O, , r , , unit vector along the line joining source and, observer V Velocity of sound in the medium., Its direction is always taken from source to, observer., , limitations of Doppler effect:, i) Doppler effect is not observed if, a) v0 vs 0 (both are in rest), b) v0 vs 0 and medium is alone in motion, direction., d) vs is perpendicular to the line of sight, ii) Doppler effect is applicable only when,, v0 v and vs v . ( v is velocity of sound), , 5. Common Cases in Doppler’s Effect, Source is moving but observer at rest., 1. Source is moving towards the observer, , 4. General expression for apparent, frequency: If v, vo , vs are the velocities of, sound, observer, source respectively and, velocity of medium is vm then apparent, frequency observed by observer when wind, blows in the direction of v (from the source to, observer )is given by, v vm v o , n' , n, v vm vS , 26, , and in opposite direction of v, , v v0 .r , n1 , v v .r n, , s, , , v, , s, , vs, , O, , v0 = 0, , v , Apparent frequency n ' n v v , S , , v vS , Apparent wavelength ' , , v , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 2., , WAVES, , Source is moving away from the observer., , 2., , When both are moving away from each other., , v, , v, vs, , vs, , v0 = 0, , O, , s, , v , n' n, , v vS , , Apparent frequency, , v vO , i) Apparent frequency n ' n v v n ' n , S , , , v vS , Apparent wavelength ' , , v , , ii) Apparent wavelength, , Source is at rest but observer is moving., 1. Observer is moving towards the source., v, vs= 0, , v0, , s, , 3., , 2., , v, s, , v v0 v v0 v , n', v v0 n, n, , v, Observer is moving away from the source, v, vs= 0, , v0, , O, , s, , v vO , Apparent frequency n ' n , , v , Apparent wavelength ' , , 4. When source and observer both are, moving, 1., , v vS , ' , ' , v , iii) Velocity of waves with respect to observer, = (v – vO), When source is moving behind observer, , O, , v v0 , Apparent frequency n ' n , , v , Apparent wavelength, , ' , , v0, , O, , s, , 4., , vs, , O, , v0, , v vO , i) Apparent frequency n ' n v v , , S , a) If vO < vS, then n ' n, b) If vO > vS, then n ' n, c) If vO = vS then n ' n, v vS , ii) Apparent wavelength ' , , v , iii) Velocity of waves with respect to observer, = (v – vO), When observer is moving behind the source, v, vs, , v0, , s, , O, , When both are moving towards each other, , v v0 , i) Apparent frequency n ' n , , v vs , , v, , s, , vs, , v0, , O, , v v0 , i) Apparent frequency n ' n v v , , S , v vS , ii) Apparent wavelength ' , , v , iii) Velocity of wave with respect to observer, = (v + v 0), NARAYANAGROUP, , a) If v0 > vs , then n ' n, b) If v0 < vs , then n ' n, c) If v0 = vs , then n ' n, v vS , ii) Apparent wavelength ' , , v , iii) The velocity of waves with respect to, , observer, , v vO , 27
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , Crossing, 1., , Some Typical Cases of Doppler’ Effect, , Moving sound source crosses a stationary 1., observer, v, , s, , v, , vs, , vs, v0= 0, Before, crossing, , M ov i ng car t ow ar ds w al l : When a car is, , moving towards a stationary wall as shown in, figure. If the car sounds a horn, wave travels, toward the wall and is reflected from the wall., When the reflected wave is heard by the driver,, it appears to be of relatively high pitch, if we, wish to measure the frequency of reflected sound., , After, crossing, , Apparent frequency before crossing, , v , n 'Before n , , v vS , , v, , Apparent frequency n ' After, , v , n, , v vS , n 'Before, , v vS , , 1, Ratio of two frequencies n ', After, v vS , Change in apparent frequency, n 'Before n ' After, , v, v , n, , , v vs v vs , , 2v , nv 2 s 2 , v vs , , Imaginary source, vc, , Observer, , vc, , Echo (sound), , Here we assume that the sound which is reflected, by the stationary wall is coming from the image, of car which is at the back of it and coming, towards it with velocity vC. Now the frequency, of sound heard by car driver be given as, , v vc , ', ', ndirect, n; nreflected, n, , v vc , No.of beats, , n ' n 'reflected n 'direct , , 2vc n, v vc, , Case (i): If the observer is at rest in between, 2nvS, source and wall as shown, v, Moving observer crosses a stationary source, , If vs v then n 'Before n ' After , 2., , v, , v, , v 0= 0, , vs, , s, 0, , v0, , v0, , vs, , Direct, , Reflected, , Sound, , Sound, , 0, source, , vs= 0, Before, crossing, , After, crossing, , Apparent frequency before crossing, v vO , n 'Before n , , v , Apparent frequency after crossing, v vO , n ' After n , , v , n 'Before v vO, , Ratio of two frequencies n ', v vO, After, Change in apparent frequency, 2nvO, n 'Before n ' After , v, 28, , observer, , , , , , , , v, , Image of, source, , wall, , v, , , , 1, n1direct , n, n ; nreflected , v vs , v vs , , No.of beats n ' n 'reflected n 'direct 0, Case (ii): If the source is in between observer, and wall, vs, , v0= 0, , Direct, Sound, observer source, , , , v, , vs, Reflected, Sound, , wall, , , , Image of source, , , , v, , , , 1, n1direct , n, n ; nreflected , v, , v, , s, v vs , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , 3. Transverse Doppler’s effect, , No. of beats n ' n 'reflected n 'direct, , i), , 2 vs , v v , , n , n n 2, 2 , v vs v vs , v vs , 2nvS, If vs v then n ' , v, Note: This method of images for solving problems, of Doppler effect is very convenient but is used, only for velocities of source and observer which, are very small compared to the speed of sound, and it should not be used frequently when the, reflector of sound is moving., 2. Moving target: Let a sound source S and, observer O are at rest (stationary). The frequency, of sound emitted by the source is n and velocity, of waves is v ., Moving traget, , S, , O, , Source and observer, are at rest, , If a source is moving in a direction making an, angle w.r.t. the observer.., A, , B, , C, , vs, , , vs cos, , , , 90°, , vs, , vs cos, , O, , The apparent frequency herd by observer O at, rest, nv, At point A: n ' , v vS cos , As source moves along AB, value of , increases, cos decreases, n ' goes on, decreasing., , At point C:, 90 , cos cos 90 0, n ' n, At point B: The apparent frequency of sound, , vr, , nv, , becomes n '' v v cos , A target is moving towards the source and, s, observer, with a velocity vT. Our aim is to find ii) When two cars are moving on perpendicular, out the frequency observed by the observer, for, roads: When car-1 sounds a horn of frequency, the waves reaching it after reflection from the, n, the apparent frequency of sound heard by carmoving target. The formula is derived by, v v2 cos 2 , applying Doppler equations twice, first with the, 2 can be given as n ' n v v cos , target as observer and then with the target as, , 1, 1 , source., The frequency n ' of the waves reaching surface, S, v, Car-1, of the moving target (treating it as observer) will, , v, v vT , v cos , v, v cos , be n ' , n, v , Car-2, Now these waves are reflected by the moving, target (which now acts as a source). Therefore, the apparent frequency, for the real observer O 4. Rotating source/observer: Suppose that a, source of sound/observer is rotating in a circle, v, v vT, will be n '' , n ' n '' , n, of radius r with angular velocity (Linear, v vT, v vT, velocity vS r ), i) If the target is moving away from the observer,, i) When source / observer at rest at centre of circle, v vT, n, ', , n, and observer / source is rotating in a circle then, then, v vT, the line of sight is perpendicular to the direcii) If target velocity is much less than the speed, tion of motion of observer / source and hence, 2v , no doppler effect. n1 n, of sound, (vT << v), then n ' 1 T n , for, O, v , , 1, , 1, , 2, , 1, , 2, , 1, , 2, , 2, , 2, , 2vT, approaching target and n ' 1 , v, , receding target, NARAYANAGROUP, , , n for, , , vs = 0, v0, , v0 = 0, v0, , vS, , vS, O, , O, , 29
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, , ii) When source is rotating, , 6. Uses of Doppler effect:, , a) Towards the observer heard frequency will, nv, be maximum i.e., nmax v v, S, b) Away from the observer heard frequency will, nv, be minimum and nmin v v, S, vS, a, r, , vS, , v, , c) Ratio of maximum and minimum frequency, nmax v vS, , nmin v vS, , iii) When observer is rotating, vO, a, r, vO, v, , a) Towards the source heard frequency will be, maximum, v v0 , i.e., nmax n , , v , b) Away from the source heard frequency will, , 5., , v v0 , be minimum and nmin n , , v , c) Ratio of maximum and minimum frequency, nmax v v0, , nmin v v0, Doppler shift in RADAR: A microwave beam, is directed towards the aeroplane and is received, back after reflection from it. If ‘v’ is the speed, of the plane and ‘n’ is the actual frequency of, the microwave beam then the frequency of the, microwave beam then the frequency received by, cv, 1, moving plane n , n, c , Now the plane act as a moving source, the frecv, 11, quency of the wave from it is n , n, cv , ( c is velocity of microwave), 2nv, Change in frequency n , c, By measuring n , the speed ‘v’ can be obtained., , 30, , It is used in, a) SONAR, b) RADAR (Radio detection and ranging used, to determine speed of objects in space), c) To determine speeds of automobiles by traffic police. The technique is applied in the airports to guide the air crafts., d) To determine speed of rotation of sun., e) In Astrophysics, it is applied in the study of, the saturn’s rings and in the study of binary satrs., Here the doppler’s shift in the frequency of light, from the atronomical objects is measured., f) Accurate navigation and accurate target bombing techniques., g) Tracking earth’s satellite., h) In medicine, it is applied to study the velocity of blood flow in different parts of the body, and the moment of the fetus in the woomb using, ultra sound. The conditions of heart beat can be, inferred by “echocardiogram” generated from, this technique., W.E-40 When a train is approaching the observer,, the frequency of the whistle is 100 cps while, when it has passed the observer, it is 50 cps., Calculate the frequency when the observer, moves with the train., nv, Sol. In case of approaching of source, 100 , v vS, while in case of recession of source,, nv, 50 , v vS, Which on simplification gives, 200, n, 66.67 Hz, 3, W.E-41: A car approaching a crossing at a speed, of 20 m/s sounds a horn of frequency 500Hz, when at 80m from the crossing. Speed of, sound in air is 330 m/s. What frequency is, heard by an observer 60 m from the crossing, on the straight road which crosses car road, at right angles ?, Sol. The situation is as shown in figure, 80m, , (car) S, , , 60m, 100m, (observer), O, , cos , , napp, , 80 4, Apparent frequency is, 100 5, , , , , , v, 330, , n, 500 , 4, v vS cos , 330 20 , , 5, , = 525.5 Hz, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , W.E-42: A whistle of frequency 540 Hz rotates, in a circle of radius 2 m at a linear speed of, 30m/s. What is the lowest and highest, frequency heard by an observer a long, distance away at rest with respect to the, centre of circle. Take speed of sound in air, as 330 m/s. Can the apparent frequency be, ever equal to actual ?, Sol. Apparent frequency will be minimum when the, source is at N and moving away from the, observer., L, vs, M, C, , O, , K, , N, , Apparent frequency will be maximum when, source is at L and approaching the observer., , v, 330 , n 'max , n, 540 594 Hz, v vS, 330 30 , Further when source is at M and K, angle, between velocity of source and line joining, source and observer is 90° (i.e., line of sight, is, perpendicular, to, vs ), or, vScos =vScos90° = 0 ., So, there will be no Doppler effect., W.E-43: A source of sound is moving along a, circular orbit of radius 3 m with an angular, velocity of 10 rad/s. A sound detector located, far away from the source is executing linear, simple harmonic motion along the line BD, with amplitude BC = CD = 6 m. The, frequency of oscillation of the detector is, , 5 / rev /sec. The source is at the point A, when the detector is at the point B. If the, source emits a continuous sound wave of, frequency 340 Hz, find the maximum and, the minimum frequencies recorded by the, detector [velocity of sound = 330 m/s]., , Further more source is moving on a circle, its speed, vs r = 3 × 10 = 30m/s and as detector is, executing SHM, vD A2 y 2, , 10 62 02 60m / s i.e., detector is at C., So n ' will be maximum when both move, v vD , towards each other. n 'max n , with vD, v vS , = max i.e., the source is at M and detector at C, and moving towards B, so, 330 60 , n 'max 340 , 442 Hz, 330 30 , Similarly n ' will be minimum when both are, moving away from each other i.e.,, , v vD , n 'min n , wit h vD = max i.e., t he, v vS , source is at N and detector at C but moving, 330 60 , towards D, so nmin 340 , 255 Hz, 330 30 , , Echo (additional), The sound reflected by an obstacle which is, heard by a listener is called an echo., Persistence of hearing is the minimum interval, of time between two sound notes to distinguish, them., Persistence of hearing is 0.1s, A person is at a distance ‘d’ from a reflected, surface (a wall, mountain etc). The person, sounds a horn and hears its echo at the end of a, time ‘t’. If V is the velocity of sound in air then., Vt, d, 2, d, S, D, , N, A, 3m, , B, , C, , Reflector, , D, , M, , Sol. Time, , period, , of, , circular, , mot ion, , T 2 / 2 /10 is same as that of, SHM i.e., T=(1/n) / 5 , so both will, complete one periodic motion in same time., NARAYANAGROUP, , To hear a clear echo, the minimum distance of, the obstacle,, V 0.1 V, d, , , min, 2, 20, -1, If V = 330 ms then dmin = 16.5m, If V = 340 ms-1 then dmin = 17 m, 31
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, 9., , C .U .Q, , CHARACTERISTICS OF PROGRESSIVE, WAVE, 1. When a wave is travelling in a medium, in that, process, the following is/are transporting from, one particle to other, 1) energy, 2) momentum, 10., 3) both 1 & 2, 4) length, 2. A plane progressive wave cannot be, represented by, 11., 1) y = a sin t kx , t x, 2) y = asin 2 , T , , 12., , 2, Vt x , , 4) y A log x B log x, The speed of wave of time period T and 13., propagation constant K is, , 3) y = a sin, , 3., , 2, TK, 1, T, 2), 3), 4), TK, 2, TK, K, The phase change between incident and, SPEED OF A TRAVELLING WAVE, reflected sound wave from a fixed wall is, 14. At any instant a wave travelling along the, 3) 3 , 4) 2 , 1) 0, 2) , string is shown in figure. Here, if point A is, The phase change between incident and, moving upward, the true statement is, reflected sound wave from a free end is, 1) 0, 2) , 3) 3 , 4) / 2, B, Figure shows the shape of a string, the pairs, C, A, O, of points which are in opposite phase is, , 1), 4., , 5., , 6., , A, , E, , 8., , 32, , x, , 1) the wave is travelling to the right, 2) the displacement amplitude of wave is equal, to displacement of B at this instant, 1) A and B 2) B and C 3) C and E 4) A and E, 3) at this instant, C also directed upward, During propagation of longitudinal plane wave, 4) 1 and 3, in a medium the two particles separated by a 15. Transverse waves are produced in a long, distance equivalent to one wavelength at an, string by attaching its free end to a vibrating, instant will be/have, tuning fork. Figure shows the shape of a part, of the string. The points in phase are, 1) in phase, same displacement, B, 2) in phase, different displacement, G, A, C, 3) different phase, same displacement, 4) different phase, different displacement, F, D, The equation of a progressive wave is Y=a, E, sin ( t-kx), then the velocity of the wave is, 1) A and D, 2) B and E, 1) k , 2) k/ , 3) /k, 4) a , 3) C and F, 4) A and G, C, , 7., , When a progressive wave is propagating in a, medium, at a given instant, two particles which, are separated by three wave lengths will, have....., 1) Different displacement in same direction, 2) Different displacement in opposite direction, 3) Same displacement in opposite direction, 4) Same displacement in same direction, The speed of sound in a medium does not, change with the change of, 1) frequency, 2) wave length, 3) pressure, 4) density, Phase difference between a particle at a, compre-ssion and a particle at the next, rarefaction is, 3) , 4) /4, 1) Zero 2) /2, One similarity between sound and light waves, is that., 1) both can propagate in vacuum, 2) both have same speed, 3) both can show polarization, 4) both can show interference, When a body is undergoing undamped, vibration, the physical quantity that remains, constant is, 1) amplitude, 2) velocity, 3) acceleration, 4) phase, , B, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , 16. A pulse in a rope approaches a solid wall and, it gets reflected from it, , The wave pulse after reflection is best, represented by _____, 1), , 2), , 3), , 4), , 17. A transverse wave is travelling along a string, from left to right. The figure below represents, the shape of the string at a given instant. At, this instant the points have an upward velocity, are (here X-wave displacement, Y-particle, displacement), y, C, B, , D, E, , x, , A, H, , F, G, , 1) D,E,F, 2) A,B,H 3) B,D,F 4) A,E,H, 18. A metal string is fixed between rigid supports., It is initially at negligible tension. It’s Young’s, modulus is ‘Y’ density is and coefficient of, linear expansion is . It is now cooled through, a temperature ‘t’, transverse waves will, move along it with a speed., Yt, , , t, , Yt, , , , 2) Y 3) , 4) t, Y, PRINCIPLE OF SUPER POSITION,, INTERFERENCE AND STATIONARY, WAVES ON STRETCHED STRINGS., 19. The interference phenomenon can take place, 1) in transverse wave, 2) in longitudinal wave, 3) in electromagnetic waves, 4) in all waves, 20. For superposition of two waves, the following, is correct, 1) they must have the same frequency and, wavelength, 2) they must have equal frequencies but may have, unequal wavelengths, 3) they must have the same wave-length, but, may have different frequencies, 4) they may have different wavelength and, different frequencies, 1), , NARAYANAGROUP, , 21. At a certain instant a stationary transverse, wave is found to have maximum kinetic, energy. The appearance of the string at that, instant is a, 1) sinusoidal shape with amplitude A/3, 2) sinusoidal shape with amplitude A/2, 3) sinusoidal shape with amplitude A, 4) straight line, 22. When stationary waves are set up, pick out, the correct statement from the following, 1) all the particles in the medium are in the same, phase of vibration at all times and distances, 2) the particles with an interval between two, consecutive nodes are in phase, but the particles, in two such consecutive intervals, are of opposite, phase, 3) the phase lag along the path of the wave increases as the distance from the source increases, 4) only antinodes are in same phase, 23. In a stationary wave along a string the strain, is, 1) zero at the antinodes, 2) maximum at the antinodes, 3) zero at the nodes, 4) maximum at the nodes, 24. In a stationary wave, 1) phase is same at all points in a loop, 2) amplitude is same at all points, 3) energy is constant at all points, 4) temperature is same at all points, 25. A wave is represented by an equation;, Y = A cos kx sin t , then, 1) it is a progressive wave with amplitude A, 2) it is a progressive wave with amplitude A, cos kx, 3) it is a stationary wave with amplitude A, 4) it is a stationary wave with amplitude A cos, kx, 26. In a stationary wave, 1) pressure change is maximum at nodes, 2) pressure change is maximum at antinodes, 3) pressure change is minimum at nodes, 4) amplitude is zero at all points, 27. A wire in sonometer experiment is vibrating, in the third overtone. There are, 1) two nodes, two antinodes, 2) three nodes, three antinodes, 3) four nodes, three antinodes, 4) five nodes, four antinodes., 28. is maximum wavelength of a transverse, wave that travels along a stretched wire, whose two ends are fixed. The length of that, wire is, 1) 2 , 2) , 3) /2, 4) 3 /2, 33
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, 29. In the sonometer experiment, a wire of density, ‘ ’ and radius ‘a’ is held between two bridges, at a distance ‘L’ apart. Tension in the wire is, ‘T’ the fundamental frequency of the wire will, be, 1), , 1 a2, 2L T , , 2), , 3), , 1, T, 2L a2, , 1, T, 4) 2 L a 2 , , 37. The relation between the intensity (I) of a wave, and on the distance (r) from a line source is, 1, , 38., , 1 T, 2L a2, , 39., 30. For a stretched string of given length, the, tension ‘T’ is plotted on the X-axis and the, frequency ‘f’ on the Y -axis. The graph is, 1) rectangular hyperbola, 2) straight line through the origin, 3) parabola, 4) straight line not through the origin, 31. The equation of a stationary wave in a, medium is given as y sin t cos kx . The, length of a loop in fundamental mode is, , , 2, K, 1), 2), 3), 4), 2K, K, K, , 32. A stretched string of length l , fixed at both, ends, can sustain stationary waves of, wavelength , correctly given by( P is number, of loops), l2, , 2, , 41., , 42., , 2l, , p, 3) 2lp 4) p, 1) 2 p 2) , 2l, 33. A knife-edge divides a sonometer wire into, two parts. The fundamental frequencies of the, t wo par t s ar e n 1 and n2. The fundamental, frequency of the sonometer wire when the, knife-edge is removed will be, n1n2, 1, 1) n1 + n2 2) n1 n2 3) n1n2 4) n n, 2, 1, 2, , SOUND AND VELOCITY OF SOUND, 34. Pitch of sound primarily depend upon, 1) intensity, 2) frequency, 3) quality, 4) overtone, 35. Quality of sound primarily depends upon, 1) intensity, 2) frequency, 3) shape of the source 4) wave form, 36. It is possible to recognize a person by hearing, his voice even if he is hidden behind a solid, wall. This is due to the fact that his voice, 1) Has a definite pitch, 2) Has a definite quality, 3) Has a definite loudness, 4) Can penetrate the wall, 34, , 40., , 43., , 1) I r 1 2) I r 2 3) I r 4) I r 2, Loudness of sound primarily depend upon, 1) intensity 2) frequency, 3) shape of the source 4) overtone, According to Laplace correction, the, propagation of sound in gas takes place under, 1) isothermal condition 2) isobaric condition, 3) isochoric condition 4) adiabatic condition, The velocity of sound is not affected by change, in, 1) temperature, 2) medium, 3) pressure4) wavelength, Velocity of sound in air at the given, temperature, 1) decreases with increase in pressure, 2) may increase on decrease with pressure, 3) is independent of the variation in pressure, 4) varies directly as the square root of pressure, If the temperature of atmosphere is increased, the following character of sound waves is, effected, 1) amplitude, 2) frequency, 3) velocity, 4) wavelength, Velocity of sound is, 1) directly proportional to absolute, temperature, 2) directly proportional to square root of, absolute temperature, 3) inversely proportional to absolute temperature, 4) inversely proportional to T, , ORGAN PIPES, 44. A closed pipe has certain frequency. Now its, length is halved. Considering the end, correction, its frequency will now become, 1) double, 2) more than double, 3) less than double, 4) four times, 45. The fundamental frequency of a closed organ, pipe is ‘n’. Its length is doubled and radius is, halved. Its frequency will become nearly., 1), , n, 2, , 2), , n, 3, , 3) n, , 4) 2n, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , 46. When an air column of length ‘l’ open at both 56. In closed pipes, the positions of antinodes are, ends produces its fundamental note, the, obtained at ____, wavelength is given by, 3 5, , 1) , ,, 2) 0, , , 1) l 2) 2l 3) 4l, 4) 3l, 4 4 4, 2, 47. In the case of closed end organ pipe, 3) , 2, 3, 4) 2, 4, 6, 1) the maximum possible wavelength is same as, 57. An open pipe of length ' l ' vibrates in, that of open end organ piple, fundamental mode. The pressure variation is, 2) the maximum possible wavelength is less than, maximum at, that of open end organ pipe, 1) 1/4 from ends 2) the middle of pipe, 3) the maximum possible wavelength may be less, 3) the ends of pipe 4) at 1/8 from ends of pipe, than that of open end organ pipe, BEATS, 4) the maximum possible wavelength is greater, 58., Beat, phenomenon, is physically meaningful, than that of open end organ pipe, only,, if, 48. In the case of standing waves in organ pipe,, 1) 1 2 1 2, 2) 1 2 1 2, y, the value of x at the open end is, 1, 1, 3) 17, 4) 1 2 , 1) >0, 2) <0, 3) =0, 4) =10, 2, 2, 49. The harmonics formed in air column in an, 59. Beats are produced by the superposition of, organ pipe closed at one end are, two waves of nearly equal frequencies. Of, 1) only odd, 2) only even, the following, the correct statement is, 3) both odd and even 4) neither odd nor even, 1) all particles of the medium vibrate simple, 50. A tube with both ends closed has same set of, harmonically with frequency equal to the, natural frequency as, difference between the frequencies of component, 1) one end closed organ pipe, waves, 2) both end open organ pipe, 2) the frequency of beats changes with the, 3) vibratory string fixed at both ends, location of the observer, 4) vibratory string fixed at one end, 3) the frequency of beats changes with time, 51. The frequency of the sound emitted by an, 4) amplitude of vibration of particle at any point, organ pipe will increase if the air in it is, changes simple harmonically with frequency, replaced by, equal to one half of the difference between the, a) hot air, b) moist air, c) hydrogen, component waves, 1) a is true, 2) a,b are true, 60., When beats are formed by two waves of, 3) b,c are true, 4) a,b,c are true, frequencies n1 and n2, the amplitude varies, 52. An empty vessel is filled partially with water, with frequency equal to, natural frequency, 1) n1-n2, 2) 2(n1-n2), 1) increase, 2) decrease, 3) (n1-n2)/2, 4) (n1+n2)/2, 3) remains unchanged 4) insufficient data, 53. End correction in a closed organ pipe of 61. Two wires are producing fundamental notes, of same frequency. The change in which of, diameter ‘d’ is, the following factors of one wire does not, 1) 0.6d, 2) 1.2d, 3) 0.3d, 4) 2.4d, produce beats between them, 54. If oil of density higher than that of water is, 1) stretching force, 2) diameter of the wire, used in place of water in a resonance tube, its, 3) material of the wire, frequency will, 4) amplitude of the vibrations, 1) increase, 2) decrease, 62. Beats are the result of, 3) remain the same, 1) diffraction 2) destructive interference, 4) depend upon the density of the material of, 3) constructive and destructive interference, tube, 4) superposition of two waves of nearly equal, 55. If 1 , 2 and 3 are wave lengths of the waves, frequencies, giving resonance with fundamental ,first and, second over tones of a closed organ pipe .the 63. To hear beats, it is essential that the two sound, waves in air should, ratio of wave lengths 1 : 2 : 3 is, 1) be travelling in opposite directions, 2) be travelling in the same direction, 1 1, 1) 1:2:3 2) 1: :, 3) 1:3:5, 4) 5:3:1, 3) have slightly different amplitude, 3 5, 4) have slightly different wavelengths, NARAYANAGROUP, , 35
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, 64. When the two tuning forks of nearly same, frequency are vibrated to produce beats, then, the velocity of propagation of beats will be, 1) less than that of sound, 2) depend upon the relative frequency, 3) more than that of sound, 4) equal to that of sound, 65. A certain number of beats are heard when, two tuning forks of natural frequencies n1 and, n2 are sounded together. The number of beats, when one of the forks is loaded, (1992), 1) increases, 2) decreases, 3) remains same, 4) may increase or decrease, 66. The frequency of sound reaching a stationary, listener behind a moving source is, 1) lower than source frequency, 2) higher than source frequency, 3) zero, 4) same as the frequency of the source, , DOPPLER EFFECT, 67. Doppler shift in frequency does not depend, upon, 1) the frequency of wave produced, 2) the speed of the source, 3) distance between source and observer, 4) the speed of the observer, 68. An observer is moving away from a source at, rest. The pitch of the note heard by the, observer is less because, 1) the pitch of the source decreases, 2) the velocity of sound in air increases, 3) wave length of the wave becomes small, 4)wavelength of the wave remains, unchanged but observer receives less number, of waves, 69. Doppler effect is not applicable to, 1) sound Waves, 2) light Waves, 3) radio Waves, 4) matter Waves, 70. In Doppler effect, when a source moves, towards a stationary observer, the apparent, increase in frequency is due to, 1) increase in wavelength of sound received by, observer, 2) decrease in wavelength of sound received, by obzerver, 3) increase in number of waves received by, observer in one second, 4) decrease in number of waves received by, observer in one second, 36, , 71. When a source moves away from stationary, observer with velocity v then apparent, change in frequency is n1 . When an observer, approaches the stationary source with same, velocity v then change in frequency is n2 then, 1) n1 n2, 2) n1 n2, n1, 3) n1 n2, 4) n 1, 2, 72. The graph between distance between source, and observer and apparent frequency in the, case of Doppler’s effect will be, 1), 2), S, , S, , n, , n, 3), , 4), S, , S, , n, n, 73. A source of sound moves towards a stationary, listener. The apparent pitch of the sound is, found to be higher than its actual value. This, happens because, 1) wavelength of sound waves decreases, 2) wavelength of sound waves increases, 3) the number of waves received by the listener, increases, 4) the number of waves received by the listener, decreases, , ASSERTION & REASON, In each of the following questions, a statement, is given and a corresponding statement or, reason is given just below it. In the, statements, mark the correct answer as, 1) Both Assertion and Reason are true and, Reason is correct explanation of Assertion., 2) Both Assertion and Reason are true but, Reason is not the correct explanation of, Assertion., 3) Assertion is true but Reason is false., 4) Assertion is false but Reason is true.., 74. Assertion (A): In transverse wave particle, velocity is perpendicular to wave velocity., Reason(R): In a longitudinal wave particle, velocity is along the direction of propagation, of wave., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 75. Assertion (A): When a sound wave is, propagating through the medium the, acceleration of a particle is directly, proportional to its displacement., Reason(R): Wave velocity depends on the, properties of the medium and is independent, of time and position., 76. Assertion (A):When a transverse wave, propagates along a string it transports energy, in two forms : kinetic energy and potential, energy., Reason(R): A transverse pulse reflected from, a fixed end gets inverted., 77. Assertion (A):When a wave is transmitted, from one medium to another, the frequency, does not change., Reason(R): A wave transports energy and, momentum along with matter., 78. Assertion (A):Mechanical waves require, material medium for their propagation., Reason(R): Transverse waves cannot be, generated with in the liquids., 79. Assertion (A):The wave reflected from a, denser medium has a phase change of from, the incident wave., Reason(R): The rate of energy transfer by a, wave is proportional to the frequency., 80. Assertion (A): A pulse traveling in one rope, is reflected at the boundary with another rope., If the reflected pulse is not inverted, then the, transmitted pulse is longer than the original, pulse., Reason(R): The speed of a transverse wave, in an elastic stretched string is doubled if, extension in the string is quadrapled., 81. Assertion (A): When a sound wave is, propagating through the medium, the pressure, fluctuations are /2 out of phase with the, displacements., Reason(R): When a sound wave is, propagating through the medium maximum, pressure deviation occurs at the positions of, zero displacement and no pressure change, occur at the positions of maximum, displacements., NARAYANAGROUP, , WAVES, 82. Assertion (A): The speed of sound in moist, air is more than that in dry air., Reason(R): Speed of sound is independent of, change in pressure at constant temperature., 83. Assertion (A): The Doppler formula is not, symmetric with respect to the speed of source, and the speed of observer., Reason(R): Doppler effect is applicable for, both mechanical as well as electromagnetic, waves., 84. Assertion (A): In propagation of sound waves,, pressure amplitude is proportional to the, displacement amplitude., Reason(R): Suppose the source and observer, both are stationary and wind is blowing in a, direction from source to observer, then the, observer detects an apparent increase in, frequency., 85. Assertion (A): In stationary wave, all the, particles of the medium between two nodes, vibrate in same phase with same frequency, but with different amplitude., Reason(R): In stationary wave, every particle, of the medium vibrates with its own energy, and it does not share or transmit it to any, other particle., 86. Assertion (A): In the phenomena of, superposition, each wave propagates as if the, other wave were not present., Reason(R): The superposition of waves is, valid only when the amplitude of the wave is, much less than the wave length and velocity, of the wave is much longer than the particle, velocity., 87. Assertion (A):A standing wave can be, produced even if the component waves have, different amplitude., Reason(R): Only periodic waves can produce, interference., 88. Assertion (A): Displacement node is a, pressure antinode., Reason(R): Standing waves can only be, transverse., 89. Two points P and Q have a phase difference, of when a traveling sine wave passes, through the region., Assertion (A): P and Q move in opposite, directions., Reason(R): P oscillates at half the frequency, of Q., 37
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WAVES, 90. Assertion (A): A transverse wave is traveling, in a string. Equation of the wave is equal of, shape of the string at an instant t., Reason(R): At a point beats frequency of n, Hz is observed. It means, at that point, zero, intensity is observed 2n times per second., 91. Assertion (A): In case of longitudinal, stationary wave, compressions and, rarefractions are obtained in place of nodes, & antinodes respectively., Reason(R): Two plane waves, one, longitudinal and the other transverse having, same frequency and amplitude are traveling, in medium in opposite directions with same, speed forms a stationary wave., 92. Assertion (A): If all the particles of a string, are oscillating in same phase, the string is, resonating in its fundamental tone., Reason(R): A sound wave is propagating, through the medium, when the particle moves, in the opposite direction to the propagation, of the wave, it is in a region of rarefaction., 93. Assertion (A): In longitudinal progressive, waves, when the particle is at the mean, position, it is a region of maximum, compression or rarefaction., Reason(R): In longitudinal progressive waves,, compressions and rare fractions travel, forward along the wave., 94. Assertion (A): In progressive waves, the, minimum distance between two particles, vibrating in phase is equal to the wavelength., Reason(R): In progressive waves, all, particles have the same amplitude, frequency, and time period., 95. Assertion (A): The intensity of sound, increases with increase in the density of the, medium., Reason(R): The sound heard is more intense, in carbon dioxide in comparison to air., 96. Assertion (A): The velocity of sound increases, with increase of temperature., Reason(R): Sound wave travel through longer, distances during night than during day., 97. Assertion (A): The velocity of sound is, generally greater in solids than in gases at, NTP., 38, , JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, Reason(R): The ratio of modulus of elasticity, to density for solids is much greater than that, for gases., 98. Assertion (A): Beats can be observed by two, light sources as in sound., Reason(R): To observe beats by two light, sources, the phase difference between the, two sources should change regularly., 99. Assertion (A): When a vibrating tuning fork, is placed on a sonometer, a large sound is, heard., Reason(R): It is due to resonance., 100. Assertion (A): When the star approaches the, earth, the spectral lines are shifted towards, the blue end of the spectrum., Reason(R): It is due to Doppler effect., 101. Assertion (A): The change in frequency due, to Doppler effect does not depend on, separation between the source and the, observer., Reason(R): When a listener and sound source, are moving with the same velocity in the same, direction, the apparent frequency as heard, by listener increases., 102. Assertion (A): The Doppler effect can be, observed in case of light and sound, but, Doppler effect formula for light differs from, that for sound., Reason(R): Light does not require a material, medium for its transmission whereas sound, requires medium for its transmission., 103. Assertion (A): In a standing wave, the, particles at nodes always remain in rest., Reason(R): In a standing wave, all the, particles cross their mean positions together., 104. Assertion (A): In a travelling wave, energy, is transmitted from one region of space to, other but in a standing wave the energy of, one region is always confined that region., Reason(R): In a travelling wave, the phases, of nearly particles are always different,, whereas in a standing wave all particles, between two successive nodes move in phase., 105. Assertion (A):Solids can support both, longitudinal and transverse waves, but only, longitudinal waves can propagate in gases., Reason(R): Solid possess elasticity of length, as well as shape, but gases possess only Bulk, modulus., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , WAVES, , 106. Assertion (A): A driver under water is unable, to hear the sound produced in air., Reason(R): The sound produced in air, undergoes reflection at the water surface., 107. Assertion (A): We need two ears to locate, the direction from which the sound is coming., Reason(R): The phase difference between the, sound reaching the two ears depends on the, direction and distance of source., 108. Assertion (A): The sound waves do not exhibit, the property of polarization., Reason(R): The sound waves do not have the, transverse nature., 109. Assertion (A): Two pulses are traveling along, a uniform string in opposite directions. One, of them is inverted pulse when compared with, the other in shape. There is no point clamped, between the pulses. Then they will have, different speeds., Reason(R): In a uniform string under a given, tension, wave speed of any pulse is same., 110. Two waves in a string (all in SI units)are, y1 0.6sin 10t 20 x and y2 0.4sin 10t 20 x ., Mass per unit length of the string is 102, , kg, m, , ., , Assertion (A): Stationary wave can be formed, by their superposition but net energy transfer, through any section will be non-zero., Reason(R): Their amplitudes are unequal., 111. Assertion (A): When a longitudinal pressure, wave is reflected at the open end of an organ, pipe, the compression pressure wave pulse, becomes rarefraction pressure wave pulse, during the reflection., Reason(R): The phase of the wave changes, by when reflected at the open end., 112. Assertion (A): An open organ pipe of certain, length have the same fundamental frequency, as closed organ pipe of half the length, Reason(R): In the case of open organ pipe,, at both the ends antinodes are formed, while, in the closed organ pipe at one end antinode, and at the other end node is formed, 113. Assertion (A) When a traveling wave is sent, along a particular string by oscillating one, end, the speed of the wave will increase if we, increase the frequency of oscillations., Reason(R): If a traveling wave sent along a, particular string by oscillating its one end, it, is the wave length of the wave that decreases., NARAYANAGROUP, , MATCH THE FOLLOWING, 114. Match the following :, List – I, List – II, In a stretched string % Change in frequency, a) Length increases by 2% e) Decreases by 4%, b) Radius increases by 4% f) Increases by 4%, c) Tension increases by 2% g)Decreases by 4%, d) Density decreases by 2% h)Increases by 8%, I) Changes by 8%, The correct match is, 1) a-g, b-e, c-h, d-f 2) a-h, b-i, c-g, d-h, 3) a-e, b-g, c-f, d-h, 4) a-f, b-i, c-e, d-g, 115. Match the following :, List – I, List – II, a) Resonance e)Law of conservation, of energy, b) Reflection, f) Doppler effect is due, to change in wave length, c) Source is in g) Doppler effect is due, motion, to number of waves, reaching the observer, d) Observer is h) Special case of, in motion, forced vibrations, i) Reverberation, The correct match is, 1) a-e, b-h, c-g, d-i, 2) a-f, b-g, c-e, d-h, 3) a-g, b-h, c-e, d-f, 4) a-h, b-i, c-f, d-g, 116. I represent intensity of sound wave, A the, amplitude and r the distance from the source., Then the match the following two comlumns., Column - I, Column - II, a) Intensity due to a, p) Proportional to r 1/ 2, point source., b) Amplitude due to a q) Proportional to r 1, point source., c) Intensity due to a line r) Proportional to r 2, source, d) Amplitude due to a, s) Proportional to r 4, line source, 1) a-r,b-q,c-q,d-p, 2) a-p,b-q,c-r,d-s, 3) a-q,b-q,c-s,d-r, 4) a-s,b-p,c-q,d-r, 117. Transverse waves are produced in a stretched, wire. Both ends of the string are fixed. Let, us compare between second overtone mode(in, numerator) and fifth harmonic mode(in, denominator). Match the following two, columns., Column - I, Column - II, (a) Frequency ratio, (p) 2/3, (b) Number of nodes ratio, (q) 4/5, (c) Number of antinodes ratio (r) 3/5, (d) Wavelength ratio, (s) 5/3, 1) a-r,b-p,c-r,d-s, 2) a-q,b-p,c-s,d-s, 3) a-r,b-s,c-q,d-p, 4) a-s,b-p,c-r,d-p, 39
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , WAVES, 118. A wave travels from a denser medium to rarer, medium, then match the following two, columns., Column - I, Column - II, (a) speed of wave, (p) will increase, (b) wavelength of wave, (q) will decrease, (c) amplitude of wave (r) will remain constant, (d)frequency of wave (s) may increase or decrease, 1) a r; b p; c r; d s 2) a p; b p; c p; d r, 3) a r; b s; c q;d p 4) a s; b p; c r;d p, 119. y-x graph of a transverse wave at a given, instant is shown in figure. match the following, two columns., y, , B, , x, A, , Column - I, Column - II, (a) velocity of particle A, (p) positive, (b) acceleration of particle A (q) negative, (c) velocity of particle B, (r) zero, (d) acceleration of particleB (s) cen’t tell, 1) a-r,b-p,c-r,d-s, 2) a-p,b-p,c-p,d-r, 3) a-r,b-s,c-q,d-p, 4) a-s,b-p,c-s,d-r, 120. Match the following, Column-I, Column-II, A) Laplace equation P) humidity, p, d, , B) Newton equation Q), C) Speed of, longitudinal wave, depends on, , R) Temperature, S) isothermal process, T), , 1), 2), 3), 4), 40, , A, P, Q, P,Q, Q,R, , B, Q, S,T, R,S, P,Q,R, , P, d, , C, S,T, P,R, T, R,S, , 121. A tuning fork ‘P’ of frequency 280 Hz produces, 6 beats/s with unknown tuning fork ‘Q’, Column-I, Column-II, A) P is waxed and P) Frequency of ‘Q’, number of beats, is 286 Hz, decreases, B) Q is filed and, Q) Frequency of, numbered beats, ‘Q’ is 274 Hz, decreases, C) P is filed and R) Frequency of‘Q’, number of beats, is 272 Hz, remains same, D) ‘Q’ is filed and S) Frequency ‘Q’ is, number of beats, 288 Hz, A, B, C D, 1), Q, R, S, R, 2), P, P, Q Q, 3), P,Q, Q,R, R,S P, 4), R,S, Q,R, S, R, 122. Match the following, Column-I, Column-II, A) Beats, P) Ratio of harmonics is, 1:2:3, B) open organ pipe Q) Transverse stationary, waves, C) string stretched at R) Superposition of, both ends, sound waves of nearly, equal frequencies, D) closed organ pipe S) longitudinal stationary, waves, T) Interference in time, A, B, C, D, 1) P,T, P,S, P,Q, S, 2) Q,R S,T, R,S,T Q, 3) S,T, Q,R,T P, Q, 4) Q, P,Q, R,S, T, 123. A string of length 1m stretched at both ends, vibrating with frequency 300 Hz which is 3, times the fundamental frequency, Column-I, Column-II, A) Number of loops, , 1, 3, , P) m, , B) Number of antinodes Q) 200 Hz, C) Distance between, R) 1st overtone, two successive, antinodes, D) 2nd harmonic, S) 3, A, B, C, D, 1) Q, R,S, P,R, Q,S, 2) T, Q,R P, S,T, 3) P,Q, R,S, P,R,T Q, 4) S, S, P, Q,R, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , MORE THAN ONE OPTION IS CORRECT, 124. A string of length l is stretched along the xaxis and is rigidly clamped at x=0 and x=l., Transverse vibrations are produced in the, string. For nth harmonic which of the following, relations may represent the shape of the string, at any time, n x , a) y 2Acost cos , , l , n x , b) y 2Asint cos , , l , n x , c) y 2Acost sin , , l , n x , d) y 2Asint sin , , l , 1) c only, 2) c and d only, 3) a only, 4) a, b, c and d, 125. In case of superposition of waves (at x=0),, y1=4sin(1026 t) and y2=2sin(1014 t), a) the frequency of resulting wave is 510 Hz, b) the amplitude of resulting wave varies at, frequency 3 Hz, c)the frequency of beats is 6 per second, d) the ratio of maximum to minimum intensity, is 9, The correct statements are, 1) a, d, 2) b, d, 3) a, c, d, 4) all, 126. In case of stationary sound waves in air the, correct statement(s) is a/are, A) each air particle vibrates with the same, amplitude, B) amplitude is maximum for some, particles and, minimum for some other particles, C) the particles do not execute periodic, motion, D) the particles do not vibrate at all, 1) A, 2) B, 3) C 4) B & C, 127. The tension in a stretched string fixed at both, ends is changed by 2%, the fundamental, frequency is found to get changed by 15 Hz., a) wavelength of the string of fundamental, frequency does not change, b) velocity of propagation of wave changes, by 2%, c) velocity of propagation of wave changes, by 1%, d) original frequency is 1500 Hz, 1) c only correct, 2) c and d are correct, 3) a, c and d are correct 4) b and d are correct, NARAYANAGROUP, , WAVES, 128. A wave y = Acos( t – kx) passes through a, medium. If v is the particle velocity and a is, the particle acceleration then,, a) y, v and a all are in the same phase, b) y lags behind v by a phase angle of /2, c) a leads y by a phase angle of , 3, d) v leads a by a phase angle of, 2, 1) a only correct, 2) b and c are correct, 3) a, b, c are correct, 4) b and d are correct, 129. The equation of the standing wave in a string, clamped at both ends, vibrating in its third, harmonic is given by, y 0.4sin 0.314 x cos 600t , , where, x and y are in cm and t in sec., a) the frequency of vibration is 300 Hz, b) the length of the string is 30 cm, c) the nodes are located at x 0 , 10 cm, 30 cm, 1) Only a is true, 2) a, b are true, 3) b, c are true, 4) a, b, c are true, 130. The equation of a stationary wave in a string, is y 4mm sin 3.14m 1 x cos t ., Select the correct alternative(s) :, a) the amplitude of component waves is 2 mm, b) the amplitude of component waves is 4mm, c) the smallest possible length of string is 0.5 m, d) the smallest possible length of string is 1.0 m, 1) a, c are correct, 2) b, c are correct, 3) a, d are correct, 4) all are correct, , CUQ - KEY, 1) 3, 7) 1, 13) 1, 19) 4, 25) 4, 31) 2, 37) 2, 43) 2, 49) 4, 55) 2, 61) 4, 67) 3, 73) 1, 79)3, 85)3, 91)3, 97)1, 103)3, 109)4, 115)4, 121)2, 127)3, , 2) 4, 8) 3, 14) 2, 20) 4, 26) 1, 32) 4, 38) 1, 44) 3, 50) 2, 56) 1, 62) 4, 68) 4, 74) 2, 80)4, 86)2, 92)2, 98)1, 104)2, 110)1, 116)1, 122)1, 128) 4, , 3) 1, 9) 4, 15) 4, 21) 4, 27) 4, 33) 4, 39) 4, 45) 1, 51) 4, 57) 2, 63) 4, 69) 4, 75) 2, 81)1, 87)3, 93)3, 99)1, 105)1, 111)1, 117)1, 123)4, 129)4, , 4) 2, 10) 1, 16) 3, 22) 2, 28) 3, 34) 2, 40) 3, 46) 2, 52) 1, 58) 2, 64) 4, 70) 2, 76)2, 82)2, 88)3, 94)2, 100)1, 106)1, 112)2, 118)2, 124)2, 130)3, , 5) 1, 11) 3, 17) 1, 23) 4, 29) 4, 35) 4, 41) 3, 47) 4, 53) 3, 59) 4, 65) 4, 71) 3, 77)3, 83)2, 89)3, 95)3, 101)3, 107)1, 113)4, 119)4, 125)1, , 6) 1, 12) 4, 18) 1, 24) 1, 30) 3, 36) 2, 42) 3, 48) 3, 54) 3, 60) 3, 66) 1, 72) 4, 78)2, 84)3, 90)2, 96)3, 102)1, 108)1, 114)1, 120)2, 126)2, 41
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 4, , LEVEL - V, , π, , given as y1 = 10 sin 3π t + ;, 3, , , SINGLE ANSWER QUESTIONS, PROGRESSIVE WAVES, 1., , Earthquakes generate sound waves inside, Earth. Unlike a gas, Earth can experience both, transvers (S) and longitudinal (P) sound waves., Typically, the speed of S, waves is about 4 km/, s. A sesimograph records P and S waves from, an earthquake. The first P waves arrive 3.0, min before the first S waves (figure). Assuming, the waves travel in a straight line, how far away, does the earthquake occur?, , (Y, , earth, , 2., , 3., , 5., , = 12.8 × 1010 pa, ρ earth = 2000kg / m3 ), , a) 1900 km, b) 1440 km, c) 1800 klm, c) 1200 km, Two particles of medium disturbed by the wave, propagation are at x1 = 0 and x2 = 1 cm. The, respective displacement (in cm) of the particles, can be given by equations y1 = 2 sin 3 p t, , 6., , π, , y 2 = 2 sin 3πt − wave velocity can be, 2, , (A) 6 cm/s (B) 7 cm/s (C) 8 cm/s (D) 9 cm/s, The following fig. show a snapshot of a, vibrating string at t = 0 . The particle P is, , observed moving up with velocity 20 3 cm/s., The tangent at P makes an angle of 600 with, the x-axis the equation of the wave is, , 7., , 0, , 1.5, , P 60°, 3.5, , 5.5, , between two points x1 =, x, 7.5 (× 10–2 m), , π x 3π , , +, A) y = 0.4 sin 10π t −, cm, 2, 4 , , π x 3π , , +, B) y = 0.4 sin 10π t +, cm, 2, 4 , , π x 3π , , +, C) y = 0.4 cos 10π t +, cm, 2, 4 , , π x 3π , , +, D) y = 0.4 cos 10π t −, cm, 2, 4 , , 72, , y2 = 5 sin 3π t + 3 cos 3π t then what is the, ratio of their amplitudes, a) 1 : 2, b) 2 : 1 c) 1 : 1 d) 3:2, A plane progressive wave has frequency 25Hz, and amplitude 2.5 ×10−5 m and initial phase is, zero propagates along the negative x-direction, with a velocity of 300 m/s. The phase, difference between the oscillations at two, points 6m apart along the line of propagation, is:, π, π, a) π, b), c) 2π, d), 2, 4, A 100 Hz sinusoidal wave is travelling in the, positive x–direction along a string with a, linear mass density of 3.5 × 10–3 kg m–1 and, a tension of 35 N. At time t = 0, the point x = 0, has zero displacement and the slope of the, string is π / 20 . Then select the wrong, alternative:, a) velocity of wave is 100 m/s, (b) angular frequency is (200 π) rad/s, c)Amplitude of wave is 0.025 m, d) propagation constant is (4π ) m −1, Equation of a stationary and a travelling waves, are as follows y1 = a sin kx cos ωt and, , y2 = a sin ( ωt − kx ) . The phase difference, , (in 10–3 m), v, 4, , 2 2, , The equation of displacement of two waves are, , in the standing wave, , 8., , π, 3π, and x2 =, is φ1 ,, 3k, 2k, , ( y1 ), , and is φ2 in, , φ1, travelling wave ( y2 ) , then ratio φ is, 2, a) 1, b) 5/6, c) 3/4, d) 6/7, The equation of a wave disturbance is given, π, , , , as y = 0.02 sin 2 + 50πt cos(10πx) where x and y, , , , , are in metres and t in seconds. Choose the, correct statement(s), a) the wavelength of wave is 0.2 m, b) displacement node occurs at x = 0.15 m, c) displacement antinode occurs at x = 0.3 m, d) the speed of constituent waves is 0.2 m/s, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , LONGITUDINAL/PRESSURE WAVES, , WAVE SPEED, 9., , Two vibrating strings of same length, same, cross section area and stretched to same, tension are made of material with densities, ρ and 2 ρ . Each string is fixed at both ends., If V1 and V2 are speeds transverse waves in, the strings with densities ρ and 2 ρ, V1, respectively, then V is:, 2, , 1, 2, 10. When an elastic cord (for which the tension is, proportional to the extension) of length l is, stretched by 2l the speed of the transverse, wave on it in V. If it is further stretched by, another l , the speed of the transverse wave, on it will be, a) 1/2, , b) 2, , c), , 2, , d), , 3, a) 2 V b) 2 2 V c) 3 V d) 2 V, , , 11.. A uniform rope of length l is hung from the, ceiling under its own weight the speed of the, transverse pulse on it at a distance x from, the ceiling will be proportional to, , b) ( l − x ) c) x, d) l − x, x, 12. A 100 m long rod of density 10.0 x 104 kg/m3, and having Young’s modules Y = 1011 Pa, is, clamped at one end. It is hammered at the, other free end. The longitudinal pulse goes to, right end, gets reflected and again returns to, the left end. How much time, the pulse take to, go back to initial point., a)0.1 sec b) 0.2 sec c) 0.3 sec d) 2 sec, 13. A rope hangs from a rigid support. A pulse is, set by jiggling the bottom end. We want to, design a rope in which velocity υ of pulse is, independent of z, the distance of the pulse from, fixed end of the rope. If the rope is very long, the desired function for mass per unit length, µ (z) in terms of µ0 (mass per unit length of, the rope at the top (z = 0) is given by, a), , gz, , −, a) µ (z) = µ e v2, 0, , b) µ ( z ) = µ e, 0, , +, , g , c) µ ( z ) = µ0 log e 2 z d) µ ( z ) = µ e, 0, υ , NARAYANAGROUP, , gz, v2, , 14. The frequency of a man’s voice is 300 Hz and, its wavelength is 1 meter. If the wavelength, of a child’s voice is 1.5 m, then the frequency, of the child’s voice is:, a)200 Hz b) 150 Hz c) 400 Hz d) 350 Hz, 15. A sound wave of frequency 440 Hz is passing, through air. An O 2 molecule (mass = 5.3 x 1026, kg) is set in oscillation with an amplitude of, 10-6 m. Its speed at the mean position of, oscillation is :, b) 17.0 × 10 −5 m / s, a) 1.70 × 10 −5 m / s, c) 2.76 ×10 −3 m / s, d) 2.77 × 10 −5 m / s, 16. Figure shown is a graph, at a certain time t, of, the displacement function S ( x, t ) of three, sound waves 1,2 and 3 as marked on the curves, that travel along x-axis through air. If P1, P2, and P3 represent their pressure oscillation, amplitudes respectively, then correct relation, between them is:, λ1/2, S, , λ1, , 1, , X, 3, 2, λ3, , a) P1 > P2 > P3, , b) P3 > P2 > P1, , c) P1 = P2 = P3, d) P2 > P3 > P1, 17. A sound wave of wavelength 40cm travels in, air. If the difference between the maximum, and minimum pressures at a given point is, 1.0 × 10 −3 N / m 2 and the bulk modulus of air, is 1.4 × 105 N / m 2 , the amplitude of vibration, of the particles of the medium is nearly, a) 1.0 × 10 −10 m, b) 2.2 × 10 −10 m, c) 3.3 × 10 −10 m, d) 4.4 × 10 −10 m, 18. The figure represents the displacement y, versus distance x along the direction of, propagation of a longitude wave. The pressure, is maximum at position marked, V, R, , P, , S, P, , υ2 , + , gz , , , Q, , R, , x, , a)P, , b)Q, , c)R, , d)S, 73
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 19. Two sound waves move in the same direction, in the same medium. The pressure oscillation, amplitudes of the waves are equal but the wave, length of the first wave is double that of, second. Let the average power transmitted, across a cross section by first wave be P 1 and, that by second wave be P 2. Then, a) P1 = P2 b) P1 =4P2 c) P2=2P1 d) P2=4P1, 20. For a sound wave travelling towards +x, direction, sinusoidal longitudinal displacement, ξ at a certain time is given as a function of x., If Bulk modulus of air is B = 5 × 105 N / m 2 , the, variation of pressure excess will be:, ξ, 10–4 m, 0, , 0.1, , 0.2, , x(m), , 0.3, , 21. S1 & S2 are two coherent sources of sound, having no initial phase differene. The velocity, of sound is 330 m/s. No minima will be formed, onthe line passing through S 2 and, perpendicular to the joining S1 and S 2 , if the, frequency of both the source is:, a) 50Hz, b) 60 Hz c) 70 Hz d) 80 Hz ., 22. Three sinusoidal waves have the same, a a, frequency, but their amplitudes are a, ,, 2 3, and their phase angles are 0, π / 2 and π, respectively. The equation of the resultant, wave obtained by the superposition of these, three, waves, is, given, by, y =, , –4, , 10 m, , 5, 6, , follow Y1 = A cos ( ωt − kx ), , + 2π × 10 Pa, 2, , 0.1, , 0, , x(m), 0.2, , Y2 =, , A, π, , cos ωt − kx + , 2, 2, , , Y3 =, , A, cos ( ωt − kx + π ), 4, , – 2π × 10 Pa, 2, , Pex, + 5π × 10 Pa, 2, , 0.1, , x(m), 0.2, , – 5π × 10 Pa, 2, , Y = Y1 + Y2 + Y3 + Y4 such as, , Y = A1 cos (ωt − kx + φ ) then………..(symbols, , Pex, + 2π × 10 Pa, 2, , have their usual meanings), 0.2, , 0, , c), , A, 3π , , cos ωt − kx +, , 8, 2 , , and their resultant wave is calculated as, Y4 =, , 0, , b), , a sin ( kx − wt + θ ) . Then tanθ =, , (A) 4 / 3 (B) 3 / 4 (C) 1 / 3 (D) 1/ 4, 23. Four waves are described by equations as, , Pex, , a), , INTERFERENCE, , 0.1, , x(m), , 1, a) A =, , 5A, 8, , 1, φ = tan −1 , 4, , 2 5A, 8, , 1, φ = tan −1 , 3, , – 2π × 10 Pa, 2, , 1, b) A =, , Pex, + 5π × 10 Pa, 2, , 3 5A, 8, , 1, φ = tan −1 , 2, , 1, d) A =, , 4 5A, 8, , φ = tan −1 (1), , x(m), , 0.1, –5π × 10 Pa, 2, , 74, , 0.2, , 0, , d), , 1, c) A =, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 24. A string of length 0.4 m and mass 10–2 kg is, tightly clamped at its ends. The tension in the, string is 1.6 N. Identical wave pulses are, produced at one end at equal intervals of time, ∆t . The minimum value of time interval, which, allows constructive interference between, successive pulses, is:, (a) 0.05 s (b) 0.10 s (c) 0.20 s (d) 0.40 s, , STATIONARY WAVES, 25. A wire of length ‘l’ having tension T and radius’, r’ vibrates with fundamental frequency “f’., Another wire of the same metal with length, ‘2l’ having tension 2T and radius 2 r will, vibrate with fundamental frequency:, f, f, 2, d), a) f, b) 2f, c), 2 2, 2, 26. A string of length 1.5 m with its two ends, clamped is vibrating in fundamental mode., Amplitude at the centre of the string is 4 mm., Distance between the two points having, amplitude 2 mm is:, a) 1m, b) 75cm c) 60cm d) 50 cm, 27. A 75 cm string fixed at both ends produces, resonant frequencies 384 Hz and 288 Hz, without there being any other resonant, frequency between these two. Wave speed for, the string is :, a)144m/s b) 216 m/s c) 108 m/s d) 72 m/s, 28. A string of length l is fixed at both ends. It is, vibrating in its 3rd overtone with maximum, l, amplitude ‘a’. The amplitude at a distance, 3, from one end is :, a) a, , b) 0, , c), , 3a, 2, , d) a / 2, , 29. A chord attached about an end to a vibrating, fork divides it into 6 loops. When its t tension, is 36N. The tension at which it will vibrate in, 4 loops when attached to same tuning fork is:, a) 24 N, b) 36 N, c) 64 N d) 81 N, 30. An aluminium wire of length 60cm is joined to, a steel wire of length 80 cm and stretched, between two fixed supports., 80 cm, , 60 cm, , Steel, , Aluminium, , The tension produced is 40 N. The cross sectional areas of the steel and, aluminiumWires are 1.0 mm2 and 3.0 mm 2, respectively., NARAYANAGROUP, , The densities of steel and aluminium are, 7.8 g / cm3 and 2.6 g / cm3 respectively. The, frequency of first overtone of this composite, wire with the joint as a node is nearly, a) 180Hz b) 240Hz c) 360Hz d) 480Hz, 31. The vibrations of string of length 60cm fixed, both ends are represented by the equations, , y = 4sin (π x /15 ) cos ( 96π t ) where x and y, are in cm and t in s. The maximum displacement, at x = 5 cm is, a) 2 3 cm b) 4 cm, c) zero, d) 4 2 cm, 32. A stretched string of length 1m fixed at both, ends, having a mass 5 × 10−4 kg is under a, tension of 20N. It is plucked at a point situated, at 25 cm from one end. The stretched string, could vibrate with a frequency of, a) 512 Hz b) 100 Hz c) 200 Hz d) 256 Hz, 33. A piano wire having a diameter of 0.90 mm is, replaced by another wire of the same length, and material but with a diameter of 0.93 mm., If the tension of the wire is kept the same,, then the percentage change in the frequency, of the fundamental tone is nearly, a) +3%, b) +3.3 % c) -3.3% d) -3%, , ORGAN PIPES, 34. An open pipe is suddenly closed at one end, with the result that the frequency of third, harmonic of the closed pipe is found to be, higher by 100 Hz than the fundamental, frequency of the open pipe. The fundamental, frequency of the open pipe is, (A) 200 Hz, (B) 300 Hz, (C) 240 Hz, (D)480 Hz., 35. An open pipe is in resonance in 2 nd harmonic, with frequency f1. Now one end of the tube is, closed and frequency is increased to f2 such, that the resonance again occurs in nth, harmonic. Choose the correct option:, (A) n = 3, f2 =, , 3, f, 4 1, , (B) n = 3, f2 =, , 5, f, 4 1, , (C) n = 5, f2 =, , 5, f, 4 1, , (D) n = 5, f2 =, , 3, f, 4 1, 75
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 36. A closed organ pipe of length L and an open, organ pipe contain gases of densities ρ1 and, , ρ2 respectively. The compressibility of gases, are equal in both the pipes. Both the pipes are, vibrating in thier first overtone with same, frequency. The length of open organ pipe is :, 1/2, , 4L ρ , , 1, (A) 3 ρ , 2, , 1/2, , 4L ρ , , (B) 3 ρ2 , 1, , (C), , 40. A source of oscillations S is fixed to the, riverbed of a river with stream velocity υ . Two, receivers R1 and R2 are fixed also to the, riverbed. If the source generates frequency, f s , the frequencies received by receivers, , R1 and R2 are respectively f1 and f 2 then, , L, 4L, (D), 3, 3, , V, Streamflow, , BEATS, , R1, , 37. There is a set of four tuning forks, one with, lowest frequency vibrating at 550 Hz. By using, any two tuning forks at a time, the following, beat frequencies are heard: 1,2,3,5,7,8. The, possible frequencies of the other three forks, are:, a) 552, 553, 560, b) 557, 558, 560, c) 552, 553, 558, d) 551, 553, 558, 38. A vibrating string of certain length l under a, tension T resonates with a mode, corresponding to the first overtone of an air, column of length 75 cm inside a tube closed at, one end. The string also generates 4 beats per, second when excited along with a tuning fork, of frequency n. Now when the tension of the, string is slightly increased the number of beats, reduces to 2 per second. Assuming the velocity, of sound in air to be 340 m/s, the frequency n, of the tuning fork in Hz is, (A) 344, (B) 336, (C) 117.3 (D) 109.3, , DOPPLER EFFECT, 39. In the figure shown a source of sound of, frequency 510Hz moves with constant velocity, Vs = 20m / s in the direction shown. The wind, is blowing at a constant velocity Vw = 20m / s, towards an observer who is at rest at point B., Corresponding to the sound emitted by the, source at initial position A, the frequency, detected by the observer is equal to (speed of, sound relative to air = 330 m/s), y, , a) 510Hz, , V, , b) 500 Hz, V, A, , y, , (A) f1 = f 2 = f s, , R2, , (B) f1 > f s , f 2 > f s, , (C) f1 < f s , f 2 < f s, (D) f1 > f s , f 2 < f s, 41. An observer is moving along positive x-axis, from the origin. One tuning fork moves away, from the observer while other moves towards, it at the same speed and with same natural, frequency f. The velocity of sound in air is V., If observer hears the beat of frequency f0 ,, find the speed of the tuning fork moving away, from observer relative to the observer, Vf, V ( f − f0 ), (A) 2 f, (B), 2 f0, 0, (C), , V ( f − f0 ), f0, , (D), , (A), , 242, 252, , (C), , Vf 0, 2f, 42. A siren placed at a railway platform is emitting, sound of frequency 5 kHz. A passenger sitting, in a moving train A records a frequency of 5.5, kHz, while the train approaches the siren., During his return journey in a different train, B he records a frequency of 6.0 kHz while, approaching the same siren. The ratio of the, velocity of train B to that train A is, (B) 2, , 5, 6, , (D), , 11, 6, , 43. A wall is moving with velocity u and a source, u, in the same, of sound moves with velocity, 2, direction as shown in the figure. Assuming that, the sound travels with velocity 10u. The ratio, of incident sound wavelength on the wall of, the reflected sound wavelength by the wall, is, equal to, , x, , u, , c) 525 Hz, S, , d) 550 Hz, 76, , S, , a) 9:11, , b) 11:9, , c) 4:5, , d) 5:4, , u/2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , MULTIPLE ANSWER QUESTIONS, 44. A wave is represented by the equation, y = (0.001 mm) sin [(50 s-1)t + (2.0 m-1)x], (A) The wave velocity = 100 m/s, (B) The wavelength = 2.0 m, (C) The frequency = 25/ π Hz, (D) The amplitude = 0.001 mm, 45. An electrically maintained tuning fork vibrates, with constant frequency and constant, amplitudes. If the temperature of the, surrounding air increases but pressure, remains constant, the sound produced will, have, (A) Larger wavelength (B) Larger frequency, (C) Larger velocity, (D) Larger time period, 46. As a wave propagates :, (A) the wave intensity remains constant for a plane, wave, (B) the wave intensity decreases as the inverse of, the distance from the source for a spherical wave, (C) the wave intensity decreases as the inverse, square of the distance from the source for a, spherical wave, (D) total intensity of the spherical wave over the, spherical surface centered at the source remains, constant at all times, 0.8, , 47. Y (x, t) = [(4 x + 5t )2 + 5] represents a moving, pulse where x and y are in meters and in t, second. Then, (A) pulse is moving in negative x-direction, (B) in 2 s it will travel a distance of 2.5 m, (C) its maximum displacement is 0.16 m, (D) pulse is moving in positive x-direction, 48. An air column in a pipe, which is closed at one, end, will be in resonance with a vibrating, tuning fork of frequency 264 Hz, if the length, of the column in cm is, (A) 31.25 (B) 62.50 (C) 93.75 (D) 12.5, 49. Velocity of sound in air is 320 m/s. a pipe closed, at one end has a length of 1 m. Neglecting end, corrections, the air column in the pipe can, resonate for sound of frequency:, (A) 80 Hz (B) 240Hz (C) 320 Hz (D) 400 Hz, 50. Standing waves can be produced :, (A) on a string clamped at both ends, (B) on a string clamped at one end and free at, the other, (C) when incident wave gets reflected from a, wall, (D) when two identical waves with a phase, difference of p are moving in the same direction, NARAYANAGROUP, , WAVES, 51. Choose the correct option (s) regarding beats, (A) beats are periodic variations in the intensity of, sound, (B) to produce beats two sound waves of nearly, equal frequencies travel in same direction, (C) One loud sound followed by faint sound form, one beat., (D) beats can heard if difference between two, frequencies is small and not more than ten., 52. Two source of intensities I 0 and 4I 0 are used, for interference experiment. Due to super, position the resulting intensity can be, (A) 9I0, (B) 4.5I 0 (C) 5I0, (D) 4I 0, 53. A listener is at rest with respect to the source, of sound. A wind starts blowing along the line, joining the source and the observer. Which of, the following quantities do not change ?, (A) Frequency, (B) Velocity of sound, (C) Wavelength, (D) Time period, 54. Two identical straight wires are streched so, as to produce 6 beats/sec when vibrating, simultaniously. On changing the tension, slightly in one of them, the beat frequency, remains unchanged. Denoting by T1 and, , T2 the higher and lower initial tension in, the strings, then it could be said that while, making the above changes in tension :, (A) T2 was decreased (B) T2 was increased, (C) T1 was decreased (D) T1 was increased, , COMPREHENSION TYPE QUESTIONS, Passage-1, A string of mass m is fixed at both ends. It is excited, to vibrate in its fundamental mode, the equation of, the stationary wave being y = A sin kx sin (ωt + π / 4 ), 55. The amplitude of oscillation of a point on the, string which is at a distance of one – third of, the length of the string from one of the fixed, ends is, , A, A, 3A, c), d), 2, 2, 2, 56. The minimum time after t = 0 at which the, kinetic energy of the string is maximum is, a) A, , b), , π, ω, , b), , a), , π, 4ω, , c), , 3π, 4ω, , d), , 2π, 3ω, 77
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 57. The maximum kinetic energy of the string is, 1, 2 2, a) mA2ω 2, b) mA ω, 2, 1, 1, 2 2, 2 2, c) mA ω, d) mA ω, 4, 8, , B = 5 × 105 N / m 2 ), , Two waves are described by the equations :, y1 = A cos(0.5πx − 100πt), and y 2 = A cos(0.46πx − 92πt), here x and y are in m and t is in s., 58. The number of maxima heard in one second, will be, (A) 4, (B) 3, (C) 2, (D) 1, 59. The speed of the higher frequency waveform, is, (A) 200m/s (B) 180m/s (C) 140m/s (D)120m/s, 60. Find the number of times y1 + y2 becomes zero, per second, at x=0., (A) 96, (B) 48, (C) 192, (D) 100, Passage-III:, A string whose ends are tied to the walls are, separated by a distance of 120 cm. waves, produced on such a string travel back and forth, between the walls and standing waves are set up., It is found that the points on the string at which, displacement amplitude is 3.5 mm are separated, by 15 cm, 61. The string is oscillating in, a) 3rd harmonic, b) 4th harmonic, c) 2nd harmonic, d) 1st harmonic, 62. The maximum displacement amplitude is, a) 3.5 mm, b) 3.5 2mm, 3.5, 2, , mm, , d) D) 7mm, , Passage IV, In an organ pipe (may be closed or open) of 99, cm length standing wave is setup, whose equation, is given by longitudinal displacement, 2π, ξ = ( 0.1mm ) cos, ( y + 1cm ) cos 2π ( 400 ) t, 80, Where y is measured from the top of the tube in, centimeters and it in t second., , y, , 78, , 2π, ( y + 1cm ) cos 2π ( 400t ), 80, 2π, 2, b) Pex = (125π N / m ) cos ( y + 1cm ) sin 2π ( 400t ), 80, 2π, c) Pex = ( 225π N / m 2 ) sin ( y + 1cm ) cos 2π ( 200t ), 80, 2π, d) Pex = ( 225π N / m 2 ) cos ( y + 1cm ) sin 2π ( 200t ), 80, , a) Pex = (125π N / m 2 ) sin, , 65. Assume end correction approximately equals, to (0.3) x (diameter of tube), estimate teh, approximate number of moles of air present, inside the tube ( Assume tube is at NTP, and, at NTP, 22.4 litre contains 1 mole), 10π, 10π, (B), (A), 36 × 22.4, 18 × 22.4, 10π, 10π, (C), (D), 72 × 22.4, 60 × 22.4, , Passage - V, Two trains A and B are moving with speeds 20 m/, s and 30 m/s respectively in the same direction on, the same straight track, with B ahead of A. The, engines are at the front ends. The engine of trains A, blows a long whistle., Intensity, , Passage II, , c), , 63. The upper end and the lower end of the tube, are respetively, a) open-closed, b) closed-open, c) open-open, d) closed-closed, 64. Equation of the standing wave in terms of, excess pressure is ___ (Bulk modulus of air, , f1, , f2, , Frequency, , Assume that the sound of the whistle is composed, of components varying in frequency from f1 = 800, Hz to f2 = 1120 Hz, as shown in the figure., The spread in the frequency (highest frequency –, lowest frequency) is thus 320 Hz. The speed of, sound in still air is 340 m/s., 66. The speed of sound of the whistle is, (A) 340 m/s for passengers in A and 310 m/s for, passengers in B., (B) 360 m/s for passengers in A and 310 m/s for, passengers in B., (C) 310 m/s for passengers in A and 360 m/s for, passengers in B., (D) 340 m/s for passengers in both the trains., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 67. The distribution of the sound intensity of the, whistle as observed by the passengers in train, A is best represented by, , (B) P0 +, , Intensity, , (A), , f2, , Intensity, , f2 Frequency, , f2, , Frequency, , Intensity, , f1, , (D), , Frequency, , Intensity, , f1, , (C), , ∆ P0, ∆P, , P0 − 0, 2, 2, , (C) P0 − ∆ P0 , P0 − 2 ∆ P0, , f1, , (B), , 71. The maximum and minimum pressure at the, closed end of the pipe are respectively, (A) P0 + ∆ P0 , P0 − ∆ P0, , f1, f2 Frequency, 68. The spread of frequency as observed by the, passengers in train B is, (A) 310 Hz (B) 330 Hz (C) 350 Hz (D) 290 Hz., , Passage - VI, The air column in a pipe closed at one end is made, to vibrate in its second overtone by a tuning fork of, frequency 440Hz. the speed of the sound in air is, 330 m/s. end corrections may be neglected. Let, P0 denote the mean pressure at any point in the, pipe and ∆P0 the maximum aplitude of pressure, variation then, 69. The amplitude of pressure variation at the, middle of the column is, , ∆p0, ∆p0, ∆p0, ∆p0, (B), (C), (D), 2, 3, 2, 3, 70. The maximum and minimum pressure at the, open end of the pipe are respectively, (A), , P0 P0, ,, 2 2, , (A) Po,Po, , (B), , (C) 2 P0 , 2 P0, , (D) P0 ,, , NARAYANAGROUP, , P0, 2, , (D) P0 , P0, , Passage - VII, A long wire PQR is made by joning 2 wires PQ, and QR of equal radii. PQ has a length 4.8 m and, mass 0.06 kg, QR has length 2.56 m and mass, 0.20kg. Wire PQR is under tension of 80 N. A, sinusoidal wave pulse of amplitude 3.5 cm is sent, along the wire PQ from end P. No power is, dissipated during propagation of wave pulse., 72. Find the speed of wave in wire PQ:, (A) 80 m/sec, (B) 75 m/sec, (C) 60m/sec, (D) 70m/sec, 73. Find the time taken by wave pulse to reach, from P to R, (A) 0.10sec, (B) 0.12 sec, (C) 0.14sec, (D) 0.16sec, 74. Find the amplitude of reflected wave pulse, after the incident wave pulse crosses the, joint Q :, (A) 1.5 cm, (B) 1.25cm, (C) 1.75cm, (D) 2.0cm, , MATRIX MATCHING QUESTIONS, 75. The figure shows a string at a certain moment, as a transverse wave passes through it. Three, particles A, B and C of the string are also, shown. Match the physical quantities in the, left column with the description in the column, on the right., A, , B, C, , Column A Column B, (A) Velocity of A, (P) Downwards, if the, wave is travelling, towards right, (B) Acceleration of A (Q) Downwards, if the, wave is travelling, towards left, (C) Velocity of B, (R) Downwards, no, matter which way, the wave is travelling, (D) Velocity of C, (S) Zero., 79
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 76. Match the following :, Column A, (A) Stiff string has N segments, (B) Open pipe of smaller diameter, (C) Closed pipe, (D) String one end connected to rigid, boundry and other end is free, Column B, (P) All harmonic of, , v, exist, 2l, , (Q) Fundamental frequency, ν =, 2l, (R) Wave length =, N, , v, 2l, , Column-I, , (S) Odd harmonics are only available, ( l = length of string / pipe, v = velocity of wave), 77. Match the appropriate condition in Column-I, with Column-II (n = 0, 1, 2, 3, 4………..), Column – I, (A) Constructive interference, (B) Destructive interference, (C) Maximum intensity (D) Minimum intensity, 78. A source of sound in moving along a circular, orbit of radius 3 m with angular velocity of, 10 rad s–1. A sound detector located for away, is executing linear SHM with amplitude 6 m, on line BCD as shown. The frequency of, 5, , , detector for oscillation is per second. The, π, source is at A when detector at B at t = 0., Source emits a continuous sound wave of, frequency 340 Hz. (velocity of sound = 330 ms–, 1, ). Match the column A with B. (T is time, period of oscillation)., N, M, , A, , B, , C, , D, , P, , Column A, (A) The frequency of sound recorded by the, detector at t = 3T/4, (B) The frequency of sound recorded by the, detector at t = T/4, (C) The ratio of the time period of source and the, detector, (D) Maximum velocity of detector maximum, velocity of source, 80, , Column B, (P) 255 Hz (Q) 1 : 1 (R) 442 Hz (S) 2 : 1, 79. T1 and T2 are higher and lower tensions in, two identical strings stretched to produce n, beats per second when vibrating, simultaniously. The beat frequncy remains, unchanged when tension is slightly changed, in one of them. If the beat frequency remains, unchanged, match the options of the two, columns :, Column-II, , (A) T1 is, , (P), , Increased, , (B) T2 is, , (Q), , Decreased, , (C) [(T 1 – T2)/2] is, (when T2 is constant), (D) [(T1 – T2)/2 is, (when T2 is constant), , 80. Three stars x,y and z have slightly different, temperatures Tx , Ty and Tz respectively. All, stars are receding from the earth with speeds, Vx , V y and Vz repectively relative to the earth., They radiate the maximum energy at the same, wavelength of the light. Match the options of, the following columns :, Column-I, , Column-II, , (A), , Vy > Vx, , (P), , Tx > Ty > Tz, , (B), , Vx > Vz, , (Q), , Tx < Ty < Tz, , (C), , Vx = Vz, , (R), , (D), , Vy > Vz, , (S), , ASSERTION & REASON QUESTIONS, (A) Statement – 1 is True, Statement – 2 is True;, Statement – 2 is a correct explanation for, Statement – 1., (B) Statement – 1 is True, Statement – 2 is True;, Statement – 2 is NOT a correct explanation, for Statement – 1., (C) Statement – 1 is True, Statement – 2 is False., (D) Statement – 1 is False, Statement – 2 is True., 81. STATEMENT – 1 : The basic of Laplace, correction was the exchange of heat between, the region of compression and rarefaction in, air is not possible. because, STATEMENT – 2 : Air is a bad conductor of, heat and velocity of sound in air is large., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, 82. STATEMENT – 1 : The velocity of sound in, air, at constant temperature, does not depend, on the ambient pressure., because, STATEMENT – 2 : This is a consequence of, the fact that the velocity of sound is a function, , P, but as P increases, ρ increases, ρ, by the same factor at constant temperature., STATEMENT – 1 :speed of sound in air was, found wrong because, he assumed process as, isothermal., because, STATEMENT – 2 : Flow of sound wave in a, medium is very fast. Quick process suppress, heat exchange, hence this process must be, adiabatic in nature., STATEMENT – 1 : If an observer places his, ear at the end of a long steel pipe, he can hear, two distinct sounds, when a workman hammers, the other end of the pipe., because, STATEMENT – 2 : Longitudinal as well as, transverses wave can be propagated in steel., STATEMENT – 1 : Soldiers are asked to, break steps while crossing the bridge to avoid, resonance situation., because, STATEMENT – 2 : When frequency of two, oscillating system are equal, their amplitude, of vibration become very high., STATEMENT – 1 : When standing waves are, produced in a closed organ pipe, the pressure, at the closed end is a constant., because, STATEMENT – 2 : The closed end, corresponds to a node and hence the pressure, is constant., STATEMENT – 1 : In the case of stationary, wave, a person hear a loud sound at the nodes, as compared to the antinodes. because, STATEMENT – 2 : In a stationary wave all, the particles of the medium vibrate in phase., Statement – 1: In case of beats, intensity of, sound at some positions in space remains, maximum and at others, it remains minimum, Statement – 2: Beat are formed due to, superposition of sound waves of unequal, frequencies., , of the ratio, 83., , 84., , 85., , 86., , 87., , 88., , NARAYANAGROUP, , WAVES, 89. Statement – 1: Two tuning forks having, frequency 410 Hz and 524 Hz are kept close, and made to vibrate. Beats will not be heard., Statement – 2 : Sound waves superimpose only, when the frequencies of superposing waves are, equal or nearly equal., 90. Statement-1 : Two sound waves of equal, intensity I produced beats. The maximum, intensity of sound produced in beats is 4I., Statement-2 : If two waves of amplitudes a1, and a 2 superpose, the maximum amplitude of, the resultant wave = a1 + a2., , INTEGER TYPE QUESTIONS, 91. Transverse wave is propagating in a string., Tension in the string becomes twice to the, initial tension. Simultaneously, area of crossaction of the string is increased so that there, is no change in speed of the wave. Initial crosssection area is A0 . Final cross-section area is, 92., , 93., , 94., , 95., , KA0 . Find the value of K, A cylindrical tube, open at both ends, has a, fundamental frequency ν . The tube is dipped, vertically in water so that half of its length is, inside the water, find ratio of new fundamental, frequency to old fundamental frequency ?, A guiter string is 90 cm long and has a, fundamental frequency of 124 Hz. It is pressed, at a distance 10x cm from one end to produce, a fundamental frequency of 186Hz. Find the, value of x., A 20 cm long string, having a mass of 1.0 g, is, fixed at both the ends. The tension in the string, is 0.5 N. The string is set into vibrations using, an external vibrator of frequency 100 Hz. Find, the separation (in cm) between the successive, nodes on the string., When two progressive waves y1=4sin(2x- 6t), π, , and y 2 = 3sin 2x − 6t − are superimposed., , , 2, , Find amplitude of resulting wave., 96. The average power transmitted across a crosssection by two sound waves moving in the, same direction are equal. The wave lengths, of two sound waves are in the ratio of 1:2, then, find the ratio of their pressure amplitudes., 81
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 97. The string fixed at both ends has standing, wave nodes for which distance between, adjacent nodes is x1. The same string has, another standing wave nodes for which, distance between adjacent nodes is x2. If l is, , x2, l, the length of the string, then x = ( l + 2 x ) ., 1, 1, What is the difference in number of the loops, in the two cases?, 98. Two vibrating strings of the same material but, of lengths L and 2L have radii 2r and r, respectively. They are stretched under the, same tension. Both the strings vibrate in their, fundamental modes, the one of length L with, frequency, , ν1 and the other with frequency, 2, , 99. The ends of a stretched wire of length L are, fixed at x = 0 and x = L. In one experiment the, πx , , displacement of the wire is y1 = A sin , L , sinω t and energy is E 1 and in other, experiment its displacement is y2 = A sin, 2πx , , sin(2ω t) and energy is E2 = mE1 .Find, L , , the value of ‘m’, 0.8, , 100. Y (x, t) = [(4x + 5t)2 + 5] represents a moving, pulse in a material medium, where x and y are, in metres and t in second. The maximum, displacement of medium particle in, transporting energy of pulse is, (N X 0.08) m.Find N., 101. Transevers wave is propagating in a string., Tension in the string is increased to twice the, initial tension. Simultaceonly, area of, crossection of the string is increased so that, there is no change in speed of the wave. Intial, crossection area is, , A0 Final crossection area, , is KA0 , Value of K is, 82, , inclination 30o with the direction of, propagation of the wave. After some time, interval its inclination Changes to 60o with, direction of propagation. Potential energy of, this small element is initially U 0 and finally it, is K . U 0 find the value of K, 103. A stretched string is fixed at both ends. When, it is vibrating in the fundamental mode the, maximum kinetic energy of the string is E1 and, amplitude at anti-node is A1 . When it is, vibrating in the 3rd harmonic, the maximum, kinetic energy of the string is E2 and amplitude, at anti-node is A2 . If E1 = E2 Find A1 / A2, , ν1, , ν 2 . Find the ratio ν, , 102. A transverse wave is travelling in a string at, any moment a small element ' dx ' is at, , 104. A motor car moving away from a cliff with a, velocity of 90 kmph sounds the horn and the, echo is heard after 20 seconds. Assuming the, velocity of sound in air to be 335 ms −1 , if the, distance between the car and cliff when the, horn is sounded is (P+ 0.1)Km, then P value is, , LEVEL-V - KEY, SINGLE ANSWER QUESTIONS, 1) B 2) A 3) A 4) C 5) A 6)D 7) D, 8) D 9) C 10) A 11) B 12) B 13) A 14) A, 15) C 16) B 17) B 18) C19) A 20) D, 21) A 22) B 23) C 24) B 25) C 26) A 27) A, 28) C 29) D 30) C31) A 32) C 33) C 34) A, 35) C 36) A 37) D 38) A 39)C 40) A 41) D, 42) B 43) A, MULTI ANSWER QUESTIONS, 44) C,D, 45) A,C, 46) A,C,D, 47) A,B,C 48) A,C, 49) A,B,C, 50) A,B,C 51) A,B,C,D 52) A,B,C,D, 53) A,D, 54) B, C, COMPREHENSION TYPE QUESTIONS, 55) D 56) C 57) C 58) A 59) A 60) A 61) B, 62) B 63) A 64) A 65) B 66) B 67) A 68) A, 69) A 70) A 71) A 72) A 73) C 74) A, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, MATRIX MATCHING QUESTIONS, 75) A-P; B-R; C-Q; D-S, 76) A-P,Q,R; B-P,Q,R; C-S ; D-S, 77) A-Q,R; B-P,S; C-Q,R; D-P,S, 78) A-R; B-P; C-Q; D-S, 79) A-Q; B-P; C-Q; D-P, 80) A-Q; B-P; C-S; D-P, ASSERTION & REASON QUESTIONS, 81) A 82) A 83) A 84) B 85) A 86)D 87) C, 88) D 89)C 90)A, INTEGER ANSWER TYPE QUESTIONS, 91) 2 92) 1 93) 6 94) 5 95) 5 96) 1, 97) 2 98) 1 99) 4 100) 2 101) 2 102) 9, 103) 3 104) 3, , LEVEL-V - HINTS, SINGLE ANSWER QUESTIONS, 01. Vs = 4km / sec, VP =, , y, 12.8 × 1010, =, ρ, 2000, , = 8000m / sec = 8km / sec, , l, l, − = 3min = 3 × 60sec, Vs Vp, l l, − = 3 × 60 ; l = 1440 km, 4 8, , 02. ∆φ =, , 2π, ( ∆x ), λ, , π 2π, =, (1) ⇒ λ = 4cm ; v = f λ = 6cm−1, 2 λ, dy dy, 03. v p = −v , = tan 60 = 3, dx dx, , v p = −v 3 ⇒ P is moving along upward direction, ⇒ wave must be travelling along negative x, direction, The, equation, of, the, wave, is, , y = A sin (ωt − Kx + φ ), 20 3, A = 4 × 10−3 m, v =, = 20cms −1, 3, , 0.2, λ = 4 ×10−2 m = 4cm ; f = 4 ×10−2 5 Hz, NARAYANAGROUP, , WAVES, ⇒ ω = 2π f = 10π, , K=, , 2π 2π π, =, = cm−1, λ, 4, 2, , at t = 0, x = 0, y = 2 2 × 10−3 m =, , 2 2, cm, 10, , ⇒, , 2 2, × 4 × 10−1 sin (φ ), 10, , ⇒, , 2 2, 1, π 3π, × 4 × 10−1 sin (φ ), ⇒ φ or, 10, 4, 4, 2, , taking at x = 1.5, t = 0 y = 0 ⇒ φ, , ,, , 3π, 4, , π x 3π , , ⇒ y = 0.4 sin 10π t −, +, , 2, 4 , , , 04. y2 = 5 sin 3π t + 3 cos 3π t , π, π, , , = 5 1 + 3 sin 3π t + = 10 sin 3π t + , 3, 3, , , , So, A1 = 10 and A2 = 10, 05. ∆φ =, , 2π, × ∆x, λ, , T, 35, v, 06. v = µ = 3.5 × 10−3 = 100 m / s , λ = = 1 m, f, , ∴, , k=, , 2π, = (2π )m −1, λ, , and, ω = v × k = (200π) rad / s, at x = 0, vp = maximum, , particle velocity = wavevelocity × slope, , ∴, , A=, , slope × wave velocity, ω, , π , × (100), 20, = , = 0.025 m, 20π, , 07. x1 and x2 are in successive loops of std. waves., So φ1 = π and, , π 7π φ1 6, 3π, φ2 = K ( ∆x ) = K , −, = =, =, 2 K 3K 6 φ2 7, 08. From the given expression for y :, amplitude A = 0.02 n, angular frequency ω = 50π rad / s, 83
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , v = 8 m/s, For constructive interference between successive, , 2l, v, , (2)(0.4) = 0.10 s, 8, (After two reflections, the wave pulse is in same, phase as it was produced, since in one reflection, its phase changes by p and if at this moment next, identical pulse is produced, then constructive, interference will be obtained.), , pulses : Dtmin=, , 25. f =, , =, , 1 T, 2l µ, , If radius is doubled and length is doubled, mass, per unit length will become four times, hence, f1 =, , 1, 2T, f, =, 2 × 2l 4 µ 2 2, , 26. λ = 2l = 3m, Equation of standing wave ( As x=0 is taken as a, node), y = 2 A sin kx cos ω t ; y= A as amplitude is 2A, A = 2A sin kx, 2π, π, 1, x = ⇒ x1 = m and, D, 6, 4, 2π, π π, .x = + ⇒ x2 = 1.25m ⇒ x2 − x1 = 1m, D, 2 3, , nv, − − − − − (1), 27. 384 =, 2l, mv, 288 =, − − − − − ( ii ), 2l, n 4, from equation (i) and (ii) = , m 3, So, n = 4From equation (i), , 4v, 10v, =, 2× 3 / 4, 3, v = 144 m/s, 28. For a string vibrating in its nth, 384 =, , overtone ( ( n = 1) hormonic ), , l, For, 3, = − a. , cos ωt, 2, , , , 4π, = a.sin 3 cos ωt, , l, 3a, i.e, at x = ; the amplitude is, 3, 2, , 29. For waves along a string : υα T, , ⇒ Dα T, , Now, for 6 loops : 3D 1 = L ⇒ D 1 = L / 3, & for 4 loops : 2D 2 = L ⇒ D 2 = L / 2, ⇒, , D1 2, =, D2 3, , 30. n =, , 1, 2l, , ⇒ T2 =, , 9, 9, × T1 = × 36 = 81N, 4, 4, , T, µ, , 31. So y will be maximum when cos ( 96π t ) =max=1, , ymax = 2 3cm at x = 5, 32. At 25 cm, there will be antinode., So wire will vibrate in two loops, , v=, , 2 T ×l, or v =, 2l M, , T, 20, =, Ml, 5 × 10−4 × 1, , = 4 ×104 Hz = 200 Hz, 33. v =, , 1, T, 1, or vα, Dl π d, D, , v1 , 30 , 100, = −3.2, Now v − 1 × 100 = 31 − 1 ×100 = −, , , 31, , , , 34. Fundamental frequency of open pipe is f0 =, , th, , ( n + 1) π x , y = 2 A, cos ωt, L, , , NARAYANAGROUP, , v, 2l, , , Third harmonic of the closed pipe fc = 3 4l , v, , , , Given: 3, , v, v, =, + 100 or,, 4l 2l, , , , v, = 200 Hz, 2l, 85
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 35. f1 =, , 43. λi = Wavelength of the incident sound, , v, ; (2nd harmonic of open pipe), l, , =, , Here, n is odd and f2 > f1, It is possible when n = 5, , f1 = Frequency of teh incident sound, , , Because with n = 5; f2 = 4 l = 4 f1, , 5 v, , 36., , u, 2 = 19u, f, 2f, , 10u −, , v, f2 = n 4l ; (nth harmonic of closed pipe), , , fc = f0, , 5, , V , V , 3 c = 2 0 , 4L , 2 L0 , , 4 V0 , 4 ρ1 , on solving L0 = 3 V L = 3 L ρ , c, 2, 37. To get beat frequency 1,2,3,5,7,8, it is possible, when other three tuning fork have frequencies 551,, 553, 558, etc..,, 38. With increases in tension, frequency of vibrating, string will increases, since number of beats are, decreasing. Therefore frequency of vibrating string, or third harmonic frequency of closed pipe should, be less than the frequency of tuning fork by 4., V , 340 , ∴ 3 + 4 = 3, + 4 = 344 Hz, 4l , 4 × 0.75 , , 39. Apparent frequency, n' = n, , (u + v, , ( u + vw ), , w, , =, , 10u − u, 18, f =, f = fr =, u, 19, 10u −, 2, Frequency of the reflected sound, =, , λr = Wavelength of the reflected sound, =, , 10u + u 11u, 11× 19 u, =, × 19 =, ., fr, 18 f, 18, f, , λi 19u, 18 f, 9, =, ×, =, λr 2 f 11× 19u 11, , MULTI ANSWER QUESTIONS, 44. (C,D); ω = 50 ;, , γP, and ν = f λ, ρ, 46. (A, C, D); For a plane wave intensity (energy, crossing per unit area per unit time) is constant at, all points., , 45. (A,C); ν =, , − vs cos 600 ), , V + V0 , V − V0 , f1 − f 2 = f 0 ; V − V f − V + V f = f0, s , s , , , , f 0 2 f (Vs + V0 ), =, ;, f, V, 42., , 2, , 1, , 2, , (Vs − V0 ) =, , But for a spherical wave intensity at a distance r, from a point source of power P (energy transmitted, per unit time) is given by, S, , Vf0, 2f, , v + v1 , v + v2 , f1 = 5.5 = f 0 , ; f2 = 6 = f0 , , v , v , v - velocity of sound, v1 - velocity of A ; v2 - velocity of B, v2, On solving v = 2, 1, , 86, , 1, , 510 ( 330 + 20 ), , 330 + 20 − 20 cos 600, 350, = 510 ×, = 525 Hz, 340, , 41., , ω, 2π, = 2, A = 0.001 mm . v =, λ, K, , l, S, , r, , I=, , P, or I, 4 πr 2, , α, , 1, r2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 47. (A, B, C); The shape of pulse x = 0 and t = 0, would be as shown, in figure (A)., y (0, 0) =, , 0.8, = 0.16 m, 5, , v, , , f = n 4l n = 1, 3, 5 ….., , –x 0 x, t = 0 (a), , v, , From the figure it is clear that ymax = 0.16 m, Pulse will be symmetric, (Symmetry is checked about ymax) if at t = 0, Y(x) = y(–x), From the given equation, 0.8 , 16 x 2 + 5 , , 0.8 at t = 0, y (– x) =, 16 x 2 + 5 , y ( x) =, , or y(–x) = y(–x), Therefore, pulse is symmetric., Speed of pulse, At t = 1 s and x = – 1.25 m, y, 0.16m, , –x, , x = –1.25m, t = 1s, , 0.16m, , x=0, t=0, , x, , value of y is again 0.16 m, i.e., pulse has traveled a, distance of 1.25 m in 1 second in negative xdirection or we can say that the speed of pulse is, 1.25 m/s and it is traveling in negative x-direction., Therefore, it will travel a distance of 2.5 m in 2, seconds. The above statement can be better, understood from figure (B)., Alternate method, If equation of a wave pulse is y = f (ax ± bt), The speed of wave is, , b, in negative x direction for, a, , y = f (ax + bt) and positive x direction for y = (ax –, bt). Comparing this from given equation we can, find that speed of wave is, , 5, 4, , = 1.25 m/s and it is, , traveling in negative x-direction., NARAYANAGROUP, , nv, , (1)(330), , 0.16m, , y, , v, , f = n 4l where, n = 1, 3, 5 …; ∴ l = 4 f, , , For n = 1, l1 = 4 × 264 × 100 cm = 31.25 cm, For n = 3, l3 = 3l1 = 93.75 cm, For n = 5, l5 = 5l1 = 156.25 cm, 49. (ABC); For closed pipe., , y, , and, , 48. (AC); For closed organ pipe,, , 320, , For n = 1, f1 = 4l = 4 × 1 = 80 Hz, For n = 3, f3 = 3f1 = 240 Hz, For n = 5, f5 = 5f1 = 400 Hz, ∴ correct options are (a), (b) and (d)., 50. (ABC); Standing waves can be produced only, when two similar type of waves (same frequency, and speed but amplitude may be different) travel in, opposite directions., 51. (ABCD); Beats are periodic variations in the, intensity of sound when two sound wages of nearly, equal frequencies travel in the same direction. One, loud sound followed by a faint sound from one beat, and the number of beats formed per second is called, beat frequency., 52. (ABCD) I = I1 + I 2 + 2 I1I 2 cos φ, I max = 9 I 0 I min = 9 I 0, all intensities between I max and I min can be obtained, 53. (A,D) If there is no relative motion between source, and sound the frequency remains same., 54. (B,C) T1 > T2 and so, v1 > v2, it implies f1 > f 2 and f1 − f 2 = 6 Hz, On increase of T1 , f1 will increase or, , ( f1 − f 2 ), , will increase.On decrease of T1 , f1 will decrease, but now, , ( f 2 − f1 ), , may be equal to 6 Hz.On, , increase of T2 , f 2 will increase and now again, , ( f 2 − f1 ) may become equal to 6 Hz., COMPREHENSION TYPE QUESTIONS, Passage- I, 55. Amplitude = A sin Kx ; Kx =, ∴ =, , 2π x 2π l π, =, × = ., λ, 2l 3 3, , A 3, 2, 87
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , Passage-VI1, 0.06 1, =, 4.8 80, 0.2, Mass per unit length of QR, m2 =, 2.56, , 72. Mass per unit length of PQ, m1 =, , T, , Speed of wave in wire PQ , v1 = m = 80m / sec, 1, 73.. Speed of wave in wire, , QR , v2 =, , 4.8 2.56, +, = 0.14 seconds, v1, v2, , 74. Reflected amplitude is given by, , Ar =, , ( 32 − 80 ) × 3.5, v2 − v1, Ai =, = −1.50cm, v2 + v1, ( 32 + 80 ), , MATRIX MATCH TYPE QUESTIONS, 76. (A) – (p), (q), (r), (B)–(p),(q),(r), (C) – (s), (D)–(s), Fixed points of the string are nodes and free end is, antinodes Wave length corresponds to N segments, when both end are fixed λ =, , 2l, nv, fn =, n, 2l, , with n = 1,, , 2, 3, In closed end organ pipe closed end will be, node while free end will be antinode., 79. f1 − f2 = n Hz, , f1 > f 2 or V1 > V2 or T1 > T2, If T1 is increased, f1 will increase or ( f1 − f 2 ), will increase and then f1 > f 2 ≠ n Hz i.e. beat, frequency changes. But if T1 is decreased, f1 will, decrease or ( f1 − f 2 ) ≠ nHz . So (A)Q., In the same way, , If Tx > Ty > Tz , λx > λz, As these wavelength appear equal on reaching, ground, λx has increased more than λ y and so, on, Hence Vx > Vy > Vz or Vx > Vz etc., , ASSERTION & REASON QUESTIONS, , T, = 32m / sec, m1, , Time taken by pulse to reach from P to R, , =, , 80. λx , λ y and λz are actual wavelengths at which, maximum energy is radiated., From Wien’s law λxTx = λyTy = λzTz, , (T1 − T2 ) / 2 =, , T1 T2, − =, 2 2, , 81. (A); According to Laplace, the changes in pressure, and volume of a gas, when sound waves propagates, through it, are not isothermal, but adiabatic. A gas, is a bad conductor of heat. It does not allow the, free exchange of heat between compressed layer,, rarefied layer and surrounding., 84.. (B); Two sound of heard because of different sound, speed in air and steel., 87 (C); The person will hear the loud sound at nodes, than at antinodes. We know that at anti–nodes the, displacement is maximum and strain is minimum, while at nodes the displacement is zero and strain, is maximum. The sound is heard due to variation of, pressure. Further, P = – E(dy/dt), where E is, elasticity and dy/dt is strain. As strain is maximum, at nodes, hence there is maximum variation of, pressure and loud sound is heard., 82. (D) ;In beats resultant amplitude is function of time., 89. (C) ; Number of beats = 524 − 410 = 14, Due to persistence of hearing we can hear more, than 10 beats per second two waves of same, frequencies and amplitude superimpose give, stationary waves., 90. The correct choices is (a). When two waves of, amplitudes a 1 and a 2 superpose to produce beats,, the resultant amplitude of the maximum of intensity, is l = a1 + a2., , INTEGER ANSWER TYPE QUESTIONS, T, T, =, is constant A ∝ T, µ, ρA Q?, , T1, , − K decreases, f1 will decrease, then, 2, , , 91. 2; v =, , ( f 2 − f1 ) may become equal to nHz . Similarly if, , 92. 1; Open tube f1 =, , T2 is increased, f 2 will increase and now, , ( f 2 − f1 ), , may be equal to n Hz. But if T2 is, , decreased, f 2 will decrease and ( f 2 − f1 ) ≠ nHz ., NARAYANAGROUP, , f2 =, , Close tube, , V, 2L, , V, L, 4 ;, 2, , f2, =1, f1, 89
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, v1 L2, 124, 93. 6; v = L ⇒ L2 = 186 × 90 = 60cm, 2, 1, 94. 5; l = 20m ; m = 1 gm; T = 0.5 N; f = 100 Hz, µ=, , 1×10−3, 1, = ×10−2 ;, 20 × 10−2 2, , λ = 0.1, , ;, , V=, , 0.5, = 10m / s, 0.5 × 10−2, , λ, = 0.05m = 5cm ., 2, , 99. Energy E ∝ (amplitude)2 (frequency)2, Amplitude (A) is same in both the cases, but, frequency 2ω in the second case is two times the, frequency (ω) in the first case., Therefore, E2 = 4E1, 100. The shape of pulse x = 0 and t = 0, would be as shown, in figure (A)., y (0, 0) =, , 95.. 5 ; A = A12 + A 22 + 2A1A 2 cos φ, 96.. P =, , 0.8, = 0.16 m, 5, y, , 1, ρω 2 A2 sV, 2, , 0.16m, , λ1 1 f1 ω1 2, since λ = 2 , f = ω = 1, 2, 2, , since P1 = P2 , ω1 A1 = ω2 A2, , –x 0 x, t = 0 (a), , Pressure amplitude, P0 = B0 Ak, , A1 k1 , , A2 k 2 , , ( Po )1 / ( Po )2 = , , 101. v =, , A λ , 1 2, = 1 2 = , = 1, A2 λ1 2 1 , , 2, , 97. let no. of loops formed in first case =n ; x1 n = 1, Let no. of loops formed in second ; case = (n+k), , x2 ( n + k ) = l ;, , T, T, =, µ, µA, , 1 ∂y , 102. P.E = T . .dx ; K=9, 2 ∂x , 104. d= (V 2t 2 − L2 ) / 4, T, , x2, l, =, ,k = 2, x1 l + kx1, , L, , R, , d, , On comparing we get k=2, 98. Fundamental frequency is given by V=, , 1 T, (with, 2l µ, , LEVEL-VI, , both the ends fixe(D)), ∴ Fundamental frequency v ∝, , 1, l µ, , [for same tension in both strings], where µ = mass per unit length of wire, = ρ.A (ρ = density) = ρ(πr2), Or, , v1, ∴v, 2, 90, , µ ∝ r;, , l, ∴v∝, rl, , r2 l2 r 2L , = r l = = l, 1 1 2r L , , SINGLE ANSWER QUESTIONS, 1., , The amplitude of a wave disturbance, propagating in the positive x–direction is, given by y =, , 1, 1, y=, at, 2 at t = 0 and, 1+ x, 2 + x2 − 2 x, , t = 2s , where x and y, , are in meter. Assuming, that the shape of the wave disturbance does, not change during the propagation, the speed, of the wave is, (A) 0.5 m/s, (B) 1 m/s, (C) 1.5 m/s, (D) 2 m/s, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, 2., , WAVES, , S1 and S2 are two coherent currents sources, of radiations separated by distance 100.25 λ ,, where λ is the wave length of radiation. S1, leads S2 in phase by π / 2 . A and B are two, points on the line joining S 1 and S 2 as shown in, figure. The ratio of amplitudes of sources S1, and S2 are in the ratio 1 : 2. The ratio of, , 6., , A string has a linear mass density ' µ ' and a, length L = 3m. Its two ends are D =2m apart., Two blocks of mass M = 8 kg each aree, suspended from the string as shown in the, figure. The time taken by a wave pulse to, travel from point A to point B is, , IA , , intensity at A to that of B I is, , , A, , S1, , S2, , B, , , , T1, , T2, B, , A, , B, , 100.25λ, , (A) ∞, 3., , 1, (B), 9, , (C) 0, , (D) 9, , M, , The ends of a stretched wire of length L are, fixed at x = 0 and x = L. In one experiment the, 7., , 2πx , , 4., , y2 = A sin L sin 2 ωt and energy is E2., , , Then:, (A) E 2 = E1, (B) E 2 = 2E1, (C) E2 = 4E1, (D) E 2 = 16E1, A transverse sinusoidal wave moves along a, string in the positive x–direction at a speed of, 10 cm/s. The wavelength of the wave is 0.5 m, and its amplitude is 10 cm. At a particular, time t, the snap–shot of the wave is shown in, figure. The velocity of point P when its, displacement is 5 cm is, , µ, L µ, L g, L µ, B), C), D), g, 2 g, 2 µ, 3 g, A rope of mass ‘m’ and length ‘L’is, suspended vertically. If a mass ‘M’is, suspended from the free end of the rope, the, time taken by a transverse wave pulse, generated at the bottom to travel to the top is, , A) L, , πx , , displacement of the wire is y1 = A sin L , , sin ωt and energy is E 1 and in other, experiment, its, displacement, is, , M, , A) 2 mg ( M + m − m ) B) mg ( M + m − m ), L, , C) 2 mg ( M − m ) D) mg ( M − m ), An object of specific gravity ρ is hung from a, thin steel wire. The fundamental frequency, for transverse standing waves in the wire is, 300 Hz. The object is immersed in water, so, that one half of its volume is submerged. The, new fundamental frequency (in Hz) is :, 1, , 8., , y, , L, , L, , L, , 12, , P, , x, , 2ρ , (B) 300 2ρ – 1 , , , , 2ρ , , 2ρ – 1 , , (C) 300 2ρ – 1 , , 5., , (A), , 3π, ĵ m / s, 50, , (B) −, , 3π ˆ, jm/s, 50, , (C), , 3π, î m / s, 50, , (D) −, , 3π, î m / s, 50, , The displacement y of a particle executing, periodic, motion, is, given, by, 2, y = 4 cos ( t ) sin (1000t ) . This expression may be, considered to be a result of the superposition, of waves :, (A) two, (B) three (C) four, (D) five, , NARAYANAGROUP, , 9., , 12, , 2ρ – 1 , (A) 300 2ρ , , , , (D) 300 2ρ , , , , , Two vibrating strings of the same material, but of lengths L and 2L have radii 2r and r, respectively. They are stretched under the, same tension. Both the strings vibrate in their, fundamental modes, the one of length L with, frequency v1 and the other with frequency v2., v1, , The ratio v is given by:, 2, (A) 2, (B) 4, (C) 8, , (D) 1, 91
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 10. In the experiment for the determination of, the speed of sound in air using the resonance, column method, the length of the air column, that resonates in the fundamental mode with, a tuning fork is 0.1 m. When this length is, changed to 0.35 m, the same tuning fork, resonates with the first overtone. The end, correction is, (A) 0.012 m, (B) 0.025 m, (C) 0.05 m, (D) 0.024 m, 11. A massless rod of length l is hung from the, ceiling with the help of two identical wires, attached at its ends. A block is hung on the, rod at a distance x from the left end. In this, case, the frequency of the 1 st harmonic of the, wire on the left end is equal to the frequency, of the 2nd harmonic of the wire on the right., The value of x is, , 14. A train moves towards a stationary observer, with speed 34 m/s. The train sounds a whistle, and its frequency recorded by the observer is, f1. If the train’s speed is reduced to 17 m/s,, the frequency registered is f2. If the speed of, f1, , sound is 340 m/s then the ratio f is :, 2, (A), , 18, 19, , (B), , 1, 2, , (C) 2, , (D), , 19, 18, , 15. A simple harmonic oscillator of frequency ‘f’, is attached to the end of a cord that has a, linear mass density ' µ ' and is under a tension, ‘T’. The power that must be provided to the, cord by the oscillator to generate a sinusoidal, wave of amplitude ‘A’, angular frequency, ' ω ' and speed ‘v’ is, A) µω A v, 2, , 2, , µω 2 A2 v, B), 2, , µω 2 A2 v 2, 3µω 2 A2 v, D), C), 2, 2, 16. Two wires of different linear mass densities, are joined, consider the junction to be at x = 0., x, , l, l, l, l, (B), (C), (D), 2, 3, 4, 5, 12. A police car moving at 22 m/s chases a, motorcyclist. The police man sound his horn, at 176 Hz, while both of them move towards a, stationary siren of frequency 165 Hz. If the, motorcyclist does not observe any beats, the, speed of motor cycle is, (velocity of sound is 330ms −1 ), , (A), , Police car, , Motorcycle, , An incident wave y1 = Ai sin ( ωt − k1x ) is, travelling to the right from the region x ≥ 0 ., At the boundary the wave is partly reflected, and partly transmitted. If Ai , A r a n d At, respectively represent the incident reflected, A, , r, and transmitted amplitudes then A = (, t, , given k2 is the wave number of transmitted, wave), 2k1, k1 − k 2, (A) k − k, (B) 2k, 1, 2, 1, k1 + k2, 2k1, (D), 2k1, k1 + k2, 17. Two loud speakers are being compared, and, one is perceived to be 32 times louder than, the other. The difference in intensity levels, between the two when measured in decibels., is, (A) 60, (B) 40, (C) 50, (D) 30, 18. There are three source of sound of equal, intensities with freqency ( n-1) ,n, ( n+1) Hz., The beat freqency heard if all the sources are, switched on simultaneously is, (A) 2, (B) 3, (C) 4, (D) 0, , (C), , 22m/s, 176Hz, , V, , Stationary siren, (165 Hz), , (A) 33 m/s (B) 22 m/s (C) zero (D) 11 m/s, 13. A string of length 0.4 m and mass 10–2 kg is, tightly clamped at its ends. The tension in the, string is 1.6 N. Identical wave pulses are, produced at one end at equal intervals of time, ∆t . The minimum value of ∆t , which allows, constructive interference between, successive pulses, is :, (A) 0.05 s (B) 0.10 s (C) 0.20 s (D) 0.40 s, 92, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 19. The temperature at which the fundamental, freqncy of an organ pipe is independent of, small variation in temperature in terms of the, coefficient of linear expansion (α ) of the, material of the tube is, (A), , 1, α, , (B), , 1, 3α, , (C), , 1, 2α, , (D), , 1, 4α, , 20. The figure shows the location of a source and, detector at time t = 0. the source and, detector are moving with velocities, υ = 5$i m / s and υ = 10 $j m / s respectively.., s, , D, , The frequency of signals received by the, detector at the moment when the source, crosses the origin is ( the frequency of source, is 100 Hz.velocity of sound 330 ms −1 ), , S, 100m, 100m, , (C), , 6000, 6000, Hz,, Hz, 30 − 2 2, 30 + 2 2, , 6000, 6000, Hz,, Hz, 30 − 4 2, 30 + 4 2, 22. A whistle of frequency f 0=1300Hz is dropped, from a height H = 505 m above the ground. At, the same time, a detector is projected, upwards with velocity υ = 50 ms −1 along the, same line. If the velocity of sound is, V=300ms-1, The frequency detected by the, detector after t = 5s is (g=10ms-2), (A) 1600 Hz, (B) 1500 Hz, (C) 1700 Hz, (D) 1800 Hz, 23. 2) Two coherent sources S1 and S 2 at a, distance interference effect at point P,O is, the middle point of S1S 2 and origin of the, coordinate system, as shown, such that, ∠POX = θ . Find the coordinate of source S1, when sources are rotated about point O so, that no interference effect is observed at P:, (D), , D, , P, , (A) 97Hz (B) 47Hz (C) 90Hz (D) 60Hz, 21. A boy is moving along a circular track in, anticlockwise sense. A children train moves, along a square path with centres of circular, track and square both coinciding, as shown in, the figure. the train as well as the boy start, from points B and A such that points O, A and, B always lie on the same radial line. The, velocity of listener is 11 m/s. The train, continuously whistles at frequency 300 Hz., during one such complete rotation the, maximum and minimum frequency heard by, the boy. are respectively ( Take velocity of, sound 330 m/s), C, , B, , S1, 0, , θ, , S2, , d, d, d, d, (A) − sin θ , cos θ (B) sin θ , cos θ, 2, 2, 2, 2, d, d, d, d, sin θ ,, cos θ, (C) cos θ , sin θ (D), 2, 2, 2, 2, 24. How long will it take sound waves to travel, the distance/between the points A and B if the, air temperature between them varies linearly, , from T1 to T2 ? (T2 > T1 ) and α =, , γR, M, , ., , l, , D, a, a/4, , A) α (T − T ) T2 − T1 , 2, 1, 4l, , B) α (T − T ) T2 − T1 , 2, 1, E, , 2l, , 9000, 9000, Hz ,, Hz, (A), 30 − 2 2, 30 + 2 2, , C) α (T − T ) T2 − T1 , 2, 1, , 9000, 9000, Hz ,, Hz, (B), 30 − 4 2, 30 + 4 2, , D) 2α (T − T ) T2 − T1 , 2, 1, , NARAYANAGROUP, , l, , 93
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 25. Figure shows two snapshots of medium, particles within a time interval of 1/60 s. Find, the possible time periods of the wave, Y, 1 1, 0.866, , 2, , X, , 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15, 1, , 1, A) 10 ( 6n + 1), , 1, B) 10 ( 5n + 1), , 1, C) 10 ( 6n − 1), , 1, D) 20 ( 6n + 1), , P, , S1, , path S1 , there is a zone of hot air having, temperature 4 times, the normal temperature,, and width d. What should be minimum, frequency of sound, so that minima can be, found at P?, S1, d, 4T, 0, , P, S2, , 94, , V0, d, , B), , 4V0, d, , C), , V0, 2d, , D), , S1 as shown in figure.Find the distance x such, that the intensity at P is equal to the intensity, at O., x, , 2λ, , 26. Airport authority has made the regula, tions that maximum allowable intensity level, detected by a microphone situated at the end, of 1630 m long runway can be 100 dB. An, aeroplane when flying at a height of 200 m, produces an intensity level of 100 dB on, ground. While taking off, this aeroplane makes, a angle 300 with horizontal. Find the maximum, distance this aeroplane can cover on the, runway, so that the regulations are not violated, (assume no reflection)., A) 1200 m. B) 1230 m C) 1430 m D) 1530 m., 27. Sound from two coherent sources S1 and S 2, are sent in phase and detected at point P, equidistant from both the sources. Speed of, sound in normal air is V0 , but in some part in, , A), , 28. Two coherent narrow slits emitting wave length, λ in the same phase are placed parallel to, each other at a small separation of 2 λ , the, sound is detected by moving a detector on the, screen S at a distance D (>> λ ) from the slit, , 2V0, d, , O, , S2, D, , A) D, , S, D, , B) 3D, , C) 3D D), 3, 29. Radio waves coming at angle α to vertical, are received by a ladder after reflection from, a nearby water surface and also directly. What, can be height of antenna from water surface, so that it records a maximum intensity (a, maxima) (wavelength = λ ), , h, , A), , α, , λ, λ, λ, λ, B), C), D), 2 cos α, 2 sin α, 4 sin α, 4 cos α, , MULTIPLE ANSWER QUESTIONS, 30. A wave equation which gives the, displacement along the y-direction is given, by: y = 10–4 sin (60 t + 2x) where x and y are in, metres and t is time in seconds. This, represents a wave:, (A) traveling with a velocity of 30 m/s in the, negative x-direction, (B) of wavelength π m, (C) of frequency 30/ π hertz, (D) of amplitude 10–4 m traveling along the, negative x-direction, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 31. A wave is represented by the equation;, y = A sin (10 π x + 15 π t + π /3), where x is in meters and t is in seconds. The, expression represents:, (A) a wave traveling in the positive x-direction, with a velocity 1.5 m/s., (B) a wave traveling in the negative x-direction, with a velocity 1.5 m/s, (C) a wave traveling in the negative x-direction, with a wavelength 0.2 m, (D) a wave traveling in the positive x-direction, with a wavelength 0.2 m, 32. A transverse sinusoidal wave of amplitude a,, wavelength λ and frequency f is traveling on, a stretched string. The maximum speed of, any point on the string is, , v, , where v is the, 10, , 34. A wave disturbance in a medium is described, π, , by y(x, t) = 0.02 cos 50πt + 2 cos (10 π x),, , , where x and y are in metre and t is in second, (A) a node occurs at x = 0.15 m, (B) an antinode occurs at x = 0.3 m, (C) the speed of wave is 5 ms–1, (D) the wavelength is 0.2 m, 35. Two tuning forks P and Q are vibrated, together. The number of beats produced are, represented by the straight line OA in the, following graph. After loading Q with wax, again these are vibrated together and the, beats produced are represented by the line, OB. If the frequency of P is 341 Hz, choose, the correct option(s), , speed of propagation of the wave. If a = 10–3, m and v = 10 m/s, then λ and f are given by :, , 103, Hz, (C) f =, 2π, , 4, , (D) f = 10 Hz, , 33. A triangular pulse is moving with speed 2 cm/, s along a rope (kept along x-axis) whose one, end is free at x = 0 as shown in the figure., Choose the correct option regarding this, pulse., 1cm, A, , O, 2cm, , (A), , X=0, , B, 1cm, , 1cm, , 2cm, 0.5 cm, , A, 3, Beats, , (A) λ = 2 π × 10–2 m (B) λ = 10–3 m, , , , B, , 2, 1, 0, , 1, , 2, , t(s), , (A) 341 Hz (B) 338 Hz (C) 344 Hz, (D) on waxing Q the no. of beats decreases, 36. In a large room, a person receives direct, sound from source at 120m away. He also, receives waves from same source after, reflection from 25 m high ceiling at a point, half ways between them. Find wavelength(s), from which two sound waves interfere, constructively, , C, , 2cm x=0 At t=1s, , 2cm, , (B), 2cm x=0 At t=1s, , (C) particle speed (between A to O) at t = 0 s is, 1 cm/s, (D) particle speed (between A to O)at t = 0 s is, 2 m/s., NARAYANAGROUP, , S, Source, (A) 10 m, , P, Detector, (B) 5m, , (C) 7 m, , (D), , 10, m, 3, 95
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 37. A sound wave of frequency f travels, horizontally to the right. It is reflected from a, large vertical plane surface moving to left, with a speed v. The speed of sound in medium, is C :, (A) The number of wave striking the surface per, second is f, , (c + v), c, c (c – v), , (B) The wavelength of reflected wave is f (c +v), , (c + v ), , (C) The frequency of the reflected wave is f ( c – v ), , 39. y (x, t) = 5 sin [ ω t - x/5], (A) Not a traveling wave, (B) A traveling wave with speed v = 10, (C) The wave is traveling in +ve x direction, (D) The wave is traveling in - ve x-direction, 40. A traveling wave is of the form y (x,t) = A cos, (kx - wt) + B sin (kx -wt), which can also be, written as y (x,t) = D sin (kx - wt - f) where, (A) D = A + B, (B) D = AB, 2, 2, 2, (C) D = A + B, (D) D = A - B, 41. Consider the snapshot of a wave traveling in, positive x-direction, , (D) The number of beats heard by a stationary, 2vf, , listener to the left of the reflecting surface is c – v, 38. Energy density E (energy per unit volume) of, the medium at a distance r from a sound source, vary according to the curve shown in figure., Which of the following are possible?, Rectangular, hyperbola, E, , A, , B, (A) The particle A is moving in -ve y-direction and, particle B is moving in +y-direction, (B) The particle B is moving in -ve y-direction and, particle A is moving in +y-direction, (C) Both are moving in the +ve y-direction, (D) Both are moving in -ve y-direction, , Passage : II, r, , A) The source may be a point isotropic source., B) If the source is a plane wave source then the, medium particles have damped oscillations., C) If the source is a plane wave source then power, of the source is decreasing with time, D) Density of the medium decreases with distance, r from the source, , COMPREHENSION TYPE QUESTIONS, Passage : I, A traveling wave on stretched string can be, understood by the function y = f(x -vt). Here v is, the wave speed ‘x’ is co-ordinate of point and ‘y’, is its instantaneous displacement. To describe the, wave completely, we must specify the function f., If the wave moves in negative x-direction y (x, t) =, f(x + vt) and if it moves in positive x-direction y (x,, t) = f(x - vt). The general relation for a traveling, wave must satisfy the relation, , d 2f, 1 d2 y, = 2 . 2 , if, 2, dx, v dt, , plane wave exists. The particle velocity and wave, velocity are related by Vpa = - (slope) (wave, velocity ). Answer the following questions, 96, , Superposition of waves results in maximum and, minimum of intensities such as in case of standing, waves. This phenomenon is called as interference., Another type of superposition result in, interference in time which is called as beats. In this, case waves are analyzed at a fixed point as a, function of time. If the two waves are of nearby, same frequency are superimposed, at a particular, point, intensity of combined waves gives a, periodic peak and fall. This phenomenon is beats., If ω 1 and ω 2 are the frequencies of two waves, then by superimposed y = y1 + y2, we get at x = 0,, , ω − ω2 ω1 + ω2 , y = 2 A cos 1, .t, .t sin, 2 2 , , Thus amplitude frequency is small and fluctuates, slowly. A beat i.e., a maximum of intensity occurs,, also intensity depends on square of amplitude. The, beat frequency is given by, ωbeat =| ω1 − ω2 |, Number of beats per second is called as beat, frequency. A normal ear can detect only upto 10, Hz of frequency because of persistence of ear., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, 42. If two sound sources of frequency difference, 25 Hz are sounded together. Then which of, the following is correct ?, (A) A normal human ear will hear 25 Hz beat, frequency, (B) A normal human ear will hear only 10 Hz, beat frequency, (C) A normal human ear cannot detect this, frequency difference, (D) A normal human ear can hear maximum of, the two frequency sounded together, 43. The phenomena of beats can take place for, (A) Only transverse waves, (B) Only longitudinal waves, (C) Both longitudinal & transverse waves, (D) For sound waves only, 44. The frequency of beats produced in air when, two sources of sound are activated, one, emitting wavelength 32 cm, other 32.2 cm is, (Take Vsound = 350 m/s), (A) 14, (B) 18, (C) 7, (D) 10, , Passage -III, A narrow tube is bent in the form of a circle of, radius R, as shown in the figure. Two small holes S, and D are made in the tube at the positions right, angle to each other. A source placed at S, generates a wave of intensity I 0 which is equally, divided into two parts: one part travels along the, longer path, while the other travels along the, shorter path. Both the part waves meet at the point, D where a detector is placed., , WAVES, 46. If a minima is formed at the detector then, the, magnitude of wavelength of the wave, produced is given by, 3, , 4, , (A) 2πR (B) πR (C) πR (D) None of these, 2, 5, 47. The maximum intensity produced at D is, given by, (A) 4I0, (B) 2I0, (C) I 0, (D) 3I0, , Passage IV, Observe O is ahead by L from source S which, are moving along same line with velocities V0 and, VS respectively. The speed of sound is V.., The source emits a wave pulse that reaches the, obsever in time t1 ., , L, S, , SI, , O, , At time t=T, the source reaches at S | . It is obvious, that the observer will not be at O this time. The, source emits a wavepulse at this time to reach the, observer in time t2 , which is measured from t=0., 48. Find the time t1, L, (A) V + V, 0, , VL, (B) (V − V ) V, 0, 0, , V0 L, (C) (V − V ) V, 0, 0, , L, (D) (V − V ), 0, , 49. Find the time t2, R, , (A), , S, , (C), , D, , 45. If a maxima is formed at a detector then, the, magnitude of wavelength of the wave, produced is given by, πR, 2, , (A) πR, , (B), , πR, (C), 4, , (D) all of these, , NARAYANAGROUP, , L + (V − V0 ) T, , (V − Vs ), , L + (V − Vs ) T, , (V − V0 ), , (V − Vs ), (B) V − V .T, (, 0), (D), , L − (V − Vs ) T, , (V − V0 ), , 50. Two pulses are emitted by sources at S, and S | . What is the time lag by which, observer observe them ?, , V − Vs , (A) V + V T, , 0 , , V + Vs , (B) V − V T, , 0 , , V + Vs , (C) V + V T, , 0 , , V − Vs , (D) V − V T, , 0 , 97
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, , Passage V, , MATRIX MATCHING TYPE QUESTIONS, , A train approaching a hill at a speed 40 km/hr, sounds a whistle of freqency 580 Hz, when it is a, distance of 1 km from the hill. A wind with speed, 40 km/hr is blowing in the direction of motion of, the train. (Velocity of sound in air = 1200 km / hr ), 51. The frequency of the whistle as heard by an, observer on the hill is, (A) 599Hz (B) 590Hz (C) 610Hz (D) 620Hz, 52. The distance from the hill at which the echo, from the hill is heard by the driver is, 31, 29, 29, 30, km (B), km (C), km (D), km, (A), 29, 31, 30, 29, 53. The frequency of echo is, (A) 599Hz (B) 590Hz (C) 610Hz (D) 620Hz, , 57. Column I represents the wave speeds in four, cases while column II represents the, expressions for speeds. Match column I and, II so that the wave speed in column I matches, the corresponding expression in column II., Column – I, Column – II, , Passage VI, ., , A source is moving across a circle given by the, equation x 2 + y2 = R 2 with constant speed, υs =, , 330π, m / s in clockwise sense. A detector is, 6 3, , stationary at the point ( 2R, 0 ) w.r.t. the centre of, the circle. The frequency emitted by the source is, f s . ( velocity of sound 330 ms −1 ), 54. The coordinates of the source when the, detector records maximum frequency is, R R 3, , R 3 R, , (A) 2 , 2 , , , , (B) 2 , 2 , , , , (C) ( R , 0 ), (D) ( 0, − R ), 55. The coordinates of the source when the, detector detects mimimum frequency is, , , R 3, , (B) 2 , 2 , , , , 6 3, , , , 6 3, , 6 3 , , 98, , (R), , B, ρ, , (D)Transverse wave in, , (S), , Y, ρ, , Fixed, end, A, , (p), , 9π2 µc 2 2, A, 4 L, , (q), , 9π2 µc 2 2, A sin2 ωt, 4 L, , L, Fixed, end, , Free, end, A, , Free, end, , , , Free, end, A, , 9π2 µc 2, , 2, 2, A (r) 16 L A sin ωt, , (C), , (C) f s π − 3 3 , f s π + 3 3 , , , , 6 3 6 3 , (D) f s π − 3 , f s π + 3 , , , , , (C)Longitudinal wave in a gas, , L, , (B) f s π − 12 3 , f s π + 12 3 , , , , 6 3 , , T, µ, , (B), , 6 3 , , , (Q), , (A), , (A) f s π − 6 3 , f s π + 6 3 , , , , , , (B)Longitudinal wave in a liquid, , Fixed, end, , (C) ( R, 0 ), (D) ( 0, − R ), 56. The values of the maximum and minimum, frequencies. are respectively, 6 3 , , γP, ρ, , a stretched string, 58. The diagrams in Column A show transverse, sinusoidal standing/travelling waveforms on, stretched strings. In each case, the string is, oscillating in a particular mode, and, its shape, and other characteristics are shown at time t, = 0. The maximum amplitude (in all the cases), is A, the velocity of the waveform on the, string is c, the mass per unit length of the, string is µ and the frequency of vibration is f, (angular frequency = ω ). The kinetic energy, of the string (of length L) is represented by, the functions in Column B. Match the correct, entries in Column B., Column A, Column B, , R 3 R, , (A) R / 2, 2 , , , , (A)Longitudinal wave in solid rod (P), , L, , A, , (D), , L, Travelling wave, , π2 µc 2 2, A sin2 ωt, (s), 4 L, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , 59. Two sources of sound S1 and S2, emitting, waves of equal wave length 40 cm are placed, with separation of 10 cm between them., Amplitudes of waves emitted are 6 mm and 8, mm respectively. Four observers A, B, C, D, lie on line joining them as shown., A, 5cm, , S1, C, 2.5cm, , S2, , D, 5cm, , B, 7cm, , 10cm, , Column – I, (A) Resulting amplitude received by A, (B) Resulting amplitude received by B, (C) Resulting amplitude received by C, (D) Resulting amplitude received by D, Column – II, (P) 12.95 cm (Q) 14 cm (R) 10 cm (S) 7 cm, 60. Match the Following:, In column I the location of observer, wall and, source with their corresponding uniform non, zero speeds ( observer is stationary) are, indicated and in column II corresponding, information are given. Match them., Column I, , (A), , O, , (B), , S, , S, , O, W, , W, , (C), , S, , (D), , O, W, , O, , S, , c) I R = 34 µW / m 2, , d) I R = 4 µW / m 2, , INTEGER ANSWER TYPE QUESTIONS, 62. A train approaching a railway crossing at a, speed of 120 km/hr sounds a short whistle at, frequency f 0 = 640 Hz .when it is 300 m away, from the crossing. The speed of sound in air is, 340 m/sec. A person standing on a road, perpendicular to the track at a distance 400 m, from the crossing, hears frequency f1 . The, value of f1 − f 0 = 10 x . Find the value of x ?, 63. A closed organ pipe is vibrating in, fundamental frequency. There are two points, A and B in the organ pipe as shown, at a, distance AB = L n . Ratio of maximum, pressure variation at point A to point B is, 2 / 3 find value of n., , B, W, , Column II, (p) Beats must be detected, (q) Beats may be detected, (r) Wavelength of sound wave after reflection, from wall may decrease to a value smaller than, actual value with stationary source., (s) Wavelength of sound wave after reflection, from wall increase to a value greater than actual, value with stationary source., 61. Two identical speakers emit sound waves of, frequency 103 Hz uniformly in all directions., The audio output of each speaker is, 9π /10 mW . A point ‘P’ is at a distance 3 m, from the speaker S1 and 5 m from speaker, , S 2 . Resultant intensity at P is I R . Match the, items in Column I with the items in column II., NARAYANAGROUP, , Column - I, i) If the speakers are incoherent, then, ii) If the speakers are driven coherently and in phase, at P, iii) If the speakers are driven coherently and out of, phase by 1800 at P, then, iv) If the speaker S 2 is switched off, then, Column-II, a) I R = 64 µW / m 2, b) I R = 25 µW / m 2, , L/n, A, , 64. A wire having a linear mass density, 5.0 × 10−3 kg / m is streched between two rigid, supports with a tension of 450N. The wire, resonates at a frequency of 420Hz. The next, higher frequency at which the same wire, resonates is (420 + 10x) Hz. Length of wire is, 2.1 m. Value of x is nearly., 65. A source emitting sound of frequency is, placed in front of a wall at a distance of 2m, from it. A detector is also placed in front of, the wall at some distance from it. Find the, minimum distance (in meter) between the, source and the detector for which the, detector detects a maximum of sound speed, of sound in air = 360 m/sec, 99
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, 66. A source of sound S and a detector D are, placed at some distance from one another. A, big cardboard is placed near the detector and, perpendicular to the line SD as shown in, figure. It is gradually moved away and it is, found that the intensity changes from a, maximum to a minimum as the board is, moved through a distance of 20 cm. The, frequency of the sound emitted is found out to, be "70 x " Hz . Find the value of x ?, (Velocity ofsound in airis336 m s-1), , 72. A closed and an open organ pipe of same, length are set into vibrations simultaniously, in thier fundamental mode to produce 2 beats., The length of open organ pipe is now halved, and of closed organ pipe is double. Now find, the number of beats produced., 73. The displacement Vs time graph for two, waves A and B which travel along the same, string are shown in the figure. Their intensity, IA, , ratio I is, B, A, , y, 3, , S, , D, , 67. Calculate the frequency of beats produced in, air to the nearest integer value when two, sources of sound are activated, one emitted a, wavelength of 32 cm and the other of, 32.2cm.The speed of sound in air is, 350 ms-1., 68. Two tuning forks with natural frequencies of, 340 Hz each move relative to a stationary, observer. One fork moves away from the, observer, while the other moves towards him, at the same speed. The observer hears beats, of frequency 3 Hz. The speed of the tuning, fork is found out to be " 6 / n " ms −1 find n?, ( given (velocity of sound 340 ms −1 ), 69. An observer standing at a railway crossing, receives frequency of 2.2 kHz and 1.8 kHz, when the train approaches and recedes from, the observer. The velocity of the train is 10n, then find n. [The speed of the sound in air is, 300 m/s], 70. nth harmonic of a closed organ pipe is equal, to mth harmonic of a open pipe. First over tone, frequency of the closed organ pipe is also, equal to first overtone frequency of the open, organ pipe. Find the value of n, if m=6., 71. Two sound sources are moving away from a, stationary observer in opposite directions with, velocities V1 and V2 (V1 > V2 ) . The frequency, of both the sources is 900Hz. V1 and V2 are, both quite less than speed of sound V=300m/, sec. Find the value of (V1 − V2 ) so that beat, frequency observed by observer is 6 Hz. (in, m/sec), 100, , 0, -2, , 1 2 3 4 5 6 7 8 9 10 11 12, , t, , B, , LEVEL-VI - KEY, SINGLE ANSWER QUESTIONS, 1) A 2) B 3) C 4) A 5) B 6) B, 7) A 8) A 9) D 10) B 11) D 12) B, 13) B 14) D 15) B 16) B 17) C 18)A, 19) C 20) A 21) A 22) B 23) A 24) C, 25) A 26)B 27) A 28) C 29) D, MULTIPLE ANSWER QUESTIONS, 30) A,B,CD 31) B,C, 32) A,C, 33) A,C, 34) A,B,C,D 35) C,D, 36) A,B,D, 37) A,B,C,D 38) A,B,C,D, COMPREHENSION TYPE QUESTIONS, 39) C 40) C 41) A 42) C 43) D 44) C, 45) D 46) A 47) B 48) D 49) C 50) D, 51) A 52) B 53) D 54) B 55) D 56) A, MATRIX MATCHING QUESTIONS, 57) A-S; B-R; C-P; D-Q, 58) A-Q; B-R; C-S; D-P, 59) A-R; B-R; C-P; D-Q, 60) A-Q,R; B-P,S; C-P; D-P,R, 61) (i ) → c; (ii ) → a; ( iii ) → d ; (iv ) → b, INTEGER ANSWER QUESTIONS, 62) 4 63) 3 64)7 65) 3 66) 6 67) 7, 68) 4 69) 3 70) 9 71) 2 72) 7 73) 1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , LEVEL-VI - HINTS, , 6., , SINGLE ANSWER QUESTIONS, 1., , 1, , or x =, (A) At t = 0 , y =, 1+ x2, , 1− y, , 2, or, ( x − 1) = y, , B, , A, , or, x = 1 +, , 1− y, = x2, y, , M, , M, , Speed of the wave, θ, , (B) ; For interference at A : S2 is behind of S1 by a, distance of 100λ +, , T1cosθ, , T1, , ∆x x2 − x1, 1, v=, =, =, = 0.5m/s, ∆t, t2 − t1 2 − 0, , 2., , T2, , 1, 1, y=, =, 2, 2 + x − 2 x 1 + ( x − 1) 2, , At t = 2s,, , ∴, , T1, , 1− y, = x1, y, , T2, , T1sinθ, , λ, (equal to phase difference, 4, , π, π, ). Further S2 lags S1 by . Hence the waves, 2, 2, , T1 cos θ = Mg ; T1 sin θ = T2 ⇒ T2 = Mg tan θ, , from S 1 and S2 interfere at A with a phase difference, of 200.5 π + 0.5 π = 201 π = π ., Hence the net amplitude at A is 2a – a = a, For interference at B : S 2 is ahead of S 1 by a distance, , L 3, = m, 4 4, , θ, , λ, π, of 100λ + . (equal to phase difference ), 4, 2, , Further S 2 lags S 1 by, , 3., , 4., 5., , π, ., 2, , 1 L 1, D = m, 2 2 4, , Hence waves from S 1 and S 2 interfere at B with a, phase difference of, 200.5 π − 0.5 π = 200π = 0 π ., Hence the net amplitude at A is 2a + a = 3a, (C); Energy E α (amplitude)2 (frequency)2, Amplitude (A) is same in both the cases, but, frequency 2w in the second case is two times the, frequency ( ω ) in the first case., Therefore, E2 = 4E1, ∂y, , ⇒ tan θ =, , t AB =, , =, , Mg, =, 8µ, , g, µ, , L µ, L, ; t AB =, 2 g, 2v, , ∂y, , (A) v P = ∂t = −v ∂x is positive and can only be, along y–axis., (B) Given : y = 4 cos 2 ( t ) sin (1000t ), = 2 [1 + cos 2t ] sin (1000t ), , = 2sin1000t + 2sin1000t ⋅ cos 2t, = 2 sin1000t + sin1002t + sin 998t, , Thus the periodic motion consists of three, components., NARAYANAGROUP, , 1, T, ⇒v= 2, µ, 8, , L-x, A, 7., , x, M, 101
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, mxg, L, ⇒ velocity of pulse at A is given by, , Tension at A = T = Mg +, , ( M + mx ) g, dx, =, v=, dt, , L, µ, , dx, =, mx, M+, g, L, , ⇒, , 9., , g, dt, µ, , 2L , mx , g, g =, t, M +, m, L 0, µ, 2L, m, , (, , L, mg, , t=2, , 8., , M, Lg, , (, , M +m− M, , ), , 3v, , = 4(0.35 + ∆l ), , Solving this equation, we get ? l = 0.025m, 11. (D), v = λf, , ), , (A) ; The diagrammatic representation of the given, problem is shown in figure. The expression of, fundamental frequency is v =, , v1 l2 r2 2 Lr, T, ; v = l = r = L2r = 1, 2, πr ρ, 2, 1, 1, , 1, ∴ ⇒ v = 2l, , v, 4(0.1 + ∆l ), , M +m − M, , T, , 10. (B) ; Let ∆l be the end correction., Given that, fundamental tone for a length 0.1 m =, first overtone for the length 0.35 m., , on integrating, , t=, , 1, , (D) ; Fundamental frequency is given by v = 2l µ, (with both the ends fixed), , T1, x, , T2, , 1 T, 2l µ, , Mg, T, , T, , T1, µ, , ρw=1g/cm3, , T = mg = vρ ' g ; f =, , T ' = mg −, , 1, f '=, 2l, f', =, f, , vρ ' g, µ, , V, ρw g, 2, , v, vρ ' g − ρ w g, 2, µ, 2 ρ '− ρ w, 2ρ ', , ρ', ρ, ρ w = specific gravity =, , 2ρ −1, f ' = 300, Hz, 2ρ, , 102, , 1, 2l, , Water, , =, , T2, µ, , 2Lf, T1, Lf ; T = 4, 2, , T1 + T2 = Mg ; T2 = Mg/5, StB = 0 ⇒ Mg × l = (Mg/5) × L; l = L/5, 12. (B); The motorcyclist observers no beats, so the, apparent frequency observed by him from two, source must be equal, 330 − V , 330 + V , f1 = f 2 ∴ 176 , = 165 , 330 − 22 , 330 , ∴ v = 22 m/s, , 13. (B) Mass per unit length of the string,, m=, , 10 –2, =, 0.4, , 2.5 × 10−2 kg/m, , ∴ Velocity of wave in the string,, T, , 1.6, , v = m = 2.5 × 10 –2 ; v = 8 m/s, For constructive interference between successive, pulses :, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, v, γ RT, 19. f = , l = l0 1 + α ( T − T0 ) ; v =, 2l, M, ⇒ we have to find the temperature at which, , f ( T ) = f ( T0 ), , 21., , B1, , C, vT, , γ RT0, M, 2l0, , G, , 45, vT, , γ RT, M, =, 2l0 1 + α (T − T0 ) , , B, , 0, , A1, , D, , θ, a, , O, , E, T, = 1 + α ( T − T0 ), T0, 1/ 2, , T − T0 , 1 +, , T0 , , , And velocity of boy and train around ‘O’ is same, , = 1 + α ( T − T0 ), , 1 T − T0 , 1, 1+ , = 1 + α (T − T0 ) ⇒ T0 =, 2 T0 , 2α, 20., 90-θ, D2, 100m, , 1, , S, , d, , O, , 100m, , 100m, D1, , cosθ =, , d vs, 5, 1, = =, =, r v 330 66, , 2, v − v0 sin θ , 1 , sin = 1 − = 1 ⇒ f D = f , , v − vs cos θ , 66 , , 330 − 10 , = 100 , = 97 Hz, 330 − 5 / 66 , 104, , aω 44, =, = 22ms −1, 2, 2, , Just before the turn at ‘C’ vT component along, the line joining boy and train is away from boy just, after turn at C, the component is towards boy ⇒, He hears max and min at the turn., velocity of train at, aω, sec 2 45 = 22 × 2 = 44ms −1, 2, , v, v + vs cos 45, , f min = f, , = 300 ×, , Time taken by ‘S’ to travel from S to 0 = 20sec, In this time detector goes from D1to D2, The signal received by detector at t = 20sec is not, that produced by source at ‘O’ but produced from, a position S 1 prior to ‘O’, r d, =, v vs, , a, 11, ⇒ aω = 44, sec 2 θω ; ω =, 2, a/4, , velocity of train at B =, , C=, , θ, , S, , d a, , a, BB1 = Tanθ ⇒ vT = Tanθ , dt 2, 2, , vT =, , r, , F, , 330, 30, = 300 ×, 330 + 22 2, 30 + 2 2, , v, 30, = 300 ×, v − vs cos 45, 30 − 2 2, 22. As sound takes finite time to travel, so the sound, received at t = 5 sec should have been emitted, earlier., f max = f, , Let t1 be the time at which source emits sound, which is detected at t = 5 sec., Then the time duration for which sound moves is, t t − t1, Let ‘h’ be dist moved by detected, h = ut −, , ( ), , 1 2, 1, gt ; H = g t1, 2, 2, , 2, , (, , − h = v t − t1, , NARAYANAGROUP, , )
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JEE-ADV PHYSICS-VOL- VII, 505 −, , ( ), , 1, g t1, 2, , 2, , − ut +, , WAVES, , (, , 1 2, gt = v t − t1, 2, , 25. The wavelength of the wave, , ), , λ = (12 − 0 ) = (14 − 2 ) = 12 units, , ( ) − 125 = 300 (5 − t ), (t ) − 60 (t ) + 224 = 0 ⇒ t = 4sec, , Therefore, in time 1/60 s, the distance moved by, the wave is, , v + vs , f = f0 , , v − vs , , d, λ, , Velocity of wave: v = 1/ 60 = 6 + nλ 60, (, ) , , , 1 2, , 2, , 1, , 1, , d = 2 + n (λ ) =, , 1, , v0 = u − gt = 0 ; vs = gt1 = 40ms −1, = 1300 ×, , λ, T=, v, , 300, = 1500 Hz., 300 − 40, , 23. No interference effect is observed if S1| S 2| is, perpendicular to OP ( ∆x = 0 ), d, |, x-coordinate of S1 = − sin θ, 2, d, d, |, ); y-coordinates of S 2 = cos θ, 2, 2, 24. For linear variation of temperature, we can write, temperature at a distance x from point A is, , (since OS1 =, |, , λ, + nλ ; n ∈ J, 6, , λ, , =, , =, , λ, , + nλ 60, 6, , 26. Since aeroplane is producing intensity level of 100, dB at a distance of 200 m from it, and this is also, the maximum allowable sound level so as not to, violate the regulations, the maximum distance of, the plane from micro phone has to be 200 m. The, diagram that follows shows the situation exactly., Let the aeroplane leave the runway, before x metres, of the location of micro phone. Then, PM is the, shortest distance between the microphone and the, aeroplane., , T2 − T1, x, l, Thus velocity of sound at this point is given as, Tx = T1 +, , 900, , T −T , v = α T1 + 2 1 x, l , , 0, , 30, dx, , dx, T −T , = α T1 + 2 1 x ⇒ α T1 + T2 − T1 x, dt, l , l , , 1, , ∫, 0, , dx, T −T , α T1 + 2 1 x, l , , 1, , ⇒t =, , PM = x sin 300 =, , x, 2, , x, ⇒ x = 400 m, 2, So, the required distance is 1630-400=1230 m., , 0, , 1, , , x = t, 0, , 2l, 2l, T2 − T1 t =, ;, , α T2 − T1 , α (T2 − T1 ) , , NARAYANAGROUP, , x, , ⇒ PM = 200 =, , = ∫ dt, , , 2l, T2 − T1, T1 +, α (T2 − T1 ) , l, , M, , = dt, , Integrating the above expression within proper, limits, we get, , 1, 10 ( 6n + 1), , Microphone, , 505 − 5 t1, , 27. Path difference = ( µ − 1) d, , µ=, , V0, =, V', , γ R (T0 ), m0, γ R ( 4T0 ), m0, , =, , 1, 2, , 105
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, d, 1 , |Path difference| = | − 1 d |=, 2, 2 , , Frequency f =, , λ d, For minima, path difference = ( 2n + 1) =, 2 2, d, V, V ( 2n + 1), =λ =, ;, f ; f =, ( 2n + 1), d, , V, d, 28. When detector is at O, we can see that the path, difference in the two waves reaching O is, d = 2λ thus at O detector receives a maximum, sound. When it reaches p and again there is a, maximum sound detected at P the path difference, between two waves must be ∆ = λ . Thus from, the figure the path difference at P can be given as, , ∆ = S1P − S 2 P ; S1Q, , f min =, , = 2λ cos θ, P, , Q, θ, S1 d = 2λ S2, , O, , ω 30, =, Hz., 2π π, , Wavelength l =, , 2π, =m, k, , Further, 60 t and 2x are of same sign. Therefore,, the wave should travel in negative x-direction., ∴ All the options are correct., 31. (BC) ; ω = 15 π , k = 10 π, Speed of wave,, , v=, , Wavelength of wave l =, , S, , And we have point P, path difference ∆ = λ , thus, , ∆ = 2λ cos θ = λ, 1, π, cos θ = ; θ =, 2, 3, Thus the value of x can be written as x = D tan θ, , π , = D tan = 3D, 3, 29. Total path difference = AB + BC + λ / 2 = λ, for maxima, h sec α cos 2α + h sec α = α / 2, α, h sec α ( 2 cos α ) = α / 2 ; h =, sec α, 2, , MULTIPLE ANSWER QUESTIONS, 30. (ABCD), y = 10–4 sin (60t + 2x), A = 10–4 m, w = 60 rad/s, k = 2m–1, Speed of wave v =, , 106, , ω, = 30 m/s, k, , 2π 2π, =, = 0.2 m, k 10π, , 10 π x and 15 π t have the same sign. Therefore,, wave is traveling in negative x-direction., ∴ correct options are (B) and (C)., 32. (A, C) ;, Maximum speed of any point on the string = a ω, = a(2 π f), v, , 10, , ∴ = 10 = 10 = 1 (Given : v = 10 m/s), 1, , ∴ 2 π af, , D, , ω, = 1.5m/s, k, , = 1 ; ∴f =, 2πa, a = 10–3 m (Given), ∴ f=, , 1, 103, =, 2π × 10–3 2π, , Hz, , Speed of wave v = f λ, , 103 –1 , ∴ (10 m/s) = 2π s λ, , , –2, ∴ λ = 2 π × 10 m, 33. (AC); Superposition of two displacement is added, y = y1 + y2, Hence Answer is (A); v p = v w ×, Hence, v p = 2 ×, , dy, dx, , 2, = 1 cm / s, 2, , For the particle between A and O., 34. (ABCD) ; It is given that, y(x, t) = 0.02 cos (50 π t + π /2) cos (10 π x), ≅ A cos( ω t + π /2) cos kx, Node occurs when kx =, 10 π x =, , π 3π, ,, etc., 2 2, , π 3π, ,, ⇒ x = 0.05 m, 0.15 (option a), 2 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- VII, , WAVES, , Antinode occurs when kx = π , 3 π , etc., 10 π x, = π , 3π, ⇒ x = 0.1 m, 0.3 m (option b), Speed of the wave is given by, v=, , ω 50π, =, = 5 m/s, k 10π, , (option c), , Wavelength is given by, 2π, , 2π, , , l = k = 10π = 5 m = 0.2 m, , 1, , (option d), , 35. (C D ) nQ = 341 ± 3 = 344 Hz or 338 Hz, on waxing Q the no. of beats decreases, 36. (ABD) ; SCP = 130 m, ∆x = 10 m ; ∆x = n λ, λ=, , 10, 10 10, ; λ = 10,5, , & so on., n, 3 4, , 37. (A, B, C,D); The number of waves encountered, by the moving plane per unit time is given by, n=, =, , distance travelled, wavelength, , c+v c, v, = 1 + = f, λ, λ, c, , v, , 1 + , c, , , (option a), , The stationary observer meets the frequency f ’ of, the incident wave and receives the reflected wave, of frequency f” emitted by the moving platform as, f′, , f (1 + v c ), , f (c + v ), , f”= 1 – v c = 1 – v c = (c – v), c, , (option c), , c c – v, , Wavelength, l” = f ′′ = f c + v (option b), , , Beat frequency= f”–f, 1+ v c, , , , f (1 + v c ), , = (1 – v c) – f, , (1 + v c) f, , Intensity , I = 2π 2 n 2 a 2 ρ v, Since, intensity I α ρ (density of medium) and, density I is decreasing with distance, therefore, the, density ρ also decreases with, distance, from the source. Hence, option (d) is also correct., , COMPREHENSION TYPE QUESTIONS, Passage:II, 42. (C) Beats will be detected by the ear only if beat, frequency fb = f1 - f2 < 10 Hz, 43. (D) When two sound waves of nearly equal, frequency travel in same direction produces beats., , 2vf, , = f 1 – v c – 1 = (1 – v c) = c – v, , , 38. (A,B,C,D); Due to propagation of a wave the, energy density at a point is given by E=I/v, Where I is intensity at that point and v is wave, propagation velocity., It means energy density E is directly proportional, to intensity I., If power emitted by a point source is P then intensity, at a distance r from it is equal to, NARAYANAGROUP, , P, 1, or I α 2, 2, 4π r, r, Hence, the shape of the curve between I and r will, also be same as that given in figure of the question., Hence option (a) is correct., If the source is a plane sound source then intensity, at every point in front of the source will be same if, damping does not take place. But if damping takes, place then the amplitude of oscillation of medium, particles decreases with distance. Hence, the, intensity decreases with the distance from the, source. In that case, the curve between I and r, may have the same shape as shown in the figure, given in the question. Hence option (b) is also, correct., If the source is a plane wave source, intensity at, every point of the source will be the same., But if power of the source is decreasing with time, then intensity will also decrease with time. But at, an instant, intensity at every density at every point, in front of source will also be same, though it will, decrease with time. Hence, option(c) is wrong., I=, , 44. (C) f1 =, f2 = =, , V 350 × 100, =, = 1093.75, λ, 32, , 350 × 100 × 10, = 1086.95, 322, , fb = f1 - f2 = 7Hz, , Passage:III, 45. (D) Path difference produced is, ∆x =, , 3, π, πR − R = πR, 2, 2, 107
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, For maxima : ∆x = n λ, , Passage: V, , πR, , , , 46. (A) For minima = πR = (2n − 1), , λ, 2, , Here I1 = I2 =, , =, , I1 + I2, , ), , 2, , I0, given, 2, , I, , I , , 1− x, 1, x, =, +, 40 1240 1160, , 29, km, 31, , v − v + v0 , , , , m, 53. fecho = f v − v, m, , , =, , 2, , 580 × 1240 1200 − 40 + 40 580 × 1240 1200, ×, =620 Hz., , =, 1200 1200 − 40 , 1200, 1160, , Passage VI 54,55,56, , Passage IV, 48. Total distance travelled by wave pulse in, , A, , t1 = L + V0t1, This distnce is also given by Vt1, So, Vt1 = L + V0t1 ;, , L, t1 =, V − V0, , 49. Total distance travelled by wave pulse in time t 2, '), , = ( L − VsT ) + V0t2, , Actual travel time = ( t2 − T ), The above distance is also given by V ( t2 − T ), , V ( t2 − T ) = ( L − VS T ) + V0t 2, L + (V − V S )T, (V − V0 ), , 3R, , R, , (because observe moves by V0t1 in time t1 ), , t2 =, , ⇒x=, , Here f is the frequency that incidents on hill, , 0, 0, ∴ Imax = 2 + 2 = 2I0, , , , (measured from S, , 1200 + 40, , ⇒ the distance travelled by train = (1 − x ) km, , 2, 2, 2, 2πR, πR, πR, πR,....., 3, 5, 7, , (, , , , 52. Let ‘x’ be the distance from hill where echo is heard, , Thus, the possible values of l are, , 47. (B) ; Maximum intensity, Imax =, , v+v, , , , m, 51. f = f 0 v + v − v = 580 , = 599 Hz., 1200 + 40 − 40 , m, s , , , ∴ n λ = π R ⇒ λ = n ,n = 1,2,3,..., , O, , 600, 600, , C, 2R, 3R, , B, Frequency detected by detector is maximum when, source is directly towards it and minimum when it, is directly away from it., From the geometry of diagram frequency detected, is maximum when source is at ‘A’ minimum when it, is at B, Time taken by source =, , 3R, v, , distance travelled by source = vs ., , 50. When source approaches stationary obeserver, V , f1 = f , > f (but consant), V − VS , When source recedes from stationary, V , observer f 2 = f V + V < f, , S , (but constant), 108, , (2R,O), , 330π, , 3R, v, , 3, , ⇒ angle swept out = 6 3 . 330 = π / 6, ⇒ angle from horizontal =, , π π π, − =, 3 6 6, , R 3 R, , ⇒ coordinates of source 2 , 2 , , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- VII, SR-MAIN-CHEM-VOL-II, , WAVES, n T, 2l µ, , 64. 7 ; v =, , ⇒ 420 =, , , , n T, 2l µ, , , , n +1 T, ; Solving x = 7, µ, 2l, , 420 + 10 x =, , v, , , , 69. From the relation, f ’ = f v ± v , we have, s , , 300, , , , 2.2 = f 300 – v , T , , , 65. 3, and 1.8, , S, , …(1), , , , , , 300, , = f 300 + v , T , , , …(2), , Here, vT = vs = velocity of source/train, Solving Eqs. (1) & (2), we get vT = 30 m/s ; n = 3, x, , V , , V , , 70. n 4 L = m 2L , c, 0, V , , -------------(1), , V , , also, 3 4 L = 2 2 L ----------(2), c, 0, , D, 1, , , x2 2, 2 22 + − x = 2 ⇒ x = 3m, 4, , , λ, = 20 ;, 4, , 66., , l = 80 cm ; f =, , λ1, S1, , D, , Lc 3, From equation (2) L = 4, o, v 336, =, × 100, λ 80, , L 6 3 9, n, = 2 c = = =, m, Lo 4 2 6, , ?, 4, , = 420 Hz, , n = 9 if m = 6, , 350, 350, −, ≈7, 67. 7; ∆f =, 0.32 0.322, , 71., , 68. Given f1 – f2 = 3 Hz, , , v, , , , , , 340, , , , v, , , , , , =3, 340, , , , or 340 340 – v – 340 340 + v , s , s , , , , =3, , –1, , vs , , vs , or 340 1 – 340 – 340 1 + 340 = 3, , , , , , as vs, << 340 m/s, Using binomial expansion, we have, , , vs , , , , vs , , , , , , , , , , 340 1 + 340 – 340 1 – 340 = 3, ∴, , 110, , 2 × 340 × vs, = 3;, 340, , 300 , f1 = 900 , , 300 + V1 , , = 900 − 3V1, , likewise, f 2 = 900 − 3V2, , or f v – v – f v + v , s , s , , , , , From equation(1), , ∴ vs = 1.5 m/s, , given f 2 − f1 = 6, , 3 (V1 − V2 ) = 6 ; V1 − V2 = 2m / sec, 72. f 0 − f c = 2 ; V − = 2 ;, 2 L 4L , 1, , 1, , V /L =8, , in second case,, f 0| − f c| =, , V V, 7V 7, −, = (8 ) = 7, ;, L 8L, 8L 8, , I1 a12 f12, (3)2 (8)2, =, =, 73. (1) I a 2 f 2 (2)2 (12)2 = 1, 2, 2 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , GEOMETRIC OPTICS (RAY OPTICS), SYNOPSIS, Introduction, , , , , , , , , , , , , Nature has endowed the human eye (retina) with, the sensitivity to detect electromagnetic waves within, a small range of the electromagnetic spectrum., Electromagnetic radiation (Wavelength from, 400 nm to 750 nm) is called light. It is mainly through, light and the sense of vision., , Light travels along straight line with enormous speed., The speed of light in vacuum is the highest speed, attainable in nature. The speed of light in vacuum is , c 2.99792458108 ms1 ., 3108 ms1, The wavelength of light is very small compared to, , the size of ordinary objects that we encounter, commonly (generally of the order of a few cm or, larger). A light wave can be considered to travel, from one point to another, along a straight line joining, them. The path is called a ray of light, and a bundle, of such rays constitutes a beam of light., The phenomena of reflection, refraction and, dispersion of light are explained using the ray picture, of light. We shall study the image formation by plane, and spherical reflection and refracting surfaces, using, the basic laws of reflection and refraction. The, construction and working of some important optical, instruments, including the human eye are also, explained., , Reflection of Light : When a light ray strikes, the boundary of two media such as air and glass, a, part of light is turned back into the same medium., This is called reflection of light., Normal, Incident, ray, , Reflected, ray, , I, , , , In case of reflection at the point of incidence ‘O’,, the angle between incident ray and normal to the, reflecting surface is called the angle of incidence, (i). The angle between reflected ray and normal to, the reflecting surface is called angle of reflection, (r)., The palne containing incident ray and normal is, called plane of incidence., , Laws of reflection : The incident ray, the, reflected ray and the normal to the reflecting surface, at the point of incidence, all lie in the same plane., The angle of incidence is equal to the angle of, reflection i r, , Types of reflections, Regular reflection:When the reflection takes, place from a perfect smooth plane surface, then, the reflection is called regular reflection (or) specular, reflection., In this case, a parallel beam of light incident will, remain parallel even after reflection as shown, in the figure., , In case of regular reflection, the reflected light, ray has large intensity in one direction and, negligibly small intensity in other direction., Regular reflection of light is useful in, determining the property of mirror., Diffused reflection: If the reflecting surface is, rough (or uneven), parallel beam of light is reflected, in random directions. This kind of refletion is called, diffused reflection., , r, O, , NARAYANAGROUP, , 111
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, As shown in the above figure if the reflecting surface, is rough, the normal at different points will be in, different directions, so the rays that are parallel, before reflection will be reflected in random, directions., We see non-luminous objects by diffused reflection. , , Important points regarding reflection, , , Laws of reflection are valid for all reflecting surfaces, either plane or curved., , From the above figure i r , But i r, Hence angle of deviation in the case of reflection is, , 2i, By keeping the incident ray fixed, the mirror is, rotated by an angle ‘ ’, about an axis in the plane, of mirror, the reflected ray is rotated through an, angle ‘ 2 ’., M, M, , , ir, , ir, , Incident ray, , ir, , Incident ray, O, , O, , , , Reflected ray, , , 1, , , , If a light ray is incident normally on a reflecting, surface, after reflection it retraces its path i.e., if, i 0 then r 0, , , M, , Reflected ray, , Vector form of law of reflection:, n̂, , ê1, , ê 2, i, , r, , , , , , In case of reflection of light frequency, wavelength, and speed does not change. But the intensity of, light on reflection will decreases., If the reflection of light takes place from a denser, medium, there is a phase change of rad., , If ê1 is unit vector along the incidnet ray ê 2 is the, unit vector along the reflected ray n̂ is the unit, vector along the normal then,, , eˆ 2 eˆ1 2 eˆ1.nˆ nˆ, , ˆ and Rˆ are vectors of any magnitude along, If ˆI, N, incident ray, the normal and the reflected ray, , respectively then, , ˆ ˆI N, ˆ N., ˆ ˆI Rˆ ˆI. N, ˆ Rˆ 0, R., , , , , , This is because incident ray, reflected ray and the, , normal at the point of incidence lie in the same plane., Deviation of a ray due to reflection: The, angle between the direction of incident ray and, reflected light ray is called the angle of deviation, ( )., O, , Reflection from Plane Surface, When you look into a plane mirror, you see an image, of yourself that has three properties., The image is up right., The image is the same size as you are, The image is located as far behind the mirror as, you are infront of it. This is shown in the figure(b)., A, , M, , M, I, , O, I, , Point object, (a), , A1, , B1, , B, , Extended object, (b), , 1, , , 112, , , , A plane mirror always form virtual image to a, real object and vice versa and the line joining, object and image is perpendicular plane mirror, as shown in figure (a)., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , , v, , , , u, , 45°, , M, , O, , (a), , Every object has its own field of view for the given, mirror. The field of view is the region between the, extreme reflected rays and depends on the location, of the object infornt of the mirror. If our eye lies in, the filed of view then only we can see the image of, the object other wise not. This is illustrated in figure., , (b), , The graph between image distance (v) and object, distance (u) for a plane mirror is a straight line as, shown in figure (b)., The ratio of image height to the object height is called, lateral magnification (m). Thus in case of plane, mirror ‘m’ is equal to one., , The principle of reversibility states that rays retrace, their path when their direction is reversed. In, accordance with the principle of reversibility object, and image positions are interchangable. The, points corresponding to object and image are, called conjugate points., This is illustrated in figure., O, , I, , I, , Field, of view, , M, I, O, , A plane mirror produces front - back reversal, rather than left - right reversal. It must be kept in, mind that the mirror produces the reversal effect, in the direction perpendicular to plane of the, mirror. The figure (a) shows that the right handed, co-ordinate system is converted into left handed, co-ordinate system., M, y, , O, , y1, z, , z, , (a), , , , , , A mirror whatever may be the size, it forms the, complete image of the object lying infront of it. Large, mirror gives more bright image than a smaller one., It is seen that the size of reflector must be much, larger than the wavelength of the incident light, otherwise the light will be scattered in all directions., The angle between directions of incident ray and, reflected or refracted ray is called deviation ( )., A plane mirror deviates the incident light through , angle 180 2i where ‘i’ is the angle of, incidence. The deviation is maximum for normal, incidence, hence max 1800 ., , x, , x, , (b), , 1, , 1, , (a), , i.e., the image formed by a plane mirror left is turned, into right and vice versa with respect to object as, shown in figure (b)., M, , (b), , When the object moves infront of stationary, mirror, the relative speed between object and, its image along the plane of the mirror is zero, and in perpendicular to plane of mirror relative, speed is twice that of the object speed., , VIO y 0 and VIO x 2v x, M, V, , y, , V, , Vy, , V, , , , , It is noted that, generally anti - clock wise deviation, is taken as positive and clock wise deviation as, negative., NARAYANAGROUP, , O, , Vx, , V, , x, , I, , 113
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , If an object moves towards (or away from) a plane, mirror at speed v, the image will also approach (or, recede) at the same speed v, and the relative velocity, of image with respect to object will be 2v as shown, in figure (a). If the mirror moved towards (or away, from) the stationary object with speed v, the image, will also move towards (or away from) the object, with a speed 2v, as shown figure (b)., , MM 2 , , V v, , Rest, , I, , O, , 2v, , I, , v, , , , , , 3D d , 4, , 4, ... (ii), , , , 2D 2d, {From (i) and (ii)], 4, , , , 2 D d D d, , 4, 2, , a) A person of height ‘h’ can see his full image in a, , h, mirror of minimum length l , 2, , D d, , Width of the mirror = M1M 2, MM 2 MM1, , M, , M, O, , and MM 2 D , , , , If two plane mirrors inclined to each other at an, angle , the number of images of a point object, formed are determined as follows, , b) A person standing at the centre of room looking, towards a plane mirror hung on a wall, can see the, whole height of the wall behind him if the length of, the mirror is equal to one-third the height of the, wall., The minimum width of a plane mirror required for a, person to see the complete width of his face, , I2, , M2, , O, , I3, , , , M1, I1, I4, , is D d / 2 , where, D is the width of his face, and d is the distance between his two eyes., , , If, , 360, is even number (say m) Number of images, , , formed n m 1, for all positions of objectes, in between the mirrors., M M1, , M2, , , , bisector of mirrors., , D, d, , , , 1, 1, , MM1 D D d , 2, 2, , MM1, , 114, , D d, , 4, , .... (i), , 360, is odd integer (say m) number of images, , formed n m , if the object is not on the bisector, of mirrors. n m 1 , if the object is on the, If, , If, , 360, is a fraction (say m). The number of images, , , formed will be equal to its integer part i.e.,, , n m ., Ex: If m=4.3, the total number of images, , n 4.3 4, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-1: A point source of light S, placed at a, 360, θ, Even, , Position of, the object, , Number of, images (n), , Any where, , m 1, , Odd, , Symmetric, , m 1, , Asymmetric, , M, , Any where, , [m], , m=, , Fraction, , distance L in front of the centre of a mirror of, width d, hangs vertically on a wall. A man, walks in front of the mirror along a line, parallel to the mirror at a distance 2L from it, as shown in figure. Find the greatest distance, over which he can see the image of the light, source in the mirror., S, , d, All the images lie on a circle whose radius is, equal to the distance between the object ‘O’, L, and the point of intersection of mirrors C. If, 2L, is less more number of images on circle with Sol: The ray diagram will be as shown in figure., large radius., G, C, , I4, , I3, M2, , , , A, , , , O, , E, , B, , , , I1, , M1, , If the objects is placed in between two parallel, mirrors 00 , the number of images formed is, infinite but of decreasing intensity in according with, 2, , I r ., , , , , , , H, , S, , , C, , I2, , D, , F, , I, J, , HI = AB =d, DS = CD = d/2, Since, AH = 2AD, GH = 2CD = 2, , d, d, 2, , Similarly IJ = d, GJ = GH + HI +IJ =d +d + d =3d, W.E-2: A ray of light travelling in the direction, , If ‘ ’ is given n is unique but if ‘n’ is given is not, 1 ˆ, unique. Since same number of images can be formed, i 3jˆ is incident on a plane mirror.., 2, for different ., After reflection, it travels along the direction, The number of images seen may be different from, 1 ˆ, number of images formed and depends on the, i 3jˆ . The angle of incidence is, position of the observer relative to object and mirror., 2, When a light ray vector incident on a mirror, only Sol: Let angle between the directions of incident ray, the component vector which is parallel to normal, and reflected ray be ,, of the mirror changes its sign without change of its, 1, 1, magnitude on reflection. It is noted that a mirror, cos ˆi 3jˆ . ˆi 3jˆ, can reflects entire energy incident on it, hence, 2, 2, the magnitude of reflected vector is same as that, 1, cos , of incident vector. Incident vector corresponding, 1200, 2, to an object and reflected vector corresponds to, an image. This vector may be position, velocity, or acceleration., , Example: If a plane mirror lies on x-z plane, a, , , , , , , , , , , , , , , , light vector 2iˆ 3jˆ 4kˆ on reflection, becomes 2iˆ 3jˆ 4kˆ ., NARAYANAGROUP, , 115
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , Reflection from Curved Surface, , W.E-3: A plane mirror is placed at origin parallel, to y-axis, facing the positive x-axis. An object , starts from (2m, 0, 0) with a velocity of, , 2iˆ 2ˆj m/s. Find the relative velocity of, , A curved mirror is a smooth reflecting part (in any, shape) of a symmetrical curved surface such as, spherical, cylindrical or ellipsoidal. In this chapter, we consider a piece of spherical surface only., , image with respect to object., 2iˆ 2ˆj, , M, , P P, , C, , 2iˆ 2ˆj, , Concave mirror (a), , Sol:, , (2, 0, 0)I, , O(2, 0, 0), , along normal 4iˆ The relative velocity image with, respect to object along plane of mirror = 0. Hence, the relative velocity of image with respect to object, , 4iˆ, W.E-4: A reflecting surface is represented by the, , 2L x , sin ,0 x L . A ray, L , , A, , travelling horizontally becomes vertical after, reflection. The coordinates of the point(s), where this ray is incident is, Sol: A horizontal ray becomes vertical after, reflection., , C, , B, Concave mirror, (a), , | ||||, , ||, , x, , dy, x, 2cos, dx, L, , 2L, 3L, sin / 3 , , , L 3L 2L 3L , ,, , , 3 & 3 , , , , , , , y, , 116, , P, , f, , F, , C, , B, Convex mirror, (b), , Sign Convertion : To derive the relevant, , |, ||||, , ||||, , |||||||||||||||||||, |||||||||, ||||||, ||||||, |||||, |, |, |, |, || | | |, |, |, |, ||||, ||| ||, |, , 2 900 450, 1 2cosx / 2, x L/3, , P, , R, , , , tan , , A, , R, , F, f, , y, , , Convex mirror (b), , If the reflection take place from the inner surface,, the mirror is called concave and if its outer surface, it is convex as shown in the figure. In case of, thin spherical mirror, the centre ‘C’ of the sphere, of which the mirror part is called the centre of, curvature of the mirror. P is the centre of the, mirror surface, is called the pole. The line CP, produced is the principal axis, AB is the aperture, means the effective diameter of the light, reflecting area of the mirror. The distance CP is, radius of curvature (R). The point F is the focus, and the distance between PF is called focal, length (f) and it is related to R as f R / 2 ., , The relative velocity of image with respect to object, , equation Y , , C, , , , formula for reflection by spherical mirrors and, refraction by spherical surfaces, we must adopt, a sign convection for measuring distance. In this, book, we shall follow the Cartesian sign, convention. According to this convention all, distances measured from the pole of the mirror., The distance measured in the same direction as, the incident light are taken as positive and those, measured in the direction opposite to the direction, of light are taken as negative., The heights measured one side with respect to, principal axis of the mirror are taken as positive, and the heights measured other side are taken as, negative., Acute angles measured from the normal (principal, axis) in the anti-clock wise sense are positive, while, that in the clock wise sense are negative., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, opposite direction of light. Similarly with the same, reason focal length of convex mirror is positive., The same sign convention is also applicable to virtual, object by treating that imaginary light rays from that, object., , +ve, light, +ve, +ve, P, , P, , ve, , C, , Relation between F and R, , ve, , ve, , Concave mirror, (a), , , , F, , ve, , a) Concave mirror b) Convex mirror, , Convex mirror, (b), , f , , Paraxial Approximation : Rays which are, close to the principal axis or make small angle, , 10 with it i.e. they are nearly parallel to, 0, , the axis, are called paraxial rays. Accordingly, we set cos 1,sin and tan ., , This is known as paraxial approximation or first, order theory or “Gaussion” optics. In spherical, mirrors we restrict to mirror with small aperture, and to paraxial rays., , Focal Length of Spherical Mirrors, We assume that the light rays are paraxial and, they make small angles with the principal axis., A beam of parallel paraxial rays is reflected from , a concave mirror so that all rays converge to a, point F on the principal axis is called principal, focus of the mirror and it is real focus., A narrow beam of paraxial rays falling on a convex, mirror is reflected to form a divergent beam which, appears to come from a point ‘F’ behind the mirror., Thus a convex mirror has a virtual focus ‘F’., The distance between focus (F) and pole (P) is, called the focal length ‘f’. Concave mirror is , also called as converging mirror. They are used, in car head lights, search lights and telescopes., Convex mirror is also called as diverging mirror., Convex mirror gives a wider field of view than, a plane mirror and concave mirror, convex, mirrors are used as rear view mirrors in, vehicles., , C, , F, , P, , Converging mirror, (a), , P, , |||||||||||||||||||, |||||||||, ||||||, |||||, ||||||, ||||, ||, , , , The focal length of mirror is independent on medium, in which it placed and wavelength of incident light., To a plane mirror focal length ‘f’ is infinite, (as R ), Rules for Image formation: In general,, position of image and its nature [i.e., whether it is, real or virtual, erect or inverted, magnified or, diminished] to an object depend on the distance of, the object from the mirror. Nature of the image can, be obtained by drawing a ray diagram. In case of, image formation unless stated object is taken to be, real, it may be point object or extended., A ray parallel to principal axis after reflection from, the mirror passes or appear to pass through its focus, F., , C, , F, (a), , , F, , C, , C, , F, , C, , (b), , F, , C, , (b), , A ray through or directed towards the centre of, curvature C, after reflection from the mirror, retraces, its path., , Diverging mirror, (b), , According to Cartesian sign convention with real, object the focal length of concave mirror is negative,, because the distance PF (P to F) is measured in, , F, , A ray passing through or directed towards focus,, after reflection from the mirror becomes parallel to, the principal axis (by principle of reversibility), , (a), , b Diverging mirror, , NARAYANAGROUP, , R, 2, , |||||||||||||||||||||||, ||||||||, |||||, |||||, |||||, ||||, |, , F, , ||||||||||||||||||||||||||||||||, ||||||, |||||, |||, |||||, , C, , C, , F, (a), , ||||||||||||||||||, ||||||, |||||||||, ||||, ||||||, ||||, ||||, , +ve, , light, , F, , C, , (b), , 117
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, A ray striking at pole P is reflected, symmetrically back in the opposite side., , i, i, , F, , i, i, , P, , (a), , ||||||||||||||||||||||||||||||||, ||||||, |||||, |||, |||||, , , , P, , a) Concave mirror b) Convex Mirror, Position of the object, , Ray diagram, , Image details, , F, , F, , At Infinity, , C, , C, , P, , Real, inverted, very, Small, at F, , P, , Real, inverted,, diminished between, F and C, , (b), , The Mirror Equation: Figure (a) shows the, ray diagram considering two rays and the image, A1B1 (in this case real image) of an object AB, formed by a concave mirror., A, , M, 1, , B, B, , C, , F, A, , P, , v, , u, (a), , 1 1 1, , u v f, , F, , C, I, , A, , O, , h, B, , 1, , O, , Between and c, , B1, , F, h1, 1, A, (b), , P, , , , f R , , 2 , , This relation is known as Gauss’s formula for a, spherical mirror. It is valid in all other situations, with a spherical mirror and also for a convex, mirror. In this formula to calculate unknown,, known quantities are substituted with proper, sign., , At C, , C, , P, , Real, inverted,, equal, at C, , P, , Real, inverted,, enlarged, beyond C, , F, I, , O, C, Between F and C, , F, , O, At F, C, , Real, inverted, very, large at infinity, , P, , F, , Image Formation by Spherical Mirrors, , , , 118, , Image details, , Ray diagram, |||| ||||||||||||||||||||| ||, | |||||, , |, , Position of the object, , At Infinity, , C, , F, , | || ||||||, , P, , Virtual, erect, very, small at F, , | | | || |, , , , F O, , || |||, , , , Virtual, erect,, enlarged behind, the mirror, , Between F and P, , Infront of mirror, , O, , ||||||||| |||||| |||||||||||||||| | |, , , , From the ray diagrams we understand that, To a real object in case of concave mirror the, image is erect, virtual and magnified when the, object is placed between F and P. In all other, positions of object the image is real and, inverted., To a real object the image formed by convex, mirror is always virtual, erect and diminished, no matter where the object is., A concave mirror with virtual object behaviour, is similar to convex mirror with real object and, convex mirror with virtual object behaviour, similar to concave mirror with real object., By principle of reversibility a convex mirror can, form real and magnified image to a virtual object, which is with in the focus and virtual images, when virtual object beyond the focus. i.e., the, convex mirror can form real and virtual images, to virtual object. A concave mirror with virtual, object always forms real images., If the given mirror breaks in to pieces, each piece, of that mirror has own principal axis, but, behaviour is similar to that of main mirror with, less intensity of image., , P, , I, , F, , Virtual, erect, diminished, between P and F, , | || ||, | | | |||, , , , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , Magnification :The size of the image relative to, , Longitudianl magnification:, , the size of the object is another important quantity, to consider. Hence we define magnification. It is, noted that magnification does not mean that the, image is enlarged. The image formed by optical, system may be larger than, smaller than or of the, same size of the object., Lateral magnification: The ratio of the, transverse dimension of the final image formed by, an optical system to the corresponding dimension, of the object is defined as transverse or lateral or, linear magnification (m). Hence it is the ratio of, the height of image ( h1 ) to the height of the object, (h). From the figure., , However, if the one dimensional object is placed, with its length along the principal axis. The ratio of, length of image to length of object is called, , A, , magnification can be expressed as, mL , , v 2 v1 , u 2 u 1 , , Where v1 and v 2 are image positions, corresponding to u1 and u 2 positions., For small objects m L , , dv, du, , 1 1 1, We have , v u f, , h, B1, B, , longitudinal magnification m L . Longitudinal, , 1, , h, , F, , , , , P, , A1, , Lateral magnification m , , In case of small linear objects , dv v , mL m2, du u , 2, , A1B1 h1, , AB, h, , here h and h1 will be taken positive or negative in, accordance with the accepted sign convention., In triangles A1B1P and ABP , we have, , B1A1 B1P, , , with sign convention this becomes, BA, BP, , Areal magnification:If a two dimensional, object is placed with its plane perpendicular to, principal axis, its magnification is called a real or, superficial magnification. If m is the lateral, magnification and mA is the areal magnification., mA , , Here negative magnification implies that image is, inverted with respect to object, while positive, magnification means that image is erect with respect, to object. i.e., m is negetive means for real object,, real image formed and for virtual object virtual image, is formed. m positive means for real object virtual, image formed and for virtual object real image is, formed., Ex: If m 2 , means, if the object is real, image, is real, inverted, magnified and mirror used is, concave., , area of image ma mb, , m2, area of object, ab, mb, , 1, h1 v , , so that, m h v, u , h, h, u, , NARAYANAGROUP, , dv du, 0, v2 u 2, , b, , a, , ma, , O, , principal, axis, , I, , Overall magnification:In case of more than, one optical component, the image formed by first, component will act as an object for the second and, image of second acts as an object for third and so, on, the product of all individual magnifications is, called over all magnifications., m0 , , I, I, I, I, 1 2 n, O O1 O 2, On, , m1 m2 mn, 119
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , Newton’s Formula :In case of spherical mirror a), if the object distance x1 and image distance, , x 2 are measured from focus instead of the pole, , b), , c), of the mirror. Then mirror formula reduces to a, simple form called the Newton’s formula., d), 1 1 1, e), reduces to, v u f, 1, 1, 1, , , f x 2 f x1 f, , , , R u , , m 1, , V1 V0, , u=R, , m 1, , V1 V0, , f u R, , m 1, , V1 V0, , u f, , m 1, , V1 V0, , m 1, V1 V0, u0, Relation between object and image velocity, given above is also valid for convex mirror. In, convex mirror speed of image slower than the, object whatever the position of the object may, Which on simplification gives x1x 2 f 2, be. Above relation is not true in terms of, acceleration of object and image., (Newton’s Formula) ( f x1x 2 ), Motion of the object Transverse to the, Motion of Object in front of Mirror Along, Principal Axis, the Principal Axis, If the object moves transverse to principal axis, When the position of the object changes with, then the image also moves transverse to principal, time on the principal axis relative to the mirror,, axis., the image position also changes with time, relative to the mirror. Hence to know the relation, between object and image speed we use the, mirror equation., , Consider the diagram. In a mirror, hi v, cons tan t m, h0 u, 1 1 1, , v u f, Differntiatae with respect to time, we get, , , 1 dv 1 du, . . 0 (or), v 2 dt u 2 dt, , Power of Curved Mirror : Every optical, , v du, v, dv, ., (or) V1 .V0, , u dt, u , dt, 2, , dh i, dt m or V1 mV0, dh 0, dt, , 2, , Where v1 velocity of image with respect to mirror, and v 0 is the velocity of object with respect to, mirror along the principal axis. Here negative sign, indicates the object and image are always moving, opposite to each other. In concave mirror depending, on the position of the object image speed may be, greater or lesser or equal to the object speed., , instrument have power, it is the ability of optical, instrument to deviate the path rays incident on, it. If the instrument converges the rays parallel, to principal axis its power is said to be positive, and if it diverges its power is said to be negetive., , F, , Converging mirror, P = +ve, (a), , F, , Diverging mirror, P = ve, (b), , For a mirror Power ‘P’, 120, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , P , , GEOMETRIC OPTICS, v, , 1, 100, or P , f metre, f cm, , v, Plane mirror, , S.I unit of power is dioptre D m1, For concave mirror (converging mirror) P is positive, and for convex mirror (diverging mirror) power is, negative., , , 2f, f, O, , 45° 1, B, u, , , , , , 1 1 1, , v u f, , 45°, 1, f, , f, , 2f, , u, , 1 1 1, 1 u f, (or) , v u f, v, uf, , f, 1, , f, u, , 1 1, 1, , v u, f, 1 f u, f, , or v f, v, fu, 1, u, In case of convex mirror, the graph between u, and v is hyperbola as shown in figure (b), Since convex mirror form only virtual image., , For virtual image, , C, , 1, f, , Since, , v, , 1, v, , 45°, , f, , u, , For real image, , 1, 1, and, to a concave mirror is, v, u, shown in figure (a), , O, , f 2f, , O, , between, , 1, f, , 2f, , Virtual image, , 1 1, Graph to Mirrors: The graph, V U, , 1, v, A, , Real image, , 1, u, , , , 1 1 1, f, or v , f, v u f, 1, u, , 1 1, 1, For all real image , v u, f, Graph in Spherical Mirror :In a spherical, 1, 1 1, , 1 1 1, v, u f, mirror: , v u f, This is a straight line equation with slope -1., v v, v v, This is represented by the line AB., 1 or 1, u f, u f, 1 1, 1, For virtual image, , v u, f, m, m, m, 1 1 1, , v u f, 1, This is a straight line equation with slope +1., v, v, v, This represents line BC., f, f, 1, The graph between 1/v and 1/u to a convex, Fig.(a), Fig.(b), Fig.(c), mirror as shown in figure (b)., Since convex mirror always form virtual image Concave mirror: If the objects is real,, to a real object., For real image, u=-ve, v=-ve, f=-ve,, 1, 1, 1 1 1 1, v, v, , , m 1or m 1, v u f, v u f, f, f, This is a straight line equation with slope +1., Graph as shown in figure (a), U-V Graph in Curved Mirror :In case of For virtual image, u=-ve, v=+ve, f=-ve, concave mirror, the graph between u and v is, v, m 1, Graph as shown in figure (b), hyperbola as shown in figure., f, , NARAYANAGROUP, , 121
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , Convex mirror: Since convex mirror always, form virtual image to a real object, u=-ve, v=+ve, Sol: 1 1 2 , Here v 70 cm ,, u v R, f=+ve,, 1 2 1, v, R 160cm , m 1, graph as shown in figure (c)., u R v, f, From the above graph it is observed that for, 1, 2, 1, 15, , , , v 0, m 1. i.e., when an object is very near, u 160 cm 70 cm, 560 cm, to pole of the mirror u 0 , then the curved, 560, u , cm 37 cm, mirror behaves like a plane mirror., 15, W.E-5: A reflecting surface is represented by the, The image is at a distance of 37 cm in front of, the mirror., 2, 2, 2, equation x y a . A ray travelling in, W.E-7: A point source of light is located 20 cm in, negative x-direction is directed towards, front of a convex mirror with f=15 cm., positive y-direction after reflection from the, Determine the position and nature of the, surface at point P. Then co-ordinates of point, image point., P are, y, 1 1 1, Sol: , u v f, ||||, ||||| |||||||||||, ||||||, ||, | | ||, , | | || | | | |, , |||||||| |||| |||||||||, , ||, , ||, || ||, , x, , |, |||||, , ||||||||||||||||||||||||, |, , The ray diagram is as shown., y, , x, , y, , ||||||||||||||||, |||||, , ||||||||| |||||, , a, 45°, , |||, |||||||, |||||, , 45°, 45°, P, , ||, |||||||, |||||, , || | | |, , |||||||, |||||||| |||||||||||, |||||, , x, , ||||||||||||||||||||||||||, |||||, , x, , a, a, and y , 2, 2, , a a , P , , , 2 2, W.E-6: A point light source lies on the principal, axis of concave spherical mirror with radius, of curvature 160 cm. Its image appears to be, back of the mirror at a distance of 70 cm from, mirror. Determine the location of the light, source., , 122, , 1 1 1, 1, 1, 35, , , , v f u 15cm 20cm 300cm, , 1, 7, , v 60 cm, , |||||, , |||||, |||||, , |||| |||| || ||||||, , Sol:, , Here u 20 cm, f 15 cm, , v 8.6 cm, Also v is positive, the image is located behind the, mirror., W.E-8: An object is 30.0 cm from a spherical, mirror, along the central axis. The absolute, value of lateral magnification is 1/2. The, image produced is inverted. What is the focal, length of the mirror?, Sol: Image inverted, so it is real u and v both are, negative. Magnification is 1/2, therefore,, , v, , u, , given, u 30 cm, v 15cm, 2, , Using the mirror formula,, , 1 1 1, , v u f, , 1, 1, 1, 1, , We have, , f 15 30 10, f 10 cm, , Since focal length is negative the given mirror, is concave., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , u 50 cm; v 10 cm, angles to the principal axis of a mirror of radius, 1, 1, 1, 50, of curvature 60 cm such that its image is, , , i.e., f 12.5cm, So, 10 50 f, 4, virtual, erect and has a length 6cm. What kind, of mirror is it and also determine the position, So R 2f 212.5 25 cm, of the object?, W.E-11: A concave mirror of focal length 10 cm, Sol: Since the image is virtual, erect and of a smaller, and a convex mirror of focal length 15 cm, size, the given mirror is ‘convex’ (concave, are placed facing each other 40 cm apart. A, mirror does not form an image with the said, point object is placed between the mirrors, on, description)., their common axis and 15 cm from the, concave mirror. Find the position, nature of, R, f 30 cm, Given R 60 cm, the image, and over all magnification, 2, produced by the successive reflections, first, Transverse magnification,, at concave mirror and then at convex mirror., I, 6, 3, v 3, Sol:, According, to given problem, for concave mirror,, m Further m , u = -15cm and f = -10 cm., O 10, 5, u 5, 1, 1, 1, 3u, , , i.e., v 30 cm, So , v, v 15 10, 5, i.e., concave mirror will form real, inverted and, 1 1 1 5 1, 1, , enlarged image I1 of object O at a distance 30, Using , v u f 3v u 30, cm from it, i.e., at a distance 40-30=10 cm from, 5 3, 1, convex mirror., , u 20 cm, M, 3u, 30, M, f = +15, f = 10, 40cm, Thus the object is at a distance 20 cm (from the, pole) in front of the mirror., I, O, P, I, W.E-10: An object is placed infront of a convex, P, 15cm, 15cm, mirror at a distance of 50 cm. A plane mirror, is introduced covering the lower half of the, M, convex mirror. If the distance between the, M, object and the plane mirror is 30 cm, if it is, For convex mirror the image I1 will act as an, found that there is no parallax between the, object and so for it u=-10 cm and f=+15 cm., images formed by the two mirrors. What is, the radius of curvature of the convex mirror?, 1, 1, 1, , , i.e, v 6 cm, v 10 15, 50 cm, So final image I2 is formed at a distance 6 cm, behind the convex mirror and is virtual as shown, 20 cm, O, P, in figure., I, F, Sol:, 30 cm, Over all magnification, M, m1 m2 26 /10 6 / 5, 30 cm, negative indicates final image is virtual w.r.t., As shown in figure the plane mirror will form, given object., erect and virtual image of same size at a distance, of 30 cm behind it. So the distance of image Refraction of Light :When a beam of light is, travelling from one medium to another medium,, formed by the plane mirror from convex mirror, a part of light gets reflected back into first, will be PI = MI - MP But as MI = MO, PI=, medium at the inferface of two media and the, MO-MP = 30-20 = 10 cm., remaining part travels through second medium, Now as this image coincides with the image, in another direction. The change in the direction, formed by convex mirror, therefore for convex, of light take place at the interface of two media., mirror,, , W.E-9: An object of length 10 cm is placed at right, , 2, , 2, , 2, , 1, , 1, , 1, , ||||||||||||||||||||||||||||||||||, , NARAYANAGROUP, , 123
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, Deviation or bending of light rays from their original, path while passing from one medium to another is, called refraction., (or), The phenomenon due to which light deviates, from its initial path, while travelling from one, optical medium to another optical medium is, called refraction., Refraction of light is due to change in speed, of light passes from one medium to another, medium., In case of refraction of light frequencey, (colour) and phase do not change. But, wavelength and velocity will change., Note: When light passes from one medium to another, medium, the colour of light is determined by its, frequency not by its wavelength., Refraction of light at plane surface:, , , , , , , , , , , , , , Angle of emergence (e): The angle which the, emergent ray makes with the normal is called, the angle of emergence., Laws of Refraction:, Incident ray, refracted ray and normal always, lie in the same plane., The product of refractive index and sine of angle, of incidence at a point in a medium is constant,, sin i = constant, , 1 sin i1 2 sin i2, If i1 i and i2 r then, 1 sin i 2 sin r;, This law is called snell’s law., According to Snell’s law,, , 2 , sin i, constant , for any pair of medium, sin r, 1 , and for light of given wavelength., i, Note:, The ratio between sine of angle of incidence, O, Air, to sine of angle of refraction is commonly called, Medium, , r, (glass), as refractive index of the material in which angle, of refraction is situated with respect to the, refracted, medium in which angle of incidence is situated., light ray, When light ray travells from medium 1 to medium, Air, e emergent ray, sin i 2, 2 then sin r 1 2 refractive index of, Incident ray: A ray of light, traveling towards, 1, another optical medium, is called incident ray., medium (2) with respect to medium (1), Point of incidence: The point (O), where an, incident ray strikes on another optical medium, Vector form of Snell’s law:, is called point of incidence., 1 eˆ1 nˆ 2 eˆ 2 nˆ , Normal: A perpendicular drawn at the surface, of seperation of two media on the point of, n̂, incidence, is called normal., Angle of incidence (i): The angle which the, incident ray makes with normal, is called angle, , of incidence., ê1, Refracted ray: A ray of light which deviates, , from its path on entering in another optical, medium is called refracted ray., ê 2, Angle of refraction(r): The angle which the, refracted ray makes with normal, is called the, angle of refraction., There ê1 unit vector along incident ray, Angle of deviation due to refraction( ): It is, ê2 unit vector along refracted ray, the angle between the direction of incident light, n̂ unit vector along normal incedence point, ray and refracted light ray., Note: Let us consider a ray of light travelling in, Emergent ray: A ray of light which emerges, situation as shown in fig., out from another optical medium as shown in, Applying Snell’s law at each interface, we get, the above figure is called emergent ray., 1, , , , , , , , , , 124, , 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , , i, , 1, r1, , r1, , 2, , r2, , r2, , 3, , Condition for no refraction : When an, incident ray strikes normally at the point of, incidence, it does not deviates from its path.i.e.,, it suffers no deviation., optically rare, medium (air), , 4, , r3, , optically denser, medium (glass), , 1 sin i 2 sin r1 ; 2 sin r1 3 sin r2, 3 sin r2 4 sin r3 ; It is clear that, 1 sin i 2 sin r1 3 sin r2 4 sin r3, , (or) sin i constant, Note: When light ray travells from medium of refractive, index 1 to another medium of refractive index, , i, , 1= , 2= , r, , sin i1 sin i2 sin i1 sin i2, , , , V1, V2, , , When a light travels from optically rarer, medium to optically denser medium obliquely:, , incident light ray, , 1 2 , From snell’s law,, sin i sin r , sin i sin r, , i r, Hence,, the ray passes without any deviation at, O, the boundary., optically denser r = ir, Note: Because of the above reason a transperant solid, medium (glass), is invisible in a liquid if their refractive indices, refracted, are same., light ray, Refractive Index :, a) it bends towards normal., Absolute refractive Index ( m ):, b) angle of incidence is greater than angle of, The absolute refractive index of a medium is the, refraction., ratio of speed of light in free space (C) to speed, When a ray of light travels from opticallly, of light in a given medium (V)., denser medium to optically rarer medium, veloctiy of light in free space (C), obliquely, , velocity of light in a given medium (V), optically rarer, medium (air), , i, , , , optically denser, medium (glass), , In this case angle of incedence (i) and angle of, refraction(r)are equal and i r 0., If the refractive indices of two media are equal, , 2 then, 1 sin i1 2 sin i2, , , , optically rare, medium (air), , incident, light ray, , It is a scalar., It has no units and dimensions., i, , optically rare, medium (air), , optically denser, medium (glass), , r = r i, , refracted, light ray, , a) it bends away from the normal at the point of, incidence., b) angle of refraction is greater than angle of, incidence., c) angle of deviation r i ., NARAYANAGROUP, , From electronmagnetic theory if 0 and 0 are the, permitivity and permeability of free space, and, are the permitivity and refractive index of the, given medium, , , , C, , V, , 1, 0 0, , , r r, 1, 0 0, , 125
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , , where r & r are the relative permittivity and , permeability of the given medium., For vaccum of free space, speed of light of all, wavelengths is same and is equal to C., So,For all wavelengths the refractive index of, , Relative Refractive Index: When light passes., from one medium to the other, the refractive index, of medium 2 relative to medium 1 is written as, and is given by, 1 2, , 2 v1 , ... 1, , , 1 v 2 , refractive index of medium 1 relative to medium, 1 v 2 , ... 2 , 2 is 2 1 and 2 1 v , 2, 1, , From eq. (1) & (2), 1, i.e., 1 2 . 2 1 1, 1 2 , 2 1, W.E-12:The refractive index of glass with respect, to water is 9/8. If the velocity and wavelength, of light in glass are 2 108 m/s and, 4000 A 0 respectively, find the velocity and, wavelength of light in water., 1, , C, 1., C, For a given medium the speed of light is different, for different wavelengths of light, greater will, be the the speed and hence lesser will be, refractive index., , free space is , , , , R V , So in medium V R, Note: Actually refractive index, , varies, , with according to the equation A , , , B, ., , , (where A & B are constants), For a given light, denser the medium lesser will, be the speed of light and so greater will be the, Sol:, refractive index., Example : Glass is denser medium when, compared to water, so glass water ., , 2 , , w, , vw , , The refractive index of water w 4 / 3, w, , The refractive index of glass g 3 / 2, , g , , g v w, vw, 9, , , ;, 8 2 108, w v g, , 9 2 108, 2.25 108 m / s., 8, , g , , g w , , , , g , w , , w g , g, w , , 9, 9 4000, , Foa a given light and given medium, the, w ; w , 4500A 0 ., refractive index is also equal to the ratio of, 8 4000, 8, wavelength of light in free space to that in the W.E-13: The wavelength of light in vacuum is, medium., . When it travels normally through glass, of, thickness ’t’. Then find the number of waves, , C f, vaccum vaccum, of light in ‘t’ thickness of glass (Refractive, V fmedium medium, index of glass is ), (when light travells from vaccume to a medium, Sol: Number of waves in a thickness ‘t’ of a medium, frequency does not change), of refractive index is, Note: If C is velocity of light in free space 0 is, thickness, t, , , number, of, waves, wavelength of given light in free space then, wavelength m, velocity of light in a medium of refractive index, , But m , C, , is Vmedium ., t, wavelength of given light in a medium of, number of waves = , , 0, Where 0 is the wavelength of light in vacuum., refractive index is medium , , , , , 126, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-14: When light of wavelength in vacuum Note: If in a given time t, light has same optical path, , length in different media, and if light travels a distance, travels through same thickness ‘t’ in glass and, d1 in a medium of refractive index 2 in same time, water, the difference in the number of waves, is ______. (Refractive indices of glass and, t, then 1d 2 2 d 2 ., water are g and w respectively.), .), Note: The difference in distance travelled by light in, vaccum and in a medium in the same interval of, Sol: We know number of waves of a given light in a, time is called optical path difference due to that, t, medium., medium of refractive index is , , x 1 d, x A ' B ' AB d d, Difference in number of waves, Note: A slab of thickness d and refractive index is, t, g w , , kept in a medium of refractive index ' . If, the two rays parallel to each other passes through, where g and w are the refractive indicies of glass, such a system with one ray passing through the slab,, and water respectively., then path difference, Optical Path ( x): Consider two points A and, Between two rays due t o slab will be, B in a medium as shown in figure. The shortest, , , distance between any two points A and B is called, x 1 d ., geometrical path. The length of geometricial path is, ' , independent of the medium that surrounds the path, AB.When a light ray travels from the point A to, 2, point B it travels with the velocity c if the medium is Note: The optical phase change (optical path, vacuum and with a lesser velocity v if the medium is, difference), other than vaccum. Therefore the light ray takes, more time to go from A to B located in a medium., The optical path to a given geometrical path in a W.E-15: The optical path of a monochromatic, light is the same if it goes through 4.00 m of, given medium is defined as distance travelled by, glass are 4.50 m of a liquid. If the refractive, light in vacuum in the same time in which it travels a, index of glass is 1.5, what is the refractive, given path length in that medium., index of the liquid?, , Sol: When light travells a distance ‘x’ in a medium, of refractive index , the optical path is x, d, B, A, Given 1 x1 2 x2 1.5 4.00 2 4.50, A1, , vaccum, , B, , 1, , d, , AB = real path or geometrical path, A1B1 = optical path, If the light travels a path lenght ‘d’ in a medium, d, at speed v, the time taken by it will be , v, So optical path length,, c, d , , x c t c d as , v, v, , Therefore optical path is times the geometrical, path. As for all media 1, optical path length is, always greater than actual path length., , NARAYANAGROUP, , 1.5 4.00, 1.333, 4.50, W.E-16: Find the thickness of a transparent, plastic plate which will produce a change in, optical path equal to the wavelength of the, light passing through it normally. The, refractive index of the plastic plate is ., Sol: When light travel a distance x in a medium of, refractive index , its optical path = x, , 2 , , Change in optical path = x x 1 x ., This is to be equal to , But 1 x , The thickness of the plate x , , , 1, 127
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-17:Consider slabs of three media A, B and C, , Principle of Reversibility of Light, , arranged as shown in figure R.I. of A is 1.5 , and that of C is 1.4. If the number of waves in, A is equal to the number of waves in the, combination of B and C then refractive index, of B is:, , According to principle of reversibility, if a ray, of light travels from X to Z along a certain path, it, will follow exactly the same path, while travelling, from Z to X. In other words the path of light is, reversible., N, , A, , Monochromatic, light, , Normal, , B, , C, , x, , 2x, , X, Medium A, (Rarer medium), P, , Sol: N A N B N C, , Q, , y, , Medium B, (Denser medium), , x A xB xC, , , A B C, , 90°, Z, Plane mirror, , M, , Figure shows a ray of light XY travelling through, medium ‘A’, such that it travels along YZ, while, travelling medium ‘B’. NM is the normal at point Y,, such XYN is the angle of incidence and MYZ is, the angle of refraction., sin XYN, a b , ....(1), sin MYZ, If a plane mirror is placed at right angles to the, path of refracted ray ‘YZ’, it found that light, retraces back its path. Now ray ZY acts as, incident ray and YX as refracted ray, such that, MYZ is angle of incidence and XYN, is the angle of refraction., , x A A x B B xC C, , , 0, 0, 0, , 3 x 1.5 x B 2 x 1.4, B 1.7, W.E-18: Two parallel rays are travelling in a, , 4, . One of, 3, the rays passes through a parallel glass slab, 3, of thickness t and refractive index 2 . The, 1, 1, sin XYN, 2, , , path difference between the two rays due to b a sin MYZ b a sin MYZ sin MYZ, sin XYN, the glass slab will be, sin XYN, 1, 2 3/ 2 , t, Comparing (1) and (2) a b , 1 t , Sol: x 1 t , b a, 1 4 / 3 8, Thus, the refractive index of medium ‘b’ with, W.E-19: A light ray travelling in a glass medium, respect to ‘a’ is equal to the reciprocal of, is incident on glass - air interface at an angle, refractive index of medium ‘a’ with respect to, of incidence . The reflected (R) and, medium ‘b’., transmitted (T) intensities, both as function W.E-20: A light ray is incident normally on a glass, slab of thickness ‘t’ and refractive index, of , are plotted. The correct sketch is, ‘ ’as shown in the figure. Then find time, 1), 2), taken by the light ray to travell through the, 100%, 100%, slab., T, T, Intensity, , Intensity, , medium of refractive index 1 , , R, , R, , , , , 90°, , 3), , , , , , 90°, , 4), 100%, T, , Intensity, , Intensity, , 100%, , R, , , T, , R, , , 90°, , , , , , 90°, , Sol: (3) After total internal reflection, there is no refracted, ray., 128, , Sol:, , t, , , , From the figure distance travelled by the light, ray through the slab is ‘t’, dis tan ce travelled, Velocity of light in glass , time, c, t, t, , , time , time, c, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-21: A light ray is incident on a plane glass, slab of thickness ‘t’ at an angle of incidence, ‘i’ as shown in the figure. If ‘ ’is the, refractive index of glass. Then find time taken, by the light ray to travel through the slab., , 1.5 , , air, , 4500, or med air , med, 1.5, 1.5, 0, , med 3674 A, , W.E-23: Monochromatic light falls at an angle of, incidence ‘i’ on a slab of a transparent material., Refractive index of this material being ‘ ’for, the given light. What should be the relation, Sol:, between i and so that the reflected and the, t, refracted rays are mutually perpendicular?, d, r, Sol: In the given figure let r is the angle of reflection and, r ' is the angle of refraction. According to the given, As shown in the above figure distance travelled by, condition, considering the reflected and the, the light ray through the slab is ‘d’. From the figure, refracted rays to be perpendicular to each other,, t, t, cos r , d , d, cos r, r, i, Dis tan ce travelled, 90°, Velocity of light in glass through the glass, r, time, i, , O, , c, d, d, , ; time , time, c, time , , From the figure r 900 r 1800, So, r 900 r, , t, 2t, , cos r c c 2 sin 2 i, , r 900 i [i = r, by law of reflection], According to Snell’s law, 1 sin i sin r , , W.E-22: Light of wavelength 4500 A0 in air is, , sin i sin 900 i , , incident on a plane boundary between air and, sin i cos i, tan i i tan 1 , another medium at an angle 300 with the plane, boundary. As it enters from air into the other, medium, it deviates by 150 towards the normal. W.E-24: A ray of light is incident at the glass-water, interface at an angle i as shown in figure, it, Find refractive index of the medium and also, emerges finally parallel to the surface of water,, the wavelength of given light in the medium., then the value of g would be, i, Air, , 30°, , Sol:, , r 15°, , Air, , Other, medium, , r, , w , , 4, 3, , r, Water, Glass, i, , Angle of incidence i 90 30 60 . As the ray, bends towards the normal, it deviates by an angle Sol: Applying Snell’s law ( sin i constant), i r 150 (given), at first and second interfaces, we have, 0 Applying Snell’s law, r 45, 1 sin i1 2 sin i2 ; But, 1 glass ,i1 i, 2 air and i2 900, air sin i med sin r ; 1 sin 600 sin 450, 1, 0, In terms of wavelengths,, g sin i 1 sin 90 or g , 0, , 0, , 0, , sin i, , NARAYANAGROUP, , 129
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W0.E-25: A light beam is travelling from region I W.E-27: The x-z plane separates two media A and, to region IV (Refer figure). The refractive, B of refractive indices 1 1.5 and 2 2 . A, , index in regions I, II,III and IV are, ray of light travels from A to B. Its directions, n0 n0, n0, in the two media are given by unit vectors, n0 , , and , respectively. The angle of, , , 2 6, 8, u1 aiˆ bjˆ and u2 ciˆ djˆ . Then, incidence for which the beam just misses Sol:, y, entering region IV is, 1 n0, 2, , I, , , 2 n0, 6, , II, 1, , IV, 90° n0, 8, , III, 2, , i, ai b j, , i, , x-z plane, r, , 0, , 0.2m, , 0.6m, , Sol: As the beam just misses entering the region IV, the, angle of refraction in the region IV must be 900 ., Application of Snell’s law successively at different, interfaces gives, n0 sin , , n0, n, n, sin 1 0 sin 2 0 sin 90 0, 2, 6, 8, , sin , , 1, 1, or sin 1, 8, 8, , W.E-26: A ray of light passes through four, , ci d j, , y i, , a, a, so sin i , b, a 2 b2, c, c, and tan r d ,sin r 2, c d2, tan i , , 3 , , a, , , , , , c, , , , 1 sin i 2 sin r ; 2 2 2 2 2 2 , a b , c d , But as aiˆ bjˆ and ciˆ djˆ are unit vectors so, , 3, a 4, transparent media with refractive indices, a 2 b 2 c 2 d 2 1 ;Hence 2 a 2c, so c 3, 1 , 2 , 3 and 4 as shown in the figure. The, W.E-28: A ray of light is incident on the surface, surfaces of all media are parallel. If the, of a spherical glass paper-weight making an, emergent ray CD is parallel to the incident, angle with the normal and is refracted in, ray AB, we must have, the medium at an angle . Cal-culate the, deviation., 1, 4, 2, 3, Sol: Deviation means the angle through which the, incident ray is turned in emerging from the, medium. In Figure if AB and DE are the incident, D, and emergent rays respectively, the deviation, will be ., B, , A, , C, , C, , , B, , Sol: Applying Snell’s law at B and C,, , B, , D, iC, C, , sin i constant or 1 sin iB 4 sin iC, But AB CD ; iB iC or 1 4, 130, , , , , A, O, , , D, E, , 4, , 1, , iB, A, , , , Now as at B ; i and r , So from Snell’s law, 1sin sin ....(1), Now from geometry of figure at D, i , So sin 1sin .....(2), Comparing Eqs. (1) and (2) , Now as in a triangle exterior angle is the sum of, remain-ing two interior angles, in BCD ,, , 2 , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-29: A ray of light falls on a transparent sphere, , , , C, , A, , , , B, , 60°, , Sol: Deviation by a sphere is 2 (i - r), Here, deviation 600 2 (i-r) or i-r = 300;, r = i 300 600 300 300, , , , , , 1, (real depth), , , 1 , , apparent shift 1 x, , , Diagram shows variation of apparent depth with, real depth of the object., app depth, , , , real depth realdepth, , relative, , 1 , , , apparent depth , , sin i sin 60 0, , 3, sin r sin 30 0, , Apparent Depth, , 1, , Hence the apparent shift OI 1 x, , , If the observer is in other than air medium of, refractive index ., Then apparent depth, , with centre at C as shown in figure. The ray, emerges from the sphere parallel to line AB., The refractive index of the sphere is, , 1, , , Case(1) : Object in denser medium and observer, Slope tan 1, in rarer medium., , , When object ‘O’ is placed at a distance ‘x’ from, Real depth, A in denser medium of refractive index as Note: If two objects O and O separated by ‘h’ on, 1, 2, shown in figure. Ray OA, which falls normally, normal, line, to, the, boundary, in a medium of, on the plane surface, passes undeviated as AD., , ., refractive, index, These, objects, are observed, Ray OB, which ‘r’(with normal) on the palne, from, air, near, to, normal, line, of, boundary., The, surface, bends away from the normal and passes, distance between the images I1 and I 2 of, as BC in air. Rays AD and BC meet at ‘I’ after, extending these two rays backwards. This ‘I’ is, h, the virtual image of real object ‘O’ to an observer, O1 and O2 is ., in rarer medium near to transmitted ray., AB, sin i tan i , ......(i), AI, air, AB, sin r tan r , .......(ii), , I, AO, O, Dividing eq. (i) and (ii), 1, , 1, , h, D, air, , i, , A, , x, , I, , i, r, , O, , C, , I2, O2, , O1O2 h I1 I 2 , , Note: Apparent depth of object due to composite slab, x, , x1 x2 x3, x, , , , x, is a , , 3, 1, 2, 1, , r, , B, , h, , , 1, , 2, , 2, , x3, , 4, , sin i AO, sin i, O, , ; According to Snell’ law , Note: If there are ‘n’ number of parallel slabs which, sin, AI, sin r, are may be in contact or may not with different, AO, AO x, refractive indices are placed between the, , AI , , AI, , , observer and the object, then the total apparent, The distance of image AI is called apparent depth, shift, , , , 1, 1 , 1 , or apparent distance. The apparent depth xapp is, s 1 x1 1 x2 1 xn, 1 , 2 , n , xreal, , x, given by i.e., app, Where x1, x2 ---- xn are the thickness of the slabs, , and 1 , 2 ....n are the corresponding refractive, x, The apparent shift OI AO AI x , indices., , NARAYANAGROUP, , 131
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , Object in rarer medium and observer in, denser medium : When the object in rarer, medium (air) at a distance’y’ from boundary and, an observer near to normal in denser medium of, refractive index ‘ ’. By ray diagram in figure, it is observed that the image is virtual, on same side, to boundary and its distance from the boundary is , times the object distance., Since 1 image distance is more than object, distance., I, r, , y, , air, , O, i, , y, , B, , A, r, , , , AB, AB, ,sin r tan r , AO, AI, According to Snell’s law 1.sin i sin r, sin i tan i , , AB, AB, , , AI . AO, AO, AI, , rarer, A, Denser, O, I, AO = x, AI = rel.x, , Application, Normal shift due to glass slab :When an, object is placed on normal line to the boundary, of slab whose thickness is ‘t’ and refractive index, ‘ ’. On observing this object (real) from other, side of the slab, due to refraction, the image of, this object shift on the normal line. This shift, value is called normal shift. This shift is towards, the slab, if the slab is denser relative to the, surroundigs and shift is away from the slab, if, the slab is rarer relative to the surrounds. Then, the Normal shift, , 1, OI 1 , , rel, , , 1, OI 1 t, for 1, normal shift, ., ,, , Therefore apparent height of object (AI) x, real height of object (AO), i.e. yapp . yreal Apparent shift AI AO, Apparent shift 1 y ., If the object is in other than air medium of , refractive index 1 . Then apparent height, , rel (real height) ; i.e., ya = 1 y, , , , 1, , 1, , ', , O, I, , rarer, , , , 1 , , shift x 1 , , real , Note: When convergent beam of rays passing from, rarer to denser medium as shown in the figure., Real image is formmed in denser medium which, is far to boundary than that of virtual object., shift real 1 x, 132, , Eye, , I O, , moving towards the plane boundary of a denser, medium., air, , , v, , , , v, O, , x, , Observer, , x, , xap x, Differentiating the above equation with, respective to time, we get, Vap V, To an observer in the denser medium, the object, appears to be more distant but moving faster. If, the speed of the object is v, then the speed of the, image will be v ., (b) Simillarly to an observer in rarer medium, and object in denser medium, the image appears, to be closer but moving slowly., air, , x, AI , rel, , 1, , ' , , Relation between the velocities of object, and image : The figure shows an object O, , Denser, , A, , 1, , Eye, , O I, , , , Apparent shift 1 1 y, , , Diagram shows variation of apparent height with, real height of the object., , slope tan 1, ', Note: When convergent beam of rays passing from, denser to rarer medium as shown in the figure., Real image is formed in rarer medium which, nearer to boundary than that of virtual object., , AO = x,, , 1 , t 1 t, , , , Observer, , I, v, , , , , v, O, , xap , , x, , , Differentiating the above equaion with, V, respective to time, we get Vap , , If the speed of the object is v. Then the speed of, v, the image will be ., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-30: In a tank, a 4cm thick layer of water, 4, , floats on a 6 cm thick layer of an, 3, , , organic liquid 1.5 . Viewing at normal, incidence, how far below the water surface, does the bottom of tank appear to be?, 6, 4, h1 h2, Sol: d AP 1.5 4 / 3 7 cm, 1, 2, , 3h, h, , 2h, , Sol:, , QP = QR = 2h, P, , W.E-31: An object is placed in front of a slab, 1.5 of thickness 6 cm at a distance 28, , i, , h, , i, , r, , i, 2h R, , Surface 1, 1=1, , Silvered face, surface 2, , O, 2 = 1.5, , Q, , T, , N, , M, K, , Sol:, , 2h, , S, , cm from it. Other face of the slab is silvered., Find the position of final image., , 2h, , i 450 ; ST = RT = h = KM = MN, , 28cm, , so KS =, , 2, , h 2 2h h 5, , 6cm, , h, 1, sin i sin 450, 5, , By the principle of reversibility of light, we can sin r , , , ; , h 5, 5, sin r 1/ 5, 2, say if light rays are coming from the mirror and, W.E-33: A person looking through a telescope, passing through the slab, the mirror will shift, 2 cm towards right for observer in front of the, focuses the lens at a point on the edge of the, slab. The position of the object from shifted, bottom of an empty cylindrical vessel. Next, mirror = 32 cm., he fills the entire vessel with a liquid of, refractive index , without disturbing the, telescope. Now, he observes the mid point of, the bottom of the vessel. Determine the radius, to depth ratio of the vessel., Sol: After the vessel is filled with the liquid, light, ray starting from the mid point O of the bottom, of the vessel as OA, after refraction goes along, CA., mirror, , So, the position of the image formed by shifted, mirror will be 32 cm behind it. Hence, position, of the image from surface 2 is 30 cm left to it, and 36 cm left of surface 1., W.E-32: An observer can see through a pin-hole, the top end of a thin rod of height h, placed, as shown in figure. The beaker height 3h and, its radius h. When the beaker is filled with a, liquid upto a height 2h, he can see the lower, end of the rod. Find the refractiveindex of the, liquid., NARAYANAGROUP, , A, , D, i, H, , C, , B, , O, R, , Applying Snell’s law, for the refraction at, point A,, , sin i 1, ... (i), sin r , OB, , From A O B , s in i O A , , R, R, , 2, , H2, 133
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, CB, , and from AC B, sin i CA , From Eq(i), , , 2, , R2 H2, 4R 2 H 2, , or 2 1 , , 2R, 2, , 4R H, , 2, , 4R 2 4H 2, 4R 2 H 2, , , , 3H 2, 4R 2 H 2, 1/ 2, , 4R 2 H 2, 1, R 1 4 2 , , , i.e.,, or, 3H 2, 2 1 H 2 2 1 , W.E-34: A diverging beam of light from a point, source S having divergence angle falls, symmetrically on a glass slab as shown. The, angles of incidence of the two extreme rays, are equal. If the thickness of the glass slab is, t and its refractive index is n, then the, divergence angle of the emergent beam is, S, i, , , , i, , 1, Normal shift t 1 , , 1, 1 , , Hence t 1 5 t 1 , 5; t = 15 cm, 1.5 , , Since the plate has given values of refractive, index (1.5) and thickness (15 cm), if its position, is shifted, it will not change the values of normal, 1, shif t 1 . It implies that the object will, , appear to be at the same distance 25 cm and its, apparent position remains the same., W.E-36: An air bubble is trapped inside a glass, cube of edge 30 cm. Looking through the face, ABEH, the bubble appears to be at normal, distance 12 cm from this face and when seen, from the opposite face CDGF, it appaears to, be at normal distance 8 cm from CDGF. Find, refractive index of glass and also the actual, position of the bubble., B, A, , n, , C, D, , t, , F, E, Sol: Divergence angle will remain unchanged because, H, G, in case of a glass slab every emergent ray is, parallel to the incidnet ray. However, the rays Sol: Let the actual distance of the bubble from the, face ABEH is ‘x’ then its actual distance from, are displaced slightly towards outer side. (In the, the face CDGF is 30 - x., figure OA BC and OD EF ), Actual dis tan ce of bubble, S, 1 , from refracting surface, A, D, , Apparent dis tan ce of bubble, But, t, from refracting surface, , The apparent distance of bubble from refracting, B, E i, C i, F, x, x, W.E-35: An observer looks at an object kept at a, surface ABEH ;12 , .....(1), , , distance 30 cm in air. If a rectangular glass, The apparent distance of bubble from refracting, plate 1.5 is placed between the observer, 30 x, 30 x, and the object with its thick-ness along the, ;8 , surface CDGF , .....(2), , , line of observation, the object appears to the, observer to be at a distance 25 cm. Find the, x 30 x, 20 ,, thickness of glass plate. Position of the glass, Adding Eqs. (1) and (2) , , , plate is now shifted (i) from object towards, we get 1.5, observers (ii) from observer towards the, object. How does it change the apparent, x, 12 or x =12, position of the object as seen by the observer?, From Eq. (1);, , 1.5, Sol: As a glass plate is placed between the observer, so, x, =, 18, cm, and the object, the object appears to be at a, This actual position of bubble is at normal, distance 25 cm in-stead of 30 cm. It implies that, distance 18 cm from face ABEH or at a distance, normal shift due to refraction at the glass plate, 12 cm from face CDGF., is 30 - 25 = 5 cm., , 134, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , , GEOMETRIC OPTICS, , Concept of Refraction in a medium of, variable refractive index :In the previous, case, we have assumed that the refractive index, of slab is constant., This is not true in atmosphere. The atmosphere, becomes thinner as we move up. When we move, up form the earth refractive index decreases. We, assumed the situation where refractive index, changes only in one direction, medium can be, considered as a collection of large number of, the thin layers. Let refractive index be a function, of y i.e., f (y) then the medium can be, considered as to be made up of large number of, thin slabs placed parallel to x-axis and optical normal, at any interface is parallel to y-axis. Similarly if, f(x) then slabs are parallel to y-axis and optical, normal at any interface is parallel to x-axis., y, , y, = f(y), = f(x), , O, , x, fig (a), , x, , O, , fig (b), , Examples of variable refractive index, , , From the snell’s law : 1. Sin0 sin 1, [Since = 1 at x = 0], Geometrically, relate the slope of this tangent to the, dy, tan 2 , angle of incidence i.e, dx, dy, Substitute from equation (1) and determine, dx, as a function of y. Integrate and obtain an expression, of y as a function of x., W.E-37:Find the variation of Refractive index, assuming it to be a function of y such that a, ray entering origin at grazing incident follows, a parabolic path y = x 2 as shown in fig:, , = f (y). Now the medium can be divided into, thin slices parallel to x axis and optical normal, parallel to y axis. Let 0 be the angle of incident, in the variable medium at point (-0,0). And is, the angle made by tangent with normal parallel, to y axis at any point (x,y) on the trajectory., Normal, , f(y), , , , , , , y, , x, 90°, , Sol:Draw a tangent at any point (x,y) which makes, an angle wiht optical normal parallel to y axis., From the Snell’s law: 1., , y, , x, , Sin 90 sin sin , , air, , Trajectory of light ray when f(y), , , , Trajectory of light ray when = f (x), , From the snell’s law:1. Sin 0 sin .... 1, Geometrically, relate the slope of this tangent to, dy, tan 90 .. 2 , the angle i.e, dx, Substitute for from equation (1) and determine, dy, as a function of y. Integrate and obtain an, dx, expression of y as a function of x., Les us Consider a Situation = f (x). Now, the medium can be divided into thin slices, parallel to y axis and optical normal parallel to, x-axis., , NARAYANAGROUP, , Geometrically,, , 1, 1 , , , dy, dy, tan 90 ; cot 2, dx, dx, y, , , x, 90° air, , dy, 2x, dx, [By Differentiating the above equation], cot 2x [From equation(2)], Hence,, 2, Given that, y x , , , , 1, cos ec 1 cot 2 1 4 x 2, sin , 135
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-38: A ray of light is incident on a glass slab, at grazing incidence. The refractive index of, the material of the slab is given by, , 1 y . If the thickness of the slab is d,, determine the equation of the trajectory of the, ray inside the slab and the coordinates of the, point where the ray exits from the slab. Take, the origin to be at the point of entry of the, ray., Sol: Draw a tangent at any point (x,y) on the trajectory, which makes an angle with optical normal parallel, to y axis f y , From the Snell’s law:, , From the snell’s law:, 1, ... 1, 1 ay , , n0 sin 90 n0 1 ay sin sin , y, , , x, 90° air, , dy, tan 90 .... 2 , dx, [Slope of tangent] From equations (1) and (2) we, get, dy, dy, ay , dx, dx, ay , , Geometrically,, , y, xmax 2000 m Ans., a, W.E-40: The refraction index of an anisotropic, , y, , x2, , , x, 90° air, , Sin90 sin , , 1., , 1, 1, , .... 1, , 1 y, , medium varies as 0, , y, , y, a, , x 4 y 2 d y d and y = d, W.E-39: Due to a vertical temperature gradient, in the atmosphere, the index of refraction, varies. Suppose index of refraction varies as, , x, , O, , x, , The ray will exit at point (x,y) where,, , where, , 0 x a. A ray of light is incident at the origin, just along y-axis (shown in figure). Find the, equation of ray in the medium., , Geometrically,, dy, dy, tan 90 , cot ... 2 , dx, dx, From equation (1) and (2) we get,, 1, dy, dy, x2, y 2 1 dx, 2 y x y, ;, dx, o, 4, y2 o, , x 1 ,, , Sol: Draw a tangent at any point (x,y) which makes, an angle with optical normal parallel to x axis., From the Snell’s law:, , , , , , 0 sin 90 0 x 1 sin ; At x 0, 0 , sin , , 1, .....(1), x 1, , n n0 1 ay , where n0 is the index of, refraction at the surface and ‘a’, = 2.0 106 m 1 . A person of height h = 2.0 m, stands on a level surface. Beyond what, distacne will he not see the run way?, dy, tan ....(2) ; [Slope of tangent], dx, From equations (1) and (2) we get,, dy, 1, dx, , dy , y 2 x Ans., dx, x, x, , Geometrically,, , Sol:Draw a tangent on the trajectory at any point (x,y), which makes an angle with optical normal., 136, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-41: A vessel of depth H is filled with a nonhomogenous liquid whose refractive index, , y, . What is the, H, , , , , varies with y as 1 , , Applying Snell’s law at interfaces P and Q, we get, , 1.sin 600 B .sin, , , 3, A Btmax .1, or, 2, 2, , 1, 3, t max A , apparent depth as seen by an observer from, , 2 , B, above?, Sol: Let us consider a thin layer of thickness dy at a W.E-43: A ray of light travelling in air is incident, depth of y., at a grazing angle on a large transparent slab, y, of thickness t 2.0m . The point of incidence, dy, H, is the origin., The medium has a variable refractive index(y), , Apparent thickness of this elementary layer, , given by, , dy, dy , dH ; 1 y, H, , y ky 1, , Where y is in m and k 0.25m 1, , 1, , Q(x0, y0), , H, , , , Apparent, , depth, , H dH, 1, , 1, , t = 2.0m, , 0, , H, , P(x, y), , O(0, 0), , (a) Express a relation between the angle of, incidence and the slope of the trajectory m,, in terms of the refractive index at that point, , dy, H ln 2, y, 0 1, H, , , =, , y ., air as shown in fig.2.65. If refractive index of Sol: (a) If be the angle made by the light ray with, the positive X axis, then slope of the light ray at, glass slab is given by A Bt where A and, B are constants and ‘t’ is the thickness of slab, any point P x, y will be m tan , measured from the top surface. Find the, Evidently, i / 2, maximum depth travelled by ray in the slab., Assume thickness of slab to be sufficiently, or, i / 2 ., .... (i), , WE-42: A ray of ligth enters into a glass slab from, , large., , 60°, Air, , Slab, , Sol: The path of ray is curved as shown in figure. As, it travels successively into denser layers, it bends, away from normal and TIR takes place at depth, , , where angle of incidence approaches, 2, , Now, refractive index for air is 1.0., Applying Snell’s law, for refraction at point O, and P, we have,, , sin 90 0 y sin i 1 i sin 1 1 , , ., , sin i, 1, , Now, from (i) cot i cot / 2 , , tan m, Slope m , , 60°, , cos i, 1 sin 2 i, , sin i, sin i, , Air, , P, , 2, , , Q, , 1, m, 1, 1 2 1 , 1, 2, sin i, i sin 1/ , , Slab, , NARAYANAGROUP, , 137
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , sin i r , Note: The later shift x t cos r can also be, incident on the slab at an angle of incidence ‘i’, , , and passing through a slab of thickness ‘t’. After, expressed as follows., two refractions at the boundary, the ray emerge, t, t, x, sin i r , sin i cos r cos i sin r , parallel to the incident ray. The perpendicular, cos r, cos r, distance between incident ray direction and, , , emergent ray direction is called lateral shift or, cos i, , sin i, , , x, t, 1, lateral displacement (x), on simplification., 2, 2 , , , , sin, i, , , Ai, O, W.E-44: A parallel sides glass slab of thickness 4, cm is made of a material of refractive index, r i-1, , Q, t, 3 . When light is incident on one of the, r, x, parallel faces at an angle of 600 , it emerges, M, P i, i, from the other parallel face. Find the lateral, displacement of the emergent beam., From the figure, the distance PQ is called lateral, displacement (or) lateral shift, i, PQ, A, From the traingle PQO, sin i r , OP, , Lateral Shift : In figure consider a ray AO, , PQ x OP sin i r , x = OP sin i r , , r, , t, , Sol:, , .....(1), , OM, OM, t, , OP , , ....(2), OP, cos r cos r, from (1) and (2), , B, , By Snell’s law, , , sin i r , x t , , cos r , for small angle of incidence sin i i,, , , , x, (Lateral shift), /2, i (Angle incidence), , xmax , 138, , t, , sin c t r c, cos r 2 , , C, , sin i, ,, sin r, , 3, , sin 600, sin r, , 3, 1, ,sin r ; Hence, r 300, 2 sin r, 2, t, sin i r , Lateral shift , cos r, , 1, r, x t i r ti 1 Lateral shift x 1 ti, i, , , Fig. Plot of lateral shift versus angle of incidence, , Note: When i (grazing incidence), 2, , E, , 3, , sin r and cos r 1, sin i r i r , , O, , D, r, , , , But cos r , , t, , i-r, , , , sin 300 , 4, 4, 4 tan 300 , 0 , 3, cos 30 , Examples of Refraction :Visibility of two, images of an object :, When an object in a glass container filled with a, liquid is viewed from outside at a level higher, than of liquid, there will be two images one due, to refraction through liquid and another due to, refraction through glass., Twinkling of stars : Due to fluctuations in, refractive index of atmosphere, the refraction, becomes irregular and the light some times, reaches the eye and some times it does not. This, gives rise to twinkling of stars., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , Critical Angle and total Internal Reflection : Note: Critical angle depends on nature of pair of, C, , , , , Consider a point object O placed in a optically, R, denser medium as shown in the figure. Rays of, media., Greater, D ratio greater will be the critical, light travel from O in all possible directions., When light is refracted at the surface into the, angle, rarer medium, it bends away from the normal., For glass - air :, Therefore, as the angle of incidence increases, the, 3, 2, angle of refraction also increases till for a certain, D , R 1, c sin 1 , c 420, 2, 3, angle of incidence, the angle of refraction is 900 and, light is refracted along the surface separting the two For water - air :, media. The corresponding angle of incidence is, 4, 3, D , R 1 c sin 1 , c 490, called the critical angle C ., 3, 4, When light is incidence at any point beyond P, For glass - water :, that is when the angle of incidence is greater than, 3, 4, 8, D , R , c sin 1 , c 630, the critical angle i C , then no light is refracted,, 2, 3, 9, and the entire incident light is reflected into the same For diamond - air :, medium. This phenomenon is known as the total, 1 , 0, internal reflection., D 2.5, R 1, c sin 1 , , c 24, 2.5 , Note: In case of total internal reflection as all (100%), incident light is reflected back into the same, medium i.e. there is no loss of intensity. This is, why images formed by total internal reflection, are much brighter than that formed by mirrors, and lenses., Note: Image due to total internal reflection is real,, Expression for critical angle C :, lateral and inverted with respect to object., According to Snell’s law, at critical angle of, Deviation of light under total internal, incidence, reflection: The figure shows a light ray, R, travelling from denser to rarer medium at an, D .sin C R .sin 90 0 , sin C , D, angle i, less than the critical angle ., c, , , V, , sin C R D D, D VR R, For R 1 , , , , , , D, , , , 1, s in , , rarer (air), , r, C, , ), C, i(<, , Condition for total internal reflection :, For total internal reflection to take place light, must be propagating from denser to rarer, medium., Ex: Ray from water to air, glass to water., Total internal reflection will take place only if, angle of incidence is greater than critical angle., i.e., 1 R , i C with C sin , D, , NARAYANAGROUP, , , Denser(), , The deviation of the light ray is given by, r i, Since sin i sin r , therefore, , sin 1 sin i i, This is a non linear equation. The maximum value, of ‘ ’ occurs when i c , and is equal to, , C ,, 2, , i.e., , max , , , c, 2, 139
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , If the light incident at an angle i c , as shown in, the figure then the angle of deviation is given by, 2i. The maximum value of occurs when, i c and is equal to max 2 c ., , i, , air, , , , , , 2, 1 , h, , , 1, sin C , tan C , , , , , Medium (), i, , , r h tan C , , , i, , , , 2 1 , 1, , 3h, For water 2C 980 and r , 7, , h2, 2 1, Total Internal reflection in Prisms:, The critical angle of ordinary glass is very nearly, 420 . If light is incident inside a prism at an angle, greater than 420 , then the light will be totally, internally reflected. This is achieved by taking, a right prism ( 900 prism) so that the other angles, of the prism are 450 each., Deviation through 900 :, d) Area of the base : A , , The variation of ‘ ’ with the angle of incidence ‘i’ , is plotted in figure., , C, , C, 2, i, , C, , , , Looming : This effect occurs when the density , of air decreases much more rapidly with, increasing height than it does under normal, conditions. This situation sometimes happens, in cold regoins particularly in the vicinity of, the cold surface of sea or of a lake. Light rays, starting from an object S (say a ship) are curved, downward and on entering the eye the rays, appear to come from S , thus giving an, impression that the ship is floating in air., , S, , E, , , , deviated, , Deviation through 1800 :, , S, , Field of vision of fish : - A fish at a depth ‘h’, from the surface of water of refractive index can, see the outer world through an inverted cone with, N, , N, r, , c, , c, deviated, , h, c c, fish, , , , , 140, , Vertex angle = 2C, Radius of the circular base of the cone formed on, surface of water is given by, , , , Erecting prism (No Deviation Prism):, Herethe rays of light are incident parallel to the, base. After refraction they are incident on the, hypotenuse face of the prism at an angle greater, than the critical angle ( 420 ). Hence total internal, reflection takes place and the rays emerge parallel, to the base., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-46: A ray of light travelling in a transparent, medium falls on a surface separating the, medium from air at an angle of incidence 450 ., The ray undergoes total internal reflection., If is the refractive index of the medium with, respect to air, select the possible value(s) of, from the followng., , 45°, 45°, However, on emergence the rays are inverted., Therefore, this prism is used for making an, inverted image erect., Sol: For total internal reflection to occur,, W.E-45: A ray of light travelling in a rarer medium, i critical angle, C or sin i sin C, strikes a plane boundary between the rarer, medium and a denser medium at an angle of, 1, 1, 1, incidence ‘i’ such that the reflected and the or sin 450 or, or 2 or 1.414, , 2 , refracted rays are mutually perpendicular., Another ray of light of same frequency is W.E-47: A liquid of refractive index 1.5 is poured, incident on the same boundary from the side, into a cylindrical jar of radius 20 cm upto a, of denser medium. Find the minimum angle, height of 20cm. A small bulb is lighted at the, of incidence at the denser-rarer boundary so, centre of the bottom of the jar. Find the area, that the second ray is totally reflected., of the liquid surface through which the light, of the bulb passes into air., Sol:Let S be the small bulb at the bottom of the jar., ir, The bulb is in the denser medium. The light from, the bulb is incident on the water - air interface., Light emerges for incident angles less than the, r, critical angle. The locus of all the points on the, surface of the water that produces an angle of, incidence C is a circle of radius r., Sol: Figure shows incidence of a ray at the rarer-denser, boundary such that reflected and refrated rays are, mutually perpendicular., i.e., r 900 r1 1800 . or, A, D, B, 1, , [r=i,, law, r1 900 r 900 i, reflection], Apply Snell’s law at the boundary,, , of, , C, , C, , R sin i D sin r1, , R sin i D sin 900 i D cos i, , D, or tan i, R, sin c , , 1, 1, , , R, D / R D, R D, , C, , h, , --- (1), --- (2), , S, , In the DBS, tan C , , r r h tan C; r , ;, h, , h, , 2 1, , The area of the surface through which the light, passes,, , Using equation (1),, 2, , 1, sin c , cot i c sin 1 cot i , tan i, , NARAYANAGROUP, , 0.20 , A r , 0.1005m 2, 2, 1.5 1, 2, , 141
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-48: An optical fibre is made of glass fibre of, , 1, ... (2), sin , From (1) and (2), considering the limiting case,, 1, 2, cos sin or 450 , cos 450, Hence refractive index 2, The least value of so that the light entering, the rod does not emerge from the curved surface, is 2 ., or , , refractive index 1.68. The outer coating of the, glass fibre is made of material of refractive, index 1.44. What is the range of angles of the, incident rays with the axis of the pipe for, which total internal reflection inside the, optical fibre takes place?, Sol: We know that maximum launching angle in case, of optical fibre as i max sin 1 22 12 , i max sin 1, , , , 2, , 1.68 1.44 , , 2, , , , W.E-50: A rectangular glass slab ABCD of, , i max sin 1 0.8653 590551, This is the maximum value of i. All light rays, with angle of incidence between 00 and 590551, will undergo total internal reflection., W.E-49: Light is incident at an angle on one, planar end of a transparent cylindrical rod of, refractive index so that the light entering, the rod does not emerge from the curved, surface of rod irrespective of the value of ., Sol: The angle incidence at the curved surface of the, , refractive index n1 is immersed in water of, refractive index n 2 n1 n 2 . A ray of light is, incident at the surface AB of the slab as, shown. The maximum value of the angle of, incidence max , such that the ray comes out, only from the other surface CD, is given by, n2, n1, , A, , D, , max, , 0, cylindrical rod is given to be 90 ., Light entering the rod at angle of incidence , on one planar end will not emerge from the, , B, , curved surface if angle of incidence 900 , is greater than the ciritical angle. Refractive, Sol:, index is denoted by ., , C, , n2, n1, , A, , D, , r2, r1, n2, , B, , C, , 0, , r1 r2 90 ;, , r1 900 r2, , r1 max 900 r2 min, , and r2 min C, (for total internal reflection at AD), , , 1, sin C, , or , , , , 1, or sin 900 , , 1, ; As a limiting case,, cos , , 1, cos , , .... (1), , According to Snell’s law, , If , rod,, 142, , sin , sin , , , as a limiting case on planar end of, 2, , n2, 1 n 2 , Where sin C n or C sin n , 1, 1, , r1 max 900 C, Now applying Snell’s law at face AB:, n1, sin max, sin max, sin max, , , , 0, n 2 sin r1 max sin 90 C cos C, , n1, 1 n 1, or sin max n cos C max sin n cos C , 2, 2, , n, n , , max sin 1 1 cos sin 1 2 , , n1 , n 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-51: What will be the minimum angle of, incidence such that the total internal reflection, occurs on both the surfaces?, 1 , , 1 sin 2 i, Critical angle is determined from, 1, 1, sin C , , 1 sin 2 i, W.E-53: Light is incident normally on face AB of, a prism as shown in figure. A liquid of, refractive index is placed on face AC of, the prism. The prism is made of glass of, refractive index 3/2. The limits of for which, total internal reflection takes place on face, AC is, , 3, , 2 , , A, , 2, , i, 2 2, , Sol:, , i, B, , 3 , , 3, , Critical angle at A C1, sin C1 , , ---(2), , 2 sin 2 r 2 cos2 r 1 sin 2 i, , 2, , 2 2, , 3 , , 1, ; cos r 1, , From equations 1 & 2, sin 900 r , , 1, 2, 1, , , ; C1 450, 2, 2, 2, , Liquid, , Critical angle at B C2, , A, , 60°, , 30° C, , 3, 3, , C2 600, 2, 2, 90°, Minimum angle of incidence for total internal, reflection to occur on both the slabs should be, B, 0, 0., , i, , 60, 60, min, Sol:Critical angle between glass and liquid face is, W.E-52: A ray of light incident on the horizontal, 2, surface of a glass slab at an angle of incidence, sin C , 3, 3, ‘i’ just grazes the adjacent vertical surface, 2, after reflection. Compute the critical angle, and refractive index of glass., sin C 2 , , A, , i, , B, , Liquid, r, F, , Sol:, , C, , D, , A, , 60°, 30°, 90°, , C, , E, , By Snell’s law of refraction, at the horizontal, surface, 1 sin i sin r; sin r sin i ---(1), Apply Snell’s law at vertical surface,, 1, sin c 1 sin 900 ; sin C , , From figure in BFD, r C 900 1800 ; c 900 r, NARAYANAGROUP, , B, , Angle of incidence at face AC is 600, For the TIR to takes place at the face AC, the, angle of incidence at the face AC, i C, 600 c or sin 600 , , 2, 3 3, or , 3, 4, 143
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-54:What is the value of the refractive index, , A, , for a 900 450 450 prism which is used to, deviate a beam through 900 by total internal, reflection?, , i, , 45°, , 45°, B, , 90°, , or sin i sin c, , 45°, , Sol: At face AB, the ray of light suffers no deviation, if the incident normally, , C, , i = 45°, , 0, or sin 45 , , 1, 1, 1, , , , 2 , , of for which 2or 1.414, Hence, the rays for which 1.414 will get TIR ;, , A, , For green and blue 1.414 , so they will suffer, TIR on face AC only red comes from this face., , 45°, , W.E-56: White light is incident on the interface, B, , C, , At face AC, the light ray undergoes TIR, so angle, of incidence at the face AC is greater than critical, angle, 450 C;sin 450 sin C;, , 1, 1, ; 2, 2 , , of glass and air as shown in the figure. If green, light is just totally internally reflected then, the emerging ray in air contains:, , Air, , Green, , W.E-55: A beam of light consisting of red, green, and blue colours is incident on a right angle, Glass, prism. The refractive indices of the material, White, of the prism for the red, green and blue, wavelengths are 1.39, 1.44 and 1.47, respectively. The colour of light that comes out, 1 1 , of the prism is, Sol: Critical angle C sin , , Wavelength increases in the sequence of, VIBGYOR. According to Cauchy’s formula, refractive index ( ) decreases as the wavelength, increases. Hence, the refractive index will increase, , 45°, , Sol: The colours for which i c , will get total internal, reflection i c, 144, , in the sequence of ROYGBIV. The critical angle C, will thus increases in the same order VIBGYOR., For green light the incidence angle is just equal to, the critical angle. For yellow, orange and red the, critical angle will be greater than the incidence angle., So, these colours will emerge from the glass air, interface., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-57: A rectangular block of glass is placed on, a printed page lying on a horizontal surface., Find the minimum value of the refractive, index of glass for which the letters on the page, are not visible from any of the vertical faces of, the block., , O, , P, , Plane mirror, , R, , i, , Sol: A.d , r, , R.d h, and its image in mirror is, , , , O, A, , , Paper, , B, , same distance i.e., , h, ., , , Sol: The situation is as shown in the Fig, Light will not, emerge our from the vertical face BD if, at it i C or sin i sin C ; sin i , , 1, ..... (1), , , So the apparent distance is, , 2h, , , W.E-59: A cubic container is filled with a liquid, whose refractive index increases linearly from, top to bottom. Which of the following, represents the path of a ray of light inside the, liquid ?, , But from Snell’s law at O., 1 sin sin r and OPR ,, 0, r i 900 , r 900 i So sin sin 90 i , , 1), , 2), , 3), , 4), , cos i sin / , 2, , But sin i 1 cos2 i 1 sin / --(2), Substituting the value of sin i from eq (2) in (1), , sin 2 1 2, 1, , 1 sin 2, 2, , 2, 2, Now as sin max 1; 2, 2, , Hence min 2, , W.E-58: A plane mirror is placed at the bottom of, a tank containing a liquid of refractive index, . P is a small object at a heigth h above the, mirror. An observer O, vertically above P,, outside the liquid, observes P and its image in, the mirror. The apparent distance between, these two will be, , NARAYANAGROUP, , Sol: Since, the refractive index is changing, the light, cannot travel in a straight line in the liquid as, shown in options (3) and (4)., Intitially it will bend towards normal and after, reflecting from the bottom it will bend away from, the noraml so (1) is correct answer., , Refraction at Spherical surfaces and by, Lenses, , , A part of a sphere of refracting material is called, a spherical refracting surface., 145
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , curvature R. The rays are incident from a medium, of refractive index 1 to another of refractive index, , Some important terms related to, spherical refracting surface are given, below:, , 2 . As before, we take like in curved mirrors the, aperture of the surface to be small compared to, other distances involved. Hence NM will be taken, to be nearly equal to the length of the perpendicular, from the point N on the principal axis., , A, Radius of, curvature, Principal axis, , P, Pole, , C, , tan , , (Centre of curvature), Denser medium, , Rarer medium, B, , Now for NOC, 'i ' is the exterior angle., , , , The point ‘P’ in the figure is the pole (P), , , , The centre of the sphere of which the refracting, surface forms a part is called the centre of curvature, (C) of the spherical refracing surface., , , , Therefore, i , Similarly r NCM NIM , Now by Snell’s law 1.sin i 2 .sin r or for small, , The radius of the sphere of which the refracting, surface forms a part is called the radius of curvature, of the spherical refracting surface (R), , , , MN, MN, MN, , tan , , tan , OM, MC, MI, , angles 1.i 2 .r substituting i & r, we get, , 1 2 , , The diameter of the spherical refracting surface is, called its aperture. In the figure, the line joining A, and B is the aperture of the spherical refracting, surface., , 1 2 2 1 , , , , The line joining the pole and centre of curvature, and extends on either side of the surface is called, the principal axis., , 1, , 1, 2 2, OM MI, MC, Here OM, MI and MC represent magnitude of, distances, applying sign convertions., , , , Sign convention:All the distances are measured, , OM u, MI v, MC R, , from the pole of the spherical refracting surface., , 2 1 2 1, , v u, R, , , , The distance measured in the direction of the incident, light are taken as positive., , , , , , Refraction at Sperical surfaces :Consider, , This is the Gaussion’s relation for a single spherical, refracting surface. Though above relation is derived, for a convex surface and for a real object and real, image, it is equally valid for all other conditions., , refraction at a spherical interface between two, transparent media. The normal at the point of, incidence is perpendicular to the tangent plane to, the spherical surface at that point and therefore,, , passes through its centre of curvature., 1, , N, , 2, , i, r, , O, , , u, , M, , R, , I, , V, , Figure shows the geometry of formation of image I, of an object ‘O’ on the principal axis of spherical, surface with centre of curvature C, and radius of, 146, , , , If the object or image itself is present at a refracting, surface, refraction at that surface is not considered., , , , It is note that with respect to real object convex, refracting surfaces can form real image (for distant, object) as well as virtual image (for nearer object),, where as concave refracting surface forms only, virtual image., , , C, , if we move in the direction of light, 1 is the, refractive index of the medium which comes before, the boundary and 2 is the refractive index of the, medium which comes after the boundary., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, As the object position changes with time, the image, position also changes., , Magnification, , , Lateral magnification or transverse, magnification:, , 1, , Hence , , 2, , 2, , h0, i, , r, , C, , v , vi 1 v 0, 2 u , , I, hi, , v i lo ngitudinal magnification v0, , v, , u, , vi m L v0, , From figure, the lateral magnification is, mt , , hi, h0, , On applying proper sign convertion we get direction, of motion of image., , , From Snell’s law:, 1i 2 .r (for small angles), h0, h, 2 i, u, v, Thus lateral magnification, , , , dh i, 2 v h i, mt , dt, 1 u h 0 dh 0, dt, , h i 1 v , ., h 0 2 u , , Longitudinal magnification at refracting, curved surface: If a small object of length ‘du’, , Velocity of image is, , is placed on the axis, produces an image of length, ‘dv’ along the axis of the refracting surface, then, longitudinal magnification, , V, , mL , , dv, 1, ; Since 2 1 2, du, v u, R, , dv 1 v 2, ., On differentiating,, du 2 u 2, 2, , mL , , 1 v, 2 u 2, , 2 2 , Longitudinal magnification mL m ., 1, , Where ‘ m t ’ is transerverse magnification., , Motion of Object, , , Along Perpendicular ( Transverse) to the, Principal Axis If the object moves transverse, to the principle axis with the speed V0 . If m is the, magnification, then, , Therefore 1, , mt , , 2 dv 1 du, , , 2, 0 , 22 v1 21 v 0 0, 2, v dt u dt, v, u, , Along the Principlal Axis :, Since, , i, , , , , , , 2, 1, , v , u .V, , , , 0, , or Vi mt V0, , , Principal FOCI :, 2 1 2 1, , v u, R, If the object at infinity i.e., u , , In the equation, , , 1, , 2, , 1, 2, F2, , f2, fig (a), , O, F1, , f2, fig (b), , 2, 1, 0 2, v, R, , 2 1 2 1, , v u, R, , NARAYANAGROUP, , 147
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 2R, From figure, it is clear that v f 2 f 2 , 2, 1, , , , i.e. The position of image corresponding to the, object at infinity, is called the second principal focus, of the refracting surface. This is shown in fig (a), Similarly if v , i.e., the object is so placed that, the refracting rays becomes parallel to the principal, axis, then, , W.E-60: A small object stuck on the surface of a, glass sphere, , viewed from the, diametrically opposite position. Find, transverse magnification., Sol: Refreaction will take place at side II of the, surface, , 2 1 2 1, , v u, R, , 2 1 2 1, , u, R, , II, , I, , 1R, From figure, it is clear that u f1 f1 , 2, 1, i.e. The position of the object, whose image is, formed at infinity to known as the first principal focus, of the refracting surface. This is shown in figure(b)., , O, , I, , P, , C, , Here, 1 1.5, 2 1 and u 2R, , f1 1, Hence f , 2, 2, , 1, 1.5, 1 1.5, , , v 2R , R, , It is easy to see that first focal length f1 for spherical, refracting surface is not equal to the second focal, length f 2 ., , 1 0.5 1.5 0.5, , , , or v 4R, v R 2R, 2R, negative sign indicates that the image is formed, to the left of refracting surface as shown., , Further, , , , , 1.5 is, , 1 v 1.5 4R , Magnification, m u 1 2R ;, , , 2, , 2 1 2 1, , v u, R, , 2 R, 1R, , 1, v 2 1 u 2 1 , , , , f 2 f1, 1, v u, , Power of Refracting Spherical surface:, Any distance divided by the refractive index of the, space in which it is measured is called a reduced, distance. The refracting power of a spherical surface, is defined as the reciprocal of the reduced focal, length. If f1 and f 2 are first and second principal, focal length of refracting surface., f1, f2, The ratios and are the reduced focal lengths., 1, , W.E-61: A solid glass sphere with radius R and an, index of refraction 1.5 is silvered over one, hemisphere. A small object is located on the, axis of the sphere at a distance 2R to the left, of the vertex of the unsilvered hemisphere., Find the position of final image after all, refraction and reflections have taken place., Sol: The ray of light first gets refracted then reflected, and then again refracted. For first refraction and, then reflection the ray of light travels from left to, right while for the last refraction it travels from right, to left. Hence, the sign convention will change, accordingly., , 2, , ||||, ||, , |||, ||||, , Then refracting power, , 2 1 , 1R , Power (P) = R f1 , , 2, 1 , 148, , ||||||||||||||||, ||||||||, |||||, |, , 1, , f, f , 2 As 1 2 , f1, f2 , 1, 2 , , I3, , I2, , O, , P, , m 3, , 2R, , 1.5R, , R, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 2 1 2 1, , with, v u, R, proper sign conventions,, 1.5 1.0 1.5 1.0, , , we have, v1 , v 2R, R, , First: refraction:Using, , Second: reflection: Using, , 1 1 1 2, , v u f R, , Sol: In case of refraction from curved surface, 2 1 2 1, , , u, R, Here, 1 1.5, 2 1;R 5cm and u 3cm ;, 1 1.5 1 1.5, So 3 5 , i.e., v 2.5 cm, , , W.E-64: One end of a cylindrical glass 1.5 , is given the shape of a concave refracting, surface of radius 10 cm. An air bubble is, situated in the glass rod at a point on its axis, such that it appears to be at distance 10 cm, from the surface and inside glass when seen, from the other medium. Find the actual, location of air bubble., , with proper sign conventions,, 1 1, 2, we have, v R, 2, , Third:, , R, 2, Again, , v2 , , refraction:, , using, , 2 1 2 1, , with reversed sign convention,, v u, R, we have, , 1.0, 1.5, 1.0 1.5, , , v3 1.5R, R, , Sol:, , 2 1 2 1, , u, R, , O, , I, , P, , C, , or v3 2R i.e., final image is formed on the vertex, Here 1 1.5, 2 1, v 10cm, R 10cm, of the silvered face., W.E-62: A point object is placed at the centre of a 1 1.5 1 1.5 1 1.5 1 1 1, ;, glass sphere of radius 6 cm and refractive, 10 u, 10, 20 u 10 20, 20, index 1.5. The distance of the virtual image, u 30 cm, from the surface is., Hence the air bubble is actually located at a distance, Sol: A point object is at the centre of a glass sphere, 30 cm from the surface and inside glass., of radius 6 cm., W.E-65: A transparent thin film of uniform, , O, , The rays from the object fall normally on the surface, of the sphere and emerge undeviated. When drawn, backwards, they meet at O. The image will be, formed at the centre O itself., W.E-63: An air bubble in glass 1.5 is, situated at a distance 3 cm from a convex, surface of diameter 10 cm as shown in figure., At what distance from the surface will the, bubble appear?, A, = 1, , = 1.5, , C, , O, , I, , P, , thickness and refractive index n1 1.4 is, coated on the convex spherical surface of, radius R at one end of a long solid glass, cylinder of refractive index n 2 1.5 , as, shown in figure. Rays of light parallel to the, axis of the cylinder traversing through the film, from air to glass get focused at distance f1 from, the film, while rays of light traversing from, glass to air get focused at distacnce f 2 from, the film. Then, the magnitudes of f1 , f 2 are, n1, , Air, , n2, , Sol: (1,3) For air to glass, , 1.5 1.4 1.5 1.5 1.4, , , f1, R, R, , f1 3R, , For glass to air, 3cm, 5cm, , NARAYANAGROUP, , 1, 1 .4 1.5 1 1 .4, , , f2, R, R, , f 2 2R, 149
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-66: A spherical solid glass paper weight of, diameter 6 cm has a small air bubble at a, distance of 1.5 cm from the centre. If the air, bubble be viewed from the side to which it is, nearest along the line joining the bubble and, the centre, find where will it appear., Sol: Fig shows the relevant situation. O is the air, bubble which acts as the object and P is the pole., The light ray travelling from the object passes from, glass to air., Glass, , O, , , 1, , 0 R V x, 1, V1 x, , Here, I1 behaves like object for pole P2 (Plane, mirror): for this I 2 is the image from refraction, through the plane mirror., Again for pole P1 , I 2 behaves as an object, then, apply, , Air, 2 = 1, , 1 = 3/2, , C, , , , 2 1 2 1, , V u, R, (for pole P1 ), , P, , 1 3/ 2 and 2 1, 3, cm, R 3cm and v ?, Also, u , 2, 2 1 2 1, , Using Eq., v, u, R, 1, 3/ 2 1 3/ 2 , , We have,, v 3/ 2cm 3cm , , , , 1, , , 0 R , V3 x 2t , , V3 , , , , 1, 3/ 2 1 3/ 2 , , v 3/ 2cm 3cm , 1, 1, 1, , or, , v 6cm 1cm, or, v 1.2cm, , or,, , x 2t , , , (This is the final image formed by the, combination), Real and Virtual Object :If a surface is, incident with a divergent beam, it means a real, object is placed infront of the surface at the, position where the rays are diverging as shown, in figure (a), Diverging beam, , Converging beam, , Real object, (a), , Virtual object, (b), , Thus, the bubble appears at a distance of 1.2 cm, from the surface P inside the sphere., , Concept of Slab when one surface is Silvered, , , Applying formula of spherical surface for pole P1, , , I1, , P1, , air O, , , , P2, , x 2t , , , , I3, I2, , x - t, , , , , , x, t, , 2 1 2 1, , v, u, R, 150, , , , If a surface is incident with a converging beam, it, means a virtual object is placed behind the surface, at the position where the rays appear to converge, as shown in figure (b)., In case of image formation unless stated object is, taken to be real, it may be point object denoted by, dot ( ) or extended and is denoted by an arrow, ( ), Real and Virtual Image : The optical image, is a point, where the rays of light either intersect, or appear to intersect., If the real rays after reflection or refraction actually, converge at a point, the image is said to be real as, shown in figure (a)., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , , GEOMETRIC OPTICS, , If the rays do not actually converge but appear to, do so. The image is said to be virtual. This is as, shown in the figure (b)., M, , M, I, , O, , O, I, , Real object, Virtual object, , I, , Convexo, - concave, , Concavo, - convex, , 1, , Fig. (c) Meniscus, , O1, , O, I, , Note: While calling the name of the lens we called first, the shape of the surface which has more radius of, curvature is to be considered., , Virtual object, Real object, , (a), , , , (b), , Note: The eye sees a virtual image as well as it sees a, real image. A real image containing light energy,, hence it can be seen on a screen and can be, photographed. On the other hand, a virtual image , can not be received on a screen., , Refraction by Lenses, , A thin lens with refractive index greaterthan that of, surroundings behaves as a convergent or convex, lens if its central portion is thicker than marginal, one. i.e. it converge if parallel rays incident on it., If the central portion of a lens (with L m ) is, thinner than marginal, it diverges parallel rays and, behaves as divergent or concave lens., , Note: A thin lens is a lens in which the thickness of the, lens is small compared to the object distance (or), material with two refracting surfaces such that atleast, the image distance or either of the two radii of, one is curved and refractive index of its material is, curvature of the lens., different from that of the surroundings., If the curved surface (or surfaces) of a lens are In case of thin spherical lenses: Optical, centre (or) pole O is a point for a given lens through, spherical, the lens is called spherical lens and if its, which any ray passes undeviated., thickness is small the lens is called thin., , , , Lense theory :A lens is a piece of transparent, , , , Here we shall limit ourselves to thin spherical lenses., , , Different types of spherical lens are shown, in figure (a) and (b), , O, , c2, , c1, , O, , c1, , c2, , F2, R1, R1, , R2, , Bi convex, lens, , R, , R, , , , R2, , R1, , R2, , R, , If the lens has two spherical surfaces, there are two, centres of curvature C1 and C2 and, , Equl convex Plano convex, lens, lens, , correspondingly two radii of curvature R1 and R 2 ., , Fig. (a) Convex Lens, , , , Principal axis C1C2 is a line passing through, optical centre and centres of curvature of two, refracting surfaces. It is perpendicular to the lens., , , Bi concave, , Equi concave, , Plano, concave, , Fig. (b) Concave Lens, , NARAYANAGROUP, , A lens has two surfaces and hence two focal points., First focal point ( F1 ) is an object point (real in case, of a convex lens and virtual for concave) on the, princiapl axis for which image is formed at infinity., , 151
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , F1, , Note: While using the sign convention it must be kept in, mind that, to calculate an unknown quantity the, known quantities are substituted with sign in a given, formula., , F1, , (a), , Longitudinal distances are measured from optical, centre and are taken to be positive if in the direction, of light propagation and negative if opposite to it., , (b), , Second focal point ( F2 ) is an image point on the, principal axis for which object lie at infinity., , Incident, R1, light, , C2, F2, , F2, , (a), , ., , Incident, light, , R2, , R1, , R2, , C1 F, , F C1, (a), , C2, (b), , (b), , For convex lens as shown in fig.(a), , , , , The distance between optical centre of a lens and, the point where the parallel beam of light converges, or appears to converge. i.e., second principal focal, point ( F2 ), is called focal length f., To a lens, if the media on the two sides is same,, then first principal focal distance is equal to second, , R1 OC1 is +ve; R 2 OC2 is -ve, f OF is +ve, For concave lens as shown in fig. (b), , R1 OC1 is -ve; R 2 OC2 is +ve, , pricipal focal distance. i.e., f1 f 2 ., Note: We are mainly concerned with the second focus, F2 because wherever we write the focal length ‘f’’ , measures second principal focal length., , , Focal plane: It is a plane passing through the, , , , Aperture: In referance to a lens, aperture is the, , principal focus and perpendicular to the principal , axis., , f OF is -ve, , Rules for image formation: In order to, locate the image and its nature by a lens graphically, following rules are adopted., A ray parallel to the principal axis after refraction, passes through the principal focus for convex lens, and appears to diverge from focus for concave lens., , effective diameter of its light transmitting area. The, intensity of image formed by a lens depends on, square of aperture. i.e., I apeture , , , 2, , 152, , F2, , Sign convention:Whenever and where ever, possible, rays of light are taken to travel from left, to right., , , , F2, , Transverse distance measured above the principal, axis are taken to be positive while those below it , negative., , (a), , (b), , A ray passing through the first focus F1 becomes, parallel to the principal axis after refraction., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , F1, , dv du , dv du, 2 0 or 2 2 0, 2, v u , v, u, , F1, , 2, , Therefore, m L , (a), , , , (b), , dv v 2 v , 2 m2, du u, u, , Magnification, , So, longitudinal magnification is proportional to the, square of the lateral magnification., , Lateral magnification: Magnification , , Angular magnification of lens:The ratio of, , produced by a lens is defined as the ratio of the, size of image to that of the object. Here the sizes, being measured perpendicular to principal axis., , the slopes of emergent ray and corresponding, incident ray with principal axis is called the angular, magnification., , h, II1, v, 1, 1, OO h 0 u, , mt , , 1, , 2, , O, , I, , 1, , O, h0, , , , I, h1, , , , O, , I, u, , tan 2, Angular magnification tan , 1, , 1, , v, , When we apply the sign convention, for erect (and, virtual) image formed by a convex or concave lens, ‘m’ is positive, while for an inverted (and real) image,, m is negative., , Note: When several lenses are used co-axially, the total, magnification, , m m1 m2 .... mn ., , From the ray diagrams it is clear that, , Note: Linear magnification for a lens can also be, expressed as, , , form a real image for a real object when the object, is placed beyond focus., , I v f v, f, m , , O u, f, f u, , , , , , Longitudinal magnification:, , Longitudinal magnification is defined as the ratio of, infinitesimal axial length (dv) in the region of the image , to the corresponding length (du) in the region of the, object., , , NARAYANAGROUP, , When the object comes with in the focus, then a, virtual image is formed for the real object., The real image formed is always inverted while, virtual image is always erect., , Regarding concave lens: A concave lens, always form virtual image for a real object., , dv, du, , , , The image formed by a concave lens is always erect, and diminished in size., , 1 1 1, , v u f, , , , A concave lens can form a real image as well as, virtual image if the object is virtual., , Longitudinal magnification mL , , On differentiating equation, , Regarding convex lens: A convex lens will, , 153
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , Lens formula : Lens formula is a relation, connecting focal length of the lens with the object, distance and image distance. The formula is, 1 1 1, , f u, , Note: The above formula is valid for convex as well as, concave lenses and it is independent of nature of, the image (real or virtual), , Position of, the object, , Image, details, , Ray diagram, , P, , At Infinity, I, , F, , Virtual,, erect,, diminished, at F, , Note: To solve the problems, the above equation can, also be expressed as follows, v, , , , uf, vf, vu, ,u , ,f , uf, f v, uv, , a) Convexlens & b) Concave lens, , Position, of the, object, , Ray diagram, , F2, , At Infinity, , F1, , Infront of, mirror, , P, O F I, , Virtual,, erect,, diminished, between, F and P, , Image details, , Real inverted,, diminished at F, , I, Between, and 2F O 2F F, , Real, inverted,, F I 2F diminished, between F, and 2F, , O, At 2F, , F 2F Real, inverted,, I, equal, at 2F, , 2F, , Between, 2F and F 2F O F, , Real, inverted,, F 2F I enlarged,, between 2F, and infinity, , O, F 2F, , At F, F, , Between, F and P, , 154, , P, I FO, , F, , Real, inverted,, enlarged, at infinity, Virtual, erect,, enlarged and, on the side, of the object, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , Lens Maker’s formula and Lens formula :, , , In case of image formation by a lens, the incident, ray is refracted at first and second surface, respectively. The image formed by the first surface, acts as object for the second., Consider an object O is placed at a distance u from, a convex lens as shown in figure. Let its image is I1, after refraction through first surface. So from the, formula for refraction at curved surface., R2, R1, u2, , u1, , F, , u1, , F, , O, u, , 2 1 2 1, , v u, R, For first surface, 2 1 2 1, , v1 u, R1, , I2, , v, , I1, , v1, , Which is known as Lens-maker’s formula and, 1 1 1, For a lens it becomes which is known, v u f, as the “lens - formula” or “Gauss’s formula” for a, lens., Though we derived it for a real image formed by a, convex lens, the formula is valid for both convex as, well as concave lens and for both real and virtual, images., Note:The lens maker’s formula is applicable for thin, lenses only and the value of R1 and R 2 are to be, put in accordance with tthe Cartesian sign, convention., 1, 1, , and, Graphs:, v, u, 1, 1, in, Convex lens: The graph between and, v, u, case convex lens is as shown in figure., , 1, f, , .... (1), , The image I1 is acts as object to second surface,, and form final image I2, For second surface, 1 2 1 2, , , .... (2), v v1, R2, So adding (1) and (2) equation, we have, , o, , 1, 1 1, 1 , r 1 , , v u, R1 R 2 , , 2, L, with r or , , 1, M, If object is at infinity, image will be formed at the, focus, i.e. for u , v f , so that above equation, , NARAYANAGROUP, , C, , 450, , 450 1, u, B, , 1, v, , 450, 1, f, , 1, u, , For real image:, , 1 , 1 1 2 1, or v u 1 R R , , 1 1, 2 , , 1, 1, 1 , becomes f r 1 R R , 1, 2 , , A, , 1, f, , 1, 1 , 1 1, 1 2 1 , , v u, R1 R 2 , , , , 1, v, , 1, 1, 1, 1, 1 1, , ;, , v u f, v, u f, It is a straight line with slope - 1, for virtual image, 1, 1, 1, 1 1 1, , ;, , , v, , u, f, , v u f, It is a straight line with slope +1 Hence AB line, when the image is real. BC line when the image is, virtual., 1, 1, Concave lens:The graph between and in, v, u, case of concave lens as shown in figure., Since concave lens only from virtual image., 1, 1, 1, 1 1 1, , ;, , v u, f, v u f, It is a straight line with slope +1., 155
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , Determination of the Focal length of a, convex lens (or) Size of the object by, Convex lens: The graph between v and u is, “LENS DISPLACEMENT METHOD”., hyperbola to convex lens as shown in figure., , U and V Graph, , , v, 2f, f, o, , v, Reailimage, u, f 2f, Virtual image, , 2f, f, o, , Fig. (a), , , , Fixed, Screen, , Fixed, Object, , f, , 2f, , u, , If the distance ‘d’ between an object and screen, is greater than four times the focal length of a, convex lens, then there are two positions of lens, between the object and screen. This method is, called displacement method and is used in the, laboratory to determine the focal length of, convex lens., If the object is at a distance u from the lens, the, , Fig. (b), , Concave lens: The graph between v and u is, hyperbola to concave lens as shown in figure., In case of thin convex lens if an object is placed at, a distance x1 from first focus and its image is formed, at a distance x 2 from the second focus., , distance of image from the lens v d u , so, , From properties of triangles,to the left of the lens, h1 h 2, , x1 f1 To the right of the lens, h1 h 2, x1 x 2, , , From, above, two, equations, f2 x 2, f1 f 2, , x1x 2 f1f 2, , For f1 f 2, , , , , x1 x 2 f is called Newton’s formula or lens user, formula. This relation can also prove by using lens, formula., Lenses with Different Media on either, Consider a lens made of a material with refractive, index with a liquid a on the left and a liquid b ., , 1 1 1, , v u f, , 1, 1 1, , du u f, , i.e., u 2 du df 0, , So that u , , 2, , side, , from lens formula, , , , 2, Now there are three possibilities., If d 4f , u will be imaginary, so physically no, position of lens is possible., d, If d 4f , in this u 2f so only one position, 2, is possible and in this v 2f . That is why the, minimum separation between the real object and, real image is 4f., , If d 4f , u1 , , I, a b, , The governing equation for this system is, b a a b , , , , v, u, R1, R2, 156, , d d d 4f , 2, , and, , d d d 4f , , for these two positions of, 2, the object real images are formed for, u2 , , O, , d d d 4f , , u u1 , v d u 1 , , d d d 4f , , For u u 2 , v 2 d u 2 , , 2, , u2, , d d d 4f , 2, , u1, , i.e., for two positions of the lens object and image, distances are interchangeble as shown in the figure., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, n = 1.5, n = 1.2, , R = 14 cm, , 1 , 1, , , So the magnification for the both positions of the Sol: PT 1.5 1 14 0 1.2 1 0 14 , 1, 0.5 0.2, 1, object are related as m1 m, PT , , , 2, 14, 14, 20, i.e., m1.m2 1, , m1m 2 , , I1 I 2 I1I 2, . , 1, O O O2, , f 20 cm, , ;, , 1, 1, 1, , , v 40 20, , Therefore O I1I2 where I1 & I2 are the sizes, 1, 1, 1, 1, , , , ; v 40 cm, of images for two positions of the object and O is, v 20 40 40, size of the object., W.E-71: An object is 5 m to the left of a flat screen., It means that the size of object is equal to the, A converging lens for which the focal length, geometric mean of the two images. This method of, is 0.8 m is placed between object and screen., measuring the size of the object is useful when the, (a) Show that for two positions of lens form, object inaccessible., images on the screen and determine how far, If ‘x’ is the distance between the two positions of, these positions are from the object? (b) How, x, do the two images differ from each other?, the lens. Then f m m, screen, 1, 2, f = 0.8m, B, W.E-69: A biconvex lens of focal length 15 cm is, in front of a plane mirror. The distance, A, between the lens and the mirror is 10 cm. A, A, small object is kept at a distance of 30 cm from Sol:, B, the lens. The final image is, (5-u), 1, , 1, , u, , 1 1 1, , v u f, 1, 1, 1, 1, 1, , or, 1.25, We have,, 5 u u 0.8 5 u u, , (a) Using the lens formula,, , Sol:, , O, , I, , u 5 u 1.25u 5 u , 6 cm, , 10 cm, , From figure, the image is real and at a distance, of 16 cm from the mirror, W.E-70: A bi-convex lens is formed with two thin, plano-convex lenses as shown in the figure., Refractive index ‘n’ of the first lens is 1.5 and, that of the second lens is 1.2. Both the curved, surface are of the same radius of curvature, R=14 cm. For this bi-convex lens, for an, object distance of 40 cm, the image distance, will be, NARAYANAGROUP, , or 1.25u 2 6.25u 5 0; u =4m and 1m, Both the values are real, which means there exist, two positions of lens that form images of object, on the screen., 5 4, v, (b) m , ; m1 4 0.25 and, , u, m2 , , 5 1 4.00, 1, , Hence, both the images are real and inverted,, the first has magnification -0.25 and the second, -4.00., 157
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-72: A point object is placed at a distance of, , In the diagram if fl and f m are respective the focal, lengths of lens and mirror. Then, 1 1 1 1 2 1, , , F f l f m fl, fl fm, In terms of focal powers of lens and mirror, P Pl Pm Pl 2 Pl Pm, , 12 cm on the axis of a convex lens of focal, length 10 cm. On the other side of the lens, a, convex mirror is placed at a distance of 10, cm from the lens such that the image formed, by the combination coincides with the object, itself. What is the focal length of convex, mirror?, 1, 1, 1, , Sol: For convex lens, , v 12 10, i.e., v 60 cm ; i.e., in the absence of convex, , 1 1 , 2, with Pl 1 R R and Pm R, 2, 1, 2 , , mirror, convex lens will form the image I1 at a , distance of 60 cm behind the lens. Since, the mirror, is at a distance of 10 cm from the lens, I1 will be at , a distance of 60 10 50 cm from the mirror, i.e.,, , MI1 50 cm, , I2, O, , 10 cm, , When the plane surface of plano convex, lenss is silvered., , I1, , M, , L, , Here Pl and Pm are substitued with sign., The system will behave as, a concave mirror if ‘P’ is positive and, The system will behave as, a convex mirror if “P” is negative., The replacement with the mirror is due to overall, reflection of given rays., , , , 50 cm, , , , , , 12 cm, 60 cm, , Now as the final image I2 is formed at the object O, itself, the rays after reflection from the mirror, retraces its path, i.e., rays on the mirror are incident, normally, i.e., I1 is the centre of the mirror, so that, and, , R MI1 50 cm, , hence, , F R / 2 50 / 2 25cm, , Lens with one Silvered surface, , , When the back surface of a convex lens is, , silvered., The rays are first refracted by lens, then refracted, from the silvered surface and finally refracted by, lens, so that we get two refractions and one, reflection., R1, , R2, , R1, =, , O, , I2, , 158, , I1, , R1, , I1, , +, , I2, , I3, , P 2 Pl Pm, , 2 1, 1 , P 2. , 0 , R, R , Since 1 , ‘P’ is positive, the system behaves as, R, a concave mirror with focal length 2 1, When curved surface of a plano convex lens, is silvered., , , , , , , , R2, , O, , R2, , Then, the focal power of the given lens is ( Pm 0 ), , R2, , +, , Then, the focal power of the given lens is, , 2 1 2 2 , , R, R R, Since ‘P’ is positive, the system behaves a, R, concave mirror with focal length, 2, P 2 Pl Pm , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-73: A pin is placed 10 cm in front of a convex W.E-74: A biconvex thin lens is prepared from, glass of refractive index 3/2. The two boundlens of focal length 20cm made of material, ing surfaces have equal radii of 25 cm each., having refractive index 1.5. The surface of the, One of the surfaces is silvered from outside to, lens farther away from the pin is silvered and, make it reflecting. Where should an object be, has a radius of curvature 22cm. Determine, placed before this lens so that the image cointhe position of the final image. Is the image, cides with the object., real or virtual?, Sol: As radius of curvature of silvered surface is 22 cm, Sol:Here, R1 25cm, R2 25cm and 3 / 2, so, Image coincides wit h object , hence, u v x (say), R 22, fM , 11cm and hence,, 1, 2, 1, 2, 3 2, 2, 2, , , 2 1 , F FL FC, 2 25 25, 1, 1, 1, PM , , , D, 1, 4, 1 1 1 1 1 4, 0.11 0.11, fM, , ,by using ; , F 25, v u F, x x 25, Further as the focal length of lens is 20cm,, i.e., 0.20 m, its power will be given by:, x 12.5cm, 1, 1, PL , D, Hence, the object should be placed at a distance, f L 0.20, 12.5cm in front of the silvered lens., Lens maker’s formula-Special Cases, It relates the focal length of the lens to the re, fractive index of material of the lens and the radii of, curvature of the two surfaces., I, O, , 10 cm, , The formula is, 11 cm, , 1 1 , 1 lens, , 1 , f medium R1 R2 , , where lens is the absolute refractive index of, Now as in image formation, light after passing, through the lens will be reflected back by the curved, mirror through the lens again, , P PL PM PL 2 PL PM, i.e., P , , 2, 1, 210, , , D, 0.20 0.11 11, , material of the lens, medium is the absolute refractive index of the medium in which the lens is placed., R1 and R2 are the radii of curvature of two surfaces of the lens., If the lens is placed in vacuum then, , 1 1 , 1, lens 1 , f, SO the focal length of equivalent mirror, R1 R2 , The lens makers’ formula is applicable for thin lenses, 1, 11, 110, F , m, cm, only and the value of R1 and R2 are to be put in, P, 210, 21, accordance with the Cartesian sign convention., i.e., the silvered lens behaves as a concave mirror, Note: For convex lens, of focal length (110/21) cm. So for object at a, distance 10 cm in front of it,, , 1, 1, 21, , , v 10, 110, i.e., v 11cm i.e., image will be 11 cm in, front of the silvered lens and will be real as, shown in figure., NARAYANAGROUP, , C1, R1, , O, , C2, R2, , 159
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , Convex Lens, , W.E-76: A concave lens of glass, refractive index, , For convex lens R1 is +ve and R2 is -ve so the, , 1.5, has both surface of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as lens, Sol: When glass lens is immersed in a medium, its, , 1 1 , 1, lens makers formula is f 1 R R , 1, 2 , 1, 2, For equiconvex lens 1 , f, R, Note: For concave lens, , C1, , C2, , O, , refractive index is m g ., a, m, , g , , a, , g 1.50 6, , , By lens makers’ formula, m 1.75 7 , , 1 1 1, 1, 6, 1 1, m g 1 or 1, , , f, R1 R2 f 7 R R , , 1 1 2 , 7R, or f , 3.5 R, f 7 R , 2, Hence, the given lens in medium behaves like, For concave lens R1 is -ve and R2 is +ve so, convergent lens of focal length 3.5R, the, lens, makers, formula, is WE-77: A hollow equi convex lens of glass will, be have like a glass plate, 1 1 , 1, 1 , Sol:Hollow convex lens is as shown in figure, f, R1 R2 , 1 1 , 1, 1, 2, g 1 0, For equiconcave lens 1 , f1, R1 R2 , f, R, Note: For converging meniscus, as R1 R2 , R1, , or, , R2, , 1 1 , 1, 1 , if R R , 1, 2, f, R1 R2 , Note: For diverging meniscus, , R1, , R4, R2 R3, , , 1 1 , 1, 1 , if R R , 1, 2, f, R1 R2 , Note: For plano convex lens, , f1, , f2, , Hollow glass lens, , or f1 , similarly f 2 , Therefore, a hollow equi convex lens of any mate1, 1, rial will behave like a glass plate., 1 R2 , f, R, W.E-78: The diagram shows a concavo -convex, W.E-75: What is the refractive index of material, lens. What is the condition on the refractive, of a plano-convex lens, if the radius of curvaindices so that the lens is diverging?, ture of the convex surface is 10cm and focal, length of the lens is 30cm?, u, u, Sol:According to lens-maker’s formula, 1, , 3, , 1 1 , 1, 1 , f, R1 R2 , Here f 30cm, R1 10cm and R2 , 1, 1 1, 1 , so, 30, 10 , , i.e., 3 3 1 or 4 / 3, 160, , The refractive index of the lens is 2, Sol:, , 3 1 1 2 2 3 1 2, , , , 3 2, 2, v u, 2R, R, 1 2, , 2 3 2 2 1 2 2 3, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-79: The magnification of an object placed in, , 1 1 1, 1 1 1, front of a convex lens of focal length 20cm is Sol: From we have , v u f, b a f, +2. To obtain a magnification of -2, the object, ab, will have to be moved a distance equal to, 1, or f , Sol:When magnification is +2 then the image is vira b, tual. Both the image and the object are on the same, Further AC 2 BC 2 AB 2, side of the lens., 2, or a 2 c 2 b 2 c 2 a b , u x; v 2 x; f 20, 1 1 1, we have 1 1 1 or, v u f, 2 x x 20, x 10cm ., To have a magnification of -2 the image must be, real., u y, v 2 y and f 20, , or a 2 b 2 2c 2 a 2 b 2 2ab, ab c 2, , Using, , , , 1 1 1, , or y 30cm, 2 y y 20, , C, , W.E-80: Two point sources S1 and S 2 are 24cm, apart. Where should a convex lens of focal, length 9 cm be placed in between them so that, the images of both sources are formed at same, place?, Sol: In this case one of the image will be real and, other virtual. Let us assume that image of S1 is, real and that of S 2 is virtual., , B, , A, , y x 20cm, , c2, Substituting this in Eq. (1) we get f , ab, W.E-82:Convex lens has a focal length of 10cm., a) Where should the object be placed if the, image is to be 30cm from the lens on the same, side as the object?, b) What will be the magnification?, f = +10, , f =9 cm, I, , S1, , S1, , I, , x, , S2, , y, 24 - x, , Applying, , 1 1 1, 1 1 1, For S1 : 1, y x 9, y x 9, , 1, 1, 1, , 2, y 24 x 9, Solving eqs. (1) and (2) , we get x 6cm, for S 2 : , , W.E-81: An object placed at A OA f . Here,, f is the focal length of the lens. The image is, formed at B. A perpendicular is errected at 0, and C is chosen such that BCA 900 . Let, OA a, OB b and OC c . Then the value of, f is, NARAYANAGROUP, , Sol:, , S2, , 30cm, , a) In case of magnifying lens, the lens is, convergent and the image is erect, enlarged,, virtual , between infinity and object and on the, same side of lens as shown in figure. So here, f 10cm and v 30cm and hence from, , 1 1 1, lens-formula, , v u f, 1, 1 1, , i.e u 7.5cm, 30 u 10, So the object must be placed in front of lens at a, distance of 7.5cm (which is <f) from it., , we have, , I v 30, 4, b) m , O , , u, , 7.5, , i.e, image is rect, virtual and four times the size of, object., 161
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-83: In the figure, light is incident on the thin, lens as shown. The radius of curvature for both, the surface is R. Determine the focal length, of this system., , u2, , lens. In the second case, the magnification is negative, the image is real, inverted and on the other, side of the lens as the object. If b is the object distance, then u b and v 2b . Hence, 1, 1, 1, , , b 45cm, 2b b 30, Thus the object has to be moved through a dis-, , u3, , tance of 45 15 30cm away from the lens., W.E-85:The distance between the object and the, real image formed by a convex lens is d. If, the linear magnification is m, find the focal, length of the lens in terms of d and m., Sol: The convex lens formula for a real image is, 1 1 1, i , v u f, Where no sign conventions are to be used. Multiu, u, 1 u, u f, plying we get 1 or 1 , v, f, m f, f, , Sol: For refraction at first surface,, 2 1 2 1, , , i , R, v1 , For refraction at 2nd surface, 3 2 3 2, , , ii , v2 v1, R, u3, , or m u f f or u , , u1 u2, I2, , v1, , I1, , 1 m f, m, , ii , , Multiplying (i) by v we get, v v, v, 1 or 1 m or v f 1 m iii , u f, f, Now u v d . Using (ii) and (iii) we have, , 1 m f f 1 m, d, , , Adding equations (i) and (ii) we get, m, 3 3 1, 3 R, md, , or v2 , v2, R, 3 1, which gives f 1 m 2, , , Therefore, focal length of the given lens system, W.E-86: A concave lens of focal length f forms, 3 R, an image which is n times the size of the obis , ject. What is the distance of the object from, 3, 1, the lens in terms of f and n?, W.E-84:The linear magnification of an object, 1 1 1, placed on the principal axis of a convex lens, , Sol:, The, concave, lens, formula, is, where, of focal length 30cm is found to be +2. In orv u f, der to obtain a magnification of -2, by how, no sign conventions are to be used. Thus, much distance should the object be moved?, u, u 1, u v, , Sol: In the first case, the magnification is positive, 1 or 1 n , v, f, n, f u, , which implies that the image is erect, virtual and on, the same side of the lens as the object. If a is the, 1 n , or u , object distance then u a and v 2a . From, f, n , the lens formula, we have, Power of A Lens :The power of lens is the, 1 1 1, 1, 1, 1, measure of its ability to produce convergence or, or, , , a 15cm, v u f, 2a a 30, divergence of a parallel beam of light. The power P, of a lens is defined as the tangent of the angle by, So the object is at a distance of 15cm from the, v2, , 162, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , which it converges or diverges a beam of light falling, at unit distant from the optical centre., , wave length of light it will have many focal lengths, 1, as 1, f, Hence it will form many images as there are different ' s .According to given diagram number of images formed by lens is 4., , , , h, , , h, , F, , O, , f, , I, , u3, (a), , tan , , (b), , h, ;, f, , if h 1, tan , , u2, , 1, f, , u3, , 1, f, If lens is placed in a medium other than air of, , , P, , refractive index . Then power, f, The S.I unit of power is diopter (D) and 1D 1m1, , u1, , As per definition power (P) = tan , , , , , , , , , 1, 100, i.e. P f in m f in cm D, A convex lens converge the incident rays. Due to, this reason, the power of a convex lens is taken as, positive. On the other hand, a concave lens diverge, the incident rays. Therefore its power is taken as, negative., , Some important points regarding lens:, , , , 1 1 , 1, lens 1 1, fa, R1 R2 , If fl is the focal length of lens immersed in a liquid, 1, 1 lens, 1 , 1 2 , then f , R R, l, 2 , liquid, 1, , 1 , 2, , fR, , NARAYANAGROUP, , Lens immersed in a liquid:, , f a is the focal length of a lens placed in air, then, , R, , If a lens made a number of layers of different, refractive indices as shown in figure, for a given, , u2, , is immersed in a liquid of refractive index liquid ,if, , fV, , , , u1, , If a lens made of material of refractive index lens, , white light, , Filling up of a lens:, , If a lens is made of two or more materials and are, placed side by side as shown in below, then there, will be one focal length and hence one image, , I2, , Every part of a lens forms complete image. If a, portion of lens is obstructed, full image will be, formed but the intensity will be reduced., The focal length of a lens depends on its refractive, 1, index i.e 1 , so the focal length of a given, f, lens is different for different wave lengths and, maximum for red and minimum for violet whatever, the nature of the lens as shown in figure., , v, , u4, , , , fl, lens 1, , f a lens, , 1, , , liquid, , , Depending upon the values of lens and liquid , we, have three cases, 163
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, lens liquid, , liquid, , L, , lens liquid, liquid, liquid, , lens liquid, liquid, liquid, , Sol: According to lens-maker’s formula,, , 1 1, , 1, 1 with L, M, f, R1 R2 , , liquid, , Initially,, , lens liquid, L, liquid, , liquid, , lens liquid, liquid L liquid, , lens, liquid, , , liquid, L, liquid, , c , , (b), , A, 1, 2, , ; R1 10cm, 3, G, 3, 2, , , , and R2 10cm, So, , (a), , , , Case(a): If lens liquid , then fl and f a are of same, sign and f l f a i.e. the nature of lens remains unchanged, but it’s focal length increases and hence, power of lens decreases., Case(b): If lens liquid then fl ,ie. the lens behaves as a plane glass plate and becomes invisible, in the medium., , 1 2 1, 1 , 2, 1 , , ie f 15cm, , f 3 10 10 , 30, ie. the air lens in glass behaves as divergent lens of, focal length 15cm., When the liquid of 2 is filled in the air cavity,,, , , , L, 2 4, , , M 1.5 3, , 1 4 1, 1 2, 1 , , f ' 3 10 10 30, f ' 15cm ie. the liquid lens in glass will behave as, a convergent lens of focal length 15cm., If the two radii of curvature of a thin lens are not, equal, the focal length remains unchanged, where, the light is incident on either of two surfaces., , So that now, , Case(c): If lens liquid , then fl and f a have oppo- , site sign and the nature of lens changes ie. a convex, lens diverges the light rays and concave lens converge the light rays., W.E-87: A glass convex lens of refractive index , (3/2) has a focal length equal to 0.3m. Find, the focal length of the lens if it is immersed in, water of refractive index (4/3)?, Sol: As according to lens-makers’ formula, 3, , 1, , fw, f, 2, f 1.2 m, , ; w , w, 3, fa g, 0.3 , , 1, , 2 1, w, , 4, , 3, , , g 1, , Cutting of a lens:, If an equi convex lens of focal length ‘f’ is cut, into two equal parts along its principal axis, as, shown in figure , as none of , R1 and R2 will, change, the focal length of each part will be equal, that of initial lens, but intensity of image formed, by each part will reduced to half., R1 = R2 = R, f, f, , f, , W.E-88: As shown in figure a spherical air lens, of radii R1 R2 10cm is cut in a glass, , 1.5 cylinder. Determine the focal length, and nature of air lens. If a liquid of refractive, index 2 is filled in the lens, what will happen, to its focal length and nature?, , (a), , , , (b), , If an equi convex lens of focal length ‘f’ is cut, into two equal parts transverse to principal axis,, as shown in figure, the focal length of each part, will become double that of initial value, but, intensity of image remains same., R1 = R2 = R, f, , 2f, , 2f, , , , = 1.5, = 1.5, , A), , =1, , = 1.5, =2, , = 1.5, (a), , 164, , (b), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 1 2 1, , 1, f, R, For each part of cut lens, , For original lens, , , , 1 1, , 2, f', R, From (1) and (2) we get f ' 2 f, a) On removing a part of lens with out disturbing, remaining part, the principal axis position of the, remaining part is same as earlier, but intensity of, image is reduced, A, , 1, 1, 1 1, 2, 1, , , ,, , Fnet f , Pnet , and, Fnet 2 f 2 f Fnet 2 f, f, If an equi convex lens of focal length ‘f’ is divided, into two equal parts along its principal axis as shown, in figure, the focal length of each part is f . If these, two parts are put in contact in different combinations, as shown in figure, , f (ve), , f (ve), , , , A, Principal, axis of, A and B, , C2, , C1, , C2, , B, , f (ve), , C1, , 1, 1 1, For the first combined f f f, net, , B, , f, 2, Pnet , 2, f, For the second combination as shown in figure, first, part will behave as convergent lens of focal length, f while the other divergent of same focal length, (being thinner near the axis), so in this case,, 1, 1 1, , Fnet f f ' ; Fnet , Pnet 0, , b) If given lens is cut along the principal axis and, the separation between them increase in a direction transverse to principal axis, each part has own, principal axis., , f net , , Principal axis of A, A, , A, , , B, , B, Principal axis of B, , If the equi convex lens of focal length ‘f’ is divided, into two equal parts transverse to the principal axis, as shown in figure, the focal length of each part is, 2f. If these two parts are put in contact in different, combinations as shown in figure, 2f, , 2f, , , , 2f, , (or), , NARAYANAGROUP, , 2f, , , , 2f, , (or), , , , , , f (ve), , 2f, , 1 1 1, , v u f, On differentiating above equation, we get, 1 dv 1 du, 2. 2, 0, v dt u dt, To a lens, , 2, , v, (or) Vi V0 where Vi velocity of image with, u, , respect to lens, V0 =velocity of object with respect, 2, , f , 2, to lens. ; i.e. Vi m .V0 , .V0, u f , , If an object is moved at constant speed towards a, convex lens from infinity to focus, the, image will move other side of the lens slower in the, beginning and faster later on away from the lens. If, the object moves from F to optical centre, the image moves with greater speed on the same side of, object from infinity to towards lens., 165
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-89: A point object O is placed at a distance, of 30 cm from a convex lens of focal length, 20cm cut into two halves each of which is displaced by 0.05cm as shown in figure. Find, the position of the image? If more than one, image is formed, find their number and distance between them?, , f1, , O, , f2, , I, , I1, , First lens of focal length f1 from the image I1 of, the object ‘O’ at a distance v1 from it., O, 20.05cm, 30cm, , f =20cm, , Sol: Considering each part as separate lens with, u 30cm and f 20cm , from lens formula, , 1 1 1, we have 1 1 1, v u f, v 30 20, ie. v 60cm, I1, L1, , O, , Q, , P, , L2, =30cm, , f =20v, , I2, V =60cm, , So each part will form a real image of the point, object at 60cm from the lens as shown in figure. As, there are two pieces, two images are formed. Now, in similar triangles OI1 I 2 and OL1 L2 , I1 I 2 OP u v , , , L1 L2 OQ, u, 90, 2 0.05 0.3cm, 30, So the two images formed are 0.3cm apart., , ie I1 I 2 , , , 166, , 1 1 1, 1, v1 u f1, , Now the image I1 will act as an object for second, lens and the second lens forms image I at a distance ‘v’ from it, then, 1 1 1, , 2, v v1 f 2, So adding (1) and (2) equations we have, 1 1 1 1, 1 1 1, or , v u f1 f 2, v u F, 1 1 1, so F f f, 1, 2, i.e., the combination behaves as a single lens of, equivalent focal length “F” given by, 1 1 1, , F f1 f 2, This derivation is valid for any number of thin lenses, in contact co-axially., 1 1 1 1, 1, ........., F f1 f 2 f3, fn, , In terms of power Pnet P1 P2 P3 .....Pn, Here focal length values are to be substituted with, sign., Note:If the two thin lens are separated by a distance, 1 1 1, d, , ‘d’ , then, and, F f1 f 2 f1 f 2, , Pnet P1 P2 dP1 P2, Note:If the medium on either side of the lens is air, and the medium between the lens is one having, Combination of Lenses :, refractive index , we can imagine that the rays, Consider two thin lenses kept in contact as shown, emerging from the first lens are incident on the, in figure. Let a point object ‘O’ is placed on the, second lens as if they have traversed a thickness, axis as shown in figure., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , d, in air.., , Hence, , 1, 1, 1 d / , , , F, f1 f 2, f1 f 2, , d , P p1 p2 P1 P2, , Note:If two thin lenses of equal focal length but of opposite nature are pair in contact, the resultant focal, length of t he combination will be, 1 1 1, , F f f, Note: If f1 and f 2, , , 0 i.e. F and P 0, , are focal lengths of two lenses, , L1 and, , L2 are separated by a distance ‘d’ on, common principal axis and ‘F’ is the equivalent focal length of the system., d, f1, , L1, , f2, , L2, , Then, i) The distance of equivalent lens from second lens, Fd, L2 is f towards the object if the value is posi1, tive and away from the object if the value is negative, ii) The distance of equivalent lens from the first lens, Fd, L1 is f away from the object, if the value is posi2, tive and towards the object if the value is negative., It is note that F, f1 and f 2 are to be substituted, according to sign convention., Note:A plane glass plate is constructed by combining a plano-convex lens and a plano-concave, lens of different materials as shown in figure., ( C is the refractive index of convergence lens,, , D is the refractive index of divergent lens and, R is the radius of curvature of common interface)., NARAYANAGROUP, , C, , D, , by lens maker’s formula, 1, 1 C 1, 1, C 1 , , 1, fC, R, R , 1, 1 D 1, 1, D 1 , , 2, fD, R, R , Now as the lenses are in contact,, , and, , 1, 1, 1 C D , R, , , , ;F , F fC f D, R, C D , , As C D , the system will act as lens. The system will behave as convergent lens if C D (as, its focal length will be positive) and as divergent, lens if C D (as F will be negative), Note:Two convex lenses made of materials of refractive indices 1 & 2 , they are placed as, shown in figure, the gap between them is filled, with a liquid of refractive index liquid .This combination is placed in air then, liquid, , R1, , R2, , R1, , air, , R2, , air, 1, , 2, , The system is equal to combination of three thin, lenses in contact so, 1 1, 1, 1, , , F f1 fliquid f 2, , 1 1 , 1, where f 1 1 R R , 1, 1, 2 , 1, fliquid, , 1, 1 , liquid 1 , , R2 R1 ' , 167
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-90:Two thin lenses, when in contact, produce, 1, 1, 1 , a combination of power +10+diopter. When, 2 1 , , , f2, R, ', R, ', 1, they are 0.25m part, the power reduces to +6, 2 , diopter. The focal length of the lenses are.......m, If the effective focal length F of the combination is, and ............m, +ve then the combination behaves like converging, lens, if F is -ve then the combination behaves like Sol: When the lenses in contact,, diverging lens., P P1 P2 or P1 P2 10 1, Note: If two convex lenses made of materials of reWhen lenses have d separation,, fractive indices 1 & 2 are kept in contact and, PP, the whole arrangement is placed in a liquid of reP1 P2 dP1 P2 P or P1 P2 1 2 6, 4, fractive index liquid then this is equivalent to comPP, bination of two lenses kept in contact in a medium., or 10 1 2 6 or P1P2 16 2 , 4, 1, 1, 1, From (1) and (2) . we get P1 8D, P2 2 D, In this case F f m f m , 1, 2, 1, 1, f1 0.125m, f 2 0.5m, 1, 1, 1, 1 , 8, 2, 1 , where f m , , liquid R1 R2 , 1, W.E-91:Two plano-concave lenses of glass of refractive index 1.5 have radii of curvature of, 1, 1, 1, 1 , 20 and 30 cm . They are placed in contact with, , 1 , , , , , curved surfaces towards each other and the, f 2 m liquid, R1 ' R2 ' , space between them is filled with a liquid of, Note:If parallel incident ray on first lens emerges, refractive index (4/3). Find the focal length of, parallel from the second lens, then f e , the system., Sol: As shown in figure the system is equivalent to, 1 1 1, d, , d f1 f 2, combination of three thin lenses in contact., f1 f 2 f1 f 2, 1 1 1 1, (i) If both the lenses are convex, then d f1 f 2, ie. F f f f, 1, 2, 3, , 1, , 2, , 3, , O, , f1, , But by lens-maker’s formula, , f2, , (ii) If second lens is concave, then, , d f1 f 2 f1 f 2, , 1 3 1 1 , 1, 1 , f1 2 20 , 40, 1 4 1, 1, 5, 1 , f 2 3 20 30 180, 1 3 1, 1 , 1, 1 , , , , f 3 2 30 , 60, , f2, f1, , 168, , 1, 1, 5, 1, , , F, 40 180 60, i.e., the system will behave as a divergent lens of, focal length 72cm., , So, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-92:Two thin symmetrical lenses of different W.E-93: Two thin converging lenses are placed, on a common axis, so that the centre of one of, them coincides with the focus of the other. An, object is placed at a distance twice the focal, length from the left-side lens. Where will its, image be? What is the lateral magnification?, The focal length of each lens is f., , nature and of different material have equal, radii of curvature R 15cm .The lenses are, put close together and immersed in water, 4, , w . The focal length of the system in, 3, , water is 30cm. The difference between refractive indices of the two lenses is, , Sol: Let f1 and f 2 be the focal lengths in water.., Then, , f, , I1, , Sol:, , O, , 1 1 , 2 , 1 1, 1 1, , 1 , , 1 1, f1 w, f1 w, R R , R , , 2f, , 2 , 1 2, , 1 2 , f 2 w R , Adding Eqs. (1) and (2) we get, , f, , f / 2 1, f, v, ; m2 2 , , 2, u2, f, 2, Therefore, final image (real) is formed at a distance, f/2 right side of the second lens wiht total lateral, magnification,, v , , 1 1 2 1 2 , , f1 f 2, w R, But the given system is equal to combination of, two lens kept in contact in liquid so, , 1, 1, m m1 m2 1 , 2, 2, , Refraction through Prism, , , R, 1 2 w ; substituting the values we, 60, get 1 2 , , f, , The image formed by frist lens will be at distance 2f, with lateral magnification m1 1 .For second lens, this image will behave as a virtual object. Using the, 1 1 1, 1 1 1, lens formula, we have, , v u f, v f, f, , 1 1 , 1 2, , 1 , f2 w R R , , 1 1 1, 1 2 1 2 , , or, , F f1 f 2, 30, w R, , f, , 4 1 1, , 3 60 3, , Prism is a transparent medium bounded by any, number of surfaces in such a way that the surface, on which light is incident and the surface from which, the light emerges are plane and non-parallel as, shown in figure., , W.E-93: A converging lens of focal length 5.0cm, first refracting, surface, second refracting, surface, , is placed in contact with a diverging lens of, focal length 10.0cm. Find the combined focal length of the system., Sol: Here f1 5.0cm and f 2 10.0cm, Therefore, the combined focal length F is given, 1 1 1, 1, 1, 1, by F f f 5.0 10.0 10.0, 1, 2, F 10.0cm ie. the combination behaves as, , a converging lens of focal length 10.0cm., , NARAYANAGROUP, , Incident, ray, , A, , refracting, edge, , emergent, ray, , Principal section, of prism, , The plane surface on which light is incident and, emerges are called refracting faces., 169
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , , , The angle between the faces on which light is , incident and from which it emerges is called, refracting angle or apex angle or angle of prism (A)., The two refracting surfaces meet each other in a, line called refracting edge., A section of the prism by a plane perpendicular to, the refracting edge is called principal section, , Minimum Deviation, From the equation i1 i2 A , the angle of, deviation depends upon angle of incidence i1 ., If we determine experimentally, the angle of deviation corresponding to different angles of incidence, and then plot a graph by taking angle of incidence, (i) on x-axis , angle of deviation on y-axis, we, get the curve as shown in figure., Y, , , , , min, A, , O, , (a), c, , , , Angle of deviation means the angle between, emergent and incident rays. While measuring the, deviation value in anticlock wise direction is taken, as positive and clock wise direction is negative., , i1, , i1, , X, , i2, , i1, , sin i, , s in i, , sin i, , 1, 2, by snell’s law sin r sin r sin r, 1, 2, , A min , sin , , 2, , , , A, sin, 2, , p, , m, , A min , sin , , 2 , , A, sin, 2, , Note:Deviaiton produced by small angled prism for, small angle, from equation above, i1 i2, = deviation angle, , , ; i1 r1 , i2 r2 But i i A, Note:If refractive index of the material of the prism, 1, 2, r1 r2, is equal to that of sorroundings, no refraction at, its surfaces will takes place and light will pass r1 r2 A; r1 r2 A But r1 r2 A, For a prism immersed in a medium of refractive, through it undeviated ie. 0 ., index m, vii) Generally we use equilateral or right angled or, Isosceles prism., , , 1 A p 1 A, Determination of Refrative index of ma m , terial of the prism for minimum deviation, Note:There are two values of angle of incidence for, X, same angle of deviation:, M, , N1, i1, Q, P, , r1, , r2, , R, , i1, , Z, , S, , Q, P, , r1, , r2, , i2, R, , N, Y, , 170, , , , c, , i2, , N, Y, , X, , N2, , , , Z, , S, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , When a light ray is incident at an angle i1 at the Note: Since the law of reversibility always true, then for, an angle of incidence i sin 1 sin A C ,, surface (xy), it emerges at an angle i2 from the, the ray grazes at the other surface., surface (zx) with a deviation angle .As the path, of light is reversible, therefore if angle of incidence Condition of grazing emergence: If a ray, can emerge out of a prism, the value of angle of, is i2 , at the face (xy), then the angle of emergence, incidence i1 for which angle of emergence, will be i1 , with the same angle of deviation , i2 900 is called condition of grazing emergence., Note:, In this situation as the ray emerges out of face, i) For a given material of prism, wave length of light, XZ, i.e., TIR does not take place at it., and angle of incidence. When the angle of prism, increases angle of deviation also increases as, A., ii) With increase in wavelength , deviation decreases, ie. deviation for red is least while maximum for vio 1 , let as 1 , , iii) When a given prism is immersed in liquid, the, angle of deviation changes as r 1, , , Maximum deviation:, Deviation of ray will be maximum when the, angle of incidence is maximum i.e i 900 ., Therefore the maximum deviation, max 90 i2 A, , Q, , i2, c, , But as in a prism r1 r2 A; r1 A r2, So r1 A C ie. r1 A C 2 , Now from snell’s law at face XY, we have, 1sin i1 sin r1, But inview of equation (2), sin i1, sin A C , sin r1 sin A C ;, , , , A, 900, , r2 C 1, , r2, , To find the angle of emergence in this case let us, apply Snell’s law at second surface., a sin r2 1, , , , sin i 2 , As i1 900 , r1 C, Also r1 r2 A, C r2 A, So, r2 A C, , sin A C 1sin i2, 1, , i2 sin sin A C , Maximum deviation is, , sin i1 sin A C , ie. sin i1 sin A cos C cos A sin C , , ie sin i1 sin A , , i.e sin i1 , , , , , 2, , 1 sin, , 2, , C cos A sin C , , , 1 sin A cos A, , , 1, (as sin C ), , 1, or i1 sin , , 1, or i1 min sin , , , , , , 2, , 1 sin A cos A, , , , , 2 1 sin A cos A 3 , , i.e light will emerge out of prism only if angle of, , max 90 i2 A, , incidence is greater than i1 min given by Eq. (3), . In this situation deviation will be given by, , max 900 sin 1 sin A C A, , i1 900 A with i1 given by Eq.(3), , 0, , NARAYANAGROUP, , 171
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , , Condition of no emergence:, The light will not emerge out of a prism for a, values of anlge of incidence if at face AB for, , Note:Normal incidence- grazing emergence: If the incident ray falls normally on the prism and grazes from, the second surface, then, , i1 max 900 at face AC, , r2 C 1, C, , A, P, i1, , r1, , r2, , , R, i2, , a) i1 r1 0, i2 900 and r2 C A, , Q, , 1 1 , b) A C sin c)Deviation d 90 C, , Now for Snell’s law at face AB, we have, 0, Note:Grazing, incidence, - grazing emergence: If the, 1 sin 90 sin r1, incident ray falls on the prism with grazing incidence and grazes from the second surface, then, 1 1 , ie. r1 sin ; or r1 C 2 , , From eq. (1) and (2) ; r1 r2 2C 3, , , B, , C, , C, , C, , However in prism ; r1 r2 A 4 , So from eq. (3) and (4) ; or A 2C 5 , A, A, A 1, C or sin sin C sin , 2, 2, 2 , , (i) i1 i2 900 (ii) r1 r2 C, , (iii)Angle of prism A 2 C, , A , i.e. cos ec 6 , (iv) Deviation d 180 2C 180 A, 2 , , i.e., A ray of light will not emerge out of a prism W.E-94: An equilateral glass prism is made of a, (what ever be the angle of incidence) if, material of refractive index 1.5. Find its angle, of minimum deviation., A, cos ec Sol: A 600 , 1.5, ?, A 2 C ,, i.e, if, min, 2, A min , (or) cot 2 A / 2 1, sin , , 2 , , Note:Limiting Angle : In order to have an emergent, , Substituting, A, ray, the maximum angle of the prism is 2C ,, sin , 2, where C is the critical angle of the prism w.r.t, 600 min , the surrounding medium 2C is called the limitsin , , 2, , , ing angle of the prism., 600 min , 1.5 , 0, 0, 1.5sin, 30, , sin, Note: If the angle of incidence at first surface i is, , , , 60 , 2, , , such that, sin, , 2 , , a) If i sin 1 sin A C , the ray grazes at the, 60 0 min , other surface., sin , 1.5 0.5000 0.75, 2, 1, , , b) If i sin sin A C , then the ray emerges, 0, 60 min, out of a prism from the other surface., 48035', 1, 2, c) If i sin sin A C , the ray under go TIR, 0, 60 min 97010 ' min 37 010 ', at the other surface., 172, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-95:A prism of refracting anlge 40 is made of W.E-98: A ray of light passing through a prism, a material of refractive index 1.652, Find its, angle of minimum deviation., Sol: A 400 , 1.652, ?, Substituting in A 1 40 1.652 1, , having 2 suffers minimum deviation. It, is found that the angle of incidence is double, the angle of refraction within the prism. What, is the angle of prism., Sol: As the prism is in ‘the position of minimum devia-, , 40 0.652 2.6080, W.E-96: A ray of light is incident normally on one, of the faces of a prism of apex angles 300 and, , tion m 2i A with r A / 2, According, to, given, , refractive index 2 . The angle of deviation, of the ray is......degree, Sol: Apply Snell’s law of refraction at P, , m 2 A A A and hence from, , A, 300, 60, , 300, N, B, , N1, P, , r, , , C, , sin 300, 1, , sin r, 2, or sin r 2 , , i 2r A as r A / 2, , , , sin A / 2], sin A / 2 , , 2 sin, 0, , 1, 1, , sin 450, 2, 2, , or r 450, r 300 450 300 150, Deviation of ray =150, , r1 r2 A when light undergoes refraction, through a prism., 0 r2 A; r2 A, When the emergent light grazes the second surface , r2 becomes the critical angle C , 1, 1, , i.e. C A and , sin C sin A, , NARAYANAGROUP, , i.e 2 , , sin A, or , sin A / 2, , A 1, A, A, A, or , 2sin cos i.e, cos , 2, 2, 2, 2, 2, , A, 1 , cos 1 450 ie. A 900, 2, 2, W.E-99: A ray of light is incident at an angle of, 600 on one face of prism of angle 300 . The, ray emerging out of the prism makes an angle, of 300 with the incident ray. Show that the, emergent ray is perpendicular to the face, through which it emerges and calculate the, refractive index of the material of the prism., Sol: According to given problem,, A 300 , i 600 and 300, and as in prism ,, A, , W.E-97: A ray of light is incident normally on, one of the refracting surfaces of a prism of, refracting angle A,. The emergent ray grazes, the other refracting surface. Find the refractive index of the material of prism., Sol: For normal incidence on one of the refracting, faces of the prism, i1 0 and r1 0 . But, , problem,, , 0, , 30, , 300, i2 00, , 0, , i2 60, , 0, , r2 0, , B, , C, , i1 i2 A ; 300 600 i2 300 ; i.e i2 00, So the emergent ray is perpendicular to the face, from which it emerges., Now, as, i2 0, r2 0;, But, as, r1 r2 A, r1 A 300, , So at first face 1sin 600 sin 300 ie 3, , 173
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-100:A ray of light undergoes deviation of 300, when incident on an equilateral prism of refractive index 2 . What is the angle subtended, by the ray inside the prism with the base of, the prism?, Sol: Here 300 and A 600 . So if the prism had, been in minimum deviation., , , , sin 300 600 / 2 , sin 600 / 2 , , , , sin 450, 1, , 2 2, 0, sin 30, 2, , And as of the prism is given to be, , 2, , A, P, , 600, , b) For maximum deviation, i1 900 so that, 2, r1 C sin 1 420 , But as in a prism, 3, , r1 r2 A so r2 A r1 600 420 180, Now applying Snell’s law at the second face,, 3, sin r2 sin i2 , i.e., sin180 sin i2, 2, ie. i2 sin 1 1.5 0.31 sin 1 0.465 280, W.E-102: Monochromatic light falls on a right, angled prism at an angle of incidence 450 ., The emergent light is found to slide along the, face AC. Find the refractive index of material of prism., , R, i1, , i1, , , , A, , i2, , i2, , 90, 45, , Q, 0, , 60, , 0, , 60, , B, , C, B, C, The prism position is in the minimum diviation Sol: Since the emergent light slides along the face, AC , angle of emergence is 900 , as shown . It, position implies that, implies that angle of incidence ray of the ray, 0, A 60 , 0, that falls on face AC is equal to the critical angle, r1 r2 r , 30, 2 2 , C r2 C 1, Therefore, angle subtended by the ray inside the, From the prism theory, we know, prism, with, t he, surface, AB ,, A, 900 r 900 300 600 and as base also, 45, subtends an angle of 600 with the face AB,the, 90, r2, r1, ray inside the prism is parallel to the base, ie., the angle subtended by the ray inside the prism, C, B, with base is zero., 0, 0, r1 r2 A 90 r2 90 r1 2 , W.E-101: A 600 prism has a refractive index of, From the equations (1) and (2) 900 r1 C, 1.5. Calculate (a) the angle of incidence for, minimum deviation, (b) the angle of emer sin 900 r1 sin C or cos r1 sin C, gence of light at maximum deviation., Sol: (a) As the prism is in the position of minimum of, 1, 1, sin C cos r1 , 0, 0, But, deviation, r A / 2 60 / 2 30 , so that at, , , Applying Snell’s law at the boundary, either, face sin i 1.5sin 30 0 0.75, or, 1, 0, i sin 1 0.75 490, AB, 1sin 45 sin r1 1 2, , Note: In this situation angle of emergence is equal to, , angle of incidence = 490 and deviation, m 2i A 2 49 60 380, 174, , , , 1, 2 1 or 2 3 / 2 1.5 1.5, 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-103: The refractive index of a prism is 2. This, prism can have what maximum refracting, angle?, Sol: Critical angle, , or, , 3, , sin i1, 3, sin 600, 0 or sin i1 , sin 30, 2, , i1 600, ii) Rotation of DCE about C for final emergent, ray to have minimum deviation:, , 1, 1, C sin 1 sin 1 300, 2, , , BD, , If A 2 C the ray does not emerge from the prism., , E, , So, maximum refracting angle can be 600 ., , W.E-104: For an equilateral prism, it is observed, that when a ray strikes grazingly at one face, it emerges grazingly at the other. Its refractive index will be, 0, Sol: i1 i2 90 , r1 r2 , , , , A, 300, 2, , sin i1, 2, sin r1, , , , W.E-105: Two identical prisms of refractive index 3 are kept as shown in figure. A light, ray strikes the first prism at face AB. Find,, , B, , D, , 60, , 60, , E, , C, , i) The angle of incidence, so that the emergent, ray from the first prism has minimum deviation, ii) Through what angle of prism DCE should be, rotated about C so that the final emergent ray, also has minimum deviation., , B, , D, 60, , 60, Sol: 60, , A, , 30, 60, , 60, C, , i) At minimum deviation r1 r2 300, sin i1, For Snell’s law, sin r, 1, NARAYANAGROUP, , C, The figure displays the position in which net deviation suffered by the ray of light is minimum. This is, achieved when the second prism is rotated, anticlockwise by 600 about C., Dispersion by A Prism :, When white light passes through a prism it splits, up into different component colours. This phenomenon is called dispersion and arises due to, the fact that refractive index of prism is different for different wave lengths. So different wave, lengths in passsing through a prism are deviated, through different angles and as 1 , violet is deviated most while red is least deviated, giving rise to display of colours known as spectrum. The spectrum consists of visible and invisible regions., , 60, , 60, , A, , A, , E, , Whitelight, , , , A, , R, , r, , Screen, R, O, GY, IB, V, , In visibile spectrum the deviation and the refractive index for the yellow ray are taken as the, mean values. If the dispersion in a medium takes, place in the order given by “VIBGYOR” it is, called normal dispersion. If however, the dispersion does not follow the rule “VIBGYOR”,, it is said to be anamalous dispersion. A medium, which brings about dispersion is called dispersive medium. Prism that separated light accordance to wavelength are known as dispersive, prisms. Dispersive prism are mainly used in, spectrometers to separate closely adjacent spectral lines. Prisms made of glass used in the vis175
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, ible region for dispersion. Dispersion also occurs, in U.V and I.R regions, but materials used for the, dispersion are different., , , Angular dispersion, The difference in the angles of deviations of any, pair of colours is called angular dispersion for, those two colours. If the refractive indices of violet, red and yellow are indicated by v , R and y ., The deviation y corresponding to yellow colour, is taken as mean deviation. The deviations, v , R and y can be written as, , v v 1 A, R R 1 A, and y y 1 A, Angular dispersion for violet and red, , , , v R v R A, Thus the angular dispersion depends on the nature of the material of prism and upon the angle, of the prism.. In general the angular dispersion, means we consider angular dispersion of violet, and red i.e the total angle through which the visible spectrum is spread out., Dispersive Power :, Dispersive power indicates the ability of the material of the prism to disperse the light rays. It is, the ratio of angular dispersion of two extreme, colours to their mean, deviation, Angular dispersion, , Mean deviation, v R, v R , , , 2 , But the mean colour of red and violet colours is, R, y, yellow colour, so v, 2, v R, So, , y, y, , but depends on material of prism., The dispersive power more precisely expressed, with reference to C, D and F Fraunhoffer’s lines, in the solar spectrum. The C,D and F lines lies, in the red , yellow and blue regions of the spec0, , 0, , trum and their wavelengths are 6563 A , 5893 A, 0, , and 4861 A respectively. Then the dispersive, F C, power may be expressed as 1, D, F C, 2, It is noted that a single prism produces both deviation and dispersion simultaneously. However if, two prisms (crown and flint) are combined together, we can get deviation without dispersion or dispersion without deviation. The dispersive power of flint, glass prism is greater than that of crown glass prism, for same refracting angle . i.e the angular separation of spectral colours in flint glass is more than, crown glass. If two prisms of prism angles A and, A ' and refractive indices and ' respectively, are placed together then the Total deviation, , Where D , , 1 2 y 1 A ' y 1 A ', and total dispersion, , 1 2 V R A 'v 'R A ', Deviation without Dispersion Or, achromatic Prism :, , , , where y is the deviation for yellow light, , , , d, v R, , y 1 1, , It is seen that the dispersive power is independent of the angle of prism and angle of incidence,, 176, , An achromatic prism is a combination of two, appropriate prisms so constructed that it shows, no colours. Flint glasses have higher dispersive, power than crown glass. Hence, it is possible to, combine two prisms of different materials and, specified angles such that ray of white light may, pass through the combination without dispersion,, though it may suffer deviation. Such a combination is called achromatic combination., i.e 0 and 0, , v R A 'v 'R A' 0, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , v R A, , , , 1, , y, , , , y, , 1, , 'v 'R A, , , ' 1, , GEOMETRIC OPTICS, In this case as the dispersion produced by flint, glass prism is opposite to crown glass prism., Therefore the net angular dispersion, C f, , ' 1 0, y, , y, , i.e C C f f 0, In this case as the deviation produced by flint, prism is opposite to crown prism. Therefore the, C f, net deviation, , V R A 'V 'R A ' or , , , V R A, , , , y 1 A ' y 1 A ', , , , , , , , y, , 'v ' R , , 1 , , 'V 'R A ', , ' 1, , ' 1, y, , y, , V R A, , W.E-106:A beam of white light passing through a, , 'v 'R A ', , hollow prism gives no spectrum why?, Sol: Light travels from air to air in case of hollow, prism. No refraction and no dispersion occur., , V R , y, , y, , C C f f, , 1, , ' 1, , y 1, , , , C f, , C f, , , , Dispersion without Deviation OR Direct, Vision Prism :, , air, , air, , air, , If the angles of crown and flint glass prism are, so adjusted that the deviation produced for the, mean rays by the first prism is equal and oppoThe glass slabs forming the prism are very thin, site to that produced by the second prism, then, and permit the rays to pass undeviated., the final beam will be parallel to the incident, beam. Such combination of two prism will proHence a hollow prism gives no spectrum., duce dispersion of the incident beam without W.E-107:White light is passed through a prism, deviation., of angle 50 . If the refractive indices for red, and blue colours are 1.641 and 1.659 respec=0, tively. Calculate the angle of dispersion beR, tween them., A, Y, , V, , Whitelight, , A, i.e 0 and 0, y 1 A ' y 1 A ' 0, , , , y, , 1, , v R , , v R A , , ' 1 ' ' A ' 0, ' ' , y, , v, , v, , R, , C f, ie. 0, C, f, NARAYANAGROUP, , R, , Sol: As for small angle of prism 1 A, , b 1.659 1 50 3.2950 and, r 1.641 1 50 3.2050 so, b r 3.2950 3.2050 0.090 0, , W.E-108:The refractive indices of flints glass, prism for C,D and F lines are 1.790, 1.795, and 1.805 respectively. Find the dispersive, power of the flint glass prism., Sol: C 1.790, v 1.795 and F 1.805, , , , F C 1.805 1.790 0.015, , , 0.1887, v 1, 1.795 1, 0.795, 177
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 1) The greater deviation, 2) no deviation, from glass of refractive index 1.54 is combined, 3) same deviation as before, with another prism P2 made from glass of re4) total internal reflection, fractive index 1.72 to produce dispersion with- Sol: Figure (a) is part of an equilateral prism of figout deviation. What is the angle of the prism, ure, b) as shown in figure which is a magnified imP2 ?, age of figure, Sol: In case of thin prism 1 A , when two, c) Therefore, the ray will suffer the same deviation, in figure(a) and figure (c), prisms are combined together., , W.E-109: A thin prism P1 with angle 40 and made, , 1 2 1 A ' 1 A ', For producing dispersion without deviation, , 0 , ie. ' 1 A ' 1 A or, 1.54 1 0, A' , 4 30, 1.72 1, So the angle of the other prism is 30 and opposite to the first., W.E-110: A crown glas prism of refracting angle, 80 is combined with a flint glass prism to obtain deviation without dispersion. If the refractive indices for red and violet rays for, crown glass are 1.514 and 1.524 and for the, flint glass are 1.645 and 1.665 respectively, find, the angle of flint glass prism and net deviation., Sol: The condition for deviation without dispersion, , is v R A 'v 'R A ', , 1.524 1.514 80 0.080, , A' , , 0.02, 1.665 1.645, For crown glass , For flint glass , , 40, , 1.514 1.524, 1.519, 2, , 1.645 1.665, 1.655, 2, , The net deviation ' 1 A ' 1 A ', , = 0.159 80 0.655 40 1.530, W.E-111: A given ray of light suffers minimum, deviation in an equilateral prism P. Addtional, prism Q and R of identical shape and of the, same material as P are now added as shown in, figure. The ray will suffer, Q, , P, , 178, , R, , Q, P, , R, , , , , , (a), , (b), , (c ), , W.E-112: Calculate (a) the refracting angle of a, flint glass prism which should be combined, with a crown glass prism of refracting angle, 60 so that the combination may not have deviation for D line and (b) the angular, seperation between C and F lines,given that, the refractive indices of the materials are as, follows:, , C, , D, , F, , Flint, 1.790 1.795 1.805, Crown, 1.527 1.530 1.535, Sol: Let A1 and A2 be the refracting angles of the, flint and crown glass prisms respectively. 1and 2, be th refractive indices for the D line of flint and, crown glasses respectively. (a) If 1 and 2 be the, angles of deviations due to the flint and crown glass, prisms respectively, then for no deviation of D line, , 1 2 0; A1 1 1 A2 2 1 0, 1, A1, 2, , A2, 1 1 , , The negative sign indicates that A1 and A2 are oppositely directed., A1 1.530 1 , 0.530, 0, , 40, ; A1 6 , 0, 6, 0.795, 1.795 1 , , b) Angular dispersion due to the flint glass prism, A1 F C 40 1.805 1.790 0.060, Angular dispersion due to the crown glass prism, , A2 F C 60 1.535 1.527 0.048, Net angular dispersion = 0.048 0.060 0.012, The negative sign indicated that the resultant dispersion, is in the direction of the deviation produced by the, flint prism., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , , , , Optical Instruments : Optical instruments are, used primarily to assist the eye in viewing the object., Optical instruments are classified into three groups,, they are, a) visual instruments, Ex: microscope, telescope, b) photographing and projecting instruments, Ex: cameras, c) analysing and measuring instruments, Ex: spectrometer, Optical instruments such as telescope and microscopes have one object lens and one eye lens., The lens towards the object is called objective, and the lens towards eye is called eye piece., Single lens forms images with defects (aberrations). If the eye is placed near to the eye lens it, will not recieve marginal rays of the eye lens., This reduces the field of view and the intensity, is not uniform in the field of view, the central, part being brighter than the marginal part., So in designing telescopes and microscopes for, practical purposes, combination of lenses are, used for both objective and eye lenses to minimize aberrations. A combination of lenses used, as an eye lens is known as eyepiece. In any eyepiece that lens near to the objective is called, field lens and the lens near to the eye is called, eye lens. The field lens increase the field of view, and the eye lens acts as a magnifier. We consider two eyepieces namely, Ramsden’s eyepiece and Huygen’s eyepiece., The Eye:The light enters the eye through a, curved front surface, called cornea and passes, through the pupil which is the central hole in the, iris. The size of pupil can change under control, muscles. The cornea-lens-fluid syst em, isequivalent to single converging lens., The light focused by the lens on retina which is, a film of nerve fibres. The retina contains rods, and cones which sense the light intensity and, colour respectively. The retina transmit electrical signals to the brain through optic nerve., The shape (curvature) and focal length of the, eye lens may be adjusted by the ciliary muscles., The image formed by this eye lens is real, inverted and diminished at the retina., The size of the image on the retina is roughly, proportional to the angle subtended by the object on the eye. This angle is known as the vi- , sual angle.Therefore it is known as the angular, size., , NARAYANAGROUP, , GEOMETRIC OPTICS, h, , , , , h, , X, , , , I, , , , , Y, , , , , , I, , When the object is distant, its visual angle and, hence image at retina is small and object looks, smaller., When the object is brought near to the eye its visual, angle and hence size of image will increase and, object looks larger as shown in figure (b), Optical instruments are used to increase this visual, angle artificially in order to improve the clarity., Eg : Microscope, Telescope, When the eye is focussed on a distant object, , 0 the ciliary muscles are relaxed so that the, focal length of the eye-lens has maximum value, which is equal to its distance from the retina., When the eye is focussed on a closer object ( , increases) the ciliary muscles of the eye are strained, and focal length of eye lens decreases. The ciliary, muscles adjust the focal length in such a way that, the image is again formed on the retina and we see, the object clearly. This process of adjusting focal, length is called accomodation., If the object is brought too close to the eye the, focal length cannot be adjusted to form the image, on the retina.Thus there is a minimum distance for, the clear vision of an object., The nearest point at which an object is seen clearly, by the eye is called the near point of the eye and, distance of near point from the eye is called the, least distance of distinct vision, It is equal to, 25cm for normal eye and it is denoted by D., The farthest point from an eye at which an object is, distinctly seen is called far point and for a normal, eye it is theoretically at infinity., Deffects of Vision: Our eyes are marvellous, organs that have the capability to interpret incoming, electromagnetic waves as images through a complex, process. But over eye may develop some defects, due to various reasons. Some common optical, defects of the eye are a)myopia b) hypermetropia, c) presbyopia, Myopia: The light from a distant object arriving, at the eye lens may be converged at a point infront, of the retina. This defect is called Myopia (or), 179
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, shortsightedness.In thi s defect, the far point of, the eye is at a distance lesser than infinity, and distant, objects are not clearly visible., , N.P.= Near Point of defected eye., If the objective is placed at D=25cm=0.25m, P, , , , , , This defect is rectified by using spectacles having, divergent lens (concave lens) which forms the image, of a distant object at the far point of defected eye., From lens formula, 1, 1, 1, , P, F .P dis tan ce of object f, Where F.P= Far point of the defective eye. If the, object is at infinity, 1, 1, Power of lens (p)= , f F .P, Hypermetropia: (or) Long-sightedness., The light from an object at the eye lens may be, converged at a point behind the retina. This defect, is called, In this type of defect, near point is at a distance, greater than 25cm and near objects are not clearly, visible., , Presbyopia: The power of accomodation of, , eyelens may change due to the decreasing, effectiveness of ciliary muscles.So, far point is lesser, than infinity and near point is greater than 25cm, and both near and far objects are not clearly visible., This defect is called presbyopia.This defect is, rectified by using bifocal lens., Astigmatism: This defect arises due to imperfect, spherical nature of lens, the focal length of eye lens, in two orthogonal directions becomes different, eye, cannot see objects in two orthogonal directions, clearly simultaneously. This defect is remedied by, cylindrical lens in a particular direction., W.E-113: A person cannot see distinctly any object placed beyond 40cm from his eye. Find, the power of lens which will enable him to, see distant stars clearly is?., Sol: The person cannot see objects clearly beyond, 0.4m., so his far point = 0.4m distance of object = ., He should use lens which forms image of distant object u at a distance of 40cm infront, of it., , 1, 1, 1, , p ; P 10 2.5D, 0.40 f, 4, W.E-114: A far sighted person cannot focus distinctly on objects closer than 1m. What is the, power of lens that will permit him to read from, a distance of 40cm?, Sol: As near point is 1m and distance of objects is, 0.40m, both in front of lens., 1 1 1 1, 1, P , , P 1.5D, f v u 1 0.40, , This defect is rectified by using spectacles having, convergent lens(i.e convex lens) which forms the, image of near objects at the near point of the, defected eye (which is more than 25cm), 1, 1, 1, , , P, N .P. dis tan ce of object f, , 180, , 1 1, 1 , , , , f 0.25 N .P. , , , , Simple Microscope :, To view an object with naked eye, the object must, be placed between D and infinity. The maximum, angle is subtended when it is placed at D., , Magnifying power of simple microscope:, The magnifying power or angular magnification, of a simple microscope is defined as the ratio of, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, magnifying power will decrease., The maximum possible magnifying power of a, simple microscope for a defect-free image is about, 4., As we use single lens in microscope, the image, formed by a single lens possesses several defects, like spherical aberration and astigmatism, at larger, magnifications the image becomes too defective., For higher magnifying power , we cannot use simple, microscope, this is because, at larger magnifications, the image becomes too defective.So we use, compound microscope for higher magnifying power., Simple magnifier is an essential part of most of, optical instruments such as microscope or telescope, in the form of an eye piece., W.E-115: A graph sheet divided into squares each, of size 1mm 2 is kept at a distance of 7cm from, a magnifying glass of focal length of 8cm. The, graph sheet is viewed through the magnifying lens keeping the eye close to the lens. Find, (i) the magnification produced by the lens, (ii), the area of each square in the image formed, (iii) the magnifying power of the magnifying, lens.Why is the magnification found in (i) different from the magnifying power?, , visual angle with instrument to the maximum visual, angle for unaided eye when the object is at least , distance of distinct vision., M, , visual angle with instrument, max imum visual angle for unaided eye, , M, , , 0, , Case(1):When the final image is formed at far point, (or) When the fimal image is formed at infinity, , D, f, As here u is maximum, magnifying power is minimum and as in this situation parallel beam of light, enters the eye, eye is least strained and is said to be, normal, relaxed and unstrained., Case(2):When the final image is formed at near point, (or) When the final image is formed at D, v D, u is ve, In this case u f , V ; So M , , 1, 1 1 1 1 1, , ; , f, D u u f D, , 1 1, D, Sol: i) u 7cm; f 8cm; v ?, M D D ; M D 1 , f , f D, , 1 1 1, , For a lens,, As the minimum value of v(=D) in this situation u is, f v u, minimum and magnifying power is maximum and, 1 1 1 1 1 1 1, eye is under maximum strain., ; ; v 56cm, 8 v 7 v 8 7 56, Note: If lens is kept at a distance ‘a’ from the eye, v 56, then D is replaced by D a , 8, Magnification, M , u 7, ii) Each square is of size 1mm 2 ie. its length and, Da, Da, M D 1 , ;, M, , , , breadth are each to 1mm. The virtual image formed, f, f , has linear magnification 8. So its length and breadth, Some important points regarding, are each equal to 8mm, The area of the image of, microscope:, each square= 8 8mm 2 64mm2, iii) Magnifying power of the magnifying glass ie., D, D, M, , 1, , ;, M, , ,, simple microscope., As D, , so M D M , f, f, D, 25, m 1 1, 4.125 D 25cm , D, D, f, 8, As M D 1 ; M , so smaller the focal, f, f, The magnification found in i is different from the, length of the lens greater the magnifying power of, magnifying power because the image distance in, the simple microscope., i is different from the least distance of distinct vi With increasing wave length of light used, focal, length of microscope will increase and hence, sion D., NARAYANAGROUP, , 181
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , are 4.17cm and 5cm respectively. (b) As in case of, simple magnifier MP=(D/u). So MP will be, minimum when, , W.E-116: If the focal length of a magnifier is 5cm, calculate, a) the power of the lens, b) the magnifying power of the lens for relaxed, and strained eye., Sol:As power of a lens is reciprocal of focal length in, P, , umax 5cm, , 1, 1, diopter 20 D, , 2, 5 10 m 0.05, , i.e MP min , , , 25, D, 5 M , 5, f, , , and, , will, , MP, , b) For relaxed eye, MP is minimum and will be, , u min 25 / 6 cm, , D 25, , 5, f, 5, While for strained eye, MP is maximum and will be, , MP max , , MP , , MP 1 , , W.E-117: A man with normal near point (25cm), reads a book with small print using a magnifying glass, a thin convex lens of focal length, , 5cm., a) What is the closest and farthest distance at, which he can read the book when viewing, through the magnifying glass?, b) What is the maximum and minimum magnifying powe r possible using the above simple, microscope?, Sol: a) As for normal eye far and near points are , and 25cm respectively, so for magnifier vmax , , 1 1 1, , v u f, ie u , , vmin 25cm, , 5, 25, , , 4.17cm, 5 / 25 1, 6, , and u will be maximum when vmax , ie u max , , 5, 5cm, 5 / 1, , so the closest and farthest distances of the book, from the magnifier (or eye) for clear viewing, 182, , when, , , 25, D, 6 1 , 25 / 6 , f, , , A simple magnifying lens is not useful where, large magnification is required. A highly magnified image must be produced in two stages. A, compound microscope is used for that purpose., , Magnifying power:, M, , Visual angle with instrument, , , Max. visual angle for unaided eye 0, , v D , M , u ue , Where u is the object distance for the objective, lens, v is image distance for the objective lens,, u e is the object distance for the eye piece., i.e M m0 me, , f, f / v 1 so u will be minimum when, , ie. u min, , maximum, , Compound Microscope, , D, 1 5 6, f, , and vmin 25cm. . However , for a lens as, , be, , The length of the tube L v ue, , Case(i): If the final image is formed at infinity (far point):, In this case ue f e, , v D, M with L v f, , e, u fe , A microscope is usually considered to operate, in this mode unless stated other wise. In this mode, ue is maximum and hence magnifying power is, minimum., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , Note:When the object is very close to the principal fo- , cus F0 of the objective, the image due to the objective becomes very close to the eyepiece. Then, , For a given microscope magnifying power for, normal setting remain practically unchanged if field, LD, and eye lens are interchanged as M f f, 0 e, , replace u with f0 and v0 with L so the expression, , , In an actual compound microscope each of the, objective and eye piece consists of a combination, L D, M, , , of several lenses instead of a single lens to eliminate, , , for magnifying power., f 0 fe , the aberrations and to increase the field of view., In low power microscopes, the magnifying power, Case-ii: If the final image is formed at D (Near, is about 20 to 40, while in high power microscopes,, point):, the magnifying power is about 500 to 2000., W.E-118: A microscope consists of two convex, In this case, for eye piece Ve D, ueis ve, lenses of focal lenghts 2cm and 5cm placed, 1, 1, 1, 20cm apart. Where must the object be placed, , , so that the final virtual image is at a distance, D ue f e, of 25cm from the eye?, Sol: For t he eyepiece, focal length, 1 1 D, , 1, , , , i.e u, f f e 5cm ; v ve 25cm , u ue ? subD, fe ; m mo me, e, stituting in, fD, v D, 1, 1, 1 1 1 1, M D 1 with LD v e, , ; , fe D, u fe , f v u 5 25 ue, In this situation as ue is minimum magnifying power, is maximum and eye is most strained., When the object is very close to the principal focus, F0 of the objective,the image due to the objective, becomes very close to the eyepiece. Then replace, u with f0 and v with L so the expression for magnifying power., , L D, M D 1 , f0 , fe , , , , , Some important points regarding, compound microscope:, As magnifying power of a compound microscope, is negative, the image seen is always truly inverted., For a microscope magnifying power is minimum, when final image is at a nd maximum when final, image is at least distance of distinct vision D, i.e and, , M max, , v D, 1 , u, fe , , NARAYANAGROUP, , 1, 1 1 6, , 25 5 25, ue, 25, cm, 6, object for the eyepiece is to be at a distance of, 25, cm to its left., 6, ue , , But v0 ue 20cm where ue , v0 20 ue 20 , , 25, cm, 6, , 25 95, , cm, 6, 6, , For the objective, v v0 , , 95, cm, 6, , f f 0 2cm; u ?, , 1 1 1 1 6 1, ; , f v u 2 95 u, 1 6 1, 83, 190, , , 2.29cm, ;u , u 95 2, 190, 83, The object is to be placed at a distance of 2.29cm, to the left side of the objective., 183
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , W.E-119: Find the magnifying power of a com- W.E-121: A compound microscope has a magnifying power 30. The focal length of its eyepiece, pound microscope whose objective has a focal, is 5cm. Assuming the final image to be at the, power of 100Dand eye piece has a focal power, least distance of distinct vision (25cm),, of 16D when the object is placed at a distance, calculatethe magnification produced by obof 1.1cm from the objective. Assume that the, jective., final image is formed at the least distance of, Sol:, In case of compound microscope,, distinct vision (25cm), Sol: The magnifying power of a compound microscope, M m0 me 1, when the final image forms at the least distance of, And in case of final image at least distance of disdistinct vision,, tinct vision,, v D, D, m 0 1 , me 1 2 , u, fe , fe , , To find v0 ; power of the objective p0 100 D ., D, so, from eqs. 1 and 2 , M m0 1 f , Focal length of the objective,, e , , 1, 1, 100, f f0 , , m, cm 1cm, Here M 30; D 25cm and f e 5cm, p0 100, 100, 25 , 30, 5, u u0 1.1cm; v v0 ?, So, 30 m 1 ie m0 , 5, 6, , 1 1 1, Negative sign implies that image formed by objecFor a lens, , tive is inverted., f v u, W.E-122: A compound microscope is used to en1 1, 1 1 1 1 0.1, , ; , , large an object kept at a distance 0.03m from, 1 v0 1.1 v0 1 1.1 1.1, its objective which consists of several convex, lenses in contact and has focal length 0.02m., v0 11cm, If a lens of focal length 0.1m is removed from, Power of the eyepiece, pe 16 D ;focal length of, the objective, find out the distance by which, the eyepiece of the microscope must be moved, the eye piece., to refocus the image?, 1, 1, 100, Sol: If initially the objective forms the image at disfe , cm 6.25cm, m, pe 16, 16, 1 1 1, Least distance of distinct vision, D 25cm, tance v1 . v 3 2 ie v1 6cm, 11 , 25 , m , 1 , 10 5 50, 1.1 , 6.25 , W.E-120: In a compound microscope, the object, is 1cm from the objective lens. The lenses are, 30 cm apart and the intermediate image is, 5cm from the eye piece. What magnification, is produced?, Sol: As the lenses are 30cm apart and intermediate, image is formed 5cm in front of eye lens,, ue 5cm and v L ue 30 5 25cm, Now as in case of compound microscope,, M m0 m e , , v D, , u ue , , here u 1cm and D 25cm, 25 25 , So M 125, 1 5, Negative sign implies that final image is inverted., 184, , 1, , Now as in case of lenses in contact, 1 1 1, 1 1 1, ........... or , F f1 f 2, F f1 F ', 1, , 1, , 1, , with F ' f f ........, 2, 3, So if one of the lenses is removed, the focal length, of the remaining lens system, 1, 1, 1, 1 1, , , ie F ' 2.5cm, F ' F f ' 2 10, , This lens will form the image of same object at a, distance v2 such that, 1 1, 1, , , v2 3 2.5 ie v2 =15cm, So to refocus the image, eyepiece must be moved, by the same distance through which the image, formed by the objective has shifted ie. 15-6=9cm, away from the objective., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-123: The focal lengths of the objective and, the eyepiece of a compound microscope are, 2.0cm and 3.0cm respectively. The distance between the objective and the eyepiece is 15.0cm., The final image formed by the eyepiece is at, infinity. Find the distance of object and image produced by the objective, from the objective lens., Sol: As final image is at infinity, the distance of intermediate image from eye lens ue will be given, by, 1, 1, 1, , ue f e ie ue f e 3cm, and as the distance between the lenses is 15.0cm ,, the distance of intermediate image (formed by objective) from the objective will be, v L ue L f e 15 3 12cm, and if u is the distance of object from objective,, 1 1 1, ie u 2.4cm, 12 u 2, So object is at a distance of 2.4cm in front of ob, jective., W.E-124: The focal lengths of the objective and, the eye piece of a compound microscope are, 2.0cm and 3.0cm respectively. The distance between the objective and the eyepiece is 15.0cm., The final image formed by the eyepiece is at, infinity. The two lenses are thin. Find the distance, incm, of the object and the image produced by the objective, measured form the objective lens, are respectively., Sol: The eyepiece forms the final image at infinity. Its, object should therefore lie at its focus., ‘F’ denotes focus of eyepiece . ‘I’ denotes image, formed by the objective lens which serves as object for eyepiece. It should be at 3cm from eyepiece., f, uc, , B, , F, A, U0, , 12cm, V0, , 3cm, , 15cm, , v0 for objective lens = 15-3-12cm(1), , , , 1 1, 1, or, v0 u0 f 0, , NARAYANAGROUP, , 1 1 1, 1, 1 1 5, , 12 u0 2, u0 12 2 12 or, , u0 2.4cm, From objective lens u0 2.4cm (to left), v0 12cm (to right), , Telescopes : A microscope is used to view the, objects placed close to it. To look at distant objects such as star, a planet or a cliff etc, we use, another optical instrument called telescope,, which increases the visual angle of distant object., The telescope that uses a lens as an objective is, called refracting telescope. However, may telescopes use a curved mirror as an objective such, telescopes are known as reflecting telescopes., There are three types of refracting telescopes in, use., i) Astronomical telescope, ii) Terrestrial telescope, iii) Galilean telescope, , Astronomical Telescope :, Magnifying power (M):, Magnifying power of a telescope is given by, Visual angle with instrument , M, , Visual angle for unaided eye 0, h, From the above figure, 0 f, 0, , h, , u, h, , f, ;M e 0, , ue, ue, and, 0 h , , f0 , The length of the tube L f 0 ue, , Case-i If the final image is at infinity (far, point): In this case, for eyepiece ve ,, ue ve, 1, 1, 1, , , ue f e, , Hence ue f e, f0, Hence M f and L f 0 fe, e, Usually a telescope is operated in this mode un185
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, less stated other wise. In this mode ue is maximum,, hence magnifying power is minimum, while length, of tube is maximum., This case is also called normal adjustment because, in this case eye is least strained are relaxed., , 1 , become 2 times of its initial value., m , , , , Case-ii:If the final image is at D (Near point):, , have large magnifying power f0 must be as large as, , In this situation for eyepiece ve D, 1, 1, 1, 1, 1, ie, , D ue f e ue f e, , MD , , f0, fe, , fe , , 1 D , , , , fe , 1 D , , , , fe D, In this case length of the tube LD f 0 f D, e, , , , , , In this situation ue is minimum , hence magnifying, power is maximum while the length of the tube is, minimum and eye is most strained., , , Some important points regarding, astronomical telescope:, , , , In case of telescope if object and final image are at, infinity and total light entering the telescope leaves, it, parallel to its axis., f, A, magnifying power 0 0, fe Ae, , , , , , , , f0 , As magnifying power for normal setting as to, fe , , , , , , practically possible and fe is small. This is why in a, telescope, objective is of large focal length while, eyepiece of smaller focal length., Larger aperture of objective helps in improving the, brightness of image by gathering more light from, the distant object. However it increase aberrations, particularly spherical., If a fly is sitting on the objective of a telescope and, we take a photograph of distant astronomical object, through it, the fly will not be seen but the intensity, of the image will be slightly reduced as the fly will, act as obstruction to light and will reduce the, aperture of the objective., A telescope produces angular magnification, whereas a microscope produces linear, magnification. The image due to a telescope appears, to be near to the eye increasing the visual angle., Terrestrial Telescope:The magnifying power, and length of telescope for relaxed eye will be, , f, f, where A0 and Ae are the apertures of objectives, M 0 1 0 , L f f 4 f, 0, e, fe, fe , and eyepiece., As magnifying power is negative, the image seen in The magnifying power and the length of telescope, astronomical telescope is truly inverted i.e left is, for image at D will be, turned right with upside down simultaneously., Dfe, f , f , However as most of the astronomical objects are, M D 0 1 e , LD f 0 4 f , symmetrical this inversion does not effect the, D fe, fe D , observations., For given telescope, magnifying power is minimum, W.E-125: An astronomical telescope has an anwhen final image is at infinity (Far point) and, gular magnification of magnitude 5 for dismaximum when it is at least distance of distinct, tant objects. The separation between the obvision (Near point) ie., jective and eye piece is 36cm and the final, f0 , f0 , fe , image is formed at infinity. Determine the foM min and M max 1 , cal length of objective and eye piece., fe D , fe , In case of a telescope when the final image is at Sol: For final image at infinity,, , now if field and eye lenses are interchanged, f, f, M 0 and L f f, 5 0 1, , 0, e, f0 , fe, fe, magnifying power will change from f to, e, and 36 f 0 f e ii , fe , 1, Solving these two equations , we have, ie it will change from m to ie will, f, m, , 0, f 30cm and f 6cm, 0, , 186, , e, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , W.E-126: A telescope has an objective of focal, , Diameter of moon 3.8 108, , length 50cm and an eyepiece of focal length, Im, age, diameter, 1, 5cm. The least distance of distinct vision is, 25cm. The telescope is focused for distinct vi- Diameter of moon= 3.8 108 Image diameter, sion on a scale 2m away from the objective., 3.8 108 0.92 102 m = 3.946 106 m 3496km, Calculate, i) For normal adjustment, the distance between the, a) magnification produced, two lenses f 0 f e 100 5 105cm, b) separation between objective and eye piece,, ii) For the final image at 25cm form the eye, the distance, Sol: Given f 0 50cm and f e 5cm, between the two lenses, For objective, Dfe , 1, 1, 1, 200, 25 5 , , , f0 , v0 , cm, 100 , 104.2cm, v0 200 50, 25 5 , 3, D fe , v, 200 / 3 1, W.E-128: In an astronomical telescope, the focal, m0 0 , , lengths of the objective and the eye piece are, u0, 200, 3, 100cm and 5cm respectively. If the telescope, For eyepiece:, is focussed on a scale 2m from the objective,, 1, 1 1, the final image is formed at 25cm from the eye., , 25 ue 5, Calculate (i) the magnification and (ii) the distance between the objective and the eyepiece, ve, 25, 25, 6, ue cm and me , Sol: f 0 100cm ; f e 5cm, ue 25 / 6 , 6, To find the image distance due to objective, a)Magnification, m m0 me 2, u0 2m 200cm; v0 ?, b) Seperation between objective and eyepiece., 1 1 1 1, 1, 1, 200 25 425, ;, , For, a, lens, L v0 ue , , , 70.83cm, f u v 100 v0 200, 3, 6, 6, 1, 1, 1, 1, W.E-127: A telescope objective of focal length 1m, , , , ; v0 200cm, forms a real image of the moon 0.92cm in, v0 100 200 200, diameter. Calculate the diameter of the moon, 200, v, taking its mean distance from the earth to be, 1, Magnifying of objective, m0 0 , v 200, 4, ., If, the, telescope, uses, an, eyepiece, 38 10 km, To find the object distance for the eyepiece, of 5cm focal length, what would be the disve 25cm , ue ?, tance between the two lenses for (i) the final, image to be formed at infinity (ii) the final, 1 1 1 1, 1, 1, image(virtual) at 25 cm form eye., For a lens f v u ; 5 25 u, e, Sol: f 0 1m, 1, 1 1 6, 25, object distance from the objective, , ue cm, ue, 25 5 25, 6, =distance of the moon from the earth, 5, 8, Magnification, of, the, eyepiece,, 3.8 10 km 3.8 10 m, image distance from the objective, ve 25 6, m, , , 6, e, = focal length of the objective =1m, ue, 25, image size = image diameter= 0.92 10 2 m, i), Magnification, of, the, eyeobject size = object diameter, piece, mo me 1 6 6, ie diameter or moon=?, ii) Distance between the objective and the eyepiece, Object diameter Object dis tan ce, 25, We know that Im age diameter Im age dis tan ce, v0 ue 200 , 204.2cm, 6, NARAYANAGROUP, , 187
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GEOMETRIC OPTICS, , JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , W.E-129: A tower 100m tall at a distance of 3km W.E-131: An astronomical telescope consisting of, , an objective of focal lenght 60cm and eyepiece, is seen through a telescope having objective, of focal length 3cm is focused on the moon so, of focal length 140cm and eyepiece of focal, that the final image is formed at least distance, length 5cm. What is the size of final image if, of distinct vision ie 25cm from the eye piece., it is at 25cm from the eye?, Assuming the angular diameter of moon is, Sol: For objective lens, 0, 1, 1, 1, 1 / 2 at the objective, calculate (a) angular, , , ie, v, , 140, cm, , f, 0, v 3 105 140, size and (b) linear size of image seen through, v, 140, 14, the telescope., 104 and as final image, so m0 , 5, Sol:, As final image is at least distance of distinct, u 3 10, 3, vision,, is at least distance of distinct vision, so for eye lens,, we have, f 0 f e 60 , 3, M, , , , , 1, 1, , , , 22.4, 1, 1 1, , 25, f, D, 3, 25, , , , e, , u , 25 ue 5 ie e, 6, , Now as by definition M , so the angular, ve, 25, me , 6, 0, ue 25 , so, size, of, image, , 6 , 0, 1, 14, M, 0 22.4 11.20, 4, m, , m, , m, , , , 10, , 6, and hence,, 0, e, 2, 3, , , 11.2 0.2rad, I , 180, But as m , O, And if I is the size of final image which is at least, I m O 28 104 100 102 28cm, I , , distance, of, distinct, vision, Negative sign implies that image is inverted., 25 , W.E-130: The diameter of the m oon is, i.e I 25 25 0.2 5cm, 3.5 103 km and its distance from the earth, C. U. Q, 3.8 105 km. It is seen through a telescope having focal lengths of objective and eye-piece, as 4m and 10cm respectively. Calculate (a), REFLECTION, magnifying power of telescope(b) angular size 1. A bird flying high up in air does not cast, of image of moon, shadow in the ground because, Sol: For normal adjustment, 1) the size of the bird is smaller than sun, f 0 4 100, 2) the size of the bird is smaller than earth, a) M f 10 40, 3) light rays fall almost normally on the bird, e, 4) none of the above, b) L f 0 f e 400 10 410cm 4.10m, 2. A plane mirror reflects a beam of light to form, c) As the angle subtended by moon on the objeca real image. The incident beam is, 1) parallel, 2) convergent, 3.5 103 3.5, , 10 2 rad, tive of telescope 0 , 5, 3), divergent, 4) any one of the above, 3.8 10, 3.8, 3. When an object is placed between two parallel, , mirrors, then number of images formed are, and as M , the angular size of final im1) 2, 2) 4, 3) 8, 4) infinite, 0, 4. If a number of images of a candle flame are, age, seen in a thick mirror, then, 3.5, 102 0.3684 rad, M 0 40 , 1) The first image is the brightest, 3.8, 2) The second image is the brightest, 1800, 3) The last image is the brightest, 210, i.e 0.368 , 4) The image are equally bright, , 188, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 5., , 6., , GEOMETRIC OPTICS, , If two plane mirrors are inclined at angle to 13. In cold countries, the phenomenon of looming, takes place, because refractive index of air, each other as shown, than angle of deviation, 1) decreases with height, of incident ray is, 2) increases with height, 3) does not change with height, 4) become infinity at the surface, , 14. A ray of light passes through four transpar, ent media with refractive indices, , , 1 , 2 , 3 and 4 as shown in figure. The, surfaces of all media are parallel. If the, 1) 360 2 2) 360 2 3) 180 2 4) 180 2, emergent ray is parallel to the incident ray,, A real, inverted and equal in size image is, we must have, formed by, 1) a concave mirror 2) a convex mirror, , , , , 3) a plane mirror 4) none of these, The rear - view mirror of a car is, 1) Plane, 2) Convex, 3) Concave, 4) None, If a spherical mirror is immersed in a liquid., it’s focal length will, 1) increase, 2)decrease, 3) remain unchanged, 1) 1 2 2) 2 3 3) 3 4 4) 4 1, 4) depend on the nature of liquid, A train is approaching towards a stationary 15. Rays of light fall on a glass slab 1 as, person with a velocity v . The train emits a, shown in the figure. If at A is maximum and, light signal. The signal will reach the stationat B it is minimum, then what will happen to, ary person with a velocity., these rays?, 1) c, 2) c v, 3) c v, 4) c 2 v 2, 1, , 7., 8., , 9., , REFRACTION, , A, , 2, , 3, , 4, , B, , 10. Light of frequency n, wave length , travelling with a velocity v enters into a, glass slab of R.I then frequency, wave, 1) they will tilt towards A, length and velocity of the wave in glass slab, 2) they will tilt towards B, respectively are, 3) they will not deviate, v, n, v, 4) there will be total internal reflection., v, n , 1) , , 2) n, , 3) n, , 4) , , v, 16., A hunter desires to shoot a fish whose image, , , could be seen through clear water. His aim, 11. Absolute refractive index of a material, should be, depends upon, 1) Above the apparent image of fish, 1) nature of material, 2) Below the apparent image of fish, 2) nature, wavelength and size of material, 3) In the line of sight of fish, 3) density, temperature, wavelength of material, 4) Parallel to the surface of water, 4) nature, temperature, wavelength of material, 17. A rectangular solid piece is placed in a liquid, 12. If a ray of light takes t1 and t 2 times in two, whose refractive index is the same as that of, media of absolute refractive indices 1 and, the solid, 1) The sides of the solid will appear to be bent, 2 respectively to travel samed i s t a n c e ,, inward, then, 2) The sides of the solid will appear to be bent, 1) 1t1 2 t 2, 2) 1t 2 2 t1, outward, 3) The solid will not be seen at all, 3) t1 1 t 2 2 4) t1 2 t 2 1, 4) The solid will appear as in air, NARAYANAGROUP, , 189
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 18. A plane glass plate is placed over various, 1 2, 2 1, coloured letters. The letter which appears to, 1), 2), 4 3, 3 4, be raised least, 1) violet 2) yellow 3) red, 4) green, 19. As temperature of medium increases the, 4 3, 3 4, critical angle, 3), 4), 1) Increases 2) Decreases, 1 2, 2 1, 3) Remains same 4) first increases then decreases, 20. A ball coated with ‘lamp black’ put in a glass 27. The relation between refractive indices, tank containing water appears silvery white, , 1 , 2 . if the behaviour of light ray is as, due to, shown in fig., 1) Refraction 2) Diffraction, , 3) Interference 4) Total internal reflection, 21. In an optical fibre, 2, 1, 1) Core region is transparent, cladding is opaque, 2) Core region is opaque, cladding is transparent, 3) Both core and cladding regions are transperent, 1) 2 1, 2) 2 1, 4) Both core and cladding regions are opaque, 22. In an optical fibre during transmission of light, 3) 2 ; 1, 4) 2 1 ; 2, 1) Energy increases 2) Energy decreases, 28. If parallel beam of light falls on a convex, 3) No loss of propagation of energy takes place, lens. The path of the rays is shown in fig. It, 4) Light partially reflects and refracts, follows that, , LENSES & THEIR COMBINATION, 23. The focal length of a lens depends on, , 1) colour of light, , , 2) radius of curvature of the lens, 3) material of the lens 4) all the above, 24. fB and fR are focal lengths of a convex lens, for blue and red light respectively and FB and, FR are the focal lengths of the concave lens, 1) 1 2, 2) 1 2, for blue and red light respectively. We must, 3) 1 2, 4) 1 2, then have, 29. A converging lens is used to form an image, 1) fB > fR and FB FR, on a screen. When the upper half of the lens, 2) fB < fR and FB FR, is covered by an opaque screen., a) half of the image will disappear, 3) fB > fR and FB FR, b) no part of image will disappear, 4) fB < fR and FB FR, c) Intensity of the image will increase, d. Intensity of the image will decrease, 25. The graph drawn with object distance along, 1) a,c are true, 2) a,d are true, abscissa and image distance along ordinate, 3), b,c, are, true, 4) b,d are true, measuring the distance from the convex lens, 30., A, convex, lens, is, placed, in contact with amirror, is, as shown. If the space between them is filled, 1) Straight line, 2) Parabola, in water, its power will, 3) Circle, 4) A hyperbola, 26. A convex lens is used to form a real image of, the object shown in the following figure:, 1, , 1 2, 4 3, Then the real inverted image is as shown, in the following figure:, 190, , 2, , 1) decrease, 2) increase, 3) remain unchanged, 4) can increase or decrease depending, on the focal length., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 31. A real image is formed by a convex lens, when, L, it is cemented with a concave lens, again the, real image is formed. The real image, 1) shifts towards the lens system, L, 2) shifts away from the lens sytem, 3) remain in its original position, 1) F1, 2) F2, 3) F1 F2 4) F1 F2, 4) shifts to infinity, 32. Parallel rays of light are focussed by a thin, LENS MAKER’S FORMULA, convex lens. A thin concave lens of same focal 38. Lens maker's formula is applicable to, length is then joined to the convex lens and the, 1) Thin lenses and paraxial rays which subtend very, result is that, small angles with the principal axis, 1) the focal point shifts away from the lens by a, 2) Thick lenses and paraxial rays which subtend, small distance, very small angels with the principles axis, 2) the focal point shifts towards the lens by a small, distance, 3) Thin lenses and for marginal rays, 3) the focal point of lens does not shift at all, 4) Thick lenses and for marginal rays, 4) the focal point shifts to infinity, 39. A spherical air bubble in water will act as, 33. A double convex lens of focal length f is cut, 1) a convex lens, 2) a concave lens, into 4 equivalent parts. One cut is, 3) Plane glass plate 4) Plano-concave lens, perpendicular to the axis and the other is 40. A liquid of refractive index 1.6 is introduced, parallel to the axis of the lens. The focal, between two identical plano-convex lenses in, length of each part is, two ways P and Q as shown. If the lens ma, 1) f/2 2) f, 3) 2f, 4) 4f, terial has refractive index 1.5, the combina, two, 34. If a lens of focal length f is divided into, tion is, equal parts and both pieces are put in, P, Q, contact as shown in fig. The resultant focal, length of combination is, 2, , 1, , L2, , L2, , L2, , L1, L1, L1, 1) convergent in both 2) divergent in both, 1) f, 2) 2 f, 3) 3 f, 4) 4 f, 3) convergent in Q only 4) convergent in P only, 41., If a convex lens is dipped in a liquid whose, 35. If a lens of focal length f is divided into two, refractive index is equal to the refractive index, equal parts and both pieces are put in c o n of the lens, then lens acts like a, tact as shown in fig. The resultant focal length, 1) concave lens, of combination are, 2) plane parallel glass plate, 3) plano convex lens, 4) plano concave lens, 1) 0,f, 2) f , f ,0 3) 2 f , f ,0 4) f , f /2, , PRISM, 36. If we added half part of each convex and con- 42 Recognize the prism (s) among the given, cave lens of focal length f as shown, the refigures., sultant focal length will be, a), , b), , 1) 0, 2) , 3) f, 4) 2 f, 37. In the figure given below there are two convex lenses L1 and L2 having focal lengths, F1 and F2 respectively. The distance between, L1 and L2 will be, NARAYANAGROUP, , c), , 1) b and c, 3) only b, , d), , 2) c, a and b, 4) a, b, c, and d, 191
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 43. The refractive index of a material of a prism, of angles 450 – 450 – 900 is 1.5. The path of, the ray of light incident normally on the hypotenuse side is shown in, A, , A, , 45, , 45, , 45, , 90, , 90, 4), C, , C, , 3) 2 1 3, 45, , B, , B, , A, , 45, , 3), , C, , A, , 45, , B, , 2), , 45, , 90, , 90, 1), , 3) the emergent ray is bent towards the edge of, the prism, 4) the emergent ray is bent towards the base of, prism, 49. Three prisms 1, 2 and 3 have A 60 , but, refractive indices are 1.4, 1.5, 1.6 and their, angles of deviation are 1 , 2 , 3 respectively.., Then, 1) 3 2 1, 2) 1 2 3, 4) 1 2 3, , DISPERSION, , 50. When white light enters a prism, its gets splits, into its constituent colours. This is due to;, 44. In the given figure,the angle between reflected, 1) high density of prism material, ray is equal to :, 2) because is different for different wavelength, 3) diffraction of light, 4) interference of light, A, 51. When a white light passes through a hollow, prism, then, 1) There is no dispersion and no deviation., 2) Dispersion but no deviation., 3) Deviation but no dispersion., 1) A, 2) 2 A, 3) 3 A, 4) 4 A, 4) There is dispersion and deviation both, 45. An equilateral prism is placed on a horizontal, 52., Which, one of the following does not exhibit, surface. A ray PQ is incident onto it. For, dispersion, minimum deviation, B, , R, Q, , S, , C, , 1), , R1, , R2, , 2) R1, , R2, , P, , 1) PQ is horizontal 2) QR is horizontal, 3) R1, 3) RS is horizontal 4) Any one will be horizontal, R2, R2 4) R1, 46. A prism produces a minimum deviation in a, light beam. If three such prisms are combined, 53. In dispersion without deviation., 1) The emergent rays of all the colours are parallel, the minimum deviation produced will be, to the incident ray., 1) 4 2) 2, 3) , 4) 0, 2) Yellow coloured ray is parallel to the incident, 47. When a ray of light is refracted by a prism such, ray, that the angle of deviation is minimum, then, 3) Only red coloured ray is parallel to the incident, 1) the angle of emergence is equal to the angle of, ray, incidence, 4) All the rays are parallel, but not parallel to, 2) the angle of emergence is greater than the angle, the incident ray., of incidence, 54. In the visible region the dispersive powers, 3) the angle of emergence is smaller than the angle, and the mean angular deviations for crown, of incidence, glass and, flint glass prism are ,, 4) the sum of the angle of incidence and the angle, 1, 1, and, d,, d, respec-tively., The condition for, 0, , of emergence is equal to 90, getting, dispersion, with, zero, deviation when, 48. If a small angled prism, made of glass is, the, two, prisms, are, combined, is. (2001), immersed in a liquid of refractive index 1 and, a ray of light is made incident on it, then, 1) d 1d 1 0 2) 1d d 1 0, 1) its deviation will be zero, 2, 2) it will suffer total reflection, 3) d 1d 1 0, 4) d 2 1d 1 0, 192, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 65. Long -sighted people who have lost their, 55. In the achromatic prism, we have, spectacles can still read a book by looking, 1) deviation without dispersion, through a small (3-4mm) hole in a sheet of, 2) dispersion without deviation, paper, 3) refraction without deviation, 1) Because the fine hole produces an image of, 4) deviation and dispersion, the letters at a longer distance, 56. The angular dispersion will be maximum in the, 2) Because in doing so, the distance of the, following pairs of colours is, object is increased, 1) Yellow and green 2) Red and blue, 3) Because in doing so, the focal length of the, 3) Green and red, 4) Blue and orange, eye lens is effectively decreased, DEFECTS OF EYE, 4) Because in doing so , the focal length of the, eye lens is effectively increased, 57. When objects at different distances are seen, by the eye, which of the following remians OPTICAL INSTRUMENTS MICROSCOPES, constant, 66. For which of the following colour, the magnify1) The focal length of the eye lens, ing power of a microscope will be maximum, 2) The object distance from the eye lens, 1) White colour, 2) Red colour, 3) The radii of curvature of the eye lens, 3) Violet colour, 4) Yellow colour, 4) The image distance from the eye lens, 67. The magnifying power of a simple microscope, 58. Near and far points of healthy human eye, can be increased, if we use eye piece of, respectively are, 1)Higher focal length 2)Smaller focal length, 1) 0 and 25 cm, 2) 0 and infinity, 3)Higher diameter 4)Smaller diameter, 3) 25cm and 100 cm 4) 25 cm and infinity, 59. The ability of eye to focus on both near and 68. The angular magnification of a simple microscope can be increased by increasing, far objects is called, 1) Focal length of lens 2) Size of object, 1) Presbyopia, 2) Myopia, 3) Aperture of lens, 4) Power of lens, 3) Hypermetropia 4) Power of accommodation, 60. Loss of the eye to focus on both far and near 69. When the length of a microscope tube, increases, its magnifing power, objects with advancing age is, 1) decreases, 2) increases, 1) Astigmatism, 2) Presbyopia, 3) does not change 4) can’t say, 3) Myopia, 4) Hypermetropia, 70., In a compound microscope the image produced, 61. The image of an object formed on the retina of, by the objective is, the eye is, 1) real enlarged and errect, 1) virtual and inverted 2) virtual and erect, 2) real enlarged and inverted, 3) real and erect, 4) real and inverted, 3) Virtual enlarged and erect, 62. Myopia occurs due to, 4) Virtual, enlarged and inverted, 1) Increase in the focal length of eye lens, 71. The magnifing power of a compound2) Decrease in the distance between retina and lens, microscope increases when, 3) Decrease in focal length of eye lens, 1) the focal length of objective lens is increased, 4) Increase in the distance between retina and lens, and that of eye lens is decreased, 63. For a myopic (short-sighted) eye, rays from, 2) the focal length of eye lens is increased and, far distant objects are brought to focus at a, that of objective lens in decreased, point, 3) focal lengths of both objective and eye-piece, 1) on the retina, 2) Behind the retina, are increased, 3) In between eye lens and retina, 4) focal lengths of both objective and eye-piece, 4) At any position, are decreased, 64. In the case of hyper metropia, OPTICAL INSTRUMENTS : TELESCOPES, 1) the image of a near object is formed behind the 72. The optical instrument with zero power is, retina, 1) microscope, 2) telescope, 2) the image of a distant object is formed infront of, 3) eyepiece, 4) all the above, the retina, 73. The image formed by the telescope in normal, 3) a concave lens should be used for correction, adjustment position is at, 4) a bifocal lens should be used for correction, 1) D, 2) 2D, 3) F, 4) Infinity, NARAYANAGROUP, , 193
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 74. If the telescope is reversed i.e. seen from the, objective side, 1) Object will appear very small, 2) Object will appear very large, 3) There will be no effect on the image formed by, the telescope, 4) Image will be slightly greater than the earlier one, 75. In an astronomical telescope the focal lengths, of objective and eyepiece should respectively, be, 1) large and small 2) small and large, 3) equal, 4) too small are too large, 76. The magnifying power of an astronomical, telescope can be increased if we, 1) increase the focal length of the objective, 2) increase the focal length of the eye-piece, 3) decrease the focal length of the objective, 4) decrease the focal length of the objective and, at the same time increase the focal length of the eye, piece, 77. The final image in an astronomical telescope, is, 1) real and errect, 2) virtual and inverted, 3) real and inverted 4) virtual and errect, 78. A photograph of the moon was taken with telescope. Later on, it was found that a housefly, was sitting on the objective lens of the telescope. In photograph, 1) The image of housefly will be reduced, 2) There is a reduction in the intensity of the image., 3) There is an increase in the intensity of the image., 4) The image of the housefly will be enlarged., 79. In Gallilean telescope, the final image formed, is, 1) Real, erect and enlarged, 2) Virtual, erect and enlarged, 3) Real, inverted and enlarged, 4) Virtual, inverted and enlarged, A:, B:, C:, D:, E:, 194, , 80. Assertion : Radius of curvature of a convex, mirror is 20 cm. If a real object is placed at, 10 cm from pole of the mirror, image is formed, at infinity., Reason : When object is placed at focus, its, image is formed at infinity., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 81. Assertion : For a prism of refracting angle, 600 and refractive index, deviation is 300 ., , 2 , minimum, , Reason : At minimum deviation light ray inside, the prism passes parallel to the base of the, prism., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 82. Assertion : Image formed by concave lens is, not always virtual., Reason : Image formed by a lens is real if the, image is formed in the direction of ray of light, with respect to the lens., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 83. Assertion : Minimum deviation for a given, prism does not depend on the refractive index, of the prism., Reason : Deviation by a prism is given by, , i1 i2 A and does not have the term, ., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 84. Assertion : Critical angle of light while passing, from glass to air is minimum for violet colour., Reason : The wavelength of violet light is, greater than that of other colours., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 85. Assertion : Different colours of light have, same velocity in vacuum, but they have, ASSERTION & REASON, different velocities in in any other transparent, If both assertion and reason are true and, medium., reason is a correct explanation of the assertion., Reason : v c / , where symbols have, If both assertion and reason are true but the, standard meanings. For different colours,, reason is not a correct explanation of assertion., If assertion is true but reason is false., refractive index, of transparent medium, If assertion is false but reason is true., has different values. Therefore, v is different., Both assertion and reason are false., 1) A, 2) B, 3) C, 4) D 5) E, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 86. Assertion : The minimum distance between an 92. Assertion : A concave mirror of focal length f, object and its real image formed by a convex, in air is used in a medium of refractive index, lens is 2f., 2. Then the focal length of mirror in medium, becomes double., Reason : The distance between an object and, its real image is minimum when its, Reason : The radius of curvature of a mirror, magnification is, - 1., is double of the focal length., 1) A, 2) B, 3) C, 4) D 5) E, 1) A, 2) B, 3) C, 4) D 5) E, 93. Assertion : When monochromatic light is, 87. Assertion : A lens has tow principal focal, incident on a surface separting two media, the, lengths which may differ., reflected and refracted light both have the, same frequency as the incident frequency., Reason : Light can fall on either surface of, Reason : The frequency of monochromatic, the lens. The two principal focal lengths differ, light depends on media., when medium on the two sides have different, refractive indices., 1) A, 2) B, 3) C, 4) D 5) E, 94. Assertion : The images formed by total, 1) A, 2) B, 3) C, 4) D 5) E, internal reflections are much brighter than, 88. Assertion : The twinkling of star is due to, those formed by mirrors or lenses., reflection of light., Reason : There is no loss of intensity in total, internal reflection., Reason : The velocity of light changes while, 1) A, 2) B, 3) C, 4) D 5) E, going from one medium to the other., 95. Assertion : The blue colour of sky is on, 1) A, 2) B, 3) C, 4) D 5) E, account of scattering of sum light., 89. Assertion : In an electromagnetic wave, the, Reason : The intensity of scattered light varies, elecric field E is much larger than magnetic, inversely as the south power of wavelength of, field B., the light., 1) A, 2) B, 3) C, 4) D 5) E, Reason : The electromagnetic waves get, deflected in perpendicular electric field but, not in a perpendicular magenetic field., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 90. Assertion : If a convex lens of glass is, immersed in water its power decreases., Reason : In water it behaves as a concave, lens., 1) A, , 2) B, , 3) C, , 4) D, , 5) E, , 91. Assertion : For observing traffic at our back,, we prefer to use a convex mirror., Reason : A convex mirror has a much larger, field of viwe than a plane mirror or a concave, mirror., 1) A, , 2) B, , NARAYANAGROUP, , 3) C, , 4) D, , 5) E, , C.U.Q - KEY, 1) 1, 7) 2, 13) 1, 19) 1, 25) 4, 31) 2, 37) 3, 43) 1, 49) 1, 55) 1, 61) 4, 67) 2, 73) 4, 79) 2, 85) 4, 91) 1, , 2) 2, 8) 3, 14) 4, 20) 4, 26) 4, 32) 4, 38) 1, 44) 2, 50) 2, 56) 2, 62) 3, 68) 4, 74) 1, 80) 4, 86) 4, 92) 4, , 3) 4, 9) 1, 15) 3, 21) 3, 27) 3, 33) 3, 39) 2, 45) 2, 51) 1, 57) 4, 63) 3, 69) 2, 75) 1, 81) 2, 87) 1, 93) 3, , 4) 2, 10) 2, 16) 2, 22) 3, 28) 3, 34) 1, 40) 3, 46) 3, 52) 3, 58) 4, 64) 1, 70) 2, 76) 1, 82) 2, 88) 4, 94) 1, , 5) 1, 11) 4, 17) 3, 23) 4, 29) 4, 35) 4, 41) 2, 47) 1, 53) 2, 59) 4, 65) 3, 71) 4, 77) 2, 83) 3, 89) 3, 95) 1, , 6) 1, 12) 2, 18) 3, 24) 4, 30) 2, 36) 2, 42) 1, 48) 4, 54) 3, 60) 2, 66) 3, 72) 2, 78) 2, 84) 2, 90) 3, , 195
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 7., , LEVEL - I (C.W), REFLECTION, 1., , 2., , Two plane mirrors are at 450 to each other. If, an object is placed between them, then the, number of images will be, 1) 5, 2) 9, 3) 7, 4) 8, Fig. Shows a plane mirror onto which a light, ray is incident. If the incidenting light ray is, turned by 100 and the mirror by 200 as shown,, then the angle turned by the reflected ray is, , 10°, , 8., , reflected ray, , A 4.5 cm needle is placed 12 cm away from a, convex mirror of focal length 15 cm. The, position of image and the magnification, respectively are, 5, 1) 3.33 cm,, 2) 6.7 cm, 1.8, 7, 5, 3) 0.15 cm, 1.8, 4) 6.7 cm,, 9, An object is placed 20 cm from the surface of, a convex mirror and a plane mirror is set so, that virtual images formed by two mirrors, coincide. If plane mirror is at a distance of 12, cm from object, then the focal length of convex, mirror is, 1) 5 cm 2) 10 cm 3) 20 cm, 4) 40 cm, , REFRACTION, , 30°, , 9., , 20°, , 3., , 4., , 5., , 6., , 196, , 1) 300 clockwise, 2) 300 anticlock wise, 3) 500 clockwise, 4) 500 anticlock wise, A small object is placed 10 cm infront of a plane, mirror. If you stand behind the object 30 cm, from the mirror and looks at its image, the, distance focussed for your eye will be, 1) 60 cm 2) 20 cm 3) 40 cm 4) 80 cm, A concave mirror gives an image three times, as large as the object placed at a distance of, 20 cm from it. For the image to be real, the, focal length should be, 1) 10 cm 2) 15 cm 3) 20 cm, 4) 30 cm, An object is 20 cm from a convex mirror of, focal length 10cm. The image fromed by the, mirror is, 1) Real and at 20 cm from the mirror, 2) Virtual and at 20 cm from the mirror, 20, 3) Virtual and at, cm from the mirror, 3, 20, 4) Real and at, cm from the mirror, 3, An object is placed at 10 cm infront of a, concave mirror of radius of curvature 15 cm., The position of image(v) and its magnification, (m) are, 1) v = 30 cm; m = 3 (real, inverted), 2) v = 20 cm; m = 3 (virtual, erect), 3) v = 10 cm; same size (real, inverted), 4) v = 10 cm; same size (virtual, erect), , If i j represents refractive index when a light, ray goes from medium i into j, then, 2 1 3 2 4 3 is equal to, , 1, 3) , 4) 4 2, 1 4, 10. The refractive index of glass with respect to, 9, water is . If the velocity and wavelength of, 8, light in water are 2.25 108 ms 1 and 5400 A0 ,, then the velcoity and wavelength of light in, glass are, 1) 2 108 ms 1 ; 4800 A0 2) 1 108 ms 1 , 6075 A0, , 1) 3 1, , 2) 3 2, , 3) 2 108 ms 1 ; 6075 A0 4) 1 108 ms 1 , 4800 A0, 11. A ray of light passes normally through a slab, , 1.5 of thickness ‘t’. If the speed of light, in vacuum is ‘c’, then time taken by the ray to, go across the slab is, t, 3t, 2t, 4t, 1), 2), 3), 4), c, 2c, 3c, 9c, 12. The angle of incidence on the surface of a, diamond of refractive index 2.4, if the angle, between the reflected and refracted rays is, 900 is, 1 1 , 1) tan 1 2.4 , 2) sin , , 2.4 , 1 1 , 3) tan , , 2.4 , , 1 1 , 4) cos , , 2.4 , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 13. A bird in air is at a height ‘y’ from the surface, of water. A fish is at a depth ‘x’ below the, surface of water. The apparent distance of fish, from the bird is (The refractive index of water, is ), , y, x, x, 2) x y 3) y, 4) y, , , , 14. A ray of light incident on a transparent block, at an angle of incidence 600 . If of block is, 1) x , , REFRACTION THROUGH, SPHERICAL SURFACES, 19. An air bubble in glass 1.5 is situated at, a distance 3 cm from a convex surface of, diameter 10 cm as shown. The distance from, surface at which the image of bubble appears, is, = 1.5, , = 1, , 3 then the angle of deviation of the refracted, ray is, 2) 250, 3) 300, 4) 450, 1) 150, 15. A fish looking up through the water see the, outside world contained in a circular horizon., 4, If the refractive index of water is and fish, 3, is 12 cm below the surface, the radius of this, circle is (in cm), 36, 36, 2) 4 5, 3), 4) 36 7, 1), 5, 7, 16. When a light ray is refracted from one medium, into another, the wavelength changes from, 4500 A0 to 3000 A0 . The critial angle for a ray, from second medium to first medium is, 2, , 2, , , 1 , 1) sin 13 , , , , 1 , 2) cos 3 , , , 3, , 2 , , , 1, 1 , 3) tan 2 , 4) tan , , 5, , 17. The fig shows a mixture of blue, green, red, colours incident on a right angled prism. The, critical angles of the material of prism for red,, green and blue colours are 460 , 440 , 430, respectively, The arrangement will separate, B, G, 45°, , R, , 1) Red from Green and Blue, 2) Blue from Green and Red, 3) Green from Red and Blue, 4) All the colours, 18. A ray of light is incident at angle of 600 on a, , , , , , 3 cm thick plate 3 . The shift in the, path of the ray as it emerges out from the plate, is (in cm), 1) 1, 2) 1.2, 3) 0.5, 4) 1.8, NARAYANAGROUP, , 3cm, , 1) 2.5 cm 2) 5 cm, , 3) 4 cm, , 4) 1.5 cm, , LENSES AND THEIR COMBINATION, 20. Two thin lenses of powers 2D and 3D are, placed in contact. An object is placed at a, distance of 30 cm from the combination. The, distance in cm of the image from the, combination is, 1) 30, 2) 40, 3) 50, 4) 60, 21. A symmetrical double convex lens is cut into, two equal parts along a plane perpendicular, to the principal axis. If the power of the orginal, lens is 4D, the power of one of the two pieces, is, 1) 2D, 2) 3D, 3) 4D, 4) 5D, 22. A parallel beam of monochromatic light falls, on a combination of a convex lens and a, concave lens of focallengths 15 cm and 5 cm., If the light emerges parallel from the concave, lens, the distance between the lenses is, 1) 20 cm 2) 3 cm 3) 10 cm, 4) 45 cm, 23. Two thin lenses when in contact produce a, combination of power + 10 D. When they are, 0.25 m apart, the power is reduced to + 6 D., The power of the lenses in diopters are, 1) 1 and 9 2) 2 and 8 3) 4 and 6 4) 5,5, 24. A beam of light converges at a point P. Now a, convex lens of focal length 20 cm placed in, the path of the convergent beam 12 cm from, P. The point at which the beam converges now, is, 1) 7.5cm right side of the lens, 2) 7.5 cm left side of the lens, 3) 15.2 cm right side of lens, 4) 15.2 cm left side of lens, 197
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GEOMETRIC OPTICS, , LENS MAKERS FORMULA, 25. The radius of curvature of the convex surface, of a thin plano - convex lens is 15 cm and the, refractive index of its material is 1.6. The, power of the lens is, 1) +1 D 2) -2 D, 3) +3 D, 4) +4 D, 26. A double convex lens is made of glass which, has refractive index 1.55 for violet rays and, 1.50 for red rays. If the focal length of violet, rays is 20 cm, the focal lenght of red rays is, 1) 9 cm 2) 18 cm 3) 20 cm, 4) 22 cm, 27. The refractive index of the material of a double, convex lens is 1.5 and its focal length is 5 cm., If the radii of curvatures are equal, the value, of radius of curvature is (in cm), 1) 5, 2) 6.5, 3) 8, 4) 9.5, 28. A diverging meniscus lens of 1.5 refractive, index has concave surfaces of radii 3 and 4, cm. The position of image if an object is placed, 12cm infront of the lens is, 1) -24 cm 2) -8 cm, 3) 8 cm, 4) 24 cm, , REFRACTION THROUGH PRISM, 29. A prism has a refracting angle of 600 . When, placed in the position of minimumm deviation,, it produces a deviation of 300 . The angle of, incidence is, 1) 300, 2) 450, 3) 150, 4) 600, 30. Light falls at normal incidence on one face of, a glass prism of refractive index 2 . Then, the angle of emergence when the angle of the, prism is 450, 1) 450, 2) 600, 3) 750, 4) 900, 31. If a light ray incident normally on one of the, faces of the prism of refractive index 2 and, emergent ray just grazes the second face of, the prism, then the angle of deviation is, 1) 00, 2) 300, 3) 600, 4) 900, 32. A ray of light passes through an equilateral, prism such that the angle of incidence is equal, to angle of emmergence and the latter is equal, 3, to th of the angle of the prism. The angle of, 4, deviation is, 1) 450, 2) 390, 3) 200, 4) 300, 33. A ray of light incident on an equilateral prism, shows minimum deviation of 300 . The speed, of light through the prism is, 1) 2.121108 ms 1, 2) 1.50 108 ms 1, 3) 1.25 108 ms 1, 4) 1.75 108 ms 1, 198, , JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, 34. A thin prism of 40 angle gives a deviation of, 2.40 . The value of refractive index of the, material of the prism is, 1) 1.6, 2) 1.7, 3) 1.8, 4) 1.9, , DISPERSION BY A PRISM, 35. A thin prism P1 of angle 40 and refractive, index 1.54 is combined with another thin prism, P2 of refractive index 1.72 to produce, dispersion without deviation. The angle of P2, is, 1) 40, 2) 5.330, 3) 2.60, 4) 30, 36. A crown glass prism with refracting angle 60, is to be achromatised for red and blue light, with flint glass prism. Angle of the flint glass, prism should be ( Given for grown glass, r 1.513 , b 1.523 , for flint glass, , r 1.645 , b 1.665 ), 1) 1.50, 2) 3.30, 3) 2.40, 4) 4.50, 37. If the ratio of dispersive powers of the, materials of two prisms be 2 : 3 and ratio of, angular dispersions produced by them is 1: 2 ,, then the ratio of mean deviation produced by, them is, 1) 4 : 3, 2) 3: 4, 3) 1: 3, 4) 3 :1, 38. Dispersive power of the material of a prism is, 0.0221 . If the deviation produced by it for, yellow colour is 380 , then the angular, dispersion between red and violet colours is, 1) 0.650 2) 0.840, 3) 0.480, 4) 1.260, , DEFECTS OF THE EYE, 39. A person can see clearly upto 1m. The nature, and power of the lens which will enable him to, see things at a distance of 3 m is, 1) concave, 0.66 D 2) convex, 0.66 D, 3) concave, 0.33 D 4) convex, 0.33 D, 40. The far point of a myopic eye is 10 cm from, the eye. The focal length of a lens for reading, at normal distance ( 25 m ) is, 1) 8.35 cm, 2) 16.7 cm, 3) 35.4 cm, 4) .32.7 cm, 41. A person can see clearly objects between 15, and 100 cm from his eye. The range of his, vision if he wears close fitting spetancles, having a power of 0.8 diopter is, 1) 5 to 500 cm, 2) 12 to 250 cm, 3) 17 to 500 cm, 4) 17 to 250 cm, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , OPTICAL INSTRUMENTS, ( MICROSCOPES ), 42. The focal length of convex lens is 10 cm. Its, magnifying power when it is used as a, magnifying glass to form the image at (i) near, point and (ii) far point is, 1) 3.5; 2.5 2) 2.5;3.5 3) 2.5;1.5 4) 1.5; 2.5, 43. A magnifying glass is made of a combination, of a convergent lens of power 20 D and, divergent lens of power 4D. If the least, distance of distinct vision is 25 cm. The, magnifying power is, 1) 7, 2) 5, 3) 3, 4) 4, 44. Four lenses A, B, C and D power 100 D ,, 50 D , 20 D and 5 D . Which lenses will you, use to design a compound microscope for best, magnification ?, 1) A and C 2) B and D 3) C and D 4) A and B, 45. The objective lens of a compound microscope, produces magnification of 10. In order to get, an over all magnification of 100 when image, is formed at 25 cm from the eye, the focal, length of the eye lens should be ( in cm ), , 25, 4) 9, 9, 46. A compound microscope has a magnifying, power of 100 when the image is formed at, infinity. The objective has a focal length 0f 0.5, cm and the tube length is 6.5 cm. Then the, focal length of the eye-piece is, 1) 2 cm 2) 2.5 cm 3) 3.25 cm 4) 4 cm, , 1) 4, , 2) 10, , 49. The magnifying power of terrestrial telescope, is 25 when it is in normal adjustment and the, length of the telescope is 124 cm. If the focal, length of the errecting lens is 5 cm, the focal, lengths of the objective and the eye-piece are, respectively., 1) 50 cm, 2 cm, 2) 50 cm, 2.5 cm, 3) 100 cm, 4 cm, 4) 100 cm, 5 cm, , LEVEL - I (C. W ) - KEY, 1) 3, 7) 4, 13) 3, 19) 1, 25) 4, 31) 3, 37) 2, 43) 2, 49) 3, , 47. The focal lengths of the eyepiece and the, objective of an astronimical telescope are 2, cm and 100 cm respectively. The magnifying, power of the telescope for normal adjustment, and the length of the telescope is, 1) 50; 102 cm, 2) 100; 204 cm, 3) 25 ; 62 cm, 4) 75 ; 125 cm, 48. The magnifying power of an astronimical, telescope is 5, the focal power of its eye piece, is 10 diopters. The focal power of its objective, ( in diopters ) is, 1) 1, 2) 2, 3) 3, 4) 4, , 2) 1, 8) 1, 14) 3, 20) 4, 26) 4, 32) 4, 38) 2, 44) 1, , 3) 3, 9) 3, 15) 3, 21) 1, 27) 1, 33) 1, 39) 1, 45) 3, , 4) 2, 10) 1, 16) 4, 22) 3, 28) 2, 34) 1, 40) 2, 46) 3, , 5) 3, 11) 2, 17) 1, 23) 2, 29) 2, 35) 4, 41) 3, 47) 1, , 6) 1, 12) 1, 18) 1, 24) 1, 30) 4, 36) 2, 42) 1, 48) 2, , LEVEL - I (C. W ) - HINTS, 1., 2., , 3), , OPTICAL INSTRUMENT, ( TELESCOPES ), , NARAYANAGROUP, , GEOMETRIC OPTICS, , 3., , 4., , 5., , 6., , 360, 1, , (i) When the incident ray is fixed and mirror rotates, through angle 200 clock wise then reflected ray, rotates clock wise through 400 angle., (ii) when mirror (M) in fixed and incident ry rotates, through angle 100 clockwise then reflected ray, rotates through angle 100 anticlockwise., Total angle turned by the reflected ray, 400 clockwise + 100 anti clockwise, 300 ( clockwise ), For a plane mirror, Object distance = Image distance, n, , 1 11 , , 1, f u m , m 3; u 20 cm, , 1 1 1, , f v u, (Apply necessary sign convention), 1 1 1, ;, f v u, m, , v, u, , 199
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 7., 8., , 1 1 1, ; mv, f v u, u, For plane mirror,, object distance = image distance, For convex mirror v 24 20 4cm, , u 20cm then, , 1 1 1, , f v u, , (Apply necessary sign convention), , j , , j, i, , g , , g, w, , 9., , i, , 10., , w, , g Cw w, , , w C g g, , x, x = thickness, c, 12. tan i, 13. Apparent distance of fish from bird, real depth, y, , , 15. r , 16., , r, , Sini, ;d i r, Sinr, h, , 18. , , 1 1.5 1 1.5, , , v 3 5 , , 100 1 1 1, , 20. p p1 p 2 ; f , p, f v u, (Apply necessary sign convention), 21. Focal length is doubled, power is halved, 22. d f1 ~ f 2, 23. p p1 p2 p ' p1 p2 d p1 p2, , 1 1, , f, R, , 200, , , , , , Sin i2, Sin r2, , 1, ; r2 A d i i A, 1, 2, sin r2, , 32. D 2i A, , AD, 2 ; C, A, C med, sin, 2, , sin, 33., , , , 1 1 A1 2 1 A2, b r Ac b ' r ' A f, , A ngular d ispersion, D, 38. Angular Dispersion v r y, , 37. , , 1, 19. 2 1 2, v, u, R, , 25., , A D, 2, , 31. r1 0;, , 36., , S in i, t, sin (i-r), ; shift , S in r, cosr, , 1 1 1, , f v u, , 1 1 1 1 1, 1, 1 ;, , f, R1 R2 f v u, , 30. r1 0 r2 A 450, , 35., , 1, , r, sin c d, , 24., , 28., , 1 1 , 1, 1 , f, R1 R2 , (Apply with sign convention), , 34. 1 A, , 2 1, , d , , 27., , f R V 1, , fV R 1, , 29. i , , 11. time , , 14. , , 26., , 1 1 1, , 39. u 3m v 1m, f v u, 1 1 1, , 40. u 25cm v 10cm, f v u, 41. p , , 1 1 1 1, ; , f, f v u, , 42. M 1 , , D, D, ;M , f, f, , D, 43. P p1 p2 ; M 1 , f, p, , 100, f, , f0 D , 44. M f 1 f , e , e , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , , D, 45. M M 0 M e ; M M 0 1 , fe , , , GEOMETRIC OPTICS, 7., , LD, , 46. M f f, 0 e, f0, 47. M f ; L f 0 f e, e, , 8., , pe, 48. M p, 0, , An object is placed at a distance 2f from the, pole of a convexmirror of focallength f. The, linear magnification is, 1, 2, 3, 1), 2), 3), 4) 1, 3, 3, 4, An object ‘O’ is placed infront of a small plane, mirror M1 and a large convex mirror M 2 of, focal length f. The distance between ‘O’ and, M1 is x and the distance between M1 and, , M 2 is y. The image of ‘O’ formed by M1 and, M 2 coincide. The magnitude of ‘f’ is, , f0, 49. M f ; L f 0 4 f f e, e, , M1, , LEVEL - I (H.W), REFLECTION, 1., , 2., , 3., , 4., , 5., , 6., , A ray reflected successively from two plane, mirrors inclined at a certain angle undergoes, a deviation of 3000 . The number of, observable images, 1) 60, 2) 12, 3) 11, 4) 5, If a plane mirror is rotated in its ownplane, through an angle of 200 keeping the incident, ray direction fixed, then the angle through, which the reflected ray turns is, 1) 400, 2) 00, 3) 200, 4) 100, A man runs towards a mirror at a rate of, 6ms 1 . If the mirror is at rest, his image will, have a velocity (with respect to man), 1) + 12 ms 1, 2) 6ms 1, 3) 6 ms 1, 4) 12ms 1, A real image formed by a concave mirror is, 4.5 times the size of the object. If the mirror, is 20 cm from the object, its focallength is, 90, 120, 150, 180, 1), cm 2), cm 3), cm 4), cm, 11, 11, 11, 11, A point object is placed at a distance of 30, cm from a convex mirror of focal lenght 30, cm. The image will be formed at, 1) Infinty, 2) Focus, 3) Pole, 4) 15 cm behind the mirror, An object is placed at 5 cm infront of a, concave mirror of radius of curvature 15 cm., The position of image (v) and its, magnification (m) are, 1) v = 15 cm; m = 3 (virtual, erect), 2) v = 5 cm; same size (virtual, erect), 3) v = 5 cm; same size (real, inverted), 4) v = 15 cm; m = 3 (real, inverted), , NARAYANAGROUP, , O, x, y, , M2, , x 2 y2 x 2 y2, x 2 y2, 1) x y 2), 3), 4), 2y, 2y, xy, REFRACTION, 9., , The refractive indices of glass and water are, 3, 4, and, respectively. The refractive index, 2, 3, of glass with respect to water is, 8, 9, 5, 1) 2, 2), 3), 4), 9, 8, 3, 10. The velocity of light in glass whose refractive, index with respect to air is 1.5 is 2 108 ms 1 .In, a certain liquid the velocity of light is found to, be 2.5 108 ms 1 . The refractive index of the, liquid with respect to air is, 1) 0.64 2) 0.80, 3) 1.20, 4) 1.44, 11. The optical path of monochromatic light is the, same if it goes through 2 cm of glass or x cm, of ruby. If the refractive index of glass is 1.510, and that of ruby is 1.760, then x is, 1)1.716cm 2)1.525cm 3)2.716cm 4) 2.525cm, 12. The reflected and refracted rays are observed, to be perpendicular to each other, when ray of, light is incident at an angle of 600 on a, transparent block. The refractive index of the, block is, 2, 3, 1, 1), 2), 3), 4) 3, 3, 2, 2, 201
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 13. In a lake, a fish rising vertically to the surface, of water uniformly at the rate of 3ms 1, observes a bird diving vertically towards the, water at a rate of 9ms 1 vertically above it. The, actual velocity of the dive of the bird is, 4, , water , 3, , 1) 9.2ms 1 2) 4.5ms 1 3) 9ms 1 4) 3.2ms 1, 14. If the angle of incidence is twice the angle of, refraction in a medium of refractive index, ' ' then the angle of incidence is, , , 1 , 1 , 1 , 1) Cos, 2) Sin, 3) 2Sin, 4) 2Cos, 2, 2, 2, 2, 15. A glass cube of edge 1 cm and 1.5 has a, sopt at the centre. The area of the cube face, that must be covered to prevent the spot from, being seen is ( in cm 2 ), , , 1) 5 2) 5, 3), 4), 5, 5, 16. The velocities of light in two different media, are 2 108 ms 1 and 2.5 108 ms 1 respectively,,, the critical angle for these media is, 1, , 1 1 , 1) S in , 5, , 1 4 , 2) S in , 5, , 1 1 , 1 1 , 3) S in , 4) S in , 2, 4, 17. White light is incident on the interface of glass, and air as shown in the figure. If green light is, just totally internally reflected then the, emergent ray in air contain, , Green, Glass, White, , Air, , 1) Yellow, orange, red 2) Violet, indigo, blue, 3) All colours, 4) All colours except green, 18. A ray of light is incident on a glass plate. The, light ray travels distance of 5 cm inside the, glass plate before emerging out of the glass, plate. If the incident ray suffers a deviation of, 300 , the perpendicular distance between, incident and the emergent ray is, 1) 5 cm 2) 2.5 cm 3) 7.5 cm 4) 10 cm, 202, , REFRACTION THROUGH, SPHERICAL SURFACES, 19. Light from a point source in air falls on a, spherical glass surface ( 1.5 and radius of, curvature 20 cm). The distance of the light, source from the glass surface is 100 cm. The, position where image is formed is, 1) 50 cm, 2) 100 cm, 3) 125 cm, 4) 25 cm, , LENSES AND THEIR COMBINATION, 20. A divergent lens produces an image of, magnification 0.5 when the object distance is, 10 cm. The focal power of the lens (in diopters), 1) + 4, , 2) - 4, , 3) + 10, , 4) - 10, , 21. A symmetrical biconvex lens of focal length, ‘f’ is cut into four identical pieces along its, principal axis and to the perpendicular to, principal axis. The focal length of one of four, pieces is, f, f, 2), 3) 2 f, 4) 4 f, 4, 2, 22. A converging lens of focallength 20 cm and a, diverging lens of focallength 5 cm are placed, coaxially at distacne ‘d’ apart. If a parallel ray, of light is incident on the converging lens and, it emerges out of the diverging lens as a, parallel ray, then the value of ‘d’ will be, 1) 30 cm 2) 25 cm 3) 15 cm, 4) 50 cm, 23. Two lenses of power -15 D and +5 D are in, contact with each other. The focal length of, the combination is, 1) +10 cm 2) -20 cm 3) -10 cm, 4) +20 cm, 24. A beam of light converges at a point P. Now a, concave lens of focal length -16 cm is placed, in the path of the convergent beam 12 cm from, P. The point at which the beam converges now, is, 1) 6.86cm right side of the lens, 2) 6.86cm left side of the lens, 3) 48cm right side of the lens, 4) 48cm left side of the lens, , 1), , LENS MAKERS FORMULA, 25. The radius of curvature of convex surface of, plano convex lens is 10 cm and its focal length, is 30 cm, then the refractive index of the, material of the lens is, 1) 3, 2) 1.5, 3) 1.66, 4) 1.33, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 26. Focal length of a lens is 0.12 m and refractive, index is 1.5. Focal length of the same lens for, blue colour is 0.1m. Then refractive index of, the lens for blue colour is, 1) 1.51 2) 1.25, 3) 1.49, 4) 1.6, 27. The focal length of a biconvex lens is 20 cm, and its refractive index is 1.5. If the radii of, curvatures of two surfaces of lens are in the, ratio 1:2, then the larger radius of curvature, is (in cm), 1) 10, 2) 15, 3) 20, 4) 30, 28. The radii of curvature of the two surfaces of, a lens are 20 cm and 30 cm and the refractive, index of the material of the lens is 1.5. If the, lens is concavo convex then the focal length, of the lens is, 1) 24 cm 2) 10 cm 3) 120 cm 4)24cm, , REFRACTION THROUGH PRISM, 29. The angle of a prism is 300 . The rays incident, at 600 at one refracting face suffer a deviation, of 300 . The angle of emergence is, 2) 300, 3) 600, 4) 900, 1) 00, 30. Light falls on a prism grazing along first, surface of a prism and the emergent ray is, normal to the 2nd face of the prism. If D is, angle of deviation then the refracting angle, of the prism is, D, 1) 90 -2D 2) 90 -D 3) 90 4) 180 2D, 2, 31. A ray of light is incident normally on one of, the refracting surfaces of prism of refracting, angle 600 . The emergent ray grazes the other, refracting surface. The refractive index of the, material of the prism is, 1) 1.155 2) 1.75, 3) 1.33, 4) 1.66, 0, 32. A ray of light is incident at 60 on a prism of, refracting angle 300 . The emerging ray is at, angle 300 with incident ray, the value of ' ', of prism is, 3, 3, 2), 3) 3, 4) 2 3, 4, 2, 33. When light rays are incident on a prism at an, angle of 450 , the minimum deviation is, obtained. If the refractive index of the, material of the prism is 2 , then the angle, of prism will be, 1) 300, 2) 450, 3) 600, 4) 900, , 1), , NARAYANAGROUP, , GEOMETRIC OPTICS, 34. A thin prism deviates an incident ray by 3.20 ., If the refractive index of the prism is 2.6 then, the angle of prism is, 1) 10, , 2) 20, , 3) 30, , 4) 40, , DISPERSION BY A PRISM, 35. A crown glass prism and a flint glass prism, are combined to produce dispersion without, deviation. Mean refractive indices of crown, and flint glass are respectively 1.5 and 1.6., Ratio of angle of crown glass prism to that of, flint prism is, 1) 1.06 2) 0.9375 3) 1.2, 4) 1.5, 36. A crown glass prism of angle 50 is to be, combined with a flint glass prism in a such a, way that the dispersion is zero.The refractive, indices for violet and red lights are 1.523 and, 1.514 respectively for crown glass and for flint, glass are 1.632 and 1.614, then the angle of, the flint glass prism is, 1) 100, 2) 2.50, 3) 20, 4) 5.450, 37. In an achromatic combination of two prisms,, the ratio of the mean deviations produced by, the two prisms is 2 : 3 , the ratio of their, dispersive power is, 1) 2 : 3, 2) 3: 2, 3) 1:1, 4) 4 : 9, 38. The angles of minimum deviations are 530 and, 510 for blue and red colours respectively, produced in an equilateral glass prism. The, dispersive power is, 1), , 51, 26, , 2), , 1, 26, , 3), , 1, 52, , 4), , 1, 51, , DEFECTS OF THE EYE, 39. The near point of a hypermetropic person is, 50 cm from the eye. The power of the lens, required to enable the person to read clearly, a book held at 25 cm from the eye is, 1) 2 D, 2) 4 D, 3) 8 D, 5) 1 D, 40. A person wears glasses of power 2.5 D . Is, the person far-sighted or near-sighted ? The, far point of the person without glasses is, 1) long-sighted, 40 cm, 2) near-sighted, 40 cm, 3) near-sighted, 20 cm, 4) long-sighted, 20 cm, 203
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 41. A long sighted person has a least distance of, distinct vision of 50 cm. He wants to reduce to, 25 cm. He should use a, 1) concave lens of focal length 50 cm, 2) convex of focal length 25 cm, 3) convex lens of focal length 50 cm, 4) concave lens of focal length 25 cm, , OPTICAL INSTRUMENTS, ( MICROSCOPES ), 42. The maximum magnification that can be, obtained with convex lens of focal length 2.5, cm is ( the least distance of distanct vision is, 25 cm ), 1) 10, 2) 0.1, 3) 62.5, 4) 11, 43. A convergent lens of power 16D is used as a, simple microscope. The magnification, produced by the lens, when the final image is, formed at least distance of distinct vision is, 1) 6, 2) 4, 3) 7, 4) 5, 44. A compound microscope is of magnifying, power 100. The magnifying power of its, eyepiece is 4. Find the magnification of its, objective., 1) 25, , 2) 20, , 3) 15, , 4) 30, , 45. The magnification produced by the objective, lens of a compound microscope is 25. The focal, length of eye piece is 5 cm and it forms find, image at least distance of distinct vision. The, magnifying power of the compound microscope, is, 1) 19, , 2) 31, , 3) 150, , 4) 150, , 46. The length of the tube of a compound, microscope 15 cm. The focal length of, objective and eye lenses are 1 cm and 5 cm, respectively. The magnifying power of, microscope for relaxed vision is, 1) 50, , 2) 75, , 3) 25, , 4) 100, , 48. The astronomical telescope has two lenses of, focal powers 0.5 D and 20 D. Its magnifying, power will be, 1) 40, 2) 30, 3) 20, 4) 8, 49. The objective of a terrestrial telescope has, focal length of 120 cm and diameter 5 cm. The, focal length of the eye piece is 2 cm. The, magnifying power of telescope for distant, object is, 1) 12, 2) 24, 3) 60, 4) 300, , LEVEL - I ( H. W ) - KEY, 1) 3, 7) 1, 13) 2, 19) 2, 25) 4, 31) 1, 37) 2, 43) 4, 49) 3, , 204, , 3) 4, 9) 3, 15) 4, 21) 3, 27) 4, 33) 3, 39) 1, 45) 3, , 4) 4, 10) 3, 16) 2, 22) 3, 28) 3, 34) 2, 40) 2, 46) 2, , 5) 4, 11) 1, 17) 1, 23) 3, 29) 1, 35) 3, 41) 3, 47) 3, , 6) 1, 12) 4, 18) 2, 24) 3, 30) 2, 36) 2, 42) 4, 48) 1, , LEVEL - I ( H. W ) - HINTS, 1., 2., 3., 4., 5., 6., 7., 8., , OPTICAL INSTRUMENT, ( TELESCOPES ), 47. The magnifying power of an astronomical, telescope for relaxed vision is 16 and the, distance between the objective and eyelens is, 34 cm. The focal length of objective and, eyelens will be respectively are, 1) 17 cm, 17 cm, 2) 20 cm, 14 cm, 3) 32 cm, 2 cm, 4) 30 cm, 4 cm, , 2) 2, 8) 2, 14) 4, 20) 4, 26) 4, 32) 3, 38) 2, 44) 1, , 360, D 360 2 n 1, If the plane mirror is rotated in its own plane, then, the direction of reflected ray does not change., Relative velocity = 2v, 1 11 , , 1 m 4.5 ; u 20 cm, f u m , 1 1 1, , f v u, , 1 1 1, ; mv, f v u, u, 1 1 1, ; mv, f v u, u, Let ‘v’ be the image distance of convex mirror., for plane mirror, Object distance = Image distance, , x yvvxy, For convex mirror u x y, v x y, 1 1 1, Then , f v u, 9., , w, , g , , g, w, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 10., , g, , l , , l C g, , g Cl, , 11. 1 x1 2 x2, 12. tan i, 13. Apparent distance of bird as seen by fish, h xy, , 14. , , dh dx, dy, dy, , , 9 3 , dt dt, dt, dt, , Sini, i, ; r, Sinr, 2, , 15. r 2 1 ; A r 2, 1, C, r, sin c Cd, 17. Yellow, Orange, Red colours undergo TIR., 18. Shift = Distance [ sin (i-r)], r, , d , , d ir, , 1, ; C r2 A ; as r1 0 , SinC, 32. d i1 i2 A ; i2 0 ; r2 0 ; r1 A, , 31. , , , , 300, , 2 1 2 1 1.5 1 1.5 1, , 19., ; v 100 20 , v, u, R, 1 1 1, 100, , 1 ; p , 20., f um , f, , Sin i1, Sin r1, , Sin i1, A, 33. Sin r ; r , 2, 1, , 34. 1 A, 35., , h, , 16., , GEOMETRIC OPTICS, , 36., , 1 1 A1 2 1 A2, v R A v ' v ' A ', , 1, 37. D, m, , 38., , , , DV DR, D DR, ; where Dm v, Dm, 2, , 1 1 1, 39. u 25cm ; v 50cm ; , f v u, 40. f = farpoint =, , 100, p, , 22. d f1 ~ f 2, , 1 1 1, 41. u 25 cm ; v 50cm ; , f v u, , 100, 23. p p1 p2 ; f , p, , 42. M 1 , , D, f, , 24., , 1 1 1, , f v u, , 43. f , , 25., , 1, 1, 1, f, R, , 44. M M 0 M e, , 26., , f b 1, , fb 1, , 1 1 , 1, 27. f 1 R R , 1, 2 , 28., , 1 1 , 1, 1 , f, R1 R2 , , 29. d i1 i2 A, 30. d i1 i2 A, , NARAYANAGROUP, , 100, D, ; M 1, p, f, , , D, 45. M M 0 M e ; M M 0 1 f , , e , LD, 46. M f f, o e, f0, 47. M f ; L f 0 f e, e, pe, 48. M p, 0, f0, 49. M f, e, 205
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 5., , LEVEL - II (C.W), REFLECTION, 1., , Two plane mirrors are arranged at right angles, to each other as shown in figure. A ray of light, is incident on the horizontal mirror at angle, ‘ ’. For what value of the ray emerges, parallel to the incoming ray after reflection, from the vertical mirror ?, , Two plane mirrors A and B are aligned parallel, to each other, as shown in the figure. A light, ray is incident at an angle 300 at a point just, inside one end of A. The plane of incidence, coincides with the plane of the figure. The, maximum number of times the ray undergoes, relfections ( including the first one ) before it, emerges out is, , 2 3m, , , , B, , 0.2 m, , 30°, A, , 2., , 1) 600, 2) 300, 3) 450 4) all of these, An object moves with 5 m/s towards right while, the mirror moves 1 m/s towards the left as, shown. Then the velocity of image., , 6., , object, 5 m/s, , 3., , 1) 7 m/s towards left, 2) 7m/s towards right, 3) 5 m/s towards right 4) 5 m/s towards left, Two mirrors labelled L1 for left mirror and L2, for right mirror in the figure are parallel to, each other and 3.0 m apart. A person standing, , 7., , 1, , u f , 2) b , , f , 8., , 4., , 206, , 1m, , 1) 2.0 m from the person, 2) 4.0 m from the person, 3) 6.0 m from the person, 4) 8.0 m from the person, A ray of light is incident at 500 on the middle, of one of the two mirrors arranged at an angle, of 600 between them. The ray then touches, the second mirror, get reflected back to the, first mirror, making an angle of incidence of, 1) 500, 2) 600, 3) 700, 4) 800, , 2), , u f 2, 1) b , , f , , L2, , 2m, , x1, x1 x2, 3) x, 4) x1 x2, 2, 2, A short linear object of length ‘b’ lies along, the axis of a concave mirror of focal length ‘f’, at a distance ‘u’ rom the pole of the mirror., The size of the image is approximately equal, to, , 1) x1 x2, , 1.0 m from the right mirror L2 looks into, this mirror and sees a series of images. The, second nearest image in the right mirror is, situated at a distance, L1, , 1) 28, 2) 30, 3) 32, 4) 34, With a concave mirror, an object is placed at a, distance x1 from the principal focus, on the, principal axis. The image is formed at a, distance x2 from the principal focus. The focal, length of the mirror is, , 9., , 1/ 2, , f , 2) b , , u f , , 2, , f , 4) b , , u f , A rod of length 10 cm lies along the principle, axis of a concave mirror of focal length 10 cm, is such a way that the end closer to the pole is, 20 cm away from it. The length of the image is, 1) 5 cm, 2) 10 cm 3) 15 cm 4) 20 cm, A car is fitted with a convex side view mirror, of focal length 20 cm. Asecond car 2.8 m behind, the first car is overtaking the first car at, relative speed of 15 m/s. The speed of the, image of the second car as seen in the mirror, of the first one is :, 1, 1, m / s 2) 10 m/s 3) 15 m/s 4), m/s, 1), 15, 10, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 10. The velocity of image w.r.t ground in the below, figure is, , 5 m/s, O, , f = 30 cm (Mirror is at rest, given velocity, w.r.t ground), , 20 cm, , 1) 45 m/s and approaches the mirror, 2) 45 m/s and moves away from the mirror, 3) 60 m/s and approaches the mirror, 4) 60 m/s and moves away from the mirror, 11. A square wire of side 3.0 cm is placed 25 cm in, front of a concave mirror of focal length 10, cm with its centre on the axis of the mirror, and its plane normal to the axis. The area, enclosed by the image of the wire is, 1) 7.5cm 2 2) 6.0 cm 2 3) 4.0 cm 2 4) 3.0 cm 2, 12. An object is moving towards a concave mirror, of length 24 cm. When it is at a distance of 60, cm from the mirror its speed is 9 cm/s, the, speed of its image at that instant is, 1) 4 cm/s towards the mirror, 2) 9 cm/s towards the mirror, 3) 4 cm/s away from the mirror, 4) 9 cm/s away from the mirror, , REFRACTION, 13. A monochromatic light passes through a glass, slab 1.5 of thickness 9 cm in time t1 . If, it takes a time t2 to travel the same distance, , through water , , is, , 4, . The value of t1 t2 , 3, , 1) 5 1011 sec, 2) 5 108 sec, 3) 2.5 1010 sec, 4) 5 1010 sec, 14. A glass slab of thickness 4cm contains the, same number of waves as 5 cm of water when, both are traversed by the same monochromatic, 4, light. If the refractive index of water is , then, 3, that of glass is, 1), , 5, 3, , 2), , NARAYANAGROUP, , 5, 4, , 3), , 16, 15, , 4), , 3, 2, , 15. When light of wavelength 4000 A0 in vacuum, travels through the same thickness in air and, vacuum the difference in the number of waves, is one. Find the thickness air 1.0008 ., 1) 0.5 mm 2) 1mm 3) 18 cm 4) 24 cm, 16. The refractive index of denser medium with, respect to rarer medium 1.125. The difference, between the velocities of light in the two media, is 0.25 108 m / s . Find the velocities of light in, the two media and their refractive indices., , c 3 10 m / s ., 8, , 1) 2.0 108 m / s; 2.25 108 m / s;1.500;1.333, 2) 2.5 108 m / s; 2.25 108 m / s;1.500;1.333, 3) 2.0 108 m / s; 2.25 108 m / s;1.333;1.500, 4) 2.5 108 m / s; 2.0 108 m / s;1.500;1.333, 17. A ray of light is travelling from medium ‘A’, into a rarer medium B. The angle of incidence, is 450 and the angle of deviation is 150 . The, refractive index of medium A w.r.to B is, , 1, 3, 2, 3, 2), 3), 4), 2, 2, 3, 2, 18. The X-Z plane separates two media A and B, of refractive indices A 1.5 and B 2 . A, ray of light travels from A to B, its direction in, two media are given by unit vectors., u A ai b j and u B ci d j respectively,,, 1), , then, a 4, a 3, b 4, b 3, , 2) , 3) 4) , c 3, c 4, d 3, d 4, 19. A cube of side 15 cm is having an air bubble., The bubble appears at 6 cm from one face and, at 4 cm from opposite face. The refractive, index of cube is, 5, 3, 2, 2, 1), 2), 3), 4), 2, 2, 3, 5, 20. Refractive index of a rectangular glass slab, is 3 . A light ray incident at an angle 600, is displaced laterally through 2.5 cm. Distance, travelled by light in the slab is, 1) 4 cm, 2) 5 cm, 3) 2.5 3 cm 4) 3 cm, , 1), , 207
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 21. A beaker contains water up to a height h1 and 25. A point source of light is placed at the bottom, of a water lake. If the area of the illuminated, kerosene of height h2 above water so that the, circle on the surface is equal to 3 times the, square of depth of the lake, the refractive, total height of ( water + kerosene ) is h1 h2 ., index of water., Refractive index of water is 1 and that of, , , , 1 3) 1 4) 1, 1) 1 2), kerosene is 2 . The apparent shift in the, 3, 4, 3, position of the bottom of the beaker when, 26. A ray of light from a denser medium strikes a, viewed from above is :, rarer medium at an angle of incidence ‘i’ if the, reflected and refracted rays are mutually, , , 1, 1 , 1) 1 h1 1 h2, perpendicular to each other then the critical, , 1, , 2 , angle is, , , 1, 1 , 1) sin 1 tan i , 2) cos 1 tan i , 2) 1 h2 1 h1, , 1, , 2 , 3) cot 1 tan i , 4) cosec1 tan i , , , 1, 1 , 27. A prism of R.I 1.5 is immersed in water of, 3) 1 h2 1 h1, 4, , 1, , 2 , R.I as shown in the figure. For the total, 3, , , 1, 1 , internal, reflection the correct choice is, 4) 1 h1 1 h2, , 1, , 2 , 22. Light ray is travelling from a denser medium, B, A, into a rarer medium. The velocity of light in, , 8, the denser and rarer medium is 2 10 m / sec ., and 2.5 108 m / sec . The critical angle of the, two media is, 5, 1) sin , 4, 1, , 4, 2) sin , 5, , C, , 1, , 8, 8, 2) sin , 9, 9, 1 1 , 1 3 , 8, 8, 3) sin , 4) sin , 3) sin , 4) sin , 2, 5, 9, 9, 23. Light takes time t1 to travel a distance x1 in 28. A light ray is incident at an angle 450 on, parallel sided glass slab and emerges out, vacuum and the same light takes time t2 to, grazing the vertical surface. The refractive, travel a distance x2 in a medium. The critical, index of the slab is, angle for that medium is, 3, 5, 3, 5, 1), 2), 3), 4), 1 x2 t 2 , 1 x1t 2 , 2, 2, 2, 2, 1) sin x t , 2) sin x t , 11 , 21, 29. The critical angle for refraction from medium, 1 to air is 1 and that from medium 2 to air, 1 x1t1 , 1 x2t1 , 3) sin x t , 4) sin x t , is 2 . If medium 2 is denser than medium, 22, 12, 1 . Find the critical angle for refraction from, 24. An under water swimmer looks upward at an, medium 2 to medium 1., unobstructed overcast sky. The vertex angle, does the sky appear to subtend at the eye of, 1 sin 2 , 1 sin 1 , swimmer is (critical angle for water air, 1) sin sin , 2) sin sin , interface is C), , 1 , , 2 , 1) C, 2) C/2, 3) 2C, 4) 3C, 3) sin 1 sin 2 , 4) sin 1 sin 1 , 208, , 1) sin , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 30. A transparent solid cylindrical rod has a, 2, refractive index of, . It is surrounded by, 3, air. A light ray is incident at the midpoint of, one end of the rod as shown in the figure., The incident angle for which the light ray, grazes along the wall of the rod is, , , 3, 1, 1) sin 2 , , , , 1 2 , 2) sin , , 3, , 1 1 , 1 1 , 3) sin , 4) sin , , 2, 3, 31. A ray of light refracts from medium 1 into a, thin layer of medium 2, crosses the layer and, is incident at the critical angle on the interface, between the medium 2 and 3 shown in the, figure. If the angle of incidence of ray is ,, the value of is, , 1=1.6, 1, 2, , 2=1.8, , 3, , 3=1.3, , 1 8 , 1) sin , 9, 1 13 , 3) sin , 16 , , 1 13 , 2) sin , 18 , 1 8 , 4) sin , 13 , , 32. A spherical convex surface of radius of, curvature R separates air a 1 from glass, , g 1.5 . The centre of curvature is in the, glass. A point object P placed in air is found to, have a real image Q in the glass. The line PQ, cuts the surface at a point O and PO OQ ., The distance PO is equal to, 1) 5 R, 2) 3 R, 3) 2 R, 4) 1.5 R, a, , NARAYANAGROUP, , 33. A denser medium of refractive index 1.5 has a, concave surace with respect to air of radius of, curvature 12 cm. An object is situated in the, denser medium at a distance of 9 cm from the, pole locate the image due to refraction in air, 1) A real image at 8 cm, 2) a virtual image at 8 cm, 3) A real image at 4.8 cm, 4) A virtual image at 4.8 cm, 34. The human eye can be regarded as a single, spherical refractive surface of curvature of, cornea 7.8 mm. If a parallel beam of light, comes to focus at 3.075 cm behind the, refractive surface, the refractive index of the, eye is, 2) 1.72, 3) 1.5, 4) 1.61, 1) 1.34, 35. A glass sphere 1.5 of radius 20 cm has, small air bubble 4 cm below its centre. The, sphere is viewed from outside and along, vertical line through the bubble. The apparent, depth of the bubble below the surface of, sphere is (in cm), 4) 30., 1) 13.33 2) 26.67 3) 15, 36. A ray of light falls on a transparent sphere with, centre at C as shown in figure. The ray, emerges from the sphere parallel to line AB., the refractive index of the sphere is, , A, , REFRACTION THROUGH, SPHERICAL SURFACES, , , , GEOMETRIC OPTICS, , 1), , 2, , C, , B, , 60°, 2), , 3, , 3), , 3, 2, , 4), , 1, 2, , LENSES AND THEIR COMBINATION, 0, , 1, 37. The sun subtends an angle of on earth., 2, The image of sun is obtained on the screen, with help of a convex lens of focal length 100, cm. The diameter of the image obtained on, the screen will be, 1) 18 cm 2) 1 mm 3) 50 cm 4) 8.73 mm, 38. An object is placed first at infinity and then at, 20 cm from the object side focal plane of a, convex lens. The two images thus formed are, 5 cm apart. The focal length of the lens is, 1) 5 cm, 2) 10 cm 3) 15 cm 4) 20 cm, , 209
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 39. The image of a square hole in a screen, illuminated by light is obtained on another, screen with the help of converging lens. The, distance of the hole from the lens is 40 cm. If, the area of the image is nine times that of the, hole, the focal length of the lens is, 1) 30 cm 2) 50 cm 3) 60 cm 4) 75 cm, 40. A plano convex lens of focal length 30 cm has, its plane surface silvered. An object is placed, 40 cm from the lens on the convex side. The, distance of the image from the lens is, 1) 18 cm 2) 24 cm 3) 30 cm 4) 40 cm, 41. The graph shows the variation of magnifictaion, m produced by convex lens with image, distance v . The focal length of the lens is used, is :, , m, , b, V, , C, , b, b, bc, c, 2), 3), 4), c, ca, a, b, 42. A convex lens of focal length f is placed, somewhere in between the object and a, screen. The distance between object and, screen is x . If magnification produced is m ,, the focal length of the lens is, mx, mx, 1) m 1 2, 2) m 1 2, , 1), , , , , , m 1, , , , 2, , , , m 1, , 2, , x, 4), x, m, m, 43. The distance between an object and the screen, is 100 cm. A lens produces an image on the, screen when placed at either positions 40 cm, apart. The power of the lens is, 1) 3D, 2) 5D 3) 7D 4) 9D, 44. Three lenses in contact have a combined focal, length of 12 cm. When the third lens is, 60, removed, the combined focal length is, cm., 7, The third lens is, 1) A converging lens of focal length 30 cm, 2) A converging lens of focal elngth 60 cm, 3) A diverging lens of focal length 30 cm, 4) A diverging lens of focal length 60 cm, 3), , 210, , 45. Arrange the following combinations in the, increasing order of focal length, a) Two plano convex lenses of focal lengths 20 cm, and 30 cm in contact, b) Two convex lens of focal lengths 20 cm and 10, cm in contact, c) Two convex lenses of focal length 25 cm, separated by 5 cm, 1) a, b, c 2) b, a, c 3) a, c, b 4) c, a, b, , LENS MAKER’S FORMULA, 46. A plano convex lens acts like a concave mirror, of 28 cm focal length when its plane surface, is silvered and like a concave mirror of 10 cm, focal length when its curved surface is, silvered. The refractive index of the material, of the lens is, 1) 1.50, 2) 1.55, 3) 1.60 4) 1.65, 47. A thin equiconvex lens has focal length 10 cm, and refractive index 1.5. One of its faces is, now silvered, and for an object placed at a, distance u infront of it, the image coincides, with the object. The value of u is, 1) 10 cm 2) 5 cm, 3) 20 cm 4) 15 cm, 48. Four lenses are made of glass and the radius, of curvature of each face is given below., Which will have the greatest focal power., 1) 10 cm convex and 15 cm convex, 2) 5 cm convex and 10 cm concave, 3) 15 cm convex and plane, 4) 20 cm convex and plane, 49. A thin liquid convex lens is formed in glass., 4, and that of, Refractive index of liquid is, 3, 3, glass is . If ‘f’ is the focal length of the, 2, liquid lens in air, its focal length and nature in, the glass is, 1) f, convex, 2) f, concave, 3) 2f, concave, 4) 3f, concave, 50. A thin converging lens of refractive index 1.5, has a focal power of 5 D. When this lens is, immersed in a liquid, it acts as a diverging, lens of focal length 100 cm. The refractive, index of the liquid is, 1), , 11, 6, , 2), , 9, 5, , 3), , 5, 3, , 4) 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 51. The focal lengths of a lens are in the ratio 8 : 3, when it is immersed in two different liquids, refractive indeces 1.6 and 1.2 respectively., The refractive index of the material of the lens, is, 1) 1.25, 2) 1.5, 3) 1.8, 4) 2, 52. A double convex lens of focal length 30 cm is, made of glass. When it is immersed in a liquid, of refractive index 1.4, the focal length is found, to be 126 cm. The critical angle between glass, and the liquid is, 1 3 , 1) sin , 4, , 53., , 54., , 55., , 56., , 57., , 1 4 , 2) sin , 5, , 1 7 , 1 7 , 3) sin , 4) sin , 13 , 8, A plano convex lens has a thickness of 6cm., Its radius of curvature is 25 cm. When its, curved surface is kept on a horizontal surface, and viewed from the top, its bottom appears, to be raised by 2 cm. The focal power of lens, is, 1) 1 D, 2) 4 D, 3) 2 D, 4) 3 D, Two plano-concave lenses of glass of, refractive index 1.5 have radii of curvatures, 20 cm, 30 cm. They are placed in contact with, curved surface towards each other and the, space between them is filled with a liquid of, 4, refractive index . The focal length of the, 3, system is, 1) 48 cm 2) 72 cm 3) 12 cm 4) 24 cm, The power of a double convex lens of radius, of curvature R each is Y. The power of a plano, convex lens of same material and of radius of, curvature 2R is, Y, Y, 1), 2), 3) 2 Y, 4) 4 Y, 4, 2, A thin glass ( refractive index 1.5 ) lens has, optical power of 8 D in air. Its optical power, in a liquid medium with refractive index 1.6, will be, 1) 25 D, 2) 1 D, 3) 25 D 4) 1D, The refractive index of a lens material is , and focal length f . Due to some chemical, changes in the material, its refractive index, has increased by 2%. The percentage, decrease, in focal length for 1.5 will be, 1) 4%, 2) 2%, 3) 6%, 4) 8%, , NARAYANAGROUP, , GEOMETRIC OPTICS, , REFRACTION THROUGH A PRISM, 58. The refractiveindex of a prism for, amonochromatic light is 2 and its refracting, angle is 600 . For minimum deviation the angle, of incidence will be, 1) 300, 2) 450, 3) 600, 4) 750, 59. The minimum deviation produced by a hollow, prism filled with a certain liquid to be 300 . The, light ray is also found to be refracted at angle, 300 . The refractive index of the liquid is, 3, 3, 4), 2, 2, 60. Under minimum deviation condition in a prism,, if a ray is incident at an angle 300 , the angle, between the emergent ray and the second, refracting surface oft he prism is, 1) 00, 2) 300, 3) 450, 4) 600, 61. The angle of deviation by prism is, , 1), , 2, , 180, , 0, , 2), , 3, , 3), , 2 A . Its critical angle will be, , A, 1 , 1) sin tan , 2, , , A, 1 , 2) sin cot , 2, , , A, A, 1 , 1 , 3) cos cot , 4) cos tan , 2, 2, , , 62. ACB is right-angled glass prism of, refractiveindex 1.5. A, B and C are, , 600 , 300 and 900 respectively. A thin layer of, liquid is on the AB. for a ray of light which is, incident normally on AC to be totally reflected, at AB, the refractive index of the liquid on AB, should be, 1) 1.5, 2) 1.4, 3) 1.3, 4) 1.2, 63. A beam of monochromatic light is incident on, one face of the equilateral prism the angle of, incidence being 550 . If the angle of emergence, is 460 , then the angle of minimum deviation, is, 1) 410, 2) 410, 3) 410 4) 410, 64. The maximum refractive index of a prism which, permits passage of the light, through it when, the refractin angle of the prism is 900 , is, 1), , 3, , 2), , 2, , 3), , 3, 2, , 4), , 3, 2, 211
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 65. The refractive index of the material of prism, is 2 and its refracting angle is 300 . One of, the refracting surfaces of the prism is made a, mirror in wards. A beam of monochromatic light, enters the prism from the other surface and, the ray retraces from the mirrored surface., The angle of incidence is, 1) 300, 2) 450, 3) 600, 4) 00, 66., , 67., , 68., , 69., , OPTICAL INSTRUMENTS (MICROSCOPES), 72. The two lenses of a compound microscope are, of focal lengths 2 cm and 5 cm. If an object is, placed at a distance of 2.1 cm from the, objective of focal length 2 cm the final image, forms at the least distance of distinct vision of, a normal eye. Find the distance between the, objective and eyepiece, 1) 46.17 cm 2) 42 cm 3) 4.17 cm 4) 40 cm, DISPERSION BY A PRISM, 73. The separation L between the objective, A glass prism deviates the red and blue rays, f 0 0.5 cm and the eye piece f e 5 cm , through 100 and 120 respectively. A second, of a compound microscope is 7.0 cm. Where, prism of equal angle deviates them through, should a small object be placed so that the eye, 80 and 100 respectively. Their dispersive, is least strained ?, powers are in the ratio, 3, 2, 1, 1) 0.5 cm 2) cm, 3) cm 4) cm, 2) 9 :11, 3) 3: 2 4) 1:1, 1) 11: 9, 2, 3, 3, A parallel beam of white light falls on a convex 74. The focal lengths of objective and eyepiece of, lens. Images of blue, yellow and red light are, a compound microscope are 5 cm, 6.25 cm, formed on other side of the lens at a distance, respectively. When an object is placed infornt, of 0.20 m 0.205 m and 0.214 m respectively.., of the objective at a distance of 6.25 cm, the, The dispersive power of the material of the, final image is formed at least distance of, lens will be, distinct vision. The length of microscope is, 1) 22.5cm 2) 25.0cm 3) 30.0cm 4) 31.25 cm, 619, 9, 14, 5, 1), 2), 3), 4), 75. The magnifying power of a compound, 1000, 200, 205, 214, microscope is 20 and the distance between its, The refractive indices of crown glass prism for, two lenses is 30 cm when the final image is at, C, D and F lines are 1.527 , 1.530 and 1.535, the near point of the eye If the focal length of, respectively. The dispersive power of the, eye-piece is 6.25 cm, the focal length of, crown, glass, prism, is, objective is, 1) 0.01509 2) 0.05109 3) 0.02108 4) 0.03402, 1) 2.5 cm 2) 3.5 cm 3) 4.5 cm 4) 5.0 cm, White light is passed through a prism of angle 76. The focal length of objective and eye-piece of, 50 , If the refractive indices of red and blue, a compound microscope are 1 cm and 5 cm, colours are 1.641 and 1.659 respectively, the, respectively. The microscope magnification is, angle of dispersion between them is, equal to 50. If the distance between two lenses, 0, 0, 0, 0, is increased by 2 cm then the magnification is, 1) 0.08, 2) 0.06, 3) 0.09 4) 0.1, 1) 31, 2) 60, 3) 16, 4) 83, DEFECTS OF THE EYE, , 70. A person cannot see an object lying beyond, 80 cm, where as a normal person can easily, see the object distant 160 cm. The focal length, and nature of the lens used to rectify this defect, will be, 1) 160 cm, cancave, 2) 160 cm, convex, 3) 60 cm, concave, 4) 60 cm, convex, 71. The near point of a person is 50 cm and the, far point is 1.5m. The spectacles required for, reading purpose and for seeing distant objects, are respectively., 2, 2, 1) 2 D , 3 D, 2) D, 2 D, , 3, 2, , , 3) 2 D , 3 D, , , 212, , , , 2, , , 4) 3 D , 2 D, , , OPTICAL INSTRUMENT ( TELESCOPES ), , 77. The focal length of obejective and eyelens of, a astonomical telescope are respectively 20, cm and 5 cm. Final image is formed at least, distance of distinct vision. The magnfying, power will be, 2) 4.0, 3) 4.8, 4) 4.0, 1) 4.8, 78. A telescope consisting of an objective of focal, length 60 cm and single eye lens of focal length, 5 cm is focused on a distant object in such a, way that parallel rays emerge from the eye, piece. If the object makes an angle of 20 at, the objective, the angular width of the image, is, 1) 100, 2) 240, 3) 500, 4) 600, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 79. Four convergent lenses have focal lengths 100, cm, 10 cm, 4 cm and 0.3 cm, for a telescope, with maximum possible magnification we, choose the lenses of following focal lengths, 1) 10 cm, 0.3 cm, 2) 10cm, 4cm, 3) 100 cm, 4 cm, 4) 100 cm, 0.3 cm, 80. Opera glass have a minimum length of 20 cm, and a magnifying power of 5 when viewing, distant objects. The focal lengths of lenses, used are, 1) 25 cm, 5 cm, 2) 25 cm, 5cm, 10 , 50 , 3) cm, cm 4) 15 cm, 10 cm, 5, 3 , 81. A telescope has an objective lens of focal, length 200 cm and an eye piece with focal, length 2 cm. It is used to see a 5m tall building, at a distance of 2 km. The height of the image, of the building formed by the objective lens is, 1) 5 mm 2) 10 mm 3) 1 mm 4) 2 mm, 82. The focal length of the objective of an, astronomical telescope is 1 m and it is in, normal adjustment. Initially the telescope is, focussed to a heavenly body. If the same, telescope is to be focussed to an object at a, distance of 21 m from the objective, then, identify the correct choice, 1) eye piece should be displaced by 2 cm away, from the objective., 2) eye piece should be displaced by 2 cm towards, the objective, 3) eye piece should be displaced by 5 cm towards, from the objective, 4) eye piece should be displaced by 5 cm away, from the objective, , 2) 1, 8) 1, 14) 1, 20) 2, 26) 1, 32) 1, 38) 2, 44) 3, 50) 3, 56) 2, 62) 4, 68) 1, 74) 3, 80) 2, , NARAYANAGROUP, , 3) 3, 9) 1, 15) 1, 21) 1, 27) 2, 33) 4, 39) 1, 45) 2, 51) 4, 57) 3, 63) 2, 69) 3, 75) 4, 81) 1, , 4) 3, 10) 1, 16) 1, 22) 2, 28) 1, 34) 1, 40) 2, 46) 2, 52) 4, 58) 2, 64) 2, 70) 1, 76) 2, 82) 4, , 5) 2, 11) 3, 17) 1, 23) 4, 29) 1, 35) 2, 41) 4, 47) 2, 53) 3, 59) 1, 65) 2, 71) 1, 77) 1, , , , 1., , , 2., , 6) 4, 12) 3, 18) 1, 24) 3, 30) 3, 36) 2, 42) 1, 48) 1, 54) 2, 60) 4, 66) 2, 72) 1, 78) 2, , The incident and reflected ray make the same angle, ‘ ’ with the vertical. Therefore they are parallel, for any value of ‘ ’, Take as + direction. v1 vm vm v0, , v1 1 1 5, 3., , v1 7 m / s and direction towards left., Mirror L2 prouduces image I1 at 1m from it., , Mirror L1 produces image I at 2m from it., I is at 2 2 1 m from L2 and acts as object, for L2 to produce I 2, distance of I 2 from the person is 6 m., 10, 10, , 4., , 50 50, , 70, , 60, , 5., 6., 7., 8., , LEVEL - II ( C. W ) - KEY, 1) 4, 7) 4, 13) 1, 19) 2, 25) 2, 31) 3, 37) 4, 43) 2, 49) 4, 55) 1, 61) 1, 67) 3, 73) 3, 79) 4, , LEVEL - II ( C. W ) - HINTS, , 9., , 40 20, 1, x, 0.2, tan 30 , , x, 3 0.2, 3, 1 1, , u x1 f ;, u v, fu, b, b, v, ; u1 u and u2 u , u f, 2, 2, u1 20 find v1 ; u2 30 find v2, 1 1 1, ; here f 20 cm ; u 280 cm, f u v, , v2 , vel obj, 2 , u , 10. For velocity component along the principle axis, , vel image , , v2, Vom II, u2, apply mirror equations :, 1 1 1, 1, 1, 1, , , v u f, v 20 30, v 60, v 60cm m , 3, u 20, shortcut solution :, , V1M II, , , , 213
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , m, , 30, f, , 3, f u 30 20 , , v mu 3 20 60 cm, since mirror is at rest ( given that ) :, 2, , 60 , V1M II , 5 45 m / s, 20 , , VIG VIM VMG 45 0 45 m / s, that means image appraoches the mirror with, velocity 45 m/s, 1, 1 1, , , 11., 10 25 v, Areal magnification, Area of theimage, v2, = 2 = Area of the object, u, , 1, 1 1, 1 dv 1 du, , ; now 2 2 , 24 60 v, v dt u, dt, x, 13. t1 t2 1 2 , c, 1, 14. and t n, , t air , 15. nair nvac 1, vac , , 12., , 1, 16. , and Vr Vd 0.25 108 ms 1, vel, sin i, 17. d r i ; A B , sin r, sin i, 18. A B , sin r, dr, 19. d ; d r 15 cm ; d a 6 4 cm, a, sin i, x, 20. , ; sin i r , sin r, l, 1, 21. S t 1 , , Vd r, 22. sin C V , r, d, , Cd, dist, 23. sin C C but C , time, r, , 24., 214, , , , 1, sin C, , 25. r , 26., , d, , h, , 2 1, , r tan i ;, , r, , d , , 1, sin c, , 1, 27. i c and i ; sin sin c , w g, 28. , , sin i, and r 900 c, sin r, , 1, cos c 1 sin 2 c ; sin c , 29. 2 air sin ; 2 air sin 2 ; 2 1 sin , 30. sin 12 22, , r 3, 1 sin , 31. sin c ; sin c, air, d, 2, 32., 33., 34., 35., 36., 37., 38., , 2 1 2 1 1.5, 1, 1.5 1, , , , ;, v, u, R, PO PO, R, 2 1 2 1 1 1.5 1 1.5, , , ; , v, u, R, v 9, 12, 2 1 2 1, 2, 1 1, , 1, ;, v, u, R, 30.75 , R, 2 1 2 1, , v, u, R, 2 1 2 1, , v, u, R, D f ( in rad ), u1 ; v1 f ; u 2 f 20 , v 2 ? ,, Given v2 v1 5 ;, , 39., 40., 41., , 42., , 1 1 1, , f u v, , uv, v2, Areal magnification 2 9 v 3u ; f , uv, u, 1, 2, 2 1 1 1, , , , F f L 30 ; F v u, 1 1 1 v, v, ;, m 1 ; m 1...... i , f u v f, f, 1 b, y mx c...... ii ; Slope , f c, v, m 2 ;, u v x 1 ;, u, 1 1 1, 3 , f, u v, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, mx, Solving 1 , 2 , 3 , f m 1 2, , , , 43. f , , D 2 x2, 4, , 1 1, 1, , 1 1 1 1, 44. f f f f ; f 2 cm ; f1 f 2 60, 1, 2, 3, 7, 1 1 1, 1 1 1, d, 45. F1 f f and F f f f f, 1, 2, 1, 2, 1 2, , 46., , 1 1, , p, R, R2, 1, 1, , , 1, 1, 1, , , , 1, , 1, 55. f R R ; p2, 1, 2 , R3, 56., , 57., , 1 2 1, , ; F1 28 cm, F1, R, 1 2 1 2, , ; F2 10 cm, F2, R, R, , 1 1 1 1, 47. F f f f ;, 1, 2, 3, 1 1 1, 1 2 1 2 1 2, , , ; then , F v u, F, R, R, R, , 48., , 1 1 , 1, 1 P, f, R1 R2 , , 49., , f 1 lens 1, , f, lens, , 1, , , liq, , , 50., , f 1 lens 1, , f, lens, , 1, , , liq, , , 51., , GEOMETRIC OPTICS, , f 1 lens 1, , f, lens, , 1, , , liq, , , 1 lens 1, , 1, f lens , 52., 1 ; and sin c, , , liq, , , 1, 1, 53. t t ( find ) ; p 1, R, , 54. For diverging meniscus lens Rconcave Rconvex, , 1 1 1 1 1, 1, 1 ; , f, R1 R2 u v f, NARAYANAGROUP, , f med Pair, 1, , lens, f air Pmed lens 1, liq, 1, 2 , 100 2, 1 ;, , f, R, f, , 100 , 100, f, 1, , A Dm , sin , , 2 , , , 58., A, sin, 2, A Dm , sin , , 2 , , , 59. a 2r and ;, A, sin, 2, 60. D 2i A, 61. d 180 2 A 2i A, A, , sin 90 , sin i, 2, , 1, , , ;, A, sin r, sin c, sin, 2, 62. i1 r1 00 ; r2 00 but r2 c, , g, 1, and , w, sin c, 63. D i1 i2 A; Dm D, , , 1, sin c, 65. r2 0 find r1 from r1 r2 A ; sin i1 sin r1, , 0, 64. A 2C 90 ; , , 1, d1 d 2 , 66., 2, f R fV fY, 67. w , fV f R, w, , w, , 68., , F C , F C, , 2, , , , 1, , 215
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 69. B R A, 70., , 1 1 1, , f v u, , LEVEL - II (H.W), REFLECTION, 1., , 100, 71. p , f, 72., , 1 1 1, and L v0 ue, f v u, , Two plane mirrors are inclined at angle ‘ ’ as, shown in figure. If a ray parallel to OB strikes, the other mirror at P and finally emerges, parallel to OA after two reflections then is, equal to, A, , 1 1, 1, 73. L v0 f e ; v u f, 0, 0, 0, 1 1, 1, 74. Find e from u v f ; ve D, e, e, e, , P, , O, , 2., , 1 1, 1, find v0 from u v f ; L v0 e, 0, 0, 0, 1 1, 1, 75. Find ue from u v f, e, e, e, , 3., , ve 25 cm f e 6.25 cm, , mirror M 2 will be ( in cm , , D, find v0 from v0 e L find me 1 f, e, , 0, , 79. f0 should be maximum and fe should be minimum., , 15cm, , 4., , 5., , f0, 80. L f 0 f e and m f, e, , 81., , 1, 1 1, v, I, , m 0 , but, f 0 v0 u0, u0 O, , 82. u1 ; v1 f 0 1m, 1 1 1, u2 21m; find v2, f v2 u2, 216, , M2, , M1, , 1, 1 1, find f0 from f u v, 0, 0, 0, , f0 , fe , 77. M f 1 D , , e , f0, 78. M f, e, , , , 5cm, , v0, find m0 from m0 me m ; m0 u, 0, , LD, 76. M f f, 0 e, , B, , 1) 900, 2) 600, 3) 450, 4) 300, Aplane mirror is approaching you at 10 cm/s., You can see your image in it. The image will, approach you with a speed of, 1) 5 cm/s 2) 10 cm/s 3) 15 cm/s 4) 20 cm/s, Two plane mirrors parallel to each other and, an object O parallel between them. Then the, distance of the first three images from the, , 6., , 1) 5, 10, 15, 2) 5, 15, 30, 3) 5, 25, 35, 4) 5, 15, 25, Two vertical plane mirrors are inclined at an, angle of 600 with each other. A ray of light, travelling horizontally is reflected first from, one mirror and then from the other mirror. then, the resultant deviation is, 1) 600, 2) 1200, 3) 1800 4) 2400, If an object is placed between two plane, mirrors a distance 2b, apart, the object is, situated at mid point between mirrors, the, position of nth image formed by the one of the, mirrors with respect to the object is, 1) nb, 2) 2nb, 3) 3nb, 4) 4nb, With a concave mirror, an object is placed at, a distance 9cm from the principal focus, on the, principal axis. The image is formed at a, distance 16cm from the principal focus. The, focal length of the mirror is, 1) 12 cm 2) 11 cm 4) 40 cm 4) 30 cm, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 7., , An infinitely long rod lies along the axis of a, concave mirror of focal length ‘f’. The near, end of the rod is at a distance u f from the, mirror. Its image will have a length., , uf, uf, f2, f2, 1), 2), 3), 4), u f, u f, u f, u f, 8. A 2 cm high object is placed on the principal, axis of a concave mirror at a distance of 12, cm from the pole. If the image is inverted,, real and 5 cm high. The location of the image, and the focal length of the mirror is, 1) 30 cm, 8.6 cm, 2) 8.6 cm 30 cm, 3) 30 cm, 10 cm, 4) 10 cm, 30 cm, 9. At what distance from a convex mirror of focal, length 2.5 cm should a boy stand so that his, image has a height equal to half the original, height ?, 1) 2.5 m from the mirror 2) 5 m from the mirror, 3) 7.5 m from the mirror 4) 10 m from the mirror, 10. The velocity of image w.r.t tround in the below, figure is, f = 20 cm, , 5 m/s, O, , GEOMETRIC OPTICS, 3, . Then, 2, the correct thickness of glass plate that will, permit the same number of wavelengths as, that by an 18 cm long column of water is, , 14. The refractive index of glass plate is, , 4, , w , 3, , 1) 12 cm 2) 16 cm, , 15. The wavelength of light in vacuum is 5000 A0 ., When it travels normally through diamond of, thickness 1.0 mm find the number of waves of, light in 1.0 mm of diamond. ( Refractive index, of diamond = 2.417 ), 1) 4834 waves, 2) 5834 waves, 3) 4384 waves, 4) 6834 waves, 16. If the refractive index of diamond is 2.4 find, the velocity of light in diamond., , c 3 10 m / s , 8, , 1) 1.25 108 m / s, (Mirror is at rest, given velocity, w.r.t ground), , 30 cm, , 1) 10 m/s moving downwards, 2) 10 m/s moving upwards, 3) 20 m/s moving downwards, 4) 20 m/s moving upwards, 11. A rectangular wire of length 2.0 cm, breadth, 1.5 cm is placed 25 cm in front of a concave, mirror of focal length 10 cm with its centre on, the axis of the mirror and its plane normal to, the axis. The area enclosed by the image of, the wire is, 1) 7.5cm 2 2) 6.0 cm 2 3) 4.0 cm 2 4) 3.0 cm 2, 12. An image of a candle on a screen is found to, be double its its size. When the candle is, shifted by a distance 5 cm then the image, becomes triple its size. Then the nature and, radius of curvature of the mirror is, 1) concave, 60 cm, 2) convex, 60 cm, 3) concave, 30 cm, 4) convex 30 cm, , REFRACTION, 13. The same colour of light takes t1 sec and t2, sec to travel the same distance ‘x’ in two media, ‘A’ and ‘B’ respectively. Refractive index of, medium ‘A’ w.r.t to ‘B’ is, xt1, t2, t2, t1, 1) t, 2) xt, 3) t, 4) t, 2, 1, 1, 2, NARAYANAGROUP, , 3) 18 cm 4) 24 cm, , 2) 2.25 108 m / s, , 3) 1.5 108 m / s, , 4) 4.5 108 m / s, 17. Refractive index of water with respect to air, is 2 . A light ray is incident on the surface at, an angle 600 travelling through water. The, deviation of light ray is, 1) 300, , 2) 1200, , 3) 00, , 4) 600, , 18. If i denotes a unit vector along an incident, ray r a unit vector along the refracted ray, into a medium of refractive index ‘ ’ and n a, unit vector normal to the boundary of the, media directed towards the incident medium,, then the law of refraction can be written as, 1) i.n r .n, , , 3) i n r n , , 2) i n n r, , , , 4) i n r n, , , , 19. A small air bubble is inside a transparent cube, of side length 24 cm and of refractive index, 4, . If the apparent distance air bubble from a, 3, face is 9 cm then its apparent distance from, opposite face is, 1) 6 cm, 2) 8 cm, 3) 9 cm 4) 12 cm, 217
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 20. A ray of light is incident upon a parallel sided, transparent slab of thickness 9 cm at an angle, of incidence 600 . If the angle of refraction is, 300 , the lateral displacement of the light ray, is, 2, cm, 1) 3 cm 2) 3 3 cm 3) 3 cm 4), 3, 21. A vessel of depth ‘d’ filled with a liquid of, refractive index 1 up to half its depth and, the remaining space is filled with a liquid of, refractive index 2 . The apparent depth while, seeing normal to the free surface of the liquid, is, , 1, 1 , 1) d , 1 2 , , vacuum and takes t2 sec to travel 10x cm in, a medium. The critical angle corresponding to, the media is, 1 t 2 , 2) sin 10t , 1, , 1 10t 2 , 1 t1 , 3) sin t , 4) sin 10t , 1 , 2, 24. A fish looks upwards at an unobstructed, overcast sky. What total angle does the sky, appear to subtend ? ( Take refractive index of, water as 2 ), 1) 1800, 2) 900, 3) 750, 4) 600, 25. In a swimming pool, a person is viewing outside, objects by keeping an eye at a depth h inside, water. If the critical angle for water is ‘ c ’,, then the value of the diameter of the circle of, view for outside objects will be, 1) 2h sin c, 2) 2h cos c, , 3) 2h tan c, 218, , 45°, 45°, , 2) d 1 2 , , d 1 1 , d, 4) 1 2 , 3) 2 , 2, 1, 2 , 22. The critical angle of light going from medium, A into medium B is . The speed of light in, medium A is v. The speed of light in medium B, is, v, v, 1), 2) v sin 3), 4) v tan , sin , tan , 23. Light takes t1 sec to travel a distance x cm in, , 1 10t1 , 1) sin t , 2 , , 26. A ray of light from a rarer medium strikes a, denser medium at angle of incidence 600 . The, reflected and refracted rays are perpendicular, to each other. The refractive index of denser, medium and angle of deviation respectively, are, 1) 3 ;300 2) 2 ; 450 3) 3 ; 600 4) 2 ;300, 27. A light ray is incident perpendicularly to one, face of a 900 prism and is totally internally, reflected at the glass air interface. If the angle, of reflection is 450 , we conclude that the, refractive index n, , 4) 2h cot c, , 45°, , 1, 1, 2) n 2 3) n , 4) n 2, 2, 2, 28. Word ‘Newton’s printed on a paper and is, placed on a horizontal surface below a cubical, glass. The minimum value of refractive index, of a cubical glass for which letters are not, visible from any vertical faces, of the glass, is, ( Critical angle = 450 ), 1) n , , 1) 3, 2) 0.5, 3) 1, 4) 2, 29. The critical angle for refraction from glass to, air is 300 and that from water to air is 370 ., Find the critical angle for refraction from glass, to water, 1 5 , 1) sin 2) 5103' 3) 610 2 ', 4) 6303', 6, 30. The refractive index of the core of an optical, fibre is 2 and that of the cladding is 1 . The, angle of incidence on the face of the core so, that the light ray just under goes total internal, reflection at the cladding is, 1 1 , 1) sin , 2, , 2) sin 1 22 12, , 3) sin 1 2 1, 4) sin 1 12 22, 31. When a ray of light enters from one medium, to another then its velocity in second medium, becomes doubled. The maximum value of, angle of incidence so that total internal, reflection may not take place will be, 1) 600, 2) 900, 3) 300, 4) 1800, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , REFRACTION THROUGH, SPHERICAL SURFACES, 32. A mark is made on the surface of a glass, sphere of diameter 10 cm and refractive index, 1.5. It is viewed through the glass from a, portion directly opposite. The distance of the, image of the mark from the centre of the, sphere will be, 1) 15 cm 2) 17.5 cm 3) 20 cm 4) 22.5 cm, 33. In a medium of refractive index 1.6 and having, a convex surface with respect to air has a, point object in it at a distance of 12 cm from, the pole. The radius of curvature is 6 cm., Locate the image as seen from air, 1) A real image at 30 cm, 2) A virtual image at 30 cm, 3) A real image at 4.28 cm, 4) A virtual image at 4.28 cm, 34. Parallel rays are incident on a transparent, sphere along its one diameter. After, refraction, these rays converge at the other, end of this diameter. The refractive index for, the material of sphere is, 1) 1, 2) 1.5, 3) 1.6, 4) 2, 35. A mark on the surface of a glass sphere, 1.5 is viewed from a diametrically, opposite position. It appears to be at a, distance 10 cm from its actual position. The, radius of the sphere is, 1) 5 cm, 2) 10 cm 3) 15 cm 4) 25 cm, 36. A ray incident at an angle of incidence 600, enters a glass sphere of refractive index, 3 . This ray is reflected and refracted, at the farther surface of the sphere. The angle, between reflected and refracted rays at the, surface is, 1) 900, 2) 600, 3) 700, 4) 400, , LENSES & THEIR COMBINATIONS, 1, , 2, , 0, , 37. The sun subtends an angle of, at earth., The image of the sun is obtained on a screen, using a converging lens of focal length 1.5 cm., The diameter of the image will be, 1) 0.13mm 2) 0.9 mm 3) 1.8 mm 4) 0.6 mm, 38. A convex lens forms an image of a distant, object at distance of 20 cm from it. On, keeping another lens in contact with the first,, 40, if the image is formed at a distance of, cm, 3, from the combination, then the focal length of, the second lens is, 1) 20 cm 2) 40 cm 3) 40 cm 4) 13.33, NARAYANAGROUP, , 39. A slide projector gives magnification of 10. If, it projects a slide of 3 cm 2 cm on a screen,, the area of image on screen is :, 1) 6000 cm 2 2) 600 cm 2 3) 3600 cm 2 4)12000 cm 2, 40. The radius of curvature of a thin planoconvex, lens is 10 cm and the refractive index of its, glass is 1.5. If the plane surface is silvered,, then it will behave like a, 1) concave mirror of focal length 10 cm, 2) concave mirror of focal length 20 cm, 3) convex mirror of focal length 10 cm, 4) convex mirror of focal length 20 cm, 41. The graph between object distance u and, image distance v for a lens given below. The, focal length of the lens, v, (cm), , +10.0, , u (cm), , 10.0, , 1) 5 0.1, 2) 5 0.05, 3) 0.5 0.1, 4) 0.5 0.05, 42. In the displacement method a conves lens is, placed in between an object and a screen. If, the magnificaiton in the two position are m1, and m2 m1 m2 , and the distance between, the two positions of the lens is x , the focal, length of the lens is, x, x, 2) m m, 1) m m, 1, 2, 1, 2, x, , x, , 3) m m, 4) m m 2, 1 2, 1 2, 43. A convex lens forms a real image 4 cm long, on a screen. When the lens is shifted to a new, position without disturbing the object or the, screen, again real image is formed on the, screen which is 16 cm long. The length of the, object is, 1) 8 cm, 2) 10 cm 3) 12 cm 4) 6 cm, 44. A convex lens of focal length 50 cm, a concave, lens of focal length 50 cm and a concave lens, focal lens 20 cm are placed in contact. The, power of this combination in diopters will be, 1) 4.67D 2) 5D 3) 3.21D 4) 3D, 2, , 219
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GEOMETRIC OPTICS, 45. Arrange the following combinations in the, increasing order of focal length, a) Two plano convex lenses of focal lengths 15, cm and 30 cm in contact, b) Two convex lens of focal lengths 40 cm and, 50 cm in contact, c) Two convex lenses of focal length 20 cm, separated by 5 cm, 1) a, b, c 2) b, a, c 3) a, c, b 4) c, a, b, , LENS MAKER’S FORMULA, 46. The radius of curvature of the convex surface, of a planoconvex lens is 12 cm and its, refractive index 1.5. If the plane face of the, lens is silvered, then the distance from the, lens at which parallel rays incident on its, convex surface converge is, 1) 12 cm 2) 18 cm 3) 24 cm 4) 30 cm, 47. An equiconcave lens having radius of, curvature of each surface 20 cm has one, surface silvered. If the refractive index of the, lens is 1.5, then the magnitude of the focal, length is, 1) 2.5 cm 2) 0.4 cm 3) 0, 4) 5 cm, 48. If R1 and R2 are the radii of curvature of, double convex lens made of same material,, the lens with more focal length is, 1) R1 20cm, R2 10cm 2) R1 R2 20cm, 4) R1 R2 5cm, 3) R1 R2 10cm, 49. A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature, R. On immersion in a medium of refractive, index 1.75, it will behave as a, 1) Convergent lens of focal length 3.5 R, 2) Convergent lens of focal length 3.0 R, 3) Divergent lens of focal length 3.5 R, 4) Divergent lens of focal length 3.0 R, 50. A thin equiconvex lens is made of glass of, refractive index 1.5 and its focal length is 0.2, m. If it acts as a concave lens of 0.5 m focal, length when dipped in a liquid, the refractive, index of liquid is, 17, 15, 13, 9, 1), 2), 3), 4), 8, 8, 8, 8, 51. A convex lens of focal length 0.15m is made, 3, of refractive index . When it is placed in, 2, liquid, its focal length increases by 0.225 m ., Then the refractive index of the liquid is, 7, 5, 9, 3, 1), 2), 3), 4), 4, 4, 4, 2, 220, , JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, 52. A diverging lens of focal length 10 cm having, refractive index 1.5 is immersed in a liquid of, refractive index 3. The focal length and, nature of the lens in liquid is, 1) 10 cm, convergent 2) 10 cm divergent, 3) 18 cm, convergent 4) 72 cm, divergent, 53. A plano convex lens a thickness of 4 cm. Its, radius of curvature is 20 cm, When its curved, surface is kept on a horizontal surface and, viewed from the top, its bottom appears to, be raised by 1 cm. The focal length of the, lens, 1) 40 cm 2) 50 cm 3) 60 cm 4) 70 cm, 54. Two equi convex lenses each of focal lengths, 20 cm and refractive index 1.5 are placed in, contact and space between them is filled with, 4, . The, water of refractive index, 3, combination works as, 1) converging lens of focal length 30 cm, 2) diverging lens of focal length 15 cm, 3) converging lens of focal length 15 cm, 4) diverging lens of focal length 40 cm, 55. If R1 and R2 are the radii of curvature of a, double convex lens. The largest power will, be for, 1) R1 , R2 10cm 2) R1 10cm, R2 , 3) R1 10cm, R2 10cm 4) R1 5cm, R2 5cm, 56. A thin convergent glass lens 1.5 has a, power of 5 D . When this lens is immersed, in a liquid of refractive index it acts as a, diverging lens of focal length 100 cm. The, value of must be, 5, 4, 5, 6, 2), 3), 4), 1), 3, 3, 4, 5, 57. The refractive index of a material of a plano5, concave lens is . Its radius of curvature is, 3, 0.3 m. Focal length of the lens in air is, 1) 0.45 m 2) 0.6 m 3) 0.75 m 4) 1m, , REFRACTION THROUGH A PRISM, 58. The angle of minimum deviation measured, with a prism is 300 and the angle of prism is, 600 . The refractive index of prism material, is, 3, 4, 1) 2, 2) 2, 3), 4), 2, 3, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 59. When light of wavelength is incident on an, DISPERSION BY A PRISM, equilateral prism kept in its minimum, deviation position, it is found that the angle 66. Two small angled prisms A and B deviate the, blue rays by 7 0 and 90 and the red rays by 50, of deviation equals the angle of the prism, itself. The refractive index of the material of, and 7 0 respectively. Which prism has a, the prism is, greater dispersive power ?, 3, 1) Prism A, 2) Prism B, 1), 2) 3, 3) 2, 4) 2, 3) same for both Prism A & B 4) none of these, 2, 67., The refractive index of the material of the, 0, 60. A ray incident at 15 on one refracting, prism for violet colour is 1.69 and that for red, surface of prism of angle 600 , suffers a, is 1.65. If the refractive index for mean colour, deviation of 550 . The angle of emergence is, is 1.66, the dispersive power of the material, of the prism is, 1) 950, 2) 450, 3) 300, 4) 1000, 1) 0.66, 2) 0.06, 3) 0.65 4) 0.69, 61. A prism of critical angle 450 is immersed water, 68., The, refractive, indices, of, flint glass prism for, of critical angle 500 . The critical angle of prism, C, D and F lines are 1.790, 1.795 and 1.805, inside water will be sin 70 0 0.94 , respectively. The dispersive power of the flint, glass prism is, 1) 700, 2) 900, 3) 1300 4) 1000, 1) 0.01587 2) 0.01887 3) 0.01187 4) 0.01387, 62. A glass prism of refractive index 1.5 is placed, in water of refractive index 1.33. The minimum 69. Refracting angle of a prism is 2 radians., Refractive indices of its material for violet and, value of the angle of the prism so that it will, red are respectively 1.62 and 1.5 Dispersion, not be possible to have any emergent ray is, produced by it is, 1) 1500, 2) 1250, 3) 1650 4) 1800, 1) 0.24, 2) 0.06, 3) 1.66 4) 1.12, 63. A certain prism is that to produce minimum, deviation of 380 . It produces a deviation of, DEFECTS OF THE EYE, 440 when the angle of incidence is either 420 70. A man cannot see clearly the objects beyond, or 620 . The refractive index of material of, a distance of 20 cm from his eyes. To see, prism is, distant objects clearly the kind of lenses and, 1) 1.51, 2) 1.33, 3) 1.62 4) 1.732, its focal length must be, 64. The maximum value of index of refraction of, 1) 100 cm, convex, 2) 100 cm concave, a material of prism which allows the passage, 3) 20 cm convex, 4) 20 cm concave, of light through it when the refracting angle of 71. A short sighted person can see objects most, prism is A is, distinctly at a distance of 16 cm. If he wears, 2 A, 2 A, spectacles at a distance of 1 cm from the eye,, 1) 1 sin, 2) 1 cos, 2, 2, then their focal length to see distinctly at a, distance of 26 cm, A, A, 3) 1 ta n 2, 4) 1 c o t 2 2, 1) 25 cm, convex, 2) 25 cm, concave, 2, 3) 37.5 cm, convex, 4) 37.5 cm, concave, 65. The prism shown in the figure has one side, 0, silvered. The angle of the prism is 30 and, OPTICAL INSTRUMENTS, , ( MICROSCOPES ), 2 . If the incident ray retraces its initial, path the angle of incidence is, 72. The two lenses of a compound microscope are, A, of focal lengths 2 cm and 5 cm. If an object is, placed at a distance of 2.1 cm from the objetive, 30°, D, of focal length 2 cm the final image forms at, the final image forms at the least distance of, i, r, C, distinct vision of a normal eye. Find the, magnifying power of the microscope, 2, B, 1) 20, 2) 6, 3) 120, 4) 60, 1) 500, 2) 450, 3) 600, 4) 750, NARAYANAGROUP, , 221
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 73. If the focal lengths of objective and eye lens, of a microscope are 1.2 cm and 3 cm, respectively and the object is put 1.25 cm, away from the objective lens and the final, image is formed at infinity, then magnifying, power of the microscope is, 1) 150, 2) 200, 3) 250, 4) 400, 74. The focal lengths of the objective and eyepiece of a compound microscope are 1 cm and, 5 cm respectively. An object is placed at a, distance of 1.1 cm from the objective has its, final image formed at least distance of distinct, vision, then the magnifying power is, 1) 20, 2) 30, 3) 50, 4) 60, 75. A compound microscope has an objective of, focal length 2.0 cm and an eye piece of focal, length 6.25 cm separted by 15 cm. If the final, image is formed at the least distance of distinct, vision (25 cm ), the distance of the object from, the objective is, 1) 1.5 cm 2) 2.5 cm 3) 3.0 cm 4) 4.0 cm, 76. The focal lengths of the objective and eyepiece of a compound microscope are 2 cm and, 3 cm respectively. The distance between the, objective and eye-piece is 15 cm. The final, image formed is at infinity. The distances in, cm of object and image from objective are, 1) 2.4 and 12, 2) 2.4 and 15, 3) 2.4 and 3.0, 4) 2.3 and 12, , 80. A Galilean telescope measures 9 cm from the, objective to the eyepiece. The focal length of, the objective is 15 cm. Its magnifying power, is, 1) 2.5, 2) 2/5, 3) 5/3, 4) 0.4, 81. A small telescope has an objective lens of focal, length 140 cm and eye piece of focal length, 5.0 cm. The telescope is used to view a 100 m, tall tower 3 km away. The height of the image, of the tower formed by objective lens is, 14, 11, 17, 8, cm 2) cm 3), cm 4) cm, 1), 3, 3, 3, 3, 82. The magnification produced by an astonomical, telescope for normal adjustment is 10 and the, length of the telescope is 1.1 m. The, magnification when the image is formed at half, distance of the distinct vision D 25 cm , 1) 6, 2) 14, 3) 18, 4) 16, , LEVEL - II ( H. W ) - KEY, 1) 2, 7) 4, 13) 4, 19) 3, 25) 3, 31) 3, 37) 1, 43) 1, 49) 1, 55) 4, 61) 1, 67) 2, 73) 2, 79) 1, , OPTICAL INSTRUMENT, ( TELESCOPES ), 77. The magnifying power of an astronomical, telescope for normal adjustment is 10 and the, length of the telescope is 110 cm. Find the, magnifying power of the telescope when the, image is formed at the least distance of distinct, vision for normal eye., 1) 14, 2) 48, 3) 28, 4) 52, 78. The eyepiece of a refracting telescope has, f 9 cm. In the normal setting, separation, betweeen objective and eyepiece is 1.8 m. Find, the magnification, 1) 20, 2) 19, 3) 18, 4) 21, 79. Four lenses of focal length 15 cm, 20 cm, 150, cm and 250 cm are available for making an, astronomical telescope. To produce at the, largest magnification, the focal length of the, eyepiece should be, 1) + 15 cm, 2) + 20 cm, 3) + 150 cm, 4) 250 cm, 222, , 2) 4, 8) 1, 14) 2, 20) 2, 26) 1, 32) 1, 38) 3, 44) 2, 50) 2, 56) 1, 62) 2, 68) 2, 74) 4, 80) 1, , 3) 3, 9) 1, 15) 1, 21) 3, 27) 2, 33) 2, 39) 2, 45) 3, 51) 2, 57) 1, 63) 1, 69) 1, 75) 2, 81) 1, , 4) 4, 10) 1, 16) 1, 22) 1, 28) 4, 34) 4, 40) 1, 46) 1, 52) 1, 58) 1, 64) 4, 70) 4, 76) 1, 82) 3, , 5) 2, 11) 3, 17) 4, 23) 1, 29) 1, 35) 1, 41) 2, 47) 4, 53) 3, 59) 2, 65) 2, 71) 4, 77) 1, , 6) 1, 12) 1, 18) 3, 24) 2, 30) 2, 36) 1, 42) 2, 48) 2, 54) 3, 60) 4, 66) 1, 72) 3, 78) 2, , LEVEL - II ( H. W ) - HINTS, 1., 2., 3., 4., 5., , 3 1800, 2u 20 cm / sec, I1 is the object for M 1 to get I 3 ., d 360 2i, The distance of the images formed by one mirror, with respect to the object are 2b, 4b ...., 2nb, 6. f x 1 x 2, 1 1 1 1 1 1, 7. f v u ; f v Length of image v2 v1, 2, 1, 1, , 1, , 1, , m, , v, h, i, u h0, , m, , f, 20 2, , f u 20 30 , , ; f uv, v 1 1 1 1, , 9. m ;, u 2 f u v, 10. Applying mirror equation :, 8., , v mu 2 30 60, For the velocity component perpendicular to the, principle axis, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, dm, , VIM m Vom h0, dt, 5 10 m / s ;[ since, h0 0 ], VIM , 2, , , 11., 12., , 13., 14., 15., 16., 17., 18., 19., 20., 21., 22., , 23., , VIG VIM VMG 10 10 10 m / s, m / s ( moving downwards ), Areal magnification, Area of theimage, v2, = 2 = Area of the object, u, Sinnce the image is formed on screen it is real., Real object and real image implies concave mirror, f, f, applying m , or 2 f u ...... 1, f u, f, after shifting 3 f u 5 ...... 2 , , , [ why u 5 ?, why not u 5 : In a concave mirror,,, are size of real image will increase, only when the, real object is brought closer to the mirror. In doing, so, its x coordinate will increase ], from (1) & (2) we get f 30 cm or R 60 cm, time, 1, t n and , 1, t n and , 1, , vel, sin i, d i r and sin r, i n i n sin i ;, r n r n sin r, RD1 RD2, , , AD1 AD2, t sin i r , x, cos r, d1 d 2, d, da , d1 d 2 , ;, 1 2, 2, r Cd, , sin c, d Cr, m ed, C, air, air, C med, , d , , t air, , d , t , m ed, , 1, , vertexangle 2c, sin c, Vertex angle = 2c, r h tan C, tan i; i r, 1, 1, i c sin i sin c , 2, 2 , 1, 1, , , 2, sin c sin 450, , 24. , 25., 26., 27., 28., , NARAYANAGROUP, , GEOMETRIC OPTICS, , a, a, 29. sin g Ca sin 30 ; sin w Ca sin 37 , g, , g, , 30. r c 90 sin r cos c, , , sin c 2 ; sin i2 1 sin r, 1, a, V1, 31. sin c V, 2, 1 1 1.5 1 1.5, , 32. 2 1 2, ; , v, u, R, v 10, 5, 2 1 2 1 1 1.6 1 1.6, , , 33., ; , v, u, R, v 12, 6, 1 2 1 2 1, , 34. 2 1 2, ;, v, u, R, 2R , R, 35. As the mark is viewed from the diametrically, opposite position, refraction takes place at side II, of the surface (the mark being on side I as shown), II, 10cm, C, , Mark, , P, , Here 1 1.5, 2 1;u 2R, 1 1 1.5 1 1.5, ;, Using 2 1 2, 2R , R, u, R, 1 0.5 1.5, 0.5, , , , or 4R, R 2R, 2R, Negative sign indicates that the image is formed to, the left of refracting surface as shown in Figure., Further, it is given that the image of mark is at a, distance 10 cm from the object., Hence: 4R 2R 10, R 5 cm, P, 60°, , 36., , Q, 60°, , r, , r, , r, C, , S, , R, , sin i sin 600 1, sin i, 0, sin r , , , , 2 r 30, sin r, 3, PC QC;, CPQ PQC r 300, Angle between reflected ray QR and refracted ray, QS at the other face =1800 r 600, 1800 300 600 900, 37. Diameter of the image D f ( in radian ), , , , 223
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 38., , 1 1 1, 1 1 1 1, 1 ;, 2, f1 u v1, f1 f 2 u v 2, 1, 1 1, 2 1 , f 2 v2 v1, , area of the im age, 2, 39. m area of the object, R, 40. F 2 1, , 1 1 1 f v u, ;, , , 41., f v u f 2 v2 u 2, , v, u, 42. m1 , m2 ; v m1 1 f , u m2 1 f, u, v, x, x v u f m1 m2 f , m1 m2, 43. OJ I1I 2, 44. p p1 p2 p2, p is +ve convex and -ve for concave, 1 1 1, 45. f f f, 1, 2, 1 1 1, 1, 1, 1 1, 0; , 46. f f f f ;, fm, f, R, , , m, 1 1 1, then ; (apply with sign convention), F v u, R, 47. F 4 2, , , 1 1 , 1, 48. f 1 R R ;, 1, 2 , 1 lens 1, , ens, f, 49- to 52., 1, , liq, , 1, 1 1, , ; t t 1 , f, R, , 1, 2 1, 2, 54. f 1 1 R ; f 2 1 R, , 1, 2, 53., , 55., , 1 1 , 1 1 , p 1 ; p R R , 1, 2 , R1 R2 , , f1 1, , , 56. f, 1, 1, 1, 1, 1 , 57., f, R, , 58., 59., , A Dm , sin , , 2 , , , A Dm ;, A, sin, 2, A, 2 cos , 2, , 60. d i1 i2 A, 224, , 61. denser rarer sin c, 62. For no emergent ray from a prism Amin 2C, , 1, sin C and g, , w, 63. i1 i2 A ;, , A , sin , , 2 , , A, sin, 2, , 1, 64. A 2C ; , sin c, , sin i1, 65. A r1 r2 ; r2 00 and sin r, 1, 1, 66. w d, y, V R, 67. w 1, Y, F C, 68. w 1, D, 69. V R A, 1 1 1, ; u , v 20 cm, 70., F v u, 71. v 16 1 15 cm, 1, , 1 1, , u 26 1 25 cm and f v u, 1 1 1, 72. Use for both objective and eyepiece, f v u, v v, M 0 e, u0 u e, 1, 1 1, v0 D , 73. f v u and M u f , 0 e , 0, 0, 0, 1 1 1, 1 1 1, 74. u v f ( find v0 ); u v f ( find ue ), 0, 0, 0, e, e, e, v0, D, m0 ; me 1 ; M m0 me, u0, fe, 1 1 1, 1 1, 75. u v f ( find v0 ) ; u v L ( find u0 ), e, e, e, 0, 0, 1 1, 1, 76. L v0 ue and u v f, 0, 0, 0, , 77., , M , , f0, fe, , fe , , 1 , , D , , , f0, 78 & 79. m f and L f 0 f e, e, f0, 80. L f 0 f e and m f, e, h0 hi, 81. u f v0 f 0 , 0, 0, f0, L, 82. M f ; L f 0 fe f e , m 1, e, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, 7, , LEVEL - III, , equation x 2 y 2 a 2 . A ray travelling in, negative x-direction is directed towards, positive y-direction after reflection from the, surface at point P. Then co-ordinates of point, P are, , REFLECTION, 1., , A plane mirror is placed at origin parallel to, y-axis, facing the positive x-axis. An object, starts from (2m,0,0) with a velocity of, 2iˆ 2 ˆj m / s . The relative velocity of image, , , , A reflecting surface is represented by the, , y, , , , with respect to object is along, , 2., , 1) Positive x - axis, , 2) Negative x - axis, , 3) Positive y - axis, , 4) Negative y - axis, , The objective and eye piece of an astronomical, telescope are double convexlenses with, refractive indes 1.5. When the telescope is, adjusted to infinity the seperation between the, lenses is 16 cm. If the space between the, lenses is now filled with water and again, 8., telescope is adjusted for infinity. Then the, present separtion between the lenses is, 1) 8 cm, , 3., , 4., , 5., , 6., , x, , 2) 16 cm, , 3) 24 cm, , 4) 32 cm, , 2) 8.6 cm, 30 cm, , 3) 30 cm, 10 cm, , 4) 10 cm, 30 cm, , A thin plano - convex lens acts like a concave, mirror of focal length 0.2 m when silvered from, its plane surface. The refractive index of the, material of the lens is 1.5. The radius of, curvature of the convex surface of the lens, will be, , 2) 0.6a,0.8a , , a a, , , 2, 2, , , a a, 4) , , 2 2, A ray of light travelling in the direction, , 3) , , 1 ˆ, i 3jˆ is incident on a plane mirror.., 2, , , , , , After reflection, it travels along the direction, , A 2 cm high object is placed on the principal, axis of a concave mirror at a distance of 12, cm from the pole. If the image is inverted, real, and 5 cm high. The location of the image and, the focal length of the mirror is (NCERT), 1) 30 cm, 8.6 cm, , 1) 0.8a,0.6a , , 1 ˆ, i 3jˆ . The angle of incidence is, 2, 1) 300 2) 450, 3) 600, 4) 750, REFRACTION, , , , 9., , , , Let the x-y plane be the boundary between two, trasparent media. Medium 1 in Z 0 has a, , 2 and medium 2 with, z 0 has a refractive index of 3 . A ray of, light in medium 1 given by the vector, , A 6 3iˆ 8 3 ˆj 10kˆ is incident on the plane, of separation. The angle of refraction in, 1) 0.4 m 2) 0.2 m, 3) 0.1 m, 4) 0.75 m, medium 2 is, A concave mirror has a focal length 20 cm., 1) 450, 2) 600, 3) 750, 4) 300, The distance between the two positions of the, 10. A ray of light entering from air to glass, object for which the image size is double of, 1.5 is partly reflected and partly, the object size is, refracted. If the reflected and refracted rays, 1) 20 cm 2) 40 cm 3) 30 cm, 4) 60 cm, are at right angles to each other, the angle of, An object is placed on the principal axis of a, refraction is, concave mirror of focal length 10 cm at a, 2, 2, distance of 8 cm from the pole. Find the position, 1, 1, sin, sin, , , , , 1), 2), and the nature of the image., 13 , , , 13 , 1) 40 cm, virtual, 2) 40 cm, real, 2 , 1 , 1 3 , 3) 20 cm, virtual, 4) 20 cm, real, 3) sin , 4) sin , , , refractive index of, , 13 , , NARAYANAGROUP, , 13 , , 225
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 11. A ray of light is incident on a rectangular plate, thickness ‘t’ and refractive index 2 3/ 2 ., at an angle of incidence 600 . The light ray, The path difference between the two rays due, suffers a deviation which is 25 % of the angle, to the glass slab will be, of incidence. The refractive index of the glass, 1) 4t / 3 2) 3t / 2, 2) t /8, 4) t / 6, will be, 17. A plane mirror is placed at the bottom of a tank, 1) 3, 2) 2, 3) 3/ 2, 4) 1.5, containing a liquid of refractive index . P is, 12. There is a water film formed on a glass block., a small object at a heigth h above the mirror., A light ray is incident on water film from air at, An observer O, vertically above P, outside the, an angle 600 with the normal. The angle of, liquid, observes P and its image in the mirror., The apparent distance between these two will, incidence on glass slab is g 1.5, w 4 / 3 , be, 3 3, 1) sin 8 , , , 1, , O, , 1 , 2) sin , , 3, 1, , , , , , 1 4 3, 1 9 3, 3) sin 9 , 4) sin 16 , , , , , 13. A ray of light travels in the way as shown in, the figure. After passing through water, the ray, grazes along the water air interface. The value, , of g interms of ‘i’ is w 4 / 3, , Water, , glass(g), , i, , P, , Plane mirror, , 1) 2 h 2), , 2h, , , 3), , , 2h, 1, 4) h 1 , 1, , , 18. A cubic container is filled with a liquid whose, refractive index increases linearly from top to, bottom. Which of the following represents the, path of a ray of light inside the liquid?, , (1), , (2), , 1, 3, 4, 2), 3), 4) sin i, sin i, 4sin i, 3sin i, 14. A layer of oil 3 cm thick is floating on a layer, (3), (4), of coloured water 5 cm thick. The refractive, index of the coloured water is 5/3 and the, apparent depth of the two liquids is 36/7cm. 19. A light ray travelling in a glass medium is, Then the refractive index of the oil is, incident on glass - air interface at an angle of, 7, 5, 7, 5, incidence . The reflected (R) and transmitted, 1), 2), 3), 4), 4, 4, 5, 3, (T) intensities, both as function of , are, 15. ‘n’ transparent slabs of refractive index 1.5, plotted. The correct sketch is, each having thickness 1 cm, 2 cm,----n cm are, arranged one over the other. A point object is, 100%, 100%, seen through this combination from top with, T, T, (1), (2), perpendicular light. If the shift of the object, R, R, by combination is 5 cm. Then the value of ‘n’, 0, 0, , 90°, 90°, is...., 1) 5, 2) 4, 3) 3, 4) 6, 100%, 100%, 16. Two parallel rays are travelling in a medium, T, T, (3), (4), of refractive index 1 4/ 3 . One of the rays, R, R, passes through a parallel glass slab of, 0, 0, 90°, 90°, 226, , Intensity, Intensity, , Intensity, , Intensity, , 1), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 20. Find the variation of refractive index assuming 23. The refraction index of an anisotropic medium, it to be a function of y such that a ray entering, varies as 0 x 1 , where 0 x a. A, origin at grazing incidence follows a parabolic, ray of light is incident at the origin just along, path y = x 2 as shown in fig:, y-axis (shown in figure). Find the equation of, y, ray in the medium., y, , a, , x, 90°, , x, , O, , 1), , 1 4x 2, , 2), , 1 2x 2, , 3) 1 x 2, 4) 1 8x 2, 21. A ray of light is incident on a glass slab at, grazing incidence. The refractive index of the, , 1) y 2 x, , 2) y 4 x, , 3) y x, 4) y 3 x, 24. A vessel of depth H is filled with a nonmaterial of the slab is given by 1 y ., homogenous liquid whose refractive index, If the thickness of the slab is d, determine the, y, , equation of the trajectory of the ray inside the, varies with y as 1 , . What is the, H, , slab and the coordinates of the point where the, ray exits from the slab. Take the origin to be, apparent depth as seen by an observer from, at the point of entry of the ray., above?, 1) H log e 2, 2) H log e 3, 1) x 2 d, 2) x d, , 3) H log e 5, 4) H log e 7, 3) x d, 4) x 4 d, 22. Due to a vertical temperature gradient in the 25. A ray of ligth enters into a glass slab from air, as shown .If refractive of glass slab is given, atmosphere, the index of refraction varies., Suppose index of refraction varies as, by A Bt where A and B are constants, and, ‘t’ is the thickness of slab measured from, n n0 1 ay , where n0 is the index of, the top surface. Find the maximum depth, refraction at the surface and a = 2.0 106 m 1 ., travelled by ray in the slab. Assume thickness, A person of height h = 2.0 m stands on a level, of slab to be sufficiently large., surface. Beyond what distance will he not see, 60°, the runway?, Air, , y, , h = 2m, , 1) 2000 m, 3) 2500 m, , NARAYANAGROUP, , x, , 1, 3, 1) B A 2 , , , , 2) 3000 m, 4) 3500 m, , 3) A B 2 , , , , 1, , 3, , , , 3, , 1, , 3, , 2) B A 2 , , , 4) B2 A 2 , , , , 227
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 26. A ray of light travelling in air is incident at a 30. A transparent thin film of uniform thickness, grazing angle on a large transparent slab of, and refractive index n1 1.4 is coated on the, thickness t 2.0m . The point of incidence, convex spherical surface of radius R at one, is the origin. The medium has a variable, end of a long solid glass cylinder of refractive, refractive index(y) given by y ky 1, index n 2 1.5 , as shown in figure. Rays of, , 1, light parallel to the axis of the cylinder, Where y is in m and k 0.25m, traversing through the film from air to glass, Q(x , y ), get focused at distance f1 from the film, while, rays of light traversing from glass to air get, P(x, y), t=2.0m, focused at distacnce f 2 from the film. Then, 0, , 0, , n1, , O(0, 0), Page 142 - fig 3, , (a)Express a relation between the angle of incidence, and the slope of the trajectory m, in terms of the, refractive index at that point, , 2 1, 1) sin 1 1 , , , , y ., , 2 1, 2) sin1 1 , , , , Air, , n2, , A) f1 3R, , B) f1 2.8R, , C) f 2 2R, , D) f 2 1.4 R, , 1) A,D 2) B,C, 3) A,C, 4) B,D, , LENS, 4) sin 1 1 , 31. A pin is placed 10 cm front of a convex lens of, , focal length 20 cm and refractive index 1.5., REFRACTION AT SPHERICAL SURFACE, The surface of the lens farther away from the, pin is silvered and has a radius of curvature is, 27. A glass sphere 1.5 of radius 20 cm has a, 22 cm. How far from the lens is the final image, small air bubble 4 cm below its centre. The, formed, sphere is viewed from outside and along a, 1) 10 cm 2) 11 cm, 3) 12 cm, 4) 13 cm, vertical line through the bubble. The apparent, depth of the bubble below the surface of sphere 32. Two plano concave lenses of glass of refractive, is ( in cm), index 1.5 have radii of curvature 20 cm and 30, 1) 13.33 2) 26.67, 3) 15, 4) 30, cm respectively. They are placed in contact, 28. A ray of light is incident on a glass sphere of, with the curved surface towards each other and, refractive index 3/2. What should be the angle, the space between them is filled with a liquid, of incidence so that the ray which enters the, of refractive index 5/2. The focal length of the, sphere does not come out of the sphere is, combination is (in cm), 1, 1, 1) 6, 2) - 92, 3) 108, 4) 12, 1) tan 2 / 3, 2) sin 2 / 3, 33., The, two, surfaces, of, a, biconvex, lens, has same, 3) 900, 4) cos 1 1/ 3, radii of curvature. This lens is made of glass, 29. There is a small air bubble inside a glass, of refractive index 1.5 and has a focal length, 10 cm in air. The lens is cut into two equal, sphere 1.5 of radius 10 cm. The bubble, halves along a plane perpendicular to its, is 4 cm below the surface and is viewed, principal axis to yield two plano-convex lenses., normally from the outside. The apparent depth, The two pieces are glued such that the convex, of the bubble is, surfaces touch each other. If this combiantion, 1) 3 cm below the surface, lens is immersed in water of refractive index, 2) 5 cm below the surface, 4/3 its focal length (in cm) is, 3) 8 cm below the surface, 1) 5, 2) 10, 3) 20, 4) 40, 4) 10 cm below the surface, 2 1, 3) sin 1 1 , , , , 228, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 34. The effective focal length of the lens, combination shown in figure is - 60 cm. The, radii of curvature of the curved surfaces of the, plano-convex lenses are 12 cm each and, refractive index of the material of the lens is, 1.5. The refractive index of the liquid is, , GEOMETRIC OPTICS, curved surfaces of the lenses then the focal, length of the combination is, R, 2R, 1) , 2) , 1, 2, 1, 2, , R, R, 3) 2 , 4) , 1, 2, 1, 2, 40. Image of the sun is formed by a binconvex lens, of focal length f. The image is a circular patch, of radius r and is formed on the focal plane of, =1.5, =1.5, the lens. Choose the correct statement from, the following, 1) The area of the image is r 2 and it is directly, proportional to f, 1) 1.33 2) 1.42, 3) 1.53, 4) 1.60, 2) The area of the image is r 2 and it is directly, A convex lens is in contact with concave lens., The magnitude of the ratio of their focal length, proportional to f 2, is 2/3. Their equivalent focal length is 30 cm., 3) The intensity of the image will increase if f is, Their individual focal lengths are, increased, 1) -75 cm, 50 cm, 2) - 10 cm, 15 cm, 4) If the lower half of the lens is covered with black, 3) - 50 cm, 75 cm, 4) - 15 cm, 10 cm, paper, the area of the image will become half, Two converging glass lenses ‘A’ and ‘B’ have 41. A plano - convex lens of refractive index 1.5, focal lengths in the ratio 2 : 1. The radius of, and radius of curvature 30 cm is silvered at, curvature of first surface of lens ‘A’ is 1/4th of, the curved surface. Now this lens has been, the second surface where as the radius of, used to form the image of an object. The, curvature of first surface of lens ‘B’ is twice, distance from this lens the object be placed, that of second surface. Then the ratio between, in order to have a real image of the same size, the radii of the first surfaces of A and B is, of the object is, 1) 5 : 3 2) 3 : 5, 3) 1 : 2, 4) 5 : 6, 1) 20cm 2) 30cm, 3) 60cm, 4) 10cm, Diameter of a plano - convex lens is 6 cm and 42. A converging lens forms a real image I of an, its thickness at the centre is 3 mm. Then the, object on its principle axis. A rectanglular slab, focal length of the lens, if the speed of light in, of refractive index and thickness x is, the material of lens is 2 108 m / s, introduced between I and the lens. Image I will, 1) 7.5 cm 2) 15 cm, 3) 45 cm, 4) 30 cm, move, Two thin symmetrical lenses one converging, 1) Towards the lens by 1 x, and other of diverging nature are made from, different material have equal radii of curvature, 1, R = 15 cm, the lenses are put in contact and, 2) Away from the lens by 1 x, , immersed in water w 4 / 3 . The focal, 3) Away from the lens by 1 x, length of the system in water is 30 cm. Then, the difference between refractive indices of the, 1, two lenses is, 4) Towards the lens by 1 x, , 1, 1, 1, 3, 43. Image of an object at infinity is formed by a, 1), 2), 3), 4), 2, 4, 3, 4, convexlens of focal length 30 cm such that the, A plano convex lens fits exactly into a plano, size of the image is 12 cm. If a concave lens of, concave lens. Their plane surfaces are parallel, focal length 20 cm is placed in between the, to each other. If the lenses are made of, convexlens and the image, at a distance 26 cm, from the convexlens, size of the new image is, different materials of refractive indices 1 and, 1) 2.5 cm, 2) 2.0 cm, 2 and ‘R’ is the radius of curvature of the, 3) 1.025 cm, 4) 1.05 cm, liquid (1), , 35., , 36., , 37., , 38., , 39., , NARAYANAGROUP, , 229
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 44. An object 2.4 m infront of a lens forms a sharp 49. A ray of monochromatic light is incident on one, image on a film 12cm behind the lens. A glass, refracting face of a prism of angle 750 . It, plate of 1 cm thick of =1.5 is interposed, passes through the prism and is incident on, between lens and film with its plane faces, the other face at the critical angle. If the, parallel to film. The distacne (from lens), refractive index of the material of the prism is, should object shifted to be in sharp focus on, 2 , the angle of incidence on the first face of, film is (AIEEE-2012), the prism is,, 1) 7.2 m 2) 2.4 m, 3) 3.2 m, 4) 5.6 m, 1) 300, 2) 450, 3) 600, 4) 00, 45. A biconvex lens of focal length 15 cm is in front, 0, of a plane mirror. The distance between the 50. The light ray is incident at an angle of 60 on, lens and the mirror is 10 cm. A small object is, a prism of angle 450 . When the light ray falls, kept at a distance of 30 cm from the lens. The, on the other surface at 900 , the refractive, final image is, index of the material of the prism and angle, 1) Virtual and at a distance of 16 cm from the mirror, of deviation ‘d’ are given by, 2) Real and at a distance of 16 cm from the mirror, 2) 1.5, d 150, 1) 2, d 300, 3) Virtual and at a distance of 20 cm from the mirror, 4) Real and at a distance of 20 cm from the mirror, 3, 3, , d 150, , d 300 4) , 46. A bi-convex lens is formed with two thin plano3) , 2, 2, convex lenses as shown in the figure., Refractive index ‘n’ of the first lens is 1.5 and 51. When a glass prism of refracting angle 600 is, immersed in a liquid its angle of minimum, that of the second lens is 1.2. Both the curved, surface are of the same radius of curvature, deviation is 300 . The critical angle of glass with, R=14 cm. For this bi-convex lens, for an object, respect to the liquid medium is, distance of 40 cm, the image distance will be, 1) 300, 2) 450, 3) 600, 4) 500, 52. Angle of prism is ‘A’ and its one surface is, n=1.5, n=1.2, silvered. Light rays falling at an angle of, incidence 2 A on first surface return back, through the same path after suffering, reflection at second silvered surface., Refractive index of the material of the prism, is, R = 14cm, 1) 2SinA 2) 2CosA 3) 1/ 2CosA 4) tan A, 53., A ray of light incident normally on an isosceles, Page 145 - fig 1, right angled prism travels as shown in the, 1) -280.0 cm 2) 40.0 cm 3) 21.5 cm 4)13.3 cm, figure. The refractive index of the prism must, REFRACTION THROUGH PRISM, be greater than, 0, 47. The angle of minimum deviation for a 75, prism of dense glass is found to be 450 when, in air and 150 when immersed in certain liquid., The refractive index of the liqud is, 1), , 3, 2, , 2), , 3, 2, , 3), , 3, 2, , 4), , 3, , 1) 2, 2) 3, 3) 1.5, 4) 2, 54. The refracting angle of a prism is A and the, 48. A certain prism of refracting angle 600 and of, refractive index of the material of the prism is, refractive index ‘2’ is immersed in a liquid of, A, cot . The angle of minimum deviation of, refractive index 2 , then the angle of, 2, minimum deviation will be (NCERT), the prism is, 1) 300, 2) 450, 3) 600, 4) 750, , , 1) 2A 2) 2A 3) A 4) A, 2, 2, 230, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRIC OPTICS, , 55. A ray of light incident on the hypotenuse of a 61. An optician while testing the eyes finds the, right angled prism after travelling parallel to, vision of a patient to be 6/12. By this means, the base inside the prism is incident on second, that, 1) The person can read the letters of 6 inches from, refracting surface. If ' ' is the refractive index, a distance of 12 m, of material of prism, the maximum value of the, 2) The person can read the letters of 12 inches from, base angle for which light is totally reflected, 6 cm, from the hypotenuse is, 3) The person can read the letters from 6m which, 1 1 , 1 1 , the normal eye can read from 12 m, sin, tan, 1), 2), , , , , 4) The focal length of eye lens had become half, that of the normal eye, 1 1 , 1 1 , sin, cos, 3), 4), , , , OPTICAL INSTRUMENTS, , , 56. The principle section of a glass prism is an 62. The length of a compound microscope is 14, cm and its magnfying power when final image, isosceles triangle ABC with AB = AC, the face, is formed at near point is 25. If the focal length, AC is silvered. A ray of light is incident, normally on the face AB and after two, of eyepiece is 5 cm. The distance of object from, reflections, it emerges from the base BC, the objective and the focal length of objective, perpendicular to the base. Angle of BAC of, lens are, the prism is, 59, 59, 59, 59, 1) 300, 2) 360, 3) 180, 4) 720, 1), cm, cm, 2), cm, cm, 25, 31, 31, 25, 57. The ratio of angle of minimum deviation of a, 3) 3cm,2cm, 4) 4cm,1cm, thin prism in air and when dipped in water will, 63. The focal length of objective and eye lens of a, 3, 4, , and, , , , , be a g, a w, , microscope are 4 cm and 8 cm respectively. If, 2, 3, , the least distacne of distinct vision is 24 cm, 1, 1, 3, 4, and object distance is 4.5 cm from the objective, 1), 2), 3), 4), lens, then the magnfying power of the, 8, 4, 4, 1, microscope will be, DISPERSION, 58. Two prisms A and B are in contact with each, 1) 18, 2) 32, 3) 64, 4) 20, other have angular dispersions of 20 and 40 64. The magnifying power of a microscope with an, respectively. The dispersive power of ‘A’ is, objective of 5mm focal length is 400. The, 0.002. If the combination produces dispersion, length of its tube is 20 cm. Then the focal length, without deviation, the dispersive power of ‘B’, of the eye - piece is, is, 1) 2.23 cm, 2) 160 cm, 1) 0.001 2) 0.004, 3) 0.002, 4) 0.006, 3) 200 cm, 4) 0.1 cm, 59. Two prisms A and B have dispersive powers, of 0.012 and 0.018 respectively. The two 65. In an astronomical telescope in normal, prisms are in contact with each other. The prism, adjustment, straight black line of length L is, 0, drawn on the objective. The eye piece forms a, ‘A’ produces a mean deviation of 1.2 , the, real image of this line. The length of this image, mean deviation produced by ‘B’ if the, is . The magnificaiotn of the telescope is, combination is achromatic is, 1) 3.60 2) 0.80, 3) 0.40, 4) 1.80, L, L, L, L 1, 60. Two prisms A and B in contact with each other., 2) 1, 3) 1, 4), 1), , , , 1, The dispersive power of A and B is 0.04 and, 0.03 respectively. The angular dispersion 66. Focal length of the objective of a terrestrial, telescope is 80 cm and it is adjusted for parallel, produced by ‘A’ is 80 . The angular dispersion, rays, then its magnfying power is 20. If the, produced by the combination, if the, focal length of erecting lens is 20 cm, then full, combination does not produce a net deviation, length of telescope will be (in cm), is, 1) 84, 2) 100, 3) 124, 4) 164, 1) 140, 2) 80, 3) 60, 4) 20, NARAYANAGROUP, , 231
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, , 67. The focal lengths of the lenses of an, sin i, astronomical telescope are 50 cm and 5 cm. 11) , sin r, The length of the telescope when the image is, formed at the least distance of distinct vision, sin i, sin 600, , , and then g w , 12) a w, is (in cm), sin r, sin r, 275, 325, sin i, sin i, 1) 45, 2) 55, 3), 4), and g w , 13) a g , 6, 6, sin r, sin r, LEVEL - III - KEY, R.d, 14) , 1) 2, 2) 4 3) 1 4) 2 5) 1 6) 1, A.d, 7) 3, 8) 1 9) 1 10) 3 11) 3 12) 1, 1, 13) 1, 14) 3 15) 1 16) 2 17) 2 18) 1, 15) Shift = t 1 ; Calculate for ‘n’ blocks, 19) 3, 20) 1 21) 1 22) 1 23) 1 24) 1, , 25) 1, 26) 1 27) 2 28) 3 29) 1 30) 3, , , 31) 2, 32) 4 33) 4 34) 4 35) 4 36) 4, x 2 1 t, 16), 37) 4, 38) 3 39) 1 40) 2 41) 1 42) 2, 1 , 43) 1, 44) 4 45) 3 46) 2 47) 3 48) 1, h, h h, 2h, 49) 2, 50) 4 51) 2 52) 2 53) 1 54) 2, 17) A.d= d ; d , 55) 4, 56) 2 57) 4 58) 2 59) 2 60) 4, , , , 61) 3, 62) 1 63) 2 64) 1 65) 1 66) 4 18) refractive index is chaning, the light cannot travel in, 67) 4, a straight line in the liquid, 19) After total internal reflection no refracted ray, LEVEL -III - HINTS, 1, dy, 1) x-coordinate inverted, tan 90 , 20) sin and, , dx, 2) M f / f ; L f f, o, , e, , o, , e, , 3), , 1 1 1, V, M and apply , f v u, u, , dy, dy, cot 2 , 2x, dx, dx, , , 4), , 1 1 1 1 2 1, , , f, fl f m fl, fl f m, , fm 0 , , 6), , 1 1 1, apply , f v u, , y, ; y a 2, x , a, 1, 1, 8) A i 3 j and B , 2, 2, , A.B AB cos, sin i, 9) , sin r, sin i, 10) , and i r 900, sin r, , 7), , 232, , , , x2, x4 y 2 d, 4, 1, dy, 22) sin 1 ay dx tan 90 , y, , a, 2, , sin 450 , , , , 1 4x2, , y, x, dy, dy, 1/ 2, y 1/ 2 dx, dx, y, 0, 0, , 1 1 1, , f v u, , apply, , 1 cot 2 , , 1, 1, 1, dy, cot , 21) sin 1 y and, dx, , 1, R, ., fl 2 1, , 5), , 1, cos ec , sin , , , , i 3 j, , , , dy, dy, ay , dx, dx, ay , x2, 23) sin , , y, xmin 2000m, a, 1, , , , 1, ;, x 1, , dy, 1, tan , dx, x, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , dx, y2 x, x, 24) Apparent thickness of small element ' dy ' is ' dh ', , dy , , H, dy, dy, dy, 1, , H ln 2, h , y, y, = 1 ;, 0 1, a, H, 0, 25) 1 sin 60 0 sin, , tmax , , 26) sin i , , , ;, 2, , GEOMETRIC OPTICS, 1 1 1, 35) f f f, 1, 2, , 1 1 , 1, 36) f 1 R R , 1, 2 , 37) , , 3, A Bt max, 2, , 2, 2 Rt t 2 y 2 t 4 ; 2 Rt y R , , 1, 3, A, , B, 2 , , 1, 1, i sin 1 , , , , M tan , , 1 sin 2 i, sin i, , 2 1, M, , 1, 1, 1 2 1 ; i, 2, sin 1 , sin i, , 2 1 2 1, , , v, u, R, 3, 28) on second surface = 180 r 60 900, 2, refraction is more than 900, 1, 29) 2 1 2, v, u, R, 1.5 1.4 1.5 1.5 1.4, , 30) For air to glass ; f , R, R, 1, , y2, 2t, , 1 1 , 1, 1 , f, R1 R2 , , 1 1 1 1, 38) f f f f, g, l, g, 1, , 1, , 1 , , 39) f 1 R R , l, 1, 2 , 1 1 1 1 1 2 1, 1 R, , and f 2, 41) f f f f ;, fl, R, l, m, l, m, , 1, 42) S t 1 , , , 27), , 43), , 1 1 1, , f v u, , 1, 44) x t 1 , , , 1 1 1, for lens, 45) u v for mirror ;, f v u, 46) apply two times ;, , 1 1.4 1.5, f1 3R and glass to air ; f R 2R, 2, 1 1 1 1 1 2 1, R, 31) f f f f ; f f f f m 2, l, m, l, l, m, , Co, ; 2R t t y 2, Cm, , 2 1 2 1, , v, u, R, , A D, 2, A, sin, 2, , sin, , 47), , , , sin, , A D, , 1 1 , 1, 2, , 32) f 1 R R , 48), A, sin, 1, 2 , 2, 33) Perpendicular to principle axis focal length does not, sin i, change, 49) a g , apply for two surfaces, sin r, 1 1 1 1, 34) f f f f, sin i, 1, 2, 3, 50) , and d 1 , sin r, NARAYANAGROUP, , 233
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 51), , L, , g , , 1, sin c, , LEVEL - IV, , 52) one face is silvered r2 0 and i2 0, 53) deviation is 1800, , COMPREHENSION - I, A point object O is placed in front of a concave, mirror of focal length 10 cm. A glass slab of, refractive index 3 / 2 and thicikness 6 cm, is inserted between the object and mirror., , gm 2, , A D , sin , , 2 , , , 54), A, sin, 2, , 55) , , 6cm, , sin i, sin r, O, , 56) i2 0 and r2 0, , d1 1, 57) d , 1, g, 2, , x, 32cm, , 58) d1 d 2 0 ; 1 1 1 2 1 2 0, 59), , w1 w2, , 0, f1 f 2, , w1 w2, 60) f f 0, 1, 2, , 61), 62), , 1., , 2., , 1, 1, 1, , p, f . p dis tan ce of object f, 1, 1 1, , L ve ue find ve, f e ve ue find ue ;, , 1) 11 cm, virtual, , 2) 17 cm, real, , 3) 14 cm, real, , 4) 20 cm, virtual, , Find the position and nature of the final image, when the distance x shown in figure is 20cm., 1) 17cm, virtual, , 2) 17cm, real, , 3) 12cm, virtual, , 4) 15cm, virtual, , COMPREHENSION - 2, Consider the situation in the figure. The, bottom of the pot is a reflecting plane mirror,, S is a small fish, and T is a human eye., Refractive index of water is , , 1, 1 1, , f e ve ue find f e, , 63), , Find the position and nature of the final image, when the distance x shown in figure, is 5 cm., , 1, 1 1, 1, 1 1, , , find, u, ;, e, f e ve ue, f e ve ue find f e, m mo me , , T, H, , vo ve, uo ue, H, , fo, 64) M f, e, fo, l, 65) M f ; m , L, e, fo, 66) M f ; L f o 4 f f e, e, , fo , fe , 67) M f 1 D , , e , 234, , S, H/2, , 3., , At what distance from itself will the fish see, the image of the eye by direct observation?, 1, , 1) H , 2, , , 1, , 2) H , 2, , , H 1, , , 2 2, , , 1 , 4) H , , 2 , , 3), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 4., , 5., , 6., , GEOMETRIC OPTICS, , At what distance from itself will the fish see or COMPREHENSION - 4, observe the image of eye by observing, A thin equiconvex lens of refractive index 3/2, through mirror is, is placed on a horizontal plane mirror as shown, in figure. The space between the lens and the, 3, , 1, , mirror is filled with a liquid of refractive index, H, , , H, , , 1) , 2) , , , 2, , 2, , 4/3. It is found that when a point object is placed, 15 cm above the lens on its principal axis, the, 3 , 3, , object coincides wiht its own image., 3) H , 4) 2 H , , 2 , 2, , At what distance from it self will the eye see, the image of the fish upon direct observation?, , 1 , 1) H 1 , , 2 , , , 1 , 2) 2 H 1 , , 2 , , , 3 , 1) 2 H , , 2 , , , 2) H 2 , , , 3 , 3) H 1 , , 2 , , , 3 , 4) H 1 , , 2 , , The radius of curvature of the convex surface, is, 1) 10 cm 2) 15 cm 3) 20 cm, 4) 25 cm, , 1 , 1 , 2, H, 1, , H, 10. If another liquid is filled instead of water, the, 3), 4) , , , , 2 , 2 , object and the image coincide at a distance 25, cm from the lens. Calculate the refractive, At what distance from itself will the eye see, index of the liquid., the image of the fish by observing from the, 1) 1.6, 2) 2.6, 3) 2.8, 4) 3.2, mirror?, , , 3, , , , , , 9., , LEVEL - IV - KEY, 1) 2, 6) 2, , 2) 1, 7) 2, , COMPREHENSION - 3, A glass shere of radius 2R and refractive index 1), ‘n’ has a sherical, 2R, R, O, , 7., , When viewer is on left side of the hollow, sphere, what will be the shift in position of the, object?, n 1, n 1, 1) n 1 R , left, 2) n 1 R , right, , , , 2 n 1, 2n 1, 3) 2n 1 R , left, 4) n 1 R , left, , , , 8. When viewer is on right side of the hollow, sphere, what will be the apparent change in, position of the object?, n 1, n 1, 1) 3n 1 R , toward left 2) 3n 1 R , toward left, , , , , , n 1, n 1, 3) 3n 1 R ,toward right 4) 3n 1 R , toward right, , , , , NARAYANAGROUP, , 3) 1, 8) 2, , 4) 1, 9) 1, , 5) 1, 10) 2, , LEVEL - IV - HINTS, The normal shift produced by a glass slab is,, 1 2, x 1 t 1 6 2cm, 3, i.e., for the mirror, the object is placed at a distanced, (32-) = 30cm for it., Applying mirror formula, , 1 1 1, , v u f, or, v = -15 cm, when x 5cm; the light falls on the slab on its return, journey as shown. But the slab will again shift it by, a distance., , l, x, , 15cm, x 2cm , hence the final real image is formed at, a distance (15+2) = 17cm from the mirror., 235
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRIC OPTICS, 2), , When x 20 cm; this time also the final image is at, a distance of 17cm from the mirror but it is virtual, as shown., , 2R, Origin, , Origin for first, reflection, , R, O, , , , , , , , l, S, , 3), , x, , Fish obersving eye :, , H, Fish, , 4), , Direct observation, H, H1 H, 2, 1, , H1 H , 2, , Fish observing image of eye by mirror,, Hence, distance of the eye image from fish,, H, H2 H H , 2, 3, , H2 H , 2, , H, , Fish, , H, , H + H, , 5), , 6), , 7), , 236, , Eye observing fish :, Direct observation, , H, 1 , H1 H , H 1 , , 2, 2 , Eye observing image of the fish, , H H, 3 , H2 H , H 1 , , 2, 2 , , S2, , 1 n , n, 1, , , v R 2 R , , H, H/2, , S1, , Origin for second, reflection, , 8), , 2R , Which on solving for v yields v , , n 1 , Image is on the right of refracting surface S., Shift = Real depth - Apparent depth, 2 R n 1, R , R, , n 1 n 1, When the viewer is on the right, two refractions, take place at surface, n 1, n, 1, For refraction at surface : v 2 R R, , 1, 2nR, 2n 1, which on solving for v1 yields, The first lies to the left of and acts as object for, refraction at the second surface. We have to shift, the origin of Cartesian coordinate system to the, vertex of . The object distance for the second, surface is, 1, n, 1 n, , , 2 4n 1, v2, 4n 1 , 2 R ; v2 , R, , R, 3, n, , 1, , , , 2n 1 , On solving we get, The minus sign shows that image is virtual and lies, 2 4n 1 R n 1, to the left of 3n 1 3n 1 ., v1 , , , , , , , , , , 9&10), The light retraces its path if it is incident normally, on a mirror. The ray after refraction through the, lens and the liquid are parallel. We will apply the, general thin lens equation with parameters., 3, 4, n1 1, n2 , n3 , u 15cm and v , H, 2, 3, n3 n1 n2 n1 n3 n2 , H/, H/2, , , u, R, R, H/2, On solving for R, we get R = 10cm, similarly when, Viewer on the left of hallow sphere: Single refraction, second liquid is filled, we have 1.6., takes place at surface S. From the single surface, refraction equation, we have, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, 5., , LEVEL-V, SINGLE CORRECT ANSWER TYPE, 1., , 2., , A concave mirror is placed on a horizontal, table, with its axis directed vertically upwards., Let O be the pole of the mirror and C its centre, of curvature. A point object is placed at C. It, has a real image, also located at C. If the mirror, is now filled with water, the image will be, (A) real and will remain at C, (B) real and located at a point between C and , (C)virtual and located at a point between C and O, (D) real and located at a point between C and O, A plane mirror of length 8 cm is moving with, speed 3 m/s towards a wall in situation as, shown in figure.A source of light S gives narrow, and parallel light on to the mirror, then the, speed of light spot formed on the wall is., Wall, S = Source of light, , 6., , The origin of x and y coordinates is the pole, of a concave mirror of focal length 20 cm. The, x-axis is the optical axis with x > 0 being the, real side of mirror. A point object at the point, (25 cm, 1 cm) is moving with a velocity 10 cm/, s in positive x-direction. The velocity of the, image in cm/s is approximately, (A) – 80 i + 8 j, (B) 160 i + 8 j, (C) – 160 i +8 j, (D) 160 i – 4 j, As the position of an object (u) reflected from, a concave mirror is varied, the position of the, image (v) also varies. By letting the u changes, from 0 to the graph between v versus u, will be, v, , v, , u, , (A), , (A) 8 m/sec (B) 6 m/sec, , u, , (B), , v, , v, , 3m/s, , (C) zero, , (D) 16 m/sec, , 8cm, , 3., , A linear object is placed along the axis of a, mirror as shown in figure. If ‘f’ is the focal, length of the mirror then the length of image, is., , u, , (C), 7., , (A) (2f) / 3 (B) f, , u, , (D), , The graph shows variation of v with change in, u for a mirror. Points plotted above the point P, on the curve are for values of v, v, , (A) Smaller then f, , 3/2f, , (C) f / 3, , 2f, , 4., , (B) Smaller then 2f, , (D) None, , P, , A point source is situated at a distance x < f, from the pole of the concave mirror of focal, length f. At time t = 0, the point source starts, moving away from the mirror with constant, velocity. Which of the graphs below represents, best, variation of image distance | v | with the, distance x between the pole of mirror and the, source., , (C) Larger then 2f, 45°, , 8., , u (D), , Larger than f, For a concave mirror, if virtual images formed,, the graph between m and u is of the form, m, , m, , V, , V, , (A), (A), , (B), , (B), x0, , f, , x, , (C), , x0, , f, , m, , (C), , (D), x0, , f, , NARAYANAGROUP, , x, , f u, , u, , x, , V, , V, , 1, , x0, , f, , x, , m, , (D), , 1, u, , u, 237
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 9., , Following figure shows the multiple reflections, of a light ray along a glass corridor where the, walls are either parallel or perpendicular to, one another. If the angle of incidence at point, P is 30°, what are the angles of reflection of, the light ray at points Q, R, S and T, respectively, R, , 14. Two plane mirrors A and B are aligned parallel, to each other, as shown in the figure. A light, ray is incident at an angle 30° at a point just, inside one end of A. The plane of incidence, coincides with the plane of the figure. The, maximum number of times the ray undergoes, reflections (including the first one) before it, emerges out is (2002; 2M), 2 3m, B, , T, Q, S, , P, , 0.2m, , (A) 30°, 30°, 30°, 30° (B) 30°, 60°, 30°, 60°, (C) 30°, 60°, 60°, 30° (D) 60°, 60°, 60°, 60°, 10. A point object is placed mid-way between two, plane mirrors distance ‘a’ apart. The plane, mirror forms an infinite number of images due, to multiple reflection. The distance between, the nth order image formed in the two mirrors, is, (A) na, , (B) 2na, , (C), , na, 2, , f x, f, , (B), , f, x, , (C), , f, x, , (D), , f2, x2, , 13. A point source of light B, placed at a distance, L in front of the centre of a plane mirror of, width d, hangs vertically on a wall. A man walks, in front of the mirror along a line parallel to, the mirror at a distance 2 L from it as shown., The greatest distance over which he can see, the image of the light source in the mirror is, (2000; 2M), , B, , d, , A, , (A) 28, (B) 30, (C) 32, (D) 34, 15. A linear object is placed along the axis of a, mirror as shown in fig. If ‘f’ is the focal length, of the mirror then the length of image is, , (D) 3na, , 11. Two plane mirrors are at right angles to each, other. A man stands between them and combs, his hair with his right hand. In how many of, the images will be seen using his right hand, (A) None (B) 1, (C) 2, (D) 3, 12. The focal length of a concave mirror is f and, the distance from the object to the principal, focus is x. The ratio of the size of the image, to the size of the object is, (A), , (A), , d, 2, , 3, , 2f, 3, , B) f, , f, f, D), 3, 2, 16. A ray of light is incident on a plane mirror along, a vector ˆi ˆj kˆ . The normal to the mirror at, , C), , the point of incidence is along ˆi ˆj . Then unit, vector along the reflected ray is, 1 ˆ ˆ ˆ, i jk, (A), 3, , , , , , (B) , , 1, 3, , , , , , , , (i j k), , , , , 1 , 1, (i j k ), ( i j k ), (D) , 3, 3, 17. A ray of light passes through four transparent, media with refractive indices 1, 2 , 3 and 4, as shown in the figure, the surfaces of all media, are parallel. If the emergent ray CD is parallel, to the incident ray AB, we must have, , (C) , , 1, , B, , (C) 2d, , 238, , f, , 2f, , (B) d, , (D) 3d, , A), , 2, , L, , 2L, , 30°, , A, , 2, , 3, , D, , C, , 4, , (A) 1 2, (B) 2 3, (C)\ 3 4, (D) 4 1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, , 18. A diverging beam of light from a point source, S h av i n g d i v er gen ce an gl e , falls, symmetrically on a glass slab as shown. the, angles of incidence of the two extreme rays, are equal. If the thickness of the glass slab is, t and the refractive index n, then the, divergence angle of the emergent beam is, , 2, , , I, , 1, , n, , , , , , , , /2, , O, , 2, , , , 1, , (D), /2, , O, , (D) 2 sin1(1 / n), , , , /2, , 2, , , , (C), , 19. The x-z plane separates two media A and B, with refractive indices 1 and 2 respectively.., A ray of light travels from A to B. Its directions, in the two media are given by the unit vectors, , , 1, , (B), O, , (B) , , (C) sin1(1/ n), , I, , , , (A), , S, , (A) zero, , 2, , , , 1, /2, , O, , 22. A ray of light is incident at the glass-water, interface at an angle i it emerges finally, parallel to the surface of water, then the value, of g would be (2003; 2M), Air, , 4, , (A) 3 sin i, , , r, , , , 1, , r A a i b j and r B i j respectively, , where î and ĵ are unit vectors in the x and y, directions. Then., (A) 1a 2, (B) 1 2a, (C) 1b 2, (D) 1 2b, 20. A cubical block of glass of refractive index n1, is in contact with the surface of water of, refractive index n2. A beam of light is incident, on vertical face of the block (see figure). After, refraction, a total internal reflection at the base, and refraction at the opposite vertical face,, the ray emerges out at an angle . The value, of is given by :, , (B) sin i, Water, Glass, , (C), , 4, 3, , (D) l, , 23. White light is incident on the interface of glass, and air as shown in the figure. If green light, is just totally internally reflected then the, emerging ray in air contains., , Air, , Green, , (2004; 2M), , (A) yellow, orange, red, (B) violet, indigo, blue, (C) all colours, , Glass, , (D) all colours except green, 24. A container is filled with water ( = 1.33) upto, a height of 33.25 cm. A concave mirror is placed, 15 cm above the water level and the image of, an object placed at the bottom is formed 25, cm below the water level. The focal length of, the mirror is (2004; 2M), White, , n2, , (C) sin <, , (B) tan <, 1, n12, , n2, , 2, , (D) tan <, , n2, , n12 n 2 2, 1, n12, , n 22, , 21. A ray of light travels from a medium of, refractive index to air. Its angle of incidence, in the medium is i, measured from the normal, to the boundary, and its angle of deviation is, . is plotted against i which of the following, best represents the resulting curve, NARAYANAGROUP, , (A) 10 cm, 15cm, , (A) sin <, , , , 2, , 33.25cm, , n1, , n12, , (B) 15 cm, 25cm, , = 1.33, , (C) 20 cm, , 1, O, , (D) 25 cm, 239
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 25. A ray of light, travelling in a medium of, refractive index 2, is incident onto the flat, surface of a slab, the angle of incidence being, 37°. The refractive index of the material of, the slab varies with depth (x) from the top, surface as (x) R(2 x) , where x is in ‘m’, and R is 1 m–1. The maximum depth reached, by the ray is (take sin 37 0.6 ), , =2, 37°, , x, , (A), , 8, m, 5, , (B), , 6, m, 5, , (C), , 3, m, 5, , (x), , 4, , container :, 3, , A) 8 cm B) 10.5 cm C) 12 cm D)14 cm, 30. An isosceles prism of angle 120º has a, refractive index of 2 . Two parallel, monochromatic rays enter the prism parallel, to each other in air as shown. the rays, emerging from the opposite faces, , (A) are parallel to each other, (B) are diverging, 120°, , (C) make an angle 30º with each other, , 4, (D) m, 5, , 26. A light beam is travelling from Region I to, Region IV (refer figure). The refractive index, in Region I, II, III and IV are, n0 ,, , 29. How much water would be filled in a container, of height 21cm, so that it appears half filled to, the observer when viewed from the top of the, , n0 n 0, n, ,, and 0 , respectively. The angle of, 2 6, 8, , (D) make an angle 60º with each other., 31. A given ray of light suffers minimum deviation, in an equilateral prism P. Additional prisms Q, and R of identical shape and of the same, material as P are now added as shown in the, figure. The ray will now suffer, , incidence for which the beam just misses, entering Region IV is (Figure), Region I, , Region II, , Region III, , Region IV, , n0, 2, , n0, 6, , n0, 8, , n0, 0, , 3, , 0.2m, , 0.6m, , 1, , 1, , 1, , , , 1 , 1 , 1 , 1 , (A) sin 4 (B) sin 8 (C) sin 4 (D) sin 3 , , , , , 27. A ball is dropped from a height of 20 m above, the surface of water in a lake. The refractive, index of water is 4/3. A fish inside the lake, in, the line of fall of the ball, is looking at the ball., At an instant, when the ball is 12.8 m above, the water surface, the fish sees the speed of, the ball as [Take g = 10 m/s2], (A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.23 m/s, , P, , Q, , R, , (A) greater deviation (B) no deviation, (C) same deviation as before, (D) total internal reflection, 32. A ray of light strikes a plane mirror at an, angle of incidence 45º as shown in the figure., After reflection, the ray passes through a, prism of refractive index 1.50, whose apex, angle is 4º. The angle through which the mirror, should be rotated if the total deviation of the, ray is to be 90º is., 45°, , 28. A ray of light in the direction i 3 j is, , , , , , incident on a plane mirror. After reflection it, 1, travels along the direction i 3 j . The, 2, angle of incidence is (IIT-2013), (A) 300, (B) 450, (C) 600, (D) 750, , , , 240, , 88°, , 90°, 4°, , , , (A)1º clockwise, (C) 2º clockwise, , (B) 1º anticlockwise, (D) 2º anticlockwise, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 33. Two beams of red and violet colours are made, to pass separately through a prism (angle of, the prism is 60°). In the position of minimum, deviation, the angle of refraction will be, (A) 30° for both the colours, (B) greater for the violet colour, (C) greater for the red colour, (D) equal but not 30° for both the colours, 34. Two identical thin isosceles prisms of refracting, angle ‘A’ and refractive index are placed, with their bases touching each other. Two, parallel rays of light are incident on this system, as shown. The distance of the point where the, rays converge from the prism is, h, h, (A), (B), A, A, , 2h, , h, h, (D), ( 1) A, ( 1) A, 35. A thin prism P1 with angle 4° and made from, glass of refractive index 1.54 is combined with, another thin prism P2 made from glass of, refractive index 1.72 to produce dispersion, without deviation. The angle of the prism P2, is: (1990; 2M), (A) 5.33° (B) 4°, (C) 3°, (D) 2.6°, 36. A ray of light is incident on an equilateral, glass prism placed on a horizontal table. For, minimum deviation which of the following is, true?, (2004; 2M), , GEOMETRICAL OPTICS, 38. A point object is placed at the center of a glass, sphere of radius 6 cm and refractive index 1.5., The distance of the virtual image from the, surface of the sphere is (2004; 2M), (A) 2 cm (B) 4 cm (C) 6 cm (D) 12 cm, 39. A ray of light is incident on a glass sphere of, 3, refractive index . What should be the, 2, minimum angle of incidence so that the ray, whcih enters the sphere does not come out of, the sphere, 2, , 2, , 1, , , , 1 , 1 , 1 , A) tan B) sin C) 900 D) cos , 3, 3, , , 3, 40. Which one of the following spherical lenses, does not exhibit dispersion? The radii of, curvature of the surfaces of the lenses are as, given in the diagrams, , (C), , (A) PQ is horizontal, Q, , R, , (B) QR is horizontal, S, , P, , (C) RS is horizontal, (D) Either PQ or RS is horizontal, , 37. A spherical surface of radius of curvature R,, separates air (refractive index 1.0) from glass, (refractive index 1.5). The centre of curvature, is in the glass. A point object P placed in air is, found to have a real image Q in the glass. The, line PQ cuts the surface at a point O and, PO = OQ. The distance PO is equal to :, (1998; 2M), (A) 5 R, (B) 3 R, (C) 2 R, (D) 1.5 R, NARAYANAGROUP, , R1, , R2, , (A), , (B), , R, , (D), , R, , , , R1 R2, , (C), , R, , R, , , , ., , 41. A biconvex lens of focal length 15cm is in front, of a plane mirror. The distance between the, lens and the mirror is 10cm. A small object is, kept at a distance of 30cm from the lens. The, final image is, (A) Virtual and at a distance of 16cm from mirror, (B) Real and at distance of 16cm from the mirror, (C) Virtual and at a distance of 20cm form the, mirror, (D) Real and at a distance of 20cm from the mirror, 42. The graph between the lateral magnification, (m) produced by a lens and the distance of the, image (v) is given by, m, , m, , (A), , (B), O, , O, , v, , m, , v, , m, , (C), , (D), O, , v, , O, , v, 241
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 43. A bi-convex lens is formed with two thin planoconvex lenses as shown in the figure., Refractive index n of the first lens is 1.5 and, that of the second lens s 1.2. Both the curved, surfaces are of the same radius of curvature, R = 14 cm. For this bi-convex lens, for an, object distance of 40 cm, the image distance, will be, n = 1.5, , n = 1.2, , (A) –280.0 cm, (B) 40.0 cm, (C) 21.5 cm, , R = 14 cm, , (D) 13.3 cm, , 44. A diminished image of an object is to be, obtained on a screen 1.0 m from it. This can, be achieved by appropriate placing :(95; 2M), (A) a concave mirror of suitable focal length, (B) a convex mirror of suitable focal length, (C) a convex lens of focal length less than 0.25 m, (D) a convex lens of suitable focal length, 45. A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature, R. On immersion in a medium of refractive, index 1.75, it will behave as a (1999; 2M), (A) convergent lens of focal length 3.5 R, (B) convergent lens of focal length 3.0 R, (C) divergent lens of focal length 3.5 R, (D) divergent lens of focal length 3.0 R, 46. The size of the image of an object, which is at, infinity, as formed by a convex lens of focal, length 30 cm is 2 cm. If a concave lens of focal, length 20 cm is placed between the convex, lens and the image at a distance of 26 cm from, the convex lens, calculate the new size of the, image (2003; 2M), (A) 1.25 cm, (B) 2.5 cm, (C) 1.05 cm, (D) 2 cm, 47. A convex lens is in contact with concave lens., The magnitude of the ratio of their focal length, is, , 2, . Their equivalent focal length is 30 cm., 3, , What are their individual focal lengths?, (2005; 2M), (A) –75, 50, (B) –10, 15, (C) 75, 50, (D) –15, 10, 242, , 48. A thin biconvex lens of focal length f is used, to form a circular image of the sun on a screen, placed in its focal plane. The radius of the, image formed on the screen is r. It can be, concluded that, (B) r 2 f 2, (A) r 2 f, (C) if the focal length of the lens is doubled keeping, its aperture constant, the brightness of the image, will increase., (D) if half of the lens is covered, the area of the, 1 2, image will become r, 2, 49. The image of an object , formed by a plane, convex lens at a distance of 8 m behind the, lens, is real and is one-third the size of the, object. The wave length of light inside the lens, 2, is times the wavelenght in free space. The, 3, radius of the curved surface of the lens is, (IIT-2013), (A) 1m, (B) 2m, (C)3m, (D) 43, 50. A thin lens of focal length f and its aperture, has a diameter d . It forms an image of, intensity I. Now the central part of the aperture, d, upto diameter, is blocked by an opaque, 2, paper. The focal length and image intensity, would change to, f I, I, 3f I, 3I, ,, A) ,, B) f ,, C), D) f ,, 2 2, 4, 4 2, 4, 51. A point object is placed at a distance of 25 cm, from a convex lens of focal length 20 cm. If a, glass slab of thickness t and refractive index, 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness, t is, A) 10 cm B) 5 cm, C) 20 cm D) 15 cm, 52. The focal lengths of the objective and the eye, piece of a compound microscope are 2.0 cm, and 3.0 cm respectively. The distance between, the objective and the eye piece is 15.0 cm. The, final image formed by the eye piece is at, infinity. The two lenses are thin. The distance, in cm of the object and the image produced by, the objective, measured from the objective, lens, are respectively :, (1995; 2M), (A) 2.4 and 12.0, (B) 2.4 and 15.0, (C) 2.0 and 12.0, (D) 2.0 and 3.0, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, 53. An eye specialist prescribes spectacles having, combination of convex lens of focal length 40, cm in contact with a concave lens of focal length, 25 cm. The power of this lens combination in, diopters is (1997; 2M), (A) + 1.5 (B) – 1.5 (C) + 6.67 (D) – 6.67, 54. In a compound microscope, the intermediate, image is :, (2000; 2M), (A) virtual, erect and magnified, (B) real, erect and magnified, (C) real inverted and magnified, (D) virtual, erect and reduced, , MULTIPLE ANSWER QUESTIONS, 55. A plane mirror M is arranged parallel to a wall, W at a distance from it. The light produced, by a point source S kept on the wall is reflected, by the mirror and produces a light spot on the, wall. The mirror moves with velocity n towards, the wall. Then., , (A) The spot of light will move with the speed n on, the wall., (B) The spot of light will not move on the wall., (C) As the mirror comes closer, the spot of light, will becomes larger and shift ways from the wall, with speed larger then n., (D) The size of the light spot on the wall remains, the same, 56. What is the relative velocity of the image in, mirror (1) with respect to the image in the, mirror (2) in situation as shown in fig., Mirror (1), , , A), , 2V, B) 2V sin , sin , , C), , 2V, V, D), sin 2 , sin , , v, Object, Mirror (2), , 57. A student performed the experiment of, determination of focal length of a concave, mirror by u–v method using an optical bench, of length 1.5 meter. The focal length of the, mirror used is 24 cm. The maximum error in, the location of the image can be 0.2 cm. The 5, sets of (u, v) values recorded by the student, (in cm) are : (42, 56), (48, 48), (60, 40), (66,, 33), (78, 39). The data set(s) that cannot come, from experiment and is (are) incorrectly, recorded, is (are), (A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39), NARAYANAGROUP, , GEOMETRICAL OPTICS, 58. A ray of light traveling in a transparent medium, falls on a surface separating the medium from, air at an angle of incidence 45°. The ray, undergoes total internal reflection. If n is the, refractive index of the medium with respect, to air, select the possible value (s) of n from, the following : (1998; 2M), (A) 1.3, (B) 1.4, (C) 1.5, (D) 1.6, 59. A ray of light traveling in water is incident on, its surface open to air. The angle of incidence, is , which is less than the critical angle. Then, there will be, (A) Only a reflected ray and no refracted ray., (B) Only a refracted ray and no reflected ray., (C) A reflected ray and a refracted ray and the, angle between them would be less than 180º – 2q., (D)A reflected ray and a refracted ray and the angle, between them would be greater than 180º – 2q, 60. An equiconvex lens forms the image of an, object. A plot of the position of image (v) versus, the position of object (u) is shown in the, adjacent figure. If the maximum error in the, measurement of length is 0.1 cm, the focal, length of the lens is, (A) 5 cm ± 0.05 cm, 11cm (B) 10 cm ± 0.05 cm, 10cm, (C) 5 cm ± 0.1 cm, v, –11cm –10cm, u, , (D) 10 cm ± 0.05 cm, , 61. Which of the following form(s) a virtual and, erect image for all positions of the real object?, (1996; 2M), (A) Convex lens, (B) Concave lens, (C) Convex mirror, (D) Concave mirror, 62. An object ‘O’ is kept infront of a converging, lens of focal length 30cm, behind the lens a, palne mirror kept at distance 15cm, f = 30cm, , O, , 15cm, , 15cm, , A) final image is formed at 60cm from the lens, towards right of it, B)final image is at 60cm from lens towards left of it, C) the final image is real D) the final image is virtral, 243
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, , COMPREHENSION TYPE QUESTIONS, Comprehension for Q. No. 63 – 65, A point object O is placed at a distance of 0.3 m, from a convex lens of focal length 0.2m. It is then, cut into two halves each of which is displaced by, 0.0005 m as shown in figure., L1, O, , 2 x 0.0005 m, , L2, , 0.3 m, , 63. Image will be formed from the lens at a, distance of, (A) 30 cm (B) 40 cm(C) 50 cm (D) 60 cm, 64. Number of images found is, (A) 3, (B) 2, (C) 1, (D) 4, 65. Separation between the images is, (A) 0.2 cm, (B) 0.4 cm, (C) 0.3 cm, (D) 0.6 cm, , MATRIX MATCH TYPE, Each question contains statements given in, two column which have to be matched., Statements (a, b, c, d) in Column A have to be, matched with statements (p, q, r, s) in Column, B. The answers to these questions have to be, appropriately bubbled as illustrated in the, following example., 69. A plane mirror is tied to the free end of an, ideal spring. The other end of the spring is, attached to a wall. The spring with mirror is, held vertically to the floor, can slide along it, smoothly. When the spring is at its natural, length, the mirror is found to be moving at a, speed of V with respect to ground frame. An, object is moving towards the mirror with speed, 2V with respect to ground frame. Then, Match, the following ., Mirror, V, , Comprehension - II, , r = 60cm, , r = 20cm, , 66. The focal length of the combination has the, magnitude, (A) 8.6 cm, (B) 7.5 cm, (C) 1.5 cm, (D) 15 cm, 67. The combination behaves like, (A) a concave mirror (B) a convex mirror, (C) a concave lens, (D) a convex lens, 68. A small object is placed on the principal axis, of the combination, at a distance of 30 cm in, front of the mirror. The magnification of the, image is, (A) , 244, , 1, 3, , (B), , 3, 4, , (C) 5, , (D) , , 1, 3, , 2V, , Wall, , The convex surface of a thin concavo–convex lens, of glass of refractive index 1.5 has a radius of, curvature of 20 cm. The concave surface has a, radius of curvature of 60 cm. The convex side is, silvered and placed on a horizontal surface as, shown in the figure., , Column – A, (A) Speed of image with respect to ground frame, when spring is at natural length, (B) Speed of image with respect to mirror when, spring is at natural length, (C) Speed of image with respect to object when, spring is at natural length, (D) Speed of image with respect to ground frame, when spring is at maximum compressed state, Column – B, (P) V, (Q) O, (R) 2V, (S) 3V, 70. An optical component and an object S placed, along its optic axis are given in, Column – I. The distance between the object, and the component can be varied. The, properties of images are given in Column II., Match all the properties of images from, Column – II with the appropriate components, given in Column I. Indicate your answer by, darkening the appropriate bubbles of the, 4 × 4 matrix given in the ORS. [2008], NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, Column – I, , GEOMETRICAL OPTICS, , Column – II, , S, , (A), , (D) 2 3, , (p) Real image, , (s) , , 2, , 1, , (t), , 2, , 1, , 3, , S, , (q) Virtual image, , (B), S, , (C), , (r) Magnified image, , S, , (D), , (s) Image at infinity, , 71. Match the following :, Column – I, Column – II, (i) Converging system (A) convex lens, (ii) Diverging system, (B) concave lens, (iii) Virtual Image is, (C) concave mirror, formed by, (iv) Magnification < 1 (D) convex mirror, is possible with, 72. Two transparent media of refractive indices, , 1 & 3 have a solid lens shaped transparent, material of refractive index 2 between them, as shown in figures in figures in Column – II., A ray traversing these media is also shown in, the figures. In Column – I different, relationships between 1 , 2 and 3 are given., Match them to the ray diagrams shown in, Column – II., Column – I, Column – II, , 3 2 1, , (A) 1 2, , (p), , (B) 1 2, , (q) 3 2, , (C) 2 3, , (r) 3 2, , NARAYANAGROUP, , 1, , 1, , 3, , STATEMENT TYPE, 73. STATEMENT – 1, The formula connected u, v and f for a spherical, mirror is valid only for mirrors whose sizes are very, small compared to their radii of curvature. [2007], because, STATEMENT – 2, Law of reflection are strictly valid for plane surfaces,, but not for large spherical surfaces., (A)Statement – 1 is True, Statement – 2 is True;, Statement – 2 is a correct explanation for, Statement – 1., (B) Statement – 1 is True, Statement – 2 is True;, Statement – 2 is NOT a correct explanation for, Statement – 1., (C) Statement – 1 is True, Statement – 2 is False., (D) Statement – 1 is False, Statement – 2 is True., 74. Assertion : When an object is placed between, two plane parallel mirrors, then all the images found, are of equal intensity., Reason : In case of plane parallel mirrors, infinity, images are possible., 75. Assertion :The mirrors used in search lights are, parabolic and not concave spherical., Reason : In a concave spherical mirror the image, formed is always virtual., 76. Assertion : The size of the mirror affect the nature, of the image., Reason : Small mirrors always forms a virtual, image., 77. Assertion : Different colours travel with different, speed in vacuum., Reason : Wavelength of light depends on refractive, index of medium., 78. Assertion : Critical angle of light passing from glass, to air is minimum for violet colour., Reason : The wavelength of blue light is greater, than the light of other colours., 79. Assertion : The images formed by total internal, reflections are much brighter than those formed by, mirrors or lenses., Reason : There is no loss of intensity in total internal, reflection., 245
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 80. A ssertion : There is no dispersion of light refracted, through a rectangular glass slab., Reason : Dispersion of light is the phenomenon of, splitting of a beam of white light into its constituent, colours., 81. Assertion : All the materials always have the same, colour, whether viewed by reflected light or through, transmitted light., Reason : The colour of material does not depend, on nature of light., 82. Assertion : The frequencies of incident, reflected, and refracted beam of monochromatic light incident, from one medium to another are same, Reason : The incident, reflected and refracted rays, are coplanar, 83. Assertion : A beam of white light gives a spectrum, on passing through a hollow prism., Reason : Speed of light outside the prism is, different from the speed of light inside the prism., 84. Assertion : If the angles of the base of the prism, are equal, then in the position of minimum deviation,, the refracted ray will pass parallel to the base of, prism., Reason : In the case of minimum deviation, the, angle of incidence is equal to the angle of emergence., 85. Assertion : Dispersion of light occurs because, velocity of light in a material depends upon its, colour., Reason : The dispersive power depends only upon, the material of the prism, not upon the refracting, angle of the prism., 86. Assertion : The refractive index of a prism, depends only on the kind of glass of which it is, made of and the colour of light, Reason : The refractive index of a prism depends, upon the refracting angle of the prism and the angle, of minimum deviation, , ASSERTION – REASONING TYPE, Read the assertion and reason carefully to, mark the correct option out of the options, given below:, (a) If both assertion and reason are true and the, reason is the correct explanation of the, assertion., (b) If both assertion and reason are true but, reason is not the correct explanation of the, assertion., 246, , (c) If assertion is true but reason is false., (d) If the assertion and reason both are false., (e) If assertion is false but reason is true., 87. Assertion : The stars twinkle while the planets do, not., Reason : The stars are much bigger in size than, the planets., 88. Assertion : The air bubble shines in water., Reason : Air bubble in water shines due to, refraction of light, 89. Assertion : The refractive index of diamond is, , 6, , and that of liquid is 3 . If the light travels from, diamond to the liquid, it will totally reflected when, the angle of incidence is 30o., Reason :, 90., , 91., , 92., , 93., , , , 1, ,, sin C, , where is the refractive index, , of diamond with respect to liquid., Assertion :We cannot produce a real image by, plane or convex mirrors under any circumstances., Reason : The focal length of a convex mirror is, always taken as positive., Assertion : Within a glass slab, a double convex, air bubble is formed. This air bubble behaves like, a converging lens., Reason : Refractive index of air is more than the, refractive index of glass., Assertion : The focal length of lens does not, change when red light is replaced by blue light., Reason : The focal length of lens does not depends, on colour of light used., Assertion : If a plane glass slab is placed on the, letters of different colours all the letters appear to, be raised up to the same height., Reason : Different colours have different, wavelengths., , INTEGER TYPE QUESTIONS, 94. Image of an object approaching a convex, mirror of radius of curvature 20m along its, opt ical axis is obser ved t o move fr om, , 25, m, 3, , 50, m in 30 seconds. what is the speed of, 7, the object in km per hour, , to, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, , 95.A light ray I is incident on a plane mirror M., The mirror is rotated in the direction as shown, in the figure by an arrow at frequency, , 9, rev/, , , sec. The light reflected by the mirror is, received on the wall W at a distance 10 m from, the axis of rotation. When the angle of, incidence becomes 37°, the speed of the spot, (a point) on the wall is ( x 102 ) m / s then the, value of x is., , 100 A uniform, horizontal beam of light is incident, upon a quarter cylinder of radius R = 5cm, and, has a refractive index, , 2, 3, , for a distance ‘xR’ from the cylinder is, unilluminated. Find the value of x., , I, , R, 37°, , M, O, , 10 m, , W, , 96. A balloon is rising up along the axis of a concave, mirror of radius of curvature 20 m. A ball is, dropped from the balloon at a height 5 m from, the mirror when the balloon has velocity 20, m/s. The speed of the image of the ball formed, by concave mirror after 4 seconds is (20) n, in (m/s), then n is [Take : g = 10 m/s2], 97. A large glass slab 5 / 3 of thickness 8cm, is placed over a point source of light on a plane, surface. It is seen that light emerges out of, the top surface of the slab from a circular area, of radius R cm. what is the value of R?, 98.AB and CD are two slabs. The medium between, the slabs has refractive index 2. The minimum, angle of incidence at Q, so that the ray is, totally reflected by both the slabs is, , , , then x, x, , m25, ratio m is, 50, B, 45°, , P, =1, , O E, Q, = 1.514, , 60°, A, , D, , 102.A ray of light travelling in air is incident at, grazing angle (incidence angle = 90°) on a, medium whose refractive index depends on the, depth of the medium. The trajectory of the light, in the medium is a parabola, y = 2x2. Find the, refractive index at a depth of 1 m in the, medium., , 2, , B, , 3, , D, , depth of medium, , Q, , 99.Figure shows an irregular block of material of, refractive index 2 . A ray of light strikes the, face AB as shown in the figure. After refraction, it is incident on a spherical surface CD of, radius of curvature 0.4 m and enters a medium, of refractive index 1.514 to meet PQ at E. The, distance OE in meter upto two places of, decimal is (2.02) n, then n is (2004; 2M), NARAYANAGROUP, , 2, , C, , y, , =2, , P C, , xR, , 101 The focal length of a thin biconvex lens is 20cm., When an object is moved from a distance of, 25 cm in front of it to 50cm, the magnification, of its image changes from m25 to m50 . The, , is (2005; 2M), A, , . A patch on the table, , 1m, , , x, , 103.A thin convergent lens produces the images of, a certain object on a screen. The height of, image is 4cm. Without changing the distance, between the object and the screen, the lens is, shifted, and it is found that the height of the, second sharp image is 9m. Determine the, height of the object (in cm), 247
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, , LEVEL-V - KEY, 1. (D) 2.(C) 3.(B) 4.(A) 5.(C) 6.(A), 7.(C) 8.(B) 9.(C) 10.(D) 11.(B) 12.(B), 13.(D) 14.(B) 15.(B) 16.(B) 17.(D) 18.(B), 19.(A) 20.(A) 21.(A) 22.(B) 23.(A) 24.(C), 25.(D) 26.(B) 27.(C) 28.(A) 29.(D) 30.(C), 31.(C) 32.(B) 33.(A) 34.(C) 35.(C) 36.(B), 37.(A) 38.(C) 39.(C) 40.(C) 41.(B) 42.(C), 43.(B) 44.(C) 45.(A) 46.(B) 47.(D) 48.(B), 49.(C) 50.(B) 51.(D) 52.(A) 53.(B) 54.(C), 55.(B, D), 56.(B), 57.(C, D), 58.(C, D), 59.(C), 60.(A), 61.(B, C), 62.(B,C), COMPREHENSION, 63.(D) 64.(B) 65.(C) 66.(B) 67.(A) 68. (A), MATRIX MATCH TYPE, 69. (a – q), (b – p), (c – r), (d – r), 70. (A) – (p), (q), (r), (s), (B) – (q),, (C) – (p), (q), (r), (s),, (D) – (p), (q), (r), (s), 71. (i) -AC; (ii) - BD; (iii) -ABCD; (iv)- ABCD, 72. (A) – p,r; (B) – q,s,t; (C) – p,r,t ; (D) – q,s, STATEMENT TYPE QUESTIONS:, 73.(C) 74.(D) 75.(C) 76.(D) 77.(E) 78.(C), 79.(A) 80.(B) 81.(D) 82.(B) 83.(D) 84.(A), 85.(B) 86.(C) 87.(B) 88.(C) 89.(E) 90.(E), 91.(D) 92.(D) 93.(E), INTEGER TYPE QUESTIONS, 94.(3) 95.(5) 96. (4) 97.(6) 98.(3), 99.(3) 100.(1)101.(6)102.(6), , 1, 1, 1, , , 1 1 1, 1 1 2, 1, , ; v 3f f ; , v u f, v, f, 3f, 3f, 2 , , V = 3f, 7., , 8., 9., , Length of image = 3f – 2f = f., , 1 1 1, ., v u f, 1 1 1, By mirror formula ., v u f, f, (2); Using formula m , ., f u, , (3); Using, , (3), R, 30°, 60°, Q, , 30°, 30°, 30°, 30°, , Image of, , 3f, 2, , S, P, , 10. (2), III, II, I, order order order, , M, , M, , I3 ', , I 2 ' I1 ', , a/2, , I1, , a/2, , a/2, , I3, , I2, , a/2, 3a/2, 5a/2, , a, , 3a/2, 5a/2, , 11. (2); In two images man will see himself using left, hand., 12. (2), I, f, , ; where u = f + x ;, O f u, , I, f, , O, x, , , , 13. (D); The ray diagram will be as follows :, G, C, A, , d, , , , D, , H, , S, I, , E, L, , J, , r, , f, , HI = AB = d ; DS = CD =, 3f/2, 2f, 248, , I, II, III, order order order, , O, , LEVEL-V - HINTS, SINGLE CORRECT ANSWER TYPE, 1. (D); The image will be real and between C and O., 3.. (B); The image of object at 2f will be formed at 2f., So the object and image will coincide at 2f., , T, , 30°, 30°, 60° 60°, , d, 2, d, , Since, AH = 2AD ; GD = 2CD = 2 = d, 2, Similarly, IJ = d, GJ = GH + HI + IJ = d + d + d = 3d, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, , 1 sin i1 = 2 sin i2, Here, 1 = glass, i1 = i; 2 = air = 1 & i2 = 90°, , 14. (B), i 2 3m, , 1, , d, 30°, 30°, , 0.2m, , g sin i = (1)(sin 90°) or g = sin i, 1, , 23. (A); Critical angle C = sin–1 , d = 0.2 tan 30° =, , 0.2, 3, , ;, , 2 3, l, = 0.2 3 = 30, d, , Therefore, maximum number of reflection are 30., 1 1 1, mirror formula, 15. (B), f v u, , 1, 1, 1, 1, 2 1, 1, , , , , ;, f V 1 3 f / 2 V1 3 f f, 3f, , V1 3 f (Image distance of right end ), 1 1, 1, 1, 1 1, 1, 2, , ; 2 , ;, f v 2f v, 2f f, 2f, v 2 2 f, Image lenghth =f (Image distance of left end), , , , , , 16. (B); The incident ray ˆi ˆj kˆ ; the reflected, ray can be found by reversing the sign of the, component along the normal., 17. (D); According to Snell’s Law, sin = constant, which gives 1 4, 18. (B); Each ray emerges without any deviation., 19. (A); 1a 2, , Wavelength increases in the sequence of, VIBGYOR. According to Cauchy’s formula, refractive index (m) decreases as the wavelength, increases. Hence the refractive index will increase, in the sequence of ROYGBIV. The critical angle, qC will thus increase in the same order VIBGYOR., For green light the incidence angle is just equal to, the critical angle. For yellow, orange and red the, critical angle will be greater than the incidence angle, will be greater than the incidence angle. So these, colours will emerge from the glass air interface., Hence, the correct option is (A)., 24. (C); Distance of object from mirror, = 15+, , Distance of image from mirror = 15+, , For the mirror,, , 25, =33.8 cm, 1.33, , 1, 1, 1 1, 1, 1, =, ;, =, +, , f, f, v u, 33.8, 40, , f, = –18.3 cm, , Most suitable answer is (C)., 25. (D); 2sin 37 (2 x)sin 90, , , , x, , 4, m, 5, , 26. (B) In order that the beam not enter region IV, the, angle should be less than a critical value so that:, , ai b j, , A, , n 0 sin C , , B, i j, , 20. (A), 21. (1), For refraction 90 i ; For T.IR. 2i, 22. (B), Air, , 33.25, = 40 cm, 1.33, , n0, 1, sin 90 i.e., C sin 1 , 8, 8, , (B) is correct., 27. (B), v = 12m/s, 12.8m, , 90° 2, , 4/3, , r, , v 2 10 (20 12.8) 12 m / s, , r, Water, Glass, , v, , 1, i, , Applying Snell’s law ( sin i= constant), at 1 and 2, we have, NARAYANAGROUP, , u, , Due to refraction at the water surface, , 2, 1, dv, , 4, , dv 2 dv, or dt 3 12 16 m / s or, , , dt 1 dt, 249
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, , 33. (1); In the position of minimum deviation r , , , 28, , A, 2, , irrespective of colour, 34. (3), , (A), , h, , Let angle between the direction of incident ray, and reflected ray be , cos , , 1, 1, i 3j, i 3j, 2, 2, , , , cos , , , , Let angle between the direction of incident ray, and reflected ray be , 1, 1, i 3j, i 3j, 2, 2, , cos , , , , h, , , h, , h, tan , x, , h, , ; x tan ( 1)A, , 35. (C), Deviation d = (m – 1) A, Given, dnet = 0, (m1 – 1)A1 = (m2 – 1)A2 or, , , , cos , , , x, , , , 1, 0 48. (A), ;, 2 120, , h, , , , , , 1, 0, ;, 2 120, , (1 1), , (1.54 1), , A2 = ( 1) A1 = (1.72 1) (4°) = 3°, 2, correct option is (C)., 36. (B); During minimum deviation the ray inside the, prism is parallel to the base of the prism in case of, an equilateral prism., Hence, the correct option is (B), 37. (A), , 29. (D), , 1 = 1.0, , 1 = 1.5, , P, O, , Q, +ve, , , , 21, 2, , Let us say PO = OQ = X, h, , Applying, , h 21, , ; h 14cm, 2, 30. (C); Each ray deviates towards the other by 15º., 31. (C); The effective situation is unchanged., , Substituting the values with sign, 1.5 1.0, 1.5 1.0, , =, X X, R, , (Distances are measured from O and are taken as, positive in the direction of ray of light), , , P, , Q, , R, , 32. (B); 1 A = 2º, mirror should be rotated by 1º anticlockwise., 250, , 2 1, , , = 2 1, v, u, R, , 2.5, 0.5, =, X = 5R, X, R, , 38. (C); When the object is placed at the center of the, glass sphere, the rays from the object fall normally, on the surface of the sphere and emerge, undeviated., Hence, the correct option is (C)., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, , 39. (C), , (1.5 1) (1.2 1), , R, R, , =, r, r, , r, , 1 1 1, , v u f, , r, , ;, , 1, 1, 1, , , v 40 20, , r, , For TIR , rmin C sin i sin r, , f = 20 cm, , u 40, , v = 40 cm, , 44. (C); Image can be formed on the screen if it is real., Real image of reduced size scan be formed by a, concave mirror or a convex lens as shown in figure., , sin imin sin rmin ; imin 900, 1, , 1, , 1 , , 40. (C); f 1 R R ,, 1, 2 , , O, , F, , C, , 1, For no dispersion d 0 Þ, f, , 1m, , 1, 1 , d , , 0 Þ R1 R 2, R1 R 2 , f, , 41. (B); First image, , l, , f = 15cm, 0, , O, , I2, , 2f, , 2f, , f, , I1, 1m, , 30cm, , 10cm, , 20cm, , 1 1 1, 1, 1, 1, , , , ;, v u t, v 30 15, v 30, image in formed 20cm behind the mirror, Second image, by plane mirror will be at 20cm, infront of plane mirror, , For third image, , A diminished real image is formed by a convex lens, when the object is placed beyond 2f and the image, of such object is formed beyond 2f on other side., Thus, d > (2f + 2f) or 4f < 0.1 m or f<0.25 m, 45. (A), R1 = –R, R2 = + R, mg = 1.5 and mm = 1.75, 1, 1 g, 1 , f = 1 R R , 2 , m, 1, , 1 1, 1, , v 10 15, , R1, , R2, , 1 1 1 3 2 5, , , = 6 cm, v 10 15, 30, 30, Hence, final image is real & formed at a distance, of 16cm from mirror., , 42. (3); By formula m , , f v, ., f, , 1 1 1, 43. (B); f f f =, 1, 2, 1 , 1 1, 1, ( 2 1) , , R , R , , = (1 1) , , NARAYANAGROUP, , Substituting the values, we have, 1, , 1.5, , 1, , 1, , 1, , =, f = 1.75 1, 3.5R, R R , , Therefore, in the medium it will behave like a, convergent lens of focal length 3.5 R. It can be, understood as, mm > mg, the lens will change its, behaviour., 251
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 51. (D), , 46. (B), , f = 20cm, I1, , I2, , O F, , Image formed by convex lens at I1 will act as a, virtual object for concave lens. For concave lens, 1, 1 1, 1 1, 1, =, (or) =, (or) v = 5 cm, f, v u, v 4, 20, , As size of the image at I1 is 2 cm. Therefore, size, of image at I2 will be 2 × 1.25 = 2.5 cm., 47. (D); Let focal length of convex lens is +f, then, 3, 2, , length of convex lens would be – f., From the given condition,, , 1, 2, 1, 1, = f 3f 3f, 30, , f = 10 cm, , Therefore, focal length of convex lens = +10 cm, and that of concave lens = –15 cm, correct answer is (D)., 48. (B), , 20cm, 25cm, , 1 , Shift , t 5cm ; t 15cm, , 52. (A); Since, the final image is formed at infinity, the, image formed by the objective will be at the focal, point of the eye-piece which is 3.0 cm. The image, formed by the objective will be at a distance of, 12.0 cm (= 15.0 cm – 3.0 cm) from the objective., If u is the distance of the object from the objective,, we have, E, , O, , O, 12.0m, , f, , , V=, , u, , 3m, , 15.0m, , r, , 1 1 1, 1, 1, 1, , = , u v f, u 12.0 2.0, , r f tan ; r 2 f 2, , (1.20)(2.0), , 24.0, , u = 12.0 2.0 10.0 = 2.4 cm, , 49. (C), , 53. (B); The focal length of combination is given by, 1, F, 24m, , 8m, , 1, 1, 1 , , , F F1 F2 , , 1, 1, , 40 25, , F=–, , or, , P=–, , 1 1 1, 1 1 1, , , , , f, R, 2R, f, R, 2R, , 3, 2, , , 1 , P , =, F (m) , , , 3 1 1, , R 3m, 24 2R, , 50. (B) Intensity Aperture , 252, , – 1.5, , 54. (C); The ray diagram is as follows :, B, F0 A, , 1 1 1, 1 1 1, f v u, 8 24 2R, , , 200, 2, cm = – m, 3, 3, , Power of the combination in dioptres,, , , 3, a , m 2, , , =, , A A O, , E, Fe, , B, B , , 2, , From the figure it is clear that image formed by, objective (or the intermediate image) is real,, inverted and magnified., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, , 55. (B, D) Light keeps falling on the same spot., 56. (B), v cos , , 59. (C), , v sin , , q, r, , , , q, , q, , Component of object velocity, , 60. (A), , v cos , , 1 1 1, , f v u, , , , , , df dv du, , , f 2 v2 u 2, , 1 1 , df f 2 2 2 .1cm .05cm, v u , 61. (B, D), v sin , Component of image velocity from mirror-1, v cos , , v sin , O I, , , , For a lens :, Component of image velocity from mirror-2, , Here the perpendicular component of velocity in, mirror(1) will change. Clearly relative velocity of, image (1) w.r.t Image (2) is 2v sin , 1 1 1, , v u f, , 57. (C, D), , 1, 1 1, =, f, v u, 1, , 1, v, , 1, , = f u, For a concave lens, f and u are negative, i.e., v, will always be negative and image will always be, virtual., i.e.,, , The data set (66, 33) does not satisfy the mirror, equation. So, options (C) and (D) are correct., , MULTIPLE CORRECT ANSWER TYPE, 58. (C, D); For total internal reflection to take place :, Angle of incidence, i > critical angle, qc, or sin i > sin qc or sin 45° >, or, , 1, 2, , >, , 1, or n >, n, , 2, , 1, n, , or n > 1.414, , Therefore, possible values of n can be 1.5 or 1.6, in the given options., NARAYANAGROUP, , O, , For a mirror :, , I, , 1, 1 1, =, f, v u, , Here, f is positive and u is negative for a convex, mirror. Therefore, v is always positive and image, is always virtual., 253
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JEE-ADV, PHYSICS-VOL- I, SR-MAIN-CHEM-VOL-II, , GEOMETRICAL OPTICS, 62. (B,C) First for lnes, 1, 1, 1, , 30 V1 15 ; V1 30cm, , For plane mirror object is real at distance 45cm, from mirror its image is at distance 45cm behind, the mirror., Again for lens image from mirror act as real object, at distance 60cm from lens . Lens makes its real, image at distance 60cm, , 2 1 (As light bends + towards normal), r C & A, (D) 2 1 ; 3 2, As light bends away from normal s = B,D, (E) 2 3 As no deviationof light, , 3, , COMPREHENSION TYPE, 63. (D), , 1, 1, 1, , , v = 60 cm, v 0.2 0.3, , v, 60, 2 depuration = 0.3 cm, 65. (C); m , v 30, , 2, , 1, , 2 1 As light bend away from normal t C & B, , STATEMENT TYPE, 73. (C); Laws are universal, where as mirror formula, is based on paraxial approximation., , 68. (A) (Q. No. 66 to 68), , INTEGER TYPE, , 1 1 1, Use the equation f f f ......., 1, 2, , to find the equivalent power of the combination,, with proper sign. Then use the equation of the lens/, mirror as appropriate., , MATRIX MATCH TYPE, 69. (a – q), (b – p), (c – r), (d – r), velocity of image w.r.t. ground = 0, velocity of image w.r.t. mirror = v, velocity of image w.r.t. object = 2v, 70. (A) – (p), (q), (r), (s), (B) – (q), (C) – (p), (q), (r),, (s), (D) – (p), (q), (r), (s), 71. (i) – AC; (ii) – BD; (iii) – ABCD; (iv) – ABCD, 72. (A) – p,r; (B) – q,s,t; (C) – p,r,t ; (D) – q,s, , 93. (3) R 20m, f 10m, For mirror,, , 1, 1, 1, 1 1 1, , , ;, V U f 25 / 3 U1 10, , 1, 1, 3, 1, , , U1 10 25, 50, 1, , 1, , U1 50cm, , 1, , and 50 / 7 U 10, 2, , U 2 25cm, , U, 25, 5, So, speed t 30 m/sec = m / sec and in, 6, , 5 18, 3km / hr, 6 5, , km/hr , 94. (5), , , , (A), , 3 2 1, , 2 2, , As there is no deviation. As the light bends towards, normal in denser medium 2 2 p A & C, (B) As light bends away from normal, , 3, , 2, , 1, , , , 2 1 and 3 2 q B & D, , y, , 10, , tan , , 10, d, 10 dy, sec 2 , 2, y, dt, y dt, , dy y 2 sec 2 d, , dt, 10, dt, 2, , (C) 2 3 (As no deviation), , , , 2, , 40 5 18, , m / s 500 m / s, 3 4 10, , 40, , 3, , 254, , 2, , 1, , for 370 ,y= &, 3, d, 9, 2 18 rad / sec, dt, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- I, , GEOMETRICAL OPTICS, 98.(3), , 95.(4), , Y, , P(x, y), , , 5m, , , O, , y 2x, , Velocity of object after 4 sec = 20 m/s downwards, 2, , VI /m , , v, V0 / m ; VI /m m 2 V0/ m, u2, , VI /m (2)2 20, VI /m 20 4 So,, , n = 4., , 96.(6), , X, , 2, , dy, dy, 4x, dx, dx, , 2 2.............(1), y 1, , from Snell’s law, 1 sin 90 sin ; sin(90 ) 1, 1 dy, tan 2 1 .............. (2), ;, dx, from (1) and (2) equations ; 3, cos , , 99.(3) Applying Snell’s law on face AB., (1) sin 45° = 2 sin r sin r =, 8cm, , tan C , , C, , C, , R, ................(i), 8, , 5, sin C 1.sin 900, 3, 3, sin C ; C 370, 5, 3 R, , ; R 6cm, 4 8, 97.(3) Critical angle at 1 and 2, 1 2, , 2 2, , 1 , 1 , C1 = sin–1 = sin–1 , = 45°, , 2, 2, 3, , = 60°, 2 , , Therefore, minimum angle of incidence or total, internal reflection to take place on both slabs should, be 60°, imin = 60°, NARAYANAGROUP, , i.e., ray becomes parallel to AD inside the block., Now applying,, 2 1, , , = 2 1 on face CD,, v, u, R, 1.514, 2, , OE, , , =, , 1.514 4, 0.4, , Solving this equation, we get, 101.(6) When object distance is 25, 1, 1, 1, 1 1 1, , = , v 25 20, v u f, v 100, , 4, u 25, When object distance is 50, 1, 1, 1, 100, , , u , cm, v 50 20, 3, v 100cm, m25 , , m25 4, , 6, m50 2, 3, f, m25 20 25 30, , , 6, f, m50, 50, 20 50, , 100, 2, m50 3 ;, 50, 3, , 3 3, , 3 , C2 = sin–1 = sin–1, 2 , , 1, or r = 30°, 2, , Alternate:, , 4 v, 9 v', 102. h u ; h u ' .......(i ), 0, 0, v' u 4 9 1, , h02 36 h0 6, h0 h0, u' v, 255
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , WAVE OPTICS (PHYSICAL OPTICS), SYNOPSIS, In geometrical optics, light is represented as a, ray which t ravels in a straight line in a, homogeneous medium. The phenomenon like, Interference and Dffraction cannot be explained, on the basis of particle nature of light. These, phenomenon can only be explained on the basis, of wave nature of light. This part of optics is, called physical optics or wave optics. The wave, theory of light was presented by Christian, Huygen. It should be pointed out that Huygen did, not know whether the light waves were, longitudinal or transverse and also how they, propagate through vaccum. It was then explained, by Maxwell by introducing electromagnetic, wave theory in nineteenth century., , b2, b2, 1 or, 1, l, l, Note: The object interacting with light may be a, mirror, a lens, a prism, an aperture (pin hole), a, slit and a straight edge., , WAVE FRONT, , Geometrical optics, 1., 2., , In geometrical optics, light is assumed to be, travelling in a straight line. This property is, known as rectilinear propagation., By using rectilinear propagation of light, laws, of reflection, refraction, total internal reflection, etc. are explained geometrically., , 1., , Physical optics or wave optics:, 1., 2., 3., , In physical optics, light is considered as a wave, Huygen’s wave principle and principle of, superposition are used to explain interference, and diffraction, Electromagnetic wave nature of light is used to, explain the concept of polarisation., , Condition for applicability of geometrical, optics and wave optics: When the size of the, , i), , object incteracting with light, is much larger than, the wavelength of light , we can apply, geometrical optics., When the wavelength of light is comparable to, or less than the size of the object interacting with, light, we can apply wave optics., If ‘b’ is the size of the object interacting with, light, ‘l’ is the distance between the object and, the screen and ‘ ’ is the wavelengh of light then,, The condition for applicability of geometrical, , b2, 1, l, ii) The condition for applicability of wave optics, is, optics is, , NARAYANAGROUP, , According to wave theory of light, a source of, light sends out disturbance in all directions. In, a homogeneous medium, the disturbance, reaches to all those particles of the medium in, phase, which are located at the same distance, from the source of light and hence at any instant,, all such particles must be vibrating in phase with, each other., The locus of all the particles of the medium,, which at any instant are vibrating in the same, phase, is called the wavefront., Depending upon the shape of the source of light,, wavefront can be of the following types, Spherical wavefront: A spherical wavefront, is produced by a point source of light. It is, because, the locus of all such points, which are, equidistant from the point source, is a sphere, (a), , (a), , (b), , (c), , 2. Cylindrical wavefront: When the source of, , 3., , light is linear in shape (such as a slit), a, cylinderical wavefront is produced. It is, because, all the points, which are equidistant, from the linear source, lie on the surface of a, cylinder (b)., Plane wavefront: A small part of a spherical, or a cylindrical wavefront originating from a, distant source will appear plane and hence it is, called a plane wavefront (c)., Huygen’s Principle: Every point on the, wave front becomes a source of secondary, disturbance and generates wavelets which, spread out in the medium with the same velocity, as that of light in the forward direction only., 1
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WAVE OPTICS, , , , The envelope of these secondary waves at any, instant of time gives the position of the new wave, front at that instant., The wave front in medium is always, perpendicular t o the direct ion of wave, propagation., , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , Principle of superposition of waves:, , If two or more waves meet at a place, simultaneously in the same medium, the particles of, the medium undergo displacements due to all the, waves simultaneously. The resultant wave is due to, the resultant displacement of the particles., Principle of superposition of waves states that when, B, two or more waves are simultaneously impressed, , i, on the particles of the medium, the resultant, A, displacement of any particle is equal to the sum of, D, r, , displacements of all the waves. (or), C, “When two or more waves overlap, the resultant, displacement at any point and at any instant is the, AB is width of incident beam, vector sum of the instantaneous displacements that, CD is width of refracted beam, would be produced at the point by individual waves,, width of incident beam cos i, if each wave were present alone”., , width of refracted beam cos r, If y1 , y2 ,......... yn denote the displacements of ‘n’, The Doppler Effect:, waves meeting at a point, then the resultant, When any source emitting light (like sun, moon, star,, displacement is given by y y1 y2 .... yn ., atom etc) is approaching or receding from the, observer then the frequency or wavelength of light a) Superposition of coherent waves: Consider, appears to be changing to the observer. This, two waves travelling in space with an angular, apparent change in frequency or waveelength of, frequency . Let the two waves arrive at some, light is called Doppler effect in light., point simultaneously. Let y1 and y2 represent the, Blue Shift: When the distance between the, displacements of two waves at this point., source and observer is decreasing (i.e. the source, is approaching the observer) then frequency of, y1 A1 sin(t 1 ) & y2 A2 sin(t 2 ), light appears to be increasing or wavelength, Then according to the principle of, appears to be decreasing i.e. the spectral line in, superposition the resultant displacement at the point, electromangetic spectrum gets displaced, is given by,, towards blue end, hence it is known as blue shift., Red Shift: When the distance between the y y1 y2 or y A1 sin(t 1 ) A2 sin(t 2 ), source and observer is increasing (i.e. the source, A1 (sin t cos 1 cos t sin 1 ), is receding from the observer) then frequency of, A2 (sin t cos 2 cos t sin 2 ), light appears to be decreasing or wavelength, appears to be increasing i.e. the spectral line in, A cos .sin t A sin .cos t, electromangetic spectrum gets displaced, A sin(t ), towards red end, hence it is known as red shift., 1, , 2, , i), , v V, , ( where V is the speed, v C, of source and C is the speed of light), W. E-1 What speed should a galaxy move with, respect to us so that the sodium line at 589.0, nm is observed at 589.6 nm?, V, ;, Sol., C, 0.6 , 8 0.6 , 5, 1, V c , 3 10 , 3.06 10 ms, 589.0 , 589.0 , Therefore. the galaxy is moving away from us, with speed 306 km/s., , Doppler Shift,, , 2, , where A cos A1 cos 1 A2 cos 2 .......(1), and Aisn A1 sin 1 A2 sin 2 ........(2), Here A and are respectively the amplitude and, initial phase of the resultant displacement, Squaring and adding equations (1) & (2), we, get, , A A12 A22 2 A1 A2 cos(1 2 ), A12 A22 2 A1 A2 cos .....(3), Where 1 2 , phase difference between the, two waves., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , Dividing equation (2) by equation (1), we get, , A sin 1 A2 sin 2, tan 1, ........(4), A1 cos 1 A2 cos 2, Since the intensity of a wave is proportional to, square of the amplitude, the resultant intensity I of , , the wave from equation (3) may be written as, I I1 I 2 2 I1 I 2 cos .........(5), , where I1 and I 2 be the intensities of the two waves., It can be seen that the amplitude (intensity) of the , resultant displacement varies with phase difference, of the constituent displacements., Case I : When 1 2 0, 2 , 4 ...... 2n, where n 0,1, 2,.........., cos 1, A A1 A2, , from (3), , , , , , and I I1 I 2, from (5), Hence the resultant amplitude is the sum of the two, individual amplitudes. This condition refers to the, constructive interference., Case II: When 1 2 ,3 ,5 ..... (2n 1), , where n 1, 2,3,....... ; cos 1, A | A1 A2 | and I | I1 I 2 |, Hence the resultant amplitude is the difference of , the individual amplitudes and is referred to as, destructive interference., , Interference:, The variation in intensity occurs due to the, redistribution of the total energy of the interfering, waves is called interference., Interference of light is a wave phenomenon., The source of light emitting wave of same frequency, and travelling with either same phase or constant, phase difference are called Coherent Sources., Ex: Two virtual sources derived from a single source, can be used as Coherent Sources., The source producing the light wave travelling with, rapid and random phase changes are called, Incoherent Sources., Ex: 1. Light emitted by two candles., 2. Light emitted by two lamps., , Conditions for Steady Interference, The two sources must be coherent., Two sources must be narrow., Two sources must be close together., NOTE: The two sources must be mono chromatic,, otherwise the fringes of different colours overlap, and hence interference cannot be observed., , Young’s Double Slit Experiment, Young with his experiment measured the most, important characteristic of the light wave i.e, wavelength ( l ), Young’s experiment conclusively established the, wave nature of light., , b) Supersposition of incoherent waves:, Incoherent waves are the waves which do not, maintain a constant phase diference. The phase of, the waves fluctuates irregularly with time and, independently of each other. In case of light waves, the phase fluctuates randomly at a rate of about, 108 per second. Light detectors such as human eye,, photographic film etc, cannot respond to such rapid, changes. The detected intensity is always the, average intensity, averaged over a time interval , which is very much larger than the time of, fluctuations. Thus, , s1s2=d, When source illuminates the two slits, the pattern, observed on the screen consists of large number of, equally spaced bright and dark bands called, “interference fringes”, , I av I1 I 2 2 I1 I 2 cos . The average a) Bright fringes :, Bright fringes occur whenever the waves from, value of the cos over a large time interval will be, S1 and S2 interfere constructively. i.e. on reach, zero and hence I av I1 I 2, ing ‘P’, the waves with crest (or trough) superimpose, This implies that the superposition of incoherent, at the same time and they are said to be in phase., waves gives uniform illumination at every point and, The condition for finding a bright fringe at ‘P’ is that, is simply equal to the sum of the intensities of, S 2 P S1 P n ,, the component waves., NARAYANAGROUP, , 3
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, Where n 0, 1, 2, 3, .... and n is called the, order of bright fringe. Hence for nth order bright, fringe, the path difference is, d sin n, y , d n n, D, n D, d, Where yn is the position of nth maximum from O., The bright fringe corresponding to n = 0, is called, the zero - order fringe or central maximum. It means, it is the fringe with zero path difference between, two waves on reaching the point P., The bright fringe corresponding to n = 1 is called, first order bright fringe i.e., if the path difference, between the two waves on reaching ‘P’ is ., Similarly second order bright fringe n 2 is located, where the path difference is 2 and so on., yn , , 2 , From I 4 I 0 cos , 2, , For maximum intensity cos, , , 1, 2, , , i.e. 0, , , 2 .........., 2, (or) Phase difference between the waves, c), 2 n with n = 0, 1, 2, 3 ........., The corresponding path difference, x n, Hence I max 4 I 0 ., , b) Dark fringes :, Dark fringes occur whenever the waves from, S1 and S 2 interfere destructively. i.e., on, reaching ‘P’ one wave with its crest and another, wave with its trough superimpose. Then the, phase difference between the waves is and, the waves are said to be in opposite phase., Destructive interference occurs at P, if S1P and, , S2P differ by a odd integral multiple of ., 2, Thus the condition for finding dark fringe at P is, , that S 2 P S1 P (2n 1) ., 2, Where n 1, 2, 3,........... , and n is called, order of dark fringe. Hence for nth order dark, 4, , fringe, the path difference, d sin (2n 1), , , 2, , , y , 2n 1 D, d n (2n 1) yn , , 2, D, 2 d, Where yn is the position of nth minima from O., The first dark fringe occurs when, S 2 P S1 P , , , .This is called first order dark (n = 1), 2, , 3, second, 2, order dark fringe (n = 2) occurs and so on., , fringe and similarly for S2P - S1P =, 2 , From I 4 I 0 cos , 2, , For minimum intensity cos, , , 0, 2, , , 3 5, , , ......, 2, 2, 2, 2, (or) , 3, 5 ......, (or) (2n 1) with n = 1, 2, 3 ......, The corresponding path difference,, , x (2n 1), 2, Hence I min O, i.e.,, , Fringe width ( ):, The distance between two adjacent bright (or, dark) fringes is called the fringe width. It is, denoted by ., The nth order bright fringe occurs from the, n D, d, th, The (n 1) order bright fringe occurs from the, , central maximum at yn , , ( n 1) D, d, The fringe separation, is given by, , central maximum at yn 1 , , D, d, In a similar way, the same result will be obtained, for the dark fringes also., D, Fringe width, , d, Thus fringe width is same every where on the screen, yn 1 yn , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , and the width of bright fringe is equal to the width, of dark fringe., D, bright dark , d, d) The locus of the point P lying in the xy-plane, , I max, , I min, , iii), , , , , I1 , , I2, , I1 , , I2, , phase difference =, , , , , 2, , 2, , 2, , , , A1 A2 , 2, A1 A2 , , 2, (path difference)., , , such that S2P - S1P = x (path difference) is a, 2, constant, is a hyperbola. If the distance D is very, , x, , large compared to the fringe width, the fringes will, be very nearly straight lines., iv) Since , Re d voilet , as red voilet, Note:, v) In YDSE, if blue light is used instead of red light, Constructive Interference, then decreases ( B R ), i) a) If the phase difference is 2n (even vi) If YDSE is conducted in vaccum instead of air,, multiples of ). Where n = 0, 1,2, 3,......, then increases ( vaccum air ), i.e. when 0, 2 , 4 ........2n, vii) In certain field of view on the screen, if n1 fringes, , are formed when light of wavelength 1 is used and, b) If the path difference x 2n (even multiples, 2, n2 fringes are formed when light of wavelength 2, of half wavelength)., is used, then, i.e when x 0, , 2 ........n, n D, The amplitude and intensity are maximum., y, = constant n = constant, d, Amax A1 A2 , n12 n2 2 (or) n11 n2 2, , ii), , , , , , 2, , A1 A2 , , 2, , viii) The distance of nth bright fringe from central, n D, n, maximum is ( yn )bri , Note: If A1 A2 a then Amax 2a, d, The distance of mth dark fringe from central, If I1 I 2 I 0 then I max 4 I 0, maximum is, Destructive Interference, (2m 1) D (2m 1), ( ym ) dark , , , a) If the phase difference 2n 1 (odd, 2, d, 2, multiples of ) where n = 1, 2, 3....., The distance between nth bright and mth dark, i.e. when ,3 , 5 ....... 2n 1 , fringes is, (2m 1), b) If the path difference x 2n 1 / 2 ( odd, ( yn )bri ( ym ) dark n , , 2, multiples of / 2 ), ix) When white light is used in YDSE the inteference, 2n 1 , 3 5, , patterns due to different component colours of white, i.e. when x , , ........, light overlap (incoherently). The central bright fringes, 2 2 2, 2, for different colours are at the same position., The ampitude and Intensity are minimum., Therefore, the central fringe is white. For a point, Amin A1 A2 , b, S, P, , S, P, , P, for, whih, where, 2, 1, 2, 2, 2, I min I1 I 2 A1 A2 , b ( 4000 A0 ) represents the wavelength for the, blue colour, the blue component will be absent, Note: If A1 A2 a then Amin 0, and the fringe will appear red in colour., I, , 0, If I1 I 2 I 0 then min, , I max , , I1 I 2, , , , NARAYANAGROUP, , , , 5
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , xiii) Missing wavelengths in front of one slit, r, where, in YDSE:, 2, r ( 8000 A0 ) is the wavelength for the red colour,,, P, S, the fringe will be predominantly blue., y, Thus, the fringe closest on either side of the central, d, O, white fringe is red and fathest will appear blue. After, S, a few fringes, no clear fringe pattern is seen., x) To know maximum number of possible maxima on, D, the screen, Suppose, P, is, a, point, of, observation in front of, n, slit S 1 as shown in figure. Path difference, If d sin n (or )sin , d, between the two waves from S1 and S2 is, n, d, x S 2 P S1P D 2 d 2 D, 1, n, As sin 1,, d, , 1/2, Therefore the maximum number of complete, d2 , , d2 , d2, D 1 2 D D 1 2 D , maxima on the screen will be 2( n) 1, 2D, D , 2D , n n, d2, As sin 1 ,, Ex: If d 3 then sin , , , x, , .........(1), 3 3, 2D, n can take values, 3, 2, 1,0,1, 2,3, But for missing wavelengths, intensity will be zero., i.e., the corresponding path difference,, Maximum number of maxima is 7., xi) Fringe visibility (or) band visibility (V) :, , x (2n 1), ...... (2), It is the measure of contrast between the bright and, 2, dark fringes, From equations (1) and (2), I max I min, d2, , Fringe visibility, V I I, (2n 1), max, min, 2D, 2, Slightly farther away where S 2Q S1Q , , 1, , 2, , where I max I1 I 2 2 I1I 2, and I min I1 I 2 2 I1 I 2, , 4 I1I 2, 2 I1I 2, V 2( I I ) ( I I ), 1, 2, 1, 2, V has no unit and no dimensional formula., Generally, 0 < V < 1., Fringe visibility is maximum, if I min 0, then, V=1, , d2, (2n 1) D, By putting n 1, 2,3,..., the wavelengths at P are, Missing wavelength, , , d2 d2 d2, ,, ,, ,......., D 3D 5D, In the above case, if bright fringes are to be, formed exactly opposite to S1 then, , , , d2, d2, n , 2D, 2 Dn, By putting n = 1, 2, 3 ,,, the possible wavelengths, at P are, , For poor visibility , I max I min , then V = 0, i.e., if V = 1, then the fringes are very clear and, contrast is maximum and if V = 0, then there will, d2 d2 d2, , ,, ,, ,...., be no fringes and there will be uniform illumination, 2 D 4 D 6D, i.e., the contrast is poor., xiv) Lateral displacement of fringes:, xii) When one slit is fully open and another one is, To determine the thickness of a given thin sheet, partially open then the contrast between the, of transparent material such as glass or mica,, fringes decreases. i.e., if the slit widths are, that transparent sheet is introduced in the path of, unequal, the minima will not be completely dark., one of the two interfering beams. The fringe, 6, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , pattern gets displaced towards the beam in whose, path the sheet is introduced. This shift is known, as lateral displacement or lateral shift., , To get central zero fringe at P, s1 p s2 p, , be introduced in one of the beams of interference,, then, ( 1)tD, 1) the lateral shift y r, d, ( 1)t, 2) the number of fringes shifted n r, , Due to the presence of transparent sheet, the phase, difference between the interfering waves at a given, 2, ( 1)t ., point is given by , , If YDSE is performed with two different colours of, light of wavelengths 1 & 2 but by placing the same, transparent sheet in the path of one of the interfering, waves then n11 n2 2 ., , S1P t t S2 P, , where n1 and n2 are the number of fringes shifted, , t, P, , S1, , y, , , d, S2, , O, , d), , D, T, , The optical path from S1 to P ( S1P t ) t., , e), , The optical path from S 2 to P S 2 P., , S 2 P S1P ( 1)t, , with wavelengths 1 & 2 ., Since 1, this implies S 2 P S1 P hence the vi) When two different transperent sheets of thickness, t1 , t2 and refractive index 1 , 2 are placed in the, fringe pattern must shift towards the beam from S1., paths, of two interfering waves in YDSE, if the central, y, bright fringe position is not shifted, then, But S 2 P S1P d sin d , where ‘y’ is the, D, (1 1)t1 ( 2 1)t2 ., lateral shift., ( 1)t d, , y, D, , Important Concepts :, , , D, , ( 1)t ( 1)t, d, , (or) Thickness of sheet, yd, y, t, , ( 1) D ( 1), From the above it is clear that, For a given colour, shift is independent of order of, the fringe i.e. shift in zero order maximum = shift in, 9th minima (or) shift in 6th maxima = shift in 2nd, minima. Since the refractive index depends on, wavelength hence lateral shift is different for different, , colours., lateral shift, The number of fringes shifted = fringe width, , Formation of colours in thin films :, a) Interference due to reflected light, , Lateral shift ( y ) , , a), , b), , c), , y ( 1)t, n , (or) n ( 1)t, , , Therefore, number of fringes shifted is more for, shorter wavelength., If a transparent sheet of thickness ‘t’ and its, relative refractive index r (w.r.t. surroundings) , , NARAYANAGROUP, , Reflected system :, Path difference between the rays Qa and QRSb., (PD) = QRS in medium - QN in air, P.D 2t cos r This is the path lag, due to reflection on film additional path lag of / 2, exists. (stoke’s theorem), Total path difference 2 t cos r , , , 2, , Condition for maximum, 2 t cos r , , , n, 2, 7
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , WE-3: Two slits are made one millimetre apart, , and the screen is placed one metre away. When, For all values of n, 2, blue-green light of wavelength 500 nm is used,, the fringe separation is, is equal to 1, 2, 3............... n., Condition for Minimum, D, , , Sol. Fringe separation, , 2 t cos r 2n 1, d, 2, 2, Given, D 1m, 500 nm 5 107 m and, 2 t cos r n for values of n = 0,1,2,3... n, d 1mm 1 103 m, = 0 gives the central minima., For normal incident i o r, 1 5 107, m, Fringe separation, , , , 1103, 2 t n for dark ; 2 t 2n 1 for bright., 2, 5 104 m 0.5 mm, Transmitted system, Interference of two rays Rc and Td. By symmetry WE-4: In YDSE, the two slits are separated by, it can be concluded that the path difference between, 0.1 mm and they are 0.5 m from the screen., the rays in 2t cos r ., The wavelengh of light used is 5000 Å . What, But there would not be any extra phase lag because, is the distance between 7th maxima and 11th, either of the two rays suffers reflection at denser, minima on the screen?, surface., Sol. Here, d = 0.1 mm = 10-4 m,, 2, , t, cos, r, , n, , Condition for maxima :, D 0.5 m, 5000 Å 5.0 107 m, Condition for minimum :, , , (2 11 1) D 7 D, 2 t cos r 2n 1, x ( X 11 ) dark ( X 7 )bright , , 2, 2d, d, OR 2 t cos r 2 n 1, , , , , , , , , If YDSE is conducted with white light,, , , , 7 D 7 5 107, x , , 2d, 2 104, , Central fringe is always achromatic (white), When path difference is small, then some coloured, fringes are obtained on two sides of the central, 8.75 10 3 m, fringe. The outer edge of the fringe is violet and, 8.75 mm, inner edge is red., The fringe width is different for different colours, WE-5: In Young’s double slit experiment, The number of fringes obtained is less than that with, interference fringes 10 apart are produced on, monochromatic light source., the screen, the slit separation is ( 589 nm), WE-2: Light waves from two coherent sources, D, having intensity ratio 81 : 1 produce, , interference. Then, the ratio of maxima and Sol. The fringe width,, d, minima in the interference pattern will be, The angualr separation of the fringes is given by, I1 A12 81, , , , Sol. Given,, , I 2 A22 1, D d, A1 9, A 1 or A1 9 A2 .... (1), 2, , I max ( A1 A2 )2, I ( A A )2, min, 1, 2, From Eq. (i), we get, I max (9 A2 A2 )2 (10)2 25, , , , I min (9 A2 A2 ) 2 (8) 2 16, 8, , 0, Given, 1 , , , rad, 180, , 589 nm, , 589 180 109, , , , 0.0337 mm, , d, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , WE-6: In Young’s double slit experiment, the WE-10: In a Young’s experiment, one of the slits, wavelength of red light is 7800 Å and that of, is covered with a transparent sheet of, blue light is 5200 Å. The value of n for which, thickness 3.6 103 cm due to which position, nth bright band due to red light coincides with, of central fringe shifts to a position orginally, (n + 1)th bright band due to blue light, is, occupied by 30th fringe.If 6000 Å, then, find the refractive index of the sheet., nR B 5200 2, n D n D, Sol. The position of 30th bright fringe,, Sol. R R B B, or n 7800 3, d, d, B, R, 30 D, nd, y30 , Now position shift of central fringe is, Therefore 2 of red coincides with 3rd of blue., d, WE-7: Young’s double slit experiment is made in, 30 D, D, y0 , a liquid. The 10th bright fringe in liquid lies, ; But we know, y0 ( 1)t, d, d, where 6th dark fringe lies in vacuum. The, 30 D D, refractive index of the liquid is approximately, ( 1)t, d, d, D, Sol. Fringe width d . When the apparatus is, 30 , 30 (6000 10 10 ), immersed in a liquid, and hence, (refractive index) times., , , , is reduced, , D, D, 10 (5.5) or10 (5.5), d, d, , , , ( 1) , , t, , , , (3.6 10 5 ), , 0.5, , 1.5, WE-11: The maximum intensity in the case of n, identical incoherent waves each of intensity, W, W, 2 2 is 32 2 the value of n is, m, m, Sol. I = n I0, 32 = n 2, n = 16, WE-12: Compare the intensities of two points, , , located at respective distance, and, from, 4, 3, the central maxima in a interference of YDSE, ( is the fringe width), , , 10, , or 1.8, 5.5, WE-8: In Young’s double slit experiment, how, many maximas can be obtained on a screen, (including the central maximum) on both sides, of the central fringe if 2000 Å and, d 7000 Å?, 2, 2 d 2 d D , Sol. For maximum intensity on the screen, x , Sol. , , =, , , D 4d , , D 4, (n)(2000) n, n, , , d sin n or sin d ;, 2 , , (7000), 3.5, , I 4 I 0 cos 2 , 4, 2, maximum value of sin 1, 4, 2, n 3, 2, 1, 0,1, 2,3 ; 7 maximas., 2 , Similarly , I 4I0 cos2 , I0, WE-9: In a double slit experiment the angular, 3, 23 , width of a fringe is found to be 0.20 on a, required ratio = 2 :1, screen placed I m away. The wavelength of, light used in 600 nm. What will be the angular WE-13: In Young’s double slit experiment, intensity at a point is (1/4) of the maximum, width of the fringe if the entire experimental, intensity. Angular position of this points is, apparatus is immersed in water? Take, I, , , refractive index of water to be 4/3., I I max cos 2 ; max I max cos 2 , Sol:, Sol. Angular fringe separation,, 4, 2, 2, , , , , 1 , ord ; In water, d , cos or , d, , , 2 2, 2 3, 2 2 , 1 3, , , or , .x where x d sin , 3 , 4, , , , 3, 3, 0, 0, d sin ,sin , , sin 1 , or 0.2 0.15, 3, 3d, 3d , 4, 4, or, , NARAYANAGROUP, , 9
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , WE-14: In Young’s double slit experiment the y, co-ordinates of central maxima and 10th, maxima are 2 cm and 5 cm respectively. When, the YDSE apparatus is immersed in a liquid, of refractive index 1.5 the corresponding y coordinates will be, Sol. Fringe width . Therefore, and hence , will decrease 1.5 times when immersed in liquid., The distance between central maxima and 10th, maxima is 3 cm in vacuum. When immersed in liquid, it will reduce to 2 cm. Position of central maxima, will not change while 10th maxima will be obtained, at y = 4 cm., WE-15: In YDSE, bi-cromatic light of, wavelengths 400 nm and 560 nm are used. The, distance between the slits is 0.1 mm and the, distance between the plane of the slits and the, screen is 1m. The minimum distance between, two successive regions of complete darkness, is:, Sol. Let nth minima of 400 nm coincides with mth minima, of 560nm, then, , First order maxima:, S 2 P S1 P (or ), , x 2 9 2 x , , or, , x 2 9 2 x Squaring both sides, we, get x 2 9 2 x 2 2 2 x . Solving this,, we get x 4 . Second order maxima:, S 2 P S1 P 2 ; (or) x 2 9 2 x 2 (or), x 2 9 2 ( x 2 ) Squaring both sides, we get, , x 2 9 2 x 2 4 2 4 x, 5, Solving, we get x 1.25, 4, Hence, the desired x coordinates are,, x 1.25 and x 4 ., 400 , 560 , WE-17: Two coherent light sources A and B with, (2n 1) , (2m 1) , or, 2 , 2 , separation 2 are placed on the x-axis, symmetrically about the origin. They emit light, 2n 1 7 14, ....., of wavelength . Obtain the positions of, 2m 1 5 10, maxima on a circle of large radius, lying in, i.e., 4th minima of 400 nm coincides with 3rd minima, the x-y plane and with centre at the origin., of 560 nm. Location of this minima is,, , (2 4 1)(1000)(400 109 ), 14 mm, 2 0.1, Next 11th minima of 400 nm will coincide with 8th, Sol:, minima of 560 nm., Location of this minima is,, Y1 , , (2 11 1) (1000)(400 109 ), 42 mm, 2 0.1, Required distance Y2 Y1 28 mm ., WE-16: An interference is observed due to two, coherent sources S1 placed at origin and S2, placed at (0,3 , 0) . Here is the wavelength, of the sources. A detector D is moved along, the positive x-axis. Find x-coordinates on the, x-axis (excluding x = 0 and x ) where, maximum intensity is observed., Sol: At x = 0, path difference is 3 . Hence, third, order maxima will be obtained. At x , path, difference is zero. Hence, zero order maxima is, obtained. In between first and second order, maxima will be obtained., Y2 , , 10, , , , For P to have maximum intensity, d cos n, n, 2 cos n cos 2 where n is integer, For n 0, 900 , 2700, n 1, 600 ,1200 , 2400 ,3000, n 2, 00 ,1800, So, positions of maxima are at, , 00 , 600 ,900 ,1200 ,1800 , 2400 , 2700 a n d, 3000 ; i.e., 8 positions will be obtained., Short cut : In d n then number of maximum, on the circle is 4n.Not e: For minima;, , x (2n 1), 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , WE-18: Two coherent point sources S1 and S2, , vibrating in phase emit light of wavelength, . The separation between the sources is 2 ., , Sol: According to the question, Intensity of ray AB, I1 =, I0, and Intensity of ray AB ,, 5, , Consider a line passing through S 2 and, perpendicular to the line S1 S 2 . Find the, position of farthest and nearest minima, S2, 2, S1, D, , P, , , The farthest minima has path, 2, difference / 2 while nearest minima has path, difference (3 / 2) . For the nearest minima., , Sol: xmin (2n 1), , S1 P S 2 P , , , 3, ; [as maximum path difference is 2 ], 2, , (2 ) 2 D 2 D , , 2, 3, 3, (2 ) 2 D 2 D , 2, 2, , , 9 2, 3, D2 2 D, 4, 2, 9 7, 7, 3 D 4 , , D , 4, 4, 12, For the farthest minima,, , S1 P S2 P , 2, , 4 2 D 2 D , 2, 4 2 D 2 , , 2, 15, D2 D D 4 / 4 , 4, 4, WE 19: A ray of light of intensity I is incident, on a parallel glass slab at a point A as shown., It undergoes partial reflection and refraction., At each reflection 20% of incident energy is, reflected. The rays AB and A’ B’ undergo, interference. The ratio Imax / Imin is, , 16 I 0, 81, , I max ( I1 I 2 ) 2 , I0 ,, 125, 125, I I, I min ( I1 I 2 ) 2 0 , max 81 ., 125 I min, WE 20: In a YDSE experiment if a slab whose, refractive index can be varied is placed in front, of one of the slits, then the variation of, resultant intensity at mid-point of screen with, ' ' will be best represented by ( 1)., [Assume slits of equal width and there is no, absorption by slab], I2 , , I0, (1), , B, B', , A, , A', , C, , NARAYANAGROUP, , (2), , , , , =1, , 42 D2 , , I, , I0, , =1, , I0, , I0, , (4), , (3), , =1, , , =1, , Sol. x ( 1)t ;, For 1, x 0, I = maximum = I0 ; As increases path, , difference x also increases.; For x 0 to ,, 2, intensity will decrease from I0 to zero., , Then for x to , intensity will increase from, 2, zero to I 0 ., Hence option 3 is correct, 11
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , Diffraction, , WE 21: Consider the optical system shown in fig., The point source of light S is having , wavelength equal to . The light is reaching, screen only after reflection. For point P to be, 2nd maxima, the value of would be , ( D d & d ), , , , , , 12d 2, 1), D, , 6d 2, 3d 2, 2), 3), D, D, (8d ) 3d, Sol: a. At P, x , ;, D, , 24d 2, 4), D, , , , , 2, , For 2nd maxima, x 2 ; , , 24d, 2, D, , , , 3d, , , , S1, 2d, P, S, , , 6d, , O, Central, maxima, , S2, , Fresnel Diffraction, , , 12d 2, , D, WE 22: Two coherent point sources S1 and S 2, vibrating in phase emit light of wavelength , . The separation between them is 2 as, shown in figure. The first bright fringe is , formed at ‘P’ due to interference on a screen, placed at distance ‘D’ from S1 ( D ), then, OP is, , , , , , 1), , 3 D 2) 1.5 D, , 4) 2 D, 2D, 1, , Sol: x d cos ; cos , d 2 2, x, 600 tan 60 x 3D, D, 12, , The bending of light around edges of an obstacle, on the enchroachment of light within geometrical, shadow is known as “diffraction of light”, Diffraction is a characteristic wave property., Diffraction is an effect exhibited by all electromagnetic waves, water waves and sound waves, Diffraction takes place with very small moving, particles such as atoms, neutrons and electrons, which show wavelike properties., When light passes through a narrow aperture some, light is found to be enchroached into shadow, regions., When slit width is larger, the enchroachment of light, is small and negligible., When slit width is comparable to wavelength of light, the enchroachment of light is more, If the size of obstacle or aperture is comparable, with the wavelength of light, light deviates from, rectilinear propagation near edges of obstacle or, aperture and enchroaches into geometrical shadow., Diffraction phenomenon is classified into two types,, a) Fresnel diffraction b) Fraunhoffer diffraction, , 4), , The source or screen or both are at finite distances, from diffracting device (obstacle or aperture), In Fresnel diffraction, the effect at any point on the, screen is due to exposed wave front which may be, spherical or cylindrical in shape., Fresnel diffraction does not require any lens to, modify the beam., Fresnel diffraction can be explained in terms of “half, period zones or strips”, , Fraunhoffer Diffraction:, The source and the screen are at infinite distance, from diffracting device (aperture or obstacle)., In Fraunhofer diffraction the wave front meeting the, obstacle is plane wave front., Fraunhofer diffraction requires lenses to modify the, beam., , Diffraction Due to Single Slit, , , Diffraction is supposed to be due to interference of, secondary wavelets from the exposed portion of, wavefront from the slit., , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , Whereas in interference, all bright fringes have same, intensity. In diffraction, bright bands are of, decreasing intensity., , 1, times. With while light,, , the central maximum is white and the rest of the, diffraction bands are coloured., , decrease and becomes, , , , Interference and diffraction bands, If N interference bands are contained by the width, of the central bright., 2 D ND, D , ;, , width N N , , a, d, d , , i), , therefore width of the slit, , Condition for minimum intensity is, , a sin n n 1, 2,3,....., , ii), , Where ‘a’ is the width of the slit, is the angle of, diffraction, Condition for maximum intensity, , a sin 2n 1, , , n 1, 2,3,....., 2, , The intensity decreases as we go to successive, maxima away from the centre, on either side. The, width of central maxima is twice as that of secondary, maxima., , a 2d, , N, WE-23: A parallel beam of light of wavelength, 500 nm falls on a narrow slit and the resulting, diffraction pattern is observed on screen 1 m, away. It is observed that the first minimum is, at a distance of 2.5 mm from the centre of the, screen. Find the width of the slit., , y, 2.5 103, , , radian, D, 1, Now, a sin n, Since is very small, therefore sin ., , Sol: , , or a , , n 1 500 109, , m, , 2.5 103, , 2 104 m 0.2 mm, , O, , y, , , P, D, , T, For first minia a sin , a, , y, D, ( sin tan ) y , D, a, , 2 D, a, Note: If lens is placed close to the slit, then D = f., Hence ‘f’ be the focal length of lens, then width, , Width of central maxima w 2 y , , 2f, of the central maximum w , ., a, Note: If this experiment is performed in liquid other, than air, width of diffraction maxima will, NARAYANAGROUP, , WE-24: A screen is placed 50 cm from a single, slit, which is illuminated with 6000 Å light,, If distance between the first and third minima, in the diffraction pattern is 3.00 mm, what is, the width of the slit?, Sol: In case of diffraction at single slit, the position, of minima is given by a sin n . Where d is, the aperture size and for small :, sin ( y / D ), D, y, a n , i.e., y (n ), a, D, , So that , y3 y1 , hence, a , , D, D, (3 ) (2 ) and, a, a, , 0.50 (2 6 107 ), 2 104 m, 3, 3 10, , 0.2 mm, , 13
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , secondary minimum is called half angular width of, the central maximum and it is given by, , (provided is small), a, If the screen is placed at a distance D from the, slit, then the linear spread of the central, maximum is given by, D, y D , a, It is, in fact, the distance of first secondary, minimum from the centre of the screen. It follows, that as the screen is moved away (D is, increased), the linear size of the central maximum, i.e., spread distance, when D Z F ,, y = a (size of the slit), Setting this condition in the above equation, we, have, , WE-25: In a single slit diffraction experiment, first minimum for 1 660 nm coincides, with first maxima for wavelength 2 ., Calculate 2 ., Sol: Position of minima in diffraction pattern is given, by; a sin n, For first minima of 1 , we have, 1, .....(i), a, The first maxima approximately lies between first, and second minima. For wavelength 2 its, position will be, 3, 3, a sin 2 2 sin 2 2, ...... (ii), 2, 2a, The two will coincide if,, 1 2, or, sin 1 sin 2, , a sin 1 (1)1, , or, , sin 1 , , 1 32, , or, a, 2a, 2, 2, 2 1 660 nm 440 nm, 3, 3, WE-26: Two slits are made one millimeter apart, and the screen is placed one meter away., What should the width of each slit be to, obtain 10 maxima of the double slit pattern, within the central maximum of the single, slit pattern., , Sol: We have a (or) , a, (a = width of each slit), , , 10 2, d, a, d 1, a 0.2 mm, 5 5, , , a2, ZF, Z, , or, F, , a, It follows that if screen is placed at a distance, beyond Z F , the spreading of light due to, diffraction will be quite large as compared to, the size of the slit. The above equation shows, that the ray -optics is valid in the limit of, wavelength tending to zero., WE-27: For what distance is ray optics a good, approximation when the aperture is 3 mm, wide and the wavelength is 500 nm?, Sol: For distance Z Z F ,, ray optics is the good appropriate, , a, , Fresnel distance Z F , , Limit of resolution:, , , The Validity of Ray Optics:, The distance of the screen from the slit, so that, spreading of light due to diffraction from the, centre of screen is just equal to size of the slit,, is called Fresnel distance. It is denoted by ZF., The diffraction pattern of a slit consists of, secondary maximum and minima on the two, sides of the central maximum. Therefore, one, can say that on diffraction from a slit, light, spreads on the screen in the form of central, maximum. The angular position of first, 14, , a 2 (3 103 )2, , 18 m, , 5 107, , The smallest linear or angular separation, between two point objects at which they can be, just separately seen or resolved by an optical, instrument is called the limit of resolution of, the instrument., , Resolving Power:, , , The resolving power of an optical instrument is, reciprocal of the smallest linear or angular, separation between two point objects, whose, images can be just resolved by the instrument., 1, , Resolving power = Limit of resolution, The resolving power of an optical instument is, inversely propotional to the wavelength of light, used., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , , , WAVE OPTICS, , Diffraction as a limit on resolving power:, , POLARIZATION, , All optical instuments like lens, telescope, , microscope, etc, act as apertures. Light on passing, through them undergoes diffraction. This puts the , limit on their resolving power., , The properties of light, like interference and, diffraction demonstrate the wave nature of light., Both longitudinal and transverse waves can exhibit, interference and diffraction effects., The properties like polarization can be exhibited, only by transverse waves., The peculiar feature of polarized light is that human, eye cannot distinguish between polarised and, unpolarised light., As light is an electromagnetic wave, among its, electric and magnetic vectors only electric vector is, mainly responsible for optical effects., The electric vector of wave can be identified as a, “light vector”, Ordinary light is unpolarised light in which electric, vector is oriented randomly in all directions, perpendicular to the direction of propagation of, light., The phenomena of confining the vibrations of electric, vector to a particular direction perpendcular to the, direction of propagation of light is called, “Polarization”. Such polarised light is called linearly, polarised or plane polarised light., The plane in which vibrations are present is called, “plane of polarization.”, , Rayleigh’s criterion for resolution:, , , , , The images of two point objects are resolved when , the central maximum of the diffraction pattern of, one falls over the first minimum of the diffraction, pattern of the other., , , Resolving Power of a Microscope:, , , , , T he resolving power of a microscope is defined as, the reciprocal of the smallest distance d between , two point objects at which they can be just resolved, , when seen in the microscope., 1 2 sin , Resolving power of microscope = =, d 1.22, , Clearly, the resolving power of a microscope , depends on:, i) the wave length of the light used, ii) Half the angle of the cone of light from each, , point object., iii) the refractive index of the medium between, the object and the objective of the microscope, , (a), , Resolving Power of a Telescope:, , , , , The resolving power of a telescope is defined as, the reciprocal of the smallest angular separation, ' d ' between two distant objects whose images, can be just resolved by it., Resolving power of telescope, , (b), , Polarized light, , 1, D, , d 1.22 , , Clearly, the resolving power of telescope depends, on : (i) the diameter (D) of the telescope objective, (ii) The wavelength of the light used., WE- 28: Assume that light of wavelength 6000 Å, is coming from a star. What is the limit of, resolution of a telescope whose objective has, a diameter of 100 inch?, Sol: A 100 inch telescope implies that, a = 100 inch = 254 cm. Thus if,, 6000 Å 6 105 cm then,, 1.22, , 2.9 10 7 radians, a, , NARAYANAGROUP, , 15
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , , , , , , , , , Plane polarised light can be produced by different WE-29: When light of a certain wavelength is, incident on a plane surface of a material at a, methods like, glancing angle 300, the reflected light is found, i. Reflection, ii. Refraction, to be completely plane polarized determine., iii. Double refraction iv. Polaroids., a) refractive index of given material and, Polarization by Reflection, b) angle of refraction., The ordinary light beam is incident on transparent Sol: a) Angle of incident light with the surface is 300., surface like glass or water. Both reflected and, The angle of incidence = 900 - 300 = 600. Since, refracted beams get partially polarised., reflected light is completely polarized, therefore, The degree of polarization changes with angle of, incidence takes place at polarizing angle of, incidence., incidence p ., At a particular angle of incidence called “polarising, angle” the reflected beam gets completely plane, p 600, polarised. The reflected beam has vibrations of, Using Brewster’s law, electric vector perpendicular to the plane of paper., tan p tan 600 3, The polarising angle depends on the nature of, b) From Snell’s law, reflecting surface., , Brewster’s Law: When angle of incidence is, , , , , equal to “polarising angle” the reflected and, refracted rays will be perpendicular to each other., Brewster’s law states that “ The refractive index of, a medium is equal to the tangent of polarising angle, ”., p, , Ordinary, light, , air, , or sin r , , Plane, polarized light, , i, , , glass, , Partially, polarized light, , , , , , The refractive index of the medium changes with, wavelength of incident light and so polarising, angle will be different for different wavelengths. , The complete polarization is possible when incident, light is monochromatic., sin p, sin p, sin p, , , , tan p, , 0, sin(90 p ) cos p, sin r, , , , From Brewster’s law, = tan p ., , , , If i= p , the reflected light is completely polarised , and the refracted light is partially polarised., , , , If i< p or i> p , both reflected and refracted rays , get partially polarised., , , , For glass p = tan-1(1.5) 570, , , , For water p = tan (1.33) 53, -1, , 16, , 0, , , , 3, , sin 600, sin r, , 3 1 1, 0., , ,, 2, 3 2 r 30, , Polarisation by Refraction, , , i = p, , sin i, sin r, , The unpolarised light when incident on a glass, plate at an angle of incidence equal to the, polarising angle, the reflected light is completely, plane polarised, but the refracted light is partially, polarised., The refracted light gets completely plane, polarised if incident light is allowed to pass, through number of thin glass plates arranged, parallel to each other. Such an arrangement of, glass plates is called “pile of plates”., , Polarisation by Double Refraction, (Additional), Bartholinus discovered that when light is incident, on a calcite crystal two refracted rays are, produced. It is called “double refraction” or, “birefringence”, An ink dot made on the paper when viewed, through calcite crystal two images are seen due, to double refraction. On rotating the crystal one, image remains stationary and the other image, rotates around the stationary image., The rotating image revolves round the stationary, image in circular path., The stationary image is formed due to ordinary, ray and revolving image is formed by, extraordinary ray., A plane which contains the optic axis and is, perpendicular to the two opposite faces is called, the principal section of crystal., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , , , , , , , The ordinary ray emerging from the calcite crystal, obey the laws of refraction and vibrations are, perpendicular to the principal section of the crystal., The extra ordinary ray does not obey the laws of, refraction and the vibrations are in the plane of, principal section of crystal., Both ordinary and extraordinary rays are plane, polarised., Polaroid: Polaroid is an optical device used to, produce plane polarised light making use of the, phenomenon of “selective absorption”., More recent type of polaroids are H-polaroids., H-polaroids are prepared by stretching a film of, polyvinyl alcohol three to eight times to original, length., , Effect of polarizer on natural light:, , WAVE OPTICS, , Effect of Analyser on plane polarized light:, When unpolarized light is incident on a polarizer,, the transmitted light is linearly polarized. If this light, further passes through analyser, the intensity varies, with the angle between the transmission axes of, polarizer and analyser., Malus states that “the intensity of the polarized light, transmitted through the analyser is proportional to, cosine square of the angle between the plane of, transmission of analyser and the plane of, transmission of polarizer.” This is known as Malus, law., , A0, , If one of waves of an unpolarized light of, intensity I 0 is incident on a polaroid and its, vibration amplitude A0 makes an angle with, the transmission axis, then the component of, vibration parallel to transmission axis will be, A0 cos while perpendicular to it A0 sin . Now, as polaroid will pass only those vibrations which, are parallel to its transmission axis, the intensity, I of emergent light wave will be, , Therefore the intensity of polarized light after, , I0, COS 2, 2, Where I 0 is the intensity of unpolarized light. The, amplitude of polarized light after passing through, A, analyser is A 0 cos ., Transmission axis, 2, A0 sin , I, Case (i) : If 00 axes are parallel then I 0, 2, 0, Case (ii): If 90 axes are perpendicular, then, A0 cos , I 0., I, Case (iii):If 1800 axes are parallel then I 0, 2, 2, 2, I KA0 cos (or), 0 axes are perpendicular then, Case, (iv):, If, 270, I I 0 cos2 [as I 0 KA02 ] In unpolarized light, all, I = 0 Thus for linearly polarized light we obtain, values of starting from 0 to 2 are equally, two positions of maximum intensity and two, positions of minimum (zero) intensity, when we, probable, therefore, rotate the axis of analyser w.r.t to polarizer by an, 2, I0, I0, 2, 2, I I 0 cos I , cos d , angle 2 . In the above cases if the polariser is, 2 0, 2, rotated with respect to analiser then there is no, change in the outcoming intensity., I, I 0, Note: In case of three polarizers P1, P2 and P3: If, 2, 1 is the angle between transmission axes of P1, Thus, if unpolarized light of intensity I 0 is incident, and P2, 2 is the angle between transmission axes, on a polarizer, the intensity of light transmitted, of P2 and P3. Then the intensity of emerging light, I, through the polarizer is 0 . The amplitude of, from P3 is, 2, I, A0, I 0 cos 2 1 cos 2 2 ., 2, polarized light is, ., 2, NARAYANAGROUP, , passing through analyser is I=, , 17
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , WE-30: Unplarized light falls on two polarizing, sheets placed one on top of the other. What, must be the angle between the characteristic, directions of the sheets if the intensity of the, tramsitted light is one third of intensity of the, incident beam?, Sol: Intensity of the light transmitted through the first, polarizer I1 I 0 / 2, where I0 is the intensity of the, incident unpolarized light., Intensity of the light transmitted through the second, polarizer is I 2 I1 cos 2 where is the angle, between the characteristic directions of the, polarizer sheets., But I 2 I 0 / 3 (given), 2, I 2 I1 cos , , I0, I, cos 2 0, 2, 3, , 2, cos 2 / 3 cos, 3, WE-31: Unpolarized light of intensity 32 Wm-2, passes through three polarizers such that the, transmission axis of the last polarizer is, crossed with the first. If the intensity of the, emerging light is 3 Wm-2, what is the angle, between the transmission axes of the first two, polarizers? At what angle will the transmitted, intensity be maximum?, Sol: If is the angle between the transmission axes, of first polaroid P1 and second P2 while , between the transmission axes of second, polaroid P2 and third P3, then according to given, problem., 900 or (900 ).....(1), 2, , According to given problem, I 3 3W / m 2, So, 4(sin 2 )2 3 i.e., sin 2 ( 3 / 2) or, 2 600 , i.e., 300 ., WE-32: Discuss the intensity of transmitted light, when a polaroid sheet is rotated between two, crossed polaroids?, Sol: Let I0 be the intensity of polarised light after passing, through the first polariser P1. Then the intensity of, light after passing through second polariser P2 will, be, I I 0 cos 2 , where is the angle between, pass axes of P1 and P2. Since P1 and P2 are, crossed the angle between the pass axes of P2, and P3 will be ( / 2 ). Hence the intensity, of light emerging from P 3 will be, , , , I I 0 cos 2 cos 2 , 2, , , 1, , I 0 cos 2 sin 2 ( I 0 / 4)sin 2 2, Therefore, the transmitted intensity will be, maximum when / 4 ., , C. U . Q, INTERFERENCE, 1., , 2., , 3., , Now if I 0 is the intensity of unpolarized light, incident on polaroid P1 , the intensity of light, transmitted through it,, 1, 1, W, I1 I 0 (32) 16 2 ......(2), 2, 2, m, Now as angle between transmission axes of, polaroids P1 and P2 is , in a accordance with, Malus law, intensity of light transmitted through, P2 will be, I 2 I1 cos 2 16 cos 2 ..........(3), And as angle between transmission axes of P2, and P3 is , light transmitted through P3 will be, I 3 I 2 cos 2 16 cos 2 cos 2 .........(4), 18, , 4., , 5., , A plane wave front falls on a convex lens. The, emergent wave front is, 1) Plane, 2) Cylindrical, 3) Spherical diverging 4)Spherical converging, When two light waves meet at a place, 1) their displacements add up, 2) their intensities add up, 3) both will add up, 4) Energy becomes zero, The following phenomena which is not, explained by Huygens’ construction of wave, front is, 1) refraction, 2) reflection, 3) diffraction, 4) origin of spectra, A wavefront is an imaginary surface, where, 1) phase is same for all points, 2) phase changes at constant rate at all points along, the surface., 3) constant phase difference continuously changes, between the points, 4) phase changes all over the surface, Huygen’s wave theory is used, 1) to determine the velocity of light, 2) to find the position of the wave front, 3) to determine the wavelength of light, 4) to find the focal length of a lens., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 6., 7., , 8., , In a Laser beam the photons emitted are, 15. The necessary condition for an interference, 1) same wavelength 2) coherent, by two sources of light is that:, 1) two light sources must have the same wavelength, 3) of same velocity 4) All the above, 2) two point sources should have the same, The amplitudes of two interfering waves are 4, amplitude and same wavelength, cm and 3 cm respectively. If the resultant, 3) two sources should have the same wavelength,, amplitude is 1 cm then the interference, nearly the same amplitude and have a constant, becomes, phase angle difference, 1) constructive, 4) the two point sources should have a randomly, 2) Destructive, varying phase difference, 3) Both constructive and destructive, 16. For the sustained interference of light, the, 4) given data is insufficient, necessary condition is that the two sources, Two coherent waves are represented by, should, y1 =a1 cos t and y2 =a2 sin t. The resultant, 1) have constant phase difference only, intensity due to interference will be, 2) be narrow, 1) a 12 a 22, 2) a 12 a 22, 3) be close to each other, 4) of same amplitude with constant phase, 3)(a1 – a2), 4)(a1+ a2), difference, Two light waves are represented by, 17., When interference of light takes place, y 1 a sin t and y 2 a sint . The phase, 1) Energy is created in the region of maximum, of the resultant wave is, intensity, , , , 2) Energy is destroyed in the region of maximum, 1) 2, 2), 3), 4), intensity, 3, 2, 4, 3) Conservation of energy holds good and, Laser light is considered to be coherent, energy is redistributed, because it consists of, 4) Conservation of energy doesn’t hold good, 1) many wavelengths, 18., Which of the following is conserved when light, 2) uncoordinated wavelengths, waves interfere, 3) coordinated waves of exactly the same, 1) momentum, 2) amplitude, wavelength, 3), energy, 4) intensity, 4) divergent beams, Two waves having the same wave length and 19. The path difference between two interfering, waves at a point on screen is 70.5 times the, amplitude but having a constant phase, wave length. The point is, difference with time are known as, 1) Dark, 2) Bright, 1) identical waves, 2) incoherent waves, 3) Not possible, 4) Green in colour, 3) coherent waves, 4) collateral waves, 20. Interference is produced with two coherent, Light waves spreading from two sources, sources of same intensity. If one of the, produce steady interference only if they have, sources is covered with a thin film so as to, 1) congruence, 2) coherence, reduce the intensity of light coming out of it to, 3) same intensity, 4) same amplitude, half, then, Four different independent waves are, 1) Bright fringes will be less bright and dark fringes, represented by, will be less dark, a) y1=a1 sin 1t, b) y2=a2 sin 2t, 2) Bright fringes will be more bright and the dark, fringes will be more dark, , c) y3=a3 sin 3t, d) y4=a4sin( 4 t+ 3 ), 3) Brightness of both types of the fringes will remain, the same, The sustained interference is possible due to, 4) Dark region will spread completely, 1) a & c 2) a & d, 3) c & d, 21. For constructive interference between two, 4)not possible with any combination, waves of equal wavelength, the phase angle, Interference fringes in Young’s double slit, should be such that, experiment with monochromatic light are, , , 1) always equispaced, 1) cos 2 1, 2) cos 2 0, 2, 2, 2) always unequally spaced, , , 3) both equally and unequally spaced, 3) cos 2 1, 4) cos 2 infinite, 4) formed by a portion of the wave front., 2, 2, , , , 9., , 10., , 11., , 12., , 13., , 14., , WAVE OPTICS, , , , NARAYANAGROUP, , , , , , 19
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , 22. Two coherent waves each of amplitude ‘a’ 30. In young’s double slit experiment the slits are, traveling with a phase difference when, of different length and widths. The amplitude, superpose with each other the resultant, of the light waves is directly proportional to, intensity at a given point on the screen is, the, 2, 2, 1) length of the slit 2) distance between the slits, 1) a (1 cos ), 2) 4a (1 cos ), 3) area of the slits, 4) width of slits, 3) 2a 2 (1 cos ), 4) (1 cos ), 31. In a double slit experiment, instead of taking, 23. In the set up shown, the two slits S1 and S2, slits of equal widths, one slit is made twice as, are not equidistant from the slit S., wide as the other. Then, in the interference, pattern., 1) The intensities of both the maxima and the minima, S, increase, S, O, 2) The intensity of the maxima increases and the, S, minima has zero intensity., 3) The intensity of the maxima decreases and that, The central fringe at O is then, of the minima increases., 1) always bright, 2) always dark, 4) The intensity of the maxima decreases and the, 3) either dark or bright depending on the position, minima has zero intensity., of S., 4) neither dark nor bright 32. When the width of slit aperture is increased, 24. In young’s experiment of double slit, the, by keeping ‘d’ as constant in Young’s, number of times the intensity of the central, experiment, bright band greater than the individual, 1) Fringe width will increase, intensity of the interfering waves, 2)Fringe width will decrease and then increase, 1) 2, 2) 4, 3) 6, 4) 16, 3)Fringe width first increases then decreases, 25. A young’s double slit experiment uses a, 4) Gradually the fringes will be merge, monochromatic source. The shape of the, 33., When viewed in white light, soap bubble show, interference fringes formed on the screen is, colours because of, a, 1) Interference, 2) Scattering, 1) straight line, 2) parabola, 3), Diffraction, 4) Dispersion, 3) hyperbola, 4) circle, 34., When, petrol, drops, from, a vehicle fall over, 26. The contrast in the fringes in any interference, rain water on road surface colours are seen, pattern depends on :, because of, 1) fringe width, 2) wave length, 1) Dispersion of light 2) Interference of light, 3) intensity ratio of the sources, 3) Scattering of light 4) Absorption of light, 4) distance between the sources., 35., Coherent light is incident on two fine parallel, 27. If monochromatic red light is replaced by, slits S1 and S2 as shown in fig. If a dark fringe, green light the fringe width becomes, occurs at P, which of the following gives, 1) increase, 2) remain same, possible phase differences for the light waves, 3) we cannot say, 4) decrease, arriving at P from S1 and S2?, 28. Interference was observed in interference, chamber, when air was present. Now the, chamber is evacuated, and if the same light, S, is used, a careful observer will see, P, 1) no interference, S, 2) interference with central bright band, 3) interference with central dark band, 1) 2, 4, 6 …, 2) 1/2, 5/2, 9/2 …, 4) interference in which breadth of the fringe, will be slightly increased., 3) , 3, 5 ….., 4) 1/2, 3/2, 5/2 …., 29. In young’s experiment with white light central 36. In young’s double slit experiment, the distance, fringe is white. If now a transparent film is, of the n-th dark fringe from the centre is, introduced in the upper beam coming from the, 2d , D , top slit, the white fringe, 1) n 2d , 2) n , , , , 1) moves down ward, 2) moves upward, D , 3) remains at the same place, D, 4d, 4) totally disappears, 3) 2n 1, 4) 2n 1, 2d, D, 1, , 2, , 1, , 2, , 20, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 37. When a thin film of thickness t is placed in the, path of light wave emerging out of the slit, then, increase in the length of optical path will be, , 4), 1) 1 t 2) 1 t 3) t, t, 38. If the Young’s double slit experiment is, performed with white light, then, 1) the central maximum will be dark, 2) there will not be completely dark fringe, 3) the fringe next to the central will be red, 4) the fringe next to the central will be violet, 39. Which of the following decides about the, contrast between bright and dark fringes in an, interference experiment?, 1) wavelength, 2) distance between two coherent sources, 3) fringe width, 4) intensity ratio, 40. If torch is used in place of monochromatic light, in Young’s experiment, what will happen?, 1) Fringe will occur as from monochromatic source, 2) Fringe will appear for a moment and then it will, disappear, 3) No fringes will appear, 4) Only bright fringe will appear, 41. At a finite distance from the source, a point, source of light produces, 1) spherical wave front, 2) plane wavefront, 3) cylindrical wavefront, 4) both spherical and plane wavefronts, 42. Nature of wave front depends on, 1) shape of source, 2) distance of source, 3) both 1 and 2, 4) none of these, 43 If Young’s double slit apparatus is shifted from, air to water, then, 1) Fringe width decreases, 2) Fringe width increases, 3) Fringe width remains same, 4) Fringe system disappears, 44. In Young’s double slit experiment the phase, difference between the waves reaching the, central fringe and fourth bright fringe will be, 1) zero 2) 4, 3) 6, 4) 8, 45. Instead of using two slits as in young’s, experiment, if we use two separate but identical, sodium lamps, which of the following occur, a) uniform illumination is observed, b) widely separate interference, c) very bright maximum, d) very minimum, 1) a only 2) a, b only 3) c, d only 4) b, d only, NARAYANAGROUP, , WAVE OPTICS, 46. Alternate bright and dark fringes appear in, Young’s double slit experiment due to the, phenomenon of, 1) Polarisation, 2) Diffraction, 3) Interference, 4) Dispersion, , DIFRACTION OF LIGHT, 47. The bending of light about corners of an, obstacle is called, 1) Dispersion, 2) Refraction, 3) Deviation, 4) Diffraction, 48. To observe diffraction, the size of an obstacle, 1) Should be of the same order as wave length, 2) Should be much larger than the wave length, 3) Has no relation to wave length, 4) May be greater or smaller than the wave length, 49. In diffraction pattern, 1) The fringe widths are equal, 2) The fringe widths are not equal, 3) The fringes can not be produced, 4) The fringe width may or may not be equal, 50. Sun light filtering through a tree leaves often, makes circular patches on the ground because, 1) The sun is round, 2) The space through which light penetrates is, round, 3) Light is transverse in nature, 4) Of diffraction effects, 51. In studying diffraction pattern of different, obstacles, the effect of, 1) full wave front is studied, 2) portion of a wave front is studied, 3) waves from two coherent sources is studied, 4) waves from one of the coherent source is, studied., 52. Both light and sound waves produce, diffraction. It is more difficult to observe the, diffraction with light waves because., 1) Light wave do not require medium, 2) Wavelength of light waves is far smaller, 3) Light waves are transverse, 4) Speed of light is far greater, 53. In Young’s double slit experiment, 1) only interference occurs, 2) only diffraction occurs, 3) both interference and diffraction occurs, 4) polarisation occurs, 54. Light travels in a straight line because, 1) it is not absorbed by atmosphere, 2) its velocity is very high, 3) diffraction effect is negligible, 4) due to interference, 21
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WAVE OPTICS, 55. One of the following statements is correct. Pick, out the one, 1) Diffraction can not take place without, interference, 2) Interference will not take place with out, diffraction., 3) Interference and diffraction are the result of, polarization, 4) The fringe width in Young’s double slit, experiment does not depends on the wave length., 56. Diffraction of light is, 1) the bending of light at the surface of separation, when it travels from rarer medium of denser medium, 2) the bending of light at the surface of, separation when it travels from denser medium to, rarer medium, 3) encroachment of light into the geometrical, shadow of the obstacle placed in its path, 4) emergence of a light ray grazing the, surface, of separation when it travels from denser to rarer, medium, 57. Pick out the correct statements, 1) diffraction is exhibited by all electromagnetic, waves but not by mechanical waves, 2) diffraction cannot be observed with a plane, polarized light, 3) the limit of resolution of a microscope decreases, with increase in the wavelength of light used, 4) the width of central maximum in the diffraction, pattern due to single slit increases as wavelength, increases, 58. A lens of focal length f gives diffraction pattern, of Fraunhoffer type of a slit having width a. If, wavelength of light is , the distance of first, dark band and next bright band from axis is, given by, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, 62. The surface of crystals can be studied using, 1) diffraction of visible light, 2) diffraction of x-rays, 3) interference of sound waves, 4) refraction of radio waves, 63. The diffraction bands observed in the case of, straight edge producing diffraction effects are, 1) equally spaced like the interference bands, but with less contrast, 2) unequally spaced with increasing width as we, move away from the edge of geometric shadow, 3) unequally spaced with decreasing width as we, move away from the edge of geometric shadow, 4) equally spaced like the interference bands but, with more contrast, 64. A we move away from the edge into the, geometrical shadow of a straight edge, the, intensity of illumination, 1) Decreases, 2) Increases, 3) Remains unchanged, 4) Increase and then decreases, 65. In Fresnel’s diffraction, wavefront must be, 1) spherical, 2) cylindrical, 3) plane, 4) both 1 and 2, , RESOLVING POWER, , 66. The resolving power of human eye is, 1) 1', 2) 10, 3) 10', 4) 5', 67. Resolving power of a telescope increases with, 1) Increase in focal length of eye piece, 2) Increase in focal length of objective, 3) Increase in aperture of eye piece, , 4) Increase in aperture of objective, a, , 2) f, 3) af, 4) af, 1) f, 68. To increase both the resolving power and, , a, magnifying power of a telecscope, 59. The class of diffraction in which incident and, diffracted wave fronts are planar is called, 1) Both the focal length and aperture of the objec, 1) Fresnel diffraction, tive has to be increased, 2) Fraunhoffer diffraction, 2) The focal length of the objective has to be in, 3) Huygen’s diffraction, creased, 4) Newton’s diffraction, 3) The aperture of the objective has to be increased, 60. Neutron diffraction pattern is used to, 4) The wavelength of light has to be decreased, determine, POLARISATION, 1) Density of solids, 69. Waves that cannot be polarised are, 2) Atomic number of elements, 3) Crystal structure of solid, 1) Longitudinal, 2) Transverse, 4) Refractive index of liquid, 3) Electromagnetic, 4) Light, 61. Geometrical shadow is formed due to the 70. Human eye, phenomenon of, 1) Can detect polarised light, 1) Diffraction of light 2) Polarisation of light, 2) Can not detect polarisation of light, 3) Interference of light, 3) Can detect only circularly polarised light, 4) Rectilinear propagation of light, 4) Can detect only linearly polarised light, 22, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 71. Polarisation of light was first successfully, explained by, 1) Corpuscular theory, 2) Huygens’ wave theory, 3) Electromagnetic wave theory, 4) Planck’s theory, 72. Plane of polarisation is, 1) The plane in which vibrations of the electric, vector takes place, 2) A plane perpendicular to the plane in which, vibrations of the electric vector takes place, 3) Is perpendicular to the plane of vibration, 4) Horizontal plane, 73. In the propagation of polarised light waves, the, angle between the plane of vibration and the, plane of polarization is, 1) 00, 2) 900, 3) 450, 4) 1800, 74. Transverse wave nature is established by, 1) Interference, 2) Diffraction, 3) Polarization, 4) All the above, 75. Choose the correct statement., 1) the Brewster’s angle is independent of, wavelength of light., 2) the Brewster’s angle is independent of the nature, of reflecting surface, 3) the Brewster’s angle is different for different, wavelengths, 4) Brewster’s angle depends on wavelength but not, on the nature of reflecting surface., 76. The polarising angle for glass is, 1) same for different kinds of glass, 2) different for different kinds of glass, 3) same for lights of all colours, 4) varies with time, 77. When an unpolarised light is polarized, then, the intensity of light of the polarized wave, 1) remains same 2) doubled 3) halved, 4) depends on the colour of the light., 78. Unpolarising light falls on two polarizing sheets, so oriented that no light is transmitted. If a, third polarizing sheet is placed between them;, not parallel to either of the above two sheets, in question, 1) no light is transmitted, 2) some light is transmitted, 3) light may or may not be transmitted, 4) certainly 50% light is transmitted., 79. When light falls on two polaroid sheets, one, observes complex brightness then the two, polaroids axes are, 1) Mutually perpendicular 2) Mutually parallel, 3) Angle between their two axes is 450, 4) None of the above, NARAYANAGROUP, , WAVE OPTICS, 80. Polaroid are used, 1) to eliminate head light glare in automobile, 2) in production of 3-D motion pictures, 3) in sun glasses, 4) all the above, 81. A diffraction pattern is obtained using a beam, of redlight. What happens if the red light is, replaced by blue light, 1) no change, 2) diffraction bands become narrower and crowded, together, 3) bands beome broader and farther apart, 4) bands disappear, 82. In a diffraction pattern the width of any fringe, is, 1) directly proportional to slit width, 2) inversely proportional to slit width, 3) Independent of the slit width, 4) None of the above, 83. Yellow light is used in a single slit of diffraction, experiment with slit width 0.6 mm. If yellow, light is replaced by X-rays then the observed, pattern will reveal, 1) that the central maximum is narrower, 2) more number of fringes, 3)less number of firnges 4)no diffraction patterns, 84. A beam of light AO is incident on a glass slab, ( = 1.54) in the direction shown. The reflected, ray OB is passed through a Nicol prism. On, viewing through a Nicol prism, we find on, rotating the prism that, B, , A, , 0, , 33, , O, , 0, , 33, , Glass slab, , 1) the intensity is reduced down to zero and remains, zero, 2) the intensity reduces down somewhat and rises, again, 3) there is no change in intensity, 4) the intensity gradually reduces to zero and then, again increases, 85. A star is going away from the earth. An, observer on the earth will see the wavelength, of light coming from the star, 1) decreased, 2) increased, 3) neither decreased nor increased, 4) decreased or increased depending upon the, velocity of the star, 23
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WAVE OPTICS, 86. Red shift is an illustration of, 1) low temperature emission, 2) high frequency absorption, 3) Doppler effect, 4) unknown phenomenon, 87. If the shift of wavelength of light emitted by a, star is towards violet, then this shows that star, is, 1) stationary, 2) moving towards earth, 3) moving away from earth, 4) Information is incomplete, 88. When there is a relative motion of an observer, from a source of light, the apparent change in, its wavelength is termed as, 1) Raman effect, 2) Seebeck effect, 3) Doppler’s effect, 4) Gravitational effect, 89. In the context of Doppler effect in light, the, term red shift signifies, 1) decrease in frequency, 2) increase in frequency, 3) decrease in intensity, 4) increase in intensity, 90. As we change the colour of light from Red to, Blue, which of the following is correct for the, polarizing angle and critical angle of glass?, 1) the former increases, the latter decreases, 2) the former decreases, the latter increases, 3) the former increases, the latter increases, 4) the former decreases, the latter decreases, 91. Atmospheric refraction is due to, 1) changing pressure in the atmosphere, 2) varying density of atmosphere, 3) varying temperature of the atmosphere, 4) both (2) and (3)., 92. Which of the following phenomenon is not, common to sound and light waves, 1) Interferenec, 2) Diffraction, 3) Polarisation, 4) Reflection, 93. Polarisation can be produced by, 1) Reflection, 2) Refraction, 3) Scattering, 4) All of the above, 94. An unpolarised light is incident on a surface, separating two transparent media of different, optical densities at the polarizing angle. Then, the reflected ray and refracted ray are, 1) parallel to each other, 2) perpendicular to each other, 3) inclined to each other making an angle 450, 4) none of the above, 24, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, 95. The intensity of the polarized light transmitted, through the analyzer is given by, 1) Brewster’s law, 2) Malus Law, 3) Fresnel’s assumptions 4) law of superposition, 96. Statement A: In the interference pattern the, intensity is same at all points in a brightband, Statement B: In Young’s double slit, experiment, as we move away from the central, maximum, the third maximum always comes, before the third minimum., 1) Both A and B are true 2) Both A and B are false, 3) A is true but B is false4) A is false and B is true, 97. A light of wavelength is incident on an object, of size b. If a screen is at a distance D from, the object. Identify the correct condition for, the observation of different phenomenon, a) if b 2 D , Fresnel diffraction is observed, b) if b 2 D , Fraunhoffer diffraction is, observed, c) b 2 D , Fraunhoffer diffraction is, observed, d) b 2 D , the approximation of geometrical, optics is applicable, 1) a, b and d are true 2) a,c and d are true, 3) a and c are true, 4) a and d are true, , DIRECTIONS Q.NO: 98 TO 127, In each of the following questions, a statement, of Assertion (A) is given followed by a, corresponding statement of reason (R) just, below it. Of the statement mark the correct, answer., 1) Both ‘A’ and ‘R’ are true and ‘R’ is the, correct explanation of ‘A’, 2) Both ‘A’ and ‘R’ are true and ‘R’ is not the, correct explanation of ‘A’, 3) ‘A’ is true and ‘R’ is false, 4) A’ is false and ‘R’ is true, 98. Assertion (A) : In Young’s double slit, experiment the band width for red colour is, more, Reason (R) : Wavelength of red is small, 99. Assertion (A) : Thin films such as soap bubble, or a thin layer of oil on water show beautiful, colours when illuminated by sunlight, Reason (R) : The colours are obtained by, dispersion of light only, 100. Assertion (A) : When tiny circular obstacle is, placed in the path of light from some distance,, a bright spot is seen at the centre of the shadow, of the obstacle., Reason (R) : Destructive interference occurs, at the centre of the shadow., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 101. Assertion (A) : Coloured spectrum is seen, when we look through a fine cotton cloth, Reason (R) :It is due to the diffraction of white, light on passing through fine slits., 102. Assertion (A) : Diffraction is common in sound, but not common in light waves, Reason (R) : Wavelength of light is more than, the wavelength of sound, 103. Assertion (A) : We cannot observe diffraction, pattern from a wide slit illuminated by, monochromatic light, Reason (R) : In diffraction pattern, all the, bright bands are not of the same intensity., 104. Assertion (A) : Transverse wave nature of, light is proved by polarisation, Reason (R) : According to Maxwell, light is, an electromagnetic wave but not mechanical, wave, 105. Assertion (A) : Coloured spectrum is seen when, we look through a cloth, Reason (R) : Diffraction of light takes place, when light is travelling through the pores of, cloth, 106. Assertion (A) : Young’s double slit experiment, can be performed using a source of white light., Reason (R) : The wavelength of red light is, less than the wavelength of other colors in white, light., 107. (A) : The unpolarised light and polarized light, can be distinguished from each other by using, Polaroid., (R) : A Polaroid is capable of producing plane, polarized beams of light., 108. Assertion (A) : Illumination of the sun at noon, is maximum because, Reason (R): The sun rays are incident almost, normally, 109. Assertion (A): The phase difference between, any two points on a wave front is zero, Reason (R): Light from the source reaches, every point of the wave front at the same time, 110. Assertion (A) : In Young’s double slit, experiment white light is used and slits are, covered with red and blue filters respectively., The phase difference at any point on the screen, will continuously change and uniform, illumination is produced on the screen, Reason (R): Two independent sources of light, would no longer act as coherent sources, NARAYANAGROUP, , WAVE OPTICS, 111. Assertion (A) : In interference pattern intensity, of successive fringes due to achromatic light, is not same, Reason (R): In interference, only redistribution, of energy takes place, 112. Assertion (A): Light from two coherent sources, is reaching the screen. If the path difference, at a point on the screen for yellow light is, 3 / 2 , then the fringe at the point will be, coloured., Reason (R): Two coherent sources always, have constant phase relationship, 113. Assertion (A): No interference pattern is, detected when two coherent sources are very, close to each other. (i.e separation almost zero), Reason (R): The fringe width is inversely, proportional to the distance between the two, slits, 114. Assertion (A): In Young’s double slit, experiment interference pattern disapperars, when one of the slits is closed, Reason (R): Interference occurs due to, superimposition of light wave from two, coherent sources, 115. Assertion (A): The maximum intensity in, interference pattern is four times the intensity, due to each slit of equal width., Reason (R): Intensity is directly proportional, to square of amplitude., 116. Assertion (A): The fringe obtained at the, centre of the screen is known as zeroth order, fringe, or the central fringe, Reason (R): Path difference between the, waves from S 1 and S 2 , reaching the central, fringe (or zero order fringe) is zero, 117. Assertion (A) : If the phase difference between, the light waves emerging from the, slits of the Young’s experiment is -radian,, the central fringe will be dark, Reason (R) : Phase difference is equal to, , 2 / times the path difference., 118. Assertion (A) : At sunrise or at sunset the sun, appears to be reddish while at mid-day the sun, looks white, Reason (R) : Scattering due to dust particles, and air molecules, 25
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WAVE OPTICS, 119. Assertion (A) : If the whole apparatus of, Young’s experiment is immersed in liquid, the, fringe width will decrease., Reason (R) : The wavelength of light in water, is more than that in air, 120. Assertion (A) : The soap film in sun light is, colourful, Reason(R):Thin films produce interference of, light, 121. Assertion (A): Coloured spectrum is seen, when we look through a cloth, Reason (R): Diffraction of light takes place, when light is travelling through the pores of, cloth, 122. Assertion (A) : Radio waves diffract, pronouncedly around the sharp edges of the, buildings than visible light waves., Reason (R) : Wave length of radio waves is, comparable to the dimension of the edges of, the building., 123. Assertion (A) : When an unpolarised light is, incident on a glass plate at Brewster angle,, the reflected ray and refracted ray are, mutually perpendicular, Reason (R) :The refractive index of glass is, equal to sine of the angle of polarisation., 124. Assertion (A) : If two waves of same amplitude, produce a resultant wave of same amplitude,, then the phase difference between them may, be 120., Reason (R) : The resultant amplitude of two, waves is equal to algebraic sum of amplitude, of two waves., 125. Assertion (A) : Although the surfaces of a, goggle lens are curved, it does not have any, power., Reason (R) : In case of goggles, both the, curved surfaces have equal radii of curvature., 126. (A) : For best contrast between maxima and, minima in the interference pattern of Young’s, double slit experiment, the intensity of light, emerging out of the two slits should be equal, (R) : The intensity of interference pattern is, proportional to square of amplitude., 26, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, 127. (A) : In Young’s double slit experiment the, fringes become indistinct if one of the slits is, converd with cellophane paper., (R) : The cellophane paper decreases the, wavelength of light., , MATCHING TYPE QUESTIONS, 128. Match list A and list B accurately, LIST - A, LIST - B, a) spherical wave e) linear source, front, b) plane wave front f) point light source, c) cylindrical wave g) at infinite, front, distance, 1) ( a, f ); ( b, g); ( c, e) 2) ( a, f ); ( b, e); ( c, g), 3) ( a, g ); ( b, f); ( c, e) 4) ( a, e ); ( b, g); ( c, f), 129. Match the following, PART-A, PART-B, D, a) achromatic light d), d, b) monochromatic, e) distance between, light, two successive, bright bands, c) fringe width, f) distance between, two successive dark, bands, g) central fringe is, always bright, h) central fringe is, always achromatic, 1) a g b e , f , g , c e , f , g, 2) a g, h b h, g c d , e, f, 3) a e , f , g b g c e , f , g, 4) a e, b h, c g , h, 130. Match the following, PART-A, PART-B, a) Polarisation, e) All types of waves, b) interference, f) longitudinal waves, c) diffraction, g) transverse waves, d) reflection, h)only with, transverse waves, i) stationary waves, produced in, stretched strings, 1) a g b e, f , g , i c e, f , g d e, f , g, 2) a h, g b f , g c g d h, 3) a e, f , g b g c e, f , g d g, 4) a e b h, i c g , h d e, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 131. Match the following :, List-I, a) Silver lining of, mountains, b) Rectilinear, propagation light, c) Polarization, d) Pile of plates, , List-II, e) polarization by, refraction, f) transverse nature, of light, g) diffraction, h) ray optics, , 1) a h, b g , c f , d e, 2) a g , b h, c e, d f, 3) a f , b h, c h, d e, 4) a g , b h, c f , d e, 132. Match the following, List-I, List-II, a) coherent,, e) Malus law, monochromatic, highly unidirectionally, b) I I 0 cos 2 , , f) Polariod, , c) Selective, g) Spherical wave, absorption is, front, exhibited by, d) Fresnel diffraction h) LASER, 1) a h, b e, c f , d g, 2) a g , b h, c e, d f, 3) a h, b g , c e, d f, 4) a g , b h, c f , d e, 133. Match the following, List-I, List-II, a) Interference, e) Thamos young, b) Polarisation by, f) Bartholinus, reflection, c) Diffraction, g) Grimaldi, d) Polarisation by, h) Malus, refraction, 1) a e, b g , c f , d h, 2) a h, b f , c g , d e, 3) a e, b h, c g , d f, , WAVE OPTICS, 134. In Young’s double slit experiment, what will be, the effects of the following, Column-I, (A) A thin translucent, plate is inserted in, front of one of the, slits., , Column-II, (p) Fringe width changes., , (B) A thin transparent, (q) Fringe width remains, glass plate is inserted, unaltered., in front of one of the, slits., (C) The entire set up is, immersed in water., , (r) Brightness of fringe, changes., , (D) Both slits are covered (s) Brightness of fringe, remains unaltered., with translucent paper., (t) Fringe width and, brightness are directly, related., , 1) A – q,r, B – q, s, C – p, r, D – q, r, 2) A – q,r, B – q, s, C – p, s, D – q, r, 3) A – p,r, B – q, s, C – p, s, D – q, r, 4) A – q,s, B – q, t, C – p, s, D – q, r, , C.U.Q - KEY, 1) 4 2) 2 3) 4 4) 1 5) 2, 8) 2 9) 2 10) 3 11) 3 12) 2, 15) 3 16) 4 17) 3 18) 3 19) 1, 22) 3 23) 3 24) 2 25) 3 26) 3, 29) 2 30) 3 31) 1 32) 2 33) 1, 36) 3 37) 1 38) 3 39) 4 40) 3, 43) 1 44) 4 45) 1 46) 3 47) 4, 50) 4 51) 2 52) 2 53) 3 54) 3, 57) 4 58) 2 59) 2 60) 3 61) 4, 64) 1 65) 4 66) 1 67) 4 68) 4, 71) 3 72) 1 73) 2 74) 3 75) 3, 78) 2 79) 2 80) 4 81) 2 82) 2, 85) 2 86) 3 87) 2 88) 3 89) 1, 92) 3 93) 4 94) 2 95) 2 96) 1, 99) 3 100)3 101)1 102)3 103)2, 106)3 107)1 108)1 109)1 110)1, 113)1 114)1 115)2 116)1 117)2, 120)1 121)1 122)1 123)3 124)3, 127)3 128)1 129) 2130)1 131)4, 134) 2, , 6) 4 7) 2, 13) 4 14) 1, 20) 1 21) 3, 27) 4 28) 2, 34) 2 35) 3, 41) 1 42) 3, 48) 1 49) 2, 55) 1 56) 3, 62) 2 63) 3, 69) 1 70) 2, 76) 2 77) 3, 83) 4 84) 4, 90) 1 91) 4, 97) 2 98) 3, 104)2 105)1, 111)1 112)4, 118)1 119)3, 125)1 126)2, 132)1 133)3, , 4) a h, b g , c f , d e, , NARAYANAGROUP, , 27
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, 8., , LEVEL - I (C. W), INTERFERENCE, 1., , 2., , 3, , The displacements of two interfering light, waves are y1 4sin t and y2 3cos t ., The amplitude of the resultant wave is ( y1, and y2 are in CGS system), 1) 5cm, 2) 7cm, 3) 1 cm, 4) zero, Two coherent sources of different intensities, send waves that interfere. The ratio of, maximum to minimum intensity is 25. The, intensity ratio of the sources is, 1) 25 : 1 2) 5 : 1, 3) 9 : 4, 4) 625 : 1, Two sources of intensity 2I and 8I are used, in an interference experiment. The intensity, at a point where the waves from two sources, superimpose with a phase difference of (a) zero, (b) / 2 and c is, 1) 18 I ,10 I , 2 I, , 1) 1.33 0.63 mm, , 9., , 10., , 3) 2 I , I ,, , 1) I 2I0 1 cos , 3) I , 5., , 1 cos , , 1 cos , , I0, 2I0, In Young’s double slit experiment, the constant, , , phase difference between two sources is ., 2, The intensity at a point equidistant from the, slits in terms of maximum intensity I0 is, 6., , 7., , 28, , 11., , 2) I I0 1 cos , 4) I , , 1) I0, 2) I0 / 2, 3) 3I0 / 4 4) 3I0, The path difference between two interfering, waves at a point on the screen is / 6 from, central maximum. The ratio of intensity at this, point and that at the central fringe will be, 1) 0.75 2) 7.5, 3) 85.3, 4) 853, In a Young’s double slit experiment, 12 fringes, are observed to be formed in a certain region, of the screen when light of wavelength 600 nm, is used. If the light of wavelength 400 nm is, used, the number of fringes observed in the, same region of the screen will be, 1) 12, 2) 18, 3) 24, 4) 8, , 2), , 0.63, mm, 1.33, , 0.63, , 2) 5 I , 4 I , I, , I, 4) 2 I ,10 I ,18 I, 2, 4. The intensity of interference waves in an, interference pattern is same as I0 . The, resultant intensity at the point of observation, will be, , A double slit apparatus is immersed in a liquid, of refractive index 1.33. It has slit separation, of 1mm and distance between the plane of slits, and screen 1.33 m. The slits are illuminated, by a parallel beam of light whose wavelength, in air is 6300 A0 . The fringe width is, , 12., , 13., , 14., , 3) 1.33 2 mm, 4) 0.63mm, The fringe width at a distance of 50cm from, the slits in young’s experiment for light of, wavelength 6000A 0 is 0.048cm. The fringe, width at the same distance for 5000A 0 will, be, 1) 0.04cm 2) 0.4cm 3) 0.14cm 4)0.45cm, In young’s double slit experiment the two slits, are illuminated by light of wavelength 58900A, and the distance between the fringes obtained, on the screen is 0.20. If the whole apparatus is, immersed in water then the angular fringe, width will be, if the refractive index of water is, 4/3, 1) 0.300 2) 0.150, 3) 150, 4) 300, A plate of thickness t made of material of, refractive index is placed in front of one of, the slits in a double slit experiment. What, should be the minimum thickness t which will, make the intensity at the centre of the fringe, pattern zero?, , 1) 1, 2) 1 , 2, , , 3) 2 1, 4) 1, In Young’s double slit experiment the, separation between the slits is halved and the, distance between the slits and screen is, doubled . The fringe width is, 1) unchanged, 2) halved, 3) doubled, 4) quadrupled, The maximum number of possible interference, maxima for slit separation equal to twice the, wavelength in Young’s double slit experiment, is, 1) infinite 2) five, 3) three, 4) zero, Two identical coherent sources produce a zero, order bright fringe on a screen. If is the band, width, the minimum distance between two, points on either side of the bright fringe where, the intensity is half that of maximum intensity, is, 1) /2, 2) /4, 3) /3, 4) /6, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 15. In Young’s double slit experiment, the 8th, POLARISATION, maximum with wavelength 1 is at a distance 23. The angle of incidence at which reflected light, is totally polarised for reflection from air to, d1 from the central maximum and the 6th, glass (refractive index n) is, maximum with wavelength 2 is at a distance, 1) sin 1 n , 2) sin 1 1/ n , d 2 from central maximum. Then, d1 / d 2 is, equal to, 3) tan 1 1/ n , 4) tan 1 n , 24. A light ray is incident on a transparent medium, 4 2 , 4 1 , 3 2 , 3 1 , 1) 3 2) 3 3) 4 4) 4 , of = 1.732 at the polarising angle. The angle, 1 , 2 , 1 , 2 , of refraction is, DIFFRACTION, 1) 600, 2) 300, 3) 450, 4) 900, 16. First diffraction minima due to a single slit 25. A ray of light in air is incident on a glass plate, diffraction is at 300 for a light of, at polarising angle of incidence. It suffers a, wavelength 6000 A0 . The width of slit is, deviation of 220 on entering glass. The angle, of polarization is, 2) 1.2 106 m, 1) 1 106 cm, 1) 900, 2) 560, 3) 680, 4) Zero, 3) 2 106 cm, 4) 2.4 10 6 m, 26., The, critical, angle, for, total, internal, reflection, 17. In a single slit diffraction, the width of slit is, 0, for, a, substance, is, 45, ., The, polarizing, angle for, 0.5 cm, focal length of lens is 40cm and, 0, wavelength of light is 4890 A0 . The distance, this substance is tan 54 44' 2, of first dark fringe is, 1) 460161 2) 540441 3) 460441 4) 540161, , 5, , 5, 2) 4 10 m, 1) 2 10 m, 27. Unpolarized light of intensity I0 is incident on, a polarizer and the emerging light strikes a, 3) 6 10 5 m, 4) 8 10 5 m, second polarizing filter with its axis at 450 to, 18. Angular width of central maxima is / 2 ., that of the first. The intensity of the emerging, When a slit of width ‘a’ is illuminated by a light, beam, a=, of wavelength 7000 A0 then, I0, I, I, 1) 9 109 m, 2) 8.9 107 m, 2) 0, 3) I0, 4) 0, 1), 2, 4, 3, 3) 9 107 m, 4) 9.8 107 m, 28. The axes of the polariser and analyser are, RESOLVING POWER, inclined to each other at 600 . If the amplitude, 0, of polarised light emergent through analyser, 19. The sun subtends an angle of 1/ 2 on earth., is A. The amplitude of unpolarised light, The image of sun is obtained on the screen, incident on polariser is, with the help of a convex lens of focal length, 100 cm the diameter of the image obtained on, A, the screen will be, 2) A, 3) 2A 4) 2 2A, 1), 2, 1) 18 cm 2) 1 mm, 3) 50 cm 4) 8.73 mm, 29. Unpolarised light of intensity I is incident on, 20. The limit of resolution of microscope, if the, a polarizer and the emerging light strikes a, numerical aperture of microscope is 0.12, and, the wavelength of light used is 600 nm, is, second polarizing filter with its axis at 450 to, that of the first. Determine, 1) 0.3 m 2) 1.2 m 3) 2.5 m 4) 3 m, a) the intensity of the emerging beam and, 21. The least resolvable angle by a telescope using, b) its state of polarization, objective of aperture 5 m is nearly, 0, I, 4000 A , 1) and parallel to second filter, 4, 1, 1, 1, 1, minute 3), sec 4), sec, I, 1) 0 2), 50, 50, 50, 500, 2) and perpendicular to second filter, 4, 22. Wavelength of light used in an optical, I, instrument are 1 4000A0 and 2 5000A0 ,, 3) and parallel to second filter, then ratio of their respective resolving powers, 8, (corresponding to 1 and 2 ) is, I, 4) and perpendicular to second filter, 1) 16 : 25 2) 9 : 1, 3) 4 : 5, 4) 5 : 4, 8, , , , NARAYANAGROUP, , , , 29
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , HUYGEN’S PRINCIPLE, 30. A parallel beam of width ‘a’ is incident on the, surface of glass slab 3 / 2 at an angle ‘i’, and the angle of refraction in glass is ‘r’. The, width of the refracted parallel beam will be, 1) equal to a, 2) less than a, 3) more than a, 4) exactly 2a/3, 31. When a parallel beam of monochromatic light, suffers refraction while going from a rarer, medium into a denser medium, which of the, following are correct?, a) the width of the beam decreases, b) the width of the beam increases, c) the refracted beam makes more angle with the, interface, d) the refracted beam makes less angle with the, interface, 1) a, c true 2) b, d true 3) a, d true 4) b, c true, 32. A parallel beam of light in incident on a liquid, surface such that the wave front makes an, angle 300 with the surface and has a width of, 3 m, the width of the refracted beam is ___, , 8. , 11., , 2), , 3m, , 11, m, 3, , 3), , 4), , 2) 3, 8) 4, 14) 2, 20) 4, 26) 2, 32) 4, , 3) 1, 9) 1, 15) 2, 21) 3, 27) 2, , 4) 1, 10) 2, 16) 2, 22) 4, 28) 4, , 5) 2, 11) 3, 17) 2, 23) 4, 29) 1, , LEVEL - I ( C. W ) - HINTS, 1., , A a12 a22 2a1a2 cos , , 2, , I m ax, , I m in, , 3., , I I1 I 2 2 I1 I 2 cos , , 4., , I I1 I 2 2 I1 I 2 cos , , , , , I1 , , I2, , I1 , , I2, , , , , 2, , 2, , I I 0 I 0 2 I 0 cos , 2 I o 1 cos , 5., 6., 7., 30, , , , I I 0 cos 2 ; I I 0 cos 2 , 2, 4, , 2, I I 0 cos 2 ; , x, , 2, n11 n2 2, , D, d, , 1 t 2n 1 / 2, , , , D, 12. , d, 1, 10. , , 2 , 14. I I 0 cos , d tan x, 2, 15. Position of nth maxima from central maxima is given, n D, by xn , d, 81D, For 8th maxima x8 d, 1, 62 D, and for 6th maxima x6 d, 2, , 16., , 11, m, 3, , 17., , 6) 1, 12) 4, 18) 4, 24) 2, 30) 3, , 18., , LEVEL - I ( C. W ) - KEY, 1) 1, 7) 2, 13) 2, 19) 4, 25) 2, 31) 4, , 9. , , 13. d sin n, , ( a L 3 ), 1) 3 m, , D, , d, , 19., , d1 n11 4 1 , , Now, x6 x8 ; , d 2 n22 3 2 , n, x , sin ; a sin n , a , d b, sin , Y, a sin n a. n ; But = D = f, D, Y, n f, a. n Y , D, a, , w, a sin n ; a sin w 2 2, D f , 1.22 , , 20. X 2 sin where sin is numarical, aperture, 1.22, 1, 21. , 22. R , a, , 0, 23. Tani p ; i p r 90, , I p r 90, 24. Tan I p, 0, 25. i + r = 90 r = i – d, I0, 1, I0, 2, 26. , Tani 27. I , I 2 cos , sin c, 2, 2, A0, A, cos , 28. A1 A0 cos 45 , ; A2 , 2, 2, Wi cos i, I, 2, 29. I 2 0 cos , 30. W cos r, 2, r, Wi cos i, 31. W cos r , i r, r, Wi cos i, 32. 1 sin i 2 sin r , W cos r, r, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , LEVEL - I (H.W), , WAVE OPTICS, 8., , INTERFERENCE, 1., , 2., , 3., , 4., , 5., , 6., , 7., , The displacements of two interfering light, waves, are, y1 2sin t, and, , , , y2 5sin t the resultant amptitude is, 3, , 4) 29 cm, 1) 39 cm 2) 39 cm 3) 7 cm, The intensity ratio of two waves is 9 : 1. If, they produce interference, the ratio of, maximum to minimum intensity will be, 1) 4 : 1 2) 2 : 1, 3) 9 : 1, 4) 3 : 2, Two beams of light having intensities I and 4I, interfere to produce a fringe pattern on a, screen. The phase difference between the, beams is / 2 at point A and at point B. then, the difference between the resultant intensities, at A and B is, 1) 2I, 2) 4I, 3) 5I, 4) 7I, The maximum intensity in Young’s double slit, experiment is I 0 . What will be the intensity of, light in front of one of the slits on a screen, , where path difference is ?, 4, I, 3, I, 1) 0, 2) I 0, 3) I 0, 4) 0, 2, 4, 4, In Young’s double slit experiment, we get 60, fringes in the field of view of monochromatic, light of wavelength 4000A o . If we use, monochromatic light of wavelength 6000A o ,, then the number of fringes obtained in the same, field of view are, 1) 60, 2) 90, 3) 40, 4) 1.5, The separation between successive fringes in, a double slit arrangement is x. If the whole, arrangement is dipped under water, what will, be the new fringe separation? [The wavelength, of light being used is 5000 Å], 1) 1.5 x 2) x, 3) 0.75 x 4) 2x, In the Young’s double slit experiment, a mica, slip of thickness t and refractive index is, introduced in the ray from first source S1. By, how much distance fringes pattern will be, displaced?, (d=distance between the slits and D is the, distance between slits and screen), d, D, 1) ( 1)t, 2) ( 1)t, D, d, d, D, 3), 4) ( 1), ( 1) D, d, , NARAYANAGROUP, , 9., , In young’s double slit experiment, the 10th, maximum of wave length 1 is at a distance of, y1 from the central maximum. When the, wavelength of the source is changed to 2 , 5th, maximum is at a distance of y2 from its central, y1, maximum. Then, y2 is, 21, 22, 1, 2, 1) , 2) , 3) 2, 4) 2, 2, 1, 2, 1, Two coherent monochormatic light sources are, located at two vertices of an equilateral, triangle. If the intensity due to each of the, source independently is 1Wm 2 at the third, vertex. The resultant intensity due to both the, sources at that point (i.e at the third vertex) is, (in Wm 2 ), , 1) zero, , 2), , 2, , 3) 2, , 4) 4, , DIFFRACTION, 10. Light of wavelength 6000A o is incident on a, single slit. The first minimum of the diffraction, pattern is obtained at 4 mm from the centre., The screen is at a distance of 2 m from the slit., The slit width will be, 1) 0.3 mm 2) 0.2 mm 3) 0.15 mm 4) 0.1 mm, 11. A plane wave of wavelength 6250Å is incident, normally on a slit of width 2 102 cm . The width, of the principal maximum on a screen distant, 50 cm will be, 1) 312.5 102 cm, 2) 312.5 104 cm, 3) 312 cm, 4) 312.5 105 cm, 12. The distance between the first and the sixth, minima in the diffraction pattern of a single slit, is 0.5 mm. The screen is 0.5 m away from the, slit. If the wavelength of light used is 5000, Å.Then the slit width will be, 1) 5 mm 2) 2.5 mm 3) 1.25 mm 4) 1.0 mm, , RESOLVING POWER, 13. The diameter of an objective of a telescope,, which can just resolve two stars situated at, angular displacement of 10-4 degee, should be, ( 5000 A0 ), 1) 35 mm 2) 35 cm 3) 35 m, 4) 3.5 cm, 14. A telescope is used to resolve two stars, separated by 4.6 10 6 rad. If the wavelength, of light used is 5460 A0 , what should be the, aperture of the objective of the telescope?, 1) 0.448 m 2) 0.1448 m 3) 1.1448 m 4) 0.011 m, 31
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, 15. Two point sources distant 0.1 meter away, viewed by a telescope. The objective is, covered by a screen having a hole of 1 mm, width. If the wavelength of the light used is, 6500 A0 , then the maximum distance at which, the two sources are seen just resolved, will be, 4., nearly, 1) 125.0 m 2) 164 m 3) 131 m 4) 144 m, 16., , 17., , 18., , 19., , 20., , 1., 2, 3., , I 4 I 2 1 4 I cos , 5I 4I I I1 I2 5 I I 4 I, Corresponding phase difference will be, 2 , 2 , ( x ) , POLARISATION, , 4 2, Two polaroids are kept crossed to each other., , 2 , Now one of them is rotated through an angle, ; I I 0 cos , or, 2 4, 2, of 450 . The percentage of incident light now, transmitted through the system is, I, I 0 cos 2 0, 1) 15% 2) 25%, 3) 50%, 4) 60%, 4 2, The amplitude of polarised light transmitted, through a polariser is A. The amplitude of 5. As, x n11 n2 2 n11 n2 2, unpolarised light incident on it is, n 60 4000, n2 1 1 , 40, , 1) A / 2 2) A / 2 3) 2A, 4) 2A, 2, 6000, Unpolarised lgiht of intensity 32 W/m2 passes 6. When the arrangement is dipped in water,, through a polariser and analyser which are at, x, , 3, , x 0.75 x, an angle of 300 with respect to each other. The, 4/3 4, intensity of the light coming from analyser is, 7. For a path difference ( 1)t , the shift is, 2) 12 W / m 2, 1) 16 3 W / m 2, D, x ( 1)t, 3) 16 W / m 2, 4) 14 W / m 2, d, The critical angle of a transparent crystal is, 2 D , 1 D , 600. Then its polarizing angle is, 8. y2 5 , ; y1 10 , , , d , d , 1 2 , 9. at the third vertex path diff = 0, hence intensity is, 1) tan , 2) sin 1 ( 2), , max., 3, 10. From, a sin n ,, 1 1 , , 1, 4) cot ( 2), 3) cos , x, , n D 1 6000 1010 2, a n or a , 2, , D, x, 4 103, When an unpolarised light of intensity I 0 is, 4, a 3 10 0.3 mm, , incident on a polarising sheet, the intensity of, 11. Here, 6250 Å = 6520 x 10-10 m, the light which does not get transmitted is, a 2 102 cm 2 104 m, 1) 1/ 2 I 0 2) 1/ 4 I 0, 3) zero, 4) I 0, D 50 cm 0.5 m, LEVEL - I ( H. W )-KEY, 2 D, 1) 2, 2) 1 3) 2 4) 1 5) 3 6) 3, Width of central maxima , 7) 2, 8) 1 9) 4 10) 1 11) 1 12) 2, a, 13) 2, 14) 2 15) 1 16) 2 17) 4 18) 2, 10, 2 6250 10 0.5, 19) 1, 20) 1, , 312.5 10 3 cm, 4, 2, , 10, LEVEL - I ( H. W )-HINTS, 5 D, 12. Distance between first and sixth minima x , A a12 a22 2a1a2 cos , a, 2, 1.22, , I max I 1 I 2 , 13. , , 2, a, I min, I1 I 2 , 14. The aperture (a) of the telescope is given as, 1.22, Here, I1 I , I2 4I ,1 / 2,2 , a, , Resultant intensity I I I 2 I I cos , 1, , 32, , I 4 I 2 I 4 I cos / 2 5 I, Resultant intensity I2 I1 I 2 2 I1 I 2 cos 2, , 1, , 2, , 1 2, , 1, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 15. , , WAVE OPTICS, 5., , 1.22, a, 2, , I, I0 I , I0 , 0 2, , (cos, 45, ), , 0, I, 16., 2, , , 2 2, 4, , I 25% of I 0, A0, I, 2, A, 17., 18. I 0 cos , 2, 2, 1, 19. , and tan 20. I I 0 cos 2 , sin c, I0, Intensity of polarised light =, 2, I0 I0, Intensity of untransmitted light I 0 , 2 2, , LEVEL -II (C.W), INTERFERENCE, 1., , 2., , In Young’s double slit experiment the intensity, of light at a point on the screen where the path, difference is is K. The intensity of light at, , a point where the path difference is, [ is, 3, the wavelength of light used] is, 1) K/4, 2) K/3, 3) K/2, 4) K, In Young’s double slit experiment, Let be, the fringe width and I0 be the intensity at the, central bright fringe. At a distance ' x ' from, the central bright fringe, the intensity will be, 1), , 3., , 4., , x, I 0 cos , , x, I 0 cos 2 , , , , 2), , x, I 0 cos 2 , , I0, x , cos 2 , , 4, , , 3), 4), In Young’s double slit experiment the fringe, pattern is observed on a screen placed at a, distance D. The slits are illuminated by light, of wavelength . The distance from the central, point where the intensity falls to half the, maximum is, D, D, D, D, 1), 2), 3), 4), 3d, 2d, d, 4d, In a double slit experiment, the slit separation, is 0.20 cm and the slit to screen distance is, 100 cm. The positions of the first three minima,, if wavelength of the source is 500 nm is, 1) 0.125cm, 0.375cm, 0.625cm, 2) 0.025 cm , 0.075 cm , 0.125 cm, 3) 12.5cm, 37.5cm, 62.5cm, 4) 1.25cm, 3.75cm, 6.25cm, , NARAYANAGROUP, , In a Young’s double slit experiment, the fringes, are displaced by a distance x when a glass plate, of refractive index 1.5 is introduced in the path, of one of the beams. Then this plate is replaced, by another plate of the same thickness, the, shift of fringes is 3 / 2 x . The refractive index, of the second plate is, 1) 2.25 2) 2.0, 3) 1.75, 4) 1.25, 6. A double slit experiment is performed with light, of wavelength 500 nm. A thin film of thickness, 2 m and refractive index 1.5 is introduced in, the path of the upper beam. The location of, the central maximum will, 1) remain unshifted, 2) shift downward by nearly two fringes, 3) shift upward by nearly two fringes, 4) shift downward by 10 fringes, 7. In Young’s double slit experiment, an, interference pattern is obtained on a screen, by a light of wavelength 6000 A0 coming from, the coherent sources S 1 and S 2 . At certain, point p on the screen third dark fringe is, formed. Then the path difference S 1 p S 2 p, in micron is, 1) 0.75 2) 1.5, 3) 3.0, 4) 4.5, 8. In double slit experiment fringes are obtained, using light of wavelength 4800 A0 One slit is, covered with a thin glass film of refractive, index. 1.4 and another slit is covered by a film, of same thickness but refractive index 1.7. By, doing so, the central fringe is shifted to fifth, bright fringe in the original pattern. The, thickness of glass film is, 1) 2 x 10 -3 mm, 2) 4 x 10 -3 mm, 3) 6 x 10 -3 mm, 4) 8 x 10 -3 mm, 9. In Young’s double slit experiment, 5th dark, fringe is obtained at a point. If a thin, transparent film is placed in the path of one of, waves, then 7th bright fringe is obtained at the, same point. The thickness of the film in terms, of wavelength and refractive index will, be, 1.5, 2.5, 1) 1 2) 1.5 1 3) 2.5 1 4) 1, 10. The Young’s double slit experiment is, performed with blue light and green light of, wavelengths 4360, and 5460, A0, A0 respectively. If y is the distance of 4th, maxima from the central one, then, y, , 5460, , b, 1) yb y g 2) yb y g 3) yb yg 4) y 4360, g, , 33
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , 11. In double slit experiment, the distance between 17. A beam of light of wavelength 600 nm from a, two slits is 0.6 mm and these are illuminated, distant source falls ona single slit 1.00 mm wide, and the resulting diffraction pattern is, with light of wavelength 4800 A0 . The angular, observed on a screen 2m away. Then distance, width of dark fringe on the screen at a distance, between the first dark fringes on either side, 120 cm from slits will be, of the central fringe is, 2) 6 104 radian, 1) 8 104 radian, 1) 1.2 mm 2) 2.4 mm 3) 3.6 mm 4) 2.4 cm, 4, , , 4, RESOLVING POWER, 3) 4 10 radian, 4) 16 10 radian, 12. Fig shows a double slit experiment, P and Q 18. A person wants to see two pillars distant 11, km, separately. The distance between the, are the two coherent sources. The path lengths, pillars must be approximately, PY and QY are n and (n + 4) respectively, where n is whole number and is wavelength., 1) 3m, 2) 1m, 3) 0.25 m 4) 0.5 m, Taking the central bright fringe as zero, what 19. Two point white dots are 1 mm apart on a black, is formed at Y?, paper. They are viewed by eye of pupil, diameter 3mm. Approximately, what is the, Y, n, maximum distance at which these dots can be, resolved by the eye? [Take wavelength of light, P, = 500 nm], [AIEEE-2005], 1) 6m, 2) 3 m, 3) 5 m, 4) 1 m, (n + 4), , POLARISATION, , Q, , 1) First Bright, 2) First Dark, 3) Fourth Bright, 4) Second Dark, 13. White light is used to illuminate two slits in, Young’s double slit experiment. Separation, between the slits is b and the screen is at a, distance d (>> b) from the slits. Then, wavelengths missing at a point on the screen, directly infront of one of the slits are, , b2 b2, b2 b 2, ,, ,, 1), 2), d 3d, d 4d, , b2 b2, b2 b2, ,, ,, 3), 4), 2d 3d, 2d 4d, , DIFFRACTION, st, , 20. A horizontal beam of vertically polarized light, of intensity 43 W/m2 is sent through two, polarizing sheets. The polarizing direction of, the first is 600 to the vertical, and that of the, second is horizontal. The intensity of the light, transmitted by the pair of sheets is (nearly), 1) 8.1 W/m2, 2) 7.3 W/m2, 3) 6.4 W/m2, 4) 3.8 W/m2, 21. Unpolarised light of intensity 32Wm 2 passes, through three polarisers such that the, transmission axis of the last polariser is, crossed with first. If the intensity of the, emerging light is 3Wm 2 , the angle between, the axes of the first two polarisers is, 1) 450, 2) 600 3) 300, 4) zero, 22. Two polaroids are oriented with their, transmision axes making angle of 300 with, each other. The fraction of incident un, polarised light is transmitted, 1) 37% 2) 37.5% 3) 3.36% 4) 36.5%, , 14. The l diffraction mininum due to single slit, diffraction is , for a light of wave length, 5000A0 . If the width of the slit is 1 104 cm, then the value of , 1) 300, 2) 450, 3) 600, 4) 150, 15. Light of wavelength 5000 1010 m is incident, normally on a slit. The first minimum of the, diffraction pattern is observed to lie at a, distance of 5mm from the central maximum on, a screen placed at a distance of 3m from the 23. The polaroids P1 , P2 & P3 are arranged, slit. Then the width of the slit is, coaxially. the angle between P1 and P2 is 370 ., 1) 3 cm 2) 0.3 cm 3) 0.03 cm 4) 0.01 cm, The angle between P2 and P3 is, if intensity of, 16. A small aperture is illuminated with a parallel, emerging light is one quarterth of intensity of, beam of 628nm . The emergent beam has, unpolarized light, an angular divergence of 20 . The size of the, 1) cos 1 54 , 2) cos 1 45 , aperture is, 1, 1, 1) 9 m 2) 18 m, 3) 27 m 4) 36 m, 3) cos 5 4 2, 4) cos 4 5 2, , , , 34, , , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 24. A ray of light is going from air to glass such, that the reflected light is found to be completely, plane polarized. Also the angle of refraction, inside the glass is found exactly equal to the, angle of deviation suffered by the ray. The, refractive index of the glass is, 2) 2, 3) 3, 4) 4/3, 1) 1.5, 25. A plane polarized beam of intensity I is, incident on a polariser with the electric vector, inclined at 30o to the optic axis of the polariser, passes through an analyzer whose optic axis, is inclined at 30o to that of polariser. Intensity, of light coming out of the analyzer is, E, P, A, , 30, , 0, , 0, , 30, , 1) (9/16)I 2) (3/4)I 3) (1/4)I 4), , , , , , 3/2 I, , LEVEL - II (C. W ) - KEY, 1) 1, 7) 2, 13) 1, 19) 3, 25) 1, , 2) 3, 8) 4, 14) 1, 20) 1, , 3) 4, 9) 4, 15) 3, 21) 3, , 4) 1, 10) 3, 16) 4, 22) 2, , 5) 3, 11) 1, 17) 2, 23) 4, , 6) 3, 12) 3, 18) 1, 24) 3, , Y, n D, 15. a sin n ; a n a , D, Y, n, rad, 16. a sin n ; , a 180, D, 17. Y , ; w = 2Y, a, 0, 1 , 18. Resolving power of eye = , 60 , 1, , d, , , 60 180 11000, 1.22 y, I, 2, , 19. sin , 20. I 2 0 cos , d, D, 2, I0, I, 2, 2, 21. I1 ; I 2 I1 cos 2 ; I 3 0 cos sin , 2, 2, I2 3, I0, 22. I1 ; I 2 I1 cos 2 ; Fraction I 8, 2, 0, I, 1, 2, 2, 23. I .cos 1 cos 2, 2, 24. i 90 r , r d , d i r , tan i, 25. Use malus law, , LEVEL - II (H. W), , LEVEL - II (C. W ) - HINTS, 1., , INTERFERENCE, , K I 0 cos2 (1), , x , 2 x, I I 0 cos 2 , 2 , 3, 3 , 2, 2, , x, 2. I I 0 cos ;, 2, , n D, 2, 3. I I 0 cos, ; y, 2, d, D, D, y x. =, d, 4 d, n D, 4. y , ; n=1,2,3,4 … 5. x 1 t, d, , 6. 1 t n, 7. x 2n 1, 2, , 5 2 1 t, 8. n 2 1 t ;, , D, 9. x 2n 1, 10. n11 n2 2, 2d, , 11. , 12. x n, d, D, 13. x 2n 1, 14. a sin n, 2d, NARAYANAGROUP, , 1., , In Young’s double slit experiment the intensity, of light at a point on the screen where the path, difference is K. The intensity of light at a, point where the path difference is, , 2., , , [ is the, 6, , wavelength of light used] is, 1) K/4, 2) K/3, 3) 3K/4, 4) K, In a Young’s double slit experiment, D equals, the distance of screen and d is the separation, between the slits. The distance of the nearest, point to the central maximum where the, intensity is same as that due to a single slit is, equla to, D, D, D, 2D, 2), 3), 4), d, 2d, 3d, d, With two slits spaced 0.2 mm apart and a screen, at a distance of 1 m, the third bright fringe is, found to be at 7.5 mm from the central fringe., The wavelength of light used is, 1) 400 nm 2) 500 nm 3) 550 nm 4) 600 nm, , 1), 3., , 35
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WAVE OPTICS, 4., , The central fringe of the interference pattern, produced by the light of wavelength 6000 A is, found to shift to the position of 4th dark fringe, after a glass sheet of refractive index 1.5 is, introduced. The thickness of glass sheet would, be, 1) 4.8 m 2) 4.2 m 3) 5.5 m 4) 3.0 m, 5. In Young’s double slit inteference experiment, the wavelength of light used is 6000 Å . If the, path difference between waves reaching a point, P on the screen is 1.5 microns, then at that point, P, 1) Second bright band occurs, 2) Second dark band occur, 3) Third dark band occur, 4) Third bright band occur, 6. When a mica plate of thickness 0.1 mm is, introduced in one of the interfering beams, the, central fringe is displaced by a distance equal, to 10 fringes. If the wavelength of the light is, 6000 Å, the refractive index of the mica is, 1) 1.06 2) 1.6, 3) 2.4, 4) 1.3, 7. In Young’s experiment inteference bands are, produced on the screen placed at 1.5 m from, the two slits 0.15 mm apart and illuminated by, light of wavelength 6000 Å . If the screen is, now taken away from the slit by 50 cm the, change in the fringe width will be, 2) 2 103 m, 1) 2 104 m, 3) 6 10 3 m, 4) 7 103 m, 8. When a thin transparent plate of Refractive, Index 1.5 is introduced in one of the, interfearing beams produces 20 fringes shift., If it is replaced by another thin plate of half, that thickness and of refractive index 1.7, the, number of fringes that undergo displacement, is, 1) 23, 2) 14, 3) 28, 4) 7, 9. In young’s double slit experiment one of the, slits is wider than other, so that amplitude of, the light from one slit is double of that from, other slit. If I m be the maximum intensity, the, resultant intensity I, when they interfere at, phase difference is given by, Im , I, 2 , 1) m (4 5 cos ), 2), 1 2 cos , 3 , 2, 9, Im , Im , 2 , 2 , 3), 1 4 cos 4), 1 8cos , 5 , 2, 9 , 2, 10. In young’s doulbe slit experiment, the two slits, act as coherent sources of waves of equal, amplitude A and wavelength . In another, experiemnt with the same arrangement the two, 36, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, slits are made to act as incoherent sources of, waves of same amplitude and wavelength. If, the intensity at the middle point of the screen, in the first case is I1 and in the second case, I1, I 2 , then the ratio I is, (AIEEE 2011), 1), 2, 1) 4, 2) 2, 3) 1, 4) 0.5, 11. A mixture of light, consisting of wavelength, 590 nm and an uknown wavelength, illuminates, Young’s double slit and gives rise to two, overlapping interference patterns on the, screen. The central maximum of both light, coincide. Further, it is observed that the third, bright fringe of known light coincides with the, 4th bright fringe of the unknown light. The, wavelength of the unknown light is (AIE 2009), 1)393.4 nm, 2)885.0 nm, 3)442.5 nm, 4)776.8 nm, 12. In Young’s experiment using monochromatic, light, the fringe pattern shifts by a certain, distance on the screen when a mica sheet of, refractive index 1.6 and thickness 2 micron is, introduced in the path of one of the interfering, waves. The mica sheet is then removed and, the distance between the slits and the screen, is doubled. It is found that the distance, between successive maxima now is the same, as the observed fringe shift upon the, introduction of the mica sheet. The wavelength, of light is, 1) 5762 Å 2) 5825 Å 3) 6000 Å 4) 6500 Å, , DIFFRACTION, 13. Plane microwaves are incident on a long slit, having a width of 5.0 cm. The wavelength of, microwaves if the first diffraction minimum is, formed at 300 is, 1) 2.5 cm 2) 5 cm, 3) 7.5 cm 4) 10 cm, 14. A screen is placed 50 cm from a single slit,, which is illuminated with 6000 A0 light. if, distance between the first and third minima in, the diffraction pattern is 3.0 mm, then the width, of the slit is, 1) 0.1 mm 2) 0.2 mm 3) 0.4 mm 4) 0.8 mm, 15. A slit of width ‘d’ is placed infront of a lens of, focal length 0.5 m and is illuminated normally, with light of wavelength 5.89 107 m. The first, diffraction minima on either side of the central, diffraction maximum are separated by, 2 103 m . The width of the slit is, 2) 2.94 10 4 m, 1) 1.47 10 4 m, 3) 1.47 10 7 m, 4) 2.92 10 7 m, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , POLARISATION, , , , ( 1)t (2n 1) 5.Path difference = (2n 1), 16. Un polarised light passes through a polariser 4., 2, 2, and analyser which are at an angle of 450 with, , D, respect to each other. The intensity of 6. Shift = ( 1)t, 7. , , d, polarised light coming from analyser is 5W /, m2. The intensity of unpolarised light incident, , 8. Shift = ( 1)t, on polariser is, , 1) 5 3 W / m 2, 2) 10 W / m 2, 9. I I1 I 2 2 I1 I 2 cos I1 4 I 2 , I max 9 I 2, 3, 10. For coherent sources, 3) 20 W / m 2, 4) 5 W / m 2, 4, I1 4 I 0 cos 2 / 2 4 I 0 For incoherent sources, 17. A beam of ordinary light is incident on a system, I1, of four polaroids which are arranged in, 2, I, , I, , I, , 2, I, ;, , succession such that each polaroid is turned, 2, 0, 0, 0, I, 2, through 300 with respect to the preceding one., The percentage of the incident intensity that 11. 31 42, emerges out from the system is approximately, 3, 3, 1770, 1) 56% 2) 6.25% 3) 21%, 4) 14%, 442.5 nm, 2 1 590 ; , 4, 4, 4, 18. Two polaroid sheets are placed one over the, other with their axes inclined to each other at, D, .... (1), an angle . If only 12.5% of the intensity of 12. y d ( 1)t, the light incident on the first sheet emerges, When the distance between the plane of slits and, out from the second sheet, the value of is, screen is changed from D to 2D, then, 1) 300, 2) 600, 3) 450, 4) 900, 2D, D, 2 D ( ), 19. An unpolarised light is incident on a plate of, , ;, ( 1)t , ... (2), d, d, d, refractive index 3 and the reflected light is, 1, found to be completely plane polarised. The, ( 1)t, angles of incidence and refraction are, 2, respectively, 13. a sin n, 1) 600, 300, 2) 300, 600, 14. Position of first minima on a single slit diffraction, pattern is given by, 3 0, 3) Sin 1 1 , 450, 4) Tan1 , ,30, d sin n, 3, 2 , y, For small value of , sin , LEVEL - II ( H. W ) -KEY, D, 1) 3, 2) 3 3) 2 4) 2 5) 3 6) 1, y.a, n D, 7) 2, 8) 2 9) 4 10) 2 11) 3 12) 3, n or y , , D, a, 13) 1, 14) 2 15) 2 16) 3 17) 3 18) 2, 19) 1, Distance between third order minima and first, order minima will be, LEVEL - II ( H. W )- HINTS, (3 1)( D) 2 D, y y3 y1 , , 1. K I 0 cos2 (1), a, a, , , , 2 D 2 f , I I 0 cos 2 2 , w 2y , , 15., 6, , a, a, , , 2, I0, I 0 4 I 0 cos 2, 2. I 4 I 0 cos, I, , cos 2 , 16., 2, 2, 2, 2, 2, yd, n 1, I, , x ; But x , 2, , , 17. I o cos where n= number of polaroids, 3, , D, 2, yd , D, , , y, I0, 2, D 3, 3d, 18. I cos 19., 2, n D, y, , 3., tan i p ; r 900 i p, d, NARAYANAGROUP, , 37
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , source, h is Planck’s constant, e is charge on, electron and m is mass of electron, then fringe, width is given as, , LEVEL - III, INTERFERENCE OF LIGHT, 1., , hD, , A ray of light of intensity I is incident on a, parallel glass slab at a point A as shown. It 7., undergoes partial reflection and refraction. At, each reflection 25% of incident energy is, reflected. The rays AB and A B undergo, interference. The ratio Imax / Imin is, , hd, , 2hD, , 2hd, , 1), 2), 3), 4), 2meV d, meV d, 2meV D, meV D, Two identical narrow slits S1 and S2 are, illuminated by light of wavelength from a, point source P. If, as shown in the diagram, above the light is then allowed to fall on a, screen, and if n is a positive integer, the, condition for destructive interference at Q is, that, , P, , S1, , l1, l2, , S2, , l3, Q, , l4, , 1) l1 l2 2n 1 / 2, 2., , 1) 4 : 1 2) 8 : 1, 3) 7 :1, 4) 49 : 1, Two coherent sources of intensity ratio , Imax Imin, interfere, then I I, is, max, min, , 4., , 5., , 6., , 38, , 3) l3 l3 l2 l4 n, 8., , 2, 2 , , 2 , 2), 3), 4) 1 , 1 , 1 , 1 , Monochromatic green light of wavelength 550, nm illuminates two parallel narrow slits, 7.7 m apart. The angular deviation of third, order (for m = 3) bright fringe a) in radian and, b) in degree, 1) 21.6, 12.40, 2) 0.216, 1.240, 0, 3) 0.216, 12.4, 4) 216, 1.240, A source emitting wavelength 480 nm and 600, nm is used in YDSE. The separation between, the slits is 0.25 mm. the interference is, observed 1.5 m away from the slits. The linear, separation between first maxima of two, 9., wavelengths is, 1) 0.72 mm 2) 0.72 cm 3) 7.2 cm 4) 7.2 mm, In the Young’s double slit experiment,, maximum number of bright bands observed, (inclusive of the central bright band) is found, to be 11. If is the wavelength of the, monochromatic light used, the distance, between the slits is, 1) 5, 2) 6, 3) 10, 4) 11, 10., In a double slit experiment, interference is, obtained from electron waves produced in an, electron gun supplied with voltage V. If is, wavelength of the beam, D is the distance of, screen, d is the spacing between coherent, 1), , 3., , 2) l3 l4 2n 1 / 2, 4) l1 l3 l2 l4 2n 1 / 2, Fig., here shows P and Q as two equally intense, coherent sources emitting radiations of, wavelength 20m. The separation PQ is 5m, and, phase of P is ahead of the phase of Q by 90o., A, B and C are three distant points of, observation equidistant from the mid-point of, PQ. The intensity of radiations of A, B, C will, bear the ratio, B, , C, , P, , Q, , A, , 1) 0 : 1 : 4 2) 4 : 1 : 0 3) 0 : 1 : 2 4) 2 : 1 : 0, Two coherent sources separated by distance, d are radiating in phase having wavelength ., A detector moves in a big circle around the, two sources in the plane of the two sources., The angular position of n = 4 interference, maxima is given as, n, , 4, , d, , , , 1) sin-1 d 2) cos-1 d 3) tan-1 4, 4) cos-1 4d, In a double slit experiment, the separation, between the slits is d and distance of the, screen from slits is D. If the wavelength of, light used is and I is the intensity of central, bright fringe, then intensity at distance x from, central maximum is, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 15. What happens to the fringe pattern when the, Young’s double slit experiment is performed, 1) I cos, in water instead of air?, 1) Shrinks, 2) Disappears, 3) I cos2, 3) Unchanged, 4) Enlarged, 11. The polaroids are placed in the path of 16. Two periodic waves of intensities I1 and I2 pass, through a region at the same time in the same, unpolarized beam of intensity I 0 such that no, direction. The sum of the maximum and, light is emitted from the second polaroid. If a, minimum intensities is:, third polaroid whose polarization axis makes, 2, an angle with the polarization axis of first, 1) I1 I 2, 2) 2 I1 I 2 , polaroid, is placed between these polariods, 2, then the intensity of light emerging from the, 4), I, , I, 3), I, , I, 1, 2, 1, 2, last polaroid will be, 17., What, is, the minimum thickness of a soap, I0 2, I0 2, 2) 4 sin 2, 1) 8 sin 2, bubble needed for constructive interference in, , , reflected light if the light incident on the film, I0 2, has wavelengths 900 nm? Assume the, 3) cos , 4) I 0 cos 4 , 2, , refractive index for the film is 1.5, 12. Two coherent point sources S1 and S 2, 1) 100 nm 2) 150 nm 3) 200 nm 4) 250 nm, vibrating in phase emit light of wavelength . 18. Two identical coherent sources are placed on, a diameter of a circle of radius R at separation, The separation between the sources is 2 ., x (<<R) symmetrically about the centre of the, The smallest distance from s2 on a line passing, circle. The sources emit identical wavelength, through S 2 and perpendicular to s1s2 , where, each. The number of points on the circle, a minimum of intensity occurs is, with maximum intensity is ( x 5 ), 7, 15, , 3, 1) 20, 2) 22, 3) 24, 4) 26, 1), 2), 3), 4), 19. In a YDSE shown in Fig a parallel beam of light, 12, 4, 2, 4, is incident on the slits from a medium of, 13. In Young’s double slit experiment S1 and S2 are, refractive index n1. The wavelength of light in, two slits. Films of thickness t1 and t2 and, this medium is 1 . A transparent slab of, refractive indices 1 and 2 are placed in front, thickness ‘t’ and refractive index n3 is put in, of S1 and S 2 respectively. If 1t1 2t2 , then, front of one slit. The medium between the, the central maximum will, screen and the plane of the slits is n2. The phase, difference between the light waves reaching, point “O”. (symmetrical, relative to the slits), is, 2, , 2 xd , , , D , , , , xd, , , , , D , , xd , 2) I sin 2 D , , , , xd, , , 4) I sin2 D , 2, , 2, , , , , , , , , , S1, , t1, , S2, , t2, , 1) Not shift, 2) Shift towards S2 irrespective of amounts of t1 and t2, 3) Shift towards S2 irrespective of amounts of t1 and t2, 4) Shift towards S1 if t2 t1 and towards S2 if t2 t1, 14. A monochromatic beam of light is used for the, formation of fringes on a screen by illuminating, the two slits in the Young’s double slit, interference experiment. When a thin film of, mica is interposed in the path of one of the, interfering beams, 1) the fringe-width increases, 2) the fringe-width decreases, 3) the fringe pattern disappears, 4) fringe-width remains the same but the pattern, shifts, NARAYANAGROUP, , 2, 1) n (n3 n2 )t, 1 1, , 2, 2) (n3 n2 )t, 1, , 2 n1 n3 , 3) n n 1 t, 2 1 2, , , 2 n1, 4) n3 n2 t, 1, 39
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , 20. In Young’s double slit experiment, the 10th, shows that the two beams have same intensity., bright fringe is at a distance x from the central, The ratio of intensity of the two beams I A & I B, fringe. Then, 1) 1: 3, 2) 3 :1, 3) 3 :1, 4) 1: 3, a) the 10th dark fringe is at a distance of, 27. An analyser is inclined to a polariser at an angle, 19 x / 20 from the central fringe, of 300 . The intensity of light emerging from, b) the 10th dark fringe is at a distance of, 21x / 20 from the central fringe, 1, the analyser is th of that is incident on the, c) the 5th dark fringe is at a distance of x/2, n, from the central fringe., polariser. Then n is equal to, d) the 5th dark fringe is at a distance of 9 x / 20, 1) 4, 2) 4/3, 3) 8/3, 4) 1/4, from the central fringe., 28. When a beam of light wavelength is incident, 1) a,b,c only, 2) b,c,d only, on the surface of a liquid at an angle , the, 3) a,d only, 4) a,b,c,d only, reflected ray in completely polarized. The, DIFFRACTION, wavelength of light in the liquid medium is, , , , 21. If I 0 is the intensity of the principle maximum, 3), 4), 1) Tan 2), in the single slit diffraction pattern then, with, Tan, Cos, Sin, doubling the slit width, the intensity becomes, LEVEL - III - KEY, 1, I, 1) I 0, 2) I 0 / 2, 3), 4) 4I 0, 1) 4 2) 2 3) 3 4) 1 5) 1 6) 1 7) 4, 0, 2, 8) 4 9) 2 10) 3 11) 1 12) 1 13) 4 14) 4, 22. Light of wavelength 6000 A 0 from a distant, source falls on a slit 0.5mm wide. The distance, 15) 1 16) 2 17) 2 18) 1 19) 1 20) 3 21) 1, between two dark bands on each side of the, 22) 4 23) 4 24) 2 25) 1 26) 1 27) 3 28) 2, central bright band of the diffraction pattern, observed on a screen placed at a distance 2m, LEVEL - III - HINTS, from the slit is, 2, I max I1 I 2 , I, 9I, 1) 1.2nm 2) 2.4nm 3) 3.6nm 4) 4.8mm, , ; I1 : I 2 , 23. The ratio of radii of Fresnel’s fourth to ninth 1. I min I1 I 2 , 4, 64, , , zone is, 2, I1, 2, 1) 1 : 4 2) 4 : 0, 3) 9 : 4, 4) 2 : 3, ; I max I1 I 2, 2., I, , I, , I, min, 1, 2, I2, 24. A parallel beam of wavelength 4500 A0, passes through a long slit of width 2 104 m ., , dy, The angular divergence for which most of the 3. d 4. D n 5. d sin n, light is diffracted is (in 10 5 radian), h, D , , , 6., ,, 2, 5, 3, , 2mVe, d, 1), 2), 3), 4), 3, 4, 4, 3, 2, 2, , , , , , , x, I, , I, cos, , , 8., ,, 0, POLARISATION, , 2, 25. A polariser and an analyser are oriented so, 2, 10. I I 0 cos, that the maximum amount of lights is 9. d sin n, 2, transmitted. Fraction of its maximum value is, I, , the intensity of the transmitted light reduced, I 0 cos 2 , x 2n 1, 11., 12., when the analyzer is rotated through (intensity, 2, 2, 0, 0, 0, of incident light = Io ) a) 30 b) 45 c) 60, t, t, 13. Shift = 1, 14. Shift = 1, 1) 0.375 I0, 0.25 I0, 0.125 I0, , , 2) 0.25 I0, 0.375 I0, 0.125 I0, , 2, 2, 3) 0.125 I0, 0.25 I0, 0.0375 I0, 15. , 16., I1 I 2 I1 I 2, , 4) 0.125 I0, 0.375 I0, 0.25 I0, 26. A polaroid examines two adjacent plane, d, polarised beams A and B whose planes of 17. 2 t cos r (2n 1) 2 for minimum thickness, polarisation are mutually perpendicular. In the, , first position of the analyser, beam B shows, t, 150nm, 4 cos r, zero intensity. From this position a rotation 300, , , , , , , , 40, , , , , , , , , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 18. Path diff at p is ; 2( x / 2 cos ) x cos , For intensity to be maximum, x n ; x cos n, n, 1 ; x 5 ; n = 1,2,3,4,5, x, Therefore in all for quadrants these can be, 20 maximum, 19. Optical path difference between the waves, n3 n2 t, Phase difference 2 (n3 n2 )t ;, , 3., , 2 (n3 n2 )t, n1 , , D , 20. ybright yn n , , d , D, ydark yn 2n 1, ; Where n = 1,2,3,4, 2d, 2, , sin , 21. I I 0 , Where is path difference for, , principal maxima, =0 Hence, intensity will remain 4., same (=I0) Increasing the width of the slit will make, the central peak more narrow., 2D, 22. Required distance =, d, 23. ra : rb a : b, n, rad, 24. a sin n ; , a 180, 5., I A cos B, I, 2, 25. I 0 cos , 26. I cos , 2, B, A, I0, 2, 27. I cos , 28. tan i, 11 2 2, 2, , LEVEL - IV, 1., , 2., , Huygen’s principle of secondary wavelets can, 6., be used to, a) deduce the laws of reflection of light, b) deduce the laws of refraction of light, c) explain the transverse nature of light waves, d) predict the location of a wavefront as time, passes, 1) a, b only 2) a, c only 3) a, b, d only 4) b, c only, Following statements which are true for light, waves but not for sound waves are/is, (I) The speed of waves is greater in vacuum, than in a medium, (II) Waves of different frequencies travel with, different speeds in a medium, , NARAYANAGROUP, , (III) Waves travel with different speeds in, different media., 1) (I) only, 2) (I) and (III), 3) (II) and (III), 4) (I), (II) and (III), If white light is used in Young’s double-slit, experiment., a) bright white fringe is formed at the centre, of the screen, b) fringes the different colours are observed, on both sides of central fringe clearly only in, the first order., c) the first order violet fringe’s are closer to, the centre of the screen than the first order, red fringes, d) The first order red fringes are closer to the, centre of the screen than the first order violet, fringes, 1) Only a and d are true 2) Only a and b are true, 3) Only a,b and c are true 4) All are true, In a double slit experiment, instead of taking, slits of equal widths, one slit is made twice as, wide as the order. Then in the interference, pattern, the intensity., a) of maxima will increases, b) of maxima will decrease, c) of minima will increase, d) of minima will decrease, 1) a,b only 2) b, c only 3) a, c only 4) c, b only, In Young’s double slit experiment for, producing interference pattern, the fringe, width depends on, i) wave length, ii) distance between the two slits, iii) distance between the screen and the slits, iv) distance between source and the slits, 1) i only 2)i, ii only 3)i, ii and iii 4) i, ii and iv, Both in interference and diffraction, phenomena, alternate dark and bright fringes, are obtained on screen, i) generally fringe width is same in interference, and not same in diffraction, ii) the central fringe in interference has, maximum brightness and the intensity, gradually decreases on either side, iii) in interference the intensity of all bright, fringes in same, iv) both the phenomena are produced from, same coherent sources, 1) i only 2) i and ii 3) i,ii and iv 4) i, ii and iii, 41
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, 7., , When light is incident on a glass block at, 1) d sin i ndB, 2) 2d cos i ndB, polarizing angle, 3) 2d sin i ndB, 4) d cos i ndB, a) reflected ray is plane polarized, b) reflected and refracted rays are 11. In an experiment, electrons are made to pass, through a narrow slit of width ‘d’ comparable, perpendicular, to their de-Broglie wavelength. They are, c) reflected and refracted rays are partially, detected on a screen at a distance ‘D’ from, polarized, the slit (see figure), d) refracted ray is partially polarised, 1) a, c and d are correct 2) a, b and d are correct, 3) b, c and d are correct 4) a, b and c are correct, , 8., , Consider the following statements A and B, identify the correct answer, (A) Polarized light can be used to study the, helical structure of nucleic acids, (B) Optic axis is a direction and not any, particular line in the crystal., 1) A and B are correct, 2) A and B are wrong, 3) A is correct and B is wrong, 4) A is wrong and B is correct, , COMPREHENSION -I, Wave property of electrons implies that they, will show diffraction effects. Davission and, Germer demonstrated this by diffracting, electrons from crystals. The law governing the, diffraction from a crystal is obtained by, requiring that electron waves reflected from, the planes of atoms in a crystal interfere, constructively (see figure) (AIEEE 2008), Incoming, Electrons, , i, , Outgoing, Electrons, , The following graphs that can be expected to, represent the number of electrons ‘N’ detected, as a function of the detector position ‘y’ (y=0, corresponds to the middle of the slit) is, y, , N, , 1), , d, , 2), , N, , y, , y, , 3), , N, , d, , d, , 4), , N, , d, , COMPREHENSION -II, , A Young’s double slit experiment is conducted, with slit separation 10mm, where the screen is, d, 2m away from the slits. If wavelength of light, used is 6000 A0 , answer the following, Crystal Plane, 12. Fringe width in mm is, 9. Electrons accelerated by potential V are, 1) 0.12 2) 0.24, 3) 0.36, 4) 0.48, 0, diffracted from a crystal. If d 1A and 13. distance of 4th dark band from central fringe, in mm is, i 300 , V should be about, 1) 0.14 2) 0.28, 3) 0.42, 4) 0.56, ( h 6.6 10 34 Js , m 9.1 1031 kg ,, 14 If the wavelength is increased by 1000A0 , and, placed the whole apparatus in water of, e 1.6 1019 C ), refractive index 4/3, the new fringe width in, 1) 2000 V 2) 50 V, 3) 500 V 4) 1000 V, mm is, 10. If a strong diffraction peak is observed when, electrons are incident at an angle ' i ' from the, 1) 0.210 2) 0.105, 3) 0.315, 4) 0.420, normal to the crystal planes with distance ‘d’, LEVEL - IV - KEY, between them (see figure) de-Broglie, 1) 3 2) 4 3) 3 4) 3 5) 3 6) 4 7) 2, wavelength dB of electrons can be calculated, 8), 1 9) 2 10) 2 11) 4 12) 1 13) 3 14) 2, by the relationship (n is an integer), 42, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, 5. In Young’s double slit experiment intensity at, , LEVEL - V, , 1, , SINGLE CORRECT ANSWER TYPE, 1., , Two beams of light having intensities I and 4I, interfere to produce a fringe pattern on a, screen. The phase difference between the, beams is, , 2., , 3., , 4., , , at point A and, 2, , , , at point B. Then, , the difference between resultant intensities at 6., A and B is: (2001; 2M), (A) 2 I, (B) 4 I, (C) 5 I, (D) 7 I, In a double slit experiment instead of taking, slits of equal widths, one slit is made twice as, wide as the other, then in the interference, pattern:, (2000; 2M), (A) the intensities of both the maxima and the, minima increases, (B) the intensity of the maxima increases and the, minima has zero intensity, (C) the intensity of maxima decreases and that of, 7., minima increases, (D) the intensity of maxima decreases and the, minima has zero intensity, In a Young’s double slit experiment, 12 fringes, are observed to be formed in a certain segment, of the screen when light of wavelength 600 nm, is used. If the wavelength of light is changed, to 400 nm, number of fringes observed in the, same segment of the screen is given by :, (2001; 2M), (A) 12, (B) 18, (C) 24, (D) 30, In a Young’s double slit experiment, the, separation between the two slits is d and the, wavelength of the light is . The intensity of, light falling on slit-1 is four times the intensity, of light falling on slit 2. Choose the correct, choice(s),, (A) if d = , the screen will contain only one, maximum, , (B) if d ,at least one more maximum, 2, (besides the central maximum) will be observed on, 8., the screen, (C) if the intensity of light falling on slit 1 is reduced, so that it becomes equal to that of slit 2, the, intensities of the observed dark and bright fringes, will increase, (D) if the intensity of light falling on slit 2 is reduced, so that it becomes equal to that of slit 1, the, intensities of the observed dark and bright fringes, will increase., , NARAYANAGROUP, , a point is 4 of the maximum intensity.., , Angular position of this point is:(2005; 2M), , , , , (A) sin–1 d (B) sin–1 2d , , (C) sin 3d , , , (D) sin 4d , , –1 , , –1 , , In the Young’s double slit experiment using a, monoclromatic light of wavelenght , the path, difference (in terms of an integer n), corresponding to any point having half the peak, intensity is, (A) 2n 1, , , 2, , (B) 2n 1, , , 4, , , , (D) 2n 1, 8, 16, The YDSE apparatus is as shown in figure, below. The condition for point ‘P’ to be a dark, fringe is [ = wavelength of light used], , (C) 2n 1, , Screen, S1, l3, , l1, , S, , P, l2, , S2, , l4, , (A) l1 l3 l2 l4 n, (B) l1 l2 l3 l4 n, (C) l1 l3 l2 l4 2n 1, , , 2, , , 2, A screen is at a distance D=80cm from a, diaphragm having two narrow slits S1 and S 2, , (D) l1 l2 l3 l4 2n 1, , which are d 2mm apart. Slit S1 is covered, by a transparent sheet of thickness t1 2.5 m, and slit S 2 is covered by another sheet of, thickness t2 1.25 m as shown in figure. Both, 43
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , sheets are made of same material having 10. In a standard YDSE the region between screen, and slits is immersed in a liquid whose, refractive index 1.4 . Water is filled in the, space between the diaphragm and screen. A, 5 T, refractvie index varies with time as l , monochromatic light beam of wavelength, 2 4, 0, =5000A, is, incident, normally, on, the, , 5, until it reaches a steady state value of . A, diaphragm. Assuming intensity of beam to be, 4, uniform calcualte ratio of intensity at C to the, glass plate of thickness 36 m and refractive, maximum intensity of interference patern, index 3/2 is introduced in front of one of the, obtained on the screen? 4 3 , slits. Find the velocity of central maximuma, when it is at ‘O’?(take d=2mm and D=1m), (A) 2 103 ms 1, (B) 3 10 3 ms 1, S1, , t1, C, , S2, , 3, 2, 8, 5, (B), (C), (D), 4, 3, 9, 7, In the given figure, S is a monochromatic point, 2, , (A) 2 , (B), (C), (D) , source emitting light of wavelength, 3, 3, 500nm . A thin lens of circular shape and 12. In an interference arrangement similar to, Young’s double-slit experiment, the slits S1 and, focal length 0.1m is cut into two identical halves, S2 are illuminted with coherent microwave, L1 and L2 by a plane passing through, sources, each of frequency 106 Hz. The sources, daimeter. The two halves are symmetrically, are synchronized to have zero phase, placed about the central axis SO with gap of, difference. The slits are separated by a, 0.5mm. The distance along the axis the from, distance d = 150.0 m. The intensity I (q) is, S to L1 and L2 is 0.15m, while that from L1, measured as a function of q, where q is defined, as shown. If I0 is the maximum intensity, then, and L2 to ‘O’ is 1.3m. The screen at O is, I(q) for 0 < q < 90° is given by : (1995; 42), normal to SO. If the third maxima occurs at, (A) I(q) = I0/2 for q = 450, point A on the screen find the disntance OA?, (B) I(q) = I0/4 for q = 90°, A, (C) I(q) = I0 for q = 0°, (D) I(q) is constant for all values of q, L1, 13. In a YDSE bi-chromatic light of wavelengths, 0.5mm, 400 nm and 560 nm are used. The distance, S, O, between the slits is 0.1 mm and the distance, between the plane of the slits and the screen, L2, is 1 m. The minimum distance between two, successive regions of complete darkness is :, (2004; 2M), 1.3m, 0.15m, (A) 4 mm (B) 5.6 mm (C) 14 mm(D) 28 mm, (A) 1 mm (B) 2 mm (C) 1.5 mm(D) 2.5 mm, , (A), , 9., , 44, , t2, , (C) 4 103 ms 1, (D) 5 10 3 ms 1, 11. In the ideal double-slit experiment, when a, glass-plate (refractive index 1.5) of thickness, t is introduced in the path of one of the, interfering beams (wavelength ), the, intensity at the position where the central, maximum occurred previously remains, unchanged. The minimum thickness of the, glass-plate is :, (2002; 2M), , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 14. Intensity observed in an interference pattern, intensity, I I 0 sin 2 . At, 30o ,, I 5 0.002 . The percentage error in angle if, , P, B, x, , 2, , d, , I 0 20 w / m is, 2, , (A) 4 3 10 %, (C), , 4 3, 10 2 %, , , 4, 2, (B) 10 %, , , (D) 3 10 2 %, , MULTIPLE ANSWER QUESTIONS, , O, , O, , A, D, , (A) d min D, , D, , (B) d min , , D, 2, , d, 15. White light is used to illuminate the two slits, (D) x d min, (C) x min, 2, in young’s double slit experiment. The, seperation between the slits is b and the screen 18. Light travels as a, (A) Parallel beam in each medium, is at a distnace d(d>>>b) from the slits. At a, (B) Convergent beam in each medium, point on the screen directly in front of one of, (D) divergent beam in each medium, the slits, certain wavelength are missing. Some, (D) divergent beam in one medium and convergent, of these missing wavelengths are, beam in other medium., b2, 2b 2, b2, 2b 2 19. The phases of the light wave at c,d,e and f are, (A) , (B) , (C) , (D) , d, d, 3d, 3d, c , d , e , f respectively. It is given that, 16. Two point monochromatic and coherent, c f , then, sources of light of wavelength are each, placed as shown in the figure below. The initial, (A) c can not be equal to d, phase difference between the sources is, (B) d can not be equal to e, zero.(Assume D>>>d). Select the correct, statement(s)., (C) d f is equal to c e , , D, S1, , S2, , O, , d, Screen, , (D) d c is not equal to f e , 20. Speed of light is, (A) same in medium-I and medium-II, (B) Larger in medium-I than in medium-II, (C) Larger in medium-II than in medium-I, (D) Different at b and d, , COMPREHENSION TYPE, , 7, Comprehension-I:, , the point ‘O’ will be minima., In the arrangement shown in figure, slits S3 and S4, 2, are having a variable separation Z. Point ‘O’ on the, (B) If d , only one maxima canbe observed on, screen is at the common perpendicular bisector of, screen, S1S2 and S3S4 . (Assume D>>>d)., (C) If d 4.8 ,then a total 10 minimas would be, there on screen., S1, S3, 5, (D) If d , , then Intensity at ‘O’ would be, 2, Z, minimum., O, d A, O, 17.The minimum value of ‘d’ so that there is a dark, fringe at ‘O’ is d min . For the value of d min the, S3, distance at which the next bright fringe is, S2, D, D, formed is x. Then, , (A) If d , , NARAYANAGROUP, , 45
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , Comprehension-III:, D, the intensity meausred at ‘O’, The figure shows a surface XY separating two, 2d, transparent media, medium-I and medium-II. The, 2 D, lines ab and cd represents wavefronts of a light wave, is, is I 0 . The intensity at ‘O’ when Z , travelling in medium-I and incident on XY. The lines, d, ef and gh represent wavefronts of light wave in, (A) I 0, (B) 2I 0, (C) 3I 0, (D) 4I 0, medium-II after refraction the, 22. The minimum value of Z for which the intensity, b, at ‘O’ is zero is, d, 3 D, D, D, D, (A), (B), (C), (D), 2d, 2d, 3d, d, medium I, 23. If a hole is made at O ' on AO ' O and the slit, a, c, X, Y, S 4 is closed then the ratio of the maximum to, h, f, the minimum observed on screen at ‘O’ if O ' S3, medium II, D, is equal to, is, e, 4d, g, (A) 1, (B) Infinity (C) 34, (D) 4, , 21. When Z , , Comprehension-II:, , ASSERTION AND REASON TYPE, , A monochromatic beam of light falls on Young’s, double slit experiment apparatus as shown in figure., A thin sheet of glass is inserted in front of lower slit, S 2 ( 600nm is wavelength of source), 27., , S1, , , , d, , , O, d << D, , 28., , S2 t, , 24. The central bright fringe can be obtained, (A) At ‘O’, (B) At ‘O’ or below ‘O’, (C) at ‘O’ or above ‘O’, (D) Any where on the screen, 25. If central bright fringe is obtained onscreen at, ‘O’ then, (A) 1 t d sin , , 29., , (B) 1 t d cos, , t, d, (D) 1 sin , 26. The phase difference between central maxima, and 5th minima is, , 3, , (A), (B) 9, (C), (D) 8 , 6, 2, 6, (C) t d, , 46, , 30., , (A) A is true and R is true and R is the correct, explanation of A., (B) A and R are true but R is not the correct, explanation of A, (C) A is true, R is false, (D) A is false, R is true., Assertion(A): For best contrast between maxima, and minima in the interference pattern of Young’s, double slit experiment the intensity of light emerging, out of the two slits should be equal., Reason(R): The intensity of interference pattern, is proportional to the square of the amplitude., Statement-I: In Young’s double slit experiment, interference pattern disaapears when one of the slits, is closed, Statement-II: Interference is observed due to, superposition of light waves from two coherent, sources, Statement-I : In Young’s double slit experiment,, the two slits are at distance ‘d’ apart. Interference, pattern is observed on a screen at a distance D, from the slits. At a point on the screen which is, directly opposite one of the slits, a dark fringe is, observed.Then the wavelength of wave is, proportional to the square of the distance between, the two slits., Statement-II: For a dark fringe, intensity is zero., Statement-I : Fringe width depends on the, refractive index of the medium., Statement-II: Refractive index changes optical, path of ray of light forming fringe pattern., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 31. Assertion(A): Thin films such as soap bubble or a 34. In Column-I the effect on fringe pattern in, YDSE is mentioned when the changes, thin layer of oil on water show beautiful colours, when illuminated by white light., mentioned in Column-II are made. Match the, Reason(R): The colours are obtained by, entries of Column-I with entries of Column-II., dispersion of light, Column-I, 32. Assertion(A): The film which appears bright in, A) Angular Fringe width changes, reflected system will appear dark in the transmitted, B) Fringe width [linear seperation between two, system and vice-versa., consecutive fringes], Reason(R): The condittions for film to appear, C) Angular Fringe width remains same, bright or dark in the reflected light are just reverse, D) The fringe pattern disappear, to those in the transmitted light, Column-II, STATEMENT TYPE QUESTIONS, P) Screen is moved away from the plane of the, (A) Statement-I is true and Statement-II is true, slits, and Statement-II is the correct explanation of, Q) Wavelength of light used is decreased., Statement-I., R) The seperation between the slits is increased, (B) Statement-I and Statement-II are true but, S) The width of the source slit is increased, Statement-II is not the correct explanation of, T) The source slit is moved closer to the double, Statement-I, slit plane., (C) Statement-I is true, Statement-II is false, (D) Statement-I is false, Statement-II is true. 35. In Young’s double slit experiment, the point, source ‘S’ is placed slightly off the central axis, MATRIX MATCHING TYPE, as shown in figure. if 500nm , then match, 33. Monochromatic light source having wavelength, the following, is used in YDSE. The separation between, the slits is ‘d’ and separation between slit and, screen is D (>>>d) as shown in figure. The slits, P, width are different and individual intensities, due to upper and lower slits at the screen are, S1, S, x = 10mm, 4I 0 and 9I 0 respectively. In column I, intensity, 2mm, of the interference pattern on the screen at, O, certain points are given and in Column II the, 20mm, distance between two points on the screen, which are having that intensity is given. Match, the entries of Column I with the enteries of, Column II., , S, , d, , 1m, 2m, Column-I, (A) Nature and order of interference at the point, P, (OP=10mm), (B) Nature and order of interference at point ‘O’, , O, , D, , Column-I, 1, th of maximum intensity, A), 25, B) Intensity is 25I 0, 19, th of maximum, C) Intensity is, 25, Intensity, 7, th of maximum, D) Intensity is, 25, Intensity, NARAYANAGROUP, , Screen, , Column-II, D, P), d, D, Q), 3d, 2 D, R), d, S), , 3 D, d, , (C) If a transparent paper (refractive index, 1.45 ) of thickness t 0.02mm is pasted on, , S1 the nature and order of interference at P.., (D) After inserting transparent paper in front of slit, S1 ,the nature and order of interference at ‘O’, Column-II, P) Bright fringe of order 80., Q) Bright fringe of order 262., R) Bright fringe of order 62., S) Bright fringe of order 280., 47
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , 36. An inteference is observed due to two coherent 40. In a modified Young’s double slit experiment,, a monochromatic uniform and parallel beam, sources A and B separated by a distnce 4, of light of wavelength 6000A0 and intensity, along Y-axis, where is the wavelength of light, . A detector ‘D’ is moved along the positive X 10 , 2, wm is incident normally on two circular, axis. Find the total number of maxima, , observed on the X-axis excluding the points, apertures A and B of radii 0.001m and 0.002m, x 0 and x ?, respectively. A perfectly transparent film of, thickness 2000A0 and refractive index 1.5 for, A, the wavelength of 6000A0 is placed in front of, aperture A as shown in figure. Calculate the, power (in microwatt) received at the focal spot, 4, F of the lens. The lens is symmetrically placed, with respect to the apertures. Assume that, 10% of the power received by each aperture, B, xn, goes in the original direction and is brought to, D, the focus, 37. Two coherent light sources each of wavelength, are separated by a distance 3 . Find the, total number of minima formed on the line AB, which runs from to ?, A, P, , B, , F, , B, , 3, , Q, , LEVEL-V - KEY, A, , 38. In Young’s double slit experiment interference, bands are produced on the screen placed at, 1.5m from two slits 0.15 mm apart and, illuminated by light of wavelength 6000A0 . If, the screen is now taken away from the slits by, 50cm then find the change in fringe width (in, mm)?, , INTEGER TYPE QUESTIONS, , SINGLE OPTION TYPE QUESTIONS, 1.B 2.A 3.B 4.A 5.C 6.B 7.C, 8.A 9.A 10.B 11.A 12.C 13.D 14.C, 15. A,C 16. A,B,C,D 17. B,D 18.A 19.C, 20.B 21. B 22. D 23.C 24.D 25.A 26.B, 27.B 28.A 29.B 30.B 31.C 32. A, 33. A P, R, S ; B P, R, S ;, , 39. In YDSE, the source is red light of wavelength, C P , Q , R , S ; D P , Q, R , S, 7, 7 10 m . When a thin glass film of refractive, index 1.5 at this wavelength is put in the path 34. A Q, R ; B P , Q, R ;, of one of the interfering beams, the central, C P , S , T ; D Q, R , S , T, , 3, bright fringe shifts by 10 m to the position, 35. A S ; B P ; C Q ; D R, previously occupies by the 5th bright fringe., Find the thickness of glass plate in, 36. 3, 37.6 38.2 39.7 40.7, micrometers?, 48, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , LEVEL-V - HINTS, , sin =, , SINGLE CORRECT ANSWER QUESTIONS, 1., , I( ) = I1 + I2 + 2 I1 I 2 cos , , correct answer is (C)., , …(1), , Here,I1 = I and I2 = 4I, , At point A, , , IA = I + 4I = 5I, IB = 1 + 4I – 4I = I, , At point, IA – IB, , B, =, = 4I, , I max, , Im cos 2 , 2, 2, , 6., , , =, 2, , , , , , , , I1 ~ I 2, , , , , , , 2n 1, 2, , x , , 7., , 2, , , , 2 4, , 1, cos , 2, 2, , , , , , 2n 1 2n 1, 2, 2, 4, For to obtain dark fringe, , 2, , 2. In interference we know that Imax = I1 I 2 and, Imin =, , , , or q = sin–1 3d , 3d, , , Under normal conditions (when the widths of both, path difference x 2n 1, 2, the slits are equal), , I1» I2 = I (say) \ Imax = 4I and Imin = 0, l1 l3 l2 l4 2n 1, When the width of one of the slits is increased., 2, Intensity due to that slit would increase, while that 8. Path difference at ‘C”, of the other will remain same. So let :, x t1 g 1 t2 g 1 g t1 t2 t1 t2 , ( > 1), I1 = I and I2 = I, , 2, , Imin = I 1 > 0, Intensity of both maxima and minima is increased., D, d, , Fringe width, w =, , 4., , Condition for maxima is d sin = n , If d = , sin = n Possible value is , only one maxima will be obtained, (A) is correct. (B) is correct., , with, , , , I1 I 2, , , , n11 n2 2, , 2, , & I min , , , , I1 I 2, , , , but when I1 = I2, Imax = 4I2 and Imin = 0, (C) and (D) are wrong., , , cos, , lmax, 4, , = lmax cos2, , , 2, , , , 1, , =, or =, 3, 2, 2, 2, , 2 2 , = 3 x …(1) where x= d sin , , , , Substituting in Eq. (1), we get, NARAYANAGROUP, , 25, 1, m ., 400 16, , 2, Phase difference x, , , , , I1 = 4I2, , l = lmax cos2 2 , , , x , , 2, , I max 9I 2 , I min I 2, , 5., , , , 1.4, , x t1 t2 g 1 2.5 1.25 3 1, 4, , , , 2 2.5, 42 , , = 1.25 1 1.25 , 40 40, 40 , , 3., , Imax , , x t1 t2 g 1, , 2, , Imax= I 1 > 4 I and, , Then,, , 2, 2 4, 1, x , 106, 10, a a, 1 5000 10 3 16, , 106 106 , , , 30 10 7 3 106 3, intensity at ‘C’, , , 2 4I cos 6 4I 34, , I C 4 I 0 cos 2 , , 2, , 0, , 0, , I C 3I 0, maximum intensity I max 4 I 0, IC, 3I, I, 3, 0 C , I max 4 I 0, I max 4, 49
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, 9., , x S2 P S1P liq g l t, A, S1, P1, P2, , S, , , , d, , yd , x l , g l t, D , For central maxima x 0, , O, , S2, a = 0.15m, , 0.3m, (v), , yd , O l , g l t, D , , D = 1m, , The images of S due to both parts of lenses are, formed at S1 and S 2 whose distance ‘v’ from, the lens is give n by, , g l tD, d l, , T, D 1 t, D 4 T t, 4, y , , 5 T d 10 T , d , 2 4 , , 1 1, 1, 10 100 150 100 50 10, , , , , , , v 0.1 0.15 1 15, 15, 15 3, 3, 0.3m, 10, , From similar les SS1S 2 and SPP, 1 2, , The time when y becomes zero is, , d, 0.3 0.15, , 3, 0.5 10, 0.15, , O, , 0.45, 0.5 103 1.5 10 3 m, 0.15, For third maximum, , D 4 T t 0 4 T 0 T 4sec, , d, , D 4 T t, d 10 T , , Speed of central maxima, , 3 D 3 500 109 1, yn , , 109 106, d, 1.5 103, 3, , yn 10 m y3 1mm, , V, , dy Dt d 4 T , , dt, d dT 10 T , , V, , tD , 1, 4 T 10 T , , , d, , 10., , , P, , tD 4 T 10 T , , , 2, d , 10 T , , , y, , S1, d = 2mm, O, S2, , , , 6 Dt, tD 4 T 10 T V , , , 2, d 10 T 2 , d 10 T , , Central maxima is at ‘O’ at T 4sec, , t, D = 1m, , Path difference, , x S2 P liq g l t S1P liquid , 50, , y, , 3 5 T , D t, 2 2 4 , y , 5 T , d , 2 4 , , 1 1 1, 1, 1, 1, , , f v u, 0.1 v 0.15 , , v, , , , l S2 P S1 P air g l t, , yn, , V, , 6 1 36 106, 2 10 3 10 4 , , 2, , , , 3 36 103, 36, , V 3 103 ms 1, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 11. Path difference due to slab should be integral, multiple of l or x = n , , 2n 1, , n, , or 1 t n , n 1, 2,3 or t = 1, for minimum value of t, n = 1, , 2, 2 , =, ( x) = (d sin q), , , , , 14. I I 0 sin 2 , , c, , =, , =, , 3 108, = 300 m and d = 150 m, 106, , 2 , 1, , , (150) =, 300, 2, , , , 2, , , % error in angle, d, 1, dI, 100 , , I 2 cot , , , , =, 2, 4, , I0, , 2 , , (ii) For 0 = 90° d = 300 (150)(1) = p, , , , , =, 2, 2, , , , and I(q) = 0, , (iii) For q = 0°, d = 0 or, , , =0, 2, , , , or, , 2n 1, 2m 1, , 7 14, = ..., 5 10, , MULTIPLE ANSWER TYPE, , (2 4 1)(100)(400 10 6 ), 2 0.4, , = 14 nm, , Next 11th minima of 400 nm coincides with 8 th, minima of 560 nm, then, 400 , , p, , S1, , S2, , From the figure:, S 2 p 2 S1 P 2 S1S 22 ; S 2 p 2 S1 P 2 S1 S22, , S2 p S1P S2 p S1 P S1S22, , 560 , , (2n – 1) 2 = (2m – 1) 2 , , , , , , NARAYANAGROUP, , d, , b, , i.e., 4th minima of 400 nm coincides with 3rd minima, of 560 nm., Location of this minima is,, Y1 =, , 6 1 1, 6 1 1 1, , 0.1 , 5, 5 10, 3, 3, , 3 1, 3, d, 4 3, , , 0.04, 100 , 102 %, 25 , , , , I(q) = I0, (option c), 13. Let nth minima of 400 nm coincides with mth 15., minima of 560 nm, then, 400 , 560 , (2n – 1) 2 = (2m – 1) 2 , , , , , , 1, 100, , , 0.002, 6, d, 100 , 100 , o, , 2 5cot 30, , , I(q) = I0 cos2 4 2 (option a), , , or, , dI, I 0 2sin cos , d, , dI I 0 2sin cos d, , 2 cot d, I, I 0 sin 2 , , (i) For q = 30°, , = v, , = 42 mm, , Required distance = Y2 – Y1 = 28 mm, Hence, the correct option is (D)., , , 12. The intensity of light is I(q) = Io cos 2 , 2, , , , 14, , (2 11 1)(100)(400 10 6 ), 2 0.1, , , , t = 1 1.5 1 = 2 , , where, , 7, , or 2m 1 = = …, 5 10, th, i.e., 11 minima of 400 nm coincides with 8th, minima of 560 nm., Location of this minima is, + Y2, =, , S 2 p S1P , , S1 S22, b2, , S2 P S1P d , S2 P S1P 2d, 51
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, Missing wavelengths means dark fringes,hence path, b2, b2 , difference x 2n 1 , 2n 1 d, 2 2d, , Given d d min, , d 2 xd, , on solving, then, D, D, b, , b2, b2, For n 1 1 , For n 2 2 , d, 3d, b2, For n 3 3 , and so on., 5d, 16. Path difference at ‘O’ is x d , which is maxima, , d, , X, , medium I, , i, , a, , c, , Y, f, , r, , e, , h, , medium II, , g, 7, d, , , the point ‘O’ will be minima., (A) If, 2, 18. The wavefronts, in both the media are parallel., The light will be a parallel beam., (B) If d , The point ‘O’ will be maxima, (C) If d 4.8 ,then a total 10 minimas can be, x d min, observed on screen 5 above ‘O’ and 5 below ‘O’, 19. Points c&d are on the same wavefrnts hence, which corresponding to, c d .Similarly points e and f are on the same, 3 5 7 9, x , , , , ,, wavefront hence e f ., ., 2, 2, 2, 2, 2, Hence d f c e, 5, d, , (D) If, , then ‘O’ will be minimum and 20. From the diagram, it clearly indicates that on, 2, entering medium-II, the ray bend towards normal, hence intensity is minimum., and hence medium-II is density medium., 17. There is a dark fringe at ‘O’ is the path difference, Hence speed of light is larger in medium I than in, x ABO AO ' O 2, medium-II., , d2, x 2 D d 2 D 2 D 1 2 2 D, D, 2, , 2, , COMPREHENSION TYPE, Comprehension-I:, 21. intensity at ‘O’ is proportional to intensity at S3, , 1, , d 2 2 , , , d2, x 2 D 1 2 1 2 D 1 , 1, 2, D , , 2D, , , , 2, , x 2 D , , , , 2 , and at S 4 . I k cos, 2 where k is constnat,, , is the phase difference., , 2, , d, d, D, , d2 , 2, 2, 2D, D, 2, , , , D, 2, There bright fringe is formed at P if the path, difference, d min , , 2 Z z, , where is the fringe width, 2 , , when Z , , D, 2d, , 2, , 2, 2, x 2 d 2 x d 2 xd, , , , , 2 D 2D, 2D, , 52, , Z, , D , , d , , , D, 2d 2, , 4 k 12 ;, , I I 0 k cos 2 , , x1 AO ' P ABP , x1 D D 2 x 2 D 2 d 2 D 2 x d , , , , When Z ' , , ', , 2 D, d, , k 2I0, , Phase difference, , 2 Z ' 2, 2 D, , d, 2, 2 D, 2d, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , Required intensity at ‘O’, , Comprehension - II:, , 2, 2 I 0 cos 2 , 2, 2, , , , I ' k cos 2 ', , 2, , 2 I 0 1, , , 25. Total path difference x 1 t d sin . For, , I ' 2I0, 22. Intensity at ‘O’ os zero if, , , , central maxima x 0 hence 1 t d sin , 26. Phase difference, , , , 2 2, , 2n 1 2 5 1 9 ; 9, , 2 Z, Z ; Z D, 2, d, , ASSERTION & REASON TYPE, , 23. Intensity at S3 is, 2, 0, , 2, 0, , 27. When intensity of light emerging out from two slits, is equal then t he intensity of minima., , 2, 0, , I S3 A A 2 A cos , , I min , , 2 D 2, D , , , d , , Where , 4d 2, 4d D, , , , 2 A02 1 cos 2 A02 2cos2 , 2 , , , 2 4A cos 4, , IS3 4A02 cos2 , I S3 4 A02 , , 2, 0, , 2, , 1, 2 A02, 2, , , , Minimum intensity at O , , I max, I min, , , , , , , , 2, , , , , 2 1 , , , 2 1 , , , , 2 A0 A0, , , , , , 2 1, 2 1, , , , 4, , 3 2 1.414 3 2.828 , , 2, , , , 2 1 2 2, , , , 2 1, , 2, , 3 2 2 , 2, , I max, 2, 5.828 34 ; I 34, min, NARAYANAGROUP, , 2, , 0 or absolutre dark. It, , , , S 2 P S1 P , , , , 2, , For dark fringe x S2 P S1 P 2n 1, , 2, , , 2 1, 2 1, , , S1 S22, d2, , S2 P S1 P 2 D, , 2, , , 2, , d2, 2, 2n 1, d 2n 1 D d 2 , 2D, 2, Intensity of dark fringe is zero., 1, D, but , , d, 31. The beautiful colours are obtained due to, interference of light reflected from the upper and, lower surfaces of thin film. As the conditions for, constructive and destructive interference are, dependent on wavelength of light. Hence coloured, interference fringes are observed. A is true and R, is false, 32. For reflected system of the film, the consition for, , 30. , , 2, , 2, 2 1 , , 2 1 , , , , , , , , S2 P S1P S2 P S1P S1S22, , Maximum intensity at O ' A0 2 A0, , , , , I I, , S 2 P 2 S1 P 2 S1S 22, , Amplitude of wave at S3 : AS3 2 A0, , , I max A0, , I min A0, , , , , provides a better contrast., A & R are true but R isnot the correct, explanation of A, 28. When one of the slits is closed, there appears general, illumination from a single source. Interference does, not take place., 29. Path difference, , I S3 2 A02 2 A02 cos , , , , , , , 24. Any where on the screen because there is no, relation b/n & ., , 2, , maxima is 2 t cos r 2n 1 2 . While the, 53
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, maxima for transmitted system of film is, 2 t cos r n and is reverse for minima. A & R, are true and R is the correct explanation of A, , 19 I 0 13I 0 2 6 I 0 cos , 6 I 0 12 I 0 cos cos , , MATRIX-MATCHING TYPE, 33. I1 4 I 0 ; I 2 9 I 0, I max , , I1 I 2, , , , 4 I 0 9I 0, , , , 2, , I 0 2 3 , , 2, , I max I 0 25 25 I 0, I min , , (A), , , , I1 I 2, , , , 2, , 5 7 11, , , , ,, ,, 2n , 3 3 3 3, 3, , path difference, , , 2, , , , x, 2, , , , I 0 2 3 I 0, , , , 2, , I0, , , 5 7 11, , , ,, , 2, 6 6 6 6, , I min I 0, , 1, , x 2n , 3 2, , , 1, th of maximum intensity:, 25, , For n 0 y0 , , I, , 1, 1, I max 25I 0 I 0, 25, 25, , 1, th of maximum intenxity gives the intensity of, 25, minima. Hence teh distance b/n such, points, , is equal to fringe width Yn , n 1 y1 , , 6I0 1, , 12 I 0 2, , n D, d, , D, 2 D, ; n 2 y2, ;, d, d, , 1 D, , yn 2n , 3 2d, , , D, 6d, , 1 D 5 D, , , For n 1 y1 2 , 3 2d, 6d, , 1 D 11 D, , , For n 2 y2 4 , 3 2d, 6d, , Required seperation, , y2 y 0 , , , 2 D, 11 D 5 D, ; y2 y1 , , 3d, 6d, 6d, , 6 D 3 D, , 6d, 3d, , 3 D, 4 D, ; n 4 y4, ;, 7, d, d, th of maximum, (D) Intensity is, 25, The distnace b/n two successive minima points, 7, D 2 D 3 D 4 D, , I 25I 0 7 I 0, ,, ,, ,, , , , , , d, , 25, d, d, d, , , n 3 y3 , , (B) Intensity is 25I 0, This represnts the maximum intensity. The distance, b/n the points on the screen. Which area having, maximum intensity, D 2 D 3 D, , d , d , d , , , , , (C) Intensity is, I, , 19, th of maximum intensity, 25, , 19, 25 I 0 19 I 0 I I1 I 2 2 I1 I 2 cos , 25, , 19 I 0 4 I 0 9 I 0 2 36 I 0 cos , 54, , I I1 I 2 2 I1I 2 cos 7 I 0 13I 0 12 I 0 cos , 1, 2 4 8, cos , ,, , , , 2, 3 3 3, , d 2 d 3 d, ,, ,, , , 3d 3d 3d, 34. Angular fringe width distance b/n slits, Required seraration, , , , D distance b/n slits & screen, d D, The arrangement for YDSE is shown in figure., Wavelength of light used, Let the size of source and separation between, Fringe width, , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , source slit and double slit plane is S. Then, , for, , s , interference fringes to be observed ., S d, If this condition is not satisfied then interference, pattern produced by different parts of the source, overlap and fringe pattern disappears., Screen, S, , x = SS 2 S 2 P SS1 S1P , , SS 2 SS1 S2 P S1P , d sin 0 d sin , , dy0 dy, x D D, 1, 2, , 20 2 20 10, , 0.04 0.1 0.14mm, 1000, 2000, A) For bright fringe:, , x , , S1 S2, S, , D, , For p: As ‘D’ is increased, angular fringe width, , x n n , , x, 0.14, 14 102, , , 500 109 103 5 107 103, , , n 2.8 102 n 280 A S, remains same as it is independnet of D. As, d, B) At the centre (origin) ‘O’, D, d y0 20 2, D increase,fringe width increases , x, , 0.04mm, d, 1000, D1, p B, C, For bright fringe x n, For q: When is decreased, angular fringe width, x 0.04 10 3 4 10 5, ( ) and fringe width decreases and fron the, n , , 500 10 9 5 107, condition as decreases and from the condition, n 80 ;, BP, n 0.8 102 80, s , as decreases this condition would be C) Due to transparent paper, the change in optical path, S d, is, failed and fringe pattern disappears., 1 t 1.45 1 0.02 0.45 0.02, q A, B, D, 0.009mm, For r: ‘d’ is increasing and both decreases, path, difference at P:, and fringe pattern disappears. r A, B , D, , , For s: As the source slit wodth increases, the, s , would be violated at some instant, S d, and fringe pattern disappears but there is no effect, on and . s C , D, For t:As the distance b/n source slit and double slit, ‘S’ is decreasing then fringe pattern disappears but, and has no effect. t C , D, , condition, , 35., , x 0.14 0.009 0.131mm, 1, , n, , x1 131 106 131 106 1310, , 262, , , 5, 500 109, 5 10 7, , C Q, n 262, D) Due to transparent paper the path difference at ‘O’, , x 0.04 0.009 0.031mm, n, , x 0.031 103 31 106, , , 500 109, 5 107, , n, , 310, 62 ; n 62 ; D R, 5, , P, S1, , S, , x1, D1, , Path difference, NARAYANAGROUP, , 10mm, , 0 , d, , y0 = 2mm, , O, , INTEGER TYPE QUESTIONS, , x2, S2, , D2, , 36. To obtain maxima the path difference, AD BD n, AD xn n, 55
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , AD n xn 1, 2, , 2, , AD AB BD, 2, , 2, , From le ABD,, 2, , n xn 16 2 xn2, , 2, , 2, n, , 2, , n x 2n xn 16 x, , S, , 2, n, , P, , d, 3d, , 16 n , , 2, , 2n xn 16 n , 2, , 2, , xn, , 38. D2 D1 50cm ; D1 1.5m 150cm, , For n 0 x0 , 15, 16 1 , n 1 x1 , 7.5, , 2, 2 , 12, 16 4 , n 2 x2 , 3, , 4, 4 , 7, 16 9 , n 3 x3 , , 6, 6 , 16 16 , n 4 x4 , 0, 2 4 , So the total no.of maxima points observed are 3., 37. Let us take any general point R on the line AB,, For any position of R on line AB,, PQR is a le ., B, P, , O, (midpoint), Q, , R, A, , PQ QR PR [In any triangle sum of two sides, is more than the third side], PR QR 3, , As PR QR represrnt the path difference at any, , 56, , D, , 2n, , 6000Å 6000 108 cm ; d 0.15mm, Change in fringe width 2 1, 50 6000 108, D2 D1 , d, 0.15 10 1, 20 104, , 2 101 cm 2mm 2mm, 3, 10, 39. Given shift of central bright fringe 5th order bright, fringe, t, , D, 5 D, 1 t , d, d, , 5, 5 7 10 7 5 7 107, , , 0.5, 1 1.5 1, , t 7 106 m t 7 m, 40. The power transmitted through A, 10 , 2, , PA 10% of 0.001, , , 10 10, , 106 106 watt, 100 , The power transmitted through ‘B’ is, 10 , 2, , PB 10% of 0.002 , , , 10 10, , 4 106 4 106 watt, 100 , Let be the phse difference t=introduced by film, , , 2, 2, 2, x, 1 t , 0.5 2000 10 10, , , 6000 1010, , , redian. The power received at F, 3, , point on AB it can occurs at, , , , 3 5, , ,, only.., 2 2 2, SO 3 minimas below the mid point ‘O’ and 3, minimas above ‘O’, Hence total no.of minimas formed is 6., , P PA PB 2 PA PB cos , 106 4 106 2 106 4 10 6 , , 1, 2, , 5 106 2 106 7 106 7 W, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , 4. ABC is a spherical wavefront centred at ‘O’, symmetric about BE is incident on slits S1 and, , LEVEL - VI, SINGLE CORRECT ANSWER TYPE, 1., , WAVE OPTICS, , If two coherent sources are placed at a, distnace 3 from each other, symmetric to the, centre of the circle as shown in figure, Find, the number of fringes shown on the screen,, placed along the circumference?, , S2 ., , S1S 2 4 ;, , BS1 3 ;, , BO 6 ;, , S1E 128 and is the wavelength of, incident light wave. A mica sheet of refractive, index 1.5 is pasted on S 2 . Find the minimum, value of thickness of mica sheet for which, central fringe forms at E?, A, , S1, , S2, , S2, , 3, , B, , 2., , (A) 16, (B) 12, (C) 8, (D) 4, In the adjacent diagram,CP represents a, wavefront and AO and BP, the corresponding, two rays. Find the condition on for, constructive interference at P between the ray, BP and reflected rays OP, , S1, , O, , E, , C, , 31, 15, 5, 7, (B), (C), (D), 8, 8, 8, 8, 5. A glass plate of refractive index 1.5 is coated, with a thin layer of thickness t and refractive, O, Q, index 1.8. light of wavelength travelling in, , air is incident normally on the layer. It is partly, d, reflected at the upper and lower surfaces of, C, the layer and the two reflected rays interfere., If 648nm , then find the least value of ‘t’, P, for which the rays interfere constructively?, A, (A) 30 nm (B) 60 nm (C) 90 nm (D)120 nm., B, 6. In a Young’s experiment, the upper slit is, covered by a thin glass plate of refractive, 3, , cos, , , cos, , , (A), (B), index. 1.4, while the lower slit is covered by, 2d, 4d, another glass plate, having the same thickness, 4, , as the first one but having refractive index 1.7., (D) sec cos , (C) sec cos , Interference pattern is observed using light of, d, d, wavelength 5400 Å. It is found that the point P, 3. In standared YDSE slits were moved apart, on the screen, where the central maximum (n, symmetrically with relative velocity v, calculate, = 0) fall before the glass plates were inserted,, the rate at which fringes pass a point at a, distance x from the centre of the fringe system, 3, now has the original intensity. It is further, formed on a screen at ' y ' distance away from, 4, observed that what used to be the fifth, the double slits if wavelength of light is , and, maximum earlier lies below the point P while, distance between the slits is ‘b’. Assuming, the, sixth minima lies above P. Calculate the, y b & b , thickness of glass plate. (Absorption of light, 2xv, xv, xv, by glass plate may be neglected). (1997; 5M), (A), (B), (C), (D) Zero, y, 2 y, y, (A) 9.3 m (B) 6.2 m (C) 8.5 m (D) 5.8 m, NARAYANAGROUP, , (A), , 57
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , In a young’s double slit experiment, two, COMPREHENSION TYPE, wavelengths of 500 nm and 700 nm were used. Comprehension-I.V:, What is the minimum distance from the central, In Young’s experiment, the source is red light of, maximum where their maximas coincide, wavelength 7 × 10–7 m. When a thin glass plate of, again? Take D/d = 103. Symbols have their, refractive index 1.5 at this wavelength is put in the, usual meanings. (2004- 4M), path of one of the interfering beams, the central, (A) 6.5mm (B) 3.5 mm (C)10.2 mm (D) 5.9mm, bright fringe shifts by 10–3 m to the position, MULTIPLE ANSWER QUESTIONS, previously occupied by the 5th bright fringe., 8. A transparents slab of thickness ‘t’ and, (1997C, 5M), refractive index is inserted in front of upper, 11. Find the thickness of the plate., slit of YDSE apparatus. The wavelength of, (A) 7 106 m, (B) 9 1010 m, light used is . Assume that there is no, (C) 4 106 m, (D) 6 1010 m, absorption of light by the slab. Mark the, correct statments., 12. When the source is now changed to green light, (A) The intensity of dark fringes will be zero, if slits, of wavelength 5 × 10–7 m, the central fringe, are identical., shifts to a position initially occupied by the 6th, (B) The change in optical path due to insertion of, bright fringe to red light. Find the refractive, plate is t ., index of glass for green light., (C) The change in optical path due to insertion of, (A) 1.4, (B) 1.2, (C) 1.6, (D) 1.3, plate is 1 t ., 13. Also estimate the change in fringe width due, to the change in wavelength., (D) For making intensity zero at centre of screen, (B) 6.22 105 m, (A) 6.22 105 m, 5, the thickness can be 2 1 ., (C) 5.71 105 m, (D) 5.71 105 m, , , , 7., , 9. In a YDSE with two identical slits, when the Comprehension-II-I:, In YDSE arangement shown in figure, fringes are, upper slit is covered with a thin, perfectly, tranparent sheet of mica, the intensity at the, seen on screen using monochromatic source S, centre of the screen reduces to 75% of the, having wavelength 3000 (in air). S1 and S 2 are, Å, initial value second minima is observed to be, two, slits, separate, by, d 1mm and D 1m . Left, above this point and third maxima below it, which of the following can not be a possible, of slits S1 a nd S 2 medium of refractive index, value of phase difference caused by the mica, 1 2 is present & to the right of slits S1 and S 2, sheet, 5, 13, , 11, 3, (A), (B), (C), (D), medum of refractive index 2 is present. A, 3, 3, 3, 3, 2, 10. In Young’s double slit experiment, the two slits, thin slab of thickness ‘t’ is placed in front of S1 ., are covered by slabs of same thickness but, refractive index 1.4 and 1.7. The distance, The refractive index of slab 3 varies with distance, between slits and screen is 1m and distance, from its starting face as shown in figure., between slits is 1mm and wavelength of, y, coherent source used is 4000 and the central, x, , Å, S, fringe shifts to the 3rd bright fringe position,, , , Screen, S, then, d/2 t, , (A) Shift will be towards slab of R.I.-1.7 by 1.2, 3, O, mm., d/2, (B) Shift will be towards slab of R.I.-1.4 by 1.2, 1, S, mm, x, 0, t, (C) Slabs are of thickness 4 m, D = 1m, D = 1m, (D) Slabs are of thickness 4, Å, 2, , 1, , 1, , 3, , 3, , 2, , 58, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 14. In order to get central maxima at the centre of Comprehension-IV-IV:, A double slit apparatus is immersed in a liquid of, the screen, the thickness of slab required is, refractive index 1.33. It has slit separation of 1 mm, (A) 1 m, (B) 2 m (C) 0.5 m (D) 1.5 m, and distance between the plane of slits and screen, 15. I f t he t hick ness of t he slab is select ed 1 m ,, is 1.33 m. The slits are illuminated by a parallel beam, of light whose wavelength in air is 6300 Å., then the position of the central maxima will be, (1996; 3M), y coordiante , 20. Calculate the fringe width., (A) 0.95 mm, (B) 0.36 mm, 1, 1, 1, 1, (C), 0.63, mm, (D) 0.56 mm, (A) mm (B) mm (C) mm (D) mm, 3, 3, 6, 6, 21. One of the slits of the apparatus is covered by, a thin glass sheet of refractive index 1.53. Find, 16. Fringe width on the screen is, the smallest thickness of the sheet to bring the, (A) 0.4mm (B) 0.1mm (C) 0.2mm (D) 0.3mm, adjacent minimum at the centre of the screen, Passage-II:, the axis., In the figure, light of wavelength 5000 is, (A) 2.132 m, (B) 2.512 m, Å, (C) 6.521 m, (D) 1.579 m, incident on the slits (in a horozontally fixed place)., Here d 1mm and D 1m . Take origin ‘O’ and Comprehension-V-VI:, The Young’s double slit experiment is done in a, XY plane as shown in figure. The screen is released, medium of refractive index 4/3. A light of 600 nm, from rest from the initial position as shown., wavelength is falling on the slits having 0.45 mm, separation. The lower slit S2 is covered by a thin, , glass sheet of thickness 10.4 m and refractive, = 5000A, d, index, 1.5. The interference pattern is observed on, S1, S2, X, a, screen, placed 1.5 m from the slits as shown in the, O, figure., (1999; 10M), Y, 22. Find the location of central maximum (bright, D, fringe with zero path difference) on the y-axis., , y, Screen, , 17. The velocity of central maxima at t 5sec is, (A) 50ms 1 along Y-axis, (B) 50ms 1 along negative Y-axis, (C) 25ms 1 along Y-axis, (D) 3 108 ms 1 along Y-axis, 18. Velcoity of 2nd maxima w.r.t central maxima at, t 2sec is, (A) 8 cms 1 i 20 ms 1 j (B) 8 cms 1 i, (C) 2 cms 1 i, , (D) 86 ms 1 i, , 19. Acceleration of 3rd maxima wrt 3rd maxima on, the other side of central maxima at t 3sec is, (A) 0.02ms2 i, , (B) 0.03ms 2 i, , (C) 10 ms 2 i, , (D) 0.6 ms 2 i, , NARAYANAGROUP, , S1, S, , O2, S2, , (A)4.33mm (B)2.56mm (C)3.26 mm (D)5.16 mm, 23. Find the light intensity of point O relative to, the maximum fringe intensity., (A), , 3, I max, 2, , (B), , 5, 3, 7, I max (C) I max (D) I max, 4, 4, 5, , 24. Now, if 600 nm light is replaced by white light, of range 400 to 700 nm, find the wavelengths, of the light that form maxima exactly at point, O. [All wavelengths in the problem are for the, given medium of refractive index, dispersion], (A) 650 nm, 6.32 nm, (C) 452 nm, 3.53 nm, , 4, . Ignore, 3, , (B) 350 nm, 4.33 nm, (D) 650 nm, 433 nm, 59
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , Comprehension-VI-III:, , 29. Assertion(A): Interference pattern is made by using, blue light instead of red light, the fringes, The arrangmeent for a mirror experiment is shown, becomes narrower., in figure. S is a point source of frequency, Reason(R):In Young’s double slit experiment,, 14, 6 10 Hz . D and C represnt the two ends of a, D, mirror placed horizontally and LOM represets the, , fringe, width, is, given, by, the, relation, screen., d, 30. Assertion(A): In YDSE, the fringes become, L, indistinct if one of the slits is covered with cellophane, paper., S, Reason(R): The cellophane paper decreases the, wavelength of light., 1mm, D Mirror C, O, 31. Assertion(A): If the whole apparatus of YDSE is, immersed in a liquid, then the fringe width will, 5cm, 5cm, 190cm, decrease, Reason(R): The wavelength of light in water is, more than that of air., M, , 25. Determine the width of the region where the, fringes will be visible, (A) 4cm, (B) 6cm (C) 2cm (D) 3cm, 26. Find the fringe width of the fringe pattern?, (A) 0.05cm, (B) 0.25cm, (C) 0.01cm, (D) 0.1cm, 27. Calculate the number of fringes, (A) 10, (B) 20, (C) 30, (D) 40, , ASSERTION AND REASON TYPE, , STATEMENT TYPE QUESTIONS, (A) Statement-I is true and Statement-II is true, and Statement-II is the correct explanation of, Statement-I., (B) Statement-I and Statement-II are true but, Statement-II is not the correct explanation of, Statement-I, (C) Statement-I is true, Statement-II is false, (D) Statement-I is false, Statement-II is true., , (A) A is true and R is true and R is the correct 32. Statement-I: In YDSE, if separation between the, explanation of A., slits is less than wavelength of light used, then no, (B) A and R are true but R is not the correct, interference pattern is observed., explanation of A, Statement-II: For interference pattern to be, (C) A is true, R is false, observed, light sources have to be coherent., (D) A is false, R is true., 33. Statement-I: In YDSE, if intensity of each source, 28. Statement-I: For the situtation shown in figure, two, is I 0 then minimum and maximum intensity is zero, identical coherent sources of light produce, interference pattern on the screen. The intensity of, and 4I 0 respectively.., minima nearest to S1 is not exactly zero., Statement-II: In YDSE, energy conservation is, not followed., Statement-II: Minimum intensity is zero when, interfering waves have same intensity at the location, MATRIX MATCHING TYPE, of superposition, 34. Column-I shows four situations of standard, young’s double slit arrangement with the, S1, P, screen placed far away from the slits S1 and, , S 2 . In each of these cases S1 P0 S2 P0 ;, S2, Screen, 60, , S1 P1 S 2 P1 4 and S1 P2 S2 P2 3 where, is the wavelength of the light used. In the, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, cases, B,C,D a transparent sheet of refractive, index and thickness ‘t’ is pasted on slit S 2 ., The thickness of the sheets are dfferent in, different cases. The phase difference between, the light waves reaching a point P on the screen, from the two slits is denoted by p and the, intensity by I p . Match the Column-I with, Column-II, valid for that situtation., (2009-JEE), Column-I, Column-II, , WAVE OPTICS, over one or both of the slits singularly or in, conjunction,causing the interference pattern to, be shifted across the screen from the original, pattern. In Column-I, how the strips have been, placed is mentioned whereas in Column-II, order of the fringe at point P on the screen that, will be produced due to the placement of the, strips as shown. Correctly match both the, Columns., Film, A, B, C, Thickness(in m ), 5, 1.5 0.25, Refractive index, 1.5, 2.5 2, , p2, p1, , S2, , p0, , A) S, , 1, , P) P0 0, , P, ( 1)t =, S2, , Slit II, , p2, , , 4, , Screen, , p1, p0, , B) S, , Q) P1 0, , 1, , ( 1)t =, S2, , p2, , , 2, , p1, p0, , C) S, , 1, , ( 1)t =, S2, , 3, 4, , R) I P1 0, , p2, p1, p0, , D), , S1, , S) I P0 I P1 , , T) I P2 I P1 , 35. A double slit interference pattern is produced, on a screen as shown in figure. Using, monochromatic light of wavelength 500nm., Point P is the location of the central bright, fringe that is produced when light waves arrive, in phase without any path differene. A choice, of three strips A,B,C of tranparent materials, with different thickness and refractive indices, is available as shown in table.These are placed, NARAYANAGROUP, , Column-I, (A) Only strip B is placed over slit-I, (B) Strip A is placed over slit-I and strip C is placed, over slit-II, (C) Strip A is placed over slit-I and strips B and C, are placed over slit-II in conjnction, (D) Strips A and C are placed over slit I in, conjunction and strips B is placed on slit-II, Column-II, (P) First bright, (Q) Fourth dark, (R) Fifth dark, (S) Central bright, 36. For the situtation shown in figure below, match, the entries of Column-I with that of Column-II, Assume thickness of films to be very small, compared to wavelength of incident light., Reflected system, Air, 1, 2, , Film-1, Film-2, , Air, Transmitted System, , 61
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, Column-I, Column-II, A) 1 2 P) Film 1 appears shiny, from the reflected, system, B) 1 2 Q) Film1 appears dark, from the reflected, system., C) 1 2 R) Film 1 appears shiny, from the transmitted, system, D) 1 2 S) Film 1 appears dark, from the transmitted, system, , INTEGER TYPE QUESTIONS, 37. In YDSE, the intensity of light at a point on, the screen is I for a path difference . The, intensity of light at a point where the path, difference becomes, , , I, is . Find the value, 3, P, , of P?, 38. In YDSE, the slits separation is 0.6mm and the, separation between slit and screen is 1.2m. The, wavelength of light used is 4800 . The, Å, angular fringe width is P 10-4 radian. FInd the, value of P?, 39. In YDSE, the wavelength of red light is, 7.5 10-5 cm and that of blue light is 5 10-5 cm ., , the screen only after reflection. For point ‘P’, , Kd 2, , ., 2, D, Where K is an integer. Find value of K?, {Assume D>>>d and d>>> ], , to be 2nd maxima, the value of, , S, , 3d, D, , LEVEL-VI - KEY, SINGLE OPTION TYPE, 1.B, 2.B, 3.C, 4.A, 5.C, 6.A, 7.B, 8.A,C,D, 9.A,C 10. A,C, Comprehension-I: 11. A 12. C 13.C, Comprehension-II: 14.A 15.C 16. B, Comprehension-III: 20.C 21.D, Comprehension-VI: 22.A 23.C, 24.D, Comprehension-V: 25.C 26.A 27.D 28.A 29.A, 30.C 31.C 32.B 33.C, MATRIX-MATCHING TYPE, 34. A P , S ; B Q ;, , th, , Find the value if ‘n’ for which n 1 blue, bright band coincides with nth red bright band?, 40. In YDSE, the slits have different widths. As a, result, amplitude of waves from two slits are A, and 2A respectively. If I 0 be the maximum 1), intensity of the interference pattern then the, intensity of the pattern at a point where the, phase difference between waves is is given, I, by 0 5 4 cos . Where P is in in integer.., P, Find the value of P?, 41. Consider the optical system shown in figure., The point source of light S is having, wavelength equal to . The light is reaching, 62, , P, , d, , C T ; D R, S ,T, 35. A R ; B R ;, C S;D P, 36. A Q, R ; B Q, S ;, C P , S ; D P , Q, R , S, 37.4 38.8 39.2 40.9 41.6, , LEVEL-VI - HINTS, SINGLE ANSWER TYPE QUESTIONS, Path dirfference at any point ‘P’ on the, circumference, P, , B, , P, r2, , , S1, , A, , , O, , S2, , C, , , S1, , x S1 P S 2 P, , , , S2, , O, 3, , 2, , S1 P R , , r1, , R, 3, , 2, , 3, cos , 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 3, cos x S1P S2 P, 2, x 3 cos For maxima at ‘P’, x n 3 cos , , S11 S 2 1 t S 2 E S11S1 S1 E, , S2 P R , , 2., , 1 2, cos 0, , ,1 [ maximum value of cos 1 ], 3 3, In each quadrant there are 3 maximas. Hence in 4, quadrants there willbe 12 maximas. Hence 1, 2, minimas in between so there are 12 fringes, PR=d, PO d sec , and, CO PO cot 2 d sec cos 2, Optical path difference between the two ras is, x C O P O d sec d sec cos 2 , , , , , , D 2 d 2 O S11 O S1 D, , 1 t OS2 OS1 , , n, cos n 0,1, 2,3, , 3, , , , , , , , O S 12 O S 2 1 t , , , , D2 d 2 D, , , , d 2 12, , 1 t 2 D 1 2 D , D , , , , , , , , d2 , d2, 2 D 1 , , D, , , , D, , D, 2, , , 2 , 2D, , , 2D , , , 16 2, d2 3 , 1 t 2 , 1 t 2 , 2 128, 2D 2 , 1, 1, 31 31, t 2 , t, , 2, 16 2, 16, 8, , , 2, , Path difference between the two rays is, , 31, 8, Conducting for costructive reference, , Thickness of mica sheet t , 2n , n 0,1, 2,3..., Condition for constructive interference should be, ....., x n , n 0,1, 2,3...., , d sec 1 cos 2 , cos , , , d , 2, , 2 cos , 2, 2 cos , , 5., , 2 t 2n 1, , , n 1, 2,3, , 2, , = 1.8, R, , 6., P, , A, , t, , = 1.5, , , , , , 4, , , For least value of ‘t’, n = 1 tmin 2 1 1, 4, , , 4d, O, , t 2n 1, , tmin , , 648 109, , 4, 4 1.8, , tmin 90 109 m tmin 90 nm, 1 = 1.4 and 2 = 1.7 and let t be the thickness, of each glass plates., Path difference at O, due to insertion of glass plates, will be, , B, , 4., , Let S11 and S 21 are the points on the wavefront, , = 1.8, , where perpendiculars can be drawn from S1 and, , S2 ., For central fringe formed ar E the path difference, should be zero., NARAYANAGROUP, , t, , = 1.5, , x, , = ( 2 – 1)t, = (1.7 – 1.4)t = 0.3 t …(1), 63
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, Now, since 5th maxima (earlier) lies below O and, 6th minima lies above O., This path difference should lie between 5l and 5l +, , S1P S 2 P t t x S1P S2 P 1 t, S1, , , 2, , th, , O, , So, let, where, , x, , = 5l = + D …(2), , x, , , <, 2, , , , th, , S2, , 1 t ., For Intensity to be zero at ‘P’ we have, , 2, ., )…(3), x, , , I(f) = Imax cos2 2 , 3, 4, , , = cos 2 , -2 , , x 2n 1, 3, I and since, 4 max, , 31, , = 5 +, , 9., , …(4), , 6, , 31, l = 0.3 t(Dx = 0.3 t), 6 6, , 7., , n1, 2, 7, n1 1 D = n2 2 D or n = =, 5, 2, 1, d, d, , This implies that 7th maxima of l1 coincides with 5th, maxima of l2 . Similarly 14th maxima of l1 will, coincide with 10th maxima of l2 and so on., nD, \ Minimum distance = 1 1, d, , = 7 × 5 × 10–7 × 103 = 3.5 × 10–3 m = 3.5 mm, , MULTIPLE ANSWER QUESTIONS, 8., , Solution:, , Path difference at ‘P’, , x S1P t air tmed S1Pair, 64, , I 4 I 0 cos 2, , cos 2, , , 2, , 75 , 2 , 4I0 , 4 I 0 cos, 2, 100 , , 3, , 3, 3, ; cos, , , 2, 4, 2, 4, 2, , 5 7 11, ,, ,, ,, ......, 2, 6 6 6 6, , (31)(5400 10 10 ), , t = 6(0.3) =, m, 1.8, or t = 9.3 × 10–6 m = 9.3 mm, Let n1 bright fringe corresponding to wavelength l1, = 500 nm coincides with n 2 bright fringe, corresponding to wavelength l2 = 700 nm., , 2n 1 , , 1 t t , 2, 2 1, , , 3, 5, Then t 2 1 , 2 1 , 2 1 , , , , 3, Imax = Imax cos2 2 , 4, , From Eqs. (3) and (4), we find that D =, i.e., x, , 2, , When there is no absorption of light by slab then, path difference at P earlier is S1P S 2 P 0 . so, change in optical path due to inserion of slab is, , 2, 2, x =, (5 + x ) = (10 +, , , , Intensity at O is given, , 6 Minima, 5 Maxima, , Due to the path difference Dx, the phase difference, at O will be, , or, , 1, , , , 5 7 11, ,, ,, ,, ......., 3 3 3 3, , the posibel value of , , 11 13 17, ,, ,, 3, 3, 3, , D, 1 t ., d, On the side of the slit where the slab is placed. On, placing the two slabs of smae thickness, the shift, should be, , 10.Shift due to the introduction of slab is, , D, 2 1 t ., d, Since three bright fringes shift is seen,, D, 3 D, 2 1 t , d, d, , t, , 3 D, 3 4000 1010, , 2 1 1 7 1.4 , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , WAVE OPTICS, , 3 4 107, , 4 106 4 m, 0.3, t 4 m, , Shift, , wgreen, = 0.143 × 10–3 m, Dw = wgreen – wred = (0.143 – 0.2) × 10–3 m, Dw, = –5.71 × 10–5 m, , Comprehension-II-I:, , D, 1, 2 1 t 3 0.3 4 10 6 1.2 103, d, 10, Shift = 1.2 mm, , 3, , , , 3, , COMPREHENSION TYPE, Comprehension - I-V:, 1, , 11. Path difference due to the glass slab,, Dx = (m – 1)t = (1.5 – 1)t = 0.5t, Due to this slab, 5 red fringes have been shifted, upwards, Therefore, , 0, , Area A 3 dx , , t, , , x, , t, , 1 3 t 2t, 2, , 14. Path difference, t, , , x 1SS2 2 S2 P 1SS1 2 S1 P 3 2 dx , , 0, , , t, , t, , 1 SS 2 SS1 2 S2 P S1 P 3 dx 2 dx, 0, , = (5)(7 × 10–7 m), t = thickness of glass slab = 7 × 10–6 m, 12. Let m¢ be the refractive index for green light then, = 6 Red, x, , x = 5 red or 0.5t, , ( 1 )t, , = 6 Red, =, , (6)(7 10 ), 7 10 6, , = 0.6, , 1.6, 13. In part (i), shifting of 5 bright fringes was equal to, 10–3 m. Which implies that, 5wred = 10–3 m [Here w = Fringe width], , ωgreen, ω red, , O, , NARAYANAGROUP, , 3, 0 2t t, 2, , 3t t, 6, , t 2 m, 2 2 t 2 10 m , 15. From the above equation, O 1 106 , , 3 yd, 0.5t, 2 D, , 3 yd, , 0.5t 1 106, 2 D, , 0.5 10 6 1 106 0.5 10 6, , green, , = , red, , wgreen = wred , red, , 2 1, , 2, , 1, y mm, 3, , D, or w µ l, d, , green, , 2 1 103 , , 10 6 2 D 106 1, 1, y, , 10 3, 3, 2 3 d 3 110, 3, , 103, wred =, m = 0.2 × 10–3 m, 5, , Now since w =, , In order to get central maxima at the centre ‘O’ of, screen is x 0, , 1 106 2t , , 7, , ( 1), , 0, , 16. , 5 107 , = (0.2 × 10 ) 7 107 , , , –3, , D 3000 1010 1 2, , 2 104 m, 2 d, 3 110 3, , 0.2mm, , 65
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , Comprehension - IV:, 20. Given — = 1.33, d = 1 mm, D = 1.33 m, , Comprehension-III-II:, S1, 1, D1 D gt 2, 2, , v = gt, , S2, , O, , X, , l = 6300 Å, Wavelength of light in the given liquid —, , Y, , ' =, P0, , x, , P, , 6300, , Å » 4737 Å = 4737 × 10–10 m, 1.33, , Fringe width, w =, , D, d, , 17. At any instant of time ‘t’, the situtation is shown is, (4737 10 10 m)(1.33 m), figure., w=, = 6.3 × 10–4 m, (1 103 m), The central maxima always lies along Y-axis at P0 . 21. Let t be the thickness of the glass slab., Its velcoity at any time ‘t’ is v gt along positive, Path difference due to glass slab at center O., Y-axis., µ glass, , 1.53, , v gt 10 5 50ms 1, x = µ liquid 1 t = 1.33 1 t, 18. Path difference corresponding to P., or x = 0.15 t, dx, x d sin d tan ; x D1, For 2nd maxima, O, , xd, 2 D1, x 2 1 2 ; x , D, d, , 1 2, , Location of central maxima is O, D gt , 2, , , , Now, for the intensity to be minimum at O, this path, , 2 D, 1, Location of 2nd maxima is d , D , , , nd, Velocity of 2 maxima wrt central maxima,, , 2, 2, V rel , 0 gt i gt i, d, d, 2 5000 1010 10 2i, Vrel , 2 102 ms 1.i, 1 10 3, , V rel 2 cms 1 .i, 3 D1 1 , 19. Location of 3 maxima d , D , , , rd, Location of 3 maxima on the other side is, rd, , 3 D 1, 1, d , D , , , , , 3, 3 g, 3 3, a r el , g , g i , g, d, d, d, , d, , x =, , , or 0.15 t, 2, , t = 15790 Å or t = 1.579 mm, , Comprehension - VI:, 22, , Given l = 600 nm = 6 × 10–7 m, d = 0.45 mm = 0.45 × 10–3 m ; D = 1.5 m, Thickness of glass sheet, t = 10.4 mm, = 10.4 × 10–6 m, Refractive index of medium, mm =, , 4, 3, , And refractive index of glass sheet, mg = 1.5, Let central maximum is obtained at a distance y, below point O. Then Dx1 = S1P – S2P =, , yd, D, , S1, , , i, , , , 6 6 5000 1010 10 , a rel , gi , i, d, 110 3, , , 2, arel 3102 ms2 i ; arel 0.03 ms i, 66, , , 2, 4737, =, Å, 2, , difference should be equal to, , 1, , O, P, , y, , S2, , , , , , , , , , g, Path difference due to glass sheet Dx2= 1 t, m, , Net path difference will be zero when Dx1= Dx2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , , WAVE OPTICS, , , , , D, , or, , yd, D, , y=, , 1.5, 10.4 10 6 (1.5), 1, , ; y = 4.33, 3, 4 3 0.45 10, , g, g, = 1 t y= 1 t d, m , m, , Substituting the values, we have, , × 10–3 m, , g, , 23. A t O, Dx1 = 0 and Dx2 = 1 t, m, , , Net path difference, Dx = Dx2, Corresponding phase difference, Df or simple f =, , 2, , Substituting the values, we have, 1.5, , , , 13 , , 13 , , Now, I(f) = Imax cos2 2 ; I = Imax cos2 6 , 3, 4, , g, , 24. At O : path difference is Dx = Dx2 = 1 t, m, , , For maximum intensity at O, Dx = nl (Here n = 1, 2, 3, …), 1.5, , x x x, ,, ,, … and so on Dx = 4 3 1, 1 2 3, , , 1.5, , , , (10.4 × 10–6 m) = 4 3 1 (10.4 × 103 nm), , , Maximum intensity will be corresponding to, l = 13000 nm,, , 1300, 1300, 1300, nm,, m,, nm …., 3, 2, 4, , = 1300 nm, 650nm, 433.33 nm, 325 nm nm …., The wavelengths in the range 400 to 700 nm are, 650 nm and 433.33 nm., , Comprehension - III:, , P2, , 25., S, , P1, O, , 1mm, B, , C 3 108, , 5 10 7 m, D 6 1014, D, Fringe width , d, D S1 A 200cm 2m ;, d SS1 2mm 2 103 m, , I = Imax, , l=, , AP1 AS1, AS BS1 200 0.1, , AP1 1, , 2cm, BS1 BD, BD, 10, Width of teh region wher fringes oberved is, ; 1 2 2cm, PP, 1 2 AP2 AP1 4 2 2cm PP, 26. Wavelength of light used is, , , , f = 6 107 4 3 1 = 3 , , , , , , AS1 BS1, 200 0.1, , 4cm AP2 4cm, BD, 5, Similarly, BCS1 and S1P1 A a re similar les, , AP2 , , or we can say y = 4.33 mm, , 2, . Dx, , , AP2 AS1, 1, , BS1 BD ; BS1 BS 1mm 10 cm, , D, , S1, , C, A, , Fringes will be observed in the region between P1, and P2 because the reflected rays lie only in this, region., From similar les : BDS1 and S1P2 A ., NARAYANAGROUP, , 5 10 7 2, 5 104 m, 3, 2 10, 5 102 cm 0.05cm, 27. No of fringes formed (N), PP, 2, 200, 1 2 , , 40, , 0.05, 5, , , , ASSERTION & REASON TYPE, 28. At the location of minima, the two waves have, different intensities hence minimum intensity is not, equal to zero., 29. b R hence b R hence fringes becomes, , D, ., d, A & R are true and R is the correct explanation, 30. When one of the slits is covered with cellophane, paper, the intensity of light imerging out from the slit, decreases. Now the two interfering beam have, different intensities or amplitudes. Hence intensity, at minima will not be zero and fringe will become, indistinct. A is true, R is flase., narrower. Fringe width , , D, 1, , , d, , , air hence air thus air, A is true and R is false., , 31. , , 67
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , WAVE OPTICS, , STATEMENT(S) TYPE, 32. path difference x d sin xmax d , if, , d then xmax so maxima can be present, and interference pattern cannot be observed., Statement-II is true,but not explaining Statement-I, 33. In YDSE, law of converstion of energy is obeyed., , MATRIX-MATCHING TYPE, 34. A), I P1 I1 I 2 2 I1I 2 cos, , , 2, , , , , 2, , , I P1 I 0 I 0 0 2 I 0, , 1, I0 I0 2I0 , 2, , I P2 2 I 0 I 0 I 0, , B), , 2 2 , , , , 4 , 2, , 2 2, , , 0, P1 1 t , 4 4 4 , , 2 2 2 , , P2 1 t , , , , , 3 4 3 , 12 , 6, , , I P1 I1 I 2 2 I1I 2 cos 00 I 0 I 0 2 I 0 4 I 0, , , I P2 I1 I 2 2 I1 I 2 cos , , 6 , 3, I 0 I 0 2 I 0 , , 2 , , , , I P2 2 3 I 0, C) P0 1 t , , BQ, , 2 2, , , , 2 , , 2 2, , P1 1 t , , 4 2 4 , , 2 , , , 4 , 2, 2 2, , P2 1 t , , 3 2 3 , , 68, , I P0 I1 I 2 2 I1I 2 cos , , I0 I0 2I0 1 2I0 2I0 0, , I P1 I 1 I 2 2 I 1 I 2 cos , 2, I0 I0 0 2 I0, , , I P2 I1 I 2 2 I1 I 2 cos , 3, , 1, 3 I 0 ( I P2 I P1 ), 2, 2 3 2 3, , , , D) P0 1 t , , 4 , 2, 2 3 2, , P1 1 t , , , 4 4 4 , , 2 2, , , 4, , , P0 0 A P, S, , P0 1 t , , , , 2 , , , 6 , 3, , I0 I0 2I0 , , 2 , I P2 I1 I 2 2 I1 I 2 cos , , 3 , , I P1 I P2 , , , , 2 3 2, , P2 1 t , , , 3 4 3 , , 5 2 5, , , , 12 , 6, I P1 I1 I 2 2 I1 I 2 cos , , 3, I P0 I1 I 2 2 I1 I 2 cos , 2, 2I0 2I0 0, , , 2I0, , , 5 , I P2 I1 I 2 2 I1 I 2 cos , , 6 , , 3, , , I 0 I 0 2 I 0 cos 2 I 0 2 I 0 , , 6, , 2 , , , , , , I P2 2 3 I 0 I P0 I P1 ; I P2 I P1 ;, 35. By using 1 t n we can find the value of, n.i.e., order of fringe produced at P, is that particular, strip has been placed over any of the slit. If two, strips are used in conjunction, path difference due, to each is added to get net path difference created., If two strips are used over different slits, their path, differences are subtracted to get net path difference., For A : When strip B is placed on slit-I alone then, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, nB, , B 1 t B 2.5 11.5 106, , , , 500 10 9, , B, , 1.5 1.5 10 6, , 4.5, 5 10 7, Order of fringe is 4.5 i.e., 5th dark Film A:, nA, , 1 t A 1.5 1 5 106, A, , , 500 10 9, , 5 1.5 10 6, 5, 5 10 7, Film C: Order of fringe, nA , , nC , , 500 109, , 1 2.5 108, 0.5, nC 0.5, 5 10 7, For B: When strip A is palced over slit-I and strip, C is placed on slit-II then, Net order of fringe: 5 0.5 4.5 i.e., 5th dark, For C: When strip A i splaced over slit-I and strips, B and C are placed over slit-II then, , , Net order of fringe 5 4.5 0.5 5 5 0, i.e., central bright fringe, For D: When slits A&C are placed over slit-I and, stipr B is placed over slit-II then Net order fringe, , 5 0.5 4.5 5.5 4.5 1, i.e.,1st bright, 36. A) For 1 2 , the two waves in the reflected, system differ by optical phase difference of and, hence due to destructive interference, the film, appears to be dark. Intransmitted system, the two, waves are in phase and hence due to constructive, interference the film appears to be shiny., A Q, R, Similarly we can go for the other conditions., , INTEGER TYPE QUESTIONS, 37. For path difference phase difference 2, 2 , I I 0 cos 2 , I 0 cos 2 , I 0 1 I 0, 2, 2 , , , , , path differece phase difference, 3, 2 2, 1 , , 3, 3, , For, , NARAYANAGROUP, , 1, I 1 I 0 cos 2 , 2, , , 2 2 , 2 , I 0 cos , I 0 cos , , 3 2 , 3, , I1 , , I0 I I, P4, 4 4 P, , 38) Angular fringe width is, , , D d, , 4800 1010 48 108, , 0.6 103, 6 104, 8 104 P 104 P 8, D, D, 39) n1 1 n2 2 n1 1 n2 2, d, d, n11 n22 ; nb b nR R, , , , C 1 tC 2 1 0.25 106, , , WAVE OPTICS, , n 1 5 105 n 7.5 105, n 1 1.5n 0.5n 1 n , , 1, 0.5, , 10, 2n2, 5, 40) As the amplitude are A and 2A then the ratio of, intensities is 1:4, n, , I max I 0 I1 I 2 2 I1 I 2 I 4 I 2 2 I, I0, 9, Intensity at any point:, I0 9I I , , I 1 I1 I 2 2 I1 I 2 cos , I 1 I 4 I 2 I 4 I cos , 1, I 1 5I 4 I cos ; I I 5 4cos , , I0, I, 5 4 cos 0 5 4 cos , 9, P, P 9, 8d 3d, 41) At P: Path difference x , D, I1 , , For 2nd maxima x 2 , , 24d 2, 12d 2, ;, D, D, , 6d 2 kd 2, , , k 6, 2, D, D, * * *, , 69
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , ELECTRO STATICS & CAPACITORS, SYNOPSIS, Charge and its properties, , , , , , , , , , , , , , , , , , , 70, , Study of characteristics of electric charges at, rest is known as electrostatics., Electric charge is the property associated with, a body or a particle due to which it is able to, produce as well as experience the electric and, magnetic effects., Charge is a fundamental property of matter and, never found free., The excess or deficiency of electrons in a body, gives the concept of charge., There are two types of charges namely positive, and negative charges., The deficiency of electrons in a body is known, as positively charged body., The excess of electrons in a body is known as, negatively charged body., If a body gets positive charge, its mass slightly, decreases., If a body is given negative charge, its mass, slightly increases., Charge is relativistically invariant, i.e. it does, not change with motion of charged particle and, no change in it is possible, whatsover may be, the circumstances. i.e., qstatic qdynamic, Charge is a scalar. S.I. unit of charge is, coulomb(C)., One electrostatic unit of charge, 1, (esu) =, coulomb., 3 109, One electromagnetic unit of charge, (emu) = 10coulomb, Charge is a derived physical quantity with, dimensions [AT]., Quantization of Charge : The electric charge, is discrete. It has been verified by Millikan’s, oil drop experiment., Charge is quantised. The charge on any body is, an integral multiple of the minimum charge or, electron charge, i.e if q is the charge then, q ne where n is an integer, and e is the, charge of electron = 1.6 1019 C ., , , , The minimum charge possible is1.6 1019 C ., , , , If a body possesses n1 protons and n2 electrons,, then net charge on it will be n1 n2 e,, i.e. n1 e n2 e n1 n2 e, , Law of conservation of charge, The total net charge of an isolated physical, system always remains constant,, i.e. q q q constant., In every chemical or nuclear reaction, the total, charge before and after the reaction remains, constant., This law is applicable to all types of processes, like nuclear, atomic, molecular and the like., Charge is conserved. It can neither be created, nor destroyed. It can only be transferred from, one object to the other., Like charges repel each other and unlike charges, attract each other., Charge always resides on the outer surface of a, charged body. It accumulates more at sharp, points., The total charge on a body is algebric sum of, the charges located at different points on the body., Electrification: A body can be charged by, friction, conduction and induction., By Friction: When two bodies are rubbed, together, equal and opposite charges are, produced on both the bodies., By Conduction: An uncharged body acquiring, charge when kept in contact with a charged body, is called conduction. Conduction preceeds, repulsion., By Induction: If a charged body is brought, near a neutral body, the charged body will attract, opposite charge and repel like charge present in, the neutral body. Opposite charge is induced at, the near end and like charge at the farther end., Inducing body neither gains nor loses charge., Induction always preceeds attraction., Repulsion is the sure test of electrification., , , 1, 1, Induced charge q q 1 where K is, K, Dielectric constant, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-1: Can two similarly charged bodies attrack, each other?, Sol: Yes, when the charge on one body (q1) is much, greater than that on the other (q2) and they are, close enough to each other so that force of, attraction between q1 and induced charge on the, other exceeds force of repulsion between q1 and, q2, Coulomb’s Law: ‘The force of attraction or, repusion between two stationary electric charges, is directly proportional to the product of, magnitude of the two charges and is inversely, proportional to the square of the distance, between them and this force acts along the line, joining those two charges’, , , F, , 1, q1q2, 4 0 r r 2, , 0 - permittivity of free space or vacuum or air.., , r - Relative permittivity or dielectric constant, of the medium in which the charges are situated., , , 0 8.857 10, , and, , 12, , farad, C2, ,, 2 or, metre, Nm, , 1, 9 109 N m, 4 0, , 2, , /C, , 2, , Permitivity of Medium: Permitivity is the, , Relative permitivity of a medium is defined as, the electrostatic force ( F0 ) between two charges, in air to the force (F) between the same two, charges kept in the medium at same distance., Dielectric constant (or) Relative permitivity, Pemitivity of the medium, K, Permitivity of free space, It has no units and no dimesions, Hence, the mathematical form of inverse square, law is given as, 1 q1q2 1 1 q1q2, F, , 4 r 2, K 4 0 r 2, For force or vacuum or air K=1 and for a good, conductor like metals, K , Conclusion : 1) The introduction of a glass, slab between two charges will decrease the, magnitude of rorce between them., 2) The introduction of a metallic slab between two, charges will decrase the magnitude of force to, zero., Note:1 When the some charges are separated by the, some distance in two different media,, 1 1 q1q2, F1 , K1 4 0 r 2 --------(1), 1, 1 q1q2, and F2 K 4 r 2 ------(2), 2, 0, , measure of degree of the medium which resist, from (1) and (2), F1 K1 F2 K 2, the flow of charges, In SI. for medium other than free space, the Note:2 When the same charges are separated by, different distance in the same medium, 1, 1, K, , constant 0, so that we can write the, Fd2 = constant (or) F1d12 F2 d 22, 4 , Note : 3 If different charges are at the same separation, equation for the force between the charges as, F 1 q11q21, F0 , 1 q1q2, , , r, in a given medium, F, F q1q2, F 0, 4 r 2, Note : 4 If the force between two charges in two, r is known as the relative permitivity of the, different media is the same for different, medium. It is a constant for a given medium and, separations., it charges separated by a medium decreases, 1 1 q1q2, compared with the force between the same, F, K 4 0 r 2 = constant, charges in free space separated by the same, distance., Kr2=constant or K1r12 K 2 r22, Relative permitivity r is also known as, If the force between two charges separated by a, dielectric constant K of the medium or specific, distance ' r0 ' in vacuum or air is same as the, inductive capacity., force between the same charges separated by a, Relative permitivity of a medium is defined as, distance ‘r’ in a medium., the ratio of permitivity of the medium to, r, permitivity of free space (or) air, Kr 2 r02 r 0, K, (or), NARAYANAGROUP, , 71
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Here K is dielectric constant of the medium., Forces between multiple charges :, The effective distance ‘r’ in medium for a Force on a charged particle due to a number of, point charges is the resultant of forces due to, r0, ., distance r0 in vaccum =, individual point charges, K, , , Similarly, the effective distance in vaccum for a, F F1 F 2 F 3 ....., dielectric slab of thickness ‘ x ’ and dielectric W.E-2: Two point sized identical spheres carrying, constant K is xeff f K, , Coulomb’s Law in Vector Form, , , , , , F12 , , 1 q1q2 , , r 12 and , F21 F12, 4 0 r22, , charges q1 and q2 on them are seperated by a, certain distance. The mutual force between, them is F. These two are brought in contact, and kept at the same separation. Now, the, 2, , F1 q1 q2 , 1 . Then, , ., force, between, them, is, F21, F12, F, F, 4q1q2, q1, q2, Here F12 is force exerted by q1 on q2 and F21 is Sol: When charges seperated by certain distance the, force is given by, force exerted by q2 on q1, 1 q1q2, Suppose the position vector of two charges q1, Then F 4 r 2 ------------(1), , , 0, and q2 are r1 and r2 , then electric force on, When charges brought in contanct and kept at, charge q1 due to q2 is,, the same distance the force is given by, , , F1 , , q1q2 , 1, r1 r2, , , 4 0 r r 3, 1, 2, , , , 2, , 1 q1 q2 , F , ----------(2), 4 0, 4r 2, , , , 1, , Similarly, electric force on q2 due to charge q1, , 2, , F 1 q1 q2 , , , 1, q1q2 , from (1) and (2) ; , F2 , r r1, F, 4q1q2, , , , 3 2, 4 0 r r, is, 2, 1, W.E-3: Consider three charges q1,q2 and q3 each, equal to q at the vertices of an equilateral, Here q1 and q2 are to be substitued with sign., , , triangle of side ‘ l ’ what is the force on any, r1 x1i y1 j z1k and r2 x2 i y2 j z2 k where, charge due to remaining charges., co- Sol : The forces acting on the charge ‘q’ are, x1 , y1 , z1 and x2 , y2 , z2 are the, ordinates of charges q1 and q2 ., , , , , , , Limitations of Colulomb’s Law, , , , , , , , , Coulomb’s law holds for stationary charges only, which are point sized., This law is valid for all types of charge, distributions., This law is valid at distances greater than, 10 15 m., This law obeys Newton’s third law., This law represents central forces., This law is analogous to Newton law of, gravitation in mechanics., The electric force is an action reaction pair, i.e, the two charges exert equal and opposite forces, on each other., The electric force is conservative in nature., Coulomb force is central., Coulomb force is much stronger than, , , , 36, gravitational force. 10 Fg FE, , 72, , , , q, , q, , F1 , , q, , 1 q2, 4 0 l 2, , 1 q2, F2 , 4 0 l 2, , , clearly F1 F2 F, The resultant force is, F 1 F 2 F 2 2 FF cos 600, , 3F 3, , 1 q2, 4 0 l 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-4: A particle of mass ‘m’ carrying a charge, q1 is moving around a fixed charge q2, along a circular path of radius ' r ' find time, period of revolution of charge q1, Sol: Electrostatic force on -q1 to +q2 will provide, the necessary centripetal force, Hence, , Kq1q2 mv 2, , ; v, r2, r, , Kq1q2, mr, , 16 3 0 mr 3, 2 r, , T, q1q2, v, W.E-5: Two identical small charged spheres each, having a mass ‘m’ hang in equilibrium as, shown in fig. The length of each string is ' l ', and the angle made by any string with vertical, is .Find the magnitude of the charge on each, sphere., So l :The forces acting on the sphere are tension in, the string T, force of gravity ‘mg’ and repulsice, , force Fe., l, , , , Let v is the volume of each ball, then mass of, each ball is m v ; When balls are in air, T cos mg ; T sin F, F mg tan vg tan ---------(1), When balls are suspended in liquid. The, F, 1, coulumbic force is reduced to F , and, K, apparent weight = weight - upthrust ;, W 1 vg vg, According to the problem, angle is, uncharged-Therefore, , F 1 W 1 tan vg vg tan ------(2), F, , K, 1, , F, Test charge: That small positive charge, which, does not influence the other charges and by the, help of which we determine the effect of other, charges, is defined as test charge., From (1) and (2) ;, , Linear charge density is defined as the, , l, , T, T T cos, T cos mg ----(1), Kq 2, T sin, F, F, T sin Fe 2 ---(2), x, r, W, W, From (1) and (2), 2, Kq, mg tan , Fe m g tan ;, r2, 1, q2, from fig r 2l sin ; 4 2l sin 2 mg tan , , 0 , , charge per unit length., , q 16 0 l 2 mg tan sin 2 , W.E-6: Two identical balls each having density, are suspended from a common point by, two insulating strings of equal length Both, the balls have equal mass and charge. In, equilibrium, each string makes an angle , with the vertical. Now both the ball are, immersed in a liquid. As a result, the angle , does not change. The density of the liquid is, . Find the dielectric constant of the liquid., Sol:, , Surface charge density is defined as the, , dq, dl, where dq is the charge on an infinitesimal length, dl., Units of are Coulomb / meter (C/m), Examples:-Charged straight wire, circular, charged ring, , , , charge per unit area., dq, ds, where dq is the charge on an infinitesimal surface, area ds. Units of are coulomb / meter 2 (C /, , , , m2 ) ., Examples:-Plane sheet of charge, conducting, sphere., , Volume charge density is defined as charge, l, T, F, W, , , , l, , l, , T, , T T cos, T sin, x, air, W, , NARAYANAGROUP, , F, , F, W, , , , l, , per unit volume., T T cos+ Vg, , T sin, x, Liquid, W, , F, , dq, dv, where dq is the charge on an infinitesimal volume, , , , 73
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, element dv. Units of are coulomb / meter 3, , , , (C/ m3 ), Examples:- Charge on a dielectric sphere etc.,, Charge given to a conductor always resides on, its outer surface., If surface is uniform then the charge distributes, uniformly on the surface., In conductors having nonspherical surfaces, the, , W.E-8: A thin fixed ring of radius ‘a’ has a, positive charge ‘q’ uniformly distributed over, it. A particle of mass ‘m’ having a negative, charge ‘Q’ is placed on the axis at a distance, of x x a from the centre of the ring., , Show that the motion of the negatively, charged particle is approximately simple, harmonic. Calculate the time period of, , oscillation., surface charge density will be larger when Sol: The force on the point charge Q due to the element, the radius of curvature is small, dq of the ring is, The working of lightening conductor is based, 1 dqQ, dF , on leakage of charge through sharp point due to, 4 0 r 2 along AB, high surface charge density., For every element of the ring, there is, W.E-7: A ring of radius R is with a uniformly, symmetrically situated diametrically opposite, distributed charge Q on it. A charge q is now, element, the components of forces along the axis, placed at the centre of the ring. Find the, will add up while those perpendicular to it will, increment in tension in the ring, cancel each other. Hence, net force on the charge, Sol: Consider an element of the ring. Its enlarged view, is as shown. For equilibrium of this segment,, Q is -ve sign shows that this force will be, we can write., towards the centre of ring., , , F, , d, , q, , F dF cos cos dF, , Q, d/2, T, , d/2, T, d, q, , d , F 2T sin , , 2 , Here F is the repulsive force between q and, elemental charge dQ, Q, , , dQ 2 R Rd , , , The electric outward force on element is, 1 qdQ, 4 0 R 2, From the above three equations, we can write, F, , 74, , , , 1, x, , r 4 0, , Qdq , r 2 , , so,, F , , 1, Qx, 4 0 3, r, , Qqx, , 1, , dq 4 , , 0, , a, , 2, , x, , 3, 2 2, , 1, , , , ----(1), , 1, , (as r a 2 x 2 2 and r a 2 x 2 2 ), As the restoring force is not linear, the motion, will be oscillatory. However, if x a , then, F , , 1 Qq, x kx, 4 0 a 3, , Qq, with k 4 a 3, 0, , 1 q QRd, d , 2T , , 2, 4 0 R 2 R, 2 , , i.e., the restoring force will become linear and, so the motion is simple harmonic with time, period, , sin , , T, , for small angle , , 4 0 ma 3, 2, m, 2, 2, , k, qQ, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-9: A point charge q is situated at a distance, , Electric Field: The space around electric, , ‘r’ from one end of a thin conducting rod of, length L having a charge Q (uniformly, distributed a long its length). find the, magnitude of electric force between the two., L, r, dE, Sol:, P, dx, x, Consider a small element of the rod of length, dx, at a distance ‘x’ from the point charge q., Trating the element as a point charge, the force, between ‘q’ and charge element will be, 1 qdQ, Q, dF , dQ dx, 4 0 x 2 ; But,, L, , charge upto which its influence is felt is known, as electric field., Electric field is a conservative field., Intensity of Electric Field: The intensity of, electric field or electric field strength E at a point, in space is defined as the force experienced by, unit positive test charge placed at that point”., The intensity of electric field is also ofted called, as electric field strength., Consider an electric field in a given region., Bring a charge q0 to a given point in that field, without disturbing any other charge that has, produced the field., , L et F be the electric force experienced by q0, , 1 qQdx, So, dF 4 Lx 2, 0, F dF , , , , 1 qQ r L dx, 4 0 L r x 2, , 1 qQ 1 r L, 1 1, 1 , , , , , , 4 0 L x r, 4 0 r r L , , 1, qQ, 4 0 r r L , Lines of Force: Line of force is an imaginary, path along which a unit +ve test charge would, tend to move in an electric field., Lines of force start from +ve charge and end at, –ve charge., Lines of force in the case of isolated +ve charge, are radially outwards and in the case of isolated, –ve charge are radially inwards., The tangent at any point to the curve gives the, direction of electric field at that point., Lines of force do not intersect., Lines of force tend to contract longitudinally, and expand laterally., F, , , , , , , , , , and it is found to be proportional to q0, , , , , is, F q0 F Eq0 . Here, F, E, proportionality constant called electric field, strength, , F, E, q0, Electric field strength is a vector quantity. Its, direction is the direction along which a free, positive charge experiences the force in the, electric field., The S.I unit of elctric field strength is newton, per coulomb (NC-1). It can also be expressed in, volt per metre (Vm-1)., , Electric field internsity due to an isolated, point charge : Consider a point charge ‘Q’, placed at point A as shown. Let us find the, , electric field E at a point P at a distance ‘r’’, from charge Q. Imagine a positive test charge, , q0 P. The charge Q produces a field E at P.., q0, Q, , The force applied by Q and q0 is given by, 1 Qq0, 4 0 r 2 . This acts along Ap., According to definition, F, , 1 Q, E E, rˆ, q0, 4 0 r 2, F, , (A) Radially outward (B) Radially inward, , –, , +, , (C), NARAYANAGROUP, , +, , +, , (D), , If ‘q’ is positive, E is along AP and if ‘q’ is, negative E will be along PA ., 75
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, If the charge ‘q’ is in a medium of p is in medium, of permititivity , and dielectric constnat K,, , , , , K the intensity of electic field in a, 0 , , medium (Emed) is given by, , Emed , , 1 Q, 4 r 2, , Emec , , E free space, , x, , (r – x), q1, q2, , Let P be the null point where Enet 0, , E1 E2 0 (due to those charges), , , or E1 E2 and E1 E2, , , , 1 q1, 1, q2, , , 2, 4 0 x, 4 0 r x 2, q1, q2, or x 2 r x 2, , , , r, q2, 1, q1, , Case 2 : If the charges are unlike, the neutral point , will be outside the charge on the lime joining, them., r, q1, q2, x, q1, q2, In this case x 2 r x 2, , , x, , , , 76, , F Eq, , F Eq, , Motion of a charged particle in a uniform, electric field :, a) A charged body of mass ‘m’ and charge ‘q’ is, initially at rest in a uniform electric field of, intensity E. The force acting on it, F Eq ., Here the direction of F is in the direction of field, if ‘q’ is ve and opposite to the field if ‘q’ is, ve ., The body travels in a straight line path with, F Eq, , initial, uniform acceleration, a , m m, velocity, u 0 ., At an instant of time t., Eq , Its final velocity, v u at , t, m , 1 2 1 Eq 2, Displacement s ut at , t, 2, 2 m , , Momentum, P mv Eq t, Kinetic energy,, , 1 2 1 E 2q 2 2, mv , t, 2, 2 m , When a charged particle enters perpendicularly, into a uniform electric field of intensity E with a, velocity ‘v’ then it describes parabolic path as, shown in figure., K .E , , +, +, q, , +, , +, , +, , +, , +, , u, y, , r, , q2, 1, q1, , If instead of a single charge, field is produced, by no.of charges, by the principle of super, , On solving we get, , E, , +, , In the case of a system of charges if the net, electric field is zero at a point, it is knwon as, null point., Application : Two point (like) charge q1 and, q2 are separated by a distance ‘r’ and fixed, We, can locate the point on the line joining those, charges where resultant or net field is zero., , Case 1: If the charges are like, the neutral point will, be between the charges., , on solving we get, , E, , K, , NULL POINT OR NEUTRAL POINT, , x, , position resultant electric field intensity, , E E1 E 2 E 3 ...., If q0 is positive charge then the force acting on it, is in the direction of the field., If q0 is negative then the direction of this force, is opposite of the field direction., , x, , Along the horizontal direction, there is no, acceleration and hence x ut ., Along the vertical direction, acceleration, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , Oblique projection of charged particle in, an uniform elctric field (Neglecting, gravitational force) : Consider a uniform, , F Eq, , (here gravitational force is not, m m, considered), a, , electric field E in space along Y-axis. A negative, charged particle of mass ‘m’ and charge ‘q’ be, projected in the XY plane from a point ‘O’ with, a velocity u making an angle with the X-axis., (Neglecting gracitational force)., , 1 Eq 2, Hence vertical displacement, y , t, 2 m , 2, , 1 qE x qE 2, , , x, 2 m u 2mu 2 , At any instant of time t, horizontal component, of velocity, vx u, vertical componet of velocity, y, , , , , Ej, , u, , , Eq , v y at , t, m , , O, , Initial velocity of the particle is, , u u cos iˆ u sin ˆj, Force acting on the particle is, , , F qE (along-ve Y axis), , 2 2 2, , E q t, m2, Two charges +Q each are separated by a, distance 'd'. The intensity of electric field at, the mid point of the line joining the charges is, zero., W.E-10 : Two charges +Q each are placed at the, two vertices of an equilateral triangle of side, a. The intensity of electric field at the third, vertex is, E E, Sol:, v v vx2 v y2 u 2 , , E1 E 2 E 2 2 EEC os , 2, , a, , qE ˆ, , a, j, m, Velocity of the particle after time ‘t’ is, , v u at ; v u cos iˆ u sin at ˆj, , If the point of projection is taken as origin, its, position vector after time ‘t’ is, , r xiˆ yjˆ where x=(ucos ) t, , a, , 1, y u sin t at 2, 2, If the charged particle is projected along the xaxis, then 00, , 2, , 2 E 2 E Cos, 2 E 2 1 cos , , +Q, , a, , +Q, , 1, Q, , ; E= 3 4 a 2, 0, 2, W.E-11: Two charges +Q, -Q are placed at the, two vertices of an equilateral triangle of side, ‘a’, then the intensity of electric field at the, third vertex is, , Sol :E1 = 2E cos = E ( 1200 ), 2, , = 2 ECos, , Eq ˆ, v uiˆ , tj, m, 1 Eq 2, t, 2 m, Direction of motion of particle after time ‘t’, makes an angle with x-axis, where, , Here x ut and y , , E, , , E, a, , +Q, , 1, , a, , Q, , E1 = 4 a 2 ., 0, NARAYANAGROUP, , Eqt, mu, A charged particle of charge Q is projected, with an initial velocity u in a vertically upward, electric field making an angle to the horizontal., Then, If gravitational force is considered, , Net force mg F mg Eq, tan , , 120°, a, , x, , –Q, , 77
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Eq, m, The negative sign is used when electric field is, in upward direction where as positive sign is, used when electric field is in downward, direction for positively charged projected , particle., , Net acceleration = g , , a. Time of flight , , 2u sin , EQ, g, m, , b. Maximum height, , , , A sphere is given a charge of 'Q' and is, suspended in a horizontal electric field. The, angle made by the string with the vertical is,, EQ , tan 1 , , mg , , The tension in the string is E Q , Hence effective acceleration, , , , mg , , 2, , 2, , F, Eq , geff g 2 , m, m, Time period of oscillation is given by, , u 2 sin 2 , EQ , , 2 g , , m , , , T 2, , u 2 sin 2, c. Range , EQ, g, m, , , , 2, , l, 2, g eff, , 1, Eq , g2 , , m , , 2, , W.E-12: An infinite number of charges each ‘q’, Intensity of electric field inside a charged hollow, are placed in the x-axis at distances of, conducting sphere is zero., 1,2,4,8...meter from the origin. If the charges, A hollow sphere of radius r is given a charge Q., are alternately positive and negative find the, Intensity of electric field at any point inside it is, intensity of electric field at origin., zero., Sol: The electric field intensities due to positive, Intensity of electric field on the surface of the, charges and due to -ve charges the field intensity, Q, is towards the charges, 1, , Q, , sphere is 4 r 2, 0, , E4, , r, , E2, , E1, , Intensity of electric field at any point outside the, sphere is (at a distance 'x' from the centre), , E3, q, , q, , q, , q, , x=1, , x=2, , x=4, , x=8, , Q, , 1 Q, 4 0 x 2, , x=0, x, , Time period of oscillation of a charged body, , , The bob of a simple pendulum is given a +ve, charge and it is made to oscillate in a vertically, upward electric field, then the time period of, , oscillation is, , , T, , l, 2, EQ, g, m, , E, , then the time period is given by, , E, , l, , 2, g, , EQ, m, , 1, Q, Q 4, , 4 0 1 4 0 5, 1 4 , , , E, , 4 Q, 5 4 0, , E, , Q, 5 0, , E, , mg, , 78, , Q , 1 1 1, , 1 2 2 2 ........ , , 4 0 2 4 8, , Since the expression in the bracket is in GP with, 1 1, a common ratio = 2 , 2, 4, E, , mg, , In the above case, if the bob is given a -ve charge, , T, , The resultant intensity at the origin, E E1 E3 E4 , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-13: A point mass ‘m’ and charge ‘q’ is, connected with a spring of negligible mass, with natural length L., Initially spring is in, natural length. Now a horizontal uniform, electric field E is switched on as shown. Find, a) The maximum separation between the mass, and the wall, b) Find the separation of the point mass and, wall at the equilibrium position of mass, c) Find the energy stored in the spring at the, equilibrium position of the point mass., E, , m, , wall, , K, , q, , R, E, m, , q, L, , mg, , As collision with the wall is perfectly elastic,, the block will rebound with same speed and as, now is motion is oppisite to the acceleration,, it, will come to rest after travelling same distance, L in same time t. After stopping it will beagain, accelerated towards the wall and so the block, will execute oscillatory motion with ‘spain’ L, and time period, , 2mL, L, qE, Sol: At maximum separation, velocity of point mass, However, as the restoring force F(=qE) when, is zero. From work energy theorem,, the block is moving away from the wall is, constance and not proportional to displacement, Wspring W field 0, x, the motion is not simple harmonic., 1 2, W.E-15:, Six charges are placed at the vertices of, qEx0 kx0 0 (x0 is maximum elongation), 2, a regular hexagon as shown in thg figure., The electric field on the line passing through, 2qE, 2qE, x0 , ; separation = L , point O and perpendicular to the plane of the, K, k, T 2t , , b), , At, , equilibrium, qE, Eq Eq kx x , k, qE, separation = L , k, , position., , figure at a distance of x a from O is, a, +Q, , +Q, , –Q, , O, , –Q, , 2, , 1 2 1 qE , q2E 2, +Q, –Q, U, , kx, , k, , c), , , 2, 2 k , 2k, Sol: This is basically a problem of finding the electric, W.E-14: A block having mass ’m’ ad carge ‘q’ is, field due to three dipoles. The dipole moment, resting on a frctionless plane at distance L, of each dipole is P Q 2a , from the wall as shown inf fig. Discuss the, motion of the block when a uniform electric, KP, Electric field due to each dipole will be E 3, field E is applied horizontally towards the wall, x, assuming that collision of the block with the, The direction of electric field due to each dipole, wall is perfectly elastic., is as shown below:, Sol: The situation is shown in fig. Electric forece, , , Enet E 2 E cos 60 0 2 E, F qE will accelerate the block towards the, a, wall producing an acceleration, +Q, –Q, F qE, 1 2, E, a , L at, 60°, , , Qa, Qa, 1, 2, m m, 2, , , +Q, E –Q, 2, , 3, 60°, 4 0 x 0 x, 2L, 2mL, E, , i.e., t , +Q, –Q, a, qE, NARAYANAGROUP, , 79
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , W.E-16: The field lines for two point charges are, , A, , shown in fig., , D B, , C, , E, , iii. What are teh sing of qA and qB ?, iv. If qA and qB are separated by a distance, , , , Q, d, , Due to this dq, electric field at centre of arc C is, given as, dq, dE , 4 0 R 2, The electric field component dE to this segment, dE sin which is perpendicular to the angle, bisector gets cancelled out on integration., The net electric field at centre will be along angle, bisector which can be calculated by integrating, dEcos within limits from / 2 to / 2, Hence net elctric field strength at centre C is, dq , , i. Is the field uniform?, ii. Datermine the ratio qA / qB ., , 10, , Consider a polar segment on arc of angular width, d at an angle from the angular bisector XY, as shown. The length of elemental segment is, Rd . The charge on this element dq is, , , , 2 1 cm, find the position of neutral, point., Sol: i. No, ii. Number of lines coming from or coming to a, charge is proportional to magnitude of charge,, q A 12, so q 6 2, B, , Ec dE cos , /2, , , , iii. qA is positive and qB is negative, iv. C is the other neutral point., v. For neutral point EA = EB, , Q, , 4 0 R 2, , qA, 1, 1 qB, , 2, 4 0 1 x , 4 0 x 2, , A, , , , EB, , B, , l, , C, , EA, , X + +, +, +, +, +, , Sol:, , , , cos d, , / 2, , Q, /2, sin / 2, 2 , 4 0 R , , Ec , , 2Q sin / 2 , 4 0 R 2, , for a semi circular ring . So at centre, 2Q sin / 2 2Q sin / 2 , 2Q, , 2, 2, 2, 4 0 R , 4 0 R , 4 0 R 2, ElECTRIC FIELD STRENGTH DUE TO A, UNIFORMLY CHARGED ROD, , Ec , , At an axial point :, , + + +, , L, , , d, , R, dEsin, , R, , C, Y, , dE, , 80, , /2, , Q, sin / 2 sin / 2, 4 0 R 2, , 2, , l x qA, x q 2 x 10 cm, , , B, ELECTRIC FIELD STRENGTH DUE TO A, CHARGED CIRCULAR ARC AT ITS, CENTRE, W.E-17. Consider a circular arc of radius R which, subtends an angle at its centre. Let us, calculate the electric field strength at C., , Q, cos d, 4 0 R 2, / 2, , , , dEcos, , P, + + + + + + + ++ r, Consider a rod of length L, uniformly charged, with a charge Q. To calculate the electric field, strength at a pont P situated at a distance ‘r’ from, one end of the rod, consider an element of length, dx on the rod as shown in the figure., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , L, , , , r, dx, , P, , x, , Ep , , dE, , Charge on the elemental length dx is dq , , Q, dx, L, , dq, Qdx, , 2, 4 0 x, 4 0 Lx 2, The net electric field at point P can be given by, integrating this expression over the length of the, rod., , E p dE , , , , Q, Q, dx , Lx 4 0, 4 0 L, 2, , r, , rL, , , r, , Ep , , Q 1, 1 , Q, , , , , 4 0 L r r L 4 0 r r L , At an equatorial point : To find the electric, field due to a rod at a point P situated at a, distance ‘r’ from its centre on its equatorial line, dE, , dE cos, , , P, , dE sin, r, + + + + + + + +, , P, r, , +++ +, , , + + +, dx, , L, (a), , (b), , Consider an element of length dx at a distance, ‘x’ from centre of rod as in figure (b). Charge on, Q, the element is dq dx ., L, The strength of electric field at P due to this point, charge dq is dE., dq, dE , 4 0 r 2 x 2 , , E p dE cos , , L, 2, , 1, Qdx, r, , , 2, 2, x , r 2 x 2 4 0, , L r, , L, , 2, , NARAYANAGROUP, , r, , 2, , x2 , , 3/ 2, , Q, Q, cos d , sin , , 4 0 Lr, 4 0 Lr, , 1 x, 1, Substituting tan r sin, , x, 2, , x r2, , , , Q , , Q , x, , EP , , , 4 0 r , 4 0 L x 2 r 2 L ;, , 2, , L, 2, , , , 1, , 2, , L, r2 , 4, , , , Q , 2, 2, , 2, 4 0 r L 4r , ELECTRIC FIELD DUE TO A UNIFORMLY, CHARGED RING :, The intensity of electric field at a distance, ‘x’ meters from the centre along the axis:, Consider a circular ring of radius ‘a’ having a, charge ‘q’ uniformly distributed over it as shown, in figure. Let ‘O’ be the cetnre of the ring ., , The component dEsin will get cancelled and, net electric field at point P will be due to, a), integration of dE cos only.., Net electric field strength at point P can be given, as, , , L, 2, , Qr, r sec 2 d, Q, r sec 2 d, , ;, 4 0 L r 3 sec3 , 4 0 Lr r 3 sec3 , , , , rL, , Ep , , , , dx, , x, r, x r tan ; On differentiation; dx r sec 2 d, , 1, dx, x2, , Q 1 , Ep , 4 0 L x r, , , , From the diagram tan , , dE , , rL, , Qr, 4 0 L, , L, 2, , Ep , , dx, , A, a, , O, , , x, , , , (x, 0), p, , dE2, X, dE1, , dx B, , Consider an element dx of the ring at point A., The charge on this element is given by, q, qdx, , dq dx charge density dq dx, 2 a 2 a, The intensity of electric field dE1 at point P due, to the element dx at A is given by, 1 dq, dE1 , 4 0 r 2, The direction of dE1 is as shown in figure. The, component of intensity along x-axis will be, 81
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 1 dq, cos dE1 cos , 4 0 r 2, The component of intensity along y-axis will be, 1 dq, sin dE1 sin , 4 0 r 2, Similarly if we consider an element dx of the, ring opposite to A which lies at B, the component, of intensity perpendicular to the axis will be, equal and opposite perpendicular to the axis will, be equal and opposite to the component of, intensity perpendicular to the axis due to element, at A. Hence they cancel each other. Due to, symmetry of ring the component of intensity due, to all elements of the ring perpendicular to the, axis will cancel., So the resultant intensity is only along the axis, of the ring. The resultant intensity is given by, 1 dq, cos , E, 4 0 r 2, E, , 1, qdx, x, (where cos x / r ), 2, , 4 0 2 ar r, , E, , 1, qx, , , 4 0 2 a , , 1, , a, , 2, , x, , 3, 2 2, , , , dx, , r 3 a 2 x 2 3/ 2 , , , 1 qx, 1, E, 2 a, 4 0 2 a a 2 x 2 3/ 2, E, , 1, qx, 4 0 a 2 x 2 3/ 2, , At its centre x = 0, Electric field at centre is zero., By symmetry we can say that electric field, strength at centre due to every small segment on, ring is cancelled by the electric field at centre, due to the element exactly opposite to it. As in, the figure the electric field at centre due to, segment A is cancelled by that due to segment B., Thus net electric field strength at the centre of a, uniformly charged ring is Ecentre 0 ., ELECTRIC FIELD STRENGTH DUE TO A, UNIFORMLY SURFACE CHARGED DISC, Consider a disc of radius R, charged on its, surface with a charge density ., 82, , Let us find electric field strength due to this disc, at a distance ‘x’ from the centre of disc on its, axis at point P as shown in figure., Consider an elemental ring of radius ‘y’ and, width dy in the disc as shown in figure. The, charge on this elemental ring dq can be given as, dq 2 ydy, {Area of elemental ring ds= dy 2 ydy }, dy, P, x, , dE, , Electirc field strength due to a ring of radius Y,, charge Q at a distance x from its centre on its, axis can be given as, Qx, E, 3/2, 4 0 x 2 y 2 , Due to the lemental ring electric field strength, dE at point P can be given as, xdq, 2 y dyx, dE , , 3/, 2, 3/ 2, 4 0 x 2 y 2 , 4 0 x 2 y 2 , Net electric field at point P due to whole disc is, given by integrating above expression within the, limits from 0 to R, R, , E dE , 0, , x, , 4 0, , 0, , 4 0 x 2 y 2 , , 3/ 2, , R, , R, , , , 2 xydy, , 2 ydy, , x, , 2, , y2 , , 3/ 2, , 2 x 1 , , , , 4 0 x 2 y 2 , 0, , , , x, 1 2, , 2 0 , x R2 , Electric field strength due to a uniformly charged, disc at a distance x from its surface is given as, E, , E, , , , x, 1 2, , 2 0 , x R2 , , , If we put x = 0 we get E 2, 0, Electric dipole: A system of two equal and, opposite point chargesfixed at a small distance, constitutes an electric dipole. Electric dipole is, analogous to bar magnet or magnetic dipole in, magnetism. Every dipole has a characteristic, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , property called dipole moment, which is similar, to magnetic moment of a bar magnet. If 2a is the, distance between the charges +q and –q, then, electric dipole moment is p = q.2a., 2a, –q, , +q, , P, , Dipole moment is a vector quantity and its, direction is from negative charge to positive, charge as shown., ELECTRIC FIELD AT ANY POINT DUE, TO A DIPOLE : We know that the electric, field is the -ve gradiant of potential. In polar, form if V is the potential at r,θ the electric, field will have two components radial and, transverse components which are represented, by Er & Eθ respectively.., E, , , E, , Field at a point on the axial line : ( 00 ), 2p, Eaxial , 4pe0 r3, Field at a point on the equitorial line ( 90 0 ), p, E eq u ito ria l , 4 p e0 r 3, The direction of E at any point is given by, psin q, 4pe0 r 3, Eq, 1, , tan f , E r 2p cos q tan f tan q, 2, 4pe0 r 3, tan 1 1 / 2 tan , Note : Electric dipole placed in an uniform electric, field experiences torque is given by, pE sin in vector form p E, , Er, , +q, , qE, , P, , 2a sin, qE, , –q, , E, , , –q, , O, , +q, , V , p cos q 1 , Then E r , , r , 4pe0 r r 2 , V, , , E r r, , 2 p cos q , , Er , 3, 1, , V, , , , , 4 p e0r, E r , The tranverse component of electric field, 1 V 1 psin q , , E , r 4pe0 r 2 , r , , Eq , , p s in q, 4 pe0r3, , E E2 E 2R, E, , p 2 sin 2 q, , 4pe r , 3, , 2, , 0, , , , 4p2 cos2 q, , 4pe r , 3, , 2, , 0, , p, 4cos2 q sin2 q, 3, 4pe0r, p, 1 3cos2 q , E, , 3, 4pe0 r , E, , NARAYANAGROUP, , The torque on the dipole tends to align the dipole, along the direction of electric field., The net force experienced by it is zero., Note : The potential energy of dipole in an electric, field is, V1, , V2, +q, , , –q, E, , 2a sin, , U= – pE cos θ ., , In vector form U p.E, if q 0o ; t 0 and U pE, if q 90o ; t pE and U 0, if q 180o ; t 0 and U pE, , , So, if p is parallel to E then, potential energy, is minimum and torque on the dipole is zero,, and the dipole will in stable equilibrium., , , If p is anti parallel to E then, potential energy, is maximum and again torque is zero, but it is in, unstable equilibrium, 83
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Note : Work done in rotating a dipole in electric W.E-18: An electric dipole of dipole moment p is, kept at a distance r from an infinite long, field from an initial angle θ1 with field to final, charged wire of linear charge density as, angle θ2 with field is, shown. Find the force acting on the dipole ?, W pE(cos q1 cos q2 ), Note : Force on dipole in non-uniform electric field:, +, P, The force on the dipole due to electric field is, +, given by F U (Force = negative potential, +, r, +, energy gradient)., +, If the electric field is along r , we can write, , , , d , F (p.E), Sol :Field intensity at a distance r from the line of, dr , If p and E are along the same direction we can, , E, charge, is, d, dE , 2 0 r, write F , (pEcos q) or F p ., , dr, dr , dE, The force on the dipole is F p dr, OSCILLATORY MOTION OF DIPOLE, , IN AN ELECTRIC FIELD, , l , pl, p , 2, 2p 0 r 2, 2p 0 r , , When dipole is displaced from its position of, equilibrium. The dipole will then experience a, torque given by pE sin , For small value of , pE ----------(1), Where negative sign shows that torque is acting, against increasing value of , Also,, I ,, Where, I = moment of inertia and, angular acceleration., , , , d 2, , Here the net force on dipole due to the wire will, be attractive., DISTRIBUTED DIPOLE: Consider a half, ring with a charge +q uniformly distributed and, another equal negative charge –q placed at its, centre. Here –q is point charge while +q is, distributed on the ring. Such a system is called, distributed dipole., +q, , d I d ---------(ii), , dt 2 , , dt 2, dt , , Hence, from eqs (i) and (ii), we have, , I, , =, –q, , If p net 2, , pE, I, , I, , where T is the time period of, pE, oscillations., 84, , dp cos q ;, , , , 2qR, , , 2qR, sin / 2, , , 0, , If the arrangement is a complete circle,, , pnet 0 ., 2, , d 2 pE, , 0, dt 2, I, On comparing above equation with standard, equation of SHM., , T 2, , d, –d, , f/2, , Eq (iii) can be written as, , pE, d, , 2 y 0 , we have ; 2 , 2, I, dt, , R, , R, , The net dipole moment is pnet , , d 2, d 2 pE, d 2, , , pE, , , , , or, ----(iii);, dt 2, dt 2, I, dt 2, This equation represents simple harmonic motion, (SHM). when dipole is displaced from its mean, position by small angle, then it will have SHM., , 2, , dq, , 2, , FORCE BETWEEN TWO SHORT, DIPOLES Consider two short dipoles, a), , seperated by a distance r. There are two, possibilities., If the dipoles are parallel to each other., r, , P1, E2, , P2, E1, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , b), , ELECTRO STATICS & CAPACITORS, Intensity of electric field at a point P that lies on the, axis of the ring at a distance x from its centre is, , 1 3p1p2, F 4p , r4, 0, Asthe, force, is positive, it is repulsive. Similarly, if p1 || p2 the force is attractive., If the dipoles are on the same axis, P1, , P2, , E2, , E, , 1, 4 0, , qx, , x, , 2, , R2, , , , 3, , 2, , x, , , where cos 2 2 , a x , , Where R is the radius of the ring. From the above, expression E = 0 at the centre of the ring., , E1, , r, , 1 6p1p 2, F 4p , dE, r4, 0., 0, E will be maximum when, dx, As the force is negative, it is attractive., Differentating E w.r.t x and putting it equal to, Quadrapole: We have discussed about, R, elecric dipole with two equal and unlike point, x, zero, we, get, and, 2, charges separated by a small distance. But in, 2 1 q , some cases the two charges are not concentrated, Emax , at its ends. (Like in water molecule) consider a, 3 3 4 0 R 2 , situation as shown in the figure. Here three, Electric field due to a Charged Spherical, charges –2q, q and q are arranged as shown. It, Conductor (Spherical Shell ), can be visualised as the combination of two, ‘q’ amount of charge be uniformly distributed, dipoles each of dipole moment p = qd at an angle, over a spherical shell of radius ‘R’, between them. The arrangment of two electric, q, dipoles are called quadrapole. As dipole moment, Surface charge density, , 4 R 2, is a vector the resultant dipole moment of the, When point ‘P’ lies outside the shell :, system is p | 2p cos q / 2 ., q, q, 1, q, E, 2, –q, 4 0 r, , –2q, , =, –q, This is the same expression as obtained for, q, q, electric field at a point due to a point charge., Few other quadrapoles are also as shown in the, Hence a charged spherical shell behave as a, following figures., point charge concentrated at the centre of it., +q, , +q, , E, –2q, , +q, , –2q, , 1 .4 R 2, q, .R 2, , , , E, , ;, 4 0, r2, 4 r 2, 0 r 2, , , When point ‘P’ lies on the shell : E , , +q, , 0, , –q, , 2q, , –q, , +q, , +q, , –q, , –q, , Electric filed at the axis of a circular, uniformly charged ring, dq, , dE sin, a2 x2, , Q, , a, , , , , E, , E, , 1, r2, , r1 = R, , dE, , , , dE cos, dE cos, , dE, dE sin, , NARAYANAGROUP, , When Point ‘P’ lies inside the shell: E = 0, , y, x, , Distance from the centre, Note : The field inside the cavity is always zero this, is known as elctro static shielding, 85
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Electric filed due to a Uniformly, charged non – conducting sphere, , , , Electric field intensity due to a uniformly , charged non-conducting sphere of charge Q,, of radius R at a distance r from the centre of , the sphere, q is t he amount of charge be uniformly, distributed over a solid sphere of radius R., , q, , 4 3, = Volume charge density, R, 3, , , When point ‘P’ lies inside sphere :, .r, 1 Qr, E, , for r R E 3 , , 4 0 R 3, , 0, , , , When point ‘P’ lies on the sphere:, . R, 1 q, E, , 3 0, When point ‘P’ lies outside the sphere:, E, , 1 q, . R3, E, , 4 0 r 2 ;, 3 0 r 2, , , , , , , E, , 1, r2, , R, , Electric field due to a uniformly charged disc, with surface charge density of radius at a , distance x from the centre of the disc is, , 20, , , 1 , , , , , x 2 R 2 , x, , If Q is the total charge on the disc, then, , E, 1, 2 , 4 0 R , 2Q, , , , 86, , 2Vq, m, , 1 2, mv or v , 2, , The work done in moving a charge of q coulomb, between two points separated by p.d. V2 V1 is, , , , x 2 R 2 , , The work done in moving a charge from one, point to another point on an equipotential surface, is zero., A hollow sphere of radius R is given a charge Q, the potential at a distance x from the centre is, x, R, , 1, , Q, , The potential at a distance when x>R is 4 . x ., 0, R, x, , , , x, , Electric Potential: Work done to bring a unit, , , , Potential due to a group of charges is the, algebraic sum of their individual potentials., i.e. V V1 V2 V3 . . . . . ., Two charges +Q and -Q are separated by a, distance d, the potential on the perpendicular, bisector of the line joining the charges is zero., When a charged particle is accelerated from rest, through a p.d. ' V ' , work done,, , 1, Q, . x R , 4 0 R, , d, , Electric Field due to a charged Disc:, , E, , 1 Q, 4 0 r, , q V2 V1 ., , E, , Er, , , , W Vq , , E, , 4 0 R 2 ;, , +vely charged body is considered to be at higher, potential and -vely charged body is considered, to be at lower potential., Electric potential at a point is a relative value, but not an absolute value., Potential at a point due to a point charge, , , , positive charge from infinite distance to a point , in the electric field is called electric potential, at that point ., W, , it is given by V , q, It represents the electrical condition or state of, the body and it is similar to temperature., , A sphere is charged to a potential. The potential, at any point inside the sphere is same as that of, the surface., Inside a hollow conducting spherical shell,, E=0, V 0 ., Relation among E, V and d in a uniform electric, dV, V, field is E d (or) E , dx, Electric field is always in the direction of, decreasing potential ., The component of electric field in any direction, is equal to the negative of potential gradient in, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, AP r a cos q AP PM ., , that direction., , V V V , E , i, j, k, y, z , x, , , , Hence potential at P due to charge +q situated at, , An equipotential surface has a constant value, of potential at all points on the surface ., For single charge q, E, , v2, , E, , E, , v1, v1, , v2, , q, , v1, , v2, , E, , E, , 1, q, ., 40 r a cos , Similarly potential at P due to charge -q at A is, q, 1, V2 , 40 r a cos ., Hence the total potential at P is, V= V1+V2, q, q, V, , 4pe0 ra cosq 4pe0 ra cosq, B is V1 , , E, , , , , Electric field at every point is normal to the, equipotential surface passing through that point, No work is required to move a test charge on, unequipotential surface., Zero Potential Point Two unlike charges Q1, and -Q2 are seperated by a distance ‘d’. The net, potential is zero at two points on the line joining, them, one (x) in between them and the other (y), outside them, , Q, Q, Q1, Q, 2 and 1 2, y dy, x dx, , V, V, , q, 4pe0, , , , 1, 1, , , , ra cosq ra cosq , , q 2a cosq , 4 pe0 r 2 a 2 cos 2 q , , But r >> a r 2 a 2 cos2 r 2, p cos q, 4pe0r 2 ., Hence potential varies inversely as the square, of the distance from the dipole., V , , SPECIAL CASSES, consists of two equal and opposite charges 1) On the axial line : For a point on the axial, 2, Potential due to a dipole: An electric dipole, , seperated by a very small distance. If 'q' is the, charge and 2a the length of the dipole then electric, 2), dipole moment will be given by p = (2a)q., P, , N, , r, , +q, , –q, O, A, , a, , , , , a, , B, , M, , Let AB be a dipole whose centre is at 'O' and 'P', be the point where the potential due to dipole is, to be determined. Let r , be the position coordinates of 'P' w.r.t the dipole as shown in, figure. Let BN & AM be the perpendiculars, drawn on to OP and the line produced along PO., From geometry ON a cos OM . Hence the a), distance ,BP from +q charge is r a cos , [because PB = PN as AB is very small in, comparsion with r]., For similar reason, NARAYANAGROUP, , line 00 Vaxial p / 4pe0 r volts for a, dipole., Point on the equitorial line : For a point on the, equitorial line 90 0 . Vequitorial 0 Volts ., Equitorial line is a line where the potential is, zero at any point., Equipotential surfaces : Equipotential, surface in an electric field is a surface on which, the potential is same at every point. In other, words, the locus of all points which have the, same electric potential is called equipotential, surface., An equipotential surface may be the surface of a, material body or a surface drawn in an electric, field. The important properties of equipotential, surfaces are as given below., As the potential difference between any two, points on the equipotential surface is zero, no, work is done in taking a charge from one point, to another., 87
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ELECTRO STATICS & CAPACITORS, b), , c), , d), , e), , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , infinity is taken as zero. If potential at infinity is V0,, The electric field is always perpendicular to an, equipotential surface. In other words electric, the potential at P is (V–V0)., field or lines of force are perpendicular to the 3) The potential difference is a property of two, equipotential surface., points and not of the charge q0 being moved., No two equipotential surfaces intersect. If they, ELECTRIC POTENTIAL DUE TO A, intersect like that, at the point of intersection field, LINEAR CHARGE DISTRIBUTION, Consider a thin infinitely long line charge having, will have two different directions or at the same, a uniform linear charge density placed along, point there will be two different potentials which, is impossible., YY 1 . Let P is a point at distance ‘r’ from the line, The spacing between equipotential surfaces, charge then manitude of electri field at point P, enables to identify regions of strong and weak, , is given by E 2 r, dV, 1, 0, fields E , . So E , (if dV is constant)., dr, dr, Y, At any point on the equipotential surface, +, component of electric field parallel to the surface, +, +, is zero., +, +, In uniform field , the lines of force are straight, dS, r, l, O+, +, P, and parallel and equipotential surfaces are planes, E, +, Gaussian, +, perpendicular to the lines of force as shown in, surface, +, +, figure, +, , equipotential, surface, , Y, , We know that V r E .dr, , , Here E 2 r and E .dr Edr, 0, The equipotential surfaces are a family of, concentric spheres for a uniformly charged sphere, or for a point charge as shown in figure, equipotential, surface, , Equipotential surfaces in electrostatics are, similar to wave fronts in optics. The wave fronts, in optics are the locus of all points which are in, the same phase. Light rays are normal to the wave, fronts. On the other hand the equipotential, surfaces are perpendicular to the lines of force., Note : 1) In case of non-uniform electric field, the, field lines are not straight, and in that case, equipotential surfaces are curved but still, perpendicular to the field., 2) Electric potential and potential energy are always, defined relative to a reference. In general we, take zero reference at infinity. The potential at a, point P in an electric field is V if potential at, 88, , , So V r Edr 2 r dr, 0, , , V r , log e r C, 2 0, , Where C is constant of integration and V(r) gives, electric potential at a distance ‘r’ from the linear charge distribution, ELECTRIC POTENTIAL DUE TO INFINITE, PLANE SHEET OF CHARGE, (NON CONDUCTING), Consider an infinite thin plane sheet of positivive, charge having a uniform surface charge density, on both sides of the sheet. by symmetry , it, follws that the electric filed is perpendicular to, the plane sheet of charge and directed in out, ward direction., , , The electric field intensity is E 2, 0, Electrostatic potential due to an infinite plane, sheet of charge at a perpendicular distance r from, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, the sheet given by V r E .dr Edr, , , , dr , rC, 2 0, 2 0 , where C is constant ofintegration similarly the, electric pontential due to an infinite plane conducting plate at a perpendicular distance r from, V r , , ELECTRO STATICS & CAPACITORS, , V is constant and is equal to that on the surface, 1 q, So, V 4 R for r R, 0, , The varitaion of V with distance ‘r’ from centre, is as shown in the graph., , V, , VS , , the plate is given by V r E .dr Edr, , , rc, , where C is constant of intergration, ELECRTIC POTENTIAL DUE TO A CHARED, SPERICAL SHELL (OR CONDUCTING, SPHERE):, V r , , , , dr , 0, 0, , E, , P, q, dS, + + +, + R r +, Charged spherical, +, +, Shell, O, +, +, +, + +, Gaussian, Surface, , Consider a thin spherical shell of radius R and, having charge+q on the spherical shell., Case (i): When point P lies outside the spherical, shell. The electric field at the point is, 1 q, E, 4 0 r 2 (for r > R), , 1 q, 40 R, v 1/r, , r=R, , r, , ELECTRIC POTENTIAL DUE TO A, UNIFORMLY CHARGED, NON-CONDUCTING SOLID SPHERE:, Consider a charged sphere of radius R with total charge q uniformly distributed on it., Case (i) : For points Outside the sphere (r > R), The electric field at any point is, 1, q, E, , 2, (for r > R), 4 0 r, The potential at any point outside the shell is, V r E .dr Edr, , 1 q, 1 q, dr , C, 2, 4 0 r, 4 0 r, Where C is constant of integration, , , If r ,V 0 and C=0, , 1 q, (r > R), The potential V r E .dr Edr, 4 0 r, Case (ii) : When point P lies on the surface of spheri1 q, 1 q, , dr, , , C, cal shell then r = R, 4 0 r 2, 4 0 r, The electrostatic potential at P on the surface is, Where C is constant of integration, 1 q, V, If r , V 0 and C 0, 4 0 R, 1 q, Case (iii) : FOr points inside the sphere (r < R), V r , r R, 4 0 r, 1 qr, E, The, electric, field, is, Case (ii) : When point P lies on the surface of spheri4 0 R 3, cal shell then r = R, + E +, electrostatice potential at P on the surface is, +, +, +, 1 q, + + +, +, +, V, +, r +, 4 0 R, + +, +, +, Case (iii) : For points inside the charged spherical, + +, +, +, R, shell (r < R), the electric field E = 0, +, dV, dV E .dr Edr, 0, So we can write , dr, NARAYANAGROUP, , V r , , 89
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, v, , r, , r, , 1 qr, v dV R E dr R 4 0 R3 dr, s, r, , 1 q r2 , V Vs , , 4 0 R 3 2 R, , V, , 1 q, 1 q r 2 R2 , , , 4 0 R, 4 0 R3 2, 2 , , 1 q 3 r2 , V , , 4 0 R 2 2 R 2 , At the centre r = 0 then, 1 3q 3 1 q, Potential at centre VC 4 2 R 2 4 R, 0, 0, The variation of V with distance ‘r’ from centre, is as shown in the graph., , v, vs, vs, vs, , W.E-19: A charge Q is distributed over two concentric hollow spheres of radii ‘r’ and R ( >, r) such that the surface densities are equal., Find the potential at the common centre., Sol: If q1 and q2 are the charges on spheres of radii, ‘r’ and R respectively, then in accordance with, conservation of charge, q1 q2 Q ------(1), And according to given problem 1 2 ,, q1 r 2, q1, q2, , , i.e.,, or, ------(2), q2 R 2, 4 r 2 4 R 2, So from Eqs (1) and (2), q1 , , QR 2, Qr 2, q, , r 2 R 2 and 2 r 2 R -----(3), , Now as potential inside a conducting sphere is, equal to that at its surface, so potential at the, common centre,, , V V1 V2 , v 1/r, , 1 q1 q2 , , 4 0 r R , , Substituting the value of q1 and q2 from Eq.(3), , r<R r<R, r=R, r, , Potential of a charged ring: A charge q is, , V, , 1 Qr, Qr , 2 2 2 2 , 4 0 R r R r , , , , distributed over the circumference of ring ( either, uniformly or non-uniformly ) , then electric, potent ial at t he cent re of the ring is, 1 QR r, , 1 q, 4 0 R 2 r 2 , V, . ., 4 o R, W.E-20: If electric potential V at any point (x, y,, At distance ‘r’ from the centre of ring on its axis, z) all in metres in space is given by V = 4x2, volt. Calculate the electric field at the point, 1, q, (1m, 0m, 2m)., would be V 4 ., R2 r 2, o, Sol : As electric field E is related to potential V, Electric potential of a uniformly charged disc, through the relation, Consider a uniformly charged circular disc, dV, E, having surface charge density ., dr, Potential a at point on its axial line at distance x, dV, d, , 2, 2, from the centre is V 2 R x x , o, , , , At the centre of disc x 0, , R, V, 2 o, , , , q, For x R , V 4 x, o, , , , R, Potential on the edge of the disc is V , o, , 90, , Ex , , Ey , , dx, , , , (4x 2 ) 8x, , dV, d, (4x 2 ) 0, dy, dy, , And, E z , , , dx, , dV, d, (4x 2 ) 0, dz, dz, , So, E ˆi E x ˆj E y kˆ E z 8xiˆ, i.e., it has magnitude 8 V/m and is directed along, negative x-axis., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-21: A conducting spherical bubble of radius, r and thickness t (t >> r) is charged to a, potential V. Now it collapses to form a, spherical droplet. Find the potential of the, droplet., Sol: Here charge and mass are conserved. If R is the, radius of the resulting drop formed and is, 4 3, R 4r 2 t, 3, , densit y of soap solut ion,, R (3r 2 t)1/ 3, , Now potential of the bubble is V , or q 4 0 rV, Now potential of resulting drop is, V| , , , , 1 q, 4 0 R, , r, , 3t , , 1 q, 4 0 r, , 1/ 3, , V., , POTENTIAL ENERGY OFA SYSTEM OF, TWO CHARGES IN AN EXTERNAL, FIELD: Consider two charges q1 and q2 located, at two points A and B having position vectgors r1, and r2 respectively. Let V1 ang V2 be the potentials, due to external sources at the two points, respectively., The work done in bringing the charge q1 from, infinity to the point A is W1 q1V1, In bringing charge q2 , the work to be done not, only against the external field but also against, the filed due to q1 ., The work done in bringing the charge q2 from, , Potential Energy of System of Charges, , infinity to the point B is W2 q2V2 ., , Two charges Q1 and Q2 are separated by a, distance 'd'. The P.E. of the system of charges is, , The workdone on q2 against the field due to q1, , 1 QQ, U, . 1 2, 4 0 d, , from U=W=Vq, , 1 q1q2, is W2 4 r, where r12 is the distance, 0, 12, , between q1 and q2 ., , d, , The total work done in bringing the charge q2, against the two fields from infinity to the point, , B is, 1 q1q2, W2 q2V2 , 4 0 r12, 1 Q1Q2, 1 QQ, Q2Q3 Q3Q1 , 1 2, , , U, U, , or, The tot al work done in assembling t he, 4 0 a, a, a , 4 0, a, configuration or the potential energy of the, Q, system is, 1 q1q2, a, a, W q1V1 q2V2 , 4 0 r12, Q, a, Q, W.E-22: Charge q1 is fixed and another point, charge q2 is placed at a distance r0 from q1 on, A charged particle of charge Q2 is held at rest at, a, frictionless horizontal surface. Find the, a distance 'd' from a stationary charge Q1 . When, velocity of q2 as a function of seperation r, the charge is released, the K.E. of the charge Q2, between them (treat the changes as point, charges and mass of q2 is m), 1 Q1Q2, at infinity is 4 . d ., q1, q2, 0, If two like charges are brought closer, P.E of Sol :, r0, the system increases., According law of concervation of energy, If two unlike chargtes are brought closer, P.E of, U1 K1 U2 K 2, the system decreses., 1 q 1q 2, 1 q 1q 2 1, For an attractive system U is always NEGATIVE., 0 , mv 2, 4, , , r, 4, , , r, 2, For a repulsive system U is always POSITIVE., 0, 0, 0, For a stable system U is MINIMUM., 1, q q 1 1, q1 q 2 1 1 , mv2 1 2 ; v , , dU, 2, 4, , , r, r, 2, , 0 m r0 r , 0 0, , i.e. F , = 0 (for stable system), , Q1, Q2, Three charges Q1 , Q2 , Q3 are placed at the three, vertices of an equilateral triangle of side 'a'. The, P.E. of the system of charges is, , 3, , 1, , 2, , dx, , NARAYANAGROUP, , 91
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ELECTRO STATICS & CAPACITORS, , W.E-23: A proton moves with a speed of 7.45 x 105, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, from law of conservation of energy, , m/s directly towards a free proton originally, 1, 1, mu 2 mv 2 mg 2 .... ( 2 ), at rest. Find the distance of closest approach, 2, 2, for the two protons., from ( 1 ) and ( 2 );, Given (1 / 4 0 ) 9 10 9 m / F; mP 1.67 1027 kg, q2, u 4g , 5 86 m / s, and e = 1.6 x 10-19 coulomb., 4 o m, Sol: As here the particle at rest is free to move,, when one particle approaches the other, due to W.E-25: If an electron enters into a space between, the plates of a parallel plate capacitor at an, electrostatic repulsion other will also start, an angle with the plates and leaves at an, moving and so the velocity of first particle will, decrease while of other will increase and at, angle to the plates, find the ratio of its kinetic, closest approach both will move with same, energy while entering the capacitor to that while, velocity. So if v is the common velocity of each, leaving., particle at closest approach, by 'conservation of Sol: Let u be the velocity of electron while entering, momentum'., the field and v be the velocity when it leaves the, plates. Component of velocity parallel to the, 1, mu m m i.e., u, plates will remain unchanged., 2, And by 'conservation of energy', u cos , Hence u cos u cos , 2, 1, 1, 1, 1 e, mu 2 m2 m2 , 2, 2, 2, 4 0 r, , So, r , , 4e 2, 4 0 mu 2, , u, , as 2 , , , , And hence substituting the given data,, 4 (1.6 10 19 )2, r 9 10 9 , 10 12 m, 1.67 10 27 (7.45 105 )2, , W.E-24: A small ball of mass 2 x 10–3 kg having a, , v, , cos , , 1, 2, 2, 2, mu , 2, u, cos , , , v, cos , 1, 2, mv , 2, , W.E-26: Figure shows two concentric, conductiong shells of radii r1 and r2 carrying, uniformly distributed charages q1 and q2 ., respectively. Find out an expression for the, potential of each shell., , charge of 1C is suspended by a string of, length 0.8m. Another identical ball having, +q, the same charge is kept at the point of, r, +q, suspension. Determ ine the minimum, r, horizontal velocity which should be impacted, to the lower ball so that it can make complete, revolution :, Sol: To complete the circle at top most point T2 = 0 Sol: The potential of each sphere consists of two, points:, F, One due to its own charge, and, V, Second due to the charge on the other sphere., Using the principle of superposition, we have, q, V1 Vr1 , surface Vr2 ,inside and, 2, , 2, , 1, , 1, , F, , Mg , , q2, 4 o , , 2, , V2 g , , 92, , , , MV2, , , q 2, ...( 1 ), 4 o M, , U, , V2 Vr1 ,outside Vr2 , surface, , 1 q1, 1 q2, Hence, V1 4 r 4 r, 0 1, 0 2, 1 q1, 1 q2, and V2 4 r 4 r, 0 2, 0, 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-27: In the previous example, if the charge Sol: a) When an object is connected to earth, q1 q0 and the outer shell is earthed, then, a) determine the charge on the outer shell,, and, b) find the potential of the inner shell., , (grounded), its potential is reduced to zero. Let, q ' be the charge on A after it is earthed as shown, in fig, Q + q, –q, , Sol: a) We know that charge on facing surfaces is, equal and opposite. So, if charge on inner sphere, is q0 , then charge on inner surface of shell, should be q0 . Now, let charge on outer surface, of shell be q2 ., As the shell is earthed. So its potential should, be zero. So,, , The charge q ' on A induces q ' on inner surface, of B and q ' on outer surface of B. In, equilibrium, the charge distribution is as shown, in fig, Potential of inner sphere = potential due to, charge on A+ potential due to charge on B = 0, q', q', Q q', VA , , , 0, 4 0 a 4 0b 4 0b, a, ', or q Q , b, , kq0 k q0 kq2, , , 0 q2 0, r2, r2, r2, Hence, charge on outer surface of shell is zero., Final charges appearing are shown in fig, Potential of inner sphere:, Vshell , , b), , Q ba, +, , +, +, , kq0 k q0 , q, , 0, r1, r2, 4 0, , 1 1 , , r1 r2 , W.E-28: Consider two concentric spherical metal, shells of radii ‘a’ and b > a. The outer shell, has charge Q, but the inner shell has no, charge, Now, the inner shell is grounded. This, means that the inner shell will come at zero, potential and that electric field lines leave the, outer shell and end on the inner shell., a) Find the charge on the inner shell., b) Find the potential on outer sphere., V1 , , This implies that a charge Q a / b has been, transferred to the earth leaving negative charge, on A., Final charge distribution will be as shown in, fig.., , +, +, , +, , +, , +, +, , b, , +, Qa, b, , +, ++, , Qa, b, , +, , +, + + +, + + +, +, +, , As b>a, so charge on the outer surface of outer, shell will be, , Q b a , 0., b, , b) Potential of outer surface VB potential due, to charge on A + potential due to charge on B., VB Va ,out Vb ,both surface, , 1 q', 1 Q, , , 4 0 b 4 0 b, , a, , Q , , 1 , 1 Q Q b a , b, , , , 4 0, b, 4 0 b, 4 0b 2, NARAYANAGROUP, , 93
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ELECTRO STATICS & CAPACITORS, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , W.E-29: Two circular loops of radii 0.05 and Sol: As the electric field at the centre of a ring is zero,, 0.09m, respectively, are put such that their, axes coincide and their centres are 0.12 m, apart. Charge of 106 coulomb is spread, uniformly on each loop. Find the potential, difference between the centres of loops., q2, q1, , R2, , R1, O1, , But here, Q 2 R and r , , r2, , r1, , the particle will not come back due to repulsion if it, crosses the centre fig., 1 2, 1 qQ, 1 qQ, mv , , 2, 4 0 r, 4 0 R, , O2, , x, , , , 3R, , , , 2, , R2 2R, , q , 1 2, 1 2 R q 1 , 1 or v , , So, mv , , 2, 4 0, R 2, 2 0 m , , Sol: The potential at the centre of a ring will be due, to charge on both the rings and as every element, q , of a ring is at a constant distance from the centre, So, Vmin , , 2 0 m , so, , V1 , , 1, 4 0, , q, q2, 1, R1, R22 x 2, , , , , , C. U. Q, , CHARGE & CONSERVATION OF CHARGE, 1. Two identical metallic spheres A and B of, 4, 104 , 9 10, exactly equal masses are given equal positive, 9 10 , , , 2, 2, 5, 9 12 , and negative charges respectively. Then, , 1) mass of A > Mass of B, 1 1 , 2) mass of A < Mass of B, 9 105 2.40 105 V, 5 15 , 3) mass of A = Mass of B, 4) mass of A Mass of B, , 1 q2, q1, 2., Two, spheres of equal mass A and B are given, , similarly, V2 4 R , 2, 2, +q, and, -q charge respectively then, R1 x , 0 , 2, 1) mass of A increases2) mass of B increases, 3) mass of A remains constant, 1 198, 3 1, 105, or V2 9 10 , 4) mass of B decreases, 9 13 117, 3 A soap bubble is given a negative charge, then, V2 1.69 105 V, its radius., 1) Decreases, 2) Increases, So, V1 V2 2.40 1.69 105 71 kV, 3) Remanins unchanged, W.E-30: A circular ring of radius R with uniform, 4) Nothing can be predicted as information is, insufficient, positive charge density per unit length is, COULOMB’S LAW, located in the y - z plane with its centre at the, 4., Two, charges, are placed at a distance apart., origin O. A particle of mass ‘m’ and positive, If a glass slab is placed between them, force, charge ‘q’ is projected from the point, between them will, p 3R, 0, 0 on the negative x-axis directly, 1) be zero, 2) increase, 3), decrease, 4) remains the same, towards O, with initial speed v. Find the, smallest (non-zero) value of the speed such 5. A negatively charged particle is situated on a, straight line joining two other stationary, that the particle does not return to P?, particles each having charge +q. The, , , , direction of motion of the negatively charged, , , r, R, particle will depend on, , , q, P, 1) the magnitude of charge, , , O, 3R , 0, 0, 2) the position at which it is situated, , , , 3) both magnitude of charge and its position, , , , 4) the magnitude of +q, Q, , 94, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 6., , ELECTRO STATICS & CAPACITORS, , Four charges are arranged at the corners of a, square ABCD as shown in the figure. The force, on the positive charge kept at the centre ‘O’, is, A, +Q, , 7., , 8., , 9., , B, +2Q, , 1) zero, O, 2) along the diagonal AC, +Q, 3) along the diagonal BD2QD, C, 4) perpendicular to side AB, Two identical +ve charges are at the ends of, a straight line AB. Another identical +ve, charge is placed at ‘C’ such that AB=BC. A,, B and C being on the same line. Now the force, on ‘A’, 1) increases, 2) decreases, 3) remains same, 4) we cannot say, Two identical pendulums A and B are, suspended from the same point. Both are given, positive charge, with A having more charge, than B. They diverge and reach equilibrium, with the suspension of A and B making, angles1 and 2 with the vertical respectively.., 1) 1 2 2) 1 2, 3) 1 2, 4) The tension in A is greater than that in B, Two metal spheres of same mass are, suspended from a common point by a light, insulating string. The length of each string is, same. The spheres are given electric charges, +q on one end and +4q on the other. Which of, the following diagram best shows the resulting, positions of spheres?, 1), , 3), , +q, , +q, , , , , , +4q, , 2), , +q , , 4), +4q, , +4q, , , , 11. Figure shows the electric lines of force, emerging from a charged body. If the electric, field at ‘A’ and ‘B’ are E A and EB, respectively and if the displacement between, ‘A’ and ‘B’ is ‘r’ then, , B, A, , 1) EA EB, 3) E A , , 2) EA EB, , EB, r, , 4) E A , , , , +q, , EB, r2, , 12. Figure shows lines of force for a system of, two point charges. The possible choice for the, charges is, , q1, , q2, , 1) q1 4 C , q2 1.0 C 2) q1 1 C , q2 4 C, 3) q1 2 C , q2 4 C 4) q1 3C , q2 2 C, 13. Drawings I and II show two samples of electric, field lines, , I, , +4q, , 10. Two point charges q and 2q are placed at, a certain distance apart. Where should a third, point charge be placed so that it is in, equilibrium?, 1) on the line joining the two charges on the, right of 2q, 2) on the line joining the two charges on the, left of q, 3) between q and 2q, 4) at any point on the right bisector of the, line joining q and 2q ., NARAYANAGROUP, , ELECTRIC FIELD, , II, , 1) The electric fields in both I and II are, produced. by negative charge locat ed, somewhere on the left and positive charges, located somewhere on the right, 2) In both I and II the electric field is the same, every where, 3) In both cases the field becomes stronger on, moving from left to right, 4) The electric field in I is the same everywhere,, but in II the electric field becomes stronger on, moving from left to right, 14. An electron is projected with certain velocity, into an electric field in a direction opposite to, the field. Then it is, 1) accelerated, 2) retarded, 3) neither accelerated nor retarded, 4) either accelerated or retarded, 95
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ELECTRO STATICS & CAPACITORS, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , 15. The acceleration of a charged particle in a 23. A charged bead is capable of sliding freely, uniform electric field is, through a string held vertically in tension. An, 1) proportional to its charge only, electric field is applied parallel to the string, 2) inversely proportional to its mass only, so that the bead stays at rest of the middle of, 3) proportional to its specific charge, the string. If the electric field is switched off, 4) inversely proportional to specific charge, momentarily and switched on, 16. An electron and proton are placed in an, 1) the bead moves downwards and stops as soon, electric field. The forces acting on them are, as the field is switched on, 2) the bead moved downwards when the field is, F1 and F2 and their accelerations are a1 and, switched off and moves upwards when the field, a2 respectively then, is switched on, 3) the bead moves downwards with constant, 1) F1 F2, 2) F1 F2 0, acceleration till it reaches the bottom of the, 3) a1 a2, 4) a1 a2, string, 4) the bead moves downwards with constant, 17. The bob of a pendulum is positively charged., velocity till it reaches the bottom of the string, Another identical charge is placed at the point, 24., An electron is moving with constant velocity, of suspension of the pendulum. The time period, along x-axis. If a uniform electric field is, of pendulum, applied along y-axis, then its path in the x-y, 1) increases, 2) decreases, plane will be, 3) becomes zero, 4) remains same., 1) a straight line, 2) a circle, 18. Intensity of electric field inside a uniformly, 3) a parabola, 4) an ellipse, charged hollow sphere is, 1) zero, 2) non zero constant, 25. An electron of mass M e , initially at rest ,, 3) change with r, moves through a certain distance in a uniform, 4) inversely proportional to r, electric field in time t1 . proton of mass M p, 19. A positive charge q0 placed at a point P near, a charged body experiences a force of, also initially at rest, takes time t2 to move, repulsion of magnitude F, the electric field E, through an equal distance in this uniform, of the charged body at P is, electric field. Neglecting the effect of gravity, F, F, F, the ratio t2 / t1 is nearly equal to, 1) q, 2) q, 3) q, 4) F, 0, 0, 0, 1) 1 2) M p / M e 3) M e / M p 4) 1836, 20. A cube of side b has charge q at each of its, vertices. The electric field at the centre of 26. Dimensions of 0 are, the cube will be (KARNATAKA CET 2000), 1) M 1 L3T 4 A2 , 2) M 0 L3T 3 A3 , 32q, q, q, 1) zero 2) 2, 3) 2, 4) 2, b, 2b, b, 3) M 1 L3T 3 A , 4) M 1 L3TA2 , 21. An electron and proton are sent into an, electric field. The ratio of force experienced 27. Two point charges q and -2q are placed some, distance d apart. If the electric field at the, by them is, locatiion of q is E, that at the location of -2q, 1) 1 : 1, 2) 1 : 1840, is, (1987), 3) 1840 : 1, 4) 1 : 9.11, E, E, 22. An electron enters an electric field with its, 2) –2E, 3), 4) – 4E, 1) , 2, 2, velocity in the direction of the electric lines, of force. Then, dV, 28. E , , here negative sign signified that, 1) the path of the electron will be a circle, dr, 2) the path of the electron will be a parabola, 1) E is opposite to V 2) E is negative, 3) the velocity of the electron will decrease, 3) E increases when V decreases, 4) the velocity of the electron will increase, 4) E is directed in the direction of decreasing V, 96, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , 29. An electron moves with a velocity v in an, , , electric field E . If the angle between v and, , E is neither 0 nor , then path followed by, the electron is, 1) straight line, 2) circle, 3) ellipse, 4) parabola, 30. A charged particle is free to move in an, electric field, 1) It will always move perpendicular to the line, of force, 2) It will always move along the line of force in, the direction of the field., 3) It will always move along the line of force, opposite to the direction of the field., 4) It will always move along the line of force in, the direction of the field or opposite to the, direction of the field depending on the nature of, the charge, 31. Two parallel plates carry opposite charges, such that the electric field in the space, between them is in upward direction. An, electron is shot in the space and parallel to, the plates. Its deflection from the original, direction will be, 1) Upwards, 2) Downwards, 3) Circular, 4) does not deflect, , ELECTRIC POTENTIAL AND, POTENTIAL ENERGY, 32. Potential at the point of a pointed conductor, is, 1) maximum 2) minimum, 3) zero, 4) same as at any other point, 33. An equipotential line and a line of force are, 1)perpendicular to each other, 2)parallel to each other, 3) in any direction 4) at an angle of 450, 34. When a positively charged conductor is placed, near an earth connected conductor, its, potential, 1) always increases, 2) always decreases, 3) may increase or decrease 4) remains the same, 35. If a unit charge is taken from one point to, another over an equipotential surface, then, 1) work is done on the charge, 2) work is done by the charge, 3) work on the charge is constant, 4) no work is done, NARAYANAGROUP, , ELECTRO STATICS & CAPACITORS, 36. Electric potential at some point in space is zero., Then at that point, 1) electric intensity is necessarily zero, 2) electric intensity is necessarily non zero., 3) electric intensity may or may not be zero, 4) electric intensity is necessarily infinite., 37. When an electron approaches a proton, their, electro static potential energy, 1) decreases, 2) increases, 3) remains unchanged 4) all the above, 38. An electron and a proton move through a, potential difference of 200V. Then, 1) electron gains more energy, 2) proton gains more energy, 3) both gain same energy, 4) none of them gain energy, 39. Two charges +q and –q are kept apart. Then, at any point on the right bisector of line joining, the two charges., 1) the electric field strength is zero, 2) the electric potential is zero, 3) both electric potential and electric field, strength are zero, 4) both electric potential and electric field, strength are non - zero, 40. When ‘n’ small drops are made to combine to, form a big drop, then the big drop’s, 1) Potential increases to n1/3 times original, potential and the charge density decreases to, n1/3 times original charge, 2) Potential increases to n2/3 times original, potential and charge density increases to n1/3, times original charge density, 3) Potential and charge density decrease to, n1/3 times original values, 4) Potential and charge density increases to ‘n’, times original values, 41. A hollow metal sphere of radius 5cm is charged, such that the potential on its surface is 10V., The potential at the centre of the sphere is, 1) 0 V, 2) 10 V, 3) same as at point 5cm away from the surface, 4) same as at point 25cm from the surface, 42. The work done (in Joule) in carrying a charge, of ‘x’ coulomb between two points having a, potential difference of ‘y’ volt is, 1), , x, y, , 2), , x2, y, , 3), , y, x, , 4) xy, , 97
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 43. Two charges q and -q are kept apart. Then at, any point on the perpendicular bisector of line, joining the two charges., (2008E), 1) the electric field strength is zero, 2) the electric potential is zero, 3) both electric potential and electric field, strength are zero, 4) both electric potential and electric field, strength are non-zero, 44. Electric potential at the centre of a charged, hollow spherical conductor is, 1) zero, 2) twice as that on the surface, 3) half of that on the surface, 4) same as that on the surface, 45. Which of the following pair is related as in, work and force, 1) electric potential and electric intensity, 2) momentum and force, 3) impulse and force, 4) resistance and voltage, 46. The equipotential surfaces corresponding to, single positve charge are concentric spherical, shells with the charge at its origin. The, spacing between the surfaces for the same, change in potential, 1) is uniform throughout the field, 2) is getting closer as r , 3) is getting closer as r 0, 4) can be varied as one wishes to, 47. Four identical charges each of charge q are, placed at the corners of a square. Then at the, centre of the square the resultant electric, intensity E and the net electric potential V, are, 1) E 0, V 0, 2) E 0, V 0, , 49. Two copper spheres of the same radii, one, hollow and the other solid, are charged to the, same potential, then, 1) hollow sphere holds more charge, 2) solid sphere holds more charge, 3) both hold equal charge, 4) we can’t say, 50. Equipotential surfaces are shown in figure a, and b. The field in, 3V0, , 2V0, , V0, 9V0, , 4V0 V0, , r0, , 2r0, 3r0, , x0, , x0, , FIGURE (A), , 51., , 52., , 53., , 54., , FIGURE (B), , 1) a is uniform only 2) b is uniform only, 3) a and b is uniform 4)both are nonuniform, Due to the motion of a charge, its magnitude, 1) changes, 2) does not changes, 3) increases (or) decreases depends on its speed, 4) can not be predicted, Induction preceeds attraction because, 1) an uncharged body can attract an uncharged, body due to induction of opposite charge on it, 2) a charged body can attract an uncharged body, due to induction of same charge on it., 3) a charged body can attract an uncharged body, due to induction of opposite charge on it., 4) a charged body can attract another charged, body due to induction of same charge on it., The coulomb electrostatic force is defined for, 1) two spherical charges at rest, 2) two spherical charges in motion, 3) two point charges in motion, 4) two point charges at rest, , F, The Electric field is given by E q , here, 0, the test charge ‘q0’ should be, a) Infinitesimally small and positive, b) Infinitesimally small and negative, 1) only a, 2) only ‘b’, 3) a (or) b, 4) neither ‘a’ or ‘b’, , 3) E 0, V 0, 4) E 0, V 0, 48. Two positive charges q and q are placed at, the diagonally opposite corners of a square, and two negative charges -q and -q are placed, at the other two corners of the square. Then, at the centre of the square the resultant 55. The p.d V B Vc between two points from C, electric intensity E and the net electric, to B, 1) does not depend on the path, potential V are, 2) depends on the path, 1) E 0, V 0, 2) E 0, V 0, 3) depends on test charge, 3) E 0, V 0, 4) E 0, V 0, 4) independent of electric field, 98, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 56. Match List-I with List-II, List-I, List-II, a) proton and, e) gains same velocity, electron, in an elctric field for, same time, b) proton and, f) gains same KE in an, positron, electric field for same time., c) Deutron and, g) experience same, force in electric field, - particle, d) electron and, h) gains same KE, positron, when accelerated, by same potential, difference., , ELECTRO STATICS & CAPACITORS, 58. Match the following, List-I, List-II, a) Fluid flow, d) Temperature, difference, b) Heat flow, e) Pressure, difference, c) Charge flow, f)Potential difference, 1) a e, b d , c f, 2) a d , b e, c f, , 3) a f , b e, c d, 4) a e, b f , c d, 59. Match List-I with List-II, List-I, List-II, a) Two like charges e) the force between, 1) a h, b g , c e, d f, are brought nearer, them decreases., b) Two unlike, f) potential energy, 2) a h, b g , c f , d e, charge of some, of the system, 3) a g , b h, c e, d f, brought nearer, increases, c) When a third, g) mut ual forces are, 4) a e, b f , c g , d h, charge, of, same, not affected, 57. Match List-I with List-II, nature is placed, List-I, List-II, equidistance from, a) Electric potential, e) inversly proportional, two like charges, inside a charged, to square of the, d) When a dielectric, h) potential energy, 2, medium, is, introduced, of the system, conducting sphere, distance (r ), between two charges decreases, b) Electric potential, f) directly proportional, 1) a h, b f , c g , d e, charged sphere, outside the conducting, 2) a f , b h, c g , d e, to distance, 3) a h, b f , c e, d g, r from the centre, c) Electric field, , d) Electric field, charged sphere, , g) constant, inside the non, conducting, charged sphere, h) inversly, outside a, conducting, proportional to, distance ( r ), , 1) a f , b e, c g , d h, 2) a e, b f , c h, d g, 3) a h, b g , c e, d f, 4) a g , b h, c f , d e, NARAYANAGROUP, , 4) a g , b e, c f , d h, 60. Match the following :, a) Electric field, e) Constant, outside a conducting, charged sphere, b) Electric potential out f) directly propor, side the conducting, national to, charged sphere, distance from, centre, c) Electric field inside g) inversely propor, a non-conducting, tional to the, charged sphere, distance, d) Electric potential in h) inversely, side a charged, proportional to, conducting sphere, the square of the, distance, 99
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 66. An electric dipole placed in a nonuniform, electric field experiences, 1) a force but no torque, 2) a torque but no force, 3) a force as well as a torque, 4) neither a force nor a torque, , 1) a h, b g , c e, d f, 2) a e, b f , c h, d g, 3) a h, b g , c f , d e, 4) a g , b h, c f , d e, , DIPOLE, , 67. If Ea be the electric field intensity due to a, 61. The angle between electric dipolemoment, short dipole at a point on the axis and Er be, p and the electric field E when the dipole is, that on the perpendicular bisector at the same, in stable equilibrium, distance from the dipole, then, 1) 0, 2) / 4, 3) / 2 4) , 1) Ea Er, 2) Ea 2 Er, 62. ‘Debye’ is the unit of, 1) electric flux, 2) electric dipolemoment, 3) Er 2 Ea, 4) Ea 2 Er, 3) electric potential 4) electric field intensity, 63. The electric field at a point at a distance r 68. The electric potential due to an extremely, short dipole at a distance r from it is, from an electric dipole is proportional to, proportional to, 1, 1, 1, 1), 2) 2, 3) 3, 4) r 2, 1, 1, 1, 1, r, r, r, 1), 2) 2, 3) 3, 4) 4, r, r, r, r, 64. An electric dipole placed with its axis in the, direction of a uniform electric field 69. An electric dipole when placed in a uniform, electric field will have minimum potential, experiences, energy, if the angle between dipole moment, 1) a force but not torque, and electric field is, 2) a torque but no force, 3) , 4) 3 / 2, 1) zero 2) / 2, 3) a force as well as a torque, 70. The angle between the electric dipole moment, 4) neither a force nor a torque, and the electric field strength due to it, on, 65. An electric dipole is placed in a non uniform, the equatorial line is, electric field increasing along the +ve, 1) 00, 2) 900, 3) 1800, 4) 600, direction of X - axis. In which direction does, 71. A metallic shell has a point charge q kept, the dipole, inside its cavity. Which one of the following, diagrams correctly represents the electric, Y, lines of forces ?, q, X, , X, , 1), , 2), , 3), , 4), , q, , Y, , 1) move along + ve direction of X - axis, rotate, clockwise, 2) move along - ve direction of X - axis, rotate, clockwise, 3) move along + ve direction of X - axis, rotate, anti clockwise, 4) move along - ve direction of X - axis, rotate, anti clockwise, 100, , ASSERTION & REASON, In each of the following questions, a statement, of Assertion (A) is given followed by a, corresponding statement of Reason (R) just, below it. Mark the correct answer., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , 72., , 73., , 74., , 75., , 76., , ELECTRO STATICS & CAPACITORS, , 1) Both ‘A’ and ‘R’ are true and ‘R’ is the, correct explanation of ‘A’, 2) Both ‘A’ and ‘R’ are true and ‘R’ is not, the correct explanation of ‘A’, 3) ‘A’ is true and ‘R’ is false, 4) ‘A’ is false and ‘R’ is true, Assertion(A) : Force between two point, charges at rest is not changed by the presence, of third point charge between them., Reason(R): Force depends on the magnitude, of the first two charges and seperation, between them, Assertion (A): Electric potential at any point, on the equatorial line of an electric dipole is, zero, Reason (R): Electric potential is scalar, Assertion (A) : Electrons always move from, a region of lower potential to a region of, highe potential, Reason (R) : Electrons carry a negative, charge, Assertion(A): A metallic shield in form of a, hollow shell may be built to block an electric, field., Reason (R): In a hollow spherical shield, the, electric field inside it is zero at every point., Assertion (A): For practical purpose, the, earth is used as a reference for zero potential, in electrical circuits., Reason (R): The electrical potential of a, sphere of radius R with charge Q uniformly, Q, , distributed on the surface is given by 4 R, 0, 77. Assertion(A): Coulomb force between, charges is central force, Reason (R): Coulomb force depends on, medium between charges, 78. Assertion(A): Electric and gravitational fields, are acting along same line. When proton and, - particle are projected up veritically along, that line, the time of flights is less for proton., Reason (R): In the given electric field, acceleration of a charged particle is directly, proportional to specific charge, 79. Assertion(A): When a proton with certain, energy moves from low potential to high, potential then its KE decreases., Reason (R): The direction of electric field is, opposite to the potential gradient and work, done against it is negative., NARAYANAGROUP, , 80. Assertion(A): In bringing an electron towards, a proton electrostatic potential energy of the, system increases., Reason (R): Potential due to proton is positive, 81. Assertion(A): The surface of a conductor is, an equipotential surface, Reason (R): Conductor allows the flow of, charge, 82. Assertion (A) : A charge ' q1 ' exerts some, force on a second charge ' q 2 ' . If a third, charge ' q3 ' is brought near , the force, , 83., , 84., , 85., , 86., , 87., , 88., , exerted by q1 on q 2 does not change, Reason (R): The elecrtostatic force between, two charges is independent of presence of, third charge, Assertion (A) : A point charge 'q' is rotated, along a circle around another point charge, Q. The work done by electric field on the, rotating charge in half revolution is zero., Reason (R) : No work is done to move a, charge on an equipotential line or surface., Assertion: (A): Work done by electric force, is path independent., Reason: (R): Electric force is conservative, Assertion (A): In bringing an electron towards, a proton electrostatic potential energy of the, system increases., Reason (R): Potential due to proton is, positive., Assertion(A): Two particles of same charge, projected with different velocity normal to, electric field experience same force, Reason (R): A charged particle experiences, force, independent of velocity in electric field, Assertion(A): The coulomb force is the, dominating force in the universe, Reason (R): The coulomb force is stronger, than the gravitational force., Assertion(A): A circle is drawn with a point, , positive charge q at its centre. The work, done in taking a unit positive charge once, around it is zero, Reason (R): Displacement of unit positive, charge is zero, 89. Assertion(A): Electric potential at any point, on the equatorial line of electric dipole is zero., Reason (R): Electric potential is scalar, 101
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ELECTRO STATICS & CAPACITORS, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , 90. Assertion(A): The potential at any point due 95. Out of the following statements, to a group of ' N ' point charges is simply, A. Three charge system can not have zero, mutual potential energy, arrived at by the principle of superposition, Reason (R): The potential energy of a system, B. The mutual potential energy of a system, of two charges is a scalar quantity, of charges is only due to positive charges, 91. Assertion (A): The electrostatic potential, 1) A is wrong and B is correct, energy is independent of the manner in which, 2) A is correct and B is wrong, the cofiguration is achieved, 3) Both A and B are correct, Reason (R): Electrostatic field is conservative, 4) Both A and B are wrong, field, 96. Statement A: Electrical potential may exist, STATEMENT QUESTIONS, at a point where the electrical field is zero, 92. Statement-1:- For a charged particle moving, Statement B : Electrical Field may exist at a, from point P to point Q, the net work done by, point where the electrical potential is zero., an electrostatic field on the particle is, Statement C : The electric potential inside a, independent of the path connecting point P to, charge conducting sphere is constant., point Q., 1) A, B are true, 2) B,C are true, Statement-2:- The net work done by a, 3) A,C are true, 4) A,B,C are true, conservative force on an ojecte moving along, a closed loop is zero, 97. Statement A: If an electron travels along the, 1) Statement-1 is true, statement-2 is true,, direction of electric field it gets accelerated, Statement-2 is the correct explanation of, Statement B: If a proton travels along the, statement-1., direction of electric field it gets retarded, 2) Statement-1 is true, statement-2 is true,, 1) Both A & B are true2) A is true, B is false, Statement-2 is not the correct explanation of, 3) A is false, B is true 4) Both A & B are false, statement-1., 98. A : Charge cannot exist without mass but mass, 3) Statement-1 is false, Statement-2 is true., can exist without charge., 4) Statement-1 is true, Statement-2 is false, B : Charge is invariant but mass is variant, 93. A dielectric slab of thickness d is inserted in a, with velocity, parallel plate capacitor whose negative plate, C : Charge is conserved but mass alone may, is at x = 3d. The slab is equidistant from the, not be conserved., plates. The capacitor is given some charge., 1) A, B, C are true 2) A, B, C are not true, As ‘x’ goes from 0 to 3d:, 1) the magnitude of the electric field remains, 3) A, B are only true 4) A, B are false, C is true, the same, 99. A particle of mass m and charge q is fastened, 2) the direction of the electric field remains the, to one end of a string fixed at point O. The, same, whole system lies on a frictionless horizontal, 3) the electric potential increases continuously, plane. Initially, the mass is at rest at A. A, 4) the electric potential dicreases at first, then, uniform electric field in the direction shown, increases and again dicreases, is then switched on. Then, 94. Choose the wrong statement, A, 1) Work done in moving a charge on equipotential, E, surface is zero., l, 2) Electric lines of force are always normal to, an equipotential surface, 60, 3) When two like charges are brought nearer,, B, O, then electrostatic potential energy of the system, 1) the speed of the particle when it reaches B is, gets decreased., 2qE, 4) Electric lines of force diverge from positive, charge and converge towards negative charge., m, 0, , 102, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 2) the speed of the particle when it reaches B is, , qE, m, 3) the tension in the string when particles, Eq, ., 2, 4) the tension in the string when the particle, reaches at B is qE., 100. A conducting sphere A of radius a, with, charge Q, is placed concentrically inside a, conducting shell B of radius b. B is earthed., C is the common centre of the A and B, , reaches at B is, , B, A, , a, , Q, , C, , p) The field at a distance r from C, where, , a r b , is k, , Q, , r2, q) The potential at a distance r from C, where, Q, r, r) The potential difference between A and B, , a r b , is k, , 1 1, is kQ , a b, s) The potential at a distance r from C, where, 1 1 , a r b , is kQ r b , Choose the correct answer, 1. p and r are true, 2. q is true, 3. p,r,s are true, 4. p,q,r,s are true, 101. A block of mass m is attached to a spring of, force constant k. Charge on the block is q. A, horizontal electric field E is acting in the, direction as shown. Block is released with, the spring in unstretched position, E, q, m, , k, , smooth, , a) block will execute SHM, b) Time period of oscillation is 2 , NARAYANAGROUP, , m, k, , qE, k, d) Block will oscillate but not simple, harmonically, Choose the correct answer, 1) a and b are true, 2) d is true, 3) a,b,c are true, 4) a,b,c,d are true, 102. A charge is moved against repulsion. Then, there is, A) decreasing its kinetic energy, B) increasing its potential energy, C) increasing both the energies, D) decreasing both the energies., 1) A, B, C, D are true 2) A, B, C are true, 3) A, B are true, 4) A only true, 103. Which of the following statements are correct?, a) The electrostatic force does not depend on, medium in which the charges are placed, b) The electrostatic force between two, charges does not exist in vacuum, c) The gravitational force between masses, can be usually neglected in comparision with, electrostatic force, d) Any excess charge given to a conductor,, not always resides on the outer surfaceof the, conductor., 1) both a & c 2) only ' c ' 3) both c & d 4) all, 104. The property of the electric line of force, a) The tangent to the line of force at any point, is parallel to the directio of ' E ' at the point, b) No two lines of force intersect each other, 1) both a & b 2) only a 3) only b 4) a or b, 105. Which of the following statements are, correct., a) Electric lines of force are just imaginary, lines, b) Electric lines of force will be parallel to, the surface of conductor, c) If the lines of force are crowded, them field, is strong, d) Electric lines of force are closed loops, 1) both a & c, 2) both b & d, 3) only a, 4) all, 106. Statement(A): Negative charges always move, from a higher potential to lower potential point, Statement (B): Electric potential is vector., 1) A is true but B is false, 2) B is true but A is false, 3) Both A and B false, 4) Both A and R are true, , c) amplitude of oscillation is, , 103
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 107. Statement (A): A solid conducting sphere, E, E, holds more charge than a hollow conducting, a), b), sphere of same radius, Statement (B) : Two spheres A and B are, r, r, connected by a conducting wire. No charge, will flow from A to B, when their radii are R, and 2R and charges on them are 2q and q, V, V, respectively, c), d), 1) A is true, B is false, 2) A is false B is true, r, r, 3) Both A and B are true, Choose the correct answer, 4) Both A and B are false, 1) b and c are true, 2) a is true, 108. A ring with a uniform charge Q and radius R,, 3) a,b,c are true, 4) a,b,c,d are true, is placed in the yz plane with its centre at the, 111., Two, concentric, shells, of radii R and 2R have, origin, given charges q and – 2q as shown in figure., a) The field at the origin is zero, In a region r < R, Q, 2q, b) The potential at the origin is k, R, q, , Q, c) The filed at the point (x, 0, 0) is k 2, x, , d) The field at the point (x, 0, 0) is k, , Q, R x2, , 0 z0 , b) Simple harmonic, for all values of z0, satisfying 0 z0 R, c) Approximately simple harmonic, provided, z0<<R, d) Such that P crosses O and continues to move, along the negative z-axis towards z , Choose the correct answer, 1) a and b are true, 2) c is true, 3) a,c,d are true, 4) a,b,c,d are true, 110. A circular ring carries a uniformly distributed, positive charge. The electric field (E) and, potential (V) varies with distance (r) from the, centre of the ring along its axis as, , r, , 2R, , 2, , Choose the correct answer, 1) a and b are true, 2) c is true, 3) a,b,c are true, 4) a,b,c,d are true, 109. A positively charged thin metal ring of radius, R is fixed in the xy plane, with its centre at, the origin O. A negatively charged particle P, is released from rest at the point (0, 0, z0),, where z0>0. Then the motion of P is, a) Periodic, for all value of z0 satisfying, , 104, , R, , a) E = 0 b) E 0 c) V = 0, d) V 0, Choose the correct answer, 1) a and c are true, 2) c is true, 3) a,d,c are true, 4) a,b,c,d are true, , C.U.Q - KEY, 1) 2, 7) 1, 13) 4, 19) 2, 25) 2, 31) 2, 37) 1, 43) 2, 49) 3, 55) 1, 61) 1, 67) 2, 73) 1, 79) 2, 85) 4, 91) 1, 97) 4, 103) 2, 109) 1, , 2) 2, 8) 3, 14) 1, 20) 1, 26) 1, 32) 4, 38) 3, 44) 4, 50) 1, 56) 1, 62) 1, 68) 2, 74) 1, 80) 4, 86) 1, 92) 1, 98) 1, 104) 1, 110) 1, , 3) 2, 9)1, 15) 3, 21) 1, 27) 3, 33) 1, 39) 2, 45) 1, 51) 2, 57) 1, 63) 3, 69) 1, 75) 1, 81) 2, 87) 4, 93) 2, 99) 2, 105) 1, 111) 1, , 4) 3, 10) 2, 16) 2, 22) 3, 28) 4, 34) 2, 40) 2, 46) 3, 52) 3, 58) 1, 64) 4, 70) 3, 76) 2, 82) 2, 88) 2, 94) 3, 100) 3, 106) 3, , 5) 2, 11) 1, 17) 4, 23) 4, 29) 4, 35) 4, 41) 2, 47) 3, 53) 2, 59) 2, 65) 1, 71) 3, 77) 2, 83) 1, 89) 2, 95) 4, 101) 3, 107) 4, , 6) 3, 12) 1, 18) 1, 24) 3, 30) 4, 36) 3, 42) 4, 48) 2, 54) 1, 60) 3, 66) 3, 72) 1, 78) 1, 84) 1, 90) 2, 96) 4, 102) 3, 108) 1, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , ELECTRIC FIELD, , LEVEL - I ( C.W ), , 9., , ELECTRIC CHARGES AND, CONSERVATION OF CHARGE, 1., , 2., , One million electrons are added to a glass, rod. The total charge on the rod is, 1) 1013 C, 2) 1.6 1013 C, 3) 1.6 1012 C, 4) 1012 C, A body has a charge of 9.6 1020 coulomb. It, is, 1) possible, 2) not possible, 3) may (or) may not possible, 4) Data not sufficient, , COULOMB’S LAW, 3., , 4., , 5., , 6., , A force of 4N is acting between two charges, in air. If the space between them is completely, filled with glass r 8 , then the new force, will be, 1) 2N, 2) 5N, 3) 0.2N, 4) 0.5N, There are two charges 1c and 2 c, kept at certain seperation . The ratio of, electro static forces acting on them will be in, the ratio of, 1) 1 : 2 2) 2 : 1, 3) 1 : 1, 4) 1 : 4, Two identical metal spheres possess +60C, and –20C of charges. They are brought in, contact and then separated by 10 cm.The, force between them is, 2) 36 1014 N, 1) 36 1013 N, 3) 36 1012 N, 4) 3.6 1012 N, A charge q is placed at the centre of the line, joining two equal charges Q. The system of, three charges will be in equilibrium if q is equal, to, 1) , , 7., , 8., , Q, 2, , 2) , , Q, 4, , 3) , , Q, 4, , 4), , Q, 2, , Three charges -q, +q and -q are placed at the, corners of an equilateral triangle of side ‘a’., The resultant electric force on a charge +q, placed at the centroid O of the triangle is, 3q 2, q2, q2, 3q 2, 1), 2), 3), 4), 4 0 a 2, 4 0 a 2, 2 0 a 2, 2 0 a 2, A charge of 2C is placed at x=0 and a, charge of 32C at x=60 cm. A third charge –, Q be placed on the x-axis such that it, experiences no force. The distance of the, point from 2C is(in cm), 1) -20, 2) 20, 3) 15, 4) 10, , NARAYANAGROUP, , Two charges of 50 C and 100 C aree, separated by a distance of 0.6m. The intensity, of electric field at a point midway between them, is, 1) 50 106 V m, , 2) 5 106 V m, , 3) 10 106 V m, 4) 10 106 V m, 10. Two point charges Q and -3Q are placed some, distnace apart. If the electic field at the location, , of Q is E , the field at the location of -3Q is, , , E, E, , , 1) E, 2) E, 3) , 4) , 3, 3, 11. A mass m carrying a charge q is suspended, from a string and placed in a uniform horizontal, electric field of intensity E. The angle made, by the string with the vertical in the equilibrium, position is, mg, , m, , 1, 1) tan Eq, , 1, 2) tan Eq, , Eq, m, , 1, 4) tan mg, , 3) tan 1, , Eq, , 12. A proton of mass ‘m’ charge ‘e’ is released, from rest in a uniform electric field of strength, ‘E’. The time taken by it to travel a distance, ‘d’ in the field is, 1), , 2de, mE, , 2), , 2dm, Ee, , 3), , 2dE, me, , 4), , 2Ee, dm, , 13. An infinite number of charges each of, magnitude q are placed on x - axis at distances, of 1,2, 4, 8, ... meter from the origin. The, intensity of the electric field at origin is, q, q, q, q, 1) 3 2) 6, 3) 2, 4) 4, 0, 0, 0, 0, 14. A uniformly charged thin spherical shell of, radius R carries uniform surface charge, density of per unit area. It is made of two, hemispherical shells, held together by pressing, them with force F.F is proportional to, 1 2 2, 1 2, 1 2, 1 2, , R, , R, 1) , 2) , 3), 4), o R, o R 2, o, o, , ELECTRIC POTENTIAL AND, POTENTIAL ENERGY, 15. The p.d. between two plates separated by a, distance of 1 mm is 100 V. The force on an, electron placed in between the plates is, 1) 105 N, 2) 1.6 1024 N, 3) 1.6 1014 N, 4) 1.6 1019 N, , 105
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 16. An infinite number of charges each equal to, 1) 2.5 10 19 J, 2) 2 10 19 J, 'q' are placed along the X-axis at x = 1, x = 2,, 3) 4 10 19 J, 4) 2 10 16 J, x = 4, x = 8 ..... The potential at the point x = 0, 24. The dipole moment of the given system is, due to this set of charges is, 2q, , 1), , Q, 4 o, , 2), , 2Q, 4 o, , 3), , 3Q, 4 o, , 4), , Q, o, , 17. A, B, C are three points on a circle of radius 1, cm. These points form the corners of an, equilateral triangle. A charge 2C is placed at, the centre of the circle. The work done in, carrying a charge of 0.1 C from A to B is, 1) Zero 2) 18 1011 J 3) 1.8 1011 J 4) 54 1011 J, 18. Charges +q, -4q and +2q are arranged at the, corners of an equilateral triangle of side 0.15m., If q=1 C, their mutual potential energy is, 1) 0.4J, 2) 0.5J, 3) 0.6J, 4) 0. 8J, 19. An electron of mass ‘M’ kg and charge ‘e’, coulomb travels from rest through a potential, difference of ‘V’ volt. The final velocity of the, electron is (in m/s), 1), , 2eV, M, , 2), , 2MV, e, , 3), , 2eV, M, , 4), , 2MV, e, , 20. A charge ‘Q’ is placed at each corner of a cube, of side ‘a’. The potential at the centre of the, cube is, (2008 M), 4Q, 8Q, 4Q, 2Q, 1) a 2) 4 a 3) 3 a 4) a, 0, 0, 0, 0, 21. A uniform electric field pointing in positive xdirection exists in a region let A be the orgin B, be the point on the x-axis at x = +1 cm and C, be the point on the Y axis at y = +1cm. Then, the potentials at the points A, B and C satisfy, 1) VA < VB 2) VA > VB 3) VA < VC 4) VA > VC., 22. The electric field at the origin is along the +ve, x-axis. A small circle is drawn with the centre, at the origin cutting the axes at the points A,, B, C and D having coordinates (a, 0), (0, a),, (-a, 0), (0, -a) respectively. Out of points on, the periphery of the circle, the potential is, minimum at, B (0,a), , 1) A, , 2)B, , 3)C, , 4)D, , C ( a, 0), , E, A (a, 0), D (0, a), , l, , l, , q, , q, , 1) 3ql along perpendicular bisector of q - q line, 2) 2 ql along perpendicular bisector of q - q line, 3) ql 2 along perpendicular bisector of q - q line, 4) 0, , LEVEL - I ( C.W ) KEY, 1)2, 7) 3, 13) 1, 19) 3, , 2) 2, 8) 1, 14) 1, 20) 3, , 3) 4, 9)2, 15) 3, 21) 2, , 4) 3, 10) 3, 16) 2, 22)1, , 5) 1, 11) 4, 17) 1, 23) 2, , 6) 2, 12) 2, 18) 3, 24) 1, , LEVEL - I ( C.W ) HINTS, 1., , Q ne n is integer, , 3., , F| , , 2. Q ne n is integer, 1 qq, , F, K, , 1 2, 4. F 4 r 2, 0, 2, , 5., 6., , 7., , 1 q1 q 2 , F , 4 0, 4d 2, 1 QQ, 1, qQ, , 0, 4 0 l 2, 4 0 l 2, 2 , , 1, , q, , d, , x, , 1 q1q2, F, 4 0 r 2, , q2, 1, q1, , 8., , , 1 Q, 10. E 4 3 r, 0 r, , q, , 1, , 1, 2, 9. E 4 x 2 4 x 2, 0 1, 0 2, , 11. qE mg tan , , s, , 12., , q 1, , 1, , 1, , 1 qE 2, t, 2 m, , , , 13. E 4 2 2 2 , 2, 4, , 0 1, 2, 2, , , R 2, 14. Pressure , and Force =, 2 o, 2 o, Vq, 15. F Eq =, d, Q 1, , 1, , 1, , 1, , , , 16. V 4 1 2 4 8 , , 0 , 23. An electric dipole is along a uniform electric 17. Equipotential surface, field. If it is deflected by 600, work done by an, , , agent is 2 1019 J. Then the work done by an 18. U 1 q1q2 q2 q3 q1q3 , r2, r3 , 4 0 r1, agent if it is deflected by 300 further is, , DIPOLE, , 106, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 1 q, 20. V 4 d, 0, 21. Along the field direction potential decreases., , 22. V E d r, 23. W1 pE 1 cos and, , 1 2, mv eV, 19., 2, , 8., , W2 pE cos 1 cos 2 , 9., , 24. p1 lq p2 and PR 3ql, , LEVEL - I ( H. W ), , 1) 256 volt2), , COULOMB’S LAW, , 1., , 2., , 4., , , , , , 2, , 2, , a, b, , ELECTRIC FIELD, , 6., , An electron mass 9.1 10 31 kg is sent into an, electric, field, of, intensity, 9.1106 newton/coulomb. The acceleration, produced is, 13., 18 m, 6 m, 1.6, , 10, 1.6, , 10, 1), 2), s2, s2, , 1), , qQ, 4 0, , 1 1, a b, , , qQ 1 1 , , 4 0 a 2 b 2 , , qQ 1, , 1, , 2) 4 a b , 0, qQ 1, , 1, , 3), 4) 4 a 2 b 2 , 0, An electric cell does 5 joules of work in carrying, 10 Coulomb’s of charge around a closed, circuit. The emf of the cell is, 1) 2V, 2) 0.5V, 3) 4V, 4) 1V, 3) 1.6 10 18 m s 2, 4) 1.6 106 m s 2, The electric field at (30, 30) cm due to a charge 14. Two positive charges 12C and 10C aree, of -8 nC at the origin in NC-1 is, initially separated by 10cm. The work done in, bringing the two charges 4cm closer is, 1) 400 i j, 2) 400 i j, 1) 7.2J, 2) 3.6J, 3) 8.4J, 4) 12.4J, 15. An insulated charged conducting sphere of, 3) 200 2 i j, 4) 200 2 i j, radius 5 cms has a potential of 10V at the, surface. What is the potential at centre?, Two charges of 10 C and -90 C aree, 1) 10V, 2) zero, separated by a distance of 24 cm. Electrostatic, 3) same as that at 5 cms from the surface, field strength from the smaller charge is zero, 4) same as that at 25 cms from the surface, at a distance of, 1) 12 cm 2) 24 cm 3) 36 cm 4) 48 cm, , , , , , , , 7., , volt, , , , 1) 2 109 C 2) 2 109 C 3) 109 C 4) 109 C, 3, 3, 5., , 1, 256, , 3) 256 1019 volt, 4) 250 volt, A charge Q is divided into two parts q1 and q2, such that they experience maximum force of 10. Two electric charges of 9 C and - 3C aree, placed 0.16m apart in air. There are two points, repulsion when separated by certain distance., A and B on the line joining the two charges at, The ratio of Q, q1 and q2 is, distances of (i) 0.04m from - 3C and in, 1) 1 : 1 : 2 2) 1 : 2 : 2 3) 2 : 2 : 1 4) 2 : 1 : 1, between the charges and (ii) 0.08m from Two charges each 1c are at P 2i 3 j k m, 3 C and out side the two charges. The, potentials, at A and B are, and Q i j k m . Then the force between, 1) 0V, 5V 2) 0V, 0V 3) 5V, 0V 4) 5V, 10V, them is _____, 11. Four charges 3C, 1C, 5C and 7 C, 4, are arranged on the circumference of a circle, 1) 100N 2) 10N 3) 10 dyne 4) 100 dyne, of radius 0.5 m. The potential at the centre is, Two charges of 200 C and 200 C aree, 1) Zero, 2) 18 104 V, placed at the corners B and C of an equilateral, 4) 288 103V, 3) 18 104 V, triangle ABC of side 0.1 m. The force on a, 12. A positive point charge ‘q’ is carried from a, charge of 5 C placed A is, point ‘B’ to a point ‘A’ in the electric field of a, 1) 1800 N 2) 1200 3N 3) 600 3N 4) 900N, point charge +Q. If the permittivity of free, Two equally charged pith balls 3 cm apart repel, space is 0 , the work done in the process is, each other with a force of 4 105 newton. The, given by, +Q, A, B, charge on each ball is, , , , 3., , Two electric charges of 109 C and 109 C, are placed at the corners A and B of an, equilateral triangle ABC side 5cm.The electric, intensity at C is, 1)1800N/C 2)3600 N/C 3)900N/C 4)2700 N/C, ELECTRIC POTENTIAL AND, POTENTIAL ENERGY, If 41020 eV is required to move a charge of 0.25, coulomb between two points, the potential, difference between these two points is, , NARAYANAGROUP, , , , , , , , , , , , 107
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ELECTRO STATICS & CAPACITORS, 16. A positive charge 'Q' is fixed at a point.A, negatively charged particle of mass 'm' and, charge 'q' is revolving in a circular path of, radius 'r1' with 'Q' as the centre. The work to, be done to change the radius of the circular, path from r1 to r2 in Joules is, , Qq, 2) 4, o, , 1) 0, , 1 1, , r1 r2 , , LEVEL-I ( H. W ) KEY, , 1 Qq 1 1 , Qq 1 1 , 3) 4 4 r r 4) 4 r r , o, o 1, 2, o 2, 1, 17. Figure bellow shows a square array of charged, particles, with distance d between adjacenet, particle. What is the electric potential at point, P at the centre of the square if the electric, potential is zero at infinity ?, 4q, , - 2q, 4 0 d, q, - 4q, 3), 4), ., 4 0 d, 4 0 d, , 1) Zero 2), , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , 21. An electric dipole of moment p is placed, normal to the lines of force of electric intensity, , E , then the work done in deflecting it through, an angle of 1800 is, 1) pE, 2) 2 pE 3) 2 pE, 4) zero, , 2q, , 1) 4 2) 4 3) 4 4) 2 5) 1 6) 3 7) 1, 8) 2 9) 1 10) 2 11) 1 12) 1 13) 2 14) 4, 15) 1 16) 2 17) 3 18) 2 19)2 20) 4 21) 4, , LEVEL-I ( H. W ) HINTS, F q1q2, , 3., , F1 F2 , , q, , P, , 2q, , q, , 1000 1500 500 , 2 3 then the, given by V = , x, x , x, electric field at x = 1 m is (in volt/m), 1) 5500iˆ 2) 5500iˆ 3) 5500iˆ 4) zero, , 20. An electric dipole of moment p is placed in, the position of stable equilibrium in uniform, electric field of intensity E. It is rotated, through an angle from the intial position., The potential energy of electric dipole in the, position is, 1) pE cos , 2) pE sin , , 108, , ; Fr F1 F2 because angle, , 1 q2, 4 0 d 2, , 5. a , , , 1 Q x, 6. E 4 3 r 7., 0 r, , 9. W qV, Q, , 1, , 11. V 4, 13. em f , , 0, , r, , Q, , 1, , 19. E , , 0, , d, q2, 1, q1, , 1, , Q, , 8. E 4 a 2, 0, , 1 q1 q2 , 10. V r r , 4 0 1, 2 , q1q2 1 1 , 12. W 4 r r , 0 1, 2, 1 1 , r r , 1 2, q1q2 1 1 , 16. W 4 r r , 0 1, 2, , 1 Q, , 17. V 4, , eE, m, , q1q2, 14. W 4, 0, , W, q, , 15. V 4 R, 0, , r, , 18. V=constant and QR, , dV, 20. U p.E 21. W1 pE 1 cos , dx, , LEVEL-II (C.W), , DIPOLE, , 4) pE cos , , F , , q, , 18. The radii of two charged metal spheres are, 5cm and 10cm both having the same charge, 60mC. If they are connected by a wire, 1) A charge of 20mC flows through the wire from, larger to smaller sphere, 2) A charge of 20mC flows through the wire from, smaller to larger sphere, 3) A charge of 40mC flows through the wire from, smaller to larger sphere, 4) No charge flows through the wire because both, spheres have same charge., 19. The electric potential at a point (x, 0, 0) is, , 3) pE 1 cos , , 1 q1q2, 4 0 r 2, , between then is1200, , q, , 4., q, , 1 qq, , 1 2, 2. F 4 r 2, 0, , 1., , COULOMB’S LAW, 1., , Two charges when kept at a distance of 1m, apart in vacuum hava some force of repulsion., If the force of repulsion between these two, charges be same, when placed in an oil of, dielectric constant 4, the distance of, separation is, 1) 0.25m 2) 0.4m, 3) 0.5m, 4) 0.6m, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 2., , 3., , 4., , 5., , 6., , 7., , 8., , ELECTRO STATICS & CAPACITORS, , The excess (equal in number) number of 9. Two identical particles of charge q each are, connected by a massless spring of force, electrons that must be placed on each of two, constant k. They are placed over a smooth, small spheres spaced 3 cm apart with force of, horizontal surface.They are released when, , 19, repulsion between the spheres to be 10 N, unstretched. If maximum extension of the, is, spring is r, the value of k is : (neglect, 1) 25, 2) 225, 3) 625, 4) 1250, gravitational effect), Two small conducting spheres each of mass, q, 1, 1 q2 1, 9 10 4 kg are suspended from the same point, 1) k r r, 2) k 4 l 2 r, 0, 0, by non conducting strings of length 100 cm., 2q, 1, q, 2, They are given equal and similar charges until, 4) k r r, 3) k r r, 0, 0, 0, the strings are equally inclined at 45 each to, ELECTRIC FIELD, the vertical. The charge on each sphere is ....., 10. In the figure shown, the electric field intensity, coulomb, 6, 6, 6, 6, at r 1m, r 6m, r 9m in Vm 1 is, 1) 1.4 10 2) 1.6 10 3) 2 10 4) 1.96 10, V, Two point charges of magnitude 4 C and -9, C are 0.5m apart. The electric intensity is, zero at a distance ‘x’ m from ‘ A’ and ‘y’ m 1)-5, -1.67, +5 2) -5, 0, +5, from ‘B’. ‘x’ and ‘y’ are respectively, 10V, p, 4mc, 9mc, r (m), 3) 0,1.67,0 4) +5, 1.67, -5 0 2, 8 10, 11. Point charges of 3 109 C are situated at, 0.5m, A, B, each of three corners of a square whose side, 1) 0.5m, 1.0m, 2) 1.0m, 1.5m, is 15 cm. The magnitude and direction of, 3) 2.0m, 1.5m, 4) 1.5m, 2.0m, electric field at the vacant corner of the square, A charge +q is fixed to each of three corners, is, of a square. On the empty corner a charge Q, 1) 2296 V/m along the diagonal, is placed such that there is no net electrostatic, 2) 9622 V/m along the diagonal, force acting on the diagonally opposite charge., 3) 22.0 V/m along the diagonal, 4) zero, Then, 12. A large flat metal surface has uniform charge, density . An electron of mass m and charge, 1) Q 2q, 2) Q 2 2q, e leaves the surface at an angle at point A, with speed v , and return to it at point B. The, 4) Q 4q, 3) Q 2q, maximum, value of AB is ____, Electrical force between two point charges is, v 2 e, vm 0, v 2 m 0, v 2e, 200N. If we increase 10% charge on one of, 1), 2), 3), 4), the charges and decrease 10% charge on the, 0 m, 0 m, e, e, other, then electrical force between them for 13. ‘n’ charges Q, 4Q, 9Q, 16Q ..... are placed at, the same distance becomes, distances of 1, 2, 3 ..... metre from a point ‘0’, 1) 198 N 2) 100 N 3) 200 N 4) 99 N, on the same straight line. The electric, N fundamental charges each of charge ‘q’ are, intensity at ‘0’ is, to be distributed as two point charges seperated, Q, Q, nQ, 1) 4 n2 2) 4 n 3) Infinity 4) 4 , by a fixed distance, then the maximum to, 0, 0, 0, minimum force bears a ratio (N is even and, 14. Two point charges q1 2 C and q2 1 C aree, greater than 2), placed at distances b=1 cm and a=2 cm from, 4N 2, N2, 2N 2, N 12, the origin on the y and x axes as shown in, 1), 2), 3), 4), 2, figure. The electric field vector at point (a, b), N 1, N 1, 4N 1, 4N, will subtend an angle with the x - axis given, A particle A having a charge of 2 106 C and, y, by, a mass of 100g is placed at the bottom of a, q, P (a, b), smooth inclined plane of inclination 300. The, distance of another particle of same mass and 1) tan 1 2) tan 2, b, charge, be placed on the incline so that it may, 3) tan 3 4) tan 4, remain in equilibrium is, x, q, 1) 27 cm 2) 16 cm 3) 30 cm 4) 45 cm, O, a, 1, , 2, , NARAYANAGROUP, , 109
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 15. A non-conducting ring of radius 0.5 m carries 22. Figure shows three circular arcs, each of radius, of total charge of 1.11x10-10c distributed nonR and total charge as indicated. The net, uniformly on its circumference producing an, electric pontential at the centre of the, electric field E everywhere in space. The value, curvature is, l 0 , Q, Q, 2Q, Q, , E.dl, 1), 2), 3), 4), (l=0 being, of the integral, 2 o R, 4 o R, o R, o R, l , 23. Two identical conducting very large plates, centre of the ring) in volts is, 1) +2, 2) -1, 3) -2, 4) zero, P1 and P2 having charges 4Q and 6Q are, ELECTRIC POTENTIALAND, placed very closed to each other at separation, d. The plate area of either face of the plate is, POTENTIAL ENERGY, A. The potential difference between plates, 16. Three charges +q, -q and -q are kept at the, vertices of an equilaterial triangle of 10cm side., P1 and P2 is, The potential at the mid point in between -q, Qd, Qd, q, if q = 5 C is, 2) VP1 VP2 A, 1) VP1 VP2 A, o, o, 1) 6.4 105V, 2) 12.8 104 V, 5Qd, 5Qd, 3) 6.4 104 V, 4) 12.8 105 V, 3) VP1 VP2 A, 4) VP1 VP2 A, 17. Two charges each ‘Q’ are released when the, o, o, distance between is ‘d’. Then the velocity of, DIPOLE, each charge of mass ‘m’ each when the distance, 24. An electric dipole consists of two opposite, between them is ‘2d’ is, charges of magnitude 1C separated by a, Q, Q, Q, Q, distance, of 2cm. The dipole is placed in an, 1) 8 dm 2) 4 dm 3) 4 dm 4) 2 dm, 0, 0, 0, 0, electric filed 105 Vm 1 . The maximum torque, 18. An oil drop carrying charge ‘Q’ is held in, that the field exert on the dipole is, equilibrium by a potential difference of 600V, 1) 10 3 Nm, 2) 2 1013 Nm, between the horizontal plates. In order to hold, another drop of twice the radius in equilibrium, 3) 3 103 Nm, 4) 4 10 3 Nm, a potential drop of 1600V had to be maintained. 25. An electric dipole is formed two particles fixed, The charge on the second drop is, at the ends of a light rigid rad of length l. The, Q, 3Q, mass of each particle is m and charges are -q, 1), 2) 2Q, 3), 4) 3Q, and +q The system is suspended by a, 2, 2, 19. A body of mass one gram and carrying a charge, torsionless thread in an electric field of, 8, intensity E such that the dipole axis is parallel, 10 C passes through two points P and Q. The, to the field if it is slightly displaced, the period, electrostatic potential at Q is OV. The velocity, of angular motion is, , 1, of the body at Q is 0.2ms and at P is, , , , ml, , 2qE, , 1, , ml, , ml, , 1, , 0.028ms 1 . The potential at P is, 1), 2) 2 qE 3) 2 2qE 4) 2 4qE, 2 ml, 1) 150V 2) 300V, 3) 600V, 4) 900V, 26. Two point charges - q and +q are located at, 20. Three charges each 20 C are placed at the, points (0,0,-a) and (0,0,a) respectively. The, corners of an equilateral triangle of side 0.4m ., electric potential at point (0,0,z) is z a , The potential energy of the system is, +Q, 1) 18 106 J 2) 9J 3) 9 106 J 4) 27J, , 21. An electric field is expressed as E 2i 3j ., 45, 0, , The potential difference VA VB between, qa, q, 2Q, 1), 2), 2, two points A and B whose positions vectors are, 4 0 z, 4 0 a, given by rA i 2j and rB 2i j 3k is, 2 qa, 2qa, 1) -1 V 2) 1 V, 3) 2 V, 4) 3 V, 3) 4 z 2 a 2 4) 4 z 2 a 2, 0, 0, , , , 110, , , , , , 0, , 30, R, , 3Q, , , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 27. Two equal charges ‘q’ of opposite sign are, separated by a small distance ‘2a’. The electric, intensity ‘E’ at a point on the perpendicular, bisector of the line joining the two charges at, a very large distance ‘r’ from, the line is, + 4Q, , e, Force on electron = eE = , 0, , e, Acceleration of electron a = m , 0, It will act as projectile wit h max range, u2 u2, m 0, = , a e, , + 4Q, , 1 qa, , 2qa, , 1, , 1) 4 r 2, 0, , P1, , 2) 4 r 3, 0, , P2, , 1, , 1 qa, 1 2 qa, 3) 4 r 2 4) 4 r 3, 0, 0, , d, , 3) 1, 10) 2, 17) 1, 24) 2, , 4) 2, 11) 1, 18) 4, 25) 3, , 5) 2, 12) 2, 19) 3, 26) 3, , 6) 1 7) 3, 13) 4 14)1, 20) 4 21)1, 27) 2, , LEVEL - II ( C. W ) HINTS, 1., , t1 , , 2., , F, , q, 15. Vo K , V 0 ;, R, , V3 , and q = ne, , 3., , F = w tan , , where, , 4., , Distance of null point x , +ve for like charges, , 5., , q2, 1, 1, 2 2, 4 0, a, 4 0, , 6., , F, , 7., , N, Fmax, 2, , F min N 1 1, , 1 q1q2, 4 0 r 2, , d, Q2, 1, Q1, , Qq, , , , 2a, , , , 2, , 110, 90, q1 and q12 , q2, 100, 100, , 2, , 8. mg sin , , , , 1, 4 0, , q2, r2, , , , , E.dl Vo V, , t , , 1, q, 4 0 3a , , , 2 , 3, , 17. gain in K.E = loss in P.E, 19., , 0, , dV, 9. FC kx, 10. E , dr, 1 q, 11. E E 2 1/ 2 E , 4 0 r 2, r = length of the side, , 12. Field near metal surface E= , 0, , , , 1 q1q2, 4 0 r 2, , -ve for unlike charges, , 1, ; q1 , , , , F, , t0, , 1 q , V, , V, , 1, 2, 16. V V1 V2 V3 ;, 4 0 a / 2 , , t, k, 1 q1q2, 4 0 r 2, , Q , , Q, , E2, 14. Tan E, 1, , LEVEL - II ( C.W ) KEY, 1) 3 2) 3, 8) 1 9) 2, 15) 1 16) 4, 22) 1 23) 2, , Q, , 1, 2, n, 13. E 4 . x 2 x 2 .... x 2 , o 1, 2, n , , V1 R1 Q2, 18. V R . Q, 2 1, 2, , 1, m vQ2 v 2p q VP VQ , 2 , , 1 q1q2 q2 q3 q3q1 , 20. U 4 r r r , 0 12, 23, 13 , 1, 2, , V, , V, , , 2dx, , 3dy , , A, 21. B, , 1, 2, , 22. V V1 V2 V3, , Q, 23. VP1 VP2 A / d, o, , 24. max pE 2aqE, 25. PE sin ; I ; I PE sin , I = moment of inertia =, , ml 2, 2, , Time period 2, , I, pE, , 26. The distance of point P from charge +q is r1 z a, and from charge -q is r2 z a, , NARAYANAGROUP, , 111
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, Pot ential, , at, , P, , 1 q q, 1, 2qa, , 2, 4 0 r1 r2 4 0 z a 2, , , , is, , 6., , , , 27. Similar to B on equitorial line of a short bar magnet, , LEVEL - II ( H. W ), COULOMB’S LAW, 1., , 2., , Two equally charged identical metal spheres, A and B repel each other with a force F., Another identical uncharged sphere C is, touched to A and then placed midway between, A and B. The net force on C is in the direction, 1) F towards A, 2) F towards B, 3) 2F towards A, 4) 2F towards B, Two unlike charges seperated by a distance, of 1m attract each other with a force of, 0.108N . If the charges are in the ratio, 1: 3 ,the weak charge is, 1) 2 C, , 3., , 2) 4 C, , 5., , 4) 5 C, , Three charges each equal to 10 9 C are, placed at the corners of an equilateral triangle, of side 1m. The force on one of the charges is, 1) 9 109 N, 3) 27 109 N, , 4., , 3) 6 C, , 2) 9 3 109 N, 4) 18 109 N, , Two particles each of mass ' m ' and carrying, charge ' Q ' are seperated by some distance.If, they are in equilibrium under mutual, gravitational and electro static forces, then, Q / m (in c/Kg) is of the order of, 1) 105, 2) 10 10, 3) 10 15, 4) 10 20, There point charges + q, – q and + q are, placed at the vertices P, Q and R of an, equilateral triangle as shown., If, 1 q2, , where 'r' is the side of the, 40 r 2, triangle, the force on charge at 'P' due to, charges at Q and R is, Y, F, , 7., , Three point charges +q, +q and –q are placed, at the corners of an equilateral triangle of side, 'a'. Another charge +Q is kept at the centroid., Force exerted on Q is:, 1 2qQ, 1 6qQ, 1) 4pe, 2) 4pe, 2, 2, o a, o a, 1 8qQ, 1 14qQ, 3) 4pe, 4), 2, 4peo a 2, o a, Three charges q1 , q 2 and q 3 are placed as, shown in fig. The X-component of the force on, q1 is proportional to, Y, q3, , q 2 q3, q q, 2 cos 2) 22 23 sin a, 2, b a, b a, q q, q q, 3) 22 22 cos 4) 22 22 sin , b, a, b a, , 1), , , q1, , b, , 1) F along positive x–direction, 2) F along negative x–direction, 3) 2 F along positive x–direction, 4) 2 F along negative x–direction, 112, , X, , The breakdown electric intensity for air is, 3 106 V/m. The maximum charge that can be, held by a sphere of radius 1 mm is, 1) 0.33 C 2) 0.33 nC 3) 3.3 C, 4) 3.3 C, 9. There is a uniform electric field of strength, 103V / m along y-axis. A body of mass 1 g and, charge 106 C is projected into the field from, origin along the positive x-axis with a velocity, 10 m/s. Its speed in m/s after 10s is (neglect, gravitation), 1) 10, 2) 5 2, 3) 10 2, 4) 20, 10. The point charges 1C , 1C and 1C are, placed at the vertices A, B and C of an, equilateral triangle of side 1m. Then, (A) The force acting on the charge at A is, 9 109 N, (B) The electric field strength at A is, 9 109 NC 1, 1) A is correct but B is wrong, 2) B is correct but A is wrong, 3) Both A and B are wrong, 4) Both A and B are correct, 11. A pendulum bob of mass m carrying a charge, q is at rest in a uniform horizontal electric field, of intensity E. The tension in the thread is, 2, , 1) T Eq mg , 2, , R, , q1, , ELECTRIC FIELD, 8., , 2, , P, , X, , E m, 3) T , q g, , 2, , E, 2, 2) T mg , q, , 2, , 4) T mg Eq, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , ELECTRIC POTENTIALAND, POTENTIAL ENERGY, , DIPOLE, , 12. Four charges 108 ; 2 108 ; 3 108 and 2 108, coulomb are placed at the four corners of a, square of side 1m the potential at the centre, of the square is, 1) zero 2)360 volt 3) 180 volt 4) 360 2 volt, 13. Two metal spheres of radii R1 and R2 aree, charged to the same potential. The ratio of the, charge on the two spheres is, 1) 1, , 2), , 1, 2, , 4) R, 2, 14. Two concentric, thin metallic spherical shells, of radii R1 and R2 R1 R2 bear charges Q1 and, Q2 respectively. Then the potential at radius, , Q1 Q2, Q1 Q2, 2) R r, r, 1, , Q1, , LEVEL - II ( H. W ) KEY, 1)1 2) 1 3) 2 4) 2 5) 2 6) 2 7) 2 8) 2, 9)3 10) 4 11) 1 12) 4 13) 4 14) 2 15) 2 16) 4, 17) 3 18) 2 19) 1, , R1, , 3) R1 R2, , ‘r’ between R1 and R2 will be, , 19. The self potential energy of hydrogen chloride, whose dipole moment is 3.44 1030 C - m and, separation between hydrogen and chlorine, atoms is 1.01 1010 m is, 1) 1.036 1019 J, 2) 3.2 105 J, 3) 4.5 107 J, 4) 1.65 106 J, , 1, 4 0, Q2, , LEVEL - II ( H .W ) HINTS, F, , 3., , F1 F2 , , times, Q1, , Q2, , 3) R R 4) R R, 1, 2, 2, 2, 4., 3, 15. An electric charge 10 C is placed at the, origin (0, 0) of X-Y coordinate system. Two, 5., points A and B are situated at 2, 2 and, (2, 0) respecitvely. The potential difference 6., between the points A and B will be:, 1) 9 V, 2) zero, 3) 2 V, 4) 4.5 V, 16. A charge 2 C at the origin, 1 C at 7cm 8., and 1 C at 7cm are placed on X axis. The, 9., mutual potential energy of the system is, 1) 0.051J 2) 0.045J 3) 0.045J 4) 0.064J, 17. Four equal charges Q are placed at the four 10., corners of a square of side ' a ' each. Work, done in removing a charge Q from its centre, to infinity is, 12., 2Q2, 1) zero, 2), 4 0 a, 14., 1), , , , 3), , 2Q 2, , 4), , a, , , , Q2, , 1 q1q2, 4 0 r 2, , 1., , Fe , , 1 q2, 4 0 r 2, , 1, , qq, , 1 2, 2. F 4 r 2, 0, , 1 q2, , ; FR 3. 4 r 2, 0, , Gm 2, 1 q2, and, F, , g, 4 0 r 2, r2, , F1 F2 and angle between them is 1200, 1 q, 2, F , 4 0 r 2 , E, , where r , , a, 3, , 1, Q, ., 4 o d 2, , Eq, m, 1 q1q2, 1 q, F, . 2 ; E, ., 4 o r 2, 4 o r, , v u at where a , , F F1 F2 ; E E1 E2, V, , 1, Q, , 4 0, r, , 1 Q, , 13. V 4 R, 0, Potential is constant within the sphere and is, additive., , 2 a, , 0, 0, 1 q1q2, q 1 1, 18. The electrostatic potential V at any point 15. V , PE , ., , 16., 4 0 r, 4 o r1 r2 , (x,y,z) in space is given by V 4 x 2, 1) The y - and z - components of the electrostatic 17. Workdone = Electrostatic potential, energy at the centre of the square, field at any point are not zero, 2) The x - component of electric field intensity at, dV, 18. E , , dx, any point is given by 8 xi, 3) The x - component of electric field intensity at a 19. p 2 qa q p 3 .41 10 2 0, , , , , , point (2, 0,2) is 8i, 4) The y - and z - components of the field are, constant in magnitude., , , , NARAYANAGROUP, , 2a, , , , PE , , 1 q2, 1.036 10 19 J, 4 0 2 a, 113
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 5., , LEVEL - III, ELECTROSTATIC FORCE, 1., , A ball of mass m = 0.5 kg is suspended by a, thread and a charge q = 0.1 C is supplied., When a ball with diameter 5cm and a like, charge of same magnitude is brought close to, the first ball, but below it, the tension decreases, to 1/3 of its initial value. The distance between 6., centres of the balls is, 2) 0.51 10 4 m, 1) 0.12 10 2 m, 3) 0.2 10 5 m, 4) 0.52 10 2 m, Five point charges each +q, are placed on five, vertices of a regular hexagon of side L, The, magnitude of the force on a point charge of, value – q placed at the centre of the hexagon, (in newton) is, , 2., , q2, , 3q2, , q2, , 3), 4), 4 0 L2, 4 3 0 L2, 4 0 L2, 7., 3. Two small objects X and Y are permanently, separated by a distance 1 cm. Object X has a, charge of + 1.0 C and object Y has a charge, of - 1.0 C . A certain number of electrons, are removed from X and put onto Y to make, the electrostatic force between the two objects, an attractive force whose magnitude is 360 N., Number of electrons removed is, 1) 8.4 1013 2) 6.25 1012 3) 4.2 1011 4) 3.5 1010 8., 4. Two identical positive charges are fixed on the, y-axis, at equal distance from the origin O, A, partical with a negative charge starts on the, negative x-axis at a large distance from O,, moves along the x-axis passed through O and, moves far away from O. Its acceleration a is, taken as positive along its direction of motion., The particle's acceleration a is plotted against, its x-co-ordinate. Which of the following best, represents the plot?, 1)Zero 2), , a, , 1), , K, K, 3, 3, K 4), 2), 3), 3, 2, K, 2, The force of attraction between two charges, separated by certain distance in air is F1. If, the space between the charges is completely, filled with dielectric of constant 4 the force, becomes F2. If half of the distance between the, charges is filled with same dielectric the force, between the charges is F3. Then F1 : F2 : F3 is, 1) 16 : 9 : 4, 2) 9 : 36 : 16, 3) 4 : 1 : 2, 4) 36 : 9 : 16, Two small spheres of masses, M 1 and M 2 aree, suspended by weightless insulating threads of, lengths L1 and L2 . the sphere carry charges, 1), , Q1 and Q2 respectively. The spheres aree, suspended such that they are in level with, another and the threads are inclined to the, vertical at angles of 1 and 2 as shown below,,, which one of the following conditions is, essential , if 1 2 ., , a, x, , o, , Two equal negative charges –q each are fixed, at points (0, –a) and (0,a) on y-axis. A positive, charge Q is released from rest at the point (2a,, 0) on the x-axis. The charge Q will, 1) execute simple harmonic motion about the origin, 2) move to the origin and remain at rest, 3) move to infinity, 4) execute oscillatory but not simple harmonic, motion, In a liquid medium of dielectric constant K and, of specific gravity 2, two identically charged, spheres are suspended from a fixed point by, threads of equal lengths. The angle between, them is 90º. In another medium of unknown, dielectric constant K1, and specific gravity 4,, the angle between them becomes 120º. If, density of material of spheres is 8 gm/cc then, K1 is :, , 2), , L1, , x, , O, , M1, Q1, a, , 3), , 114, , O x, , 1, , 2, , L2, , M2, Q2, , a, x, 4), , O, , 1) M 1 M 2 but Q1 Q2 2) M 1 M 2, 3) Q1 Q2, , 4) L1 L2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , ELECTRIC FIELD, 9., , 15. Electric field on the axis of a small electric, , , dipole at a distance r is E1 and E2 at a distance, of 2r on a line of perpendicular bisector. Then, , , , , 1) E2 E1 / 8, 2) E2 E1 /16, , , , 3 ) E2 E1 / 4, 4) E2 E1 / 8, 16. A particle having charge that on an electron, and mass 1.6 x 10–30 kg is projected with an, initial speed 'u' to the horizontal from the lower, plate of a parallel plate capacitor as shown., The plates are sufficiently long and have, separation 2cm. Then the maximum value of, velocity of particle not to hit the upper plate., (E=103 V/m upwards)., , If the electric field between the plates of a, cathode ray oscilloscope be 1.2 104 N / C , the, deflection that an electron will experience if it, enters at right angles to the field with kinetic, energy 2000 eV is (The deflection assembly is, 1.5cm long.), 1) 0.34 cm2) 3.4 cm, 3) 0.034 mm, 4) 0.34 mm, 10. A electric field of 1.5 104 NC 1 exists, between two parallel plates of length 2 cm. An, electron enters the region between the plates, at right angles to the field with a kinetic energy, of Ek 2000eV . The deflection that the, u, electron experiences at the deflecting plates, E = 103V/m, is, 45°, 1) 0.34 mm 2) 0.57 mm 3) 7.5 mm 4) 0.75 mm, 6, 11. A bob of a simple pendulum of mass 40gm with, 1)2 x 10 m/s, 2) 4 x 106 m/s, 6, , 6, a positive charge 4 10 C is oscillating with a, 3) 6 x 10 m/s, 4) 3x 106 m/s, time period T1 .An electric field of intensity 17. An electric field is acting vertically upwards., A small body of mass 1 gm and charge -1 m C, 3.6 104 N/C is applied vertically upwards.Now, is projected with a velocity 10 m/s at an angle, T2, 450 with horizontal. Its horizontal range is 2m, the time period is T2 the value of T is (g =, then the intensity of electric field is :(g = 10 m/, 1, s2 ), 10m/s2), 1)0.16, 2) 0.64, 3)1.25, 4)0.8, 1) 20,000 N/C, 2) 10,000 N/C, 12. A particle of mass m and charge q is placed, 3) 40,000 N/C, 4) 90,000 N/C, at rest in a uniform electric field E and then 18. A thin copper ring of radius ‘a’ is charged with, released. The kinetic energy attained by the, q units of electricity. An electron is placed at, particle after moving a distance y is, the centre of the copper ring. If the electron, 2, 2, 2, is displaced a little, it will have frequency., 1) qEy, 2) qE y, 3) qEy, 4) q Ey, 13. Four equipotential curves in an electric field, 1, eq, 1, q, 1) 2 4 ma3, 2) 2 4 ema3, are shown in the figure. A,B,C are three points, 0, 0, in the field.If electric intensity at A,B,C are, eq, q, E A , EB , EC then, 3) 4 ma, 4) 4 ema3, C, , B, , A, , 0, , 0, , 19. A thin fixed ring of radius 1 metre has a, positive charge 1 105 C uniformly distributed, over it. A particle of mass 0.9gm and having a, 120V, 90V, 60V, 60V, negative charge of 1 10 6 C is placed on the, 1) E A EB EC, 2) E A EB EC, axis at a distance of 1 cm from the centre of, 4) E A EB EC, 3) E A EB EC, the ring. Assuming that the oscillations have, 14. A particle of mass 1Kg and carrying 0.01C is, small amplitude, the time period of oscillations, at rest on an inclined plane of angle 300 with, is, 1) 0.23s 2) 0.39s, 3) 0.49 s, 4) 0.63s, 490, NC 1 20. A sphere carrying charge 0.01 C is kept at, horizontal when an electric field of, 3, rest without falling down, touching a wall by, applied parllel to horizontal .The coefficient of, applying an electric field 100 N/C.If the, friction is, coeffcient of friction between the sphere and, 1, 3, 3, the wall is 0.2 , the weight of the sphere is, 1) 0.5, 2), 3), 4), 3, 2, 7, 1) 4N, 2) 2 N 3) 20 N, 4) 0.2 N, NARAYANAGROUP, , 115
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 21. A particle of mass 1kg and carrying positive, ELECTRIC POTENTIAL, charge 0.01 C is sliding down an inclined plane 25. Two thin rings each having a radius R are, placed at distance d apart with their axes, of angle 300 with the horizontal. An electric, coinciding.The charges on the two rings are, field E is applied to stop the particle. If the, +q, -q. The potential difference between the, coefficient of friction between the particle and, rings, 1, , , Q.R, Q 1, 1, Q 1, 1, the surface of the plane is, , E must be, 1) 4 .d 2) 2 R R d 3) 4 R R d , 2 3, , , , , E cos , 4)0, 26. Two metal sphres A and B have their capacities, in th ratio 3:4. They are put in contact with, E, each other and an amount of charge, E sin , 7 106 C is given to the combination. Next, the, 1)1260 V/m, 2)245 V/m, two spheres are separated and kept wide the, mg sin , apart so that one has no electrical infuence on, 490, the other. The potential due to the smaller, mg, 3) 140 3 V/m 4), V/m, sphere at a distance of 50m from it is, = 30, 3, 1) 540V 2) 270V, 3) 1180V 4) zero, 22. Two identical point charges are placed at a 27. A solid conducting sphere having a charge Q, separation of l . P is a point on the line joining, is surrounded by an uncharged concentric, the charges, at a distance x from any one, conducting spherical shell. The potential, charge. The field at P is E. E is plotted against, difference between the surface of solid sphere, x for values of x from close to zero to slightly, and the shell is V. The shell is now given a, less than l . Which of the following best, charge –3Q. The new potential difference, between the same surfaces will be, 1) –, represents the resulting curve?, 2V, 2) 4V, 3) V, 4) 2V, 28. A spherical charged conductor has surface, 1), 2), E, E, charge density s . The intensity of electric, field, and potential on its surface are E and V., O, O, X, l, X, l, Now radius of sphere is halved keeping the, charge density as constant. The new electric, field on the surface and potential at the centre, 3) E, 4), of the sphere are, E, 1) 4E, V 2) E, V/2 3) E, V 4) 2E, 4V, X, O, O X, l, l, 29. Two spherical conductors A and B of radii 1, mm and 2mm are seperated by a distance of 5, 23. A particle of charge q and mass m moves in, cm and are uniformly charged. If the spheres, a circular orbit of radius r about a fixed charge, are connected by a conducting wire then in the, Q . The relation between the radius of the, equilibrium condition the ratio of electric fields, at surfaces of A and B is, orbit r and the time period T is, 1) 4: 1 2) 1 : 2, 3) 2 : 1, 4) 1 : 4, Qq, Qq, 3, 3, 2, 30. A charge +q is fixed at each of the points, 1) r 16 2 m T, 2) r 16 3 m T, 0, 0, x=x0,x=3x0, x=5x0 . . . . . on the x - axis and, Qq, Qq, a, charge-q is fixed at each of the points, 2, 3, 2, 3, 3) r 16 3 m T 4) r 16 m T, 0, 0, x 2 x0 , x 4 x0 , x 6 x0 ....... . Here x0 is a, 24. A thin semicircular ring of radius ‘r’ has a, positive constant. Take the electric potential, positive charge distributed uniformly over it., at a point due to a charge Q at a distance r, The net field E at the centre ‘O’ is (AIEEE, Q, i, from it to be 4 r . Then the potential at, 2010), 0, the origin due to the above system of charges, q, q, is, 1) 2 2 r 2 j 2) 4 2 r 2 j, i, 0, 0, q, O, q log e 2 , q, q, 1), 0, 2), 3), 4), 8 0 x0 log e 2 , 4 0 x0, 3) 4 2 r 2 j 4) 2 2 r 2 j, 0, 0, 2, , 0, , 0, , 2, , 2, , 0, , 2, , 0, , 116, , NARAYANAGROUP, , 2
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 31. A non – conducting ring of radius 0.5 m carries 37. There are three uncharged identical metallic, a total charge of 1.11 x 10 –10 C distributed non, spheres 1,2 and 3 each of radius r and are, – uniformly in its circumference producing an, placed at the vertices of an equilateral triangle, electric field E everywhere in space. The value, of side d. A charged metallic sphere having, l 0, of the integral Edl ( l = 0 being centre, charge q of same radius r is touched to sphere, l , 1, after some time it is taken to the location of, of the ring) in volt is :, sphere 2 and is touched to it, then it is taken, 1) +2, 2) –1, 3) –2, 4) zero, 32. Some equipotential surfaces are shown in, far away from spheres 1,2 and 3. After that, figure. The electric field strength is, the sphere 3 is grounded, the charge on sphere, Y, 3 is (neglect electrostatic induction by, 10V 20V 30V, assuming d>> 2r), 1) 100 V/m along x-axis, , 300, 0, , 0, , 5cm, , 10cm, , 15cm, , 1) Zero, , x cm, , 2) 100 V/m along y-axis, , 2), , 3qr, 4d, , 3), , qr, 2d, , 4), , qr, 4d, , 38., , 3) 400 V/m at an angle 1200 with x-axis, 400, 4), V/m an angle 1200 with x-axis, 3, 33. A field of 100Vm–1 is directed at 300 to positve, x - axis. Find VBA if OA = 2m and OB = 4m, , P (a , b, 0), , (0 , 0, 0), Q (2a , 0, 0), R (a , b, 0), , B, , O, , 3 2 V, 3) 100 2 3 V, , 300, A, , , 4) 200 2 3 V, 2) 100 2 3 V, , 1) 100, , E, , 34. Here is a special parallelogram with adjacent, side lengths 2a and a and the one of the, possible angles between them as 60°. Two, charges are to be kept across a diagonal only., , A point charge q moves from point P to point S, along the path PQRS in a unifrom electric field, , E pointing parallel to the positive direction, of the x-axis. The coordinates of the points P,, Q, R and S are (a,b,0), (2a,0,0), (a, –b, 0) and, (0,0,0) respectively. The work done by the field, in the above process is given by the expression, 1) qaE, 2) –qaE, , 3) q( a 2 b 2 ) E, 4) 3qE a 2 b 2, The ratio of the minimum potential energy of, the system to the maximum potential energy 39. The potential at a point x (measured in m m), due to some charges situated on the x-axis is, is, 1) 3 : 7 2) 3 : 7 3) 1 : 2, 4) 1 : 4, 20, volt. The electric field, given by V x 2, 35. Two concentric spherical conducting shells of, x 4, radii R and 2R carry charges Q and 2Q, E at x = 4 m m is given by, respectively. Change in electric potential on the, outer shell when both are connected by a, 5 V, , , and in the positive x - direction, 1), 1, 3 mm, , conducting wire is : k , , , 3kQ, , 4pe0 , kQ, 3) R, , 2kQ, , 1) zero, 2) 2R, 4) R, 36. The longer side of a rectangle is twice the length, of its shorter side. A charge q is kept at one, vertex. The maximum electric potential due to, that charge at any other vertex is V, then the, minimum electric potential at any other vertex, will be, 1) 2V, 2) 3 V, 3) V / 5 4) 5 V, NARAYANAGROUP, , 2), , 10 V, and in the negative x - direction, 9 mm, , 3), , 10 V, and in the positive x-direction, 9 mm, , 4), , 5 V, and in the negative x-direction, 3 mm, 117
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 40. Two points char ges q1 and q2 (=q1/2) are placed 44. A dielectric slab of thickness d is inserted in a, at points A(0, 1) and B (1, 0) as shown in the, parallel plate capacitor whose negative plate, figure. The electric field vector at point P(1, 1), is at x=0 and positive plate is at x=3d. The, makes an angle q with the x–axis, then the, slab is equidistant from the plates. The, angle q is, capacitor is given some charge. As one goes, y, from 0 to 3d,, 1) the magnitude of the electric field remains the, same, A, q, P(1,1), 1 1 , 1 1 , 2) the direction of the electric field remains the same, 1) tan 2) tan , 2, 4, 3) the electric potential increases continuously', 4) the electric potential increases at first, then, 1, B, x, 3) tan 1 1 4) tan 0 , decreases and again increases., O, q, 45., A solid sphere of radius R is charged, 41. Figure shows three spherical and equipotential, uniformly. The electrostatic potential V is, surfaces 1,2 and 3 round a point charge q. The, plotted as a function of distance r from the, potential difference V1–V2 = V2 – V3. If t1 and, centre of the sphere. Which of the following, t2 be the distance between them. Then, best represents the resulting curve ?, 1, , 2, , 3, 2, 1, q, , 1) t1=t2, , t1, , 2) t1>t2, , 1) V, , R, , t2, , 4) t1 t2, , 3) t1<t2, , o, , 42. A half ring of radius ‘ r ’ has a linear charge, density .The potential at the centre of the, half ring is, , 1) 4, 0, , , 2) 4 2 r, 0, , 2) V, , , 3) 4 r, 0, , , 4) 4 r 2, 0, , R, o, , 3) V, , r, , 4) V, o, , R, r, , R, o, , r, , 43. The distance between plates of a parallel plate, POTENTIAL ENERGY, capacitor is 5d. The positively charged plate, is at x=0 and negativily charged plates is at 46. Along the X-axis, three charges q ,-q and q, 2, 2, x=5d. Two slabs one of conducotor and the, are placed at x = 0, x =a and x =2a, other of a dielectric of same thickness d are, respectively . The resultant electric potential, inserted between the plates as shown in figre., Potential (V) versus distance x graph will be, at x =a+r(if a ,<<r) is ( 0 is the permittivity of, +q, q, free space, DIELECTRIC, , CONDUCTOR, , O, , qa, 1) 4 r 2, 0, x, , d, , 2d, , 3d, , 4d, , V, , V, , 1), , 2), O d 2d 3d 4d 5d, , X, , O d 2d 3d 4d 5d, , X, , V, , V, , 4), , 3), O d 2d 3d 4d 5d, 118, , 5d, , X, , O d 2d 3d 4d 5d, , X, , qa 2, 2), 4 0 r 3, , q, q(a 2 / 4), 3), 4) 4 r, 3, 4 0 r, 0, 47. An electron travelling from infinity with, velocity ‘v’ into an electric field due to two, stationary electrons separated by a distance, of 2m. If it comes to rest when it reaches the, mid point of the line joining the stationary, electrons.The initial velocity ‘ v ‘ of the, electron is, 1) 16m/s, 2) 32m/s, 3) 16 2m / s, 4) 32 2m / s, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 48. Work performed when a point charge, 2 108 C is transformed from infinity to a point, at a distance of 1cm from the surface of the, ball with a radius of 1cm and a surface charge, density =108 C / cm 2, 1) 1.1 104 J, 2) 11 104 J, 3) 0.11 104 J, 4) 113 104 J, 49. A conducting sphere A of radius a, with a, charge Q, is placed concentrically inside a, conducting shell B of radius b. B is earthed. C, is the common centre of A and B., B, Q, , A, a, C, , b, , 1) The field at a distance r from C, where a r b ,, is k, , Q, r2, , 2) The potential at a distance r from C, where, a r b , is k, , Q, r2, , 3) The potential difference between A and B is, 4) The potential at a distance r from C, where, a r b , is, 50. Given figure shows an arrangement of six fixed, charged particle. The net electrostatic force F, acting on charge +q at the origin due to other, charges is, q, , 2q, , q, (0,5), , q2, 6q 2, 18q 2, 3q 2, 1), 2), 3), 4), 4 o, 4 o 91, 4 o 91, 4 o, 52. An electron travelling from infinity with, velocity 'V' into an electric field due to two, stationary electrons seperated by a distance, of 2m. If it comes to rest when it reaches the, midpoint of the line joining the stationary, electrons, the initial velocity 'V' of the, electron is (in m/sec), 1) 16, 2) 32, 3) 16 2, 4) 32 2, 53. A particle of mass m and charge q is projected, , vertically upwards. A uniform electric field E, is acted vertically downwards. The most, appropiate graph between potential energy, U (gravitational plus electrostatic) and height, h(<< radius of earth) is : (assume U to be zero, on surface of earth), U, , U, , 1), , 2), h, U, , +q, a, , B, q, , C, +q, , 2a, , 2) zero, , 7q 2, q2 3, , 3, 3), 4), 2, 2 , 2 0 a, 4 0 a 2, , 51. 2q and 3q are two charges separated by a, distance 12 cm on x-axis. A third charge q is, placed at 5 cm on y-axis as shown in figure., Find the change in potential energy of the, system if 3q is moved from initial position to a, point on X-axis in circular path:, NARAYANAGROUP, , h, , h, , a, , 6q 2, 1), 4 0 a 2, , 4), , 0, , 30 300 a, O +q, , 2a, , h, , U, , 3), , A, q, , 3q, (12,0), , (5,0), , (0,), , 54. Three charged particles are initially in, position 1. They are free to move and they, come in position 2 after some time. Let U1, and U2 be the electrostatic potential energies, in position 1 and 2. Then :, 1) U1>U2 2) U2>U1 3) U1=U2 4) U 2 U1, 55. Two identical thin rings, each of radius R,, are coaxially placed a distance r apart. If Q1, and Q2 are respectively the charges uniformly, spread on the two rings, the work done in, moving a charge q from the centre of one ring, to that of the orther is, , , , 1) zero 2) q Q1 Q2 , , , , 2 1 /, , 24pe0 R, , , , 3) q 2 Q1 Q2 / 4pe0 R , , , , 4) q Q1 / Q2 , , , , 2 1, , 24pe0 R, , , 119
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 56. The electric potential at a point (x, 0, 0) is given 61. Two opposite and equal charges, 4 x10–8 coulomb when placed 2 x 10–2 cm away,, 1000 1500 500 , 2 3 then the electric, from a dipole. If this dipole is placed in an, by V = , x, x , x, external electric field, 2, field at x = 1 m is (in volt/m), x 10–2 newton/coulomb, the value of maximum, torque and the work done in rotating it through, 1) 5500iˆ 2) 5500iˆ 3) 5500iˆ 4) zero, 1800 will be, 57. Six charges are placed at the vertices of a, regular hexagon as shown in the figure. The, 1) 32x104 Nm and 32x104 J, electric field on the line passing through point, 2) 64x104 Nm and 64x104 J, O and perpendicular to the plane of the figure, at a dist ance of x (>> a) from O is, 3) 64x104 Nm and 32x104 J, Q, , a, , Q, , Q, , 4) 32x104 J and 64 x104 Nm, 62. An electric dipole is made up of two particles, having charges 1mc , mass 1 kg and other, , Q, , with charge 1mc and mass 1 kg separated, by distance 1m. It is in equilibrium in a uniform, electric field of 20 x 103 V/m. If the dipole is, 2Qa, Qa, 3Qa, deflected through angle 20, time taken by it to, 1) x 3 2) x 3 3), 4) zero, 0 x 3, 0, 0, come again in equilibrium is, DIPOLE, 1) 2.5 p s 2) 2.5 s, 3) 5 p s, 4) 4 p, 58. A small electric dipole is placed at origin with 63. A point particle of mass M is attached to one, its dipole moment directed along positive x end of a massless rigid non-conducting rod of, axis. The direction of electric field at point, length L. Another point particle of the same, mass is attached to the other end of the rod., 2, 2 2, 0 is, The two particle carry charges + q and – q, 1) along z - axis, 2) along y - axis, respectively. This arrangement is held in a, 3) along negative y -axis 4) along negative z-axis, region of a uniform electric field E such that, the rod makes a small angle q (say of about, 59. Two electric dipoles each of dipolemoment, 0, 5 ) with the field direction (see figure). The, p 6.2 10 30 C m are placed with their axis, expression for the minimum time needed for, along the same line and their centres at a, the rod to become parallel to the field after it, distanced=108 cm . The force of attraction, is set free., between dipoles is, Q, , , , Q, , , , 1) 2.1 1016 N, 3) 2.1 1010 N, , 2) 2.1 1012 N, 4) 2.1 108 N, , A, , q, , , , E, , O, , 60. Two charges 3.2 10, , 19, , C and 3.2 10, , 19, , q, , C, , B, , placed 2.4 A0 apart form an electric dipole. It, is placed in a uniform electric field of intensity, , 4 105 V / m the work done to rotate the, electric dipole from the equilibrium position by, 1800 is, 1) 3 1023 J, 3) 12 1023 J, 120, , 2) 6 1023 J, 4) Zero, , 1) t , , p mL, 2 2qE, , 2) t , , p 2mL, qE, , 4) t 2, , 3) t 2, , p mL, 2 qE, p 3mL, 2qE, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , LEVEL - III - KEY, 1) 4, 7) 4, 13) 3, 19) 4, 25) 2, 31) 1, 37) 2, 43) 2, 49) 1, 55) 2, 61) 4, , 2) 3, 8) 2, 14) 4, 20) 4, 26) 1, 32) 3, 38) 2, 44) 2, 50) 2, 56) 2, 62) 1, , 3) 2, 9) 4, 15) 2, 21) 3, 27) 3, 33) 1, 39) 3, 45) 3, 51) 3, 57) 1, 63) 1, , 4) 2, 10) 4, 16) 1, 22) 4, 28) 2, 34) 1, 40) 1, 46) 2, 52) 2, 58) 2, , 5) 4, 11) 3, 17) 3, 23) 2, 29) 3, 35) 1, 41) 3, 47) 2, 53) 1, 59) 4, , 16. Maximum height , 6) 4, 12) 3, 18) 1, 24) 4, 30) 4, 36) 3, 42) 1, 48) 2, 54) 1, 60) 2, , LEVEL - III - HINTS, 1., , T, , 1 q1q2, mg, 4 o r 2, , 1, q1q2, 2 ,3&5. F 4 , d2, 0 r, 6. F mgTan, 1, qq, 7. F 4 1 22, d, 0 r, 8. There are three forces acting on each sphere are, (i) tension (ii) weight(w) (iii) electrostatic force of, repulsion for sphere, In equailibirum, from figure, tan 1 F1 / M 1 g, From sphere 2, in equilibirum from figure, tan 2 F2 / M 2 g, , for F1 F2, F1, F2, or 1 2 only for M g M g, 1, 2, , But, F1 F2 and then M 1 M 2, 9., , Deflection y , , 10. y , , eEx2, where K is kinetic energy.., 4 K , , eE 2, K K .E , 4k, , 17. Range , , 18., , E, , d2x, 1 qex, , 2, dt, 4 0 a3, So motion is S.H.M., qe, 1, 2 , 4 0 ma 3, m, , Qq x, m, 19. F 4 R 3 kx and T 2, k, 0, 20. mg qE, 21. mg sin cos qE sin qE cos , 1 Qq, 2, 2, 23. F 4 r 2 mr ; , T, 0, 1 q sin / 2, 24. E 4 r 2 / 2, 0, q sin / 2, E, j, 2 2 0 r 2, , , , Q 1, , 1, 1, , 2, R d2, R, V V1 V2, , V2 , , Q, 4 0, , , , 29. V K ., , r, , and Ebicector , , NARAYANAGROUP, , , , , , , , r1 , 1 q1, 26. q1 r r q ; V1 4 r, 0, 1 2, 27. Pd between the two spheres is independent of, charge on outer shell., , 12. K.E =FS K.E = qEy, dV, 13. E , dx, 14. N =mg sin + qE sin , mg sin = N qE cos, 3, , 1, , 25. V1 4 R 2 2 , R d , 0 , , 28. E , , 2 kp, , u 2 sin 2, EQ, g, m, , 1, qx, qx, , 3/ 2, 2, 2, 4 0 a x , 4 0 a3, , l, 11. T = 2 g, eff, , 15. Eaxis , , u 2 sin 2 , EQ , , 2 g , , m , , , 0, , R, , and V , 0, , Q V, Q, Q, K., K., R, 2, 2R, R, 1 Q , Q, K. K., , d 2R, 2 R , d, , kp, 2r 3, , When the two conducting spheres are connected, by a conducting wire, charge will flow from one, sphere (having higher potential) to other (having, lower potential) till both acquire the same potential., , 121
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, There fore, E , , , V, r, , , , E1 r2 2, 2 :1, E2 r1 1, 1, , q, , 1, , 1, , 1, , , , 30. V 4 x 2 x 3 x 4 x , 0 0, 0, 0, 0, , q, 1 1 1, , = 4 x 1 2 3 4 , , 0 0 , q, , log 2 , 4 0 x0, , , 31. V E .dr, 32. E , , dV, dx, , , , 33. V E.r, 34. long and short diagonal lengt hs are, p 2 q 2 2 pq cos , 1, , 47., , 2 pq, 4 0, , 1, 1 , , , , 3, 3, d l d l , , 1 2 1 q1q2, mv 4, 0, 2, r, , 48 . Potential at a distance 2cm from its centre, r 2, , 1, Q, 4 r 2, , = 4 r 4 r ' r ' 2 100, 0, 0, 0, 0, since r=1 cm and r ' =2 cm, , PD b/w the two points is equal to 200 , 0, , work done =VQ= 200 X 2108 =11104 J, 0, 49. field concept, 50. concept of force, 1, , 51. U U f Ui and U 4, , 0, , q1q2, r, , 1, e2, u, , 2, , ., 52. ui 0 , f, 4 0 d / 2 , , 3Q, , 35. V 4 2 R, 0, 1, 36. If the charges is kept at ‘A’ then maximum and, PE KE mv 2 calculate ‘ v ‘, 2, minimum potentials at D and C respectively, 53. conceptual., 37. commen potential, , 54. Particles moves in a direction where potential energy, 38. w = Fs ; W =q E.S, of the system decreased., dv, Q1, Q2, 39. E , V1 , , 55., dx, 4 0 R 4 0 2 R, 1, , 1, , Q, , 41. V1 V2 kq r r , 1, , r2 r1 , , V1 V2 r1r2, kq, , ; but, , r2 r1 t, , W12 q V2 V1 , , dV, dx, 57. concept of field, 58. Use vector representation, 59. Force of interaction, , 56. E , , t r1r2, , if P.D is constant then r2 r1 t, 42. potential due to small element ‘ p ’ at the centre, , 1 , dl , dl, r, 4 0 r , dl, dv K ., ;, 1 , , , r , r, 4 0 r, 4 0, v dv k, , 1, 1 , , , , 3, 3, d l d l , W PE2 PE1 2 PE 2 2qdE, , , 2 pq, 4 0, , dv, E inside the conductor is zero., dx, 44. The direction of E is constant., , 60., 61. pESin, , 1 Q, 45. V 4 r, 0, , 62. T 2, , I, pE, , 46. Force of interaction, , 63. T 2, , I, pE, , 43. E , , 122, , Q, , 1, 2, and V2 4 2 R 4 R, 0, 0, , 2, , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , ELECTRIC CAPACITANCE, Electric Capacity:, , , Imagine a Gaussian surface enclosing the plate as, shown., , The ratio of charge to potential of a conductor is, , x, , Q, , called its capacity. C V, Unit : farad (F), , +, x1, , Parallel Plate Capacitor: If two plates each, A, , , , +, , +, , +, , +, , +, y1, , q, , q, , or, , q, , where E is a field in the medium, , q, .... (b), A 0K, The P.D. between the two plates of the capacitor., , -, , -, , -, , -, , V E 0 d t E.t, , 0 A, 0 A, new capacity , , t, 1, d t 1 d t , k, k, , V=, , GAUSS METHOD, Let us consider a case of parallel plate capacitor, in which a medium of dielectric constant K is, partially filled as shown in figure., Then the field is uniform in air as well as in, medium but they will have different values. let, 't' be the thickness of the medium whose relative, permittivity is K. The remaining space of (d - t), thickness be occupied by air., q, , x, , 1, , +, , E , , -, , x, , +, , E.ds e0K EA e0 K, , t, , d, , +, , q, .... (a), A 0, Similarly by considering a Gaussian surface, through the medium, then by Gauss law,, , +, , d, , , , +, , E0 , , air, , -, , +, , q, , When a dielectric medium is introduced between, the plates of a parallel plate capacitor, its, capacity increases to 'k' times the original, capacity., When a dielectric slab of thickness 't' is, introduced between the plates of a parallel plate, capacitor,, +, , +, , E0 ds e0 E0 A e0, , (dielectric medium), , +, , +, , If E0 is the field in air, then from Gauss law, , capacity C 0d (air as medium),, k0 A, d, , +, , Fig. (c), , of area A are seperated by a distance 'd' then its, , C, , y, , , , , air, , , , y, , , , , , , , 1, , y, K, , , , q, q, t, d t , A 0, A 0 K, , q , t, d t , , A0 , K, , q, q, , q, V, d t t / K , A 0, Ae0, C, , t , d t , K , , , , or C , , , , When a metal slab of thickness 't' is introduced, between the plates of a parallel plate capacitor,, , o A, ., d t, ( for metal k = ), , new capacity , , Fig. (b), 142, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , , ELECTRO STATICS & CAPACITORS, , The method for the calculation of capacitance, requires integration of the electric field between, two conductors or the plates which are separated, with a potential difference Vab, , C, , , , a, , i.e. Vab E.d r, b, , or V V E.d r from this C , , , , , , , , q, Vab, , C, K1 K2 ., 2, , t, t , d t1 t2 .....tn 1 ...... n , kn , k1, , In the above case if the dielectric media are, completely filled between the plates, effective, capacity C , , When a thin metal sheet t 0 is introduced , between the plates of a parallel plate capacitor,, then capacity remains unchanged., A dielectric slab of thickness 't' is introduced, between the plates, to restore the original , capacity, if the distance between the plates is, , 1, increased by x, then x t 1 k ., , , Two dielectric slabs of equal thickness are , introduced between the plates of a capacitor as, shown in figure, then new capacity, , , , 0 A, , 0 A, t1, tn , ..... , kn , k1, , The capacity of a parallel plate capacitor is, independent of the charge on it, potential, difference between the plates and the nature of, plate material., In a capacitor, the energy is stored in the electric, field between the two plates., Capacity of a spherical conductor = 4 0 r ,, where r is the radius of the sphere., If we imagine earth to be a uniform solid sphere, then capacity of earth is 4 0 R, Where R = Radius of the earth = 6400 103 m, , Note : For the earth, R 6.4 106 m, The capacity of earth is, , 1, 6.4106 711 F, 9, 910, W.E-1: A metal slab of thickness, equal to half, A/2, A/2, the distance between the plates is introduced, If the two dielectrics are of different face areas, between the plates of a parallel plate capacitor as shown. Find its capacity., A1 and A2 but of same thickness, then capacity,,, K1, , , , d, , C 4 0 R , , q, , , C 0 K1 A1 K 2 A2 , d, If two dielectric slabs of constants k1 and k2, are introduced as shown in figure, new capacity, , , , , K2, , 2k1k2, .C, k, 1 k2 , , , , , , d, , , air, , , , , , , , , d/2, , Sol: When capacitor is partially filled with dielec0 A, , K1, , d, 2, , tric capacity, , K2, , d, 2, , For metal slab of thickness t = d/2,, , If number of dielectric slabs of same cross, sectional area ‘A’ and of t hicknesses, t1 , t2 , t3 ,........t n and const ants k1 , k2 .......kn are, introduced between the plates, effective capacity, , NARAYANAGROUP, , C, , , C, , d t (1 1 ) , k , , , 0 A, ( K for metal slab), dt, 0 A, A, 2 0 ., d, d, d, 2, 143
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , W.E-2: Two conductors carrying equal and, , W.E-4: Capacitor has square plates each of side, , opposite charges produce a non uniform, electric field along X - axis given by, , ‘l’ making an angle ' ' with each other as, shown. Then for small value of , the capacitance ‘C’ is given by, , E, , Q, (1 Bx 2 ) where, 0 A, , A and, , B, , are, , l, , , constants. Separation between the, conductors along X-axis is X. Find the, capacitance of the capacitor formed., Sol: Potential difference between the conductors is, given by V V V Edx, 0, X, , Q, , 0 A (1 Bx, , 2, , )dx, , 0, , Capacity C , , Q, , V, , d d l , l, d, 2, 2, 0 A 0 l 2, Capacity C , , l, d, d, 2, , , Q , Bx 3 , V, x, , 0 A , 3 , , or, , l, , Sol: At one side, distance between plates d,, At another side,, distance d l sin d l, Mean distance between the plates, , X, , V, , d, , X, , , O, , Q , BX3 , X, , 0 A , 3 , , 0 A, BX 2 , X 1 , 3 , , , , , W.E-3: Find the capacitance of a system of two, identical metal balls of radius a if the distance between their centres is equal to b, with, b>>a. The system is located in a uniform dielectric with permittivity K., Sol: Let q and -q be the charges on two balls. Then, , The potential difference between the balls, q, , l , 1 2d , , , , 1, , , , 0 l 2, d, , l , 1 2d , , , , Spherical condenser, 1 q q , V = Vp Vq 4 a b 0, , 0 , , , V1 Vball V V, V2 Vball V V, , 0 l 2, d, , 1, ba, q, 4 0 ab , , C, , q, V, , -q, b, , a, , a, , -Q, , r, , a, P, , b, , +Q, , q, , V1 V2 2V, b a, , 2, , , , (a) C 4 0, , E dr, , a, b a, , 2, , charged and outer sphere is earthed., , , 1, , q, , 2q, , 1, , 1 , , 4 0 K r2 dr 4 0 K a b a , a, , C, , , , q, q, , V1 V2 2q b 2a , , , 4 0 K a b a , , 2 0 K a(b a), (b 2a), , For b >> a, we can write C 2p 0 K a ., 144, , ab, , if inner sphere is, ba, , a, b, , b2, (b) C 4 0, ,, ba, If inner sphere is earthed and outer sphere is, charged., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , Cylindrical Capacitor: A cylindrical capacitor, consists of two coaxial cylinders and its capacitance, is given by, O, , Note : For a capacitor having constant potential, difference across the plates the force, 20 A 2 V 2, C2 V 2, , F, 2 0 A, d 2 2 0 A, 1, V2, 0 2 A, 2, d, In this case force depends on the separation, between the plates. Thus to change the separation, variable force is needed., F, , c, , l, , 2 0l, b, log e , a, , Where l is the length of each of cylinder and a, and b are the radii of the inner and outer, cylinders., , Force between the plates of a capacitor, Consider a parallel plate capacitor with plate, area A. Let Q and –Q be the charges on the plates, of capacitor. Let F be the force of attraction between the plates. Let E be the field between the, capacitor plates.The expression for the force can, be derived by energy method. Let the distance, between the plates be x., So electric field energy between the plates is, 1, U 0 E2 (Ax), 2, dU 1, 0 E2 A, dx 2, , CAPACITORS IN SERIES, In series combination, the capacitors are first, arranged in a series order such that the, second plate of first capacitor is connected, to the first plate of second capacitor, the, second plate of second capacitor is connected, to first plate of third capacitor and so on. And, finally the first plate of first capacitor and, second plate of last capacitor are connected, to opposite terminals of battery., Let us consider three capacitors of capacities, C1, C2 and C3 connected in series across a, source of potential difference 'V' as shown in, figure., V1, , V2, -q, , +q, C1, , X, dx, , dU 1, 0 E2 A, dx 2, (Conservative force), So the force of attraction between the plates is, 1, 2, F = 0 E A, 2, Note : For an isolated charged capacit or, , By definition F , , Q2, . This force does not depend on the, 2 0 A, separation between the plates, and so the constant, amount of force is needed to change the, separation., F, , NARAYANAGROUP, , C2, , , F, , V3, -q +q, , +q, , -q, C3, , , V, , At the moment, the system is connected to the, source, left plate of first condenser acquires, positive charge due to conduction. This inturn, will produce negative charge of equal, magnitude, on the left face of second plate of, first condenser due to induction. The process, continues for the remaining two condensers., Hence the charge acquired by all the three, capacitors will be same., As the capacitors are different, the potentials, developed across them will be different., q C1V1 C2 V2 C3 V3, q, q, q, V1 , V2 , , V3 , C1, C2, C3, But V V1 V2 V3, 145
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 1, 1, 1 , V q, , , C1, C2, C 3 , , , W.E-6: When the space between the plates of a, parallel plate condenser is completely filled, with two slabs of dielectric constants K1 and, K2 and each slab having area A and thickness, , .... (1), , If a single capacitor when connected across the, same source draws the same charge, that, d, as shown in the figure, equal to, capacitance is said to be t he equivalent, 2, capacitance of the three capacitors. If CS is the, equivalent capacitance., q, K, CS , V, K, q, V, -----(2), CS, Fig. The equivalent circuit is as shown, Substituting (2) in (1), q, q, q, q, A, , , , CS C1 C2 C3, d, K, 2, 1, 1, 1, 1, , , , A, Cs C1 C2 C3, d, 1, , 2, , 1, , , , , , , , 1, 1, In general C C, S, n, The resultant capacity of series combination is, smaller than the least capacity of the capacitors a), of the combination., In series, ratio of charges on three capacitors is b), 1 : 1 :1., The ratio of potential differences across three c), capacitors is, d), Q Q Q, 1 1 1, V1 : V2 : V3 :, :, :, :, C1 C2 C3 C1 C2 C3, P.D across first capacitor is, 1, C1, V1 , V, 1, 1, 1 , C C C , 1, 2, 3, , e), , Capacity of the upper half C1 , , 8F, A, , B, , 2K 2 o A, d, C1 and C2 may be supposed to be conencted in, series., Effective capacity, , Capacity of the lower half C2 , , C, , 12F, , Sol: Here 12 F and 12 F are short circuited. Hence, they are not charged., Take only 8 F and 8 F parallel combination. Fig., 146, , 2K1 K 2 , C1C2, A 2K1K 2 , 0 , C0 , , C1 C2, d K1 K 2 , K1 K 2 , , Here C0 is the capacity of the condenser with, air medium., , 2K1 K 2 , Effective dielectric constant K= , K1 K 2 , Capacitors are said to be connected in parallel, if the two plates of any capacitor are, connected one to positive terminal and the, other to negative terminal of the source, then, the connection is said to be parallel, connection., , 8F, , C 8 8 16 F, , 2K1 o A, d, , Capacitors in parallel, , similary we can find V2 and V3., W.E-5: The equivalent capacity between A and B, in the given circuit is, , 12F, , K2, , 2, , V, , , , , q1, C1, , , , , q2, C2, , , , , q3, C3, , Let us consider three capacitors of capacities, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , C1, C2 and C3 connected in parallel across a source W.E-7 : In the net work three identical capacitors, are connected as shown. Each of them can, ‘V’ as shown, withstand to a maximum 100 V potential, The moment capacitors are connected, charge is, difference. What is the maximum voltage that, drawn from the voltage source and this charge, can be applied across A and B so that no, is drawn along three branches and thus gets, capacitor gets spoiled., C, shared. As all capacitors are connected in, C, parallel, the potential across any of the, A, B, capacitors is same . Here charge gets shared, depending upon their capacitances for, C, maintaining same potential., Sol: Let q max be the max-charge supplied by the, battery between A and B so that no capacitor, q, q, q, V 1 2 3, gets spoiled., C1 C2 C3, For each capacitor, q1 q 2 q 3 C1V C2 V C3 V, q max CV0 C(100) 100 C, q V C1 C2 C3 , For the combination q max Cequivalent Vmax , , q, 2, C1 C2 C3, 100 C C Vmax Vmax 150 V, ... (1), 3, V, Among, 150V, potential difference across, If a single capacitor when connected to the same, parallel, combination, is 50V and the potential, source draws a charge q then that capacitor is, difference across the other capacitor is 100V., said to be the effective or equivalent capacitor, W.E-8: Calculate the capacitance of a parallel, for the three parallel capacitors., plate capacitor, with plate area A and distance, If the effective capactiance is Cp,, between the plates d, when filled with a dielectic, whose permittivity varies as, q, CP , ... (2), d, d, , , V, ( x) 0 kx 0 x ; ( x ) 0 k( d x ) x d , , , 2, , , 2, from (1) and (2), dx, , C P C1 C 2 C 3, In general CP Cn s, , , , , , , , , X, , The resultant capacity of parallel combination, is greater than the largest capacity of the Sol:, capacitors of the combination., In parallel, ratio of P.D. on three capacitors is 1, X=0, X=d, : 1 :1., The given capacitor is equivalent to two, capacitors in series. Let C1 and C2 be their, The ratio of charges on three capacitors is, capacities. Then, Q1 : Q2 : Q3 C1V : C2V : C3V C1 : C2 : C3, The charge on first capacitor is, 1, 1, l , , , , C, C1, dC1 dC2 , Q1 , Q, Consider an element of width dx at a distance x, C1 C2 C3, from the left plate. Then, similarly we can find Q2 and Q3., kx A, When n identical capacitors each of capacity C, dC1 0, for 0 x d, dx, 2, are first connected in series and next connected, k(d x) A for d x d, in parallel then the ratio of their effective, and dC2 0, 2, dx, capacities, on substituting these two values we get, Cs, C, n 2 :1, 2 Kd , 1, l, 2, KA 2 0 Kd , Cs ; Cp nC, , , n 0, n, Cp, n, C, , C, , NARAYANAGROUP, , dC, , KA, , , , 2 0, , , , 2, , , , 2 0, , , , 147
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , W.E-9: When the space between the plates of a, parallel plate condenser is completely filled, with two slabs of dielectric constants K1 and, K2 and each slab having area, , A, 2, , A, , The equivalent circuit as shown, A, 2, d, 2, , A, 2, , K1, , K2, , K3, , K1, , d, , A, 2, , K1, , d, 2, , Fig. The equivalent circuit is as shown, , d, , K1, d, , K2, , A, 2, , A/2, , and thickness, , equal to distance of seperation d as shown in, the figure., , K1, , A/2, , K2, , d, 2, , K3, A, , 1, 1, 1, , , C C3 C1 C2, , ,, , 1, 1, 1, , , KC 2K 3 C K1 K 2 C, , 1, 1, 1, , , K K1 K 2 2K 3, , W.E-11: Four identical metal plates are located, a), b), c), , 0 A, Capacity of the left half C1= K1, 2d, 0 A, Capacity of the right half C2 = K2, 2d, C1 and C2 may be supposed to be connected in, parallel then effective capacity, C = C1 + C2 , , in air at equal distance d from one another., The area of each plate is A. Find the, equivalent capacitance of the system between, X and Y., 1, , X, , 2, , 0 A K1 K 2 , d , 2 , , 3, Y, 4, , Sol: Let us give numbers to the four plates. Here X, K1 K 2 , C = C0 , where, C, is, capacity, of, , and Y are connected to the positive and negative, 0, 2 , terminals of the battery (say), then the charge, capacitor without dielectric., distribution will be as shown, K1 K 2, , 1, , d) Effective dielectric constant K , X, , , 2, 2, , , W.E-10: A parallel plate capacitor of area A, plate, , 3, , separation d and capacitance C is filled with, Y, , , three different dielectric materials having, 4, Here the arrangement can be represented as the, dielectric constants K1, K2 and K3 as shown, grouping of three identical capacitors each of, in fig. If a single dielectric material is to be, used to have the same effective capacitance, A, capacity 0 . The arrangement will be as, as the above combination then its dielectric, d, constant K is given by :, shown, Sol: Let, , A, C 0, d, , ;, , 0 A, 2 K C, C1 K1, 1, d, 2, , 0 A, 2 K C, C2 K 2, 2 ;, d, 2, 148, , C3 , , K 3 o A, , 2K 3 C, d, 2, , 2 1, 3 4, , X, (), , Y, (), , 2 3, , Now the equivalent capacitance between X and, Y is CXY , , (C C)C 2C 2 0 A, , , CCC, 3, 3d, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , W.E-12: Find equivalent capacity between X and, 1, , Y, , , , 2, X, , Types of Dielectrics :, , Y, , 3, 4, , Sol: Let us give numbers to the four plates. Here X, and Y are connected to the positive and negative, terminals of the battery (say) ,, 1, , , , , 2, X, , Y, , 3, , , , 4, , Here the arrangement can be represented as the , grouping of two identical capacitors each of, 0 A, d, , capacity, , . The arrangement will be as, , shown, 1, , 2, , , X , , Y, 4, , 2, , , Now the equivalent capacitance between X and , Y is, C XY (C C ) 2C 2, , e0 A, d, , W.E-13: Find equivalent capacity X and Y, 1, 2, , X, , 3, 4, , Y, , , , Sol: Let us give numbers to the four plates. Here X, and Y are connected to the positive and negative, terminals of the battery (say)., 1, 2, , , , , , X, , , 3, 4, , 0 A, d, , . The arrangement will be as, 2, , 1, , 3, , 2, , 3, , 4, , X, , Y, , , Now the equivalent capacitance between X and, C C , , C, , 3, , C C C, Y is CXY , 2, 2, C C , , 3, 2, , = CXY C , NARAYANAGROUP, , 3 0 A, 2 d, , 1, , q 1 q 1 , K, , , , , where K is dielectric constant., Electric field due to induced charges on the, dielectric is, , Y, , Here the arrangement can be represented as the, grouping of three identical capacitors each of, capacity, shown, , A dielectric is an insulating material in which, electrons are tightly bound to the nuclei of the, atoms., Ex: glass, mica, paper etc., There are two types of dielectrics, 1) Non-polar dielectrics, 2) Polar dielectrics, In non polar dielectrics the centre of positive, charge and centre of negative charge of each, molecule coincide, Under ordinary conditions Non-polar molecule, will have zero dipole moment., When a Non-polar dielectric is subjected to, electric field, the positive charge of each, molecule is shifted in the direction of electric, filed and negative charge in the opposite, direction., Ex: oxygen, nitrogen, In polar dielectrics the centre of positive, charges and centre of negative charges of each, molecule do not coincide., Each molecule has a permanent dipole moment., When polar dielectric is subjected to external, electric field, the electric field exerts torque, on the dipoles, tending to align them in the, direction of the field., Ex: CO2, NH3,HCl, etc., If a dielectric is charged by induction then, induced charge q1 is less than inducing charge, q. Induced charge,, , Eind or E p E0 , , E0, 1, E0 1 ., K, k, , Dielectric Strength of Air : A conducting, sphere cannot hold very large quantity of, charge. It can hold a maximum charge Q such, that the electric intensity on the surface is equal, to dielectric strength of air 3 106 Vm 1 , 1 Q, 6, 1, i.e. 4 R 2 3 10 Vm, 0, , Energy stored in a condenser : Energy, stored, U , , 1, CV, 2, , in, , a, , charged, , condenser, , 2, , 2, , , , 1, q, qV , 2, 2C, 149
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , , If a condenser is connected across a battery and, U is the energy stored in the condenser then the, work done by the battery in charging the, , condenser is 2U ( W = qV = 2U ), For a parallel plate capacitor, U , , 1, 2 , , ( Ad ), as E , , 2, 0 , 0 , , Q, , Energy density, U 2 1, , 0 E 2 ( here V is volume i.e. Ad), V 0 2, , a) When three capacitors are in series, the ratio, of energies is, , , , U1 : U2 : U3 , , , , Q 2 Q 2 Q2, 1 1 1, :, :, :, :, 2C1 2C2 2C3 C1 C2 C3, , b) When three capacitors are in parallel, the, ratio of energies is, 1, 1, 1, U1 : U2 : U3 C1V2 : C2V2 : C3V2 C1 : C2 : C3, 2, 2, 2, , , , c) Energy density ( ) = energy/ volume, , 1, 1, E 2 k 0 E2, 2, 2, (Where K in the dielectric constant of, medium between the plates), , Effect of Dielectric:, , , , , , , , , , , , 150, , 1, times the original energy.., k, A capacitor is fully charged to a potential 'v'., After disconnecting the battery, the distance, between the plates of capacitors is increased by, means of insulating handles. Potential difference, , between the plates increases. ( V C , Q remains, same, and C decreases), A capacitor with a dielectric is fully charged., Without disconnecting the battery if the dielectric, slab is removed, then some charge flows back, to the battery., , Mixed Grouping of Capacitors:, , , Number of capacitors in a row, n, , , , , desired potential, given potential, desired capacity, , Number of such rows m original capacity n, Total number of capacitors = m n ., , Coalesence of a Charged Oil Drops:, There are ‘n’ charged drops of radius ‘r’ and, charge ‘q’. The drops are merge to form a bigger, drop. If capicity of small drop is ‘C’ then, 1, , 1). capacity of bigger drop is C , , 1, n3, , C, , A parallel plate capacitor is fully charged to a 2) Potential of bigger drop is, potential V. Without disconnecting the battery, Q, nq, n2/3 q, V1 1 1/3 , n2/3V., if the gap between the plates is completely filled, C, n .C, C, by a dielectric medium,capacity increases to k, 3) Energy of bigger drop is, times the original capacity., P.D. between the plates remains same., Q2, n2 q2, n5/3 q2, Charge on the plates increases to k times the, U 1 1/3 , n5/3U., 2C, 2n .C, 2C, original charge., Energy stored in the capacitor increases to k 4) Surface charge density of bigger drop is, times the original energy., Q, nq, n1/3 q, 1, After disconnecting the battery if the gap, , , , n1/ 3 ., 2, 2/3 2, 2, 4R, 4 n .r, 4r, between the plates of the capacitor is filled by, S.No., Quantity, For, each, charged, For the big, a dielectric medium,capacity increases to k times, small drop, drop, the original capacity., 1. Radius, r, R=n1/3 r, 1, 2. Charge, q, Q=n×q, P.D. between the plates decreases to, times, k, 3. Capacity, C, C1 = n1/3 × C, the original potential., 4. Potential, V, V1 = n2/3 × V, Charge on the plates remains same., 5. Energy, U, U1 = n5/3 U, Energy stored in the capacitor decreases to, 6. Surface charge , n1/ 3 ., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , INTRODUCTION OF DIELECTRIC IN A, CHARGED CAPACITOR, A dielectric slab (K) is introduced between the, plates of the capacitor, , Charge transfered is q1 q11 or q2 q12 , CV, 1 1 CV, 1 (or) C2V2 C2V, , C1 V1 V or C2 V2 V , WE.14: A capacitor of capacitance Co is charged, 1, Capacity, K times increases, K times increases, to a potential Vo and then isolated. A small, 2, charge, K times increases, Remains constant, capacitor C is then charged from C o ,, 3, P.D, Remains constant, K times decreases, discharged and charged again, the process, 4, Electric Intensity Remains constant, K times decrease, being repeated n times. Due to this, potential, 5, Energy stored in K times increases, K times decrease, of the large capacitor is decreased to V. Find, condenser, the capacitance of the small capacitor:, The distance between the plates of condenser is Sol: When key is closed, common pot ential, increased by n times., C V, V1 o o charge left on large capacitor after, Sno Physical With battery, with battery, Sno, , Physical, quantity, , quantity, 1, , With battery, , with battery, , permanently connected disconnected, , permanently connected disconnected, , Capacity, , n times decreases, , n times decreases, , 2, , charge, , n times decreases, , Remains constant, , 3, , P.D, , Remains constant, , n times increases, , 4, 5, , Electric Intensity n times decrease, Energy stored in n times decreases, condenser, , Remainsconstant, n times increases, , Co C, , first sharing of charges Q1o Co V1, common potential after second sharing of charges, C, , 0, in V2 C C V1, 0, , ;, , V2 =, , C2o V0, ( Co + C )2, , n, REDISTRIBUTION OF CHARGE, COMMON, th sharing charges V = Co V, after, n, POTENTIAL AND LOSS OF ENERGY, n, C + C 0, o, Two capacitors of capacities C1 and C2 are, n, charged to potentials V1 and V2 separately and, V 1/ n , C , they are connect so that charge flows. Here But Vn V ; V 0 V0 ; C Co o 1, V, , C0 C , charge flows from higher potential to lower, potential till both capacitors get the same W.E-15: In the circuit shown in figure C1 1 F, potential, and C2 2 F . The capacitor C1 is charged to, a) Two capacitors are connected in parallel such, 100V and the capacitor C2 is charged to 20V., that positive plate of one capacitor is connected, After charging then are connected as shown., to positive plate of other capacitor, When the switches S1 , S2 and S3 are closed, the, charge flowing through S1 is, S1 C1, S3 C2, S2, Let V be the common potential, +, +, Then Q = Q1 + Q2 (charge conservation), , (C1 + C2) V = C1V1 + C2 V2 ; V , , C1V1 C 2 V2, C1 C 2, , V1=100, V2=20V, In this case there will be loss in energy of the Sol: When S ,S and, S, are, closed, both t he, 1 2, 3, system, capacitors are in parallel with unlike charged, 1, 1, 2, 2, plates together. So, they attain a common, U Uf Ui ; where Uf C1V C2 V, potential., 2, 2, Before closing the switches,, 1 C 1C 2, 1, 1, (V1 V2 ) 2, U i C1V12 C 2 V2 2 ; U , Charge on C1 is q1 100 1 100 C, 2 C1 C 2, 2, 2, Charge on C2 is q 2 20 2 40 C, b) If positive plate of one capacitor is connected, After closing the switches, to negative plate of other capacitor, common, potential is given by, q q, 100 40, Common potential V 1 2 , 20 V, C1 C2, 3, C1V1 ~ C2 V2, V, Now final charges q11 C1V 1 20 20 C, C1 C2, , Here charge flow takes place if V1 V2, In this case, the loss of energy, 1 C1C 2, U , (V1 V2 ) 2, 2 C1 C 2, , NARAYANAGROUP, , q12 C2 V 2 20 40 C, , The charge that flows through S1 is, q 100 20 80 C, , 151
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Application :, , which is uncharged., , a) Redistribution of charges when two conductors are connected by conduting wire, In charging a conductor, work is required to be, done. This work done is stored up as the potential energy of the conductor., Energy of a charged conductor,, 1, Q2, 2 1, U CV QV , 2, 2, 2C, When two charged bodies are connected by a, conducting wire then charge flows from a, conductor at higher potential to that at lower, potential until their potentials are equal., Let the amounts of charge on two conductors A, and B are Q1 and Q2 their capacities are C1 and, C 2 and t heir pot entials are V 1 and V 2, respectively, then Q1 C1V1 and Q2 C2 V2, Let the amount of charge after the conductors, are connected, are Q 1| and Q |2 respectively, then, Q11 C1V;Q12 C2 V, , Electric potential of inner sphere is, 1 q1, V1 , 4o r1, Electric potential of outer sphere is, 1 q1, 4o r2, potential difference between the two, conductors, V2 , , V1 V2 , , q1 1 1 , , 4o r1 r2 , , If ‘ q 2 ’ charge is on the outer shell, r2, q1, , 1 q1 q 2 , V1 , , 4 o r1 r2 , , q2, , 1 q1 q 2 , , , 4o r2 , , Charges are redistributed in the ratio of their, capacities., (since V is same), QI1 : QI2 C1 : C2, In case of spherical conductors, C 4 0 r, so, Q I1 : Q I 2 r1 : r2, , V2 , , Van De Graff Generator, , potential difference V1 V2 will remain the, , Van De Graaff generator is used to develop very, high voltages and resulting large electric fields, and used to accelerate charged particles to high, energies, 1., Principle :- Whenever a charge is given to a, metal body it will spread on the outer surface of, it. if we put a charged metal body inside the 2., hollow metal body and the two are connected, by a wire, whole of the charge of innner body, will flow to the outer surface of the hallow body., No matter how, large the charge is on the inner body., Consider a sperical conductor 1 of radius r1, , same for any value of q 2, , holding charge q1 uniformly distributed on it. it, is kept inside a hollow conductor 2 of radius r2, 152, , r1, , V1 V2 , , q1 1 1 , , 4o r1 r2 , , C. U . Q, A condenser stores, 1) potential 2) charge, 3) current, 4) energy in magnetic field, Out of the following statements, A) The capacity of a conductor is affected, due to the presence of an uncharged isolated, conductor, B) A conductor can hold more charge at the, same potential if it is surrounded by dielectric, medium., 1) Both A and B are correct, 2) Both A and B are wrong, 3) A is correct and B is wrong, 4) A is wrong and B is correct, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 3., , If an earthed plate is brought near positively, charged plate, the potential and capacity of, charged plate, 1) increases, decreases 2) decreases, increases, 3) decreases, decreases 4) increases, increases, 4. The plates of charged condenser are, connected by a conducting wire. The quantity, of heat produced in the wire is, 1) Inversely proportional to the capacity of the, condenser., 2) Inversely proportional to the square of the, potential of the condenser., 3) proportional to the length of wire, 4) independent of the resistance of the wire, 5. A capacitor works in, 1) A.C. circuits only 2) D.C. circuits only, 3) both A,C & D.C, 4) neither A.C. nor in D.C. circuit., 6. In order to increase the capacity of a parallel, plate condenser one should introduce between, the plates a sheet of (assume that the space, is completely filled), 1) Mica, 2) Tin, 3) Copper, 4) Stainless steel, 7. In a parallel plate capacitor, the capacitance, 1) increases with increase in the distance, between the plates, 2) decreases if a dielectric material is put, between the plates, 3) increases with decrease in the distance, between the plates, 4) increases with decrease in the area of the, plates, 8. When a dielectric material is introduced, between the plates of a charged condenser,, after disconnecting the battery the electric, field between the plates, 1) decreases, 2) increases, 3) does not change 4) may increase or decrease, 9. A parallel plate capacitor is charged and the, charging battery is then disconnected. If the, plates of the capacitor are moved further, apart by means of insulating handles, 1) the charge in the capacitor becomes zero, 2) the capacitance becomes infinite, 3) the charge in the capacitor increases, 4) the voltage across the plates increases, 10. The ratio of charge to potential of a body is, known as, 1) conductance, 2) capacitance, 3) inductance, 4) reactance, NARAYANAGROUP, , ELECTRO STATICS & CAPACITORS, 11. A parallel plate capacitor filled with a materail, of dielectric constant K is charged to a certain, voltage and is isolated. The dielectric material, is removed. Then, a)T he capacit ance decr eases by a fact or K, b) T he elect r ic field r educes by a fact or K, c) T he volt age acr oss t he capacit or incr eases, by a fact or K, d ) T h e ch ar ge st r or ed i n t h e cap aci t or, incr eases by a fact or K, , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , 1) a and b are true, 2) a and c are true, 3) b and c are true, 4) b and d are true, Force acting upon a charged particle kept, between the plates of a charged condenser is, F. If one of the plates of the condenser is, removed, force acting on the same particle, will become, 1) zero, 2) F/2, 3) F, 4) 2F, A condenser is charged and then battery is, removed. A dielectric plate is put between the, plates of condenser, then correct statement, is, 1) Q constant V and U decrease, 2) Q constant V increases U decreases, 3) Q increases V decreases U increases, 4) Q, V and U increase, If an uncharged capacitor is charged by, connecting it to a battery, then the amount of, energy lost as heat is, 1) 1/2QV 2) QV, 3) 1/2QV2 4) QV 2, When air is replaced by a dielectric medium, of constant K, the capacity of the condenser, 1) increases K times 2) increases K 2 times, 3) remains unchanged 4) decreases K times, If we increase the distance between two plates, of the capacitor, the capacitance will, 1) decrease 2) remain same, 3) increase 4) first decrease then increase, In a charged capacitor the energy is stored in, (r) is less than at B, 1) both in positive and negative charges, 2) positive charges, 3) the edges of the capacitor plates, 4) the electric field between the plates, A metal plate of thickness half the separation, between the capacitor plates of capacitance, C is inserted. The new capacitance is, 1) C, 2) C/2, 3) zero, 4) 2C, One plate of parallel plate capacitor is smaller, than the other, the charge on the smaller plate, will be, 1) less than other, 2) more than other, 3) equal to other, 4) will depend upon the medium between them, 153
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ELECTRO STATICS & CAPACITORS, , JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , 20. Two condensers of unequal capacities are, the battery is disconnected and the plate, connected in series across a constant voltage, separation is doubled is E1, d.c. source. The ratio of the potential, B) The energy stored in the capacitor when, differences across the condensers will be, the charging battery is kept connected and, 1) direct proportion to their capacities, the separation between the capacitor plates, 2) inverse proportion to their capacities, E1, 3) direct proportion to the square of their, is doubled is E2 . Then E value is, 2, capacities, 1), 4, 2), 3/2, 3), 2, 4) 1/2, 4) inverse proportion to the square root of their, 27. Select correct Statements, capacities, a) Charge cannot be isolated, 21. A parallel plate capacitor is first charged and, b) Repulsion is the sure test to know the, then isolated , and a dielectric slab is, presence of charge, introduced between the plates. The quantity, c) Waxed paper is dielectric in paper capacitor, that remains unchanged is, 1) Charge Q, 2) Potential V, d) Variable capacitor is used in tuning circuits, 3) Capacity C, 4) Energy U, in radio, 1) a, b only, 2) a, c only, ENERGY STORED IN A CONDENSER, 3), a,, b,, c, only, 4) b,c,d only, AND TYPES OF CAPACITORS, 28., A, variable, parallel, plate, capacitor and an, 22. The condenser used in the tuning circuit of, electroscope, are, connected, in parallel to a, radio receiver is, battery. The reading of the electroscope, 1) paper condenser 2) electrolytic condenser, would be decreased by, 3) leyden jar, 4) gang condenser, 1) increasing the area of overlap of the plates, 23. Space between the plates of a parallel plate, 2) placing a block of paraffin wax between the, capacitor is filled with a dielectric slab. The, plates, capacitor is charged and then the supply is, 3) decreasing the distance between the plates, disconnected to it. If the slab is now taken, 4) decreasing the battery potential, out then, 29. Three identical capacitors are connected, 1) work is not done to take out the slab, together differently. For the same voltage to, 2) energy stored in the capacitor reduces, 3) potential difference across the capacitor, every combina-tion, the one that stores, is decreased, maximum energy is, 4) potential difference across the capacitor, 1) the three in series 2) the three in parallel, is increased, 3) two in series and the third in parallel with it, 24. A parallel plate condenser is charged by, 4) two in parallel and the third in series with it, connecting it to a battery. The battery is 30. The magnitude of electric field E in the, disconnected and a glas slab is introduced, annular region of a charged cylindrical, between the plates. Then, capacitor, 1) potential increases, 1) is same throughout, 2) electric intensity increases, 2) is higher near the outer cylinder than near the, 3) energy decreases, inner cylinder, 4) capacity decreases, 3) varies as 1/r where r is the distance from the, 25. A parallel plate condenser is charged by, axis, connecting it to a battery. Without, 4) varies as r where r is the distance from the, disconnecting the battery, the space between, axis, the plates is completely filled with a medium, 31., Two identical capacitors are joined in parallel,, of dielectric constant k. Then, charged to a potential V, separated and then, 1) potential becomes 1/k times, connected in series i.e., the positive plate of, 2) charge becomes k times, one is connected to the negative plate of other., 3) energy becomes 1/k times, 1) the charges on the free plates are enhanced, 4) electric intensity becomes k times., 2) the charges on the free plates are decreased, 26. A par allel plat e capacit or of capacit y C o is, 3) the energy stored in the system increases, charged to a potential Vo., 4) the potential difference between the free plates, A) The energy stored in the capacitor when, is 2V, 154, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 32. Two parallel plate air capacitors are, constructed, one by a pair of iron plates and, the second by a pair of copper plates of same, area and same spacings. Then, 1) the copper plate capacitor has a greater, capacitance than the iron one, 2) both capacitors will have equal non zero, capacitances, in the uncharged state, 3) both capacitors will have equal capacitances, only if they are charged equally, 4) the capacitances of the two capacitors are, unequal even they are unequally charged, 33. Select correct statement for a capacitor, having capacitance C, is connected to a source, of constant emf E, 1) Almost whole of the energy supplied by the, battery will be stored in the capacity, if, resistance of connecting wire is negligibly small, 2) Energy received by the capacitor will be half, of energy supplied by the battery only when the, capacitor was initially uncharged, 3) Strain energy in the capacitor must increases, even if the capacitor had an initial charge, 4) Energy stored depends on type of the source, of emf, 34. Van de Graff genetor is used to :, 1) supply electricity for industrial use, 2) produce intense magnetic fields, 3) generate high voltage, 4) obtain highly penetrating X-rays, 35. A number of spherical conductors of different, radii have same potential. Then the surface, charge density on them, 1) is proportional to their radii, 2) is inversely proportional to their radii, 3) are equal, 4) is proportional to square of their radii, 36. Three charged particles are initially in, position 1. They are free to move and they, come in position 2 after some time. Let U1, and U2 be the electrostatic potential energies, in position 1 and 2. Then, 1) U1 > U2 2) U2 > U1 3) U1 = U2 4) U 2 U1, 37. An insulator plate is passed between the plates, of a capacitor. Then current, A, , B, , 1) always flows from A to B, 2) always flows from B to A, 3) first flows from A to B and then from B to A, 4) first flows from B to A and then from A to B, NARAYANAGROUP, , ELECTRO STATICS & CAPACITORS, 38. Read the following statements, a) Non polar molecules have uniform charge, distribution, b) Polar molecules have non - uniform charge, distribution, c) Polar molecules are already polarized, d) Molecules are not already polarized without, electric field in Non - polar molecules, 1)only a & b are correct 2)only c & d are correct, 3) only c is wrong, 4) all are correct, 39. The capacitance of a capacitor depends on, 1) the geometry of the plates, 2) separation between plates, 3) the dielectric between the plates, 4) all the above, , , , , 40. The electric field E between two parallel, plates of a capacitor will be uniform if, 1) the plate separation (d) is equal to area of the, plate (A), 2) the plate separation (d) greater when, compared to area of the plate (A), 3) the plate separation (d) is less when compared, to area of the plate (A), 4) 2 (or) 3, 41. For metals the value of dielectric constant (K), is, 1) One, 2) Infinity 3) Zero, 4) Two, 42. A capacitor C is connnected to a battery, circuit having two switches S 1 and S 2 and, resistors R1 and R 2 . The capacitor will be, fully charged when, R1, , S1, , S2, R2, C, , 1) both S 1 and S 2 are closed, 2) S 1 is closed and S 2 is open, 3) S 1 is open and S 2 is closed, 4) any one of the above, 43. Figure shows two capacitors connected in, series and joined to a cell. The graph shows, the variation in potential as one moves from, left to right on the branch containing, capacitors., , 1) C 1 C 2, 2) C 1 C 2, 3) C 1 C 2, 4) data insufficient to conclude the answer, 155
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 44. Two condensers of unequal capacities are, connected in parallel across a constant voltage, d.c. source. The ratio of the charges stored, in the condensers will be, 1) direct proportion to their capacities, 2) inverse proportion to their capacities, 3) direct proportion to the square root of their, capacities, 4) inverse proportion to the square of their, capacities, 45. A parallel plate capacitor is charged and then, isolated. Regarding the effect of increasing, the plate separation, select the appropriate, alternative., Charge Potential Energy, 1) decreases constant decreases, 2) increases increases increases, 3) constant decreases decreases, 4) constant increases increases, 46. A parallel plate capacitor is charged by, connecting its plates to the terminals of a, battery. The battery remains connected to, the condenser plates and a glass plate is, interposed between the plates of the capacitor,, then, 1) the charge increases while the potential, difference remains constant, 2) the charge decreases while the potential, difference remains constant, 3) the charge decreases while the potential, difference increases, 4) the charge increases while the potential, difference decreases, 47. A parallel plate capacitor is charged to a fixed, potential and the charging battery is then, disconnected. If now, the plates of the, capacitor are moved further apart, then, 1) the charge on the capacitor increases, 2) the voltage across the capacitor increases, 3) the energy stored in the capacitor decreases, 4) the capacitance increases, 48. A parallel plate air condenser is charged and, then disconnected from the charging battery., Now the space between the plates is filled, with a dielectric then, the electric field, strength between the plates, 1) increases while its capacity increases, 2) increases while its capacity decreases, 3) decreases while its capacity increases, 4) decreases while its capacity decreases, 49. When two identical condensers are connected, in series choose the correct statement, regarding the working voltage (the maximum, p.d. that can be applied to a condenser) and, the capacity, 156, , 50., , 51., , 52., , 53., , 54., , 1) working voltage increases, capacity increases, 2) working voltage increases, capacity decreases, 3) working voltage decreases, capacity increases, 4) working voltage decreases, capacity decreases, Two unequal capacitors, initially uncharged,, are connected in series across a battery., Which of the following is true, 1) The potential across each is the same, 2) The charge on each is the same, 3) The energy stored in each is the same, 4) The equivalent capacitance is the sum of the, two capacitances, Which of the following will not increase the, capacitance of an air capacitor?, 1) adding a dielectric in the space between the, plates, 2) increasing the area of the plates, 3) moving the plates closer together, 4) increasing the voltage, In a parallel-plate capacitor, the region, between the plates is filled by a dielectric slab., The capacitor is connected to a cell and the, slab is taken out. Then, 1) some charge is drawn from the cell, 2) some charge is returned to the cell, 3) the potential difference across the capacitor, is reduced, 4) no work is done by an external agent in taking, the slab out, Which of the following statements are, correct?, a) When capacitors are connected in parallel the, effective capacitance is less than the individual, capacitances, b) The capacitances of a parallel plate capacitor, can be increased by decreasing the separation, of plates, c) When capacitors are connected in series the, effective capacitance is less than the least of the, individual capacities, d) In a parallel plate capacitor the electrostatic, energy is stored on the plates, 1) (a) and (b), 2) (a) and (c), 3) (c) and (d), 4) (b) and (c), Three identical condensers are connected, together in four different ways. First all of, them are connected in series and the, equivalent capacity is C1. Next all of them, are connected in parallel and the equivalent, capacity is C 2 . Next two of them are, connected in series and the third one, connected in parallel to the combination and, the equivalent capacity is C3. Next two of, them are connected in parallel and the third, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , one connected in series with the combination, and the equivalent capacity is C4. Which of, the following is correct ascending order of the, equivalent capacities?, 1) C1 C3 C4 C2 2) C1 C4 C3 C2, 57., 3) C2 C3 C4 C1 4) C2 C4 C3 C1, 55. On a capacitor of capacitance C0 following, steps are performed in the order as given in, column I., (A) Capacitor is charged by connecting it, across a battery of emf E0, (B) Dielectric of dielectric constant K and, thickness d is inserted, (C) Capacitor is disconnected from battery, (D) Separation between plates is doubled, Column-I, Column-II, (Steps performed), (Final value of, Quantity (Symbols have, usual meaning), CE, (a) (A)(D)(C)(B), (p) Q 0 0, 2, KC0 E0, (b) (D)(A)(C)(B), (q) Q , K 1, KC0, (c) (B)(A)(C)(D), (r) C , K 1, E ( K 1), (d) (A)(B)(D)(C), (s) V 0, 2K, 1) a-p,r,s, b-p,r,s, c-r, d-q,r, 2) a-p, b-p,r c-r, d-q,, 3) a-p,s, b-r,s, c-r, d-q,, 4) a-r,s, b-s, c-r, d-q,r, 56. In the circuit, both capacitors are indentical., Column I indicates action done on capacitors, 1 and Column II indicates effect on capacitor, 2, (1), , (2), , Column-II, (p)Amount of charge, on left plate, increases, (b) Area increased (q) Potential difference, increases, (c) Left plate is earthed (r) Amount of charge, on right plate, decreases, , (d) It's plates are short, (s) None of the above, circuited, effects, 1.a-r, b-p,q, c-s, d-p,q 2.a-r, b-p, c-s, d-q, 3.a-r, b-p, c-r, d-q, 4.a-s, b-,q, c-s, d-q, The potential across a 3 F capacitor is 12 V, when it is not connected to anything. It is then, connected in parallel with an uncharged 6 F, capacitor. At equilibrium, the charge and, potential difference across the capacitor 3, F and 6 F are listed in column I. Match it, with column III., Column-I, Column-II, (a) charge on 3 F capacitor, (p) 12 C, (b) charge on 6 F capacitor, (q)24 F, (c) potential difference across 3 F (r) 8 V, capacitor, (d) potential difference across 6 F (s) 4 V, capacitor, 1) a-r, b-p, c-s, d-q 2) a-p, b-q, c-s, d-s, 3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q, 58. Some events related to a capacitor are listed, in column-I. Match these with their effect(s), in column - II, Column-I, Column-II, (a) Insertion of dielectric, (p) Eelctric field, while battery, between, plates changes, remain attached, (b) Removal of dielectric, (q) Charge present on, while battery, plates changes, is not present, (c) Slow decrease in, (r) Energy stored in, separation between, capacitor increases, plates while battery is, attached, (d) Slow increase of, (s) Work done by, separation between, capacitor agent is, plates while battery, positive, is not present, , Column-I, (a) Plates are moved, further apart, , NARAYANAGROUP, , 1) a-r, b-p,, c-p,q,s, d-q, 2) a-p, b-p,, c-r,s, d-s, 3) a-q,r, b-p,r,s c-p,q,r, d-r,s, 4) a-r, p,b-q,, c-s, d-q, 157
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 59. The effective capacity of the following, capacitors is ________, e), , 2c, 3, , battery remains connected. Due to this, inserting some physical quantities may change, which are mentioned in Column-I and the, effect is mentioned in Column-II.Match the, Column I with Column-II, , f) 2C, , A, e, , g) 3C, , K, B, , Column-I, Column-II, (a) Charge on A, (p) Increases, (b) Charge on B, (q) Decreases, 3C, (c) Potential difference, i), across A, (r) Remains constant, 2, (d), Potential, difference, 1) a g , b f , c e, d i, across B, (s) Will change, 2) a g , b h, c e, d i, 1) a-r, b-p, c-s, d-q 2) a-p,s b-q,s, c-q,s d-q,s, 3) a i, b h, c e, d g, 3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q, 62., Match the following, 4) a g , b e, c h, d i, Set -I, Set-II, 60. The circuit involves two ideal cells connected, a), Electrolyte, e) Radio circuits, to a 1 F capacitor via key K. Initially the, Capacitors, & cheap in cost, key K is in position 1 and the capacitor is, b) Paper Capacitor f) Proper Polarity, charged fully by 2V cell. The key is pushed to, high capacitance, position 2. Column I gives physical quantities, of order 103 F, involving the circuit after the key is pushed, c) Multiple, g)High frequency, from position 1. Column.II gives corresponding, Capacitor, oscillating circuits, results. Match the column-I with Column-II, d), Variable, h) Tuning circuits, C = 1mF, Capacitor, in radio & T.V, receivers, 4V, 2V, K, 1) a f , b g , c h, d e, 2) a g , b f , c e, d h, 1 2, 3) a f , b e, c g , d h, Column-I, Column-II, 4) a h, b e, c f , d g, (a) The net charge crossing the, 63. Column - I, Column - II, 4 volt cell in C is, (p) 2, A) electrical potential p) vector, (b) The magnitude of work done, 1, 2, B) energy stored in a q) CV, by 4 volt cell in J is, (q) 6, 2, (c) The gain in potential energy of, condenser, C) force between two r) scalar, capacitor in J is, (r) 8, (d) The net heat produced in, 1, 0 E 2 A, D), electric, capacity, s), circuit in J is, (s) 16, 2, 1) a-r, b-p, c-s, d-q 2) a-p, b-r, c-q, d-p, capacitor plates, 3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q, A, B, C, D, 61. Two identical capacitors A and B are, 1), r, q,r, p,s r, connected to a battery of emf E as shown in, 2), r, q,r, p,q s, figure. Now a dielectric slab is inserted, 3), q,r, p,q, r,s s, between the plates of capacitor B while, 4), p,q, r, q,r s, h), , 158, , 5C, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , ASSERTION & REASONING, , 64., , 65., , 66., , 67., , 68., , 69., , 1) Both A and R false, 2) Both A and R true and R is not correct, reason for A, 3) A is true and R is false, 4) Both A and R are true and R is correct, reason of A., Assertion (A) : The strength of electric filed, in the charged and isolated capacitor is, decreased when the dielectric slab is inserted., Reason(R): When the dielctric slab is inserted, between the plates of a charged capacitor,, electricfield produced due to induced charge,, opposite to the external field., Assertion: If temperature is increased, the, dielectric constant of a polar dielectric, decreases whereas that of a non-polar, dielectric does not change significantly, Reason: The magnitude of dipole moment of, individual polar molecule decreases, significantly with increase in temperature., Assertion: The heat produced by a resistor in, any time t during the charging of a capacitor, in a series circuit is half the energy stored in, the capacitor by that time., Reason: Current in the circuit is equal to the, rate of increase in charge on the capacitor., Assertion: A dielectric is inserted between the, plates of an isolated fully-charged capacitor., The dielectric completely fills the space, between the plates. The magnitude of, electrostatic force on either metal plate, decreases, as it was before the insertion of, dielectric medium., Reason: Due to insertion of dielectric slab, in an isolated parallel plate capacitor (the, dielectric completely fills the space between, the plates), the electrostatic potential energy, of the capacitor decreases., Assertion: If the potential difference across, a plane parallel plate capacitor is doubled then, the potential energy of the capacitor is, doubled then the potential energy of the, capacitor becomes four times under all, conditions, Reason: The potential energy U stored in the, 1, 2, capacitor is U CV , where C and V have, 2, usual meaning., Assertion: A parallel plate capacitor is, charged to a potential difference of 100V, and, disconnected from the voltage source. A slab, of dielectric is then slowly inserted between, the plates. Compared to the energy before, , NARAYANAGROUP, , 70., , 71., , 72., , 73., , 74., , 75., , the slab was inserted, the energy stored in the, capacitor with the dielectric is decreased., Reason: When we insert a dielectric between, the plates of a capacitor, the induced charges, tend to draw in the dielectric into the field, (just as neutral objects are attracted by, charged objects due to induction). We resist, this force while slowly inserting the dielectric,, and thus do negative work on the system,, removing electrostatic energy from the, system., Statement ' A ' : The energy stored gets, reduced by a factor ' K ' when the battery is, disconnected after charging the capacitor and, then the dielectric is introduced, Statement ' B ' : The energy stored in the, capacitor increases by a factor ' k ' when a, dielectric is introduced between the plates, with the battery present in the circuit, Assertion (A): A metallic sheild in form of a, hollow shell may be built to block an electric, field., Reason (R): In a hollow spherical sheild, the, electric field inside it is zero at every point., Assertion (A): When two spheres carrying, same charge but a different radii are, connected by a conducting wire, the charge, flows fromsmaller sphere to large sphere., Reason (R): Smaller sphere is at high potential, when equal charges are imparted to both the, spheres, Assertion (A): Two capacitors are connected, in par allel to a battery. If a dielectric medium, is inserted between the plates of one of the, capacitors then the energy stored in the, system will increase, Reason (R): On inserting dielectric medium, between the plates of a capacitors, its capacity, increases, Assertion (A): When a charged capacitor is, discharged through a resistor, heat is produced, in the resistor, Reason (R): In charging a capacitor, energy, is stored in the capacitor., Assertion (A): A capacitor of capacitance C, is connected across a battery of potential, difference V. The energy stored in the, 1, 2, capacitor is CV, 2, Reason (R): The energy supplied by the, 1, 2, battery is CV, 2, 159
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 76. Assertion (A): Two metal plates each of area, A form a parallel plate capacitor. Now one, plate is displaced up, then the capacitance of, capacitor decreases., Reason (R): Due to displacing one plate, the, overlapping area decreases, capacitance, A, C 0 decreases., d, 77. Assertion (A): Two plates of a parallel plate, capacitor are drawn apart, keeping them, connected to a battery. Next the same plates, are drawn apart from the same initial, condition, keeping the battery disconnected,, then the work done in both cases are same., Reason (R): Capacitor plates have same, charge in both cases and displacements of, plates in both cases are also same., 78. Assertion (A) : Two metallic plates placed, side by side form three capacitors., Reason (R) : The infinity and first face of first, plate is one capacitor, the second face of first, plate and first face of second plate forms, second capacitor and the second face of, second plate and infinity forms the third, capacitor, but the capacitance of first and third, capacitance are extremely small, 79. Statement ' A ' : The energy stored gets, reduced by a factor ' K ' when the battery is, disconnected after charging the capacitor and, then the dielectric is introduced, Statement ' B ' : The energy stored in the, capacitor increases by a factor ' k ' when a, dielectric is introduced between the plates, with the battery present in the circuit, , 160, , 2) 1, 8) 1, 14) 1, 20) 2, 26) 1, 32) 2, 38) 4, 44) 1, 50) 2, 56) 1, 62) 3, 68) 4, 74) 2, , 3) 2, 9) 4, 15) 1, 21) 1, 27) 4, 33) 3, 39) 4, 45) 4, 51) 4, 57) 2, 63) 1, 69) 1, 75) 3, , 4) 4, 10) 2, 16) 1, 22) 4, 28) 4, 34) 3, 40) 3, 46) 1, 52) 2, 58) 3, 64) 4, 70) 4, 76) 1, , CAPACITANCE, 1., , 2., , 5) 3, 11) 2, 17) 4, 23) 4, 29) 2, 35) 2, 41) 2, 47) 2, 53) 4, 59) 1, 65) 3, 71) 1, 77) 4, , 6) 1, 12) 2, 18) 4, 24) 3, 30) 3, 36) 1, 42) 2, 48) 3, 54) 2, 60) 2, 66) 4, 72) 1, 78) 1, , The capacity of a parallel plate condenser, consisting of two plates each 10 cm square, and are seperated by a distance of 2 mm is, (Take air as the medium between the plates), 2) 4.42 1012 F, 1) 8.85 1013 F, 12, , 3) 44.25 10 F, 4) 88.5 1013 F, Sixty four spherical drops each of radius 2 cm, and carrying 5C charge combine to form a, bigger drop. Its capacity is, 8, 9, , 1) 1011 F 2) 90 1011 F 3) 1.11011 F 4) 9 1011 F, 3., , 4., , A highly conducting sheet of aluminium foil of, negligible thickness is placed between the, plates of a parallel plate capacitor. The foil is, parallel to the plates. If the capacitance before, the insertion of foil was 10 F, its value after, the insertion of foil will be, 1) 20 F 2) 10 F 3) 5 F 4) Zero, Two metal plates are separated by a distance, d in a parallel plate condenser. A metal plate, of thickness t and of the same area is inserted, between the condenser plates. The value of, capacitance increases by ...... times, 1), , 5., , d t, d, , 1, t , , 1 , d, , , t, t, , , 2) 1 d 3) t d 4), , , , , , A radio capacitor of variable capacitance is, made of n parallel plates each of area A and, separated from each other by a distance d., The alternate plates are connected together., The capacitance of the combination is, 1), , 6., , C. U.Q - KEY, 1) 2, 7) 3, 13) 1, 19) 3, 25) 2, 31) 4, 37) 4, 43) 3, 49) 2, 55) 1, 61) 2, 67) 4, 73) 1, 79) 4, , LEVEL - I (C.W), , n A o, d, , 2), , n 1 A o, d, , 3), , 2n 1 A o, d, , 4), , n 2 A o, d, , The radius of the circular plates of a parallel, plate condenser is ‘r’. Air is there as the, dielectric. The distance between the plates if, its capacitance is equal to that of an isolated, sphere of radius r' is, 1), , r2, 4r ', , 2), , r2, r', , 3), , r, r', , 4), , r2, 4, , CAPACITORS IN SERIES AND IN, PARALLEL, 7., , When two capacitors are joined in series the, resultance capacity is 2.4 F and when the same, two are joined in parallel the resultant, capacity is 10 F . Their individual capacities are, 1) 7 F , 3 F, 2) 1 F , 9 F, 6, , F, ,, 4, , F, 3), 4 8 F , 2 F, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 8., , 9., , ELECTRO STATICS & CAPACITORS, , LEVEL - I (C.W ) - KEY, , Three condensers 1 F , 2 F and 3 F are, connected in series to a p.d. of 330 volt. The, p.d across the plates of 3 F is, 1) 180 V 2) 300 V 3) 60 V, 4) 270 V, The effective capacitance between the point, P and Q in the given figure is, , 1) 2, 7) 3, 13) 2, , LEVEL - I (C.W ) - HINTS, , 4F, , 1., 1) 4 F 2) 16 F P, 4F, , 4F, , 10F, , P, , 5F 5F, , 5F, , C, , 2) 3 F, , 7., , C, , X, , Y, , 8., 3) 1 F 4) 0.5 F, C, 12. The equivalent capacitance of the network 9., given below is 1 F. The value of ‘C’ is, 1.5F, Q, , P, C, , 1) 3 F 2) 1.5 F 3) 2.5 F 4) 1 F, 13. Three capacitors of 3 F , 2 F and 6 F are, connected in series. When a battery of 10V is, connected to this combination then charge on, 3 F capacitor will be, 1) 5C, 2) 10C 3) 15C, 4) 20C, , ENERGY STORED IN ACONDENSER,, TYPES OF CAPACITORS, 14. Two spheres of radii 12 cm and 16 cm have, equal charge. The ratio of their energies is, 1) 3 : 4, 2) 4 : 3, 3) 1 : 2, 4) 2 : 1, 15. A condenser of capacity 10 F is charged to, a potential of 500 V. Its terminals are then, connected to those of an uncharged condenser, of capacity 40 F. The loss of energy in, connecting them together is, 1) 1J, 2) 2.5J 3) 10J, 4) 12 J, 16. A 2 F condenser is charged to 500V and then, the plates are joined through a resistance. The, heat produced in the resistance in joule is, 1) 50 102 Joule, 2) 25 102 Joule, , 2, 3) 0.25 10 Joule, 4) 0.5 102 Joule, NARAYANAGROUP, , 1, , C1 n 3 C, , 0 A, , C, , ;k = , t, k, Due to n plates n-1 capacitors are formed, o r 2 , d, , 4 o r 1, , d , , C1C2, C1 C2, , CS , , r2, 4 r1, , ; C1 C2 CP, , Q Ceff V Q C1V, , C1 , , C1C2, C1 C2, , C 11 , , C3C4, C3 C4, , Ceff C1 C11, , 3F, , 3F, , 0 A, d, , d t , , 5., Q, 1) 10 F 2) 20 F 3) 5 F, 4) 15 F, 11. The equivalent capacity between the points 6., X and Y in the circuit with C 1 F (2007M), 1) 2 F, , C, , Q, , 2., 3) 26 F 4) 10 F, 4F, 10. The equivalent capacitance between P and Q, is, 10F 10F 10F 10F, 4., 10F, , 2) 1 3) 2 4) 4 5) 2 6) 1, 8) 3 9) 1 10) 3 11) 1 12) 2, 14) 1 15) 1 16) 2, , 10 10, 5 F, 10 10, C 1 5 5 10 F, , 10. From Left C , , 10 10, 5 F and so on, 10 10, 10 10, 5 F, finally Ceff , 10 10, 11. Ceff C1 C2, C 11 , , 12. 1.5 c, c are in parallel ;, its effective capacitance 1.5 + c, 1.5+c, 3 F , 3 F are in series, 13. In series charge constant Q Ceff V, 1, q2, 14. U =, , U, r, 2C, 1 C1C2, 2, 15. E 2 C C V1 V2 , 1, 2, , 16. Energy Stored =, , 1 2, cv, 2, , 161
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , LEVEL - I (H.W), CAPACITANCE, 1., , 9., , Two capacitors with capacitances C1 and C2, are charged to potentials V1 and V2, respectively. When they are connected in, parallel the ratio of their respective charges, is, , The charge stored in a capacitor is 20C and, the potential difference across the plates is, 500 V. Its capacity is, C1, V1, V12, C12, 2, 6, 2), 3), 4), 1), 1) 0.04F 2) 10 F 3) 2 10 F 4) 250 F, C2, V2, V22, C22, An oil condenser has a capacity of 100 F . The 10. The equivalent capacitance between P and Q, oil has dielectric constant 2. When the oil leaks, of the given figure is (the capacitance of each, out its new capacity is, capacitor is 1 F ), 1) 200 F 2) 0.02 F 3) 50 F, 4) 0.5 F, C, C, A dielectric of thickness 5cm and dielectric, P, C, C, constant 10 is introduced between the plates, O, O, of a parallel plate capacitor having plate area, C, 500 sq. cm and separation between the plates, C, 1) 2 F, 2) 0.5 F 3) 5 F 4) 0.2 F, 10cm. The capacitance of the capacitor with, dielectric slab is, 11. The resultant capacity between the points P, 12, 2, 2, and Q of the given figure is, 0 8.8 10 C / N m , , 2., , 3., , 1) 4.4 pF 2) 6.2 pF 3) 8 pF 4) 10 pF, The capacitance of a sphere of radius 10cm, situated in air is approximately, 1) 11106 F 2) 1110 9 F 3) 11 10 12 F 4) Zero, , 4., , CAPACITORS IN SERIES AND IN, PARALLEL, 5., , 6., , 7., , 8., , 162, , 4F, , 1) 4 F, , 2), , P, , 2F, , 16, F, 3, , 3) 1.6 F 4) 1 F, 4F, Q, 12. Charge ‘Q’ taken from the batteryof 12V in, the circuit is, 12V, , The ratio of the resultant capacities when, three capacitors of 2 F , 4 F and 6 F are, 3F, 6F, connected first in series and then in parallel, 1) 72 C 2) 36 C, is, 3) 156 C 4)20 C, 1) 1 : 11 2) 11 : 1 3) 12 : 1 4) 1 : 12, 4F, A condenser A of capacity 4 F has a charge 13. If 3 capacitors of values 1, 2 and 3 F are, available. The maximum and minimum values, 20 C and another condenser B of capacity, of capacitances one can obtain by different, 10 F has a charge 40 C . If they are, combinations of the three capacitors together, connected parallel, then, are respectively .... and, 1) charge flows from B to A till the charges on, 11, 6, them are equal., 2) 6 F , F, 1) 6 F , F, 2) charge flows from B to A till common poten, 6, 11, tial is reached, 3) 3 F ,1 F, 4) 4 F ,2 F, 3) charge flows from A to B till common, ENERGY STORED IN ACONDENSER,, potential is reached, TYPES OF CAPACITORS, 4) charge flows from A to B till charges on them, are equal., 14. A capacitor of 8 micro farad is charged to a, A capacitor of 30 F charged to 100 V is, potential of 1000V. The energy stored in the, capacitor is, conncected in parallel to capacitor of 20 F, 1) 8 J, 2) 12 J, 3) 2 J, 4) 4 J, charged to 50 volt. The common potential is, 15., A, condenser, is, charged, to, a, p.d., of 120 volt., 1) 75 V 2) 150 V 3) 50 V, 4) 80 V, 5, The equivalent capacity between the points, Its energy is 110 joule . If the battery is, ‘A’ and ‘B’ in the following figure will be, there and the space between plates is filled, up with a dielectric medium r 5 , its new, energy is, A, 1) 105 J 2) 2 105 J 3) 3 10 5 J 4) 5 105 J, 1) 3C, 2) C/3, 3)3/C, 4) 1/3C, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 16. The plates of a parallel plate capacitor have, an area of 90 cm 2 each and are separated by, 2 mm. The capacitor is charged by connecting, if to a 400 V supply. Then the density of the 1., energy stored in the capacitor, , , , 8.8 10 12 F / m , , o, , 1) 0.113Jm 3, , 2) 0.117 Jm 3, , 3) 0.152 Jm 3, , 4) 0.226 Jm 3, , LEVEL - I ( H.W ) - KEY, 1) 1, 7) 4, 13) 1, , 2., , 2) 3 3) 3 4) 3 5) 1 6) 3, 8) 1 9) 1 10) 2 11) 1 12) 1, 14) 4 15) 4 16) 2, , LEVEL - I ( H.W ) - HINTS, 1., 2., 3., 4., 5., 6., , 7., 8., 9., 10., , 11., 12., 13., 14., 15., 16., , q, V, C', C, K, 0 A, C, t, d t , k, C 4 0 R, 1, 1, 1, 1, , , Cs C1 C2 C3, , 3., , C, , 4., C p C 1 C2 C3, , In parallel potential constant V , , C1V1 C2V2, C1 C2, , then find charges q11 C1V , q21 C2V, C V C2V2, V 1 1, C1 C2, Capacitors are in parallel, Parallel potential constant and Q C, C1 C C ; C2 C ; C3 C C, 1, 1, 1, 1, , , Ceff C1 C2 C3, 4 4, C1 , ; C2 2 F ; Ceff C1 C2, 44, 6 3, Ceff , 4 ; Q Ceff V, 63, For maximum capacitors are parallel, For minimum capacirots are in sseries, U = 1/2 CV 2, E1 r E, 1, V, U oE 2 , E , 2, d, , NARAYANAGROUP, , LEVEL - II (C.W), CAPACITANCE, A parallel plate condenser has initially air, medium between the plates. If a slab of, dieletric constant 5 having thickness half the, distance of seperation between the plates is, introduced, the percentage increase in its, capacity is, 1) 33.3% 2) 66.7% 3) 50%, 4) 75%, When a dielectric slab of thickness 4 cm is, introduced between the plates of parallel plate, condenser, it is found the distance between, the plates has to be increased by 3cm to, restore to capacity to original value. The, dielectric constant of the slab is, 1) 1/4, 2) 4, 3) 3, 4) 1, The area of the positive plate is A1 and the, area of the negative plate is A 2 ( A 2 < A1 ) ., They are parallel to each other and are, separated by a distance d . The capacity of a, condenser with air as dielectric is, 0 A1, 0 A1, 0 A2, 0 A1 A2, 2), 3), 4) A d, 1), 2, d, d, d, The cross section of a cable is shown in fig., The inner conductor has a radius of 10 mm, and the dielectric has a thickness of 5 mm., The cable is 8 km long. Then the capacitance, of the cable is log e 1.5 0.4, 1) 3.8 F 2) 1.1 F 3) 4.8 1010 F 4) 3.3 F, , CAPACITORS IN SERIES AND IN, PARALLEL, 5., , Two condensers of capacity C and 2C are, connected in parallel and these are charged, upto V volt. If the battery is removed and, dielectric medium of constant K is put between, the plates of first condenser, then the potential, at each condenser is, V, , 6., , k, , 2V, , 3V, , 1), 2) 2 , 4), 3), k2, 3V, k2, k2, Given a number of capacitors labelled as C,V., Find the minimum number of capacitors, needed to get an arrangement equivalent to, Cnet , Vnet, C, V 2, 1) n net net2, C, V, C, V, 3) n C V, net, net, , C, V2, n, , , 2), Cnet Vnet 2, , 4) n , , Cnet Vnet, , C, V, 163
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , Two capacitors of capacities 3 F and 6 F 13. The equivalent capacity of the infinite net work, shown in the figure (across AB) is (Capacity, are connected in series and connected to, of each capacitor is 1 F), 120V. The potential difference across 3 F, is V0 and the charge here is q0 . We have, , 7., , A) q0 40 C, , B) V0 60V, , C) V0 80V, D) q0 240 C, 3 1 , 3 1, 1) A, C are correct, 2) A, B are correct, 1 F 3) 2 F 4) 2 F, 1), 2), , 3) B, D are correct 4) C, D are correct, , , , , n Capacitors of 2 F each are connected in 14. The extra charge flowing through the cell on, parallel and a p.d of 200v is applied to the, closing the key k is equal to, k, combination. The total charge on them was, C, 1c then n is equal to, C, C, 1) 3333, 2) 3000 3) 2500, 4) 25, CV, 1), 2), 4, CV, An infinite number of identical capacitors each, C, 4, of capacitance 1 mF are connected as shown, 4, 3, V, 3) CV 4) CV, in the figure. Then the equivalent capacitance, 3, 4, between A and B is, , 8., , 9., , LEVEL - II (C. W ) - KEY, , 1) 1 mF, , 2) 2 mF, , 1) 2, 7) 4, 13) 3, , 8 Capacitors, , 4) 0.75 mF, , A, , B, , 10. Two capacitors of capacites 1 F and C F, are connected in series and the combination, is charged to a potential difference of 120 V., If the charge on the combition is 80 C , the, energy stored in the capacitor C in micro joules, is :, 1) 1800 2) 1600 3) 14400 4) 7200, 11. A parallel capacitor of capacitance C is, charged and disconnected from the battery., The energy stored in it is E. If a dielectric, slab of dielectric constant 6 is inserted, between the plates of the capacitor then, energy and capacitance will become, 1) 6E, 6C 2) E, C, 3) E/6, 6C 4) E, 6C, 12. In the circuit diagram given below, the value, of the potential difference across the plates, of the capacitors are, 3F, , 7F, 13kv, , 1) 17.5 KV, 7.5 KV, 3) 5 KV, 20 KV, 164, , 2) 10 KV, 15 KV, 4) 16.5 KV, 8.5KV, , 0 A, 0 A C , ;, d t 1 1/ k , d, C C0, C % , 100%, C0, , 1., , C0 , , 2., , C, , 3., , Effective area only A2, , 4., 5., , K o 2l, ln b / a , Q = constant , CV + 2CV = KCV | 2CV |, , 6., , F , , ENERGY STORED IN A CONDENSER, TYPES OF CAPACITORS, , 12kv, , 4) 1 5) 4 6) 1, 10) 2 11) 3 12) 1, , LEVEL - II (C. W ) - HINTS, , 16 Capacitors, , 3) ½ mF, , 2) 2 3) 2, 8) 3 9) 2, 14) 1, , 0 A, A, 0, d t 1 1/ k d d ', , C, , Q, Q2, , 2 0 2 0 A, , CC , Q 1 2 V, C1 C2 , 8. Q nCV, C C, 9. C R C ......, 2 4, 10. CV, 1 1 C2V2, , 7., , q2, q2, ,U f , 2C, 2 KC, 12. By Kirchoff loop theorem, , 11. U , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, q, q, q, q, 13 0, ; V3 , V4 , 3, 7, 3, 7, 13. Between D & E effective capacitance is x, 1, 11 x, x 1, 3, 3, 14. Cnet C ; when key was open q CV, 4, 4, when key was closed 3C becomes short, circuited. Net charge on C is now q ' CV, , ELECTRO STATICS & CAPACITORS, , 12 , , q q ' q , , CV, 4, , 5., , LEVEL - II (H.W), CAPACITANCE, 1., , 2., , The capacity of a condenser A is 10 F and it, is charged to a battery of 100 volt. The battery, is disconnected and the condenser A is 6., connected to a condenser B the common, potential is 40V. The capacity of B is, 1) 8 F, 2) 15 F 3) 2 F, 4) 1 F, A parallel plate capacitor has the space 7., between its plates filled by two slabs of, thickness, , d, 2, , each and dielectric constant K1, , and K 2 . d is the plate separation of the, capacitor. The capacitance of the capacitor is, , 3., , K2, , 1), , A 0, K1 K 2 , d, , 2), , 2 A 0, K1 K 2 , d, , 3), , A 0, K1 K 2 , 2d, , 4), , 2 A 0 K1 K 2, ., d, K1 K 2, , ‘A’ and ‘B’ are two condensers of capacities, 2 F and 4 F. They are charged to potential, differences of 12V and 6V respectively. If, they are now connected (+ve to +ve), the, charge that flows through the connecting wire, is, 1) 24 C from A to B 2) 8 C from A to B, 3) 8 C from B to A 4) 24 C from B to A, Force of attraction between the plates of a, parallel plate capacitor is, 1), , q2, 2 0 A, , 2), , q2, 0 A, , 3), , 2 0 A, K1 K 2 , d, , 1), , 2), , 3), , 2 0 A K1 K 2 , , , d K1 K 2 , , 4), , 2 0 d K1 K 2 , , , A K1 K 2 , , 3), , 4), , 3) CV, 0, 0, -CV, , 10, F ., 11, , Which of the combination shown in figure will, achieve the desired result, , 2), , CV CV, ,, ,0, K, K, CV CV, ,, , CV, 4) CV,, K, K, , 4), , q2, 2 0 A2, , obtain an effective capacitance of, , 2 0 A K1 K 2 , , , d K1 K 2 , , An isolated capacitor of capacitance ‘C’ is 8., charged to a potential ‘V’. Then a dielectric, slab of dielectric constant K is inserted as, shown in fig. The net charge on four surfaces, 1,2,3 and 4 would be respectively., , q, 2 0 A, , Seven capacitors each of capacitance 2 F, are to be connected in a configuration to, , 1), , 1) 0, CV,-CV, 0, , The equivalent capacitance between ‘A’ and, ‘B’ in the adjoining figure is, A, , 3F, , B, , 9F, , 9F, , 2) 0,, , CAPACITORS IN SERIES AND IN, PARALLEL, 4., , K1, , d, , 9F, , 1), , 51, F, 30, , 2) 6 F, , 3) 30 F, , 4) 12 F, , NERGY STORED IN A CONDENSER, TYPES OF CAPACITORS, A parallel plate capacitor with plate area ‘A’, and separation ‘d’ is filled with two dielectrics 9. A capacitor 4 F charged to 50V is connected, of dielectric constants K1 and K 2 . If the, to another capacitor 2 F charged to 100V ., permittivity of free space is 0 , the, capacitance of the capacitor is given by, , NARAYANAGROUP, , The total energy of combination is, , 165
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, +, , –, , 6., 7., , 1) 13.3 103 J 2) 20 103 J, 3) 5 10 3 J, , 4) 10 103 J, , +, , 8., , –, , 10. A 4 F capacitor is charged by a 200V, battery. It is then disconnected from the, supply and is connected to another uncharged, 2 F capacitor. During this process, Loss of, energy (in J) is, 1) Zero, 2) 5.33 x 10-2, -2, 3)4 x 10, 4) 2.67 x 10-2, 11. A capacitor of capacitance C has charge Q, and stored energy W . If the charge is, increased 2Q the stored energy would be, 1) W 4, 2) W 2, 3) 2W, 4) 4W, 12. The equivalent capacitance between points M, and N is, C, , 9., , q, , 2) 3, 8) 2, , 3) 2, 9) 1, , 1., , 2., C2, , 4) 3 5) 2 6) 1, 10) 4 11) 4 12) 1, , LEVEL - II ( H. W ) - HINTS, V, , 2., , 1 1, 1, A Kk 2 A o, , , C1 o 1 , k1, ;, C C1 C2, d 2, d, , 3., , 2 A o, k2, d, Due to polarization, charge on dielectric slab, 1, would be CV 1 , K, , 4., , A, A, k1 o , k2 o , 2 ; C , 2, C1 , 2, d, d, , C2 , , C1 and C2 connected in parallel.., Charge flow Q C1 V1 Vo , where Vo common potential, t thickness of metal sheet, , 5, , 166, , 3., , c1v1 c2 v2, c1 c2, , 1., , q, , LEVEL - III, , LEVEL - II ( H. W ) - KEY, 1) 2, 7) 1, , C1V1 C2V2, 1, U C1 C2 V 2, ;, C1 C2, 2, , 12. Ceq V V 0 , M, N, , N, , M, , V, , 1 C1 C 2, 2, 1, 2, 10. U 2 C C V1 V2 ; U C1 C2 V, 2, 1, 2, 11. E Q 2, , 1, , C2, 1) Infinity 2) C1 C, 1, C1C2, C1C2, 3) C C 4) C C, 1, 2, 1, 2, , C1C2, q, q CV C C .V . Vo C, 1, 2, 1, Verify the cases individually, Ceff nc ; Q Ceff V, Resultant capacitance of 9 F , 9 F and 9 F, is in parallel to 3 F ., , The time in seconds required to produce a P.D, at 20V across a capacitor at 1000 F when, it is charged at the steady rate of 200 C / sec, is, 1) 50, 2) 100, 3) 150, 4) 200, A parallel plate capacitor of capacity 5 F and, plate separation 6cm is connected to a 1V, battery and is charged. A dielectric of, dielectric constant 4 and thickness 4 cm is, introduced into the capacitor. The additional, charge that flows into the capacitor from the, battery is, 1) 2C, 2) 3C, 3) 5C, 4) 10C, The force between the plates of a parallel, plate capacitor of capacitance C and distance, of separation of the plates d with a potential, difference V between the plates, is, 1), , 4., , CV 2, 2d, , 2 2, 2) C V, , 2d 2, , 3), , C 2V 2, d2, , 4), , V 2d, C, , Two identical capacitors are connected as, show in the figure. A dielectric slab is, introduced between the plates of one of the, capacitors so as to fill the gap, the battery, remaining connected. The charge on each, capacitor will be (charge on each condenser, is q0 ; k = dielectric constant ), I, , 2q0, 1), 1 1k, 2q0, 3), 1 k, , q0, 2), 1 1k, q, 4) 0, 1 k, , 2V, B, II, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 5., , ELECTRO STATICS & CAPACITORS, , Two identical capacitors 1 and 2 are connected, in series to a battery as shown in figure., Capacitor 2 contains a dielectric slab of, dielectric constant K as shown. Q1 and Q2, are the charges stored in the capacitors. Now, the dielectric slab is removed and the, corresponding charges are Q '1 and Q ' 2 ., Then, , 1), , Q /1 K 1, , Q1, K, , 3), 6., , 7., , 8., , Q/2 K 1, , Q2, 2K, , 2), 4), , Q/2 K 1, , 2, Q2, Q/2 K, , Q2, 2, , A capacitor of capacitance 1 F withstands, a maximum voltage of 6 kV, while another, capacitor of capacitance 2 F withstands a, maximum voltage of 4 kV. If they are, connected in series, the combination can, withstand a maximum voltage of, 1) 3 kV, 2) 6 kV 3) 10 kV 4) 9 kV, Energy ‘E’ is stored in a parallel plate, capacitor ‘C 1 ’. An identical uncharged, capacitor ‘C2’ is connected to it, kept in, contact with it for a while and then, disconnected, the energy stored in C2 is, 1) E/2, 2) E/3, 3) E/4, 4) Zero, A parallel plate capacitor has area of each, plate A, the separation between the plates is, d . It is charged to a potential V and then, disconnected from the battery. The amount, of work done in the filling the capacitor, Completely with a dielectric constant k is, 1 0 AV 2 1 , 1, 1), 2 d k 2 , 2, , 1 V 2 0 A, 2), 2 kd, 2, , 1 V 0 A, 1 0 AV 1 , 1, 4), 2, 2, d K , 2 k d, A capacitor of capacitance 10 F is charged, to a potential 50 V with a battery. The, battery is now disconnected and an additional, charge 200 C is given to the positive plate, of the capacitor. The potential difference, across the capacitor will be, 1) 50 V 2) 80 V 3) 100V 4) 60 V, 3), , 9., , B, , NARAYANAGROUP, , 10. A capacitor is filled with an insulator and a, certain potential difference is applied to its, plates. The energy stored in the capacitor is, U. Now the capacitor is disconnected from the, source and the insulator is pulled out of the, capacitor. The work performed against the, forces of electric field in pulling out the, insulator is 4U. Then dielectric constant of, the insulator is, 1) 4, 2) 8, 3) 5, 4) 3, 11. A capacitor of capacitance C is charged to a, potential difference V from a cell and then, disconnected from it. A charge +Q is now, given to its positive plate. The potential, difference across the capacitor is now, Q, Q, Q, 4) V , ifV CV, 1) V 2) V 3) V , C, 2C, C, 12. A parallel plate capacitor with plates, separated by air acquires 1 C of charge, when connected to a battery of 500V. The, plates still connected to the battery are then, immersed in benzene [ k = 2.28]. Then a, charge that flows from the battery is, 1) 1.28 C 2) 2.28 C 3) 1 / 4 C 4) 4.56 C, 13. An air capacitor with plates of area 1 m2 and, 0.01 metre apart is charged with 106 C of, electricity. When the capacitor is submerged, in oil of relative permittivity 2, then the energy, decreases by, 1) 20 % 2) 50 % 3) 60 %, 4) 75 %, 14. Three uncharged capacitors of capacities, C1,C2 and C3 are connected as shown in the, figure to one another and the point. A, B and, C are at potentials V1,V2 and V3 respectively., Then the potential at O will be, A, , C1, , 1), 3), , V1C1 V2C2 V3C3, C1 C2 C3, , V1 V2 V3 , C1 C2 C3 , , V1 V2 V3, 2) C C C, 1, 2, 3, , 4), , VV, 1 2V3, C1C2C3, , O, C2, , C3, , B, , C, , 15. In the given figure the capacitor of plate area, A is charged upto charge q. The ratio of, elongations (neglect force of gravity) in, springs C and D at equilibrium position is, , k1, 1) k, 2, , 3) k1k2, , k2, 2) k, 1, k1, 4) k, 2, , k2, , k1, , D, , C, , 167
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, , 16. If metal section of shape H is inserted in 22. A capacitor is connected with a battery and, between two parallel plates as shown in figure, stores energy U. After removing the battery,, and A is the area of each plate then the, it is connected with another similar capacitor, equivalent capacitance is, in parallel. The new stored energy in each, capacitor will be, U, , U, , 3U, , 2) U, 3), 4), 1), 2, 4, 2, A 0 A 0, A 0 a, b, , 2), 1), 23., A, parallel, plate, condenser, with, a, dielectric, of, a, b, a b, dielectric constant K between the plates has, A 0 A 0, A 0, a capacity C and is charged to a potential V, , 4), 3), a, b, a b, volts. The dielectric slab is slowly removed, 17. The equivalent capacitance C AB of the circuit, from between the plates and then reinserted., The net work done by the system in this, C, shown in the figure is, process is, C, C, C, C, B, 1, 5, 4 A, 2, 1) K 1 CV, 2) CV 2 K 1 / K, 1) C, 2) C, C, C, 2, 4, 5, C, C, C, C, 3) 2 C, 4) C, C, 3) K 1 CV 2, 4) zero, 18. A solid conducting sphere of radius 10cm is 24. A fully charged capacitor has a capacitance, enclosed by a thin metallic shell of radius, C. It is discharged through a small coil of, 20cm. A charge q=20 C is given to the inner, resistance wire embedded in a thermally, sphere. The heat generated in the process is, insulated block of specific heat capacity s and, 1) 12 J, 2) 9 J, 3) 24 J, 4) zero, mass m. If the temperature of the block is, 19. A condenser of capacity 500 F is charged, raised by T , the potential difference V, across the capacitor is, at the rate of 400 C per second. The time, required to raise its potential by 40V is, mC T, ms T, 2mC T, 2msT, 1) 50 s, 2) 100 s 3) 20 s, 4) 10 s, 1), 2), 3), 4), s, C, s, C, 20. In the figure shown the effective capacity, across P and Q is (the area of each plate is ‘a’) 25. A parallel plate capacitor of capacity 100 F, P, is charged by a battery at 50 volts. The, battery remains connected and if the plates, d, K, 2, of the capacitor are separated so that the, K, d, d, K, distance between them is halved the original, 2, distance, the additional energy gives by the, battery to the capacitor in Joules is ......, Q, 1) 125 103, 2) 12.5 103, a 0 K1, K2 K3 , a 0 K 2, K1 K 3 , 1) d 2 K K 2) 2d 2 K K , 3) 1.25 103, 4) 0.125 103, , 2, 3, , 1, 3, 26. The equivalent capacity between the points, a 0 K 3, K1 K 2 , a 0 K1 K1 K 2 , 3) 3d 2 K K 4) d 2 K K , A and B in the adjoining circuit will be, , 1, 2 , , 2 3 , 2, , 1, , 3, , 21. Two capacitors C1 2 F and C2 6 F in, series, are connected in parallel to a third, capacitor C3 4 F . This arrangement is then, connected to a battery of e.m.f.=2 V, as shown, in figure. The energy lost by the battery in, charging the capacitors is, C, C, 1, , 2, , C3, , 1) 22 106 J, 32, , 2) 11106 J, 16 , 6, 10 J, , , , , 6, 3) 3 10 J 4) 3, , , 168, , 2V, , C, , B, , C, , 1) C, , 2) 2C, , 3) 3C, , 4) 4, , C, , C, , C, , C, , C, , C, , A, C, , 27. A parallel plate capacitor with air as medium, between the plates has a capacitance of 10, F. The area of the capacitor is divided into, two equal halves and filled with two media, having dielectric constant K1 2 and K 2 4 ., The capacitance will now be, 1) 10 F 2) 20 F 3) 30 F 4) 40 F, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS & CAPACITORS, , 28. The capacity of a parallel plate condenser with 34. In the following circuit two identical, capacitors, a battery and a switch(s) are, air medium is 60 F having distance of, connected as shown. the switch(s) is opened, seperation d . If the space between the plates, and dielectric of constant K 3 are, is filled with two slabs each of thinckness d 2, inserted in the condensers. The ratio of, and dielectric constants 4 and 8, the effective, electrostatic energies of the system before, capacity becomes, and after filling the dielectric will be, 1) 160 F 2) 320 F 3) 640 F 4) 360 F, 29. In the adjoining diagram, the condenser C will, S, A, B, be fully charged to potential V if, 1) 3: 1 2) 5 : 1 V, 5, , 10, , S1, , S2, , +, –, , V, , C, , 1) S1 and S2 both are open, 2) S1 and S2 both are closed, 3) S1 is closed and S2 is open, 4) S1 is open and S2 is closed., 30. The capacity between the point A and B in, the adjoining circuit wil be, , 3) 3:5, 4) 5 : 3, 35. In the given figure a capacitor of plate area, A is charged upto charge q. The mass of, each plate is m2. The lower plate is rigidly, fixed. The value of m1 if the system remains, in equilibrium is, q2, 2) m2, 1) m2 , m, 2 0 Ag, q, m2 4) 2m2, 3), 2 0 Ag, 36. One plate of a capacitor is connected to a, spring as shown in figure. Area of both the, plates is A. In steady state; separation, between the plates is 0.8d (spring was, unstretched and the distance between the, plates was d, when the capacitor was, uncharged). The force constant of the spring, is approximately, 1, , 3, , 1), 3), , 2C1C2 C3 C1 C2 , C1C2 C2C3 C3C1, 2), C1 C2 C3, C1 C2 2C3, , C1 C2 C3 C2 C1 C3 , C1C2C3, 4), C, C, , C2C3 C3C1, C1 C2 3C3, 1 2, , 31. The capacitance C AB in the given network, 5F, , 10F, , A, , 50, 2 0 AE, 4 0 AE 2, 5F, F, 1), 2), 3, 7, E, d2, d, B, 7, 2, 3, F, 6 E, AE, 3) 7.5 F 4), 5F, 10F, 3) 0 3 4) 0 3, 50, Ad, 2d, 32. In the following circuit; find the potentials at, 37., A, capacitor, is made of a flat plate of area A, points A and B is, + –, and second plate having a stair-like structure, 10V, as shown in figure. The width of each plate, is ‘a’ and the height is ‘b’. The capacitance, 1) 10V, 0V 2) 6 V, -4V, of the capacitor is, B, 3) 4V, -6V 4) 5V, -5V A, 33. The potential difference between the points A, and B in the following circuit in steady state, 1F, will be 3F, B, , 1) 7 F, , 2), , d, , 1 F, , 3F, , 3a, , 1F, 10, , A, , 200, , 1) VAB 100 volt, 3) VAB 25 volt, NARAYANAGROUP, , 100V, , 2 A 0, 1) 3 d b , , 2), , A 0 3 d 2 6bd 2 b 2 , , C, , 2) VAB 75 volt, 4) VAB 50 volt, , A 0 d 2 2bd b 2 , , 3) 3d d b d 2b 4), , , , , 3 d b d d 2 b , , 2 A 0 d 2 2bd b 2 , 3d d b d 2b , 169
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS & CAPACITORS, 38. A parallel plate capacitor of capacitance C is, connected to a battery and is charged to a, potential difference V. Another capacitor of, capacitance 2C is similarly charged to a, potential difference 2V. The charging, battery is now disconnected and the, capacitors are connected in parallel to each, other in such a way that the positive terminal, of one is connected to the negative terminal, of the other. The final energy of the, configuration is, 3, 35, 9, 2, CV 2 4) CV 2, 1) zero 2) CV 3), 2, 6, 2, 39. Two identical capacitors, have the same, capacitance C. One of them is charged to, potential V1 and the other to V2.The negative, ends are also connected, the decrease in, energy of the combined system is, , , , 2, , 1) 1/ 4C V1 V2, , 2, , , , , , 2, , 2) 1/ 4C V1 V2, , 2, , 2, , , , 2, , 3) 1/ 4C V1 V2 4) 1/ 4C V1 V2 , 40. Consider the situation shown in the figure., The capacitor A has a charge q on it whereas, B is uncharged. The charge appearing on, the capacitor B a long 7 time after the switch, q, –, is closed is:, +, –, –, –, –, –, –, –, , +, +, +, +, +, +, , S, , 1) Zero, 2) q /2, 3) q, 4) 2q, A, B, 41. Find the capacitance of a system of two, identical metal balls of radius a if the distance, between their centres is equal to b, with, b>>a. The system is located in a uniform, dielectric with permittivity K., 2) 4 0 Ka, 1) 0 Ka, 3) 2 0 Ka, , 4) 2/3 0 Ka, , LEVEL - III KEY, 1) 2, 9) 4, 15) 2, 21) 2, 27) 3, 33) 3, 39) 3, , 2) 3 3) 1 4) 1, 10) 3 11) 3, 16) 4 17) 1, 22) 3 23) 4, 28) 2 29) 3, 34) 3 35) 3, 40) 1 41) 3, , 5) 3 6) 4 7) 3, 12) 1 13) 2, 18) 2 19) 1, 24) 4 25) 1, 30) 1 31) 1, 36) 1 37) 2, , 8) 4, 14) 1, 20) 1, 26) 2, 32) 2, 38) 2, , LEVEL - III - HINTS, 1., , dq c.dv, , dt, dt, , 2., , c, /, q1 cv ; q2 C V , c / , , 170, , d t, d, , t, K, , 3., , F, , QE CV V , , 2, 2 d , , C0 k, ; q Ceff V, 1 k, CE, ', ', ; Before the slab is removed, 5. Q1 Q2 , 2, k , C1 C and C2 kC ; Cnet k 1 C, , , ', Q2 k 1, , Q2, 2k, 6. V1 Vmax1 , V2 Vmax 2, 1, /, 2, 7. U 2 C 2 V common, 2, 8. Work done = decrease in energy, 1 0 A 2 0 Av 2 1 , E, , E, , v , 1, ie w = 1, 2, 2 d, 2d k , 9. q0 CV 500 C, 700 q, q, 500 q, q, , , , 2 A 0 2 A 0 2 A 0, 2 A 0, q, q 600 C ; V =60V, C, 2, 1 q2, 1q, U, , 4, U, , 10. U , ;, 2 C0, 2C, 2, C, 1q, 5U , ; k C 5, 2 C0, 0, Q, CV, Q, CV, ', 2, V Ed , , 11. E , ;, A 0, 2, 0, A 0, d, Q, CV, Q, = 2, V , C, 2C, 12. Q0 C0V0 ; Q CV0 KC0V0, , 4., , Ceff , , Q Q Q0 K 1 C0V0, U, /, 13. U , K, 14. q1 q2 q3, V1 V0 C1 V0 V2 C2 V0 V3 C3, C V C2V2 C3V3, V0 1 1, C1 C2 C3, x1 k2, 15. Fe k1 x1 k 2 x2 x k, 2, 1, 16. Net space between metal plates is a-b, , ; q q2 q1, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 5., , LEVEL - V, SINGLE ANSWER QUESTIONS, CHARGE AND ITS PROPERTIES, 1., , qQ, A) 8 2 r 2, 0, , The linear charge density of a thin metallic rod, varies with the distance ‘x’ from one end as, , 0 x 2 0 x l . The total charge on the, rod is, 6., l3, l3, 2 l 3, l3, B) 0, C) 0 D) 0, A) 0, 3, 4, 3, 2, The linear charge density of a uniform, semicircular wire varies with ' ' shown in the, , 2., , figure as 0 cos . If R is the radius of the, loop, the charge on the wire is, +, , +, , +, , +, , +, , +, , +, +, , +, , +, , +, , +, , +, , +, , +, , +, , +, , +, , +, , +, , +, +, , 2, , +, , +, +, , +, , +Q (fixed), , +, , , , 1, , curvature is k 4 , , , 0, , , +, , +, , + +, , +, , + +, , +, , A), , kQq sin, , , 2, , R 2, , , 2kQq cos, 2, C), R 2, 172, , B), D), , 2kQq sin, R 2, , , 2, , +, , +, +, , , 2, , 2kQq tan, , 1 Q1Q2, C) 4 , D) zero, 2, 0 d, A positive point charge +Q is fixed in space. A, negative point charge -q of mass m revolves, around a fixed charge in elliptical orbit. The, fixed charge +Q is at one focus of the, ellipse.The only force acting on negative, charge is the electrostatic force due to positive, charge.Then which of the following statement, is true., m -q, , A charge ‘+Q’ is uniformly distributed along, the circular arc of radius ‘R’ as shown in the, figure. The magnitude of the force experienced, by the point charge +q placed at the centre of, , , 1 Q1Q2, B) 4 , 2, 0 d, , +, , , , COULOMB’S LAW & PRINCIPLE, OF SUPERPOSITION, 4., , radius R charged separately to Q1 and Q2 ,, are at a distance ‘d’ between their centres, (d>2R). The force between them will be, , +, , 0 R 2, A) 2 0 R B), 2, 0 R 2, C), D) 0 R 2, 3, , +, , with ' ' (shown in figure) as 0 cos ., The total charge on the sphere is, , qQ, B) 4 2 r 2, 0, , qQ , qQ, , C) , D), 2, 2, 2 0 r 2, 4 0 r , Two metallic spheres each of mass M and, , 1 Q1Q2, A) 4 , 2, 0 d, , A) 0 R B) 30 R, , R, 20 R, 7., C), D) 20 R, 3, For the hemisphere of radius ‘R’ shown in the, figure the surface charge density ' ' varies, , 3., , A thin circular wire ring of radius ‘r’ has a, charge ‘Q’. If a point charge ‘q’ is placed at, the centre of the ring then the tension in the, wire will increase by, , R, , , , R, , 8., , A) Linear momentum of negative point charge is, conserved., B) Angular momentum of negative point charge, about fixed positive charge is conserved., C) Total kinetic energy of negative point charge is, conserved., D) Electrostatic potential energy of system of both, point charges is conserved., Find the force experienced by the semicircular, rod charged with a charge q, placed as shown, in figure. Radius of the wire is R and the line, of charge with linear charge density is passing, through its centre and perpendicular to the, plane of wire., , R 2, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, +, , +, , ELECTRIC FIELD AND LINES OF FORCE, 9., , (D), , Z, , ++ +, +, R, +, +, +, , +, , q, q, (A) 2 2 R (B) 2 R, 0, 0, q, q, (C) 4 2 R (D) 4 R, 0, 0, , (C), , Z, , X, , X, , ELECTRIC POTENTIAL &, POTENTIAL ENERGY, , Two identical point charges are placed at a 11. A conducting bubble of radius “a” and, separation of l. P is a point on the line joining, thickness t t a has a potential ‘V’. Now, the charges, at a distance x from any one, the bubble collapses into a droplet (of radius, charge. The field at P is E and E is plotted, R). The electric potential on the droplet is, against x for values of x from close to zero to, 1, 1, 1, slightly less than l. Which of the following best, a 2, 2a 3, a 3, A) V B) V C) V D)V, represents the resulting curve?, 3t , 3t , 3t , E, E, (A), (B), 12. In the following diagram a uniform electric field, of magnitude 325 NC-1 is directed along ve, y-direction. The co-ordinates of point A are (x, x, 20, -30) cm and those of point B are (40, 50)cm., O, L, O, L, The potential difference along the path shown, is, y, , (C) E, , (D) E, L, , O, , x, , C, , L, , O, , B, , x, , x, , E, , A, , 10. Two infinitely large charged planes having, uniform surface charge density + and – are, placed along x-y plane and yz plane, A) 260V B) 260V C) 130V D) 130V, respectively as shown in the figure. Then the 13. An infinite by long plate has surface charge, density . As shown in the fig. a point charge, nature of electric lines of force in x-z plane is, q is moved from A to B. Net work done by, given by :, Z, –, electric field is :, +, , X, , , , B(X1,O), (B), (A), , A(X2,O), , Z, , Z, , X, , X, , q, , (A) 2 (X1 – X2), 0, q, , (C) (X2 – X1), 0, NARAYANAGROUP, , q, , (B) 2 (X2 – X1), 0, q, , (D) ( 2 r r ), 0, 173
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 14. A conducting disc of radius R rotates about its, axis with an angular velocity . Then the, potential difference between the centre of the, disc and its edge is (no magnetic field is, present) :, (A) zero (B), , me 2R2, m R 3, eme R2, (C) e, (D), 2e, 3e, 2, , q, A) 2 ln 2, 0, , q, B) 4 ln 2, 0, , q, C) 4 ln 2, 0, , q, D) 2 ln 0.5 , 0, , 19. The diagram shows three infinitely long, uniform line charges placed on the X, Y and Z, 15. A neutral conducting spherical shell is kept, axis. The work done in moving a unit positive, near a charge q as shown. The potential at point, charge from (1, 1, 1) to (0, 1, 1) is equal to, P due to the induced charges is, Y, 3, r, , q, , C, , r', , P, , X, 2, , (A), , kq, r, , (B), , kq, r', , (C), , , , kq kq, kq, –, (D), r, r', CP, , Z, , 16. The following diagram represents a semi, circular wire of linear charge density, , (A) (l ln 2) / 2 0, (B) (l ln 2) / 0, (C) (3l ln 2) / 2 0 (D) None, 0 sin where 0 is a positive constant. 20. Two positively charged particles X and Y are, initially far away from each other and at rest., The electric potential at ‘o’ is, X begins to move towards Y with some initial, , , 1, velocity. The total momentum and energy of, take k , , the system are p and E., 4 0 , dQ, , (A) If Y is fixed, both p and E are conserved., (B) If Y is fixed, E is conserved, but not p., , (C) If both are free to move, p is conserved but not, E., (D) If both are free, E is conserved, but not p., k 0 cos , A) k 0 sin B), C)zero D) k 0 cos 21. A metal ball of radius R is placed, 2, concentrically inside a hollow metal sphere of, 17. The electric potential at the centre of, inner radius 2R and outer radius 3R. The ball, hemisphere of radius ‘R’ having uniform, is given a charge +2Q and the hollow sphere a, surface charge density is, total charge – Q. The electrostatic potential, R, 2 R, R, R, energy of this system is :, A) , B) , C) 2 D) 3 , 0, 0, 0, 0, 18. For an infinite line of charge having linear, charge density ' ' lying along x-axis, the work, done in moving a charge ‘q’ from C to A along, CA is, B, C, , a, , A, +, , 174, , a, +, , 7Q 2, (A), 240 R, , 5Q 2, (B), 160 R, , 5Q 2, (C), 80 R, , (D) None, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , RELATION BETWEEN E & V, 22. Figure shows an electric line of force which, curves along a circular arc. The magnitude of, electric field intensity is same at all points on, this curve and is equal to E. If the potential at, A is V, then the potential at B is, , (B) V – E2R sin, , A, , 2, , b, , (C) 15 volt (D) 0, , 27. Three concentric metallic spherical shells A,, B and C of radii a, b and c (a < b < c) have, surface charge densities – , + , and, – respectively. The potential of shell A is :, B, (A) 0 [a + b – c] (B) 0 [a – b + c], , E, , (A) V – ER , , a, , (A) 10 volt (B) 5 volt, , R, , , , (C) 0 [b – a – c] (D) none, 28., There are four concentric shells A, B, C and D, , (C) V + ER (D) V + 2ER sin, of radii a, 2a, 3a and 4a respectively. Shells B, 2, and D are given charges +q and –q, 23. Uniform electric field of magnitude 100 V/m in, respectively. Shell C is now earthed. The, space is directed along the line y = 3 + x. Find, potential difference VA – VC is :, the potential difference between point A (3, 1), Kq, Kq, Kq, Kq, & B (1, 3), (A), (B), (C), (D), 2a, 3a, 4a, 6a, (A) 100 V (B) 200 2 V (C) 200 V (D) 0, 29. At distance 'r' from a point charge, the ratio, 24. Electrical potential 'V' in space as a function, 1, 1, 1, of co-ordinates is given by, V = + + ., x, y, z, , Then the electric field intensity at (1, 1, 1) is, given by :, ˆ, (A) – (iˆ ˆj k), , (B) ˆi ˆj kˆ, , (C) zero, , 1 ˆ ˆ ˆ, ( i j k), (D), 3, , u, , v2, , (where 'u' is energy density and 'v' is, , potential) is best represented by :, (A), 2, u/v, , (B), 2, u/v, r, , (C), , r, , (D), , 25. A graph of the x component of the electric field, u/v, u/v, as a function of x in a region of space is shown., The Y and Z components of the electric field, r, r, are zero in this region. If the electric potential 30. Six charges of magnitude + q and –q are fixed, is 10 V at the origin, then potential at x = 2.0, at the corners of a regular hexagon of edge, length a as shown in the figure. The, m is :, electrostatic interaction energy of the charged, 20, –q, particles is :, +q, 2, , EX(N/C), , 1 2 3, , 2, , x(m), , –q, , -20, (A) 10 V (B) 40 V (C) – 10 V (D) 30 V, 26. If the electric potential of the inner metal, (A), sphere is 10 volt & that of the outer shell is 5, volt, then the potential at the centre will be :, (C), NARAYANAGROUP, , +q, , 4 5, +q, q2 3 15 , – , , 0 a 8, 4 , q2, 0 a, , 3 15 , – , , 2 , 4, , (B), , –q, , q2 3 9 , – , , 0 a 2 4 , , (D), , q2, 0 a, , 3 15 , – , , 8 , 2, 175
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, , MOTION OF CHARGED PARTICLE, IN UNIFORM & NON UNIFORM, ELECTRIC FIELD, , 34. In space a horizontal EF (E = (mg)/q) exists as, shown in figure and a mass m attached at the, end of a light rod. If mass m is released from, the position shown in figure find the angular, 31. A particle of mass ‘m’ carries an electric, velocity of the rod when it passes through the, positive charge Q and is subjected to the, bottom most position, combined action of gravity g and a uniform, horizontal electric field of strength E. It is, 0, =45, projected with speed u in the vertical plane, mg, , E, parallel to the electric field and at an angle , q, g, 2g, to the horizontal. The maximum distance the, m, (A), (B), particle travels horizontally when it is at the, l, l, m, same level as its starting point is, 3g, 5g, +q, (C), (D), u2, u2 EQ mg , l, l, B), A), mg 2, g, 35. A unit positive point charge of mass m is, 2, 2, projected with a velocity V inside the tunnel, u, EQu, 2, 2, C) mg 2 EQ mg EQ, D) mg 2, as shown. The tunnel has been made inside a, uniformly charged non conducting sphere. The, 32. The charge and mass of two particles are +Q,, minimum velocity with which the point charge, M and -q, m respectively. The particles, separated by a distance L, are released from, should be projected so that it can reach the, rest in a uniform electric field E. The electric, opposite end of the tunnel, is equal to, field is parallel to line joining both the charges, and is directed from negative to positive, R/2, charge. For the separation between particles, to remain constant, the value of L is (A) [rR2/4m ]1/2, 0, , , , , , , 1 , k , , 4 0 ., , , (B) [rR2/24m 0]1/2 (C) [rR2/6m 0]1/2, , (D) zero because the initial and the final points are, at the same potential., 36. A particle of charge – q & mass m moves in a, mMKQq, mMKQq, circle of radius r around an infinitely long line, C) E qM Qm, D) E QM qm, charge having linear charge density + . Then, 33. A bullet of mass m and charge q is fired towards, time period will be., a solid uniformly charged sphere of radius R, +, and total charge + q. If it strikes the surface of, sphere with speed u, find the minimum speed u, -q, so that it can penetrate through the sphere., r, (Neglect all resistive forces or friction acting, on bullet except electrostatic forces), M m KQq, A) E qM Qm, , M m KQq, B) E qm QM , , q, +q, m, , 176, , u, , + +, + +, +, +, + +, + +, +, +, +, R +, + + +, , +, +, +, +, , (A), , q, 2 0 mR, , (B) 4 mR, 0, , q, , (C), , q, 8 0 mR, , (D), , 3q, 4 0 mR, , (A) T = 2 r, (C) T =, , m, 2k q, , 4 2 m 3, r, (B) T =, 2k q, 2, , 1 2k q, 2 r, m, 1, , m, , 1, , (D) T = 2 r 2k q where k = 4 , 0, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 37. A point mass of charge ‘-q’ and mass ‘m’ is, released from rest from a distance ‘R’ along, the axis of a uniform circular disc of charge, ‘Q’. Assuming gravity free situation the, velocity attained by point mass when it reaches, centre of disc is (Radius of disc-R), -q, m, , +, +, +, , +, , +, , +, ++, , +, , +, , +, , + +, , +, , +, , +, Q, , , , , , 2Qq 2 1, , A) mR, 0, , , , , 1/2, , , , , 1/2, , , , Qq 2 1, , C) 2 mR, 0, , , , , , , , , Qq 3 1, , D) 2 mR, 0, , , , , , 1/2, , , , Qq 2 1, , B) mR, 0, , , 4, 2, , B), C), D) , 3, 3, 3, 41. In the diagram shown the charge +Q is fixed,, another charge +2q is projected from a, distance ‘R’ from a fixed charge. Minimum, separation between the charges, if the velocity, at this moment is half of velocity of projection, in magnitude (Assume gravity to be absent ), , A), , R, + +, , +, , +, , direction of first ball changes by 600 and, velocity magnitude is reduced by half, the, direction of velocity of second ball changes by, 900 then magnitude of charge to mass ratio of, second ball if that of first ball is ' ' is, , , , 1/2, , 0, , +Q, , , , , R, , 30, m, 2q, , +, , +, , +, , R, 38. The adjoining figure represents a semi-circular, 3R, A), B), C), D) 4R, 3R, loop of charge ‘Q’ and of radius ‘R’. The, 2, 2, minimum velocity with which a point mass ‘m’, ELECTRIC DIPOLE, of charge ‘-q’ should be projected so that it, becomes free from the influence of the loop is 42. The dipole moment of a system of charge +q, distributed uniformly on an arc of radius R, subtending an angle /2 at its centre where, another charge -q is placed is :, +, , +, , +, , +, , qR, 2qR, 2 2qR, 2qR, (B), (C), (D), A), B), , , , , 43. As shown in the figure an electric dipole lies, Qq, Qq, at a distance ' x ' from the centre of a charged, C), 2 0 mR D) 4 0 mR, ring of radius ' R ' with charge ' Q ' uniformly, 39. A charge ‘q’ of mass ‘m’ is projected from a, distributed over it . The net force acting on, long distance with speed ‘v’ towards another, the dipole is, stationary particle of same mass and charge., The distance of closest approach of the, particles is, +, , R, , (A), , +, , Qq, 0 mR, , 2Qq, 0 mR, , +, , +, , +, , +, , +, , +, , -q, , q, , A), , 2, , 2 0 mv, , 2q, 2, , B), , 3q 2, , C), , 2 0 mv 2, , x, , 0 mv 2, q2, , D), , 0 mv 2, , 2a, , aQq, A) 4 0, , R 2 2 x2, 5/2, , R2 x2 , , 40. Two balls of charges q1 and q2 in magnitude, have velocities of equal magnitude (v) and, aQq R 2 2 x 2, same direction. After a uniform electric field, is applied for a certain interval of time the C) 4 0 R 2 x 2 5/2, , , , NARAYANAGROUP, , -q, , 2, , , , aQq, B) 2 0, aQq, D) 2 0, , R 2 2 x2, 5/2, , R2 x2 , , R 2 2 x2, 5/2, , R2 x2 , , 177
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 44. Two point dipoles p k̂ &, , p, k̂ are located at (0,, 2, , 0, 0) & (1m, 0, 2m) respectively. The resultant, electric field due to the two dipoles at the point, (1 m, 0, 0) is :, 9p, , – 7p, , (A) 32 k̂, 0, , (B) 32 k̂, 0, , 7p, , (C) 32 k̂, 0, , GAUSS LAW, 48. Two infinite line charges, each having uniform, charge density , pass through the mid points, of two pairs of opposite faces of a cube of edge, ‘L’ as shown in figure. The modulus of the total, electric flux due to both the line charges, through the face, H, , (D) none of these, , 45. A dipole of dipole moment p is kept at the, centre of a ring of radius R and charge Q. The, dipole moment has direction along the axis of, the ring. The resultant force on the ring due to, the dipole is :, (A) zero, , (B), , G, , D, , kpQ, R3, , (C), , a, E, A, , L, , 2kpQ, , L, , A) CDHG is 2 , 0, , R3, , B) AEHD is 3 , 0, , L, , kpQ, , F, B, , L, , C) ABCD is 2 , D) CDHG is 3 , (D) 3 only if the charge is uniformly distributed, 0, 0, R, 49. The figure shows an imaginary cube of length, on the ring., L/2 and a uniformly charged rod of length L, 46. A point electric dipole is placed at the origin, touching the centre of the right face of the cube, ' O' and is directed along the ve x axis . At, normally. At time t = 0, the rod starts moving, a point ' P' faraway from the dipole , the, to the left slowly at a constant speed ‘v’. The, electric field is parallel to the y axis . The, electric flux (F) through the cube is plotted, against time (t). The correct graph showing, line 'OP' makes an angle ' ' with the x axis ., the variation of flux with time is, Then, A ) tan 3, , B) tan 2, , 1, 2, 47. A large sheet carries uniform surface charge, density ' ' A rod of length '2l ' has a linear, charge density ' ' on one half and on the, other half . The rod is hinged at mid point, 'O' and makes an angle ' ' with normal to the, sheet . The torque experienced by the rod is, C) 45, , D) tan , , +, , 178, , L/2, , (A) , , , , (B) , , t, , +, , l 2, l 2, cos, , cos2 , (A), (B), 2 0, 0, l 2, sin (D) zero, (C), 2 0, , L, , (C) , , t, , (D) , , -, , t, , t, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 50. A ring of radius R is placed in the plane with 53. A point charge ' q ' is placed at the centre of a, its centre at origin and its axis along the xcylinder of length ‘L’ and radius ‘R’ the electric, axis and having uniformly distributed positive, flux through the curved surface of the cylinder, charge. A ring of radius r(<<R) and coaxial with, is, the larger ring is moving along the axis with, , , , q, q, L, q, L, q, L, B) , C) , D), constant velocity then the variation of A) , 0 L2 R2 0 L2 2R2 0 L2 4R2 , 2 0, electrical flux ( ) passing through the smaller, ring with Position will be best represented by : 54. Two charges q1 and q2 are placed at A and, y, B respectively. A line of force emanates from, q1 at angle ' ' with line AB and it terminates, r, , at an angle ' ' as shown then, R, , V, , (A), , X, A, , (B), , , , , , , -q2, , +q1, , , , B, , q, , , , q, , , , q, , , , 1, 1, 1, 1, (A) sin q sin 2 (B) 2sin q sin , 2, , , 2, , , (C), , q, , (D), , , , 1, 1, 2, 1, (C) 2sin q sin 2 (D) 2sin q sin 2 , 1, , , 2, , , , 55. The potential inside a charged ball depends, only on the distance ‘r’ of the point from its, centre according to the following relation, 51. A point charge ‘q’ is placed at a distance ' l ', V Ar 2 B volts . The charge density inside, from the centre of a disc of radius ‘R’ along its, the ball will be, axis. The electric flux through the disc is, 3 0 A, 6 0 A, A), , 6, , A, B), 6, , A, C), D), , , , , 0, 0, q, l, q, l, r, r, A) 1 2 2 B) 2 1 2 2 , 56. Two very large thin conducting plates having, 0, 0, R l , R l , same cross-sectional area are placed as shown, in figure. They are carrying charges 'Q' and, , , q 1, l, q 1, l, '3Q' respectively. The variation of electric field, C) 2 2 2 2 D) 2 2 2 , as a function of x (for x = 0 to x = 3d) will be, 0, 0, R l , R l , best represented by., 52. Two point charges ' q ' and ' q ' aree, Y, separated by '2l ' . The electric flux of electric, Q, 3Q, field strength vector through a circle, perpendicular to the line joining the charges, as shown in the figure is, X, (d, 0) (2d, 0) (3d, 0), , (A), , (B), , E, , +q, , 60, , 0, , 60, , 0, , E, , -q, d, , 2d, , 3d X, , (C), , l, , q, B) , 0, , NARAYANAGROUP, , 2q, C) , 0, , 2d, , D) Zero, , 3d X, , 2d, , 3d X, , E, , 2l, , q, A) 2, 0, , 2d, , (D), E, , l, , d, , d, , 3d X, , d, , 179
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, , (A) Only induced charge on outer surface will, redistribute., (B) Only induced charge on inner surface will, redistribute., (C) both induced charge on outer and inner surface, will redistribute., a, a, a, (D) force on charge q placed inside the cavity will, C), D), A) 0, B), change., 2 0, 30, 0, 61., The following diagram shows three metal balls., 58. Two spherical, nonconducting, and very thin, Ball ‘A’ is charged to ‘Q’ coulombs and B,C, shells of uniformly distributed positive charge, are uncharged. The charges on balls B and C, Q and radius d are located a distance 10d from, when the switches S1 and S 2 are closed are, each other. A positive point charge q is placed, inside one of the shells at a distance d/2 from, respectively, the center, on the line connecting the centers, B, 2a, S, of the two shells, as shown in the figure. What, is the net force on the charge q?, , 57. Inside a ball charged uniformly with volume, density ' ' there is a spherical cavity. The, centre of the cavity is displaced with respect, to the centre of the ball by a distance a. Then, the field strength inside the cavity is, , 1, , Q, , 5a, , Q, , 5a, , d, , C, , A, , a, , d/2, , S2, , 5a, Q, , 10d, , qQ, qQ, (A) 361 d 2 to the left (B) 361 d 2 to the right, 0, 0, 362qQ, 360qQ, (C) 361 d 2 to the left (D) 361 d 2 to the right, 0, 0, 59. A point charge ‘q’ is located at the centre of, the spherical layers of uniform isotropic, dielectric with relative permitivity ‘K’. The, inside radius of the layer is equal to ‘a’ and, the outside radius is equal to ‘b’. The, electrostatic energy inside the dielectric layer, is, , 1, q2 1 1 , , A), 4 0 2 K 2 a b , C), , 1 q2 1 1 , , B), 4 0 K 2 b a , , 1, q2 1 1 , , 4 0 2 K 2 b a , , D), , 1 q2 2 1 , , 4 0 K 2 b a , , PROPERTIES OF CONDUCTORS &, CHARGE DISTRIBUTION, , 8Q 3Q, 3Q 8Q 8Q 3Q, 3Q 8Q, ,, ,, ,, ,, B), C), D), 23 23, 23 23, 23 23, 23 23, 62. In the following diagram the conducting shells, are concentric. The amount of charge that flows, through the switch(S) after closing it is, , A), , C, B, A, S, 2R, R, Q, , 4R, , -2Q, 3Q, , Q, Q, Q, C), D), 2, 3, 4, 63. In the following diagram the plates are, conducting and are held parallel,then amount, of charge that flows through the switch after, closing it is, , A) Q, , B), , Q, , -2Q, , 3Q, , 4R, , 60. The figure shows a charge q placed inside a, cavity in an uncharged conductor. Now if an, external electric field is switched on :, d, , 2d, , 4d, , c, q, , A) Q, 180, , B), , 5Q, 2, , C), , 7Q, 2, , D) 2Q, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 64. Two concentric thin conducting spherical shells 68. Four charges of 1 mC, 2 mC, 3 mC, and – 6mC, are placed one at each corner of the square of, having radii R and 2R are as shown in the, side 1m. The square lies in the x-y plane with, figure. A charge +Q is given to the shell ‘A’, its centre at the origin., and -4Q is given to the shell ‘B’. If A and B, (A) The electric potential is zero at the origin., are connected by a thin conducting wire then, (B) The electric potential is zero everywhere along, the amount of heat produced after the, the x-axis only on the sides of the square which are, connection is, parallel to x and y axis., -4Q, (C) The electric potential is zero everywhere along, Q, the z-axis for any orientation of the square in the xy plane., R, (D) The electric potential is not zero along the zaxis except at the origin., 2R, A, 69. Two fixed charges 4Q (positive) and Q, B, (negative) are located at A and B, the distance, AB being 3 m., 3Q 2, Q2, 5Q 2, Q2, +4Q, -Q, A), B), C), D), 4 0 R, 16 0 R 8 0 R, 8 0 R, 3m, B, A, (A) The point P where the resultant field due to, MULTI ANSWER QUESTIONS, both is zero is on AB outside AB., 65. Five balls numbered 1 to 5 are suspended using, (B) The point P where the resultant field due to, separate threads. Pairs (1,2), (2,4) and (4,1), both is zero is on AB inside AB., show electrostatic attraction while pairs (2,3), (C) If a positive charge is placed at P and displaced, and (4,5) show repulsion. Therefore ball 1 must, slightly along AB it will execute oscillations., be, (D) If a negative charge is placed at P and displaced, A) positively charged B) negatively charged, slightly along AB it will execute oscillations., C) neutral, D) made of metal, 70. Two identical charges +Q are kept fixed some, 66. A point charge ‘Q’ is placed at origin ‘O’. Let, distance apart. A small particle P with charge, , , q is placed midway between them. If P is given, E A , EB and Ec represent electric fields at A,B, a small displacement D, it will undergo simple, and C respectively. If coordinates of A,B and, harmonic motion if, C are respectively 1, 2,3 m, 1,1, 1 m and, (A) q is positive and D is along the line joining the, charges., 2, 2, 2 m then, (B) q is positive and D is perpendicular to the line, , , , joining the charges., A) E A E B, B) E A || EC, (C) q is negative and D is perpendicular to the line, , , , , joining the charges., C) EB 4 EC, D) EB 8 EC, (D) q is negative and D is along the line joining the, 67. A negative point charge placed at the point A, charges., is, 71. Select the correct statement : (Only force on, a particle is due to electric field), a, a, (A) A charged particle always moves along the, electric line of force., +2q, +2q, A, (B) A charged particle may move along the line of, force, (A) in stable equilibrium along x-axis, (C) A charge particle never moves along the line of, (B) in unstable equilibrium along y-axis, force, (C) in stable equilibrium along y-axis, (D) A charged particle moves along the line of force, (D) in unstable equilibrium along x-axis, only if released from rest., NARAYANAGROUP, , 181
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, , 72. Two point charges Q and – Q/4 are separated 76. A thin-walled, spherical conducting shell S of, radius R is given charge Q. The same amount, by a distance x. Then, of charge is also placed at its centre C. Which, Q, -Q/4, x, of the following statements are correct?, (A) potential is zero at a point on the axis which is, (A) On the outer surface of S, the charge density is, x/3 on the right side of the charge – Q/4, Q, (B) potential is zero at a point on the axis which is, ., x/5 on the left side of the charge – Q/4, 2R 2, (C) electric field is zero at a point on the axis which is at, (B) The electric field is zero at all points inside S., a distance x on the right side of the charge – Q/4, (C) At a point just outside S, the electric field is, (D) there exist two points on the axis where electric, double the field at a point just inside S., field is zero., (D) At any point inside S, the electric field is inversely, 73. Three point charges Q, 4Q and 16Q are placed, proportional to the square of its distance from C., on a straight line 9 cm long. Charges are placed 77. A hollow closed conductor of irregular shape, in such a way that the system has minimum, is given some charge. Which of the following, potential energy. Then, statements are correct?, (A) 4Q and 16Q must be at the ends and Q at a, (A) The entire charge will appear on its outer, distance of 3 cm from 16Q., surface., (B) 4Q and 16Q must be at the ends and Q at a, (B) All points on the conductor will have the same, distance of 6 cm from 16Q., potential., (C) Electric field at the position of Q is zero., (C) All points on its surface will have the same, Q, charge density., (D) Electric field at the position of Q is, ., (D) All points near its surface and outside it will, 40, have the same electric intensity., 74. Two infinite sheets of uniform charge density, + and – are parallel to each other as 78. Three point charges are placed at the corners, shown in the figure. Electric field at the, of an equilateral triangle of side L as shown in, the figure., 2q, –, +, +S, , –S, L, , L, , –, , (A) points to the left or to the right of the sheets is, zero. (B) midpoint between the sheets is zero., , (C) midpoint of the sheets is and is directed, 0, , (A) The potential at the centroid of the triangle is, zero., (B) The electric field at the centroid of the triangle, is zero., , towards right., , (C) The dipole moment of the system is 2 qL, , +q, , L, , +q, , , (D) The dipole moment of the system is 3 qL ., (D) midpoint of the sheet is 2 and is directed, 79. An electric dipole is placed at the centre of a, 0, sphere. Mark the correct answer, towards right., (A) the flux of the electric field through the sphere, 75. If we use permittivity e, resistance R,, is zero, gravitational constant G and voltage V as, (B) the electric field is zero at every point of the, fundamental physical quantities, then, 0, 0, 0, 0, sphere., (A) [angular displacement] = e R G V, –1, –1, 0, 0, (C) the electric potential is zero everywhere on the, (B) [Velocity] = e R G V, (C) [dipole moment] = e1R0G0V1, sphere., 1, 0, 0, 2, (D) [force] = e R G V, (D) the electric potential is zero on a circle on the, surface., 182, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 80. An electric field converges at the origin whose, (A) With the switch S open, the potential of the inner, magnitude is given by the expression, sphere is equal to that of the outer., E = 100rN/C, where r is the distance measured, (B) When the switch S is closed, the potential of, from the origin., the inner sphere becomes zero., (A) total charge contained in any spherical volume, (C) With the switch S closed, the charge attained, with its centre at the origin is negative., by the inner sphere is – q/3., (B) total charge contained in any spherical volume,, (D) By closing the switch the capacitance of the, irrespective of the location of its centre, is negative., system increases., (C) total charge contained in a spherical volume, 83. X and Y are large, parallel conducting plates, of radius 3 cm with its centre at the origin, close to each other. Each face has an area A., has magnitude3 ×10–13C., X is given a charge Q. Y is without any charge., (D) total charge contained in a spherical volume of, Points A, B and C are as shown in figure., radius 3 cm with its centre at the origin, X, Y, has magnitude 3 × 10–9 C., 81. A conducting sphere A of radius a, with charge, Q, is placed concentrically inside a conducting, B, C, A, shell B of radius b. B is earthed. C is the, common centre of the A and B., B, Q, , A, a, , Q, (A) The field at B is 2 A, 0, , b, , Q, (A) The field at a distance r from C, where, (B) The field at B is A, 0, 1 Q, (C) The fields at A, B and C are of the same, a r b is 4 r 2 ., 0, magnitude., (B) The potential at a distance r from C, where, (D) The field at A and C are of the same magnitude,, 1 Q, but in opposite directions., a r b , is 4 r ., 0, 84. A particle of mass m and charge q is thrown in, (C) The potential difference between A and B is, a region where uniform gravitational field and, electric field are present. The path of particle, 1, 1 1, Q , (A) may be a straight line (B) may be a circle, 4 0 a b , (C) may be a parabola (D) may be a hyperbola, (D) The potential at a distance r from C, where, 85. Two large, parallel conducting plates are, 1, 1 1 , placed close to each other. The inner surface, a r b is, 4 Q r b ., of the two plates have surface charge densities, 0, and . The outer surfaces are without, 82. Two thin conducting shells of radii R and 3R, charge., The electric field has a magnitude of, are shown in the figure. The outer shell carries, a charge + Q and the inner shell is neutral., , A) in the region between the plates, The inner shell is earthed with the help of a, 0, switch S., , B) in the region outside the plates, 0, 3R, R, S, , NARAYANAGROUP, , 2, C) in the region between the plates, 0, D) Zero in the region outside the plates, 183
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, , 86. The electric potential in the region of space is 90. Two infinite, parallel, non- conducting sheets, carry equal positive charge density . One is, given by: V x A Bx Cx 2 , where V is, placed on the other at distance x=a. Take, potential V = 0 at x = 0. Then:, in volts, x is in meters and A, B, C aree, constants. Then,, A) For 0 x a, potential Vx 0, , , A) E varies linearly with x B)the unit of E is N/C, , , , , , x a, B) For x a, potential Vx , C) E is in the negative x-direction, , 0, , D) the electric field E in this region is constant, , 87. A dipole is placed in x y plane parallel to, C) For x a, potential Vx x a , 0, the line y 2 x . There exists a uniform electric, , field along z axis . Net force acting on the, x, D) For x a, potential Vx , dipole will be zero. But it can experience some, 0, torque. We can show that the direction of this 91. Consider two large, identical, parallel, torque will be parallel to the line, conducting plates having surface X and Y,, facing each other. The charge per unit area on, A) y 2 x 1, B) y 2 x, X is 1 and charge per unit surface area on Y, 1, 1, y, , , x, y, , , x, , 2, C), D), is . Then,, 2, , 2, , 2, , A) 1 2 only if a charge is given to either X, 88. 10 C of charge is given to a conducting, or Y, spherical shell and a 3C point charge is, B) 1 2 0 if equal charges are given to both, placed inside the shell. For this arrangement,, markout the correct statement(s)., X and Y, A) The charge on the inner surface of the shell will, C) 1 2 if X is given a charge more than that, be 3C and it can be distributed uniformly or nongiven to Y, uniformly., D) 1 2 in all cases, B) The charge on the inner surface of the shell will 92. A particle having specific charge ‘s’ starts from, be 3C and its distribution would be uniform., rest in a region where the electric field has, constant direction but magnitude varying with, C) The net charge on outer surface of the shell will, time t as E E0t . In time t, it is observed that, be 7C and its distribution can be uniform or nonthe particle acquires a velocity v after covering, uniform., a distance x, then, D) The net charge on outer surface of the shell will, be 7C and its distribution would be uniform., 1, 1, 3, x, , E, s, t, x, , , , E0 s t 3, A), B), 0, 89. For Gauss’s law, mark out the correct, 2, 6, statement(s)., 1, 2, A) If we displace the enclosed charges (within a, C) v E0 s t 3, D) v E0 s t, Gaussian surface) without crossing the boundary,, 2, , then E and both remain same., 93. Two charges Q1 and Q2 are placed at the, points A and B having separation r lying inside, B) If we displace the enclosed charges without, , and, outside the uncharged conducting shell., crossing the boundary, then E changes but, The force on Q1 is F and that on Q2 is f. Then, remains the same., , , , C) If charge crosses the boundary, then both E, and would change., D) If charge crosses the boundary, then changes, , , , but E remains the same., 184, , A) F 0, C) f , , 1 Q1Q2, 4 0 r 2, , B) F , , 1 Q1Q2, 4 0 r 2, , D) f = 0, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 94. In a parallel plate capacitor, the potential 98. The velocity of m at a point P very close to, 1, difference between the plates is V. A particle, of mass m and charge -Q leaves the negative, Q2 at a distance h1 from the surface of the earth, plate and reaches the positive plate at distance, (if the initial velocity of ' m1 ' was zero, air drag, d in time t with a momentum p. Then, and earths magnetic field being ignored) is, B) p 2mQV, A) p mQV, Q1Q2, Q1Q2, A) 2 gh 2 h m B) 2 gh , md 2, 2md 2, 4 0 h12 m1, 0 1 1, D) t , C) t , , QV, , QV, , COMPREHENSION TYPE QUESTIONS, PASSAGE : I, The sketch below shows cross-sections of, equipotential surfaces between two charged, conductors that are shown in solid black., Some points on the equipotential surfaces, near the conductors are marked as, A,B,C,........ . The arrangement lies in air., (Take = 8.85 × 10–12 C2 /N m2 ], 0, 0.3m, , Solid, conducting, sphere, , A, , E, B C, , –30V –20V–10V, , Large, conducting, plate, , D, , 10V 20V 30V 40V, , C), , 2gh D), , 2 gh , , Q1Q2, 4 0 h1m1, , 99. The magnitude of charge Q2 at which the, velocity of ' m1 ' at an altitude h2 is zero is given, by, A) Q2 , , 0 m1ghh2, Q1, , C) Q2 , , 2 0 m1 ghh2, m ghh, Q2 0 1 2, D), Q1, 4Q1, , B) Q2 , , 4 0 m1 ghh2, Q1, , 100. At what altitude h3 will object m1 be in, equilibrium and what will be the nature of, objects in motion if it is disturbed slightly from, equilibrium., , 95. Surface charge density of the plate is equal, Q1Q2, to, A) h3 4 m g , periodic, not oscillatory, –10, 2, –10, 2, (A) 8.85 × 10 C/m (B) –8.85 × 10 C/m, 0 1, (C) 17.7 × 10–10 C/m2 (D) –17.7 × 10–10 C/m2, Q1Q2, 96. A positive charge is placed at B. When it is, , periodic and oscillatory, h3 , B), released :, 2 0 m1g, (A) no force will be exerted on it., (B) it will move towards A., Q1Q2, C) h3 4 m g , periodic and oscillatory, (C) it will move towards C., 0 1, (D) it will move towards E., Q1Q 2, 97. How much work is required to slowly move, D) h , non- periodic and non, 3, 4, , 0 m1 g, a – 1mC charge from E to D ?, (A) 2 × 10–5 J, (B) –2 × 10–5 J, oscilllatory, –5, (C) 4 × 10 J, (D) –4 × 10–5 J, PASSAGE - III, PASSAGE-II:, There is a fixed semicircular ring of radius ‘R’, A particle of mass ' m2 ' carrying a charge Q2, lying in yz plane , with centre at origin and it, is fixed on the surface of the earth . Another, is uniformly charged with charge Q A pipe is, particle of mass ' m1 ' and charge Q1 is, fixed along x axis from the origin . The inner, positioned right above the first one at an, surface of pipe is smooth and is made of, insulating material . A small ball of charge, altitude h R R is radius of earth. The, q and mass ' m ' is projected in the pipe with, charges Q1 and Q2 are of same sign . Then, negligible velocity , ball can move in the pipe., NARAYANAGROUP, , 185
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, Whole arrangement is in gravity free space., , 104. The value that fills " FB1 " is, , z, , 25, 15, B) 25, C), y, 2, 2, +, +, +, 105. The region that fills " FB2 " is, +, +, x, A) ab, B) de, C) bc, +, +, 106. The value that fills " FB3 " is, +, A) 1, B) 2, C) 3, 101. The maximum acceleration of the ball in the 107. The plot E vs x is, x, pipe is, +, , +, , +, , + +, , A), , +, , D) -50, D) dc, D) zero, , Ex, , Qq, (A), , 4 0 mR 2, , Ex, , Qq, (B), , 12 3 0 mR 2, , A), , B), x, , x, , Qq, , -3, , Qq, , (D), 6 3 0 mR 2, 12 2 0 mR 2, 102. The kinetic energy of the particle when the, acceleration is maximum is, , -2, , 0, , 1, , 3, , -3, , -2, , C), , D), x, -3, , -2, , 0, , 0, , 1, , 2, , 3, , 2, , 3, , Ex, , Ex, , (C), , 1 Qq , 2, 1 Qq 1, 1, , , (B), (A) 4 R , 3, 4 0 R 2, 0, , , 2, , 1, , 2, , 3, , -2, , 0, , -3, , 1, , x, , 108. In the V vs x curve , the potential possesses, a zero value at, A) x , , 12, m, 5, , 5, B) x m, , 13, 1 Qq , 1 , 1 Qq , 1 , 1, 1, (D), , , , , 13, 14, 4 0 R , 4 0 R , 3, 3, m, m, C) x , D) x , 5, 5, 103. Normal reaction exerted by pipe on ball when, PASSAGE : V, the ball is moving in pipe is along, The law governing electrostatics is coulomb’s, (A) Z a x is always (B) Z a x is always, law. In principle coulomb’s law can be used to, (C) Initially along y axis and then along Z axis, compute electric field due to any charge, configuration. In practice, however, it is a, (D) y axis always, formidable task to compute electric fields due, PASSAGE-IV:, to uniform charge distributions. For such cases, v(volts), Gauss proposed a theorem which states that, b, 15, q, e, E.ds 0, 10, Where ds is an element of area directed along, c d, 5, the outward normal for the surface at every, x(m), point. The integral is called electric flux. Any, 1, 2, 3, -3, -2 -1, convenient surface that we choose to evalvate, the surface integral is called the Gaussian, -10, a, surface., Suppose electric potential varies along the, 109. Gauss theorem in electrostatics states that, x axis as shown in the above figure. The potential, q, does not vary in y or z direction of the intervals, electric flux E.ds . Where S is the, shown (ignore the behaviour at the end points of, s, 0, Gaussian surface . The electric field E is due, the intervals), the field Ex has a maximum absolute, to all the charges, value " FB1 " Vm1 in the region " FB2 " . Its value, A) both inside and outside the surface S, B) outside the surface S, in the region cd is " FB3 " Vm 1 then, C) inside the surface S D) on the surface S, , (C), , 186, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , Three conducting concentric shells A,B and C, 110. The SI unit of electric flux is, of radius a , 2a and 3a are as shown in the, A) Vm 1 B) Vm2 C) NC 1m 2 D) NC 1m 2, figure . The charge on shell B is ‘Q’. The switch, 111. Consider a uniform electric field E =, ‘S’ is closed. Then after closing the switch, 3ˆ, 1 . What is the flux through a, 115., The charge on innermost shell is, 3 x 10 i NC, Q, Q, Q, Q, square of side 10 cm whose plane is parallel to, (A), (B), (C), (D), the y-z plane?, 4, 4, 2, 2, 1 2, 1 2, 116., The, charge, on, outermost, shell, is, A) 30 NC m, B) 20 NC m, Q, Q, Q, Q, C) 10 NC1m 2, D) zero, (A), (B), (C), (D), 4, 4, 2, 2, PASSAGE -VI, 117. If A , B and C are the surface charge, Figure shows concentric spherical conductors, densities on A,B and C respectively then, A,B and C with radii R, 2 R and 4 R, A : B : C is, respectively. A and C are connected by a, (A) 9 : 9 :1 (B) 9 : 9 :1 (C) 9 : 9 :1 (D) 1:1:1, conducting wire and B is having uniform charge, , ASSERTION AND REASON QUESTIONS, , Q, then, , A) Both A and R are True and R is the correct, explanation of A, B, B) Both A and R are True but R is not the, correct explanation of A, R, A, C) A is True, R is False, 2R, 4R, D) A is False, R is True, 118., Assertion(A) : An applied field will polarize the, y, polar dielectric material., Reason(R) : In polar dielectrics, each molecule, 112. The charge on conductor A is q A and that on, has a permanent dipole moment but these are, C is q C then, randomly oriented in the absence of an externally, applied electric field., Q, 2Q, 119. Assertion(A) : Total work done by non-uniform, A) q A q C Q / 3 B) q A , q C , 3, 3, electric field on a charged particle starting from rest, to any time is non-negative.(Assume no other forces, Q, Q, Q, Q, act on the charged particle.), C) q A , q C , D) q A , q C , 3, 3, 3, 3, Reason(R) : The angle between electrostatic force, and velocity of the charged particle released from, 113. The electric potential of ' A ' is, rest in non-uniform electric field is always acute., Q, Q, Q, Q, (Assume no other forces act on the charged, A) 4 R B) 12 R C) 16 R D) 8 R, particle.), 0, 0, 0, 0, 120., Assertion(A) : A metallic shield in the form of a, 114. The electric potential of ‘B’ is, hollow shell may be built to block an electric field., Q, 5Q, 3Q, 5Q, Reason(R) : In a hollow spherical conducting, A) 4 R B) 12 R C) 6 R D) 48 R, sphere , the electric field inside it is zero at every, 0, 0, 0, 0, point., PASSAGE - VII:, , 121. Assertion(A) : E in the outside vicinity of a, C, conductor depends only on the local charge density, B, and it is independent of the other charges present, A, a, 3a, anywhere on the conductor., , Reason(R), :, E in outside vicinity of a conductor, 2a, C, , 1, , S, , NARAYANAGROUP, , is given by, , , 0, 187
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 122. Assertion(A) : If Gaussian surface does not, , enclose any charge, then E at any point on the, Gaussian surface must be zero., Reason(R) : No net charge is enclosed by, Gaussian surface, so net flux passing through the, surface is zero., , STATEMENT TYPE QUESTIONS, , MATRIX-MATCHING QUESTIONS, This section contains 4 questions. Each question, contains statements given in two columns which have, to be matched. Statements (A, B, C, D) in Column, I have to be matched with statements (p, q, r, s) in, Column II. The answers to these questions have, to be appropriately bubble as illustrated in the, following example. If the correct matches are A–p,, A–s, B–q, B–r, C–p, C–q and D–s, then the, correctly bubbled 4 × 4 matrix should be as given :, , A) If both statements are TRUE and, STATEMENT 2 is the correct explanation of, STATEMENT 1., B) If both statements are TRUE but, 128. Two charges q and Kq are placed at, STATEMENT 2 is not the correct explanation, origin O and point A:. P is a point on a line, of STATEMENT1., parallel to y-axis where potential has a zero, C) If STATEMENT 1 is TRUE and, STATEMENT 2 is FALSE, value., D) If STATEMENT 1 is FALSE but, y, STATEMENT 2 is TRUE, 123. Statement - 1: The potential of an uncharged, P, conducting sphere of radius R, for a point charge q, r1, R2, located at a distance r from its cent re, , r R is, , q, 4 0 r, , O, (+q), , Statement - 2: Electric field intensity inside the, conductor is zero therefore potential at each point, Column-I, on the conductor is zero., 124. Statement - 1: Any charge will move from electric A) If OP r1 , AP r2, potential V 1 to V 2 by its own only when, and OA d , then r2, , V 1 V 2 ., , P1, , A(-kq), , x, , Column-II, p) Kr1, , is given by, , Statement - 2: Electron moves from B) If OA d , then, q) Kr1 sin , V1 2V towards V2 4V, PP1 is equal to, 125. Statement - 1: Electric field intensity at surface of, uniformly charged spherical shell is E. If the shell is, PAP1 , punctured at a point then intensity at punctured point, becomes E/2., C) If OA 6m and, r) x1 4; x2 1, 1, Statement - 2: Electric field intensity due to, K 2, then locus of, spherical charge distribution can be found out by, using Gauss Law., point P is a circle of radius and centre, 126. Statement - 1: If a point charge q is placed in front, given by x1 and x2, of an infinite grounded conducting plane surface,, respectively., the point charge will experience a force., Statement - 2: This force is due to the induced D) For some value of, s)none of these, charge on the conducting surface, which is at zero, OA (> 0) and K(>0),, potential., locus of point P will be, 127. Statement - 1: The surface of a charged conductor, a circle of radius and, is always an equipotential., centre given by, Statement - 2: Electric field lines are always, perpendicular to the equipotential surface., x1 and x2 respectively.., 188, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 129. The variation of electric field (E) with distance, INTEGER TYPE QUESTIONS, (x) from its centre is shown. The sphere has a 131. A uniform surface charge density 8 exists, positive charge., over the entire x-y plane except for a circular, hole of radius a centred at the origin. The, E, , , , , , electric field at a point P 0,0, 3a on the z-, , , axis is found to be 2 x . Find the value of, 0, X, ., R, x, 132., Two, fixed, equal, positive charges, each of, Column - II, magnitude 5 105 C are located at points A, p)non-conducting material, and B separated by a distance of 6m. An equal, and opposite charge moves towards them along, q)different, the line COD , the perpendicular bisector of, the line AB. The moving charge when it reaches, the point C at a distance of 4 m from O , has a, kinetic energy of 4 joules. The distance of the, r)non-uniform, farthest point D , in metre, where the negative, charge will reach before returning towards C, is 6 x . Find x . +q, , Column - I, A) The sphere is, made of a, B) The dielectric, constant of the, sphere and its, surrrounding are, C) Charge density, throughout the, volume of the, sphere is, A, D) Electric potential s)none of the above, 3m, at the centre, C, D, O, due to sphere is, 4m, 3m, 130. In options (A) to (C) spherical conductor is, B, hollow and neutral and in (D), it is hollow and, +q, charged. Match the columns:, 133. A thin fixed ring of radius 1 meter has a, Column - I, Column - II, positive charge 1 10 5 coulomb uniformly, A), distributed over it. A particle of mass 0.9g, p)Potential inside the, and having a negative charge of 1 106, coulomb is placed on the axis at a distance of, conductor is varying., 1 cm from the centre of the ring. The motion, B), of the negatively charged particle is approxi, , P(q), mately simple harmonic. Hence the time period, q) E inside the, conductor is zero., C), q2, , P(q), , , r) E inside the, conductor is constant., , D), P(q1), , q2, , , s) E inside the, conductor is varying., , NARAYANAGROUP, , of oscillations is, , , . Find k ., k, , 134. Two concentric spherical shells have radii 0.15, m and 0.3 m. The inner sphere is given a, charge 6 102 C . An electron escapes, from the inner sphere with negligible speed., Assuming that the region between the spheres, is vacuum, the speed with which it will strike, the outer sphere is x 10 6 ms 1 , Find the, value of x?, e, 11, 1 , for electron 1.76 10 Ckg , m, , 189
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 135. A particle having a charge of q 8.85 C is, placed on the axis of a circular ring of radius, R = 30 cm at a point P at a distance of a=40, cm from the centre of the ring. The electric, flux passing through the ring is x 105 N / C ., Find the value of x ?, 136. It is required to hold four equal point charges, , 8, each having a charge Q 1 2 2 C in, 7, , , , , , equilibrium at the corners of a square. Find, the point charge, in coulomb, that will do this if, placed at the centre of the square., 137. A ball of mass 2 kg, charge 1 C is dropped, from top of a high tower. In space electric field, exists in horizontal direction away from the, tower which varies as, , E= 5 - 2x 106 Vm1., , Find maximum, horizontal distance that the ball can go from, the tower., 138. A solid sphere of radius R has charge Q, distributed in its volume with a charge density, , kr a , where k and a are constants and r, is the distance from its centre. If the electric, field at r , , R 1, is times that at r=R , find the, 2 8, , value of a., 139. A charge Q , , 5, nC is distributed over two, 100, , concentric hollow spheres of radii r=3 cm and, R=6 cm such that the densities are equal. Find, the potential, in volt, at the common centre., 140. A point charge + Q is placed at the centroid of, an equilateral triangle. When a second charge, + Q is placed at a vertex of the triangle, the, magnitude of the electrostatic force on the, central charge is 8 N. The magnitude of the, net force on the central charge when a third, charge + Q is placed at another vertex of the, triangle is :, 141. Two very small particles 1 and 2 each of mass, 0.5kg and of charges q1 1mc and q2 1 c, 1., respectively are connected by a mass less, spring of spring constant 200Nm 1 and placed, on a horizontal rough surface. The particle 1, is fixed and 2 is free to move. Initially the, 190, , spring is in its natural length (2m) when ' q2 ' is, released. If coefficient of static friction, between 2 and horizontal surface is 0.5 the, separation between the charges when they are, in equilibrium will be (Neglect gravitational, interactions between 1 and 2) g=10ms-2., , LEVEL - V -KEY, 1) A 2) D 3) B 4) B 5) A 6) C 7) B, 8) B 9) D 10) C 11) A 12) C 13) A 14) BC, 15) C 16) C 17) A 18) A 19) B 20) B 21) A, 22) A 23) D 24) B 25) D 26) A 27) C 28) D, 29) B 30) D 31) C 32) A 33) B 34) B 35) A, 36) A 37) A 38) C 39) D 40) A 41) B 42) A, 43) D 44) B 45) B 46) B 47) C 48) C 49) C, 50) C 51) B 52) A 53) C 54) D 55) A 56) C, 57) A 58) C 59) A 60) A 61) C 62) C 63) D, 64) B, 65) C,D 66) C, 67) C,D, 68) A,C, 69) A,D 70) A,C 71) B, 72) A,B,C 73) B, C 74) A, C 75) A,C,D, 76) A,C,D 77) A,B, 78) A, D 79) A, D, 80) A, B, C 81) A, C, D 82) A, B, C, D, 83) A,C,D, 84) D 85) A,D 86) A,B,C, 87) C,D, 88) A,D, 89) B,C 90) A,B,D, 91) A,C 92) B,D, 93) A,C 94) B,D, PASSAGE-I : 95) A, 96) B 97) D, PASSAGE-II: 98) A, 99) B 100) C, PASSAGE-III: 101) C 102) A, 103) A, PASSAGE-1V :, 104) B 105) A 106) D 107) D 108) C, PASSAGE-V : 109) A, 110) D, 111) A, PASSAGE-VI : 112) D 113) C, 114) D, PASSAGE-VII : 115) B 116) A 117) C 118) A, 119) C 120) A 121) D 122) D 123) C 124) D, 125) B 126) A 127) A, 128) A-p, B-q, C-s, D-r, 129) A-p, B-q, C-s, D-r, 130) A-p,s;B-q,r; C-p,s; D-p,s, 131) 3 132) 2 133) 5 134) 8 135) 1 136) 2, 137) 5 138) 2 139) 9 140) 8 141) 2, , LEVEL - V - HINTS, dx, x, dQ dx , , 0 x 2 dx , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, l, , 2., , ELECTRO STATICS, , , 2 kq sin, kq /2, cos d , F, 2, R /2, R, , 0l 3, Q, 3., , 0 x 2 dx , 0, , , , 2kQq sin, Q, 2, but , F, R, 2, R, , d, , , , dQ R d 0 cos Rd, , T cos d/2, , T cos d/2, , d/2, , , , T, , 2, , Q, , d/2, T tan d/2, , 0 R cos d , , , , T, , 5., , , , 2, , 2, , = 0 R sin 2 20 R, , r, , dA, , 3., , 2T sin, , T, , d, q.dQ, , 2, 4 0 r 2, , q r, 4 0 r 2, , T, , 6., 2 R sin d , , +, , +, , +, , +, , The distance between the centres of charge will be >d, The force between them will be less than, +, , +, , +, , 1 Q1Q2, 4 0 d 2, , +, , +, , +, , 7, , , , 4., , +, +, +, +, +, +, , 2, sin 2 d 0 R cos 2 /2, 0, 2, , Q 0 R 2, +, , /2, , d, , +, , , 0, , Q2, , +, +, +, +, +, , Q 0 R, , qQ, 8 0 r 2, 2, , Due to induction the charge distribution will be as, shown, , dQ 2 0 R 2 sin cos d 0 R 2 sin 2 d, /2, , T, , 2, , dA 2 rRd, , 2, , d, q rd, , 2 4 0 r 2, , qQr, (2 r)4 0 r 2, , Q1, , , , 2T, , /2, , Only angular momentum x & total energy are, conserved but not kinetic energy, dF sin, , , , dF, dF dF cos 5, , , dF cos, , The force on the pt. charge q due to the differential, , kqdq k R d q, element is dF 2 , R, R2, , dF cos , , k .q R d , , NARAYANAGROUP, , R2, , cos , , d, , 8., , , , E, , , 1, 2 0 R dF = E dq E Rd , 191
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, From symmetry dF cos 0 F dF sin , , , 14. (B) eE = mew2r, R, , m e 2, Edr = e 0 r dr, , , , , , R1 sind , 20 R, 0, , ER d 1, 0, , 15. (C) Vp =, , Z –, E E, , dV , , B, , E2 E2, , A, , C, , E2 E2, , X, +, D, , 1, E, , E1, , E, , X, , E1, , 2, , , , (Figure 2), , 1 q, 11. V 4 a, q 4 0 aV, 0, equating volumes of bubble and droplet, , , , , , a, V , 3t , , 0, /2, cos / 2, 4 0, , 1, , 1 2 R 2 sin d R, sin d, , 2 0, 4 0, R, , 3, , V, , C B , V, , V, , A E.dr E.dr, 12. B, A, , A, 0.5, o, , 325 dl cos180, , 325 0.8 =260V, , 0.3, , 192, , 0, , R, 2 0, , C, , 18., , A, , x, , a, +, , +, , E, , , ,, 2 0 y For a small displacement, , , dr dy j dxi, , , , , = q E.d, , , sin d , , y, , C, , , , , , a, , E.dr 0 VB VA E.dr, , where E = 2 î, 0, , /2, , E, , C , , 13. (A) Wnet, , R, 2 0, B, , C, , B , , , , =0, , 1 dq, 2 R 2 sin d dV 4 R, 0, , , , 1 4 0 aV, 1 q, , 1, 4 0 R 4 0, 3a2t 3, , 1, 0 sin d , 4 0 /2, , 17. The area of shaded element is dA 2 rRd, , 1, 4, 4 a 2t R 3 R 3a 2t 3, 3, , Vx , , 0 sin R d , , /2, , V, , (Figure 1), , dQ, , 1 0 R sin d , 4 0, R, , Z, , E1, , me 2R 2, 2e, , ka, kq, kq kq, Vin = V =, , V, =, C, in, r, r, r, r, , q, on solving F 2 R, 0, 16. 0 sin , 9. Diagram is the solution, 10. The electric field intensity due to each uniformly, charged infinite plane is uniform. The electric field, intensity at points A, B, C and D due to plane 1,, , plane 2 and both planes are given by E1, E2 and, E as shown in figure 1. Hence, the electric lines, dQ dl , of forces are as given in figure 2., E1, , V=, , Wnet q, , , (X1 X 2 ), 20, , , , , , , dw E.dr, , , dy, j. dy j dxi , 2 0 y, 2 0 y, , , , , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, W , , ELECTRO STATICS, , q, q, q, a, ln 0.5 , ln 2, ln y2a , 2 0, 2 0, 2 0, , q, 4, +q, , -q, , , A B, D, 28., dr, C, 2 0 r, 20. (A) vA – vB = work done by electric field on +, 1 coul. charge from A to B = E Rq, q, q, q , K, , VA, vB = vA – E Rq = v – E Rq, 4(3a) 2a 4a , 21. Usystem = Uself + Uint, 19. dw Vdr, , V , , r2, , r1, , Kq, 6a, , VA , , , 1 0K 2 Q 2, V V V , 1, 2, , E, i, j, k, 29. (B) u =, = 2, 0, 22. E , 2, 2 r4, y, z , x, 23. No change in potential for the wire placed on X, 1, Q2, 0K 2 4, axis. So W = 0 and for wires on Y & Z axes is, 1 0, u, KQ, 2, r, =, = 2 2, V=, 2, 2, 2, 2r, , r, r, V, K Q, W r, ln 2 W= W + W, 2, 1, 2, 20, r1, r, W, , 2, , 4, l n , l n2, W , 20 1 , 0, , , 24. (B) E = = , , = – [Area under Ex – x curve], VB = 30 V., , t, , q, q , q, Vc K C B A , b, a , c, , 1 4 c 2 4 b 2 4 a 2 , VA , , , , 4 0 , c, b, a , , (b c a), 0, , NARAYANAGROUP, , r2, , so the correct option is B., , 3 15 , , , , 8 , 2, , 1, 1 EQ 2, t y usin t gt 2, 2, 2 m, , 2usin , u2, 1 QE 4u2 sin2 , x, sin2 , ., g, g, 2 m, g2, , 4, 2, , –, +, –, , , , 1, , 1, 1, [U1 ... U6 ] =, [6U1] = 3U, 1, 2, 2, , q q , Vc K 1 2 ? Clearly V = 10V, c, b a, , A B C, , , , 31. x ucos t , , q1 q2 , q1 q2 , 26. K 10v K , 5v, b a , b , , 27., , V, , 2, , q2, 2, 1 , 2 2, 3, kq, , , , , =, =, 0 a, a a 3 2a , , 25. (D) VB VA Ex dx, VB – 10 = .2.( 20) = 20, , u, , 30. (D) Interaction energy of system of charges is =, , 1 ˆ 1 ˆ 1 ˆ, i 2 j 2 k, x2, y, z, , = ˆi ˆj kˆ, , 1, 2, , because, , , , mg, dx, 0 then tan2 , EQ, d, , u2, 2, 2, x max , EQ mg EQ, 2, mg, 32. In order to maintain constant separation, the, particles must have the same acceleration., Assuming the system of both charges to acceleration, towards left. Applying Newton’s second law to the, left particle we get, , , , , , 193
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, -Q,m, , +Q,M, , dV , , L, E, , r dr , 2Q, Q, , 2, R 4 0 R 2 r 2, 2 0 R 2, R, , r dr, R2 r 2, , Q, r dr, Q, KQq, V, , R2 r 2, 2 , 2, QE 2 Ma ............... (1), 2, 1, 2 0 R o R r, 2 0 R, L, Under given condition the acceleration of both, Q 2 1, Q, charges should be same and should also be equal, , 2 1 R , 2, 2 0 R, to acceleration of centre of mass of both the charges., 2 0 R, , , , , , a, , Q q E, Fnet, , ............. (2), total mass, mM, , Qq, U of q , , dV , , 33. (TE)surface = (PE)centre, Velocity at centre is nearly zero then it can cross, the sphere, , 1 q.q 1 2 3 1 q , mv , , 40 R 2, 2 40 R , , 1 2, I, 2, , 35. Use conservation of energy, , PE KE at surface PE at contrect path, 36. (A) We have centripetal force equation, , R, , dQ , , Q, , 2 rdr, , 2kq, Now,,, m, , 1, where, k = 40, , 2Q r dr , R2 4 0 r, , , , Q, dr, 2 0 R 2, , Q, 2 0 R U of q at centre of disc is, , U , , Qq, 2 0 R, , Qq, 2 0 R, , , , , , 2 1 , , Qq, 1 2, mv , 2, 2 0 R, , 1, Qq, mv 2 , 1 2 1, 2, 2 0 R, , , , , , 1, Qq, mv 2 , 2 2, 2, 2 0 R, , , , 1, mv2 , 2, , Qq 2, , , , , , 2 1, , 2 0 R, , , , , 1/ 2, , , , 2Qq 2 1, v, 0 mR, , , , , , 1 2, 1 Qq, 38. 2 mv 4 R 0, 0, , The charge on the shaded part, , , , 2Q, , R, R2, Potental due to dQ at P, 194, , 2, , so v =, , P, , V, , O, , Eq l sin mg l (1 cos ) , , 37., , , , 2 1, , Applying mechanical energy conservation, , 34. WE Wg K .E, Use work energy theorem, , m, 2k q, , , , Similarly the potential due to dQ at centre of disc is, , M m KQq, L, E qM Qm, , 2 r, T=, =, v, , , , 0, , 2 0 R, , Hence from equation (1) and (2) we get, , 2k , mv 2, q, =, , r , r, , , , , , , , R, , r dr , , v, , 2Qq, Qq, , 4 0 mR, 2 0 mR, , 39. As the first particle approaches the second due to, repulsion it recedes away from it along the same l, ine of apprach. The separation between them is, minimum when their velocities are equal., Applying conservation of linear momentum, NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, to the dipole moment. (figure below), Hence, net force on ring is F = QE =, , L, L, Flux through CDHG = O 4 4 , 0, 0, , kpQ, R3, , L, L, Flux through AEHD = O 4 4 , 0, 0, , Alternate solution, Electric field due to ring at point P on its axis at, a distance x from the centre O of the ring is, E = k, , dE , ;, , dx at x 0, , Qx, (x 2 R 2 )3 / 2, , Force on dipole =, , =, , L, L, L, Flux through ABCD = 4 4 2 , 0, 0, 0, , kQ, R3, , k Qp, dE, =, dx, R3, , 49. Flux first increases linearly ( v is constant),, remains constant (when length L/2 is inside the cube), and then decreases linearly (as rod moves out)., , , 50. (C) f = E ds, since r << R so we can consider electric field is, constant throughtout the surface of smaller ring,, hence, , , , , , 46., , f= E, , , , , x, , x, 2, , (R x 2 )3 / 2, , So, the best represented graph is C., 51., , , 1 1, tan tan , 2, , , 1 1, given / 2 tan tan , 2, 2, , E, , , dA, , q, q, , , l, , 1, tan cot tan 2, 2, , , + , , DqE, , , 47., +, , (dq)E, , l 2, By symmetry 2 sin , o, 48. The charge inside the cube due to each line is L ., Flux due to horizontal wire through the faces AEDH,, BCGF is zero as the electric field produced will be, parallel to these faces., , L, But the total flux due to this line is as it is equally, 0, distributed among the remaining four faces. Similarly, flux due to vertical coire through the faces, DHGC,AEFB is zero., 196, , 1, q, dA 2 r dx E 4 2 2 d E dA cos , 0 r l, , , , 1, q, l, 2 r dx , 4 0 r 2 l 2, r2 l2, , , , ql, 2 0, , , , , , r dr, 3/2, , r2 l2 , , ql, , 2 0, , , ql, 2 0, , , , , 1, , r 2 l 2, , 1, 1, , l, R2 l 2, , ql, 2 0, , R, , , 0, , r dr, , r2 l2 , , 3/2, , r R, , , , r 0, , , q, , 2 0, , , , l, 1 , , , R 2 l 2 , , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , q, q, , , , 2, 52., 2 0 1 cos 60 , 2 0, , dV, V Ar B E dr E 2 Ar, 2 Ar 4 r 2 ., , dA, , E, , , , 53., , q, , , x, , qR 2, , 2 0, , Y, , 1, q, 4 0 R 2 x 2, , d , , Q, , 1, q, R, 2 R dx , 4 0 R 2 x 2, R2 x2, dx, 2 3/2, , R2 x , , qR 2, , 2 0, , L /2, , , L /2, , q1, , E, , 3/2, , R 2 x2 , , E, 2d, , d, , 3d X, , E, , q1, q, 1 cos 2 1 cos , 2 0, 2 0, , sin 2, , 2d, , 3d X, , d, , -q2, , q1 2 sin 2, , X, , dx, , , , , , 3Q, , (d, 0) (2d, 0) (3d, 0), , , q , L, , , 0 L2 4 R 2 , , 54., , 4 r3, 6 0 A, 3 0, , 56. Two very large thin conducting plates having same, cross-sectional area are placed as shown in figure., They are carrying charges 'Q' and '3Q' respectively., The variation of electric field as a function of x (for, x = 0 to x = 3d) will be best represented by., , dx, , E, , q, E.ds , 0, , , , 2, , , , , q2 2 sin 2 , 2, 2, , , q, q1, , , sin 2, 2 sin 1 1 sin , 2 q2, 2, 2, q2, , E, 2d, , d, , 2d, , 3d X, , 3d X, , d, , 57. Electric field at any point inside the cavity =, Electric field due to entire sphere at that point- the, field at the same point due to a solid sphere which, can tightly fit into the cavity, r1, P, r, OA 0 2, , r, , 55., , , E E1 E 2 According to Guass's Law, , , E1 , r1, 3 0, , NARAYANAGROUP, , , , E, (r1 r2 ), 3 0, 197
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, , Vectorially E , , EA , , 1, 4 0, , , EA , , 1, 4 0, , , EB , , Q, , , , , , 12 2 2 32, , i 2 j 2 k^ , , , , , , 10C – 3C, +3C, , , , –3C, , 1, 4 0, , 1, 4 0, , , EC , , 1, 4 0, , i 2 j 2 k^ , , 3 , , 14 , , Q, , , , 2, , 2, , 1 2 3, , Q, , 3, , 3, , , , 3, , , , 2, , i j k, , Q, , , , 2, , 2, , 2 2 2, , 2, , , 1 Q 2i 2 j 3k, 3, 4 0, 23 3, , , , , , 3, , i j k , , , , 2i 2 j 3k , , , , , , , E A . EB 0 A option is correct, 1 , EC EB C option is correct, 4, +, , P, , 87. Torque will be perpendicular to the line y = 2x and, it should be in xy plane, because electric field in zdirection. The lines in option (c) and (d) both are, perpendicular to y = 2x., 88. Due to induction, charge on various faces are as, shown., , Q, , , EB , , , EC , , 1 q , r, 4 0 r 3, , –, , Q, , Charge on the inner surface of shell = +3 C, Net charge on outer surface of shell =, , 3 C 10 C 7 C, Distribution of charge on inner surface would be, uniform if charge is placed at the centre, otherwise, non-uniform.On outer surface, charge would be, always uniformly distributed as displacement of, inside charges does not affect, the distribution of the outer charge., 89. crossing through Gaussian surface does not, , , , depend on the location of charge, while E depends, on it. If q crosses the boundary, then qenclosed and, , , , hence the flux and E also change., 90. 0 x a;, , x, , Vx Ex dx V 0 0 (as E = 0), x, 0, , x, , R, , x a; Vx E x dx V a , , 85., , 0, , Vx , , , , , EP , , 0 EQ , , 0, 2 0 2 0, 2 0 2 0, EQ , , , , , , , 2 0 2 0 0, , 86. E V / x B 2Cx, i.e., E varies linearly with x and is along negative xdirection. It is also clear that B has got dimensions, or units of E, i.e., NC 1 ., 200, , x , , , dx V a x a x a, , , 0, 0, a 0 , , x 0; V x , , x, , , E x d x V 0 E x , , 0 , , 0, , , Vx , 0, , , , x V 0 , x, 0, , , NARAYANAGROUP
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 1 q1q2, q1q2, 100. 4 h 2 mg h3 4 mg, 0, 3, 0, , PASSAGE-IV:, 104,105,106,107,108, V, , PASSAGE - III, 101,102,103, , b, , 15, 10, 5, , , , dE cos, , , , –2, , O, , dFx q dEx , Fx , , 1, Qqx, 2, 4 o R x 2 3/ 2, , 1, , R, , 2, , x2 , , 1, , 2, , 3 x(m), , value ' FB1 ' in the region FB2, , Ex value in cd is FB3 Vm 1 ., 25 , dV, Eab 25, dx, 1 , , 5 15 10, Ebc , , 1 2 3, , Ecd 0, , 15 5 , Ede , 10, 3 2 , , , , d , x, 0, dx R 2 x 2 3/2 , , , , , Ex Max, , E, , 1, Qqx, 2, 4 0 m R x 2 3/ 2 for a to be maximum, , a, , d, , 10, , a, , 1, dQ, cos , 2, 4 o R x 2 , , c, , –3, , dE, , q, , e, , Ex Max 25Vm 1 FB2 is ab, , 3 2, 2 5/2, R x 2x 0, 2 , , , 1 x, 3/2 , , Taking the values of Eab , Ebc , Ecd , Ede , The plot, of Ex Vs x will be as shown in ‘D’ option., For a b part V 25x 65, , , , 1, , R, , 2, , x, , 2 3/ 2, , , , , , 3x 2, , R, , 2, , x, , amax , , x R/ 2, 1, Qq (2 2), , 2, 4 o mR, 2 3 3), , 2 5/ 2, , , , R 2 x 2 3x2, , 1, QqR / 2, 4 o m 2 R 2 3/ 2, R , , 2 , , , Qq, , 6 3 o mR 2, , Applying mechanical energy conservation, 1 Qq, 1, , 4 o R 4 o, , Qq, , 1, mv 2, 2, R x, 2, , 2, , 1 2, 1 Qq , 2, KE 2 mv 4 R 1 3 , o, , , 202, , when V 0 x , , 65 13, , m, 25, 5, , PASSAGE : V, 109. (a) Conceptual, 110. (d), , =, , , E .S, , F, , = q S NC1m2, , , 111. (a), E S cos , is angle between area vector and field vector, S 10 cm 10 cm 10 2 m 2, E 3 103 NC1 E S Cos 30 NC 1m 2, PASSAGE -VI, 112. VA VC, , 1 qA qB qC 1 qA qB qC , , , , 40 R 2R 4R 40 , 4R, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, , 4q A 2qB qC q A qB qC, , 1 q Q q , 1 Q, , , , 4 0 a 2a 3a 4 0 3a, , 3q A qB Q, qA , , Q, 3, , qC , , Q, 3, , 1 q A qB qC , , , 113. VA , 4 0 R 2 R 4 R , , , , , 1, 1, 4Q 6Q Q , 4 0 12 R, , 1, 4 0, , Q , Q Q, 6 R 2 R 12 R , , 1, , 4 0, , Q , Q Q, 6 R 2 R 12 R , , 1, Q, , 2 6 1, 4 0 12 R, , ., , 5Q, , 48 0 R, , Q Q, , Q, 4, , ASSERTION AND REASON QUESTIONS, , , 118. E in outside vicinity of conductor’s surface depends, on all the charge present in the space, but expression, , E, , , ., 0, , 119. E at any point on Gaussian surface may be due to, outside charges also., , 1, , q, , 120. V0 4 r , 0 , , V0 , , induced ch arg e , r, , , , 1 q, 0, 4 0 r, , Hence, potential is constant, 121. Electron being a negative charge, will be move from, lower to higher potential, 122. Conceptual, , C, , Q, , Q, Q, . qA , 4, 4, , 1 1 1, A : B :C : :, 9 : 9 :1, 4 4 36, , PASSAGE - VII:, 115., , B, A, a, , q, , Q, Q, Q, , , C, 117. A 4 4 a 2 B , 4 4 9a 2, 4 4a 2, , 1 q A qB qC , 114. VB 4 2 R 2 R 4 R , , 0 , , , q Q Q 2 q Q Q Q, , , 3 2 3 3 3 2, 6, , 116. qC , , 1 Q Q, Q , , , , , 4 0 3R 2 R 12 R , , 1, 1, Q, 3Q , 4 0 12 R, 16 0 R, , q, , 2a, , STATEMENT TYPE QUESTIONS, , 3a, , S, , When switch is closed the inner most and outer, most shells will acquire same potential., Let the charge on outer shell be ‘q’ and that on the, inner shell be -q, the total charge on inner and outer, shells is zero., , 123. Apply the concept of electric image, 124. Since field is zero for a charged conductor on the, surface and inside it also. So the surface of a, charged conductor is always equipotential. Also,, , , for equipotential surface. Hence E.dl 0, , , E dl. Hence both are true and Statement -, , 2 is the correct explanation to Statement - 1., , VA VC, NARAYANAGROUP, , 203
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, 1, , q, , 125. (A) V P 4 r , 0 1, , Kq 0, , , , r2, , or r2 Kr1, , Enet , , ', ' , z, , 1 2, 2 0 2 0 , z a2, , (B) Now, PP1 sin POP1 r1 sin PAP1 r2, sin POP1 , , Enet , , PP1, sin POP1 K sin or r K sin , 1, , E, , (C) OP1 x, P1 A 6 x , , x 3, , r, Kr, 2 1 K, sin PAP1 r1, r1, , 'z, 2 0, , 2, , 2, , 2, , 12, , , , 2 3, 0, 0, , + + + + + + + +, +, + + + + + + + +, +, + + + + + + + +, +, + + +, + + +, +, + +, + +, + + +, + + +, + + + + ++ + + +, + + + + + + + + +, + + ++ + + + + +, + + + ++ + + ++, +, , r2 2r1 from r2 Kr1 6 x y 2 4 x 2 y 2 , 2, , z2 a2, , 8 3a , 2 0 2a , , r12 x 2 y 2 r22 6 x y 2, , On solving, x 2 y 0 4 , , , , , , 2, , Locus of point P is a circle of radius x1 4m and, centre x2 2, 0 , (D) The values of OA and K should be known, 126. E is proportional to distance upto x R , so the, sphere has uniformly distributed charge over its, volume and because of this reason, the sphere is, made of non-conducting material., The electric field is same just inside and outside of, the sphere, therefore dielectric constants of the, materials of sphere and surrounding are, same.Potential at the centre is maximum., 127. For option (B), ther e is no charge in the conductor,, , thus E inside the conductor is zero and the potential, is same as potential of the conductor and is, constant., , For other cases, E inside the conductor is nonzero and varying. Potential inside the conductor is, also varying., , MATRIX-MATCHING QUESTIONS, 128. Electric potential due to equitorial line, 129 Electric field intensity due to conducting sphere, 130. Electric field and electric potential due to spherical, conductor., , INTEGER TYPE QUESTIONS, 131., 204, , Enet , , ', Edisc, 2 0, , AO 2 OC 2, , 132. From figure AC , , AC 32 4 2 5m Similarly, BC = 5 m, 9, P.E. at C is U c 2 9 10 , , q q , AC, , K.E at C is 4 J, Total energy at C is Ec P.E K .E ., , EC 9 4 5 J, Potential energy at D is U D , given by, , UD , , , , 2 9 109 5 105, , , , 2, , r, , where AD = BD = r (say), , U D , , 45, J, r, , K.E at D = 0, , So, total energy at D is, , ED 0 , , 45 45, , J, r, r, , By Law of Conversation of Energy,, , U C KC U D K D, , , 45, 5, R, , r 9m, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , ELECTRO STATICS, 134. Gain in KE = e V2 V1 , , AD 2 AO 2, , NowOD , , 1, 1, 1 1, mv 2 e, q , 2, 4 0 a b , , OD 92 32 72 m, , x2, , OD 6 2, A, , –q, , –q, O, , D, , v 8106 ms1 x106 x=8, , C, , 135. E , , (+q), B, , 133. Let us first the force on a-q charge placed at a, distance x from centre of ring along its axis., Figure shows the respective situation. In this case, force on particle P is, , FP , , FP qE, , q, Qx, 4 0 x 2 R 2, , , , , , 1 qQx, 4 0 R3, , E , , 3/2, , Q, + ++, , ++, , R, , + ++, , 3, , 9 1 0 9 1 0 5 1 0 6, , , s k5, 5, +, +, +x, p, +, + F(–q), +, +, +, , ++, , +, +, +, +, +, +, +, +, , NARAYANAGROUP, , 106 4 , 1 , 2 5, , 106 1, E , 105 NC 1, 2 5, , F21 F24 cos 45 F25 cos 45 F23 cos90 0, , , , Q, Q, , 2, 4 0 a, 4 0 2a, , , , Q, , 2, 4 0 mR 3, T , 2, , Qq, , T , , , , 40 , 1 , 50 , , , , F2 0 F 2 x i F2 y j 0 F2 x 0, , Qq, 4 0 mR 3, , 0 .9 1 0 3 1 , , , , 2 8.85 10, , 12, , 136. Let us consider the equilibrium of charge Q placed, at 2. Then for equilibrium, , , , qQ, x, x0, 3 , 4 0 mR , which is a standard SHM equation. So, , T 2, , 8.85 106, , , , , , E 1 105 x 105 x=1, , Now, acceleration of particle is, F, 1, qQ, a, , x, m, 4 0 mR 3, , , , q , a, 1 , 2 0 , R2 a2, , E , , For small x, x<<R, we can neglect x, com, pared to R. So, , F , , 2e 1 ab , , q, , m 4 0 b a , , v2 , , (+q), , , , 2, , Qp, , , , a , 4 0 , , 2, , 2, , 1, 0 0, 2, , Q, 2q 0, 2 2, , , , , , Q 2 2 1 4q 0 q , Since Q , , 8, 1 2 2, 7, , , , Q, 2 2 1, 4, , , , , , , , 1 8, , q 1 2 2 1 2 2 , 4 7, , , , , , , , , 18, q 7 2C, 47, 137. ax , , qE, m, , , , dvx q, 5 2x , dt m, 205
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , ELECTRO STATICS, At time t , E 5 2 x 10, , v, , , , , vx2, 2, , , , q, 5x x2, m, , , , dx, , dt, , , , v, , 6, , vx2 , , 2q, 5x x2, m, , , , , , dv q, 5 2x , dx m, , 2q, 5x x 2, m, , , , , , For x to be maximum, , dx, 0 5x x2 0 x 5 m, dt, , Therefore q1 Q q2, , R2 , r2 , q1 Q 2, Q, , Q, 2, 2 , 2 , r, , R, r, , R, , , , , the potential V1 at common centre due to, charge q1 is given by, , V1 , , 1 q1, 1 r2 Q, V1 , , , 4 0 r, 4 0 r 2 R 2 r, , F23, F24, 1, , 2, , F25, , V1 , , F21, , y, , 5, q, , The potential V2 at common centre due to, , x, , 4, , charge q2 is V2 , , 3, , 138. Total charge, , 4k, Kr 4r dr a3 R , a, , a3, , 2, , r0, , r, ', , Q dV , , R, 2, , Kr , a, , r 0, , 4 k R , 4 r dr , , a 3 2 , 2, , V , , 1, QR, r R , 4 0 r 2 R 2, , V , , 1 QR r, 4 0 R 2 r 2, , a 3, , , , According to question, , 1, Q', 1 1 Q , , , , 4 0 R 2 8 4 0 R 2 , , 2, , , , V 9 109 , , , , 5, 9, 109 100, 100, 45, , V 9V ., , putting the value of Q and Q ' get a = 2, , 140. (D) R =, , 82 82 2.8.8 cos 120 = 8 N, , 139. Let q1 and q2 be the respective charges, distributed over two concentric spheres of radii r, and R such that, , 8N, , q1 q2 Q, , 1 2, , q1 r 2, , , q2 R 2, , q1, r2, q1 q2 r 2 R 2, , 1 2 1 , , q2, q2, R, R2, using (1), we get, , R2 , Q r 2 R2, Q, , This gives q2 2, 2, q2, R2, r R , 206, , 60°, , R, , As surface densities are given to be equal,, therfore, , q, q, 12 2 2, 4 r, 4 R, , 1, QR, 2, 4 0 r R 2, , Net potential at common centre, V V1 V2, , rR, , Q dV , , 1, Qr, 2, 4 0 r R 2, , 120°, 60°, 8N, , 141. Based on friction and Coulombs Law, 9 109 109, 2.25 N, 4, which is less mg acting on 2, final separation =2m only, F, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , CAPACITORS, 4., , LEVEL - V, 1., , n conducting plates are placed face to face., Distance between two consecutive plates is d., , A A A 1 , , , ..... n1 A, 2 4 8, 2 , A dielectric slab of dielectric constant k is, inserted between the first and second plates, and the assembly is charged by a battery of, emf . Find the charge stored in the assembly., Area of plates is A,, , A, , C, d, , 5., A/4, , 0 A, A) 2d 1 2n 1 2 , k, , , 0 A, B) 2d 1 2n 1 1, k, , , 0 A, , 2., , 3., , 0 A, , C) d 1 2n 1 2, D) 3d 1 2n 1 2 , k, , k, , A capacitor of capacitance 10mF is charged, up a potential difference of 2 V and then, the cell is removed . Now it is connected to, a cell of emf 4 V and is charged fully. In both, cases the polarities of the two cells are in, the same directions. Total heat produced in, the second charging process is :, (A) 10mJ (B) 20 mJ (C) 40mJ (D) 80mJ, For the circuit shown ,which of the following, statement is true ?, , V1=30V, , V2=20V, , S1 C =2pF S2 C =3pF S2, 1, 2, (a) With S1 closed V1 15V, V2 20V, (b) With S3 closed V1 V2 25V, (c) With S1 andS2 closed V1 V2 0, (d) With S3 closed V1 30V, V2 20V, NARAYANAGROUP, , d/3, R, , 15d 9vt 0 R, 6 0 R, (b) 2d 2 3dvi 9v 2 t 2, (a), 5d 3vt, (c), , A/2, , A parallel plate capacitor C with plates of unit, area and separation d is filled with a liquid of, dieletric constant K 2 .The level of liquid is, d/3 initially.Suppose the liquid level decreases, at a constant speed v, the time constant as a, function of time t is, , 6., , 6 0 R, 15d 9vt 0 R, (d) 2, 5d 3vt, 2d 3dvt 9v 2 t 2, , A dielectric slab of thickness d is inserted in a, parallel plate capacitor whose negative plate, is at x = 0 and positive plate is at x 3d . The, slab is equidistant from the plates.The, capacitor is given some charge. As one goes, from 0 to 3d (1998), (a) the magnitude of the electric field remains the, same, (b) the direction of the electric field remains of the, same, (c) the electric potential increases continuously, (d) the electric potential increases at first, then, decreases and again increases, What amount of heat will be generated in the, circuit shown after the switch S w is shifted from, position 1 to position 2., , 2CC0, 2CC0, (A) 2C C (B) C C, , , 0, 0, , C, , C, , CO, , S, 2CC0, 2 2CC0, 1, 2, (C) C 2C (D) C 2C, , , 0, 0, , 7. A parallel plate capacitor was lowered into, water in a horizontal position, with water filling, up the gap between the plates. The distance, between the plates is d . Then a constant, voltage V was applied to the capacitor. Find, the water pressure increment in the gap., Dielectric constant of water is k., w, , 0 k 1V 2, 0 k k 1 V 2, (B) p , (A) p , 2d 2, 2d 2, 0 k 1V 2, 2 0 k 1V 2, (C) p , (D) p , 4d 2, d2, 207
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, 8., , 9., , The gap between the plates of a parallel plate, capacitor is filled with glass of dielectric, constant k = 6 and of specific resistivity 100, Gm. The capacitance of the capacitor is, 4.0nF . When a voltage of 2.0kV is applied to, the capacitor, the leakage current of the, capacitor will be, (A) 2.0 106 A, (B) 3.0 10 6 A, (C) 1.5 10 6 A, (D) zero, Two parallel plate capacitors with area A are, connected through a conducting spring of, natural length l in series as shown. Plates P, and S have fixed positions at separation d. Now, the plates are connected by a battery of emf, as shown. If the extension in the spring in, equilibrium is equal to the separation between, the plates, find the spring constant k., , P, , Q, , R, , S, , l, d, , 27 A 0 2, k, , A), 8 d l 3, , 27 A 0, B) k 8 d l 3, , K, E, C, R, , , t , (A) V t E 1 exp , , RC , , (B) V t , , E, 2t , 1 exp , , , 2, RC , , , 2t , (C) V t E 1 exp , , RC , , E, t , 1 exp , , , 2, RC , 12. A capacitor of capacitance C is given a charge, Q. At t=0, it is connected to an uncharged, capacitor of equal capacitance through a, resistance R. Find the charge on the second, capacitor as a function of time., (D) V t , , 2t, , , , RC, Q, 1, , e, , 1) , , , , 2t, , , Q, RC, 1, , e, , 2) 2 , , , , t, t, , , , , , Q, RC, RC, 1, , e, Q, 1, , e, , , 3) 2 , 4) , , , , , 28 A 0, 27 A 0, k, , k, , 13. A capacitor of capacirtance C is given a charge, C), D), 8 d l 3, 8 d l 2, Q. At t=0, it is connceted to an ideal battery of, emf E through a resistance R.Find the charge, 10. The potential difference between the points A, on the capacitor at time t., and B and that between E and F of the circuit, shown in figure respectively are:, A) CEe t / CR Qe t /CR, , B) CE 1 e t / CR Qe t / CR, , 23V, 15F, , 5F, A, , 15F, E, , C) CE 1 e t / CR Qe t / CR, , D) CE 1 e t / CR , 14. A circuit consists of a source of a constant emf, E, a resistance R and a capacitor with, B, F, capacitance C connected in series. The internal, 5F, 15F, 5F, resistance of the source is negligible. At a, A) 5V, 5V, B) 10 V, 5V, moment t= 0, the capacitance of the capacitor, C) 15 V, 5V, D) 0V, 0V, is abruptly decreased to -fold. Find the, 11. In the circuit shown in the diagram, E is the, current flowing through the circuit as a function, e.m.f of the cell, connected to two resistances,, of time t., each of magnitude R and a capacitor of, t, t, , , E, E, capacitance C as shown in the diagram. If the, 1) 1 e CR, 2) 1 e CR, switch key K is closed at time t = 0, the growth, R, R, , t, t, of potential V across the capacitor will be, , , , , , , E, E, CR, CR, , , 1, 1, , e, , , 1, 1, , e, , , , , , 4), , , correctly given by, 3) R, R, 0.75F, , 0.75F, , , , 208, , , , , , NARAYANAGROUP, ,
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JEE-ADV PHYSICS-VOL- II, , MORE THAN ONE ANSWERS, 15. A parallel plate capacitor of plate area A plate, separation d is charged to potential difference, V and then the battery is disconnected. A slab, of dielectric constant K is then inserted, between the plates, of the capacitor so as, to fill the space between the plates.If Q, E and, W denotes respectively, the magnitude of, charge on each plate, the electric field between, the plates (after the slab is inserted), and work, done on the system in the process of inserting, the slab, then (1991), 0 AV, 0 KAV, (b) Q , (a) Q , d, d, , CAPACITORS, 18. The plates of a capacitor are connected to a, source of emf and the energy stored is U 0 ., When dielectric material of dielectric constant, K is introduced between the plates filling the, whole space, the energy stored becomes U., Now, (A) U U 0 (B) U U 0, (C) U U 0, (D) The electric field intensity between the plates is, less after the dielectric material is, introduced., 19. Four capacitors and two batteries are, connected as shown in the diagram. If Va and, Vb denote the potentials of the points ‘a’ and, ‘b’ then, , 0 AV 2 1 , V, 3 F, a, 1 , (c) E , (d) W , , 2d K , Kd, 5F, –, +, 11V, 16. A parallel plate capacitor has smooth square, – 14V, + 10F, plates of side “a”. It is charged by a battery, so that, the charge density becomes . After, b, 2 F, charging, the battery is disconnected. Now a, smooth dielectric slab of length a which can, (A) Va Vb 15V, (B) Vb Va 4V, fill the space between the plates is introduced, (D) Va Vb 10V, (C) Va Vb 4V, between the plates from one side between the, plates., 20. There are 10 identical capacitors each of, capacitance C., (A) If all of them are connected in series, their, +, equivalent capacitance is C/10, K, –, (B) If all of them are connected in parallel, their, equivalent capacitance is 10C., (C) If two series combinations each having 5, (A) The slab can execute SHM between the plates., capacitors are connected in parallel, their equivalent, (B) The plate can execute oscillatory motion which, capacitance is 2C/5, is not SHM, (D) If they are connected as five parallel, (C) The magnitude of the force experienced by the, combinations each one having two capacitors in, slab is constant., series, their equivalent capaitance is 5C/2., (D) The magnitude of the force experienced by the, 21. The plates of a parallel plate capacitor are, slab is not constant., charged by a battery and the battery is, 17. The plates of a capacitor are connected to a, disconnected after the charging. Now, the, source of e.m.f 320V. Between the plates, there, are two dielectric slabs each of uniform, plates are placed as shown in the figure. Then, thickness filling the whole space. One slab A, (plates are not parallel to each other), is of thickness 4mm and dielectric constant 8., +, A, B, The other slab B is of thickness 3mm and, dielectric constant 10., (A) The electric field intensity in A is 5 10 4 N / C, –, (B) The electric field intensity in B is 4 104 N / C, (A), the, surface, charge density is greater at point A, (C) The energy stored per unit area of the capacitor, (B) the surface charge density is greater at point B, is 5.7 104 J ., (C) the potential at points A and B is same., (D) The (induced) bound charge on B is more than, (D) the potential at point A is greater, in A., NARAYANAGROUP, , 209
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, , 22. Select the correct statement(s), 27. The charge on 6 F capacitor is :, (A) Capacitance of a capacitor depends on the, (A) 8c, (B) 35 c (C) 6 c (D) zero, charge on it, (B) Capacitance of a capacitor does not depend PASSAGE-2:, In Figure the plate A has 100 µC charge, while, on the charge on it, the plate B has 60 µC charge., (C) Capacitance of a capacitor does not depend, on the potential difference, A, B, (D) Capacitance of a capacitor does not depend, on the material with which the capacitor plates are, q, q, q, q, prepared., 23. In the given circuit, find the charge on, 1, , C1 , C2 , C3 , C4 and C5 . A battery is connected, as shown in the figure., A, , C1 = 2f, , C, , B, , H, , C4, 2f, , C5, I, , 4, , S2, , 28. When both switches are open, then, E, , 4f, 2f C6, , S1, , 3, , D, , C2 = 4f C3 = 10f, , V= 10v, , 2, , J, , A) q1 80 C, , B) q2 20C, , D) q4 10 C, C) q3 20 C, 29. When only switch S1 is closed, then, A) q1 10 C, , Select the correct options, (A) The charge on C1 10 C, , B) q2 60C, , C) q 3 50 C, D) q4 10 C, 30. When switch S2 is also closed, then, , (B) The charge on C2 20 C, , A) q1 10 C, , (C) The charge on C3 zero, , B) q2 60C, , D) q4 10 C, C) q 3 0 C, (D) The charge on C4 10C, PASSAGE -1:, PASSAGE-3:, In the given circuit, K1 and K 2 are open In the fig. shown the circuit is in steady state., initially and the capacitors are uncharged. After, K1 and K 2 are closed and steady state is, 10, 40, attained. Now answer the following questions., 10F, , A, , ( ), K1, 5F, , H, , 36V, 80, , C, , ( ), , G, , 9V, 4F, , B, , 3F, F, , 7V, , K2, , 6F, , E, , D, , 24. The charge on 4 F capacitor is :, (A) 8c, , (B) 35 c (C) 6 c, , (D) zero, , 25. The charge on 5 c capacitor is :, (A) 8c, , (B) 35 c (C) 6 c, , (D) zero, , 26. The charge on 3 F capacitor is :, (A) 8c, 210, , 20, , (B) 35 c (C) 27 c (D) zero, , 31. what is the potential difference across, the capacitor in steady state?, a) 5V, b) 10V, c) 20V, d) 40V, 32. If the battery is disconnected give the, capacitor charge as a function of time., a) Q 200e 3t /10, , b) Q 200e 3t /100, , c) Q 200e 3t, , d) Q 100e 3000 t, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , CAPACITORS, , 33. How long does it take for the capacitor, to, discharge until the potential difference across, it becomes 1 V?, b) 333.34s, a) 1766.1s, c) 1000s, , LEVEL - V - KEY, , d) 333.34s, , MATRIX MATCHING QUESTIONS, 34. Initially K1 and K 2 are open and the capacitors, are uncharged. After K1 and K 2 are closed, and steady state is attained. Now match the, following for the magnitude of the potential, difference between different points., 1., 9V, , A, , K1, , 4F, H, , C, , K2, 7V, , 3F, F, , (a) n 1 capacitors are in series, So,, , r, s, , Column – II, Between Hand B, Between E and D, Between E and F, Between H and G, , t, , Between E and H, , p, q, , Ceq , , D, , upto 2V =, , 2F, 5, , C3, , 10F, , 3, 20V, , upto 4V =, , C2, 10V, 2, , 4.4s wheen the plates are connected by a thin, wire. find the resistance of the wire in ohms. 4., , NARAYANAGROUP, , 1, 10 42 = 80µJ = u2 (suppose), 2, , Increase in charge = 40 – 20 = 20µC, Energy drawn from cell = 20 x 4 = 80µJ = u, (suppose), Heat produced, = u1 + u – u2, = 20 + 80 – 80, = 20 µJ, , 2F, , 36. An electric field between the plates of a 3., parallelplate capacitor of capacitance, 2.0F drops to one third of its initial value in, , ( log e 3 1.1), , 1, 10 22 = 20µJ = u1 (suppose), 2, , Energy stored in capacitor when it is charged, , 5, C1, , 0 A, 1, , 2d 2n 1 2 , k, , , V1 30V and V2 20V (Conceptual), Energy stored in capacitor when it is charged, , INTEGER TYPE QUESTIONS, 35. In the circuit shown in figure. The charge on, capacitor C2 in steady state is 10 x µC.Find the, value of x., , 2 4 ..., , 1, , 2d 2n 1 2 , k, , , q Ceq q , , 2., , A, , 0 A, , 6F, E, , Column – I, 7 volt, 2 volt, 9 volt, 0 volt, , LEVEL - V - HINTS, , 1, 1, 1, 1, , , .... , , 0 A / 2n1 s, C 0 A / 2 0 A / 4, d, d, d, , 5F, , G, , A, B, C, D, , B, , 1) A 2) B 3)D 4)A 5)B 6)A 7)B, 8)C 9) A 10) A 11)B 12)B 13)C 14) A, 15)A,C,D 16)B,D, 17)A,B,C,D, 18)A 19)C 20)A,B,C,D, 21)A,C 22)B,C,D 23)A,B,C,D 24)A 25)B, 26)C 27)D 28) A,B,C 29)B 30)C 31)C, 32)D 3)C, 34) A p, t ; B s ; C r ; D q, 35) 1 36) 2, , 5., , When S3 is closed, both capacitors attian, common potential which is average of the two., , C, , RC, , 0 A, , d x , , x k 2 x d vt, 3, k, , E 0 d E d 2d (Conceptual), 211
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, 6., , The charge on the condensers in position 1 are, , the capacitor at time t. But q = CV, , q q0 q q0, shown. Here C C C C, 0, 0, , Heance i , , 1, 1, , and q q0 , C C0 C , , dV, 1, , dt. The minus sign is because, V, CR, dV is negative., , Hence, q , , +q, –q, , (a), , q0, +q+q0, C, –q–q0, , –q–q0, C, +q+q0, , 2, , 1, , t , Integrating between limits, V V0 exp , , CR , V V0, t , exp , , R R, CR , This gives the leakage currnt i at instant t. Intitial, leakage current is obtained at t = 0, Hence initial leakage current, , CC0, C 2, and q0 C 2C, C0 2C, 0, , +q0, C, –q0 O, , C, , Heance, , C C C0 , C0 2C, , or q q0 , , b), , Hence i , , q0, –q0, CO, +q0, , –q, C, +q, , i t 0 , , 2, , 1, q0, , , , , , q0, , After the switch thrown to position 2, the charges, change as shown in (Fig - b). A charge q0 has flown, in the right loop through the two condensers and a, charge q0 through the cell, Because of the, symmetery of the problem there is no change in the, energy stored in the condensers. Thus, H (Heat product) = Energy delivered by the cell, , 9., , V0 Q0, Q, , 0, R CR k 0, , 4 109 2 103, , 1.5 106 A ., 9, 12, 100 10 6 8.85 10, Sol: Let charge on capacitors be q and separation, between plates P & Q and R & S be x at any, time. Distance between plates P & Q and R & S is, same because force acting on them is same., Q', P, , CC0 2, Q q0 , C0 2C, , R', , 7., 8., , Let Q0 be the initial charge on the capacitor and, , x, , Capacitance of capacitor PQ, C1 , , 0 A, x, , Capacitance of capacitor RS, C2 , , 0 A, x, , V0 its initial potential. Let A be the area of the, plates and be the distance between them., k 0 A, , , k 0, , A, C, If q be the charge at time t, then the leakage current, Resistance of the capacitor is R , , dq V, where V is the volt age across, i is i , dt R, 212, , S, , d, , Q2, A 0, , Capacitance C of the capacitor is C , , R, , Q, , x, , F, P , A, , F, , dq, dV V, C, ., dt, dt R, , From KVL, we have, , q, q, , , C1 C2, , or, , 0 A, 2x, At this moment extension is spring d 2 x l ., Force on plate Q towards P., q, , F1 , , q2, A2 2 A 2, 0 2 02, 2 A 0 8 Ax 0, 8x, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , CAPACITORS, , Spring force on plate Q due to extension in spring., At equilibrium, separation between plates =, extension is spring, Thus x y d 2 x l , x, , d l, 3, , …… (1), , At equilibrium, F1 F2 …… (2), From equation (1) and (2), we have, , A 2 , A 0 2, x 0 …… (3), , ky, , kx, 8x 2, 8k , From equation (1) and (3),, , q3, 345 4, , 5V, 0.75 92 3, p.d. between E and F is also 5V but in the opposite, direction., 11. (B) The given problem may be approached in a, simplified manner by the following considerations., Since the resistances R and R are in series with the, battery, the potential difference across R(which, charges the capacitor) will have maximum value E/, 2. Also, if the battery is absent, and the (charged), capacitor were to discharge, the two resistors R, and R will be connected in parallel (i.e) effective, resistance across the capacitor is (R/2) and hence, the time-constant in the cirstant in the circuit is, RC / 2 . Thus the growth of potential in the, capacitor will be given by, , p.d. between A and B =, , 3, 27 A 0 2, A 0 2, d l , k , , , 8 d l 3, 8k, 3 , E, E, t , 2t , 10. The distribution of charge is shown in figure V t 2 1 exp 2 1 exp RC , , , , , , , according to the junction rule., 12. At any instance charges on capacitors will be as, q, q –q, q, shown, 1, , 1, , 1, , +, , +, –, q2, , +, –, +, –, , –, , 1, , 3, , +, –, +, –, , –, , +, –, +, –, , 2, , +, , +, –, q2 – q3, , (Q – q), , 3, , i, , +, –, q2, , q, , Applying KVL, we get, , 23V, , dq, q Q q , , iR 0 and i , dt, C, C, 13. Similar to above question. (or) substitute t 0 and, verify the options, 14. Magnitude for on the slab is constant, which is, , Applying loop rules, For loop – 1: , , q2, q, q, 3 1 0, 5 0.75 15, , q1 3q2 20q3 0 ….(1), For loop -2 :, , , q2 q3, q, q q, q, 3 1 3 3 0, 15, 0.75, 5, 0.75, , 3q1 q2 44q3 0 …., , (2), , For loop -3 : 23 , , q2 q1 q3 q2, , 0, 5, 15, 5, , 345 7q2 q3, , …. (3), , Solving for q1 , q2 , q3, q1 , , 19 345, 13 345, 345, , q2 , , q3 , 92, 92, 92, , NARAYANAGROUP, , F, , 0 bV 2 K 1, 2d, , 1, AV 2, 2, 15. U1 CV 0, 2, 2d, Charge remains same electric potential, hence the, field b/w plates decreases by factor K, dU, , where U is, dx, energy stored in capacitor with a part of dielectric, slab in it. Here F is not directly proportional to x ,, but depends on x, , 16. Force on the slab is given by F , , 213
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, E, E, 4 103 3 103 320 or E 4 105, 17., 8, 10, , Therefore, , E, E, 5 10 4 N / C, 4 104 N / C, 8, 10, , 1 2, 4, Energy stored per unit area = cv 5.7 10 J, 2, , R is effective resistance across capacitor, 3 t, , 33. q q0 e t /, , V V0 e t / i 20e1000 t 1000s, .34. Let xV be the potential of the point E and 0V(zero), be the potential of the point C then, potential of H, is 7V. So, by charge conservation at the point H,, , 9 7 x 4 5 7 x x 6 9 x 3 0, So, potential at H 7Volt, x0, on B is greater than on A because K B K A, Also, potential at G 9Volt, 18. As battery is connected , V is constant, E is also 35. Use the concept of CR-circuit., constant, t, t, 19. P.d across 2 F capacitor, 36. q q0 e t / E 0 e t / ln 3 rc , , ln 3, 3, 4.4, 11 14 , 15V, R, 2, 3 2, 1.1 2, Va Vb 11 15 4V, , LEVEL - VI, , 20. Apply series and parallel combination of capacitors, 21. On a conductor potential is same as every point, 2, C1, A, , D,E, , SINGLE ANSWER QUESTIONS, , 2, C4, , C1G, , J,I,H, C2, 4, , 1., B, , C5, 4, , Seven capacitors each of capacitance 2 F are, connected ina configuration to obtain an, 10, F . Which of the, 11, following combination will achieve the desired, result ?, , 23., , effective capacitance of, 10v, , From the diagram , we can find the following, On C1 10 c On C4 20 c, On C2 20 c On C5 20 c, , (a), , (b), , (c), , (d), , On C6 20 c On C3 zero, 24-27. Apply KVL and solve the equations, 28 (a) q1 80 C, q3 20 C, , q2 20C, q4 80 C, , 29 When S1 is closed , q1 0 and q2 60 c so 2., that VA 0, (b) q1 0, , q2 60 C q3 60 C q4 0, , 30 When both S1 & S 2 are closed ;all charges will be, zero, (c) q1 0 q2 0, q3 0, q4 0, 31. In steady state capacitor is equivalent to open circuit, VC 20V, .32. q q0 e, 214, , t / r, , q0 200 c, , A parallel plate capacitor of capacitance C is, connected to a battery and is charged to a, potential difference V. Another capacitor of, capacitance 2C is similarly charged toa, potential difference 2V . The charging battery, is now disconnected and the capacitors are, connected in parallel to each other in such a, way that the positive terminal of one is, connected to the negative of the other. The, final energy of the configuration is (1995), (a) zero, , RC, , (b), , 3, 25, 9, CV 2 (c), CV 2 (d) CV 2, 2, 6, 2, NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 3., , CAPACITORS, , A parallel plate capacitor of area A, plate 6., separation d and capacitance C is filled with, three different dielectric materials having, dielectric constants K1 , K 2 and K 3 as shown.If, a single dielectric material is to be used to have, the same capacitance C in this capacitor then, its dielectric constant K is given by (2000), A/2, , A/2, , K1, , K2, , d/2, d, , 7., , K3, , 1, 1, 1, 1, 1, 1, 1, (a) K K K 2K (b) K K K 2K, 1, 2, 3, 1, 2, 3, K1 K 3, K2K3, 1, K1K 2, (c) K K K 2K 3 (d) K K K K K, 1, 2, 1, 3, 2, 3, , 4., , Consider the situation shown in the figure.The, capacitor A has a charge q on it whereas B is, uncharged .The charge appearing on the, capacitor B a long time after the switch is closed, is (2001), , Two identical capacitors, have the same, capacitance C.One of them is charged to, potential V1 and the other to V2 . Likely, charged plates are then connected.Then the, decrease in energy of the combined system is, (2002), 1, 1, 2, 2, 2, 2, (a) C V1 V2 , (b) C V1 V2 , 4, 4, 1, 1, 2, 2, (d) C V1 V2 , (c) C V1 V2 , 4, 4, A parallel plate capacitor is located, horizontally, so that one of its plates is, submerged into liquid while the other is over, the surface. The dielectric constant of the, liquid is equal to k. Its density is equal to ., To what height will the level of the liquid in the, capacitor rise after its plates get a charge of, surface density ?, , k 1 2, , (A) h , , 2 0 k g, , q, +, , –, , +, , –, , +, , –, , +, , –, , +, , –, , S, , 8., B, , (a) zero, (b) q / 2 (c) q, (d) q, Four capacitors and a battery of emf 24V are, connected as shown in the figure. Initially, S, is open and the capacitors are uncharged., After S is closed and steady state is attained,, the potential difference acoss the 4 f, capacitor is 7.5 V. The capacitance in f of 9., the capacitor A is, S, , A, , A X, 10F 10F, , 3F, , 8F, 6F 6F, X B, 5F, , The equivalent capacitance in F between A, and B is, (A) 3.75, (B) 8.5, (C) 10.25 (D) 15, Four capacitors and two batteries are, connected as shown in the diagram. The p.d, between the points a and b is, , 3F, 22V, , 2 F, , 4F, , 8F, , (B) 7, , NARAYANAGROUP, , (C) 8, , (D) 10, , (A) 22V, , a, , 5F, , b, , (A) 5, , 0k g, , 2 k 1 2, (D) h 0, 0k g, Seven capacitors are connected as shown in, the figure., , 2F, , 24V, , k 1 2, , (C) h , , A, , 5., , (B) h , , (B) 15V, , 7V, , 3F, , (C) 13V, , (A) 0V, 215
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, , MORE THAN ONE ANSWERS, , 210, C, 11, 120, C, (B) The final charge on C2 , 11, (C) The final charge on C2 zero, , (A) The final charge on C1 , , 10. A parallel plate air capacitor is connected to a, battery.The quantities charge, voltage, electric, field and energy associatated with this, capacitor are given by Q0 , V0 , E 0 and U 0, respectively.A dielectric slab is now introduced, 45, to fill the space between the plates with the, C, (D) The final charge on C 3 , battery still in connection. The corresponding, 11, quantites now given by Q, V, E and U are 14. A parallel plate capacitor has smooth square, plates of side “a”. It is connected to battery of, related to the previous one as (1985), emf. V. Now a smooth dielectric slab of length, (a) Q Q0 (b) V V0 c E E 0 (d) U U 0, a which can fill the space between the plates is, 11. A parallel plate capacitor is charged and the, introduced between the plates from one side, charging battery is then disconnected.If the, as shown in the figure. Now, plates of the capacitor are moved farther apart, by means of inslulting handles, +, (a) the charge on the capacitor increases, (b) the voltage across the plates increases, –, (c) the capacitance increases, (d) the electrostatic energy stored in the capacitor, (A) The slab can execute SHM between the plates., increases, (B) The plate can execute oscillatory motion which, is not SHM, 12. Three parallel plate capacitors C1 4 F ,, (C) The magnitude value of the force experienced, C2 2 F , C3 6 F with respective charges, by the slab is constant., (D) The magnitude of the force experienced by, q1 20 C , q2 10 C , q3 5C are, the slab is not constant., connected in series with a battery of emf 10V, through an open switch as shown. Now, the 15. Two parallel - plate capacitors with different, distances between the plates are connected in, switch is closed and steady state is reached., parallel to a voltage source. A point positive, C, C, C, –, –, –, +, +, +, charge is moved from a point A,that is exactly, in the middle between the plates of a capacitor, S, C1 to a point B (or a capacitor C2 that lies at, 1, , 2, , 3, , a distance from the negative plate of C2 equal, , V, , (A) The final charge on C1 30 C, , to half the distance between the plates of C1 ), During this process of moving the charge, , (B) The fial charge on C2 20 C, , +, , (C) The final charge on C2 zero, (D) The final charge on C3 15 C, 13. Three parallel plate capacitors C1 4 F ,, , C2 2 F , C3 6 F with respective charges, q1 20 C , q2 10 C , q3 5C are, connected in series through an open switch as, shown. Now, the switch is closed and steady, state is reached., –, , +, C1, , 216, , –, , +, C2, , –, , +, C3, , Sw, , A, , B, C1, , C2, –, , (A) The potential at A is greater than potential, at B, (B) The potential at A is less than potential at B, (C) The work done to move that point positive, charge from A to B is greater than zero., (D) The work done to move the point positive, charge from A to B is zero., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , CAPACITORS, , 16. A parallel - plate capacitor is filled with a 20. In the circuit shown in the figure four capaciors, dielectric upto one - half of the distance, are connected to a battery. The p.d across the, between the plates. The manner in which, the, 7 F capacitor is 6V.., potential between the plates varies is, illustrated in the figure., E, 10V, , A, , B, , 0V, , 12F, 3.9F, a, , b, (A) The electric field strength is greater in part A, 7 F, 3 F, (B) The electric field strength is greater in part B, (A) The p.d across the 12 F capacitor is 10V.., (C) Dielectric slab fills the space of part A, (D) Dielectric slab fills the space of part B., (B) The charge on the 3 F capacitor is 42 C ., 17. A capacitor of capacitance 2 F is initially, (C) The p.d across 3.9 F capacitor is 10V, connected to a battery of emf 10 volt and, (D) The emf of the battery is 30V ., steady state is reached. Now, 10V battery is, removed and another battery of emf 20V is, PASSAGE TYPE QUESTIONS, connected with like polarities together. Find, the amount of heat energy developed (after PASSAGE - 1, In the circuit diagram shown here seven, the steady state is reached), capacitors are connected to a source of emf E., (A) 104 J (B) 2 104 J (C) 3 10 4 J (D) zero, Initially, the switch S is closed and steady state, 18. Solve the above problem, if 20V battery is, attained. Now, the potential difference between, connected with unlike polarities together., the points ‘a’ and ‘b’ is 4 volt. Now, answer the, (A) 4 104 J, (B) 9 104 J, following questions:, (D) 104 J, (C) 2 104 J, i, f, e, a, 19. A conducting slab is in between a parallel plate, capacitor .The thickness of the slab is t. The, 10F, 12F, distance between the plates is d (d > t) . A, battery of emf ‘v’ volt is connected across the, 4.5F, 6F, b, plates (see diagram). The system attains, steady state. Now select the correct options., 5F, , 30F, , , , , h, , , , t, , , , , , , , 6F, g, , d, , 21. The emf E of the battery is :, (A) 46V, (B) 26V (C) 14V, 22. The charge on 4.5 F is :, , (A) 90C (B) 45C (C) 54C, (A) The electric field inside the conducting slab is zero. 23. The charge on 10 F is :, (B) The charge density induced on the surface of the, (A) 70 c (B) 45 c (C) 75 c, slab should be equal to the charge density on the, plates of the capacitor., 24. The charge on 5 F is :, (C) The electric field inside the slab always be zero,, (A) 70 c (B) 45 c (C) 75 c, whatever be the metal with which the slab is, prepared., 25. The charge on 12 F is :, (D) The capacitance of the capacitor is increased after, (A) 48 c (B) 90 c (C) 120 c, the introduction of the slab., NARAYANAGROUP, , c, , (D) 10V, (D) zero, (D) 120 c, (D) 120 c, (D) 60 c, 217
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, PASSAGE-2, , MATRIX MATCHING QUESTIONS, , Four capacitors C1 1 F , C2 2 F , 29. In the circuit diagram shown here, seven, capacitors are connected to a source of emf E., C3 3 F and C4 4 F are connected in, Initially, the switch S is closed and steady state, attained. Now the potential difference between, a network as shown in the diagram. The emf, the points ‘a’ and ‘b’ is 4 Volt. Now, match the, of the battery is E =12V and its internal, following; for potential difference between, resistance is negligible. The keys S1 and S2, different points., can be independently put on or off. Indicate, i, f, e, a, the charge on the capacitors by, 10F, 12F, , q1 , q 2 , q 3 and q 4 respectively and the potential, drops across them by V1 , V2 , V3 and V4, respectively., C1, , C3, S2, , +, , –, C2, , E, , C4, , S1, , 12V, , 4.5F, , b, , 6F, 5F, , 30F, h, , 6F, g, , c, , d, , Column – I, p, A, 12 volt, q, B, 10 volt, r, C, 14 volt, s, D, 22 volt, , Column – II, P.d Between g and b, P.d Between g and d, P.d Between g and i, P.d Between g and a, P.d Between a and d, , t, 26. Initially both the keys are open. Then the key, S1 is closed. Then the charges on the, SUBJECTIVE QUESTIONS, capacitors are, 30. Two parallel plate capacitors A and B have the, A) q1 q 2 16 C, q 3 q 4 9 C, same separation d 8.85 104 m between, the plates. The plate areas of A and B are 0.04, B) q1 q 3 16 C, q 2 q 4 9 C, 2, m 2 and 0.02m respectively. A slab of, C) q1 q 4 9 C, q 2 q 3 16 C, dielectric constant (relative permitivity ), D) q1 q 3 9 C, q 2 q 4 16 C, K 9 has dimensions such that it can exactly, fill the space between the plates of capacitor, 27. Initially key S2 is closed. Then the key S1 is, B.(IIT 1993), now closed. Then the charges on the capacitors, are, A, , A) q1 q 2 24 C,q 3 q 4 12 C, , B, , B A, , B) q1 q 2 12 C, q 3 q 4 24 C, C) q1 10.8 C, q 2 14.4 C,, 3q 3 2q 4 , 2q 4 25.2 C, D) 2q1 q 2 16.8C, 4q3 3q 4 43.2 C, 28. Initially key S2 is open. The key S1 is closed, and the capacitors are charged. If now the key, S2 be closed, the charge that will flow across, this key is, A) 2.4 C B) 0.4 C C) 0.2 C D)1.2 C, 218, , 110V, , 110V, , a) The dielectric slab is placed inside A as shown, in figure(i). Calculate the capacitance of A and the, energy stored in it., b)The battery is disconnected and then the dielectric, slab is removed from A. Find the work done by the, external agency in removing the slab from A., c)The same dielectric slab is now placed inside B,, filling it completely. The two capacitors A and B, are then connected as shown in figure (iii). Calculate, the energy stored in the system., NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, , CAPACITORS, , 31. Two square metal plates of side 1 m are kept 34. Find how the voltage across the capacitor C, 0.01 m apart like a parallel plate capacitor in, varies with time t after closing of the switch, air in such a way that one of their edges is, S w at the moment t 0 ., perpendicular to an oil surface in a tank filled, Sw, with an insulating oil.The plates are connected, to a battery of emf 55V. The plates are then, R, C, , lowered vertically into the oil at a speed of, R, 0.001 ms 1 .Calculate the current drawn from, the battery during the process., 35. In the circuit shown, a capacitor charged to a, (Dielectric constant of oil = 11,, potential difference V0 is connected to an, 12, 2, 1, 2, ), 0 8.85 10 C N m, uncharged capacitor through a resistor at, 32. Two capacitors A and B with capacities, t 0 , by closing the switch. Find current in the, 3 F and 2 F are charged to a potential, circuit as function of time., difference of 100V and 180 V respectively.The, R, plates of the capacitors are connected as shown, in the figure with one wire of each capacitor, C, C, free.The upper plate of A is positive and that, of B is negative.An uncharged 2 F capacitor, Sw, C with wires falls on the free ends to complete 36. A circuit has a section AB shown in figure. The, the circuit. Calculate ., emf of the source equals 10V , the, C, capacitor capacitances are equal to, 2 F, + 3 F, A, , 100V, , C1 1.0 F , C2 2.0 F , and the potential, 2 F –, 180V, , difference A B 5.0V . Find the voltage, across each capacitor., , , B, , a)The final charge on the three capacitors and, B, A, b)The amount of electrostatic energy stored in the, system before and after completion of the circuit., C1, C2, 33. The capacitance of a parallel plate capacitor 37. In a circuit shown in figure find the potential, with plate area A and separation d, is C, The, difference between the left and right plates of, space between the plates is filled with two, each capacitor., wedges, of, dielectric, constants, K1 and K 2 respectively (figure).Find the, C1, 1, 2, capacitance of the resulting capacitor, A, , C2, 38. Find the charge of each capacitor in the circuit, shown in figure., , K2, d, K1, , 1, , C1, , 2, , C2, NARAYANAGROUP, , 219
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, , LEVEL - VI - KEY, 1)A, 2)B, 3)D, 4)A 5)D, 6)C, 7)A, 8)A, 9)C, 10)A,D, 11)B,D, 12)A, C, D13)A,B,D, 14)B,D, 15)A,C, 16)B, C 17)A 18)B, 19)A, B, C, D 20)A,B,D 21)A 22)B 23)C, 24)D 25)A 26) D 27)D, 28)A, 30)(a) CA 2.0 109 F , U A 1.21105 J, (b) W 4.84 10 J, , (c) U 1.1 10 J, 5, , 31) 4.43 10 9 A, 32)(a) q1 90 C, q 2 210 C, q 3 150 C, (b) U f 18mJ , U i 47.4mJ, CK 1 K 2, K2, 0A, 33) K K ln K ; where C d, 2, 1, 1, V 2t / RC, 1, 2t / RC, 34) V 1 e, , 35) 0 e, 36)10V, 5V, 2, R, , 37) V1 , 38) q , , q 2 1 C2, , C1, C1 C2, , V2 , , q 2 1 C2, , C1, C1 C2, , 2 1 C1C2, C1 C2, , U decreases in potential energy, , 8., , The p.d. across the 8 F capacitor is zero., C 2, , 9., , P.D, , 5 3, 10 6, 3.75 F C , 3.75 F, 53, 10 6, across, is, 2 F capacitor, , 22 7 , , 29) A t ; B r , s ; C p ; D q, 5, , 6., , 3, 9V, 3 2, , p.d across 8 F capacitor is 22V, As the positively charged plates of 2 F and, 8 F are connected Va Vb 22 9 13V, 10. As battery is not disconnected from the capcitor,, , V V0 , E E0 , C KC0 , Q KQ0 , U KQ0, 11. C decreases, so V increases and U also increases, 12. After the switch is closed, let a charge Q be driven, by the battery. Now , new, charges on capacitors would become,, , Q1 20 Q C ; Q2 10 Q C, Q3 5 Q C, , 20 Q 10 Q 5 Q, , , 10, 4, 2, 6, , 60 3Q 60 6Q 10 2Q 120, 11Q 110 ; Q 10C, 13. Let the final charges on the capacitors be, , LEVEL - VI - HINTS, , a1 , q2 & q3, , 10 1 10 F, , q1 q2 q3, 0 ....................(1), 4 2 6, , 1., , C net , , 2., , U' , , 1 ' '2 1, 3, C V 3C V 2 CV 2, 2, 2, 2, , 3., , K, , K1 K 2, K2K3, , K1 K 2 K 2 K 3 (Conceptual), , 10 1, , 11, , q1 q2 30 ........................(2), q1 q3 15 ...........................(3), solve these equations and get the answers., 14. Magnitude of force on the slab is constant, which is, , 0 bV 2 K 1, Due to attraction with positive charge, no negative, F, charge on capacitor A will flow through the switch S., 2d, 5. P.d across the plates of, 16. In dielectric slab, electric field is lesser than that of, outside region, 7.5 4 , , A 24 7.5 , 17. When 10 volt battery is connected,, 3 6.5V, , 1, 6, 4, q1 20 C U1 2 10 100 10 J, Charge on A 17.5 2 4 7.5 65c., 2, 65, When 20V battery is connected, q2 40 C, 10 f, Capacitance of A , 6.5, 4., , 220, , NARAYANAGROUP
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JEE-ADV PHYSICS-VOL- II, 1, U 2 2 106 400 4 10 4 J, 2, q 40 20 20 C W.D by the battery =, , 20 q 20 20 106 4 104 J, , CAPACITORS, If V be the potential of the four common plates, joined by S2 , when S1 is closed., Hence, , q1 12 3.6 1 8.4 C, q 2 12 3.6 2 16.8 C, , q 3 3.6 3 10.8 C, q 4 3.6 4 14.4 C, Heat developred = U1 U 2 + W.D by the, battery 104 4 104 4 104 104 J Am. 28. Before S2 is closed, C1 and C3 are in series and, hence the total charge on their common plate is, 18. q1 20 C U1 104 J q2 40 C, Q Q 0 . Similarly for the plates joining, U 2 4 10 4 J Q 20 40 60 C, C2 andC4 . However when C2 and C4 is closed, the, (because, polarity has changed), 6, 4, charges, read just to the values calculated in problem, , , W.D by the battery = 20 60 10 12 10 J, 86. Hence total charge on the linking platesofC1 and, Heat U1 U 2 W .D 104 4 104 12 104, C3 is 8.4 C 10.8 C 2.4 C while the, 9 104 J, total, charge on the linking plates of, 19. Electric field in a conductor is always zero, C2 andC4 is 16.8 C 14.4 C 2.4 C which, 73 76, , 6, 120, , C, , , , 20. q 3.9 , shows that a charge of 2.4 C has crossed the key, 7 3 , 3 , , S2 from C2 , C4 to C1 , C3 ., 120, 10V A , p.d across 12 F capacitor , 120, 12, C Charge supplied by battery (i.e), 29. Ceff , 46, Charge on 3 F capacitor 7 6 42 C B , C5 120 C E 46V Q fg 45C Q10 75C, p.d. across, 76 , 3.9 F capacitor 6 , 20V, 3 , , 30. (a) CA 2.0 109 F , U A 1.21105 J, (b) W 4.84 105 J (c) U 1.1 105 J, , The emf of the battery 20 10 30V D , , 0 A, dQ Q CV , C x, dx, And v , 31. i , , d, , x, , , dt, dt, K, 32. (a) q1 90 C, q 2 210 C, q 3 150 C, , 120, C, 46, Charge supplied by battery (i.e), , 21-25: Ceff , , C5 120 C E 46V Q fg 45C Q10 75C, 26. Refer to the circuit network in the problem., With both keys S1 and S2 initially open, if now, , (b) U f 18mJ, CK 1 K 2, K2, 0A, 33. K K ln K ; where C d, 2, 1, 1, , 1, 2t / RC, 34. V 1 e, , 2, Sol. Let, at any moment of time, charge on the plates be, VC1C3 12 1 3, q and q respectively, then voltage, Hence q1 q 3 C C 1 3 9 C, q, 1, 3, across the capacitor, 1, C, VC2C4 12 2 4, q, , q, , , , 16, , C, Now,, from, charge, conservation,, 2, 4, Also, C2 C4, 2 4, dq, i i1 i 2 , Where i 2 , 2, 27. If initially S2 is closed, C1 and C2 are in parallel, as, dt, also C3 and C 4 . Effectively, In the loop 65146, using 0 ., , S1 is closed, the capacitance C1and C3 are in, series as also the capacitors C2 and C4, , therefore C1 C2 and C2 C4 are in series, when key S1 is closed., NARAYANAGROUP, , q , dq , i1 R 0 3, C , dt , 221
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JEE-ADV, PHYSICS-VOL- II, SR-MAIN-CHEM-VOL-II, , CAPACITORS, , 36. 10V, 5V, Let, us make the charge distribution,, shown in the figure., q, q, Now, A B C C, 1, 2, , Using eq. 1 and 2 , In the loop 25632, using 0, q, q, i1R 0 or i1R 4 , C, C, 4, , 3, i, , 6, , or, q , , i1, R, , , , +q, –q, , C, , i, 1, , 2, , 5, , A B , C1 C 2, , a, , s, , C1C 2, , Hence, voltage across the capacitor C1, , , q A B , , C 2 10V, C1, C1 C 2, , From (1) and (2),, and voltage across the capacitor, C2, dq, 2q, dq, dt, R 1 , or,, 5, 2q R, q A B , dt, C, , , , C1 5V, C, C2, C1 C2, On integrating the expression 5 between, q q q q, .B, A., suitable limits,, C, C, 2q, q, q 2 1 C 2, q 2 1 C 2, , t, dq, 1, C, t, 1 , , 2 , , C, ;, 37., C1, C1 C 2, C1, C1 C 2, q 2q R 0 dt or, 2 in R, , Let 2 1 , then using 0 in the closed, C, circuit, (Fig), q, 1, 2t/RC, , , , , V, 1, e, thus, , , +q C1 –q, C, 2, A, B, V0 2t / RC, e, 35., R, 1, 2, Sol. Let, at any moment of time, charge flown be q then, dq, current i , dt, Applying loop rule in the circuit, 0 , we get:, D, C, –q +q, C, 2, dq, 1, CV0 q q 0, dq, q, q, IR , C C, or CV 2q RC dt, 2 , 1 0 q 2 1 1 2, dt, C, C, 0, C1, C2, C1 C 2, CV0 2q, t, Hence the p.d across the left and right plate of, 2, So, In CV, RC for 0 t t, capacitors,, 0, 2t, , q 2 1 C 2, CV0 , RC, 1 , , and similarly, or, q 2 1 e , C1, C1 C 2, , , 2, , 1, , 2t, dq CV0 2 RC, V, , e 0 e 2t / RC, Hence, i , dt, 2 RC, R, , 2 , , R, , 38. q , –q, C, +q, , –(CV0–q), C, , dq, i, dt, 222, , q 2 1 C 2, , C1, C1 C 2, , 2 1 C1C2, , C1 C2, Taking benefit of the foregoing problem, the amount, of charge on each capacitor, q , , 2 1 C1C2, C1 C2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , CURRENT ELECTRICITY, v is linear velocity of the charge q, r is radius of the circular path, , SYNOPSIS, Strength of Electric Current, , 1., , The strength of electric current is defined as rate of, flow of charge through any cross section of a, conductor., The instantaneous current is defined by the, equation,, Q dQ, I Lt, , t 0 t, dt, q, Average current i , t, Ampere : If one coulomb of charge passes, through a cross-section of the conductor per, second then the current is one ampere., 1 ampere =, , 1coulom b, 1sec ond, , current is a scalar quantity., , Applications on electric current, 1., , If the current is varying with time t, then the charge, flowing in a time interval from t1 to t 2 is, t2, , q Idt, , 3., , 4., , If in a discharge tube n1 protons are moving from, left to right in t seconds and n2 electrons are moving, simultaneously from right to left in t seconds, then, the net current in any crossection of the discharge, tube is, (n1 n2 )e, (from left to right), t, here e is the magnitude of charge of electron (or), proton., DRIFT VELOCITY : Drift velocity is the, average velocity acquired by free electrons, inside a metal by the application of an electric, field which results in current., I, , J, I, , ne Ane, where, J = I/A is current density, n is number of free electrons per unit volume, e is charge of electron, The drift velocity is related to relaxation time, is, , Drift velocity vd , , eE, , m, If n particles, each having a charge q, pass, through a given cross sectional area in time t, Note : 1.The drift velocity of electrons is of the order, of 104 ms 1 ., nq, then average current is i =, t, 2. Greater the electric field, greater will be the, If a point charge q is revolving in a circle of, drift velocity vd E, radius r with speed v then its time period is, 3. The direction of drift velocity for electrons in a, metal is opposite to that of electric field applied, V, , r, E, q, W.E-1:, In a hydrogen atom, electron moves in an, T (2 r / v), orbit of radius 5 × 10-11 m with a speed of 2.2 ×, 106 m/s. Calculate the equivalent current., The average current associated with this, v, revolving charge is, .e, Sol: Current i f .e , 2 r, q, , vq, I fq , q, 2.2 106, T, 2, 2 r, =, × 1.6 × 10–19, Where f is the frequency of revolution in Hz., 2 5 1011, is the angular frequency in rad/sec, = 1.12 × 10–3 amp = 1.12 mA., t1, , 2., , 5., , NARAYANA GROUP, , vd , , 1
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-2: The current through a wire depends on, 2, i, 29, ms 1, Sol: vd , 6, 19, time as i i0 t , where i0 10 A and, nAe 10 10 1.6 10, = 12.5 × 10–4 ms–1, 4 A / s . Find the charge that crossed, , Average time taken by an electron to cross the, length of wire, , through a section of the wire in 10 seconds., Sol: i i0 t ; but i , , dq, dt, , t, , dq (i0 t )dt, t 10, , q, , 2, , Mobility ( ) : Mobility ( ) of a charge, carrier (like electron) is defined as the average, drift velocity resulting from the application of, unit electric field strength., , 10, , , t , dq q i0 t , , 2 0, , t 0, , , , = (10i0 50 ) = 300 coloumb, , Current Density ( J ) : Current density at a, point is defined as a vector having magnitude, equal to current per unit area., , I dI, J Lt, nˆ, s 0 s, ds, If the normal to the area makes an angle with, the direction of the current, then the current, density is, , dI J .ds, , SI unit of, , I, J, , dI Jds cos (or), s cos , , i.e., I J .ds, , –2, J is Am, , l, 4, 4, , vd 1.25 104 s = 3.2 × 10 s, , drift velocity, electric field ; , , | vd |, E, Mobility depends on pressure and temperature., OHM’S LAW : For a given conductor, at a, given temperature the strength of electric, current through it is directly proportional to, the potential difference applied across at its, ends”., , , , , , V, ; V = IR, R, Where R is electrical resistance of the conductor, Note :, ohm’s law is neither a basic law nor a deriavable, one, ohm’s law is just an empherical relation., Microscopically Ohm’s law is expressed as, , Dimensional formula of J is [ AL2 ], Current is the flux of current density., Relaxation time ( ) : 1. It is the time interval, between two successive collisions of electrons, with +ve ions in the metallic lattice.The , resistance of a conductor is given by, , i.e. I V I , , J nevd J E where is the, electrical conductivity of the material., The conductors which obey Ohm’s law are, called Ohmic conductors., Ex : all metals, For Ohmic conductors V – i graph is a straight, line passing through origin (metals)., , 2ml, , ne 2 A, where n = number density of electrons, T, e = electron charge, 1, V, m = mass of electron, V, T, = relaxation time., 2, W.E-3: Consider a wire of length 4m and cross, sectional area I mm2 carrying a current of, , , i, 2A. If each cubic metre of the material, i, contains 1029 free electrons, find the average, (A) Slope of the line, (B) Here tan 1 tan 2, time taken by an elctron to cross the length, So R1 R2 i.e, of the wire., tan v / i R ; T1 T2, , R=, , 1, , 2, , , , , , 2, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , The substances which do not obey Ohm’s law are, called non-Ohmic conductors., Ex: Thermistor, Electronic Valve, Semiconductor devices, gases, crystal rectifier etc.,, The V – i graph for a non – Ohmic conductor is, non-linear., , , , , , , , , (mA), , i, , i, , , X, , 1, , X, , 0, , V, , V, , Factors Effecting the Resistance of A, Conductor, , (A), i, , V, vaccum tube, , semi conductors, , 1., , i, , i, i2, , V, , i1, i, , V1, V, Thermistor, , dilH2SO4(Platinum Electrode), , 2., 0, , V, , R1 A2 r22 , , R2 A1 r12 , For small changes in area (or) radius we have, 1, 1, i.e., R (or) R 2 ;, A, r, , V, i, Neon Gas(With tungsten Electrode), , Noo o ooooo oooooooo : The circuits in which, , Ohm's law is not obeyed are called non-ohmic, circuits. The V-I graph is a curve, e.g. torch bulb,, electrolyte,, semiconductors, thermonic valves etc. as shown, by curves (a), (b), (c)., I, , I, , Semi conductor, , b), , Diode, , 3., , R A, 2r, , , R, A, r, As the temperature increases resistance of, metallic conductors increases and that of, semiconductors decreases., Conductance: The reciprocal of resistance (R), is called conductance., 1, ., R, The S.I unit of conductance is mho or siemen or, ohm-1., Resistivity: As we know, that the resistance of, the conductor is directly proportional to its, length and inversely proportional to its area of, cross section, we can write, l, l, R R, A, A, where is specific resistance or resisitivity, of the material of the conductor., , conductance, G , , V, , V, , I, , c), V, , Resistance-Definiton : The resistance of a, conductor is defined as the ratio of the potential, difference ‘V’ across the condutor to the current, ‘i’ flowing through the conductor., NARAYANA GROUP, , R l, , R, l, The resistance of a conductor is inversely, proportional to the area of cross-section (A), , For small changes in the length,, , i, , a), , The resistance of the conductor is directly, proportional to the length (l) of the conductor, i.e., R1 l1, R l (or) R l, 2, 2, , 0, , V, , V, i, The resistance of a conductor depends upon, 1) shape (dimensions) 2) nature of material, 3) impurities, 4) Temperature, The resistance of a conductor increases with, impurities., The resistance of a semi conductor decreases, with impurities., , Resistance R , , 3
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, Note:, 1. Resistitivity is the specific property of a material, but Resistance is the bulk property of a, conductor., 2. Resistivity is independent of dimensions of the, conductor such as length, area of the cross, section., 3. Resistivity depends on the nature of the material, of the conductor, temperature and impurities., 4. Resistivity of any alloy is more than resistivity, of its constituent elements., i), , R alloys, , R conductors, , ii), , metals, , 2., , a., b., c., , l 2, R l 2, V, V, 1, 1, R 2 R 2 4, A, A, r, , R, , Interms of mass of the wire R, , 4., , l2, m, , m m, , A2 r 4, For small changes in the length or radius during, the stretching, R, l, R, A, r, 2, 2, 4, ;, R, l, R, A, r, In case of a cuboid of dimensions l b h is, , and R, , 3., , 5., 6., , The alternate forms of resistance is, l2, l 2 d V, m, , 2 , V, m, A, d A2, Where d is density of material of conductor, V is volume of the conductor, m is mass of the conductor, If a conductor is streched or elongated or drawn, or twisted, then the volume of the conductor is, constant. Hence, , l, h, Rmin , bh, l b, If a wire of resistance R is stretched to ‘n’ times, its original length, its resistance becomes n2 R., If a wire of resistance R is stretched until its, 1, radius becomes th of its original radius then, n, its resistance becomes n4R., When a wire is stretched to increase its length, by x% (where x is very small) its resistance, increases by 2x %., When a wire is stretched to increase its length, by x% (where x is large) its resistance increases, Rmax , , 7., , R, , 8., , 2 , , 2x x , by , 100 ., , 9. When a wire is stretched to reduce its radius, byx% (where x is very small), its resistance, increases by 4x%., W.E-4: A rectangular block has dimensions 5 cm, × 5 cm × 10 cm. Calculate the resistance, measured between (a) two square ends and, (b) the opposite rectangular ends. Specific, resistance of the material is 3.5 105 m ., , Sol: a) Resistance between two square ends R1 , R1 , , 3.5 10 5 10 10 2, 5 5 10, , 4, , , A, , 1.4 103 , , 5 cm, , F, , h, , h, l b, , If l b h , then, , alloys, , Special Cases :, 1., , Resistance across EF, REF , , 5 cm, , 10 cm, , C, b, , F, , A, , E, , B, , D, E, , ends R2 , , l, , l, Resistance across AB, RAB , b h, b, Resistance across CD, RCD , lh, 4, , b) Resistance between the opposite rectangular, , R2 , , , A, , 3.5 10 5 5 10 2, 5 10 10 4, , 1.4 10 4 , , Conductivity: Conductivity is the measure of, the ability of a material to conduct electric, current through it. It is reciprocal of resistivity., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , 1, l, , RA, S.I unit : sieman / m : (Sm–1), For perfect insulators 0, For perfect conductors, is infinity.., , CURRENT ELECTRICITY, , , , , , , Temperature dependence of resistance:, , R1 R2 R1' R2', , For conductors i.e metals resistance increases, with rise in temperature, , R1 R2 R1 (1 1t ) R2 (1 2t ), R11 R2 2, Variation of resistance of some materials, , R t R o (1 t t 2 ) for t > 300oC, , R t R o (1 t) for t < 300oC or, , , , Rt Ro, 0, R ot / C, , Material, , R 0 = resistance of conductor at 0oC, If R t = resistance of conductor at toC And, , If, , , = temperature co-efficients of resistance, If R1 and R2 are the resistances at t1oC to t2oC, R1 1 t1, respectively then R 1 t, 2, 2, R R1, 2, R1t2 R2t1, The value of is different at different, temperatures., , 1, At a given temperature R, t, , dR , dt at t0C, , , , du, c, fo, rc, on, , Resistivity, , , , 0, , Metals, , Positive, , Increases, , Solid nonmetal, , Zero, , independents, , Semiconductor, , Negative, , Decreases, , Electrolyte, , Negative, , Decreases, , Ionized, gases, , Negative, , Decreases, , Alloys, , Small positive, value, , Almost, constant, , Variation of Resistivity with Temperature:, , , If 1 is the resistivity of a material at temperature, at temperature t 2 ,then, , for manganin and constantan, for, se, , Temp. coefficient, Variation of, of, resistance with, temperature rise, resistance (), , t1 and 2 is the resistivity of the same material, , to, rs, , Y, , The resistivity of manganin and constantan is almost, independent of temperature., Two resistors having resistances R1 and R2, at 0o C are connected in series. The condition, for the effective resistance in series in same, at all temperatures, , mic, ond, u, , 2 1 1 t 2 t1 , , W.E-5: The temperature coefficient of resistance, of platinum is 3.92 103 K 1 at 0° C. Find, the temperature at which the increase in the, resistance of platinum wire is 10% of its value, at 00C., , cto, rs, , X, Temperature, , Graph shows the variation of resistivity with, temperature for conductors, semiconductors, and for alloys like manganin and constantan., Since the resistivity of manganin and, constantan remains constant with respect to, change in temperature, these materials are, used for the bridge wires and resistance coils., NARAYANA GROUP, , Sol: R2 , t , , , , 110R1, 1.1R1 ; 3.92 103 K 1, 100, , R2 R1, 1.1R1 R1, , R1, R1, , R1 (1.1 1) 0.1R1, 0.1, , , R1, R1, 3.92 10 3, , 0, t 25.510 C ; t 2 25.51 20 45.51 C, , 5
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-6: The resistance of iron wire is 10 and, , 5 10, , 3, , C . If a current of 30A is, flowing in it at 20C , keeping the potential, difference across its length constant, if the, temperature is increased to 120C , what is, the current flowing through that wire ?, , R120 R20, R 10, 5 10 3 120, ;, 10 100, 20 120 20 , R120 15 ; But V = IR, , Sol: a R, , Here V is constant. Hence,, , or, , cos sin , , R 0 T2 T1 , sin cos , , cos 2, sin 2 ; or T2 T1 cot 2, 2, W.E-10: Figure shows a conductor of length l, having a circular cross -section. The radius, of cross-section varies linearly from a to b., The resistivity of the material is . Assuming, that b - a <<l, find the resistance of the, conductor., dx, , R 0 T2 T1 , , Sol :, I120 R20 I120 10, , ; I120 20 A, ;, ba ya, I 20 R120 30 15, tan , , 0, l, x, W.E-7: Resistance of a wire at temperature t C, is R R0 (1 at bt 2 ), Here, R0 is the temperature at 00C. Find the, temperature coefficient of resistance at, temperature t is, Sol :, 1 dR, 1, ., , [ R0 (a 2bt )], R dt R0 (1 at bt 2 ), a 2bt , , 2 , 1 at bt , W.E-8: A silver wire has a resistance of 2.1 at, 27.5C & 2.7 at 1000C. Determine the temperature coefficient of resistivity of silver., Sol: Rt = R0 (1 + αθ ), ......(1), 2.1 = R0 (1+ α 27.5), 2.7 = R0 (1+ α 100), ......(2), Solve equation (1) and (2) α=0.00390 C-1, W.E-9: V- I graph of a conductor at temperature, T1 and T2 are shown in the figure (T2 – T1) is, proportional to, , y, , a, , b, , yl - al = bx - ax, x, dy , l , l (b a) dx , dy (1), ba, dx , Resistance across the elemental disc under, dx, consideration dR (2), A, l dy, from (1) and (2) dR , 2, ba y, Resistance across the given conductor,,, b, , y b, , l, dy, l, R dR R , . 2 ; R , (b a) y a y, ab, y a, , W.E-11: A hollow cylinder of specific resistance, , inner radius R, outer radius 2R and length, l is as shown in figure. What is the net, resistance between the inner and outer, surfaces ?, Sol : Consider a ring of width ‘dr’ and radius ‘r’., 2r, R, , V, T2, T1, , Resistance accross the ring is, , , , , I, , Sol: Slope of line gives resistance, So, R1 tan R 0 1 T1 , , dR , , dr dr, , dA 2 rl, , R2 tan(90 ) cot R0 1 T2 , cot tan R 0 T2 T1 , 6, , Net resistance , , p, 2R, , R, , r (dr ) r , , l n (2), (2p rl ) 2p l , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-12: There are two concentric spheres of, , W.E-15: How many number of turns of nichrome, , radius a and b respectively. If the space, between them is filled with medium of, resistivity , then the resistance of the, intergap between the two spheres will be, (Assume b > a), Sol: Consider a concentric spherical shell of radius, x and thickness dx, its resistance is, dx, dR, dR =, 4 x 2, Total resistance, , wire of specific resistance 106 m and, diameter 2mm that should be wound on a, cylinder of diameter 5cm to obtain a resistance, of 40 ?, Sol: If R is the radius of the cylinder, r is the radius of the wire, N is the number of turns, , b, , l, R2 r2 , l, l k, R1 = 52 42 9 9, , , R, , Final Resistance, l, l, k, R2 = 52 25 25, , Percentage of change = R2 R1 100, R1, k k, , 25, 9 100 64%, , k, 9, W.E-14: If resistivity of the material of a, conductor of uniform area of cross-section, varies along its length as 0 1 x . Find, then the resistance of the conductor if its, lengths is ‘L’ and area of cross-section is ‘A’, 0, A, , 1 2, , L 2 L , , , , ρ, A, , R ', , r 2 p R N, pr2, , 106 22.5102N , = N 800, 40, 1106, , b, , dx 1 1 , R = dR R 4 x 2 4 a b , , a, , , a, W.E-13: A hollow copper cylinder is of inner, radius 4cm and outer radius 5cm. Now, hollow portion is completely filled with, suitable copper wires. Find percentage, change in its electric resistance., Sol: A hollow cylidner of inner radius ‘r’ and outer, radius ‘R’ has specific resistance ' ' . If its, length is ' l ' then its resistance, , r, , then R ' , , Thermistor: A thermistor is a heat sensitive, and non-ohmic device., This is made of semiconductor compounds as, oxides of Ni, Fe, Co etc., This will have high +ve (or) -ve temperature, coefficient of resistance., Thermistor with -ve ‘ ’ are used as resistance, themometers which can measure low temperature, of order, of 10K and small changes of in the order, –3, of 10 K., Having -ve , these are widely used in, measuring the rate of energy flow in micro wave, beam., Thermistor can also be used to serve as, thermostat., , , Resistor Colour codes, Colour Number Multiplier Tolerance(%), Black, 0, × 10°, Brown 1, × 101, Red, 2, × 102, Orange 3, × 103, Yellow 4, × 104, Green, 5, × 105, Blue, 6, × 106, Violet, 7, × 107, Gray, 8, × 108, White, 9, × 109, –, –1, Gold, –, ×10, 5%, Silver, –, ×10–2, 10%, No clour –, 20%, digit 2, tolerence, multiplier, digit 1, , L, , Sol: dR = , , dx, dx, 0 1 x ; R dR ], A, A, 0, , NARAYANA GROUP, , wire, lead, 7
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , Colour bands on a resistor: B.B.ROY of, , V1 , , Great Britain having Very Good Wife with Gold, and Silver, Resistors in the higher range are made mostly, from carbon. Carbon resistors are compact,, inexpensive and thus find extensive use in, electronic circuits., W.E-16: Suppose the colours on the resistor as, shown in Figure are brown, yellow, green and, gold as read from left to right. Using the table,, find the resistance of the resistor, , Resistances in Parallel, I1, , Brown Yellow, Green, 1, 4, ×105, >, >, , , , , , , (1.4 0.07)10 6 (1.4 0.07)M , Some times tolerance is missing from the code, and there are only three bands. Then the, tolerance is 20%., Super Conductor : There are certain metals, for which the resistance suddenly falls to zero, below certain temp. Called critical, temperature., Critical temperature depends on the nature of, material. The materials in this state are called, super conductors., Without any applied emf steady current can be, maintained in super conductors., Ex: Hg below 4.2 K or Pb below 8.2K, , Resistances In Series:, A, , B, R1, , C, R2, , I, , I, , 2., , 3., 4., 8, , V, , If resistors of resistance R1, R2, R 3 ……..are, connected in parallel, the resultant resistance R, is given by, 1, 1, 1, 1, , , , ............, R R1 R 2 R 3, , 2., , If resistances R1 and R2 are connected in parallel,, , 3., , 4., , R 1R 2, the resultant resistance. R R R, 1, 2, When resistors are joined in parallel the, potential difference across each resistor is same., But the currents are in the ratio i1 : i2 : i3:............., 1, 1, 1, = R : R : R : ............, 1, 2, 3, , When two resistances are parallel then, IR2, IR1, I1 , I2 , and, R1 R2, R1 R2, , Note:, 1., 2., , , , V, , 1., , R3, , 1., , I, , , R2, , I, , D, R3, , I2, I3, , Gold, +5%, >, , 5 , , 14 105 1 , , 100 , , R1, , A, , Yellow, Gold, Brown, Green, , Sol:, , VR1, VR2, V2 , and, R1 R2, R1 R2, , If resistors of resistances R 1, R 2, R3, ..... are, connected in series, the resultant resistance, 3., R = R1 + R 2+ R3 + ........, When resistances are connected in series, same, current passes through each resistor. But the 4., pot ential differences are in the ratio, V1 : V2 : V3 ..... = R1 : R2 : R3 ....., When resistors are joined in series, the effective, 5., resistance is greater than the greatest resistance, in the circuit., When two resistances are connected in series, then, , When resistors are joined in parallel, the, effective resistance is less than the least, resistance in the circuit., A wire of resistance ‘R’ is cut into ‘n’ equal, parts and all of them are connected in parallel,, R, equivalent resistance becomes 2 ., n, In ‘n’ wires of equal resistances are given, the, number of combinations that can be made to give, different resistances is 2n –1 ., If ‘n’ wire of unequal resistances are given, the, number of combinations that can be made to give, different resistances is 2n (If n >2)., If R s and R p be the resultant resistances of R1, and R2 when connected in series and parallel, then, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-19: Find equivalent resistance of the, network in Fig. between points (i) A and B, and (ii) A and C., 10, B, , A, 30, , C, , 7.5, , Sol: (i)The 10 and 30 resistors are connected in, parallel between points A and B. The equivalent, resistance between A and B is, 10 30, R1 , ohm = 7.5, 10 30, , (ii) The resistance R1 is connected in series with, resistor of 7.5 , hence the equivalent resistance, between points A and C is, R2=(R1+7.5) ohm =, (7.5+7.5) ohm = 15 ., W.E-20: Find potential difference between points, A and B of the network shown in Fig. and, distribution of given main current through, different resistors., 4, , R1 , , R, R, . r , ; Resistance of section PTQ, 2 r, 2, , Rr 2, , R 2 , ;, R2 , 2 r, 2, As R1 and R2 are in parallel, R2 , , Req , , So,, , R1 R2, R, 2 2, R1 R2 4, , W.E-22: Determine the current drawn from a 12V, supply with internal resistance 0.5 Ω . by the, infinite network shown in Fig. Each resistor, has 1 Ω . resistance., 1, , 1, , 1, , 12V, 0.5, , 1, , 1, , 1, , 1, , 1, , Sol: First calculate net resistance of network, 1, , 6, , I=2.7A, A, , 8, B, , 9, , X, , 1, , x, , I, , 1, , Sol: Between points A and B resistors of 4 , 6, and 8 resistances are in series and these are, in parallel to 9 resistor.., Equivalent resistance of series combinaiton is, R1 = (4 + 6 + 8 ) ohm = 18, If equivalent resistance between A and B is, R = 9 × 18 / (9 + 18) ohm = 6 , Potential difference between A and B is, V = IR = 2.7 × 6V = 16.2V, Current through 9 resistor = 16.2/9=1.8A, Current through 4 ,6 and 8 resistors =, 2.7 – 1.8 = 0.9A., W.E-21: P and Q are two points on a uniform, ring of resistance R. The equivalent resistance, between P and Q is, P, , , O, , Q, , Sol: Resistance of section PSQ, P, , S, , , O, , Q, , x=2+, , x, ; 2, ;, x+1 x 2x 2 0, , on solving, x 1 3 2.73, Total resistance = 2.73 + 0.5 = 3.23 Ω, 12, I=, =3.73A, 3.23, JOULE’S LAW: According to Joule’s law,, the current passing through a conductor produces, heat., W = vit, Now, work done, W = (iR) i t, 2, W = i2 R t = v t = v i t, R, This work is converted into energy in the, conductor., Thermal energy produced, Q = i2 Rt in Joules, , i 2 Rt, in cal., 4.2, As H i 2 , heating effect of current is common, to both A.C and D.C., Joule’s effect is irreversible., Or Q =, , T, , 10, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-23: A fuse wire with radius of 0.2mm blows, , Electrical Energy:, , , The electric energy consumed in a circuit is, defined as the total workdone in maintaing the, current in an electric circuit for a given time., Electrical Energy = Vit = Pt = i2Rt , , off with a current of 5 Amp. The fuse wire of, same material, but of radius 0.3mm will blow, off with a current of, , V2t, R, , S.I. unit of electric energy is joule, 1 K.W.H. = 36 × 105 J, , 3) 5, , Electrical Power:, , , The rate at which work is done in maintaining, the current in electric circuit. Electrical power, P, , , , W, V2, Vi i2R , t, R, , 27, Amp, 8, , 4) 5 Amp, , Sol: i2 r 3 ;, , watt (or) joule / sec, , W Pt E it i 2 Rt E 2t, , , , , J, J, J, J, RJ, H mst, Where s 4200J /Kg0C, where J is mechanical equivalent of heat., Fuse wire: A fuse wire generally prepared, from tin - lead alloy (63% tin + 37% lead). It, should have high resistivity, low melting point., Let R be the resistance of fuse wire., H=, , rL, We know that R 2, pr, (L and r denote length and radius), The heat produced in the fuse wire is, i2rL, pr 2, If H0 is heat loss per unit surface area of the, fuse wire, then heat radiated per second is =, H 0 2rL At thermal equilibrium,, H i2R , , i2 r2 L, i2 r, , H, 2, p, rL, H, , (or), 0, 0, pr 2, 2p2r 3, According to Newton’s law of cooling., H 0 K, Where is the increase in temperature of fuse, wire and K is a constant., i 2r, 2p2r3K, Here is independent of length L of the fuse, wire provided i remains constant., For a given material of fuse wire i2 r 3 ., NARAYANA GROUP, , 2), , 3, , Heat energy produced due to the electric current, , q, , 3, Amp, 2, , 1) 5, , i1 r1 , , i 2 r2 , , 2, , 5 3, Amp, 2, , 3, , 0.2 , , 0.3 , , 2, , 27, Amp, 8, If radiation losses are neglected, due to, heating effect of current the temperature of, fuse wire will increase continuously, and it, melt in time ‘t’ such that, i2 5, , , , H ms ;, , t, , I 2 Rt, ms( mp r ), J, , 2 r 4 s mp r J, 2, , ;t r 4, , I , i.e., in absence of radiation lossess, the time, in which fuse will melt is also independent on, length and varies with radius as r4., Note :, a) If resistances are connected in series , i.e.., I is, same, P R with V a R [ as V IR ], i.e.., in series potential difference and power, consumed will be more in larger resistance., However, if resistances are connected in, parallel, i.e.., V is same, 1, 1, Pa, with I a, [ as V IR ], R, R, i.e.., in parallel current and power consumed, will be more in smaller resistance. This in turn, implies that more power is consumed in larger, resistance if reistances are in series and in, smaller reistance if reistances are in parallel., b) A reistance R under a potential difference V, dissipates power., P V 2 / R , , So If the resistance is changed from R to (R/n), keeping V same, the power consumed will be, 11
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 2, , V2, V2, n, nP, R, ( R / n), i.e.., if for a given voltage, resistance is changed, from R to (R/n) , power consumed changes from, P to nP., If n equal resistances are connected in series, with a voltage source, the power dissipated will, be, , V , P A W, VS , , P1 , , c), , 2, , so, P 100 1000 250W, 200 , c) The total electrical energy consumed by an, electric appliance in a specified time is given, by,, W1h1, E, kWh, 1000, so, E 1000 (10 30) 300kWh, 1000, , V2, [ as Rs nR ], nR, And if the same resistances are connected in, parallel with the same voltage source, Ps , , V2, nV 2, Pp , , ( R / n), R, So,, , d), , Pp, , [as Rp ( R / n)], , Bulbs connected in Series:, , , n 2 i.e.., P n 2 P ., P, S, , Ps, i.e.., power consumed by n equal resistors in, parallel is n 2 times that of power consumed in, series if V remains same., As resistance of a given electric appliance, ( e.g.., bulb , heater, geyser or press ) is constant, and is given by,, 2, , V, VS, V, R S , s, I, (W / VS ) W, , P R & V R P i2Rt , , , , , , , , by , R , , , , b) The ‘ actual power ‘ consumed by an electric, appliance is given by ,, 12, , 1, 1, and I ., R, R, , i.e. The current and power consumed will be, more in smaller resistance., When the appliances of power P1, P2 , P3 .... are, in parallel, the effective power consumed(P) is, P P1 P2 P3 ........., , 2, , VS, 200 40, so, R , W, 1000, , V2, ., nR, , If Bulbs (or electrical appliances) are connected, in parallel, the potential difference across each, resistance is same. Then P , , Sol:a) The resistance of an electric appliance is given, 2, , less than the power of individual appliance., If ‘n’ appliances, each of equal resistance ‘R’, are connected in series with a voltage source, , Bulbs connected in parallel:, , W.E-24: A 1 kW heater is meant to operate at, 200 V. (a) What is its resistance ? (b) How, much power will it consume if the line voltage, drops to 100 V ? (c) How many units of, electrical energy will it consume in a month, (of 30 days) if it operates 10 hr daily at the, specified voltage ?, , i.e. effective power is, , ‘V’, the power dissipated ‘ Ps ’ will be Ps , , 2, , 2, 2, VA VA , P, W [ as R VS ]., R VS , W, , i.e. In series combination; t he potential, difference and power consumed will be more, in larger resistance., When the appliances of power P1, P2 , P3 .... are, in series, the effective power consumed (P) is, 1 1, 1, 1, , , ........., P P1 P2 P3, , W, [ as I ], V, , Where Vs and W are the voltage and wattage, specified on the appliance. So if the applied, voltage is different from specified, the ‘ actual, power consumption’ will be, , If Bulbs (or electrical appliances) are connected, in series, the current through each resistance is, same. Then power of the electrical appliance, , , , i.e. the effective power of various electrical, appliance is more than the power of individual, appliance., If ‘n’ appliances, each of resistance ‘R’ are, connected in parallel with a voltage source ’V’,, the power dissipated ‘Pp’ will be, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, PP , , , , , , CURRENT ELECTRICITY, , W.E-27: A 100W and a 500W bulbs are joined in, , V2, nV 2, , R / n R, , PP, n 2 or PP n 2 PS, PS, This shows that power consumed by ‘n’ equal, resistances in parallel is n2 times that of power, consumed in series if voltage remains same., In parallel grouping of bulbs across a given, source of voltage, the bulb of greater wattage, will give more brightness and will allow more, current through it, but will have lesser resistance, and same potential difference across it., For a given voltage V, if resistance is changed, R, , from ‘R’ to n , power consumed changes from, , R, V2, where R , then, ‘P’ to ‘nP’ P' , n, R', P , , , , V2, nV 2, , nP ., R / n R, , If t1, t 2 are the time taken by two different coils, for producing same heat with same supply, then, If they are connected in series to produce same, heat, time taken t t1 t2, If they are connected in parallel to produce same, tt, , 1 2, heat, time taken is t t t ., 1, 2, W.E-25: A lamp of 100W works at 220 volts. What, is its resistance and current capacity ?, Sol: Power of the lamp, P = 100W, Operating voltage, V = 220V, Current capacity of the lamp,, , i, , P 100, , 0.455A, V 220, , Resistance of the lamp, R V (220) 484, 2, , P, , 2, , 100, , W.E-26: A 100W – 220V bulb is connected to, 110V source. Calculate the power consumed, by the bulb., Sol: Power of the bulb, P = 100W, Operating voltage, V = 200V, Resistance of the bulb, R , , V 2 (220)2, , 484, P, 100, , series and connected to the mains. Which, bulb will glow brighter ?, Sol: Let R1 and R2 be the resistances of the two bulbs., If each bulb is connected separately to the mains, of voltage V,, then P1 , , , P1, , , , P11, , R2, , R1, , P2, , 500, , , , 5, P2 R1 (or) R 2 P1 100, If the two bulbs are in series with the mains, the, same current ‘i’ flows through each of them., Let P1 and P2 be the powers dissipated by two, bulbs, turn, P11 i2 R1 and P21 i2 R 2, P21, , , , V2, V2, and P2 , R1, R2, , , , R1, R2, , 5, , or P11 5P21, , Since 100 watt bulb dissipates more power, it, glows brighter, W.E-28: A cell develops the same power across, two resistances R1 and R2 separately. The internal resistance of the cell is, Sol: Let r be the internal resistance of the cell and E, its EMF. When connected across the resistance, R1 in the circuit, current passing through the resistance is, E, i, R1 r ;, , E , R, P1 i R1 , R1 r 1, 2, , , , 2, , E , , Similarly P2 , R 2 ; Given that P1 = P2, R 2 r , 2, , Substituting the values, we get, ;, r R1R 2, W.E-29: A 100 W bulb B1 and two 60 W bulbs B2, and B3, are connected to a 250V source, as, shown in the figure. Now W1 , W2 and W3 are, the output powers of the bulbs B1, B2 and B3, respectively. Then, B1, , B2, , B3, , V1, , Actual operating voltage, = 110 V, Therefore, power consumed by the bulb,, P1 , , (V1 )2 (110)2, , 25W., R, 484, 250V, , NARAYANA GROUP, , 13
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , W.E-30: A battery if internal resistance 4 is, V2, connected to the network of ressitances as, Sol: A bulb is essentially a resistance R , where, P, shown. What must be the value of R so that, P denotes the power of the bulb., maximum power is delivered to the network, Resistance of B1(R1) = V2 / 100, ? Find the maximum power ?, Resistance of B2 (R2) = V2 / 60, R, 2R, Resistance of B3(R3) = V2 / 60, 6R, 250, 250300, 4R, 2R, I1 = Current in B1 R R 8V 2, 1, 2, 250, 250 300, I2 = Current in B2 R R , 8V 2, 1, 2, I3 = Current in B3 = I1 as B1, B2 are in series, , E, , Sol:i) According to maximum power transfer theorem, 3R 6R, 4, 4 R 2, R ext R int, 9R, 2, , W1 output power of B1 = I12 R1, , ii), , 250300 2 V 2, W1 , , 8V2 100, 250 300 V, W2 I 22 R 2 or W2 , , 8V 2 60, , 2, , 250 300 2 V 2, W3 I22 R 3 or W3 , , 8V 2 60, , W1 : W2 : W3 15 : 25 : 64 or W1 < W2 < W3, , , , Secondary Cells (or) Storage Cells:, , , R, , Consider a device of resistance R connected to , a source of e.m.f E and internal resistance r as , E , shown. Current in the circuit is i , ., R r , , Power dissipated in the device is P = i2R, E2R, (R r)2, For maximum power dissipated in the device, , , , d E2R , dP, 0 , , 0, dR (R r)2 , dR, On simplification, we can get R = r, So, the power dissipated in an external , resistance is maximum if that resistance is equal, to internal resistance of the source supplying the, current to that device., 14, , Primary Cells: Voltaic, Leclanche, Daniel and, Dry cells are primary cells. They convert, chemical energy into electrical energy. They, can’t be recharged. They supply small currents., , i, , P, , Units of electrical energy consumed by an, electrical appliance =, Number of watts× Number of hours, 1000, It is in KWH., , CELLS, , Maximum power transfer theorem, , E, , E 2, E2, , , 4 , 4 4 , 16, , Consumption of Electrical Energy:, , 2, , r, , Pmax i R ext, 2, , Electrical energy is first converted into chemical, energy and then the stored chemical energy is, converted into electrical energy due to these, cells., These cells can be recharged., The internal resistance of a secondary cell is, low where as the internal resistance of a primary, cell is large., EMF of a Cell: The energy supplied by the, battery to drive unit charge around the circuit is, defined as electro motive force of the cell., EMF is also defined as the absolute potential, difference between the terminals of a source, when no energy is drawn from it. i.e., in the open, circuit of the cell. It depends on the nature of, electrolyte used in the cell., Unit :J/C (or) Volt, , emf of a cell depends on, a) metal of electrodes, b) nature of electrolyte, c) temperature, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , , CURRENT ELECTRICITY, , emf of the cell is independent of, , r, , a) area of plate, b) quantitiy of electrolyte, c) distance between plate, d) size of the cell, , Internal Resistance of a Cell, , , , , , , , , , It is the resistance offered by the electrolyte of, the cell., It depends on, 1, area of the electrodes used ( r ), A, nature of electrolyte , concentration( r C ), area of cross section of the electrolyte through, which the current flows and, age of the cell., Internal resistance of an ideal cell is zero., , , , ....................(B), , E V, r , V, R, , , E V, V, , , V = iR =, , ER, R r , , , E , R 1 R .... (C), , V, , , , , V, R, , E Rr, % of fractional useful energy, R , V , 100, = 100 = , E, Rr, , , , V', r, , Fractional energy lost,, E Rr, , , , % of lost energy, E 100 R r 100, , , , , internal resistance, r = V R, , , , , , Terminal Voltage:, When no current flows through the cell, the, circuit is said to be an open circuit. This is shown, in figure., , E V, i, , Fractional energy useful =, , V' , , , , , r, , , , E V , , DIFFERENT CONCEPTS WITH A CELL, , In such a case, the potential difference (p.d), across the terminals of the cell, called the, terminal voltage (V) will be equal to the emf, (E) of the cell., If an external resistance R is connected across, the two terminals of the cell, as in figure then, current flows in the closed circuit.,, , R, , i, , i, , E, , , , E, r, i, , When the cell is charging, the EMF is less than, the terminal voltage (E < V) and the direction of, current inside the cell is from + ve terminal to, the –ve terminal., , i, , EV, , V=E +ir, When the cell is discharging, the EMF is greater, than the terminal voltage (E >V) and the, direction of current inside the cell is from – ve, terminal to the +ve terminal., i, , i, , E, , V, i, R, , ........... (1), , E, and also i , ........... (2), R r, iR + ir = E, V +ir = E, V = E – ir, Lost volts: It is the difference between emf, and P.D. of a cell It is used in driving the current, between terminals of the cell., Lost volts E - V = i r, Note: Formulae related with cells, E V, ..................(A), i, r, NARAYANA GROUP, , , , EV, , V = E – i r ; Hence E V, Power delivered will be maximum when R = r., , E2, 4r, This statement in generalized form is called, ‘maximum power transfer theorem’, , So Pmax , , , 2, , Pmax = E /4r, P, R=r, R, 15
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, Here the % of energy lost and energy useful are, each equal to 50%, Back EMF: When current flows through the, electrolyte solution, electrolysis takes place with, a layer of hydrogen and this hinders the flow of, current. In the neighbourhood of both electrodes,, the concentrations of ions get altered. This, opposing EMF is called back EMF and the, phenomenon is called Electrolytic polarisation., To reduce back emf manganese dioxide (or), potassium dichromate is added to electrolyte of, cell., W.E-31: When a current drawn from a battery is, 0.5A, its terminal potential difference is 20V., And when current drawn from it is 2.0A, the, terminal voltage reduces to 16 V. Find out., e.m.f and internal resistance of the battery., Sol: We know, V = E --- Ir ; I = 0.5 A, V = 20 Volt, we have, 20 = E – 0.5r, ....... (i), I = 2 A, V = 16 Volt, we have, 16 = E-0.2r, ....... (ii), From eqs (i) and (ii), 2E – r = 40 and E – 2r = 16, Solving we get E = 21.3V, r = 2.675 ., W.E-32: An ideal battery passes a current of 5A, through a resistor. When it is connected to, another resistance of 10 in parallel, the, current is 6A. Find the resistance of the first, resistor., R2 = 10, R1, , W.E-33: When a battery is connected to the, resistance of 10 the current in the circuit is, 0.12A. The same battery gives 0.07A current, with 20 . Calculate e.m.f. and internal, resistance of the battery., Sol: We know that E Ir IR, I1r I1R1 I2 r I2 R2 ; r , , 0.07 20 0.12 10 1.4 1.2 0.2, , , 4, 0.05, 0.05, 0.12 0.07, Internal resistance r 4, e. m. f E Ir IR, 0.12 4 0.12 10 0.48 1.2 ; E = 1.68 volt., r, , GROUPING OF CELLS, 1. Electric Cells in Series: When ‘n’ identical, , , , , , , , , R1, , 5A, Sol:, , cells each of EMF ‘E’ and internal resistance, ‘r ’ are connected in series to an external, resistance ‘R’, then, total emf of the combination = n E, effective internal resistance = n r, nE, Current through external resistance i =, R nr, , , = current from one cell, r, nE, If R>> n r then i =, R, If two cells of different emf’s are in series, E1 E2, Eeq = E1 + E2 ; req = r1 + r2 ; i r r R, 1, 2, If R << n r then i =, , E1, , E2, , 6A, r1, , V, V, Current through R1 in the first case i1 = 5A, Current in the second case i2 = 6A, Effective resistance in the second case, R, , R1R2, R1 R2, , I1R1 I2, 5 6, , R1 , , RR, , 1 2, ; V I1R1 and V I2 R R, 1, 2, , R1R2, R2, I1 I2, R1 R 2, R1 R2, , 10, 5(R1 10) 60, R1 10, , 5R1 + 50 = 60, 5R1 = 10, , 16, , I2 R2 I1R1, I1 I2, , 10, 2 R1 2 ., 5, , r2, , R, , , , T.P.D across the first cell V1 = E1 - ir1, T.P.D across the second cell V2 = E2 - ir2, If one of the cell is in reverse connection, ( E1 > E2 ) then Eeq = E1 - E2, E1 E2, req = r1 + r2 ; i r r R, 1, 2, E1, , r1, , E2, , r2, , R, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , First cell is discharging then V1 = E1 - ir1, Second cell is charging then V2 = E2 + ir2, cell having less emf in charging state., , W.E-36: In the given circuit as shown below,, calculate the magnitude and direction of the, current, , WRONGLY CONNECTED CELLS, , A 2, , , , By mistake if ‘m’ cells out of ‘n’ cells are, wrongly connected to the external resistance ‘R’, (a) total emf of the combination = (n – 2m) E, (b) total internal resistance = n r, (c) total resistance = R + n r, n 2mE, (d) current through the circuit (i) =, R nr, W.E-34:Two cells A and B with same e.m.f of 2 V, each and with internal resistances, rA 3.5 and rB 0.5 are connected in series, with an external resistance R 3 . Find the, terminal voltages across the two cells., Sol: Current through the circuit, i, , , 22, 4, , , R, , r, 3, , 3.5, , 0.5, 7, , , , , i) R 3, rA 3.5 , E 2V, Terminal voltages A, VA E ir, 4, 2 3.5 0 volt, 7, , ii) rB 0.5 , R 3 , E2V, Terminal voltage at B, VB Eir, , 10V, D, , B, , 10 5, 1A, 5, Since the cell of larger emf decides the direction, of flow of current, the direction of current in the, circuit is from A to B through e, W.E-37: A voltmeter resistance 500 is used to, measure the emf of a cell of internal resistance 4 . The percentage error in the reading of the voltmeter will be, Sol: V = E – ir, E, ir, Percentage error 100 100, E, E, , E , , r, R r , , ELECTRIC CELLS IN PARALLEL, , , , , , 2, , 4, 1.9 = 2, 3.8, , NARAYANA GROUP, , – 2 = 0., , When ‘n’ identical cells each of EMF ‘E’ and, internal resistance ‘r’ are connected in parallel, to an external resistance ‘R’, then, total emf of the combination = E, r, effective internal resistance =, n, r, total resistance in the circuit = R +, n, current through the external resistance, i=, , 4, 4, , , A, 1 1.9 0.9 3.8 ., , potential difference at A, VA ir ,, , E, , r , 100 , 100, R r , , 4 , , 100 0.8%, 500 4 , , , , R =1, , C, , i=, , Sol:, , A, , 1, , Reff, , connected in series to an external resistance, R=1 ohm . The internal resistance of A is rA, =1.9 ohm and B is rB =0.9 ohm. Find the, potential difference between the terminals of, A., , 0.9, , 5V, , V V, total current in the circuit is i 1 2, , W.E-35: Two cells A and B each of 2 V are, , r = 1.9, , B, , Reff 2 2 1 5, , , , voltage, Total resist an ce, , 2, , Sol: Effective resistance of the circuit is, , 4, 2 0.5 1.714 volts., 7, , Total current through the circuit i , , e, , , , E, r, R, n, , If R >>, , , , nE, nR r, , r, E, , then i =, = current from one cell., n, R, 17
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , , , r, nE, , then i =, n, r, If two cells of emf E1 and E2 having internal, resistances r1 and r2 are connected in parallel, to an external resistance ‘R’, then, , If R <<, , i1, , r1 E1, + -, , i2, , R, , , , Condition for maximum power R , , , , E2, 4r, Condition for maximum current, R r, + =minimum, n m, Pmax mn , , i, , E2, i, , +r, 2, , i, , N, , n m , , , R, r, R r, 2 0; i.e., , (N = n x m), N m, n m, So in case of mixed grouping of cells, current, d mR r , , 0;, dm N m , , E 1 r 2 E 2r 1, , the effective emf, E =, , r 1 r2, , r1 r2, the effective internal resistance, reff r r, 1, 2, , E, Current through the circuit, i r R, eff, i = i1 + i2, E1 i R, , E2 iR, and i 2 r, i1 = r, 2, 1, Potential difference across R, i.e terminal, ER, potential of the cells is V iR R r, eff, , , , When the cell E2 is reversed in polarity then, we should use - E2 in all the above equations., Mixed Grouping: If n identical cell’s are, connected in a row and such m rows are, connected in parallel as then, 1, 2, , E,r E,r, 1, , , , R r , in the circuit will be maximum when , n m, nE mE, I max , , 2 R 2r, Total number of cells = m n, , KIRCHHOFF’S LAWS, When the circuit is complecated to find current, kirchhoff’s laws are formulated., (i) Kirchhoff`s First Law (Junction Law or, Current law) : It states that the sum of the, currents flowing into a junction is equal to the, sum of the currents flowing out of the junciton ., Or, “The algebraic sum of currents at a junction is, zero”., R4, , E,r, n, , 2, , R1, , I1, , I4, , R3, I3, , A, i, , nr, and, m, , I2 R2, , m, V, R, , , , , , , Equivalent emf of the combination Ee q = nE, Equivalent internal resistance of the combination, nr, req =, m, Main current flowing through the load, i, , 18, , nE, nmE, , nr mR nr, R, m, , Distribution of current at a junction in the circuit, I1 + I2 = I3 + I4 or I1 + I2 – I3 – I4 = 0, If we take currents approaching point A in Fig, as positive and that leaving the point as negative,, then the above relation may be written as, I1 + I2 +(–I3) + (–I4) = 0, I 0, Note: Thus, Kirchhoff’s first law is accordance with, law of conservation of charge, since no charge, can accumulate at a junction., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , (ii) Kirchoff`s Second Law (Loop Law or Potential Note:, law) : Kirchhoff’s second law states that the 1) This law represents “conservation of energy”, algebraic sum of changes in potential around any 2) If there are n meshes in a circuit, the number of, closed loop is zero., independent equations in accordance with loop, (Kirchhoff’s second law) can be expressed as, rule will be (n – 1)., Application : This is the most general case of, V 0 ., parallel grouping in which E and r of different, In terms of potential drops and emfs, the law is, cells are different and the positive terminals, expressed as iR E 0, cells are connected as shown, , Sign conventions:, , (a) The change in potential in traversing a resistance, in the direction of current is –IR while in the, opposite direction +IR as shown in the figure., , A, , R, , I, , B, , R, , A, , E1, , r1, , E2, , r2, , i1, i2, , B, i, , V A - IR = V B, V A - V B = IR, , V B + IR = V A, V A - V B = IR, , (b) The change in potential in traversing an emf, source from negative to positive is +E while in, the opposite direction –E irrespective of the, direction of current in the circuit as shown in, the figure., A I +E I B, A I +E I B, VA - E = VB, VA - VB = E, , VB + E = VA, VA - VB = E, , Example 1:, i, , A, , B, R2, , R3, , i C, , Apply the Kirchoff’s second law to the loop, ABCDA, then, E, – iR1 – iR2 – iR3 + E = 0 ; i R R R , 1, 2, 3, Example 2:, r2 B, A r1, a, i, , i, , E1 E2, R, , i, , i, , D, C, i, Apply the kirchoff’s second law to the loop, ADCBA, then, –iR – ir2 + E2 – E1 – ir1 = 0, i(r1 + R + r2) = E2 – E1 i , NARAYANA GROUP, , or i1 E1 iR ................ (1), r1 r1, , E2 iR i2 r2 0, E2 iR, ............... (2), r2 r2, Adding Eqs. (1), (2) we get, i1 i2 ( E1 / r1 ) ( E2 / r2 ) iR (1/ r1 1/ r2 ), i2 , , ( E1 / r1 ) ( E2 / r2 ), 1 R (1/ r1 1/ r2 ), W.E-38: Find the emf (V) and internal resistance, (r) of a single battery which is equivalent to a, parallel combination of two batteries of emfs, V1 and V2 and internal resistances r1 and r2, respectively, with polarities as shown in figure, i , , i, , D, , Kirchhoff’s second law in diferent loops gives, the following equations,, E1 iR i1r1 0, , or i[1 R(1/ r1 1/ r2 )] ( E1 / r1 ) ( E2 / r2 ), , R1, , i, , i, , R, , E 2 E1, r1 r2 R, , +, A, , i2, , V2, i1, , V1, , B, , Sol: EMF of battery is equal to potential difference, across the terminals, when no current is drawn, from battery (for external circuit) [Here, all the, elements in the circuit are in series], Current in internal circuit = i, Net emf, V V2, i, or i= 1, Total resistance, r1 r2, 19
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JEE-ADV PHYSICS- VOL- III, P R, , .......(5), Q S, This is the balancing condition for Wheat stone, bridge., APPLICATIONS ON WHEATSTONE BRIDGE, 1. We can compare two unknown resistances R and, P R, S from , Q S, 2. In place of resistances we can use capacitors to, form a D.C. Wheatstone bridge with four, capacitors of capacitances C1, C2, C3 and C4., C1 C3, The balancing condition will be C C, 2, 4, 3. It has been found that the bridge has the greatest, sensitivity when the resistances are as nearly, equal as possible., The bridge is most sensitive if P = Q = R = S., 4. Equivalent resistance of balanced bridge across, the ends of battery when the bridge is balanced, is given by, P Q R S, PQR S, 5. There are seven closed meshes in wheatstone’s, bridge, Application : Direction of current in an, unbalanced wheatstone’s bridge :, , CURRENT ELECTRICITY, , W.E-41: Determine the current in each branch of, the network shown in fig., B, 10, , i, , A, , 10, 1, AI, 1, I-I, , VAD VA VD i2 R i, , R S P, PQ R S, , D, F, , E, , .......(1), , 5 I I1 10I1 10 10I=0, , , , 5I1 + 15I = 10, .......(2), From equation (1) and (2), 21 4, 10, I1 = A, I= ;, 5 17, 17, 4, Current in AB branch , 17, 10 4, 6, =I I1 A, 17 17 17, Current in DB branch, 10 8 2, I 2I = - = A, 17 17 17, Metre bridge: It works on the principle of, P R, Wheatstone Bridge , Q S , Cell, , RB, R, , i, QR PS , PQ R S, , NARAYANA GROUP, , 10, , 10V, , +, , QR PS , VB VD Balanced bridge, , I, , 2I 5I , Apply KVL in ADCEFA loop, , PQ R S, , QR PS , VB VD current flows from D to B, , 10, , 10I1 5 I 2I1 5 I I1 0, , P Q R, , if QR PS , VB VD current flows from B to D, , I1, , C, , Sol: Apply KVL in loop ABDA, , i, VB VD P Q R S P Q R R S P , , , , 5, 1, I-2I, , 5, , I, , D, , VAB VA VB i1P i, , B I-I, 5, 1, , S, , R, , 10, , 10V, , C, , i2, , 10, D, , Q, , P, , C, , 5, 5, , B, i1, , 5, , A, , A, , X, J, l1, , H.R, , , , ( ), Key, , l2, , B, , G, , When the Meter bridge is balanced then, R l1, l1, , X l2 100 l1, 21
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, Where l1 is the balancing length from the left, end., Note:, 1. If resistance in the left gap increases or, resistance in the right gap decreases, balancing, point shifts towards right side., 2. If resistance in the left gap decreases or, resistance in the right gap increases, balancing, point shifts towards left., 3. If a cm, b cm are the end corrections at A and B,, R l1 a, then X l b, 2, 4. Meter bridge is more sensitive if l1 = 50 cm, 5., , The resistance of copper strip is called end, resistance, W.E-42: In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer, is x, what would be its value if the radius of, the wire AB is doubled?, R1, , R2, G, , A, , x, , C, , B, , Sol:For null deflection of galvanometer in a, metrebridge experiment,, R1 R AC, R, x, , or 1 , R 2 RCB, R 2 100 x, Since R1 / R2 remains constant, x / (100-x) also, remains constant. The value of x remains as such., Length of AC = x, W.E-43: A resistance of 2 is connected across, one gap of a metre-bridge (the length of the, wire is 100 cm) and an unknown resistance,, greater than 2 , is connected across the, other gap. When these resistances are interchanged, the balance point shifts by 20 cm., Neglecting any corrections, the unknown resistance is, Sol: Refer to the diagram Apply the conditions of, the balanced Wheatstone’s bridge for the two, cases., R1, , R2, G, , l, , 22, , 100-l, , 2, , , ..................................(i), x 100 , x 20, , ....................................(ii), 2 80 , Equations (i) and (ii) give x 3, Potentiometer: Potentiometer is an instrument, which can measure accurately the emf of a, source or the potential difference across any, part of an electric circuit without drawing any, current., a) Principle : The principle of potentiometer states, that when a constant current is passed through a, wire of uniform area of cross-section, the, potential drop across any portion of the wire is, directly proportional to the length of that portion., The principle of potentiometer require that, i) potentiometer wire should be of uniform, area of cross-section and, ii) current through the wire should remain, constant., c) Theory of potentiometer : The end of the, potentiometer wire AB are connected to a, standard cell of emf E or a source of emf E that, supplies constant current. The current through, the potentiometer wire can be varied by means, of a series resistance Rs which is adjustable., i, , + r E, , I, A, , r, , k, RS, ( ), Primarycircuit, I, , 1, , D, , Secondary circuit J, HR, + e, , B, , G, , Let r be the internal resistance of the cell of emf, E connected across the potentiometer wire of, length L and resistance R. The current through, the potentiometer wire is, E, I, r R Rs, The potential of the wire decreases from the end, A to the end B. The potential fall or potential, drop across a length l of the potentiometer wire, is, V = Current x Resistance of length l of the, R, potentiometer wire I l, L, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , R, is, L, denoted by , the potential drop across the wire, is, , , , If the resistance per unit length of the wire ,, , and balancing length l1 is obtained. Since cell, E is in open circuit so it’s emf balances on length, l1, , V I , V, is called potential drop per unit length of the, l, potentiometer wire or potential gradient of the, wire. It is given by, , 1), , , R, E, V, I , , l, r R RS L, Thus, the unknown voltage V is measured when, no current is drawn from it., When specific resistance (S) of potentiometer, wire is given then potential gradiant, , X, , resistance R1 and R2 are joined together to form, , x1 R1 L2, , x2 L1 R2, Potential gradient depends on, , the potentiometer wire, then, , a) Resistance per unit length of the potentiometer, wire ( =R/L), b) Radius of crosssection of the potentiometer, wire, when the series resistance is included in, the circuit and cell in the primary circuit is not, ideal., c) Current flowing through potentiometer wire., d) emf of the cell in primary circuit, e) Series resistance in the primary circuit, f) Total length (L) and resistance (R) of the, potentiometer wire., g) If cell in primary circuit is ideal and in the, absence of series resistance potential gradient, only depends on emf of cell in primary circuit, and length of potentiometer wire, TO DETERMINE THE INTERNAL, RESISTANCE OF A PRIMARY CELL:, e, r, , , , E , r 1 .R /, V, , Where E = emf of cell in secondary circuit, V = Terminal voltage, , E l1, E, l, , 1 1 1, ,, V l2, V, l2, COMPARISON OF EMF’S OF TWO CELLS, Let l1 and l2 be the balancing length with the, , , cell E1 and E2 respectively, then E1 xl1 and, , NARAYANA GROUP, , ( ), , R1, , K1, , Rh, , K, ( ), , e,r, , J, , A, , B, G, , E1, , 1, , E2, , , , 2, , Let E1 E2 and both are connected in series., If balancing length is l1 when cells assist each, other and it is l2 when they oppose each other, as shown then:, E1, , B, G, , E, , E1 l1, , E2 l2, , E2 xl2 , , E2, , , , , E1, , , E1 E2 xl1, , J, , A, , l1 l2 , R', i.e p.d on R, r , l2 , , , , Rh, , k, ( ), , i.e E xl1 .....(i), Now key K ’ is closed so cell E comes in closed, circuit. If the process of balancing is repeated, again keeping constant then potential difference, V balances on length l2, , i.e V xl2 .....(ii), By using formula internal resistance, , IS, IS, 2, A pr, , where A = area of cross - section of potentiometer, wire r = Radius of potentiometer wire., 2) When two wires of length L1 and L2 and, , Initially in secondary circuit key K’ remains open, , , , E1 E2 l1, , E1 E2 l2, , , , E2, , , , , , , E1 E2 xl2, E1 l1 l2, (or) E l l, 2, 1, 2, 23
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , Comparison of resistances:, , SENSITIVITY OF POTENTIO METER, , Let the balancing length for resistance R1 (when, , 1., , XY is connected) be l1 and let balancing length, , 2., , for resistance R1 R2 (when YZ is connected), be l2 . keeping X constant, , 3., , Sensitivity of potentio meter is estimated by its, potential gradient., Sensitivity is inversly proportional to potential, gradient so lower the potential gradient higher, will be the sensitivity., The best instrument for accurate measurement, of e.m.f. of a cell is potentiometer, because it, does not draw any current from the cell., Calibration of ammeter: Checking the, correctness of ammeter readings with the help, of potentiometer is called calibration of ammeter., +e -, , R2 l2 l1, , R1, l1, , , , R, A, , K, (), , Rh, , A, B, , HRB, , , , G, , + E0 1 2 3, , G, , Cold ice, , , , 10, , , , , , , , 24, , Hot sand, , The value of thermo-emf in a thermocouple for, ordinary temperature difference is very low, 6, , , , , , volt . For this the potential gradient x, , must be also very low 104 V / m . Hence a, high resistance (R) is connected in series with, the potentiometer wire in order to reduce current, in the primary circuit, The potential difference across R must be equal, to the emf of standard cell, E, i 0, i.e iR E0, R, The small thermo emf produced in the, thermocouple e xl, , iR|, iR| I, x i , e, , , L, L, where L = Length of potentiometer wire,, =resistance per unit length, l balancing, length of e and, R| = Resistance of potentiometer wire, , B, G, A, + Rh, , K2, (), , + -, , To determine thermo emf:, e r, + -, , Rh, , A +, + - E1 1, 2, 3, 1, , Then iR1 xl1 and, i R1 R2 xl2 , , K, (), , For the calibration of an ammeter, 1 resistance, coil is specifically used in the secondary circuit, of the potentiometer, because the potential, difference across 1 is equal to the current, following through it i.e V i, If the balancing length for the emf, E0, E0 is l0 then E0 xl0 x l (Process of, 0, standardisation), Let i ' current flows through 1 resistance giving, potential difference as V ' i ' 1 xl1 where l1, is the balancing length. so error can be found as, E, i i i ' i xl1 i 0 l1, l0, Here i is ammeter reading, , Calibration of voltmeter:, , , , , Checking the correctness of voltmeter readings, with the help of potentiometer is called, calibration of voltmeter., If l0 is balancing length for E0 the emf of, standard cell by connecting 1 and 2 of bidirectional key, then x E0 / l0, + -, , K1, (), , A +, + - E0, 2, V, , Rh, , B, 1, 3, , G, , RB, , + -, , K2, (), , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , , CURRENT ELECTRICITY, , The balancing length l1 for unknown potential, difference V ' is given by (closing 2 and 3), , E , V | xl1 0 l1, l0 , If the voltmeter reading is V then the error will, be V V | which may be +ve, -ve or zero, W.E-44:The length of a potentiometer wire is 1m, and its resistance is 4 . A current of 5 mA is, flowing in it. An unknown source of e.m.f is, balanced on 40 cm length of this wire, then, find the e.m.f of the source., R 5 4, Sol: x I I L 1 20 mV, E = 1 x = 0. 40 x 20 = 8 mV, W.E-45: A cell of e.m.f 2 volt and internal, resistance 1.5 is connected to the ends of, 1m long wire. The resistance of wire is, 0.5 / m . Find the value of potential gradient, on the wire., I R E R 2 0.5, , =0. 5 V/m, , L, R r L 0.5 1.5, W.E-46: In a potentiometer experiment the, balancing length with a cell is 560 cm. When, an external resistance of 10 is connected in, parallel to the cell, the balancing length, changes by 60 cm. Find the internal resistance, of the cell., Sol: Balancing length 1 560 cm, Change in balancing length (1 2 ) 60 cm, , C. U. Q, ELECTRIC CURRENT AND DRIFT, VELOCITY, 1., , 2., , 3., , 4., , 560 2 60, 2 500 cm, , 5., , W.E-47: In a potentiometer experiment when a, battery of e.m.f. 2V is included in the, secondary circuit, the balance point is 500cm., Find the balancing length of the same end, when a cadimum cell of e.m.f. 1.018V is, connected to the secondary circuit., , 6., , Sol: E , E1 1, , E2 2, 2 , , E2, 1.018, 1 , 500 254.5cm ., E1, 2, , NARAYANA GROUP, , ne 2 A, ne 2 τA, ne 2 , 2m, Among the following dependences of drift, velocity vd on electric field E, Ohm’s Law, obeyed is, 1) vd E, 2) vd E 2, 1), , Sol: X , , , 60 6, rR 1 2 r10, 1.2 ., 500 5, 2 , , If n, e, , m, are representing electron density,,, charge, relaxation time and mass of an, electron respectively then the resistance of, wire of length l and cross sectional area A is, given by, , 7., , ml, , 2), , 2mA, , 3) ne 2 A 4), , 4) vd constant, 3) vd E, A steady current is passing through a linear, conductor of nonuniform cross-section. The, net quantity of charge crossing any cross, section per second is, 1) independent of area of cross-section, 2) directly proportional to the length of the, conductor, 3) directly proportional to the area of cross, section., 4) inversely proportional to the area of the, conductor, Given a current carrying wire of non-uniform, cross section. Which of the following quantity, or quantities are constant throughout the, length of the wire?, 1) current, electric field and drift speed, 2) drift speed only, 3) current and drift speed, 4) current only, , When electric field ( E ) is applied on the ends, of a conductor, the free electrons starts, moving in direction, , , 1) similar to E, 2) Opposite to E, , 3) Perpendicular to E 4) Cannot be predicted, The drift speed of an electron in a metal is of, the order of, 1) 10–13 m/s, 2) 10–3 mm/s, –4, 3) 10 m/s, 4) 10–30 m/s, In metals and vacuum tubes charge carriers, are, 1) electrons, 2) protons, 3) both, 4) positrons, 25
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 8., , The electric intensity E, current density j and, conductivity are related as :, 1) j E 2) j E / 3) jE 4) j 2 E, 9. Electric field (E) and current density (J) have, relation, 1, 1, 2, 1) E J 1 2) E J 3) E 2 4) E , J, J, 10. Assertion : A current flows in a conductor only, when there is an electric field within the, conductor., Reason : The drift velocity of electron in, presence of electric field decreases., 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, , OHM’S LAW & FACTORS, EFFECTING RESISTANCE, 11. In an electric circuit containg a battery, the, charge (assumed positive) inside the battery, 1) always goes form the positive terminal to the, negative terminal, 2) may move from the positive terminal to the, negative terminal, 3) always goes from the negative terminal to the, positive terminal, 4) does not move ., 12. From the following the quantity which is, analogous to temperature in electricity is, 1) potential, 2) resistance, 3) current, 4) charge, 13. The flow of the electric current through a, metallic conductor is, 1) only due to electrons, 2) only due to +ve charges, 3) due to both nuclei and electrons., 4) can not be predicted., 14. For making standard resistance, wire of, following material is used, 1) Nichrome, 2) Copper, 3) Silver, 4) manganin, 15. Material used for heating coils is, 1) Nichrome, 2) Copper, 3) Silver, 4) Manganin, 16. A piece of silver and another of silicon are heated, from room temperature. The resistance of, 1) each of them increases, 2) each of them decreases, 3) Silver increases and Silicon decreases, 4) Silver decreases and Silicon increases, 26, , 17. i-v graph for a metal at temperatures t1, t2, t3, are as shown. The highest temperature is, t3, t2, i, , t1, , V, , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 1) t1, 2) t2, 3) t3, 4) t1 t2 t3, A certain piece of copper is to be shaped into, a conductor of minimum resistance. Its length, and cross sectional area should be, 1) L and A, 2) 2L and A/2, 3) L/2 and 2A, 4) 3L and A/3, When light falls on semiconductors, their, resistance, 1) decreases, 2) increases, 3) does not change, 4) can’t be predicted, With the increase of temperature, the ratio of, conductivity to resistivity of a metal conductor, 1) Decreases 2) Remains same, 3) Increases 4) May increase or decrease, The conductivity of a super conductor, in the, super conducting state is, 1) Zero, 2) Infinity, 3) Depends on temp, 4) Depends on free election, When a piece of aluminium wire of finite length, is drawn through a series of dies to reduce its, diameter to half its original value, its resistance, will become, 1) Two times, 2) Four times, 3) Eight times, 4) Sixteen times, Metals have, 1) Zero resistivity, 2) High resistivity, 3) Low resistivity, 4) Infinite resistivity, Consider a rectangular slab of length L, and, area of cross-section A. A current I is passed, through it, if the length is doubled the potential, drop across the end faces, 1) Becomes half of the initial value, 2) Becomes one-forth of the initial value, 3) Becomes double the initial value, 4) Remains Same, A metallic block has no potential difference, applied across it, then the mean velocity of, free electrons is (T = absolute temperature, of the block), 1) Proportional to T 2) Proportional to T, 3) Zero, 4) Finite but independent of temperature., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, 26. The resistance of a metal increases with, increasing temperature because, 1) The collisions of the conducting electrons, with the electrons increases., 2) The collisions of the conducting electrons, with the lattice consisting of the ions of the metal, increases, 3) The number of the conduction electrons, decreases., 4) The number of conduction electrons increase., 27. In the absence of applied potential, the electric, current flowing through a metallic wire is zero, because, 1) The average velocity of electron is zero, 2) The electrons are drifted in random direction with, a speed of the order of 10-2cm/s., 3) The electrons move in random direction with a, speed of the order close to that of velocity of light., 4) Electrons and ions move in opposite direction., , 28. A long constan wire is connected across the, terminals of an ideal battery. if the wire is, cut in to two equal pieces and one of them is, now connected to the same battery, what will, be the mobility of free electrons now in the, wire compared to that in the first case?, 1) same as that of previous value, 2) double that of previous value, 3) half that of previous value, 4) four times that of previous value, 29. Ohm’s law is not applicable for, 1) insulators, 2) semi conductors, 3) vaccum tube, 4) all the above, 30. V - I graphs for two materials is shown in the, figure. The graphs are drawn at two different, temperatures., Y, T2, , T1, , I, , , 33. Fuse wire is a wire of, 1) low melting point and low value of a, 2) high melting pointand high value of a, 3) high melting point and low value of a, 4) low melting point and high value of a, 34. Assertion : Material used in the construction, of a standard resistance is constantan or, manganin., Reason : Temperature coefficient of constantan, is very small., 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, 35. Assertion (A) : Bending of a conducting wire, effects electrical resistance., Reason (R) : Resistance of a wire depends on, resistivity of that material., 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, 36. Assertion (A) : When the radius of a copper, wire is doubled, its specific resistance gets, increased., Reason (R):Specific resistance is independent, of cross-section of material used, 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, , THERMISTOR, , , V, , X, , 1) T1 T2 cot 2q, 2) T1 T2 sin 2q, 3) T1 T2 tan 2q, 4) T1 T2 cos 2q, 31. Wires of Nichrome and Copper of equal, dimensions are connected in series in, electrical circuit. Then, 1) More current will flow in copper wire, 2) More current will flow in Nichrome wire, 3) Copper wire will get heated more, 4) Nichrome wire will get heated more, 32. At absolute zero silver wire behaves as, 1) Super conductor, 2) Semi conductor, 3) Perfect insulator, 4) Semi insulator, NARAYANA GROUP, , CURRENT ELECTRICITY, , 37. The thermistors are usually made of, 1) metals with low temperature coefficient of, resistivity, 2) metals with high temperature coefficient of, resistivity., 3) metal oxides with high temperature coefficient, of resistivity, 4) semiconducting materials having, low temperature coefficient of resistivity, 38. For a chosen non-zero value of voltage, there, can be more than one value of current in, 1) copper wire, 2) thermistor, 3) zener diode, 4) manganin wire, 27
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 45. Two metallic wires of same material and same, ELECTRIC POWER, 39. A heater coil is cut into two equal parts and, length have different diameters. When the, only one part is used in the heater. Then the, wires are connected in parallel across an ideal, heat generated becomes, battery the rate of heat produced in thinner, 1) become one fourth 2) halved, wire is Q1 and that in thicker wire is Q2. The, 3) doubled4) become four times, correct statement is, 40. Two lamps have resistance r and R, R being, 1) Q1 = Q2, 2) Q1 < Q2 3) Q1 > Q2, greater than r. If they are connected in parallel, in an electric circuit, then, 4) It will depend on the emf of the battery, 1) the lamp with resistance R will shine more, brightly, 46. There are two metalic wires of same material,, 2) the lamp with resistance r will shine more brightly, same length but of different radii. When these, 3) the two lamps will shine equal brightly, are connected to an ideal battery in series, heat, 4) the lamp with resistance R will not shine at all, produced is H1 but when connected in parallel,, 41. Two bulbs are fitted in a room in the domestic, electric installation. If one of them glows, heat produced is H2 for the same time. Then, brighter than the other, then, the correct statement is, 1) the brighter bulb has smaller resistance, 1) H1 H2, 2) H1 H2, 2) the brighter bulb has larger resistance, 3) both the bulbs have the same resistance, 4) No relation, 3) H1 H2, 4) nothing can be said about the resistance unless, 47. Two electric bulbs rated P1 watt and V volt ,, other factors are known, are connected in series, across V-volt supply., 42. Three identical bulbs P, Q and R are connected, The total power consumed is, to a battery as shown in the figure. When the, circuit is closed, P1 P2, P1 P2, 2), P1 P2 3), 4) P1 P2 , 1), P, P, P, 2, , K, , , Q, , R, , , , 1) Q and R will be brighter than P, 2) Q and R will be dimmer than P, 3) All the bulbs will be equally bright, 4) Q and R will not shine at all, 43. Figure shows three similar lamps L1, L2, L3, connected across a power supply. If the lamp, L3 fuses. The light emitted by L1 and L2 will, change as, , Power supply, , L1, , L2, , L3, , 1) no change, 2) brilliance of L1 decreases and that of L 2, increases, 3) brilliance of both L1 and L2 increases, 4) brilliance of both L1 and L2 decreases, 44. The potential difference across a conductor is, doubled, the rate of generation of heat will, 1) become one fourth 2) be halved, 3) be doubledtimes, 4) become four times, 28, , 1, , 2, , 48. In above question, if the bulbs are connected, in parallel, total power consumed is, P1 P2, 2), 2, , P1 P2, , 3) P P 4) P1 P2 , 1, 2, 49. Which of the following causes production of, heat, when current is set up in a wire, 1) Fall of electron from higher orbits to lower orbits, 2)Inter atomic collisions, 3)Inter electron collisions, 4)Collisions of conduction electrons with atoms, 50. A constant voltage is applied between the two, ends of a metallic wire. If both the length and, the radius of the wire are doubled, the rate of, heat developed in the wire, 1) will be doubled, 2) will be halved, 3) will remain the same 4) will be quadrupled, 51. A resistor R1 dissipates the power P when, connected to a certain generator. If the, resistor R2 is put in series with R1, the power, dissipated by R1, 1) Decreases, 2) Increases, 3) Remains the same, 4) Any of the above depending upon the relative, values of R1 and R2, 1), , P1 P2, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CELL-INTERNAL RESISTANCE, EMF, 52. Back emf of a cell is due to, 1) Electrolytic polarization, 2) Peltier effect, 3) Magnetic effect of current, 4) Internal resistance, 53. The direction of current in a cell is, 1) ve pole to ve pole during discharging, 2) ve pole to ve pole during discharging, , 54., , 55., , 56., , 57., 58., , 59., , 60., 61., , 3) Always ve pole to ve pole, 4) always flows from (+)ve ploe to (-) ve pole, When an electric cell drives current through, load resistance, its Back emf,, 1) Supports the original emf, 2) Opposes the original emf, 3) Supports if internal resistance is low, 4) Opposes if load resistance is large, The terminal voltage of a cell is greater than, its emf. when it is, 1) being charged, 2) an open circuit, 3) being discharged, 4) it never happens, What is constant in a battery ( also called a, source of emf) ?, 1) current supplied by it, 2) terminal potential difference, 3) internal resistance, 4) emf, From the following the standard cell is, 1) Daniel cell, 2) Cadmium cell, 3) Leclanche cell, 4) Lead accumulator, A cell is to convert, 1) chemical energy into electrical energy, 2) electrical energy into chemical energy, 3) heat energy into potential energy, 4) potential energy into heat energy, ‘n’ identical cells, each of internal resistance, (r) are first connected in parallel and then, connected in series across a resistance ( R)., If the current through R is the same in both, cases, then, 1) R = r/2 2) r = R/2 3) R = r 4) r = 0, The value of internal resistance of ideal cell is, 1) Zero 2) infinite 3) 1 4) 2 , In a circuit two or more cells of the same emf, are connected in parallel in order, 1) Increases the pd across a resistance in the circuit, 2) Decreases pd across a resistance in the circuit, 3) Facilitate drawing more current from the, battery system, 4) Change the emf across the system of batteries, , NARAYANA GROUP, , CURRENT ELECTRICITY, 62. The resistance of an open circuit is, 1) Infinity, 2) Zero, 3) Negative, 4) cann’t be predicted, 63. According to joule's law if potential difference, across a conductor having a material of specific, resistance , remains constant, then heat, produced in the conductor is directly, proportional to, 1, 1, 2) 2, 3) , 4), 1) , , 64. Internal resistance of a cell depends on, 1) concentration of electrolyte, 2) distance between the electrodes, 3) area of electrode, 4) all the above, 65. When cells are arranged in series, 1) the current capacity decreases, 2) The current capacity increases, 3) the emf increases, 4) the emf decreases, 66. To supply maximum current, cells should be, arrange in, 1) series 2) parallel 3) Mixed grouping, 4) depends on the internal and external resistance, 67. The terminal Pd of a cell is equal to its emf if, 1) external resistance is infinity, 2) internal resistance is zero, 3) both 1 and 2, 4) internal resistance is 5, 68. The electric power transfered by a cell to an, external resistance is maximum when the, external resistance is equal to ...(r internal, resistance), r, 1), 2) 2r, 3) r, 4) r2, 2, 69. Which depolarizers are used to neutralizes, hydrogen layer in cells, 1) Potassium dichromite 2) Manganese dioxide, 3) 1 or 2, 4) hydrogen peroxide, 70. Assertion : Series combination of cells is used, when their internal resitance is much smaller, than the external resistance., n, Reason : I , where the symbols have, R nr, their standard meaning,in series connection, 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, 29
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CURRENT ELECTRICITY, 71. Assertion (A) : To draw more current at low, P.d; parallel connection of cells is preferred., Reason (R) : In parallel connection, current, i, , nE, r, , , if r >> R., , 1) Both (A) and (R) are true and (R) is the correct, explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, , KIRCHHOFF’S LAWS, WHEATSTONE BRIDGE, 72. Kirchoff’s law of meshes is in accordance with, law of conservation of, 1) charge, 2) current, 3) energy, 4) angular momentum, 73. Kirchoff’s law of junctions is also called the, law of conservation of, 1) energy, 2) charge, 3) momentum, 4) angular momentum, 74. Wheatstones’s bridge cannot be used for, measurement of very ——— resistances., 1) high 2) low 3) low(or) high 4) zero, 75. In a balanced Wheatstone’s network, the, resistances in the arms Q and S are, interchanged. As a result of this :, 1)galvanometer and the cell must be interchanged, to balance, 2) galvanometer shows zero deflection, 3) network is not balanced, 4) network is still balanced, 76. If galvanometer and battery are interchanged, in balanced wheatstone bridge, then, 1) the battery discharges, 2) the bridge still balances, 3) the balance point is changed, 4) the galvanometer is damaged due to flow of high, current, 77. Wheatstone bridge can be used, 1) To compare two unknown resistances., 2) to measure small strains produced in hardmetals, 3) as the working principle of meterbridge, 4) All the above, , 30, , JEE-ADV PHYSICS- VOL- III, 78. In a wheatstone's bridge three resistances, P,Q,R connected in three arms and the fourth, arm is formed by two resitances S1,S2 connected in parallel.The condition for bridge to, be balanced will be, P, R, 1) Q S S, 1, 2, , P, 2R, 2) Q S S, 1, 2, , P R ( S1 S 2 ), 3) Q S S, 1 2, , P R ( S1 S 2 ), 4) Q 2S S, 1 2, , 79. Assertion : At any junction of a network,, algebraic sum of various currents is zero, Reason : At steady state there is, no accumulation of charge at the junction., 1) Both (A) and (R) are true and (R) is the, correct explanation of A., 2) Both (A) and (R) are true but (R) is not the, correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, , METERBRIDGE, 80. Metal wire is connected in the left gap, semi, conductor is connected in the right gap of, meter bridge and balancing point is found., Both are heated so that change of resistances, in them are same. Then the balancing point, 1) will not shift, 2) shifts towards left, 3) shifts towards right, 4) depends on rise of temperature, 81. A metre bridge is balanced with known, resistance in the right gap and a metal wire, in the left gap. If the metal wire is heated, the balance point., 1) shifts towards left, 2) shifts towards right, 3) does not change, 4) may shift towards left or right depending on, the nature of the metal., 82. In metre bridge experiment of resistances, the, known and unknown resistances are interchanged . The error so removed is, 1) end correction, 2) index error, 3) due to temperature effect, 4) random error, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 83. In a metre-bridge experiment, when the 89. If the emf of the cell in the primary circuit is, doubled, with out changing the cell in the, resistances in the gaps are interchanged, the, secondary circuit, the balancing length is, balance-point did not shift at all. The ratio of, resistances must be, 1) Doubled, 2) Halved, 1) Very large, 2) Very small, 3) Uncharged, 4) Zero, 3) Equal to unity, 4) zero, 90. The potential gradients on the potentiometer, 84. Assertion (A) : Meterbridge wire is made up, wire are V1 and V2 with an ideal cell and a, of manganin, Reason (R) : The temperature coeffiecient, real cell of same emf in the primary circuit, of resistance is very small for manganin, then, 1) Both (A) and (R) are true and (R) is the, 1) V1 V2 2) V1 V2 3) V1 V2 4) V1 V2, correct explanation of A., 2) Both (A) and (R) are true but (R) is not the 91. If the current in the primary circuit is, decreased, then balancing length is obtained, correct explanation of A., at, 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, 1) Lower length, 2) Higher length, POTENTIOMETER, 3) Same length, 4) 1/3rd length, 85. A potentiometer is superior to voltmeter for 92. Temperature coefficient of resistance ' ' and, measuring a potential because, resistivity ‘ ’ of a potentiometer wire must, 1) voltmeter has high resistance, be, 2) resistance of potentiometer wire is quite low, 1) high and low, 2) low and high, 3) potentiometer does not draw any current from, 3) low and low, 4) high and high, the unknown source of emf. to be measured., 4) sensitivity of potentiometer is higher than that 93. A series high resistance is preferable than, shunt resistance in the galvanometer circuit, of a voltmeter., of potentiometer. Because, 86. In comparing emf’s of 2 cells with the help of, potentiometer, at the balance point, the, 1) shunt resistances are costly, current flowing through the wire is taken from, 2) shunt resistance damages the galvanometer, 1) Any one of these cells., 3) series resistance reduces the current through, 2) both of these cells, galvanometer in an unbalanced circuit, 3) Battery in the primary circuit, 4) high resistances are easily available, 4) From an unknown source, 87. A potentiometer wire is connected across the 94. The sensitivity of potentiometer wire can be, increased by, ideal battery now, the radius of potentiometer, wire is doubled without changing its length., 1) decreasing the length of potentiometer wire, The value of potential gradient, 2) increasing potential gradient on its wire, 1) increases 4 times, 2) increases two times, 3) increasing emf of battery in the primary circuit, 3) Does not change 4) becomes half, 4) decreasing the potential gradient on its wire, 88. In a potentiometer of ten wires, the balance, point is obtained on the sixth wire. To shift 95. A cell of emf ‘E’ and internal resistance ‘r’, connected in the secondary gets balanced, the balance point to eighth wire, we should, against length ‘ ’ of potentiometer wire. If a, 1) increase resistance in the primary circuit., resistance ‘R’ is connected in parallel with, 2) decrease resistance in the primary circuit., the cell, then the new balancing length for, 3) decrease resistance in series with the cell whose, the cell will be, emf. has to be measured., 4) increase resistance in series with the cell whose, R , Rr, R, R , emf. has to be measured., , 4) , l, l 2) , 1) , R l 3) r, , Rr, R r, , NARAYANA GROUP, , , , , , , , , , 31
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 96. Potentiometer is an ideal instrument, because, 1) no current is drawn from the source of unknown, emf, 2) current is drawn from the source of unknown, emf, 3) it gives deflection even at null point, 4) it has variable potential gradient, 97. On increasing the resistance of the primary, circuit of potentiometer, its potential gradient, will, 1) become more, 2) become less, 3) not change, 4) become infinite, 98. If the value of potential gradient on potentiometer, wire is decreased, then the new null point will, be obtained at, 1) lower length, 2) higher length, 3) same length, 4) nothing can be said, 99. A cell of negligible internal resistance is, connected to a potentiometer wire and, potential gradient is found. Keeping the, len gt h as const an t , i f t h e radi u s of, potentiometer wire is increased four times,, the potential gradient will become (no series, resistance in primary), 1) 4 times 2) 2 times 3) half 4) constant, 100. For the working of potentiometer, the emf of cell, in the primary circuit (E) compared to the emf of, the cell in the secondary circuit (E1) is, 1) E > E1, 2) E < E1, 3) Both the above, 4) E = E1, 101. The balancing lengths of potentiometer wire, are l1 and l2 when two cells of emf E1 and E2, are connected in the secondary circuit in, series first to help each other and next to, E1, , oppose each other E is equal to (E1>E2)., 2, l1, , 1) l, 2, , l1 l2, , 2) l l, 1, 2, , l1 l2, , 3) l l, 1, 2, , l2, , 4) l, , 1, , 102. At the moment when the potentiometer is, balanced,, 1) Current flows only in the primary circuit, 2) Current flows only in the secondary circuit, 3) Current flows both in primary and secondary, circuits, 4) current does not flow in any circuit, 32, , 103. The quantity that cannot be measured by a, potentiometer is ..........., 1) Resistance, 2) emf, 3) current in the wire 4) Inductance, 104. Assertion : Potentiometer is much better than, a voltmeter for measuring emf of cell, Reason : A potentiometer draws no current, while measuring emf of a cell, 1) Both (A) and (R) are true and (R) is the, correct explanation of A., 2) Both (A) and (R) are true but (R) is not, the correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, 105. A : The emf of the cell in secondary circuit, must be less than emf of cell in primary circuit, in potentiometer., R : Balancing length cannot be more than, length of potentiometer wire., 1) Both (A) and (R) are true and (R) is the, correct explanation of A., 2) Both (A) and (R) are true but (R) is not, the correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, , C. U. Q KEY, 1) 1, 7) 1, 13) 1, 19) 1, 25) 3, 31) 4, 37) 3, 43) 2, 49) 4, 55) 1, 61) 3, 67) 3, 73) 2, 79) 1, 85) 3, 91) 2, 97) 2, 103) 4, , 2) 1, 8) 1, 14) 4, 20) 1, 26) 2, 32) 1, 38) 2, 44) 4, 50) 1, 56) 4, 62) 1, 68) 3, 74) 2, 80) 3, 86)3, 92) 2, 98) 2, 104) 1, , 3) 1, 9) 2, 15) 1, 21) 2, 27) 1, 33) 4, 39) 3, 45) 2, 51) 1, 57) 2, 63) 4, 69) 3, 75) 3, 81) 2, 87) 3, 93) 3, 99) 4, 105) 1, , 4) 4, 10) 3, 16) 3, 22) 4, 28) 1, 34) 1, 40) 2, 46) 2, 52) 1, 58) 1, 64) 4, 70) 1, 76) 2, 82) 1, 88) 1, 94) 4, 100)1, , 5) 2 6) 3, 11) 2 12) 1, 17) 1 18) 3, 23) 1 24) 3, 29) 4 30) 1, 35) 2 36) 4, 41) 1 42) 2, 47) 3 48) 4, 53) 1 54) 2, 59) 3 60) 1, 65) 3 66) 4, 71) 1 72) 3, 77) 4 78) 3, 83) 3 84) 1, 89) 2 90) 2, 95) 4 96) 1, 101)3 102) 1, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 7., , LEVEL - I (C. W), , In the circuit shown, the reading of the, voltmeter and the ammeter are, 6V, , ELECTRIC CURRENT & DRIFT, VELOCITY, 1., , A, 100, , If the electron in a Hydrogen atom makes, 6.25x1015 revolutions in one second, the, current is, 1) 1.12 mA 2) 1 mA 3) 1.25 mA 4) 1.5 mA, The current through a wire connected to a 8., condenser varies with time as i 2t 1 A, The charge transport to the condenser from, t 0 to t 5s is, 1) 5C, 2) 55C, 3) 30C, 4) 60C, A copper wire of cross-sectional area 2.0 9., , 100, , 100, V, V, , 1)4V, 0.2A 2) 2V, 0.4A 3)3V, 0.6A 4)4V, 0.04A, The resistance of a wire of 100 cm length is, 10 . Now, it is cut into 10 equal parts and, all of them are twisted to form a single bundle., Its resistance is, 1) 1 , 2) 0.5 3) 5 , 4) 0.1 , A, metallic, wire, of, resistance, 20, ohm, stretched, 3., until its length is doubled. Its resistance is, 8, 1) 20 2) 40 , 3) 80 , 4) 60 , mm2 , resistivity = 1.7 10 m , carries a, current of 1 A. The electric field in the copper 10. A wire of resistance 20 is bent in the form, wire is, of a square. The resistance between the ends, of diagonal is, 5, , , 4, 1) 8.5 10 V / m, 2) 8.5 10 V / m, 1) 10 , 2) 5 , 3) 20 , 4) 15 , 4) 8.5 102 V / m, 3) 8.5 103 V / m, 11. Resistance of each 10 are connected as, shown in the fig. The effective resistance, OHM’S LAW AND COMBINATION OF, betweeen A and G is, RESISTANCES, 2., , 4., , 5., , B, Using three wires of resistances 1 ohm, 2ohm, A, and 3 ohm, then no.of different values of, D G, C, resistances that possible are, E, F, 1) 6, 2) 4, 3) 10, 4) 8, 1) 16 , 2) 20 3) 12 4) 8 , Six resistances of each 12 ohm are connected 12. Which arrangement of four identical, as shown in the fig. The effective resistance resistances should be used to draw maximum, between the terminals A and B is, energy from a cell of voltage V, , 1), A, , B, , 2), , 6., , 1) 8 , 2) 6 3) 4 , 4) 12 , Current ‘i’ coming from the battery and, ammeter reading are, , 3), , 4, 2V, , i, , 4, 4, , A, 4, , 3 1, A, A, 8 8, 2, 3) 2 A, A, 3, , 1), , NARAYANA GROUP, , 1 1, A, A, 8 8, 1, 4) 2 A, A, 8, , 4), , 13. If four resistances are connected as shown in, the fig. between A and B the effective, resistance is, 4, , 2), , 6, , A, 1.6 , , B, , 4, , 1) 4 , , 2) 8 , , 3) 2.4 , , 4) 2 , 33
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CURRENT ELECTRICITY, , JEE-ADV PHYSICS- VOL- III, , 14. A letter 'A' is constructed as a uniform wire of 23. Four wires made of same material have, JEE, MAINS, - VOL, - VII, resistance 1 ohm/cm. The sides of the letter, different lengths and, radii,, the wire, having, are 20 cm long and the cross piece in the middle, more resistance in the following case is, is 10cm long while the vertex angle is 600. The, 1) 100cm, r 1mm 2) 50cm, r 2mm, resistance of the letter between the two ends, 1, 1, of the legs is, 3) 100 cm , r 2 m m 4) 50cm, r mm, 2, 1) 40/3 2) 80/3 3) 40 , 4) 10 , 24., Two, different, wires, have, specific, resistivities,, 15. Find the value of colour coded resistance, lengths, area of cross-sections are in the raio, shown is fig, 3:4, 2:9 and 8:27. Then the ratio of resistance, of two wires is, 16, 9, 8, 27, 2), 3), 4), 1), 9, 16, 27, 8, Green Red Orenge Silver, 25., Two, wires, made, of, same, material, have, their, 1) 520 10%, 2) 5200 1%, length are in the ratio 1:2 and their masses in, 3) 52000 10%, 4) 52000 1%, the ratio 3 : 16. The ratio of resistance of, 16. The resistance of a wire is 2 . If it is drawn, two wires is, in such a way that it experiences a longitudinal, 1) 3/4, 2) 1:2, 3) 2:1, 4) 4:3, strain 200%. Its new resistance is, 26., A, wire, of, resistance, 18, ohm, is, drawn, until its, 2) 8 , 3) 16 4) 18 , 1) 4 , 17. ‘n’ conducting wires of same dimensions but, 1, radius, reduce, th of its original radius then, having resistivities 1, 2, 3,...n are connected, 2, in series. The equivalent resistivity of the, resistance of the wire is, combination is, 1) 188 2) 72 , 3) 288 4) 388 , 2, n, 27. A piece of wire of resistance 4 is bent, n n 1, n 1, n 1, 1), 2), 3), 4) n 1, through 1800 at its midpoint and the two, 2, 2n, 2, halves are twisted together. Then the resis18. An Aluminium ( = 4 x 10-3K-1) resistance R1, tance is, and a carbon ( = -0.5 x 10-3K-1) resistance, 1) 8 , 2) 1 , 3) 2 , 4) 5 , R2 are connected in series to have a resultant, resistance of 36 at all temperatures. The 28. If three wires of equal resistance are given, then number of combinations they cany be, values of R1 and R2 in respectively are :, made to give different resistance is, 1) 32, 4 2) 16, 20 3) 4, 32, 4) 20, 16, 1) 4, 2) 3, 3) 5, 4) 2, 19. The temperature coefficient of a wire is, 0.00125°C–1. At 300 K its resistance is one ohm. 29. The effective resistance between A and B in, the given circuit is, The resistance of the wire will be 2 at, A, 1) 1154 K 2) 1100 K 3) 1400 K 4) 1127 K, 2, 2, 2, 20. The electrical resistance of a mercury column, 3, 3, 2, in a cylindrical container is ‘R’. The mercury, B, is poured into another cylindrical container with, 2, 2, 2, half the radius of cross-section. The resistance, 1) 20 2) 7 , 3) 3 , 4) 6 , of the mercury column is, 30. How many cells each marked 6V 12 A, 1) R, 2) 2R, 3) 16R, 4) 5R, 21. Four conductors of same resistance connected, should be connected in mixed grouping so that, to form a square. If the resistance between, it may be marked 24V 24 A, diagonally opposite corners is 8 ohm, the, 1) 4, 2) 8, 3) 12, 4) 6, resistance between any two adjacent corners, 31. The effective resistance in series combinais, tion of two equal resistance is ‘s’. When they, 1) 32 ohm 2) 8 ohm 3) 1/6 ohm 4) 6 ohm, 22. The resistivity of a material is S ohm meter., are joined in parallel the total resistance is p., The resistance between opposite faces of a, If s = np then the minimum possible value of, solid cube of edge 10 cm is ( in ohm), ‘n’ is, 1) S/2, 2) S/10, 3) 100S, 4) 10S, 1) 4, 2) 1, 3) 2, 4) 3, 34, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , ELECTRIC POWER & JOULES LEW 42. A resistance coil of 60 is immersed in 42kg, 32. A 25 watt, 220 volt bulb and a 100 watt,, 220 volt bulb are connected in series across, 440 volt line, 1) only 100 watt bulb will fuse, 2) only 25 watt bulb will fuse, 3) none of the bulb will fuse, 4) both bulbs will fuse, 33. There are 5 tube-lights each of 40W in a, house. These are used on an average for 5, hours per day. In addition, there is an, immersion heater of 1500W used on an, average for 1 hour per day. The number of, units of electricity are consumed ina month is, 1) 25 units 2) 50 units 3) 75 units 4) 100 units, 34. Three equal resistors connected in series, across a source emf together dissipate 10, watt. If the same resistors are connected in, parallel across the same emf the power, dissipate will be, 1) 10 watt 2) 30 watt 3) 10/3 watt 4) 90 watt, 35. Time taken by a 836 W heater to heat one, litre of water from 100 C to 400 C is, 1) 50 s, 2) 100 s 3) 150 s, 4) 200 s, 36. A lamp of 600W-240V is connected to 220V, mains. Its resistance is, 1) 96 2) 84 3) 90 , 4) 64 , 37. A 200W - 200V lamp is connected to 250V, mains. It power consumption is, 1) 300 W 2) 312.5W 3) 292 W 4) 250 W, 38. If the current in a heater increases by 10% ,, the percentage change in the power consumption, 1) 19% 2) 21%, 3) 25%, 4) 17%, 39. The power of a heating coil is P. It is cut into, two equal parts. The power of one of them, across same mains is, 1) 2P, 2) 3P, 3) P/2, 4) 4P, 40. In a house there are four bulbs each of 50W, and 5 fans each of 60W. If they are used at, the rate of 6 hours a day, the electrical energy consumed in a month of 30 days is, 1) 64 KWH 2)90.8KWH 3)72 KWH 4) 42 KWH, 41. An electric kettle has two coils. When one coil, is switched on it takes 15 minutes and the other, takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when, connected in series and in parallel to boil the, same mass of water is, 1) 9 :2, 2) 2:9, 3) 4:5, 4) 5:4, NARAYANA GROUP, , of water. A current of 7A is passed through it., The rise in temperature of water per minute, is, 1) 40 C, 2) 80 C, 3) 1.30C, 4) 120 C, 43. What is the required resistance of the heater, coil of an immersion heater that will increase, the temperature of 1.50 kg of water from, 100 C to 500 C in 10 minutes while operating, at 240V ?, 1) 25 2) 12.5 3) 250 4) 137.2, 44. A 50 C rise in the temperature is observed in, a conductor by passing some current. When, the current is doubled, then rise in tem, perature will be equal to, 2) 100 C, 3) 200 C 4) 400 C, 1) 50 C, , CELLS AND COMBINATION OF CELLS, 45. In the following diagram, the pd across 6V, cell is, 12V, , 6V, , 3, , 2, , 1) 6V, 2) 5.6V, 3) 8.2V, 4) 8.4V, 46. While connecting 6 cells in a battery in series,, in a tape recorder, by mistake one cell is, connected with reverse polarity. If the, effective resistance of load is 24 ohm and, internal resistance of each cell is one ohm, and emf 1.5V, the current devlivered by the, battery is, 1) 0.1A 2) 0.2A, 3) 0.3A, 4) 0.4A, 47. A 10m long wire of resistance 15 ohm is, connected in series with a battery of emf 2V, (no internal resistance) and a resistance of 5, ohm. The potential gradient along the wire is, 1) 0.15 Vm-1, 2) 0.45V m-1, -1, 3) 1.5Vm, 4)4.5Vm-1, 48. When a resistance of 2 ohm is placed across, a battery the current is 1A and when the, resistance across the terminals is 17 ohm, the, current is 0.25A. the emf of the battery is, 1) 4.5 V 2) 5 V, 3) 3 V, 4) 6 V, 49. A battery has six cells in series. Each has an, emf 1.5V and internal resistance 1 ohm. If, an external load of 24 is connected to it., The potential drop across the load is, 1) 7.2V 2) 0.3V 3) 6.8V, 4) 0.4V, 35
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 50. 12 cells of each emf 2V are connected in series, among them, if 3 cells are connected, wrongly. Then the effective emf. of the, combination is, 1) 18 V 2) 12 V, 3) 24 V, 4) 6 V, 51. When a battery connected across a resistor, of 16 , the voltage across the resistor is, 12V.When the same battery is connected, across a resistor of 10 , voltage, across it is 11V. The internal resistance of, the battery in ohms is, 1) 10/7 2) 20/7, 3) 25/7, 4) 30/7, , KIRCHOFF’S LAWS, WHEATSTONE, BRIDGE, 52. Six resistors of each 2 ohm are connected as, shown in the figure. The resultant resistance, between A and B is., 1) 4 2) 2 , 3) 1 4) 10 A, B, 53. In the given circuit current through the, galvanometer is, C, 6, , 1) Zero, , 3A, , 3, , A, , G, , 2) Flows from C to D, , B, 6, , 3, , 3) Flows from D to C, , D, 4) In sufficient information, 54. The potential difference between A & B in, the given branch of a circuit is, , 1, A, , 2A, , 2, 9V, , 3, 3V, , B, , 1) 6V, 2) 12V, 3) 9V, 4) 0V, 55. The resistance between A and B is, 2, , 4, A, , 8, , 2, , 2, , 2, , B, , 4, , 1) 8 , 2) 4 , 3) 3.75 4) 2 , 56. The resistance between A and B is, C, , 3, , 1, , 288, 2) 12 , 56, 9, 8, 3) , 4) , 4, 3, , A, , 5, , 1), , 36, , B, 6, , 2, , D, , 57. The value of E of the given circuit is, JEE MAINS - VOL - VII, E, 5, 1A, , 1) 10 V, , 2) 12 V, , 1A, , 4, , 10V, , 3) 14 V 4) 18 V B, C, 5V, 6, 58. In the circuit shown in the figure, the value, of Resistance X, when potential difference, between the poins B and D is zero will be, B, 21, A, , 1) 9 , 3) 6 , , 2) 8 , , 8, , X, 3, , C, , 6, , 18, D, , 4) 4 , , METRE BRIDGE, 59. When an unknown resistance and a resistance, of 4 are connected in the left and right gaps, of a Meterbridge, the balance point is, obtained at 50cm. The shift in the balance, point if a 4 resistance is now connected in, parallel to the resistance in the right gap is, 1) 66.7cm 2)16.7 cm 3) 34.6 cm 4) 14.6 cm, 60. In a meter bridge, the gaps are closed by, resistances 2 and 3 ohms. The value of shunt, to be added to 3 ohm resistor to shift the, balancing point by 22.5 cm is, 1) 1 , 2) 2 , 3) 2.5 , 4) 5 , 61. Two equal resistance are connected in the, gaps of a metre bridge. If the resistance in, the left gap is increased by 10%, the balancing point shift, 1) 10 % to right, 2) 10% to left, 3) 9.6% to right, 4) 4.8% to right, , POTENTIO METER, 62. A potentiometer having a wire of 4m length, is connected to the terminals of a battery with, a steady voltage. A leclanche cell has a null, point at 1m. If the length of the potentiometer, wire is increased by 1m, The position of the, null point is, 1) 1.5m 2) 1.25m 3) 10.05m 4) 1.31m, 63. The emf of a battery A is balanced by a length, of 80cm on a potentio meter wire. The emf of, a standard cell 1v is balanced by 50cm. The, emf of A is, 1) 2 V 2) 1.4 V, 3) 1.5 V 4) 1.6 V, 64. When 6 identical cells of no internal resistance are connected in series in the secondary circuit of a potentio meter, the balancing, length is ‘ l ’, balancing length becomes l /3, when some cells are connected wrongly, the, number of cells conected wrongly are, 1) 1, 2) 3, 3) 2, 4) 4, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 65. In a potentiometer experiment, the balancing, length with a cell is 560cm. When an external, resistance of 10ohms is connected in parallel, to the cell the balancing length changs by, 60cm. The internal resistance of the cell in ohm, is, 1) 3.6, 2) 2.4, 3) 1.2, 4) 0.6, 66. The resistivity of a potentio meter wire is, if, the area of cross section of the wire is 4cm2., The current flowing in the circuit is 1A, the, poetntial gradient is 7.5 v/m, 2) 2 × 10–6 - m, 1) 3 × 10–3 - m, 3) 4 × 10–6 - m, 4) 5 × 10–4 - m, 67. A potentiometer wire of 10m legnth and 20, Ohm resistance is connected in series with a, resistance R ohms and a battery of emf 2V,, negligible internal resistance, Potential gradient on the wire is 0.16 millivolt / centimetre, then R is ...ohms, 1) 50 2) 60 3) 230 4) 46 , , LEVEL -I (C. W ) KEY, 1) 2, 7) 4, 13) 4, 19) 4, 25) 4, 31) 1, 37) 2, 43) 4, 49) 1, 55) 2, 61) 4, 67) 3, , 2) 3, 8) 4, 14) 2, 20) 3, 26) 3, 32) 2, 38) 2, 44) 3, 50) 2, 56) 3, 62) 2, , 3) 3, 9)3, 15) 3, 21) 4, 27) 2, 33) 3, 39) 1, 45) 4, 51) 2, 57) 4, 63) 4, , 4) 4, 10) 2, 16) 4, 22) 4, 28) 1, 34) 4, 40) 2, 46) 2, 52) 3, 58) 2, 64) 3, , 5) 3, 11) 1, 17) 2, 23) 3, 29) 4, 35) 3, 41) 1, 47) 1, 53) 3, 59) 2, 65) 3, , LEVEL - I (C. W ) - HINTS, 1., , i = qf, , 2., , q idt, , 3., 4., 5., 6., 7., 8., 9., , 5, , 0, , i, A, no of combinations x = 2n, combination of resistors, combination of resistors, combination of resistors, R, R 2, eff, n, E, , R, , 1, 10. R , , l, , V A.l, A, 20, 5, 4, , NARAYANA GROUP, , 6) 1, 12) 2, 18) 3, 24) 2, 30) 1, 36) 1, 42) 3, 48) 2, 54) 1, 60) 2, 66) 1, , 10, 5, 2, Solving for effective resistence by series and, parallel combination, combination of resistors, combination of resistors, combination of resistors, R 52 103 10%, R l2, R R1 R 2 ...Rn, , R1 10, R2 10 ; RP , , 11., 12., 13., 14., 15., 16., 17., , ρ nl , , l, l, l, + 2 + .....+ n, A, A, A, A, n 1 2 3 ......... n, =1, , n n 1, , n1, S, 2, 2, 18. R1 1 = R2 2 and R1+R2 = 36 ohm., R2 R1, 19. R t R t, 1 2, 2 1, n , , 20. R , , , , l, , V A.l, A, , 21. Combination of resistors, l, 22. R , A, l, 23. R . Check the options, r, l, 24. R , A, l2, 25. R , m, 1, 26. R 4, r, R1 R2, 1, 27. R R R, 1, 2, 28. 2n-1, 29. Combination of resistors, required current, 30. Number of rows , Given current, 2, , 24A, 2m, 12A, Number of cells in each row, , , , , requried potential, Given potential, 37
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 24, 4n, 6, Total no of cells n m, , , , 2 4 8, , 31. RS = n.RP, , V2, ; V iR, P, E, 33. P , 1 K.W.H = 1 unit, t, V2, P, R, P, , S P, 34., R, PP RS, 35. W JQ P t JmsT, 32. R , , 36., 37., 38., 39., 40., 41., 42., 43., , (V ) 2, R, P, V2, P, R, P i2, l, V2, P, ; R, A, R, no. watts no. of hours, 1 unit =, 1000, t1t2, ts = t1 + t2 ; tp = t t, 1, 2, 2, 2, ,, JQ i Rt mSt i Rt, use Joule’s law, Q msT but Q i 2 RT T i 2, t, i, t 20 C, t, i, V = E + ir, n 2m E, i, R nr , , E V , , l 50 16.7, , 2, 62.5, , 2, l, 100 62.5, , 60., l 40 cm ; 3r, 3 100 l, 3 r, X, l, , 61., R 100 l, l1 L1, 62. l L l L, 2, 2, 63. v = i l, , 64. E ' n 2m E, , 1 l2 , 65. r R , 2 , iR, RA, 66. P.G , ; Resistivity , l, L, 67. v = i l, , LEVEL - I (H. W), ELECTRIC CURRENT & DRIFT, VELOCITY, , 2, , 44., 45., 46., , 1, , 1, , 2, , 2, , 2, , o, , 2, , E, R, 47. Potential gradient = r R R L, S, P, 2, 15, 0.15, 0 15 5 10, E, 48. i , Rr, nE , 49. V , R, R nr , 50. Eeq N 2m E, , 1., , 2., , =, , 38, , E V , , 1, 2, VOL, - VII, R2MAINS, 51. r V R1 V JEE, . Solve -for, E and, 1 , 2 , substitute for r, 52. to 56. Use K.V.L., 57. from K V L, P R, , 58., Q S, 4, l, x 50, , 59., ---- (1) ; 2 100 l ----- (2), y 50, , 3., , 4., , The current passing through a conductor is 5, ampere. The charge that passes through that, conductor in 5 minute is, 1) 1200C 2) 300 C 3) 1000C 4)1500C, In a hydrogen atom, an electron is revolving, with an angular frequency 6.28 rad/s around, the nucleus. Then the equivalent electric, current is ..... 1019 A, 1) 0.16, 2) 1.6, 3) 0.016 4) 16, A current of 1.6 A is flowing in a conductor., The number of electrons flowing per second, through the conductor is, 1) 109, 2) 1019, 3) 1016, 4) 1031, If an electron revolves in the circular path of, radius 0.5A0 at a frequency of 5 x 1015 cycles/, sec. The equivalent electric current is, 1) 0.4 mA 2) 0.8 mA 3) 1.2 mA 4) 1.6 mA, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , A current flows in a wire of circular cross 13. The resistance of a wire is 10 ohm. The, section with the free electrons travelling with, resistance of a wire whose length is twice and, , the radius is half, if it is made of same material, drift velocity V . If an equal current flows in, is, a wire of twice the radius, new drift velocity, 1) 20 , 2) 5 , 3) 80 , 4) 40 , is, 14., The, resultant, resistance, of, two, resistors, , , , V, V, , when connected in series is 48 ohm. The ratio, 1) V, 2), 3), 4) 2V, of their resistances is 3 : 1. The value of each, 2, 4, resistance is, OHM’S LAW AND COMBINATION OF, 1) 20 , 28 , 2) 32 , 16 , RESISTANCES, 3), 36, ,, 12, 4) 24 , 24 , , , 6. Three resistances each of 3 are connected, as shown in fig. The resultant resistance 15. The resistance of a bulb filament is 100, at a t em p erat u re o f 1000 C ., If i t s, between A and F is, temperature coefficient of resistance be, E, B, C, D, 0.005 per 0 C , its resistance will become, A, F, 200 at temperature of, 1) 3000 C 2) 4000 C 3) 5000 C 4) 2000 C, 1) 9 , 2) 2 , 3) 4 , 4) 1 , 7. Two wires made of same material have 16. The current 'i' in the circuit given aside is, 5., , I, , lengths in the ratio 1 : 2 and their volumes in, the same ratio. The ratio of their resistances, is, 1) 4 : 1 2) 2 : 1, 3) 1 : 2, 4) 1 : 4, 8. Two wires made of same material have their, electrical resistances in the ratio 1 : 4. If, their lengths are in the ratio 1 : 2, the ratio, of their masses is, 1) 1: 1, 2) 1 : 8 3) 8 : 1 4) 2 : 1, 9. There are five equal resistors., The minimum resistance possible by their, combination is 2 ohm. The maximum possible, resistance we can make with them is, 1) 25 ohm 2) 50 ohm 3) 100 ohm 4) 150 ohm, 10. An electric current is passed through a circuit, containing two wires of the same material,, connected in parallel. If the lengths and radii, of the wires are in the ratio 4/3 and 2/3, then, the ratio of the currents passing through the, wires will be, 1) 3, 2) 1/3, 3) 8/9, 4) 2, 11. A current of 1 A is passed through two, resistances 1 and 2 connected in parallel., The current flowing through 2 resistor will, be, 1) 1/3 A 2) 1 A 3) 2/3 A, 4) 3 A, 12. The colour coded resistance of corbon, resistance is (Initial three bands are red and, fourth band is silver), 1) 222. 10%, 2) 2200 10%, 3) 333 5%, 4) 33000 10%, NARAYANA GROUP, , 1) 0.1 A 2) 0.2A 2V, , B, , +, – 30, , 30, 30, , 17., , 18., , 19., , 20., , 21., , 3) 1.0A 4) 2.0 A, A, C, The combined resistance of two conductors, in series is 1 . If the conductance of one, conductor is 1.1 siemen, the conductance of, the other conductor in siemen is, 1) 10, 2) 11, 3) 1, 4) 1.1, Four conductors of resistnace 16 each are, connected to form a square. The equivalent, resistance across two adjacent corners is (in ohm), 1) 6, 2) 18, 3) 12, 4) 16, When two resistances are connected in, parallel then the equivalent resistance is 6/5., When one of the resistance is removed, then the effective resistance is 2 . The, resistance of the wire removed will be, 3, 6, 1) 3 ohm 2) 2 ohm 3) ohm 4) ohm, 5, 5, A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has, twice the diameter of a wire made of ‘A’., Then for the two wires to have the same, resistace, the ratio lB /lA of their respective, lengths must, 1) 1, 2) 1/2, 3) 1/4, 4) 2/1, If a wire of resistance ‘R’ is melted and, recasted in to half of its length, then the new, resistance of the wire will be, 1) R/4, 2) R/2, 3) R, 4) 2R, 39
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 22. When a wire is drawn until its radius decreases, by 3%. Then percentage of increase in resistance is, 1) 10% 2) 9%, 3) 6%, 4) 12%, 23. When three wires of unequal resistances are, given the number of combinations they can, be made to give different resistances is, 1) 6, 2) 4, 3) 2, 4) 8, 24. The resistance of a coi is 4.2 at 1000C and, the temperature coefficient of resistance of, its material is 0.004/0C. Its resistance at 00C, is, 1) 6.5 2) 5 , 3) 3 , 4) 2.5 , 25. You are given several identical resistors each, of value 10 and each capable of carrying a, maximum current of 1A. It is required to make, a suitable combination of these to resistances, to produce a resistance of 5 which can carry, a current of 4A. The minimum number of resistors required for this job is, 1) 4, 2) 8, 3) 10, 4) 20, 26. A wire of resistance 50 is cut into six equal, parts and they ae bundled together side by, side to form a thicker wire. The resistance of, the bundle is, 18, 9, 25, 25, 2), 3), , , 1), 4), 25, 12.5, 9, 18, 27. Three conductors of resistance 12 each are, connected to form an equilateral triangle. The, resistance between any two vertices is, 2) 2 , 3) 6 , 4) 8 , 1) 4 , 28. When three equal resistance are connected, in parallel, the effective resistance is 1/ 3 ., If all are connected in series, the effective, resistance is, 1) 9 , 2) 3 , 3) 6 , 4) 12 , 29. A technician has only two resistance coils. By, using them in series or in paralle he is able to, obtain the resistance 3,4,12 and 16 ohms. The, resistance of two coils are, 1) 6, 10 2) 4, 12, 3) 7, 9, 4) 4, 16, 30. The effective resistance between A&B in the, given circuit is, 2, , 2, , 2, , A, 3, , 2, , B, 2, , 1) 7 , 40, , 2) 2 , , 2, , 2, , 3) 6 , , 4) 5 , , 31. The effective resistance between A and B is, JEE MAINS - VOL - VII, 3 then the value of R is, A, 4, B, , 1) 2 , , 4, , 2) 4 , , 4, , 3) 6 , 4) 8 , R, 32. The effective resistance between A and B in, the given circuit is, 1) 2 2) 4 , 3) 3 4) 6 , , A, , 3, , 6, , 3, , 6, , 3, , 6, , B, , ELECTRIC POWER & JOULES LAW, 33. An electric bulb is rated 220 volt and 100 watt., Power consumed by it when operated on 110, volt is, 1) 50 watt 2) 75 watt 3) 90 watt 4)25 watt, 34. A heater coil is cut in to two parts of equal, length and only one of them is used in the, heater. The ratio of the heat produced by this, half-coil to that by the original coil is, 1) 2 : 1 2) 1 : 2, 3) 1 : 4, 4) 4 : 1, 35. If the electric current in a lamp decreases by, 5% then the power output decreases by, 1) 20% 2) 10%, 3) 5%, 4) 2.5%, 36. Two electric bulbs whose resistances are in, the ratio of 1 : 2 are connected in parallel to, a constant voltage source. The powers, dissipated in them have the ratio, 1) 1 : 2 2) 1 : 1, 3) 2 : 1, 4) 1 : 4, 37. A bulb rated 60 W -120V is connected to 80V, mains. What is the current through the bulb, 2, 5, 3, 1, 2) A, 3) A, 4) A, 1) A, 3, 3, 3, 5, 38. An electric bulb has the following specifications 100 watt, 220 volt. The resistance of, bulb, 1) 384 2) 484 3) 344 4) 584 , 39. A 200W and 100W bulbs, both meant for operation at 220V, are connected in series to, 220V. The power consumption by the combination is, 1) 46 W 2) 66 W 3) 56 W, 4) 75 W, 40. Five bulbs, each rated at 40 W-220 V are used, for 5 hours daily on 20V line. How may units, of electric energy is consumed in a month of, 30 days?, 1) 20 units 2) 25 units 3) 15 units 4) 30 units, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 41. An electric Kettle has two heating coils. When, one of them is switched on water in it boils in, 6 minutes and when other is switched on water boils in 4 minutes. In what time will the, water boil if both coil are switched on simultaneously, 1) 1.6 min 2) 2.8 min 3) 2.4 min 4) 3 min, 42. A 10 V storage battery of negligible internal, resistance is connected across a 50 resistor.., How much heat energy is produced in the, resistor in 1 hour, 1) 7200J 2) 6200J 3) 5200J 4) 4200J, , 50. If the external resistance is equal to internal, resistance of a cell of emf E. The current, across the circuit is, E, r, r, E, 1), 2), 3), 4), r, E, 2E, 2r, 51. Two cells each of emf 10V and each 1 internal resistance are used to send a current, through a wire of 2 resistance. The cells, are arranged in parallel. Then the current, through the circuit, 1) 2A, 2) 4A, 3) 3A, 4) 5A, , CELLS AND COMBINATION OF CELLS, , KIRCHOFF’S LAWS, WHEATSTONE, BRIDGE, , 43. A cell of emf 6V is being charged by 1A, current. If the internal resistance of the cell, is 1 ohm, the potential difference across the, terminals of the cell is, 1) 5V, 2) 7V, 3) 6V, 4) 8V, 44. When two identical cells are connected either, in series or in parallel across 2 ohm resistor, they send the same current through it. The, internal resistance of each cell is, 1) 2 ohm 2) 1.2 ohm 3) 12 ohm 4) 21 ohm, 45. The emf of a Daniel cell is 1.08V. When the, terminals of the cells are connected to a, resistance of 3 , the potential difference, across the terminlas is found to be 0.6V. Then, the internal resistance of the cell is, 1) 1.8 2) 2.4 3) 3.24 4) 0.2 , 46. Four cells each of emf 2V and internal, resistance 1 ohm are connected in parallel, with an external resistance of 6 ohm. The, current in the external resistance is, 1) 0.32 A 2) 0.16 A 3) 0.2 A, 4) 0.6 A, 47. A student is asked to connected four cells of, emf of 1 V and internal resistance 0.5 ohm in, series with an external resistance of, 1 ohm. But one cell is wrongly connected by, him with its terminal reversed, the current, in the circuit is, 1, 2, 3, 4, 1) A, 2) A, 3) A, 4) A, 3, 3, 4, 3, 48. Two cells of emf 1.25V, 0.75V and each of, internal resistance 1 are connected in, parallel. The effective emf will be, 1) 1 V, 2) 1.25 V 3) 2 V, 4) 0.5 V, 49. The emf of a cell is 2V. When the terminals, of the cell is connected to a resistance 4 ., The potential difference across the terminals,, if internal resistance of cell is 1 is, 3, 8, 6, 5, 1) V, 2) V, 3) V, 4) V, 5, 5, 5, 8, NARAYANA GROUP, , 52. The figure below shows current in a part of, electric circuit. The current i is, 1amp, , 2amp, , 1.3amp, 2amp, i, , 1) 1.7amp 2) 3.7 amp 3) 1.3 amp 4) 1 amp, 53. Current in the main circuit shown is, 6, , 6, , 6, , 4, 6, , 4, , 6, , 6, , 10V, 1, , 1) 1.5 A, , 2) 2 A, , 3) 0.6 A, , 4) 1 A, , 54.. Find ‘i’ for the given loop., A, 5V, , 3, , 6, 8, i, 1) A, 2) A, 6V, 5, 9, 5, 1, 1, C, 3) A, 4) 1A, B, 3V, 2, 55. The potential difference between points A, and B is, , 2V 2 5V, A, , 1A, , 1, , B, , 1) 0 V, 2) 10 V, 3) 4 V, 4) 5 V, 56. In wheat stone bridge, P and Q are approximately equal. When R is 500 , the bridge is, balanced. On interchanging P and Q, the values of R is 505 for balanching . The value, of ‘S’ is, 1) 500.5 2) 501.5 3) 502.5 4) 503.5 , 41
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 57. To balance the bridge in the circuit, the valR, ues of R is, 5, 12, G, , 1) 8 , 3) 20 , , 2) 4 , 4) 12 , , 15, , 60, , METRE BRIDGE, 58. The point in a Metre bridge is at 35.6 cm. If the, resistances in the gaps are interchanged,the new, balance point is, 1) 64.4 cm 2) 56 cm 3) 41.2 cm 4) 56.7 cm, 59. In a metre bridge expt, when the resistances, in the gaps are interchanged the balance point, is increases by 10cm. The ratio of the resistances is, 15, 12, 11, 10, 1), 2), 3), 4), 5, 8, 9, 9, 60. When an unknown resistance and a resistance, 6 are connected in the left and right gaps, of a meter bridge, the balance point is obtained at 50cm. If 3 resistance is connected, in parallel to resistance in right gap, the balance point is, 1) decrease by 25 cm 2) increase by 25 cm, 3) decrease by 16.7 cm 4) increase by 16.7 cm, 61. When un known resistance and a resistance, of 5 are used in left and right gaps of meter, bridge the balance point is 50cm. The, balanceing point if 5 resistance is now connected in seriece to the resistor in right gap, 1) 20 cm 2) 33.3 cm 3) 60 cm 4) 60 cm, 62. In a meter bridge experiement two unkonwn, resistances X and y are connected to left and, right gaps of a meter bridge and the balancing point is obtained at 20cm from right (X>Y), the new position of the null point from left if, one decides balance a resistance of 4X against, Y., 1) 114 cm 2) 80 cm 3) 53.3 cm 4) 70 cm, , POTENTIO METER, 63. In a potentiometer the balance length with, standard cadmium cell is 509 cm. The emf of, a cell which when connected in the place of, the standard cell gave a balance length of 750, cm is (emf of standard cell is 1.018V), 1) 1.5V 2) 0.5V, 3) 1.08V 4) 1.2V, 64. Two cells of emf’s E1 and E2 when placed in, series produce null deflection at a distance of, 204 cm in a potentio meter. When one cell is, reversed they produce null deflection at 36, cm if E1 1.4v then E2=, 1) 0.98 V 2) 2.47 V 3) 0.098 V 4) 98.8 V, 42, , 65. When 6 identical cells of no internal resistance, JEE, MAINS, VOL - VII, are connected in series, in the, second- arycircuit, of a poetntio meter, the balancing length is l., If two of them are wrongly connected the balancing length becomes, l, l, 2l, 2), 3) l, 4), 4, 3, 3, 66. In an experiment to determine the internal, resistance of a cell with potentiometer, the, balancing length is 165cm. When a reistance, of 5 ohm is joined in parallel with the cell the, balancing length is 150cm. The internal resistance of cell is, 4) 0.5 , 1) 2.2 2) 1.1 3) 3.3 , 67. The resistivity of a potentio meter wire is 40, x 10-8 m and its area of cross section is 8, x 10-6 m2. If 0.2A current is flowing through, the wire, the potential gradient will be, 1) 10–2 V/m, 2) 10–1 V/m, 3) 3.2 × 10–2 V/m, 4) 1 V/m, 68. The emf of a cell is Ev, and its its internal, resistance is 1 . A resistance of 4 is joined, to battery in parallel. This is connected in secondary circuit of poetntio meter. The balancing length is 160cm. If 1V cell balances for, 100cm of potentio meter wire, the emf of cell, E is, 1) 1 V, 2) 3 V, 3) 2 V, 4) 4 V, , 1), , LEVEL -I ( H. W ) KEY, 1) 4, 7) 4, 13) 3, 19) 1, 25) 2, 31) 3, 37) 1, 43) 2, 49) 2, 55) 1, , 2) 2, 8) 1, 14) 3, 20) 4, 26) 4, 32) 3, 38) 2, 44) 1, 50) 4, 56) 3, , 3) 2, 9) 2, 15) 2, 21) 1, 27) 4, 33) 4, 39) 2, 45)2, 51) 2, 57) 1, , 4) 2, 10) 2, 16) 1, 22) 4, 28) 2, 34) 1, 40) 4, 46) 1, 52) 1, 58) 1, , 5) 3, 11) 1, 17) 2, 23) 4, 29) 2, 35) 2, 41) 3, 47) 2, 53) 4, 59) 3, , 6) 4, 12) 2, 18) 3, 24) 3, 30) 3, 36) 3, 42) 1, 48) 1, 54) 2, 60) 2, , 61) 2, 67) 1, , 62) 3 63) 1 64) 1 65) 2, 68) 3, , 66) 4, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , LEVEL - I ( H. W ) - HINTS, , CURRENT ELECTRICITY, RR, 6, 19. R R 5 . If R is removed R 2, 1, , 1., , q, i, t, , 2., , i qf, , 2, , 2, , 1, , 1, , 2, , 2R, 6, 5R 6 3R R 3, 2 R 5, 2, , 2, , 2, , 2, , 2, , ne, it, n, t, e, , 3., , i, , 4., , iq f, , 6., , 1, Vd 2, r, The 3 resistances are parallel, , 7., , R l2, , 8., , l2, R, m, , 5., , 9., , 10., , R, 2, 5, , , RA, r2, , l , A, , , , 20. R , , V constant, , 21. R 2, 1, r4, , 22. R , 23. 2n, , 24. Rt R0 1 t , R, ; i p mi, n, m × n = ..............., , 25. RP , Rmax 5 R Rmin , , 2, 1, 2, 2, , i1 r l2, , i2 r l1, , iR1, 11. i2 R R, 1, 2, , 12. R 22 102 10%, R1 l1 r22, , 13., R2 l2 r12, , 26. R ' , , R, n2, , 27. R , , 2R, 3, , 28. RP , , R, and RS nR, n, , 29. R1 , , R1 R2, 14. RS R1 R2 , RP R R, 1, 2, , Solving for R1 & R2, , 16. use ohm’s law, 17. R1 + R2 = 1 ohm., , 33. P , , 1, 10, =, 1.1, =>, R, =, 1, R1, 11, 1, , R2 , , RS Rs2 4 Rs R p, 2, Rs Rs2 4 Rs Rp, 2, , 30. Using combination of resistors, 31. Using combination of resistors, 32. Using combination of resistors, , R2 R1, 15. R t R t, 1 2, 2 1, , 1, , R2 = 1 - R1 R 1 R, 2, 1, , NARAYANA GROUP, , R, ., 5, , V2, R, , 34. W JQ Q , , V2, Q R, 1 2, RJ, Q2 R1, P, , I, , 100 2 100, 35. P i 2 R P i 2 , P, I, 43
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , V2, 36. P , R, 37. R , , V '2, P', , and i , , V, R, , 52., 53., 54., 55., , K.C.L, K.C.L, K.C.L, K.C.L, , 56., , P R, , Q S, , 57., , P R, , Q S, , 58., , P 35.6 Q 64.4, , , , Q 64.4 R 35.6, , 59., , X, l, , R 100 l, , 60., , X, l, , R 100 l, , 61., , X, l, , R 100 l, , 62., , x, l, , y 100 l, , V2, 38. R , P, P1 P2, 39. P P P, 1, 2, , 40., , no. of watts no. of hours, 1000, , 41. ts t1 t2 ;, , tp , , tt1 2, t1 t2, , 2, , V, t Q 7200J, 42. Q , R, , 43. V E ir, nE, E, , 44. iS iP ; R nr R r , n, , E V, 45. r , V, , E1 l1, 63. E l 3, 2, 2, , , R, , , 2 4 8, , , , 46. i , 25, 25, r 6 1, R, 4, n, E, , 47. i , , R Nr, , E1r2 E2 r1, r1 r2, , 49. V E i r, E, 50. i , Rr, , 51. i , , 44, , 2, , 64., , 1 4 0.5, , E1 1 2, , E2 1 2, , 65. N E l1 ,, , N 2n E 4 2 1 2 1, , 48. Eeff , , JEE MAINS - VOL - VII, , 3, , N 2m E l2, , 1 l2 , 66. r R , 2 , 67. P. G , , i, A, , 1 l2 , 68. r R , 2 , , E, r, R, n, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , OHM’s LAW AND COMBINATION, OF RESISTANCES, , LEVEL - II (C. W), ELECTRIC CURRENT AND, DRIFT VELOCITY, 1., , 2., , In a hydrogen tube it is observed that through, a given cross - section 3.13 x 1015 electrons, per sec, moving from right to left and 3.12 x, 1015 protons per sec are moving from left to, right. The electric current in the discharge, tube and its direction is, 1) 1 mA towards left, 2) 1 mA towards right, 3) 1.5 mA towards right 4) 2 mA towards left, An electron of mass m, moves around the, nucleus in a circular orbit of radius ‘r’ under, the action of centripetal force ‘F’. The, equivalent electric current is, 1), , e, 2, , F, mr, , 2), , e, , , 6., , 1) 4V, , 7., , 4., , 6, , 3, , 2, , 2) 3V, , 3) 2V, 4) 5V, B, The resistance of the network between the, terminals A and B is, A, , 5, , 5, , 20, , 5, , 5, , B, 5, , 8., , 3), 3., , 4, 3A, , Fr, m, , e Fm, e Fm, 4), 2, r, , r, The current in a conductor varies with time, ‘t’ as I 3t 4t 2 . Where I in amp and t in, sec. The electric charge flows through the, section of the conductor between t = 1s and, t=3s, 100, 127, 140, 150, C 2), C 3), C 4), C, 1), 3, 4, 3, 3, A conductor has a non-uniform section as, shown in the figure. A steady current is, flowing through it. Then the drift speed of, the electrons, (M-2012), , A current of 3A flows in a circuit shown in, the figure. The potential difference between, A, A and B is, , 10, , 1) 30 2) 20, 3) 50, 4) 60, In the figure, the value of resistance to be, connected between C and D so that the, resistance of the entire circuit between A and, B does not change with the number of, elementary sets used is, R, , A, R, B, , R, , R, , R, , R, R, , R, , R, , C, , R, , R, , D, , R, , , , 9., , 5, , , , , , , , 1) R 2) R 3 1 3) 3R 4) R 3 1, The effective resistance across the points A, C, and I is, 2, , 2, 2, , B, , D, , 2, , 2, I, , A, Q, , P, , 1) 2, , 2) 1, , 2, 2, G, , 5., , 1) is constant throughout the wire, 2) varies unpredictably, 3) decreases from P & Q, 4) increases from P to Q, A current of 16 A is made to pass through a, conductor in which the number of density of, free electrons is 4 × 1028 m–3 and its area of, cross section is 10–5 m2 . The average drift, velocity of free electrons in the conductor is, (M-2012), 1) 1.6 × 10–4 ms–1, 2) 2.5 × 10–4 ms–1, 3) 6.4 × 10–4 ms–1, 4) 3.2 × 10–4 ms–1, , NARAYANA GROUP, , E, , 2, 2, , 3) 0.5 4) 5 , , 2, F, , 10. In the circuit shown below, the cell has an, emf of 10V and internal resistance of 1 ohm., the other resistances are shown in the figure., The potential difference VA VB is, E = 10V, r = 1, 4, A, , 2, , 1) 6V, , 2) 4V, , B, , 2, , 1, , 4, , 3) 2V, , 4) -2 V, 45
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 11. A uniform wire of resistance 20 having, resistance 1 / m is bent in the form of circle, as shown in fig. If the equivalent resistance, between M and N is 1.8 , then the length of, the shorter section is, M N, 1) 2 m, 2) 5 m, 3) 1.8 m 4) 18 m, 12. If the voltmeter reads 0.2 V and the ammeter, reads 0.101A, the resistance of the voltmeter, is (in ohm), 2, , A, , 1) 500, , 2) 1000, , A, 3) 200, 4) 400, 13. In the given circuit Ammeter reading is same, when both switches S 1 , S 2 are closed or, opened. The value of resistance R is, 100, A, , S1, , S2, , 50, , R, , 17. A 220 V and 800 W electric kettle and three, 220V and 100W bulbs, are connected, JEE MAINS, - VOL - in, VII, parallel. On connecting this combination with, 200 V supply, the total current in the circuit, will be, 1) 0.15 A 2) 5A, 3) 5.5A, 4) 4.55A, 18. What is the equivalent resistance of the, 2, circuit, 4V 1, – 2, , +, , 1) 6, , 2) 7, , 2, , 4, , 3) 8, 4) 9, V, 19. The temperature coefficient of resistance, of platinum is 3.92 103 K 1 at 200 C ., Find the temperature at which the increase, in the resistance of platinum wire is 10%, of its value at 200 C, 1) 40.50 C 2) 45.50 C 3) 48.50 C 4) 43.50 C, 20. Four identical resistance are joined as shown, in fig. The equivalent resistance between, points A and B is R1 and that between A and, R1, C is R2. Then ratio of R is, 2, 2, A, , 300, , A, , 1) 5 2) 5, V, 3) 5 4) 5, 15. Twelve resistances each of resistance R are, connected in the circuit as shown in fig. Net, resistance between points A and C would be, B, R, G, , R, , 5R, 3, , 2), , 7R, 6, , R, A, , R, , 46, , 2, , 2) 3:4, , R, E, , 2, , C, D, 3) 2:5, 4) 1:2, 2, 21. If the galvanometer reading is zero in the, given circuit, the current passing through re500, sistance 250 is, G, , +, –, , 12V, , 1) 0.016 A 2) 0.16 A, , +, 4V, –, , 250 x, , 3) 0.032A 4) 0.042 A, 22. The effective resistance between A and B is, 3, the given circuit is, 3, , R, , 6, , C, , R, , R, 3R, R, 4, D, 16. A resistance is made by connecting two wires, (series) of same material of radii 2 mm and 5, mm and length 8 cm and 5 cm. A potential, difference of 22V is applied to them. The, potential difference on the longer wire is, 1) 15 V 2) 18 V, 3) 16 V, 4) 20 V, , 4), , 1) 1: 5, , R, , F, R, , R, , 3) R, , B, , 1.5V, , 1) 200 2) 100 , 3) 400 4) -300 , 14. In the following diagram ammeter reading is, 4A, voltmeter reading is 20V, the value of R, is, R, , 1), , A, , 6, , 3, 6, , 1) 3 , 2) 2 , 3, 3, A, B, 3) 4 , 4) 6 , 3, 23. The equivalent resistance between points A, and B of an infinite network of resistance each, of 1 connected as shown is, A, , 1, , 1, , 1, , 1, , 1, , 1, , B, , 1), , 1 5, 2 5, 3 5, 1 7, 2), 3), 4), 2, 4, 2, 3, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 24. Equivalent resistance across A and B in the, r, given circuit is, 8r, r, r, r, r, 7, 3), 4) 6r A, O, r, r, B, 3, 25. Two resistances of 400 and 800 are connected in series with 6V battery of negligible, internal resistance. A voltmeter of resistance, 10,000 is used to measure the p.d. across, 400 . The error in the measurement of p.d., in volts approximately, 1) 0.05 V 2) 0.5 V, 3) 0.75 V 4) 5 V, 26. Copper and carbon wires are connected in, series and the combined resistor is kept at, 0°C. Assuming the combined resistane does, not vary with temperature, the ratio of the, resistances of carbon and copper wires at 0°C, is (Temperature coefficients of resisti-vity of, copper and carbon respectively are 4 × 10–3/, °C and – 0.5 × 10–3/°C) (M-2013), 1) 2, 2) 4, 3) 8, 4) 6, 27. Three resistances of equal values are, arranged in four different configurations as, shown below. Power dissipation in the, increasing order is, (E-2012), , 1), , 2r, 73r, , 2), , R, , R, , I, , R, , R, , R, , R, II, , R, I, , R, , R, R, R, , R, , III, , IV, , 1) (III) < (II) < (IV) < (I) 2) (II)<(III)<(IV)<(I), 3) (I) < (IV) < (III) < (II) 4) (I)<(III)<(II)<(IV), 28. If 400 of resistance is made by adding four, 100 resistances of tolerance 5%, then the, tolerance of the combination is [Mains-2011], 1) 5%, 2) 10%, 3) 15%, 4) 20%, , ELECTRIC POWER, 29. Two wires A and B with lengths in the ratio, of 3 : 1, diameters in the ratio of 1:2 and, resistivities in the ratio of 1:20 are joined in, parallel with a source of emf. 2V. Ratio of the, R1 / R2 is:, 1) 5 : 2 2) 2 : 5, 3) 5 : 3, 4) 3 : 5, NARAYANA GROUP, , 3, 1) 5 amp, 2, , 2), , 5 3, amp, 2, , 27, amp, 4) 5 amp, 8, 33. In a large building, there are 15 bulbs of 40, W, 5 bulbs of 100 W, 5 fans of 80 W and 1, heater of 1 kW. The voltage of electric mains, is 220 V. The minimum capacity of the main, fuse of the building will be: [Mains-2014], 1) 8 A, 2) 10 A, 3) 12 A, 4) 14 A, 34. The supply voltage to room is 120 V. The, resistance of the lead wires is 6 . A 60 W, bulb is already switched on. What is the, decrease of voltage across the bulb, when a, 240 W heater is switched on in parallel to the, bulb?, [Mains-2013], 1) zero 2) 2.9 Volt 3) 13.3 Volt 4) 10.04 Volt, 3) 5, , INTERNAL RESISTANCE AND EMF, R, , I, , 30. An electric heater operating at 220 volts boils, 5 litre of water in 5 minutes. If it is used on 110, volts, it will boil the same amount of water in, 1) 10 minutes, 2) 20 minutes, 3) 15 minutes, 4) 25 minutes, 31. Three electric bulbs of 40W, 60W and 100W, have the tungsten wire of the same diameter., Then the longer wire is used by, 1) 60W 2) 100W 3) 40W, 4) All use the same length, 32. A fuse wire with radius of 0.2 mm blows, off with a current of 5 Amp. The fuse wire, of same material but of radius 0.3 mm will, blow off with a current of, , 35. In the circuit shown here, cells A and B have, emf 10 V each and the internal resistance is, 5 for A and 3 for B. For what value of R, will the potential difference across the cell A, will be zero?, E, 5, +, A, , 1) zero, , E, 3, +, B, , 2) 1 ohm, , R, 3) 2 ohm 4) 3 ohm, 36. In the circuit of fig. with steady current, the, potential drop across the capacitor is, , 1) V, V, 3), 3, , V, 2, 2V, 4), 3, , 2), , V, , R, , V, , C, , 2V, , 2R, 47
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 37. In the circuit, the galvanometer G shows zero, deflection. If the batteries A and B have, negligible internal resistance, the value of the, resistor R will be :, 1000, G, , 12V, , 1) 100 2) 200, 38., , 39., , 40., , 41., , 42., , 43., , 48, , B, , R, , A, , 2V, , 3) 500 4) 1000, Twenty four cells each of emf 1.5V and, internal resistance 0.5 ohms are to be, connected to a 3 ohm resistance. For, maximum current through this resistance the, number of rows and number of columns that, you connect these cells is., 1) 12 cells in series 2 rows in parallel, 2) 8 cells in series 3 rows in parallel, 3) 4 cells in series 6 rows in parallel, 4) 6 cells in series 4 rows in parallel, A battery of four cells in series each having, an emf of 1.5V and internal resistance 1 , are connected in series with an ammeter, a, coil of resistance 2 and a filament lamp. If, the ammeter reads 0.5A, the resistance of, the filment lamp is, 2) 6 , 3) 2 4) 12 , 1) 4 , A 5V battery with internal resistance 2 and, a 2V battery with internal resistance 1 are, connected in parallel with unlike polarities, connected together .This combination is, connected to 10 resistor The current in the, 10 resistor is, 1) 0.27 A 2) 0.05 A 3) 0.25 A 4) 0.3 A, A voltmeter resistance 500 is used to, measure the emf of a cell of internal, resistance 4 . The percentage error in, the reading of the voltmeter will be, 1) 0.4%, 2) 0.6% 3) 0.8% 4) 1.2%, When two identical cells are connected either in series or in parallel across a 4 ohm, resistor, they send the same current through, it. The internal resistance of the cell in ohm, is, 1) 4 , 2) 2 , 3) 1 , 4) 7 , Two cells with same e.m.f. ‘E’ and different, internal resistances r1 and r2 are connected, in series to an external resistance ‘R’. The, value of R so that the p.d. across the first cell, be zero is, r r, 1) r2 r1 2) r1 r2, 3) r1 r2 4) 1 2, 2, , 44. Two conductors have the same resistance at, JEE MAINS, - VOL - of, VII, 0°C but their temperature, coefficients, resistance are 1 and 2 . The respective, temperature coefficients of their series and, parallel combinations are nearly [AIE-2010], 1 2, 2, , 1 2, 2) 1 2 , 1, 1), 2, 2, 1 2, 2 1 2, ,, 3) 1 2 , , 4) 1, 2, 2, 1, 2, , KIRCHOFF’S LAWS AND WHEAT, SHONE’S BRIDGE, 45. The electric current i in the circuit shown is, 2A, (E-2011) 3A, A, , 1) 6 A, , B, , 2A, , 2) 2 A, , 1A, C, i, , 3) 3 A, 4) 4 A, 46. In the circuit shown in the figure, the current, C, ‘I’ is, (EAM-2013), 10V, , 1) 6 A, , 2) 2 A, , 24V, , A, , P, , 2, B, , 3, , 1, , 1, , D, , 3) 4 A, 4) 7 A, 9V, 47. Four resistors A, B, C and D form a, wheatstone’s bridge. The bridge is balanced, when C = 100 . If A and B are inter changed,, the bridge balances for C = 121 . The vlaue, of D is, (E-2012), 1) 10 2) 100 3) 110 4) 120 , 48. In the circuit shown below, the ammeter, reading is zero. Then the value of the, resistance R is, ( E-2011), 12V, , 1) 50 2) 100 , 500, 3) 200 4) 400 , , +, , –, , – +, R, , METRE BRIDGE, 49. Two unknown resistrance X and Y are, connected to left and right gaps of a meter, bridge and the balancing point is obtained at, 80 cm from left. When a 10 resistance is, connected in parallel to ‘x’, the balance point, is 50 cm from left. The values of X and Y, respectively are, 1) 40 , 9, 2) 30 , 7.5, 3) 20 , 6, 4) 10 , 3, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 50. In the meter bridge experiment, the length AB, of the wire is 1 m. The resistors X and Y have, values 5 and 2 respectively. When a, shunt resistance S is connected to X, the, balancing point is found to be 0.625 m from, A. Then, the resistance of the shunt is, S, (Eam-2013), , are 4 × 10–7 m2, respectively. The potential, gradient will be equal to, [Mains-2011], 1)1 V/m 2)0.5 V/m 3)0.1 V/m 4)0.2 V/m, , LEVEL - II ( C. W ) KEY, 1) 2, 7) 2, 13) 4, 19) 2, 25) 1, 31) 3, 37) 2, 43) 2, 49) 2, 55) 3, , Y, X, , 1) 5 , , 2) 10 , , G, A, , A, , B, , J, , 3) 7.5 4) 12.5 , , POTENTIO METER, 51. The potential gradient along the length of a, uniform wire is 10 volt/m B and C are two, points at 30 cm and 60 cm in a scale fitted, along the wire. The pd between B and C is, 1) 3V, 2) 0.4V, 3) 7V, 4) 4V, 52. In the determination of the internal resistance of a cell using a potentiometer, when, the cell is shunted by a resistance “R” and, connected in the secondary circuit, the, balance length is found to be L1. On doubling, the shunt resistance, the balance length is, found to increase to L2. The value of the, internal resistance is, 2 R ( L2 L1 ), , 2 R ( L2 L1 ), , 1) ( L 2 L ), 1, 2, , 2) (2 L L ), 1, 2, , R( L2 L1 ), , R( L2 L1 ), , 2) 1, 8) 2, 14) 1, 20) 2, 26) 3, 32) 3, 38) 1, 44) 4, 50) 2, , 3) 3, 9) 2, 15) 4, 21) 1, 27) 1, 33) 3, 39) 2, 45) 4, 51) 1, , 4) 3, 10) 4, 16) 4, 22) 2, 28) 1, 34) 4, 40) 3, 46) 3, 52) 2, , 5) 2, 11) 1, 17) 4, 23) 1, 29) 3, 35) 3, 41) 3, 47) 3, 53) 1, , 6) 3, 12) 3, 18) 3, 24) 2, 30) 2, 36) 3, 42) 1, 48) 2, 54) 2, , LEVEL - II (C. W ) - HINTS, 1., , i, , n, , e, , np e, t, , q, ev, , t 2 r, , 2., , i, , 4., , vd , , t2, , 3., , q I . dt, t1, , 5., 7., 8., , i, 6., V = iR, neA, End resistors are not considered, Resitance between AB is independent of number, of sets used. Let x be the R eff between AB, vd , , R, , A, , 3) ( L 2 L ), 4) (2 L L ), 1, 2, 1, 2, 53. Figure shows a potentiometer circuit for, comparision of two resistan ces. The, balance point with a standard resistor, R 10.0 is found to be 58.3 cm, while, that with the unknown resistance X is, 68.5 cm. The value of X is, , 1, A, , C, , R, , r, , R, , B, , D, R, , Rr , , A R 2R , R r , , A, , 1, C, , x, , r, B, , D, , R, , A, , x, , R, , B, R, , A, , 1) 11.75 2) 12.55, , B, R, , G, , 9., , x R 2 R x, 2 Solve for r, The equivalent circuit is shown, , X, , 3) 9.55 4) 12.75, 54. In a experiment for calibration of voltmeter, a standard cell of emf 1.5 V is balanced at, 300 cm length of potentiometer wire. The, P.D across a resistance in the circuit is, balancedat 1.25 m. If a volt met er i s, connected across the same resistance it, reads 0.65 V. The error in the volt meter, is, 1) 0.5V 2) 0.025V 3) 0.05 V 4) 0.25V, 55. The current in the primay circuit of a, potentiometer is 0.2 A. The specific resistance, and cross-section of the potentiometer wire, , , NARAYANA GROUP, , A, , B, , 10. Apply Ohm’s law., 11. Let the resistance of shorter part MN be x., Then resistance of longer part is (20 – x) , (20 x)x, R eq , 1.8 , 20 x x, Solving we get x = 2 , So length of shorter part = 2m, 2x , , 12. 2 x (0.101) = 0.2, , , , solve for ‘x’., 49
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 13. When S1 and S2 are opened, i =, , 1.5, ..... (1), 450, , When S1 and S2 are closed,, 1.5[100 R ], i=, ....... (2), 400R 30,000, 14. iR 20, here i 4A R 5, 15. When a cell is connected across A and C, no, current flows in the arms BG and ED due to, symmetry in the arrangement. Then equivalent, circuit will be as shown in Fig.(a) and (b). The, effective resistance between A and C is, 1, 1, 1, 1, , , , Reff 2 R 3R 2 R, , , 3 2 3 8, 4, 3R, , , or Reff , ., 6R, 6 R 3R, 4, B, , R, , G, R, , R, , R, , A, , R, R, , R, , C, , H, , R, R, E, , R, , D, (a), 2R, , 44. R1 R0 (1 1 t ) ;, , 3R, A, , V2 V2A, 1, , PJEE MAINS - VOL - VII, R, l, l, V2, 33. p v i, 34. P , 32. i 2 r 3, R, e1 e2, 35. i R r r , v e1 ir1 0 7, 1, 2, 2V V V, , 36. I , 3R, 3R, Apply loop law in upper loop, V, V, R or VC ., IR V V VC 0 or VC IR , 3R, 3, 37. i = E / r + R, mnE, nE, & m n 24, 38. i , 39. i , mR nr, R nr, 2E, 2E, , and I P , 40. I s , 2r R, r 2R, r=R, E, ir, 100 100, 41. % error =, E, E, 42. when is i p ; r = R, E1 E2, 43. i R r r ; V1 E1 i r1, 1, 2, 31. P , , C, 2R, , 16. Apply Ohm’s law., , (b), , V, V, ; find Req, then i R, 17. R , eq, P, 18. Combination of resistors, R2 R1, 19. t R , 1, 20. Combination of resistors, V, 21. i R, total, 22. to 24. Combination of resistors, 25. error = V1 V11, 26. R11 R2 2, 27. P i 2 R, 28. 4R 400 20, l, V2, t ; R 2, 29. H , r, R, 2, t2, V12, V, , , t ;, 30. H , t1, V 22, R, 2, , 50, , 45., 46., 47., 48., 49., , R2 R0 (1 2 (t )), 1 1, 1, Series R R1 R2 ; Parallel R R R, 1, 2, Applying kirchoff’s first law, Using kirchoff’s law, Using wheatstone bridge principle, Applying kirchoff’s law, X, 80, , 1, Y 100 80, , 10x, 10 x 50 2, Y, 50, , 50., , X, l, , R 100 l, , 51. Potential gradient =, , V, l, , E , , 52. R r R K .l, , , 53., , E1, l IR, l, 1 ;, 1, E2, l 2 IX, l2, , V1, , l1, , 54. V l V2 0.625V, 2, 2, error = 0.65 - 0.625, 55. V i l, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , LEVEL-II (H. W), , CURRENT ELECTRICITY, 8., , ELECTRIC CURRENT AND DRIFT, VELOCITY, 1., , 2., , An electron of mass 9 x 10-31kg moves around, a nucleus in a circular orbit of radius 2A0 under the action of centripetal force 3.2N. Then, the equivalent electric current is, 32, 3, 16, 3, 2), 3), 4), 1), 3, 32, 3, 16, The current in a conductor varies with time, ‘t’ as I = 2- 0.02t ampers. The electric charge, that passes from t = 0 to t = 100 sec is, 1) 50 C, 2) 100 C, 3) 25 C 4) 75 C, , OHM’s LAW AND COMBINATION, OF RESISTANCES, 3., , 4., , 5., , 6., , 7., , Four resistances 10 , 5 ,7 and 3 are, connected so that they form the sides of a, rectangle AB, BC, CD and DA respectively., Another resistance of 10 is connected, across the diagonal AC. The equivalent, resistance between A and B is, 1) 2 , 2) 5 , 3) 7 , 4) 10 , A 3 resistor and a 6 resistor are, connected in parallel and the combination is, connected in series to a battery of 5V and a, 3 resistor. The potential difference across, the 6 resistor, 1) 2V, 2) 4V, 3) 3V, 4) 1V, You are given a wire of length 100 cm and, linear resistance of 1 ohm/cm. If it is cut into, two parts, so that when they are in parallel,, the effective resistance is 24 ohm. The, lengths of the two parts are, 1) 30cm & 70cm, 2) 60cm & 40cm, 3) 70cm & 30cm, 4) 20cm & 80cm, The resistance of a platinum wire of a, platinum resistance thermometer at the, ice point is 5 and at steam point is, 5.4 . When the thermometer is inserted, in a hot bath, the resistance of the platinum, wire is 6.2 . Find the temperature of the, hot bath., 1) 3000o C 2) 30o C 3) 300o C 4) 300 K, Three unequal resistors in parallel are, equivalent to a resistance 1 ohm. If two, of them are in the ratio 1 : 2 and if no, resistance value is fractional, the largest, of the three resistance in ohm is, 1) 4, 2) 6, 3) 8, 4) 12, , NARAYANA GROUP, , 9., , A carbon filament has resistance of 120 , at 0o C what must be the resistance of a, copper filament connected in series with, carbon so that combination has same, resistance at all temperatures, (carbon 5104 / o C, copper 4 10 3 / o C ), 1) 120 2) 15, 3) 60 4) 210 , The equivalent resistance across XY in fig., r, , 1) r, , 2) 2r, , r, , 2r, , X, , Y, r, , r, , 2r, , 3) 4r, 4) r / 2, 10. If the resistance of a circuit having 12V source, is increased by 4 , the current drops by 0.5A., What is the original resistance of circuit, 1, , 1) 4, 2) 8, 3) 16 4), 16, 11. An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and, radii of the wire are in the ratio 4/3 and 2/3,, then the ratio of the currents passing through, the wires will be, 1) 1/3, 2) 3/1, 3) 4/3, 4) 3/4, 12. When ‘n’ wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by X/Y, . Then the resistance of each wire is, 1), , xn, yn, xn, yn, 2), 3), 4), 2, 2, y (n 1), x (n 1), y (n 1) x(n 1), , 13. The equivalent resistance across A and B is, 6, 6, , 6, , 1) 2 , , 2) 4 , 6, , 8, 3) 8 , 4) 12 A, B, 14. An ammeter A is connected as shown in the, diagram. Ammeter reading is, E, , r, r, , E, 1), r, , 2E, 2), r, , r, 3), 2E, , E, 4), 2r, , r, A, r, r, , 51
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CURRENT ELECTRICITY, , JEE-ADV PHYSICS- VOL- III, , 15. A current of 7A flows thorugh the circuit as 23. The pd across terminals of a cell is found to be, 29 volt and 28 volt respectively when it delivers, shown in the figure the potential difference, a current of 1 ampere and 2 ampere, across points B and D is, B, 5, 1, respectively. The emf and internal resistance, of a cell are respectively, 7A, 1)30V, 2 2) 30V,1 3) 29V, 1 4) 28V,2 , 1) 5 V 2) 3 V, 24. The current in a circuit containing a battery, D, connected to 2 resistance is 0.9 A. When a, 3) 10 V 4) 7 V, 10, 5, resistance of 7 connected to the same, 16. If a wire is stretched to make it 0.1% longer,, battery, the current observed in the circuit is, its rsistance will :, [Mains-2011], 0.3A. Then the internal resistance of the, 1) increase by 0.2%, battery is, 2) decrease by 0.2%, 1) 0.1 2) 0.5 , 3) 1 , 4) Zero, 3) decrease by 0.05%, 4) increase by 0.05%, 25. Th e p ot en t i al d i fferen ce across t h e, terminals of a battery is 10 V when there, ELECTRIC POWER, is a current of 3A in the battery from the, 17. Two resistances R1 and R2 when connected, negative to the positive terminal. When, in series consume power equal to 25W. When, the current is 2A in the reverse direction,, connected in parallel they consume 100W. The, the potential difference becomes 15V. The, ratio of power of each is, internal resistance of the battery is, 1) 1/4, 2) 1/3, 3) 1/2, 4) 1, 2) 0.4 3) 0.6 4) 0.8, 1) 1, 18. Two electric bulbs marked 500 W, 220 V are 26. Two cells of emf 3V and 5V and internal reput in series with 110V line. The power, sistance r1 and r2 respectively are in series, dissipated in each of the bulb is, with an external resistance R. If the p.d., 125, 25, 225, 325, across 1st cell is zero, then R is, W 2), W, W, 1), W 4), 3), 4, 4, 4, 4, 5r 3r2, 2r 3r2, 3r 5r2, 4r 5r2, 19. A conductor of resistance 3 ohm is stretched, 1) 1, 2) 1, 3) 1, 4) 1, 3, 4, 3, 3, uniformly till its length is doubled. The wire, 27. A battery when connected by a resistance of, now is bent in the form of an equilateral, 16 gives a terminal voltage of 12V. and, triangle. The effective resistance between the, when, connected by a resistance of 10 gives, ends of any side of the triangle in ohms is, a terminal voltage of 11V. Then the emf of, 1) 9/2, 2) 8/3, 3) 2, 4) 1, the battery and its internal resistance, 20. The energy in kilowatt hour is consumed in, 1) 12.8 V 2) 13.7 V 3) 10.7 V 4) 9 V, operating ten 50W bulbs for 10 hours per day, 28., When a resistor of 11 ohm is connected in, in a month of 30 days is, series with an electric cell the current flow1) 1500, 2) 15000 3) 15, 4) 150, ing in it is 0.5A. Instead, when a resistor of, 21. Two electric bulbs rated 25 W - 220 V and, 5 is connected to the same electric cell in, 100 W – 220 V are connected in series to a, series, the current increases by 0.4A. The in440 V supply. Which of the bulbs will fuse?, ternal resistance of cell is, [Mains-2012], 1) 1.5 , 2) 2 , 3) 2.5 4) 6 , 1) Both, 2) 100 W 3) 25 W 4) Neither, INTERNAL RESISTANCE AND EMF 29. Two cells of emf 4V and 8V are connected to, two resister 4 and 6 as shown. If 8V cell, 22. Two batteries of different emf and internal, is short circuited. Then current through reresistances connected in series with each, sistance 4 and 6, other and with an external load resistor. The, 4, current is 3.0 A. When the polarity of one, battery is reversed, the current becomes 1.0, 4V, 6, 8V, A. The ratio of the emf of the two batteries, is, 1) 2 A, 2) 1 A, 3) 2.5 A 4) 3 A, 1) 2.5 : 1 2) 2 : 1, 3) 3 : 2, 4) 1 : 1, 52, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 30. If in the circuit shown below, the internal, resistnce of the battery is 1 andVp and VQ, are the potentials at P and Q respectively,, the potential difference between the points P, 10V, and Q is, P, , , 1) 9 V, , 2) 11 V, , 3) 7 V, , 4) 6 V, , 14, , 1) 1 A 2) 3 A, , 7, , 15V, 1, 6, , 3) 2 A 4) 5 A, , 0.5, , , 8, , 10, , KIRCHOFF’S LAWS AND WHEAT, SHONE’S BRIDGE, , 4, , Q, 4, , 2, , 14, , 36. The current i drawn from the 5 volt source, will be, 10, , 31. Voltmeter reading in the given circuit is (voltmeter is ideal), 14V.1, , 10, , 1) 0.5, , 20, , 5, , 2) 2 A I, , 10, , 1) 6 V, , 2) 8 V, , 3) 10 V, , 4) 14 V, , 20hm, , 4, , 12V, , V, , 32. For a cell the graph between the p.d(v) across, the terminals of the cells and the current (I), drawn from the cell as shown. The emf and, inernal resistance is, V, , 2, , i1, , 30V, , 1) 0.4 A, , 2) 0.6 A, , 10, , 20, , i2, 30, , 2V, 3) 1.6 A, 4) 2 A, 38. A 5V battery with internal resistance 2 , and a 2V battery with internal resistance 1 , are connected to a 10 resistor as shown in, the figure, [Mains-2008], , 1, in volt, P, 3, 2, 0, 3) 3, 4) , 3, 5V, 2V, 3, 1in amp, 10, 1, 2, 33. The minimum number of cells in mixed grouping required to produced a maximum current, P, of 1A through external resistance of 20, The, current, in, the, 10, resistor is, given the emf of each cell is 2V and internal, 1) 0.27 A P2 to P1, 2) 0.03 A P1 to P2, resistance 1 is, 1) 25, 2) 20, 3) 16, 4) 30, 3) 0.03 A P2 to P1, 4) 0.27 A P1 to P2, 34. A battery of emf ‘E’ and internal resistance, METRE BRIDGE, ‘r’ is connected to a resistor of restance ‘ r1 ’ 39. When a conductor is connected in the left gap, and known resistance in the right gap the baland Q Joules of heat is produced in a certain, ancing length is 50cm. If the wire is stretched, time ‘t’. When the same battery is connectted, so that its length increased by 20% , New balto another resistor of resistance r2 the same, ancing length is, 1)40.98cm 2)38.23cm 3)42.56cm 4)48.21cm, quantity of heat is produced in the same time, 40., In a meter bridge experiment when a resis‘t’. Then, the value of ‘r’ is (M-2011), tance wire is connected in the left gap, the, r12, r22, 1, balance point is found at the 30th cm. When, 1), 2), 3) ( r1 r2 ) 4) r1 r2, the wire is replaced by another wire, the balr2, r1, 2, ance point is found at the 60th cm. find the, 35. The emf of a cell E is 15 V as shown in the, balance point when the two wires connectd, figure with an internal resistance of 0.5 ., in series and in parallel in the left gap succesThen the value of the current drawn from the, sively, cell is, (M-2013), 1) 20 cm 2) 25 cm, 3) 23 cm 4) 30 cm, , 1), , 3, , 2, , 3) 1.5 A, 4) 3 A, 5V, 37. In the given circuit which is a part of a closed, cirucit the current i1, i2 are respcetively, , 2), , 2, , 1, , NARAYANA GROUP, , 53
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , LEVEL - II (H. W ) KEY, , 41. Shown in the figure below is a meter-bridge set, up with null deflection in the galvanometer., , 1) 1, 7) 2, 13) 2, 19) 2, 25) 1, 31) 2, 37) 2, 43) 4, , R, , 55, , G, 21cm, , 2) 2, 8) 2, 14) 2, 20) 4, 26) 1, 32) 4, 38) 3, 44) 3, , 3) 2, 9) 4, 15) 1, 21) 3, 27) 2, 33) 2, 39) 1, 45) 2, , 4) 1, 10) 2, 16) 1, 22) 2, 28) 3, 34) 4, 40) 2, , 5) 2, 11) 1, 17) 4, 23) 2, 29) 2, 35) 1, 41) 2, , 6) 3, 12) 1, 18) 1, 24) 2, 30) 1, 36) 1, 42) 1, , LEVEL - II (H . W ) - HINTS, The vlaue of the unknown resistor R is, [Mains-2008], 1) 13.75 2) 220 3) 110 , 4) 55 , , 1., , 30 E, 30 E, 30 E, 100 E, 1), 2), 3), 4), 100.5, 100 0.5, 100, 30, 45. 1 ohm resistance is in series with an, Ammeter which is balanced by 75 cm of, potentiometer wire. A standard cell of, 1.02 V is balanced by 50 cm. The ammeter, shows a reading of 1.5 A. The error in, the ammeter reading is, 1) 0.002 A 2) 0.03 A 3) 1.01 A 4) no error, 54, , q, qv, , t 2 r, , , mv 2 , , F, , , r , , , POTENTIO METER, 42. A potentiometer wire 10 m long has a, resistance of 40 . It is connected in series, with a resistanced box and a 2 v storage cell., If the potential gradient along the wire is, 0.01V/m the resistance unplugged in the box, is, 1) 760 2) 260, 3) 1060 4) 960, 43. The ratio of potential gradients is 1 : 2,, the resistance of two potentiometer wires, of same length are 2 & 4 respectively.., The current flowing through them are in, the ratio, 1) 1 : 2, 2) 2 : 1, 3) 1 : 3 4) 1 : 1, 44. The length of potentiometer wire is 100, cm and the emf of its standard cell is E, volts. It is employed to measure the emf, of a battery whose internal resistance is, 0.5 ohm. If the balance point is obtained, at l = 30 cm from positive end, the emf, of the battery is, , i, , t2, , 2., , q idt, t1, , 3., 4., 5., , Combination of resistors, Combination of resistors, Combination of resistors, , 6., , R2 R1 1 T , , 7., , 1, 1, 1, , , 1, R1 R2 R3, , R2 : R3 1: 2, 8., 9., , R11 R2 2, Combination of resistors, , 10. V = iR, 12 = ( i – 0.5) (R + 4 ), 11. V - iR, , V = constant, , i1R1 i2 R2, i1 R2, , i2 R1, R , , l, A, , X, 12. Rs R p , Y , 13. Combination of resistors, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 5., , LEVEL - III, , A, , ELECTRIC CURRENT AND DRIFT, VELOCITY, 1., , The electron of hydrogen atom is considered, to be revolving around the proton in circular, orbit of radius, , , 3), 2., , me2, , with velocity, , e2, , where, , , h, . The current I is, 2, 2, , 1), , 2, , 4 me, , 2, , 2), , 4 2 m 2 e 2, , 4), , 3) 3, 6., , 4 me, , 2, , h3, 4 2 me5, , h3, h3, In a straight conductor of uniform cross- 8., section charge q is flowing for time t. Let s, be the specific charge of an electron. The, momentum of all the free electrons per unit, length of the conductor, due to their drift, velocity only is, 2, , q, q, q, 9., 2) , 3), 4) qts, ts, ts, ts , Potential difference of 100 V is applied to, the ends of a copper wire one metre long., Find the ratio of average drift velocity and, thermal velocity of electrons at 27o C ., (Consider there is one conduction electron, per atom. The density of copper is 9.0 103 ;, Atomic mass of copper is 63.5 g., , 1), , 3., , 1) 1, , 7., 2, , h2, , Find the equivalent resistance across AB:, , N A 6.0 1023 per gram-mole, conductivity, , 2, , 2, , 2, , 2, , 2) 2, 4) 4, , 2, , B, , Two wires of the same material have length, 6cm and 10cm and radii 0.5 mm and 1.5 mm, respectively. They are connected in series, across a battery of 16V. The p.d. across the, shorter wire is, 1) 5V, 2) 13.5 V 3) 27 V, 4) 10 V, Three ammeters P,Q and R with internal, resistances r, 1.5r ,3r respectively . Q and R, parallel and this combination is in series with, P , The whole combination concted between, X and Y . When the battery connected, between X and Y , the ratio of the readings of, P,Q and R is, 1) 2:1:1 2) 3:2:1, 3) 3:1:2, 4) 1:1:1, The potential difference between the points, 3, 2, A and B is, A, , 12V, , 1, , 2, , 1) 1.50 V 2) 2.50 V, B, , 2, 3, 3) 1.00 V 4)0.50 V, The resistance of a semicircle shown in fig., between its two end faces is (Given that radial, thickness = 3 cm, axial thickness = 4 cm, inner, radius = 6 cm and resistivity = 4 106 cm ), , 4cm, , 1) 24.15 106 , 2) 7.85 107 , , 3cm, , 3) 7.85 106 , , 4) 7.85 105 , 4cm, 10. ABCD is a square where each side is a, K 1.38 1023 JK 1 ), uniform wire of resistance 1 . A point, E lies on CD such that if a uniform wire, 1) 3.67 10 6, 2) 4.3 106, of resistance 1 is connected across AE, 3) 6 105, 4) 5.6 106, and constant potential difference is, OHM’S LAW AND COMBINATION, applied across A and C, then B and E are, equipotential ., OF RESISTANCES, A, B, of copper is 5.81 107 1 ., , 4., , 56, , The sides of rectangular block are 2cm, 3cm, and 4cm. The ratio of the maximum to mini, mum resistance between its parallel faces, is, 1) 3, 2) 4, 3) 2, 4) 1, , 1), 3), , CE, 1, ED, CE 1, , ED 2, , 2), 4), , CE, 1, , ED, 2, CE, 2, ED, , D, , E, , C, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , 11. A h eati n g elemen t u si n g ni ch rome, connected to a 230 V supply draws an, initial current of 3.2 A which settles after, a few seconds to a steady value of 2.8 A., What is the steady temperature of the, heating element if the room temperature, is 27.0o C ? Temperature coefficient of, resistance of nichrome averaged over the, temperature range involved is, 1.70 104 C 1 ?, 1) 6800 C 2) 8670 C 3) 9200 C 4) 7500 C, , CELLS, KIRCHOFF’S LAW ‘S ,, WHEAT STONE BRIDGE, , 3) 0.15 A, , 4) 1.5 A, , 2) 24 J, , 3) 36 J, , 4) 48 J, , +8V, , 3, 5, , 4, , 3) r, , 2) 2r, , 4) 5r, , 14. For t he circui t shown in the fi gure,, potential difference between points A, and B is 16V. Find the current passing, through 2 , 4, i1, i1 + i2, , 9V, i2, , 1, i2, , 3V 4, B, , i1, , 2, , 1) 3.5A 2) 3A, 3) 4.5A 4) 5.5A, 15. The minimum number of cells in mixed group, ing required to produce a maximum current, of 1.5 A through an external resistance of, 30 ,given the emf of each cell is 1.5 V and, internal resistance is 1 is, 1) 30, 2) 120, 3) 40, 4) 60, 16. The p.d between the terminals A & B is, 20, , 18. In the circuit shown in figure, the potentials, of B,C and D are :, A, 1, –, +, , 2) VB 11V ;VC 9V ;VD 6V, , 2) 3V, , 3) 3.6 V, , 4) 1.8 V, , NARAYANA GROUP, , 5v, A, , 10, , 12V, , B, , –, , 6V, , C, , +, 3, , 3) VB 9V ;VC 11V ;VD 6V, D, , 19. A current of 0.10 A flows through the 25, resistor represented in the diagram to the, right. The current through the 80 resistor, is:, 80, , B, , 2v, , 0.1A, , 25, , V, , 20, , 1) 0.10 A 2) 0.20 A, 20, , 60, , 3) 0.30 A 4) 0.40 A, 20. In Wheat stone’s bridge shown in the, adjoining figure galvanometer gives no, deflection on pressing the key, the balance, condition for the bridge is :, B, R1, A, , R1 C1, R1 C2, 1) R C 2) R C, 2, 2, 2, 1, , R2, C, , G, C1, , C2, D, , Key, , R1, C1, R1, C1, 3) R R C C, 4) R R C C, 1, 2, 1, 2, 1, 2, 1, 2, 21. In the steady state, the energy stored in the, capacitor is :, i, , E1 , r 1, , a, , + –, , i, , C, , 1, C ( E1 E2 ) 2, 2, 1, 2, 2) C ( E1 E2 ), 2, , R2, , 1), , R1, , +, E2, r2, –, b, , 2, , 1) 2V, , 2, 3F, , 4, , 13. Cell A has emf 2E and internal resistance 4r., Cell B has emf E and internal resistance r., The negative of A is connected to the positive, of B and a load resistance of R is connected, across the battery formed. If the terminal, potential difference across A is zero, then R, is equal to, , 1) 3r, , 1, , P, , 4) VB 9V ;VC 6V ;VD 11V, , I, , N, , 2) 5.1 A, , Current, , 1) 12 J, , 2, , 2, , 1) 0.51 A, , +3V, , 1) VB 6V ;VC 9V ;VD 11V, , 12. A group of N cells where e.m.f. varies directly, with the internal resistance as per the, equation EN = 1.5 rN are connected as shown, in the figure. The current I in the circuit is:, 1, , 17. The energy stored in the capcitor is, , i, 2, , , 1 E1R1 E1R2 , 1 , ER, 1 1, 3) C , 4) C E2 , , 2 r1 r2 R1 R2 , 2 , r1 R1 R2 , , 10, , 57
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 22. A part of circuit in steady state along with, the currents flowing in the branches, the value, of resistances is shown in figure. Calculate, the energy stored in the capacitor., 1A, 4v 3, a, , i1, , 5, , e, , 4F, , 1) 8 101 J 2) 8 102 J, , 3v 1, b, 2A, , 1, 4, , 3, , i2f, , 2, , 3) 8 10 3 J 4) 8 104 J, 1A, 23. Equivalent resistance across A and B in the, given circuit if r = 10 , R = 20 is, R, , A, r, , 1) 7 , , 2) 14 , , r, r, , R, , 3) 35 4) 20/3 , B, 24. For a cell, the graph between the p.d.(V), across the terminals of the cell and the current, I drawn from the cell is shown in the fig. the, emf and the internal resistance of the cell is, E and r repectively., 1) E 2V, r 0.5, , 2, , V(Volt), , 2) E 2V, r 0.4, 3) E 2V, r 0.5, O, , 5, , I (in amp), 4) E 2V, r 0.4, 25. The charge developed on 4 F condenser is, 6, , 4F, , 24, , 2F, , 1) 18 C 2) 4 C, 3) 8 C 4) Zero, 4, , 10V, , 26. Find out the value of current through, 2 resistance for the given circuit., , 1) 0, , 5 10, , 2) 1.6 A, 10V, , 3) 2.4 A, , 4)3A, , 20V, , 2, , ELECTRIC POWER, JOULE’S LAW, 27. Same mass of copper is drawn into 2 wires of, 1mm thick and 3mm thick. Two wires are, connected in series and current is passed., Heat produced in the wires is the ratio of, 1) 3 : 1, 2) 9 : 1, 3) 81 : 1 4) 1 : 81, 58, , 28. Masses of three are in the ratio 1:3:5. Their, lengths are in the ratio 5:3:1. When they are, connected in series to an external source, the, amounts of heats produced in them are in the, ratio, 1) 125 : 15 : 1, 2) 1 : 15 : 125, 3) 5 : 3 : 1, 4) 1 : 3 : 5, 29. A heater coil rated at 1000W is connected to, a 110V mains. How much time will take to, melt 625 gm of ice at 00 C . (for ice L = 80, cal/gm), 1) 100s 2) 150s, 3) 200s, 4) 210s, 30. In the following circuit, 5 resistor develops, 45 J/s due to current flowing through it. The, power developed across 12 resistor is, i2, , 9, , 6, , 12, , i1, , 5, , 1) 16 W 2) 192 W 3) 36 W, 4) 64 W, 31. Two wires ' A ' and ' B ' of the same material, have their lengths in the ratio 1 : 2 and radii, in the ratio 2 : 1. The two wires are connected, in parallel across a battery. The ratio of the, heat produced in ' A ' to the heat produced in, ' B ' for the same time is, 1) 1 : 2, 2) 2 : 1, 3) 1 : 8, 4) 8 : 1, 32. An electric motor operating on 50 volt D.C., supply draws a current of 10 amp. If the, efficiency of motor is 40%, then the resistance, of the winding of the motor is, 1) 1.5 2) 3, 3) 4.5 4) 6 , 33. The resistance of a 240 V – 200 W electric, bulb when hot is 10 times the resistance when, cold. The resistance at room temperature and, the temperature coefficient of the filament, are (given working temperature of the, filament is 2000 C ), 1) 28.8 , 4.5 103 / C, 2) 14.4 , 4.5 10 3 / C, 3) 28.8 ,3.5 103 / C, 4) 14.4 , 3.5 103 / C, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, 34. A wire of length L and 3 identical cells of, negligible internal resistances are connected, in series. Due to the current, the temperature, of the wire is raisied by T in a time t. A, number N of similar cells is now connected in, series with a wire of the same material and, cross-section but the length 2L, The, temperature of the wire is raised by the same, amount T in the same time t. The value of, N is :, 1) 3, 2) 2, 3) 6, 4) 4, 35. Three bulbs with their power and working, voltage are connected as shown in the circuit, diagram to a 12 V battery. The total power, consumed by the bulbs is (ignore the internal, resistance of the battery shown), 12v 6W, 12v, 12W, 12v, , 6W, , CURRENT ELECTRICITY, 40. ‘n’ identical resistors are taken. ‘n/2’, resistors are connected in series and the, remaining are connected in parallel. The, series connected group is kept in the left gap, of a meter bridge and the parallel connected, group in the right gap. The distance of the, balance point from the left end of the wire is, 400, 400, 100n 2, 100n 2, 2), 3) 2, 4) 2, 2, 2, n 4, n 1, n 4, n 1, 41. In a metere bridge, the balance length from, left end (standard resistance of 1 is in the, right gap) is found to be 20 cm, the length of, 1, resistance wire in left gap is m and radius, 2, is 2mm its specific resistance is, 1) 10 6 ohm m, 2) 2 106 ohm m, , 1), , 3), 12v, , 1) 24 W, 2) 12 W, 3) 6 W, 4) 15 W, 36. A cell of emf 12 V and internal resistance, 6 is connected in parallel with another cell, of emf 6 V and internal resistance 3 , such, that the positive of the first cell joins the, positive of the second cell and similarly the, negative of first cell joins the negative of, the second cell. A bulb of filament resistance, 14 is connected across the combination., The power delivered to be bulb is, 1) 4.0 W 2) 3.5 W, 3) 8.5 W 4) 2.5 W, 37. A cell develops the same power across, two resistances R1 & R2 separately. The, internal resistance of the cell is, 1), , R1 R2 2), , 2R1 R2 3) R1 R2 4) R1 R2, , METRE BRIDGE, 38. A metallic conductor at 10 C connected in the, left gap of meter bridge gives balancing, length 40 cm. When the conductor is at 60 C ,, the balancing point shifts by---cm,, (temperature coefficeint of resistance of the, material of the wire is 1/ 220 / 0 C ), 1) 4.8, 2) 10, 3) 15, 4) 7, 39. When a conducting wire is connected in the, right gap and known resistance in the left gap,, the balancing length is 60cm. The balancing, length becoms 42.4 cm when the wire is, stretched so that its length increases by, 1) 10%, 2) 20%, 3) 25%, 4) 42.7%, NARAYANA GROUP, , , 106 ohm m, 2, , 4) 3 106 ohm m, , POTENTIO METER, 42. In an experiment with potentiometer to measure the internal resistance of a cell, when, the cell is shunted by 5 , the null point is, obtained at 2m. When cell is shunted by 20 , the null point is obtained at 3m. The internal, resistance of cell is, 1) 2 , 2) 4 , 3) 6 , 4) 8 , 43. A potentiometer wire of length 100cm has a, resistance 5 . It is connected in series with, a resistance and a cell of emf 2v and of, negligible internal resistance. A source of emf, 5mv balanced by 10 cm length of, potentiometer wire. The value of external, reistance is _____, 1) 540 2) 195, 3) 190 4) 990 , 44. 1 resistance is in series with an Ammeter, which is balanced by 75 cm of potentiometer, wire. A standard cell of 1.02V is balanced by, 50 cm. The Ammeter shows a reading of 1.5A., The error in the Ammeter reading is, 1) 0.002A 2) 0.03A, 3) 1.01A 4) no error, 45. An ideal battery of emf 2V and a series, resistance R are connected in the primary, circuit of a potentio meter of length 1m and, resistance 5 . The value of R to give a, potential difference of 5mV across the 10cm, of potentiometer wire is, 1) 180 2) 190 , 3) 195 4) 200 , 59
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 46. In an experiment for calibration of voltmeter,, a standard cell of emf 1.5V is balanced at 300cm, length of potentiometer wire. The P.D.across, a resistance in the circuit is balanced at 1.25m., If a voltmeter is connected across the same, resistance, it reads 0.65V. The error in the, volt meter is, 1) 0.05V 2) 0.025V 3) 0.5V, 4) 0.25V, 47. In the circuit shown in fig., the potential, difference between the points C and D is, balanced against 40 cm length of potentiometer wire of total length 100 cm. In order, to balance the potential difference between, the points D and E. The jockey to be pressed, on potentiometer wire at a distance of, , 4., , R max Pa ab a 2 4 2, , , 4., R ma in bc Pc c 2 2 2, 5., , Apply series and parallel combinations, , 6., , l1 6cm , l2 10cm,, r1 0.5 10 3 , r2 1.5 10 3, , In series combination i = constant, , l1, V1 R1, A, l A, , 1 1 2, V2 R2 l2 l2 A1 V1 V2 16V, A2, , 40cm, J, B, 10, , G, , 4, , C, , D, + –, 6v, , 1, , 1) 16 cm 2) 32 cm, , E, , Solving for V1 13.5V, , ( ), K1, , 3) 56 cm 4) 80 cm, , 7., , i, , LEVEL-III KEY, 1) 4, 7) 2, 13) 3, 19) 3, 25) 3, 31) 4, 37) 1, 43) 2, , 2) 1, 8) 4, 14) 3, 20) 3, 26) 1, 32) 2, 38) 1, 44) 2, , 3) 1, 9) 3, 15)2, 21) 2, 27) 3, 33) 1, 39) 4, 45) 3, , 4) 2, 10) 4, 16) 4, 22) 4, 28) 1, 34) 3, 40) 1, 46) 2, , 2., , I, , 6) 2, 12) 4, 18) 2, 24) 2, 30) 2, 36) 2, 42) 2, , 8., , ii1, , 2, , A, , I 1, , II, , i, , 3, , 2, B, , For the first loop 12 5i i1, For the second loop 0 7(i i1 ) i1, or, , 8i1 7i or , i1 (7 / 8)i, , 7, 47i, Therefore, we obtain 12 5i i , 8, 8, , I, q/t, , I nAevd or vd , nAe nAe, No. of free electrons per unit length of conductor, , or,, , i, , 12 8, A 2.04 A, 47, , 7, i1 2.04 A 1.79 A, 8, Thus, the p.d across A and B is, , p Nmvd, E v 3K B T, rms, me, ne, , 12 V, , 2, , e, e, ev, , , t 2r / v 2r, , vd , , 3, , i1, , N nA 1, Momentum of all the free electrons is, , 3., , V, R, , i, 5) 1, 11) 2, 17) 2, 23) 2, 29) 4, 35) 3, 41) 2, 47) 2, , LEVEL - III - HINTS, 1., , a, c, , R min , bc, ab, , a b c432, , + –, , A, , R max , , VA VB (i i1 ) 2 0.25 2 0.50V, 9., , Here, A = 4 cm 3 cm = 12 cm2, , l r = 6 3/ 2 7.5 cm, 60, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , R, , CURRENT ELECTRICITY, , l 4 106 7.5, , A, 12, , 1.5 n, 2 30, n 60, , 1.5 , , 7.85 106 ohm., 1, , 1, , 1.5 m, 2 1, m2, total no of cells; n m 2 60 120, , B, , A, , 1.5 , , 1, , 1, , 10., D, , x, , E, , (1 x) , , C, , Equivalent resistance between A and E is, , y, , x 1, , 16., , x2, For B and E to be at equal potential , we get, , RAE REC, x 1, 1 x, , , , RAB RBC, x 2 1 1, , V iR i 10 1.8v, B, , Solving x 2 1, Now, 11. R27 , , CE 1 x, , 2, ED, x, V 230, 230, , , R , I 3.2, 2.8, , R27 R0 1 27 , , R, R0 1 , , EN, 12. i r 1.5, N, e1 e2, 13. i R r r , V1 E1 ir1 0, 1, 2, , 14. VA VB 16, , 4i1 2 i1 i2 3 4i1 16.... 1, 9 i2 2 i1 i2 0, Solving eqs (1) and (2) i1 1.5 A and i2 2 A, , E1 E 2, 5, 2, , , r1, r2, 20 10, i, , 1, 1 1, 1, 1 R 1 10 , 20 10 , r1 r2 , , A, , 17., , +8V, , Current(i) P, 4, , +3V, , 1, 2, , 3F, C, , We have, VA VB i (4 1), , 8 3 i5, 5 i 5, i 1A, VA V p 4 1, , 8 VP 4; VP 4 volt, Now VC 0 . So, the energy stored in the, 1, capacitor is 3 16 24 J, 2, 18. Potential at O is zero being earthed., Applying Kirchhoff’s second law, i (1 2 3) 12 6 or i 1A, , VA VD (1 2 3) 1 6V, mE nE, , 2r 2R, n = number of cells in each row., m = number of rows., , 15. i max , , NARAYANA GROUP, , VA VB 1 1 1V, VA VC (1 2) 1 3V, Also, VA VO 12V or, , VA 12V, 61
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , P.D across 4 2 Volt, Charges on 4 F 8 C, 26. 2 resistor is in open circut so current is 0, , 40, 500 100 R, , 100, 500, , 33. Resistance of the hot bulb R2 , , 27. JQ i 2 Rt, Q R , , V 2 240 240, , P, 200, , Resistance of the bulb at room temperature, , 1, when wire is stretched, A2, , Q1 r24 34, 81.1, Q2 r14 14, 28. m1 : m 2 : m 3 1: 3 : 5, l1 : l2 : l3 5 : 3 :1, , R1 , , R2 288, , 10, 10, , , , R2 R1, R1 t, , 34. In the first case, three identical cells are, connected in seires with a wire of length L. Let, the terminal potential difference of each cell is, V and resistance of the wire is R. Then heat, developed in the wire in time t is, , (3V )2, t msT, R, where m is the mass of the wire, s-the specific, heat of its material and T is the rise in its, temperature., When N such indentical cells are connected in, series, the effective terminal otential is NV volt, and if the length of the wire is doubled, its, resistance and mass also doubled. Then heat, develope in the wire is, H, , dl 2, Q i Rt R , m, 2, dl, Q R , m, 2, Q1 l1 m2, , , Q2 l22 m1, 2, , Q1 : Q 2 : Q3 125 :15 :1, 29. JQ P t, , ( NV )2, H , t (2m)s.T, 2R, Dividing both the equations, we get, `, , J m L = P t, 80 4.2, 1000 t, 103, t 210s, , 1 625 10 3 , , 30. P i2R 192W, 31. Q , , V2, R, , Q1 R 2, l, r2, , 2 12 ., Q2 R1, l1 r2, , N2, 2 N 6, 2 9, 35 . P , , V2, R, , Eeff, 2, 36. i R , p i R, eff, 2, , E , 37. P1 i R1 , R1, R1 r , 2, , 2, , Q1 2 2 , 8, 2 , Q2 1 1, 1, , 2, , E , P2 i R2 , R2, R2 r , , 32. Input power, P = VI = 50 × 10, Power dissipated as heat = I2 R = 100 R, Efficiency =, , Out put power, Input power, , NARAYANA GROUP, , 2, , 38., , X 40 2 X 0 1 t1 2, , ;, , R 60 3, R, 3, 63
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, , X 0 1 t2 , l, , 100 l, R, 1 t1 2 100 l , , 1 t2 3 l , l 44.8, , 39., , 44. 1.02V 50cm, ?, 75cm, V, , X 60, , ;, R 40, , error 1.53 1.5 0.03, 3, 45. 5 10 , , R ' 60 57.6, , R 40 42.4, R l, , 2, , R1 R , l1 l, 100 , 100, , , l, R, , , , sl, A, , 46. 300 10 3 m 1.5V ; 1.25m ?, V 0.625V ; Error in ammeter reading, 0.625 0.65 0.025v ., V1 iR1 l1, 47. V iR l, 2, 2, 2, , X 20, , R 80, , 1, 1, R2, 2, , X, 3 2, 4 2 10 , 8, 1, S 210 6 m, 4, , V1 l1, 42. V l, 2, 2, , 1., , 2., , E , E , V1 , R 1 , V2 , R2, R1 r , R2 r , R1 R 2 r , R 2 R1 r , , , , l1, l2 r 4, , E R, 43. E ' i l R R L . l, 3 , , , 40, , 4 l2, , LEVEL - IV, , S, , X, , V, iR, l l, L, L, , 2 5, 2, 5 103 , 10 10, R51, , X, nr / 2, , 100 X 2r / n, , 41. X , , 64, , 75 1.02, 1.53 ;, 50, , Balancing point shifts by =44.8-40=4.8 ., , X 42.4, , ;, R ' 57.6, , 40., , 2 5, 5 10 3 , 10, , 5 R 3 100, R S 195 , , 3., , Instructions for Assertion & Reason Type, questions:, 1) Both (A) and (R) are true and (R) is the, correct explanation of A., 2) Both (A) and (R) are true but (R) is not, the correct explanation of A., 3) (A) is true but (R) is false, 4) (A) is false but (R) is true, Assertion : Terminal voltage of a cell is, greater than emf of cell, during charging of, the cell., Reason : The emf of a cell is always greater, than its terminal voltage., Assertion : In metrebridge experiment, a high, resistance is connected in series with the, galvanometer., Reason : As resistance increases, current, through the circuit increases, Assertion (A) : In a metrebridge ; copper wire, is connected in the left gap and silicon is, connected in the right gap, when the temp of, both wires increase, balancing point shifts to, right., Reason (R) : Temperature coefficient of, copper is -Ve and that of silicon is +Ve., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, 4., , 5., , 6., , Assertion (A) : If a current flows through a, wire of non-uniform cross-section, potential, difference per unit length of the wire in the, direction of current is same at different, points., Reason (R) : V = iR and current in the wire is, same throughout., Assertion (A) : Voltmeter is much better than, a potentiometer for measuring emf of cell., Reason (R) : A potentiometer draws no, current while measuring emf of a cell., Assertion (A):The equivalent resistance, between the points X and Y in the figure, is, 10 ., 10, B, 10, , 10, x, A, , C, , y, 10, , 10, D, , Reason (R) : According to wheatstone bridge, points A and C have the same potential., 7. Assertion : The drift velocity of electrons in a, metallic wire will decrease, if the temperature, of the wire is increased., Reason : On increasing temperature,, conductivity of metallic wire decreases., 8. Assertion (A) :The electric bulb glows, immediately when switch is on., Reason (R) : The drift velocity of electrons, in a metallic wire is very high., 9. Assertion (A) : If the length of the conductor, is doubled, the drift velocity will become half, of the original value (keeping potential, difference unchanged), Reason (R) : At constant potential difference,, drift velocity is inversely proportional to the, length of the conductor., 10. Assertion (A): If the current of a lamp, decreases by 20%, the percentage decrease, in the illumination of the lamp is 40%, Reason (R) : Illumination of the lamp is, directly proportional to square of the current, through lamp., 11. Assertion (A) : However long a fuse wire may, be, the safe current that can be allowed is, the same., Reason (R): The safe current that can be, allowed to pass through a fuse wire depends, on the radius of the wire., NARAYANA GROUP, , CURRENT ELECTRICITY, 12. Assertion (A) : The resistance of an ideal, voltmeter should be infinite., Reason (R) : The potential difference, measured by a voltmeter across a resistor is, always less than the actual potential, difference across the resistor., 13. Assertion (A) : Current is passed through a, metallic wire, heating it red. When cold water, is poured on half of its portion, then rest of, the half portion becomes more hot., Reason (R) : Resistance decreases due to, decrease in temperature and then current, through wire increases., , MATCHING TYPE QUESTIONS, 14. Match list - I with List - II, List - I, List- II, a) Ohm’s law, e) conservation of, charge, b) Joule’s Law, f) conservation of, energy, c) Kirchhoff’s I Law g) v Ri, d) Kirchhoff’s II Law h) H i 2 Rt, 1) a - h, b - g, c - e, d - f, 2) a - g, b - h, c - e, d - f, 3) a - h, b - f, c - e, d - g, 4) a - e, b - f, c - g, d - h, 15. Match list - I with List - II, List - I, List - II, a) Potentiometer, e) For measuring, current, b) Metrebridge, f)For measuring, internal resistance, c) Ammeter, g)For measuring, specific resistance, of wire, d) Voltmeter, h) For measuring, potential difference, 1) a-f, b-g, c-e, d-h, 2) a-g, b-e, c-f, d-h, 3) a-h, b-e, c-f, d-g, 4) a-h, b-f, c-e, d-g, 16. Match list - I with List - II, List - I, List - II, a) Thermistor, e) High ve ' ', b) Carbon, f) almost zero, c) Nichrome, g) either positive or, negative ' ', d) Constantan,, h) Negative ' ', and manganin, 1) a-g, b-h, c-e, d-f, 2) a-h, b-g, c-e, d-f, 3) a-e, b-f, c-g, d-h, 4) a-e, b-g, c-h, d-f, 65
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 17. Match list - I with List - II, List - I, List - II, a) Resistivity, e) Volt, b) Conductivity, f) Siemen, c) emf, g) ohm - metre, d) conductance, h) mho metre 1, 1) a-e, b-f, c-g, d-h, 2) a-f, b-e, c-g, d-h, 3) a-g, b-h, c-e, d-f, 4) a-h, b-g, c-e, d-f, 18. Three wires of same material are connected, in parallel to a source of emf. The length ratio, of the wires is 1 : 2 : 3 and the ratio of their, area of cross section is 2 : 4 : 1., Table – 1, Table - 2, (a) Resistance ratio (p) 6 : 6 : 1, (b) Current ratio, (q) 1 : 6 : 6, (c) Power ratio, (r) 1 : 1 : 6, (s) None, 1) a-r,b-q,c-p, 2) a-p,b-q,c-r, 3) a - r; b - p, c - p, 4) a-q,b-p,c-r, 19. In the figure shown, each resistance is R., a, , b, , d, , c, , Table – 1, , Table - 2, , (a) Resistance between a and b, , (p), , (b) Resistance between a and c, , (q) R, , (c) Resistance between b and d, , R, 2, , 2V 2, 3V 3, A, B, , Table - 2, (p) Zero, (q) 2 V, (r) 4 V, (s) 6 V, (t) none, 1) a - p; b - p; c - p,d-p 2) a-p,b-s,c-q, 3) a-q,b-q,c-s, 4) a-r,b-r,c-p, 21. In the potentiometer arrangement shown in, figur e, null point is obt ained at lengt h l ., E1, , R, , l, J, , E2, , Table – 1, (a) If E1 is increased, (b) If R is increased, (c) If E 2 is increased, , Table - 2, (p) l should increase, (q) l should decrease, (r) l should remain, the same to again, get the null point, 1) a-q,b-q,c-p, 2) a-r,b-r,c-q, 3) a-p,b-p,c-r, 4) a - q; b - p; c - p, 22. In the circuit shown in figure, if a resistance, R is connected in parallel with R 2, R1, , R2, , i, , 5, 8, , (r) R, (s) None, 1) a-q,b-p,c-r, 2) a - r; b - p; c - p, 3) a-p,b-q,c-s, 4) a-s,b-r,c-p, 20. Six batteries of increasing emf and increasing, internal resistance are as shown in figure., 1V 1, , Table – 1, (a) Potential of point A, (b) Potential of point B, (c) Potential of point C, (d) Potential of point D, , V, , Table – 1, (a) Main current i, (b) Power across R1, (c) Power across R 2, 1) a-p,b-p,c-p, 3) a - p, b - p, c - q, 23. In the circuit shown,, , Table - 2, (p) will increase, (q) will decrease, (r) will remain same, 2) a-q,b-q,c-q, 4) a-r,b-r,c-r, , A, , B, , 15V,1, , 10V,1, , C, 8V 6, , 5V 6, D, , 66, , 4V 4, , 3, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, Table – 1, (a) Potential difference, across battery A, (b) Potential difference, across battery B, (c) Power is supplied by, battery, (d) Power is consumed, by battery, , CURRENT ELECTRICITY, Table - 2, (p) A, (q) B, (r) 14 V, (s) 9 V, , (t) 11V, 1) a-p; b-p;c-q;d-r, 2) a-q;b-q;c-r;d-s, 3) a-r, b-t, c-p, d-q, 4) a-r;b-r;c-s;d-t, 24. Current is flowing through a wire of nonuniform cross section. Cross section of wire, A is less than the cross section of wire at B., Table – 1, Table - 2, (a) current at A, (p) is zero, (b) drift velocity of, (q) is more than at B, electrons at A, (c) electric field in, (r) is less than at B, the wire at A, (d) current density, (s) is equal to that, at A, at B, 1) a-p;b-q;c-q;d-p, 2) a-s, b-q, c-q, d-q, 3) a-q;b-r;c-p;d-q, 4) a-r;b-s;c-r;c-s, 25. In the circuit shown in figure,, R1 R 2 R 3 R ., , 27., , 28., , E, , R1, E, , E, R3, , 29., , R2, , Table – 1, Table - 2, (a) current through R1 (p) E/R, (b) current through R 2 (q) 2E/R, (c) current through R 3 (r) E/2R, (s) Zero, 1) a-p, b-p, c-p, 2) a-p;b-q;c-s, 3) a-q;b-r;c-r, 4) a-r;b-s;c-p, 26. Matrix Matching, 2, , 3, , 4, , 5, , V, , NARAYANA GROUP, , 30., , Table – 1, Table - 2, (a) Minimum current will flow, (p) 2, through, (b) Maximum current will, (q) 4, flow through, (c) Maximum power will be, (r) 3, generated across, (d) Minimum power will be, (s) 5, generated across, 1) a-p;b-p;c-q;d-r, 2) a-r;b-r;c-p;d-s, 3) a - q, b - p, c - r, d - q 4) a-s;b-s;c-r;d-q, Statement (A) : Thermistor can have only, negative temp eratu re coefficeints of, resistances., Statement (B) : Thermistors with negative, temperature coefficients of resistance are, used as resist ance th ermometers, to, measure low temperatures of the order of, 10 K., 1) both A and B are true, 2) both A and B are false, 3) A is true and B is false, 4) A is false, but B is true, Statement (A) : Resistivity of insulators is, about 1022 times the resistivity of metallic, conductors., Statment (B) : Metals like silver, copper and, aluminium have very high values of, resistivity., 1) A and B are true, 2) A and B are false, 3) A is true, B is false 4) A is false, B is true, Statement (A) : Series combination of cells is, preferred when external resistance is large, compared to internal resistance of cell., Statement (B) : Parallel combination of cells, is preferred when external resistance is small, compared to the internal resistance of each, cell., 1) A and B are true, 2) A and B are false, 3) A is true, B is false 4) A is false, B is true, Statement (A) : The difference between a, new torch light cell and an old one is due to, increase in internal resistance., Statement(B) : At 0 kelvin specific resistance, of prefect insulator is infinity., 1) Both A and B are true, 2) A is true, B is false, 3) A is false, B is true, 4) Both A and B are false, 67
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JEE-ADV PHYSICS- VOL- III, , CURRENT ELECTRICITY, 31. Statement-1: The temperature dependence of, resistance is usually given as, , A) the potentiometer wire PQ, B) the resistor R, C) the galvanometer G, R R0 1 t . The resistance of a wire, 1) A, B and C, 2) A and B only, changes from 100 to 150 when its, 3) B and C only, 4) C only, o, o, temperature is increased from 27 C to 227 C. 34. The two ends of a uniform conductor are, joined to a cell of emf E and some internal, This implies that 2.5 103 / 0 C, resistance. Starting from the mid point P of, the conductor, we move in the direction of, Statement 2 : - R R0 1 t is valid only, current and return to P. The potential V at, when the change in the temperature T is, every point on the path is plotted against the, distance covered (x). One of the following, small and R R R0 R0, best represent the resulting curve?, 1) Statement -1 is true, statement -2 is true,, V, V, Statement-2 is the correct explanation of, statement -1., 2), 1), <E, E, 2) Statement -1 is t rue, statement -2 is, true,statement -2 is not the correct explanation, of statement -1, X, X, 3) Statement - 1 is false, Statement - 2 is true, V, 4) Statement - 1 is true, Statement - 2 is false, V, 32. In the above circuit, C denotes the balance, >E, position on the potentiometer wire AB. Which, <E, 3), 4), of the following procedures can shift C, towards the end B?, X, X, , 35. In the circuit shown L1, L2, L3, and L4 are, identical light bulbs. There are six voltmeters, connected to the circuit as shown. Assume that, the voltmeters are ideal. If L3 were to burn, out, opening the circuit, which voltmeter(s), would read zero volts?, , R, , X, , C, , A, , B, , Y, G, , a) replacing the driving cell X by one with a, smaller EMF, b) adding a resistance in series with the, galvanometer G, c) increasing the resistance of the rheostat R, 1) a, b and c, 2) a and b only, 3) b and c only, 4) a and (c) only, 33. The potentiometer circuit shown is used to, find the internal resistance of the cell E. At, balance, the galvanometer pointer does not, deflect, and no current flows through, , T, R, , E, 68, , V2, , L2, , L3, , V3, , L4, , V4, , V0, , 1) none would read zero, 2) only V3, 3) only V4, 4) only V3, V4, and V5, , LEVEL-IV KEY, , V, , P, , V0, , L1, V1, , Q, G, , 1) 3, 7) 2, 13) 1, 19) 1, 25) 1, 31) 3, , 2) 3, 8) 3, 14) 2, 20) 1, 26) 3, 32) 4, , 3) 3, 9) 1, 15) 1, 21) 4, 27) 4, 33) 4, , 4) 4, 10) 4, 16) 1, 22) 3, 28) 3, 34) 2, , 5) 4, 11) 1, 17) 3, 23) 3, 29) 1, 35) 3, , 6) 1, 12) 2, 18) 3, 24) 2, 30) 1, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, 5., , LEVEL - V, MULTIPLE ANSWER QUESTIONS, 1., , 2., , 3., , 4., , A piece of copper and another of germanium, are cooled from room temperature to 80 K., The resistance of, A. each of them increases, B. each of them decreases, C. copper increases and germanium decreases, D. copper decreases and germanium increases, Read the following statements carefully, Y: The resistivity of semiconductor decreases, with increase of temperature, Z: In a conducting solid, the rate of collisions, between free electrons and ions increases with 6., increase of temperature., Select the correct statement(s) from the, following, A. Y is true but Z is false, B. Y is false but Z is true, C. Both Y and Z are true, D. Y is true and Z is the correct reason for Y, A steady current flows in a metallic conductor, of non-uniform cross-section. The quantity/, 7., quantities constant along the length of the, conductor is/are, A. current, electric field and drift speed, B. drift speed only, C. current and drift speed D. current only, Two bars of radius r and 2r are kept in contact, as shown. An electric current I is passed, through the bars. Which one of following is, correct? [2006, 3M], , A battery of internal resistance 4 is, connected to the network of resistances as, shown in figure. In order that the maximum, power can be delivered to the network, the, value of R in should be, R, , R, E, , R, , 6R, 4R, , R, , 4, , R, , 4, 8, B. 2, C., D. 18, 9, 3, The effective resistance between points P and, Q of the electrical circuit shown in the figure, is [2002, 2M], , A., , 2R, , 2R, 2R, P, , r, , r, , Q, , 2R, 2R, , 2R, , 8R R r , 2Rr, 5R, 2r, B., C. 2r 4R D., Rr, 2, 3R r, The three resistances of equal value are, arranged in the different combinations shown, below. Arrange them in increasing order of, power dissipation, , A., , A. III < II < IV, B. II > III > IV, C. IV < III < II, D. III > II > IV, 8. Six equal resistances are connected between, points P, Q and R as shown in the figure. Then,, the net resistance will be maximum between, , A. Heat produced in bar BC is 4 times the heat, produced in bar AB, A. P and Q, B. Electric field in both halves is equal, C. Current density across AB is double that of across C. P and R, BC, D. Potential difference across AB is 4 times that of, across BC, , P, , B. Q and R, D. any two points, , Q, , R, , NARAYANAGROUP, 69
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 9. A rigid container with thermally insulated walls 12. Consider a thin square sheet of side L and, thickness t, made of a material of resistivity, contains a coil of resistance 100 , carrying, p. The resistance between two opposite faces,, current 1A. Change in internal energy after 5, shown by the shaded areas in the figure is, min will be [2005, 2M], [2010], A. zero, B. 10 kJ C. 20 kJ D. 30 kJ, 10. Find out the value of current through 2, resistance for the given circuit [2005, 2M], t, , 20V, , 10V, , L, , 10, , 5, , A)directly proportional to L B)directly proportional to t, C) independent of L, D) independent of t, A. 5 A, B. 6 A, C. zero, D. 4 A, 13. Incandescent bulbs are designed by keeping, 11.Figure shows three resistor configurations, in mind that the resistance of their filament, and, R, connected, to, 3V, battery., If, the, R1 , R 2, increases with the increase in temperature. If, 3, at room temperature, 100 W, 60 W and 40 W, power dissipated by the configuration R1 , R 2, bulbs have filament resistances R 100 , R 60 and, and R 3 is P1 , P2 and P3 , respectively, then, R 40 , respectively, the relation between these, [2008, 3M], resistances is [2010], 2, , 1, 1, 1, A. R R R, 100, 40, 60, , B. R100 R 40 R 60, , C. R100 R 60 R 40, , 1, 1, 1, D. R R R, 100, 60, 40, , 1, 3V, , 1, , 1, , 1, , 1, , 3V, , 14. In the circuit shown in figure the heat produced, in the 5 resistor due to the current flowing, through it is 10 cal/s, , 1, , 1, , 1, , 1, , 1, , 4, , R2, , 6, , 5, , 1, , 1, 1, , 1, , 3V, , The heat generated in the 4 resistor is, A. 1 cal/s B. 2 cal/s C. 3 cal/s D. 4 cal/s, 15. The current i in the circuit (see figure) is, , 1, 2V, , 1, , A. P1 P2 P3, , B. P1 P3 P2, , C. P2 P1 P3, , D. P3 P2 P1, , A., , 1, A, 45, , B., , 1, A, 15, , C., , 1, A, 10, , D., , 1, A, 5, , NARAYANAGROUP, 70
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 16. In the circuit shown in the figure, the current 20. A circuit is connected as shown in the figure, through [1998, 2M], with the switch S open. When the switch is, 3, 2, 2, closed, the total amount of charge that flows, from Y to X is [2007, 3M], 9V, 8, , 8, , 4, , 3 F, 2, , 2, , 6 F, , X, , 2, , S, , A. the 3 resistor is 0.50 A, 3, 6, B. the 3 resistor is 0.25 A, Y, C. the 4 resistor is 0.50 A, D. the 4 resistor is 0.25 A, 9V, 17. In the given circuit, it is observed that the, A. zero, B. 54 C C. 27 C D. 81 C, current I is independent of the value of the, resistance R 6 . Then, the resistance values 21. A resistance of 2 is connected across one, gap of a meter-bridge (the length of the wire is, must satisfy [2001, 2M], 100 cm) and an unknown resistance, greater, R, than 2 , is connected across the other gap., R, R, When these resistances are interchanged, the, I, R, 8, balance point shifts by 20 cm. Neglecting any, R, R, correct ions, the unknown resistance is [2007,, 3M], A. R1R 2 R 5 R 3 R 4 R 6, A. 3, B. 4, C. 5, D. 6, 22. A meter bridge is set-up as shown in figure, to, 1, 1, 1, 1, , , , B. R, determine an unknown resistance X using a, R 6 R1 R 2 R 3 R 4, 5, standard 10 resistor. The galvanometer, C. R1R 4 R 2 R 3, D. R1R 3 R 2 R 4, shows null point when tapping key is at 52 cm, 18. A wire of length L and 3 identical cells of, mark. The end corrections are 1 cm and 2 cm, negligible internal resistances are connected, respectively for the ends A and B. The, in series. Due to the current, the temperature, determined value of X is [2011], of the wire is raised by T in a time t. A number, N of similar cells is now connected in series, X, 10, with a wire of the same material and crosssection but of length 2L. The temperature of, the wire is raised by the same amount T in, A, B, the same time. The value of N is [2001, 2M], A. 4, B. 6, C. 8, D. 9, A. 10.2 B. 10.6 C. 10.8 D. 11.1, 19. A 100 W bulb B1 , and two 60 W bulbs B2 and, 23. In the circuit shown P R , the reading of, B3 , are connected to a 250 V source, as shown, galvanometer is same with switch S open or, closed. Then, in the figure. Now W1 , W2 and W3 are the, 5, , 1, , 3, , 6, , 2, , 4, , output powers of the bulbs B1 , B2 and B3, respectively. Then,[2002, 2M], B1, , A. W1 W2 W3, B. W1 W2 W3, C. W1 W2 W3, D. W1 W2 W3, , P, , Q, S, , B2, R, , G, , B3, , V, , 250V, , A. IR IG ] B. IP IG C. I Q I G D. I Q I R, , NARAYANAGROUP, 71
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 24. In the given circuit, with steady current, the 29. To verify Ohm’s law, a student is provided with, potential difference across the capacitor must, a test resistor R T , a high resistance R1 , a small, be [2001, 2M], V, R, resistance R 2 , two identical galvanometers, , G1 and G 2 , and a variable voltage source V., C, V, V, The correct circuit to carry out the experiment, A. V, B., is [2010], 2, 2R, 2V, V, 2V, G, G, C., D., R, 3, 3, R, G, G, 25. Which of the following set-up can be used to, R, R, R, R, A., B., verify Ohm’s law? [2003, 2M], 1, , 1, , 2, , 1, , 2, , 2, , T, , T, , 1, , V, , V, , A, V, , A., , 2, , B., , G1, , V, , A, , G2, , G2, , RT, , RT, , C., , R2, , D., , R1, , V, , A, , V, , C., , D., A, , V, , 26. In the shown arrangement of the experiment, of the meter bridge if AC corresponding to null, deflection of galvanometer is x, what would be, its value if the radius of the wire AB is doubled?, [2003, S], A. x, , B. x / 4, , C. 4x, , D. 2x, , R1, , R2, , G, x, , A, , C, , B, , 27. For the post office box arrangement to, determine the value of unknown resistance,, the unknown resistance should be connected, B, A, between [2004, 2M] C, , A. B1 and C1 B. A and D, C. C and D, , R2, , G1, , R1, , V, , 30. A microammeter has a resistance of 100 and, full scale range of 50 A . It can be used as a, voltmeter or as a higher range ammeter, provided a resistance is added to it. Pick the, correct range and resistance combination(s), [1991, 2M], A. 50 V range with 10k resistance in series, B. 10 V range with 200k resistance in series, C. 5 mA range with 1 resistance in parallel, D. 10 mA range with 1 resistance in parallel, 31. A capacitor is charged using an external, battery with a resistance x in series. The, dashed line shows the variation of ln I with, respect to time. If the resistance is changed to, 2x, the new graph will be [2004, 2M], , D, , D. B and D, C', , S, , I, , B', , A. P, , B. Q, , C. R, , D. S, , R, Q, P, , 28. A moving coil galvanometer of resistance, t, 100 is used as an ammeter using a resistance 32. A 4 F capacitor and a resistance of 2.5M, 0.1 . The maximum deflection current in the, are in series with 12 V battery. Find the time, after which the potential difference across the, galvanometer is 100 A . Find the current in, capacitor is 3 times the potential difference, the circuit, so that the ammeter shows, across the resistor, maximum deflection [2005, 2M], [Given, ln (2) = 0.693] [2005, 2M], A. 100.1 mA, B. 1000.1 mA, A. 13.86 s B. 6.93 s C. 7 s, D. 14 s, C. 10.01 mA, D. 1.01 mA, NARAYANAGROUP, 72
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 33. Find the time constant for the given RC circuits, in correct order (in s )[2006, 3M], V, , 8, 8, 8, 8, A. 18, 4, B. 18, , 4 C. 4,18, D. 4, ,18, 9, 9, 9, 9, , 34. Capacitor C1 of capacitance 1 F and capacitor, , C2 of capacitance 2 F are separately charged, fully by a common battery. The two capacitors, are then separately allowed to discharging, through equal resistors at time t = 0 [1989, 2M], A. The current in each of the two discharging circuits, is zero at t = 0, B. The currents in the two dischargig circuits at t=0, are equal but not zero, C. The currents in the two discharging circuits at, t=0 are unequal, D. Capacitor C1 , loses 50% of its initial charge, , sooner than C2 loses 50% of its initial charge, +, PASSAGE : 1, Resistance value of an unknown resistor is calcu- – E2, , 0.4, , R1, +, – E1, , R2, , Current (A), , R1 1, R 2 2, C1 4 F, C2 2 F, , Let RA and RB be the calculated values in the two, cases A and B respectively., 35. The relation between RA and the actual value, R is, (A) R > RA (B) R < RA, (C) R = RA, D) dependent upon E and r., 36. The relation between RB and the actual value, R is, (A) R < RB, (B) R > RB (C) R = RB, D) dependent upon E and r., 37. If the resistance of voltmeter is RV = 1 k W, and that of ammeter is RG = 1W. the magnitude of the percentage error in the measurement of R (the value of R is nearly 10W) is, (A) zero in both cases, (B) non zero but equal in both cases, (C) more in circuit A, D) more in circuit B, PASSAGE : 2, In the circuit given below, both batteries are, ideal. EMF E1 of battery 1 has a fixed value,, but emf E2 of battery 2 can be varied between, 1.0V and 10.0V. The graph gives the currents, through the two batteries as a function of E2,, but are not marked as which plot corresponds, to which battery. But for both plots, current is, assumed to be negative when the direction of, the current through the battery is opposite the, direction of that battery’s emf. (direction of, emf is from negative to positive), , 0.2, , 0, , –0.2, V, where V and I be, I, 38. The value of emf E1 is, the readings of the voltmeter and the ammeter reA) 8 V, B) 6V, C) 4V, spectively. Consider the circuits below. The inter- 39. The resistance R has value, 1, nal resistances of the voltmeter and the ammeter, A) 10, B) 20, C) 30, (RV and RG respectively) are finite and non zero., 40. The resistance R2 is equal to, A) 10, B) 20, C) 30, , lated using the formula R , , 10, , E2(V), , D) 2V, , , , NARAYANAGROUP, 73
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , MATRIX MATCHING QUESTIONS, , Potential difference across rod I, (C), 41. Match the magnitude values of the potential, Potential difference across rod II, difference across the resistors in different, Ave.timetakenby free e 1 to move from Ato B, branches in the given circuit under steady, (D), Ave. timetakenby free e1 to move from B toC, state., Column II, F, G, (p) 0.5, B, C, (q) 1, (r) 2, E, (s) 4, H, , LEVEL - V - KEY, , D, , A, , 1. D 2.C 3.D 4.A 5. B 6.A 7.A 8.A, 9.D 10.C 11.C 12.C 13.D 14.B 15.C, 16.D 17.C 18.B 19.D 20.C 21.A 22.B, 23.A 24.C 25.A 26.A 27.B 28.A 29.C, 30.B,C 31.B 32.A 33.B 34.B 35) A 36) A,, 37) D 38)B 39) B 40)D, 41)A t, B r,s, C q, D p, 42) A q B s C s D r, , In the given circuit 12 wires of each of resistance 2 forms a cube structure. A battery of, emf 10 volt is connected across the points A, and G. A steady current flows through the battery. Nowmatch the column - I and column - II, for potential difference between different, points., , 10 volt, , p, , between B and F, , B, , 6 volt, , q, , between A and E, , C, , 4 volt, , r, , between A and H, , 2 volt, , s, , between A and F, , t, , between A and G, , A, , D, , LEVEL - V - HINTS, , Column II, , Column I, , 1., , 2., 42. Column I gives physical quantities of a situation in which a current i passes through two, rods I and II of equal length that are joined in, series. The ratio of free electron density (n),, resistivity () and cross-section area (A) of, both are in ratio n1 : n2 = 2 : 1, : = 2 : 1, and A1 : A2 = 1 : 2 respectively. Column II gives 3., corresponding results. Match the ratios in Column I with the values in Column II., , I, A, , 4., , II, B, , C, , Column I, Drift velocity of free electron in rod I, (A) Drift Velocity of free electron in rod II, Electric field in rod I, (B) Electric field in rod II, , 5., , Copper is metal and germanium is semiconductor, resistance of a metal decreases and that of a, semiconductor increases with decreases in, temperature., , Correct option is d , Resistivity of conductors increases with increase in, temperature because rate of collisions between free, electrons and ions increases with increase of, temperature.However, the resistivity of, semiconductors decreases with increase in, temperature, because more and more covalent, bonds are broken at higher temperatures., i ne A vd, Therefore, for non-uniform cross -section (different, values of A) drift speed will be different at different, sections.Only current (or rate of flow of charge), will be same., Current flowing through both the bars is equal Now, the heat produced is given by, or H R H BC 4H AB, H I2 Rt, The given circuit is a balanced wheatstone’s bridge, E, 4, , R, , R, R, , 6R, R, , R, 4R, , NARAYANAGROUP, 74
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, Therefore, the correct option is (a), , R, R, , 9. Q i 2 Rt 1100 5 60 30 k J, 10. Current in the respective loop will remain confirmed, in the loop itself, Therefore, current through 2 resistance = 0, Correct answer is (c), 11. Applying, , 2R, 6R, , 2R, , 4R, , 4, , E, , Thus, no current will flow across 6R of the side, CD. The given circuit will now be equivalent to, R, , P, , V2, , R 1 1, R 2 0.5, R, , 2R, , and R 1, 2R, , 4R, , P2 , 4, , 3, , , 2, , 9 P1, , 1, , 3, 2, , 3, , 2, , 2, , 18W, , 2, , 4.5W P2 P1 P3, , Correct option is c , , L L , , , i.e Ris independent of L, A, tL, t, Hence the correct option is (c), , 12. R , , 6., , For maximum power, net external resistance, = Total internal resistance, R 2, The circuit can be redrawn as follows, 2R, , 2R, , r, , r, , 2R, , 2R, 2R, , P, , Q P, , Q, 2R, , 2R, , 2R, , 2R, , 1, 1, 1, V2, 1, , , or R ; , R100 R 60 R 40, P, P, Hence the correct option is (d), 14. Since, resistance in upper branch of the circuit is, twice the resistance in lower branch. Hence, current, there will be half., , 13. R , , 2R, , 4, , 6, , 4R, , 5, 2r, , P, , Q, , 2Rr / R+r, P, , Q, , 2, , P4 1/ 2 4 , , 4R, , 7., , P i2R, , Now,, Current is same, so P R, , P2 i , , 2, , , , 5, , 10, 2cal / s, 5, , Correct options is (b), 2, r , in 15. The simplified circuit is shown in the figure., 3, third case it is r / 3 and in fourth case the net, resistance is, , In the first case it is 3r, insecond case it is, , 3r/2 RIII RII RIV R1 PIII PII PIV PI, 8., , 5, 4, 3, R PO r, R QR r and R PR r, 11, 11, 11, , 30 2V, , 2V, , 20, , 30, 30, , R PQ is max imum, NARAYANAGROUP, 75
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , NE , , 2m s T, 2R, Dividing Eq (ii) by (i) , we get, , 2V, 30, , Therefore, current i =, , 60, , N2, 2 or N 2 36or N 6, 18, , 2, 1, A, 20 10, , Correct option is c , , 19. P , , 16. Net resistance of the circuit is ., Current drawn from the battery, i, , 2, , 1A, , 2, , i2, , V2, V2, and , R 2 R 3 , 100, 60, , Now, W1 , , 250 , , 2, , R1 R 2 , , 2, , R1, , i4, , 9V, 8, , 8, i1, , 2, , V2, V2, so, R , R, P, , R1 , , 9, 1A current through 3 resistor, 9, 3, , 2, , 4, , W2 , , 250 , , 2, , R1 R 2 , , 2, , R 2 and W3 , , 250 , , 2, , R3, , i3, 2, , 2, , Potential difference between A and B is, , W1 : W2 : W3 15 : 25 : 64 or W1 W2 W3, 20. From F to X charge flows to plates a and b, , q0 q b i q0 q b 27 C, , VA VB 9 1 3 2 4V 8i1, i1 0.5A, , 3F, , 6F, , i 2 1 i1 0.5 A, Similarity potential difference between C and D, , 18C, , 18C, , 3, , VC VD VA VB i 2 2 2 , , 6, 1A, , 4 4i 2 4 4 0.5 2V 8i 3, , 9V, X, , i3 0.25A, 9C, , Therefore i 4 i 2 i3 0.5 0.25, , 17. Current I can be independent of R 6 only when, R1 , R 2 , R 3 , R 4 and R 6 form, Wheatstone bridge, , 36C, , i 2 0.25A, , a, , 3, , balanced, , R1 R 3, Therefore R R, 2, 4, 2, , V2 , 3E , , 18. In the first case, t mx T H R t , , , R, When length of the wire is doubled , resistance and, mass both are doubled, Therefore, in the second case,, , 1A, , 6, , 9V, , 27 C charge flows from Y to X, , Currect option is c , 21. R 2 100 x x, 2, , X, , R, , 100, , X, , NARAYANAGROUP, 76
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JEE-ADV PHYSICS- VOL- III, R, , Applying, , 25. Circuit is option (a) can be used to verify the, equation V IR by varying net resistance of the, circuit.No circuit is given to verify Ohm’s law, , 2, , X+20, , CUURENT ELECTRICITY, , AC, will remain unchanged, CB, 27. Total external resistance will be the total resistance, of the whole length of box. It should be connected, between A and D, Hence correct option is (b), , 80-X, , 26. The ratio, , P R, , Q S, , 2, x, R x 20, , , (i), R 10 x, 2 80 x, Solving equations (i) and (ii) we get R 3, , We have, , Correct option is a , , 28. Vab ig .G i i g S, , G, i 1 i g, S, , , Substituting the values, we get i 100.1 mA, , Correct answer is a , 22. Using the concept of balanced wheatston’s bridge,, 29. We will require a voltmeter , an ammeter a test, we have, resistor and a variable battery in verify Ohm’s law, P R, X, 10, 10 53, , , X , 10.6, volt meter which is made by connecting in parallel, Q S 32 1 48 2 , 50, with the test resistor, Further, an ammeter which is formed by connecting, Correct option is (b), 23. As there is no change in the reading of galvanometer, a low measure in parallel with galvanometer is, with switch S open or closed. it implies that bridge, required to measure the current through test resistor, is balanced current through S is zero and, The correct cption is (c), 30. To increases the range of ammeter a parallel, IR IG , I P IQ, resistance (called shunt) is required which is given, 24. In steady state condition, no current will flow, by, through the capacitor C. Current in the outer circuit, i, , 2V V, V, , 2R R 3R, , ig, S , ii, g, , , , G For option (c), , , R, V, A, , B, , V, 2R, 2V, , Potential difference between A and B, , VA V V iR VB, V, V , VB VA iR , R , 3, 3R , Note: in this problem charge stored in the capacitor, V, with, 3, positive charge on B side and negative on A side, , can also be asked, which is equal to q C, , because VB VA, , , , 50 106, S , 100 1, 3, 6 , 5 10 50 10 , To change it into voltmeter, a high resistance R is, , V, put in series where R is given by R i G, g, , 10, 100 200k, 50 106, Therefore, option (b) and (c) are correct, , For option (b) R =, , E RCt, 31. Charging current, I e, R, t, E, Taking log both sides, log I log , R RC, , NARAYANAGROUP, 77
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, When R is doubled, slope of curve decreases.Also, at t 0, the current will be less. Graph Q represents, the best, Hence the correct option is (b), 32. Given : EC 3VR 3 V VC , Here , V is the applied potential, VC , , 3, 3, V or E 1 e t / tc V, 4, 4, , e t / c , , RB R, R, 100 G 100 100 10%, R, R, , Hence percentage error in circuit B is more than, that in A., 38 to 40. (B), (B), (D) i 1 = 0.1A, E2 = 4V,, i2 = 0 As; 0.1 R1 + 0.1 R2 - E1 = 0 0.1 R2 - 4V = 0, R2 = 40., i2 = 0, , 1, Here c CR 10s, 4, , t, we get t 13.86 Correct answer is a , , 1 C1 C2 R1 R 2 18 s, , 33. CR, , C C R R 8 2 8, 2 1 2 1 2 s, C1 C 2 R1 R 2 6 3 9, , R1, , +, – E2, 4V, , substituting this value of c in Eq (i) and solving for, , i1 = 0.1A, , R2, , + 0.1A, – E1, , Now, i2 = 0.3A, i1 = -0.1 A, E2 = 8V, Now, 0.1 R1 + E1 - 8 = 0 0.1 + 3 - E1 = 0, 0.2R1 - 4 = 0, , RR , 2, 3 C1 C2 1 2 6 4 s, 3, R1 R 2 , 34. The discharging current in the circuit is, , R1 , , 4, 20 E1 = 2 + 4 = 6V.., 0.2, 0.2A, , 0.3A, , R1, , 8V, , 0.1A, , 40, , V, R, Here, V is the potential with which capacitor was 41. Distribute current by supporting symmetry given in, the problem, charged. Since, V and R for both the capacitors, are same, initial discharging current will be same 42. A) since current in both rods is same, but non-zero, Correct option is (b), v, n A, 1 2, 1 2 2 1, n, ev, A, =, n, ev, A, 1, 1 1, 2, 2 2, Further, CR C C or , v, n Al1 2 1, i i 0 e t / CR Here, i0 initial current , , c, , 1, , 2, , C1, , C2, , or C1 loses its 50% of intial charge sooner than C2, Option (d) is also correct, , R.RV, 35. RA R R R, V, , 2, , ( B) E J , , p.d . across rod I, E AB, 1, 4, p.d .across rod II E2 BC, , (D), , avg.timetakenby free e1 to move from Ato B, avg .timetaken by free e 1 to move from B toC, , , , , R, 100 1% % error in case B, R RV, , I, E, A 2 2, 1 1 2 4, A E2 2 A1 1 1, , (C ), , 36. RB = R + RG > R, 37. D) % error in case A., , RV, , RA R, 100 , 1 100, R, R RV, , , 1, , AB V2, , 1, V1 BC, , NARAYANAGROUP, 78
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , LEVEL - VI, , R, , R, , R, , MULTIPLE ANSWER QUESTIONS, 1., , A steady current flows through the conductor, of variable cross-sectional area. Now select, the correct options., I, 1, , 2., , 3., , 2, , (A) R 2 R0, 5., , 3, , (A) The value of current is same through the cross, - sections 1, 2 and 3., (B) The current density is the maximum at the cross, - sections (3), (C) The drift velocity is greater at the cross - section 6., (1), (D) The electric field is maximum at the cross section (3), Select the correct statement (s), (A) A current carrying conductor is electrically, neutral., (B) In a current carrying conductor, it is possible,, all the free electrons can have same drift velocity. 7., (C) In a current carrying conductor, the speed of, all the free electrons need not be same., (D)The electric field at any point inside the currrent, carrying conductor is non-zero., (e) The electric field just outside the current carrying, conductor is non - zero, Two conducting layers of length ' L ' each,, (A), specific resistivites 1 and 2 and cross-, , (B) R 3R0, , (D) R R0, (C) R R0 / 3, What amount of heat will be generated in a, coil of resistance R due a charge q passing, through it, if the current in the coil decreases, down to zero uniformly during a time interval, t, , 2q 2R, 3q 2 R, 4q 2R, 3q 2 R, (A), (B), (C), (D), 3t, 2t, 3t, 4t, What amount of heat will be generated in a, coil of resistance R due to a charge q passing, through it, if the current in the coil decreases, down to zero halving its value every t secs., q2 ln 2, q2 ln 2, q2 ln 2, q2 ln 2, R B), R C), R D), R, A), 3t, 4t, t, 2t, Six resistors are connected so as to form the, edges of a tetrahedron ABCD, the resistances, of opposite pairs being equal. (Note that the, resistors AB and CD do not touch each other), Find the equivalent resistance between A and, A, C., , , r3 2rr, 1 2 r3 r1 r2 , , , 2 r1 r3 r2 r3 , , r1, , r3, r2, C, , D, , r, sectional areas A1 and A2 respectively are, , , r, r, , r, r2 2rr, , , 12, 3 1, 2, r, r, , connected to a voltage source 'V ' such that (B) , 2 r1 r3 r2 r3 , current flow lines do not cross from one layer, B, to the other., r1 2r1r2 r3 r1 r2 , r3 2r1r3 r2 r1 r2 , (1) Find the total current in the system and, (C) 2 r r r r (D) 2 r r r r , describe how it is divided between the 2 layers, 1 3 2 3 , 1 3 2 3 , (2) Find the total current when a third layer of, 8. 12 wires of different resistances are, cross-sectional area A1 A2 , length L and, connected as shown. Find the equivalent, C, 10, D, resistance A and F., specific resistivity 3 is attached at one of the, 10, 10, 10, 10, ends of the given system., 8, O, 8, B, E, A dc source with internal resistance R0 is, 5, 5, 5, 5, 60, 50, loaded with three identical resistance R as, (A), (B), F, A, 3, 29, 29, shown in the figure. At what value of R will the, 40, 20, thermal power generated in this circuit be the, (C), (D), highest?, 29, 29, 2, , 3, , 4., , 1, , NARAYANAGROUP, 79
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, 9., , In the electrical network, shown in the 12. If voltage is applied between terminals 1 and, 2 when terminals 3 and 4 are open, the power, diagram, R1 5 , R2 2 , R3 3 and, liberated is P1 40W and when terminals 3, E1 2 E2 10V . Sources have neigligible, and 4 are connected, the power liberated is, internal resistance. For this network,, P2 80W . If the same source is connected to, R3, R2, the terminals 3 and 4, the power liberated in, the circuit when terminals 1 and 2 are open in, +, +, R1, P3 20W . Determine the power P4 consumed, E1, E2, in the circuit when the terminals 1 and 2 are, connected and the same voltage is applied, (A) The power generated in R1 is 6.4 J/s, between the terminals 3 and 4., (B) if the sources of e. m. f E1 and E2 were, physically inter changed, without changing the, polarities, the power generated in R1 will be the, same., (C) If the polarities of the sources E1 and E2 were, , R1, , R2, , 1, , 3, R3, 4, , 2, , both reversed, the power generatd in R1 will be 13. A storage battery with emf 2.6V loaded, the same, with an external resistance produces a, (D) The ratio of the powers generated in, current I 1.0 A . In this case the potential, R2 and R3 is 1 : 30, difference between the terminals of the, 10. In the given network, the powers generated in, battery is V 2.0V Now, the resistors R1 and R2 are P1 and P2 and the (A) the thermal power generated in the battery is 0.6 W, (B) the power developed in the external resistance, power generated in R is P. For this network,, is 2.0 W, E, R, +, (C) the thermal power generated in the battery is 0.5 W, (D) the power developed in the external resistance, R, is 4.0 W, R, 14. Two batteries of emf 1 and 2 with respective, 1, , 2, , (A) If P1 P2 , it necessarily follows that R1 R2, (B) If R R1 R2 , then P P1 P2, R1 R2, 4, (D) If R1 R2 , then it necessarily follows that, , (C) If P P1 P2 , then R , , P P1 P2, 11. Five 1 resistances are connected as shown, in the figure. The resistance in the conducting, wires (fully drawn lines) is negligible., Determine the resulting resistance R between, A and B., A, , 1, , 1, , 1, , 1, , 1, , B, , internal resistance r1 and r2 are connected in, parallel. Now, (A) the effective emf of this parallel combination is, , 1r1 2r2 , r1 r2 , (B) the effective internal resistance of this, 2r1r2, , combination is r r , 1, 2, (C) the effective emf of this parallel combination is, , 1r2 2r1 , r1 r2 , (D) the effective internal resistance of this, , r1r2, combination is r r , 1, 2, NARAYANAGROUP, 80
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 18. In a metre bridge, the wire consists of two parts, one of length 30 cm and of radius r and the, 3, 2, other of radius 2r. Where will the null point, 10V, 32V, occur if the resistance in the left and right gaps, 5, 10, are 5 and 8 , respectively? The material, 7, 3, of, the wires is the same., e, d, (A) the potential difference between the points ( a 19. When an ideal voltmeter is connected between, the [points E and F the reading of the meter is, and d) is 16V, V0 . When an ideal ammeter is connected, (B) the current through the battery of 10V is zero, (C) the current through 5 resistor is 2A, between E and F, reading is I 0 . Find the current, (D) the current through 10 resistor is 2A, I through a resistor R connected between E, B, (E) the current through 32 battery is 4A, and F., (F) the current throguh 2 resistor is 2A, (G) the potential difference between the points c, E, A, C, and e is 26V, F, (H) the potential difference between along 7, resistor is zero, D, 16. In the given circuit, the resistance of all the, resistors is equal to 2 . Find the current, 20. A network of four unequal resistors R1 , R2 , R3, flowing through each resistor under steady, and R4 are connected in a network (see, state., A, C, R, D, diagram) that contains a source of e.m.f E and, an ammeter A. The source has negligible, R, F, R, internal resistance. For this network,, G, 20v, E, R, R, R, (A) If R3 R4 , the current in the resistor R3, will be less than that shown by the ammeter, B, H, I, J, Select the correct options, (B) If R3 R4 the current in the resistor R2, (A) Current through R1 5 A, will be greater than that shown by the ammeter, (C) If the source of e.m.f and the ammeter were, (B) Current through R2 5 A, physically interchanged in the network the am, (C) Current through R3 zero, meter will show the same current., (D) Current through R4 5 A, (D) If the resistors R2 and R1 were physically, interchanged, the ammeter will show the same, (E) Current through R5 5 A, current, R, (F) Current through R6 10 A ., A, 17. ABCDFPA is a network of three batteries of, E, R, the e.m.f’s E, 12 V and 4 V respectively and, R, three resistance 2 , 4 and 6 connected, R, as shown in the diagram. An ideal ammeter, 21. A galvanometer with a coil resistance of 100, connected between F and P shows a current, shows a full-scale deflection when a current of, reading of 0.5 A. Then the value of the e.m.f., 1 mA is passed through it. What is value of the, E is, resistance, which can convert this, D, B, C, galvanometer into an ammeter showing a full, +, +, +, 4V, E, 12V, - scale deflection for a current 10A?A, resistance of the required value is available, 2, 4, 6, but it will get burnt if the energy dissipated in, it is greater then 1 W. Can it be used for the, A, F, P, A, conversion of the galvanometer described, (A) 6V, (B) 6.6V (C) 8 V (D) 5.5V, above?, 15. In the circuit diagram shown in the figure,, a, , b, , c, , 1, , 2, , 6, , 3, , 5, , 4, , 1, , 3, , 4, , 2, , NARAYANAGROUP, 81
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CUURENT ELECTRICITY, , JEE-ADV PHYSICS- VOL- III, , 22. When the modified galvanometer of previous 28. An accumulator of e.m.f 2 volt and negligible, question is connected across the terminal of a, internal resistance is connected across a, battery, it shows a current of 4A. The current, uniform wire of length 10 m and resistance, drops to 1 A when a resistance of 1.5 is, 30 . The appropriate terminals of a cell of, connected in series with the modified, e.m.f. 1.5 volt and internal resistance 1 is, galvanometer. Find the emf and the internal, connected to one end of the wire. What length, resistance of the battery., of the wire will be required to produce zero, 23. An ammeter and a voltmeter are connected in, deflection of the galvanometer? How will the, series to a battery of emf 6V. When a certain, balancing change, resistance is connected in parallel with the, (A) when a coil of resistance 5 is placed in, voltmeter, the reading of the latter decreases, n = 2 times , whereas the reading of the, series with the accumulator, ammeter increases the same number of times., (B) the cell of 1.5 volt is shunted with 5, What is the ratio of the voltmeter resistance, resistor?, to the ammeter resistance? Find the voltmeter 29. A certain thermocouple (treat it as a seat of, reading after the connection., e.m.f) which has a total resistance of 10 , has, 24. An ammeter and a voltmeter are connected, one junction in melting ice and the other, as in figures of Circuit 1 and Circuit 2, in order, junction in steam. The e.m.f. between its ends, to, as measured with a potentiometer is 4 millivolt., X, X, A, A, What would be its reading when it is connected, to a millivoltmeter which has a resistance of, 50 ohm?, 30. A 10m wire potntiometer has resistance 10, V, V, and is connected to an accumulator of 2 volt, Circuit 2, Circuit 1, and negligible internal resistance. There are, measure an unknown resisance. Justify, quantitatively which connection is to be, two resistance boxes R1 and R2 in series with, perferred., the accumulator and one can have any integral, 25. It is required to measure the resistance, values of resistance from resistance boxes. A, of a circuit operating at 120V. There is only, standard Cd-cell of 1.018 V having a sensitive, one galvanometer of current sensitivity 106, galvanometer in series with it is connected, A per division. How should the galvanometer, across R1 . How would you proceed with the, be connected in the circuit to operate as an, above arrangement to obtain a potential drop, ohm meter? What minimum resistance can be, measured with such a galvanometer if its full, 1V per mm of the potentiometer wire?, scale has 40 divisions? Construct the entire, Calculate values of R1 and R2 required. What, scale of such an ohmmeter in ohms per 5, length of this potentiometer will balance the, divisions.Neglect galvanometer resistance., 26. The resistance of a potentiometer wire 8 m long, thermo e.m.f. of iron-copper couple at 3000 C, is 8 ohm. A high resistance box and a 2-volt, which develops 17 V / o C ?, accumulator are connected in series with it., What should be the value of the resistance in 31. A five wire potentiometer is connected to an, the box, if it is desired to have a p ot e n t i a l, accumulator of e.m.f. 2.2V and 1 internal, drop of 1 micro volt per mm?, resistance. The potentiometer wire has, 27. The terminals of a cell are connected to, resistance 1 per metre. What is the, resistance R and the fall of potential across R, maximum voltage that you can measure with, is balanced against the fall of potential on a, this particular arrangement of potentiometer?, potentio-meter wire., What length of this potentiometer will balance, When R is 20 and 10 respectively, the, the e.m.f of a Daniell cell (e.m.f. = 1.18V)?, corresponding length on the potentiometer wire, What resistance in series with accumulator will, are 1.5 m and 1.2 m. Calculate the internal, be required to balance this cell exactly at the, resistance of the cell., centre of the last wire?, NARAYANAGROUP, 82
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, , 32. A potentiometer wire of length 1000 cm has a 37. If on connecting the key K, no currect flows, resistance of 10 ohms. It is connected in seres, through the segment CD, whatever potential, with a resistance and a cell of e.m.f. 2 volts, difference V exists between A and B, which of, and of negligible internal resistance. A source, the following cases will satisfy this condition?, of e.m.f. 10 millivolts is balanced against a, (A) R1 1, R2 2, R3 3, R4 4, length of 40 cm of the potentiometer wire., What is the value of the external resistance?, (B) R1 2, R2 1, R3 4, R4 2, PASSAGE - 1, Twelve str aight unifor m wires of length ‘ a’ and, r esist ance R ar e j oined t o for m t he edges of a, cube of side ‘ a’ . Cur r ent ent er s t he syst em at, one cor ner and leaves fr om a point on one of, t he edges meet ing at t he opposit e cor ner at a, distance Ka 0 K 1 from it., , 33. The eqivalent resistance of the system is, (A), , R, 10 4 K 5 K 2 , , 12, , (B), , R, 10 4 K 5K 2 , , 6, , R, R, 2, 5 2K 5K 2 , , (D) 5 2 K 5K , 12, 6, 34. The equivalent resistance is maximum, when, , (C), , 4, 2, 1, (B) K (C) K (D) K 0, 5, 5, 5, 35. The maximum value of the equivalent, resistance is, , (C) R1 1, R2 2, R3 2, R4 1, (D) R1 R2 2, R3 2 R4 4, 38. If R1 R4 1 and R2 R3 2 , the, effective resistance between the terminals A, and B will change,on closing the key K, by, (A) an increase of 0.5, (B) a decrease of 0.16, (C) an increase of 0.16, (D) a decrease of 0.5, PASSAGE - 3, In the circuit shown, the resistances are given, in ohms and the battery is assumed ideal with, emf equal to 3.0 volts., , (A) K , , 9R, 9R, (B), (A), 5, 2, PASSAGE - 2., , 9R, (C), 10, , 5R, (D), 12, , Four resistors R1 , R2 , R3 and R4 are connected, between two terminals A and B in a network as, shown in the diagram. A key K can connect the, two points (see diagram) C and D. A constant, , 50, R1, , R3, R2, , 3.0V, , 30, , PASSAGE TYPE QUESTIONS, , R4, , 50, R5 30, , 39. The resistor that dissipates maximum power, A) R1, B) R2, C) R4, D) R5, 40. The potential difference across resistor R3 is, A) 0.4 V, B) 0.6V C) 1.2V D) 1.5v, 41. The current passing through 3V battery is, A) 10mA B) 30mA C) 40mA D) 60mA, , MATRIX MATCHING QUESTIONS, potential difference VA VB V is maintained, between the points A and B, 42. Match the magnitude values of the potential, difference across the resistors in different, R, R, C, branches in the given circuit under steady, state., 1, , 2, , A, , R3, , R4, , 36. If V 25 , R1 1 , R2 2 , R3 3 and, , R4 4 the current that will flow from C to D, on connecting key K is, (A) zero, (B) infinity (C) + 1 A (D) - 1 A, , 3, , Q, 18V, , V, , 2, , M, , P, , 2, S, , N, 10, , 6, T, , O, 4, U, , NARAYANAGROUP, 83
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CUURENT ELECTRICITY, , JEE-ADV PHYSICS- VOL- III, column I with the direction of current through, galvanometer and the value of the current, through the battery in column II., A, S1, , 15, , , , 20, , S3, , 60, , 30, , , +, , , 30, , S2, , B, , 43. Match the magnitude values of the potential, difference across the resistors in different, branches in the given circuit under steady, state., , S4, , 180V, , Column - I, (A) only switch S1 is closed, (B) only switch S2 is closed, F, G, (C) only switch S3 is closed, B, C, (D) only switch S4 is closed, Column II, E, H, (p) current from A to B, D, A, (q) current from B to A, (r) current through the battery is 12.0A, In the given cirucit 8 wires of each of, (s) current through the battery is 15.6 A, resistance 2 forms a cube structure. A, (t) current through the Galvanometer is 1.2A, battery of emf 10 volt is connected across the 45. In the circuit shown, battery, ammeter and voltpoints A and G. A steady current flows through, meter are ideal and the switch S is initially, the battery., closed as shown. When switch S is opened,, Now match the column - I and column - II for, match the parameter of column I with the efpotential difference between different points., fects in column II., Column II, , Column I, A, , 10 volt, , p, , between B and F, , B, , 6 volt, , q, , between A and E, , C, , 4 volt, , r, , between A and H, , D, , 2 volt, , s, , between A and F, , t, , between A and G, , 44. Consider the circuit shown. The resistance connected between the junction A and B is 60, including the resistance of the galvanometer., The switches have no resistance when shorted, and infinite resistance when opened. All the, switches are initially open and they are closed, as given in column I. Match the condition in, , R, , R, , V, , S, , E, , A, , Column - I, (A) Equivalent resistance across the battery, (B) Power dissipated by left resistance R, (C) Voltmeter reading, (D) Ammeter reading, Column - II, (p) remains same, (q) increases, (r) decreases, (s) becomes zero., NARAYANAGROUP, , 84
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JEE-ADV PHYSICS- VOL- III, , LEVEL - VI- KEY, , CUURENT ELECTRICITY, Hence, total resistance across the voltage source is :, , 1. A, B, D 2. A, B, C, D, E, , v, , 3.i) R R R1 R2 R1 ,, 1 2, , v, , R1 R2 R2, , R1R2, , , 4., , The equivalent circuit may be drawn as in the figure., Let, us assume that e.m.f. of the cell is , then, current, , 4.B, 5.C, 6.D, 7.A, 8.A, 9. A, C, D, 10.A, B, C 11. 0.5 12.40w, 13.A, B 14.C,D 15.A, B, C, D, E, F, G, H, 16.A, B, C, D, E, F, 17.B, , i, , 20.A, B, C, D, , P i2 R / 3 , , 2, , R R / 3 , , 2, , R / 3, , 0, , dP, 0 , which yields, dR, , For P to be maximum,, , 21. 0.01, yes, R 3R0, , 22. r 0.45 2volt 23.2 volts, 25. 3M 26. 1992 27.6.67 , 28.7.5 m, 8.75m, 6.25m 29.3.33 milli volt, 30. R1 1018, , , R0 R / 3, , Now, thermal power, generated in the circuit, , v0 I 0, 19. I v I R, 0, 0, , 18.3m, , R, R, R, , 31.1.83.v, 3.218m, 2.39, , 32.70 , 33) A, 34. B, 35.C, 36).C 37) B 38) B 39).A 40).A 41). C, 42). A p,q ; B r ; C t; D s,u, 43). A t ; B r ,s ; C q; D p, 44). A p,s,t ;B q,r;C q,t;D p, 45). A q;B r;C r;D r, , LEVEL - VI - HINTS, I, nAe, We know, for a conductor during the flow of steady, current,, vd , , v Ri, ei, ei, , E, , e , A A, electric feild is maximum at cross - section (3), 3. (1) Resistance of the 2 layers are :, E, , R1 , , A2, , Resistance of the network = R0 R / 3, , 1l, R1R2, R3 Where R1 A, 1, R1 R2, , 1. I nAvd e, , A1, , 2, , V, , v, , ii), , 1, , L, L, 1 ; R2 2, A1, A2, , i, i, ,R0, , 5., , As current i is function of time, and at t 0 and, , t , it equals i0 and zero respectively, it may be, t , , represented as, i i0 1 , t , t, , t, , i0 t, t , , Thus q idt i0 1 t dt 2, , , 0, 0, 2q, t, The heat generated., , So, i0 , , t, , t, , 2, , 2q , t , 4q 2 R, H i Rdt 1 Rdt , t , 3t, 0, 0 t , 2, , NARAYANAGROUP, 85
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JEE-ADV PHYSICS- VOL- III, , CUURENT ELECTRICITY, 21. We have S , , , 103, i, G , 100, I i, 10 103, , 0.1, 0.01, 9.999, , V ', 2, , P I 2 R 10 103 , , and 4 0.01 0.49 4 0.5 2volts, 23. Before connection of the resistance,, E, i (ammeter reading) =, rR, E, R, and V (voltmeter reading) =, rR, After connection of the resistance (S) in parallel to, R, i’ (ammeter reading), ER S, E, , RS, r R S RS, r, RS, and V (voltmeter reading), , , , , E R, , ERS, r R S RS, , It is given that i ' ni and nV ' V, E R S , E, , n, r R S RS, r R or, , ER 2, E, 6, V', , 2volts, R 2 n 1 or n 1 2 1, 24. In circuit 1 , i (main current), , , RV X RA , , , RV X RA , RV X RA ., RV X RA, V’ (voltmeter reading) = p.d. across the voltmeter, , , RV X RA RV X RA , , ., RV X RA , RT X RA, i’ (ammeter reading), = current in the branch, , , , , RV X RA , , RV , RV X RA RV X RA , X RA, , V', X RA, i', X ' X, R, 100 A 100., %error , X, X, When X 0, and hence this circuit is, unsuitable for low resistance, When X 0, and hence this circuit is, suitable for high, resistance., In circuit 2,, , i' , , R S r R n r R S RS ---- (i), ERS, ER, and n r R S RS R r, or nS R r r R S RS ------- (2), S, , 2, , X’ (apparent value) =, , E R S, RS, RS, , , R S r R S RS R S, , From (i) and (ii) ,, , ERS, r R S RS, , R, n 1, , R, R , R, R 2 R 2, n, n 1 , n 1, , 0.1, 10 103 , , 10 10 2 0.1 0.999 watt<1 watt., Hence it can be used for the purpose., , , 22. 4 , and 1 , 0.01 r, 0.01 r 1.5, 4 r 1.51, or 4r 0.04 r 1.51, , 1 r 0.01, or 3r 1.47 or r 0.49, , i ', , and putting this value of S in (i) we have R = nr., R : r n :1, , R, n 1, 2, , V', , , RV X , , , RV X, RA X RV RA RA RV, RV X, , RV X , RV X, , RV X X RV RA RA RV, RV X, RV RA RARV, NARAYANAGROUP, , 88
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JEE-ADV PHYSICS- VOL- III, , X ', , V', RV X, , i ' X RV RA RA RV, , , X RA RV RA RV, R X, V, RV X , RV X, , ,% error, X ' X, X, 100 , 100, X, RV X, (numerical value), , , 100, R, 1 V, X, When X 0, 0 and when X , 1, 25. We have by ohm’s law,, 120, I , R, R, But I K by theory of galvanometer.., When I 106 A , 1 ;, K 10 6, , , CUURENT ELECTRICITY, , Rx Rr R z p RZ 0 (or), -------- (2), 2z x r P, Ry Rq R x r Rx 0 (or), --------- (3), 2x y q r, Therefore 2 p 3 y z 3 x ------- (4), ----------- (5), 2q 3 x y 3 z, ------------ (6), 2r 3 z x 3 y, Again x r x r q r z p r , (or) 2 x z 4r p q ------------ (7), substituting values of p, q, r, we find , z y --- (8), again,, z p z p r K z x q p, p 1 K y q p , , This gives , K 2 y x 8 x y 0 (or), x, y, , ------ (9), 8 2K 8 K, , V R x x r x r q K z x q p , 120, 120 106, 3M ., or R , R, 40, 7x 2 y, , V R, 2 K 2 x y , Putting 0,5,10,15, 20, 25,30,35, 40 we have, ------ (10), 2, , the entire scals as ,, ------------ (11), I x 2y, 24,12,8,6, 4.8, 4,3.43 and 3M ’ss, 26. 1992 , 7 x 2 y 4K 2 x y , V, Req R , , 27. 6.67, i, 2 x 2 y, , , 28. 7.5m; 8.75m ; 6.25m, R, 29 3.33 millivolt, 10 4 K 5K 2 , 12, 30. Put R1 1018 (one thousand times the e.m.f. of, 2, 9R, Cd-cell); adjust R2 till no deflection, in the, R is maximum when K and Rmax , 5, 10, galvaometer; R2 972,5.1m ), 19-21.Let the resistance (small but not zero) in the arm, 31. 1.833 V; 3.218 m; 2.39 ), CD be x . Assume currents i1 , i2 , i3 , i4 and ix as, 32. 70, shown in the diagram. Then applications of, 16-18, Kirchhoff’s junction law and loop laws give,, (x, +y, +z, (x – r), ), i1 i2 ix ----- (1) i3 i4 ix ----- (2), x, 6, 10 , , q, , ., , (z–p+r), , r, Z, , (x – r + q), , (z + x + q – p), (x – p), P, , (y – q + p), (x + y + z), , Ry R y q RP RZ 0 (or), 2y Z p q, , ---------- (1), , V i1R1 i1R2 i4 R4, , ----- (3), , and i1R1 Xix i3 R3 0 ----- (4), Solving these equations, for ix , we get, from (1), &(3),, , V i1 R1 R2 ix R2 -----(5), and from (2) and (3) ,, , NARAYANAGROUP, 89
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , MAGNETISM, SYNOPSIS, Magnet: A body which attracts Iron,Cobalt,, Nickel, like substances and which exhibits, directive property is called Magnet., , Types of Magnet:, i) Natural magnets: a) The magnet which is, found in nature is called a natural magnet, Eg: magnetite. Fe3O4 ., b) Generally they are weak magnets., ii) Artificial magnets:The magnets which are, artificially prepared are known as artificial, magnets. These are generally made of iron, steel, and nickel., , PROPERTIES OF MAGNETS :, 1) Attractive property : The property of, attracting pieces of iron, steel , cobalt , nickel, etc by a magnet is called attractive property. It, was found that when a magnet is dipped into, iron filings the concentrations of iron fillings is, maximum at ends and minimum at centre. The, places in a magnet where the attracting power, is maximum are called poles., 2. Directive property : If a magnet is suspended, freely, its length becomes parallel to N-S, direction. This is called directive property. The, pole at the end pointing north is called north, pole while the other pointing south is called south, pole., Magnetic poles always exist in pairs If a magnet, is broken into number of pieces, each piece, becomes a magnet with two equal and opposite, poles This implies that monopole does not exist., . The two poles of a magnet are found to be equal, in strength and opposite in nature., Unlike poles attract each other and like poles, repel each other, There can be magnets with no poles., Eg: Solenoid and toroid has properties of, magnet but no poles., , Magnetic axis and magnetic meridian, The line joining the poles of a magnet is called, magnetic axis and the vertical plane passing, through the axis of a freely suspended magnet is, called magnetic meridian, NARAYANA GROUP, , Geometrical length (L) : The actual length of, magnet is called geometric length, , Magnetic length 2l The shortest distance, , , , between two poles of a magnet along the axis is, called magnetic length or effective length. As, the poles are not exactly at the ends the magnetic, length is always lesser than geometric length of, a magnet. Effective length depends only on the, positions of the poles but not on the magnet, , Examples :, 2l, S, , N, , S, , N, , L, 2R, Magnetic length = 2l Magnetic length = 2R, Geometrical length = L Geometrical, length p R, Magnetic length is a vector quantity. its direction, is from south pole to north pole along its axis, 5, Geometrical length, 6, Pole Strength (m) : The ability of a pole to, attract or repel another pole of a magnet is called, pole strength S.I Unit : ampere - meter. Pole, strength is a scalar It depends on the area of, cross section of the pole. Its dimensional formula, is M 0 LT 0 A1, Inductive property: When a magnetic, substance such as iron bar is kept very close to, a magnet an opposite pole is induced at the, nearer end and a similar pole is induced at the, farther end of the magnetic substance.This, property is known, as inductive property., A magnet attracts certain other magnetic, substance through the phenomenon of magnetic, induction. induction precedes attraction., Repulsion is a sure test of magnetism.A pole of a, magnet attracts the opposite pole while repels, similar pole.How ever a sure test of magnetism, is repulsion but not attraction.Because attraction, can takes place between opposite poles or, between a pole, and a piece of, unmagnetized material due to induction., , Magnetic length =, , 91
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , , Magnetic Moment, Magnetic dipole and magnetic dipole, moment (M) : A configuration of two, magnetic poles of opposite nature and equal, strength separated by a finite distance is, called as magnetic dipole., The product of pole strength (either pole) and, magnetic length of the magnet is called, magnetic dipole moment or simply magnetic, moment., If ‘m’ be the pole strength of each pole and, ' 2 ' be the magnetic length, then magnetic, moment M is given by M m 2, In vector form, M 2 m, Magnetic moment is a vector whose direction, is along the axis of the magnet from south to, north pole. The S.I. unit of magnetic moment is, ampere-meter2 (A-m2) its dimensional formula, [AL2], , N, , , , S, x parts, y parts, , pole strength of each part = m/x, Length of each part = 2 / y, 2, , m, , M, , Magnetic moment of each part, M1 y x xy, , Variation of magnetic moment due to bending, of magnets, , , Variation of magnetic moment due to , cutting of magnets :, Consider a bar magnet of length ' 2 ' , pole, strength ‘m’ and magnetic moment ‘M’, When the bar magnet is cut into ‘n’ equal, parts parallel to its length, then, N, S, , When the magnet is cut into ‘x’ equal parts, parallel to its length and ‘y’ equal parts, perpendicular to its length, then, , When a bar magnet is bent, its pole strength, remains same but magnetic length decreases., Therefore magnetic moment decreases., When a thin bar magnet of magnetic moment, -shape with the, M is bent in the form of, arms of equal length as shown in figure, then, S, , N, , M/3, , M/3, N, S, , S, N, , M/3, , Pole strength of each part = m/n, ( area of cross section becomes (1/n) times, of original magnet), Length of each part = 2 (remains same), m, , M, , Magnetic moment of each part, M1 2 n n, Note: If it is cut ‘n’ times , parallel to its length then, magnetic moment of each part is, m, M, M 1 2l , , n 1 n 1, When the magnet is cut into ‘n’ equal parts, perpendicular to its length then, N, S, , Magnetic moment of, each part = M / 3, Net magnetic moment of the combination,, M, M, M, M, M1 j i j i , 3, 3, 3, 3, M, M1 , 3, When a thin magnetic needle of magnetic, moment M is bent at the middle, so that the, two equal parts are perpendicular as shown, in figure, then, N, , 2, , M/2, S, N, , Pole strength of each part =m ( area of cross, section remains same), Length of each part 2 / n, 2, M, Magnetic moment of each part, M1 m , n, n, 92, , /, , S, , M/2, , M, 2, Net magnetic moment of the combination,, , Magnetic moment of each part =, , M1 , , M, M, i j , 2, 2, , M 1 2, , M, M, , 2, 2, , NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , When a thin bar magnet of magnetic moment, M is bent into an arc of a circle subtending an, angle ' q ' radians at the centre of the circle,, then its new magnetic moment is given by, q , 2 M sin , 2, ( q must be in radians), M1 , q, y, , S, , 2, 2, ie q , R, R, q, , R, , y, , Effective length of the magnet increases. Hence, Magnetic moment increases, New magnetic moment is given by, M1 , , N2, , /2, , M2, , S2, , , , from the figure, Effective length = 2y= 2R sin, , ( q must be in radians), , Resultant Magnetic Moment due to, combination of Magnets :, , N, , , , , , q, 2, , , , M1, , S1, , N1, , When two bar magnets of moments M1 and M2, are joined so that their like poles touch each other, and their axes are inclined at an angle ' ' , then, the resultant magnetic moment of the combination, ‘M1’ is given by, , , , sin q y y R sin q , 2 , , 2 R, , New Magnetic Moment,, 2 q, M 1 m 2 y m 2 sin, q 2, q, 2 M sin , 2, M1 , q, , Mq, q , 2 sin , 2 , , M1 M12 M 22 2M1M 2 cos , ( q angle between the directions of magnetic, moments), S2, , M 2 m, , N2, , , , , S, , M1, , 0, , (180 - ) S, , , , If q , , p, radians,, 2, , , , M2, , , 2, , When two bar magnets of moments M1 and M2, are joined so that their unlike poles touch each, other and their axes are inclined at an angle ‘ ’,, then the resultant magnetic moment, , N, , i.e., if the magnet is bent in the form of quadrant, of a circle, then, M1 , , , , p, 4 2 2M, p , p, , 2 , , 2 M sin, , M1 M12M 22 2M1M 2 cos(1800 q), , [ angle between directions of magnetic moments is, (1800 - q )], , M1 M12 M 22 2M1M 2 cos , , If q p radians, i.e., if the magnet is bent in, the form of a semi circle, then, , p, 2 2M, =, M1 , N, p, p S, If q 2p radians, i.e., if the magnet is bent in, the form of a circle, then, 2 M sin p, M1 , 0, 2p, When a magnet in the form of an arc of a, circle making an angle ' q 'at the centre having, magnetic moment 'M' is straightened, then, , M1, , , , NARAYANA GROUP, , M2, , , , 2 M sin, , , , N1, , 1, , N1N2, , S1, , S2, , When two bar magnets of moments M1 and M2, (M1>M2) are placed coaxially with like poles, in contact then resultant magnetic moment,, M1=M1–M2, ( angle between directions of magnetic moments,, q1800 ), M1, , M2, , , S1, , N1 S2, , N2, 93
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, When two bar magnets of moments M1 and M2, (M1>M2) are placed coaxially with unlike poles, are in contact then resultant magnetic moment,, M1=M1+M2, ( angle between directions of magnetic moments,, q 00 ), S2, , N2, M2, , , , an equilateral triangle with unlike poles at each, corner, resultant magnetic moment is given by, , M1 2M M 2 22M M cos1200 3M, When four bar magnets of moments M, 2M, 3M, & 4M are arranged to form a square with unlike, poles at each corner, then resultant magnetic, moment is given by, 2, , 3M, , M1, S1, , S, N, , N1, , When two bar magnets of magnetic moments, M1 and M2 are placed one over the other with, like poles on the same side, then resultant, magnetic moment,, , = M1+M2 q 0, , M1, , N2, , S2, , S1, , N1, , 0, , 4M, 2M, , M, , M1 2M 2M 22M 2M cos900 2 2M, 2, , When two bar magnets of magnetic moments M1, and M2 are placed one over the other with unlike, poles on the same side, then resultant magnetic, , S2, , , , S1, , N2, S1, , N2 S2, , S2, , N1, , , , 0, M1 M12 M22 q 90 ., , M M M, , N, , M M , , p 2 , , 2, , S, , M, N, S, , S N, , N, , , , M, , When identical magnets each of magnetic, moment M are arranged to form a closed polygon, like a triangle (or) square with unlike poles at, each corner, then resultant magnetic moment, M1, = 0., In the above point , if one of the magnets is, reversed pole to pole then resultant magnetic, moment, M 1 2M, S, , N, , 3M, , 120, 2M, , N, , =, , M, , M, , , , 0, , M, , M, , N, , , , , , , M, , When three bar magnets of equal length but, moments M, 2M and 3M are arranged to form, , 4 p 2 , , M1, M2, , N, , Magnetic field :, , 0, , 2M 60, , M, 2p, , NS, , S, S, , 94, , 2, 2, , N S, , S, N, , S, M, , S, , , , 2, 1, , M, , , , , , M2, , 2, , M, , M, , N, , M , 2 , 2 p , 2 M M M M , 2p , , , , p, 2, p, 2, In the above case if the two parts are arranged, perpendicular to each other, then resultant, magnetic moment is, 1, , N, , M, , S, , M1, , When two bar magnets of magnetic moments, M1 and M2 are placed at right angles to each, other t hen resultant magnet ic moment,, S, , N, , resultant magnetic moment, M1 = M1+M2, , S1, , N2, , 2, , S, , moment, M1 = M1 M2. q 1800 , N1, , 2M, , When half of the length of a thin bar magnet of, magnetic moment M is bent into a semi circle as, shown in figure, then, , M1, , N1, , =, , N, S, , S N, , , , M2, , , , 2M, , N S, , , , Around a pole there exist a region called, magnetic field in which the influence of the pole, is felt., The space around the magnet is said to be, associated with a field known as magnetic field,, if another magnet is brought into the space, it is, acted upon by a force due to this energy., Magnetic induction is the measure of magnetic, field both in magnitude and direction., Magnetic Field Lines :, The imaginary path in which a free unit north, pole would tend to move in a magnetic field is, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , known as a magnetic line of force (or) simply (ix) When a soft iron ring is placed in magnetic field,, then most of lines of force pass through the ring, magnetic “field line”., and no lines of force pass through the space, inside the ring as shown in figure. The, phenomenon is known as magnetic screening or, N, S, shielding., , Magnetic line of force with magnetic needle, Characteristics of lines of force :, (i), , Magnetic lines of force are closed curves., B=0, Outside the magnet, their direction is from north, to south pole, while inside the magnet they are, from south to north pole. Hence they have neither, (x) If the magnetic lines of force are straight and, origin nor end., parallel, and equally spaced the magnetic field, (ii) Tangent, at any point to the line of force gives, is said to be uniform., the direction of magnetic field at that point., Magnetic Induction (or), (iii) Two lines of force never intersect each other. If, the two lines of force intersect, at the intersecting, Induction Field Strength (B), point the field should have two directions, which, Magnetic induction field strength at a point, is not possible., in the magnetic field is defined as the force, (iv) The lines of force tend to contract longitudinally, experienced by unit north pole placed at that, or length wise . Due to this property the two, point. It is denoted by ‘B’., unlike poles attract each other., If a pole of strength ‘m’ placed at a point in a, magnetic field experiences a force ‘F’, the, N, S, magnetic induction (B) at that point is given by, Magnetic lines of force between two unlike poles., (v) The lines of force tend to repel each other , laterally. Due to this property the two similar, poles repel each other., , F, i.e., F mB, m, B is a vector quantity directed away from Npole or towards S-pole., B, , N, Unit North, Pole, , N, , S, , Magnetic lines of force between two like poles, (vi) If in any point, in the combined field due to two, magnets, there are no lines of force, it follows, that the resultant field at that point is zero. Such, points are called null or neutral points., (vii)Lines of force in a field represent the strength of, the field at a point in the field. Lines of force, are crowded themselves in regions where the, field is strong and they spread themselves apart, at places where the field is weak., (viii) Lines of force have a tendency to pass through, magnetic substances. They show maximum, tendency to pass through ferro magnetic, materials., NARAYANA GROUP, , B, , B, , S, , Unit North, Pole, , S.I. Unit of B :, N, J, V s, wb, (or ), (or ) 2 (or ) 2 (or )tesla (T ), 2, Am, Am, m, m, CGS Unit of B : gauss (G) 1G = 10–4 T, , Dimensions of B :, F [ MLT 2 ], , [ MT 2 A1 ], m, [ AL], When placed in an external magnetic field, all, N-poles experience a force (F = mB) in the, direction of the field and all S-poles experience, the same force in the direction opposite to the, field., B, , N, , F = mB, B, , F = mB, , S, , 95
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , Magnetic induction at a point due to an, isolated magnetic pole :, Consider a magnetic pole of strength ‘m’ kept at, the point ‘O’. Consider a point ‘P’ at a distance, ‘r’ from ‘O’. To find the magnetic induction at, the point ‘P’, imagine a unit north pole at P., O, m, P, r, m 1, Force on unit north pole at P o, N, 4 r 2, Force on unit north pole at ‘P’ gives the, magnetic induction at that point., Magnetic induction at P is, , B, , , , , , , , , , , o m, newton/amp–metre (or) tesla (T), 4 r 2, , Types of Magnetic Field, , , , , , Couple acting on the bar magnet (or), Torque on a Magnetic Dipole, , , Uniform magnetic field: The magnetic field,, in which the magnetic induction field strength is, same both in magnitude and direction at all, points, is known as uniform magnetic field., In such a magnetic field the magnetic lines of, force are equidistant and parallel straight lines., Ex: Horizontal component of earth’s magnetic, field in a limited region., Non uniform magnetic field: The magnetic, field, in which the magnetic induction or field, strength differs either in magnitude, in direction, or both is known as non uniform magnetic field., It is represented by non-parallel lines of force , Ex: The magnetic field near the pole of any, magnet, , Magnetic flux : It is equal to the total, number of magnetic lines of force passing normal, through a given area. Its S.I. unit is weber and, C.G.S. unit is maxwell, 1 weber = 108 maxwell, , , B.A BAcos, , , , Where ‘ ’ is the angle made by magnetic field, , , B with the area n̂ , , A = area of the coil, A Anˆ, It is a scalar. Dimensional formula is , , , , ML2 T 2 I 1 ., , Magnetic Flux Density (B): The number of , magnetic flux lines passing per unit area of cross, 96, , section normal to the cross section is called, magnetic flux density., B= B / A, SI unit is weber metre-2 or tesla or NA–1m–1., Its C.G.S. unit is gauss, 1 gauss = 10-4 tesla, Its dimensional formula is [M1L0T-2A-1], It is also known as magnetic induction and, magnetic field., The relation between B and H is B0 = H in, vacuum and B = H in a material medium Where, is the absolute permeability of the medium., The force experienced by a pole of strength ‘m’, ampere meter in a field of induction B is F = m, B, , When a bar magnet of moment M and length 2l, is placed in a uniform field of induction B, then, each pole experiences a force mB in opposite, directions., B, N, 2l, mB, , mB, , , , S, , As a result the bar magnet experiences a couple, and moment of couple is developed., Moment of couple acting on the bar magnet is, C = Force x perpendicular distance between two, forces., , C m 2 B sin ( or) C = M B sin, Where is the angle between magnetic moment, and magnetic field., In vector notation C M B, When the bar magnet is either along or opposite, to the direction of magnetic field then moment, of couple=0., When the bar magnet is perpendicular to the, direction of applied magnetic field, then the, moment of couple is maximum. i.e. Cmax = MB, In a uniform magnetic field a bar magnet, experiences only a couple but no net force., Therefore it undergoes only rotatory motion., In a non-uniform magnetic field a bar magnet, experiences a couple and also a net force. So it, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , , MAGNETISM, , undergoes both rotational and translational motion, Two magnets of magnetic moments M1 and M2, are joined in the form of a (+) and this, arrangement is pivoted so that it is free to rotate, in a horizontal plane under the influence of, earth's horizontal magnetic field. If ' q ' is the, angle made by the magnetic meridian with M1, in equilibrium position, then, t1 t 2 ; i.e., M1BH sin q, tan q , , = M2BH sin 90 q ;, 2, , M2, M1, , B1 sin 450, 2, , , 0, B2 sin 30, 1, A pivoted magnetic needle of length 2 and pole, strength 'm' is at rest in magnetic meridian. It is, held in equilibrium at an angle ' q ' with BH by, pulling its north pole towards east by a string., Then tension in the string is, from the figure ,, y, cos q y cos q In equlibrium, , , = M B2 sin (750 - 300) ; , , BH, , 1, , BH, , N, , , M2, , 90–, , N2, , F, l, , M1, N1, , , , y, , l, S, , S1, , , , S2, , Two magnets of moments M1 and M2 are joined, as shown in figure and the arrangement is pivoted, so that it is free to rotate in a horizontal plane, under the influence of magnetic field B. Then, net torque acting on the system is given by, t t1 t 2 ; M1B sin q1 M 2 B sin q2, , BM1 sin q1 M 2 sin q2 , , t tension t BH ; i.e.,, , F cos q 2m BH sin q, F 2mBH tan q, b) In the above case, if the magnetic needle is, held in equilibrium at an angle ' q ' to a uniform, magnetic induction field BH by applying a force, F at a distance 'r' from the pivot along a direction, perpendicular to the field, then, (or), , BH, , B, M1, N1, , F cos q MBH sin q, , N, , , , , , S2, , r, , , , M2, , N2, , S, , S1, , Two uniform magnetic fields of strengths B1 and, B2 acting at an angle 750 with each other in, horizontal plane are applied on a magnetic, needle of moment M, which is free to move in, the horizontal plane. If the needle gets aligned, B, at an angle 300 with B1, then the ratio 1 is, B2, In equilibrium position,, , Fr cos q MBH sinq ;, , , , MBH tan q, r, , r, c) In the above case, if the force is applied at, one end which is always perpendicular to length, of the magnetic needle, then, BH, N, , M, , , , 750, 300, , NARAYANA GROUP, , F, , 2m BH tan q, , B2, , t1 t 2 ;, , F, , l, B1, , i.e., MB1 sin 300, , S, , l, , F, , t tension t BH, , 97
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , BH, N, , moment M2 is rotated through 300. Then find, the ratio of M1 & M2 ., Sol: C a q MBsin q, For first magnet, C 180 45 M 1 B sin 450, ----(1), For second magnet, C 180 30 M 2 B sin 300 ----(2), Diving equation (1) by equation (2), , , , M, 9, 135 M 1, , 2 1 , M 2 10 2, 150 M 2, , r, , Fr sin 900 MBH sin q, , F, l, , S, , 2mBH sin q, r, A magnet of moment 'M' is suspended in the, magnetic meridian with an untwisted wire. The, upper end of the wire is rotated through an angle, ' a ' to deflect the magnet by an angle ' q ' from, magnetic meridian. Then deflecting couple acting, on the magnet = MBH sin q, F, , , , MBH sin q, r, , ;, , , , , , WE - 3 : A magnetic dipole is under the influence, of two magnetic fields. The angle between the, two field directions is 600 and one of the fields, has a magnitudeof 1.2 × 10–2T. If the dipole, comes to stable equilibrium at an angle of 150, with this field, what is the magnitude of the, other field?, Sol. Here B1 = 1.2 × 10–2 T Inclination of dipole, with B1 is 1 = 150 Therefore, inclination of, dipole with B 2 is 2 600 150 450 As the, dipole is in equilibrium, therefore the torque on, the dipole due to the two fields are equal and, opposite. If M is magnetic dipole moment of the, dipole, then, , N, , N, , S, , B1, , 1 =150, , i.e., F 2m BH sin q ; F 2mBH sin q, d) In the above case, if the force applied is, always perpendicular to length of the magnetic, needle but at a distance 'r' from the pivot, then, , 0, , 45, =, 2, B2, , S, , Restoring couple developed in suspension wire, = Ca q where C is couple per unit twist of, suspension wire., In equilibrium position,, , MBH sin q Ca q, W.E-1 : When a bar magnet is placed at 90 o to, a uniform magnetic field, it is acted upon by, a couple which is maximum. For the couple, to be half of the maximum value, at what, angle should the magnet be inclined to the, magnetic field (B) ?, Sol: We know that, = MB sin , If = 90° then max = MB ..... (1), max, MBsin , 2, , ..... (2), , 60, , MB1 sin 1 MB2 sin 2, , 98, , or, , B2 , , B1 sin q1, 1.2102 sin150, , sin q2, sin 450, , ;, , 2, , 1.210 0.2588 4.39103 T, ;, , 0.707, , W.E-4: A compass needle of magnetic moment, 60A-m2, pointing towards geographical north, at a certain place where the horizontal, component of earth’s magnetic field is, 40 wb/m 2 experiences a torque of 1.2 × 10–3, Nm. Find the declination at that place., Sol. If is the declination of the place, then the, torque acting on the needle is M BH sin , , From equations (1) and (2), 2 = sin1 or sin 1 or 30, 2, W.E-2 : A bar magnet of magnetic moment M1 is, suspended by a wire in a magnetic field. The, upper end of the wire is rotated through 1800,, then the magnet rotated through 450. Under, similar conditions another magnet of magnetic, , 0, , sin , , , 1.2 10 3, 1, , q 300, , 6, M BH 60 40 10, 2, , Work done in rotating a magnetic dipole in a, magnetic field, , , The work done in deflecting a magnet from, angular position 1 to an angular position 2, wit h the field is change in PE given as, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , , , MAGNETISM, , W.E - 7 : A bar magnet has a magnetic moment, W MB cos 1 cos 2 , 2.5 J T–1 and is placed in a magnetic field of, The work done in deflecting a bar magnet through, 0.2 T. Calculate the work done in turning the, an angle from its state of equilibrium position in, magnet from parallel to antiparallel position, a uniform magnetic field is given by, relative to field direction., 0, Sol., Work, done in changing the orientation of a dipole, W MB1 cos here 1 0 , 2 , of moment M in a field B from position q1to q2 is, When it is released, this workdone converts into, rotational KE, , 1, MB(1-Cos )= I 2, 2, When a bar magnet is held at an angle with, the magnetic field, the potential energy, possessed by the magnet is U = –MB cos , When the bar magnet is parallel to the applied, field, then = 00 and potential energy is (-MB).It, is said to be stable equilibrium., When the bar magnet is perpendicular to the, applied field, then = 900 and potential energy, is zero, When the bar magnet is anti-parallel to the, applied field, then = 1800 and potential energy, is maximum i.e. U = +MB.It is said to be unstable, equilibrium., W.E-5: A magnet is suspended at an angle 600 in an, external magnetic field of 5 × 10–4 T. What is the, work done by the magnetic field in bringing it in, its direction ? [The magnetic moment = 20 Am2], Sol. Work done by the magnetic field,, W=MB cos1 cos 2 Here 1 600 and q2 00, , q1 00 and q2 1800, So, W 2 MB 2 2.5 0.2 1 J, W.E- 8 : A bar magnet with poles 25 cm apart and, pole-strength 14.4 A-m rests with its centre on, a frictionless pivot. It is held in equilibrium at, 600 to a uniform magnetic field of induction, 0.25 T by applying a force F at right angles to, its axis, 10cm from its pivot. Calculate F. What, will happen if the force is removed?, Sol. The situation is shown in figure. In equilibrium, the torque on M due to B is balanced by torque, , due to F, i.e., i.e., M B r F, Here,, , mB, B, M, , 10, , MB sin Fr sin 900 or, , F, , , , , , 10 10, , r, , = 7.8 N, , Axial line: The magnetic induction at a point, 0 , , 2Md, , on the axial line is Ba = 4 d 2 l 2 2, , , For a short bar magnet i.e. l d, , q, , 2, So, W MB (1 cos q ) 2 MB sin, 2, , q, 2, , 0, So, W cot 2 , i.e C W cot 30 3W, , 3/2, , Field of a Bar Magnet, , q1 0 and q 2 6 0 0, , q, 2, , , , 2, , 0, , and , C MB sin q 2 MB sin cos, , m 2l B sin , , , , If the force F is removed, the torque M B, will become unbalanced and under its action the, magnet will execute oscillatory motion about the, direction of B on its pivot O which will not be, simple harmonic as sin , , W MB cos q1 cos q2 and C = MB sin q, , NARAYANA GROUP, , F, , 14.4 25 102 0.25, , magnetic field requires W units of work to, turn it through 60 0 . What is the torque, needed to m aintain the needle in this, positon?, Sol. In case of a dipole in a magnetic field,, , q , , F, , ( as M = m x 2l) ; So substituting the given data,, , W.E-6 : A magnetic needle lying parallel to a, , C, , 0, , mB, , 1 , 3, 2 1 5 10 J ., , , , Here,, , 60, , 0, , W 20 5104 cos60 0 cos 0 0 , 2, , W MB cos q1 cos q2 , , given by, , 0 2M, , 3, 4 d, , then Ba = , , , The direction of magnetic induction on the axial, line is along the direction of magnetic moment., 99
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , each other, , And as U = MB, the interaction energy of the, , , system (i.e., P.E. of M 2 in the field of M1 or, , , P.E. of M1 in the field of M 2 ) will be, , , m 2M1M2, U M 2 . B1 M1 . B2 0, 4p r 3, , , [as M 2 is parallel to B1 , i.e., q 00 ].........(3), , Now as F = - (dU/dr) so force on M1 due to, , , , M 2 or force on M 2 due to M1 will be, , Equatorial line: The magnetic induction at a, point on the equatorial line at a distance d from, , , M, , 0, the centre is Be 4 (d 2 l 2 )3/ 2, For a short bar magnet i.e. l d, , then Be =, , , 0 M, , 4 d 3, , The direction of magnetic induction on the, equatorial line is in the direction opposite to, magnetic moment., , At any point in the plane of axial and, equatorial lines:, , , , , , , M1, , S O1 N, , M2, F1, , F2, , B2, , M1, , S O2 N, , S O1 N, B1, , M2, F1, , F2, , B2, , N O2 S, B1, , r, , r, , Attraction, (A), , Repulsion, (B), , , Similarly, C1 M1 B2 0, , , [as M1 is parallel to B2 i.e., q 0 ] .......(2), i.e., the magnets will not exert any couple on, 100, , F1 F2 , , 2, , M (3cos 1), 0, B = 4 , , , d3, 0, = 0 for axial line ; = 900 for equatorial line, For a short bar magnet , at two equidistant, points , one on the axial and the other on, equitorial line Ba 2 Be, Force between two magnets :When one, magnet is placed in the field of another magnet, it usually experiences a couple or force o, r, both and has potential energy. Depending on the, orientation of the magnets relative to each other,, the following situations are discussed ., When magnets are along the line joining their, centres, If the opposite poles of two magnets face each, , other as shown in Fig.(A), the field due to M1, , at the position of M 2 , i.e., at O2, will be :, , m 2M, B1 0 3 1 with q 00, [ as O2 lies, 4p r, , on the axis of M1 ], , , , So couple on M 2 due to M1 , i.e., B1 is, , , [ as M 2 is, C2 M 2 B1 0, , parallel to B1 , i.e., q 0 ]........(1), , , , , , M1, , B2, , d m0 2M1M 2 , , , dr 4p r 3 , , m0 6M1M 2, 4p r 4, , ––––––(4), , From equation (4) it is clear that interaction, force between the magnets varies as (1/r4)., When magnets are perpendicular to the line, joining their centres, If the similar poles of two magnets face each, , other as shown in Fig. (A), the field due to M1, , at the position of M 2 , i.e., at O2 will be, N, O1, S, , F1, , F2, , r, Repulsion, (A), , N, O2, S, , M2, , B1, , M1, , B2, , N, O1, S, , F1, , F2, , r, , S, O2, N, , M2, , B1, , Attraction, (B), , m M, B1 0 31 with f 900 [as O2 lies on the, 4p r, , equatorial line of M1 ], , , , Now as B1 is antiparallel to M 2 and B2 to, , M1 , i.e., q 1800 , so, , , C2 M 2 B1 0 and C1 M1 B2 0, ........... (5), i.e., the magnets will not exert any couple on, each other, , And as U M.B , the interaction energy of, , , the system (i.e.,P.E. of M 2 in the field of M1, , , or P.E. of M1 in the field of M 2 ) will be, , , m M M, U M 2 .B1 M1.B2 0 1 3 2 ........... (6), 4p r, , NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , , [as M 2 is antiparallel to B1 , i.e., q 1800 ], , Now as F = -(dU/dr), so force on M1 due to, , , , M 2 or on M 2 due to M1 will be, m 3M1M 2, d m M M , F1 F2 0 1 3 2 0, .............. (7), , dr 4p r 4p, r4, From equation (7) it is clear that interaction, force varies as (1/r4)., , MAGNETISM, induction field due to the magnet is exactly same as, that of earth’s horizontal component. These points, are called null points. If the average distance of N1, and N2 from the centre of the magnet is ‘d’ then, Bmagnet = BH (horizontal component of earth’s, magnetic field), , , Superposition of Magnetic fields, Neutral points and their location :, In the combined field due to bar magnet and, horizontal component of earth’s magnetic field, (BH) :, Earth’s magnetic field is present every where, and its horizontal component extends from, south to north. When a magnet is placed any, where, its field gets superimposed over the, earth’s field, giving rise to resultant magnetic, field. In this resultant magnetic field, there are, certain points where the resultant magnetic, induction field becomes zero. At these points,, the horizontal component of earth’s magnetic, field exactly balances the field due to the, magnet. These points are called null points, or neutral points., “The points in the magnetic field where the, resultant magnetic induction field becomes, zero are called null points”., North pole of the magnet pointing towards, geographical north : When a magnet is placed, in the magnetic meridian, with its north pole, facing geographic north, the combined magnetic, field lines due to earth and the bar magnet are, as shown in the figure., , N2, , Magnetic lines of force when northpole, of the magnet pointing towards, geographic north Results:, , , , , , , Along the axial line, on both sides, the two fields, have same direction. The magnitude of resultant, magnetic field is the sum of the magnitudes of, two fields., As we deviate from axial line, the two fields, differ in direction., On the equatorial line, the direction of the two, fields are exactly opposite to each other., At N1 and N2 on the equatorial line, the magnetic, , NARAYANA GROUP, , , , , , 3/2, , For short magnet, , BH 4.16, , 0 M, BH 4.17, 4 d 3, , North pole of the magnet pointing, towards geographic south:, When a magnet is placed in the magnetic, meridian with its north pole facing geographic, south, the field lines of the resultant magnetic, field are shown in the figure., N1, N, W, , S, , E, , O, N, S, N2, , Magnetic lines of force when north pole, of the magnet pointing towards, geographic south Results:, , , , N1, , 0, M, 4 d 2 l 2, , , , , The directions of the two fields ( horizontal, component of earth’s magnetic field and the field, due to the magnet) are exactly opposite to each, other, on the axial line., As we deviate from the axial line, the two fields, differ in direction., The directions of the two fields at all points on, the equatorial line is the same., Along the axial line, the magnetic field due to, magnet decreases in magnitude on moving away, from the centre of the magnet. There will be, points N1 and N2 situated at equal distances from, the centre of the magnet where the fields are, exactly balanced by the earth’s horizontal, component field. These points are called null, points., B, , m0 2Md, BH, 4p d 2 2 2, , At null points,, (where BH is earth’s horizontal magnetic induction, 101
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, field), , distance from the pole of the magnet where the, , m0 2M, m, BH, neutral point is formed, then BH 40 d 2, 4p d3, If the horizontal component of earth’s magnetic, If the north pole is on the table, then the, field B H at the given place is known, the, neutral point is formed towards geographic south, magnetic moment (M) of the magnet can be, side of the pole., determined by locating the neutral points., iv) If the south pole is on the table, then the, Magnet placed perpendicular to the, neutral point is formed towards geographic north, magnetic meridian : When a bar magnet is, side of the pole., placed with its axial line perpendicular to the Note: A short bar magnet is kept along magnetic, magnetic meridian with its north pole facing east, meridian with its north pole pointing north. A, of earth, the resultant magnetic field is shown in, neutral point is formed at point 'P' at distance, the figure Along a line making an angle of, 'd' from the centre of the magnet then, For short magnet,, , tan1, , 2 with east - west line, there are two, , points (N1 and N2) where the resultant magnetic, induction field is zero. Thus N1 (on the N-W, line) and N2 ( on the S-E line) are the null points., At the null point,, m M, BH 0 3 1 3 cos 2 q Where Tanq 2, 4p d, , BH, n2, d, , , , , E, , S, , m0 M, 4p d3, 3, If a very long magnet is placed vertically with, its one pole on a horizontal wooden table (or), when an isolated magnetic pole is kept in the, earth’s magnetic field, then, , 1, , BH, , BH, , S, , Be, , Be11, , At a distance 'd' on equatorial line, net megnetic, induction B net = 0, 0 M, . BH, ie Be BH , 4 d3, At a distance d/2 from the centre of the magnet, on equatorial line, the net magnetic induction is, given by, , BH 2, , , , n1, d P 2d, , d, 2, , Be1, , Bnet B|e BH , , N2, , cos q , , S, , Be, , N1, , O, , BH, , N, , N, , W, , Ba, BH, , 0 M, BH 7BH, 4 d 3, , 2 , , At a distance '2d' on equatorial line, the net, magnetic induction is given by, , M, Bnet= BH B||e BH 0, 4 2d 3, BH 7BH, , 8, 8, At a distance 'd' on axial line of the bar magnet,, the net magnetic induction is given by, Bnet = Ba + BH = 2Be + BH = 2BH + BH = 3BH., If the axis of the bar magent is rotated through, 900 clockwise at the same position then the net, magnetic magnetic induction at the same point, 'P' is, , = BH , , , , N, , BH, , BH, BH, n, B, , N, BH, , S, , n, , , , B, , A single neutral point will be formed in the, combined field on the horizontal table., If ‘m’ is the polestrength and ‘d’ is the, 102, , Bnet=, , B2a B2H =, , 5 BH Ba 2Be 2BH , , NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, BH, , N, S1, , MAGNETISM, , N1, , Bnet, , P, , Ba, , , , S, , If the axis of the magnet is rotated through 1800 at, the same position, then net magnetic induction at, the same point 'P' is Bnet = Be + BH = 2BH, Note: A short bar magnet is kept along magnetic, meridian with its south pole pointing north. A, neutral point is formed at a point 'P' at a, distance 'd' from the centre of the magnet, then, at a distance 'd' on axial line of the bar magnet, net magnetic induction,, , , , Neutral points in the combined field due, to isolated magnetic poles :, , , BH, d n1 P, Ba, S, , BH, N, , Bnet = 0 i.e., Ba = BH, , , 0 2M, BH, ., 4 d 3, , d, on axial line of bar magnet,, 2, netmagnetic induction is given by, , At a distance, , Bnet B1a BH , , 0 2M, ., BH 7BH, 4 d 3, , 2 , , At a distance '2d' on axial line of the bar magnet,, net magnetic induction is given by, , 2M, Bnet BH B||a BH 0, 4 2d 3, BH 7B H, , 8, 8, At a distance 'd' on equatorial line of the bar, magnet, net magnetic induction is, B, Bnet Be BH a BH, 2, B, 3, H BH B H, 2, 2, If axis of the magnet is rotated through 900, clockwise at the same position, then net magnetic, induction at the same point 'P' is given by, BH , , , , , , NARAYANA GROUP, , When two like magnetic poles of pole strengths, m1 and m2 (m1 < m2) are separated by a distance, ‘d’, then neutral point is formed in between the, poles and on the line joining them. Let ‘x’ be, the distance of neutral point from weaker pole, of strength m1 ., , N, , X, , m1, , B2, , (dX), n, , N, , B1, , m2, , At neutral point, B1 = B2, m m, m, m2, 0 . 21 0 ., 4p x, 4p d x2, , BH, Be, , , , , 5, B, B , BH Be a H , , 2, 2, 2 , If axis of the magnet is rotated through 1800 ,then, magnetic induction at the point 'P' is, Bnet = Ba + BH = BH + BH = 2BH, , 2, Bnet Be2 BH, , , d, , on solving, we get x , , , , m2, 1, m1, , When two unlike magnetic poles of strengths, m1and m2 (m1 < m2) are separated by a distance, ‘d’, then neutral point is formed outside and on, the line passing through the poles. It always, lies closer to weaker pole., B1, n, , B2, , N, , X, , m1, , d, , S, m2, , At neutral point, B1 = B2, m m, m, m2, 0 . 21 0 ., 4p x, 4p d x 2, on solving, we get x , , d, m2, 1, m1, , Neutral points in the combined field due, to short bar magnets :, Two short bar magnets of magnetic moments, M1 and M2 (M1 < M2) are placed at a distance, ‘d’ between their centres with their magnetic, axes oriented as shown in the figure, Then two, neutral points are formed (i) in between and (ii), outside and on the line passing through centres, of the magnets. In either case, null point is always, 103
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, closer to magnet of weaker moment., B1, , M1, , B2, x, , n2, , S1, , N1, , B1, , M2, , B2, , x, , N2, , n1, , S2, , d, , B1, , B1, , S1, x, , N2, , , , x, , n2, , M 1 and M 2 M 1 M 2 are placed along the, same line with like poles facing each other and, ‘d’ is the distance between their centres, the, distance of null point from M 1 is, , n1, B2, , N1, , S2, M2, , B2, , M1, , Case i) : If the neutral point is formed in between the, magnets, then B1 = B2, m 2M, m, 2M2, 0 . 31 0 ., 4p x, 4p d x3, on solving, we get x , , x, , 1, , 3, , 1, , Case ii) : If the neutral point is formed outside the, combination, then, , , m0 2M1 m0 2M 2, ., ., , 4p x 3, 4p d x3, , on solving , we get x , , , d, , M , 2 , M , , 1, , 3, , 1, , m1 m2 are separated by a distance d, then, the distance of the neutral point from the first, pole m1 is, x, , d, m2, 1, m1, , for like poles , , , for unlike poles , , a) For like poles the neutral point is situated in, 104, , 1/ 3, , , , Time period of Suspended Magnet in, the Uniform Magnetic Field, Principle : When a bar magnet is suspended, freely in a uniform magnetic field and displaced, from its equilibrium, it starts executing angular, SHM., Time period of oscillation and frequency of, magnet is, , 1, , Note:No null points are obtained when unlike poles, of the magnets are placed closer to each other, When t wo or more magnetic fields are, superimposed in the same region, according to, the resultant magnetic field the space in the, region gets modified., The magnetic field of induction at any point is, the resultant of all the fields superimposed at, that point., Null Point (or) Neutral Point : The point at, which the resultant magnetic field is zero is, called null point., If two poles of pole strengths m1 and m2, , d, , M2 , , 1, M1 , a) + for null point formed between the magnets., b) for null point formed outside the magnets., c) When unlike poles face each other, null point, , d, M , 2 , M1 , , between the poles, b) For unlike poles the neutral point is situated, on line joining the poles. But not in between, them., c) In either case null point is always closer to, the weaker pole., If two short bar magnets of magentic moments, , T = 2, , I, MB H and, , n, , 1, 1, , 2, T, , MBH, I, , where M magnetic moment, BH Horizontal, component of earth magnetic induction and I, moment of inertia, I m, , (l 2 b 2 ), 12, , for a thin bar magnet I , , ml 2, 12, , , , , where m is mass, l is length and b is breadth, of the magnet., For small percentage changes in moment of, T, 1 I, 100 , 100, inertia, T, 2 I, As I increases , T increases, For small percentage changes in magnetic, , , , moment T 100 2 M 100, As M increases , T decreases, , , , T, , 1 M, , Comparision of magnetic moments :, , , If two magnets of moment M1 and M2 of same, dimensions and same mass are oscillating in the, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , same field separately, then, T1, M2, , T2, M1, , T 1, , M, , , , (Bar magnets of equal size), , , , , A magnet is oscillating in a magnetic field B, and its time period is Tsec. If another identical, magnet is placed over that magnet with similar, poles together, then time period remain , unchanged., ( I| 2I and M| = 2M,, T| 2p, , , , T22 M1 M 2, , T12 M1 M 2, , |, , I, 2I, I, 2p, 2p, T), |, MB, 2MB, MB, , A magnet is oscillating in a magnetic field B , and its time period is T sec. If another identical, magnet is placed over that magnet with unlike, poles together, then time period becomes, infinite. i.e., it does not oscillate., , , I, M| M M 0; T 2p, , , , , 0, , B, , , , , , The time period of a thin bar magnet is T. It is, cut into 'n' equal parts by cutting it normal to its, length. The time period of each piece when, oscillating in the same magnetic field will be, T| , , T, n, , , , , I 1 , , , , 2, m , , , , , , , I, M , n n, , 3 &M1, 12, n, n , , , , T 1 2p, , , , I1, T, , 1, M B n, , The time period of a thin bar magnet is T. It is, cut into 'n' equal parts by cutting it along its, length. The time period of each piece remains, unchanged, when oscillating in the same field., , m 2, , ( M| M & I| n I, n, 12, n, T| 2p, , , , I|, I/n, 2p, T), |, M, MB, B, n, , a) Two magnets of magnetic moments M1 and, M2 (M1>M2) are placed one over the other., If T1 is the time period when like poles touch, each other and T2 is the time peiod when unlike, poles touch each other, then, , NARAYANA GROUP, , M1 T22 T12, 2 2, M2 T2 T1, b) If n1 and n2 are the corresponding, M1 n12 n 22, 1, frequencies, then, M2, n1 n 22, When same bar magnet used in the vibration, magnetometer at two different places 1 and 2, then, , , , T a 1 , T22, , , 2, , B, , , H, BH2 T1, When two bar magnets of moments M1 and M2, are placed one over the other such that (i) like, poles together (ii) unlike poles together and (iii), their axes are perpendicular to each other. When, vibrated in the same magnetic field, the ratio of, their time periods respectively is, , BH1, , T 2p, T1 : T2 : T3 , , , , , , T a 1 , , M , , I, 1, Ta, MB, M, 1, 1, :, :, M1 M 2, M1 M 2, , 1, , , , M M 22, 2, 1, , , , If T0 is the time period of oscillation of the, experimental magnet oscillating in B H. An, external field B is applied due to a bar magnet, in addition to BH at the point where the first, magnet is oscillating. Then its new time period, is T., T0, Br, , Then T B where Br B BH, H, , , a) If B and BH are along the same direction,, Br B BH T T0, , , b) If B and BH are in opposite directions,, Br B BH T0 T, , c) If B and BH are in opposite directions, and, also if B BH t hen Br B BH 0, T (i.e., it does not oscillate), , , d) If B and BH are perpendicular to each other,,, Br B2 BH2 T T0, , Here B , , m0 2M, (if the point is on the axial line), 4p d3, 105
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, B, , , , m0 M, (if the point is on the equatorial line), 4p d 3, , W.E - 11 : A short magnet oscillates in a vibration, magnetometer with a time period of 0.1s, where the horizontal component of earth’s, magnetic field is 24T . An upward current of, 18A is established in the vertical wire placed, 20 cm east of the magnet. Find the new time, period ?, , e) If a straight wire carries current vertically up, or down placed on the east or west or north or, m i, south side, then B 0 (From ampere’s law, 2p r, in electro magnetism), If n1 and n2 are frequencies of oscillation of the Sol. T2 B1 Where B = B =24 x 10–6T, 1, H, T1, B2, bar magnet in uniform magnetic field when B, supports BH and when B opposes BH, i, and B2 BH B BH 0, 2 r, n1, B BH, 7, then n B B (let B>BH), 4, , , 10, , 18, 6, 2, , , , 24 10, , H, , B, n 2 n 22, 12, BH n1 n 22, , , , 1, , For a bar magnet , T B (or ) n BH, H, If B1 and B2 be the earth’s magnetic induction at, two different places having angles of dip, q1 and q2 then, T1, , T2, , n1, , BH 2, B H1, , , , BH 1, , B2 Cosq 2, B1Cosq1, , T2, , 0.1, , , , 6 106 T, , 2 0.2, , 24 10 6, 2, 6 10 6, , T2 0.2s, , W.E-12: A magnet is suspended so as to swing, horizontally makes 50 vibrations/min at a, place where dip is 300, and 40 vibrations / min, where dip is 450. Compare the earth’s total, fields at the two places., Sol. n BH, , , n1, , n2, , B1Cos 1, B2Cos 2, , ie, , B1Cosq1, , 50, , 40, , B1 Cos300, , B2 Cos 450, , B, , 25, , 25 B1, 3, 1, or n B B Cosq, , , , (or) B 8 6, 2, H2, 2, 2, 16 B2, 2, 2, W.E-9: Two bar magnets placed together in a WE - 13: When a short bar magnet is kept in tan, vibration magnetometer take 3 seconds for 1, A position on a deflection magnetometer, the, vibration. If one magnet is reversed, the, magnetic needle oscillates with a frequency’f’, combination takes 4 seconds for 1 vibration., and the deflection produced is 450. If the bar, Find the ratio of their magnetic moments., magnet is removed find the frequency of, Sol. Given that, T1 = 3s and T2 = 4s, osciullation of that needle ?, M1 T22 T12 4 2 32, 16 9 25, M, , , , , or 1 3.57, M 2 T22 T12 4 2 32, 16 9 7, M2, , WE - 10 : A bar magnet makes 40 oscillations per, , Sol. n B, , , , n1, , n2, , B1, B2, , Where B1 B2 BH2 BH tan 450 BH2, minute in a vibration magnetometer. An, identical magnet is demagnetised completely, 2 BH & B2 BH, and is placed over the magnet in the, magnetometer. Calculate the time taken for, n, 2 BH, n, f, 1 , 21/ 4 n2 1/14 1/ 4, 40 oscillations by this combination. Ingore, n2, BH, 2, 2, induced magnetism., W.E-14:, Two, bar, magnets, of, the, same, length and, Sol. In the first case, frequency of oscillation,, breadth but having magnetic moments M and, 1 MB, n, 2M are joined with like poles together and, 2p, I, suspended by a string. The time of oscillation, In the second case, frequency of oscillation,, of this assembly in a magnetic field of strength, n1, 1, T1, 1 MB, 1, B is 3 sec. What will be the period of, , 2, n , n, T, 2p 2I, 2, oscillation, if the polarity of one of the, 1, 1, (or) T 2T (or) 40T 2 40T, magnets is changed and the combination is, again made to oscillate in the same field ?, 1, (or) t 2t 2 minute = 1.414 minute, 106, , 2, , NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , Sol. As magnetic moment is a vector, so when magnets, are joined with like poles together, M1 M 2 M 3M , so, T 2, , I1 I 2 , 3MB, , ........... (1), , When the polarity of one of the magnets is, reversed, M2 = M ~2M = M;, so T ' 2, , I1 I 2 , , ............ (2), , MB, , Dividing Eq. (2) by (1),, T', 3, i.e., T ' , T, , 3 T 3, , 3 sec, , Magnetic Materials, , , , , , , , , , Curie and Faraday discovered that all the, materials in the universe are magnetic to some, ext ent. These magnetic substances are, categorized mainly into two groups., Weak magnet ic materials come under, diamagnetic and paramagnetic materials. Strong, magnetic materials are Ferro-magnetic materials., According to the modern electron theory of, magnetism, the magnetic response of any material, is due to the circulating electrons in the atoms., Each circulating charge constitutes a magnetic, moment in a direction perpendicular to the plane, of circulation., In magnetic material all these magnetic moments, due to the orbital and spin motion of all the, electrons in the atoms of the material, vectorially, add up to a resultant magnetic moment. The, magnitude and direction of this resultant, magnetic moment is responsible for the magnetic, behaviour of the material., Magnetic material are studied interms of the, following physical parameters, , Intensity of Magnetising field( H ) :, Any magnetic field in which a magnetic material, is placed for its magnetizat ion is called, magnetising field., In a magnetising field the ratio of magnetising, , field Bo to the permeability of free space is, called intensity of magnetising field, , B, , , o, In air . H , or Bo o H, , o, , In a medium H , NARAYANA GROUP, , B, , , The value of H is independent of medium ., Intensity of magnetising field is a vector in the, direction of magnetic field and has unit, Wb / m 2, V s, A, , , H /m, sm m, , Dimensions AL1, , 2) Intensity of magnetisation I : When a, , magnetic material is magnetised by placing it in, a magnetising field, the induced dipole-moment, per unit volume in the specimen is called, intensity of magnetisation., , M, , , but as M mLn and V = SL, i.e I , V, m, I n i.e , intensity of magnetisation is, S, numerically equal to the induced pole-strength, per unit area of cross -section.It is a vector, quantity having direction of magnetising field, or opposite to it as shown in figure. Its unit is, (A/m) and dimensions AL1 , , Magnetic Susceptibility m : The ratio of, magnitude of intensity of magnetisation to that, of magnetising field strength is called magnetic, I, susceptibility m , H, It is a scalar with no units and dimensions. It, physically represents the ease with which a, magnetic material can be magnetised.i.e large, value of m implies that the material is more, susceptible to the field and hence can be easily, magnetised., Magnetic permeability : When a, magnetic material is placed in a magnetising, field, the ratio of magnitude of total field inside, the material to that of intensity of magnetising, field is called magnetic permeability; i.e.,, B , i.e., B H, H, It measures the degree to which a magnetic, material can be penetrated by the magnetising, field or ability of the material to allow magnetic, lines of force. It is a scalar having unit Hm 1, and dimensions MLT 2 A2 ., , Relative permeability r :, It is the ratio of magnitudes of total field inside, the material to that of magnetising field or it is, the ratio of permeability of a medium to that of, free space., 107
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , W.E-19 : An iron bar of length 10 cm and diameter, , H, , B, , , B0 0 H 0, It has no units and dimensions, r , , 2 cm is placed in a magnetic field of intensity, 1000Am –1 with its length parallel to the, direction of the field. Determine the magnetic, Relation between relative permeability, moment produced in the bar if permeability, and susceptibility :, of its material is 6.3 × 10–4 TmA–1., , Sol. we know that, 0 1 , B, I, We know B m0 H I or, m0 1 , H, H, , 6.3 10 4, , 1 , 1 = 500.6, 0, 4 10 7, I, B, c], or, m m0 1 c [as m and, Intensity of magnetisation,, H, H, I H 500.6 1000 = 5 × 105 Am–1, m, 2, (or) m 1 c, mr 1 c, magnetic moment, M = I x V I pr , 0, This is the desired result., W.E-15 : A magnetising field of 1600Am -1, produces a magnetic flux of 2.4 × 10–5 weber, in a bar of iron of cross section 0.2 cm 2 . , Calculate permeability and susceptibility of, the bar., , 2.4 10 5, Sol: Magnetic induction, B A 0.2 104 = 1.2 Wb/m2, i) Permeability, , , B, 1.2, , 7.5 10 4 TA 1m, H 1600, , , , ii) As m m 0 1 c then, Susceptibility, c , , 4, m, 1 7.5 10 1, m0, 4 10 7, , = 596.1, , W.E-16 : The permeability of substance is, 6.28×10 –4 wb/A-m. Find its relative, , permebility and suscepibility ?, 6.28 10 4, , , Sol. 4 10 7 500, 0, 1 , , 1 500 1 499, , , , W.E-17 : The magnetic moment of a magnet of, , mass 75 gm is 9×10–7 A-m2. If the density of, the material of magnet is 7.5×103 kg m–3, then, find intensity of magnetisation is, , Sol. I , , , M, V, , mass m , , Where volume, V density , , M , 9 10 7 7.5 10 3, , 0.09 A / m, m, 75 10 3, , , , WE- 18 : A magnetic field strength (H) 3×103Am–, 1, , produces a magnetic field of induction (B) of, in an iron rod. Find the relative, permeability of iron ?, 12T, , B, 12, 4 103, Sol. , H 3 103, 4 10 3, r , , 10 4, 0 4 10 7, 108, , , , 5 10 5 3.14 10 2, , 2, , 10 10 =17.70 A-m2, 2, , Electron Theory of Magnetism, i) Molecular theory of magnetism was first given, by Weber and was later developed by Ewing., ii) Electron theory of magnetism was proposed, by Langevin., iii) The main reason for the magnetic property, of a magnet is spin motion of electron. Most of, the magnetic moment is produced due to electron, spin. The contribution of the orbital revolution, is very small., , A) Explanation of diamagnetism:, i) Since diamagnetic substance have paired, electrons, magnetic moments cancel each other, and there is no net magnetic moment., ii) When a diamagnetic substance is placed in, an external magnetic field each electron, experiences radial force F = Bev either inwards, or outwards. Due to this the angular velocity,, current, and magnetic moment of one electron, increases and of the other decreases. This results, in a non-zero magnetic moment in the substances, in a direction opposite to the field., iii) Since the orbital motion of electrons in atoms, is an universal phenomenon, diamagnetism is, present in all materials. Hence diamagnetism is, a universal property., , Properties of Dia-magnetic substances, , , The substances which when placed in a external, magnetic field acquire feeble magnetism, opposite to the direction of the magnetising field, are known as dia-magnetic substances., NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , Ex: Bismuth (Bi), Zinc (Zn), Copper (Cu), Silver (Ag),, Gold (Au). Salt (Nacl), Water (H2O), Mercury, (Hg), Hydrogen (H2O) etc., When a bar of dia-magnetic substance is, suspended freely between two magnetic poles, [see figure] , then the axis of the bar becomes, perpendicular to magnetic field., , , N, , , , , , N, , S, , S, , When a dia-magnetic substance is placed in a, non-uniform field, then it tends to move towards , the weaker part from the stronger part of the field, as shown in figure., , , N, , S, , a) Magnet closely, spaced, , , , N, , S, , b) pole pieces, moved apart, , , , Dia magnetic substances acquire feeble, magnetism in a direction oppoite to magnetising, field. The intensity of magnetisation I is very, small, negative and is directly proportional to , magnetising field H as shown in figure., , NARAYANA GROUP, , I, , The magnetic susceptibility, , c I / H , , is small, , x, , O, , T, , The relative permeability is less than unity, because m r 1 c and c is negative., The origin of diamagnetism is the induced, dipole moment due to change in orbital motion of, electrons in atoms by the applied field. Diamagnetism is shown only by those substances, which do not have any permanent magnetic, moment., , B) Explanation of Paramagnetism:, , S, , , , , , H, , O, , and negative(Because I is small and opposite in, direct ion to H). This is independent of, temperature as shown in figure., , When a dia-magnetic material is placed inside, a magnetic field, the magnetic field lines become, less dense in the material., If one limb of a narrow U-tube containing a , dia-magnetic liquid is placed between the poles, of an electromagnet, then on switching the field,, the liquid shows a depression. This is shown in , figure., , N, , I, , i) Paramagnetic materials have a permanent, magnetic moment in them. The moments arise, from both orbital motion of electrons and the, spinning of electrons in certain axis., ii) In atoms whose inner shells are not completely, filled, there is a net moment in them since more, number of electrons spin in the same direction., This permanent magnet behaves like a tiny bar, magnet called atomic magnet., iii) In absence of external magnetic field atomic, magnets are randomly oriented due to the thermal, agitation and the net magnetic moment of the, substance is zero., iv) When it is placed in an external magnetic, field the atomic magnets align in the direction of, the field and thermal agitation oppose them to, do so., v) At low fields the total magnetic moment would, be directly proportional to the magnetic field B, 109
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, and inversely proportional to temperature T., , Properties of Paramagnetic substances, N, S, N, S, The substances which when placed in a magnetic, field, acquire feeble magnetism in the direction, fig(a), fig(b), of magnetising field are known as paramagnetic, are near to each other. In figure (b), the, substances., distance, between the poles is, Ex: Aluminium (Al), Platinum (Pt), Manganese (Mn),, increased. i.e., the field is stronger near the, Copper chloride (CuCl2), Oxygen (O2), solutions, poles., of salts of iron etc. are examples of paramagnetic, (vi) The intensity of magnetisation I is very small, substances., and compared to one. It follows from the relation, (ii) When a bar of paramagnetic substance is placed, mr 1 cm . The variation of mr or c with H, in a magnetic field, it tries to concentrate the, is shown in figure. As is clear from the figure,, lines of force into it as shown in figure, the variation is non-linear. The large value of, , (i), , s, , N, , n, , S, , This shows that the magnetic induction B in it is, numerically slightly greater than the applied field, H. So the permeability m is greater than one, , mr is due to the fact that the field B inside the, material is much stronger than the magnetising, field due to ‘pulling in’ of a large number of, lines of force by the material., Y, , because m B / H ., (iii) When the bar of paramagnetic material is, suspended freely between two magnetic poles,, its axis becomes parallel to magnetic field., Move over, the poles produced at the ends of, the bar are opposite to nearer magnetic poles., (iv) If a paramagnetic solution is poured in a U-tube, and if one limb is placed between the poles of, an electromagnet in such a way that liquid level, is parallel to field, then on switching the field,, the liquid rises. This is shown in figure., , N, , S, , (v) In a non-uniform magnetic field, the paramagnetic, substances are attracted towards the stronger, parts of the magnetic field from the weaker parts, of the field. The situation is shown in figure. In, figure (a), the field is stronger in the middle as, the poles, 110, , r or x, O, , H, , X, , Curie’s law : Curie law states that far away, from saturation, the suceptibility c (I/H) of, paramagnetic substance is inversely proportional, to absolute temperature, i.e.,, c, , 1, T, , or, , c, , C, T, , where C is constant and is called as Curie, constant., When a ferromagnetic material is heated, it, becomes paramagnetic at a certain temperature., This temperature is called as Curie temperature, and is denoted by TC. After this temperature, the, susceptibility varies with temperature as, c, , C|, T Tc , , where C| is another constant. For iron, Tc = 1043, K. = 7700 C, , NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , Properties of Dia, Para and Ferror Magnetic materials, DIA, , PARA, , FERRO, , 1. They are feebly repelled, by a magnet., , 1. They are feebly attracted, by a magnet, , 1. They are strongly attracted, a magnet, , 2. The net magnet moment, due to all the electrons in, the atom is zero, , 2. The net magnetic moment, atoms due to all electrons, is not zero., , 2. The net magnetic moment, in atoms is very strong., , 3. When subjected to the, magnetising field they are, feebly magnetised in, opposite direction to the, magnetising field, , 3. Magnetised feebly in the, direction of magnetising, field., , 3. Magnetized strongly in the, direction of magnetising, field., , 4. When suspended inside the 4. They align with their, magnetic field, they align, length along the direction, their length perpendicular, of magnetic field., to the magnetic field., , 4. They align with their, length along the direction, of magnetic field., , 5. Magnetic lines of force, prefer to move out of the, specimen., , 5. Few lines pass through the 5. Almost all lines prefer to, specimen., move through the, specimen., , 6. They move from stronger, part of the magnetic field, to the weaker part of the, magnetic field, , 6. They move from weaker to 6. They move from weaker to, stronger part of the, stronger part of the, magnetic field, magnetic field., , 7. r < 1, , 7. r > 1, , 7. r >>> 1, , 8. Intensity of magnetization, (I) is small and negative., , 8. I is small and positive, , 8. I is high and positive, , 9. m is small and negative, , 9. m is small and positive, , 9. m is highly positive, , 10. m is independent of, temperature., , 10. m is dependent on, temperature., , 10. m is dependent on, temperature, , 11. Doesn’t obey Curie law., , 11. Obey Curie law, , 11. Obey Curie law and at, Curie temperature they are, turned to paramagnetic, materials., , 12. Substances following, Diamagnetism are, Bismuth, Copper, lead,, silicon, water, glass etc., , 12. Substances following, paramagnetism are, Aluminum, Platinum,, Manganese, Chromium,, Calcium, Oxygen,, Nitrogen (at STP), , 12. Substances following, ferromagnetism are Iron,, Cobalt, Nickel and alloys, like alnico, , NARAYANA GROUP, , 111
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, Hysteresis: When a bar of ferromagnetic material is, magnetized by a varying magnetic field H and, the intensity of magnetization I induced is, measured. The graph of I versus H is as shown, is figure., , Elements of Earth’s Magnetism, (Terristrial Magnetism ): There are three, , , I, B, C, , , , , , , , Hysterisis, curve, , , , 112, , resolved into a horizontal component BH and, vertical component BV at any place., , When magnetising field is increased from O the, intensity of magnetisation I increases and, becomes maximum i.e at point (A)., This, maximum value is called the saturation value., When H is reduced, I reduces but is not zero, when H = 0. The remainder value OB of, magnetisation when H = 0 is called the residual, magnetism or retentivity. OB is retentivity., When magnetic field H is reversed, I reduces , and becomes zero i.e., for H = OC, I = 0. This, value of H is called the coercivity., , When field H is further increased in reverse, direction, the intensity of magnetisation attains, saturation value in reverse direction (i.e., point, D). When H is decreased to zero and changed, direction in steps, we get the part DFGA, , Properties of soft iron and steel: For soft, iron, the susceptibility, permeability and, retentivity are greater while coercivity and, hysteresis loss per cycle are smaller than those, of steel., I, , , , H, , G, F, , D, , , , field Be in the magnetic meridian may be, , A, , O, , I, H, , Soft Magnetic, Material, , elements of earth’s magnetism, (i) Angle of declination (ii) Angle of dip, (iii) Horizontal component of earth’s field., Earth’s Magnetic Field: The earth’s magnetic, , Hysterisis, curve, , H, Hard Magnetic, Material, , Permanent magnets are made of steel and cobalt, while electromagnets are made of soft iron., Diamagnetism is universal. It is present in all, materials. But it is, weak and hard to detect, if substance is para or ferromagnetic, Shielding from magnetic fields: For, shielding a certain region of space from magnetic, field, we surround the region by soft iron rings., Magnetic field lines will be drawn into the rings, and the space enclosed will be free of magnetic, field., , Geographical, meridian, , Geographical, L, north, P, , Magnetic, north, , , S, , N, , Magnetic, meridian, , M, , O, , , , BH, , BV, , BP, , R, , =dip (or) inclination = declination, Horizontal component of earth’s magnetic field, , ....... 1, , BH Be cos , , Vertical component of earth’s magnetic field, , ....... 2 , , BV Be sin , , Be , , B, , 2, H, , BV2, , , , Dividing equation (2) by equation (1), we have, BV Be sin , , tan , BH Be cos , , Geographical Meridian: A vertical plane, passing through the axis of rotation of the earth, is called the geographic meridian., Magnetic Meridian: A vertical plane passing, through the axis of a freely suspended magnet is, called the magnetic meridian., , Angle of Declination : The acute angle, , , , , , , bet ween the magnetic meridian and t he, geographical meridian is called the ‘angle of, declination’ at any place., The value of declination at equator is 17 0, Declination varies from place to place, The lines joining the places of equal declination, are called isogonal lines., The lines joining the places of zero declination, are called agonic lines., NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , Thus, one can get the true dip d without locating, the magnetic meridian., with its centre coinciding with the centre of, More about angle of dip ( ) :, earth, show that the angle of dip is related, to magnetic latitude through the relation (i) At a place on poles, earth’s magnetic field is, perpendicular to the surface of earth, i.e.,, tan 2 tan , Sol. Considering the situation for dipole, at position, 900, , W.E - 20 Considering the earth as a short magnet, , r, we have, Br, , B, , S, , r, , , , Bv Bsin 90 0 B, Further, BH = B cos 900 = 0, So, except at poles, the earth has a horizontal, component of magnetic induction field., (ii) At a place on equator, earth’s magnetic field, , N, , is parallel to the surface of earth, i.e., 00, , BH Bcos 00 B, Further BV = BH sin 00 = 0, So, except at equator, the earth has a vertical, component of magnetic induction field., (iii) In a vertical plane at an angle q to magnetic, meridian, tan 2 cot ; But From figure 900 , B'H BH cos q and B'V BV, So, tan 2 cot 900 ; i.e., tan 2 tan , So, the angle of dip d in a vertical plane making, Apparent Dip: If the dip circle is not kept in, an angle q to magnetic meridian is given by, the magnetic meridian, the needle will not show, B|, BV, the correct direction of earth’s magnetic field., d ' |V , tan, The angle made by the needle with the horizontal, BH BH cos q, is called the apparent dip for this plane. If the, , tan d BV, dip circle is at an angle q to the meridian, the, or tan d ' cos q B tan d , , , H, effective horizontal component in this place is, (a) For a vertical plane other than magnetic, BH BH cos q . The vertical component is still, meridian, q 00 and cos q 1, i.e., d1 d, Bv . If d1 is the apparent dip and d is the true, ( angle of dip increases ), dip, we have, (b) For a plane perpendicular to magnetic, B, BV, meridian, q 900, tan d1 v , BH BH cos q, tan d, tan d1 , or d1 900, , , cos 90, tan d tan d BV , , or tan d1 , This shows that in a plane perpendicular to, BH ..... (1), cos q , magenetic meridian, the dip needle will become, Now suppose, the dip circle is rotated through, vertical., an angle of 900 from this position. It will now, , 2M cos , M sin , Br 0, and B 0, 3, 4, r, 4 r 3, BV, Br, and as tan B B , so in the light of Eq. (1), H, , , make an angle 90 q with the meridian. The, effective horizontal component in this plane is, BH BH sin q . if d 2 be the apparent dip, we, 1., shall have, B, BV, tan d, tan d2 V'' , tan d2 , .........(2), BH, BH sin q or, sin q, From (1) and (2) cot 2 d1 cot 2 d2 cot 2 d, NARAYANA GROUP, , C.U.Q, MAGNETIC MOMENT AND, RESULTANT MAGNETIC MOMENT, The dimensional formula for magnetic, moment is, 1) M0L2T0A1, 2) M0L1T0A2, 3) M0L2T0A2, 4) M0L0T1A1, 113
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 2., , 3., , 4., , 5., , 6., , 7., , 8., , If two bar magnets of different magnetic, lengths have equal moments, then the pole, strength is, 1) equal for both the magnets, 2) less for shorter magnet, 3) more for longer magnet, 4) more for shorter magnet, A bar magnet of moment M is bent into arc,, its moment, 1) decreases, 2) increases, 3) does not change, 4) may change, A bar magnet is cut into two equal halves by, a plane parallel to the magnetic axis of the, following physical quantities the one which, remains unchanged is, 1) pole strength, 2) magnetic moment, 3) intensity of magnetisation 4) moment of inertia, Two magnets of magnetic moments of, M1, M2 are placed one over the other with like, poles touching, the resultant magnetic, moment is, 1) M1 + M2, 2) M1 - M2, , MAGNETIC FIELD, 9., , 10., , 11., , 12., , 4) M 12 M 22, 3) M 12 M 22, 13., A bar Magnet consists of, 1) two poles of different nature and different, strength, 2) equal poles in magnitude, 3) equal and opposite magnetic poles, 4) opposite poles, A small hole is made at the centre of the, magnet then its magnetic moment, 1) decreases, 2) increases, 14., 3) remains same, 4) depends on the nature of the magnetic material, A magnetised wire of magnetic length ‘ 2l ’,, pole strength ‘m’ and magnetic moment ‘M’, is bent at angle is ' ' radian at the centre of, the circle, then, 15., 1) Its pole strength remains same, 2) Its length decreases and becomes, , , 4l sin 2 , , , , , , , , , 3) Its new magnetic moment becomes, , , 2 M sin 2 , , , , , , , , 114, , 16., 4) All the above are correct, , S.I. unit of Magnetic flux is, 1) ampere-meter, 2) amp. m2, 3) weber, 4) weber/m2, The Source of magnetic field is, 1) isolated Magnetic pole, 2) static electric charge, 3) current loop, 4) moving light source, The earth’s magnetic field, 1) varies in direction but not in magnitude, 2) varies in magnitude but not in direction, 3) varies in both magnitude and direction, 4) is centred exactly about the centre of the earth, The electric and magnetic field lines differ in, that, 1) electric lines of force are closed curves while, magnetic field lines are not, 2) magnetic field lines are closed while electric, lines are not, 3) electric lines of force can give direction of, the electric field while magnetic lines can not, 4) magnetic lines can give direction, of magnetic field while electric lines can not., The incorrect statement regarding the lines, of force of the magnetic field B is, 1) Magnetic intensity is a measure of lines of, force passing through unit area held normal to it, 2) Magnetic lines of force form a closed curve, 3) Inside a magnet, its magnetic lines of force, move from north pole of a magnet towards its, south pole, 4) Magnetic lines of force never cut each other, Two bar magnets are placed on a piece of cork, which floats on water. The magnets are so, placed that their axis are mutually, perpendicular. Then the cork, 1) rotates, 2) moves a side, 3) oscillates, 4) neither rotates nor oscillates, When a bar magnet of magnetic moment M, is placed in a magnetic field of induction field, strength B , each pole experiences a force of, F then the distance between the South and, North pole of the magnet measured inside it, is, MB, F, FB, 1) MBF 2), 3), 4), F, MB, M, Lines of force due to earth’s horizontal, magnetic field are, 1) parallel and straight 2) elliptical, 3) concentric circles 4) curved lines, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , 25. A magnet is kept fixed with its length parallel, 17. Magnetic lines of force are, to the magnetic meridian. An identical magnet, 1) continuous, 2) discontinuous, is parallel to this such that its center lies on, 3) some times continuous and some times, perpendicular bisector of both. If the second, discontinuous, 4) nothing can be said, magnet is free to move, it will have, 18. In case of a bar magnet, lines of magnetic, 1) translatory motion only, induction, 2) rotational motion only, 1) start from the north pole and end at the south, 3) both translatory and rotational motion, pole., 4) vibrational motion only, 2) run continuously through the bar magnet and 26. There is no. couple acting when two bar, outside., magnets are placed co-axially separated by a, distance because, 3) emerge in circular paths from the middle of, 1) there are no forces on the poles., the bar, 2) the forces are parallel and their lines of action, 4) are produced only at the north pole like rays, do not coincide, of light from a bulb, 3) the forces are perpendicular to each other, 19. The total number of magnetic lines of force, 4) the forces act along the same line, originating or terminating on a pole of strength, 27. Find the wrong statement among the following., ‘m’ is, Two unlike isolated magnetic poles are at, m, some distance apart in air., 0 m, 2) , 3) m 2, 4) 0 m, 1), 1) the resultant induction at a point beween the, 4, 0, poles is B1 B2 on the line joining them, COUPLE ACTING ON THE BAR MAGNET, 2) The resultant induction is B1 B2 at any point, 20. A magnetic needle is kept in a non uniform, out side the poles on the line joining them, magnetic field. It experiences, 3) No neutral point is formed on the line joining, 1) a force and a torque, them if the pole strengths are equal., 2) a force but not a torque, 4) A neutral point is formed in between the poles, 3) torque but not a force, and nearer to weak pole on the line joining them., 4) neither a torque nor a force, 28. A magnetic field is produced and directed, 21. A magnetic field is produced and directed, along y-axis. A magnet is placed along y-axis., along y-axis. A magnet is placed along x-axis, The direction of torque on the magnet is, .The direction of the torque on the magnet is, 1) in the x-y plane, 2) along y-axis, 1) in the x-y plane, 2) along z-axis, 3) along z-axis, 4) Torque will be zero, 3) along y-axis, 4) torque will be zero, FIELD OF A BAR MAGNET, 22. A bar magnet of moment M is in a magnetic 29. The magnetic intensities at points lying at the, same distance from the magnetic pole are, field of induction B . Then the couple is, 1) same both in magnitude and direction, 1) M x B, 2) B x M, 2) same in magnitude and different in direction, 3) different in magnitude but same in direction, 3) M . B, 4) B . M, 4) different both in magnitude and direction, 23. If a bar magnet of moment is suspended in a, SUPERPOSITION OF MAGNETIC FIELDS, uniform magnetic field B it is given an angular, 30. When N-pole of the given bar magnet is placed, deflection, w.r.t equilibrium position. Then the, on a table pointing geographic north, the null, restoring torque on the magnet is, points are formed due to the superposition of, 1) MB sin , 2) M B cos , the magnetic field of the bar magnet and the, 2, earth’s magnetic field. The two null points are, 3) MB tan , 4) MB sin , located, 24. The effect due to uniform magnetic field on a, 1) on the axial line at equidistant on either sides, freely suspended magnetic needle is as follows, 2) on the equitorial line at equidistant on either, 1) both torque and net force are present, sides, 2) torque is present but no net force, 3) on the axial line only on one side of the magnet, 3) both torque and net force are absent, 4) on the equitorial line only on one side of the, 4) net force is present but no torque, magnet, NARAYANA GROUP, , 115
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MAGNETISM, 31. When S-pole of the given bar magnet is placed, on a table pointing geographical N-pole, 1) two null points are located on the axial line, at equidistant on either sides, 2) two null points are located on the equitorial, line at equidistant on either sides, 3) two null points are located on the axial line, only on one side of the magnet, 4) two null points are located on the equitorial, line only on one side of the magnet, 32. A very long magnet is held vertically with its, south pole on a table. A single neutral point is, located on the table to the, 1) East of the magnet 2) North of the magnet, 3) West of the magnet 4) South of the magnet, 33. The null points are on the axial line of a bar, Magnet when it is placed such that its south, pole points, 1) South 2) East, 3) North 4) West, 34. The null point on the equatorial line of a bar, magnet when the north pole of the magnet is, pointing, 1) North 2) South, 3) East, 4) West, 35. When the N - pole of a bar magnet points, towards the south and S- pole towards the, north, the null points are on the, 1) magnetic axis, 2) magnetic centre, 3) perpendicular division of magnetic axis, 4) N and S pole, TIME PERIOD OF SUSPENDED MAGNET IN, THE UNIFORM MAGNETIC FIELD, 36. The restoring couple for a magnet oscillating, in the uniform magnetic field is provided by, 1) horizontal component of earth’s magnetic field, 2) gravity, 3) torsion in the suspended thread, 4) magnetic field of magnet, 37. Vibration of suspended magnet works on the, principle of, 1) torque acting on the bar magnet and rotational, inertia, 2) force acting on the bar magnet and rotational, inertia, 3) both the force and torque acting on the bar, magnet, 4) neither force nor torque, 38. The factors on which the period of oscillation, of a bar magnet in uniform magnetic field, depend, 1) nature of suspension fibre, 2) length of the suspension fibre, 3) vertical component of earth’s magnetic, induction, 4) moment of inertia of the magnet, 116, , JEE-ADV PHYSICS- VOL- III, 39. The time period of a freely suspended, magnetic needle does not depend upon, 1) length of the magnet 2) pole strength, 3) horizontal component of earth’s magnetic field, 4) length of the suspension fibre, 40. A magnetic needle suspended by a silk thread, is oscillating in the earth’s magnetic field. If, the temperature of the needle is increased by, 5000C, then, 1) the time period decreases, 2) the time period increases, 3) the time period remain unchanged, 4) the needle stops vibrating, , TYPES OF MAGNETIC MATERIALS, 41. The following instrument i.e. used to measure, magnetic field, 1) Thermometer, 2) Pyrometer, 3) Hygrometer, 4) Fluxmeter, 42. A watch glass containing some powdered, substance is placed between the pole pieces, of a magnet. Deep concavity is observed at, the centre. The substance in the watch glass, is (assume poles are far), 1) iron 2) chromium 3) carbon 4) wood, 43. Permanent magnets are made from, 1) diamagnetic substances, 2) paramagnetic substance, 3) ferromagnetic substances, 4) wood, 44. Out of dia, para and ferromagnetism, the, universal property of all substances is, 1) diamagnetism, 2) paramagnetism, 3) ferromagnetism, 4) antiferromagnetism, 45. The following one is a diamagnetic, 1) Liquid oxygen, 2) Air, 3) Water, 4) Copper sulphate, 46. The following one is para-magnetic, 1) Bismuth, 2) Antimony, 3) Water, 4) Chromium, 47. Ferromagnetic ore properties are due to, 1) filled inner sub-shells, 2) vacant inner sub-shells, 3) partially filled inner sub-shells, 4) all the sub-shells equally filled, 48. The major contribution of magnetism in, substances is due to, 1) orbital motion of electrons, 2) spin motion of electrons, 3) equally due to orbital and spin motions of, electrons, 4) hidden magnets, 49. If the magnetic moment of the atoms of a, substances is zero, the substance is called, 1) diamagnetic, 2) ferromagnetic, 3) paramagnetic, 4) antiferromagnetic, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , 50. A uniform magnetic field exists in certain space 55. Curie temperature is the temperature above, in the plane of the paper and initially it is, which, directed from left to right. When a rod of soft, 1) a paramagnetic material becomes ferro, iron is placed parallel to the field-direction, the, magnetic, magnetic lines of force passing within the rod, 2) a ferromagnetic material becomes, will be represented by figure, paramagnetic, 3) a paramagnetic material becomes diamagnetic, 4) a ferromagnetic material becomes diamagnetic, 1), 2), 56. For a paramagnetic material, the dependence, of the magnetic susceptibility on the, absolute temperature T is given by, 2) constant T, 1) T, 3), 4), 1, 3) , 4) constant, T, 51. A rod of a paramagnetic substance is placed, in a non-uniform magnetic field. Which of the 57. The area enclosed by a hysteresis loop is a, measure of, following figure shows its alignment in the field, 1) retentivity, 2) susceptibility, ?, 3) permeability, 4) energy loss per cycle, 58. A material produces a magnetic field which, N, S, N, S, helps the applied magnetic field, then it is, 2), 1), 1) diamagnetic, 2) paramagnetic, 3) electro magnetic 4) all the above, 59. A material produces a magnetic field which, N, S, N, S, oppose the applied magnetic field, then it is, 3), 4), 1) diamagnetic, 2) para magnetic, 3) electro magnetic 4) ferro magnetic, 52. The relative permeability of silicon is 0.99837, and that of palladium is 1.00692, choose the 60. The permeability of a material is 0.9. The, correct options of the following, material is, 1) silicon is paramagnetic and palladium is, 1) diamagnetic, 2) para magnetic, ferromagnetic, 3) ferro magnetic, 4) non-magnetic, 2) silicon is ferromagnetic and palladium is 61. The susceptibility of a diamagnetic substance, paramagnetic, is, 3) silicon is diamagnetic and palladium is, 1) , 2) zero, paramagnetic, 3) small but negative 4) small but positive, 4) Both are paramagnetic, 62. Liquids and gases never exhibit, 53. The relative permeability is represented by, 1) diamagnetic properties, r and susceptibility is denoted by for a, 2) para magnetic properties, magnetic substance then for a paramagnetic, 3) ferro magnetic properties, substance., 4) electro magnetic properties, 1) r 1, 0, 2) r 1, 0, 63. Alnico is used for making permanent magnets, because it has, 3) r 1, 0, 4) r 1, 0, 1) High coercivity and high retentivity, 54. Two like poles of strengths m1 and m 2 aree, 2) high coercivity and low retentivity, at far distance apart. The energy required to, 3) low coercivity and low retentivity, bring them r0 distance apart is, 4) low coercivity and high retentivity, 64. A mariners compass is used, 0 m1 m 2, 0 m1 m 2, 1) 4 r, 2) 8 r, 1) to compare magnetic moments, 0, 0, 2) for determination of H, 0 m1 m 2, 0 m1 m 2, 3) for determination of direction, 3) 16 r, 4) 2 r, 0, 0, 4) for determination of dip at a place, NARAYANA GROUP, , 117
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , 65. The hysteresis cycle for the material of a 75. Relative permitivity and permeability of a, permanent magnet is, material are r and r . respectively. Which, 1) Short and wide, 2) tall and narrow, of the following values of these quantities are, 3) tall and wide, 4) short and narrow, allowed for a diamagnetic material, 66. The relation between r and is, 1) r 1.5, r 0.5 2) r 0.5, r 0.5, 1) r 1 , 2) r 1, 3) r 1.5, r 1.5 4) r 0.5, r 1.5, 3) 0 r, 4) r / 0, 76. Susceptibility is large and positive for, 67. The curie weiss law is obeyed by iron, 1) para magnetic, 2) diamagnetic, 1) at all temperatures, 3) ferro magnetic, 4) electromagnetic, 2) above the curie temperature, 77. For soft iron, in comparison with steel, 3) below the curie temperature, 1) hysteresis loss is more, 4) at the curie temperature, 2) hysteresis loss is same, 68. Which of the following quantities:, 3) hysteresis loss is less, (I) magnetic declination (II) dip is used to, 4) hysteresis loss is negligible, determine the strength of earths magnetic, 78. 1 and 2 are susceptibilities of diamagnetic, field at a point on the earths surface, 69., 70., , 71., , 72., , 73., , 74., , 118, , 1) Both I & II, 2) Neither I nor II, 3) I Only, 4) II Only, Domain formation is the necessary feature of, 1) ferro magnetism 2) paramagnetism, 3) diamagnetism, 4) electro magnetism, The magnetic force required to demagnetise, the material is, 1) retentivity, 2) coercivity, 3) energy loss, 4) hysterisis, Substances in which the magnetic moment of, a single atom is zero, 1) dia magnetic, 2) ferro magnetic, 3) para magnetic, 4) electro magnetic, Property possessed by ferro magnetic, substance only is, 1) attracting magnetic substance 2) hysterisis, 3) directional property, 4) susceptibility independent of temperature, Needles N1 , N 2 , N 3 are made of a, ferromagnetic, paramagnetic and a, diamagnetic substance respectively. A magnet, when brought close to them will 1) attract all, three of them, 2) attract N1 and N 2 strongly but repel N 3, weakly, 3) attract N1 strongly, N 2 weakly and repel N 3, weakly, 4) attract N1 strongly, but repel, N 2 , N3, weakly, The substance used for preparing electro, magnets is, 1) soft iron 2) steel 3) nickel, 4) copper, , substance at temperatures T1K and T2 K, respectively, then, 1) 1T1 2T2, 2) 1 2, 79., , 80., , 81., , 82., , 83., , 3) 1 T1 2 T2, 4) 1T2 2T1, Ferromagnetic materials have their properties, due to, 1) vacant inner subshells, 2) partially filled inner subshells, 3) filled inner subshells, 4) completely filled outer shells, When a diamagnetic liquid is poured into a Utube and one arm of the U-tube is placed, between the two poles of strong magnet with, the meniscus along the lines of the field, then, the level of the liquid in the arm where, magnetic field is applied will, 1) fall 2) rise 3) oscillate 4) remain unchanged, At Curie temperature, in ferromagnetic, materials, 1) the atomic dipoles get aligned, 2) the atomic dipoles lose alignment, 3) the atomic dipoles lose alignment, 4) magnetism is zero, A sensitive magnetic instrument can be, shielded very effectively from outside, magnetic fields by placing it inside a box of, 1) wood, 2) plastic, 3) metal of high conductivity, 4) soft iron of high permeability, The value of susceptibility for super conductor, is, 1) 0, 2) , 3) 1, 4) 1, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, , MAGNETISM, , 93. The core of electromagnet is made of soft iron,, because, a) the susceptibility of soft iron is very high, b) coercivity of soft iron is very low, 1) only a is correct, 2) only b is correct, 3) both a and b are correct, 4) both a and b are wrong, 94., The angles of dip at the poles and the equator, TERRESTRIAL MAGNETISM, respectively are, The angle of dip at a place on the earth's, 1) 300 , 600, 2) 900 , 00, surface gives, 1) direction of earth’s magnetic field, 3) 300 ,900, 4) 00 , 00, 2) horizontal component of earth's magnetic field 95. Select the correct answer., 3) vertical component of earth's magnetic field, a) When ‘n’ identical magnets are arranged, 4) location of geographic poles, in the form of closed polygon with unlike poles, A point near the equator has, nearer, the resultant magnetic moment is, zero., 1) BV BH, 2) BH BV, b) If one magnet is removed from the polygon,, 3) BV BH, 4) BV BH 0, the resultant magnetic moment becomes ‘M’., If I is the intensity of earth's magnetic field,, c) If one magnet is reversed in the polygon,, H its horizontal component and V the vertical, the resultant magnetic moment of, component, then these are related as, combination becomes 2M, 1) a, b and c are correct, 2, 2, 1) I V H, 2) I H V, 2) a and b are correct but c is wrong, 3) I H 2 V 2, 4) I 2 V 2 H 2, 3) only a is correct, 4) a, b and c are wrong, A line joining places of zero declination is, 96. Arrange the following in the descencalled, ding order of their resultant magnetic, 1) agonic, 2) isoclinic, moments consider two magnets of same, 3) isodynamic, 4) isogonal, moment, A line joining places of equal declination is, a) They are kept one upon the other with like, called, poles in contact, 1) aclinic, 2) isoclinic, b) They are kept one upon the other with, 3) isodynamic, 4) isogonal, unlike poles in contact, The needle of a dip circle when place at a, c) They are arranged in perpendicular, geomagnetic pole stays along, directions, 1) south north direction only, d) They are inclined 600 with like poles in, 2) east west direction only, contact, 3) vertical direction, 1) a, c, d, b, 2) a, b, c, d, 4) horizontal direction, 3) a, d, c, b, 4) d, b, c, a, The value of angle of dip is zero at the, 97., Among, the, following, statements:, magnetic equator because on it, A) A magnet of moment M is bent into a, 1) V and H are equal, semicircle, then its magnetic moment, 2) the value of V and H are zero, decreases, 3) the value of V is zero, B) Magnetic moment is directed parallel to, 4) the value of H is zero, axial line from south pole to north pole, Earth's magnetic field always has a horizontal, 1) A is true & B is false, component except at, 2) A is false & B is true, 1) equator, 2) magnetic pole, 3) A and B are true, 3) a latitude of 600, 4) an inclination of 600, 4) A and B are false, , 84. In a permanent magnet at room temperature, 1) magnetic moment of each molecules is, zero, 2) the individual molecules have non-zero, magnetic moments which are all perfectly, aligned, 3) domains are partially aligned, 4) domains are all perfectly aligned, 85., , 86., , 87., , 88., , 89., , 90., , 91., , 92., , NARAYANA GROUP, , 119
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MAGNETISM, 98. When a bar magnet is suspended freely in a, uniform magnetic field, identify the correct, statements, a) The magnet experiences only couple and, undergoes only rotatory motion, b) The direction of torque is along the, suspension wire, c) The magnitude of torque is maximum when, the magnet is normal to the field direction, 1) only a and c are correct, 2) only a and b are correct, 3) only b and c are correct, 4) a, b, c are correct, 99. Among the following statements:, (A) The resultant induction at a point on the, axial line of a bar magnet is parallel to, magnetic moment., (B) The resultant induction at a point on the, equatorial line is antiparallel to magnetic, moment, 1) A is true & B is false 2) A is false & B is true, 3) A and B are true, 4) A and B are false, 100. (i) Soft iron conducts electricity, (ii) Soft iron is magnetic material, (iii) Soft iron is used for permanent magnets, (iv) Soft iron is used as electro magnet, Out of the statements given above, 1) (i) and (ii) are correct, 2) (i) ,(ii) and (iii) are correct, 3) (ii) and (iv) are correct, 4) (i), (ii) and (iv) are correct, 101. Match the following:, Physical quantity, Unit, a) Magnetic moment e) Amp-m, b) Magnetic flux, f) Amp/m, density, c) Intensity of, g) N-m3/wb, magnetic field, d) Pole strength, h) Gauss, 1) a-e, b-f, c-g, d-h 2) a-g, b-h, c-f, d-e, 3) a-g, b-f, c-h, d-e 4) a-e, b-f, c-h, d-g, 102. Some physical quantities are given in the list, I the related units are given in the list II., Match the correct pairs in the lists, List-I, List-II, a) Magnetic field, e) Wb m-1, intensity, b) Magnetic flux, f) Wb m-2, c) Magnetic potential g) Wb, d) Magnetic induction h) Am-1, 1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-e, d-f, 3) a-h, b-e, c-g, d-f 4) a-f, b-g, c-e, d-h, 120, , JEE-ADV PHYSICS- VOL- III, 103. When a bar magnet is suspended in an uniform, magnetic field, then the torque acting on it, will be, List-I, List-II, a) maximum, e) 450 with the field, b) half of the, f) 600 with the, maximum value, field, 0, c) 3 / 2 times, g) 30 with the, the maximum, field, d) 1/ 2 times, h) 900 with the, the maximum, field, 1) a-h, b-g, c-f, d-e 2) a-e, b-f, c-g, d-h, 3) a-f, b-e, c-g, d-h 4) a-h, b-g, c-f, d-e, 104. Match the following, LIST - 1, LIST- 2, a) Magnetic moment d) Am 2, b) Pole strength, e) Am, c) Relative, f) weber, Wb, permeability, g), Am, H, h), m, 1) a e b d c g, 2) a g b e c d, 3) a d b e, f c g , h, 4) a f b e c d, , ASSERTION & REASON, 1) Both A and R are true and R is the correct, explanation of A., 2) Both A and R are true and R is not correct, explanation of A., 3) A is true, But R is false, 4) A is false, But R is true, 105. Assertion (A) : The net magnetic flux coming, out of a closed surface is always zero., Reason (R) : Unlike poles of equal strength, exist together, 106. Assertion (A): A magnet remains stable, If it, aligns itself with the field, Reason (R): The P.E. of a bar magnet is, minimum, if it is parallel to magnetic field., 107. Assertion (A) : To protect any instrument, from external magnetic field, it is put inside, an iron box, Reason (R) : Iron is a ferro magnetic, substance, NARAYANAGROUP
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JEE-ADV, JEE, MAINS PHYSICS- VOL - VIIIVOL- III, 108. A sser t ion(A ): T graph for a diamagnetic, material is a straight line parallel to T axis, Reason (R): This is because susceptibility of, a diamagnetic material is not affected by, temperature, 109. Assertion(A): If one arm of a U-tube, containing a dia magnetic solution is placed in, between the poles of a strong magnet with, the level in line with the field, the level of the, solution falls,, Reason(R): Diamagnetic substances are, repelled by the magnetic field, 110. Assertion(A): The ferro magnetic substances, do not obey curie's law, Reason(R) : At curie point ferro magnetic, substances start behaving as a para magnetic, substances, 111. Assertion(A): Earth's magnetic field inside a, closed iron box is less as compared to the out, side, Reason(R) : The magnetic permeability of, iron is low, 112. Assertion: Magnetic moment of an atom is, due to both, the orbital motion and spin motion, of every electron., Reason: A charged particle at rest produces, a magnetic field., 113. Assertion: Electromagnetis are made of soft, iron., Reason: Coercivity of soft iron is small., 114. Assertion: Time period of vibrations of a pair, of magnets in sum position is always smaller, than in difference position., Reason: T 2 I / MBH , where symbols, have their standard meaning, 115. Assertion: Magnetism is relativistic, Reason: When we move along with the, charge, so that there is no motion relative to, us, we find no magnetic field associated with, the charge, 116. Assertion: Steel is attracted by a magnet, Reason: Steel is not a magnetic substance, 117. Assertion: When radius of a circular wire, carrying current is doubled, its magnetic, moment becomes four times, Reason: Magnetic moment is directly, proportional to area of the loop, 118. Assertion: It is not necessary that every, magnet has one north pole and one south pole., Reason: It is a basic fact that magnetic poles, occur in pairs, 119. Assertion: Relative magnetic permeability, has no units and no dimensions, Reason: r / 0 , where the symbols have, their standard meaning., NARAYANA GROUP, , MAGNETISM, 120. Assertion: A magnet suspended freely in an, uniform magnetic field experiences no net, force, but a torque that tends to align the, magnet along the field when it is deflected, from equilibrium position, Reason: Net force mB mB 0 , but the, forces on north and south poles being equal,, unlike and parallel make up a couple that, tends to align the magnet, along the field., 121. Assertion: Basic difference between an, electric line and magnetic line of force is that, former is discontinuous and the latter is, continuous or endless., Reason: No electric lines of forces exit inside, charged conductor but magnetic lines do exist, inside magnet., 122. Assertion: The earth’s magnetic field is due, to iron present in its core., Reason: At a high temperature magnet losses, its magnetic property or magnetism., 123. Assertion: The properties of paramagnetic, and ferromagnetic substances are not, affected by heating., Reason: As temperature rises, the alignment, of molecular magnets gradually decreases., 124. Assertion: A soft iron core is used in a moving, coil galvanometer to increase the strength of, magnetic field., Reason: From soft iron more number of the, magnetic lines of force passes., , C.U.Q - KEY, 1) 1, 7) 1, 13) 3, 19) 4, 25) 3, 31) 1, 37) 1, 43) 3, 49) 1, 55) 2, 61) 3, 67) 2, 73) 3, 79) 2, 85) 1, 91) 3, 97) 3, 103) 1, 109) 1, 115) 1, 121) 1, , 2) 4, 8) 4, 14) 4, 20) 1, 26) 4, 32) 2, 38) 4, 44) 1, 50) 2, 56) 3, 62) 3, 68) 1, 74) 1, 80) 1, 86) 2, 92) 2, 98) 4, 104) 3, 110) 2, 116) 3, 122) 4, , 3) 1, 9) 3, 15) 2, 21) 2, 27) 4, 33) 3, 39) 4, 45) 3, 51) 1, 57) 4, 63) 1, 69)1, 75) 1, 81) 3, 87) 2, 93) 3, 99) 3, 105) 1, 111) 3, 117) 1, 123) 4, , 4) 3, 10) 3, 16) 4, 22) 1, 28) 4, 34) 1, 40) 2, 46) 4, 52) 3, 58) 2, 64) 3, 70) 2, 76) 3, 82) 4, 88) 1, 94) 2, 100) 4, 106) 1, 112) 3, 118) 4, 124) 1, , 5) 1, 11) 3, 17) 1, 23) 1, 29) 2, 35) 1, 41) 4, 47) 3, 53) 4, 59) 1, 65) 3, 71) 1, 77) 3, 83) 4, 89) 4, 95) 1, 101) 2, 107) 1, 113) 1, 119) 1, , 6) 3, 12) 2, 18) 2, 24) 2, 30) 2, 36) 1, 42) 1, 48) 2, 54) 1, 60) 1, 66) 1, 72) 2, 78) 2, 84) 3, 90) 3, 96) 3, 102) 2, 108) 1, 114) 1, 120) 1, , 121
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , LEVEL - I ( C.W ), MAGNETIC MOMENT AND, RESULTANT MAGNETIC MOMENT, 1., 2., , 3., , 4., , 5., , The geometric length of a bar magnet is, 24 cm. The length of the magnet is, 1) 24cm 2) 28.8cm 3) 20cm 4) 30cm, The magnetic moment of a bar magnet is, 3.6x10-3 A.m2. Its pole strength is 120 milli, amp. m. Its magnetic length is, 1) 3cm, 2) 0.3cm 3) 33.33cm 4) 3x102 cm, Two magnets have their lengths in the ratio, 2 : 3 and their pole strengths in the ratio 3 : 4., The ratio of their magnetic moment is, 1) 2 :1, 2) 4 :1, 3) 1 : 2, 4) 1 : 4, , The length of a magnet is 16 cm. Its pole, strength is 250 milli. amp. m. When it is cut, into four equal pieces parallel to its axis, the, magnetic length, pole strength and moments, of each piece are: (respectively), 1) 4 cm; 62. 5 milli Am; 250 milli amp. cm2, 2) 8 cm ; 500 milli Am; 400 milli amp. cm2, 3) 16 cm; 250 milli Am; 4000 milli amp. cm2, 4) 16 cm; 62.5 milli Am; 0.01 A.m2, A bar magnet of magnetic moment M1 is, axially cut into two equal parts. If these two, pieces are arranged perpendicular to each, other, the resultant magnetic moment is M2., M1, (2007M), Then the value of M is, 2, , 1, 4) 2, 2 2, 2, The resultant magnetic moment for the, following arrangement (non coplanar vectors), 1), , 6., , 1, , 2) 1, , 3), , M, , 600, , M, , M, , 7., , 8., , 122, , 1) M, 2) 2M 3) 3M, 4) 4M, Two magnets of moments 4Am 2 and 3Am2, are joined to form a cross (+), then the, magnetic moment of the combination is, 1) 4Am 2 2) 1Am 2 3) 7Am 2, 4) 5Am2, A magnet of magnetic moment M and length, 2l is bent at its mid-point such that the angle, of bending is 600. The new magnetic moment, is., M, M, 1) M, 2), 3) 2M 4), 2, 2, , 9., , A bar magnet of magnetic moment M is bent, in ‘ ’ shape such that all the parts are of, equal lengths. Then new magnetic moment, is, 1) M/3, 2) 2M, 3) 3 M 4) 3 3M, magnetic, 10. A thin bar magnet of length ' ' and, moment 'M' is bent at the mid point so that the, two parts are at right angles. The new magnetic, length and, magnetic moment are respectively, , , M, , , , M, , ,, 1) 2, 2M 2), 3) 2, 2 4) 2 , 2 M, 2 2, 11. The resultant magnetic moment for the following, arrangement is, M, 600, M, 600, , M, , M, , 1) M, 2) 2M, 3) 3M, 4) 4M, 12. Three magnets of same length but moments, M,2M and 3M are arranged in the form of an, equilateral triangle with opposite poles nearer,, the resultant magnetic moment of the arrangement, is, 1) 6M, , 3, , 2) zero, , M, 3) 3M, 4), 2, 13. A bar magnet of moment M is cut into two, identical pieces along the length. One piece is, bent in the form of a semi circle. If two pieces, are perpendicular to each other, then resultant, magnetic moment is, 2, , 1) M M , 2 , 2, , 3), , 2, , 2, , 2), , M M , , 2 , , 2, , 2, , M M, , , 2, MAGNETIC FIELD, , M M , , 2 , , 4), , 14. A magnetic pole of pole strength 9.2 A m. is, placed in a field of induction 50x10-6 tesla., The force experienced by the pole is, 1) 46N 2) 46x10-4N 3) 4.6x10-4N 4) 460N, 15. The magnetic induction at distance of 0.1 m, from a strong magnetic pole of strength, 1200 Am is, 1) 12x10-3 T, 2) 12x10-4 T, -3, 3) 1.2x10 T, 4) 24x10-3 T, 16. If area vector A 3i 2 j 5k m 2 flux, density vector B 5i 10 j 6k ( web / m 2 ) . The, magnetic flux linked with the coil is, 1) 31Wb, 2) 9000Wb, 3) 65Wb, 4) 100Wb, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, 17. P and Q are two unlike magnetic poles., Induction due to ‘P’ at the location of ‘Q’ is, B, and induction due to ‘Q’ at the location of, P is B/2. The ratio of pole strengths of P and, Q is, 1) 1 : 1, 2) 1 : 2, 3) 2 : 1 4) 1 : 2, 18. Two north poles each of pole strength m and, a south pole of pole strength m are placed at, the three corners of an equilateral triangle, of side a. The intensity of magnetic induction, field strength at the centre of the triangle is, m, 6m, 9m, m, 1) 0 2 2) 0 2 3) 0 2 4) 0, 4 a, 4 a, 4 a, 4 2a 2, 19. The pole strength of a horse shoe magnet is, 90 Am and distance between the poles is 6, cm. The magnetic induction at mid point of, the line joining the poles is,, 1) 10 2 T 2) Zero 3) 2 10 2 T 4) 10 4 T, 20. The force acting on each pole of a magnet, when placed in a uniform magnetic field of 7, A/m is 4.2x10-4 N. If the distance between the, poles is 10 cm, the moment of the magnet is, , 15, 2, 2), 1), , 15 Am, 2, 2, 3) 7.5 x 10-12 Am, 4) 6x10-6 Am, 21. An iron specimen has relative permeability, of 600 when placed in uniform magnetic field, of intensity 110 amp /m. Then the magnetic, flux density inside is....... tesla., 1) 18.29 x 10-3, 2) 8.29 x 10-2, 3, 3) 66 x 10, 4) 7.536 x 10-4, COUPLE ACTING ON THE BAR MAGNET, 22. A magnetic needle of pole strength 'm' is pivoted, at its centre. Its N-pole is pulled eastward by a, string. Then the horizontal force required to, produce a deflection of from magnetic, meridian (B H horizontal componet of earths, magnetic field), 1) mBcos 2) mBsin 3) 2 mBtan 4) mBcot , 23. Two identical bar magnets are joined to form a, cross. If this combination is suspended freely in, a uniform field the angles made by the magnets, with field direction are respectively, 1) 60°, 30° 2) 37°, 53° 3) 45°, 45° 4) 20°, 70°, , 24. A bar magnet of length 16 cm has a pole, strength of 500 milli amp.m. The angle at, which it should be placed to the direction of, external magnetic field of induction 2.5 gauss, so that it may experience a torque of 3 x105, Nm is, 1) , 2) / 2 3) / 3, 4) / 6, NARAYANA GROUP, , MAGNETISM, 25. A bar magnet is at right angles to a uniform, magnetic field. The couple acting on the, magnet is to be one fourth by rotating it from, the position. The angle of rotation is, 1) Sin-1(0.25), 2) 900-Sin-1(0.25), 3) Cos-1(0.25), 4) 900- Cos-1(0.25), , , , , , , ^, , 26. A bar magnet of moment M = i + j is placed, ^, , ^, , in a magnetic field induction B 3 i 4 j 4 k ., The torque acting on the magnet is, , , , , , , 1) 4 i -4 j + k, , , 27., , 28., , 29., , 30., , , , , , , , , , , , 2) i + k, , , 3) i - j, 4) i + j + k, A bar magnet of magnetic moment 1.5 J/T is, aligned with the direction of a uniform, magnetic field of 0.22 T. The work done in, turning the magnet so as to align its magnetic, moment opposite to the field and the torque, acting on it in this position are respectively., 1) 0.33J, 0.33N-m, 2) 0.66J, 06.66N-m, 3) 0.33J, 0, 4) 0.66J, 0, The work done in turning a magnet of, magnetic moment M by an angle of 900 from, the meridian is n times the corresponding, work done to turn it through an angle of 600 ,, where n is given by, 1, 1, 2) 2, 3), 4) 1, 1), 2, 4, A bar magnet of moment 4Am 2 is placed in a, nonuniform magnetic field. If the field, strength at poles are 0.2 T and 0.22 T then, the maximum couple acting on it is, 1) 0.04Nm 2) 0.84Nm3) 0.4 Nm 4) 0.44Nm, A magnet of length 10 cm and pole strength, 4x10-4 Am is placed in a magnetic field of, induction 2x10-5 weber m-2, such that the axis, of the magnet makes an angle 300 with the, lines of induction. The moment of the couple, acting on the magnet is, 1) 4x10-10 Nm, 2) 8x10-10 Nm, 3) 4x10-6 Nm, 4) 3 x10-11 Nm, , 31. A bar magnet of magnetic moment 2Am 2 is, free to rotate about a vertical axis passing, through its center. The magnet is released, from rest from east - west position. Then the, KE of the magnet as it takes N-S position is, , BH, , 25T , , 1) 25 J, , 2) 50 J 3) 100 J 4) 12.5 J, 123
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 32. A bar magnet of length 10cm and pole, strength 2 Am makes an angle 600 with a, uniform magnetic field of induction 50T. The, couple acting on it is, 1) 5 3Nm, 2) 3Nm, 3) 10 3Nm, , 4) 20 3Nm, , FIELD OF A BAR MAGNET, 33. The magnetic induction field strength due to, a short bar magnet at a distance 0.20 m on, the equatorial line is 20x10 -6 tesla. The, magnetic moment of the bar magnet is, 1) 3.2Am2 2) 6.4Am2 3) 1.6Am2 4) 16Am2, 34. The magnetic induction field strength at a, distance 0.3 m on the axial line of a short bar, magnet of moment 3.6 Am2 is, 1) 4.5 10-4 T, 2) 9 10-4 T, -5, 4) 2.6 10-5 T, 3) 9 10 T, 35. A magnet of length 10 cm and magnetic, moment 1Am 2 is placed along the side of an, equilateral triangle of the side AB of length, 10 cm. The magnetic induction at third vertex, C is, 1) 10 -9 T 2) 10 -7 T 3) 10-5 T 4) 10 -4T, 36. The length of a magnet of moment 5Am2 is, 14 cm. The magnetic induction at a point,, equidistant from both the poles is 3.2x10-5 Wb/, m2. The distance of the point from either pole, is, 1) 25 cm 2) 10 cm 3) 15 cm 4) 5 cm, 37. A pole of pole strength 80 Am is placed at a, point at a distance 20cm on the equatorial line, from the centre of a short magnet of magnetic, moment 20Am2 . The force experienced by it, is, 1) 8 x 10-2 N, 2) 2 x 10-2 N, -2, 3) 16 x 10 N, 4) 64 x 10-2 N, 38. A short bar magnet produces magnetic fields, of equal induction at two points one on the, axial line and the other on the equatorial line., The ratio of their distances is, 1) 2:1, 2) 21/2:1, 3) 21/3:1 4) 21/4:1, SUPERPOSITION OF MAGNETIC FIELDS, 39. Two short bar magnets with magnetic, moments 8Am2 and 27Am2 are placed 35cm, apart along their common axial line with their, like poles facing each other. The neutral point, is, 1) midway between them, 2) 21 cm from weaker magnet, 3) 14 cm from weaker magnet, 4) 27 cm from weaker magnet, 124, , 40. A short magnetic needle is pivoted in a, uniform magnetic field of induction 1T. Now,, simultaneoulsy another magnetic field of, induction 3T is applied at right angles to, the first field; the needle deflects through an, (EAM 2010), angle where its value is, 0, 0, 0, 1) 30, 2) 45, 3) 90, 4) 600, 41. Two magnetic poles of pole strengths 324 milli, amp.m. and 400 milli amp m are kept at a, distance of 10 cm in air. The null point will be, at a distance of ...... cm, on the line joining, the two poles, from the weak pole if they are, like poles., 1) 4.73, 2) 5, 3) 6.2, 4) 5.27, , TIME PERIOD OF SUSPENDED, MAGNET IN THE UNIFORM, MAGNETIC FIELD, 42. With a standard rectangular bar magnet, the, time period in the uniform magnetic field is 4, sec. The bar magnet is cut parallel to its length, into 4 equal pieces. The time period in the, uniform magnetic field when the piece is used, (in sec) (bar magnet breadth is small), 1) 16, 2) 8, 3) 4, 4) 2, 43. A bar magnet of moment of inertia, 1×10 -2 kgm 2 vibrates in a magnetic field of, induction 0.36 10-4 tesla. The time period of, vibration is 10 s. Then the magnetic moment, of the bar magnet is (Am2), 1) 120, 2) 111, 3) 140 4) 160, 44. Two bar magnets are placed together, suspended in the uniform magnetic field, vibrates with a time period 3 second. If one, magnet is reversed, the combination takes 4s, for one vibration. The ratio of their magnetic, moments is, 1) 3 : 1, 2) 5 : 18 3) 18 : 5 4) 25 : 7, 45. A bar magnet of length ‘l’ breadth ‘b’ mass, ‘m’ suspended horizontally in the earths, magnetic field, oscillates with period T. If ‘l’,, m, b are doubled with pole strength remaining, the same, the new period will be, 1) 8T, 2) 4T, 3) T/2, 4) 2T, 46. The time period of a suspended magnet is T0., Its magnet is replaced by another magnet, whose moment of inertia is 3 times and, magnetic moment is 1/3 of that of the initial, magnet. The time period now will be, To, T, 2) To, 3), 4) o, 1) 3To, 3, 3, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , LEVEL - I (C. W ) KEY, , 47. A magnetic needle is kept in a uniform, magnetic field of induction 0.5 x 10-4 tesla. It, makes 30 oscillations per minute. If it is kept, in a field of induction 2 x 10-4 tesla. Then its, frequency is, 1) 1 oscillation/s, 2) 60 oscillations/s, 3) 15 oscillations/min 4) 15 oscillations/s, 48. A magnet is suspended horizontally in the, earth’s field. The period of oscillation in the, place is T. If a piece of wood of the same, moment of inertia as the magnet is attached, to it, the new period of oscillation would be, 1), , T, 2, , 2) T/2, , 3) T/3, , 4), , 2T, , 49. A magnet freely suspended makes 30, vibrations per minute at one place and 20, vibrations per minute at another place. If the, value of BH at first place is 0.27 tesla. The, value of BH at other place is, 1) 0.12 T, 2) 2.1 T 3) 5.4 T 4) 0.61 T, , 1) 3, 7) 4, 13) 2, 19) 3, 25) 2, 31) 2, 37) 2, 43) 2, 49) 1, , 3) 3 104 H / m, NARAYANA GROUP, , 4) 4 10 4 H / m, , 3) 3, 9) 1, 15)1, 21) 2, 27) 4, 33) 3, 39) 3, 45) 4, 51) 2, , 4) 4, 10)2, 16) 3, 22) 3, 28) 2, 34) 4, 40) 4, 46) 1, 52) 4, , 5) 4, 11) 2, 17) 3, 23) 3, 29) 2, 35) 4, 41) 1, 47) 1, 53) 1, , 6) 2, 12) 3, 18) 2, 24) 3, 30) 1, 36) 1, 42) 3, 48) 4, 54) 2, , LEVEL-I (C. W ) HINTS, 1., 2., 3., 4., , TYPES OF MAGNETIC MATERIALS, 50. A magnet has a dimensions of, 25 cm x 10 cm x 5 cm and pole strength of, 200 milli ampm The intensity of magnetisation, due to it is, 1) 6.25A/m 2) 62.5A/m 3) 40A/m 4) 4A/m, 51. The mass of iron rod is 110g, its magnetic, moment is 20 Am2. The density of iron is 8g/, cm3. The intensity of magnetization is nearly, 1) 2x105 Am-1, 2) 2.26x106 Am-1, 6, -1, 3) 1.6x10 Am, 4) 1.4 x 106 Am-1, 52. Relative permeability of iron is 5500, then its, magnetic susceptibility will be:, 1) 5500 107, 2) 5500 107, 3) 5501, 4) 5499, 53. A specimen of iron is uniformly magnetised, by a magnetising field of 500Am 1 . If the, magnetic induction in the specimen is 0.2, Wbm 2 . The susceptibility nearly is, 1) 317.5 2) 418.5 3) 217.5 4) 175, 54. The magnetic susceptibility of a rod is 499., The absolute permeability of vacuum is, 4 10 7 H / m . The absolute permeability of, the material of the rod is, 1) 104 H / m, 2) 2 10 4 H / m, , 2) 1, 8) 2, 14) 3, 20) 1, 26) 1, 32) 1, 38) 3, 44) 4, 50) 3, , 5., , 5, geometric length, 6, M, 2l , m, , 2l , , M 1 m1 2l 1, , , M 2 m2 2l 2, Magnetic length remains same, m, 1, Pole strength m , 4, M, 1, Magnetic moment = M , 4, M1, M, , M2 M / 2, , , , , , 2, , M2, , 6., , M res , , 7., , M M 12 M 22, , 8., , M 1 M sin, , 9., , 1, M 1 m 2l , 2l , , 3M, , , 2, 1, , 1, 10. M M sin, , 2l, 3, , , , 1, , 2l 2l sin, 2, 2, , 11. M R M 12 M 22 2 M 1M 2 cos, 12. M R M 12 M 22 2 M 1M 2 cos, 13. M 1 , , 2M, , 2, , M, , ,M2 , 2, , , M 1 M 12 M 22, , 14. F mB, 125
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 15. B , , 40. B BH Tan, , 0 m, 4 d 2, , m, , 16. B. A, 17. B m, 18. Bres, , , m, B 0, 4 a / 3, 2 B where, , , , 42. T 2, , , , 2, , 0 8m, 4 d 2, M, 20. F 0 H , 2l, 21. B 0 r H, 22. MB sin , 23. MB sin ,1 2, 24. C MB sin , 25. C MB sin , , 26. M B, 27. = MB sin , W MB(cos1 cos 2 ), 28. W MB(cos1 cos 2 ), 29. Cmax MBavg, 30. C MB sin , 19. B , , 31. W MB cos 1 cos 2 , 32. C MB sin , M, 33. B 0 3, 4 d, 2M, 34. Ba 0 3, 4 d, 0 M, 35. Be , where x d 2 l 2, 4 x 3, , 0 M, 36. Be , where x d 2 l 2, 4 x 3, M, 37. Be 0 3 , F mBe, 4 d, d , B1, 2 2 , 38., B2, d1 , , 2, 43. M 4, , 45. T 2, 46. T , , 39. from weaker pole, , 1/ 3, , M2 , , , M1 , , 1, , , , , , I, T 2B, , m 2, I, l b2, , I, 12, MB, , , , , , I, M, , 47. n B, 48. T I, 49. n B, 50. I , , M m, , V, a, , 51. I , , M, ; Mass dV, V, , 52. r 1, 53. B H 0 r H ; r 1, 54. o 1 , , LEVEL - I (H.W), MAGNETIC MOMENT AND RESULTANT, MAGNETIC MOMENT, 1., , 2., d, , m 2, I, l b2, , I, 12, MB, , M 1 T22 T12, , 44., M 2 T22 T12, , 3, , x, , 126, , m, , 1, 2, 41. x 2 d x 2, , If a bar magnet of pole strength ‘m’ and, magnetic moment ‘M’ is cut equally 4 times, parallel to its axis and 5 times perpendicular, to its axis then the pole strength and magnetic, moment of each piece are respectively, m M, m M, m M, m M, ,, 1), 2) ,, 3) ,, 4) ,, 20 20, 4 20, 5 20, 5 4, Three identical bar magnets each of, magnetic moment M are arranged in the form, of an equilateral triangle such that at two, vertices like poles are in contact. The, resultant magnetic moment will be, 1) Zero, , 2) 2M, , 3), , 2 M 4) M 3, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, 3., , If two identical bar magnets, each of length, ‘l’, pole strength ‘m’ and magnetic moment, ‘M’ are placed perpendicular to each other, with their unlike poles in contact, the magnetic, moment of the combination is, 1), , 4., , 5., , 6., , M, 2, , 2) lm 2 3) 2lm 2 , , 4) 2M, , A magnetised wire of magnetic moment ‘M’, and length ‘ l ’ is bent in the form of a, semicircle of radius ‘r’. The new magnetic, moment is, M, M, M, 2Mr, 1), 2), 3), 4), l, , 2, 4, A long thin magnet of moment M is bent into, a semi circle. The decrease in the Magnetic, moment is, 2) M/2, 1) 2M/ , 3) M( -2) / , 4) M(2- )/2, A magnet of magnetic moment M is in the, form of a quadrant of a circle. If it is, straightened, its new magnetic moment will, be, , M, 2M, 4), 2, 2, 2 2, , A bar magnet of moment 'M' is bent into a, shape' '. If the length of the each part is, same, its new magnetic moment will be, 1), , 7., , MAGNETISM, , M, , M, , 2), , M, , M, , 3), , M, , 2, M, 3, 5, 2, 3, 8. Four magnets of magnetic moments M, 2M,, 3M and 4M are arranged in the form of a, square such that unlike poles are in contact., Then the resultant magnetic moment is, 1) 2 2M 2) 2M 3) 10M, 4) 2M, COUPLE ACTING ON THE BAR MAGNET, 9. A torque of 2 x 10-4 Nm is required to hold a, magnet at right angle to the magnetic, meridian. The torque required to hold it at, 300 to the magnetic meridian in N-m is, 1) 0.5 x 10-4 2) 1 x 10-4 3) 4 x 10-4 4) 8 x 10-4, 10. A bar magnet of 5 cm long having a pole, strength of 20 A.m. is deflected through 300, from the magnetic meridian. If, 1), , 2), , 320, , 3), , 4), , H = 4 A/m, the deflecting couple is, 1) 1.6 x 10-4 Nm, 2) 3.2 x 10-5 Nm, -5, 3) 1.6 x 10 Nm, 4) 1.6 x 10-2 Nm, NARAYANA GROUP, , 11. A short bar magnet placed with its axis at 300, with a uniform external magnetic field of 0.16, T experience a torque of magnitude 0.032 N, m. If the bar magnet is free to rotate, its, potential energies when it is in stable and, unstable equilibrium are respectively, 1) -0.064J, +0.064J 2) -0.032J, +0.032J, 3) +0.064J, -0.128J 4) 0.032J, -0.032J, 12. When a bar magnet is placed at 900 to uniform, magnetic field, it is acted upon by a couple, which is maximum. For the couple to be half, of the maximum value, it is to be inclined to, the magnetic field at an angle is, 1) 300, 2) 450, 3) 600, 4) 900, 13. A magnet of moment 4Am 2 is kept suspended, in a magnetic field of induction 5 10-5T . The, workdone in rotating it through 1800 is, 1) 4 10 4 J 2) 5 104 J 3) 2 10 4 J 4) 10 4 J, 14. The work done in rotating the magnet from, the direction of uniform field to the opposite, direction to the field is W. The work done in, rotating the magnet from the field direction, to half the maximum couple position is, W, W, 3W, 2 3 4), 1 3, 3), 4, 2, 2, 15. The work done in rotating a magnet of pole, strength 1 A-m and length 1 cm through an, angle of 600 from the magnetic meridian is, (H=30 A/m), 1) 9.42 x 10-8 J, 2) 3.14 x 10-8 J, -8, 3) 18.84 x 10 J, 4) 10 x 10-8 J, 16. The work done in turning a magnet normal to, field direction from the direction of the field, is 40x10-6 J. The kinetic energy attained by it, when it reaches the field direction when, released is, 1) Zero, 2) 30 x 10-6J, 3) 10 x 10-6 J, 4) 40 x 10-6 J, 17. A magnet is parallel to a uniform magnetic, field. The work done in rotating the magnetic, through 600is 8x10-5 J. The work done in, rotating through another 300 is, 1) 4 x10-5J, 2) 6 x10-5J, -5, 3) 8 x 10 J, 4) 2 x 10-5J, , 1) 2 W 2), , , , , , , , , , FIELD OF A BAR MAGENT, 18. The magnetic induction field strength at a, distance 0.2 m on the axial line of a short, bar magnet of moment 3.6 Am 2 is, 1) 4.5 104 T, 3) 9 105 T, , 2) 9 104 T, 4) 4.5 105 T, 127
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 19. A short bar magnet produces magnetic fields, of equal induction at two points on the axial, line and the other on the equatorial line. Then, the ratio of the distance is, 1/ 3, , 1/ 3, , 1) 1: 2, 2) 1/ 2, 3) 2 :2 4) 2 :1, SUPERPOSITION OF MAGNETIC FIELDS, 20. A short bar magnet of magnetic moment, 1.2Am 2 is placed in the magnetic meridian, with its south pole pointing the north. If a, neutral point is found at a distance of 20 cm, from the centre of the magnet, the value of, the horizontal component of the earth’s, magnetic field is, 1) 3×10-5 T 2) 3×10-4 T 3) 3×103T 4) 3×10-2 T, 21. A very long magnet of pole strength 4 Am is, placed vertically with its one pole on the, table.The distance from the pole, the neutral, 1/ 3, , point will be formed is BH 4 10 5 T , 1) 0.5 m 2) 0.1 m 3) 0.15 m 4) 6.66 m, TIME PERIOD OF SUSPENDED MAGNET IN, THE UNIFORM MAGNETIC FIELD, 22. A bar magnet of magnetic moment M and, moment of inertial I is freely suspended such, that the magnetic axis is in the direction of, magnetic meridian. If the magnet is displaced, by a very small angle , the angular, , 25. A bar magnet has a magnetic moment equal, to 5 x 10-5 weber x metre. It is suspended in a, magnetic field which has a magnetic induction, (B) equal to 8 10 4 tesla. The magnet, vibrates with a period of vibration equal to, 15 seconds. The moment of inertia of the, magnet is:, 1) 9 1013 kg m2, 2) 11.25 x 1013 kg m2, 3) 5.62 x 1013 kg m2 4) 0.57 x 1013 kg m2, 26. Two bar magnets are suspended and allowed, to vibrate. They make 20 oscillations /minute, when their similar poles are on the same side, and they make 15 oscillations per minute with, their opposite poles lie on the same side. The, ratio of their moments, 1) 9:5, 2) 25:7 3) 16:9, 4) 5:4, , TYPES OF MAGNETIC MATERIALS, 27. The variation of magnetic susceptibility ( ), with temperature for a diamagnetic substance, is best represented by, 1) x, T, , 3) x, , acceleration is (Magnetic induction of earth’s, horizontal field = BH ), Mθ, Iθ, MBH θ, IB H θ, 2), 3) IB, 4) MB, I, M, H, H, 23. If the moments of inertia of two bar magnets, are same, and if their magnetic moments are, in the ratio 4 : 9 and if their frequencies of, oscillations are same, the ratio of the induction, field strengths in which they are vibrating is, 1) 2 : 3, 2) 3 : 2 3) 4 : 9, 4) 9 : 4, 24. If the strength of the magnetic field is, increased by 21% the frequency of a magnetic, needle oscillating in that field., 1) Increased by 10%, 2) Decreases by 10%, 3) Increases by 11%, 4) Decreased by 21%, , 1), , 128, , 2) x, T, , T, , 4), , x, , T, , T, , 28. The magnetic induction and the intensity of, magnetic field inside an iron core of an, electromagnet are 1 Wbm2 and 150 Am1, respectively. The relative permeability of iron, is : 0 4107 Henry/m, 106, 1), 4, , 106, 2), 6, , 10 5, 3), 4, , 10 5, 4), 6, , 29. The mass of an iron rod is 80 gm and its, magnetic moment is 10 Am2. If the density of, iron is 8 gm c.c. Then the value of intensity, of magnetisation will be, 1) 106 A m, 2) 104 A m, 3) 102 A m, , 4) 10 A m, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , 30. A rod of cross sectional area 10cm 2 is placed, with its length parallel to a magnetic field of, intensity 1000 A/M the flux through the rod is, 104webers. Then the permeability of material, of rod is, 1) 104 wb Am, 2) 103 wb Am, 3) 102 wb Am, , 4) 10 wb Am, 31. A bar magnet of magnetic moment 10 Am 2 has, a cross sectional area of 2.5 104 m2 .If the, intensity of magnetisation of the magnet is, 106 A m , then the length of magnet is, 1) 0.4m, 2) 0.04cm 3) 0.04m 4) 40 cm, , 10. C M o H sin , 11. P.E M .B, 12. C sin , 13. W MB cos 1 cos 2 , 14. W MB cos 1 cos 2 , 15. W m 2l B cos1 cos 2 , 16. K . E = Work done, 17. W MB cos 1 cos 2 , 18. B , , LEVEL - I (H. W ) - KEY, 1) 3, 7) 2, 13) 1, 19) 4, 25) 1, 31) 3, , 2) 2, 8) 1, 14) 3, 20) 1, 26) 2, , 3) 2, 9) 2, 15)3, 21) 2, 27)2, , 4) 2, 10)3, 16) 4, 22) 1, 28) 4, , 5)3, 11) 1, 17) 3, 23) 4, 29) 1, , 6) 4, 12) 1, 18)3, 24) 1, 30) 1, , LEVEL - I ( H. W ) - HINTS, 1., , 3., , M R M 12 M 22, , 4., , 1, For semi circle , M , , 2M, here l r, , , 5., , 1, For semi circle , M , , 2M, , , decrease in M , M M M 1, , , M 2 M 1 sin ; , 2, 2, 1, , 2l , , M1 , , , , 2l, 5, , M, 5, , 8., , M R M 12 M 22 2 M 1 M 2 cos , , 9., , C MBH sin , , NARAYANA GROUP, , o 2M, 4 d 3, , 21. BH , , o m, 4 d 2, , 24. n B, , M R M M 2 M 1M 2 cos, , 7., , 20. BH , , 2, 2, , 2., , 6., , 2M M, 3, d13, d2, , 22. I MBH, 23. MB = constant, , m, M, m , M1 , 5, 5 4, 1, , 2, 1, , 19., , o 2M, 4 d 3, , B2, , n2, , 1 100, 1 100 , n1, , B1, , , 25. I , , MBT 2, 4 2, , M 1 n12 n22, , 26., M 2 n12 n22, 27. For diamagentic material, is - ve & independent of temperature, B, 28. r H, o, , 29. I , , M, mass, ,V , V, density, , 30. , , , AH, , 31., , 2l , , M, IA, 129
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , LEVEL - II (C.W ), , 7., , MAGNETIC MOMENT AND, RESULTANT MAGNETIC MOMENT, 1., , 2., , A magnetised wire is bent into an arc of a, circle subtending an angle 600 at its centre., Then its magnetic moment is X. If the same, wire is bent into an arc of a circle subtending, an angle 900 at its centre then its magnetic, moment will be, 3x, x, (2 2) x, x 2, 1), 2), 3), 4), 2 2, 3, 3, 3, A magnet of length 2L and moment ‘M’ is, axially cut into two equal halves ‘P’ and ‘Q’., The piece ‘P’ is bent in the form of semi circle, and ‘Q’ is attached to it as shown. Its moment, is, N, S, Q, N, S, P, , M, M 2 , M, M, 2), 3), 4) 2 , , 2, 2, A bar magnet of magnetic moment ‘M’ is bent, in the form of an arc which makes angle 600., The percentage change in the magnetic, moment is, 1) 9% Increase, 2) 9% Decrease, 3) 4.5% Decrease, 4) 4.5% Increase, , COUPLE ACTING ON THE BAR, MAGNET, 8., , 9., , 10., , 1), 3., , MAGNETIC FIELD, 4., , 5., , 6., , 130, , At two corners A and B of an equilateral, triangle ABC, a south and north pole each of, strength 30Am are placed. If the side of the, triangle is 1m. The magnetic induction at C is, 1) 3 x 10-6 T, 2) 4 x 10-6 T, -6, 3) 8 x 10 T, 4) 2 x 10-6 T, A bar Magnet of Magnetic Moment 3.0, amp.m2 is placed in a uniform Magnetic, induction field 2x 10-5 T. If each pole of the, magnet experience a force of 6x 10-4 N, the, length of the magnet is, 1) 0.5 m 2) 0.3 m 3) 0.2 m 4) 0.1 m, The magnetic induction at a distance ‘d’ from, the magnetic pole of unknown strength ‘m’ is, B. If an identical pole is now placed at a, distance of 2d from the first pole, the force, between the two poles is, mB, mB, 1) mB, 2), 3), 4) 2mB, 2, 4, , Two identical north poles each of strength m, are kept at vertices A and B of an equilateral, triangle ABC of side a. The mutual force of, repulsion between them has a magnitude of, F. The magnitude of magnetic induction at C, is, 2) F/ 3m 3) F/3m 4) 3F/m, 1) F/m, , 11., , 12., , 13., , Two magnets of magnetic moments M and, 3M are joined to form a cross +. The, combination is suspended freely in a uniform, magnetic field. In the equilibrium position, the, angle between the magnetic moment 3 M, and the field is, 1) 30o, 2) 45o, 3) 60o, 4) 90o, The rate of change of torque ‘ ’ with, deflection is maximum for a magnet, suspended freely in a uniform magnetic field, of induction B when is equal to, 1) 0o, 2) 45o, 3) 60o, 4) 90o, The couple acting on a bar magnet of pole, strength 2 Am when kept in a magnetic field, of intensity 10 A/m, such that axis of the, magnet makes an angle 300 with the direction, of the field is 80 107 Nm . The distance, between the poles of the magnet is, 2, , 1, 1) m, 2) m 3) 63.36m 4), m, , 2, 2, A bar magnet with poles 25cm apart and pole, strength 14.4 Am rests with its center on a, frictionless pivot. If it is held in equilibrium, at 600 to a uniform magnetic field on induction, 0.25 T by applying a force F at right angles to, its axis 10cm from the pivot, the value of F in, newton is (nearly), 1) 3.9N, 2) 7.8N 3) 15.6N 4) 31.2N, A bar magnet of magnetic moment M1 is, suspended by a wire in a magnetic field. The, top of the wire is twisted through 1800, then, the magnet is rotated through 450 . Under, similar conditions another magnet of magnetic, moment M2 is rotated through 300, the ratio, M1:M2 is, 1) 9 : 10 2 2) 1: 2 3) 1 : 1, 4) 1:3, Two magnets of moments M1 and M2 are, rigidly fixed together at their centres so that, their axes are inclined to each other. This, system is suspended in a magnetic field of, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, induction 'B' so that M1 makes an angles 1, and M2 makes an angles 2 with the field, direction and unlike poles on either side of, the field direction. The resultant torque on, the rigid system is, 1) B M1 sin 1 M 2 sin 2 , 2) B M1 cos 1 M 2 cos 2 , 3) B M1 sin 2 M 2 sin 1 , 4) B M1 cos 2 M 2 cos 1 , , 14. A short magnet placed with its axis making, an angle with a uniform external magnetic, field of induction B experiences a torque ., If the magnet is free to rotate, which, orientation would correspond to its stable and, unstable equilibrium., 1) 00 , 90 0 2) 0 0 , 1800, , MAGNETISM, o, M, . 2 3, 4, d, M, 3) o . 3, 4 d, , 1), , o, M, . 3 3, 4, d, , M, 4) o . 5 3, 4, d, , , , , , 19. Magnetic induction at a point on the axial line, of a short bar magnet is B towards east. If, the magnet is turned through 900 in clock wise, direction, then magnetic induction at the same, point is (Neglect earth’s magnetic field), 1) B/4 towards east 2) B/2 towards west, 3) B/2 towards north 4) B/2 towards south, 20. Two short bar magnets of equal dipole, moments ‘M’ each are fastened perpendicular, at their centers as shown in figure. The, magnitude of the magnetic field at ‘P’ at a, distance d from their common center as shown, in figure is, P, , N, , 3) 450 , 1350 4) 00 , 2700, , 450, , FIELD OF A BAR MAGNET, 15. Two short magnets each of moment 10, A-m 2 are placed in end - on position so that, their centres are 0.1m apart. The force, between them is, 1) 0.4N, 2) 0.5N 3) 0.6N, 4) 0.8N, 16. The ratio of magnetic fields on the axial line, of a long magnet at distances of 10cm and, 20cm is 12.5:1. The length of the magnet is, 1) 5cm, 2) 10cm 3) 10m 4)15 m, 17. Two short magnets AB and CD in the X-Y, plane and are parallel to X-axis and the, co-ordinates of their centres respectively are, (0,2) and (2,0). Line joining the North - South, poles of CD is opposite to that of AB and lies, along the positive X-axis. The resultant field, induction due to AB and CD at a point P(2,2), is 100×10-7 T . When the poles of the magnet, CD are reversed, the resultant field induction, is 50×10-7 T . The values of magnetic moments, of AB and CD are (in Am2), 1) 300:200, 2) 400:600, 3) 200:100, 4) 300:100, 18. Two identical bar Magnets each having, Magnetic Moment of ‘M’ are kept at a, distance of 2d with their axes perpendicular, to each other in a horizontal plane. The, Magnetic induction at midway between them, is, NARAYANA GROUP, , , , 2), , S, , N, , S, , 0 M, 4 d 3, , 0 2 2M, 4 d 3, 0 M, 0 2M, 3), 4), 3, 4 d, 2 d 3, SUPERPOSITION OF MAGNETIC, FIELDS, 1), , 2), , 21. A magnetic dipole is under the influence of, two magnetic fields having an angle of 1200, between them. One of the fields has a, magnitude 1.2 10 2 T . If the dipole comes to, stable equilibrium at an angle of 300 with this, field, then magnitude of the other field is, 1) 8.484x10 2 T, 2) 0.6 x10-2 T, 3) 4.242x10-3T, 4) 4.242x10-5T, 22. A short bar magnet is placed with its south, pole facing geographic south and the distance, between the null points is found to be 16 cm., When the magnet is turned pole to pole at, the same place then the distance between the, null points will be, 1) will be same , along the axial line, 2) will be same , along the equatorial line, 3) will be 16 21/3 , on the axial line, 4) will be 16 21/3 , on the equatorial line, 131
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MAGNETISM, 23. A bar magnet is placed with its North pole, pointing North. Neutral point is at a distance, ‘d’ from the center of magnet. The net, magnetic induction at the same distance on, the axial line of the magnet is, 1) 2BH, 2) 3BH, 3) BH, 4) 7BH, 24. A bar magnet is placed with its North pole, pointing North. Neutral point is at 12 cm., Another magnet is now placed on the first, magnet, then the neutral point is found to be, at 8 cm. The ratio of their magnetic moments, is, 1) 3:2, 2) 27:19 3) 9:4, 4) 9:5, TIME PERIOD OF SUSPENDED MAGNET IN, THE UNIFORM MAGNETIC FIELD, 25. The period of a thin magnet in a magnetic, field is 2s. It is cut into four equal parts by, cutting it along length and breadth. The, period of each of them in the same field is, 1) 1 s, 2) 2 s, 3) 3 s 4) 4 s, 26. A bar magnet suspended in magnetic meridian, executes oscillations with a time period of 2, sec in the earth’s horizontal magnetic field, of 24 microtesla. When a horizontal field of, 18 microtesla is produced opposite to the, earth’s field by placing a current carrying, wire, the new time period of magnet will be:, 1) 1s, 2) 2s, 3) 3s, 4) 4s, 27. Two bar magnets are bound together side by, side and suspended. They swing in 12s when, their like poles are together and in 16s when, their unlike poles are together, the magnetic, moments of these magnets are in the ratio, 1)27:5, 2)25:7, 3)7:25, 4)24:7, 28. A short bar magnet is oscillating in a magnetic, field and its time period is 2 seconds . If, another piece of brass of double moment of, inertia be placed over that magnet the time, period of that combination in that field is, 4) 1 2 S, 1) 2 3 S 2) 2 2 S 3) 2 S, 29. When two identical bar magnets placed one, above the other, such that they are mutually, perpendicular and bisect each other. The time, period of oscillation in a horizontal magnetic, field is 4 seconds. If one of the magnets is, removed the time period of the other in the, same field (21/4=1.189), 1)1.34sec 2) 2.34sec3) 3.36sec 4) 4.34sec, 132, , JEE-ADV PHYSICS- VOL- III, 30. A bar magnet suspended freely in uniform, magnetic field is vibrating with a time period, of 3 seconds.If the initial field strength is 2T,, then the final field strength, for which time, period becomes 4 seconds is, 1) 1.125Tesla, 2) 0.625Tesla, 3) 3.55 Tesla, 4) 0.75 Tesla, 31. A short bar magnet of magnetic moment 2Am2, and moment of inertia 6x102kgm2 is freely, suspended such that the magnetic axial line, is in the direction of magnetic meridian. If the, magnet is displaced by a very small angle (30),, the angular acceleration is –– x10–6rad/sec2, (Magnetic induction of earth's horizontal field, = 4x 10-4T)., 1) /20 2) /45 3) /60 4) /75, 32. The period of oscillation of a magnet at a, place is 4 seconds. When it is remagnetised,, so that the pole strength becomes 1/9th of, initial value, the period of oscillation in, seconds is, 1) 3, 2) 12, 3) 5, 4) 4, 33. The magnetic needle of a vibration, magnetometer makes 12 oscillations per, minute in the horizontal component of earth’s, magnetic field.When an external short bar, magnet is placed at some distance along the, axis of the needle in the same line it makes 15, oscillations per minute.If the poles of the bar, magnet are inter changed , the number of, oscillations it takes per minute is, 1) 61, 2) 63, 3) 65, 4) 67, 34. The magnetic needle of a V.M.M completes, 10 oscillations in 92seconds. When a small, magnet is placed in the magnetic meridian, 10cm due north of needle with north pole, towards south completes 15 oscillations in, 69seconds. The magnetic moment of magnet, , BH 0.3G , , is, , 1) 4.5Am 2, 2) 0.45Am 2, 3) 0.75 Am2, 4) 0.225Am 2, 35. A magnetic needle has a frequency of 20, oscillations per minute in the earth’s, horizontal field. when the field of a magnet, supports the earths horizontal field, the, frequency increases to 30 oscillations per, minute. The ratio of the field of the magnet, to that of the earth is, 1)4:7, 2)7:4, 3)5:4, 4)4:5, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , TYPES OF MAGNETIC MATERIALS, 36. A thin iron rod is cut into 10 equal parts, parallel to its length. The intensity of, magnetisation of each piece will be...., 1), , 1, th of initial value 2) 10 times initial value, 10, , 3) does not change, 4) become half, 37. The dipole moment of each molecule of, paramagnetic gas is 1.5 10 23 Am 2 .The, temperature of gas is 270 C and the number, of molecules per unit volume in it is, 2 1026 m 3 . The maximum possible intensity, of magnetisation in the gas will be ( in A/m) is, 1) 3 103 2) 4 103 3) 5 105 4) 6 104, 38. A paramagnetic sample shows a net, magnetisation of 8 A/m when placed in an, external magnetic field of 0.6T at a temperature of 4K.When the same sample is placed, in an external magnetic field of 0.2T at a temperature of 16K , the mangnetisation will be:, 1), , 42. The real angle of dip, if a magnet is suspended, at an angle of 300 to the magnetic meridian, and the dip needle makes an angle of 450 with, horizontal is, 3, , 1, 1) tan 2 , , , , 3, , LEVEL - II ( C. W ) KEY, 1) 3, 7) 4, 13) 1, 19) 3, 25) 1, 31) 2, 37) 1, 43) 1, , 6) 3, 12) 1, 18) 4, 24) 2, 30) 1, 36) 3, 42) 4, , In the arrangement magnetic moment of P is, 2, M1 , , 1) 7 x 10 T, 2) 6 x 10-5T, 3) 5 x 10-5T, 4) 9.2 x 10-5T., 40. The correct value of dip angle at a place is, 450 . If the dip circle is rotated by 450 out of, the meridian, then the tangent of the angle, of apparent dip at the place is, , NARAYANA GROUP, , 5) 4, 11) 2, 17) 1, 23) 2, 29) 3, 35) 3, 41) 2, , 2., , , , times per minute where dip is 300 . The total, magnetic field of earth at second to first, places is, 1) 1.51, 2) 1.83, 3) 1.63 4) 1.23, , 4) 1, 10) 1, 16) 2, 22) 3, 28) 1, 34) 3, 40) 4, , 1., , -5, , 1) 1, 2) 1/2, 3) 1/ 2 4) 2, 41. A compass needle oscillates 20 times per, minute at a place where dip is 450 and 30, , 3) 3, 9) 1, 15) 3, 21) 2, 27) 2, 33) 2, 39) 4, , , 2 M sin , 2, M1 , , , -5, , , , 2) 3, 8) 1, 14) 2, 20) 2, 26) 4, 32) 2, 38) 2, , LEVEL - II (C. W. ) HINTS, , 0, , sin 40.60 0.65, , 2 , , 0.4 104 T and 0.3 104 T respectively. The, resultant earth’s magnetic field is, 2) 104 T, 1) 0.5 10 4 T, 3) 2 104 T, 4) 5 104 T, , TERRESTRIAL MAGNETISM, , field BV = 6 x 10 T.. The total intensity of the, earth's magnetic field at this place is, , 3, , 1, 1, 3) tan 2 , 4) tan , 3, , , 43. At a place the value of BH and BV are, , 32, 2, A / m 2) A / m 3) 6 A / m 4) 2.4 A / m, 3, 3, , 39. The angle of the dip at a place is 40.6 and, the vertical component of the earths magnetic, , 1, 2) tan, , M, , sin, 2, 2 M and magnetic moment, , , , of Q is M 2 , , M, Resultant magnetic moment, 2, , M r M1 M 2 , 3., , M M M 2 , , , , 2, 2, , 2M sin 2 , ', M , , , Percentage change in the magnetic moment, M 'M, 100, M, m, B 0 2, 4 d, F = mB, , , 4., 5., , 133
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , 6., , 7., 8., , M = m x 2l, m, B o 2 and, 4 d, mm, m m Bm, F 0, o 2 , 2, 4 2d , 4 d, 4, 4, , B, , 0 m, 0 m1m2, , F, 2, 4 d, 4 d 2, , 9., , 21., , o 2 M cos 45, 2, 4, d3, , B1 sin 300 B2 sin 900, 2M, , 0, 22. BH Be 4 d 3, 1, , BH Ba , , 0 M, 4 d 23, , 23. Be BH and B Ba BH 2 Be BH, , 2 BH BH 3BH, , 3MB sin MB sin 90 , , 24., , 1, 300, 3, , tan , , 20. B , , o M 1, M1 M 2 , BH and o, BH, 3, 4 12, 4, 83, , I, I, 25. T 2 MB and T , M, H, , d, MB cos , d, , 26. T , , 10. C MB sin , 11. MBH sin Fr, 12. C (1800 45o ) M 1 B sin 45 0, , 1, B, , I, T1, 27 T 2 MB and T , H, 2, , M1 M 2, M1 M 2, , I, 28. T 2 MB, H, , I1, I, 1, , , I2, 3I, 3, , C (180 0 30 o ) M 2 B sin 30 0, , 13. Net torque, , B, N, , S, 1 2, , I, 29. T 2 MB, H, T1, and T , 2, , N, , S, , M 1B sin 1 M 2 B sin 180 2 , 14. In stable equlibrium PE is minimum and in, unstable equilibrium PE is maximum and , is zero in both cases, 6 M 1M 2, 15. F o, 4 d 4, 0 2Md, B, , axial, 16., 4 d 2 l 2 2, 17. BR B1 B2 ; BR B1 B2, 1, , 2, , 2, Beq2, 18. B Baxial, , 0 2 M, 0 M, 19. Ba , ; Be , 3, 4 d, 4 d 3, 134, , T1, and T , 2, , 30. T 2, , 2I, I1 M 2, M, , , , 21/ 4, I2 M1, I, 2M, 1, I, and T , B, MB, , MBH, 31. , I, , , , , , 32. T , , 1, and M m 2l, M, , 33. n1 , , 1, 2, , MBH, 1, , n2 , 2, I, , M BH B , I, , 1 M BH B , 2, I, where B is field strength due to external magnet, at the site of magnetic dipole ., , and n3 , , T1, 34. T1 9.2 S , T2 4.6 S and T , 2, , B2, B1, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , B BH, 4, BH B 5BH 1.5 10 T, , , , 9.2, , 4.6, , , , M o 2M, 1.5 104 M 0.75 Am 2, 4 d 3, , 20, 35. n B 30 , , LEVEL - II (H.W), MAGNETIC MOMENT AND, RESULTANT MAGNETIC MOMENT, 1., , BH, 5, B BH, B BH, 4, , 36. I is independent of dimensions, 37. I , , nM, V, , N, , 38. As I m H and m , , I, , S, , C, ( Curie law),, T, , N, , CH, I, CH 3 / T1 H 1 T2 , or 1 , , , T, I 2 CH 2 / T2 H 2 T1 , , S, S, , or, 2., H 2 T1 , 0.2T 4 K , M 2 M1 , 8 A / m , 0.6T 16 K , H1 T2 , , , 2, A/m, 3, , 39. BV B sin B , , BV, sin , , BV, 40. tan B and, H, , BV1, BV, tan 1 , BH BH cos 450, 1, , tan 1 , , tan , cos 45, , 2, BH 2 T12, B2 cos T1, , , , 41. B, 2, B1 cos 1 T2 2, T2, H1, 2, , , , Three identical thin bar magnets each of, moment M are placed along three adjacent, sides of a regular hexagon as shown in figure., The resultant magnetic moment of the system, is, , B2 T1 cos 1, , B1 T22 cos 2, , tan , cos , =true, dip,, , ' = dip with the field, = angle made by the meridian, , 42. tan ' , , 43. B BH 2 BV 2, NARAYANA GROUP, , N, , 1) M 2) M 3 3) M 2 4) 2M, The magnetic moment of a bar magnet is 0.256, amp.m2. Its pole strength is 400 milli amp. m., It is cut into two equal pieces and these two, pieces are arranged at right angles to each, other with their unlike poles in contact (or, like poles in contact). The resultant magnetic, moment of the system is, 3, 1) 2 256 10 Am2 2) 250 x 10-3Am2, , 256, 128, 103 Am2, 10 3 Am2, 4), 2, 2, COUPLE ACTING ON THE BAR MAGNET, 3. A bar magnet is suspended in a uniform, magnetic field in a position such that it, experiences maximum torque. The angle, through which it must be rotated from this, position such that it experiences half of the, maximum troque is, 1) 60°, 2) 30°, 3) 45°, 4) 37°, 4. If the maximum couple acting on a magnet in, a field of induction 0.2 tesla is 10 Nm, what is, its magnetic moment ?, 1) 50 Am2 2) 2 Am2 3) 5 Am2 4) 20 Am2, 5. A bar magnet of length 10 cm experiences a, torque of 0.141 N-m in a uniform magnetic, field of induction 0.4 wb/m2 , when it is, suspended making an angle 450 with the field,, the pole strength of the magnet is, 3), , 1) 5 A-m, , 2) 2.5 A-m, , 3) 10 A-m, , 4) 15 Am, 135
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 6., , A bar magnet of pole strength 2 A-m is kept in, a magnetic field of induction 4 105 wbm2, such that the axis of the magnet makes an, angle 300 with the direction of the field. The, couple acting on the magnet is found, 80 107 N-m . Then the distance between the, poles of the magnet is, 1) 20 m, 2) 2m, 3) 3cm, 4) 20 cm, 7. A magnet of magnetic moment 20 k̂ Am2 is, placed along the z - axis in a magnetic field, , ˆ T. The torque acting on the, B = (0.4jˆ 0.5k), magnet is, 1) 8 î N-m 2) 6 ĵ N-m 3) – 8 î N-m4) – 6 ĵ N-m, 8. The torque required to keep a magnet of, length 10cm at 450 to a uniform magnetic field, is 2 10 5 Nm . The magnetic force on each, pole is, 1) 0.2mN 2) 20 N 3) 0.02N 4) 2N, 9. A bar magnet of moment 40 A-m2 is free to, rotate about a vertical axis passing through, its centre. The magnet is released from rest, from east west direction. The kinetic energy, of the magnet as it takes north-south, direction is (B H = 30T ), 1) 0.6 mJ 2) 1.2 mJ 3) 2.4 mJ 4) 0.3 mJ, 10. A bar magnet of magnetic moment M is, divided into 'n' equal parts cutting parallel to, length. Then one part is suspended in a, uniform magnetic field of strength 2T and held, making an angle 600 with the direction of the, field. When the magnet is released the K.E, of the magnet in the equilibrium position is, 1), , M, J, n, , 2) Mn J, , 3), , M, J, n2, , 4) Mn2 J, , 13. Two north poles each of pole strength 8Am, are placed at corners A and C of a square, ABCD. The pole that should be placed at B, to make D as null point is, 1) North pole of pole strength 8 2Am, 2) North pole of pole strength 16 2Am, 3) North pole of pole strength 8 2Am, 14., , 15., , 16., , 17., , FIELD OF A BAR MAGNET, 11. A short bar magnet of magnetic moment, 12.8 103 Am2 is arranged in the magnetic, meridian with its south pole pointing, geographic north. If BH 0.4 gauss, the, distance between the null points is, 1) 4cm, 2) 8cm 3) 12cm 4) 16cm, 12. The magnetic field strength at a point a, distance ‘d’ from the centre on the axial line, of a very short bar magnet of magnetic, moment ‘M’ is’B’.Then magnetic induction, at a diatance 2d from the centre on the, equatorial line of a magnetic moment 8M will, be, 1) 4B, 2) B/2, 3) B/4, 4) 2B, 136, , 18., , 19., , 4) North pole of pole strength 16 2Am, Two short bar magnets of magnetic, moments 0.125 Am 2 . and 0.512 Am 2 are, placed with their like poles facing each, other . If the distance between the centres, of the magnet is 0.26m. The distance of, neutral point from the weaker magnet is, 1) 0.13 m 2) 0.2 m 3) 0.26 m 4) 0.1 m, TIME PERIOD OF OSCILLATION, A bar magnet of moment of inertia I is, vibrated in a magnetic field of induction, 0.4 104 T . The time period of vibration is, 12 sec. The magnetic moment of the magnet, is 120 Am 2 . The moment of inertia of the, magnet is (in kgm2) approximately, 1) 1728 102, 2) 172.8 104, 3) 2.1 2, 4) 1.5 102, A bar magnet has moment of inertia, 49 x102 kgm2 vibrates in a magnetic field of, induction 0.5 x104 Tesla. The time period of, vibration is 8.8sec . The magnetic moment of, the bar magnet is, (2007E), 2, 1) 350Am 2 2) 490 Am, 3) 490 Am 2 4)500Am2, A thin rod 30 cm long is uniformly magnetised, and its period of oscillation is 4 s. It is broken, into three equal parts normal to its length., The period of oscillation of each part is, 1)12 s, 2)6 s, 3)1.33 s 4)2.66 s, The moment of inertia as well as the, frequencies of two magnets are in the ratio, 3:4 the ratio of their magnetic moments is, 1) 27 : 64 2) 64 : 27 3) 9 : 16 4) 16 : 9, A magnet freely suspended in a vibration, magnetometer makes 40 oscillations per, minute at a place A and 20 oscillations per, min at a place B. If the horizontal component, of earth’s magnetic field at A is 36 106 T ,, then its value at ‘B’ is, 1) 36 106 T, 2) 9106 T, 3) 144 106 T, 4) 288106 T, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , 20. A magnetic needle pivoted through its centre, of mass and is free to rotate in a plane, containing uniform magnetic field 200 x 10-4T., When it is displaced slightly from the, equilibrium it makes 2 oscillations per second., If the moment of inertia of the needle about, the axis of oscillation is 0.75 x 10–5 kg m2, the, magnetic moment of the needle is, 1) 0.06 J/T, 2) 0.03 J/T, 3) 0.12 J/T, 4) 0.6 J/T, , PROPERTIES OF MAGNETIC, MATERIALS, 21. The magnetic susceptibility of a medium is, 0.825.Its relative permeability is, 1)1.825 2)825, 3)285, 4)1825, 22. A magnetic field strength (H) 3 x103 A m, produces a magnetic field of Induction (B)of, 12 telsa in an iron rod.The relative, permeability of iron is, 1) 105, 2) 104, 3) 103, 4) 102, 23. The magnetic moment of magnet of mass 75, gm is 9 10 7 Am 2 . If the density of the, material of magnet is 7.5 10, , 3, , kg, m3, , 24. A magnetising field of 5000 A m produces a, magnetic flux of 5105 weber in an iron rod., If the area of cross section of the rod is, 0.5 cm 2 , then the permeability of the rod will, be, 1) 1103 2) 2104 3) 3105 4) 4106, 25. A short bar magnet of magnetic moment, 20Am 2 has a cross sectional area of, 1.5104 m2 . If the intensity of magnetisation, of the magnet is 105A/m. The length of magnet, is, 1) 0.33m 2) 0.13cm 3) 1.33m 4) 1.33cm, , NARAYANA GROUP, , 3) 1, 9) 2, 15) 2, 21) 1, , 4) 1, 10) 1, 16) 4, 22) 2, , 5) 1, 11) 2, 17) 3, 23) 2, , M net M 1 M 2 M 3, , 2., , M1 M 2 , , 6) 4, 12)2, 18) 1, 24) 2, , M, 2, , M net M 12 M 22, , 3., 4., 5., 6., 7., 8., 9., , max MB, , MB sin , MB sin , M B, MB sin , K .E W MB 1 cos , , M, , W MB 1 cos , n, 0 2M, 11. BH , 4 d 3, 2M, M, 12. Ba o 3 ; Be o 3, 4 d, 4 d, 13. Fnet 0, 1, 10. M , , 14., , d, , x, , 1, , M 2 / M1 3 1, , 15. T 2, , I, MB, , 16. T 2, , I, MB, , 17. T 2, , I, MB, , 18. f , , 1, 2, , MB, I, , 19. T 2, , I, MB, , 1, 2, , MB, I, , 20. f , , LEVEL - II ( H. W ) - KEY, 2) 3, 8) 1, 14) 4, 20) 1, , 1., , then, , intensity of magnetisation will be, 1) 0.9 A m 2) 0.09 A m 3) 9 A m 4) 90 A m, , 1) 4, 7) 3, 13) 4, 19) 2, 25) 3, , LEVEL - II ( H. W ) - HINTS, , 21. r 1 , B, , 22. H & r , o, , 137
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 23. I , , M, V, , 24. B B. A, 6., , M m, , 25. I , V, A, , LEVEL - III, MAGNETIC MOMENT, 1., , 2., , A cylindrical rod magnet has a length of 5.0, cm and a diameter of 1.0 cm . It has a unifrom, magnetisation of 5.3 103 A / m What is its, magnetic dipole moment ?, 2) 10.8mJT 1, 1) 20.8mJT 1, 3) 5.8mJT1, 4) 30.8mJT 1, Find the resultant magnetic moment for the, following arrangement, M, , 7., , much should we rotate the wire in order to twist, the magnet through 450 from its original, position, 1) 2570, 2) 2520 3) 2750, 4) 1270, A magnetic dipole is under the influence of, two magnetic fields. The angle between the, field directions is 600 and one of the fields, has a magnitude of 1.2x10–2T. If the dipole, comes to stable equilibrium at angle of 150, with this field, then the magnitude of the other, field (sin150 = 0.2588), 1) 1.39x10–3T, 2) 2.39x10–3T, 3) 3.39x10–3T, 4) 4.39x10–3T, Two small magnets X and Y of dipole moments, M1 and M2 are fixed perpendicular to each, other with their north poles in contact. This, arrangement is placed on a floating body so as, to move freeely in earth's magnetic field as, shown in figure then the ratio of magnetic, moment is, B, H, , M, 900, , 1), 3), , 2), , 2M, , , , , , 2 1 M, , , , , , N2 N1, , 2 1 M, , 4) M, , 1) 1: 3, , 0, , 30, M, , COUPLE ACTING ON THE BAR, MAGNET, 3. A bar magnet with poles 25.0 cm apart and of, pole strength 14.4 Am rests with its centre, on a friction less point. It is held in equilibrium, at 600 to a uniform magnetic field of induction, 0.25 T by applying a force F at right angle to, the axis, 12 cm from its pivot. The magnitude, of the force is, 1) 15 3N 2) 75 3N 3) 3.75 3N 4) 25 3 N, 4. A magnet is suspended in the magnetic, meridian with an untwisted wire. The upper, end of the wire is rotated through 1800 to, deflect the magnet by 300 from magnetic, meridian. Now this magnet is replaced by, another magnet. Now the upper end of the, wire is rotated through 2700 to deflect the, magnet 300 from the magnetic meridian. The, ratio of the magnetic moments of the two, magnets is, 1) 3 : 4, 2) 1 : 2 3) 4 : 7, 4) 5 : 8, 5. A magnet is suspended in a uniform magnetic, field by a thin wire. On twisting the wire, through half revolution, the magnet twists, through 300 from the original position. How, 138, , 2) 2 : 3, , 3), , 3:2, , 4), , 3 :1, , 60, , S1, , 0, , S2, , FIELD OF A BAR MAGNET, 8., , Two magnets A and B are identical and these, are arranged as shown in the figure. Their, length is negligible in comparision with the, seperation between them. A magnetic needle, is placed between the magnets at point P, which gets deflected through an angle , under the influence of magnets. The ratio of, distances d1 and d2 will be, B, A, P, , d1, , 1. (2 cot ), , 1, , 3, , 1, , d2, , 2. (2 tan ), , 1, , 3, , 1, , 3. (2cot ) 3, 4. (2 tan ) 3, TIME PERIOD OF SUSPENDED MAGNET, IN THE UNIFORM MAGNETIC FIELD, 9. When a bar magnet is placed at some distance, along the axis of the magnetic needle of an, oscillation magnetometer located in earth’s, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , 10., , 11., , 12., , 13., , 14., , magnetic field, the needle makes 14, oscillations per minute. If the bar magnet is, turned so that its poles exchange their, positions, the needle makes 20 oscillations per, minute. If the magnet is completely removed,, the frequency of the needle is nearly (assume, B>BH at needle), 1) 20 oscillations/minute, 2) 15 oscillations/minute, 3) 5 oscillations/minute, 4) 10 oscillations/minute, A vibration magnetometer consists of two, identical bar magnets placed one over the, other such that they are mutually, perpendicular and bisect each other. The time, period of oscillations of combination in a, horizontal magnetic field is 4s. If one of the, magnets is removed, then the period of, oscillations of the other in the same field is, 1) 2 14 sec 2) 2 5 4 sec 3) 2 7 4 sec 4) 2 34 sec, A magnet is suspended in such a way that it, oscillates in the horizontal plane. If it makes, 20 oscillations per minute, at a place, 0, where dip angle is 30 and 15 oscillations per, minute at a place where dip angle is 600 .The, ratio of total earth’s magnetic field at the two, places is, 1) 3 3 : 8 2) 16 : 9 3 3) 4 : 9 4) 2 3 : 9, A thin rectangular magnet suspended freely, has a period of oscillation equal to T. Now it, is broken into two equal haves (each having, half of the original length) and one piece is, made to oscillate freely in the same field. If, T', its period of oscillation is T ' then, is, T, 1, 1, 1, 1), 2), 3), 4) 2, 2 2, 4, 2, A compass needle makes 10 oscillations per, minute in the earths horizontal field.A bar, magnet deflects the needle by 600 from the, magnetic meridian. The frequency of, oscillation in the deflected position in, oscillations per minute is (field due to magnet, is perpendicular to BH), 1) 5 2, 2) 20 2 3) 10 2 4) 10, Two bar magnets are placed in vibration, magnetometer and allowed to vibrate. They, make 20 oscillations per minute when their, similar poles are on the same side, while they, , NARAYANA GROUP, , MAGNETISM, make 15 osillations per minute when their, opposite poles lie on the same side. The ratio, of their magnetic moments is, (Eamcet (M) 2008, E(2009), 1) 7 : 25 2) 25 : 7 3) 25 : 16 4) 16 : 25, 15. With a standard rectangular bar magnet the, time period of a vibration magnetometer is 4, seconds. The bar magnet is cut parallel to its, length into four equal pieces. The time period, of vibration magnetometer when one piece is, used (in seconds) (bar magent breadth is, small.) is, ( E-2008), 1) 16, 2) 8, 3) 4, 4) 2, , TYPES OF MAGNETIC MATERIALS, 16. The relation between and H for a specimen, 0.4, , 12 104 henry/metre., of iron is , H, , The value of H which produces flux density, of 1 tesla will be, 2) 500 A / m, 1) 250 A / m, 3) 750 A / m, 4) 103 A / m, 17. The mass of a specimen of a ferromagnetic, material is 0.6kg. and its density is, 7.8 103 kg / m3 . If the area of hysteresis loop, of alternating magnetising field of frequency, 50Hz is 0.722 MKS units then the hysteresis, loss per second will be, 1) 277.7 105 Joule 2) 277.7 106 Joule, 3) 277.7 104 Joule 4) 277.7 103 Joule, 18. 300 turns of a thin wire are uniformly wound, on a permanent magnet shaped as a cylinder, of length 15cm. When a current 3A is passed, through the wire, the field outside the magnet, disappears. Then the coercive force of the, material is, 1) 2 kNm 1 2) 4 kNm 1 3) 5 kNm 1 4) 6 kNm 1, 19. At a temperature of 300 C , the susceptibility, of ferromagnetic material is found to be ‘ ’., Its susceptibility at 3330 C is, , 1) , 2), 3) 2 , 4) 11.1 , 2, 20. What will be the energy loss per hour in the, iron core of a transformet if the area of its, hysteresis loop is equivalent to 2500erg/cm 3 ., The frequency of alternating current is 50 Hz., The mass of core is 10 Kg and the density of, iron is 7.5gm/cm 3 ., 1) 2 102 Joule, 3) 6 104 Joule, , 2) 4 103 Joule, 4) 8 105 Joule, 139
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, 21. Find the percentage increase in the mag netic, field B when the space within the current, carrying toroid is filled with aluminium. The, susceptibility of aluminium is, 2.1105, 1) 3.1 103, 3) 2.1 103, , 1) tan 2 tan 2 1 tan 2 2, , 2) 1.1 103, 4) 2.1105, , TYPES OF MAGNETIC MATERIALS, 22. A rod of ferromegnetic material with, dimensions 10cm x 0.5cm x 2cm is placed in a, magnetising field of intensity 2x105A/m. The, magnetic moment produced due it is 6, amp m 2 . The value of magnetic induction, will be-----10–2T., 1)100.48, 2)200.28 3)50.24, 4)300.48, 23. A magnetic material of volume 30cm3 is, placed in a magnetic field of intensity 5, oersted.The magnetic moment produced due, it is 6 amp m 2 .The value of magnetic, induction will be., 1)0.2517 Tesla, 2)0.025 Tesla, 3)0.0025 Tesla, 4)25 Tesla, 24. The total magnetic flux in a material, which, produces a pole of strength mp when a, magnetic material of cross- sectional area A, is placed in a magnetic field of strength H,, will be, , , , 1) 0 AH m p, , , , 2) 0 AH, , , , , , 3) 0 m p, 4) 0 m p AH A, 25. Relative permittivity and permeability of a, material are er and mr , respectively. Which of, the following values of these quantities are, allowed for a diamagnetic material ?, (AIE 2008), 1) er 0.5, mr 1.5 2) er 1.5, mr 0.5, 3) er 0.5, mr 0.5 4) er 1.5, mr 1.5, , TERRESTRIAL MAGNETISM, 26. The angle of dip at a place is . If the dip is, measured in a plane making an angle with, the magnetic meridian, the apparent angle of, dip 1 will be, 1) tan 1 tan , 3) tan 1 tan sec , 140, , 2) tan 1 tan cos , 4) 0, , 27. If 1 and 2 be the angles of dip observed in, two vertical planes at right angles to each, other and is the true value of dip then, , 2), , c o t 2 c o t 2 1 co t 2 2, , 3), , ta n 2 , , ta n 2 1 t an 2 2, ta n 2 1 ta n 2 2, , 4) cot 2 1 cot 2 1 cos 2 2, 28. A magnet makes 10 oscillations per minute, at a place where the angle of dip is 450 and, the total intensity is 0.4 gauss. The number, of oscillations made per sec by the same, magnet at another place where the angle of, dip is 600 and the total intensity 0.5 gauss is, approximately., 1) 6Hz, , 2), , 1, Hz, 1.06 6, , 3) 6 1.06 Hz, , 4), , 1, Hz, 6, , 29. The horizontal component of earth’s magnetic, field at place is 0.36x104 weber m2 . If the angle, of dip at that place is 60 then the value of, vertical component of earth’smagnetic field, will be (in wb m2 ), 1) 6 x105 T, 3) 3.6 3 10 5 T, , 2) 6 2 x105 T, 4) 2 x105 T, , 30. An iron rod is subjected to cycles of magnetisation, at the rate of 50Hz. Given the density of the rod, is 8 103 kg / m3 and specific heat is 0.11 10–3, cal / kg0C. The rise in temperature per minute,, if the area enclosed by the B – H loop, corresponds to energy of 10–2 J is (Assume, there is no radiation losses), 1) 780C, 2) 880C 3) 8.10C 4) none of these, , LEVEL - III - KEY, 1) 1, 7) 4, 13) 3, 19) 2, 25) 3, , 2) 2, 8) 1, 14) 3, 20) 3, 26) 3, , 3) 3, 9) 4, 15) 3, 21) 3, 27) 2, , 4) 4, 10) 3, 16) 2, 22) 1, 28) 2, , 5) 1, 11) 2, 17) 1, 23) 1, 29) 3, , 6) 4, 12) 3, 18) 1, 24) 1, 30) 3, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- III, , LEVEL-III - HINTS, 1., , m, M or m MV, V, , or m M r 2 l , 2, , 5.3 103 3.14 5 103 5 102 J / T, 20.8 103 J / T 20.8mJ / T, , , , , , 2., , M 2M M, , 3., , Anticlockwise torque = clockwise torque., F 0.12 MB sin , F, , 4., , 2 1, , 3.6 0.25 3 / 2, 3.75 3N, 0.12, , C 180 30 M1 BH Sin30 1, C 270 30 M 2 BH Sin30 2 , Divide, , I, 11. T1 2 MB cos 30 3, 1, I, and T2 2 MB cos 60 4, 2, 2, , mass× length , 12. Moment of inertia =, 12, When magnet is divided into two equal halves,, mass is reduced by a factor of 2 and length is, also reduced by factor of 2. S new moment of, 1, th of the initial moment of inertial., 8, Also, magnetic moment = pole strength length, Pole strength is unchanged and the length is, halved. So, new magnetic moment is one-half, of the intial magnetic moment., , inertia is, , I', M 'B, , T ' 2, , M1 5, , M2 8, , 5., , MAGNETISM, , C 180 30 M 1BH Sin30 1, , 2, Now,, , C 45 MBH Sin45 2 , T' 1, , T 2, , C1, Divide C, 2, , 6., , MB1Sin150 MB2 Sin 450, B1 1.2 10, , 7., , 8., 9., , I /8, T, T, , , M, 4 2, B, 2, , 13. n1 , , 2, , M BH, I, , 1, 2, , ________ (1), , M 1 B Sin300 M 2 B Sin600, , 1, n2 , 2, , M1, 3, , M2, 1, , B BH Tan60, , BA BH tan , , Solving n2 2 n1, , M BH2 B 2, I, , B 3BH, , ______ (2), , _________ (3), , 2, 1, , n K ( B BH ), , n2 10 2, , 2, , n2 K ( B BH ), 2, , n3 KBH, T1, 10. T , 2, , I1 M 2, , I 2 M1, , NARAYANA GROUP, , 14. n , , 1, 2, , 20 2, , m BH, I, , M 1 M 2 BH, 2I, 141
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JEE-ADV PHYSICS- VOL- III, , MAGNETISM, , 15 2, , 21. In the absence of A1, B0 0 H and in the, presence of A1, , M1 M 2 B, , H, , 2I, , B H 0 1 H, % age increase, , M1 M 2, M1 M 2, , 4, , 3, , M 1 25, Solving M 7, 2, , 15. T 2, , T 1 2, , , , 100 2.1 10 5 100 2.1 10 3, , I, 4, MBH, , 22. I , , I /4, 4, M / 4. BH, , B 0 H I , 23. H = 50 ersted, , 0.4, , B, 12 x104 H, H, , , 5, , 17. WH VAft, , 24. BA, , B 0 H I , , 0.6, , 0.722 50, 7.8 103, , 25. for diamagnetic r 1, , 5, , 277.7 10 Joule, 18. The magnetic field at the centre of a long straight, conductor, whose cross section is in the form of a, thin half ring is, , 0 I, 1, 2R, So the force per unit length of the wire carrying, current I and placed on the axis of the first, conductor is, F BI l, B, , 0 I 2, 2R, , 19. , , M, 6, 103, , AM 1 ; I , V 30 10 6, 4, , B 0 H I , , m, Aft, d, , , , M, 6, , V 10 0.5 2 106, , H 2 105 A / M, , 16. B H, , WH , , B B0, H, 100 0, 100, B0, 0 H, , r 1, 26. tan , , V, V, 27. tan 1 H cos cos H tan ....(1), 1, V, V, sin , H sin , H tan 2 ...(2), Squaring and adding Equations (1) and (2), tan 2 , , (for unit length) 2 , 28. T , , 1, T, , 20. Total energy loss E nAVt nA, 50 2500 101 , , m, t, d, , 10, 3600, 7.5 103, , V, V, and tan 1 , H, H cos , , 1, n 2 B cos 1, ; n B and 12 1, n2 B2 cos 2, B, , 29. BH 0.36 104 T, Tan , , BV, BH, , 30. 50 60 102 d 0.11 103 4.2 l, , 6 104 Joule, 142, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , ELECTRO MAGNETIC INDUCTION, , , SYNOPSIS, , , , , n B then magnetic flux linked with the surface, , The phenomenon in which electric current is, induced by varying magnetic fields is called, electromagnetic induction., Magnetic Flux ( ) : The number of magnetic, lines of force passing normally through given, area is called magnetic flux., , A, , l, ma, nor, , , , 0, is zero i.e., 0 90 , , ii) When the plane of the surface is perpendicular, to magnetic field (or) normal drawn to the surface, , is parallel to the magnetic field n B , then, , , , , , magnetic flux linked with the surface is, maximum. i.e., max BA 00 , , B, , iii) When the flux entering the surface is opposite, When a surface of area A is placed in a uniform, magnetic field of induction B, such that the unit , vector along the normal ( n ) makes an angle ‘ ’, with direction of magnetic field then the flux, passing through it is given by, B. A BA cos , , , , If magnetic field is non uniform then B.ds, , , , The SI Unit of flux is weber (Wb)., CGS unit of flux is maxwell (Mx), , 1 weber = 1 tesla - meter2, 1 weber = 108 maxwell, Dimensional formula of the magnetic flux is, , , , , , ML2T 2 A1, Magnetic flux is a scalar, Magnetic flux can be positive, negative or zero, depending upon the angle between area vector, and field direction., When a cylinder is placed in a uniform magnetic, field as shown in the below figure, n B, n, BA, n B, , =0, C, , B, , n, , A, , B, , =BA, n B, , i) When the plane of the surface is parallel to the, direction of the magnetic field (or) normal drawn to, the surface is perpendicular to the magnetic field, NARAYANAGROUP, , to the area vector ( n ) then BA 1800 , The magnet ic flux linked with a coil, , NBA cos , , can be changed by, , a) Changining the no. of turns (N), b) Varying the magnetic field (B), c) Changing the area of the magnetic field, bounded by the coil by moving the coil into or, out of the magnetic field, d) Changing the angle made by the coil with the, direction of the field, , The change of flux due to rotation of the, coil: When the coil is rotated from an angle of, .t, 1 to an angle of 2 (both are measured w.r.t, normal) in a uniform magnetic field then the, initial flux through the coil is, i NBA cos 1, The final flux through the coil after rotation is, f NBA cos 2, , The change in the flux associated with the coil, is, f i, , NBA cos 2 cos 1 , if 1 00 and 2 900 then NBA, if 1 900 and 2 1800 then NBA, if 1 00 and 2 1800 then 2NBA, , 1
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , W.E-1: A rectangular loop of area 0.06 m2 is placed, , Where ‘ N ’ is total flux linked with the coil of N, turns., (or), , in a uniform magnetic field of 0.3T with its, plane (i) normal to the field (ii) inclined 300, to the field (iii) parallel to the field. Find the, flux linked with the coil in each case., Sol: NBA cos , , d, d, N NBA cos , dt, dt, Negative sign is in accordance with Lenz’s law., The above law is also called Neumann’s law., e, , i) 1 0.06 0.3 cos 00 0.018 weber, ii) 1 0.06 0.3 cos 600 0.009 weber, , LENZ’S LAW AND CONSERVATION, OF ENERGY, , iii) 1 0.06 0.3 cos 900 0, W.E-2: At a certain location in the northern, hemisphere, the earth’s magnetic field has, magnitude of 42T and points downwards , at 530 to the vertical. Calculate the flux, through a horizontal surface of area, , 2.5m 2 . sin 530 0.8 , , “The direction of the induced emf is always such, that it tends to produce a current which opposes, the change in magnetic flux”, Induced emf can exist whether the circuit is, opened or closed. But induced current can exist, only in the closed circuits., A metallic ring is held horizontally and a bar, magnet is dropped through the ring with its length, along the axis of the ring, as shown in figure., , 530, , B, , A, , Sol: B BA cos , 42 106 2.5 cos530 63Wb, , FARADAY’S LAWS OF ELECTRO, MAGNETIC INDUCTION, First Law : Whenever the magnetic flux linked with, an electric circuit (coil) changes, an emf is, induced in the circuit (coil). The induced emf, exists as long as the change in magnetic flux, continues., Second Law : The induced emf produced in the coil, is equal to the negative rate of change of magnetic , flux linked with it., d, dt, where flux through each turn, If the coil contains N turns, an emf appears in, every turn all these emfs are to be added. Then,, the induced emf is given by, , In both the cases net force on the magnet is, Fnet mg f, Hence net acceleration of the fall is, f, a net g, m, where f=force exerted by the induced magnetic, field of ring on the magnet., When the magnet is allowed to fall through an, open ring (or) cut ring, then, a net g , , e, , d, d, e N., N , dt, dt, 2, , S, N, , mg, , Cut ring, a) an emf is induced, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , b) No current is induced (since the ring is not closed), and hence no induced magnetic field., , c) No opposition to the motion of the magnet., d) Fnet mg, , , , , , , , e) a net g Magnet falls with an acceleration = g, When a magnet is allowed to fall through two, identical metal coils at different temperatures, then magnet falls slowly through the coil at low, temperature as its resistance is less more induced, current flows so more is the opposition., A magnet allowed to fall through a long, cylindrical pipe then the acceleration of magnet, is always less than ‘g’ and the acceleration, continuously decreases due to induced currents., But the velocity increases until the magnet moves, with acceleration. At a particular instant the, acceleration becomes zero and the magnet moves, downwards with uniform velocity, called , terminal velocity., When north pole of a magnet is moved, perpendicular to the plane of the coil as shown, in figure, then, , Induced, Current, , N, , N, , A, , (because there is no change of flux linked with, the coil), When a current carrying conductor is placed, beside a closed loop in its plane then the induced, current direction for the following are, a) Current in conductor is constant., , BO, Bi , , A, , CW, B, , I, C, I= Increasing, = Increasing, , BO - Original field, Bi - Induced field, , (B), , S, , N, , , , a) emf is induced, b) Induced current flows from A to Balong the , coil when A and B are connected through, , resistor., c) Electrons flow from B to A along the coil, , NARAYANAGROUP, , No induced current, , Ii, , B, , N, , Clockwise induced current, , No induced current, b) Current through the conductor increases as, shown., , a) emf is induced, b) Induced current flows from B to A along the, coil when A and B are connected through, resistor., c) Electrons flow from A to B along the coil, d) Hence plate A will become positively charged, and plate B becomes negatively charged., When the two magnets are moved perpendicular, to plane of coil as shown, then, S, , Clockwise induced current, , S, , A B, Capacitor, , , , d) Hence plate A will become negatively charged, and plate B becomes positively charged., The directions of induced current in coil for different, kinds of motion of magnets, , In this case, the flux through the loop due to, current carrying wire is out of the plane of the, coil., As current is increasing, the outward flux through, the coil also increases., Hence to oppose this, an inward flux is created, by the clock wise induced current., c) Current through the conductor decreases as, shown., 3
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , c) Instantaneous emf e AN cos , , , dB, dt, , If ‘A’ is changed then, a) Average induced emf, e BN cos , , A2 A1 , t2 t1 , , b) If the plane of the coil is perpendicular to, magnetic field, then 00 cos 1, In this case, the flux through the loop due to current, carrying wire is out of the plane of the coil., As current is decreasing, the outward flux through, the coil also decreases., Hence to oppose this, an outward flux is created, by the anti-clock wise induced current., , EXPRESSIONS FOR INDUCED EMF,, INDUCED CURRENT AND, INDUCED CHARGE, , , If the coil has N turns then e N, , e N, , , , , , d, dt, , d, dt, , 2 1 , dt, , d, dt, The emf is induced (or) change in flux is caused, by changing B (or) A (or) N (or) , If ‘B’ is changed then, a) Average induced emf, , As BAN cos and e , , B B1 , e AN cos 2, t2 t1 , Here B1 is magnetic field induction at an instant , , t1 B2 is magnetic field induction at an instant t2, b), If the plane of t he coil is, perpendicular t o magnet ic field, then, 00 cos 1, , B2 B1 , then e AN t t, 2 1, 4, , A2 A1 , t2 t1 , , dA, dt, If ‘ ’ is changed (i.e., if coil is rotated), a) Average induced emf, , c) Instantaneous emf e BN cos , , According to Faraday’s second law and Lenz’s, law the induced emf is given by e , , , , then e BN, , e BAN, , cos 2 cos 1 , t2 t1 , , d, cos , dt, If the coil is rotated with constant angular velocity, ‘ ’ then t and, , b) Instantaneous emf e BAN, , e BAN, , d, cos t BAN sin t, dt, , e BAN sin t, c) t 900 , if the plane is parallel to the, magnetic field then induced emf is maximum., Then Peak emf., e0 BAN , e e0 sin t, This is the principle of AC generator., , INDUCED CURRENT, If the magnetic flux in a coil of resistance R, changes from 1 to 2 in a time ‘dt’, then a, current ‘i’ is induced in the coil as i , , e, R, , N 2 1 , , d , , e N . , Rdt, dt , , Induced current is given by, i, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, Magnitude of current, i, , W.E-4: A circular coil of 500 turns of wire has an, , Induced emf, N d , , , Resistance in the circuit R dt , , INDUCED CHARGE, , , ELECTRO MAGNETIC INDUCTION, , The amount of charge induced in a conductor is, given as follows, 1 d , e, (or) I , R dt , R, dq, 1 d, 1, , , (or) dq d , dt, R dt, R, , We know, I , , enclosed area of 0.1m2 per turn. It is kept, perpendicular to a magnetic field of induction, 0.2T and rotated by 1800 about a diameter, perpendicular to the field in 0.1s. How much, charge will pass when the coil is connected, to a galvanometer with a combined resistance, of 50 ., Sol: q , , i f, R, , , , NBA NBA 2 NBA, , R, R, , 2 500 0.2 0.1, 0.4 C ., 50, d, , , W.E-5: Some magnetic flux is changed from a, i, coil of resistance 10 . As a result an induced, f, 1, q f i (or) q i, (magnitude, current is developed in it, which varies with, R, R, tim e as shown in figure. What is the, of charge), magnitude of change in flux through the coil, In general, induced charge is given by, ?, change of magnetic flux, q, , resistance, Sol: The induced charge is q , R, N, But,, Area, of, i-t, curve, gives, charge, For N turns, the induced charge is q d , R, R Area of i t curve ; qR, Induced emf is independent of total resistance, of the circuit but depends on time of change of, flux., Induced current depends on both time of change, 1, of flux and resistance of circuit, 4 0.1 10, 2wb, 2, Induced charge is independent of time but, depends on the resistance of circuit., When a magnet is moved towards a stationary, W.E-6: A long solenoid with 1.5 turns per cm has, coil(i) slowly and (ii) quickly, then, a small loop of area 2.0 cm 2 placed inside the, a) induced charge is same in both cases, b) induced emf is more in second case, solenoid normal to its axis. If the current in, c) induced current is more in second case, the solenoid changes steadily from 2.0 A to, W.E-3: The magnetic flux through a coil, 4.0 A in 1.0s. The emf induced in the loop is, perpendicular to its plane is varying according Sol: The magnetic field along the axis of solenoid is, B 0 ni where n is no. of turns per unit length., to the relation B 5t 3 4t 2 2t 5 weber.., flux through the smaller loop placed in solenoid, Calculate the induced current through the, is B A Since current in solenoid is, coil at t 2 second. The resistance of the, changing, emf induced in loop is, coil is 5 ., 1, Induced charge, q R, , f, , Sol: 5t 3 4t 2 2t 5, , q, , e, , di , d d, 0 niA ; e 0 nA , dt dt, dt , , d, 15t 2 8t 2, dt, at t 2 sec, e 78V, , 42, 4107 1.5 102 2 104 , , 1 0 , , i 5 15 4 8 2 2 i 15.6 A, , 0.75 106 V, , e , , NARAYANAGROUP, , 5
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , W.E-7: A sqaure loop of side 10cm and resistance, , Therefore, change in flux,, B f i ; 8 103 Wb, , 0.5 is placed vertically in the east-west, plane. A uniform magnetic field of 0.10T is, B, set up across the plane in the north-east, (a) Given t 0.01 s, R 5 ; e , direction. The magnetic field is decreased to, t, zero in 0.70s at a steady rate. The magnitude, e 0.8, 8 103, of current in this time-interval is., , 0.8V ; or i , 0.16 A, 0.01, R, 5, Sol: The initial magnetic flux is given by, and q it 0.16 0.01 ; 1.6 103 C, BA cos , Given, B=0.10 T, area of square loop, B, b) t 0.02s ; e t, 2, 2, 2, 10 10 100cm 10 m, , , , , , 0.1102, 2, , Wb, , , , Final flux, min 0, The change in flux is brought about in 0.70 s, The magnitude of the induced emf is, 3, , 0, 10, , , 1 mV, t, t, 2 0.7, The magnitude of current is, e, , e 103, , 2 mA, R 0.5, WE-8: A square loop ACDE of area 20cm 2 and, resistance 5 is rotated in a magnetic field, B = 2T through 1800, a) in 0.01 s and, b) in 0.02 s., Find the magnetiude of e,i and q in both the, cases., I, , , , A, , , , , , , C, , , E, , , , B, , , D, , , , Sol: Let us take the area vector S perpendicular to, plane of loop inwards. So intially dS parallel, to B and when it is rotated by 1800 , S is anti, parallel to B. Hence, initial flux passing through, the loop,, , i BS cos 00 2 20 10 4 1, , 4 103 Wb, Flux passing through the loop when it is rotated, by 1800 , f BS cos1800, 2 20 10, 6, , 4, , 1, , 4.0 103Wb, , e 0.4, 8 1 0 3, 0.08 A, ; = 0.4V ; i , 0 .0 2, R 5, , and q it 0.08 0.02 ; 16 103 C, , MOTIONAL EMF, , , The motional emf is the emf which results from, relative motion between a conductor and the, source of magnetic field., When a condutor of length l is moved with a, velocity v perpendicular to its length in uniform, magnetic field (B), which is perpendicular to, both its length and as well as its velocity, the, emf induced across its ends e=Blv, If the rod moved making an angle with its, length, then e Blv sin , , , In vector form e B. l v or l. v B, , , among B, l and v , if any two are parallel the, emf induced across the conductor is zero, W.E-9: A rectangular loop of length ‘l’ and, breadth ‘b’ is placed at a distance of x from, an infinitely long wire carrying current ‘i’, such that the direction of current is parallel, to breadth. If the loop moves away from the, current wire in a direction perpendicular to, it with a velocity ‘v’, the magnitude of the, e.m.f. in the loop is : ( 0 permeability of, free space), , , , , , , , v, , Sol:, , b, , i, , l, x, , emf = Blv =Bbv B1 B2 bv, i, 0i , ibv 1 1 0ilbv, 0 , bv ; 0 , ; 2x x l , 2 x l x , 2x 2 x l , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , FLEMINGS’S RIGHT HAND RULE, , WE-10: A horizontal magnetic field B is produced, across a narrow gap between the two square , iron pole pieces. A closed square loop of side, a, mass m and resistance R is allowed to fall, with the top of the loop in the field. The loop, attains a terminal velocity equal to :, , a, , Stretch the first three fingers of right hand such that, they are mutually perpendicular to each other. If, the fore finger represents the direction of magnetic, field and the thumb represents the direction of the, motion of the conductor, then the central finger, indicates the direction of induced current, , Motion of, conductor, , B, , a, Sol: Induced emf in the loop, when it is falling with, terminal velocity, , Field, , e Bva, , R, R, Vertically upward force experienced by loop due, to this, , e = BVa ; i , , Bva , B 2va 2, , B, a, , F = Bia ;, , ;, R , r, When the loop attains terminal velocity ‘v’, , Thumb, , Index, finger, , , , mgR, B 2va 2, ; V 2 2, Ba, R, W.E-11: A conducting wire of mass m slides, down two smooth conducting bars, set at an, angle to the horizontal as shown in figure., The seperation between the bars is l . The, system is located in the magnetic field B ,, perpendicular to the plane of the sliding wire, and bars. The constant velocity of the wire is, , Central, finger, Induced Current, A conductor of length ‘l’ measured from P to Q is, moved with a speed of ‘v’ in a uniform magnetic, field ‘B’ as shown in figure., v, y, B, , mg , , R, , Q, z, , , Here B B(k ), l l ( i ), , and v v cos i v sin j, Induced emf is, , e l. v B l ( i ). v cos i v sin j B(k ), , P, , , , B, , , , , , Sol: Along inclined plane the force acting downwards, mg sin .......(1), magnetic force acting upwards, , x, , , , , , , , , , , , Blv sin , The change in the flux in the time of ‘ t ’ is, et Blv sin t, A conductor of length ‘l’ is bent at its midpoint, and is moved along its perpendicular bisector, with a constant speed of ‘v’ in a uniform magnetic, field of strength ‘B’ as shown in figure, P, l/2, , 2 2, , Blv , Bl v, ......(2), F Bil B R l ; , , , R, From (1) and (2), , B 2l 2v, mg sin ;, R, NARAYANAGROUP, , v, , mgR sin , B 2l 2, , , , , V, , x, , l/2, , Q, , 7
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, From the figure sin , , x, l, x sin , l/2, 2, , , , Here B B(k ), v vi, and effective length of the conductor, , l 2 x( j ) l sin ( j ) Induced emf is, , , e l . v B l sin ( j ).vi B ( k ) Blv sin , , , , , , , , The change in the flux associated in time internal, of ‘ t ’ is et Blv sin t, Here the effective length between free ends of, conductor is l sin ., The emf induced across the ends of the conductor, shown in the figure is, , , , N, , w, s, Bv, , BH, n E, , v, 0 and e , , l, e=0, (B), , v, , d, 0, dt, , iii) If the wire is moving in a horizontal plane in any, direction as shown in figure(C), it will cut flux of, BV(as BH will always be parallel to area) and so, , BH, , l, , l1, , Fm, , Bv, , v, , e=Bv lv, , l2, , e BVl BV l1 sin 1 l2 sin 2 , , C , d V, BV vl, V Bvls and e , dt, , i) If a conductor is moving vertically downwards, ds , , with constant velocity v with its ends pointing, with v dt , , , east-west, it will cut the horizontal component, of earth’s field BH as shown in figure (A) and WE-12: A wire of length 2l is bent at mid point so, hence the flux linked with the area generated by, that the angle between two halves is 600. If it, the motion of the conductor, and induced emf, moves as shown with a velocity v in a magnetic, d H, field B find the induced emf., BH v y l, will be H BH ly and e , dt, , FmB, , H, , B, y, , dy , , with v y , , dt , , , W, , l, e=BHvl, (A), , A, B, F, , Sol: e=Blv. Here l = Effective length = PQ, W.E-13 : A conductor of length 0.1m is moving with, a velocity of 4m/s in a uniform magnetic field of, 2T as shown in the figure. Find the emf induced?, , ii) However in case of vertical motion, if the, ends of the conductor point north-south, both BH, and BV will be parallel to the plane of area, generated by the motion of the conductor as, shown figure (B) and hence it doesn’t cut the, magnetic lines. so, Sol: e=Blv sin 900 = (2) (0.1) (4) = 0.8 Volt, 8, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , WE-14: Figure shows a conducting rod PQ in, contact with metal rails RP and SQ, which, are 0.25m apart in a uniform magnetic field, of flux density 0.4T acting perpendicular to, the plane of the paper. Ends R and S are, connected through a 5 resistance. What is, the emf when the rod moves to the right with, a velocity of 5ms 1 ? What is the magnitude, and direction of the current through the, 5 resistance? If the rod PQ moves to the, left with the same speed, what will be the new, current and its direction ?, , ELECTRO MAGNETIC INDUCTION, The effective electric circuit can be redrawn as, shown in Fig., , A, , E, , i1, R1, , B, i2, , i, , R2, , e=Bvl, , D, , F, , C, , The resistance R1 and R2 are in parallel, so the, 1 1, 1, equivalent resistance R is given by R R R, 1, 2, From Ohms law, the total current is i , , e, R, , 1, 1 , i Blv , R1 R2 , , -, , Sol: e Blv 0.4 0.25 5 0.5V, , BlV, Current in AD is i1 R ; Current in BC is, 1, , e 0.5V, BlV, , 0.1A, i2 , R, 5, R2, As the rod ‘PQ’ moves to right as shown, the, free electrons in it experience a Lorentz force. WE-16: A rectangular loop with a slide wire of, length l is kept in a uniform magnetic field, According to F.L.H., the force is towards the, as shown in the figure. The resistance of, end ‘Q’ of rod. They move from P to Q,, slider is R. Neglecting self inductance of the, hence the end of the rod P becomes deficient of, loop find the current in the connector during, electrons VP VQ, its motion with a velocity v., Applying Flemming’s right hand rule, the current, in the rod shall flow from Q to P., V, R2, R1, l, (b) : If the rod PQ moves to the left with the, B, same speed, then the current of 0.1 A will flow, inthe rod PQ from P to Q, Sol: The equivalent circuit is, WE-15: A loop ABCD containing two resistors as, shown in figure is placed in a uniform, magnetic field B directed outwards to the, R 1R 2, R, plane of page. A sliding conductor EF of, R1 R 2, length l and of negligible resistance moves to, Blv, the right with a uniform velocity v as shown, in Fig. Determine the current in each branch., The equivalent resistance of the circuit is, Current, I , , A, , E, B, , R1, D, , B, v, , l, F, , B R2, C, , Sol: The magnetic field induction B, length l and the, velocity v of the conductor EF are mutually, perpendicular, hence the emf induced in it is, e=Blv (with end F of the rod at higher potential), NARAYANAGROUP, , , RR , R R 1 2 , R1 R2 , , Hence the current in the connector is i , , , i, , e, R, , Blv R1 R2 , , RR1 RR2 R1R2 , 9
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , Solving (1) and (2) gives :, is moving with a uniform speed v=2.0m/s in a, 18, 10, i1 103 A and i2 10 3 A, uniform magnetic field B=4.0T directed into, 55, 55, the paper. A capacitor of capacity C 10F, Current through 5 i1 i2, is connected as shown in the figure. Then, 8, 8, what are the charges on the plates A and B of, 10 3 A mA, the capacitor., 55, 55, P, W.E-19: A conducting rod MN moves with a speed, v parallel to a long straight wire which carries, B, A, a constant current i, as shown in Fig. The, v, length of the rod is normal to the wire. Find, B, the emf induced in the total length of the rod., State which end will be at a lower potential., Q, Sol: The motional emf is, p.d across the capacitor Blv 4 1 2 8V, q CV 10 8 80C, A is +Ve w.r.t. B (from fleming right hand rule), The charge on plate A is q A 80 C, , W.E-17: A conducting rod PQ of length L = 1.0m, , The charge on plate B is qB 80C, , W.E-18: Two parallel rails with negligible Sol: The magnetic field induction due to current i is, , different at different sections of the rod, because, they are at different distances from the wire., Let us, first of all, subdivide the entire length of, the conductor MN into elementary sections., Consider a section (shown shaded in the figure, (b)) of thickness dx at a distance x from the wire., As all the three, v, B and (dX) are mutually, normally to each other, so the emf induced in it, is de=Bvdx., (from N to M by Fleming’s right hand rule), , resistance are 10.0 cm apart. They are, connected by a 5.0 resistor. The circuit also, contains two metal rods having resistances of, 10.0 and 15.0 along the rails. The rods, are pulled away from the resistor at constant, speeds 4.00 m/s and 2.00 m/s respectively. A, uniform magnetic field of magnitude 0.01T, is applied perpendicular to the plane of the, rails. Determine the current in the 5.0 , resistor., , x, M, , dx, B, de, , N, , Figure (b), , Sol: In the figure R 5.0 , r1 10 , r2 15 ,, e1 Blv1 0.01 0.1 4 4 10 3V, e2 Blv2 0.01 0.1 2 2 10 3V, Applying Kircoff’s law to the left loop :, , For the rest of sections, the induced emf is in the, same sense, (i.e., from N to M), Tot al emf induced in the conductor is, , , , e de , , b, , 10i1 5 i1 i2 4 103, 15i1 5i2 4 103, 20i2 5i1 2 10, 10, , Bv dx, , Substituting for B , (1), , Right loop : 15i2 5 i1 i2 2 103, 3, , ba, , (2), , 0i, , the above equation, 2x, , gets changed to, e, , b a, , b, , 0iv dx, iv, iv, b a, e 0 ln x b or,, e 0 ln 1 a / b , 2 x, 2, 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , W.E-20 : A square loop of side a is placed in the W.E-22 : A metallic square loop ABCD is moving in its, same plane as a long straight wire carrying a, current i. The centre of the loop is at a, distance r from the wire where r a . The, loop is moved away from the wire with a, constant velocity v. The induced e.m.f. in the, loop is, , own plane with velocity v in a uniform magnetic, field perpendicular to its plane as shown in the figure., Find, a) In which sides of the loop electric field is, induced., b) Net emf induced in the loop, c)If one ‘BC’ is outside the field with remaining loop, in the field and is being pulled out with a costant, velocity then induced current in the loop., B, , A, , v, Sol: Magnetic field by the straight wire of current i at a, D, C, i, distance r is B 0, 2r, Sol: a) The metallic square loop moves in its own, flux associated with t he loop is, plane with velocity v., A uniform magnetic field is imposed, i, BA 0 a 2, perpendicular to the plane of the square loop., 2 r, AD and BC are to the velocity as well as , d 0 2 d 1 0 2 1 dr, to field applied. Hence electric field is induced, e , , ia, ia 2 , , dt, 2, dt r 2, r dt, across the sides AD and BC only., Hence the induced emf in the loop is, b) As there is no change of flux through the entire, 2, coil net emf induced in the coil is zero., a, dr, , e 0 i 2 v v , e, 2 r, dt, , c)Induced current i , Where R is the, R, W.E-21: Two conducting rings of radii r and 2r, resistance of the coil., move in apposite directions with velocities 2, and respectively on a conducting surface S., Blv, i , (Only the side AD cuts the flux), There is a uniform magnetic field of magnitude, R, B perpendicular to the plane of the rings. The, MOTIONAL EMF INDUCED IN A, potential difference between the highest points, ROTATING BAR, of the two rings is, B, x, , 2v r, , 2r, , Bin, , v, , P, , S, , Sol: Replace the induced emfs in the rings by cells, , dr, , emfs e1 B 2r 2v 4 Brv, , O, , e2 B 4r v 4 Brv The equivalent circuit is, e1 +, , 1, , 2, +, -, , s, Hence the potential difference between the, highest points of t he t wo rings is, V2 V1 e1 e2 8Brv, NARAYANAGROUP, , +, , , , e2, , -, , If a rod of length l is rotated with a constant, angular velocity ‘ ’ about an axis passing, through its end (O) and perpendicular to its length, , and if a uniform magnetic field B is present, perpendicular to it, then emf across its ends is, given by e , , 1 2, Bl , 2, 11
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , , In the above case if the rod is rotated about an axis Sol: Here each spoke of wheel act as a source of an, passing through its centre (O) and perpendicular to, induced emf (cell) and emf’s of all spokes are, tis length them emf across its ends is zero, parallel., , f=120 rev/min = 2 rev/second,, B=0.40 gauss= 0.4 104 T ,, A1 2, emf across OA is e Bl , Area swept, by each spoke per second, A r 2 f, 8, +, 1 2, o, Magnetic flux cut by each spoke per second,, emf across OB is e Bl , 8, d B, BA B r 2 f, - B, Net emf across AB is zero, dt, end ‘A’ is -ve with respect to ‘O’, Induced emf, e Br 2 f (numerically), end ‘B’ is -ve with respect to ‘O’, 22, A spoked wheel of spoke length ‘l’ is rotated about, e 0.4 10 4 0.5 0.5 2, its axis with an angular velocity ‘ ’ in a plane normal, 7, to uniform magnetic field B as shown., Rim, spokes, , The emf induced across the ends of each spoke is, e 6.29 105 volt, Induced emf in a wheel is independent of no. of, 1, e Bl 2 , with axle (centre) at higher potential., spokes., 2, Since all the spokes are parallel between axle and W.E-25: A metal rod of resistance 20 is fixed, rim, the emf induced between axle and rim is, along a diameter of a conducting ring of, radius 0.1m and lies on x-y plane. There is a, 1, e Bl 2 ., , 2, magnetic field B 50T k . The ring rotates, It is independent of number of spokes., with an angular velocity 20 rad / s about, W.E-23: A copper rod of length 2m is rotated with, its axis. An external resistance of 10 is, a speed of 10 rps, in a uniform magnetic field, connected across the centre of the ring and, of 1 tesla about a pivot at one end. The, rim. The current through external resistance, magnetic field is perpendicular to the plane, is, of rotation. Find the emf induced across its, ends, 10, 1, 1, 2, 2, 2, Sol: e Bl B 2n l Bnl, 10, 2, 2, Sol:, 10, e 3.14 110 2 2 125.6 volt, , W.E-24: A wheel with 10 metallic spokes, each 0.5m, , The equivalent circuit is, , long, is rotated with a speed of 120 rev/minute in a, plane normal to the earth’s magnetic field at the, place. If the magnitude of the field is 0.40 gauss,, what is the induced emf between the axle and the, rim of the wheel ?, , 10, 10, 10, , 1 2, 1, Bl 50 0.1 0.1 20 ;, 2, 2, Hence, the current through the external, e 5V, e, 5 1, i , A, resistance is i , R, 15 3, e, , Rim, spokes, , 12, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , MOTIONAL EMF INDUCED IN A, ROTATING DISC, Y, , The rate at which work is done by the applied force, to move the rod is,, , B 2l 2v 2, R, The rate at which energy is dissipated in the, circuit is,, Papplied Fv , , R, , B, , B, , r, , , dr, , u, B, , X, , 2, , B 2l 2 v 2, Bvl , Pdissipated i R , R, , , R, R , This is just equal to the rate at which work is, done by the applied force., W.E-27: A 0.1 m long conductor carrying a current, of 50 A is perpendicular to a magnetic field of, 1.25 mT. The mechanical power to move the, , Z b, , 2, , I, , , , A circular disc of radius ‘R’ is rotating with an, angular velocity ‘ ’ about an axis passing, through centre and plane of rotation is normal to, an uniform magnetic field of induction B. It is, equivalent to a spoked wheel with a large, number of spokes each of length ‘R’ between, conductor with a speed of 1ms 1 is, centre and rim without any air gap. The emf, induced between centre and rim is independent Sol: Power P=Fv ; P=Bilv ; l=0.1m ; i=50, of number of spokes., B 1.25 103 ; v=1m/sec ; p Bilv, So, the emf induced between centre and rim is, 1.25 103 50 0.1 1 ; 6.25 103 ; =6.25 mW, 1 2, 1, 2, e Bl BR , W.E-28: A short - circuited coil is placed in a time, 2, 2, varying magnetic field. Electrical power is, W.E-26: A copper disc of radius 1m is rotated, dissipated due to the current induced in the coil., about its natural axis with an angular velocity, If the number of turns were to be quadrupled, 2 rad/sec in a uniform magnetic field of 5 telsa, and the radius of the wire is to be halved, then, with its plane perpendicular to the field. Find, find the electrical power dissipated., the emf induced between the centre of the disc, Sol: Current is induced in the short-circuited coil due, and its rim., to the imposed time - varying magnetic field., 1, 1, 2, Sol: e Br ; e 5 2 1 1 5 volt, d, e2, 2, 2, e, P, , Power, ;, Here, where NBA, ENERGY CONSIDERATION, dt, R, l, P, R 2 wher l and r are length and radius, and, F.v, r, l R, F, m, , Q, , , , A conductor PQ is moved with a constant, velocity v on parallel sides of a U shaped, conductor in a magnetic field as shown in figure., Let R be the resistance of the closed loop., The emf induced in the rod is e=Blv, e Blv, The current in the circuit is i , R, R, As current flows in the conductor PQ from Q to, P of the conductor. So, an equal and opposite, force F has to be applied on the conductor to, move the conductor with a constant velocity v., Thus, F Fm , , NARAYANAGROUP, , B 2l 2 v, R, , 2, , r 2 d, , NBA or, of the wire. P , , l dt, , 2, , r 2 2 2 dB , N 2r 2, P, N A , or P=(constant), l, l, dt , , when r2 , , r1, then t2 4l1, 2, 2, , 2, , P 4N r l , 2 , , P1, N2, 2r 4l , P2 1, P2 16 N 2 r 2 l, 2, or P 1, 2, P1 N 4r 4l, 1, Power, dissipated, is, the, same., , , , , 13
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , W.E-29: A pair of parallel horizontal conducting WE-30: Two parallel vertical metallic bars XX 1, rails of negligible resistance, shorted at one, end is fixed on a table. The distance between, R can slide on the rails frictionlessly. The rod, is tied to a massless string which passes over a, pulley fixed to the edge of the table. A mass, m, tied to the other end of the string, hangs, vertically. A constant magnetic field B exists, perpendicular to the table. If the system is, released from rest, calculate :, B, , FM, , and YY 1 , of negligible resistance and separated, by a length ‘l ’, are as shown in Fig. The ends, of the bars are joined by resistance R1 and R2., A uniform magnetic field of induction B exists, in space normal to the plane of the bars. A, horizontal metallic rod PQ of mass m starts, falling vertically, making contact with the, bars. It is observed that in the steady state the, powers dissipated in the resistance R1 and R2, and the terminal velocity attained by the rod, PQ., R1, X, Y, , T, , B, , mg, , Q, P, i) The terminal velocity achieved by the rod., l, ii) The acceleration of the mass at the instant, when the velocity of the rod, is half the terminal, X', Y', velocity., R2, Sol: i) the velocity of rod = V, Sol: Let V0 be the terminal velocity attained by the, Intensity of magnetic field = B, rod PQ (in the steady state). If i1 and i2 be the, V, emf induced in rod (e)=BLV, currents flowing through R1 and R2 in this state,, BlV, then, current flowing through the rod PQ is, current induced in rod (i) , R, i i1 i2 (see the circuit diagram) as shown in, B 2VL2, Force on the rod F BiL , R, Net force on the system = mg - T, mg - T = ma, Fig., B 2VL2, B 2VL2, ma, but T F , Hence, mg , R, R, Applying Kirchoff’s loop rule, yields., B 2VL2, i1 R1 BV0 l and i2 R2 BV0 l, or a g , ...............(i), mR, 1, 1 , i1 i 2 B V 0 l , , For rod to achieve terminal velocity VT , a 0, , ......(i), R, R, B 2VT L2, 0 g , mR, mgR, .........(ii), B 2 L2, V, ii) Acceleration of mass when V T, 2, mgR, or V , . Put this value of V in (i), 2 B 2 L2, B 2 L2 mgR , g, a g , 2 2 or a g , mR 2 B L , 2, g, or a ...............(iii), 2, , or Terminal velocity VT , , 14, , , , 1, , 2, Given that, P1 i1 R1 , , 2, , , , B 2V02 l 2, R1, , .....(ii), , B 2V02 l 2, and P2 i R2 , R2, .....(iii), Also in the steady state, the acceleration of PQ=0, 2, 2, , mg B i1 i2 l, 1, 1 , 2 2, (or) mg B l V0 R R P1 P2, 1, 2 , [From equation (ii) and (iii)], P1 P2, The terminal velocity is V0 mg, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , Substituting for V0 in equation (ii),, 2, , 2, , Bl P1 P2 1, B2l 2 P1 P2 , P1 , , , R1 , R1 mg , mg, , P1, Similarly from equation (iii), , The net emf induced, d, dB, dB, dB , dB, e, A, A v A, v. , dt, dt, dx, dx , dt, 144 10 4 10 3 8 10 3 , , 2, 144 9 107 129.6 10 6 V, Bl P1 P2 , 1, R2 , , WE-33: A bar of mass m and length l moves on, mg, , P2, two frictionless parallel rails in the presence, WE-31: The loop ABCD is moving with velocity, of a uniform magnetic field directed into the, ‘v’ towards right. The magnetic field is 4T., plane of the paper. The bar is given an initial, The loop is connected to a resistance of 8 ., velocity vi to the right and released. Find the, velocity of bar, induced emf across the bar and, If steady current of 2A flows in the loop then, the current in the circuit as a function of time, value of ‘v’ if loop has a resistance of 4 , is, : (Given AB=30cm, AD=30 cm), C, D, v, vi, 8, l, R, B, A 37, 0, , Sol: The induced emf in the loop is e Blv, 0, , 0, , e B AD sin 37 v 4 0.3sin 37 v, Effective resistance of the circuit is, e Blv, R 4 8 12 ; Hence i , R, R, 2, , 4 0.3 sin 37 0 v, 100, v , m/ s, ;, 4 8, 3, , WE-32: A square loop of side 12cm with its sides, parallel to x and y-axes is moved with a, velocity 8 cm/s along positive x-direction in, an environment containing magnetic field, along +ve z-direction. The field has a gradient, of 103 tesla/em along -ve x-direction, (increasing along -ve x-axis) and also, decreases with time at the rate of 103 tesla/s., The emf induced in the loop is, Sol: The magnetic field in loop varies with position, ‘x’ of loop and also with time simultaneously., The rate of change of flux due to variation of ‘B’, with time is, , d, dB, A, dt, dt, , The rate of change of flux due to variation B, with position ‘x’ is, d, dB, dB dx, dB, A, A, , A, v, dt, dt, dx dt, dx, , Since both cause decrease in flux, the two effects, will add up, NARAYANAGROUP, , Sol: The induced current is in the counter clockwise, direction and the magnetic force on the bar is, given by FB ilB . The negative sign indicates, that the force is towards the left and retards, motion., F=ma, dv, ilB m., dt, Because the force depends on current and the, current depends on the speed, the force is not, constant and the acceleration of the bar is not, constant. The induced current is given by, Blv, dv, i, ; ilB m., R, dt, , dv, dv, B 2l 2, Blv , , lB, , m, ., , , , dt, , dt, v, mR, R , v, , t, , dv, B 2l 2, v, B 2l 2, t, , , dt, ln , t, v v, , ;, mR 0, mR, T, v1 , 1, t, mR, T, , v, , v, e, i, B 2l 2, The speed of the bar therefore decreases, exponentially with time under the action of, magnetic retarding force., , where T , , t, Blv Bl Tt, v1e, emf iR Blvi e T ; current : i , R, R, , 15
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , The force required to keep the arm PQ in constant, motion is IlB. Its direction is to the left., , WE-34: The arm PQ of the rectangular conductor, is moved from x=0, outwards in the uniform, magnetic field which extends from x=0 to x=b, and is zero for x>b as shown. Only the arm, PQ possesses substantial resistance r., Consider the situation when the arm PQ is, pulled outwards from x=0 to x=2b, and is then, moved back to x=0 with constant speed v., Obtain expressions for the flux, the induced, emf, the force necessary to pull the arm and, the power dissipated as Joule heat. Sketch, the variation of these quantities with distance., K, , B 2l 2 v, 0 x b : F 0 b x 2b, r, The Joule heating loss is, F, , PJ I 2 r , , EDDY CURRENTS, , P, v, , l, , x=0, , , , x=b, , x=2b, , , Sol: Let us first consider the forward motion from, x 0 to x 2b . The flux B linked with the, circuit SPQR is, , The induced emf is,, , , d B, Blv, dt, , 0 xb, , 0 b x 2b, When the induced emf is nonzero,, the current I is I , , L, , , , Blv, (in magnitude), r, , OUTWARD, K, , , , INWARD, M, , L, , K, , Flux, , , , EMF, , , , Force, , Blb, , , , Blv, , B 2l 2 v, r, B 2l 2 v 2, r, , Power, , -Blv, B 2l 2 v, r, , x=, , 16, , 0, , b, , 0 xb, , PJ 0, b x 2b, One obtains similar expressions for the inward, motion from x 2b to x 0 . One can, appreciate the whole process by examining the, sketch of various quantities displayed in Fig, , M, , L, , B 2l 2 v 2, r, , 2b, , b, , 0, , When bulk pieces of conductors are subjected, to changing magnetic flux, induced currents are, produced in them., The flow patterns of induced currents resemble, the whirling eddies in water. This effect was, discovered by Foucault and these currents are, called eddy currents (or) Foucault currents., A copper plate is allowed to swing like a simple, pendulum between the pole pieces of a strong, magnet, its motion is damped and the plate comes, to rest inthe magnetic field due to eddy currents, in the plate., If rectangular slots are made in the copper plate, area available to the flow of eddy currents is, less. So, electromagnetic damping is reduced, and the plate swings more freely., The eddy currents heat up the metallic cores and, dissipate electrical energy inthe form of heat in, the devices like transformers, electric motors and, other such devices., The eddy currents are minimized by using., laminations of metal to make a metal core. The, laminations are separated by an insulating, material like lacquer., The plane of the laminations must be arranged, parallel to the magnetic field, so that they cut, across the eddy current paths reduces the strength, of the eddy current., , Advantages :, Eddy currents are used in, a) Magnetic braking in trains., b) Electromagnetic damping., c) Induction furnace., d) Electric power meters., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , SELF INDUCTION :, L, , 0 N 2r, 2, , SELF INDUCTANCE OF A SOLENOID :, EE, If current flowing in a coil changes, the magnetic, flux linked with the coil changes. Then emf, induced in the coil is called self induced emf, and the phenomenon is called self induction., If ‘i’ is the current flowing through the coil and, ‘ ’ is magnetic flux linked with the coil, then, , i L i,, L , i, Here ‘L’ is called coefficient of self induction, of the coil or self inductance of the coil., , i L i,, L , i, Self induced e.m.f is given by, , d , di, e, L, dt, dt, Self inductance of a coil is magnetic flux linked, with the coil when unit current flows through it, (or) emf induced in the coil when current changes, in it at the rate of 1 A/sec., S.I. Unit of self inductance : Henry., Other Units : weber / ampere, volt-second/ampere,, , J / amp 2 , Wb 2 / J , voltsec 2 coul 1 . Dimensional i), i.e.,, formula of L is ML2T 2 I 2 , A coil having high self inductance is called, inductor., ii), Self induction is also known as inertia of iii), electricity as it opposes the growth or decay of , the current in the circuit., Inductance may be viewed as electrical inertia. It , is analogous to inertia in mechanics. It does not, oppose the current, but is opposes the change in , current., , , , SELF INDUCTANCE OF A FLAT, CIRCULAR COIL :, , , , Let us consider a circular coil of radius r and , containing N-turns. Suppose it carries a current, ‘i ’. The magnetic field at the centre due to this , Ni, current B 0, 2r, 2, 0 Ni 2 0 N ri, , , NBA, , N, , r, , And total flux, , , 2, 2r , , Now comparing with N B Li we get, NARAYANAGROUP, , Consider a long solenoid of length l, area of cross, section A and number of turns per unit length n, and length is very large when compared with, radius of cross section., Let I be the current flowing through the solenoid., The magnetic field inside the long solenoid is, uniform and is given by B 0 nI, Total number of turns in the solenoid of length l, is N=nl., Now, the magnetic flux linked with each turn of, the solenoid B A 0 nIA, l, , I, Total magnetic flux linked with the whole, solenoid, =magnetic flux with each turn , number of turns in the solenoid., 0 nIA nl 0 n 2 IAl ............(1), But LI LI 0 n 2 IAl from (1) & (2), , N, N2, , L 0, A, l, l, Self inductance of coil depends on, Geometry of the coil, a) Number of turns of the coil, b) The length (l ) of the solenoid,, c) The area of cross-section (A) of the solenoid,, Medium inside the coil (permeability), Nature of the material of the core of the solenoid., More is the permeability of the medium, more is, the self inductance, An inductor will have large inductance and low, resistance., Resistor opposes the current, inductor opposes, the change of current, One can have resistance without inductance, One cannot have inductance without resistance., An ideal inductor has inductance and no, resistance., When the current in the coil either increases or, decreases at a rate, then the coil can be imagined, di, to be a cell of emf e L., dt, One can have self inductance without mutual, inductance., One cannot have mutual inductance without self, inductance., , L 0 n 2 Al, , Since n , , 17
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, The direction of induced emf for different states of, current in a coil :, a) Steady current, , , di, 0, dt, , i, , i, , , , e = 0 no opposition, b) Make of circuit or increasing current, Circuit is made on, or I increasing, , e, eL, , + -, , di, dt, , (or) emf induced in one coil when current in the, other coil changes at the rate of 1 Amp/second., S.I. unit : Herry, Dimensional formula of self inductance or mutual, inductance is ML2T 2 A2, The value of mutual inductance depends on, 1) Distance between the two coils, 2) Number of turns of coils, 3) Geometrical shape of the coil, 4) Material of the core medium between the coils, 5) Orientation of the coils i.e., angle between, the axes of the coils., If the axes the parallel, then M is maximum, If the axes are perpendicular then M is minimum, , MUTUAL INDUCTANCE OF TWO, LONG COAXIAL SOLENOIDS :, , (A), , c) Breaking of circuit of decreasing of current, , , , Circuit is made off, or I decreasing, , Consider two solenoids S1 and S2 such that the, solenoid S2 completely surrounds the solenoid, S1., l, I1, , e, , I1, S1, , di, e L, dt, , - +, , S2, , (B), , MUTUAL INDUCTION, , , When current in one coil changes, magnetic flux, linked with the second coil placed near by it, also changes. The emf induced in secondary is, called mutually induced emf and the phenomenon , is called mutual induction., , Let l be length of each solenoid (or length of, primary coil) and of nearly same area of crosssection A. N1 and N2 are the total number of turns, of solenoid S1 and S2 respectively., Number of turns per unit length of solenoid S1, is, n1 , , S, , G, , , , , , 18, , E, , K, , Number of turns per unit length of solenoid S2, N, is, n2 2, l, Magnetic field inside the solenoid S1 is given, N1, I1, l, , If ‘ i p ’ is current flowing in the primary coil,, , by B1 0 n1 I1 0, , ‘ S ’ is magnetic flux linked with secondary coil, , then S i p, , Magnetic flux linked with each turn of solenoid, , S, , ip, Here ‘M’ is called coefficient of mutual induction, or mutual inductance., Induced emf in secondary coil is, e, di , d , e, M p (or) M , di p / dt, dt, dt , Mutual indictance between two coils is equal to, the magnetic flux linked in the secondary coil, when unit current passes through the primary coil, S Mi p ,, , , , P, , N1, l, , M , , S2 B1 A 0, , N1, I1 A, l, , Total magnetic flux linked with N2 turns of the, solenoid S2 is, 2 N 2 B1 A 0, 2 , , N1, I1 A N 2, l, , 0 N 1 N 2 I1 A, .............. (i), l, , But 2 M 12 I1 ................ (ii), Where M12 is the mutual inductance when current, varies in solenoid S 1 and makes magnetic, flux linked with solenoid S2,, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , from (i) and (ii) we get, ENERGY STORED IN AN INDUCTOR, NN IA, N N A Consider an ideal inductor of inductance ‘L’, M 12 I1 0 1 2 1 , M 12 0 1 2, connected with a battery. Let I be the current, l, l, inthe circuit at any instant ‘t ’, NN A, Similarly, M 21 0 1 2 , where M21 is the, l, di, e L, mutual inductance when current varies in, dt, solenoid S2 and makes magnetic flux linked with, solenoid S1., VL, It can be proved that M 12 M 21 M, E, The above equation is treated as a general result,, + -+ if the two solenoids are wound on a magnetic, This induced emf is given by, substance of relative permeability r , then the, mutual inductance is given by, dI, e L, NN A, dt, M 0 r 1 2 0 r n1n2 Al, l, -ve sign shows that ‘e’ opposes the change of, W.E-35: Two different coils have self inductance, current I in the inductor., To drive the current through the inductor against, L1 8mH and L2 2mH . The currents in, the induced emf ‘e’, the external voltage is, both are increasing at the same constant rate., applied. Here external voltage is emf of the, At a certain instant of time, the power given, battery = E, to the two coils is the same. At this moment, According to Kirchoff’s voltage law, E+e=0, the current, the induced voltage and energy, dI, E e ; E L dt, , stored in the first coil are i 1 , V 1 and U 1, respectively. The corresponding values in the, second coil are i2, V2 and U2 respectively., , Let an infinitesimal charge dq be driven through, the inductor in time dt. So, the rate of work, done by the external voltage is given by, , i1 V1, U1, Then the values of i , V and U are, 2, 2, 2, , dW, dI, dI, EI L I LI, dt, dt, dt, The total work done in establishing a current, through the inductor from 0 to I is given by, , respectively, i1 L2 2 1 v1 L1 8, Sol: i L 8 4 ; v L 2 4, 2, 1, 2, 2, U1 L2 2 1, , , U 2 L1 8 4, , I, I2 1, W dW LI dI ; W L LI 2, 2 2, 0, , W.E-36: Two coaxial solenoids are made by winding thin insulated wire over a pipe of crosssectional area A 10cm 2 and length = 20cm., If one of the solenoids has 300 turns and the, other 400 turns, their mutual inductance is, , , , 0, , 1 2, LI, 2, The work done in maintaining the current through, the inductor is stored as the potential energy (U), in its magnetic field. Hence energy stored inthe, inductor is given by, W, , 4 10 7 TmA1 , , Sol: M , , 0 N1 N 2 A, L, 7, , , , 2, , 410 3 10 4 10 10, 2 101, 2.4 104 H, , U, , 3, , M, , NARAYANAGROUP, , , , 1 2, LI, 2, , The equation U , , 1 2, LI is similar to the, 2, 19
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ELECTRO MAGNETIC INDUCTION, , JEE-ADV PHYSICS- VOL- IV, , 1 2, LdI 5 2, dI, dt , , s, expression for kinetic energy E mv . It shows that, e L, Now,, or, 2, e, 50, dt, L is analogues to mass ‘m’ and self inductance is, 10, 1, called electrical inertia., s s 0.2 s, 50, 8, The self inductance of a coil is numerically equal to, twice the energy stored in it when unit current flows, So, the desired emf can be produced by reducing, through it., the given current to zero in 0.2 second, i.e., When i=1A, L=2U, WE-40: Two different coils have self-inductances, di, , L1 16 mH and L2 12 mH . At a certain, Induced power P e i Li ., dt , instant, the current in the two coils is, 2, increasing at the same rate of power supplied, In case of solenoid L 0 n Al, to the two coils is the same. Find the ratio of, Magnetic energy stored per unit volume, i) induced voltage ii) current iii) energy stored, 1 2, in the two coils at that instant., 2, Li, B, 1, uB 2, uB u0 n 2i 2 Hence u B 2, dI, dI V1 L1 16 , 0, Sol: i) V1 L1 ;V2 L2 ; V L 12 3, Al, 2, dt, dt, 2, 2, The magnetic energy stored per unit volume, I1 V1 3, similar to electrostatic energy stored per unit, ii) P V1 I1 V2 I 2 I V 4, 1, 2, 2, 2, volume in a parallel plate capacitor u B 0 E, 2, 1, 2, 2, L I2, In both cases the energy is proportional to the, U1 2 1 1 L1 I1 4 3 , 3, , , square of field stregth., iii) U 2 1, 4, L2 I 22 L2 I 2 3 4 , W.E-37: The self-inductance of a coil having 200, 2, turns is 10 milli henry. Calculate the WE-41: The network shown is a part of the closed, magnetic flux through the cross-section of the, circuit in which the current is changing. At, coil corresponding to current of 4, an instant, current in it is 5A. Potential, milliampere. Also determine the total flux, difference between the points A and B if the, linked with each turn., current is, Sol: Total magnetic flux linked with the coil,, A, B, , N LI 102 4 103 4 105Wb, , 5A 1, , 15V, , 5H, , 1) Increasing at 1A/sec, 4 105, 7, 2) Decreasing at 1A/sec, 2 10 Wb, Flux per turn, , Sol: 1) The coil can be imagined as a cell of emf, 200, WE-38 : A coil of inductance 0.2 henry is, di , e L 5 1 5V ; Equivalent circuit is, connected to 600 volt battery. At what rate,, dt , will the current in the coil grow when circuit, A, B, is completed ?, 5A 1, 15V 5V, Sol: As the battery and inductor are in parallel, at, VA 5(1) 15 5 VB, any instant, emf of the battery and self emf in the, inductor are equal, Hence VA VB 5 15 5 25V, e, 2), The coil can be imagined as a cell of emf, dI, dI, 600V, e L, , 3000 A s 1, or, dt, dt L 0.2 H, di , e L 5 1 5V ; Equivalent circuit is, W.E-39: An inductor of 5H inductance carries a, dt , steady current of 2A. How can a 50V selfA, B, induced emf be made to appear in the inductor, 5A 1, Sol: L 5H ; e 50 V ; Let us produce the required, emf by reducing current to zero, 20, , 15V, , 5V, , VA 5(1) 15 5 VB, Hence VA VB 5 15 5 15V, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , WE-42: Calculate the mutual inductance between, two coils when a current of 2A changes to 6A, in 2 seconds and induces an emf of 20 mV in, the secondary coil, dI, Sol: e M, dt, , 20 10 3 M, , 6 2, , B2 , , 0 I 2, 2r2, , The second co-axially placed coil has very small, radius. So B2 may be considered constant over, its cross-sectional area., , (or) M = 10mH, 0 r12, 2, 2, 2 0 I 2 , , , , , , r, B, r, , , I2, or 1, 1, 2, 1 , WE-43 : If the coefficient of mutual induction of Now, 1, 2r2, 2r2 , the primary and secondary coils of an, 0 r12, induction coil is 6H and a current of 5A is, M, , cut off in 1/5000 second, calculate the emf Comparing with 1 M 12 I 2 , we get ; 12, 2r2, induced in the secondary coil., 0 r12, r12, 5, dI, M, , M, , , M, , 4, 12, Also, 21, V 15 10 V, ; e 6, Sol: e M, 2r2, r2, dt, 1/ 000, It would have been difficult to calculate the flux, WE-44 : A solenoid is of length 50 cm and has a, through the bigger coil of the nouniform field, radius of 2cm. It has 500 turns. Around its, due to the current in the smaller coil and hence, central section a coil of 50 turns is wound., the mutual inductance M 12 . The equality, Calculate the mutual inductance of the, system., M 12 M 21 is helpful. Note also that mutual, 2, inductance depends solely on the geometry., Sol: N P 500, N S 50 ; A 0.02 0.02m, WE-47: A small square loop of wire of side l is, 0 4 107 Hm 1 ,1 50 cm 0.5 m, placed inside a large square loop of wire of, 0 N P N S A, side L l . The loops are coplanar and, Now, M , l, their centres coincide. What is the mutual, 2, 7, inductance of the system ?, 4 10 500 50 0.02 , , H, Sol: Considering the large loop to be made up of four, 0.5, rod each of length L, the field at the centre, i.e.,, 789.8 107 H 78.98 H, at a distance (L/2) from each rod, will be, WE-45: A solenoidal coil has 50 turns per, I, B 4 0 sin sin , centimetre along its length and a cross4 d, sectional area of 4 104 m 2 . 200 turns of, , I, another wire is wound round the first solenoid, 2sin 45, B 4 0, i.e.,, co-axially. The two coils are electrically, 4 L / 2 , insulated from each other. Calculate the, l, mutual inductance between the two coils., L, Sol: n1 50 turns per cm ; = 5000 turns per metre, n2l 200, A 4 10 4 m 2 ; M 0 n1 n2l A, 4 107 5000 200 4 104 H 5.03 104 H, WE-46 : Two circular coils, one of smaller radius, r1 and the other of very large radius r2 are, placed co-axially with centres coinciding., Obtain the mutual inductance of the, arrangement., Sol: Suppose a current I2 flows through the outer, circular coil. The field at the centre of the coil, is, 22, , 0 8 2, I, 4 L, So the flux linked with smaller loop, , i.e., B1 , , 2 B1 S 2 , , 0 8 21 2, l, 4 L, , and hence, M , , 2, l2, l2, 2 2 0 M , I, L, L, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , WE-48: (a) A toroidal solenoid with an air core, , d B, d, NBA cos t NBA sin t, has an average radius of 0.15m, area of cross , dt, dt, section 12 104 m 2 and 1200 turns. Obtain The magnitude of induced emf is, the self inductance of the toroid. Ignore field, NBA sin t 0 sin t, variation across the cross section of the toroid., where e0 NBA is the maximum value of the, (b) A second coil of 300 turns is wound closely, on the toroid above. If the current in the, emf., primary coil is increased from zero to 2.0 A in, 0 is called the amplitude or peak value of emf., 0.05s, obtain the induced emf in the secondary, The induced emf depends upon (i) strength of, coil., the magnetic field, (ii) area of the coil, (iii) speed, 0 N1 I 0 N1 I, of rotation, and (iv) the number of turns of the, , Sol: (a) B 0 n1 I , l, 2r, coil., If f be the frequency of rotation of coil, then, 0 N12 IA, , , N, BA, , Total magnetic flux, B, 1, 0 sin 2ft, 2r, A graph plotted between and t , is a sine, 0 N12 A, curve as shown in Fig., L , But B LI, 2r, , L, , 4107 1200 1200 12 104, H, 2 0.15, , e, , 2.3 103 H 2.3mH, , d, t, 2 , where 2 is the total magnetic, dt, flux linked with the second coil., WE-49: A boy pedals a stationary bicycle at one, revolution per second. The pedals are attached, d, d NI , e N 2 BA N 2 0 1 A, to 100 turns coil of are 0.1m2 and placed in a, dt, dt , 2r, , uniform magnetic field of 0.1T. What is the, maximum voltage generated in the coil ?, 0 N1 N 2 A dI, or e , 2r, dt, Sol: 0 NBA NBA 2 f f 1, , (b) e , , 4107 1200 300 12 104 2, V, 2 0.15 0.05, =0.023 V, or e , , AC GENERATOR :, , , , , 0 100 0.1 0.1 2 3.14 1 V 6.28V, , WE-50: A coil of 800 turns and 50 cm 2 area, makes 10 rps about an axis in its own plane, in a magnetic field of 100 gauss perpendicular, to this axis. What is the instantaneous induced, emf in the coil?, , An ac generator converts mechanical energy into, electrical energy. The device used for the, purpose is called ac generator., When the coil having N turns is rotated with a Sol: A 50cm 2 50 10 4 m 2, constant angular speed , the angle between the, n 10 rps, N 800, area vector A and the magnetic field vector B is, B 100 gauss 100 104 T 102 T, at any instant t is t (assuming 00 at t=0)., Now, 0 sin t NBA sin t, The flux linked with the coil at any instant t is, , B NBA cos NBA cos t, From Faraday’s law, the induced emf for the, rotating coil of N turns is,, NARAYANAGROUP, , 800 102 50 10 4 2 10sin 20 t , or 2.5sin 20 t volt, 23
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , WE-51: A person peddles a stationary bicyle the, pedals of the bicycle are attached to a 100 turn, coil of area 0.10 m2. The coil rotates at half a, revolution per second and it is placed in a, uniform magnetic field of 0.01 T, perpendicular to the axis of rotation of the coil,, What is the maximum voltage generated in the, coil ?, Sol: Here f 0.5 Hz : N 100, A 0.1 m 2 and, from, t he, equat ion, B 0.01T, , 0 sin t NBA sin t maximum emf, , 0 NBA NBA 2 f , 0 100 0.01 0.1 2 3.14 0.5 0.314V, INDUCED ELECTRIC FIELDS :, When a conducting loop is placed in a varying, magnetic field, a varying electric field produced, in the loop, is called induced electric field., An electric field is always generated by a, changing magnetic field, even in free space, where no charges are present., Consider a conducting loop of radius R, situated, in a uniform magnet ic field B t hat is, perpendicular to the plane of the loop as shown, in the figure, , E, , E, r, , E, Bin, , E, , e, 2r, Using this result along with Faraday’s law and the, fact that B BA Br 2 for a circular loop, the, induced electric field can be expressed as, , qe qE 2r ; E , , 1 d B , 1 d, r dB, Br 2 , , , , 2r dt , 2r dt, 2 dt, The emf for any closed path can be expressed as, , the line integral of E.dl over that path. Hence,, the general form of Faraday’s law of induction, is, d , e E.dl B, dt, It is important to recognize that the induced, electric field E that appears in the equation is a, non-convervative field that is generated by a, changing magnetic field., Points to remember about induced electric field., 1) The induced electric field is produced only, by changing magnetic field and not by charged, particles., 2) One cannot define potentials w.r.t this induced, field, 3) The lines of induced electric field are closed, curves and have no starting and terminating, points., 4) As long as the magnetic field keeps on, changing, the induced electric field will be, present because this electric field is produced, only by variable magnetic field., WE-52: A uniform magnetic field of induction B, is confined in a cylinderical region of radius, R. If the field is increasing at a constant rate, dB, T / s , then the intensity of the, of, dt, electric field induced at point P, distant r from, the axis as shown in the figure is proportional, to :, E, , If the magnetic field changes with time, then an, d , emf e , is induced in the loop. The, dt, induced current thus produced implies the, presence of an induced electric field E that must, be tangential to the loop in order to provide an, B, electric force on the charge around the loop., r, The work done by the electric field on the loop, R, P, in moving a test charge q once around the, d B, loop=qe. Because the magnitude of electric force Sol: For r < R ; e E.ds ; , dt, onthe charge is qE, the work done by the electric, dB , 2 dB , field can also be expressed as qE 2r , where E.2r A , ; E.2r r , , dt , dt , 2r is the circumference of the loop. These, r dB , r, r, two expressions for the work must be equal; E , ; E ; E ; Er, 2 dt , 2, 2, therefore, we see that, , , , 24, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , WE-53: Magnetic flux linked with a stationary WE-56: The magnetic field at all points within, loop of resistance R varies with respect to time, during the tim e period T as follows:, at T t the amount of heat generated in, the loop during that time (inductance of the, coil is negligible) is, d, Sol: Give that at T t ; induced emf, E , dt, , , d, at T t ; at 0 1 a T 1 ; a T 2t , dt , , the cylindrical region whose cross- section is, indicated in the accompanying figure starts, increasing at a constant rate ' ' . Find the, magnitude of electric field as a function of r,, the distance from the geometric centre of the, region., , Sol: Case -1 : For r < R, , R, , So, induced emf is also a function of time Heat Case -2 : r = R, generated, dB, dB, T, T, E.2 r A, E.2 R R 2, ;, 2 3, a2, E2, dt, dt, H , dt ; T 2t dt ; a T, R, R, dB, R dB, 3R, 0, 0, E.2 r r 2, ; E, WE-54: A closed loop of cross-sectional area, dt, 2 dt, 2, 2 which has inductance L=10 mH and, r dB, r, R, 10 m, E, ;E , ;E r, negligible resistance is placed in a time2 dt, 2, 2, varying magnetic field. Figure shows the, variation of B with time for the interval 4 s. Case - 3 r > R ; E.2 r R 2 dB, dt E, The field is perpendicular to the plane of the, loop (given at t 0, B 0, I 0 ). The value of, the maximum current induced in the loop is, R, 1, 1, R 2 dB, R2, E, 2 Er, , E, E, , , E, , , , ;, ;, r, out, B(T), r, 2r dt, , WE-57: A wire is bent in the form of a square of, , di, A dB , dB , A, L di , dt, dt, L dt , dt , B, , A, , A, , di L dB ; I L B, 0, , ;, , I max , , 0, , t(s), A, Bmax, L, , side ‘a’ in a varying magnetic field, B B t k . If the resistance per unit length, 0, , is , then find the following., Y, , P, , 2, , , , 10, 0.1, 10 103, , ; =0.1A = 100mA, , r, , R, , di 0.1, Sol: Induced emf (e) L, dt, , l, , 2r, , B, , i)The direction of induced current, , WE-55: A magnetic field directed into the page ii) The current in the loop, , a, , S, , Q, a, R, , X, changes with time according to the expression iii) Potential difference between P and Q, B 0.03t 2 1.4 T , where t is in seconds. The Sol: i) Direction of current is closewise., field has a circular cross - section of radius R d d, BA a 2 B0 .t a 2 B, ii) e , 2.5cm. What is the magnitude and direction of, dt dt, electric field at P, when t = 3.0s and r = 0.02 m., e a 2 B0, a B0, d, R 4a , Current : i , , e, , E, ., dl, , Sol:, R, 4a, 4, , dt, e, dB, d, iii) v p i.a VQ ,where ‘e’ is the total emf, E 2 r A., r 2 0.03t 2 1.4 , 4, dt, dt, 2, e, r, r, E, 0.06t 0.06t , induced or VP VQ ia , 2 r, 2, 4, 0.02, e, e, e e, E , 0.06 3 18 10 4 N / C, .a ; VP VQ 0, or VP VQ , 2, 4a, 4, 4 4, NARAYANAGROUP, , 25
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ELECTRO MAGNETIC INDUCTION, , JEE-ADV PHYSICS- VOL- IV, , WE-58: Shown in the figure is a circular loop of, , E, , i0 , radius r connected to a resistance R. A variable, R, magnetic field of induction B e t is II. Decay of Current, established inside the coil. If the key(k) is, a) When circuit is disconnected from the battery, closed, find the electric power developed ?, and switch ‘s’ is connected to point ‘b’, the, current now beings to fall. But inductor opposes, R, di, decay of current L Ri, dt, K, t, , Where, i, is, the, current, at, any, instant, and, d, dB, d, i, , i, e, 2, t, 0, E, , , , A, ., E, , r, e, , , or, Sol:, , dt, dt, dt, L, t, , , , where, 2, 4, r, R, E 2 2 r 4 e 2t, E r 2 e t ; P R R ; at t = 0 ; P , R, i, i0, D.C. CIRCUITS, Growth and decay of current in an inductor, 0.37 i0, Resistor (L - R) circuit, I. Growth of current, t, t=, Consider a circuit shown in the diagram, i, b) At t , i 0 0.37 i0, e, c) The inductive time constant ( ) can also be, defined as the time interval during which the, current decays to 37% of the maximum current., d) For small value of ‘L’, rate of decay of current, a) When a switch S is connected to ‘a’ , the, will be large., current in the circuit beings to increase from zero, e) Current becomes zero after infinite time., to a maximum value ‘ i0 ’. The Inductor opposes, WE-59 : In the given circuit, current through the, the growth of the current., 5 mH inductor in steady state is, di, , E L Ri, 5 mH, dt, Where ‘i ’ is the current in the circuit at any, t, , , , i, , i, 1, , e, , , 0, instant ‘t ’ and, , , , 10 mH, 5, , 20 V, , Sol: 5mH, 10mH are connected in parallel, Equivalent inductance, L, , , Here, called Inductive time constant, 5 10 50, 10, R, Leq , ; mH, 5 10 15, 3, 1, , 20, b) At t , i i0 1 e 0.63 i0, 4A, Current at steady state ; I , 5, c) Thus the inductive time constant of a circuit, is defined as the time in which the current rises, As L1 and L2 are in parallel, from zero to 63% of its final value., L2 , 10 , d) Greater the value of ‘ ’ smaller will be the, I1 , , I ;, 4, rate of growth of current., 10 5 , L1 L2 , e) Current reaches i0 after infinite time., 10, 8, Amp, ; 4;, f) When current attains maximum value, Inductor, 15, 3, doesn’t work., Where i0 is the maximum current., , 26, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , WE-60: In the given circuit diagram, key K is WE-63: A cell of 1.5V is connected across an, switched on at t 0 . The ratio of current, inductor of 2mH in series with a 2 resistor.., through, the, cell, at, to, that, at, will, t , i, t 0, What is the rate of growth of current, be, , immediately after the cell is switched on., R, C, dI, dI E IR, , Sol: E L IR , therefore,, dt, dt, L, 6R, E 1.5Volt , R 2 , L 2mH 2 103 H, K, E, When the cell is switched on, I = 0, Sol: At t = 0, the branch containing L will offer infinite, resistance while the branch containing the capacitor, dI E, 1.5, , As 1 750 As 1, Hence, will be effectively a short circuit., dt L 2 103, Hence,, WE-64: A coil having resistance 15 and, E, inductance 10H is connected across a 90 Volt, (i )t 0 , R, dc supply. Determine the value of current after, Similarly, at t , L will offer zero, resistance,, 2sec. What is the energy stored in the, where as ‘c’ will be an open circuit., magnetic field at that instant., Hence, effective resistance, Sol: Give that ; R 15 , L 10 H , E 90Volt, 6 R 3R, e, Peak value of current, R, ; (i )t , 6 R 3R, 3R, E 90, L 10, I0 A 6 A also, L 0.67sec, e 3R, R 15, R 15, The required ratio ; , ; =3:1, R e, Rt, , , WE-61: An inductor of inductance L 400 mH, L, I, , I, 1, , e, , After 2sec,, 0, Now,, , , and resistors of resistance R1 4 and, L, , 3R, , R2 2 are connected to battery of emf 12V, I 6 1 e2 / 0.67 6 1 0.05 5.7 A, as shown in the figure. The internal resisEnergy stored in the magnetic field, tance of the battery is negligible. The swich, 1, 1, 2, S is closed at t 0 . The potential drop across, U LI 2 10 5.7 J 162.45 J ., 2, 2, L as a function of time is, WE-65: Calculate the back e.m.f of a 10H, 200 , E 12, dl2, I, , , , 6, A, E, , L, , R, , l, Sol: 1 R, ;, coil 100 ms after a 100V d.c supply is, 2, 2, 2, dt, 1, connected to it., E, 12, Sol:, The value of current at 100ms after the switch is, , 6A, I 2 I 0 1 e t / tc ; I 0 , closed is, R2 2, , L 400 103, tc , 0.2 ; I 2 6 1 e t /0.2 , R, 2, Potential drop across L, VL E R2 I 2 12 2 6 1 ebt ; 12e 5t, WE-62: An inductor of 3H is connected to a, battery of emf 6V through a resistance of, 100 . Calculate the time constant. What, will be the maximum value of current in the, circuit ?, Sol: Give that L 3H , E 6V , R 100 , , Time constant, , L , , Maximum Current, NARAYANAGROUP, , L, 3, , 0.03sec, R 100, E, 6, I0 , amp 0.06 amp, R 100, , t, , , T0, I I 0 1 e , Here, I 0 100 0.5 amp;, 200, , , , 0 , , L 10, , 0.05sec; t 0.1sec, R 200, , I 0.5 1 e0.1/ 0.05 0.5 1 e 2 0.4325 A, , dI, , or, dt, dI, 100 0.4325 200 L, dt, dI, Back e.m.f L 100 0.4325 200 13.5V, dt, , Now, E IR L, , 27
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ELECTRO MAGNETIC INDUCTION, , WE-66: A coil of resistance 20 and inductance, 0.5 henry is switched to dc 200 volt supply., Calculate the rate of increase of current:, a) At the instant of closing the switch and, b) After one time constant, c) Find the steady state current in the circuit., Sol: a) This is the case of growth of current in an, L - R circuit. Hence, current at time t is given, t, , L, i, , i, l, , e, 0, by, , , , , Rate of increase of current,, , , , t, , di i0 L, di i, E/R E, e ; At t 0 0 , , dt L, dt L L / R L, , di 200, , 400 A / s, dt 0.5, di, 1, b) At t L , 400 e 0.37 400 148 A / s, dt, c) The steady state current in the circuit, is, E 200, i0 , 10 A, R 20, , GROWTH AND DECAY OF CHARGE IN, A CAPACITOR - RESISTOR (C - R), CIRCUIT, I. Growth of Charge : Consider a circuit shown, , JEE-ADV PHYSICS- VOL- IV, 1, c) When t . q q0 1 0.63 q0, e, d) Thus the capacitive time constant is the time, in which the charge on the plates of the capacitor, becomes 0.63 q0, e) Smaller the value of CR, more rapid is the growth, of charge on the condenser., f) Charge on the capacitor becomes maximum, after infinite time and it is q0 EC . Then, current in the circuit becomes zero., , II. Decay of charge :, a) When the capacitor is fully charged the key is, connected to point ‘b’., b) Charge slowly reduces to zero after infinite, time., t, q, q, dq, Ri (or), R, and q q0e , c, c, dt, q, , q0, 0.37 q0, t=, , t, , q0, 0.37 q0, e, in the diagram, d) Thus capacitive time constant can also be, q, defined as the time interval in which the charge, q0, b, decreases to 37% of the maximum charge, R C, 0.63q0, K, e) Smaller the time constant, quicker is the, a, discharge of the condenser., t, t=, WE-67: In the circuit shown in figure switch S is, E, a) When the key’s is connected to point ‘a’, the, closed at time t = 0. Find the current through, charging of capacitor takes place until the, different wires and charge stored on the, potential difference across the plates of the, capacitor, at, any, time, t., condenser becomes E., 6R, b) But charge attained already on the plates, S, R 3R, V, opposes further introduction of charge, C, Sol: Calculation of equivalent time constant, q, q, dq, E Ri (or) E R, 6R, c, c, dt, R 3R, Where ‘q’ is the instantaneous charge, i is the, instaneous current in the circuit., In the circuit shown in figure, after short circuiting, t, , , the battery 3R and 6R are parallel, so their, , and q q0 1 e , 6R 3R 2R, , , combined resistance is, . Now, 6 R 3R, where q0 is the maximum charge., this 2R is in series with the remaining R., Where CR , called capacitive time, Hence, Rnet 2 R R 3R ; c Rnet C 3RC, constant, , 28, , c) At t , q , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , Calculation of steady state charge q0 :, WE-68: 4F capacitor and a resistance 2.5 M , At t , capacitor is fully charged and no, are in series with 12V battery. Find the time, current flows through it., after which the potential difference across the, P.D across capacitor = P.D across 3R, capacitor is 3 times the potential difference, V, CV, V , across the resistor. [Given ln(2)=0.693], , q0 , 3R ,, 9, R, 3, 3, , , t, Now, let charge on the capacitor at any time t be q Sol: a) Charging current i V0 e RC, R, and current through it is i1 . Then, t, Potential difference across R is, , , 3 RC, t / c, t, q q0 1 e, i.e., q q0 1 e , , RC, VR iR V0 e, 1, dq q0 t / c, q, Pot ential difference across ‘C’ is, e, 0 e 3 RC ...(1), and i1 , dt c, 3RC, t, t, , , , , RC, RC, V0 1 e , VC V0 VR ; V0 V0 e, , , q0, , but given VC 3VR , we get, 1 e t / RC 3e t / RC or 4e t / RC, , Applying Kirchhoff’s second law in loop, ACDFA, we have 6iR 3i2 R V 0, , e, , t, RC, , 4, , t, ln 4 t 2 RC ln 2, RC, , V, ...(ii), t 2.5 106 4 106 2 0.693, 3R, Applying Kirchoff’s junction law at B, we have, or t = 13.86 sec, i i1 i2 ...(iii), WE-69: In a circuit inductance L and capacitance, Solving Eqs. (i), (ii) and (iii), we have, C are connected as shown in figure and A1, V 2, V 2q0 t / tc, and A2 are ammeters. When key k is pressed, i2 , e, i1 , , to complete the circuit, then just after closing, 9R 3, 9 R 3tc, key k, the reading of A1 and A2 will be :, t, CV, V, 2q0 3 RC, , e, i.e., i2 , where q0 , 3, 9 R 3RC, A1 c R1, 2i i2 , , t, V, q0 t / tc V, q0 3 RC, i, , e, , , e, A2, L, 9 R 3tc, 9 R 3RC, R2, WE-67: A parallel - plate capacitor, filled with a, k, E, dielectric of dielectric constant k, is charged, to a potential V0. It is now disconnected, Battery, fromthe cell and the slab is removed. If it, now discharges, with time constant , Sol: At t = 0 capacitor offers zero resistance and acts, through a resistance, then find time after, like a short circuit. While inductor offers infinite, which the potential difference across it will, resistance and it acts like an open circuit., be V0?, Therefore no current flow through inductor, Sol: When slab is removed, the potential difference, branch and maximum current flows through, across capacitor increases to kV0, t, capacitor branch., , CV0 kCV0 e as q0 KCV0, Hence reading of A2 is zero and reading A1 is, , t, t, , t, 1, , , e k e ; ln t lnk, , k, , NARAYANAGROUP, , E, given by R, 1, 29
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , C. U. Q, , 8., , FARADAY’S EXPERIMENT,, INDUCED E.M.F & LENZ’S’ LAW, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 30, , When ever the flux linked with a coil changes,, then, 1) current is always induced, 2) an emf and a current are always induced, 3) an emf is induced but a current is never induced, 4) an emf is always induced and a current is, induced, when the coil is a closed one, Whenever the magnet flux linked with a coil, changes, then there is an induced emf in the, circuit. This emf lasts, 1) For a short time, 2) For a long time, 3) For ever, 4) So long as the change in the flux takes place, A magnet is moved towards the coil (i) quickly, (ii) slowly then the induced emf is, 1) Larger in case (i), 2) Smaller in case (i), 3) Equal in both, 4) Larger or smaller depending upon the radius of, the coil, The laws of electromagnetic induction have, been used in the construction of a, 1) galvanometer, 2) voltmeter, 3) electric motor, 4) electric generator, When a rate of change of current in a circuit, is unity, the induced emf is equal to, 1) Total flux linked with the coil, 2) induced charge, 3) Number of turns in the circle, 4) Coefficient of self induction, A bar magnet is dropped along the axis of, copper ring held horizontally. The acceleration, of fall is, 1) Equal to ‘g’ at the place 2) Less than ‘g’, 3) More than ‘g’, 4) Depends upon diameter of the ring and length, of the magnet, An annular circular brass disk of inner radius, ‘r’ and outer radius ‘R’ is rotating about an, axis passing through its center and, perpendicular to its plane with a uniform, angular velocity ‘ ’ in a uniform magnetic filed, of induction ‘B’ normal to the plane of the disk., The induced emf between the inner and outer, edge of the annular disk is, B( r 2 R 2 ), B( R 2 r 2 ), 1), 2), 2, 2, B(r R ), B(r R ), 3), 4), 2, 2, , A, , X, , Consider the situation shown in the figure. If, the current I in the long straight conducting, wire XY is increased at a steady rate then the, induced e.m.f.’s in loops A and B will be, 1) clockwise in A, anti clockwise in, Y, 2) anti clockwise in A, clockwise in B, B, 3) clockwise in both A and B, 4) anti clockwise in both A and B, , FLEMING’S RIGHT HAND RULE, 9., , The direction of the induced e.m.f. is, determined by, 1) Fleming’s left hand rule, 2) Fleming’s right hand rule, 3) Maxwell’s right hand screw rule, 4) Ampere’s rule of swimming, 10. A wire moves with a velocity “v” through a, magnetic field and experiences an induced, charge separation as shown. Then the, direction of the magnetic field is, , -, , +, v, , 1) in to the page, 2) out of the page, 3) towards the bottom of the page, 4) towards the top of the page, 11. An electric potential difference will be induced, between the ends of the conductor shown in, the figure, if the conductor moves in the, direction shown by, M, N, , L, , S, R, 1) P, 2) R, P, 3) L, 4) M, 12. A horizontal straight conductor when placed, along south-north direction falls under gravity;, there is, 1) an induced current form south-to-north direction, 2) an induced current from north-to-south direction, 3) no induced emf along the length of the conductor, 4) an induced emf along the length of the conductor, 13. Two circular, similar, coaxial loops carry equal, currents in the same direction. If the loops are, brought nearer, what will happen?, 1) Current will increase in each loop, 2) Current will decrease in each loop, 3) Current will remain same in each loop, 4) Current will increase in one and decrease in the other, 14. A long conducting wire AH is moved over a, conducting triangular wire CDE with a constant, velocity v in a uniform magnetic field, , B directed into the plane of the paper.., Resistance per unit length of each wire is ., Then, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, D, F, , A, , ELECTRO MAGNETIC INDUCTION, , B, , G, , r, when the loop is pulled? I, , H, , b, , v, C, , a, , E, , Force on, , Force on, , Induced, 1) a constant clockwise induced current will flow, Right side, Left side, Current, in the closed loop, a.Cunterclockwise, To the left, To the left, 2) an increasing anticlockwise induced current will, b. Counterclokwise To the right, To the left, flow in the closed loop, 3) a decreasing anticlockwise induced current will, To the right, c. Clokwise, To the left, flow in the closed loop, To the right, d. Clockwise, To the left, 4) a constant anticlockwise induced current will flow, in the closed loop, 18. The four wire loops shown figure have vertical, 15. A squar e coil ACDE with its plane vertical is, edge lengths of either L, 2 L or 3L . They will, released from rest in a horizontal uniform, move with the same speed into a region of, , , megnetic field B of length 2 L. The, uniform magnetic field B directed out of the, acceleration of the coil is, page. Rank them according to the maximum, magnitude of the induced emf greatest to least., C, D, , L, A, , E, B, B, , 2L, , 1, , 2, , 3, , 4, , 1) 1 and 2 tie, then 3 and 4 tie, 2) 3 and 4 tie, then 1 and 2 tie, 3) 4,2,3,1 4) 4 then, 2 and 3 tie and then 1, 1) less than ‘g’ for all the time till the loop crosses, 19., A rod lies across frictionless rails in a uniform, the magnetic field completely, , magnetic field B as shown in figure. The rod, 2) less than ‘g’ when it enters the field and greater, moves to the right with speed V . In order to, than ‘g’ when it comes out of the field, make, the induced emf in the circuit to be zero,, 3) ‘g’ all the time, the, magnitude, of the magnetic field should, 4) less than ‘g’ when it enters and comes out of the, V, field but equal to ‘g’ when it is within the field, 16. A conducting wire frame is placed in a magnetic, field which is directed into the plane of the 1) not change, paper. The magnetic field is increasing at a, constant rate. The directions of induced 2) increase linearly with time, currents in wires AB and CD are, C, 3) decrease linearly with time, 4) decrease nonlinearly with time, A, 20. An electron moves on a straight line path, 1) B to A and D to C, 2) A to B and C to D, YY ' as shown in figure. A coil is kept on the, B, right such that YY ' is the plane of the coil. At, 3) A to B and D to C, the instant when the electron gets closest to, 4) B to A and C to D, D, the coil (neglect self-induction of the coil), 17. A rectangular loop of wire with dimensions, Y’, shown in figure is coplanar with a long wire, c, 1) The current in the, carrying current ‘I’. The distance between the, e, coil flows clockwise, d, b, wire and the left side of the loop is r . The loop, O, O, 2), The, current, in, the, is pulled to the right as indicated. What are, coil flows anticlockwise, a, the directions of the induced current in the loop, 3) The current in the coil is zero Y, and the magnetic forces on the left and the right, 4) The current in the coil does not change the, sides of the loop when the loop is pulled?, direction as the electron crosses point O, B, , NARAYANAGROUP, , 31
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 21. In figure, there is a conducting ring having, resistance R placed in the plane of paper in a, uniform magnetic field B0 . If the ring is rotating, in the plane of paper about an axis passing, through point O and perpendicular to the plane, of paper with constant angular speed in, clockwise direction, then, , r, , 1) point O will be at higher potential than A, 2) the potential of point B and C will be different, 3) the current in the ring will be zero, 4) the current in the ring will be 2 B0 r 2 / R, 22. In the space shown a non-uniform magnetic, , field B B 1 x kˆ tesla is present. A, , , , A, , IA, , C, , P, , Q, D, , B, , 1) the rod PQ will move downward with constant, acceleration, 2) the rod PQ will move upward with constant, acceleration, 3) the rod will remain at rest 4) any of the above, 25. Three identical coils A, B and C carrying, currents are placed coaxially with their planes, parallel to one another. A and C carry current, as shown in figure B is kept fixed while A and, C both are moved towards B with the same, speed. Initially, B is equally seperated from, A and C . The direction of the induced current, in the coil B is, I, , I, , 1) same as that in coil A, 0, 2) same as that in coil B, closed loop of small resistance, placed in the, 3) zero, xy plane is given velocity V0 . The force due to, 4) none of these, A, B, C, magnetic field on the loop is, 26. Two identical conductors P and Q are placed, on two frictionless rails R and S in a uniform, y, 1) zero, V, magnetic field directed into the plane. If P is, 2) Along x direction, , moved in the direction shown in figure with a, Q, P, 3) along x direction, constant speed, then rod Q, x, 4) along y direction, 23. Two identical cycle wheels (geometrically), R, have different number of spokes connected 1)will be attracted towards P, v, from centre to rim. One is having 20 spokes, S, 2)will, be, repelled, away, from, P, and the other having only 10 (the rim and the, spokes are resistanceless). One resistance of 3) will remain stationary, value R is connected between centre and rim. 4) may be repelled away orattracted towards P, The current in R will be, SELFINDUCTIONAND MUTUALINDUCTION, 1) double in the first wheel than in the second wheel 27. An inductance stores energy in the, 2) four times in the first wheel than in the second, 1) electric filed, 2) magnetic field, wheel, 3) resistance of the coil, 3) will be double in the second wheel than that of, 4) electric and magnetic fields, the first wheel, 28. If ‘N’is the number of turns in a coil, the value, 4) will be equal in both these wheels, of self inductance varies as, 24. AB and CD are fixed conducting smooth rails, 1) N0, 2) N, 3) N2, 4) N-2, placed in a vertical palne and joined by a 29. A series combination of L and R is connected, constant current source at its upper end. PQ, to a battery of emf E having negligible internal, is a conducting rod which is free to slide on the, resistance. The final value of current depends, rails. A horizontal uniform magnetic field exists, upon, in space as shown in figure. If the rod PQ is, 1) L and R only, 2) E and L only, released from rest then,, 3) E and R only, 4) L, R and E only, 0, , B, , 32, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 30. Two coils of inductances L1, and L2 are linked 36. In the circuit shown in figure, a conducting wire, such that their mutual inductance is M, HE is moved with a constant speed v towards, 1, left. The complete circuit is placed in a uniform, 1) L1+L2 2) ( L1 L2 ) 3) ( L1 L2 ) 4) L1 L2, , 2, magnetic field B perpendicular to the plane of, 31. The coefficient of self inductance and the, the circuit inwards. The current in HKDE is, coefficient of mutual inductance have, A, , H, , K, , 1) same units but different dimensions, 2) different units but same dimensions, 1) clockwise, 3) different units and different dimensions, C, R, V, 2) anticlockwise, 4) same units and same dimensions, 3) alternating, 32. The mutual inductance between a pair of coils, B, E, D, each of ‘N’ turns is ‘M’. If a current is ‘I’ in, 4) zero, the first coil is bought to zero in a time t, then 37. In which of the following cases the emf is induced, the average emf induced in the second coil is, due to time varying magnetic field(induced field emf)?, 1) MI/t, 2) Mt/I, 3) Mt/IN, 4) It/MN, Case I A magnet is moving along the axis of a, 33. A circuit contains two inductors of selfconducting coil, inductance L1 and L2 in series. If M is the, Case II A loop having varying area (due to moving jumper) is placed in a magnetic field, mutual inductance then the effective, inductance of the circuit shown will be, Case III The resistance of the coil is changing,, which is connected to an ideal battery. Case IV, A current carrying wire is approaching a conL, L, ducting ring., 1) L1 +L 2 2) L1 +L2 2M, 1) I, II and III only, 2) I, III and IV only, 3), I,, II, and, IV, only, 4) All the four, 3) L1 +L 2 +M 4) L1 +L 2 +2M, 38. A closed conducting ring is placed in between, 34. In the circuit of figure, (1) and (2) are, two bar magnets as shown in the figure. The, ammeters. Just after key K is pressed to, pole str ength of M 1 is double that of M2. When, complete the circuit, the reading is, the two bar magnets are at same distance from, C R1, the centre of the ring, the bar magnet M1 has, 1, given a velocity 2v while M2 is given velocity, 1) maximum in both (1) and (2), L, 2, v in the direction as shown in the figure., R2, 2) zero in both (1) and (2), 1, , 2, , K, , 3) zero in (1), minimum in (2), E, 4) maximum in (1), zero in (2), 35. A pure inductor L, a capacitor C and a, resistance R are connected across a battery, of emf E and internal resistance r as shown in, the figure. Switch SW is closed at t 0 , select, the correct alternative(s)., , 1, , The direction of induced current in the ring as seen, from XX from this moment to the moment till bar, L, magnets collide is, 1) always clockwise 2) always anticlockwise, R, 3) first clockwise, and then anticlockwise, C, 4) first anti-clockwise, and then clockwise, 39. Two identical ciruclar loops of metal wire are, E, r, lying on a table without touching each other., S, Loop A carries a current which increases with, time. In response, the loop B, 1) current through resistance R is zero all the time, 1) remains stationary, 2) current through resistance R is zero at t 0 and, 2) is attracted by the loop A, t, 3) maximum charge stored in the capacitor is CE, 3) is repelled by the loop A, 4) maximum enrgy stored in the inductor is equal to, 4) rotates about its CM , with CM fixed, the maximum energy stored in the capacitor, w, , NARAYANAGROUP, , 33
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 40. A metallic square loop ABCD is moving in its, own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown, in the figure. Electric field is induced., A, , B, , 1) in AD, but not in BC, 44. An infinitely long cylinder is kept parallel to, 2) in BC, but not in AD, an uniform magnetic field B directed along, 3) neither in AD nor in BC, D, C, positive, z-axis. The direction of induced cur4) in both AD and BC, rent as seen from the z-axis will be, 41. Two circular coils can be arranged in any of, 1) clockwise of the +ve z-axis, the three situations shown in the figure. Their, 2) anticlockwise of the +ve z-axis, mutual inducatnce will be, 3) zero, 4) along the magnetic field, 45. The figure shows certain wire segments joined, together to form a coplanar loop. The loop is, placed in a perpendicular magnetic field in the, C , (A), (B), direction going into the plane of the figure. The, 1) maximum in situation (A) 2)maximum in situation (B), magnitude of the field increases with time., 3) maximum in situation (C) 4) the same in all situations, I1 and I 2 are the currents in the segments, 42. As shown in the figure, P and Q are two coaxil, ab and cd . Then,, conducting loops separated by some distance., c, d, When the switch S is closed, a clockwise cur-- 1) I I, 2) I1 I 2, 1, 2, a, b, rent I p flows in P (as seen by E) and an in- 3) I1 is in the direction ba, and I 2 is in the direction cd, duced current I Q1 flows in Q . The switch re- 4) I is in the direction ab, 1, mains closed for a long time. When S is and I 2 is in the direction dc, 46. A coil is suspended in a uniform magnetic field, opened, a current I Q2 flows in Q . Then the, with the plane of the coil parallel to the magnetic lines of force. When a current is passed, through the coil, it starts oscillating; it is very, difficult to stop. But if an aluminium plate is, placed near to the coil, it stops. This is due to, 1) development of air current when the plate is, placed, 43. The variation of induced emf e with time, 2) induction of electrical charge on the plate, 3) shileding of magnetic lines of force as aluminium, t in a coil if, is a paramagnetic material, 4) electromagnetic induction in the aluminium plate, giving rise to electromagnetic damping, 47. Which of the following units denotes the didirection I Q1 and I Q2 (as seen by E) are, 1) respectively clockwise and anticlockwise, 2) both clockwise, 3) both anticlockwise, 4) respectively anticlockwise and clockwise, , a short bar magnet is moved along axis of, the coil shown with a constant velocity is best, represented as, , , , , t, , t, (1), , 34, , (2), , mensions ML2 / Q 2 , where Q denotes the, electric charge?, 1) Wb / m2 2)henry (H) 3) H / m 2 4)weber (Wb), 48. A rod of length l rotates with a small but uniform angular velocity about its perpendicular bisector. A uniform magnetic field B exists, parallel to the axis of rotation. The potential, difference between the centre of the rod and, an end is, 1) zero, 2) 1/ 8 Bl 2 3) 1/ 2 Bl 2 4) B l 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 49. A rod of length l rotates with a uniform angu- 57. Assertion (A): Only a change in magnetic flux will, maintain an induced current in a closed coil., lar velocity about its perpendicular bisecReason (R): The presence of large magnetic flux, tor. A uniform magnetic field B exists parallel, through a coil maintains a current in the coil if the, to the axis of rotation. The potential difference, coil is continuous., between the two ends of the rod is, 58., Magnetic flux in a circular coil of resistance, 1) zero, 2) 1/ 2 Bl 2 3) Bl 2 4) 2Bl 2, 10 changes with time as shown in figure., 50. Consider the situation shown in figure . If the, Symbol indicates a direction perpendicular, switch is closed and after some time it is, to paper inwards. Match the following:, opened again, the closed loop will show, wb , , 1) an anticlockwise current-pulse, 2) a clockwise current-pulse, Table - 1, Table-2, 3) an anticlockwise current-pulse and then a clock1), At, 1s, is, induced, current, is, p), clockwise, wise current-pulse, 2), At, 5s, induced, current, is, q), anticlockwise, 4) a clockwise current-pulse and then an, 3) At 9s induced current is r) zero, anticlockwise current-pulse, 4) At 15s induced current is s) 2A, 51. A bar magnet is released from rest along the, t) None, axis of a very long, vertical copper tube. Af1) a-q; b-r; c-p; d-q 2) a-p; b-r; c-q; d-p, ter some time the magnet, 3) a-r; b-p; c-q; d-q 4) a-p; b-r; c-s; d-q, 1) will stop in the tube, 59. Three coils are placed infront of each other as, 2) will move with almost contant speed, shown. Currents in 1 and 2 are in same, 3) will move with an acceleration g 4) will oscillate, direction, while that in 3 is in opposite direction., ASSERTION & REASON, Match the following table., 1, 2, 3, 1) Both A and R are true and R is the correct, explanation of A, 2) Both A and R are true and R is not the correct, explanation of A, 3) A is true but R is false 4) A is false but R is true., 52. Assertion : Magnetic flux is a vector qunatity, Table - 1, Table - 2, Reason: Value of magnetic flux can be positive,, a) When current is, p) Current in 1 will, negative or zero, increased, increase, 53. Assertion : Lenz’a law violates the principle of, b) When current in q) Current in 2 will, conservation of energy, 2 is increased, increase, Reason: Induced emf always oppose the change, c) When current in r) Current in 3 will, in magnetic flux responsible for its production, 3 is increased, increase, 54. Assertion: When number of turns in a coil is, s) None, doubled, coefficient of self-inductance of the coil, 1) a-r; b-r; c-p,q, 2) a-p; b-p; c-q, becomes 4 times., 3) a-q; b-q; c-r, 4) a-r; b-q; c-p, Reason: This is because L N 2, 55. Assertion : The induced emf and current will be, C. U. Q - KEY, same in two identical loops of copper and 1) 4 2) 4 3) 1 4) 4 5) 4 6) 2 7) 2 8) 1, aluminium, when rotated with same speed in the 9) 2 10) 1 11) 4 12) 3 13) 2 14) 4 15) 4 16) 1, same magnetic field., 17) 4 18) 4 19) 4 20) 3 21) 3 22) 3 23) 4 24) 4, Reason: Mutual induction does not depend on the 25) 3 26) 1 27) 2 28) 3 29) 3 30) 4 31) 4 32) 1, orientation of the coils, 33) 4 34) 4 35) 2 36) 4 37) 2 38) 2 39) 3 40) 4, 56. Assertion : When two coils are wound on each, other, the mutual induction between the coils is 41) 1 42) 4 43) 2 44) 3 45) 4 46) 4 47) 2 48) 2, 49) 1 50) 4 51) 2 52) 4 53) 4 54) 1 55) 3 56) 3, maximum., Reason: Mutual induction does not depend on the 57) 3 58) 1 59) 1, orientation of the coils., NARAYANAGROUP, , 35
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , LEVEL - I (C. W), , 7., , MAGNETIC FLUX & FARADY’S, LAWS, INDUCED EMF,CURRENT, AND CHARGE, 1. A field of strength 5 104/ ampere turns /, , 2., , 3., , 4., , 5., , meter acts at right angles to the coil of 50 8., turns of area 10-2 m2. The coil is removed from, the field in 0.1 second. Then the induced e.m.f, in the coil is, 1) 0.1 V, 2) 0.2 V 3) 1.96 V 4) 0.98 V, A coil has 1,000 turns and 500 cm2 as its area., The plane of the coil is placed at right angles, 9., to a magnetic induction field of 2 X 10-5 web/, m2. The coil is rotated through 180° in 0.2, seconds. The average emf induced in the coil,, in milli volts, is:, 1) 5, 2) 10, 3) 15, 4) 20, A square loop of side 22cm is changed to a, circle in time 0.4 sec with its plane normal to a, magnetic field 0.2T. The emf induced is, 1) 6.6mv, 2) 6.6mv, 3) 13.2mv, 4) 13.2mv, A coil of 1200 turns and mean area of 500, cm 2 is held perpendicular to a uniform, , The horizontal component of the earth’s, magnetic field at a place is 3 104 T and the, dip is = tan-1(4/3). A metal rod of length, 0.25m placed in the north-south position is, moved at a constant speed of 10 cm/s towards, the east. The e.m.f induced in the rod will be:, 1) zero 2) 1 mV, 3) 5 mV 4) 10mV, A metal bar of length 1m falls from rest under, the action of gravity remaining horizontal with, its ends in east-west direction. The induced, e.m.f in it at the instant when it has fallen for, 10s is (BH = 1.7 x 10–5T and g = 10ms–2), 1) 2.5mV 2) 3.2 mV 3) 1.7mV 4) 0.5mV, A thin semicircular conducting ring of radius, R is falling with its plane vertical in a, horizontal magnetic induction B (figure). At, the position MNQ the speed of the ring is V., The potential difference developed across the, ring is :, X, , X, B, , X, X, , 36, , X, , V, , X, M, , X, Q, , 1) zero, magnetic field of induction 4 104 T . The, 2) BV R 2 / V and M is at higher potential, resistance of the coil is 20 ohms. When the, 3) RBV and Q is at higher potential, coil is rotated through 1800 in the magnetic, 4) 2RBV and Q is at higher potential, field in 0.1 seconds the average electric 10. Two thick rods AB, CD are placed parallel to, current (in mA) induced is :, each other at a distance l. their ends are, 1) 12, 2) 24, 3) 36, 4) 48, joined to a resistance R. A magnetic field of, A closed coil with a resistance R is placed in a, induction B is applied perpendicular to the, magnetic field. The flux linked with the coil, plane containing the rods. If the rods are, is . If the magnetic field is suddenly reversed, vertical, the terminal uniform velocity of the, rod PQ of mass m is given by, in direction, the charge that flows through the, coil will be, A, C, R, 1) /2R, 2) /R, 3) 2 /R, 4) zero, , MOTIONAL E.M.F, 6., , X, , N, , An aeroplane with wing span 50 m is flying, horizontally with a speed of 360 km/hr over a, place where the vertical component of the, earth’s magnetic field is 2x10-4 Wb/m2. The, potential difference between the tips of the, wings would be:, 1) 0.1V 2) 1.0V, 3) 0.2V 4) 0.01V, , P, , X, , X, , X, , X, , X, , X, , X, , X, , Q, , X, X, , D, , l, B, , 1), , mg.R, B 2l 2, , 2), , X, X, , mg.R, Bl, , 3), , mg, BlR, , 4), , mgl, BR, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 11. A conducting ring of radius ‘r’ is rolling without, slipping with a constant angular velocity , (figure). If the magnetic field strength is B and, is directed into the page then the emf induced, across PQ is, X, , X, X, , 12., , 13., , 14., , 15., , , , X, , P X, B r 2 X, 1) B r 2 2) 2 2, X, r2 B X, 3) 4B r 2 4), X, X, Q, 8, A cycle wheel with 64 spokes is rotating with, N rotations per second at right angles to, horizontal component of magnetic field. The, induced e.m.f. generated between its axle and, rim is E. If the number of spokes is reduced, to 32 then the value of induced e.m.f. will be, 1) E, 2) 2E, 3) E/2, 4) E/4, A uniform circular metal disc of radius R is, rotating about a vertical axis passing through, its center and perpendicular to its plane with, constant frequency f. If B H and B V are, horizontal and vertical components of the, Earth’s magnetic field respectively, then the, induced e.m.f between its center and the rim, is, 1) BVfR2 2) BHfR2 3) 2 BVfR2 4) Zero, A copper disc of diameter 20 cm makes 1200, r.p.m. about its natural axis kept parallel to a, uniform magnetic field of 10-2 T. The potential, difference between the centre and edge of, the disc is, 1) 6.28 x 10-3 V, 2) 62.8 x 10-3 V, -3, 3) 0.628 x 10 V, 4) 0.628 V, In an AC generator, a coil with N turns, all of, the same area A and total resitance R, rotates, with frequency in a magnetic field B. The, maximum value of emf generated in the coil, is, 1) NABR 2) NAB, 3) NABR 4) NAB, , INDUCED ELECTRIC FIELDS, 16. A flat circular coil having N turns (tightly, wounD) is placed in a time varying magnetic, field, , B B0 sin t . The outer radius of the, , coil is R. Determine the maximum value of, the induced emf in the circuit., NARAYANAGROUP, , 1) R 2 NB0 2) 3 R 2 NB0, R 2 NB0, R 2 NB0, 4), , 3, 17. A uniform but time-varying magnetic field, , 3), , B t exists in a circular region of radius a and, is directed into the plane of the paper as, shown. The magnitude of the induced electirc, field at point P at a distance r from the centre, B(t), of the circular region, P, , r, , 1) is zero 2) decreases as 1/ r, 3) increases as r, 4) decreases 1/ r 2, , a, , SELF INDUCTION & MUTUAL, INDUCTION, 18. A coil has self inductance of 0.01H. The, current through it is allowed to change, at the rate of 1A in 10 –2 s. The induced, emf is, 1) 1V, 2) 2V, 3) 3V, 4) 4V, 19. The average self-induced emf in a 25mH, solenoid when the current in it falls from, 0.2 A to 0 A in 0.01 second, is, 1) 0.05 V 2) 0.5 V 3) 500 V 4) 50 V, 20. Two inductors each of inductance L are, j oi n ed in parallel. Thei r equ i valen t, inductance is, 1) zero, 2) 2L 3) L/2, 4) L, 21. A coil of 100 turns with a current of 5A, produced a magnetic flux of 1 Wb and, each turn of the coil. The coefficient of, self induction is, 1) 10 H 2) 20 H 3) 30 H 4) 40 H, 22. In a n i n d u ct an ce c oi l t h e cu rren t, increases from zero to 6 ampere in 0.3, second by which an induced e.m.f. of 60, volt i s p rodu ced i n it. The valu e of, coefficient of self-induction of coil is, 1) 1 henry, 2) 1.5 henry, 3) 2 henry, 4) 3 henry, 37
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 23. Two coils are at fixed locations. When coil 1, has no current and the current in coil 2, increases at the rate of 15.0 A/s, the emf in, coil 1 is 25.0 mV, when coil 2 has no current, and coil 1 has a current of 3.6 A, the flux linkage, in coil 2 is, 1) 16 mWb, 2) 10 mWb, 3) 4.00 mWb, 4) 6.00 mWb, , D.C TRANSIENT CIRCUITS, 24. A coil of inductance 0.20 H is connected in, series with a switch and a cell of emf 1.6 V., The total resistance of the circuit is 4 . What, is the initial rate of growth of the current when, the switch is closed?, 1) 0.050As -1 2) 0.40As -1 3) 0.13As -1 4) 8.0As -1, 25. Two inductance coils made of different metal, wires are having the same inductance. But, their time constants are in the ratio 1 : 2., Then the ratio of their resistances is, 1) 1 : 2, 2) 1: 2 3) 2 :1 4) 2 : 1, 26. The time constant of an inductor is 1. When, a pure resistor of R is connected in series, with it, the time constant is found to decrease, to 2. The internal resistance of the inductor, is, 1), , R 1, R( 1 2 ), R 2, R( 1 2 ), 2) 3), 4), 1, 1 2, 1, 2, 2, , LEVEL - I ( C. W ) - KEY, 01) 1, 08) 3, 15) 4, 22) 4, , 02) 2, 09) 4, 16) 4, 23) 4, , 03) 2, 10) 1, 17) 2, 24) 4, , 04) 2, 11) 1, 18) 1, 25) 4, , 05) 3 06) 2 07) 2, 12) 1 13) 1 14) 1, 19) 2 20) 3 21) 2, 26) 1, , 6., , d, R, e BH lv, , 7., , e BV lv tan =, , 8., , e BH lv where v gt, , 5., , 1., , 2., 3., 4., 38, , 0 5 104, , NBA 50 2 102 102, e, , =0.11, 1, time, 0.1, 2NBA, e, t, A, e B, t, 2NBA, E, E, , i, t, R, , BV, BH, , 10. mg Bil and i , 11., 12, 13., 14., 15., , 17., , B lv T, R, , Bl 2, E , , l 2r, 2, .E.M.F is independent of no of spokes here., 1, e BR 2 where 2 f, 2, 1, e BR 2 where 2 f, 2, The emf generated would be maximum when, flux(cutting) would be maximum i.e., angle, betweena area vector of coil and magnetic field, is 00 . The emf generated is given by [as a, function of time], e NBA cos t emax NAB, , d, dB, E.dl dt ; S dt, , dB, a 2 dB, for r a ; E , dt, 2r dt, 1, induced electric field ; for r a, r, dB, r dB, E 2 r r 2, or E r, or E , dt, 2 dt, 2, or E 2 r a, , a dB, 2 dt, Therefore, variation of E with r (distance from, centre) will be as follows, di, eL, dt, di, eL, dt, LL, Lp 1 2, L1 L 2, n L i, , At r a, E , , LEVEL-I ( C. W ) - HINTS, B 0 H , , q, , 18., 19., 20., 21., 22., , di , e L , dt , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 23. Coefficient of mutual inductance M is given by, , e1, M , 2, di2 / dt i1, , ELECTRO MAGNETIC INDUCTION, 5., , 25.0 10 3 3.6 , , e1i1, 2 , , di2 / dt , 15 , , dI, , at t 0, I 0, thus we have, dt, dI V 1.6, , 8A / s, dt L 0.2, t1 R 2, 25. t R, 2, 1, , 24. V RI L, , A flip coil consists of N turns of circular, coils which lie in a uniform magnetic field., Plane of the coils is perpendicular to the, magnetic field as shown in figure. The coil, is connected to a current integrator which, me as u res t h e t ot al c h a rg e p a ss i n g, through it. The coil is turned through 1800, about the diameter. The charge passing, through the coil is, 1), , NBA, R, , 2), , NBA, 3NBA, 3), 2R, 2R, , 4), , 2NBA, R, , B in, , L, L, 2 , ;, Solve for r from the, r, Rr, above equations, , 26. 1 , , N turns, of a coil, , LEVEL - I (H. W), MAGNETIC FLUX AND, MOTIONAL EMF, , 1., , 2., , 3., , 4., , In a coil of area 10cm2 and 10 turns with 6., magnetic field directed perpendicular to, the plane and is changing at the rate of, 10 8 gauss/ second. The resistance of the, coil is 20 . The current in the coil will, be, 1) 0.5A, 2) 5A, 3) 50A 4) 5 x 10 8A, A magnetic flux of 500 micro-webers 7., p assi n g t h rough a 200 t u rn s coi l i s, reversed in 20×10 -3 seconds. The average, emf induced in the coil in volts, is :, 1) 2.5, 2) 5.0, 3) 7.5, 4) 10.0, A rectangular coil of 200 turns and area, 100 cm2 is kept perpendicular to a uniform, magnetic field of induction 0.25 tesla. If, the field is reversed in direction in 0.01 8., second, the average induced emf in the, coil is, 1) 10 6 V 2) 10 4 V 3) 10 2 V 4) zero, A coil having an area 2m 2 is placed in a, magnetic field which changes from 1Wb/, m 2 t o 4W b / m 2 i n a n i n t e rv al o f 2, second.The average e.m.f. induced in the, coil will be, 1) 4V, 2)3V, 3)1.5V, 4)2V, , NARAYANAGROUP, , R, , C, , A conductor AB of length l moves in, , xy plane wit h velocity v v0 iˆ ˆj . A, , magnetic field B B0 iˆ ˆj exists in the, , , , , , , , , , region. The induced emf is, 1) zero, , 2) 2B0lv0 3) B0lv0 4), , 2B0lv0, To measure the field ‘B’ between the poles of, an electromagnet, a small test loop of area 1, cm2, resistance 10 and 20 turns is pulled, out of it. A galvanometer shows that a total, charge of 2C passed through the loop. The, value of ‘B’ is, 1) 0.001 T 2) 0.01 T 3) 0.1 T, 4) 1.0 T, A thin circular ring of area A is perpendicular, to uniform magnetic field of induction B. A, small cut is made in the ring and a, galvanometer is connected across the ends, such that the total resistance of circuit is R., When the ring is suddenly squeezed to zero, area, the charge flowing through the, galvanometer is, 1), , BR, A, , 2), , AB, R, , 3) ABR, , 4), , B2 A, R2, 39
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ELECTRO MAGNETIC INDUCTION, , JEE-ADV PHYSICS- VOL- IV, , 9., , A short- circuited coil is placed in a time varying, SELF INDUCTION &, magnetic field. Electrical power is dissipated, MUTUAL INDUCTION, due to the current induced in the coil. If the, number of turns were to be quadrupled (four 13. A coil has an inductance of 0.05H and 100, turns and 0.02A current is passed through, times) and wire radius halved, the electrical, it. Flux linked with coil is, power dissipated would be, 1) halved 2) the same 3) doubled 4) quadrupled, 1) 102 Wb 2) 103 Wb3) 104 Wb 4) 105 Wb, MOTIONAL E.M.F, 14. A current of 2 A is increasing at the rate, 10. A conducting square looop of side L and, of 4 A/s through a coil of inductance 2 H., resistance R movesin its plane with a uniform, The energy stored in the inductor per unit, velocity v perpendicular to one of tis sides., time is, A magnetic induction B , constant in time and, 1) 2 W, 2) 1 W, 3) 16 W 4) 4 W, space, pointing perpendicular to and into the 15. The current decays from 5 A to 2 A in, plane of the loop exists every where., 0.01s in a coil. The emf induced in a coil, X, X, X, nearby it is 30V. The mutual inductance, between the coils is, The current induced in, 1) 1.0 H 2) 0.1 H 3) 0.001 H 4) 10 H, v, the loop is, 16. A varying current in a coil change from, 1) BLv / R clock wise, 10A to 0A in 0.5sec. If the average emf, ind uced in t he coil i s 220V, t he self, 2) BLv / R anticlockwise X, X, X, inductance of the coil is, 3) 2BLv / R anticlockwise 4) zero, 1) 5H, 2) 6 H 3) 11H, 4) 12 H, 11. A metal rod moves at a constant velocity in a 17. An air - cored solenoid is of length 0.3m, area, direction perpendiuclar to its length. A, of cross section 1.2 10 3 m 2 and has 2500, constant uniform magnetic field exists in space, urns. Around its central section, a coil of 350, in a direction perpendicular to the rod as well, turns is wound. The solenoid and the coil are, as its velocity. Select the correct statement, (s) from the following., electrically insulated from each other., 1) The entire rod is at the same electric potential, Calculate the emf induced in the coil if the, 2) There is an electric field in the rod, initial current of 3A in the solenoid is reversed, 3) The electric potential is highest at the centre, in 0.25s., of the rod and decrease towards its ends, 1)0.1056V, 2)1.056V, 4) The electric potential is lowest at the centre, 3)10.56V, 4) 0.01056V, of the rod and increases towards its ends, 12. A thin flexible wire of length L is connected 18. A solenoid of length 50cm with 20 turns per, centimetre and area of cross-section 40cm 2, to two adjacent fixed points and carries a, current I in the clockwise direction, as shown, completely surrounds another coaxial solenoid, in the figure. When the system is put in a, of the same length, area of cross-section, uniform magnetic field of strength B going, 25cm 2 with 25 turns per centimetre., into the plane of the paper, the wire takes the, Calculate the mutual inductance of the, shape of a circle. The tension in the wire is, x, x, system., 1) 9.7 mH 2) 7.9 mH 3) 8.9 mH 4) 6.8 mH, 19. The current in a coil is changed from 5A, to 10A in 10 -2 s. An emf of 50mV is, induced in coil near by it. The mutual, x, x, inductance of two coils is, IBL, IBL, IBL, 1) 100 H, 2) 200 H, 1) IBL, 2), 3), 4), , 2, 4, 3) 300 H, 4) 400 H, 40, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 20. A small square loop of wire of side l is placed, inside a large square loop of wire of side, , L L l . The loops are coplanar and their, centres coincide. The mutual inductance of the, system is proportional to, 1) l / L, 2) l 2 / L, 3) L / l, 4) L2 / l, , L-R AND C-R CIRCUITS (D.C.), 21. A coil is connected to a battery of 12 V, emf and negligible internal resistance., The current in the solenoid grows to 63%, of its final steady state value in, 0.3, s. If the final steady state current is 0.6, A, the inductance of the solenoid is, 1) 0.6 H 2) 6.0 H 3) 0.015 H 4) 0.15 H, 22. A coil of inductane 8.4mH and resistance, 6 is connected to a 12V battery. The current, in the coil is 1A at approximately the time, 1) 500 s, 2) 20 s, 3) 35 ms, 4) 1 ms, 23. An ideal coil of 10H is connected series with, a resistance of 5 and a battery of 5V.2s after, the connection is made, the current flowing, (in ampere) in the circuit is, 1) I e , , 2) e, , 3) e I, , 4) I e 1 , , LEVEL-I (H. W) - KEY, 01) 2 02) 4 03) 3 04) 2 05) 4 06) 1 07) 2, 08) 2 09) 4 10) 4 11) 2 12) 3 13) 4 14) 3, 15) 2 16) 3 17) 1 18) 2 19) 1 20) 2 21) 2, 22) 4 23) 4, , LEVEL-I - ( H. W ) - HINTS, 1., 2., 3., 4., 5., , 1, dB, 1, NA., 10 103 104, R, dt, 20, d, eN, dt, , i, , NA B2 B1 , NA( B B) 2 NAB, , , time, time, time, dB, e A, dt, , , 2 NBA, Q B , R, R, , , 6. l , v and B are coplanar.., e, B1 B2 NA, 7. q dt , R, R, B 0 20 104 0.01T, 2 c , 10, 2 1 BA 0 BA, , , , 8. Q , R, R, R, R, -3, = 6×10 = 6mWb, e2, P, , 9. Power, R, d , Here, e induced emf , where NBA, dt , dB , 1, e NA , ; Also, R 2, r, dt , Where R= resistance , r= radius, l=length, P, 2 4, P N 2r 2 ;, P1, 10. Net change in magnetic flux passing through the, coil is zero., Current(of emf) induced in the loop is zero, 11. A motional emf, e Blv is induced in the rod., Or we can say a potential difference is induced, between the tow ends of the rod AB,with A at, higher potential and B at lower potential. Due, to this potential difference, there is an electric, field in the rod., L, 12. L 2 R ; R , ; 2T sin d d, 2, For small angles, sin d d, , 2T d I dL B sin 900 ; I 2 R.d .B, T IRB , , ILB, ; Correct option is (c), 2, , 13. n Li, 1 2, dU, 14. U Li ; P , 2, dt, di, , Initial flux through the coil, Bi NBA, 15. e M , dt , Final flux through the coil, Bi NBA, di, When the coil is turned through 1800 its flux 16. E L dt, reverses; the angle between magnetic field, NN A, and area vector is reversed., 17. M 0 1 2 2, l, B Bf Bi NBA NBA 2 NBA, e, , NARAYANAGROUP, , 41
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 2., 0 N1 N 2 A2, l, di, 19. E M , dt, 20. Magnetic field produced by a current i in a large, square loop at its centre, i, i, B, say B K, L, L, Magnetic flux linked with smaller looop,, i, K l 2 , B.S, L, Therefore,, the, mut ual, inductance 3., 18. M , , , l2, l2, K or M , i, L, L, L Li, 21. , R V, 22. The current-time i t equation in L R circuit, is given by [Growth of current in L R circuit], M, , 4., , i i0 1 e i / tL i , , V 12, L 8.4 103, , 2 A and t L , R 6, R, 6, and i 1A ; t ?, Substituting these values in Eq. (i), we get, t 0.97 103 s or t 0.97 ms t 1ms, E, L 10, 23. I 0 1A ; 2s ; I 1 e 1 A, R, R 5, , Where i0 , , A physicist works in a laboratory where the, magnetic field is 2 T. She wears a necklace, enclosing an area 100 cm2 of field and having, a resistance of 0.1 . Because of power, failure, the field decays to 1 T in millisecond., The electric charge circulated in the necklace, assuming that the magnetic field is, perpendicular to area covered by the, necklace is, 1) 0.01 C 2) 0.001 C 3) 0.1 C, 4) 1.0 C, , MOTIONAL E.M.F, Two parallel rails of a railway track insulated, from each other and with the ground are, connected to a millivoltmeter. The distance, between the rails is one metre. A train is, traveling with a velocity of 72 kmph along, the track. The reading of the millivoltmeter, ( in m V ) is : ( Vertical component of the, earth’s magnetic induction is 2 105 T ), 1) 144, 2) 0.72, 3) 0.4, 4) 0.2, A rod PQ is connected to the capacitor plates., The rod is placed in a magnetic field (B), directed downwards perpendicular to the, plane of the paper. If the rod is pulled out of, , magnetic field with velocity v as shown in, Figure., B, P, M, v, , LEVEL - II (C. W), , N, , Q, , MAGNETIC FLUX & FARADY’S LAWS,, INDUCED EMF,CURRENT AND CHARGE, 1., , A circular coil of ‘n’ turns is kept in a uniform, magnetic field such that the plane of the coil, is perpendicular to the field. The magnetic, flux associated with the coil is now . Now, 5., the coil is opened and made into another, circular coil of twice the radius of the, previous coil and kept in the same field such, that the plane of the coil is perpendicular to, the field. The magnetic flux associated with, this coil now is, 1) , , 42, , 2) 2, , 3), , , 4, , 4), , , 2, , 1) Plate M will be positively charged, 2) Plate N will be positively charged, 3) Both plates will be similarly charged, 4) no charge will be collected on paltes., A wire is sliding as shown in Figure. The angle, between the acceleration and the velocity of, , the wire is, , R, , B, , v, , b, , 300, , 1) 300, , 2) 400, , 3) 1200, , 4) 900, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 6., , A conducting wire xy of length l and mass m 9., is sliding without friction on vertical conduction, rails ab and cd as shown in figure. A uniform, magnetic filed B exists perpendicular to the, plane of the rails, x moves with a constant, velocity of, 10., R, a, , c, , y, , x, , l, , b, , d, , mgR, B 2l, A conduct ing r od A B of lengt h l = 1 m is, moving at a velocity v = 4 m/s making an angle, 30° with its length. A uniform magnetic field, B = 2T exists in a direction perpendicular to, the plane of motion. Then :, , 1), , 7., , mgR, Bl, , 2), , mgR, Bl 2, , 3), , X, , X, , X, , X, , X, , X, 300, , A, X, , 8., , ELECTRO MAGNETIC INDUCTION, , X, , X, , mgR, B 2l 2, , , , A uniform magnetic field exists in region given, , by B 3iˆ 4 ˆj 5 kˆ. A rod of length 5 m is, placed along y – axis is moved along x – axis, with constant speed 1 m/sec. Then the, magnitude of induced e.m.f in the rod is:, 1) zero, 2) 25 volt 3) 20 volt 4) 15 volt, A conducting rod PQ of length 1 m is moving, with uniform velocity of 2 m/s in a uniform, magnetic field of 2T directed into the plane, of paper. A capacitor of capacity c 10 F is, connected as shown. Then:, X, , X, , X, , X, , X, , X, , 4), , A, , X, , B, , X, , X, , X, , X, , X, , X, , X, , X, , 1) q A 40 C , qB 40 C, , X, , 2) q A 40 C , qB 40 C, , X, , 3) q A 40 C , qB 40 C, , B, , 4) q A q B 0, , INDUCED ELECTRIC FIELD, , X, , 1) VA – VB = 8V, 2) VA – VB = 4 V, 3) VB – VA = 8V, 4) VB – VA = 4V, A wire KMN moves along the bisector of the, angle with a constant velocity v in a uniform, magnetic field B perpendicular to the plane, of the paper and directed inward. Which of, the following is correct?, , 11. A time varying magnetic field is present in a, cylindrical region of radius R as shown in the, figure. A positive charge q is taken slowly, from P to Q through POQ, the magnetic field, varies with time as B B0t (where B0 is a, constant) and directed into the plane of the, paper. If W is the workdone thenAW =, 1) Zero 2) B0, , O, , P, , Q, , R, , 3) Infinite 4) 2B0, 12. A metallic square loop ABCD is moving in its, own plane with velocity v in a uniform, magnetic field perpendicular to its plane as, shown in the figure. Electric field is induced:, 1) Effective length of the wire is 2 L sin, , , 2, , , 2, 3) The shape of KMN is immaterial, only the, end points K N are important., 4) All the above, , A, , 1) in AD, but not in BC, , 2) E.m.f induced between K and N is 2 BLVSin, , NARAYANAGROUP, , B, , 2) in BC, but not in AD, , V, D, , C, , 3) neither in AD nor in BC, 4) in both AD and BC, 43
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , SELF INDUCTION, AND MUTUAL INDUCTION, 13. The flux linked with a coil is 0.8 Wb when a 2, A current is flowing through it. If this current, begins to increase at the rate of 400 A/s, the, induced emf in the coil will be, 1) 20 V, 2) 40 V, 3) 80 V, 4) 160 V, 14. A solenoid of self inductance 1.2 H is in series, with a tangent galvanometer of reduction, factor 0.9 A. They are connected to a battery, and the tangent galvanometer shows a, deflection of 53°. The energy stored in the, magnetic field of the solenoid is, (tan 53o = 4/3 ), 1) 0.864 J 2) 0.72 J 3) 0.173 J 4) 1.44 J, 15. There are two batteries ‘A’ and ‘B’ having, same emf. A has no internal resistance and B, has some internal resistance. An inductance, is connected first to ‘A’ and the energy in the, uniform magnetic field setup inside is ‘U’. It, is now disconnected from ‘A’ and reconnected, to ‘B’. The energy stored in the uniform, magnetic field will be, 1) U, 2) > U, 3) < U, 4) zero, 16. An emf induced in a secondary coil is 10000 V, when the current breaks in the primary. The, mutual inductance is 5 H and the current, reaches to zero in 104 s in primary. The, maximum current in the primary before the, break is, 1) 0.2 A, 2) 0.3 A 3) 0.4 A, 4) 0.5 A, 17. A mutual inductor consists of two coils X and, Y as shwon in figure in which one quarter of, the magnetic flux produced by X links with Y,, giving a mutual inductance M. What will be, the mutual inductance when Y is used as the, primary?, X, , Y, , L, , 0 A N1 / N2 , L, 0 A N1 N 2 , 2), L, 3) 0 A N1 N 2 L, , N1, , 2, , 4), , 0 A N12 N 2 , , L, 19. A small coil of radius r is placed at the centre, of a large coil of radius R , where R r ., The two coils are coplanar. The mutual, inductance between the coils proportional to, 1) r / R, 2) r 2 / R 3) r 2 / R 2 4) r / R 2, 20. The coefficient of mutual inductance of two, circuits A and B is 3 mH and their respective, resistances are 10 and 4 . How much current, should change in 0.02 s in circuit A, so that, the induced current in B should be 0.0060 A?, 1) 0.24 A 2) 1.6 A 3) 0.18 A 4) 0.16 A, , D.C TRANSIENT CIRCUITS, 21. The key ‘K’ is switched on at t = 0. Then the, currents through battery at t = 0 and t , are, , 1), , 1, 1, A, A, 15 10, , 1, 1, 2, 1, 1, 2, A, A 3), A, A 4), A, A, 10 15, 15 10, 15 25, 22. The current in an L - R circuit builds upto 3/, 4th of its steady state value in 4sec. Then, the time constant of this circuit is, , 2), , 1, 3, 4, 2, sec 2), sec 3), sec 4), sec, ln 2, ln 2, ln 2, ln 2, 23. For the circuit shown, initially ‘S’ is closed, for a long time so that steady state has been, reached. Then at t = 0, ‘S’ is opened, due to, which the current decays to zero. The heat, generated in inductor is., , 1), , 1) M/4, 2) M/2, 3) M 4) 2M, 18. A long solenoid of length L, cross section A, E2L, having N1 turns has wound about its centre a 1) zero 2), 2( R r ), E, small coil of N 2 turns as shown in figure. The, E2L, E2R, 3), 4), mutual inductance of two circuit is, 2r ( R r ), 2r ( R r ), 44, , N2, , l, , 1), , L, r, R, S, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 24. A coil of inductance 1 henry and resistance, 10 is connected to a resistance less battery, of e.m.f 50V at time t=0. Calculate the ratio, of the rate at which magnetic energy is stored, in the coil to the rate at which energy is, supplied by the battery at t=0.1s, 1, 2), e 1, , 1) e, L1, , 1, 3) 1 , e, , 02) 3, 09) 2, 16) 1, 23) 3, , 03) 3, 10) 2, 17) 3, 24) 4, , 1., , BAn, , t, , 07) 2, 14) 1, 21) 1, 28) 3, , n 2 r constant, , 6., , BIl mg or B, , 7., , e Blv sin , , 8., , e B l e f f v ; leff 2 l sin 2, , el vB, , i, , b, , 06) 3, 13) 4, 20) 4, 27) 3, , 5., , B), , C), , 05) 3, 12) 4, 19) 2, 26) 3, , N B1 B2 A 1 2 1102, q, , 0.1C, R, 0.1, e Bvlu 2 105 1 20 0.4mV, Consider the force on an electron in PQ . This, electron experiences a force towards Q . Free, electron in PQ tend to move towards N . So, M will be positively charged., , , Fm v ; 900 300 1200, , R2, , a, , 04) 1, 11) 1, 18) 2, 25) 3, , LEVEL- II (C. W ) - HINTS, , 3., 4., , S, , L2, , 01) 2, 08) 4, 15) 3, 22) 4, , 2., , R1, , 25. A), V, , 1, 4), e, , LEVEL-II ( C. W ) - KEY, , Bvl, mgR, l mg or v 2 2, R, Bl, , , , , , , , , , 9., Current growth in two LR circuits (A) and, (B) is as shown in figure (C). It follows that 10. einduced Blv and q =Ceinduced Apply Right hand, rule to know direction of induced current., 1) R1 > R2; L1 < L2, 2) R1 > R2; L1 = L2, 11., In taking charge from P to Q one has to perform, 3) R1 = R2; L1 < L2, 4) R1 < R2; L1 > L2, work against the force experienced by charge, 26. In an LR circuit, current at t = 0 is 20 A. After, due to the induced electric field. The induced, 2 s it reduces to 18 A. The time constant of, electric field is perpendicular to line POQ and, the circuit is :, hence W1 = 0, 0.8, 2, 0.4, 13. Li L , 10, 10 , , , 2, 10, , , 1) ln 2) 2 3) ln 4) 2 ln , 9, 9, 9, di, eL, 0.4 400 160v, dt, 27. A choke coil has an inductance of 4 H and a, 4, 1, 1, resistance of 2 . It is connected to a battery 14. i K tan 0.9 3 1.2 A,U 2 Li 2 1.2 1.2 0.864, of 12 V and negligible internal resistance. The 15. U i2, time taken for the current to become 3.78 A, di, i1 0 , 16. e L 5 4 , is nearly, dt, 10 , 1) 8 s, 2) ½ s, 3) 2 s, 4) 4 s, 17. The mutual inductance M remains the same, 28. A coil of 40 H inductance is connected in series, whether X or Y is used as the primary., with a resistance of 8 and this combination, 0 N1 I1, , , N, A, 18., or, or, , , N, B, A, 2, 2, 2, 2 1, is connected to the terminals of a 2V battery., L, The inductive time constant of the circuit is, NN A, 2 0 1 2 I1, (in seconds), L, 1) 40, 2) 20, 3) 5, 4) 0.2, 2, , NARAYANAGROUP, , 2, , 45
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, Comparing, M2 , , 19., , 20., , 21., 22., , wit h, , 2 MI1 ,, , we, , get, , 0 N1 N 2 A, L, , 0 I, Magnetic field at the centre of a large coil E , 2R, I, 2, Magnetic flux linked with smaller coil 0 r, 2R, r2, r2, ; M 0, ; M , I, 2R, R, Induced current in B 0.006 A 6 103 A, induced emf in B=6×10-3 ×4V=24×10 -3V, dI, 3, Now, M 24 10, dt, 24 103 0.02, A 0.16 A, or dI , 3 103, V = iR, Rt, , , i i0 1 e L , i 3i0 when t = 4 sec., , , 4, , i0 1 e, , 4, , 4R, , L, , 3 i0, , R, , t, L, 27. i i0 1 e , , , , 28. t , , LEVEL - II (H. W), MAGNETIC FLUX AND, MOTIONAL EMF, 1., , A coi l of 30 t u r n s of w i r e each of 10 cm 2, , 2., , area is placed with its plane perpendicular, to a magnetic field of 0.1 T. When the, coil is suddenly withdrawn from the field,, a galvanometer connected in series with, the coil indicated that a 10 C charge, passes around the circuit. The combined, resistance of the coil and galvanometer, is, 1) 3 , 2) 30 3) 300 4)3000 , A square coil of side 0.5 m has movable, sides. It is placed such that its plane is, perpendicular to uniform magnetic field, of induction 0.2 T. If all the sides are, allowed to move with a speed of 0.1 m/s, for 4 sec outwards, average induced emf, is, 1) Zero, 2) 0.01V 3) 0.028V 4) 0.072V, A uniform magnetic field exists in a, , regiion given by B 3iˆ 4 ˆj 5kˆ . A rod of, , , , e4 / 1, , 4, , e 4 / 4, 4, 4, 2, ln 4, , , 2, 2ln 4 ln 2, 23. Energy, stored, , in, , indcut or, 2, , 3., , 1, E, 1 E, U Li02 , i0 ,U L 2, 2, r, 2 r, fraction of energy dissipated in ‘r’, U, , 24., , 1 LE 2 r, LE 2, ,, U, , 2 r2 r R, 2r ( R r ), , dU, di, L (2i ), R, 1, dt , dt L i R e L t R V e 1 1, 0, , p, 2 Vi, V L, V R, e, , V, V, 25. i0 = R R R1 R2, 1, 2, , 1 2 , , L1 L2, , L1 L2, R1 R2, , t / , 18 20 e 2 / , 26. I I 0 e, , ln, , 46, , 10, 2, , , 9, , , 2, 10 , ln , 9 , , L, R, , 4., , length 5 m along y axis moves with a, constant speed of 1 m/s along x axis ., Then the induced emf in the rod will be, 1) 0, 2) 25 V 3) 20 V, 4) 15 V, Two identical conducting rings A and B, of radius R are rolling over a horizontal, conducting plane with same speed v but, in opposite direction. A constant magnetic, field B is present pointing into the plane, of paper. Then the potential difference, between the highest points of the two, rings is, , 10, e2 / , 9, , A, , 1) 0, , B, , 2) 2BvR 3) 4BvR 4) none of these, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 5., , 6., , 1), 3), 7., , 8., , ELECTRO MAGNETIC INDUCTION, , A flexible wire loop in the shape of a circle 9. A rod of length 10 cm made up of conducting, and non-conducting material (shaded part is, has a radius that grows linearly with time., non-conducting). The rod is rotated with, There is a magnetic field perpendicular to, cosntant angular velocity 10 rad / s about point, the plane of the loop that has a magnitude, O, in constant magnetic field of 2 T as shown, inversely proportional to the distance from, in the figure. The induced emf between the, 1, point A and B of rod will be :, B, r, , , , the centre of the loop,, How does, r, 3 cm, A, the emf E vary with time?, B, 1) E t 2 2) E t 3) E t 4) E is constant, A square loop PQRS of side ‘a’ and resistance 1) 0.029 V 2) 0.1 V, 2 cm, ‘r’ is placed near an infinitely long wire, carrying a constant current I. The sides PQ 3) 0.051 V 4) 0.064 V, O, and RS are parallel to the wire. The wire, INDUCED ELECTRIC FIELD, and the loop are in the same plane. The loop, 10. A triangular wire frame ABC (each side =2m), is rotated by 180 about an axis parallel to the, is placed in a region of time variant magnetic, long wire and passing through the mid points, field having dB/dt = 3 T/s. The magnetic, of the side QR and PS. The total amount of, field is perpendicular to the plane of the, charge which passes through any point of the, triangle. The base of the triangle AB has a, loop during rotation is:, resistance 1 while the other two sides have, resistance 2 each. The magnitude of, potential difference between the points A and, B will be, 0 Ia, Ia, n 2 2) 0 n 2, 1) 0.4 V, 2) 0.6 V 3) 1.2 V, 4) None, 2 r, r, 11. A uniform magnetic field of induction B is, ia, confined to a cylindrical region of radius R., 0 Ia 2, 4) 0, The magnetic field is increasing at a constant, 2 r, 2 r, rate of dB / dt (tesla/second). A charge q of, MOTIONAL E.M.F, mass m, placed at the point P on the periphery, A rectangular loop with a sliding connector of, of the field experiences an acceleration:, lengt h l = 1.0 m is situated in a uniform, magnetic field B = 2T perpendicular to the, plane of loop. Resistance of connecter is, R, asshown in figure. The external force required, to keep the connector moving with a constant, 1 eR dB, 1), toward left, velocity v = 2 m/s is, 2 m dt, B, 1 eR dB, 6, 3, V, 2), toward right, 2 m dt, 1) 6N, 2) 4 N, P, eR dB, 3) 2 N, 4) 1 N, 3), toward left, 4) zero, m dt, A straight rod of length l is rotating about axis, SELF INDUCTION & MUTUAL, passing through O is shown. A uniform, INDUCTION, magnetic field B exists parallel to the axis of, 12., The, e.m.f., induced, in a secondary coil is, rotation. E.m.f induced between P and Q is:, 20000, V, when, the, current, breaks in the, l/5, 4l/5, primary coil. The mutual inductance is, 5H and the current reaches to zero in 104, P, Q, sec in the primary. The maximum current, O, in the primary before it breaks is (EAM8, 3, 7, 2, 2, 2, 07), Bl 2) Bl 3), Bl 4) zero, 1), 25, 10, 25, 1) 0.1A, 2) 0.4A 3) 0.6A, 4) 0.8A, , NARAYANAGROUP, , 47
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 13. A small square loop of wire of side ‘l’ is, potential difference of 300V is: (Given, placed inside a large square loop of side, log10 4 0.602 ), ‘L’ (L>>l). If the loops are coplanar and, 1)2.2sec 2)1.1sec 3)0.55sec 4)0.48secs, t h ei r ce n t re s co i n ci d e , t h e mu t u al, 19., A, resistance with an inductor of 8 H has, i n d u ct i on o f t h e sy st em i s d i re ct ly, the same time constant as it has with a, proportional to : (EAMCET 2006 ENG), 1) L/l, 2) l/L, 3) L2/l, 4) l 2 /L, condenser of capacity 2 F . The value, 14. A straight solenoid of length 1 m has 5000, of the resistance expressed in ohms is, turns in the primary and 200 turns in the, 1) 500, 2) 250 3) 4000, 4) 2000, secondary. If the area of cross section is, 20. The cell in the circuit shown in figure is, 4cm 2 , the mutual inductance will be, ideal. The coil has an inductance of 4 mH, 1)503 H 2) 503 mH 3) 503 µ H 4) 5.03 H, and a resistance of 2mΩ . The switch is, 15. Two coaxial circular loops of radius 0.5 m and, closed at t 0 . The amount of energy, 5102 mare separated by a distance 0.5 m, stored in the inductor at t 2 s is (take, and carry currents 2 A and 1A respectively. The, e 3), force between the loops due to mutual, E = 2V, induction is, L L = 4mH, R = 2m, , 4, J, 1), S, 3, 1) 2.09 108 N, 8, 8, 3, 3, 2) 10 J 3) 10 J 4) 2 103 J, 2) 1.06 106 N, 9, 3, x, 21. In the circuit, the final current through, 3) 4.18108 N, 4) 8.3105 N, 30 resistance when circuit is completed is, 16. The mutual inductance between the, rectangular loop and the long straight w i r e, 30, as shown in figure is M., 20, 1H, 1) 3 A 2) 0.1 A, b, , 3V, , B, , K, , 3)5 A 4) 0.5 A, 22. In the circuit shown in figure, switch S is closed, at time t 0 . The charge that passes through, the battery in one time constant is, , 1) M = zero, µ 0a c , C, ln 1+ , 2) M =, a, 2π b , L, R, 2, µ 0a a + c , L, eR E, ln , 3) M =, i, 1), 2) E , 2π b , R, L, EL, eL, µ 0a b , ln 1 , 3), 4), 4) M =, S, E, eR 2, ER, 2π c , 23., A, coil, of, wire, having, finite, inductance, and, 17. An infinite long straight conducting cylindrical, resistance, has, a, conducting, ring, placed, coshell of radius a is surrounded by a thin coaxial, axially within it. The coil is connected a, infinite conducing cylindrical shell of radius, b. Assuming current flows uniformly through, battery at time t 0 , so that a time dependent, the cylindrical shell and returns through the, current I1 t starts flowing through the coil., outer shell, the inductance per unit length for, this arrangement is, If I 2 t is the current induced in the ring and, 0 b , 20 b , ln, ln, 1) , 2) a , , B t is the magnetic field at the axis of the, a, , 0 b , 3) 2 ln a , , , , 0 b , 4) 4 ln a , , , L-R AND C-R CIRCUITS (D.C.), 18. A 8 F capacitor is charged by a 400V, supply through 0.1 M resistance. The, time taken by the capactor to develop a, 48, , coil dut to I1 t then as a function of time, , t 0 , the product I 2 t B t , 1) increases with time 2) decreases with time, 3) does not vary with time, 4) passes through a maximum, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , LEVEL - II ( H. W ) - KEY, 1) 3, 08) 2, 15) 1, 21) 2, , 2) 2, 09) 3, 16) 4, 22) 3, , 3) 2 4) 3 5) 4 06) 2 07) 2, 10) 2 11) 1 12) 2 13) 4 14) 3, 17) 3 18) 4 19) 2 20) 2, 23) 4, , i, 11., , , , qE, a , Direction of induced field can determined, m, , LEVEL - II ( H. W ) - HINTS, 1., , 2., 3., , 4., , d NA B1 B2 , , R, R, 3, 30 10 0.1 0 , 10 c , ; R 300, R, dA , e B, , dt , , , , , , , e v B .l ; e i 3i 4 j 5k .5 j, e 25V, 1, 2, V2 V1 B 2 R 2 BR R 2 BRv, 2, q, , , , , , , , 4, F, , , , 2, , A, , V, , V, , , , V, RT= 5 ; VAB iRAB, RT, , , d , d B, E d l , A, d t, d t, , from Lenz’s law, i2 i1 2 104 5 0 i1 , 12. e M ., ;, time, 104, 13. Flux M i BA ;, , 0 N1 N 2 A, a, Here l =1m 503 106 H, l, 0 6 M 1 M 2, 15. F , 4, d4, where M 1 n1 i1 A1 , M 2 n2 i2 A2, 14. M , , 16. d B adx ; d , , a, 3, , , , 1, , V3 V4 2 BRv But V1 V3 ; V2 V4 4 BRv, 5. Let radius of the loop is r at any time t and, in further time dt ,radius increases by dr . 17., , r, , dr, , b, , oia c b , ln , =Mi, 2, c , , 0i, , 0i, , e, , d, dr k, 2 r , dt, dt r, , k, dr, e 2 ck (constant) dt c, B r , , , , B 2l 2v, 63, 2, here RP , RP, 63, , 7., , F, , 8., , 3Bl 2, e B l x dx , , 10, 5, 4l, , 9., , 5, , 10, , e B x dx 0.051V, , NARAYANAGROUP, , 0, , b, , L, 18. RC, R, L, , C, , , , 8, , 2 10 6 ; R 2 K , , q, 19. q q0 1 e t /T ; V , C, E, 1 2, t /, 20. Use i 1 e and U Li, R, 2, 21. The current is short cicuited through inductor. In, stready state, current will pass through inductor, instead of 20 . Then current through 30 is, 3, 0.1A, :I , 30, , 22. The current at time t is given by i i0 1 e t / , Here i0 , , 7, , d, dB, A, 10. P.D in the loop E dl , dt, dt, , b, , a, , t + dt, , Then change in flux : d 2 rdr B, , oi, a dx, 2 x, , 2 x dx 2 log a ; L 2 log a , R , , t, , 4 o l 2i, L 2, l2, M, L, Mi , , E, L, and , R, R, , , , t, , 0, , 0, , q idt i0 1 et / dt, 49
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 3., E L , i R R EL, 0 2, e, e, eR, 23. The equation of I1 t , I 2 t and B t will take the, following forms, I1 t K1 1 e k 2t current growth in L-R 4., circuit, B t K 3 1 e k 2t B t I1 t , , I 2 t K 4 e k 2t, e2, dI1 , , I 2 t R and e2 M dt , , , Therefore,, the, , A 0.1 m long conductor carrying a current of, 50 A is perpendicular to a magnetic field of, 1.25 mT. The mechanical power to move the, conductor with a speed of 1ms 1 is, 1) 0.25 mW, 2) 6.25 mW, 3) 0.625 mW, 4) 1 W, A circuit ABCD is held perpendicular to the, uniform magnetic field of B 5 102 T, extending over the region PQRS and, 3directed into the plane of the paper. The, circuit is moving out of the field at a uniform, speed of 0.2ms-1 for 1.5s. During this time,, the current in the 5 resistor is, , product, , P, , Q, , I 2 t B t K 5 e k 2 t 1 e k 2 t , , The value of this product zero at t 0 and t . 1) 0.6 mA from B to C, Therefore, the product will pass through a, maximum value., 2) 0.9 mA from B to C, , LEVEL - III, MAGNETIC FLUX & FARADY’S LAWS,, INDUCED EMF,CURRENT & CHARGE, 1., , 0.3m, , C, , D, S, , R, , , , I, , 2, , passes through a loop of area 5.0 cm is placed, flat on xy plane is, 1)750nWb 2)-750nWb 3) 360nWb 4) -360nWb, , a, , 1) 0.23 mV 2) 0.46 mV, , b, v, , MOTIONAL E.M.F, A straight conducting rod of length 30cm 6., and having a resistance of 0.2 ohm is allowed, to slide over two parallel thick metallic rails, with uniform velocity of 0.2 m/s as shown in, the figure. The rails are situated in a, horizontal plane if the horizontal component, of earth’s magnetic field is 0.3 104 T and, a steady current of 3 A is induced through, the rod. The angle of dip will be :, , 3) 0.16 mV 4) 0.32 mV, A rectangular loop with a sliding conductor, of length l is located in a uniform magnetic, field perpendicular to the plane of the loop., The magnetic induction is B . The conductor, has a resistance R1 and R2 , respectively.., Find the current through the conductor, during its motion to the right with a constant, velocity v ., A, D, R1, , 3, 1 1 , 1) tan 2) tan , , 4, 3, 1 1 , 1, 3 4) tan , 3) tan, 3, 1, , , , 50, , 5, 0.2ms1, , B, , induced emf in the rod, where v 5ms-1 ,, a 1cm , b 100cm ., , 4, by B 40i 15k 10 T . The magnetic flux, , 2., , B, , 3) 0.9 mA from C to B 4) 0.6 mA from C to B, 5. Figure shows a copper rod moving with, velocity v parallel to a long straight wire, carrying current 100A . Calculate the, , A magnetic field in a certain region is given, , , , 0.5m, , A, , R1 R1 R2 , , Bl v, R1 R1R2, , B, , Blv R1 R2 , , 2, , 2), , v, , R2, , Blv R1 R2 , , 1), , V =0.2m/s, , R, , 3) R R, 1, , 2 R R1 R2 , , C, 2, , 4), , Bl v, R1R2 R R1 R2 , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , A conductor of length l and mass m can slide 12. The coil of an ac generator rotates at a, frequency of 60 Hz and develops an inducecd, without any friction along the two vertical, conductors connected at the top through a, emf of 120 V (rms). The coil has an area of, capacitor. A uniform magnetic field B is set, 3.0 x10-3 m2 and consists of 500 turns. The, up to the plane of paper. The acceleration, magnitude of the magnetic field in which the, C, of the conductor, coil rotates is, 1) 0.30 T 2) 3.00 T 3) 0.03 T 4) 30 T, 13. A conducting loop of radius R is present in, a uniform magnetic field B perpendicular to, 1) is constant 2) Increases, the plane of ring. If radius R varies as a, 3) decreases 4) cannot say, function of time t as R R0 t 2 . The emf, 8. A wheel has three spokes and is in a uniform, induced in the loop is, magnetic field perpendicular to its plane,, with the axis of rotation of the wheel, parallel to the magnetic field. When the, wheel rotates with a uniform angular, R, velocity , the emf induced between the, center and rim of the wheel is ‘e’. If another, wheel having same radius but with six, spokes is kept in the same field and rotated, 1) 2 Bt R0 t 2 Clockwise, with a uniform angular velocity ‘ /2’, the, emf induced between the center and the rim, 2) 2 Bt R0 t 2 Anticlockwise, will be, 1) e, 2) e/2, 3) 2e, 4) e/4, 2, 3) 4 Bt R0 t Anticlockwise, 9. A metal rod of length 1 m is rotated about, one of its ends in a plane at right angles to, 2, 4) 4 Bt R0 t Clockwise, a uniform magnetic field of induction, 2.5 103Wbm2 . If it makes 1800 rpm, then the 14. A conducting rod is free to slide along a pair, induced emf between its ends approximately, of conducting rails, in a region where a, is, uniform and constant (in time) magnetic field, 1) 0.24 V 2) 0.12 V 3) 0.36 V 4) 0.48 V, is directed into the plane of the paper, as the, 10. A bicycle is resting on its stand in the eastdrawing illustrates. Initially the rod is at rest., west direction and rear wheel is rotating at, There is no friction between the rails and the, 100 revolution per minute. Length of each, rod. What happens to the rod after the switch, spoke is 30 cm, and vertical component of, is closed? If any induced emf develops, be sure, earth’s magnetism is 1.5 105 tesla. If the, to account for its effect., , 6, emf induced in the spokes is 3 10 V , the, Rod, B (into paper), angle, of, di p, wi ll, be, :, , 7., , 1, , 3, , 1, , 1 , , , , 1, , 1, , 1) tan 4 2) tan 3) tan 1 3 4) tan 3 , , , 3, 11. Find the linear speed the bicycle required to, power its head light by a generator, whose, rubber shaft presses against the wheel of cycle, of radius 0.33 m, turns at an angular speed of, 44 times as great as the angular speed of the, tire itself. The coil consists of 75 turns, has, an area of 2.6 X 10-3 m2, and rotates in a 0.10, T magnetic field. When the peak emf being, generated is 6.0 V., 1) 2.5 m/s 2) 5 m/s 3) 2.3 m/s, 4) 4.6 m/s, NARAYANAGROUP, , , Swltch, Conducting rail, , 1) The rod accelerates to the right, its velocity, increasing without limit., 2) The rod does not move., 3) The rod accelerates to the right for a while, and then slows down and comes to a halt., 4) The rod accelerates to the right and eventually, reaches a constant velocity at which it continues, to move., 51
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 15. A rectangular loop has a sliding connector PQ 19. A wire is bent to form the double loop shown in, Figure. There is a uniform magnetic field, of length l and resistance R and it is moving, directed into the plane of the loop. If the, with a speed v as shown. The set-up is placed, magnitude of this field is decreasing, the, in a uniform mangetic field going into the plane, current will flow from, of the paper. The three currents I1 , I 2 and, P, l, a, b, I are, c, d, 1) a to b and c to d, 1), 2), , Blv, 2 Blv R, I1 I 2 , ,I , R, R, Blv, 2 Blv, I1 I 2 , ,I , 3R, 3R, , 3) I1 I 2 I , , Blv, R, , v, , R, , R, , 3) a to b and d to c, , l, l1, , 4) I1 I 2 , , Q, , 2) b to a and d to c, , l2, , Blv, Blv, ,I , 6R, 3R, , INDUCED ELECTRIC FIELD, 16. A specially uniform magnetic field of 0.080 T, is directed into the plane of the page and, perpendicular to it, as shown in the, accompanying figure. A wire loop in the plane, of the page has constant area 0.010 m2. The, magnitude of the magnetic field decreases at, a constant rate of 3.0 X 10 - 4 T/s. The, magnitude and direction of the induced emf is, , 4) b to a and c to d, , SELF INDUCTION AND MUTUAL, INDUCTION, 20. Two coils have self-inductance L1 = 4 mH and, L2 = 1 mH respectively. The currents in the, two coils are increased at the same rate. At, a certain instant of time both coils are given, the same power. If I1 and I2 are the currents, in the two coils, at that instant of time, respectively, then the value of (I1/I2) is:, 1) 1/8, 2) 1/4, 3) 1/2, 4) 1, 21. The length of a wire required to manufacture, a solenoid of length l and self-induction L is, (cross-sectional area is negligible), , 4 Ll, 0 Ll, 0 Ll, 3., 4., 1) 3.0 X 10 V clockwise, 0, 4, 2, 2) 3.0 X 10 - 6 V anticlockwise, 22. The inductance L of a solenoid of length l ,, 3) 2.4 X 10 - 5 V anticlockwise, 4) 8.0 X 10 - 4 V clockwise, whose windings are made of material of, 17. A magnetic field induction is changing in, density D and resistivity , is (the winding, magnitude in a region at a constant rate dB/, resistance is R), dt. A given mass m of copper drawn into a, 0 Rm, 0 lm, 0 lm, 0 R 2 m, wire and formed into a loop is placed, 1) 4 l D 2) 4 r D 3), 4) 2 R D, perpendicular to the field. If the values of, 4 l D, specific resistance and density of copper are 23. Two concentric and coplanar coils have radii, and respectively, then the current in, a and b a as shown in figure. Resistance, the loop is given by :, of the inner coil is R. Current in the outer coil, 4 m dB, m dB, m dB, 2 m dB, 1) dt 2) 4 dt 3) dt 4) dt, is increased from 0 to i , then the total charge, circulating the inner coil is, 18. Magnetic flux linked with a stationary loop, of resistance R varies with respect to time, during the time period T as follows:, b, at T t the amount of heat generated in, 0 iab, 0ia 2, a, the loop during that time (inductance of the, 1), 2), 2, R, 2, Rb, coil is negligible) is, ib, 0iab b 2, aT, a 2T 2, a 2T 2, a 2T 3, 3), 4) 0, 1), 2), 3), 4), 2 R, 2a R, 3R, 3R, R, 3R, -6, , 52, , 1., , 2 Ll, 0 2., , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 24. The inner loop has an area 5 104 m2 and a 28. A coil of inductance 300 mH and resistance, 2 is connected to a source of voltage 2V.., resistance of 2 . The larger circular loop, The, current reaches half of its steady state, is fixed and has a radius of 0.1m. Both the, value, in (nearly), loops are concentric and coplanar. The, 1), 0.05, s 2) 0.1 s, 3) 0.15 s 4) 0.03 s, smaller loop is rotated with an angular, velocity rad s-1 about its diameter. The 29. In the circuit shown, X is joined to Y for a, long time and then X is joined to Z. The total, magnetic flux with the smaller loop is, heat produced in R2 is, R, 2, , L, , 1) 2 106 weber, , 1A, , Z, , 1A, , X, Y, LE 2, 1., 2 R12, E, R, 2, 2, 3) 109 cos t weber 4) zero, LE, LE, LE 2 R2, 2., 3., 4., 25. A circular wire loop of radius R is placed in, 2 R22, 2 R1 R2, 2 R13, the x-y plane centred at the origin O. A square 30. A condenser, in series with a resistor is, loop of side a a R having two turns is, connected through a switch to a battery of, negligible internal resistance and having an, places with its centre at z 3R along the, emf of 12 V. One second after closing the, axis of the circular wire loop, as shown in, switch, the condenser is found to have a, figure. The plane of the squre loop makes an, potential difference of 6 V. The time constant, angle of 450 with respect to the z-axis. If the, of the system is, mutual inductance between the loops is given, 1, 1, a2, 3) log e 2 4) log e , 1) 2 s 2) log 2, by P0/ 2 , then the value of P is, 2, e, 2 R, z, 31. The time constant of the circuit shown is, 45, , 2) 109 weber, , 1, , 0, , R, a, , R, 3R, , 1) 3, , R, , 3) 7, , 1) RC, , 2) 5, 4) 9, , y, O, , x, , D.C TRANSIENT CIRCUITS, , R, , 27. An inductance L and a resistance R are, connected in series to a battery of voltage V, and negligible internal resistance through a, switch. The switch is closed at t = 0. The, charge that passes through the battery in one, time constant is [e is base of natural, logarithms], VL, VL, eL, eR 2V, 1), 2), 3), 4), 2, R, eR, VR, L, NARAYANAGROUP, , 4C/3, , 3) 3RC 4) 4RC, 32. In the circuit shown below. the key K is closed, at t 0 . The current through the battery is, V, , 26. A coil of some internal resistance ‘r’ behaves, like an inductance. When it is connected in, series with a resistance R1, the time constant, is found to be 1. When it is connected in, series with a resistance R2, the time constant, is found to be 2. The inductance of the coil, is, 1 2 ( R1 R2 ), ( 2 1 ), ( R1 R2 ), ( 2 1 ), 1) ( ) 2) ( R R ) 3) ( ) 4) ( R R ), 2, 1, 1 2, 1, 2, 2, 1, 1, 2, , 2) 2RC, , R1, , L, R2, , VR1 R2, , 1), 2), , 2, 1, , 2, 2, , at t 0 and, , R R, , V R1 R2 , V, at t 0 and, at t , R2, R1 R2, , V, 3) R at t 0 and, 2, , 4), , V, at t , R2, , VR1 R2, R12 R22, , at t , , V R1 R2 , V, at t 0 and, at t , R1 R2, R2, 53
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , LEVEL-III - KEY, 01) 2, 08) 2, 15) 2, 22) 1, 29) 1, 1., , 02) 4, 09) 1, 16) 1, 23) 1, 30) 2, , , B.A, , 03) 2, 10) 1, 17) 2, 24) 3, 31) 2, , 04) 1, 11) 3, 18) 4, 25) 3, 32) 2, , 05) 2, 12) 1, 19) 3, 26) 1, , 06) 3, 13) 3, 20) 2, 27) 3, , 07) 1, 14) 4, 21) 3, 28) 2, , LEVEL-III - HINTS, , 40i 15, k 10 4 5 10 4 k 750 nWb, , , , , , , 2., , , , Only veritical lines of earth’s magnetism will, cut when the rod is allowed to slide over the rails., Let the vertical component of earth’s magnetic field, is Bv ., Induced, Bv l 0 Bv l, , R, R, 5, BV 10 T Given BH 0.3 10 4 T, , current i , , 1 Bv , The angle of dip is given by tan B , H, , 3., , P Fv BIlv 1.25 10 3 50 0.1 1W, , 6.25 103 W 6.25mW, An iterating alternating, P EI Blv I, 4., , 5., , Blv, 5 102 0.3 0.2, I, , A 0.6mA, ;, R, 5, Area and flux are decreasing. So, current flows, to increase the flux. Clearly, current should be, clockwise. So, it flows from B to C through 5 ., Let there be an element dx of rod at a distance, x from the wire., emf developed in the element, dE Bdxv, 2I , dE 0, dxv, 4 x , I, , E , , 0 Iv b dx 0 Iv, b, , log e, , a, 2, x, 2, a, , E , , 4 107 100 5, 100, log e, 2, 1, , 54, , I , , 7., , Blv R1 R2 , Blv, , R1R2, R1R2 R R1 R2 , R, R1 R2 , , Let v is the velocity of conductor at any time, then, induced emf: e Blv ............. (i), charge on capacitor : q Ce CBlv, , dq, dv, CBl, dt, dt, mdv, for conductor mg IBl , dt, dv mdv, dv, mg, mg CB 2l 2, , , , dt, dt, dt m CB 2l 2, This is the acceleration of conductor which is, constant., dA, 1, Br 2, 8. e B, dt, 2, In, every, spokes, same emf developed, and, , cycle spokes is equivalent to parallel, combination of cells., 1, 2, 9. B l, 2, 10. The bicycle is resting in east -west direction,, therefore horizontal magneti field lines of earth’s, magnetisation will cut across the spokes of the, wheel. Let this horizontal field of earth’s, magnetism is BH . The induced emf between the, , Current in circuit : I , , axle and the rim of the wheel is e , , 1, BH l 2, 2, , BH 2 10 5 T, , given, , Bv 1.5 10 5 T . Angle of dip, , B , tan 1 v , BH , , e.m. f, 1 , 11. v r coil where coil , NAB, 44 , , 2Vr .m. s, NA, 4.6 10 V 0.46mV, Induced emf Blv . R is internal resistance of, d, d, B R2 , 13. e , seat of emf, i.e., of rod, dt, dt, 4, , 6., , R1R2, Total resistance of circuit R R R, 1, 2, , 12. B , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , 15. A moving conductor is equivalent to a battery of, 1 i, L, 1, 20. i ; 1 2 , emf vBl ( motion emf ), L i2 L1 4, Equivalent circuit ;, I I1 I 2, 0 N 2 A, Applying Kirchhoff’s law, 21. L , l, I1R IR vBl 0 i , If x is the length of the solenoid with r as radius,, then x 2 rN , A r 2, I 2 R IR vBl 0 ii , Adding Eqs. (i) and (ii), x2 r 2 , x , , , L, , , N, , 0, 2vBl, 2 2 , , 2 r , , 4 r l, 2IR IR 2vBl ; I 3R, 4 Ll, 0 iR 2, 0 a 2, x , B, M, , , 3/, 2, ; B ACos 45 ;, 0, a 2 R2, i 27 / 2 R 2, , , , , , 2 A, dB, L, , , N, . If x is the length, 22., For, a, solenoid,, , 2, , 4, , 6, 0, 16. e A, 10 3 10 3 10, l, dt, of the wire and a is the area of cross-section,, 17. Let r is the radius of the loop. The magnetic flux,, x, at a time t, through the loop is given by, then R , and m axD, a, d, 2 dB, 2, r ( ), B r ; e , dt, dt, x, Rm, Rm , axD, x , (the numerical valu of induced emf), a, D, Induced current in the loop is given by, 2, x L 0 N A , e r 2 dB, i , ....(i), , , Also, x 2 rN , N , l , 2 r , R, R dt, where R is the resistance of the wire let a is the, 2, 2, Rm, x r, corss-section area and l the length of the wire., L 0 , 0, , 4 l D, 2 r l, l, R, , , Then, .......(ii), 0ia 2, , 0 i , 2, a2, , , , a, , , q, , 23., ; 0; f , , 2b, R i, 2 b, where l 2 r ; m a 2l, On putting the value of a 2 in eq. (ii), we get, 0ia 2, 0ia 2, f i , ; So, q , 2b, 2 Rb, m dB, i, 24., Magnetic, field, due, to, larger, loop, 4 dt, , , , 18. Given that at T t , induced emf, E , , d d, at T t , dt dt , , at 0 1 a T t ; a T 2t , So, indeced emf is also a function of time., Heat generated in time T is, , , , 01 4 107 1, , T 2 106 T, 2r, 2 0.1, Now, magnetic flux linked withthe smaller loop, , , , NBA cos t 1 2 106 5 104 cos t, 25., , B, , 0 iR 2, , a, , 2, , 3/ 2, , R2 , , E2, a2 T, a 2T 3, 2, 0 a 2, H , dt , T, , 2, t, dt, , , , M 7/2 2, 0 R, R 0, 3R, i 2 R, 19. By Lenz’s law, clockwise current is induced in, both loops. Greater the area, large will be the, induced emf Outer loop has greater area., 26. 1 L ; 2 , R1 r, , ; B ACos 45 , , T, , NARAYANAGROUP, , L, R2 r, 55
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, R, t , , 27. q idt i0 1 e L dt, , , R, , t, L, 28. i i0 1 e , , , , 2., , 3., , E, 29. The current in L for steady state R, 1, Energy stored in L , , 1 2 1 E2, LI 0 L 2, 2, 2 R1, , ; t = 1 second, 1, CV CV 1 e ; e, 2, , , 30. q q0 1 e, , t, , 4., , Rc, , t, , Rc, , t, , Rc, , 0, , 5., , 1, 1, ln e ; RC , Rc, ln 2, 6., R, 3R, 3R 4C, 1, , 2 RC, 31. R R , ; i RC , 2, 2, 2, 3, 32. At t 0 , inductor behaves like an infinite, ln 2 , , V, resistance so at t 0, i R, 2, and at t , inductor behaves like a conducting, V R1 R2 , V, wire i R , R1 R2, eq, , Assertion (A): Lenz’s law is a violation of the law, of conservation of energy., Reason (R): Induced emf always oppose the, change in magnetic flux that has produced them., Assertion(A): An inductor in a D.C. circuit, opposes both a steady current and a changing, current, Reason(R): Induced emf is generated only when, the flux linked with the inductor remains, unchanged., Assertion(A): The possibility of an electric bulb, fusing is higher at the time of switching on and, off., Reason(R): Inductive effects produce a large, current at the time of switch-on and switch-off., Assertion(A): L/R and CR both have same, dimensions., Reason(R): L/R and CR are time constants, Assertion(A): When a charged condenser, discharges through a resistor, the time taken for, half the charge to be lost is always same,, irrespective of the initial value of the charge., Reason(R): The rate of decay of charge in aCR, circuit is a linear function of time., , MATCH THE FOLLOWING, 7., , A square loop is placed near a long straight, current carrying wire as shown, match the, following table., , LEVEL - IV, ASSERTION & REASON, , 1., , 56, , Instructions for the Assertion & reason type, questions:, 1) Both Assertion and Reason are true and, Reason is the correct explanation of Assertion, 2) Both Assertion and Reason are true and, Reason is not the correct explanation of, Assertion, 3) Assertion is true but reason is false, 4) Assertion is false but reason is true., Assertion (A): Whenever the magnetic flux, linked with a closed coil changes there will be, an induced emf as well as an induced current., Reason (R): According to Faraday, the induced, emf is inversely proportional to the rate of, change of magnetic flux linked with a coil., , Table - 1, Table - 2, a) If current is increased p) Induced current in, loop is clockwise, b) If current is decreased q) Induced current in, loop is anticlockwise, c) If loop is moved away r) Wire will attract, from the wire, the loop, d) If loop is moved to, s) Wire will repel the, wards the wire, loop, 1) a-q,s; b-p,r; c-q,s; d-p,r, 2) a-p,r; b-q,s; c-p,r; d-q,s, 3) a-q,s; b-p,r; c-q,r; d-p,q, 4) a-q,s; b-p,r; c-p,r; d-q,s, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, , GRAPHS, 8., , An equilateral triangular loop ADC having, some resistance is pulled with a constant, velocity v out of a uniform magnetic field, directed into the paper. At time t 0 , side DC, of the loop is at edge of the magnetic field., , If e denotes the induced emf in L and i the, current flowing through the circuit at time t ,, then which of the following graphs correctly, represents the variation of e with i ?, , 1), , A, , 2), , e, , e, , i, , D, , i, , 3), , C, , 4), e, , V, , e, , The induced current (i) versus time (t) graph, i, i, will be as, 11. In an LR circuit connected to a battery, the, rate at which energy is stored in the inductor, e, i, is plotted against time during the growth of, curretn in the circuit. Which of the following, 1), 2), t, i, best represents the resulting curve?, i, Rate, , 3), t, 9., , Rate, , 4), t, , The current through the coil in figure(i) varies, as shown in figure (ii). Which graph best, shows the ammeter A reading as a function of, time?, , i1, i2, , 2), Time, , Time, , Rate, , Rate, , 3), , 4), , i1, , Time, , S, , A, , 1), , Time, , t, 12. In the circuit shown, switch k2 is open and, , R, , i2, 1), , 2), , i2, , t, , t, , switch k1 is opened and switch k2 is, simultaneously closed. The variation of, inductor current with time is, K2, , i2, , i2, 3), , t, , 4), , 10. Switch S of the circuit shown in figure is closed, at t=0, , R, , NARAYANAGROUP, , i, 1) Et, 0, t0, , S, L, , E, , K1, , t, , Et0, 3) t0, , t0, , i, t0, , i, 2) Et, 0, t, 0, t, i, Et0, 4) t0, t, , t0, , t, , t0, , t, 57
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETIC INDUCTION, 13. Switch S of the circuit shown in figure is closed, at t 0 . If e denotes the induced emf in L and, I is the current flowing through the circuit at, time t , which of the following graphs is/are, correct?, , B, , t=0, , S, , 1), R, , V, , 900, , 2), , i, , i, , L, , t, e, , t, , i, , 1), , 2), O, , 3), O, , t, , 4), , i, , t, , i, , t, , t, , 16. A flexible conducting wire in the form of a, circle is kept in a uniform magnetic field with, its plane normal to the field. Radius of that, circle changes with time as shown. Then which, O, O, t, t, of the following graphs represents the, 14. Two infinitely long conducting parallel rails, variation of induced emf with time, are connected through a capacitor C as shown, R R0 , t t0 ; R R0 t , t0 t 2t0 :, in figure. a conductor of length l is moved, e, , i, , 3), , 4), , with constant speed v0 . Which of the following, graph truly depicts the variation of current, through the condctor with time?, , e, , e, , 1), , 2), t, , V0, , l, , 2t, , t, , 3) e, , 2t, , t, , t, , 2t, , 4) e, i, , 1), , t, , 2), , 2t, , 17. A wire loop is placed in a region of time, varying magnetic field which is oriented, t(Time), t(Time), orthogonally to the plane of the loop as shown, 3), 4), in figure. The graph shows the magnetic field, Current, Current, variation as the function of time. Assume the, I(t), I(t), positive emf is the one which drives a current, t(Time), t(Time), in the clockwise direction and seen by the, 15. The figure shows an isosceles triangle wire, observer in the direction of B. Which of the, following grapsh best represents the induced, , frame with angle equal to . The frame starts, emf a function of time., 2, B, entering into the region of uniform magnetic, field B with constant velocity v at t 0 . The, longest side of the frame is perpendicular to, the direction of velocity. If ‘i’ is the, instantaneous current through the frame then, t, t, t, t, choose the alternative showing the correct, variation of ‘i’ with time., Current, I(t), , Current, I(t), , 1, , 58, , 2, , 3, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , 1) e, , ELECTRO MAGNETIC INDUCTION, , 2) e, t1, , e, , t2 t3, 1), , t1, , t2 t3, , e, 2), t, , 3) e, , 4), , 3) e, , e, , t, , 4) e, i, , t2 t3, , t1, t2 t3, , t1, , t, , t, , 18. The current is an induction coil varies with 20. A flexible wire bent in the form of a circle is, time t , according to the graph shown in figure., place in a uniform mgnetic field, Which of the following graphs shows the, perpendicularly to the plane of the coil. The, induced emf (e) in the coil with time., radius of the coil changes as shown in Figure., The graph of magnitude of induced emf in the, coil is represented by, i, , y, O, , c, , t, , e, , b, , O, , 1, , r, , e, , 1), , a, , d, , 2), t, , t, , 3) e, , t, , y, , 4) e, , 2), e, , t, , 19. A and B are two coils placed closely as shown., The current in coil A varies as shown in the, graph., , e, A, , O, , 1, , 2, , x, , e, , 3), , e, , O, , t, , y, , 2, , x, , 1 2, t, , x, , 1, , t, , y, 4) e, , O 1 2, t, , B, , x, , O, , LEVEL - IV - KEY, , t, Then which of the following graphs gives the, best representation of variation of induced, emf in coil B?, NARAYANAGROUP, , y, , 1), , i, t, , x, , 2, , 01) 3, 05) 1, 09) 1, 13) 3,4, 17) 3, , 02) 4, 06) 3, 10) 3, 14) 3, 18) 3, , 03) 4, 07) 4, 11) 1, 15) 4, 19) 3, , 04) 1, 08) 2, 12) 1, 16) 3, 20) 2, 59
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, 4., , LEVEL - V, SINGLE ANSWER QUESTIONS, ELECTRO MAGNETIC INDUCTION, 1., , 2., , an initial velocity v0 to the right and is released, at t=0. The velocity of the bar as a function of, time is given by, , A helicopter has metallic blades with a length, of 3 m extending outward from a central hub, and rotating at 2 rev/s. If the vertical, component of the earth’s magnetic field is, 50 T , then the emf induced between the, blade tip and the centre hub is, (A) 4.95 V, (B) 2.83 mV, (C) 5.66 mV, (D) 11 mV, In the figure shown AB is a plastic rod moving, with a constant velocity v, W is an infinite long, wire carrying current I. The end of the rod, which is at higher potential is, , , , , , B0, , , , , , R, , , , , , , , , , , , , mRt, B02l 2, , , , , , , , , , , V0, , , , , l, , , , , , , , , , (B) v e , 0, , , , , , B02l 2t, mR, , B 2l 2t, 0, e 2mR, , 2B02l 2t, e mR, , V, , A, , , , (A) v0 , , 5., , a, , A conducting bar of mass m and length l, moves on two frictionless parallel conducting, rails in the presence of a uniform magnetic field, B0 -directed into the paper. The bar is given, , (C) v, (D) v, 0, 0, In a cylinder region of radius R, a uniform, magnetic field is there which is increasing with, time, according as B B0t 2 . A positive point, , B, , I, b, , , , , charge q is released from rest at P OP , W, , 3., , at t=0 [the instant the field is switched on], , (A) A, (B) B, (C) No emf will be induced in the rod, (D) None of the above, A non-conducting disk of radius R is rotating, about its own axis with constant angular, velocity in a perpendicular uniform, magnetic field as shown in figure. The emf, induced between centre and rim of disk is, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , R, , , , , P O , , , , , The force experienced by, the point charge at, t=1s, is ( R = 2m ), (A) qB0 , anti-clockwise (B) qB0 , clockwise, 6., , 60, , R, , 2, , B R 2, (A), 2, , (B) B R 2, , (C) Zero, , B R 2, (D), 3, , (C) 2qB0 , anti-clockwise(D) 2qB0 , clockwise, Three circuits are arranged in a region of, magnetic field as shown in the figure. In the, cylindrical region of magnetic field shown,, magnetic field is increasing with time. Rank, the circuits for the thermal energy dissipated, to resistor R2 from the least to the greatest, R1, , R2, C1, , R1, , R2, C2, , R1, , R2, , C3, , (A) C2 , C1 , C3, , (B) C3 , C1 , C2, , (C) C1, C2 , C3, , (D) None of these, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, , MULTIPLE ANSWER QUESTIONS, 7., , Two parallel resistance less rails are connected PASSAGE - 1, by an inductor of inductance L at one end as, A conductor of mass m and length l is sliding, shown in figure. A magnetic field B exists in, smoothly an two vertical conducting rails AB and, the space which is perpendicular to the plane, CD as shown in figure. The top ends of two, of the rails. Now a conductor of length l and, conducting rails are joined by a capacitor of, mass m is placed transverse on the rail and, capacitance C. The conductor is released from rest, given an impulse J towards the rightward, when it is very close to AC i.e., y 0 . A uniform, direction. Then select the correct option (s) :, , L, , I, , a) Velocity of the conductor is half of the initial, velocity after a displacement of the conductor, 2, , 3J L, 4 B 2l 2 m, b) Velocity of the conductor is half of the initial, velocity after a displacement of the conductor, d, , 3J 2 L, d, B 2l 2 m, c) Current flowing trough the inductor at the instant, when velocity of the conductor is half of the initial, , 8., , COMPREHENSION TYPE QUESTIONS, , magnetic field B0 perpendicular to plane of figure, existing. Neglect the resistance of rails and, connecting wires. Take acceleration due to gravity, to be g., , , , , , , , , , , , , , , , , , , , , , , , , , , , , A, , B, , l, , , , , , , , C , , , , , , , , , , , , , , , , , , , , , D, , ., , Y, , Based on above information answer the following, questions:, 9. Mark the correct statement about the motion, 3J 2, of conductor, velocity is i , 4 Lm, (A)It is falling down with constant acceleration g, d) Current flowing through the inductor at the instant, (B) It is falling down with constant acceleration but, when velocity of the conductor is half of the initial, not equal to g, 3J 2, (C) It is fallig down with increasing acceleration, velocity is i , mL, (D) It is falling down with decreasig acceleration, In the figure shown, the wires PQ, and, P, Q, 10. Charge on the capacitor as a function of y is, 1 1, 2 2, are made to slide on the rails with same speed, given by, , 1, of 5cms . In this region a magnetic field of 1, 2mgy, T exists. The electric current in the 9, (A) CB0l, m CB02l 2, resistance is :, , , P1, , P2, , , , , , , , , , , , , , , , , , , , 2, , , , 9, , , , , , , , , , , , , , , , 4cm, , , , , , , , , , , , , , 2, , Q1, , , , Q2, , a) Zero if both wires slide towards left, b) zero if both wires slide in opposite directions, c) 0.2 mA if both wires move towards left, d) 0.2 mA if both wires move in opposite directions, NARAYANAGROUP, , (B) CB0l, , (C), , CB0, , m CB02l 2, 2mgy, mgy, mCB02l 2, le, mgy, , 2mgy, m CB02l 2, , e, (D) CB0l, m CB02l 2, 61
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, , 14. The maximum current in the inductor is :, 11. Current in the circuit is, (A) constant, 3V 3C, 3C, (B) increasing with time, a) 0, b) V0, 2, L, L, (C) decreasing with time, (D) First increases, then reaches a maximum value,, 3C, C, c) 2V0, d) V0, and then starts decreasing to attain a constant value, L, L, finally, 15. Potential difference across capacitor of, PASSGAE -2, capacitance C when the current in the circuit, A thin non-conducting ring of mass m , radius a ,, is maximum is, carrying a charge q can rotate freely about its own, V0, 3V0, axis which is vertical. At the initial moment the ring, a), b), was at rest and no magnetic field was present. At, 4, 4, instant t = 0, a uniform magnetic field is switched, 5V0, 7Vo, on which is vertically downwards and increase with, c), d), 4, 4, time according to the law B B0t . Neglecting, 16. Potential difference across capacitor of, magnetism induced due to rotational motion of the, capacitance 3C when the current in the circuit, ring. Now answer the following questions., is maximum is :, 12. The angular acceleration of the ring and its, V0, V0, direction of rotation n as seen from above: if, b), a), E is induced e.m.f, 4, 4, Eq, 5V0, 7V, a), , anticlockwise, c), d) o, 2ma, 4, 4, Eq, PASSGAE -3, , anticlockwise, b), ma, Switches S1 , S2 remain open and switch S3 remains, 2Eq, c), clock wise, closed for long time such that capacitor becomes, ma, fully charged and current in inductor coil becomes, Eq, maximum, Now switches S1 , S2 are simultaneously, d), clock wise, ma, 13. The power developed by the forces acting on, closed and S3 is simultaneously opened at t 0 ., the ring, as a function of time:, Assume that battery and inductor coil are ideal, 2 2, 2 2, 2 2, 2 2, E q, E q, 2E q, 2E q, t b), t c), t d), t, a), S1, 2m, m, m, m, PASSAGE -3, -1F, 100, Two capacitors of capacitance C and 3C are, S2, charged to potential difference V0 and 2V0, respectively, and connected to an inductor of, 10mH, inductance L as shown in figure. Initially the current, in the inductor is zero. Now, the switch S is closed., 100, V0, 2V0, + –, + –, S3, , C, , 3C, S, , 10V, , L, , 62, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, Required induced emf,, , 17. Charge (in C ) on capacitor as a function of, time is :, , e, , 4 3 , a) q 10 2 sin 10 t , 4 , , , Bl 2, 2, , 50 106 4 9, 2.83 mV, or e , 2, , 4 3 , b) q 10sin 10 t , 4 , , , 2., 3., , 2 3 , c) q 10sin 10 t , 4 , , , 4., , 2 3 , d) q 10 2 sin 10 t , 4 , , , by, I , , 18. Maximum current in the inductor coil at any, times t > 0, 1, , a), , 1, A, 10, , b), , c), , 1, A, 5, , d) 1A, , 5 2, , As the rod is of plastic so no free electrons are, present and hence no emf will develop in the rod, As the ring is non-conducting, no motion of charge, carriers occur and hence there is no induced emf, Let at any time t, velocity of rod be v, then emf, developed across its ends is e B0vl . Due to this, induced emf, a current will establish in circuit given, , B0, , A, , F, , V, I, , 19. Find time t 0 , when energy store in inductor, would be minimum for the first time?, , , 4, a) 10 s, 4, c), , 3, 104 s, 4, , , 4, b) 10 s, 2, , For rod,, , d) 10 4 s, , v, , dv, B02l 2, m IB0l , v, dt, R, , COMPREHENSION TYPE QUESTIONS, 9. (B) 10. (A), (A) 15. (C), 19.(C), , 11. (A) 12. (B) 13. (B) 14., 16 (C) 17. (A) 18.(A), , LEVEL- V - HINTS, 1., , The situation is as shown in the figure below., B, l, , , NARAYANAGROUP, , t, , dv, B02l 2, B 2l 2t, , , dt, 0, v, mR, v v0e mR, v0, 0, , LEVEL - V- KEY, SINGLE ANSWER QUESTIONS, 1. (B) 2. (C) 3. (C) 4. (B) 5. (C) 6. (B), MULTIPLE ANSWER QUESTIONS, 7. (A), (C) 8. (B), (C), , B0vl, ., R, , 5., 6., , At P, E B0t , so force, F qB0t in anticlockwise (direction form Lenz’s law), Due to changing B, flux linked with cir5cuit changes, and hence a current induces in the circuit., The direction of this induced current is in anticlockwise direction., In C3 , R2 is shorted, so no current flows through, R2 and hence energy dissipated is zero., Under the assumption that induced emf is same for, both C1 and C2 , induced current in C1 would be, less than that in C2 , as a result, thermal energy, dissipated through R2 in C1 would be less than, that in C2 ., So, C 3 C1 C 2 ., 63
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, , P 7 - 9:, Let us say that in time t the conductor falls down by, E 2q 2, y and acquires a velocity v, then at this instant,, P, t, 12., m, induced emf is, Loss in energy of capacitor = Energy store in inductor, e B0vl [with Q at higher potential], , C, , A, , 1, 1, 1, 1, 2, VC02 3C 2V0 4CV 2 LI, 2, 2, 2, 2, 3, 3V0, I V0, 2, L, mg, IB0l, P13-14:, B, D, dI, Charge on the capacitor at this instant is,, 0, When current is maximum, dt, q Ce B0Cl v, e.m.f across L=0 so potential difference across, the current in the circuit at this instant is,, the capacitor will be same., dq, dv, From the law of conservation of charge on plate, I, B0Cl, B0Cl a, 2 and 3., dt, dt, where a is the acceleration of the conductor at this, 5, instant., 3CV CV 6CV0 CV0 V V0, 4, Writing Newton’s 2nd law equation for conductor,, 10C, , Y, , mg IB0l ma, mg, a, ,, m CB02l 2, , +, , 2, , mgt, mgt, y, v, and, 2 m CB02l 2, m CB02l 2, , , , 15., , , , (1/10)A, , After solving above equations, we could have q as, a function of y as I B0Cl a, As a is constant, I is also constant., r dB dB, a, B0 E B0, 10. E , 2 dt dt, 2, , 1 Q02 1 Q 2 1 2, , Li Q0 10 2C, 2 C 2 C 2, 1, Q, , 10 4 and Q 0, [Capacitor is, LC, 2, discharging], 3 , , q 10 2 sin 104 t , , 4 , , , , , , , , , , , , , , , , , , , , , , , , , d dF .a dqE .a d Ea dq Eaq, , I, , , , , , Eaq, Eq, , , 2, I ma ma, , –, , 16., , 1 2, 1 Q2 1 2, 2, 1, Limax , Li imax , A, A, 2, 2 C 2, 10, 5 2, , 17. When energy store in inductor is minimum at that, time energy store in capacitor is maximum qmax, first time when, 3 3, 3, 104 t , , t , 104, 4, 2, 4, , Eq, anticlockwise, ma, , 11. Power P .w Eaq t Eaq Eq t, ma, 64, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, 3., , LEVEL-VI, SINGLE ANSWER QUESTIONS, EMI, 1., , A rectangular loop of wire with dimensions, shown in figure is coplanar with a long wire, carrying current I. The distance between the, wire and the left side of the loop is r. The loop, is pulled to the right as indicated. What are, the directions of the induced current in the loop 4., and the magnetic forces on the left and the right, sides of the loop when the loop is pulled?, r, , A coil having N turns is wound tightly in the, form of a spiral with inner and outer radii a, and b respectively. When a current I passes, through the coil, the magnetic field at the, centre is., µ 0 NI, b, µ 0 NI, b, ln, (C), 2b a a, , 2µ 0 NI, a, 0 IN, b, (D) b a ln a ., , , , (B), , (A), , A rectangular loop with a sliding conductor of, length l is located in a uniform magnetic field, perpendicular to the plane of the loop. The, magnetic induction is B. The conductor has a, resistance R. The sides AB and CD have, resistances R1 and R2 , respectively. Find the, current through the conductor during its motion, to the right with a constant velocity v ., , b, , I, , X, , X, , X, , X, , X, , X, , A, X, , a, , Induced current Force on left side Force on, right side, A. Counterclockwise To the left, To the left, B. Counterclockwise To the right, To the left, C. Clockwise, To the right, To the left, D. Clockwise, To the left, To the right, 2. There is a uniform magnetic field B in a circular, region of radius R as shown in figure whose, magnitude changes at the rate of, , dB, . The 5., dt, , e.m.f. induced across the ends of a circular, concentric conducting arc of radius R1 having, an angle as shown OAO ' is, A., ×× ××, × × × ×R × ×, × × × × × ×× × × ×, × × × × ×× × × × ×, × × × × × ×× × × × × ×, × × × × A × ×× ×, × × × × × ×× × × × × ×, × ×× × × ×× ×, ×× × × × ××××, ×× × × ××××, × ×, , X, , X, R1, , R1, , 2 dB, R, 2, dt, , X, , X, , X, , X, , X, , X, , X, , X, , X, , R2, , X, X, , D, , C, , X, , X, , X, , Blv R1 R2 , A., R1 R1 R2 , , X, , X, , X, , 2, , B., , Bl v, R1 R1R2, , Blv R1 R2 , Bl 2v, C., D., R1R2 R R1 R2 , R1R2 R R1 R2 , PQ is an infinite current-carrying conductor., AB and CD are smooth conducting rods on, which a conductor EF moves with constant, velocity V as shown in figure. The force needed, to maintain constant speed of EF is, P, , A, , R, , C, F, , I, , V, a, , 2 dB, R, C., 2, dt, , b, Q, , D. None of these, , X, , X, , E, , B., , X, , V, , X, , 2 dB, R1, 2, dt, , O', , NARAYANAGROUP, , B, , B, , D, , 1 0 IV b , ln , A., VR 2, a , , 2, , 2, , 0 IV b 1, ln , B. , a VR, 2, 2, 0 IV b V, ln , C. , a R, 2, V 0 IV b , ln , D. , R 2, a , , 2, , 65
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, 6., , AB is a resistanceless conducting rod which 8., forms a diameter of a conducting ring of radius, r rotating in a uniform magnetic field B as, shown in figure. The resistors R1 and R2 do, not rotate. Then the current through the, resistor R1 is, X X XX X XX X XX X X, R1, X X XX X XX X XX X X, A, , B, , X X XX X XX X XX X X, R2, X X XX X XX X XX X X, , B r 2, A., 2 R1, , B r 2, B., 2 R2, , A copper rod is bent into a semi-circle of radius, a and at ends straight parts are bent along, diameter of the semi-circle and are passed, through fixed, smooth and conducting ring O, and O ' as shown in figure. A capacitor having, capacitance C is connected to the rings. The, system is located in a uniform magnetic field, of induction B such that axis of rotation OO ', is perpendicular to the field direction. At initial, moment of time (t=0), plane of semi-circle was, normal to the field direction and the semi-circle, is set in rotation with constant angular velocity, . Neglect the resistance and inductance of, the circuit. The current flowing through the, circuit as function of time is, O, , O, , B r 2, B r 2, R1 R2 D., C., 2 R2 R2 , 2 R1R2, 7., , X, , A conducting rod PQ of mass m and length l, is placed on two long parallel (smooth and, conducting) rails connected to a capacitor as, shown. The rod PQ is connected to a non, conducting spring of spring constant k, which, is initially in relaxed state. The entire, arrangement is placed in a magnetic field, perpendicular to the plane of figure., Neglect the resistance of the rails and rod., Now, the rod is imparted a velocity v0 towards, right, then acceleration of the rod as a function, of its displacement x is given by, ×, , ×, , ×, , ×, , P, , ×, , ×, , ×, , 9., , ×, , 66, , ×, , ×, , × Q×, , ×, , ×, , ×, , A., , kx, m, , B., , kx, m B 2 l 2C, , C., , kx, m B 2l 2C, , D., , kx, B 2l 2 c, , B, C, , A., , 1 2 2, a CB cos t, 4, , B., , 1 2 2, a CB cos t, 2, , C., , 1 2 2, a CB sin t, 4, , D., , 1 2 2, a CB sin t, 2, , ×, , v0, , 1, , A conducting wire of length l and mass m is, placed on two inclined rails as shown in figure., A current I is flowing in the wire in the direction, shown. When no magnetic field is present in, the region, the wire is just on the verge of, sliding. When a vertically upward magnetic, field is switched on, the wire starts moving up, the incline. The distance travelled by the wire, as a function of time t will be, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, to thickness dx of the wire, , 1 IBl, , 2g t 2, A. , 2 m, , , Rails, , B., , l, I, , , , 1 IBl, 1, , , 2 g sin t 2, , 2 m cos , , , C., D., , 1. (D), 7. (C), , B, , 1 IBl, , 2 g sin t 2, , 2 m, , , a, , b, , , , 1 IBl cos 2, , 2 g sin t 2, , 2 m cos , , , a, , SINGLE ANSWER, 2. (B) 3. (C) 4. (C) 5. (A) 6. (A), 8. (B) 9. (D), , Induced emf Blv . R is internal resistance of seat, of emf, i.e., of rod, , I , , r, A, , NI, b, dx 0 NI, b, log e xa ; B 20 b a loge a, x 2 b a , , , , Total resistance of circuit = R , , LEVEL- VI - HINTS, , 1., , 0 NI dx 0 NI, , 2 ba x, 2 b a , , Hence correct option is (c), 4., , LEVEL- VI - KEY, , NI, , On integrating we get, b, , , , , , NI dx, , 0, 0, dB 2 b a x 2 b a , , R1R2, R1 R2, , Blv R1 R2 , Blv, , R1R2, R1R2 R R1 R2 , R, R1 R2 , , B, b, , 5., , b, , 0 I, BVdx, 2, , x, a, , Induced emf BVdx , a, , V, , I, , 0 IV b , Induced e.m.f. = 2 In a , , D, , C, , As the flux decreases, to maintain flux, current in, the loop is clockwise. Force on DA due to the long, wire is towards left while on BC is towards right., 2., , Required emf, , I, 3., , dq 1 2 2, a CB cos t , dt 2, , (c) Let us consider a thickness dx of wire. Let it be, at a distance x from the centre O., 6., N, Number of turns per unit length , ba, number of turns in thickness dx , , N, dx, ba, , Small amount of magnetic field produced at O due, NARAYANAGROUP, , E2, Power dissipated , R, E2, Also, power = FV F , VR, 1 0 IV b , F, In , VR 2, a , , 2, , The equivalent diagram is, , R1, , R2, e, 67
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JEE-ADV PHYSICS- VOL- IV, , ELECTRO MAGNETISM & EMI, The induced emf across the centre and any point, on the circumference is, , 1, B r, e Bl 2 , 2, 2, , 2 BA, , 2, , Total flux linked with the circuit is, , B r, Current through R1 2 R, 1, 7., , Flux linked withthis part is, , 1, a 2 B cos t BA, 2, , 2, , Induced emf in the circuit,, , Let the velocity of rod be v when it has been, displaced by x . Due to motion of rod an emf will, be induced in rod given by e Bvl , due to this, induced emf, charging of the capacitor takes place, as a current, flows in the circuit [for very small time], a result of this current, the rod experiences a, magnetic force given by IBl ., ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, , ×, , ×, , ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, , ×, , ×, , ×, , P, , Q, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, v0, ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , From Newton’s second law,, , e, , d 1, a 2 B sin t , dt 2, , Since resistance of the circuit is negligible, therefore,, potential difference across the capacitor is equal to, induced emf in the circuit., , Charge on the capacitor at time t is q=Ce, 1, a 2CB sin t , 2, But current I , 9., , The front view of the arrangement is shown in figure, IBl sin, , dv , , IBl kx ma a , dt , , , I, , 8., , mg sin, , d, d, dv, Q C Bvl CBl , dt, dt, dt, , kx, a, 2 x, 2 2, mB l C, , dq 1 2 2, a CB cos t , dt 2, , IBl cos, , mg cos, , , mg, , From initial condition, mg sin mg cos, , tan , , Which also shows that rod is performing SHM, , ma IBl cos mg sin N, , When the copper rod is rotated, flux linked with, the circuit varies with time, , N mg cos IBl sin , , Therefore, anemf is induced in the circuit., , IBl, IBl sin 2 , a, cos q 2 g sin , m, m cos , , At time t, plane of semi-circle makes angle t with, the plane of rectangular part of the circuit. Hence,, component of the magnetic induction normal to, plane of semi-circle is equal to B cos t ., Flux linked with semicircular part is, , 1, 1 a 2 B cos t, 2, , , , IBl cos 2, 2 g sin , m cos , , Now,, , s, , 1 2 1 IBl cos 2, , at , 2 g sin t 2, 2, 2 m cos , , , Let area of rectangular part of the circuit be A., , 68, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , ALTERNATING CURRENT, SYNOPSIS, , , I or E, , Introduction : When a resistor is connected, across the terminals of a battery, a current is, established in the circuit. The current has a, unique direction, it goes from the positive, terminal to the negative terminal via the external, resistor. The magnitude of the current also, remains almost constant. This is called direct, current (dc)., If the direction of the current in a resistor or in, any other element changes alternately, the, current is called an alternating current (ac). In, this chapter, we shall study the alternating, current that varies sinusoidally with time., , ALTERNATING CURRENT(A.C.), •, , •, , •, , Electric current, which keeps on changing in, magnitude and direction periodically is defined, as alternating current., It obeys Ohm’s law and Joule’s heating law., It is produced using the principle of, electromagnetic induction., Graphical representations for alternating, quantities can be represented in the form of the, following graphs., , t, , Square form of ac, , ADVANTAGES OF ALTERNATING, CURRENT OVER DIRECT, CURRENT, •, •, •, •, , DISADVANTAGES OF ALTERNATING, CURRENT OVER DIRECT CURRENT, •, •, , ALTERNATING VOLTAGE (A.V), •, , •, , The voltage, which changes in magnitude and, direction with respect to time is defined as, alternating voltage., The alternating voltage in general use is, sinusoidal voltage. It is produced by rotating a, coil in a uniform magnetic field with uniform, angular velocity., , •, , , , •, , •, t, , , ac is transmitted more by the surface of the, , conductor. This is called skin effect. Due to, this reason that several strands of thin insulated, wire, instead of a single thick wire, need be, used., For electrorefining,electro - typing, electroplating, only dc can be used but not ac., , The value of current or voltage in an ac circuit, at any instant of time is called its instantaneous, value., Instantaneous current,, I I 0 sin wt (or) I I 0 sin(w t f), Instantaneous voltage,, E E0 sin w t (or) E E 0 sin(wt f), Where (w t f) is called phase, I or E, , sinusoidal form of ac, , ac is more dangerous than dc., , INSTANTANEOUS VALUE OF, CURRENT OR VOLTAGE (I or E), , •, , I or E, , The cost of generation of ac is less than that of dc., ac can be conveniently converted into dc with the, help of rectifiers., By supplying ac at high voltages, we can, minimise transmission losses or line losses., ac is available in a wide range of voltages., These volatages can be easily stepped up or, stepped down with the help of transformers., , I or E, , +, , +, , , E 0 or I 0, Positive half cycle, +, , +, , , , , O, , t or , , T/4, , _, , T/2, , Triangular form of ac, , NARAYANA GROUP, , T, , Negative, half cycle, , 69
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , AMPLITUDE OF A.C. ( PEAK, VALUE) (I0) or ( I m ):, , c), , It is the maximum value of A.C. The value of A.C., becomes maximum twice in one cycle., Note: Average value of a function F t over a period, of T is given by, T, , , , T, , dt, , FREQUENCY OF A.C. (f), •, , It is the number of cycles completed by A.C. in, one second., , 1, F t dt, T 0, , •, , It is the time taken by A.C. to complete one cycle., f = 1/ T, , MEAN SQUARE VALUE OF A.C. < I2 >, , 0, , 1 cos 2 t 1, :, 2, 2, sin 2t 0 ; cos 2t 0, , •, , 2, Eg:- sin t , , •, , The value of current at any instant ‘t’ is given by, , •, , I I0 sin wt ., The average value of a sinusoidal wave over, one complete cycle is given by, , < I2 > =, , It is the square root of the average of squares, of all the instantaneous values of current over, one complete cycle., T, 2, I rms, , , , I avg , , , , dt, , , , dt, , 0, , 0, , For half cycle:, T, , I , , T, , 2, , I, , 0, , 0, , T, , , , 0, , sin t dt, T, , 2, , dt, , , 2, , dt, , 0, , 2I 0, 0.636 I 0, , , Similarly, , Note: a), , Eavg , , (+), I or E, A, , B, t, , 2 Eo, 0.637 E0 63.7% E0, , Between A and B,, Iavg = 0, Eavg = 0, , (-), , (+) A, , b), , I or E, (-), , 70, , t, , , , 0, T, , 2, , wt.dt, , 0, , T, , I 02, T, , T, , , 0, , T, 1 cos 2wt , I 2 sin 2wt , , dt 0 t , , 2, 2T , 2w 0, , , , •, , I0, I02, 0.707 I 0, ; I rms , 2, 2, It is equal to that direct current which produces, same heating in a resistance as is produced by, the A.C. in same resistance during same time., MEAN SQUARE VALUE OF A.C. < I2 >, , •, , < I2 > =, , 0, , I avg 63.7% of I 0, , •, , 2, , , , 2, , Idt, , I0 .sin, , .dt, , 0, , 0, , T, , 2, , dt, , I .dt I 0 sin wt.dt, 0, , T, , I, , T, , 0, T, , I02, 2, , R.M.S. VALUE (Irms) or EFFECTIVE, VALUE (I) or VIRTUAL VALUE OF A.C., , AVERAGE VALUE OF A.C. < I >, , T, , B, t, , C, , TIME PERIOD OF A.C. (T), , F t dt, 0, , Between A and C, or, C and B, Iavg 0, Eavg 0, , (-), , T, , F t Favg , , (+), I or E, A, , Between A and B,, Iavg = 0, Eavg = 0, , I02, 2, , FORM FACTOR, rms value, Form factor = average value over half cycle, Form factor , , I rms, E, rms, I avg, Eavg, , We know that I rms , , I0, 2I 0, and Iave , 2, p, , B, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, I0, , Form factor , , 2, , , , ALTERNATING CURRENT, , p, p, , 1.11, 2I0 2 2, , Note :, •, ac ammeter and voltmeter read the r.m.s value, i.e., effective value of alternating current and, voltage respectively., •, ac can be measured by using hot wire ammeters, or hot wire voltmeters because the heat generated, is independent of the direction of current., •, , ac produces the same heating effects as that, of dc of magnitude i = irms, , •, , ac is more dangerous than dc of same voltage., 100V ac means Erms 100V , E0 100 2 V, , •, , 100V dc is equivalent to Erms, ac can be produced by the principle of, electromagnetic induction., , POWER IN ac CIRCUITS :, In dc circuits power is given by P = VI. But in, ac circuits, since there is some phase angle, between voltage and current, therefore power is, defined as the product of voltage and that, component of the current which is in phase with, the voltage., Thus P = EI cosf , where E and I are r.m.s., values of voltage and current., Power factor:The quantity cosf is called, power factor., a) In stantaneous power : Suppose in a circuit, E = E0 sin wt and I I 0 sin wt f then, Pinstantaneous = EI = E0I0 sin wt sin wt f, b) Average power (True power) : The average, of instantaneous power in an ac circuit over a, full cycle is called average power. Its unit is, watt i.e., T, , Pavg , , W, , t, , T, 2, Average power over complete cycle,, W E0 I 0 cos X, , Pavg , , , , W E0 I 0, , cos , T, 2, , E0 I 0, 2, , 2, , cos Erms I rms cos , , c) Apparent or virtual power : The product, of apparent voltage and apparent current in an, electric circuit is called apparent power. This, is always positive., EI, Papp E rms I rms 0 0, 2, W.E-1: You have two copper cables of equal length, for carrying current. One of them has a single, wire of area of across section A, the other has, ten wires each of cross section area A/10., Judge their suitability for transporting ac and, dc., Sol: For transporting d.c.., both the wires are equally, suitable, but for transporting a.c., we prefer wire, of multiple strands.ac is transmitted more by the, surface of the conductor. This is called skin effect, .Due to this reason that several strands of thin, insulated wire, instead of a single thick wire,, need be used., W.E-2: If the voltage in an ac circuit is represented, by the equation., V 220 2 sin 314t volt calculate (a), , peak and rms value of the voltage, (b) average, voltage, (c) frequency of ac., Sol: (a) As in case of ac,, , V V0 sin t ; The peak value, V0 220 2 311V and as in case of ac., Vrms , , V0, 2, , ;Vrms 220V ; (b) In case of ac, , T, , P.dt P.dt, 0, , , , T, , dt, , 0, , T, , T, , Vavg , , ; W P.dt, 0, , (c) As 2 f , 2 f 314, , 0, , T, , W E0 I 0 cos sin 2 tdt , 0, , NARAYANA GROUP, , 2, 2, V0 311 198.17V, , , , T, , E0 I 0, sin sin 2tdt, 2, 0, , i.e, f , , 314, 50 Hz, 2, 71
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , W.E-3: A current is made of two components a dc, , •, , component i 1 = 3A and an ac component, i 2 4 2 sin t ., , Find the reading of hot wire, , •, , ammeter?, , Sol: i i1 +i2 3 4, , 2 sin w t, , T, , 2, irms, , , | alternating emf |, , Z = | alternating current |, , T, , , , 2, , i dt, , , 0, T, , 3 4, , , , 2, , 2 sin wt dt, , , , , , 1, i 9 24 2 sin wt 32sin2 wt dt, T 0, irms = 5A, W.E-4: If a direct current of value a ampere is, superimposed on an alternating current, I b sin t flowing through a wire, what is, the effective value of the resulting current in, the circuit?, , dc, , I, , I, a, , ac, , b, , •, , •, , •, •, , •, , , , t, , t, , •, , I I dc I ac a b sin t, •, 1, 2, , T 2 , 1, I dt , T, 2, , , 1 a b sin t 2 dt, 0T, , , , , , T 0, , dt , 0, , , I eff, , =, , RMS value of alternating voltage, RMS value of AC, , ADMITTANCE(Y): Reciprocal of impedance, of a circuit is called admittance of the circuit., 1, admittance (Y) =, Z, S.I. Unit:ohm-1 i.e. mho or siemen., PHASE: The physical quantity which, represents both the instantaneous value and, direction of A.C. at any instant is called its phase., It is dimensionless quantity and its unit is Radian, Phase Difference: The diffrerence between, the phases of current and voltage is called Phase, difference., If alternat ing emf and current are, , E E0 sin t 1 and i i0 sin t 2 , , =?, , Sol: As current at any instant in the circuit will be,, , So,, , peak value of alternating voltage, peak value of AC, , T, , 0, , T, , =, , 0, , dt, 2, rms, , RESISTANCE (R) It is the opposition offered, by a conductor to the flow of direct current., IMPEDANCE (Z) It is the opposition offered, by a conductor to the flow of alternating current., , then phase difference is 1 2, The quantity varies sinusoidally with time and, can be represented as projection of a rotating, vector, is called as phasor., A diagram, representing alternating emf and, current (of same frequency) as rotating, vectors (Phasors) with phase angle between, them is called as phasor diagram., , 1, , 1 T 2, 2, 2, 2, I, , i.e, eff a 2b sin t b sin t dt , T 0, , But as, 1, T, , , , T, , 0, , sin t dt 0 and, 1/ 2, , So, I eff, 72, , 1 , , a 2 b2 , 2 , , , 1, T, , , , T, , 0, , sin 2 t dt , , 1, 2, , •, , In the above figure , OA and OB represent two, rotating vectors having magnitudes E0 and I0 in, anti clock wise direction with same angular, velocity ‘ w ’., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , •, •, •, , OM and ON are the projections of OA and OB •, on Y-axis respectively., •, OM = E and ON = I, represent the instantaneous, values of alternating emf and current., BOA f represents the phase angle by which, current I0 leads the alternating emf E0., The phasor diagram, in a simple representation, is, , instantaneous emf is E=E 0 sin t, instantaneous current I I 0 sin t , , , , , where ; Current leads emf by, 2, 2, , , Y, , Y, , E, , E and I, , •, , ALTERNATING CURRENT, , I, t, , O, , I, I, , E, , , , , t, , X, , O, , , E=E 0 sin t ; I I 0 sin t where 2, Current lags emf by /2, or, emf leads current by /2, , X, , O, , •, , WE-5: Use a phasor diagram to represent the sine, waves in the following Figure., , A.C THROUGH A RESISTOR, A pure resistor of resistance R is connected, across an alternating source of emf, , •, , R, , Sol:The phasor diagram representing the sine waves, is shown in figure.The length of each phasor, represents the peak value of the sine wave., , ~, E=E0 sin t, , The instantaneous value of alternating emf is, E = E0 sin t, The instantaneous value of alternating current, , •, •, , is I , , E0, Peak value of current, I 0 , R, , •, Note:If e.m.f (or voltage) in A.C. is E = E 0 sin t, , E and I, , and the current I = I0sin ωt+φ Where phase, difference φ is Positive if current leads,Negative, if current lags and zero if current is inphase with, the emf (or voltage)., E, , E E0, , sint I 0 sin t, R R, , Phasor diagrams:, , •, , E.m.f, , I,E, , Y, , Current, , O, , I, , , , 2, 3, , 2, , t, , 2, , O, t, , I, , E, , X, , O, , NARAYANA GROUP, , •, , emf and current will be in phase 0 , , •, , emf and current have same frequency, 73
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, •, •, •, , Peak emf is more than peak current, The value of impedance (Z) is equal to R, and reactance (X) is zero, Apart from instantaneous value, current in the, circuit is independent of frequency and, decreases with increase in R (similar to that in, dc circuits)., slope, , R=constt., , f =constt, , I0, 2, Irms, , I, , Irms, , or f, , R, , •, , Instantaneous power Pi Eo I o sin 2 t, Average power over time ' T ' sec =, , Pavg Erms I rms cos Erms . I rms, , •, , A pure inductor of inductance L is connected, across an alternating source of emf E, , •, , The instantaneous value of alternating emf, is E = E0 sin t ...........(1), dI, The induced emf across the inductor = L., dt, which opposes the growth of current in the, circuit. As there is no potential drop across the, circuit, so, dI , , dI, E L. 0 or L. E, , dt , dt, dI E0, , sin t ; On integrating, dt, L, E0, , , cos t I 0 sin t , L, 2, , , 2, , 2, t, , , , , , 2, , INDUCTIVE REACTANCE (XL), •, •, , •, •, , •, , The opposition offered by an inductor to the flow, of ac is called an inductive reactance., The quantity L is analogous to resistance and, is called reactance of Inductor represented by, XL ., It allows D.C. but offers finite impedance to the, flow of A.C., Its value depends on L and f., Inductance not only causes the current to lag, behind emf but it also limits the magnitude of, current in the circuit., I0 , , E0, E, L 0 X L ,, L, I0, , X L L 2 fL X L f ; X L f curve, X L L curve, Y, , .......... 2 , , XL, , The instantaneous value of alternating current, is, , , I I0 sin t , 2, , , 3, , 2, , LL, , ~, E=E0 sin t, , 74, , VL, , E.m.f, , O, , •, , •, , Phasor diagram, , , , L, , I , , , 2, , emf by a phase angle of, , 2, Erms, , R, , A.C THROUGH AN INDUCTOR, , •, , , 2, The alternating current lags behind the, and current is, , Current, , power factor cos cos 0 1, , •, , E 0 E0, L X L, , From equation 1 & 2, Phase difference between alternating voltage, , I, E, , •, , •, , •, , 1, R, , POWER, , •, , Where Peak value of current, I0 , , Y, , slope = tan = 2L, XL, , , , X, f, , •, , slope = tan = 2f, , , X, L, , For dc, f 0 X L 0, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , •, , ALTERNATING CURRENT, q=CE=CE0 sin t, , For ac, high frequencies, X L , dc can flow easily through inductor.., Inductive reactance in terms of RMS value, is, , E, X L L rms, Irms, , I, , •, , Power supplied to inductor, •, , , , I I 0 sin t ...............(2), 2, , , The instantaneous power supplied to the inductor, is, , , , PL iv i0 sin t v0 sin t , 2, , , where peak value of current,, , i0 v0, •, sin 2 t , 2, So, the average power over a complete cycle is, , , , I,E, , E.m.f, , 3, , O , , /2, , 2, , , , 2, , When an alternating emf is applied to a •, capacitor, then alternating current is constituted, in the circuit. Due to this, charge on the plates, and electric field between the plates of capacitor •, vary sinusoidally with time., At any instant the potential difference between, the plates of a capacitor is equal to applied, emf at that time., , 2, , CAPACITIVE REACTANCE (XC), The resistance offered by a capacitor to the flow, of ac is called capacitive reactence., 1, is analogous to resistance and, C, is called reactance of capacitor represented by, , The quantity, , XC, I0 , , C, , •, , E0, E, E, 1, 1, XC , , 0 rms, C 2 fC I 0, I rms, 1 , , , C , , It is the part of impedance in which A.C. leads, the A.V. by a phase angle of, , ~, E=E0 sin t, , •, •, , A capacitor of capacity C is connected across, an alternating source of emf, The instantaneous value of alternating emf is, E = E0 sin t ...................(1), Let q be the charge on the capacitor at any instant., Accoding to kirchhoff’s loop rule, E, , q, 0 q CE0 sin t, C, , NARAYANA GROUP, , t, , VC, , A.C THROUGH A CAPACITOR, , •, , IC, , Current, , cycle is zero., T hus, t he average power supplied to an, inductor over one complete cycle is zero., , •, , , ., 2, , PHASOR DIAGRAM, , Since the average of sin 2t over a complete, , •, , E0, 1 , , , C , , current leads the emf by an angle, , 90, , i0 v0, sin 2 t 0, 2, , I0 , , From equation 1 & 2, , i0v0 cos t sin t , Pavg Erms . I rms cos 0, , dq, , , C E0 cos t I 0 sin t ............... 2 , dt, 2, , The instantaneous value of alternating current, is, , , ., 2, , 1, 1, , ., C 2fC, , •, , Its value is Xc , , •, •, •, , Its value depends on C and f., It bypasses A.C. but blocks D.C., It is produced due to pure capacitor or induced, charge., , •, , X C f curve, X C C curve, 75
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, = const, , C = const, XC, , XC, , WE-8: An inductor of 1 henry is connected across, a 220 v, 50 Hz supply. The peak value of the, current is approximately., Sol: Peak value of current, , or f, C, 2 Erms, 2 Erms, E, 2 220 , , Note: Resistance, Impedance and Reactance have the i0 0 , , 0.99 A, 2 fL, XL, L, 2 50 1, same units and Dimensional Formulae., i.e. SI unit is ohm; Dimensional Formula is WE-9: A capacitor of 2 F is connected in a radio, circuit. The source frequency is 1000 Hz. If, ML2T 3 A2 , the current through the capacitor branch is, Power supplied to capacitor:, 2 mA then the voltage across the capacitor is, •, The instantaneous power supplied to the, capacitor is, 1, I, VC IX C I , , Sol:, Pc iv i0 cos t v0 sin t , C 2 fC, iv, 2 103, i0 v0 cos t sin t ; 0 0 sin 2 t , , 0.16V, 2, 2 103 2 106, So, the average power over a complete cycle is, A.C THROUGH LR SERIES CIRCUIT, zero, •, LR circuit consists of a resistor of resistance, since sin 2t 0 over a complete cycle., R and an inductor of inductance L in series, Pavg Vrms . I rms cos Vrms . I rms cos 90 0, with a source of alternating emf, no power is consumed in a purely, •, The instantaneous value of alternating emf is, capacitive circuit., E E 0 sin t, WE-6: An alternating voltage, , E 200 2 sin 100t volt is connected to a, , L, , R, , 1 F capacitor through an ac ammeter. What, will be the reading of the ammeter?, , VL, , VR, , Sol: Comparing E 200 2 sin 100t with, , E E0 sin t ; E0 200 2V and 100 rad / s , XC , , 1, 1, , 104 , 6, C 100 10, , I rms , , Erms, E0, 200 2, , , 20 mA, Z, 2XC, 2 104, , WE-7: Find the maximum value of current when, a coil of inductance 2H is connected to 150V,, 50 cycles / sec supply., Sol: Here L 2H, Erms 150 V , f 50 Hz, X L L L 2f = 2 2 3.14 50 = 628 ohm, RMS value of current through the inductor ,, Irms , , Erms 150, , 0.24A, XL 628, , Maximum value (or peak value) of current is, given by I r m s , or, 76, , I0, 2, , I0 2 Irms 1.414 0.24 0.339A, , E, ~, , •, •, , The potential difference across the inductor is, given by, VL IXL, .....(1), The potential difference across the resistor,, .....(2), VR IR, , current I lags the Voltage VL by an angle of ,, 2, Therefore, the resultant of VL and VR is, 2, , OC OA 2 OB2 or E VR VL, , 2, , y, C, , B, , XL, (or), VL, , E, , , VR (or) R, , A, , x, I, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , Using equations ( 1 ) and (2), we get, 2, , 2, , E I2R 2 I2 X L I R 2 X L, where X L L is the inductive reactance., •, E, or I , .....(3), 2, R2 XL, E, 2, 2, 2, 2 2, I, Z LR ; Z LR R X L R L , •, The effective opposition offered by LR circuit, to ac is called the impedance of LR circuit., •, Let be the angle made by the resultant of VL, and VR with the X-axis, then from figure, we get, , AC OB VL IX L, tan , , , , OA OA VR, IR, X L L, or tan , , R, R, Note: In series LR circuit, emf leads the current or, the current is said to lag behind the emf by an, angle , Current in L-R series circuit is given by, , I, , The current leads emf by an angle, , The potential difference across the resistor,, VR = IR, .....(ii), The emf and current are in phase when ac flows, through resistor., , Phasor diagram., I, VR (or) R, , A, , VC, (or), XC, , E, , B, , •, , ALTERNATING VOLTAGE, Let an alternating source of emf E =E0 sin w t is, connected to a series combination of a pure, capacitor of capacitance (C) and a resistor of, resistance (R) as shown in figure (a), , C, , In figure VC is represented by OB along negative, Y - axis and the current I is represented along, X - axis., VR is represented by OA along X - axis., The resultant potential difference of VC and VR is, represented by OC., Also, the emf and current are in phase when ac, flows through the resistor. So, VR is represented, by OA along X-axis., Therefore, the resultant potential difference of, VC and VR is represented by OC and is given, by, OC OA 2 OB2, , VR, , ~, E, , •, , Let I be the r.m.s value of current flowing through, the circuit. The potential difference across the, , NARAYANA GROUP, , 2, , or E VR VC, , 2, , Using equations (i) and (ii), we get, 2, , R, , VC, , X, , , , -Y, , E, E, 0 sin(w t f ), Z LR Z LR, , C, , , when ac, 2, , flows through capacitor., , O, , (or) I I 0 sin( w t f ), Note:, •, •, Z LR R 2 L2 2 R 2 L2 4 2 f 2 ., Thus Z LR increases with the frequency of ac, so, •, ZLR is low for lower freqeuncy of ac and high, for higher frequency of ac, •, •, The phase angle between voltage and current, increases with the increase in the frequency of, ac, •, C-R SERIES CIRCUIT WITH, •, , capacitor,, VC IX C .....(i), , 2, , E I2R 2 I2XC I R 2 XC, E, E, or I , , 2, 2, R XC, Z CR, From the above equations of I and E, we have, 2, 1 , 2, 2, 2 , , R, , , Z CR R XC, , Cw , Where Z CR is the effective opposition offered, 77
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, by the CR circuit to ac, which is the impedance of, CR circuit., Let be the angle made by E with X-axis, AC VC IX C, tan , , , OA VR, IR, XC, I, , or tan , R CR, In series CR circuit, emf lags behind the current, or in other words, the current is said to lead the, emf by an angle given by the above equation., Current in C-R series circuit is given by, E, E, 0 sin( wt f ), I, Z CR Z CR, (or) I I0 .sin(wt f ), Note:, •, , •, , •, , •, , The current I lags VL by an angle / 2 ., , •, , The P.D across capacitance is VC I . X C ., , •, , The current I leads VC by an angle / 2 ., , •, , The voltage VL and VC are represented by OB, and OC respecitvely., Y, B, VL, , VC, -Y, , The resultant P.D of VL and VC is, , V VL VC I X L X C , , The resultant potential difference of VC and VR is, represented by OC For very high frequency (f), of ac. Z R and for very low frequency of ac, •, Z, Phase angle between voltage and current is, given by, As, , f, , 1, 1, , C R 2 fC R, , increases, phase angle decreases., , •, , L - C SERIES CIRCUIT WITH, ALTERNATING VOLTAGE, •, , Let an alternating source of emf E E0 sin t is •, connected to the series combination of a pure, capacitor of capacitance (C) and an inductor of •, inductance (L) is shown in fig., L, , C, , I, C, , The resultant potential difference of VC and VR is, represented by OC Impedance of CR circuit., 1, 2, Z CR R 2 X C R 2 2 2, C, 1, R2 2 2 2, 4 f C, , tan , , X, , O, , 1 , , I L , IZLC, C , , From the above equations, Impedance of, L -C circuit is, 1 , , Z LC L , C , , , If L , , 1, i.e, X L X C then VL VC potential, C, , difference V VL VC ., Now current lags behind voltage by / 2 ., If L , , 1, then VL VC resultant potential, C, , difference V VC VL, Now current leads emf by / 2 ., , VL, , VC, , If L , ~, E, , •, , Let I be the rms value of current flowing in the, circut, , •, , The P.D across ‘L’ is VL I . X L, , 78, , 1, 1, 0, then Z L , C, C, , E, , Z, In L - C, circuit, the phase difference between, voltage and current is always / 2 ., , Current I , , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , WE-11: When 100 volt dc is applied across a coil,, , Power factor cos cos / 2 0 ., So, power consumed in L - C circuit is, P Vrms I rms cos 0, In L - C circuit no power is consumed., Note:, •, , In L - C, circuit, the impendence Z L , , 1, C, , E, ., Z, So, the impedence and current varies with, frequency., , Current I , , •, , At a particular angular frequency, L , , 1, C, , a current of 1 amp flows through it; when 100, V ac of 50 Hz is applied to the same coil, only, 0.5 amp flows. Calculate the resistance and, inductance of the coil., Sol: In case of a coil, i.e, L - R circuit,, V, I with Z R2 X L 2 R2 L 2, Z, So when dc is applied, 0, so Z R, V, V 100, 100, and hence I , i.e, R , R, I, 1, and when ac of 50 Hz is applied., V, V 100, I , i,e. Z , 200, Z, I 0.5, but Z R 2 2 L2 , i.e, 2 L2 Z 2 R 2, 2, , E, and current I becomes maximum I 0 and, Z, resonance occurs., E, At resonance Z 0 and I 0 0 ., Z, , Resonant angular frequency 0 , , 1, LC, , 1, ., 2 LC, WE-10: A 0.21 H inductor and a 12 ohm, resistance are connected in series to a 220 V., 50 Hz ac source. Calculate the current in the, circuit and the phase angle between the, current and the source voltage ., Sol: Here, Resonant frequecny f 0 , , X L L 2 fL 2 50 0.21 21 , 2, , 2, , 2, , 2, , Z R X L 12 21 144 4348, V, 220, Z 4492 67.01 ; I Z 67.02 3.28 A, , X , 21 , tan 1 L tan 1 , , R , 12 , The current will lag the applied voltage by an, 1 21 , angle tan , ., 12 , , NARAYANA GROUP, , i.e, 2 fL 2002 1002 3 104, 3 102, 3, , H 0.55H, 2 50, , WE-12: A 10µF capacitor is in series with a 50, resistance and the combination is connected, to a 220V, 50 Hz line. Calculate (i) the, capacitive reactance, (ii) the impedance of the, circuit and (iii) the current in the circuit., Sol: Here, C 10F 10 10 6 10 5 F, R 50 ohm, E rms 220V, 50Hz ,, (i) Capacitive reactance,, L, , XC , , 1, 1, , C 2C, , , , 1, 2 3.14 50 10 5, , = 318.5 , , (ii) Impedance of CR circuit., 2, , ZCR R 2 X C , , 2, , 50 318.5 , , (iii) Current, I rms , , E rms, 220, , Z CR 322.4, , 2, , 322.4, , = 0.68A, , WE-13: A coil has an inductance of 0.7 H and is, joined in series with a resistance of 220 ., When an alternating e.m.f of 220 V at 50 cps, is applied to it, then the wattless component, of the current in the circuit is, X, L 2 50 0.7, , 1, Sol: tan L , R, R, 220, , 45, Z R 2 X L2 2202 2202, , 220 2, Wattless component of currnet I v sin , , , Ev, 220, 1, sin 45 , , 0.5 A, Z, 220 2, 2, 79
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , A.C THROUGH LCR SERIES CIRCUIT, •, , •, , •, , •, •, , Similarly if XC = 0 then Z R 2 X L 2 i.e., expression for series RL circuit., R, Also, cos , Z, Case (i) : If X L X C then is +ve. In this case, the current lags behind the emf by a phase, , •, , B, VL, (VL-VC), or, (XL-XC), O, , D, , L, E, , , , X, VR or R, , A I, , VC, C, , XL XC , , R, , , , 1 , angle Tan , , -Y, , •, , Since VL and VC are in opposite phase, so, their resultant (VL –VC) is represented by OD, (Here VL > VC), The resultant of VR and (VL–VC) is given by OL., The magnitude of OL is given by, , OL , , 2, , OA OD, , 2, , I R2 XL XC , , 2, , ; VR VL VC , , 2, , E, 2, R2 X L XC , I, Impedance (Z) of LCR circuit is given by, , 2, , Case (ii) : If X L X C then is -ve. In this case, the current leads the emf by a phase angle, X XL , Tan 1 C, , R, , , , Case (iii): If X L X C then is 0. In this case, the current and emf are in phase., •, , If X L X C , then the circuit will be inductive, , •, , If X L X C , then the circuit will be, capacitive, , •, , If X L X C , then the circuit will be purely, resistive., The LCR circuit can be inductive or, capacitive or purely resistive depending on, , Z, , Z R2 X L X C , , 80, , E, , 2, I, A circuit containing pure inductor ofinductance (L),, 1 , , 2 ;, 2, 2, R L , R X L X C , pure capacitor of capacitance (C) and resistor, , C , of resistance (R), all joined in series, is shown, in figure., •, Let be the phase angle between E and I, then, Let E be the r.m.s value of the applied alternating, from Phasor diagram, emf to the LCR circuit., V VC IX L IX C, X XC, tan L, , L, VR, IR, R, 1 , , L C , I, , tan , R, Current in L-C- R series circuit is given by, E E, I 0 sin( w t f ), The potential difference across L,, Z, Z, (or), I, , I, .....(i), VL IX L, 0 .sin( wt f ), •, If XL and XC are equal then Z = R i.e.,, The potential difference across C,, expression for pure resistance, VC IX C .....(ii), circuit., The potential difference across R,, If XL = 0 then Z R 2 X C 2 i.e., expression, .....(iii), VR IR, PHASOR DIAGRAM, for series RC circuit., , Y, , •, , , , E, , 2, , •, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , the value of frequency of alternating source WE-14: In a circuit L, C and R are connected in, series with an alternating voltage source of, of emf., frequency f. The current leads the voltage, •, At some frequency of alternating source,, by 45 . The value of C is :, X L X C and for some other frequency, X L X C ., Sol: As current leads the voltage by 45 ,, There exists a particular value of frequency, XC X L, where X L X C (This situation is, tan 45 1, tan , R, explained under resonance of LCR series, X C X L R or X C X L R, circuit ), Note:Relation between applied pd & pd’s across, 1, 1, L R C , the components in L - C - R circuit, or, L R , C, L, C, R, L, C, R, 1, C, 2 f 2 fL R , V, V, V, V, V, V, WE-15: In a series LCR circuit, the voltage across, ~, the resistance, capacitance and inductance is, E, V, 10V each. If the capacitance is short circuited, then the voltage across the inductance w i l l, For ‘dc’, For ‘ac’, be, V = VR + VL + VC, Sol: As VR VL VC ; R X L X C, V IZ, Z R ; V IR 10 volt, (only before steady state), When capacitor is short circuited,, L, , C, , L, , R, , I R2 X L X C , , 2, , ; , , V 2 = VR 2 + VL - VC , , C, , R, , IR 2 IX L IX C , , 2, , Z R 2 X L2 R 2 R 2 R 2, , 2, , V, 10, , I, R 2 R 2, where VL IX L I L and VC IX C , C, Potential drop across inductance, and VR = IR, 10 R 10, I ' XL I 'R , , volt, R 2, 2, Note: Rules to be followed for various, 200, combinations of ac circuits, mH . a capacitance, WE-16 : An inductance of, •, Compute effective resistance of the circuit as R, , •, Calculate the net reactance of the circuit as, 10 3, 1, F and a resistance of 10 are, of, X X L X C where X L L , X C , ., , C, connected in series with an AC source of 220, •, Resistance offered by all the circuited elements, V, 50 Hz. The phase angle of the circuit is, to the flow of ac is impedance ( Z ), 200, 200 103, 0.2, 2, mH , H, H, Sol: Here, L , Z R 2 X 2 R 2 X L X C , , , , E0, 103, I, , C, , F , R 10 ; Ev 220V , n 50 Hz, •, Calculate the peak value of current as 0, Z, , •, The phase difference between emf & current can, 0.2, X L L 2 nL 2 50 , 20 , be known by constructing an ac triangle as, , X, tan f , R, 1, 1, , XC , , , 10, X, Z, sin f , C 2 nC 2 50 103, X, Z, R, X L X C 20 10 1 , co s f , tan , Z, , ;, f, 4, R, 10, R, NARAYANA GROUP, , New current, I ' V / Z , , 81
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , WE-17: In a series LCR circuit, R 200 , the, voltage and the frequency of the main supply, is 220 V and 50 Hz. respectively. On taking, out the capacitance from the circuit, the current, lags behind the voltage by 30 .On taking out, the inductor from the circuit, the current, leads the voltage by 30 . The power, dissipated in the LCR circuit is, Sol: Here, R 200, Ev 220V, , Z R2 X C X L , , 2, , 2, , 100, , 3 , 90 21.3 , 9, , 2, , Power dissipated / cycle Ev I v cos , 2, , E R E , E0 v v R, Z Z Z , , XL, R, , 3, , 15, , , , 3 0.744W, 2 21.3 , X, WE-19: A 750 Hertz - 20 volt source is connected, In C - R circuit, tan 30 C, R, to a resistance of 100 ohm, an inductance of, 0.1803 henry and a cpacitance of 10 F , all in, X, X, L C or X L X C, series. What is the tim e in which the, R, R, resistance (Thermal capacity 2 joule/ C ), In L - C - R circuit, if is the phase difference, between voltage and current, then, will get heated by 10C ?, X XC, 0, Sol: Here, v 750 Hz , Ev 20V , R 100 , tan L, , 0 0, R, 200, L 0.1803 H , C 10 F 105 F , t ?, i.e., current and voltage are in the same, 10C , thermal capacity 2J/C, phase., , In L -R circuit, tan 30 , , Average power Ev Iv cos , , 220 , , , Ev2, 0, R, , 2, , 242 W, 200, WE-18: An LCR circuit has L = 10 mH. R = 3, ohm and C 1 F connected in series to a, source of 15 cos t volt. What is average, power dissipated per cycle at a frequecny that, is 10% lower than the resonant requency?, Sol: Here, L 102 H , R 3, C 106 F, Resonant frequency,, 0 , , 1, 1, , 10 4 rad/s, 2, 6, LC, 10 10, , XC , , 1, 1, , 21.2 , C 2 750 105, , Z R2 X L X C , , 2, , 2, , 1002 850 21.2 835 , Power dissipated Ev I v cos , 2, E R 20 100, Ev v , 0.0574W, 2, Z Z , 835 , , Heat produced in resistance 2 10 20J, If t is the required time, then, 20, 20, , 348s, P 0.0574, , Actual frequency, 90% 0, , P t 20 t , , 9 103 rad/s, , RESONANT FREQUENCY, , X L L 9 103 10 2 90 , XC , 82, , X L L 2 vL 2 3.14 750 0.1803, , 1, 1, 1000, , , , 3, 6, C 9 10 10, 9, , Electrical Resonance Series L-C-R Circuit, Electrical resonance is said to take place in a, series LCR circuit, when the circuit allows, maximum current for a given frequency of, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , alternating supply, at which capacitive reactance Note: Series LCR circuit at resonance admit, becomes equal to the inductive reactance., maximum current at particular frequencies, so they, The current (I) in a series LCR circuit is given by, can be used to tune the desired frequency or filter, unwanted frequencies. They are used in transmitters, E, E, and receivers of radio, television and telephone, I , 2, Z, 1 .....(i), , 2, carrier equipment etc., R L , , , C , , RESONANCE IN L - C CIRCUIT :, , From the above equation (i), it is clear that, current I will be maximum if the impedance (Z) a), of the circuit is minimum., b), At low frequencies, L L 2 f is very small c), 1, 1, and C C 2 f is very large., , 1, At high frequencies, L is very large and, is, C, very small., 1, i.e., For a particular frequency (f0), L , C, X L X C and the impedance (Z) of LCR circuit, is minimum and is given by Z = R., Therefore, at the particular frequency ( f0 ), the, current in LCR circuit becomes maximum. The, frequency ( f0 ) is known as the resonant, frequency and the phenomenon is called, electrical resonance., Again, for electrical resonance (XL–Xc) = 0., i.e. XL = XC, 1, 1, 2 , or L , C, LC, 1, 1, 2f 0 , or , LC, LC, 1, , or f 0 , , At resonance ,, Net reactance X = 0, , X L XC, Impedance Z = 0, , E0, , Z, 1, , d), , peak value of current I0 , , e), , Resonant frequency f 0 , , f), , Voltage and current differ in phase by, , g), , Power factor cos f 0, , 2 LC, p, 2, , RESONANCE IN L - C - R CIRCUIT :, a), b), c), d), , At resonance,, Net reactance X = 0, X L XC, Impedance Z = R ( minimum ), E0 E0, peak value of current I0 ( maximum but, Z R, not infinity ), , 1, , e), , Resonant frequency f 0 , , f), g), h), i), , Voltage and current will be in phase, power factor cos 1, Resonant frequency is independent of value of R., A series L - C - R circuit behaves like a pure, resistive circuit at resonance., , .....(ii), 2 LC, This is the value of resonant frequency., The resonant frequency is independent of the •, resistanace R in the circuit. However, the, sharpness of resonance decreases with the, increase in R., Series LCR circuit is more selective when, resistance of this circuit is small., , 2 LC, , HALF POWER FREQUENCIES AND, BAND WIDTH., The frequencies at which the power in the circuit, is half of the maximum power (The power at, resonance) are called half power frequencies., Pmax, , Y, R3>R2>R1, , I0, , P, , R1, , R3, , NARAYANA GROUP, , Pmax, 2, , R2, , I, , O, , P=, , f0, , ., X, , 1, , 2, , 3 V, , f, , 83
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84, , = tan–1, I lags E by , , 8) Lead / lag, , L, R, , E0, I0 = Z, , 7) Phase diff. between, E&I, , 6) Peak value of current, , Z = R2 + (L)2, , 5) Impedance, , 1, C, , –1, RC, , I lags E by , , = tan–1, , E0, I0 = Z, , Z = R2 +, , 2, , R, , R, –1, C, , I = I0 sin (t + ), , I = I0 sin (t – ), , X = XC =, , E = E0 sin t, , E = E0 sin t, , X = XL = L, , R-C circuit, , R-L circuit, , 4) Net reactance, , 3) Resistance, , 2) Resulting current, , 1) Input emf, , Parameter, , 1, C, 1, C, , 2, , If XL > XC , I Lags E, by 90°, If XL < XC , I leads E, by 90°, If XL = XC , E and I are, in phase, , = 90°, , E0, I0 = Z, , Z = L –, , X = L –, , R, , XL – XC, , If XL > XC , I lags E, by , If XL > XC , I lags E, by , If XL > XC , I lags E, by , , = tan–1, , E0, I0 = Z, , 1, C, , 1, C, Z = R2 + L –, , X = L –, , R, , I = I0 sin (t ), , , I = I0 sin t 2, 0, , E = E0 sin t, , L-C-R circuit, , E = E0 sin t, , L-C circuit, , Table for values of different parameters for different components applied to ac, , 2, , ALTERNATING CURRENT, JEE-ADV PHYSICS- VOL- IV, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, •, , •, , •, , ALTERNATING CURRENT, , The current in the circuit at half power frequecies, (HPF) is 1 2 or 0.707 or 70.7% of maximum, current (current at resonance)., There are two half power frequencies, , 1, VL, VC 0 L, Q factor V or V R or CR, R, R, 0, , Q factor , , 1 called lower half power frequency. At this, frequency the circuit is capacitive., , WATTLESS CURRENT:, , 3 called upper half power frequency. It is, , In an ac circuit ,, , greater than 2 . At this frequency the circuit is, inducitve., , For series resonant circuit it can be proved, , Pav 0, i.e.., in resistanceless circuit the power, consumed is zero, Such a circuit is called the, wattless circuit and the current flowing is, called the wattless current., Or, The component of current which does not, contribute to the average power dissipation, is called wattless current., , R / L , , wattless current = I rms sin , , QUALITY FACTOR (Q - FACTOR) OF, •, SERIES RESONANT CIRCUIT., , CHOKE COIL:, , Band width : The difference of half power, frequencies 1 and 2 is called band width, , and, •, , •, , 3 1 ., , The characteristic of a series resonant circuit is, determined by the quality factor Q factor of, , •, , •, , •, the circuit., It defines sharpness of i v curve at resonance, when Q factor is large, the sharpness of •, resonance curve is more and vice - versa., R=0, Q- factor =infinity, R = Very low, Q- factor = large, R = low, Q- factor = normal, R = High, Q- factor = low, , i, , V0, V, Resonance curve, , •, , 1 L, R C, , Q factor also defined as follows, Q factor 2 , , Maximum energy stored, energy dissipation, , Choke coil (or ballast) is a device having high, inductance and negligible resistance., It is used to control current in ac circuits and is, used in fluorescent tubes., The power loss in a circuit containing choke coil, is least., In a dc circuit current is reduced by means of a, rheostat.This resutls in a loss of electrical energy, I 2 R per sec., Iron core, Starter, Coil of Cu wire, , ~, Choke coil, , •, , •, •, , Choke, coil, L, R, Application of choke coil, , ~, , It consists of a copper coil wound over a soft, iron laminated core. This coil is put in series, with the circuit in which current is to be reduced., Soft iron is used to improve inductance (L) of, the circuit., The inductive reactance or effective opposition, of the choke coil is given by X L L 2 vL, , , , 2 Maximum energy stored, , T Mean power dissipated, , •, , Resonant frequency 0, , Band width, , , •, , , , NARAYANA GROUP, , R 0 cos 0 so, , For an ideal choke coil r 0 , no electric cnergy, is wasted, i.e., average power P = 0., In actual practice choke coil is equivalent to a, R L circuit., 85
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, •, •, , •, , •, , •, , •, , Choke coil for different frequencies are made by WE-21 :An electric bulb has a rated power of 50, using different substances in their core., W at 100 V. If it is used on an AC source of, For low frequency L should be large thus iron, 200 V, 50 Hz, a choke has to be used in series, core choke coil is used. For high frequency ac circuit,, with it. This choke should have an inductance, L Should be small, so air cored choke coil is used., of, The choke coil can be used only in ac circuits not in Sol: Here, P 50W , V 100volt, dc circuits, because for dc frequency v 0 . Hence, P 50, V 100, I , 0.5 A, R , 200, X L 2 vL 0., V 100, I 0.5, Choke coil is based on the principle of wattless, Let L be the inductance of the choke coil, current., E, E, 200, I v v or Z v , 400 , E, Z, Iv, 0.5, The current in the circuit I with, Z, Now X L Z 2 R 2 4002 200 2, 2, 2, ., Z R r L , L 100 2 3, The power loss in the choke, pav Vrms I rms cos 0, , r, r, r, , 0, as cos Z 2, 2 2, L, r L, WE-20: An ideal choke coil takes a current of 8, ampere when connected to an AC supply of, 100 volt and 50 Hz. A pure resistor under the, same conditions takes a current of 10 ampere., If the two are connected to an AC supply of, 150 volts and 40 Hz. then the current in a, series combination of the above resistor and, inductor is, Sol: For pure inductor,, XL , , E0 100 25, , , 8, 2, Iv, , L , , 25, 25, 25, 1, ;L , , , H, 2, 2 2 2 50 8, , V 100, , 10, I, 10, For the combination, the supply is 150 v, 40, Hz, R, , X L L 2 40 , , 1, 10, 8, , Z X L2 R 2 102 10 2 10 2ohm, Iv , , 86, , Ev, 150, 15, , A, A, Z 10 2, 2, , L, , 200 3 200 3 200 3 2 1.732, , , , 1.1H, , 2 v, 100, 3.14, , LC OSCILLATIONS, A capacitor (C) and an inductor (L) are, connected as shown in the figure. Initially the, charge on the capacitor is Q, , Q2, Energy stored in the capacitor U E , 2C, The energy stored in the inductor, UB = 0., The capacitor now begins to discharge through, the inductor and current begins to flow in the, circuit. As the charge on the capacitor decreases,, 1 2, U E decreases but the energy U B LI in the, 2, magnetic field of the inductor increases. Energy, is thus transferred from capacitor to inductor., When the whole of the charge on the capacitor, disappears, the total energy stored in the electric, field in the capacitor gets converted into, magnetic field energy in the inductor. At this, stage, there is maximum current in the inductor., Energy now flows from inductor to the capacitor, except that the capacitor is charged oppositely., This process of energy transfer continues at a, definite frequency (v). Energy is continuosly, shuttled back and forth between the electric field, NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , in the capacitor and the magnetic field in the, inductor., *, If no resistance is present in the LC circuit, the, LC oscillation will continue infinitely as shown., +q0, -q0, , *, , t, , However in an actual LC circuit, some resistance *, is always present due to which energy is, dissipated in the form of heat. So LC oscillation, will not continue infinitely with same amplitude *, as shown., +q0, , *, , t, -q0, , 1 V, , tells us the potential, C q , difference required to store a unit charge, F, In a mechanical oscillation K tells us the, x, external force requred to produce a unit, displacement of mass, In L - C oscillations current is the analogous, quantity for velocity of the mass in mechanical, oscillations, In L - C oscillations energy stored in capacitor, is analagous to potential energy in mechanical, oscillations, In L - C oscillations energy stored in inductor is, analogous to kinetic energy of the mass in, mechanical oscillations, In L - C oscillations maximum charge on, capacitor q 0 is analogous to amplitude in, mechanical oscillations, As Vmax = A in mechanical oscillations,, I 0 q0 0 in L- C oscillations, , In L - C oscilations, , Let q be the charge on the capacitor at any *, di, time t and, be the rate of change of current., dt, *, Since no battery is connected in the circuit,, q, di, dq, Analogies between Mechanical and Electrical Quantities, L. 0, but i , c, dt, Mecha nical System, Electrical System, dt, from the above equations, we get, Mass m, Inductance L, , q, d 2q, d 2q 1, L 2 0 2 , q0, C, dt, dt, LC, The above equation is analogus to, , d2 x, 2 x = 0, 2, dt, , (differential equation of S.H.M), 1, 1, 2, , Hence on comparing , LC, LC, 1, 1, 2f , f , LC, 2 LC, The charge therefore oscillates with a frequency, 1, f , and varies sinusoidally with time., 2 LC, , *, *, , COMPARISON OF L - C OSCILLATIONS, WITH SHM : The L - C oscillations can be, compared to S.H.M of a block attached to a, spring, 1, In L - C oscillations 0 , LC, In Mechanical oscillations 0 , is the spring constant, , NARAYANA GROUP, , K, where K, m, , Force constant k, Displacement x, Velocity v = dx/dt, Mechnical energy, , Reciprocal capacitance 1/C, Charge q, Current I = dq/dt, Electromagnetic, energy, , Energy of LC Oscillations: Let q0 be the, initial charge on a capacitor. Let the charged, capacitor be connected to an inductor of, induct ance L. LC ciruit will sust ain an, 1, ), oscillations with frequency ( 2 f , LC, At an instant t, charge q on the capacitor and the, current i are given by; q(t ) q0 cos t ;, i q0 sin t, Energy stored in the capacitor at time t is, , 1, 1 q 2 q02, U E CV 2 , , cos2 (t ), 2, 2 C 2C, Energy stored in the inductor at time t is, 1, U M Li 2, 2, , , 1 2 2 2, q2, Lq0 sin t 0 sin 2 (t )( 2 , 2, 2C, , 1, ), LC, , Sum of energies, 87
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , q02, q2, (cos 2 t sin 2 t ) 0 \, 2C, 2C, As q0 and C, both are time independent, this •, sum of energies stored in capacitor and induc, tor is constant in time. Note that it is, equal to the initial energy of the capacitor., UE UM , , TRANSFORMER, •, •`, •, •, , A transformer works on the principle of mutual, induction., It is a static device that is used to increase or, decrease the voltage in an AC circuit., On a laminated iron core two insulated copper, coils called primary and secondary are wound., Primary is connected to an alternating source of •, emf, By mutual induction, an emf is induced in, the secondary., , VOLTAGE RATIO:, •, , •, •, •, , If V1 and V2 are the primary and secondary •, voltages in a transformer, N1 and N2 are the, number of turns in the primary and secondary, •, V1, N1, , coils of the transformer, then, ., V2, N2, In a transformer the voltage per turn is the same, in primary and secondary coils., The ratio N2/N1 is called transformation ratio., The voltage ratio is the same as the ratio of the, number of turns on the two coils., , CURRENT RATIO:, •, , If the primary and secondary currents are I1 and, I 2 respectively, then for ideal transformer •, N, V2 I1, 2, , V1 I 2, N1, , •, •, •, •, •, •, •, , 88, , ., , In an ideal transformer the ampere turns are the •, same in primary and secondary coils., If N s N P voltage is stepped up, then the, transformer is called step - up transformer., •, If N s N P voltage is stepped down, then the, transformer is called step - down transformer. •, In step - up transformer, VS VP and I S I P, •, In step - down transformer, VS VP and I S I P, •, Frequency of input a.c is equal to frequency of, output a.c, •, Transformation of voltage, is not possible with, •, d.c, EFFICIENCY OF TRANSFORMER ( ), Effeiciency is defined as the ratio of output •, power and input power., •, output power, Efficiency , input power, , Pout, Vs is, i.e., % P 100 V i 100, in, P P, , For an ideal transformer Pout Pin so, 100% (But efficiency of practical transformer, lies between 70% - 90 %), For practical transformer Pin Pout Plosses, P, , out, So P 100, in, , P PL 100, Pout, 100 in, Pin, Pout PL , In an ideal transformer the input power is equal to, the output power., V1I1 = V2 I2, The efficiency of an ideal transformer is 100%., LOSSES IN A TRANSFORMER:, The losses in a transformer are divided in to two, types. They are copper losses and iron losses., The loss of energy that occurs in the copper coils, of the transformer (i.e. primary and secondary coils), is called ‘copper losses’. These are nothing but joule, heating losses where electrical energy is converted, in to heat energy., The loss of energy that occurs in the iron core of, the transformer (i.e. hysteresis loss and eddy current, loss) is called ‘iron losses’., MINIMIZING THE LOSSES IN A, TRANSFORMER:, The core of a transformer is laminated and each, lamination is coated with a paint of insulation to, reduce the ‘eddy current’ losses., By choosing a material with narrow ‘hysteresis, loop’ for the core, the hysteresis losses are, minimized., , , Uses of transformer:, A transformer is used in almost all ac operations,, e.g, In voltage regulators for TV, refrigerator, computer,, air conditioner etc., In the induction furnaces., Step down transformer is used for welding, purposes., In the transmission of ac over long distnace., Step down and step up transformers are used in, electical power distribution., Audio Frequency transformers are used in, radiography, television, radio, telephone etc., Radio frequency transformers are used in radio, communication., NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , WE- 22: A transformer having efficiency 90% is, working on 100 V and at 2.0 kW power. If the, current in the secondary coil is 5A, calculate, (i) the current in the primary coil and (ii), voltage across the secondary coil., 1., 9, , I s 5A , E p 100V,, 10, (i) E p I p 2kW 2000W, , Sol: Here 90% , , Ip , , 2000, Ep, , (ii) , , or I p , , 2000, 20A, 100, , Output power E s I s, , Input power, Ep Ip, , 2., , or E s Is E p I p, 3., , 9, 2000 1800W, 10, , Es , , 1800 1800, , 360 volt, Is, 5, , WE-23:A step up transformer operates on a 230 4., V line and a load current of 2 ampere. The, ratio of the primary and secondary windings, is 1 : 25. What is the current in the primary ?, Sol: Using the relation, NP, I, N I, S ; IP S S, NS IP, NP, , 5., , Here N p / Ns 1 / 25 (or) Ns / N p 25 / 1 = 25, and IS = 2A, Current in primary, I P 25 2 50A, , SKIN EFFECT:, •, •, •, •, , •, , A direct current flows uniformly throughout the, 6., cross section of the conductor., An alternaitng current, on the other hand, flows, mainly along the surface of the conductor. This, effect is known as skin effect., When alternating current flows through a, conductor, the flux changes in the inner part of, the conductor are higher., Therefore, the inductance of the inner part is, higher than that of the outer part. Higher the 7., frequency of alternating current, more is the skin, effect., The depth upto which ac current flows through a, wire is called skin depth ., VR2, R, R VR2, PR, , VR rated voltage, PR rated power , NARAYANA GROUP, , 8., , C. U. Q, INSTANTANEOUS, PEAK,R.M.S &, AVERAGE VALUES OF A.C AND A.V, In an ac circuit the current, 1) is in phase with the voltage, 2) leads the voltage, 3) lags the voltage, 4) any of the above depending on t he, circumstances, The average e.m.f during the positive half, cycle of an a.c. supply of peak value E0 is, 1) E0 / 2) E0 / 2 3) E0 / 2 4) 2 E0 / , Alternating current is transmitted to distant, places at, 1) high voltage and low current, 2) high voltage and high current, 3) low voltage and low current, 4) low voltage and high current, In case of a.c circuit, Ohm’s law holds good, for, a) Peak values of voltage and current, b) Effective values of voltage and current, c) Instantaneous values of voltage and current, 1) only a is true, 2) only a and b are true, 3) only c is true, 4) a, b and c are true, In case of AC circuits the relation V = i Z,, where Z is impedance, can directly applied to, 1) peak values of voltage and current only, 2) rms values of voltage and current only, 3) instantaneous values of voltage and current, only, 4) both 1 and 2 are true, Alternating current can not be measured by, direct current meters, because, 1) alternating current can not pass through an, ammeter, 2) the average value of current for complete cycle, is zero, 3) some amount of alternating current is, destroyed in the ammeter, 4) peak value of current is zero, The r.m.s. value of potential due to, superposition of given two alternating, potentials E1 = E0 sin t and E2 = E0 cos t, will be, 1) E0, 2) 2E0, 3) E 0 2 4) Zero, If the instantaneous values of current is, I 2 cos(t ) A in a circuit, the r.m.s. value, of current in ampere will be, 1) 2, 2) 2, 3) 2 2, 4) zero, 89
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 9., , If a capacitor is connected to two different A.C., generators, then the value of capacitive, reactance is, 1) directly proportional to frequency, 2) inversely proportional to frequency, 3) independent of frequency, 4) inversely proportional to the square of, frequency, 10. In general in an alternating current circuit, 1) the average value of current is zero, 2) the average value of square of the current is, zero, 3) average power dissipation is zero, 4) the phase difference between voltage and, current is zero, , A.C ACROSS R-L,R-C,L-C &, L-C-R SERIES CIRCUIT, 11. The magnitude of induced e.m.f in an LR, circuit at break of circuit as compared to its, value at make of circuit will be, 1) less, 2) more, 3) some times less and some times more, 4) nothing can be said, 12. The emf and current in a circuit are such that, E = E0 sin t and I = I0 sin ( t ) . This AC, circuit contains., 1) R and L 2) R and C 3) only R 4) only C, 13. The correct graph between the resistance of, a conductor with frequency is, 1) Y, , 2), , R, , 16. The phase angle between current and voltage, in a purely inductive circuit is, 1) zero, 2) , 3) / 4, 4) / 2, 17. Ratio of impedence to capacitive reactance, has, 1) no units 2) ohm, 3) ampere 4) tesla, 18. An inductor coil having some resistance is, connected to an AC source. Which of the, following have zero average value over a, cycle, 1) induced emf in the inductor only, 2) current only 3) both 1 and 2 4) neither 1 nor 2, 19. The current does not rise immediately in a, circuit containing inductance, 1) because of induced emf, 2) because of high voltage drop, 3) both 1 and 2 4) because of joule heating, 20. In an AC circuit containing only capacitance, the current, 1) leads the voltage by 180º, 2) lags the voltage by 90º, 3) leads the voltage by 90º, 4) remains in phase with the voltage, 21. A bulb is connected first with dc and then ac, of same voltage.Then it will shine brightly with, 1) AC, 2) DC, 3) Equally with both, 4) Brightness will be in ratio 1/14, 22. A capacitor of capacity C is connected in A.C., circuit. If the applied emf is V = V0 sin t ,, then the current is, , Y, , 1) I , , R, , V0, sin t, L, , 2), , I, , V0, , , sin t , C, 2, , , , , , 3) I V0 C sin t 4) I V0C sin t 2 , , , 23. At low frequency a condenser offers, 1) high impedance, 2) low impedance, Y, Y, 3), 4), 3) zero impedance, R, R, 4) impedance of condenser is independent of, frequency, f, f, X, X, 24. Statement ( A ) : The reactance offered by, 14. Same current is flowing in two alternating, an inductance in A.C. circuit decreases with, circuits. The first circuit contains only, increase of AC frequency., inductance and the other contains only a, Statement ( B ) : The reactance offered by a, capacitor. If the frequency of the e.m.f. is, capacitor in AC circuit increases with increase, increased, the current will, of AC frequency., 1) increase in first circuit and decrease in the, 1) A is true but B is false, other, 2) Both A and B are true, 2) increase in both circuits, 3) A is false but B is true, 3) decrease in both circuits, 4) Both A and B are false, 4) decrease in first circuit and increase in the 25. Statement ( A ) : With increase in frequency, other, of AC supply inductive reactance increases., 15. When an a.c source is connected across a, Statement ( B ) : With increase in frequency, resistor, of AC supply capacitive reactance increase, 1) The current leads the voltage in phase, 1) A is true but B is false, 2) The current lags behind the voltage in phase, 2) Both A and B are true, 3) The current and voltage are in same phase, 3) A is false but B is true, 4) The current and voltage are out of phase, 4) Both A and B are false, f, , 90, , X, , f, , X, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , 26. In an A.C circuit having resistance and 32. In series L - C - R resonant circuit, to increase, the resonant frequency, capacitance, 1) L will have to be increased, 1) emf leads the current, 2) C will have to be increased, 2) current lags behind the emf, 3) LC will have to be decreased, 3) both the current and emf are in phase, 4) LC will have to be increased, 4) current leads the emf., 27. Select the correct options among the 33. If in a series L - C - R ac circuit, the voltages, across R, L, C are V1 ,V2 ,V3 respectively., following: In an R-C circuit, Then the voltage of applied AC source is, a) instantaneous A.C is given by, always equal to, I = I0sin ( wt ), b) the alternating current in the circuit leads, 1) V1 +V2 +V3, 2) V12 (V2 V3 )2, the emf by a phase angle ., 3) V1 -V2 -V3, 4) V12 (V2 V3 ) 2, 2, 2, c) Its impedance is R ( c), 34. In non-resonant circuit, the nature of circuit, d) Its capacitive reactance is c, for frequencies greater than the resonant, 1) a, b are ture, 2) b, c, d are true, frequency is, 3) c, d are true, 4) a, c are true, 1) resistive, 2) capacitive, 28. If the frequency of alternating e.m.f. is f in L3) inductive, 4) both 1 and 2, C-R circuit, then the value of impedance Z 35. The phase difference between voltage and, will change with log (frequency) as, current in an LCR series circuit is, 1) increases, 1) zero always, 2) / 4 always, 2) increases and then becomes equal to, 3) , 4) between 0 and / 2, resistance, then it will start decreasing, 36. In an LCR a.c circuit at resonance, the, 3) decreases and when it becomes minimum, current, equal to the resistance then it will start increasing, 1) Is always in phase with the voltage, 4) go on decreasing, 2) Always leads the voltage, 29. An inductance and resistance are connected, 3) Always lags behind the voltage, in series with an A.C circuit. In this circuit, 4) May lead or lag behind the voltage, 1) the current and P.d across the resistance lead 37. An inductance L and capacitance C and, P.d across the inductance by /2, resistance R are connected in series across, 2) the current and P.d across the resistance lags, an AC source of angular frequency . If, behind the P.d across the inductance by angle, 1, 2 , then, /2, LC, 3) The current across resistance leads and the, 1) emf leads the current, P.d across resistance lags behind the P.d across, 2) both the emf and the current are in phase, the inductance by /2, 3) current leads the emf, 4) the current across resistance lags behind and, 4) emf lags behind the current, the P.d across the resistance leads the P.d across 38. Consider the following two statements A and, the inductance by /2, B and identify the correct answer., 30. An LCR circuit is connected to a source of, A) At resonance of L - C - R series circuit, the, alternating current. At resonance, the applied, reactance of circuit is minimum., voltage and the current flowing through the, B) The reactance of a capacitor in an A.C, circuit will have a phase difference of, circuit is similar to the resistance of a, capacitor in a D.C. circuit, 1) / 4, 2) zero, 3) , 4) / 2, 1) A is true but B is false, 31. The incorrect statement for L-R-C series, 2) Both A and B are true, circuit is, 3) A is false but B is true, 1) The potential difference across the resistance, 4) Both A and B are false, and the appleid e.m.f. are always in same phase, 39., Choose the wrong statement of the following., 2) The phase difference across inductive coil is, 0, 1) The peak voltage across the inductor can be, 90, less than the peak voltage of the source in an, 3) The phase difference between the potential, LCR circuit, difference across capacitor and potential, 2) In a circuit containing a capacitor and an ac, 0, difference across inductance is 90, source the current is zero at the instant source, 4) The phase difference between potential, voltage is maximum, difference across capacitor and potential, 0, difference across resistance is 90, NARAYANA GROUP, , 91
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ALTERNATING CURRENT, , JEE-ADV PHYSICS- VOL- IV, , 3) When an AC source is connected to a 48. A step up transformer is connected on the primary, side to a rechargable battery which can deliver a, capacitor,then the rms current in the circuit gets, large current. If a bulb is connected in the, increased if a dielectric slab is inserted into the, secondary, then, capacitor., 1) the bulb will glow very bright, 4) In a pure inductive circuit emf will be in phase, 2) the bulb will get fused, with the current., 3) the bulb will glow, but with less brightness, 40. The essential difference between a d.c., 4) the bulb will not glow, dynamo and an a.c. dynamo is that, 49. The ratio of primary voltage to secondary, 1) a.c. has an electromagnet but d.c. has a, voltage in a transformer is ‘n’. The ratio of, permanent magnet, the primary current to secondary current in, 2) a.c. will generate a higher voltage, the transformer is, 3) a.c.has slip rings but the d.c. has a commutator, 1) n, 2) 1/n, 3) n2, 4) 1/n2, 4) a.c. dynamo has a coil wound on soft iron, 50. In a step down transformer, the number of, but the d.c. dynamo has a coil wound on copper, turns in the primary is always, 41. The unit of impedence is, 1) greater than the number of turns in the, secondary, 1) ohm, 2) mho, 3) ampere 4) volt, 2) less than the number of turns in the secondary, 42. The power factor of a.c. circuit having L and, 3) equal to the number of turns in the secondary, R connected in series to an a.c. source of, 4) either greater than or less than the number of, angular frequency is given by, turns in the secondary, R, L, R, R 2 2L2, 51. The transformer ratio of a step up transformer, 1), 2), 3), 4), is, R, L, R 2 2 L2, R, 1) greater than one, 2) less than one, 43. The capacitor offers zero resistance to, 3), less, than, one, and, some, times greater than one, 1) D.C. only, 2) A.C. & D.C., 4) greater than one and some times less than one, 3) A.C. only, 4) neither A.C. nor D.C., 52. A stepup transformer develops 400V in, 44. Power factor is defined as, secondary coil for an input of 200V A.C. Then, 1) apparent power/true power, the type of transformer is, 2) true power/apparent power, 1) Steped down 2) Steped up 3) Same, 3) true power (apparent power)2, 4) Same but with reversed direction, 4) true power x apparent power, 53. Assertion(A) : If changing current is flowing, through a machine with iron parts, results in, TRANSFORMER, loss of energy., 45. The core of a transformer is laminated so that, Reason(R): Changing magnetic flux through, 1) energy loss due to eddy currents may be, an area of the iron parts causes eddy currents., reduced, 1)Both A and R are individually true and R is, 2) rusting of the core may be prevented, the correct explanation of A, 3) change in flux may be increased, 2)Both A and R are individually true but R is, 4) ratio of voltage in the primary to that in the, not the correct explanation of A, secondary may be increased, 3)A is true but R is false, 46. A step up transformer is used to, 4)Both A and R are false, 1) increase the current and increase the voltage 54. Transformers are used in, 2) decrease the current and increase the voltage, 1) d.c circuits only, 2) a.c. circuits only, 3) increase the current and decrease the voltage, 3) Both a.c and d.c circuits 4) Integrated circuits., 4) decrease the current and decrease the voltage 55. The magnitude of the e.m.f. across the, secondary of a transformer does not depend, 47. A transformer changes the voltage, on, 1) without changing the current and frequency, 1) The number of the turns in the primary, 2) without changing the current but changes the, 2) The number of the turns in the secondary, frequency, 3) The magnitude of the e.m.f applied across the, 3) without changing the frequency but changes the, primary, current, 4)The resistance of the primary and the, 4) without changing the frequency as well as the, secondary, current, 92, , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , capacitive reactance equals inductive reactance, 56. For an ideal transformer ratio of output to the, input power is always, 1, 4) current will be leading voltage if LC, 1) greater than one, 2) equal to one, 3) less than one, 4) zero, 64. The value of current in two series L C R, 57. Consider the following two statements A and, circuits at resonance is same when connected, B and identify the correct answer., across a sinusoidal voltage source. Then, A) In a transformer a large alternating, 1) both circuits must be having same value of, current at low voltage can be transformed into, capacitance and inductance, a small alternating current at high voltage, 2) in both circuits ratio of L and C will be same, B) Energy in current carrying coil is stored in, 3) for both the circuits X L / X C must be same at, the form of magnetic field., that frequency, 1) A is true but B is false, 4) both circuits must have same impedance at, 2) Both A and B are true, all frequencies, 3) A is false but B is true, 4) Both A and B are false, 65. When an AC source of emf e E0 sin 100t , 58. Statement ( A ) : Flux leakage in a, is connected across a circuit, the phase, transformer can be minimized by winding the, difference betwen the emf e and the current, primary and secondary coils one over the, , other., in the circuit is observed to be ahead,If, i, Statement ( B ) : Core of the transformer is, 4, made of soft iron, the circuit consists possibly of R C or, 59. Statement (A ) : In high current low voltage, R L or L C in series, find the relationship, windings of a transformer thick wire is used, between, the two elements:, to minimize energy loss due to heat produced, 1) R 1k , C 10 F 2) R 1k , C 1 F, Statement ( B ) : The core of any transformer, is laminated so as to reduce the energy loss, 3) R 1k , L 10 H 4) R 1k , L 1H, due to eddy currents, 66. An AC voltage source of variable angular, 60. Statement ( A ) : Step up transformer, frequency and fixed amplitude V0 is, converts low voltage, high current to high, connected in series with a capacitance C and, voltage, low current, Statement (B) : Transformer works on both, an electric bulb of resistance R (inductance, ac and dc, zero). When is increased, 61. To reduce the iron losses in a transformer, the, 1) the bulb glows dimmer, core must be made of a material having, 2) the bulb glows brighter, 1) low permeability and high resistivity, 3) total impendance of the circuit is unchanged, 2) high permeability and high resistivity, 4) total impendance of the circuit increases, 3) low permeability and low resistivity, ASSERTION & REASON, 4) high permeability and low resistivity, 1) Both Assertion and Reason are true and, 62. Maximum efficiency of a transformer depends, Reason is the correct explanation of, on, Assertion., 1) the working conditions of technicians., 2) Both Assertion and Reason are true but, 2) weather copper loss =1/2 x iron loss, Reason is not the correct explanation of, 3) weather copper loss = iron loss, Assertion., 4) weather copper loss =2 x iron loss, 3) Assertion is true but Reason is false, 63. For a LCR series circuit with an A.C. source, 4) Assertion is false but Reason is true, of angular frequency , 67. Assertion (A): The average value of, 1, , , 1) circuit will be capacitive if, <sin 2t is zero., LC, Reason (R): The average value of function, 2) circuit will be inductive if , , 1, LC, , 3) power factor of circuit will be unity if, NARAYANA GROUP, , F t over a period T is F t , , 1, T, , T, , F t dt, 0, , 93
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 68. Assertion (A): If current varies sinusoidally, the average power consumed in a cycle is, zero., Reason (R): If current varies sinusoidally the, average power consumed is zero, 69. Assertion (A) : The power consumed in an, electric circuit is never negative, Reason (R) : The average power consumed, in an electric circuit is P =, , 1) 4, 8) 2, 15) 3, 22) 4, 29) 2, 36) 1, 43) 4, 50) 1, 57) 2, 64) 3, 71) 4, , V2, = I2 R, R, , 70. Assertion (A): The inductive reactance limits, the current in a purely inductive circuit in the, same way as the resistance circuit., Reason (R): The inductive reactance is, directly proportional to the inductance and to, the frequency of the varying current., 71. Assertion (A) : An ac emf which oscillates, symmetrically about zero, the current it, sustains also oscillates symmetrically about, zero., Reason (R): In any circuit element, current is, always in the phase with voltage, 72. Assertion (A): A lamp is connected in series, with a capacitor and ac source connected, across their terminals consequently current, flow in the circuit and the lamp will shine., Reaosn(R): capacitor block dc current and, allow ac current, 73. Assertion (A): An electric lamp is connected, in series with a long solenoid of copper with, air core and then connected to AC source. If, an iron rod is inserted in solenoid the lamp, will become dim., Reason (R): If iron rod is inserted in solenoid,, the induction of solenoid increases., 74. An inductor, capacitor and resistance, connected in series. The combination is, connecte across AC source., Assertion (A): Peak current through each, remains same, Reason (R) : Average power delivered by, source is equal to average power consumed, by resistance., 75. Assertion (A): when frequency is greater than, resonance frequency in a series LCR circuit,, it will be an inductive circuit., Reason (R): Resultant voltage will lead the, current, 76. Assertion (A): Maximum power is dessipated, in a circuit (through R) in resonance, Reason (R) : At resonance in a series LCR, circuit, the voltage across indcutor and, capacitor are out of phase., 94, , C. U. Q - KEY, 2) 4, 9) 2, 16) 4, 23) 1, 30) 2, 37) 1, 44) 2, 51) 1, 58) 4, 65) 1, 72) 1, , 3) 1, 10) 1, 17) 1, 24) 4, 31) 3, 38) 1, 45) 1, 52) 2, 59) 2, 66) 2, 73) 1, , 4) 2, 11) 2, 18) 3, 25) 1, 32) 3, 39) 4, 46) 2, 53) 1, 60) 1, 67) 4, 74) 2, , 5) 4, 12) 1, 19) 3, 26) 4, 33) 4, 40) 3, 47) 3, 54) 2, 61) 2, 68) 4, 75) 1, , 6) 2, 13) 1, 20) 3, 27) 1, 34) 3, 41) 1, 48) 4, 55) 4, 62) 3, 69) 1, 76) 1, , 7) 1, 14) 4, 21) 3, 28) 3, 35) 4, 42) 2, 49) 2, 56) 2, 63) 3, 70) 2, , C. U. Q - HINTS, 67., , sin 2 t 1/ 2, 1 2, 68. P im R, 2, 69. I is scalar in Joules heating effect is independent, an direction of current., V, V, 70. I X & i R X L L 2 vL, L, , , 2, , In capcitor current leads the voltage by, 2, 1, 72. X C , 2 fC, 71. In inductor current lags the voltage by, , for dc f 0 then X L , for ac f 0 then X C finite, 73. L r, more voltage is present across inductor, so less voltage across bulb, 74. In series current is same, inductor and capacitor, does not consume power, 75. At resonance X L X C and frequency, , 1, 1, If f f0 then X L X C , so it, 2 LC, will be an inductive circuit. AC current must lag, AC voltage., 2, 76. At resonance P I max, R and VL and VC are out, f0 , , of phase. I max is due to Z min R which is due, to out of phase of VL and VC ., , NARAYANA GROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 9., , LEVEL-I (C.W), INSTANTANEOUS, PEAK,R.M.S &, AVERAGE VALUES OF A.C AND A.V, 1., , 2., , A steady P.D. of 10V produces heat at a rate, 'x' in resistor. The peak value of A.C. voltage, which will produce heat at rate of x/2 in same, resistor is, , 1) 5 V, 2) 5 2 V 3) 10 V, 4) 10 2 V, The r.m.s. value of an a.c. of 50 Hz is 10 A. 10. An alternating voltage of E 200 2 sin(100t)V, The time taken by the alternating current in, is connected to a condenser of 1 F through, reaching from zero to maximum value and the, an A.C. ammeter. The reading of the ammeter, peak value of current will be, will be, 1) 2 102 sec and 14.14 A 2) 110 2 sec and 7.07 A, 1) 10 mA 2) 40 mA 3) 80 mA 4) 20 mA, 3) 5 10 3 sec and 7.07 A 4) 5 10 3 sec and 14.14 A 11. The inductance of a coil is 0.70 henry. An A.C., source of 120 volt is connected in parallel with, An inductor has a resistance R and inductance, it. If the frequency of A.C. is 60Hz, then the, L. It is connected to an A.C. source of e.m.f, current which is flowing in inductance will be, EV and angular frequency , then the current, 1) 4.55 A 2) 0.355 A 3) 0.455 A 4) 3.55 A, I in the circuit is, v, , 2, 2, EV, EV, E, EV EV , , 2) V 3), 4), R L , R 2 2 L2, L, R, , , , The peak voltage of 220 Volt AC mains (in Volt), is, 1) 155.6 2) 220.0, 3) 311 4) 440.0, The peak value of A.C. is 2 2A . It’s apparent, value will be, 1) 1A, 2) 2A, 3) 4A, 4) zero, Alternating current in circuit is given by, I I 0 sin 2 nt . Then the time taken by the, current to rise from zero to r.m.s. value is equal, to, 1) 1/2n, 2) 1/n, 3) 1/4n, 4) 1/8n, Using an A.C. voltmeter the potential, difference in the electrical line in a house is, read to be 234 volt. If the line frequency is, known to be 50 cycles/second, the equation for, the line voltage is, 1) V = 165 sin(100 t ) 2) V = 331 sin(100 t ), 3) V = 220 sin(100 t ) 4) V = 440 sin(100 t ), A mixer of 100 resistance is connected to, an A.C. source of 200V and 50 cycles/sec. The, value of average potential difference across, the mixer will be, 1) 308V 2) 264V 3) 220V 4) zero, , 1), , 3., , 4., , 5., , 6., , 7., , A.C ACROSS PURE RESISTOR,, INDUCTOR & CAPACITOR, 8., , TRANSFORMER, 12. A transformer steps up an A.C. voltage from, 230 V to 2300 V. If the number of turns in the, secondary coil is 1000, the number of turns in, the primary coil will be, 1) 100, 2) 10,000 3) 500, 4) 1000, 13. The transformer ratio of a transformer is 5. If, the primary voltage of the transformer is 400, V, 50 Hz, the secondary voltage will be, 1) 2000 V, 250 Hz, 2) 80 V, 50 Hz, 3) 80 V, 10 Hz, 4) 2000 V, 50 Hz, 14. A step-up transformer works on 220V and gives, 2 A to an external resistor. The turn ratio between, the primary and secondary coils is 2:25., Assuming 100% efficiency, find the secondary, voltage, primary current and power delivered, respectively, 1) 2750 V, 25 A, 5500 W 2) 2750 V, 20 A, 5000 W, 3) 2570 V, 25 A, 550 W 4) 2750 V, 20 A, 55 W, , A.C ACROSS L-R, L-C &, L-C-R SERIES CIRCUITS, 1, , 15. A coil of self - inductance H is connected, , in series with a 300 resistance. A voltage of, 200V at frequency 200Hz is applied to this, combination. The phase difference between, the voltage and the current will be, 1) tan, , 1 4 , , , 3, , 2) tan, , 1 3 , , , 4, , 3) tan, , 1 1 , , , 4, , 4), , 5, tan 1 , 4, , The equation of an alternating voltage is 16. A condenser of 10 F and an inductor of 1H, E=220 sin(t / 6) and the equation of the, are connected in series with an A.C. source of, frequency 50Hz. The impedance of the, current in the circuit is I=10 sin( t / 6) ., combination will be (take 2 10 ), Then the impedance of the circuit is, 1) zero, 2) Infinity 3) 44.7 4) 5.67 , 1) 10 ohm 2) 22 ohm 3) 11 ohm 4) 17 ohm, , NARAYANAGROUP, , 95
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , 17. A 100 km telegraph wire has capacity of 23. The voltage time (V - t) graph for triangular, 0.02 F / km , if it carries an alternating current, wave having peak value V0 is as shown in, of frequency 5 kHZ. The value of an, figure., inductance required to be connected in series, so that the impedence is minimum., 1) 50.7mH 2) 5.07mH 3) 0.507mH 4) 507mH, 18. In an LCR series circuit the rms voltages, across R, L and C are found to be 10 V, 10 V, and 20 V respectively. The rms voltage across, the entire combination is, The rms value of V in time interval from t 0, 1) 30 V, 2) 1 V, 3) 20V, 4) 10 2 V, T, to, is, 19. In the circuit shown, a 30V d.c. source gives a, 4, current 2.0 A as recorded in the ammeter A, V0, V0, V, and 30V a.c. source of frequency 100Hz gives, 1), 2) 0, 3), 4) 2Vo, a current 1.2A. The inductive reactance is, 3, 2, 2, R, , L, , LEVEL-I (C.W) - KEY, 1) 4, , 4) 2, , 5) 4, , 6) 2, , 7) 4, , LEVEL-I (C.W) - HINTS, 1, T, ,t , 4, f, , 1., , i0 2irms , T , , 2., , i, , 3., , V0 2 .Vr .m.s . 2 200 311 volt, , 4., , I rms , , 5., , t, , 6., , E E0 sin t ; voltage read is r.m.s. value, , 7., , E0 2 234V 331 volt, and t 2 n t 2 50 t 100 t, Thus, the eqn of line voltage is given by, V = 331 sin(100 t ), For one complete rotation, average voltage is zero, , 8., , Z, , Z, , 1) only capacitor 2) both inductor and resistor, 3) either capacitor, resistor and inductor or only, 9., capacitor and resistor, 4) only resistor, 96, , 3) 3, , 8) 2 9) 3 10) 4 11) 3 12) 1 13) 4 14) 1, 15) 1 16) 1 17) 3 18) 4 19) 2 20) 3 21) 4, 22) 3 23) 1, , A, , 1) 10 ohm 2) 20 ohm 3) 5 34 ohm 4) 40 ohm, 20. A choke coil has negligible resistance. The, alternating potential drop across it is 220 volt, and the current is 5mA. The power consumed, is, 5, 220, W, W, 1) 220 , 2), 1000, 5, 3) zero, 4) 2.20 x 5W, 21. In an A.C. circuit, the instantaneous values of, e.m.f. and current are E = 200 sin 314t volt, and I sin(314t / 3) ampere then the, average power consumed in watts is, 1) 200, 2) 100, 3) 0, 4) 50, 22. In a black box of unkown elements (L, C or R, or any other combination) an AC voltage, E E 0 sin( t ) is applied and current in the, circuit was found to be i i0 sin(t /4) ., Then the unknown elements in the box may, be, , 2) 3, , E0, 2, , R X L2, , , X L L, , I0, 2, , T, 1, , 4 4f, , E0, I0, , v2, v2 x, v, x, 1 v1 , R, R 2, 2, , in the second case Vrms= V1, , V0 =, , 2 V1, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, E, , E, , ALTERNATING CURRENT, , C, , rms, 0, 10. Irms X 2, C, 11. XL = 2 fl 6.28 60 0.70 263.76, , V, 120, , 0.455 A, I, X L 263.76, , LEVEL-I (H.W), INSTANTANEOUS, PEAK,R.M.S &, AVERAGE VALUES OF A.C AND A.V, 1., , For a given AC source the average emf during, the positive half cycle, 1) depends on E0, 2) depends on shape of wave, 3) both 1 and 2, 4) depends only on peak value of E0, , 2., , An alternating emf given by V V0 Sin t has, peak value 10 volt and frequency 50 Hz. The, 1, s is, instantaneous emf at t , 600, , n s Vs, 12. n V, p, p, Vs, 13. Frequency remains same. V 5, p, , Es N s i p, 14. E N i , P Esis, p, p, s, 15. tan , , 2 fL, R, , 1, , f, , ,, , 2 LC, , 1, , , , 16. Z 2 fL 2 fC , , , 17. , , 3., , , , 1, 1, 1, L 2 , C (2 n) 2 C, LC, , 18. V VR2 VL VC , , 3) I , , 2, , V 30, , 15, I, 2, 30, 25, When a.c. source, Z , 1.2, , 4., , 19. When d.c. source, R , , X L (25) 2 (15) 2 625 225 20, 20. Average power is zero, 1 200, , cos 60, 21. Pavg Irms Erms cos , 2, 2, 50W, 22. Here current leads the voltage. So, there is, reactance which is capacitive, X X C X L or X X C alone besides R, 23. Ans : (a), , 5., 6., , 2, , Sin 150 t , , NARAYANAGROUP, , , , 7., , E0, , 2), , 1 , , R 2 L , , C , , , E0, 1 , , 2 R 2 2 L 2 2 , C , , , T 4, , , 2 , t dt , 4V0 0, V0, , T, , T 4 , 3, , dt , , , 0, , , 4) I 20 2 Sin 75 t , , The voltage of an A.C. source varies with time, according to the equation V 50sin100 t cos100 t ,, where 't' is in sec and 'V' is in volt. Then, 1) The peak voltage of the source is 100 V, 2) The peak voltage of the source is 100 / 2V, 3) The peak voltage of the source is 25 V, 4) The frequency of the source is 50 Hz, The form factor for a sinusoidal A.C. is, 1) 2 2 : 2) : 2 2 3) 2 : 1 4) 1 : 2, At resonance the peak value of current in LC-R series circuit is, , 3), , , , V2, , 20, , 1) E0/R, , V t 4V t, V 0 0, T, T, 4, , Vrms , , 4) 1V, 1) 10 V, 2) 5 3V 3) 5 V, The equation of A.C. of frequency 75Hz, if it’s, RMS value is 20A is, 1) I 20Sin 150 t , 2) I 20 2 Sin 150 t , , 2, , 4), , 2, , E0, 2R, , In an AC circuit, the rms value of the current,, I rms is related to the peak current I 0 as, 1) I rm s , , 1, I0, , , 3) I rms , , 2I0, , 2) I rms , , 1, I0, 2, , 4) I rms I 0, 97
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 8., , 9., , A voltmeter connected in an A.C circuit reads 17. The transformer ratio of a transformer is 10:1., If the primary voltage is 440V, secondary emf, 220V. It represents,, is, 1) peak voltage, 2) RMS voltage, 1) 44 V 2) 440V 3) 4400 V 4) 44000 V, 3) Average voltage, 4) Mean square voltage, A.C, ACROSS L-R, L-C & L-C-R, If the instantaneous current in a circuit is given, , SERIES CIRCUITS, by I 2cos t A, the rms value of the, 18. The frequency at which the inductive reactance, current is, of 2H inductance will be equal to the capacitive, 2) 2 A, 3) 2 2 A 4) zero, 1) 2 A, reactance of 2 F capacitance (nearly), 1), 80Hz 2) 40 Hz 3) 60Hz 4) 20Hz, 10. The time taken by an AC of 50 Hz in reaching, 19. In a series LCR circuit R 10 and the, from zero to its maximum value will be, 1) 0.5 s, 2) 0.005 s 3) 0.05 s 4) 5s, impedance Z 20 . Then the phase, 11. A generator produces a voltage that is given, difference between the current and the voltage, by V=240 sin 120t V, where t is in second. The, is, 1) 60o, 2) 30o 3) 45o, 4) 90o, frequency and r.m.s. voltage are respectively, 20. In an L-C-R series circuit,, 1) 60Hz and 240V, 2) 19Hz and 120V, 3) 19Hz and 170V, 4) 754Hz and 170V, R 5, X L 9, X C 7 . If applied voltage, in, the circuit is 50V then impedance of the, A.C ACROSS PURE RESISTOR,, circuit in ohm will be, INDUCTOR & CAPACITOR, 1) 2, 2) 3, 3) 2 5, 4) 3 5, 12. A 220 V, 50 Hz AC supply is connected across, a resistor of 50 k . The current at time t 21. In an AC circuit the potential differences, across an inductance and resistance joined in, second, assuming that it is zero at t 0, is, series are respectively 16 V and 20 V. The total, potential difference across the circuit is, 1) 4.4sin 314t mA 2) 6.2sin 314t mA, 1) 20 V, 2) 25.6 V 3) 31.9 V 4) 53.5 V, 22., Current, in, an, ac circuit is given by, 3) 4.4sin 157t mA 4) 6.2sin 157t mA, i 3sin t 4cos t then, 13. A resistance of 20 is connected to a source, 1) rms value of current is 5 A, of alternating current rated 110 V, 50 Hz. Then, 2) mean value of this current in one half period will, the time taken by the current to change from, be 6/, its maximum value to the r.m.s. value is, 3) if voltage applied is V Vm sin t then the, 1) 2.5 103 sec, 2) 2.5 102 sec, circuit must be containing resistance and, 3) 5 103 sec, 4) 25 10 3 sec, capacitance, 14. A condenser of capacity 1pF is connected to, 4) if voltage applied is V Vm sin t , the circuit may, an A.C source of 220V and 50Hz frequency., contain resistance and inductance, The current flowing in the circuit will be, 23. A fully charged capacitor C with initial, 1) 6.9 x 10-8A 2) 6.9A 3) 6.9 x 10-6A 4) zero, charge q0 is connected to a coil of self, 1000, inductance L at t 0 . The time at which the, 15. In a circuit, the frequency is f , Hz and, 2, energy is stored equally between the electric, the inductance is 2 henry, then the reactance, and the magnetic fields is, will be, , LC 2) 2 LC 3) LC, 1), 4) LC, 4, 1) 200 2) 200 3) 2000 4) 2000, TRANSFORMER, 16. The transformer ratio of a transformer is 10:1., The current in the primary circuit if the secondary, current required is 100 A assuming the, transformer be ideal, is, 1) 500 A 2) 200 A 3) 1000 A 4) 2000 A, 98, , LEVEL-I (H.W) - KEY, , 1) 3, 8) 2, 15) 3, 22) 3, , 2) 3 3) 2 4) 3 5) 2 6) 1 7) 2, 9) 2 10) 2 11) 3 12) 2 13) 1 14) 1, 16) 3 17) 3 18) 1 19) 1 20) 2 21) 2, 23) 1, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 22.. Ans : (c), , LEVEL-I (H.W) - HINTS, 2 E0, T, , T /2, , sin t dt , , 2 E0, , , 4, 3, , i 5 sin t cos t 5 sin t , 5, 5, , , 1, s, 600, , rms value is, , 1., , Eav , , 2., , V 10sin 100 t ;, , 3., , i i0 sin t 2 irms sin 2 ft , , 4., , V0 2.Vr .m.s ., , 5., , Form factor , , 9., , , , 0, , t, , rms value, avg value over half a cycle, , Mean value can not be decided., Here current leads voltage so, it is RC circuit, 23. As initially charge is maximum, q q0 cos t, i , , I, Irms 0, 2, , T, 1, , 4 4f, 11. V = Vm sin t compare to given equation, we get, Vm = 240 and 120, , q cos t , 1, 2, L q0 sin t 0, 2, 2C, , 120, , 19 H 2 and, 2 6.28, , voltage =, , Vm, 2, , , , 240, 170V, 2, , t , , i0, 2, , 1., , 1000, 2 2000, 2, , 18. f , , 1, 2 LC, , 10, 5, 10, 20 2, 2 A 2), 2 A 3), A, A 4), , , , , , 2., , A 100 resistance is connected in series with, a 4H inductor. The voltage across the resistor, is VR 2sin 1000t V . The voltage across the, inductor is, , R, Z, , 20. Impedance, Z = R+XC+XL, , , , , , LC, 4 4, , A.C ACROSS L-R, L-C &, L-C-R SERIES CIRCUITS, , N S Vs, 17. N V, p, p, , , , t, , The average current of a sinusoidally varrying, alternating current of peak value 5A with initial, phase zero, between the instants t = T/8 to t =, T/4 is ( Where 'T' is time period), 1), , NS I p, 16. N I, p, s, , 19. cos , , tan t 1, , INSTANTANEOUS, PEAK,R.M.S &, AVERAGE VALUES OF A.C AND A.V, , E, rms, Xc, , 15. X L L 2 fl 2 , , , 4, , 2, , LEVEL-I (H.W), , 2 Erms, E0, , R, R, , 13. E E0 cos t , i i0 cos 2 ft but i , 14. irms, , 1, LC, , But, , , 12. i i0 sin t, , 2 f ; i0 , , dq, q0 sin t, dt, , 1 2 q2, Given Li , 2, 2C, , 10. t , , f , , 5, 2, , , , 5i 7 j 9 j 5i 2 j, , | Z | 5 4 9 3, , , , 1) 80 sin 1000t 2 , , , , , , , , 2) 40 sin 1000t 2 , , , , 3) 80 sin 1000t 2 , , , , 4) 40 sin 1000t 2 , , , , , , , , , , , , , , , , , , , , 21. Vrms 162 202 656 25.6V, NARAYANAGROUP, , 99
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 3., , The reading of voltmeter and ammeter in the 8., following figure will respectively be, , A, XC = 4, , V, , The figure shows variation of R, XL and XC, with frequenc f in a series L, C, R circiut. Then, for what frequency point, the circiut is inductive, , 90V, , XL, , XL = 4, , 4., , R = 45, 1) 0 and 2A, 2) 2A and 0V, 3) 2V and 2A, 4) 0V and 0A, In the following circuit, the values of current, flowing in the circuit at f = 0 and f = will 9., respectively be, , XC, , 0.01H, , 10–5F 25, , R, , A, , B, , C, , f, , 1) A, 2) B, 3) C, 4) All points, A constant voltage at different frequencies is, applied across a capacitance C as shown in the, figure. Which of the following graphs correctly, depicts the variation of current with frequency, C, A, , 200V, , 5., , A.C. generator, , 1) 8A and 0A, 2) 0A and 0A, 3) 8A and 8A, 4) 0A and 8A, In the series L-C-R circuit figure the voltmeter, and ammeter readings are, 400V, , I, , I, , 1), , 2), , 400V, , , , , , V, I, , I, , R = 50, , L, , C, , A, , 3), 100V 50Hz, , 6., , 7., , 100, , 4), , , , , 10. In a series L C R circuit R 200 and the, voltage and the frequency of the main supply, is 220 V and 50Hz respectively. On taking out, the capacitance from the circuit the current, lags behind the voltage by 300 . On taking out, the inductor from the circuit the current leads, the voltage by 300 . The power dissipated in, the L C R circuit is, VR, 1) 305 W 2) 210 W 3) zero, 4) 242 W, 1) 2VR 2) VR, 3), 4) 5VR, 2, 11. In a series resonant LCR circuit, the voltage, A 220V, 50Hz a.c. generator is connected to, across R is 100V and R 1 k with C 2 F ., The resonant frequency is 200 rad/s. At, an inductor and a 50 resistance in series., resonance, the voltage across L is, The current in the circuit is 1.0A. The P.D., 1) 2.5 10 2 V 2) 40 V 3) 250 V 4) 4 103V, across inductor is, 1) 102.2V 2) 186.4V 3) 213.6V 4) 302V, , 1) V=100 volt, I=2A 2) V=100 volt, I = 5 A, 3) V=1000 volt, I=2A 4) V=300 volt, I = 1 A, The potential difference between the ends of, a resistance R is VR, between the ends of, capacitor is VC = 2VR and between the ends of, inductance is VL =3VR. Then the alternating, potential of the source in terms of VR will be, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , LEVEL-II (C.W) - KEY, 1) 1, 8) 3, , 2) 1, 9) 2, , 3) 1 4) 2 5) 1, 10) 4 11) 2, , 6) 1, , 7) 3, , LEVEL-II (C.W) - HINTS, T /4, , 1., , , i , , , idt, , T /8, T /4, T /8, , dt, , V0 R, , , , , VL V0 L sint and V0 L XLi, R, 2, , , 2., , i, , 3., , E, Irms rms 2A ; Vrms Irms (X L XC ) 0, R, , 5., , I , , E, , Z, , I r .m.s . , , E, , 1 , R 2 2 f L , , 2, f C , , , 1., , 2, , V r .m . s . V r .m . s . 100, , , 2A, Z, R, 50, , V VR2 VL VC , 6., , LEVEL-II (H.W), INSTANTANEOUS, PEAK,R.M.S &, AVERAGE VALUES OF A.C AND A.V, , circuit is at resonance, , 4., , ALTERNATING CURRENT, 1, 11. At resonance, L , C, current flowing through the circuit, V, 100, I R , 0.1A, R 1000, So, voltage across L is given by, I, VL I X L I L, but L , , 0.1, I, , 250V, VI , C 200 2 106, , 2, , A.C ACROSS L-R, L-C &, L-C-R SERIES CIRCUITS, , V S VB VC VL VR i 2VR j 3VR j, , 2., , VR i VR j , V 2VR, , 7., , I, , E, 220, , I=, , Z=220, Z, Z, , 3., , Z 2 R 2 X L2 X L Z 2 R 2, L, , 1, Z 2 R2, , , L , , 1, 2 f, , Z 2 R 2 0.68 H, , 8., , VL LI 2 0.5 0.68 1 213.6 V, At A : XC > XL ; At B : XC = XL ; At C : XC < XL, , 9., , For capacitive circuits X C , i , , 1, C, , V2, 242W, R, , NARAYANAGROUP, , In an LR circuit, R = 10 and L = 2H. If an, alternating voltage of 120V and 60Hz is, connected in this circuit, then the value of, current flowing in it will be _______ A (nearly), 1) 0.32, 2) 0.16, 3) 0.48, 4) 0.8, The equation of an alternating current is, I 50 2 sin400t A, then the frequency and, the root mean square value of the current are, respectively., 1) 200Hz, 50 A, 2) 400Hz, 50 2A, 3) 200Hz, 50 2A, , 4., , V, V C i , XC, , 10. The given circuit is under resonance as X L X C, Hence, power dissipated in the circuit is, , P, , An alternating current 'i' is given by, i i0 sin 2 (t / T 1 / 4) . Then the average, current in the first one quarter time period is, 2i, I, I, 3I, 1) 0, 2) 0, 3) 0, 4) 0, , , 2, , , 5., , 4) 500Hz, 200A, , 360, Hz contains a 1 F, 2, capacitor and a 20 resistor. The inductor, must be added in series to make the phase, angle for the circuit zero is, 1) 7.7 H 2) 10 H, 3) 3.5 H 4) 15 H, A resistor R and capacitor C are connected in, series across an AC source of rms voltage 5, V. If the rms voltage across C is 3 V then that, across R is, 1) 1V, 2) 2 V, 3) 3 V, 4) 4 V, , A circuit operating at, , 101
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 6., , 7., , 8., , An LCR series circuit containing a resistance, TRANSFORMER, of 120 has angular resonance frequency 13. The efficiency of a transformer is 98%. The, primary voltage and current are 200 V and 6A., 4 105 rad S 1 . At resonance the voltage, If the secondary voltage is 100 V, the, across resistance and inductance are 60V and, secondary current is, 40V respectively. Then the values of L and C, 1) 11.76 A 2) 12.25 A 3) 3.06 A 4) 2.94 A, are respectively., 1) 0.2 mH ,1/ 32 F 2) 0.4 mH ,1/16 F, LEVEL-II (H.W) - KEY, 3) 0.2 mH ,1/16 F 4) 0.4 mH ,1/ 32 F, 1) 1 2) 2 3) 1 4) 1 5) 4 6) 1 7) 3, The natural frequency of an LC - circuit is, 8) 2 9) 4 10) 2 11) 4 12) 2 13) 3, 1,25,000 cycles per second. Then the capacitor, C is replaced by another capacitor with a, LEVEL-II (H.W) - HINTS, dielectric medium of dielectric contant k. In, T /4, this case, the frequency decreases by 25 kHz., o idt, The value of k is, , i, , T /4, 1., 1) 3.0, 2) 2.1, 3) 1.56, 4) 1.7, dt, , o, In the given figure, the instantaneous value of, E, E, alternating e.m.f. is e = 14.14 sin t . The, I , 2., Z, 2, reading of voltmeter in volt will be, R 4 2 f 2 L2, XL, 300, , 3., , A, , V, R, , XC, , 200, , 1) 141.4 2) 10, 3) 200.0 4) 70.7, A coil of inductance 0.1H is connected to 50V,, 100Hz generator and current is found to be, 0.5A. The potential difference across, resistance of the coil is, 1) 15V, 2) 20V, 3) 25V, 4) 39V, 10. The voltage of A.C. source varies with time, according equation. V = 120 sin 100t cos, 100 t. Then the frequency of source is, 1) 50Hz 2) 100Hz 3) 150Hz 4) 200Hz, 11. The current in a coil of self inductance 5 henry, is increasing according to i = 2 sin 2 t . The, amount of energy spent during the period when, current changes from 0 to 2 amperes is, 1) 10J, 2) 5J, 3) 100J, 4) 2J, 12. In an AC circuit the voltage applied is, E E0 sin t . The resulting current in the, , and I 0 50 2, , 9., , , , circuit is I I 0 sin t 2 ., , , , 3) P , 102, , E0 I 0, 2, , I rms , , 4) P 2 E0 I 0, , I0, 2, , 50 A, , 1, C, , 4., , L , , 5., , E ER2 EC2, At resonance, , 6., , i, , 7., 8., , V VL, , X L, , R XL ; L, , 9., , 1, LC, , n1, n1, C2, Kc, ;n C c ; n K, 2 LC 2, 2, 1, Reading of voltmeter in rms value, , n, , E, , The power, , consumption in the circuit is given by, E0 I 0, 1) P , 2) P= zero, 2, , I 50 2 sin 400 t ; I = I0 sin t, Comparing two equations, we get, 2 f 400 ; f = 200Hz, , 1, , 14.14, 10V, 2, , E, 50, ; 0.5= Z 100, Z, Z, 2, 2, 2 2 , then R =, 78, Z R L, I, , Now VR VLR2 VL2 39V ; VR2 VL2 VLR2 , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , 4., 10. V = 120sin 100 t cos100 t, = 60 x 2 sin100 t.cos100 t, = 60sin 200 t and 200, n 100cps, 11. Energy E = 1/2LI2, When I = change of current from minimum to, maximum. = 2–0 = 2J., 12. For given circuit current is lagging the voltage by 5., / 2 so circuit is purely inductive and there is no, power consumption in the circuit. The work, done by battery is stored as magnetic energy in, the inductor., Vs I s, 98 100 is, 13. V I 100 200 6, p p, , 2., , 3., , An AC voltage source of variable angular, , 6., , frequency and fixed amplitude V0 is, connected in series with a capacitance C and, an electric bulb of resistance R (inductance, zero). When is increased, 1) The bulb glows dimmer, 7., 2) The bulb glows brignther, 3) Total impedance of the circuit is unchanged, 4) Total impedance of the circuit increases, In an A.C circuit the instantaneous values of, current and voltage are I 120 sin t ampere, and E 300sin t / 3 volt respectively.., 8., What will be the inductive reactance of series, LCR circuit if the resistance and capacitive, reactance are 2 ohm and 1 ohm respectively?, 1) 4.5 ohms 2) 2 ohms 3) 2.5 ohms 4) 3 ohms, A pure resistive circuit element 'x' when, connected to an A.C. supply of peak voltage, 100 V gives a peak current of 4 A which is in, phase with the voltage. A second circuit, element ‘y’ when connected to the same AC, supply also gives the same value of peak, current but the current lags behind by 900 . If, 9., the series combination of 'x' and 'y' is, connected to the same supply. R.M.S. value, of current is, 1), , 5, A, 2, , 2) 2A, , NARAYANAGROUP, , 100 2 V , 40 Hz supply, the current through, the circuit will be, 1) 10 A, 2) 12.5 A 3) 20 A, 4) 25 A, A circuit containing resistance R1 , Inductance, , L1 and capacitance C1 connected in series, resonates at the same frequency 'n' as a, second combination of R2 , L2 and C2 . If the, two are connected in series. Then the circuit, will resonates at, 1) n, , LEVEL-III, 1., , An ideal inductor takes a current of 10 A when, connected to a 125 V, 50 Hz AC supply. A pure, resistor across the same source takes 12.5 A., if the two are connected in series across a, , 3) 1/2 A, , 4), , 2, A, 5, , 2) 2n 3), , L2C2, L1C1, , 4), , L1C1, L2C2, , An AC source of variable frequency is applied, across a series L-C-R circuit. At a frequency, double the resonance frequency. The, impedance is 10 times the minimum, impedance. The inductive reactance is, 1) R, 2) 2R, 3) 3R, 4) 4R, A 20V, 750 HZ source is connected to a series, combination of R = 100 , C = 10 F and, L = 0.1803 H. Calculate the time in which, resistance will get heated by 10o C . (If thermal, capacity of the material = 2 J / o C ), 1) 328 sec 2) 348 sec 3) 3.48 sec 4) 4.32 sec, An AC source of angular frequency is fed, across a resistor R and a capacitor C in series., The current registered is I. If now the, frequency of source is changed to / 3 (but, maintaining the same voltage), the current in, the circuit is found to be halved. The ratio of, reactance to resistance at the original, frequency is, 3, 5, 3, 5, 2), 3), 4), 5, 3, 5, 3, An LCR circuit has L = 10 mH, R = 3 , and C, = 1 F connected in series to a source of, , 1), , 15 cos t volt. The current amplitude at a, frequency that is 10% lower than the resonant, frequency is, 1) 0.5 A, 2) 0.7 A 3) 0.9 A 4) 1.1 A, 103
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, , 10. In the given circuit, R is a pure resistor, L is a 16. The potential difference across a 2H inductor, pure inductor, S is a 100V, 50 Hz AC source,, as a function of time is shown in figure. At, time t = 0, current is zero., and A is an AC ammeter. With either K1 or, Current t = 2 second is, K alone closed, the ammeter reading is I. If, 2, , VL(volt), , the source is changed to 100 V, 100 Hz, the, ammeter reading with K1 alone closed and, , 10, , with K 2 alone closed will be respectively.., K1, , K2, , 2, , S, R, , L, , A, , 11., , t(s), , 4, , 1) 1A, 2) 3A, 3) 4A, 4) 5A, 17. For the circuit shown in the figure the rms value, of voltages across R and coil are E1 and E2,, respectively., , 1) I , I / 2 2) I , 2 I, 3) 2 I , I, 4) 2 I , I / 2, A capacitor has a resistance of 1200 M and, L,r, R, capacitance of 22 F . When connected to, Resistor, Coil, an a.c. supply of frequency 80 hertz, then the, alternating voltage supply required to drive a, current of 10 virtual ampere is, e = E sint, 1) 904 2V 2) 904V 3) 904 / 2V 4) 452V, e =E, A 120V, 60Hz a.c. power is connected 800, The power (thermal) developed across the, non-inductive resistance and unknown, coil is, capcitance in series. The voltage drop across, the resistance is found to be 102V, then voltage, E E12, E E12 E22, 1), 2), drop across capacitor is, 2R, 2R, 1) 8V, 2) 102V 3) 63V, 4) 55V, 2, A 100 V a.c source of frequency 50 Hz is, 2, E E1 , E, , connected to a LCR circuit with L = 8.1, 3), 4), 2R, 2R, millihenry, C 12.5 F and R 10ohm , all, connected in series. What is the potential 18. A bulb is rated at 100 V, 100 W, it can be, treated as a resistor. Find out the inductance, difference across the resistance?, of an inductor (called choke coil) that should, 1) 100 V 2) 200 V 3) 300 V 4) 450 V, A coil has an inductance of 0.7H and is joined, be connected in series with the bulb to, in series with a resistance of 220 . When an, operate the bulb at its rated power with the, alternating e.m.f. of 220V at 50 c.p.s. is applied, help of an ac source of 200 V and 50 Hz, to it, then the wattless component of the, , current in the circuit is, H 2) 100 H 3) 2 H 4) 3 H, 1), 3, , , 1) 5 ampere, 2) 0.5 ampere, 19. In the circuit diagram shown,, 3) 0.7 ampere, 4) 7 ampere, Two alternating voltage generators produce, X C 100 , X L 200 & R 100 . The, emfs of the same amplitude E0 but with a phase, effective current through the source is, , C, difference of . The resultant e.m.f.is, 3, 200V, R, 0, , 12., , 13., , 14., , 15., , rms, , , , 1) E 0 sin t , 3, , , 3), 104, , , , 2) E 0 sin t , 6, , , , , 3E 0 sin t 4), 6, , , , , 3E 0 sin t , 2, , , L, , 1) 2 A, , 2) 2 2 A, , 3) 0.5 A, , 4) 0.4 A, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, C1C2, L1C1 L2C2 ; Lnet L1 L2 ; Cnet C C, 1, 2, , LEVEL-III - KEY, 1) 2 2) 1 3) 2 4) 1 5) 1 6) 4 7) 2, 8) 1 9) 2 10) 1 11) 2 12) 3 13) 1 14) 2, 15) 3 16) 4 17) 2 18) 4 19) 2, , LEVEL-III - HINTS, 1., , CC , Lnet Cnet ( L1 L2 ) 1 2 ; L C L C, net net, 2 2, C1 C2 , 6., , In R C circuit, the impedance is, Z R2 , , 10 R 2 R 2 (2o L 1/ 2o C ) 2, , 1, ;, C2, , minimum impedance Zmin = R, , 2, , o2 LC 1 ------- (1), , As increases, Z decreases., Since, Power , , Z 2 R 2 ( L 1/ C )2, , 1, , therefore the bulb, impedance, , 1, 3R ------- (2), 2o C, , 2o L , , glows brighter., 2., , I 120sin t , E 300sin t / 3, Clearly, / 3 ,, R, 1, Now, cos cos 60 Z 2 R, Z, 2, , 7., , from(1), , 1, R, 2o C, , from(2), , X C 2o L 3R R 4 R, , XC , , As R 2, Z 2 2 4; X C 1, 2, , 2, , 2, , 2, , 1, 21.2, 2nc, , Z R 2 ( X L X C )2 835, , 2, , Now X L X C Z R 4 2 12, , IV Ev / z 0.0239 A, , X L X C 12 2 3, , 2, , IV Rt (ms) t , , X L X C 2 3 1 3.464, Taking + value, X L 1 3.464 4.465 , 3., , XL , , 4., , ', at frequency / 3, X C , , 4/ 2, 2A, 2, , V, 1, V, L, R 10, ,, 8, iL, iR, , For 40 Hz, 100 2 V supply, i, , n, , V, R 2 X L2, , , , V, R 2 4 2 f 2 L2, , 1, 1, , 2p L1C1 2p L2 C2, , NARAYANAGROUP, , (ms), 2, IV R, , at frequency , X C 1/ C, , I, , V, 2, , R X, , For 50 Hz and 125 V supply, X L L , , 5., , 8., , o, , =25 ; R o 25 ; Z= R 2 X 2 ;, C, Io, Io, , I 01 o / Z 4 / 2 A; I r .m.s. I 01 / 2 , , XC R, , 9., , cv , , 2, C, , ;, , I, , 2, , 3, 3X C, C, , V, , XC, 3, , ;, R 9X, R, 5, 2, , 2, C, , 90, 1, 90, x, cv0 , =9000 rad/s, 100, LC, 100, E0, , i0 , , 1 , , R2 L , C , , , 2, , 10. In the second case induction reactance becomes 2, times thus current through L when K2 is closed, i, . But current through R when K1 is, 2, closed does not change, , becomes, , 105
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, 11. f = 80Hz, IV = 10A, EV, EV, , Current through R, I R , R 12 108, EV, Current through C I C X 2 fC EV, C, 2 80 22 10 6 EV, , = 352 10 5 EV, , (102 ), , IV2 I R2 I C2, , EV2, (352 105 EV )2, (12 108 ) 2, , , , 220 L, 22, 0.7 2 50, 2, 2 2, (R L ), 7, , , , 220 (0.7 2 50), 220, (2202 2202 ), , , , 220 220 1, 0.5, 220 2 (2) 2, , 15. E1 = E0 sin t; E 2 E0 sin(t / 3), E = E2 + E1, = E0 sin(t / 3) E0 sin t, = 2 E0 sin(t / 6) cos( / 6), , 1, , , EV2 , 1.2 104 , 16, 144 10, , , =, , 3E0 sin(t / 6), , 4, , EV2 , , 100 10, EV 904 volt, 1.2, , 16. e L, , 12. V 2 VR2 VC2, 2, C, , 2, , di, e dt L i2 i1 , dt, , e dt area of le for t 0 to 2 sec., 17. Draw the phasor diagram., , 2, R, , V V V, , E 2 E12 E 22 2E1E 2 cos ., , VC2 (120) 2 (102) 2, VC = 63V, , E2, E2, , 13. Here, E 100V , v 500 Hz, , E, , L 8.1 103 H , C 12.5 106 F , R 10 , , , X L L 2 vL 1000 8.1 10 3 25.4, , E1, , I, , , 1, , XC , C 1000 12.5 106, , Thermal power developed in coil is, P E 2 cos I and, , 103, , 25.4, 12.5, , I, , Z R2 X L X C , , 18. Resistance of bulb is R , , 2, , 2, , 102 25.4 25.4 10, Ev 100, , 10 A, Z, 10, Potential difference across R I v R, Iv , , 10 10 100V, 14. Watt less component of, E, A.C. = IV sin V sin , Z, 220, L, , , R2 L22, R2 L22, , E1, EE, E 2 E12 E 22, P 1 2 cos , R, R, 2R, , Rated current is, In ac, I rms , , 100, 1A, 100, , Vrms, ; Z 200, Z, , 2, 1002 L 200 2 L2 30000 and, , L, , 19. I R , , L0.7 2 50, , 100 100, 100, 100, , 30000, , 100 , , 2, , , , 3, , , henry.., , 200, V 200, 2A, , 2 A ; I LC , X, R 100, L XC, , I 22 22 2 2 A, , as I R inphase with V, , I LC lages behind V by 2, 106, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENT, Ferromagnetic core, , LEVEL-IV, , R, , AC, , 1., , At resonance, VL and VC are both very much, greater than the applied potential, V itself., The quality factor for an LCR circuit in, resonance is given by Q=, , XL, . In practice,, R, , Q 200 has been achieved., L, 200H, 10, , 0.10V, 1.0 MHz, , (a) At resonance, the capacitor has been adjusted, for, 1 200×10-6 µF, , 2. 0.00013µF, , a), 3. 0.0013µF, 4. 0.0013F, (b) At resonance, the potential difference across, the inductance is, 1) 1.3 V 2) 13 V 3) 0.3 V 4) none of these, b), (c) The potential across the capacitance at, resonance is, 1) 13 V 2) > 13 V 3) < 13 V 4) none of these, (d) The Q factor is, c), VL, VC, VC, VL, 1. V, 2. V, 3., 4., V, V, C, L, (e) choose the right statement., 1. VL +VC can be greater than Vapplied, 2. VL +VC =Vapplied, 3. VL +VC <Vapplied, 2., , 1°Coil, , 2°Coil, , The primary coil is connected to a source of, alternating (AC) current. The secondary coil, is connected to a resistor such as a light bulb., The AC source produces an oscillating voltage, and current in the primary coil that produces, an oscillating magnetic field in the core, material. This in turn induces an oscillating, voltage and AC current in the secondary coil., Student collected the following data comparing, the number of turns per coil (N), the voltage, (V) and the current (I) in the coils of three, transformers, Primary Coil Secondary coil, N1 V1 I1, N 2 V2 I2, Transformer 1 100 10V 10 A 20 20 V 5 A, Transformer 2 100 10V 10 A 50 5 V 20 A, Transformer 3 200 10V 10 A 100 5 V 20 A, The primary coil of a transformer has 100 turns, and is connected to a 120V AC source. How, many turns are in the secondary coil if there is, a 2400 V across it, 1) 5, 2) 50, 3) 200, 4) 2000, A transformer with 40 turns in its primary coil, is connected to a 120 V AC source. If 20 W of, power is supplied to the primary coil, the power, developed in the secondary coil is, 1) 10 W 2) 20 W 3) 80 W 4) 160 W, One of the following is a correct expression, for R, the resistance of the load connected to, the secondary coil (pick the correct one), V10, 1) I, 10, , N 20, , N10, , , , , , V10, 2) I, 10, , N10., , N 0, 2, , , , , , V10, 3) I, 10, , N10, , N 20, , , , , , V10, 4) I, 10, , N10, , N 0, 2, , , , , , 2, , 2, , 4. none of these, d). A 12 V battery is used to supply 2.0 mA of, A physics lab is designed to study the transfer, current to the 300 turns in the primary coil of, of electrical energy from one circuit to another, a given transformer. What is the current in the, by means of a magnetic field using simple, secondary coil if N 2 150 turns, transformers. Each transformer has two coils, 1) zero, 2) 1.0 mA 3) 2.0 mA 4) 4.0 mA, of wire electrically insulated from each other, but wound around a common core of, LEVEL-IV - KEY, ferromagnetic material. The two wires are, 1) a) 2 b) 3 c) 1 d) 3,4 e) 4, close together but do not touch each other., 2) a) 4 b) 2 c) 1 d) 4, , NARAYANAGROUP, , 107
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 7., , LEVEL - V, SINGLE ANSWER QUESTIONS, 1., , 2., 3., , 4., , The value of current in two series L C R, circuits at resonance is same when connected, across a sinusodial voltage source. Then:, A. both circuits must be having same value of, capacitance and inductor, B. in both circuits ratio of L and C will be same, C. for both the circuits X L / X C must be same at, that frequency, D. both circuits must have same impedance at all, frequencies, The series RLC circuit in resonance is called:, A. Selector circuit, B. rejector circuit, C. amplifier circuit, D. oscillator circuit, In a series R-L-C circuit, the frequency of the, source is half of the resonance frequency. The, nature of the circuit will be, a) capacitive, b) inductive, c) purely resistive, d) selective, The graphs given below depict the dependence, of two reactive impedances X 1 and X 2 on the, frequency of the alternating e.m.f. applied, individually to them. We can then say that, , 8., , The r.m.s. value of an ac of 50 Hz is 10 amp., The time taken by the alternating current in, reaching from zero to maximum value and the, peak value of current will be, A) 2 × 10–2 sec and 14.14 amp, B) 1 × 10–2 sec and 7.07 amp, C) 5 × 10–3 sec and 7.07 amp, D) 5 × 10–3 sec and 14.14 amp, The voltage time (V-t) graph for a triangular, wave having peak value Vo is as shown in, figure. The rms value of V is, V, V0, T, , T/4, , t, , 2T, , V0, , A), 9., , Vo, 3, , B), , Vo, 2, , C), , Vo, , D), , 2, , Vo, 3, , The average value for the saw-tooth voltage, of peak value of V0 as shown in figure is, V0, V, , T/2, , T, , 3T/2, , t, , X1, , Impedance, , Impedance, , V0, , Frequency, , 5., , X2, Frequency, , A. X 1 is an inductor and X 2 is a capacitor, B. X 1 is a resistor and X 2 is a capacitor, C. X 1 is a capacitor and X 2 is an inductor, D. X 1 is an inductor and X 2 is a resistor, In which of the following electrical appliances, will AC fail to function where DC in normally, used?, A. electric light, B. voltmeter, C. solenoid for electromagnet D. a cathode ray tube, , RMS & Average value of Alternating current, 6., , The instantaneous value of current and emf, in an AC circuit are I , , V0, V0, 2V, V, b), c) 0, d) 0, 3, 2, 3, 2, 10. An alternating voltage is given by:, e e1 sin t e2 cos t . Then the root mean, square value of voltage is given by:, a), , e1e 2, e12 e 22, A) e e B) e1e2 C), D), 2, 2, 2, 11. If i t , 0 t T then rms value of current is, 2, 1, , 2, 2, , T2, A), C), D), 2, 5, 5, 12. The rms current value of a semicircular wave, which has a maximum value i0 is, T2, , T2, , T2, B), 2, , 1, sin314t amp and, 2, , Y, , , , E 2 sin 314t V , respectively. The, 6, , , phase difference between E and I (with respect, to I) will be, A) , 108, , , , , , rad B) rad C), rad D) rad, 6, 3, 6, 3, , -i0, , A., , i0, 3, , +i0, , B., , 2, i0, 3, , C., , i0, 2, , X, , D. 2i0, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 13. The rms and the average value of the voltage, wave shown in figure are, V, 4, 2, 0, , 1, , 2, , t, , 3, , -2, , 16. Calculate the reading which will be given by a, hot-wire voltmeter if it is connected across the, terminals of a generator whose voltage, waveform is represented by, v 200sin t 100sin 3t 50sin 5t, A. 110V B. 162V C. 200V D. 220V, 17. The current ‘i’ in an inductance coil varies with, time ‘t’ according to following graph, i, , -4, , 32, V;1V, 3, 11, V;3V, C., 3, , 11, V;1V, 3, 32, V;3V, D., 3, , A., , B., , 14. If i1 i01 sin t , i2 i02 sin t , then, i3 , , t, , (0, 0), , Which one of the following figures shows the, variations of voltage in the coil, V, , i1, , V, , A), , B), (0, 0), , i2, 2, 01, , t, , (0, 0), , V, , t, , V, , 2, 02, , i i sin t, , A., , C), , , , B. i01 i02 sin t , 2, , , , , , , , D. , , i021 i022 2i01 i02, , , cos sin t , , i02 sin , 1, where Tan i i cos , 01 02, , , B, , A, , C, , T, Time, , Vm, , 2, V, Vm and m, , 2, Vm, 2, C. Vm and, , 2, , A., , NARAYANAGROUP, , 2, , 3, , Phase, , Vm, V, and m, , 2, Vm, V, and m, D., 2, 2, , B., , t, , y, , 10, , V, , , , D), , 18. Find the rms and average value of the waveform shown in figure., , 20, , 15. The average and effective values for the, waveshape shown in the figure are:, , 0, , t, , (0, 0), , i021 i022 2i01 i02 cos sin t , , C., , (0, 0), , 2T, , t, , A. 8.5, 10 B. 10.3, 20 C. 15.2, 15 D. 2.6, 5, 19.. Determine the rms value of a semi-circular, current wave which has maximum value of a., A. 2.515 a B. 1.815 a C. 0.615 a D. 0.816 a, 20. An electric bulb is designed to operate at 12, volts DC. If this bulb is connected to an AC, source and gives normal brightness, what, would be the peak voltage of the source ?, A. 37 V B. 17 V, C. 18 V D. 10 V, 109
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 21. The current in a discharging LR circuit is given, , I, , I, , t / , , where is the time constant of, by i i0 e, the circuit. Calculate the rms current for the, period t 0 to t ., A., , i0, e, , e, , 2, , , , 1, , B., , 2, , i0, e, , e, , 2, , i0, e, , e 2, , , , , , , , 24. The resonance point in X L f and X C f, curves is, , 2, , , , D., , D), , 1, , , , 2, i0 e 2, 3, e, 2, 22. The voltage across a pure inductor is, represented by the following diagram. Which, one of the following diagrams will represent, the current, , C., , C), , XL, P, , R, , S, , Q, , f, , XC, , V, , A. P, , B. Q, , C. R, , D. S, , AC across L-R, L-C,L-C-R circuits, t, , i, , i, , A), , B), , t, , i, , t, , i, , C), , D), , t, , t, , 23. A constant voltage at different frequencies is, applied across a capacitance C as shown in, the figure. Which of the following graphs, Signal, Generator, , , , C, , V, , A, , Correctly depicts the variation of current with, frequency ?, I, , I, , A), , B), , , 110, , , , 25. When 100 volt DC is applied across a solenoid,, a current of 1.0 amp flows in it. When 100 volt, AC is applied across the same coil, the current, drops to 0.5 amp. If the frequency of the AC, source is 50 Hz, the impedance and inductance, of the solenoid are, A) 200 ohm and 0.55 henry, B) 100 ohm and 0.86 henry, C) 100 ohm and 1.0 henry, D) 100 ohm and 0.93 henry., 26. A coil having an inductance of 1 / henry is, connected in series with a resistance of, 300 . If 20 volt from a 200 cycle source are, impressed across the combination, the value, of the phase angle between the voltage and, the current is :, 1 5, 1 4, 1 3, 1 4, A) tan, B) tan, C) tan, D) tan, ., 4, 5, 4, 3, 27. In a circuit containing an inductance of zero, resistance, the current leads the applied a.c., voltage by a phase angle at, A) 90o, B) -90o, C) 0o, D) 180o, 28. In RLC circuit, at a frequency , the potential, difference across each device are VR max =, 8.8 V, VL max 2.6 V and VC max 7.4 V ., The composed potential difference, VC VL max across inductor and capacitor, is, A) 10 V B) 7.8 V C) 7.4 V D) 4.8 V, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 29. The natural frequency of the circuit shown in, the figure is, C, , C, , 34. In the a.c. circuit shown in figure, the supply, voltage has a constant r.m.s. value but variable, frequency f. Resonance frequency is, 1, H, 2, , L, , A), , 1, F, 2, , L, , 1, 1, B), 2 LC, LC, , C), , 2, D) none, LC, , 30. If the phase difference between voltage and, current is / 6 and the resistance in the circuit, is 300 , then the impedance of the circuit, will be, A) 40 B) 20 C) 50 D) 13 , 31. In the circuit as shown in the figure, if value of, R 60 , then the current flowing through the, condenser will be, R, , L, , C, , 15V, , 20V, , 10V, , , A) 0.5 A B) 0.25 A C) 0.75 A D) 1.0 A, 32. The power in ac circuit is given by P =, ErmsIrms cos .The vale of, LCR circuit at resonance is:, , cos, , in series, , 1, 1, A) zero, B) 1, C), D), 2, 2, 33. A generator with an adjustable frequency of, oscillation is connected to resistance, R =, 100 , inductances, L1 = 1.7 mH and L2 = 2.3, mH and capacitances, C1 4 F , C2 2.5 F, and C3 3.5 F . The resonant angular, frequency of the circuit is, L1, , , A) 10 Hz B) 100 Hz C) 1000 Hz D) 200 Hz, 35. In a LCR circuit capacitance is changed from, C to 2C. For the resonant frequency to remain, unchanged, the inductance should be change, from L to, A) 4L, B) 2L, C) L/2, D) L/4, 36. In an a.c. circuit V and I are given by, V = 100 sin 100t volts;, , , I 100 sin 100t mA ., 3, , , , The power dissipated in the circuit is, A) 104 watt B) 10 watt C) 2.5 watt D) 5 watt, 37. A series combination of R, L, C is connected, to an a.c. source. If the reistance is 3 and, the reactance is 4 , the power factor of the, circuit is, A) 0.4, B) 0.6, C) 0.8, D) 1.0, 38. If a current I given by I0 sin(t / 2) flows, in an ac circuit across which an ac potential of, E 0 sin(t ) has been applied, then the power, consumption P in the circuit will be, A) E0 I0 / 2 B) E0 I0 / 2 C) EI / 2 D) Zero, 39. In a series CR circuit shown in figure, the, applied voltage is 10 V and the voltage across, capacitor is found to be 8V. Then the voltage, across R, and the phase difference between, current and the applied voltage will, respectively be, 4, A) 6V, tan–1 , 8V, VR, 3, , R, , , , C1, , C2, , C3, , L2, , A) 0.5 rad/s, , B) 0.5 104 rad / s, , C) 2 rad/s, , D) 2 104 rad / s, , NARAYANAGROUP, , , 10V, , 3, B) 3V, tan–1 , 4, 5, C) 6V, tan–1 , 3, 1 4 , D) 3V , tan , 3, 111
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 40. An inductor of inductance L and resistor of, resistance R are joined in series and, connected by a source of emf V and frequency, . Power dissipated in the circuit is, , ( R 2 2 L2 ), A), V, , V2R, B), ( R 2 2 L2 ), , V, R 2 2 L2, C), D), ( R 2 2 L2 ), V2, 41. Power loss in AC circuit will be minimum when, A) Inductance is high, resistance is high, B) Inductance is low, resistance is high, C) Inductance is low, resistance is low, D) Inductance is high, resistance is low, 42. In the circuit, as shown in the figure, if the, value of R.M.S current is 2.2 ampere, the, power factor of the box is (E=220V), 100, , C, , 1/ Henry, Box, , , 1, 1, 3, B) 1, C), D), 2, 2, 2, 43. The impedance of a series RL circuit is same, as the series RC circuit when connected to the, same AC source separately keeping the same, resistance. The frequency of the source is, , A), , 1, 1, R, 1, B), C), D), LC, 2 LC, L, RC, 44. A current source sends a current, A), , i i0 cos t . When connected across an, unknown load gives a voltage output of,, , 112, , , , , Load, , , v = v0 sin (t + /4), , i0 cos t, , v v0 sin t / 4 across that load. Then, voltage across the current source may be, brought in phase with the current through it, by:, , a) Connecting an inductor in series with the load, b) Connecting a capacitor in series with the load, c) Connecting a capacitor in parallel with the load, d) Connecting a capacitor in parallel with the load, 45. A combination of elements is enclosed in a, black box and the voltage and currents are, measured across this black box. The, expression for applied voltage; and the current, flowing in it is V V0 sin t ;, i 2 2 sin t / 4 where 100 rad/, sec. Then the wrong statement is:, , , , Black, box, , a) There must be a capacitor is the black box, b) Power factor of circuit = 0.707, c) There must be a resistor in the box, d) There must be an inductor in the box, 46. A high impedance AC voltmeter is connected, in turn across the inductor, the capacitor and, the resistor in a series circuit having an AC, source of 100 V(rms) and gives the same, reading in volts in each case. This reading is:, a) 100 V b) 141 V c) 150 V d) 200 V, 47. In a black box of unknown elements (L, C or, R or any other combination) an AC voltage, , E E0 sin t is applied and current in the, circuit, was, found, to, be, i i0 sin t / 4 . Then the unknown, elements in the box may be:, Z, , , A) only capacitor B) inductor and resistor both, C) either capacitor, resistor and inductor or only, capacitor and resistor, D) only resistor, 48. A given alternating current has an rms value, of 5.6 ampere. If this current flows in a curcuit, containing 10 of resistance in series with, 20 of inductive reactance, the power, consumed in the circuit will be, A) 313.6 W B) 940.8 W C) 627.2 W D) 168 W, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 49. In an a.c.circuit, V & I are given by, V = 100 sin (100 t) volt., , 54. If the reading of the voltmeters vary with time, , , as: V1 20sin t and V2 20 cos t , 6, , then the unknown circuit element x is a:, , , , I 100sin 100t mA, 2, , The power dissipated in the circuit is:, A) 1 watt B) 10 watt C) zero, D) 5 watt, 50. In R-L-C series circuit, we have same current, , V1, , at angular frequencies 1 and 2 . The, resonant frequency of circuit is, , X, R, , 12, 22, A., B., C. 12 D. 1 2, 2, 1, 51. A choke coil of resistance R and inductance L, is connected to an A.C. source of frequency f, , , A. pure (or ideal) inductor B. practical inductor, C. pure (or ideal) capacitor D. practical capacitor, , and maximum voltage V0 . Then, the average, power dissipated in the choke is proportional, to:, A. f 2, , B. f 2, , C. f 1, , D. f 0, , 52. When two A.C. generators of emfs V1 and V2, and same frequency connected in series, the, emf across A and B is ( phase angle, difference between the generators):, , A., , 2, V, Vm and m, , 2, , B., , C., , V, 2, Vm and m, , 2, , D., , v1, , Vm, 2, , and, , Vm, 2, , and f 2 of AC source, the current amplitude, , B, , v2, , Vm, V, and m, , 2, , 55. In a series LCR circuit, at the frequencies f1, , falls to, , A, , V2, , 1, of the current amplitude at, 2, , resonance. Then the value of f 2 f1 is, , V V, A. 1 2, 2, , B. V1 V2, , R, R, R, R, B., C., D. 2, 2L, L, L, L, 56. In the figure, which of the phasor diagram, represents RLC circuit driven at resonance?, , A., , D. V12 V22, V12 V22 2VV, 1 2 cos , 53. At resonance of the given series R-L-C, circuit:, C., , V1, R, , VL, , L, , V2, , C, , VR, , VL, , A., , VR, , B., , V3, , VC, , VC, , V, VL, , , C., , 2, , NARAYANAGROUP, , VL, VR, , A. V 3 V1 V2 V32 B. V3 0, C. V1 0, , VR, , VC, , D., VC, , D. V2 0, 113
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 57. In LCR circuit at resonance current in the, circuit is 10 2 A. If now frequency of the, source is changed such that now current lags, by 45º than applied voltage in the circuit., Which of the following is correct, A. Frequency must be increased and current after, the change is 10 A, B.Frequency must be decreased and current after, the change is 10 A, C. Frequency must be decreased and current is, same as that of initial value, D.The given information is insufficient to conclude, anything, 58. A pure resistive circuit element X when, connected to an sinusoidal AC supply of peak, voltage 200V gives a peak current of 5A which, is in phase with the voltage. A second circuit, element Y, when connected to the same AC, supply also gives the same value of peak, current but the current lags behind by 900. If, the series combination of X and Y is connected, to the same supply, the rms value of current is, A., , 10, 2, , A, , B., , 5, 2, , A, , C., , 5, A, 2, , D. 5A, , 59. An A.C. circuit contains a resistor 'R', an, inductor 'L' and a capacitor 'C' connected in, series. When it is connected to an A.C., generator of fixed output voltage and variable, frequency, the current in the circuit is found, , to be leading the applied voltage rad,when, 4, the frequency is f1 . When the frequency of, the generator increased to f 2 the current is, found to be lagging behind the applied voltage, , by rad. The resonant frequency of the circuit, 4, is, f1f 2, 2f1f 2, f f, A. f f, B. 1 2 C. f f, D. f1f 2, 2, 1, 2, 1, 2, 60. When an AC source of emf E = E0 Sin (100t), is connected across a circuit, the phase, difference between the emf E and the current, , I in the circuit is observed to be , as shown, 4, in the figure. If the circuit consists only of R–, C(or)R–L(or) L–C series. Possible values of, the elements of the circuit are (IIT-scre-2003), 114, , i, , e, , t, , A. R = 1K , C = 10 F, B. R = 1K , C = 1 F, C. R = 1K , L = 10H D. R = 1K , L = 1H, 61. A radio tuner has a frequency range from, 500kHz to 5 MHz. If its LC circuit has an, effective inductance of 200 H , what is the, range of its variable capacitor? (Take 2 10 )., , A. 2.5 pF to 250 pF B. 5.0 pF to 500 pF, C. 7.5 pF to 750 pF D. 10 pF to 1000 pF, 62. In a series LCR circuit the frequency of a 10V,, AC voltage source is adjusted in such a, fashion that the reactanace of the inductor, 1., measures 15 and that of the capacitor 11, If R= 3 , the potential difference across the, series combination of L and C will be:, A. 8V, B. 10V, C. 22V, D. 52V, 63. A resistor of resistance 100 is connected to, an AC source e (12 V )sin(250 ps 1 )t. Find, the energy dissipated as heat during t 0 to, t 1.0 ms., A. 0.61104 J, B. 0.61104 J, C. 2.61104 J, D. 2.61106 J, 64. A lamp consumes only 50% of peak power in, an a.c. circuit. What is the phase difference, between the applied voltage and the circuit, current, , , , , A., B., C., D., 6, 3, 4, 2, 65. Voltage and current for a circuit with two, elements in series are expressed as follows :, , , v t 170 sin 6.28 t volts, 3, , , , i t 8.5sin 6.28 t amp, 2, , , What are the values of the elements?, A. R 27.32 , C 25.92 mF, B. R 17.32 , C 15.92 mF, C. R 7.32 , C 5.92 mF, D. R 10.32 , C 5.92 mF, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 66. When an ac source of e.m.f. e E0 sin 100 t , is connected across a circuit, the phase, difference between the e.m.f. e and the current, , i in the circuit is observed to be , as shown, 4, in the diagram. If the circuit consists possibly, only of RC or LC in series, find the, relationship between the two elements, i or e, , 71. An, , AC, , source, , producing, , emf, , 0 cos 100 s 1 t cos 500 s 1 t , , e, , i, , 70. A circuit consisting of a capacitor and a coil, in series is connected to the mains. Varying, the capacitance of the capacitor, the heat, power generated in the coil was increased, n 1.7 times. How much (in per cent) was, the value of cos changed in the process?, A. 80%, B. 25%, C. 50%, D. 30 %, , is, , connected in series with a capacitor and a, resistor. The steady-state current in the, circuit is found to be, , , , A. R 1k , C 10 F B. R 1k , C 1 F, C. R 1k , L 10 H D. R 1k , L 1 H, 67. The figure shows variation of R, X L and X C, with frequency f in a series L, C, R circuit., Then for what frequency point, the circuit is, inductive, XC, , XL, , R, , AB, , f, , C, , A. A, B. B, C. C, D. All points, 68. Which of the following plots may represent the, reactance of a series LC combination, , i i1 cos 100 s 1 t 1 , i2 cos 500 s 1 t 2 , A. i1 i2, B. i1 i2 C. i1 i2, D. the information is insufficient to find the relation, between i1 and i2 ., 72. In the circuit shown in the figure, the ac, source gives a voltage V 20 cos 2000 t ., Neglecting source resistance, the voltmeter, and ammeter reading will be, , , , 6, , 5mH, , 4, , A, , Reactance, , a, , V, , c, b, , A. 0V, 0.47A, C. 0V, 1.4A, PASSAGE-1, , B. 1.68V, 0.47A, D. 5.6V, 1.4A, B, , Frequency, C=2F, , d, , A. a, B. b, C. c, D. d, 69. An inductor-coil, a capacitor and an AC source, of rms voltage 24 V are connected in series., When the frequency of the source is varied, a, maximum rms current of 6.0 A is observed. If, this inductor coil is connected to a battery of, emf 12 V and internal resistance 4.0 , what, will be the current?, A. 2 A, B. 1.5 A C. 0.5 A D. 2.5 A, NARAYANAGROUP, , 5F, , A, , R=2 i1, O, +, CL, -, , +, –2t, V =3e, - C, i2 R=3, , C, , L=4H, , D, , 73. The current iL is, 2 t, a) 2 2 1 e A, , 2 t, b) 2 2 1 e A, , 2 t, c) 3 2 1 e A, , 2 t, d) 2 3 1 e A, , 115
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 74. The variation of current in the inductor with, time can be represented as :, i2 (A), , i2 (A), , 4, , 4, , A) 2, , B) 2, t, , C), , i2 (A), , 4, , 4, , 2, , D) 2, , VCD, , 28, , 28, , C) 12, , D) 12, t, , t, i2 (A), , VCD, , PASSAGE-2, An ac generator G with an adjustable, frequency of oscillation is used in the circuit,, as shown., R = 100, L1 = 1.6mH, , t, , t, , 75. The potential difference across inductor VL, is:, a) 8e2tV b) 9e 2 tV c) 16e 2 tV d) 18e 2tV, 76. The variation of potential difference across A, and C with time can be represented as :, VAC, , VAC, , 8, , 4, , A), , 1, , B) 2, t, , 12, VAC, , VAC, 28, , 17, , C) 12, , D), t, , t, , 77. The potential difference across AB VAB is:, 1 3t, A) 8e2tV B) e V C) 17e 2tV D) 16e 2 tV, 2, 78. The variation of potential difference across C, , and D VCD with time can be expressed as:, VCD, , VCD, , 28, , 28, , A) 12, , B) 12, t, , 116, , t, , t, , G, , S, , C1 = 3F, , C2 =, 4.5F, , C3 = 2.5F, , L2 = 2.4mH, , 79. Current drawn from the ac source will be, maximum if its angular frequency is, A) 105 rad/s, B) 104 rad/s, C) 5000 rad/s, D) 500 rad/s, 80. To increase resonant frequency of the circuit,, some of the changes in the circuit are carried, out. Which changes would certainly result in, the increase in resonant frequency ?, A) R is increased, B) L1 is incresed and C1 is decreased, C) L2 is decreased and C2 is increased, D) C3 is removed from the circuit, 81. If the ac source G is of 100 V rating at resonant, frequency of the circuit, then average power, supplied by the source is, A) 50 W, B) 100 W, C) 500 W, D) 1000 W, 82. Aver age ener gy st or ed by t he induct or L 2, (source is at resonance frequency) is equal to, A) zero, B) 1.2mJ, C) 2.4 mJ, D) 4 mJ, 83. Thermal energy produced by the resistance, R in time duration 1 ms, using the source at, resonant condition, is, A) 0J, B) 1m J, C) 100m J, D) not possible to calculate from the given, information, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , LEVEL-V - KEY, 1) C, 8) D, 15) C, 22) D, 29) A, 36) C, 43) B, 50) C, 57) A, 64) B, 71) C, 78) B, , 2) A, 9) D, 16 )B, 23) B, 30) B, 37) B, 44) B, 51) D, 58) C, 65) B, 72) D, 79) C, , 3) A, 10) D, 17) D, 24) C, 31) B, 38) D, 45) D, 52) C, 59) D, 66) C, 73) B, 80) D, , 4) C, 11) C, 18) C, 25) A, 32) B, 39) A, 46) A, 53) C, 60) A, 67) C, 74) A, 81) B, , 5) C, 12)B, 19)D, 26) D, 33) B, 40) B, 47) C, 54) D, 61) A, 68) D, 75) C, 82) B, , 6) A, 13) A, 20) B, 27) B, 34) C, 41) D, 48) A, 55) A, 62) A, 69) B, 76) A, 83) D, , 7) D, 14) D, 21) A, 28) D, 35) C, 42) A, 49) C, 56) C, 63) C, 70) D, 77) C, , LEVEL-V - HINTS, 1., , X L X C at resonance, , 3., , 6., , , e1, e2, e12 e22 sin t, cos t, 2, 2, 2, , e1 e2, e1 e22, , 1 e2, e e12 e 22 sin t where tan e, , 1, , erms , , emax, , e12 e22, , , , 2, , 2, , , , e12 e22, 2, , 1 T 2, T2, i, dt, , W, , T 0, 5, , i rms, , or x 2 y 2 i02 or y 2 i02 x 2, , The series RLC circuit at resonance selects that, current out of many currents whose frequency is, equal to its natural frequency, hence called as, ‘acceptor’ or ‘selector’ circuit., , 1, 2i0, , 2, irms, , , i0, 2, , y dx, , i0, , 2, , idt, dt, , i dt, dt, , 1, Xc , and X L L, C, , 13. iave , , At res , X C X L The circuit capacitive., , 14. Apply principle of superposition, , I, , 1, 2, , , , ; irms , , , , , , , , ; E 2 sin 314t 6 , , sin 314t, , , , 15. V , ave, , Vm sin d , 0, , , , 0, , Veff , , , 2, , , , 1, t , , , 5 103 sec, 2 2 2 f 200, 16Vo2, T, 2, for 0 t ; V T 2, 4, , , , t, , , 16Vo2 T / 4 2, 2 t dt, T 0, V2, 2, V 0 T / 4 , o, T/4, 3, , dt, , , , , , Vm, 2, , 16. Since hot wire voltmeter reads only rms value we, will have to find rms value of the given voltage., Considering one complete cycle,, , I Io sin t, , 4V, V t o, T, , 2Vm, , , , , Let I = 0 when t = 0., , t , , , , Vm2 sin 2 d , , Irms 10A Io 10 2 A 14.14 Amp, , Io Io sin t, , , , , , e12 e 22 sin t cos cos t sin , , , , , So phase difference is , 6, , 8., , 2V0, t V0, T, , 12. The equation of a semicircular wave is, , with respect to current, the emf is lagging., , 7., , V, , 10. e e1 sin t e 2 cos t, , 11., , X, L 1 . For both circuits, XC, , 2., , 9., , v dt ,, dt, , Vmean , , ; Vrms , , 2, o, , 2, t, , , V, V, o, 3, 3, , vrms , , vrms , , , , 1, 2, , 1, 2, , 1, 2, , 2, , v, , 2, , d, , 0, , 2, , 200 sin 100 sin 3 50 sin 5 , , 2, , d, , 0, , 2, , 200 sin 100sin 3 50, , 2, , sin 2 5 , , 2, , 0, , 2 200 100 sin sin 3, , 0, , NARAYANAGROUP, , 117
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 2 100 50 sin 3 sin 5, , x 2 y 2 a 2 or y 2 a 2 x 2, , 2 50 200sin 5 sin )d, , 26, 250, , , , , , x2, 1 a 2, 1 2, 2, a dx x dx , a x, , 2a a, 2a, 3, , , , a, , , , a, , 1 3 a3, a 3 2a 2, 3, , a a , 2a , 3, 3, 3, I rms 2a 2 / 3 0.816 a ., , 1 T, 1 T, 10 , y dt 10 t dt, , T 0, T 0, T , , 1 T, 10, 5t, 1, 10.dt .t.dt 10t , , , 0, T , T, T, T, , 1 a 2, a x 2 dx, 2a a, , 2, or I rms , , vrms 26250 162V ., 18. The slope of the curve AB is BC/AC = 20/T. Next,, consider the function y at any time t. It is seen that, DE / AE BC / AC 10 / T, or y 10 / t 10 / T or y 10 10 / T t, This giveus the equation for the function of one, cycle., Yav , , 1 a 2, y dx, 2a a, , I rms , , 1 200 2 1002 502 , , , , 2, 2 2, 2, 2 , , i0, V0, ; VDC , 2, 2, , 20. irms , , 2 T, , 15, , 12 1.4 16.8 17, , 0, , y, , , 2, , B, , 20, , 21., , D, 10, , 2, , irms, , i dt, 0, , dt, 0, , A, , EC, , y, , t, , T, Time, , 25. A solenoid consists of inductance and resistance., When 100 V dc is applied, 0 Z R, , 2T, , Z, , Mean square value, , , , Vrms, 100, R, 100 , Irms, 1, , When 100 V, 50 Hz ac is applied,, , 2, , T, 1 T, 10 , y 2 dt 10 t dt, 0, 0, , T, T , , Z, , 1 T, 100 2 200 , .t dt, 100 2 t , , 0, T, T, T , , 1, 100t 3 100t 2, 100t , , T, 3T 2, T, , T, , , 0, , 700, 3, , or rms value 10 7 / 3 15.2 ., 19. The equation of a semi-circular wave (shown in, figure) is, y, , Vrms 100, , 200 , I rms, 0.5, 2, , Z2 R2 XL2 2002 100 X2C, XL 100 3 2fL 100 3, L, , -a, , +a, 0, , x, , 100 3, 0.55H, 2 50, , 26. XL 2fL 2 200 , , 1, 400 , , , R 300, , tan , , 118, , , , XL 400 4, , , R 300 3, , 4, tan1 , 3, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 27. tan , , ALTERNATING CURRENTS, , XL XC XL 0, , , R, 0, , 90o, , 35. fo , , 1, 1, 1, ; f 'o , =, 2 LC, 2 L ' C ', 2 L ' 2C, , As per the question f 'o fo ; L ' , , E, , 100, volt., 2, , 36. Erms , Irms , I, , Current lags behind the applied voltage by a phase, angle of 90o or current leads the applied voltage, by a phase angle of 90o ., 28. In series R-L-C circuit, VL and VC are always, oppsoite in phase., Vc VL 7.4 2.6 4.8 V, , f, , 30., , , , 2 L eqCeq, , cos , , 1, C, 2 2L , 2, , , , 1, , R R, 1, Z R, , Leff L1 L2 1.7mH 2.3mH 4mH, , Resonance frequency,, , 34., , V , X , tan 1 C tan 1 C , R , VR , 8, 4, tan 1 tan 1 , 6, 3, , 1, L eff Ceff, , , , 1, 3, , 4 10 10 10 6, , 10 4, 0.5 10 4 rad / s, 2, 1, 2 LC, , , , 1, 1 1, 2, . 10 6, 2 2, , NARAYANAGROUP, , 40., , V, , P VIcos V , , 2, , 2 2, , R, , , , 2, , R 2L2, , R L, , P, , 41., , 4 F 2.5 F 3.5 F 10 F, , fo , , 39. V 2 VR2 VC2 VR 6 volt, , 1, 0.25 A, 4, , 33. Ceff C1 C2 C3, , , , Z R 2 X2 3 2 4 2 5, , R, 3, 300, or, , or Z 20 , |Z|, 2, | Z|, , VR IR 15 I 60 I , , , , 37. x 4 , R 3 , , 2 LC, , 31. I = current flowing through condenser/capacitor., R, L, C are connected in series. So same current, flows through R., , 32. cos , , 100 100, , , 10 3 cos, 3, 2, 2, 100 100, 1, , 103 2.5 watt, 2, 2, , ErmsIrms cos , , R 3, 0.6, Z 5, Eo, I, , o cos 0, 38. P ErmsIrms cos , 2, 2, 2, , C1C2, C2 C, , , C1 C2 2C 2, 1, , 100, 100, , 10 3 A , , mA , 2, 2, 3, , Power factor cos , , 29. L eq L L 2L, Ceq , , L, 2, , V 2R, R2 2L2, V, , I, 2, , R L , , 2, , Power loss=I2R , , V 2R, R 2 L , , 2, , , , V2, 2L2, R, R, , For minimum power loss, resistance should be low, and inductance shold be high., 220, 100 , R = Z, 42. Z , 2.2, 1, X C X L 100 100, , Power factor of box, , 1000Hz, , R, , =, , 2, , R X, , 2, C, , , , 100, 2, , 100 100, , 2, , , , 1, 2, 119
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 43. The impedance in RL circuit is same as the RC, circuit:, , , R 2 2 L2 R 2 , , 2 , , 1, LC, , or, , 1, 2 L2, , f , , 1, 2 LC, , , , 44. v v0 sin t / 4 v0 cos t , 4, , , But we cannot obtain a leading phase difference of, / 4 if we use only capacitor, (phase difference of / 2 ), or only (inductor and, resistor) (Phase difference of, / 2 ). or only resistor (phase difference of 0)., 48. Inductance does not consume power, 2, , 2, , Pav I rms R 5.6 10 313.6 W, , 49. P Vrms I rms cos , , , 100 100, , 103 cos900 = 0, , 2 2, , , 1, 1, 50. 1 L C 2 L C, 1, , I, , 2, , 1, 02, LC, 52. Apply parallelogram law, 53. At resonance VL and VC are equal in magnitude, with phase difference 1800, , Solving 12 , , /4, V, , Since V lags current, an inductor canbring it in phase, with current., 45. Since current leads voltage by 450 , there mustbe, a resistor and a capacitor. We, can say nothing about inductor surely. It may or, may not be present., 46. It is condition of resonance then only potential on, each are equal and 100V., 100 V, , V0, 100 V, , 1 , , R 2 L , , C , , , 100 V, , 47. If we have all R, L and C then I vs.E will be, IXC, , , , V0, 2R, , 1, Solving, 1 L C R, 2 L , , E, IXL, , ..... (i), , 1, R, 2C, , ...... (ii), , 1, LC, , ...... (iii), , To obtain a leading phase differene of /4 :, , Adding, 12 , , If X L X C and we use all R, L and C in the circuit,, then the resultant graph will be:, , Subtracting and using (iii) 2 1 , , I(XC Xi), Resultant, , IR, , E, , Which can give a leading phase difference of / 4 :, Similarly, if we have only resistance and capacitor, then we can obtain a phase, difference of / 4 (leading) for suitable values of, 120, , 2, , 1, , IR, , I, X C and R., , , , , , , , , , 20sin t , 54. V2 20 cos t 6 , 3, , , , The element x is practical capacitor, ir, 55. i , (say at angular frequencies 1 and 2 ), 2, , R, L, , R, 2L, 56. At resonance, VL = VC also VL and VC are in, opposite phase. A series LCR circuit containing a, resistance of 120 has an angular resonance, frequency. At resonance, voltage across resistance, and inductance are 60 V and 40 V., 57. At resonance X L X C in LCR circuit at, resonance current, V, I mms rms 10 2, ...... i , R, f 2 f1 , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, When current lags by, Vrms, , I rms , , , : R X , 4, , Vrms, , 10 A, 2, 2, 2R, R X, E, 58. Initially 5 0 (for X) and E0 200V, R, E, 5 0 (for Y), XL, In the series combinat ion of X and Y, E0, i, i0 , irms 0, 2, 2 ;, R XL, 2, 1, 1, 2f1 L, 2f 2 L , 2f 2C, 59. tan 2f1C, and tan , 4, R, 4, R, , 1, Resonant frequency , 2 LC, 60. From the graph current leads emf, hence the circuit, should be R-C, 1, 61. Resonant frequency , 2 LC, 62., , 10, , i, , 32 15 11, , 2, , 2A, , VL VC i X L X C 2 15 11 8V, 63. Pave Erms irms cos , Energy dissipated Pave time, 1, P Ppeak cos , 2, 1, 1, , Ppeak Ppeak cos cos , 2, 2, 3, , ALTERNATING CURRENTS, In LCR circuit, Z R 2 X L X C , 1 , , R L , , C , , , , 1, , and , CR, 6, 66. From the graph current leads emf. Hence the circuit, , 1, should be RC. For the circuit tan , 4 CR, 24, 69. Initially at resonance 6 , R, Resistance of coil R 4 , 12, 12, , 1.5 A, Later with 12V battery i , Rr, 44, 70. Average power, , 65. tan , , P , , V 2R, Z2, , NARAYANAGROUP, , 2, , 2, , V 2R, , For C C1 , P1 , , , 1 , R L , C1 , , , 2, , 2, , V 2R, , For C C2 , P2 , , , 1 , R L , C2 , , Accoriding to problem,, P2 P1, , 2, , 2, , V 2R, , , , V 2R, , , , 2, , , 1 , L, , , R2, , , C2 , , 2, , , 1 , L, , , R2, , , C1 , , or, 2, , , 1 , L , R2, 2, , , C1 , , , 1 , 2, ..........1, L C R , , 2, , Power factor, cos , , R, Z, , R, , cos , , 2, , , 1 , 2, L C R, 1, , 64. P V0i0 cos , , , , 2, , R, , cos 1 , , 2, , , 1 , 2, L C R, 2, 2, , , 1 , R2, 2, L , , R, , .......... 2, C1 , cos 2 1, , Similarly,, 2, , , 1 , R2, 2, , L, , , R, , .......... 3, , C1 , cos 2 2, Form eqns. (1), (2) and (3), we get, 2, , , 1 , 2, L , , R, , C, , , 1 , 1, L , R2 , C2 , , , 2, , 121
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 7., , LEVEL-VI, SINGLE ANSWER QUESTIONS, RMS & AVERAGE VALUE OF AC, 1., , If a direct current of value of ‘a’ ampere is, superimposed on an alternating current, I b sin t flowing through a wire, what is the, effective value of the resulting current in the, circuit?, a, , dc, I bo, , I, o, , R=50, , (a) 10, , t, , 8., , 3., , 4., , The secondary coil of an ideal step down, transformer is delivering 500 watt power at, 12.5 A current. if the ratio of turns in the, primary to the secondary is` 5:1; then the, current flowing in the primary coil will be:, a) 62.5 A b) 2.5 A c) 6 A, d) 0.4 A, In a step-up transformer the turn’s ratio is 10., If the frequency of the current in the, primary coil is 50 Hz then the frequency of, the current in the second ary coil will be, a) 500 Hz b) 5 Hz, c) 60 Hz d) 50 Hz, A power transformer is used to step up an, alternating emf of 220 volt to 11 kv to transmit, 4.4 kw of power. If the primary coil has 1000, turns, what is the current in the secondary?, a) 4A, b) 0.4A, c) 0.04A d) 0.2A, , 6., , An alternating voltage of 200 volt, at 400, cycles/sec is applied in a circuit containing an, inductance of 0.01 henry in series with a, resistance of 22.8 ohms. The voltage across, the inductance is, a) 148.2 volt, b) 392.4 volt, c) 74.1 volt, d) 196.2 volt, If the readings V1 and V3 are 100 volt, each then, , (c) 1000, , (a), , 3, henry, , , (b) henry, , (c) 3 henry, 9., , (d) 200, , (d) 3 henry, , An inductor X L 2 a capacitor X C 8 , and a resistance 8 are connected in series, with an ac source. The voltage output of A.C, source is given by v =10 cos 100 t . The, instantaneous p.d. between A and B, when it, is half of the voltage output from source at, that instant will be:, xL=2, , xC=8, , A, , 8, , B, , , , a), , 24, 24, 7, 5, volts b), volts c), volts d), volts, 7, 5, 24, 24, , 10. A resistor of resistance 100 is connected to, , reading of V2 is:, L, , R, , V1, , V2, , C, , an AC source of emf 12V sin 250s 1 t ., , V3, , find the energy dissipated as heat during t=0, to t=1.0ms, , ~, 200 volt, 50 Hz, , a) 0 volt b) 100 volt c) 200 volt, d) cannot be determined by given information, NARAYANAGROUP, , (b) 100, , If the power factor is 1/2 in a series RL circuit, with R = 100 . If AC mains, 50 Hz is used,, then L is, , AC across L-R, L-C, L-C-R of circuit, 5., , 1, H, , , , V=300V, , a.c, , t, , L, 1, F, 4, , b2, a2, b2, b 2 (d) a 2 , a) a2 b2 (b) a 2 (c), 2, 2, 3, , 2., , In the a.c. circuit shown in the figure. The, supply voltage has a constant r.m.s, value V,, but variable frequency f. Resonance, frequency in hertz is, , a) 0.61 10 J, , b) 0.61 10 J, , c) 2.61104 J, , d) 2.61106 J, 123
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 11. In the given AC circuit, which of the following, is incorrect:, xL=10 xC=10 R=10, , a) Voltage across resistance is lagging by 900 than, the voltage across capacitor, b) Voltage across capacitor is lagging by 1800 than, voltage across inductor, c) Voltage across inductor is leading by 900 than, voltage across resistance, d) resistance of the circuit is equal to impedance, reactance of circuit, 12. In the series circuit shown in the figure the, voltmeter reading will be(all the meters are, ideal)., , A, , 300V, V, , R, , L, , C, , X, , , , X, , -1 , C , b) tan R , , , , [Given cos 1 0.6 530 ], , = 100 sin (100t + /2), , 300V, V, , , , , -1 , C , c) tan R 2, d), 2, , , 15. In the circuit current through source will be, , , , V, , X, , -1 , C , a) tan R 2, , , , 200V, , , a) 300 V b) 200 V c) 100 V d) 600 V, 13. In the circuit diagram shown,, X C 100, X L 200 & R 100 ., The, effective current through the source is:, , XL=40, , R=30, , , , V 10 10 2 sin 100 t 450 , , (A), , 1, 2, , sin 100 t 80 , 3 5, , (B), , 1, 2, , sin 100 t 80 , 5 5, , (C), , 1, 2, , sin 100 t 980 , 3 5, , (D), , 1, 2, , sin 100 t 980 , 5 5, , 16. In figure below if Z L Zc and reading of, ammeter is 1 A., Find value of source voltage V., ZL, , ZC, , (A) 80 volt, , C, 200V , , L, , 1, H, , , R, , R=80, , L, , I1, , (D) None, , V, 30Hz, , 17. As shown in figure value of inductive, reactance X L will be if source voltage is, 100 volt, R1=20, , L, , (C) 100 volt, , , A, , , , a) 2A, b) 2 2A c) 0.5A, d) 0.4 A, 14. In the given circuit assuming inductor and, source to be ideal, the phase difference, between current I1 and I2 :, , (B) 60 volt, , XL1=30, , R2=10 XC1=40 , , I=2A, XL, , , , , I2, , V, , XC, , 124, , R, , (A) 40 , (C) 50 , , (B) 30 , (D) Can have any value, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 18. The power factor of the circuit shown in the, figure is, , III) Readings in A and V1 are always in phase, which of these statements are/is correct, , XC=20, , R=20, , V1, , XL=100, 40, , , , , , 220V, 50Hz, , (A) 0.4, (B) 0.2, (C) 0.8, (D) 0.6, 19. What will be the reading of the voltmeter, across the resistance and ammeter in the, circuit shown in the figure?, R=100, , A, , V, 100V, , V, 100V, , V, , , 220V, 50Hz, , (A) 300 V, 2A, (B) 800 V, 2A, (C) 100 V, 2A, (D) 220 V, 2.2A, 20. In an L-R-C series circuit the current is given, by i I cos t . The voltage amplitudes for, the resistor, inductor and capacitor are VR ,VL, and VC respectively.., (a) The instantaneous power into the resistor, is pR VR I cos 2 t ., (b) The instantaneous power into the inductor, is pL VL I sin t cos t ., (c) The instantaneous power into the capacitor, is pC VC I sin t cos t ., (d) pR pL pC equals total power p, supplied by the source at each instant of time., A. (a), (c), (d) are correct, B. (b), (c) are correct, C. (a) is correct, D. (a), (b), (c), (d) are correct, 21. The diagram shows a capacitor C and a, resistor R connected in series to an ac source., V1 and V2 are voltameters and A is an ammeter, consider now the following statements, I) Readings in A and V2 are always in phase, II) Reading in V1 is ahead in phase with reading, in V2, NARAYANAGROUP, , C, , R, , V2, , A, , A. I only, B. II only, C. I and II only, D. II and III only, 22. Two impedances Z1 and Z 2 when connected, separatley across a 230V, 50Hz supply, consumed 100 W and 60 W at power factors, of 0.5 lagging and 0.6 leading respectively. If, these impedances are now connected in series, across the same supply, find :, (a) total power absorbed, (b) the value of the impedance to be added in, series so as to raise the overall power factor, to unity., B. 19 W ,95 , A. 19 W , 295 , C. 99 W ,195 , D. 75W ,195 , 23. In the LCR circuit shown in figure, XC=20, , R=10 XL=10, , , V 200 2 sin t, , (1) current will lead the voltage, (2) rms value of current is 20 A, , 1, 2, (4) voltage drop across resistance is 100 V, A. (1) and (3) are correct, B. (1) and (4) are correct, C. (2) and (3) are correct, D. (3) and (4) are correct, (3) power factor of the circuit is, , MULTI ANSWER QUESTIONS, 24. In an AC series circuit, the instantaneous, current is zero when the instantaneous voltage, is maximum. Connected to the source may be, (a) pure inductor, (b) pure capacitors, (c) pure resistor, (d) combination of an inductor and a capacitor, 125
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 25. Which statement(s) is False for the series, resonant condition, (A) current maximum and phase difference between, E and i is / 2, (B) current maximum and phase difference between, E and i is zero, (C) voltage maximum and phase difference between, E and i is zero, (D) voltage maximum and phase difference between, E and i is / 2, 26. An alternating e.m.f. of frequency, 1 , , , 2 LC is applied to a series LCR, circuit. For this frequency of the applied e.m.f., A. The circuit is at resonance and its impedance is, made up only of a reactive part, B. The current in the circuit is in phase with the applied, e.m.f. and the voltage across R equals this applied, emf, C. The sum of the p.d’s across the inductance and, capacitance equals the applied e.m.f. which is 1800, ahead of phase of the current in the circuit, D. The quality factor of the circuit is, L / R or 1/ CR and this is a measure of the, voltage magnification (produced by the circuit at, resonance) as well as the sharpness of resonance, of the circuit., 27. An alternating voltage (in volts) varies with, time t (in seconds) as V = 200 sin (100 t ), (A) The peak value of the voltage is 200 V, (B) The rms value of the voltage is 220 V, (C) The rms value of the voltage is 100 2 V, (D) The frequency of the voltage is 50 Hz, 28. An AC source rated 100 V (rms) supplies a, current of 10 A (rms) to a circuit. The average, power delivered by the source, (A) must be 1000 W, (B) may be 1000 W, (C) may be greater than 1000 W, (D) may be less than 1000 W, 29. In an L-R circuit, the value of L is (0.4 / ), henry and the value of R is 30 ohm. If in the, circuit, an alternating emf of 200 volt at 50, cycles per second is connected, the impedance, of the circuit and current will be, (A) 50 ohm, (B) 60 ohm, (C) 2 ampere, (D) 4 ampere., 126, , 30. A circuit has three elements, a resistance of, 11 , a coil of inductive reactance 120 and, a capacitive reactance of 120 in series and, connected to an A.C. source of 110 V, 60 Hz., Which of the three elements have minimum, potential difference?, (A) Resistance (B) Capacitance (C) Inductor, (D) All will have equal potential difference, 31. An inductor 20 103 henry, a capacitor, 100F and a resistor 50 are connected in, series across a source of emf V 10sin 314t ., (a) The energy dissipated in the circuit in 20, minutes is 951 J., (b) If resistance is removed from the circuit, and the value of inductance is doubled, then, the variation of current with time in the new, circuit is 0.52 cos (314 t), A. Both (a) and (b) are correct, B. Both (a) and (b) are false, C. Only (a) is correct D. Only (b) is correct, 32. What will be the reading of the voltmeter, across the resistance and ammeter in the, circuit shown in the figure?, 100, , V, 100V, , A, , V, 100V, , V, , , 200V, 50Hz, , (A) 200 V (B) 300 V (C) 3 A, (D) 2 A, 33. A 50 electric heater is connected to, 100 V, 60 Hz ac supply., (A) The peak value of the voltatge is 100 V, (B) The peak value of the current in the circuit is, 2 2A, , (C) The rms value of the voltage is 100 V, (D) The rms value of the current is 2 A, 34. In a series LCR circuit, 1, H, 2, , 1, F, 2, , R, , , (A) the voltage VL across the inductance leads the, current in the circuit by a phase angle of / 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , (B) the voltage VC across the capacitance lags, behind the current by a phase angle of / 2, (C) the voltage VR across the resistance is in phase, with the current, (D) the votage across the series combination of L,, C and R is V = VL + VC + VR., 35. In a series LCR circuit with an AC source, (Erms = 50 V and = 50 / Hz ), R = 300 ,, C 0.02mF, L 1.0H . Which of the following, , is correct?, (A) the rms current in the circuit is 0.1 A, (B) the rms potential difference across the capacitor, is 50 V, (C) the rms potential difference across the capacitor, is 14.1 V, (D) the rms current in the circuit is 0.14 A, 36. In the figure shown R 100 L , C, , 2, H and, , , 8, F are connected in series with an a.c., , , source of 200 volt and frequency ‘f’. V1 and, , V2 are two hot-wire voltmeters. If the readings, of V1 and V2 are same then:, V1, R, , V2, , L, , A) The current leads the emf in the circuit, B) The circuit is more inductive than capacitive, C) To increase the rate at which energy is, transferred to the resistive load, L should be, decreased., D) To increase the rate at which energy is, transferred to the resistive load, C should be, decreased., 38. In the given AC circuit, which of the following, is incorrect:, XL=10 XC=10, , R=10, , , = 100sin (100t + /2), , A) Voltage across resistance is lagging by 900 than, the voltage across capacitor, B) Voltage across capacitor is lagging by 1800 than, voltage across inductor, C) Voltage across inductor is leading by 900 than, voltage across resistance, D) Resistance of the circuit is equal to reactance of, circuit, 39. In the circuit shown, resistance R 100 ,, 1, inductance L H and capacitance, , 4, F are connected in series with an ac, , source of 100 volt and frequency ‘f’. If the, readings of the hot wire voltmeters V1 and, c, , , , A) f=125 Hz, B) f=250 Hz, C) Current through R is 2A, D) V1 V2 1000 volt, 37. Graph shows variation of source emf V and, current i in a series RLC circuit, with time., , V2 are same then:, V1, R, , V2, , L, , Vori, V, , , , i, t, , NARAYANAGROUP, , A) f = 125 Hz, B) f = 250 Hz, C) Current through R is 1A, D) V1 V2 =500 volt, 127
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 40. Choose correct statement of capacitance, increases from zero (0) to infinity ., Box–1, C, , XL, , R, , V1, , A, , , V,fHz, , (A) Current increases from 0 (zero) to maximum, then decreases to zero, (B) Reading of voltmeter first increases and it will, be maximum when X L X C, (C) Power factor of circuit first increases then, decreases, (D) V1 may be greater than V, V1 may be equal V,, , V1 maybe less than V, where V1, is reading of voltmeter and V is source voltage., 41. A box P and a coil Q are connected in series, with an AC source of variable frequency. The, emf of the source is constant at 10 V. Box P, contains a capacitance of 1F in series with a, resistance of 32 . Coil Q has a selfinductance 4.9 mH and a resistance of 68, in series. The frequency is adjusted so that, the maximum current flows in P and Q. At this, frequency, (a) The impedance of P is 77, (b) The impedance of Q is 85, (c) Voltage across P is 7.7V, (d) Voltage across Q is 9.76V, 1mF, C, , 32, , 68W, , R1, BoxP, , L, R2, Coil Q, , , A. Only (a), (c) are correct, B. Only (a), (d) are correct, C. Only (c), (d) are correct, D. (a), (c), (d) are correct, 128, , 42. A series LCR circuit containing a resistance, of 120 has angular resonance frequency, 4 105 rads 1 . At resonance the voltages, across resistance and inductance are 60 V and, 40 V respectively., 1, F, (a) The value of L and C are 0.2 mH,, 32, (b) 8 105 rad/s, the current lags the voltage by 450, (c) 6 105 rad/s, the current lags the voltage by 450, A. (a), (c) are correct B. (a), (b) are correct, C. (a), (b), (c) are correct, D. (a), (b), (c) are wrong, 43. A current of 4 A flows in a coil when connected, to a 12 V DC source. If the same coil is, connected to a 12 V, 50 rad/s AC source a, current of 2.4 A flows in the circuit., (a) The inductance of the coil is 0.06H, (b) The inductance of the coil is 0.08H, (c) When 2500 F capacitor is connected in, series with the coil, then power developed in, the circuit is 17.28 W, A. (a), (c) are correct, B. (b), (c) are correct, C. (a), (b), (c) are correct D. Only (b) is correct, 44. In the given series R-L-C circuit, R 100 ,, L 103 H , C 0.1F , V0 200V, Bulb, R, , 6L, C, , C, 3L, , R, , , V= 200sin cot, , (a) The resonant frequency is 15924 Hz, (b) The current at resonance is 1 A, (c) The power dissipated in the circuit at, resonance is 100 W, A. (a), (b), (c) are correct B. (a), (b), (c) are wrong, C. Only (a), (b) are correct D. Only (b), (c) are correct, , 45. An alternating emf of frequency f = 50 Hz,, peak voltage V0 21 volt is applied to a series, circuit of resistance R 20ohm , an inductance, L 100mH and a capacitor of C 30F ., (a) The maximum current is 3A, (b) The phase difference between current and, applied voltage is 750, (c) The current i as a function of time ‘t’ is, i 3sin 314t 750 , A. Only (a), (b) are correct B. Only (b), (c) are correct, C. (a), (b), (c) are correct, D. (a), (b), (c) are wrong, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 46. A resistor R is connected in series with a coil., The system is subjected to an A.C supply of, peak voltage V0 . If the peak voltages dropped, across the resistor R and the coil are V1 and, V2 respectively, (a) The power dissipated in the coil is, , V02 V12 V22, 2R, (b) The power dissipated in the circuit is, V02 V12 V22, 2R, , COMPREHENSION QUESTIONS, PASSAGE-1:, A 100 resistance is connected in series with a, 4H inductor. The voltage across the resistor is,, VR = (2.0V) sin (103t)., 49. Find the expression of circuit current, (A) 2 102 A sin103 t , (B) 2 103 A sin102 t , (C) 2 103 A sin103 t , (D) 2 102 A sin 102 t , , Coil., R, V1, , V2, , A. Only (a) is correct B. Only (b) is correct, C. (a), (b) are wrong D. (a), (b) are correct, 47. For the AC circuit shown, the reading of, ammeter and voltmeter are 5A and, 50 5 volts respectively, then, –, , 20mH, 10, , 50F, , –, , –, , A. average power delivered by the source is 250W, B. rms value of AC source is 50 volts, C. voltage gain is 2 D. frequency of ac source is, 48. In the circuit shown in fig. If both the lamps, L1 and L2 are identical., 500F, , L1, L2, , 10mH, , , , N, 200 V, 500 Hz, , A. their brightness will be same, B. L1 will be brighter than L2, C. as the frequency of supply voltage is increased,, brightness of L1 will increase and that of L2 will, decrease, D. only L2 will glow because the capacitor has, infinite resistance, NARAYANAGROUP, , 50. Find the inductive reactance, (A) 2 × 103 ohm, (B) 3 × 103 ohm, 3, (C) 4 × 10 ohm, (D) 5 × 103 ohm, 51. Find amplitude of the voltage across the, inductor., (A) 40 V (B) 60 V (C) 80 V (D) 90 V, PASSAGE-2:, If various elements, i.e., resistance, capcitance and, inductance which are in series and having values, 1000 , 1F and 2.0 H respectively. Given emf, as, V 100 2 sin1000 t volts, 52. Voltage across the resistor is, (A) 70.7 Volts, (B) 100 Volts, (C) 141.4 Volts, (D) 270.7 Volts, 53. Voltage across the inductor is, (A) 70.7 Volts, (B) 100 Volts, (C) 141.4 Volts, (D) 270.7 Volts, 54. Voltage across the capacitor is, (A) 70.7 Volts, (B) 100 volts, (C) 141.4 Volts, (D) 270.7 Volts, PASSAGE-3:, One application of L-R-C series circuits is to high, pass or low pass filters, which filter out either the, low or high frequency components of a signal. A, high pass filter is shown in figure Where the output, voltage is taken across the L-R where L-R, combination represents and inductive coil that also, has resistance due to the large length of the wire in, the coil., , Vs, , C, R, , L, Vout, 129
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 55. Find the ratio for Vout / Vs as a function of the, angular frequency of the source, R 2 L2, , R 2 2L2, , 1 , , R 2 L , , C , , , (A), , 2, , 1 , , R 2 L , , C , , , (B), , 2, , R 2 2L, 1 , , R 2 C , , L , , , (C), , 2, , (D) 1, , 56. Which of the following statement is correct, when is small in the case of Vout / Vs, R, R, (C) RL, (D), L, C, 57. Which statement is correct in the limit of large, frequency is reached ? (for Vout / Vs ), , (A) RC (B), , (B) RC (C) RL, , (A) 1, , (D), , R, L, , PASSAGE-4:, In A.C. source peak value of A.C. is the maximum, value of current in either direction of the cycle. Root, moon square (RMS) is also defined as the direct, current which produces the same heating effect in, a resistor as the actual A.C., 58. A.C. measuring instrument measures its, (A) rms value, (B) Peak value, (a), (C) Average value, (D) Square of current, 59. Current time graph of different source is given, which one will have R.M.S. value V0, V0, , A), , 2V0, , T/2, T, , V0, , 3T 2T, 2, , 2T, , T, , B), , T/2, 2V0, , 2V0, , 3T, 2, , 4V0, , C), , (D), , V=2V0 sin r, , 60. Average voltage for the given source is, 2V0, , T/2, , (A) V0, 130, , (B) 2V0, , T, , 3T 2T, 2, , (C), , V0, 2, , (D), , 3V0, 2, , PASSAGE-5:, An alternating voltage of 260 volt and, = 100 radian/second, is applied in an LCR series, circuit where L = 0.01H, C = 4 104 F and, R 10 ., 61. Find the power supplied by the source:, A) 1000 W B) 6760 W C) 3380 W D) 3000 W, 62. Find the resonance frequency of the circuit (in, hertz):, 25, 250, 40, 200, A), B), C), D), , , , , 63. The power sullpied by the source at the above, resonance frequency is:, A) 1000W B) 6760W C) 3380W D) 3000W, PASSAGE-6:, A steady 4A flows in an inductor coil when, connected to a 12V source as shown in figure. If, the same coil is connected to an ac source of 12V,, 50 rad/s, a current of 2.4A flows in the circuit as, shown in figure2. Now after these observations, a, 1, F is connected in, capacitor of capacitance, 50, series with the coil and with the same AC source, as shown in figure 3., , 12V, , , , , , 12V, 50 red/sec. 12V, 50 red/sec., (Figure 1), (Figure 2), (Figure 3), , 64. The inductance of the coil is nearly equal to, A) 0.01 H B) 0.02 H C) 0.04 H D) 0.08 H, 65. The resistance of the coil is:, A) 1, B) 2, C) 3, D) 4, 66. The average power supplied to the curcuit, after connecting capacitance in series is, approximately equal to:, A) 24 W B) 72 W C) 144 W D) 18.2 W, PASSAGE-7:, In a series L-R circuit, connected with a sinusoidal, ac source, the maximum potential difference across, L and R are respectively 3 volts and 4 volts., 67. At an instant the potential difference across, resistor is 2 volts. The potential difference in, volt, across the inductor at the same instant, will be:, a) 3cos 300, b) 3cos 600, c) 6 cos 450, d) 6, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 68. At the same instant, the magnitude of the, potential difference in volt, across the ac, source will be, A) 3cos 67 0 B) 5sin 37 0 C) 4 cos 97 0 D) 0, 69. If the current at this instant is decreasing the, magnitude of potential difference at that, instant across the ac source is, A) increasing, B) decreasing, C) Constant, D) Can’t be said, PASSAGE-8:, A constant voltage at a frequency of 1 MHz is, applied to an inductor in series with variable, capacitor; when capacitor is 500 pF, the current, has its maximum value, while it is reduced to half, when capacitance is 600 pF. Find, 70. Resistance (R), (A) 30 (B) 20 (C) 40 (D) 50 , 71. The inductance L, (A) 0.05 mH (B) 0.5mH (C) 0.005mH (D) 5 mH, 72. Q factor of an inductor, (A) 10.4 (B) 20.8 (C) 5.2, (D) 9.4, PASSAGE-9:, When 1A passed through three coils A, B, C in, series the voltage drops are respectively 6, 3 and, 8 volt on direct current source and 7, 5 and 10 volt, on Alternating current source, 73. Power factor of coil B, will be, (A) 0.6, (B) 0.8, (C) 0.7, (D) None, 74. Power dissipated in coil C, (A) 10 watt (B) 6 watt (C) 5 watt (D) 8 watt, 75. Power factor of whole circuit when alternating, current flow, (A) 0.6, (B) 0.8, (C) 0.78 (D) 1, PASSAGE-10:, A series circuit connected across a 200 V, 60 Hz, line consists of a capacitive reactance 30 non, inductive resistor of 44 and a coil of inductive, reactance 90 and resistence 36 as shown in, the diagram., , 78. The power dissipated in the inductance coil is, A. zero, B. 320 W C. 144 W D. 160 W, PASSAGE-11:, The maximum values of the phasors (currents and, voltage) in A.C. circuits can be treated as vectors, rotating with an angular frequency equal to the, angular frequency of the rotor of the generator. If, , the phase difference between two phasors A1 and, , A2 is , the resultant phasor is:, , f, , The rms value of y f t is, 1, , yrms, , T, 2, 2, f t dt , , , 0, , T, , , , , , Average value, T, , The average value of y f t is y , av, , ydt, 0, , T, , Using the above concepts, answer the following, questions., 79. The currents i1 and i2 in A.C. circuit are given, , , , , , , , , as: i1 4 sin t 3 and i2 4 sin t 3 , , , , , , , , , i3, R1=44, , i2, XL=90, R2=36, , 76. The potential difference across the coil is, A. 100 V B. 194 V C. 97 V D. Zero, 77. The power used in the circuit is, A. 320 W B. 144 W C. 160 W D. 96 W, NARAYANAGROUP, , b, A1, , A A12 A22 2 A1 A2 cos , , and the phase of A with respect to A1 is, A2 sin , tan 1, A1 A2 cos , RMS value, , XC=30, , 200V , 60Hz, , w, , A2, , i1, , The current i3 can be given as:, 2, , , , , , A. 4 3 sin t 3 , , , , , , B. 2 3 cos t 3 , , C. 4sin t , , D. 4cos t , , , , , , 131
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 80. The rms value of i3 is, A. 2 6, B. 6, C. 3 2, D. 2 2, 81. The average value of i in i t graph (semicircular) is, , 86. Current versus time graph across the inductor, wil be, iL, , iL, t, , A., , i, , t, , B., , 2, , iL, , iL, , t, , , 2, , 4, , , B., 2, , A. , , , D., 2, , , C., 3, , PASSAGE-12:, A series R-L-C circuit has R=100 ohm., L=0.2, 1, mH and C F . The applied voltage, 2, V 20sin t . Then, , VR max, , , ,, 0, 82. At resonant frequency, VL max, A. 2, B. 5, C. 3, D. 4, 83. When the current lags the applied voltage by, 450 , the value of is approximately, A. 5 10 5 rad / s, B. 3 105 rad / s, C. 4 105 rad / s, D. 4 1010 rad / s, 84. Referring the above question, the equation of, the current is, , B. 0.2 sin t tan 0.3 , C. 0.3cos t tan 1 0.3 , D. 0.3cos t tan 0.3 , 1, , PASSAGE-13:, The potential difference across a 2H inductor as a, function of time is shown in figure. At time t = 0,, current is zero, 85. Current at t = 2 s is, VL(volt), 10, t(s), , 132, , B. 3 A, , , , , , , , , , , , , , , , , , , , L, , , , , , , , , , C, , 87. Find the rms current, B. 5105 A, , C. 4105 A, D. 7105 A, 88. Find the energy dissipated in 50 sec., A. 6.12106 J, B. 8.12105 J, , 1, , A. 1 A, , t, , PASSAGE-14:, In the given arrangement the square loop of area, 10 cm2 rotates with an angular velocity about, its diagonal. The loop is connected to a inductance, of L 100 mH and a capacitance of 10 mF in, series. The lead wires have a net resistance of, 10 . Given that B = 0.1 T and 63 rad/s,, , A. 6105 A, , A. 0.2 sin t tan 1 0.3 , , 2, , t D., , C., , 4, , C. 4 A, , D. 5 A, , C. 5.12 105 J, D. 8.1210 6 J, 89. If the current is in phase with voltage, what, should be the frequency of rotation of the coil., A. 31.6 rad/s, B. 29.5 rad/s, C. 25.6 rad/s, D. 20.5 rad/s, PASSAGE-15:, A 20 V 5 watt lamp is used in ac main 220 V and, frequency 50 c.p.s., 90. Capacitance of capacitor, to be put in series, to run the lamp, A. 2m F B. 4m F, C. 6m F, D. 8mF, 91. Inductance of inductor, to be put in series to, run the lamp., A. 2.53H B. 5H, C. 7.5H D. 9H, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 92. What pure resistance should be included in, place of the above passive elements so that, the lamp can run on its rated voltage ?, A. 120 B. 240 C. 800 D. 720 , PASSAGE-16:, In the circuit, , shown in the figure, , R 50 , E1 25 3 volt and E2 25 6, , sin t volt where 100 s 1 . The switch s is, closed at time t = 0, and remains closed for 14, minutes, then it is opened., 93. Find the amount of heat produced in the, resistor, , , E2, , Transformer Primary Coil Secondary Coil, 1, 2, 3, , R, , E1, , The primary (10) coil is connected to a sorce of, alternating (AC) current. The secondary (20) coil, is connected to resistor such as a light bulb. The, AC source produces an oscillating voltage and, current in the primary coil that produces an, oscillating magnetic field in the core material. This, in turn induces an oscillating voltage and AC current, in the secondary coil, Students collected the following data comparing, the number of turns per coil (N) , the voltage (V), and the current (I) in the coils of three transformers., , S, , A. 64000 J B. 56000 J C. 63000 J D. 75000 J, 94. If total heat produced is used to raise the, temperature of 3 kg of water at 200 C , what, would be the final temperature of water ?, A. 150 C B. 250 C C. 450 C D. 750 C, 95. Find the value of the direct current that will, produce same amount of heat in the resistor, in same time as combination of DC source and, AC source has produced Specific heat of water, = 4200 J/kg-0 C., A. 1.23 A B. 1.22 A C. 2.24 A D. 3.25 A, PASSAGE-17:, A physics lab is designed to study the transfer of, electrial energy from one circuit to another by means, of a magnetic field using simple transformers. Each, transformer has two coils of wire electrically, insulated from each other but wound a round a, common core of ferromagnetic material. The two, wires are close together but do not touch each other, Ferromagnetic core, , R, , AC, , 100 10V 10A 200 20V 5A, 100 10V 10A 50 5V 20A, 100 10V 10A 100 5V 20A, , 96. The primary coil of a transformer has 100 turns, and is connected to a 120 V AC source. How, many turns are in the secondary coil if there’s, a 2400 V acorss it, A. 5, B. 50, C. 200, D. 2000, 97. A transformer with 40 turns in its primary coil, is connected to a 120 V AC source. If 20 W of, power is supplied to the primary coil, how much, power is developed in the secondary coil?, A. 10 W B. 20 W C. 80 W D. 160 W, 98. Which of the follwing is a correct expression, for R, the resistance of the load connected to, the secondary coil, , V10 N 20 , A. I N , 10 10 , , V10 N 20 , B. , , I10 N10 , , V10 N10 , C. I N , 10 20 , , V10 N10 , D. , , I10 N 20 , , 2, , 2, , 99. A 12 V battery is used to supply 2.0 mA of, current to the 300 turns in the primary coil of, a given transformer. What is the current in the, secondary coil if N2 = 150 turns, A. zero, B. 1.0 mA C. 2.0 mA D. 4.0 mA, PASSAGE-18:, In the circuit shown in figure:, R1 10, L , , 3, H , R2 20, 10, , and, , C, , 3, mF.., 2, , Current in L R1 circuit is I1 in C R2 circuit is, 10 Coil, , NARAYANAGROUP, , 20 Coil, , I 2 and the main current is I, 133
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, C, , R2, , R1, , L, , , V 200 2 sin (100t)V, , 100. Phase difference between I1 and I 2 is, A. 00, B. 900, C. 1800, D. 600, 101. At some instant current in L R1 circuit is 10, A. At the same instant current in C-R2 branch, will be, A. 5 A, B. 5 2A C. 5 6 A D. 5 3A, 102. At some instant I1 in the circuit is 10 2A ,, then at this instant current I will be, A. 20 A, B. 10 2A C. 20 2A D. 25 A, , MATRIX MATCHING QUESTIONS, 103. Column I, A. In case of series L-C-R circuit, at resonance., B. Only resistor in an a.c. circuit., C. Only inductor in an a.c. circuit., D. Only capacitor in an a.c. circuit., Column II, P. Current in the circuit has same frequency as that, of source voltage., Q. Voltage lags the current by / 2 ., R. Current lags the voltage by / 2 ., S. Reactance of the circuit is zero., T. Current is in phase with applied voltage., 104. Column I, A. For square wave having peak value v0., B. For sinusoidal wave having peak value v0., C. Current leads the voltage by / 2 ., D. Wattless current., Column II, P. v0 > vrms > vav., Q. In a pure inductance., R. vav = vrms = v0, S. In a pure capacitance., 105. Figure shows a series LCR circuit connected, to a variable frequency 200 V source. L = 5 H,, C 80 F and R 40 ., L, , Column I, (A) The impedance of the circuit at resonance (in, ohm), (B) The current amplitude at resonance (in A), (C) The rms potential drop across the inductor at, resonance (in volt), (D) The rms potential drop across the resistor at, resonance (in V), Column II, (P) 1250 V (Q) 200 V (R) 40 V (S) 5 2 A, 106. In L-C-R series circuit suppose r is the, resonance frequency, then match the following, columns:, Column I, (A) If r, (B) If r, (C) If 2r, (D) If r, Column II, (P) current will lead the voltage, (Q) voltage will lead the current, (R) XL 4XC, (S) current and voltage are in phase, (T) None, 107. In column I, variation of current with time t is, given in figures. In column II root mean square, current irms and average current are given., Match the column I with corresponding, quantities given in column II., Column I, i, i0, 0, , (A), , i, i0, T/2, , T, , t, , 0, , (B), , -i0, , 0, , i, i0, , T/2, , t, , T, , (D), , 0, , -i0, , Column II, (P) irms , , io, 3, , (R) Avreage current for positve half cycle is, R, , 134, , t, , T/2 T, , (Q) Average current for positive half cycle is io, , C, , , , t, , T, , -i0, , i, i0, , (C), , T/2, , io, 2, , (S) Full cycle averge current is zero., (T) irms io, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, Column II, imax, , 108. Column I, L, , R, , ALTERNATING CURRENTS, , (a), , 111. All voltmeters are ideal and reading of, voltmeters V1 and V2 are given by, , volt and V2 4 volt in all cases. Match the, following:, Column II, Column I, , (p), , , , 0, , O, , R, , w, , imax, , C, , (b), , V3, , (q), , , R C, , 0, , O, , V2, , L, , R, , (p) V3 5 Volt, , (A), , (r), , , C, , V1, , w, , imax, , L, , (c), , V1 3, , , , imax, , L, , (d), , , , (s), , V3, , , , , 109. In the series R-L-C a.c circuit, at resonance, Pmax - maximum power dissipation, , i0 - maximum current, f - frequency of the supply, (Given 2 10 ), Column I, Column II, (a) power-factor, (p) 0, Pmax, (b) i 2 R, (q) 0.1, 0, (c) 4 f 2 LC, , V1, , (q) V3 1 Volt, , , V3, V1, , (d) X L X C, (s) 0.5, 110. An LCR series circuit has a current which lags, behind the applied voltage by . The voltage, across the inductance has a maximum value, equal to twice the maximum of the voltage, , C, , (C), , V3, V1, , V2, , L1, , L2, , (D), , 0, , (r) V3 7 Volt, , , , (b) Reactance of inductor if 450, (d) Reactance of circuit if 600, Column II, (p) 20 (q) 20 3 (r) 40, , V2, , L, , across the capacitor. EL 30sin 100t . If, R 20 , then match the items given in, Column-I with that in Column-II, Column I, (a) Reactance of capacitor if 450, , NARAYANAGROUP, , R, , C, , (B), , (r) 1, , (c) Impedance of circuit if 45, , V2, , (s) Current is lagging in, phase from applied voltage, (t) Applied voltage is, lagging in phase from, current, , , , (s) 20 2, 135
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 112. A resistance R and inductance L and a, capacitor C all are connected in series with an, a.c. supply. The resistance R is 16 and for, a given frequency, the inductive reactance of, L is 24 and capacitive reactance of C is 12, . If the current in the circuit is 5 amp, then, match the following, Column I, Column II, (A) Potential difference across R (P) 100 V, (B) Potential difference across L, (Q) 60 V, (C) Potential difference across C (R) 120V, (D) The voltage of A.C supply, (S) 80 V, 113. Four different circuit components are given in, each situation of column I and all the, components are connected across an ac source, of same angular frequency w 200 rad/s., The information of phase difference between, the current and source voltage in each situation, of column I is given in column II. Match the, circuit components in column I with, corresponding results in column II, COLUMN - I, , 115. Referring to the given circuit, match column-I, with column - II, 10mH, , , V=10 sin t (Volt), , COLUMN -I, A) For 8000 rad / s B) For 10000 rad / s, C) For 10500 rad / s D) For 10000 rad / s, COLUMN -II, p) Peak current in the circuit is less than 0.1A, q) Voltage across the combination and the current, are in same phase, r) Voltage across the combination leads the current, s) Current through the circuit leads the voltage, across it, 116. In column I some ac circuits with meter reading, are given and in column II some circuit, quantities are given., Match the entries of column I with the entries, of column II, COLUMN - II, COLUMN - I, , 10, , V1, , A), B), C), D), , 500F, , 1k, , V2, , A), , 5H, 4H, , 3F, , , 200V, V1=158.11V; V2=200V, , p) VR 150 V, , V1, , 5H, , COLUMN - II, , B), , , p) The magnitude of required phase difference is, 2, , q) The magnitude of required phase difference is, 4, r) The current leads in phase to source voltage, s) The current lags in phase to source voltage., 114. In a series RLC circuit,, COLUMN -I, COLUMN -II, A) If C decreases, p) Z may increase, B) If L increases, q) Z may decrease, C) If R decreases, r) Resonant, frequency increases, D) If P.D across L and C s) Resonant, are increases in magnitude frequency decreases, 136, , 100 1F, , q) VL 50 V, , , 250V, V1=150V, , V3, V2, , V1, , C), , D), , , 250V, V1=158.11V; V2=291.6V;, V3=150V, , r) VC 250 V, , s) Power factor, , 250V, , of the circuit 3/5, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 117. Referring to Fig. A and B match column - I, with column - II, 1mH, , 1mH, , 1 0.1F, , 1 0.1F, , , V=V0 sin t, , , V=V0 sin t, , D) In fig. B, for 10 rad/s, COLUMN -II, p) Current through the inductor and current through, the capacitor are in same phase, q) Current through the inductor and current through, the capacitor are in opposite phase, r) Current I through the source is zero, s) Current I through the source is non-zero, 118. Referring to the given circuit, match column I with column - II, 4, , L, , B1(bulb), , , , C, , B2(bulb), , , B3(bulb), , , , (Assume the same value of applied voltage, amplitude in each case and also a negligible, internal resistance of the source), COLUMN -I, A) Brightness of bulb B1 will increase if, B) Brightness of bulb B2 will increase if, C) Brightness of bulb B3 may increase or decrease, if, D) Brightness of bulb B3 will be maximum if, NARAYANAGROUP, , 1, LC, q) Frequency is increased, r) space between the plates of capacitor is filled, with a meterial of high dielectric constant (assuming, the capacitor to be an air capacitor initially), s) A meterial of high permeability is inserted in the, inductor along its axis (assuming the inductor to be, an air inductor initially), , p) , , R, , COLUMN -I, A) In fig. A, for 105 rad/s, B) In fig. A, for 104 rad/s, C) In fig. B, for 105 rad/s, , C, , COLUMN -II, , STATEMENT TYPE QUESTIONS, a), , Statement-I is true, statement-II is true;, statement-II is a correct explanation for, statement-I, b) Statement-II is true, statement-I is true;, statement-II is a not a correct explanation, for statement-I, c) Statement-I is true, statement-II is false, d) Statement-I is false, statement-II is true, 119. Statement-I: The D.C and A.C both can be, measured by a hot wire instrument., Statement-II: The hot wire instrument is based, on the principle of magnetic effect of current., 120. Statement-I: The electrostatic energy stored in, capacitor plus magnetic energy stored in inductor, will always be zero in a series LCR circuit driven, by ac voltage source under condition of resonance., Statement-IIThe complete voltage of ac source, appears across the resistor in a series LCR circuit, driven by ac voltage source under condition of, resonance., 121. Statement-I : The r.m.s. value of alternating current, is defined as the square root of the average of I2, during a complete cycle., Statement-II : For sinusoidal a.c., , I0, ., 2, 122. Statement-I : In series LCR circuit resonance can, take place., Statement-II : Resonance takes if inductive, reactance and capacitive reactance are equal with, phase difference 1800 ., 123. Statement-I : In series LCR resonance circuit,, the impedance is equal to the ohmic resistance., Statement-II : At resonance, the inductive, reactance is equal and opposite to the capacitive, reactance., ( I = I0 sin wt ) Irms , , 137
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ALTERNATING CURRENTS, 124. Statement-I : In series LCR circuit, the resonance, occurs at one frequency only., Statement-II : At this frequency, inductive, reactance is equal to capacitive reactance., 125. Statement-I : For series RLC network, current, of circuit in region (1) leading and in region (2), lagging over applied voltage., Statement-II : Overall nature of circuit in region, (1) is inductive while in region (2) is capacitive., I, (1), , (2), , f1 f, , 126. Statement-I : In a tube light choke coil is used to, limit the current and at starting induced emf across, tube light is higher than supply voltage., Statement-II : Choke coil having higher inductive, reactance than resistance, power factor of choke, coil is more., 127. Statement-I : Power loss in ideal choke coil will, be zero., Statement-II : Ideal choke coil has zero power, factor., 128. Statement-I : KVL rule can also be applied to, AC circuit ., Statement-II : Varying electrostatic field is nonconservative, , INTEGER TYPE QUESTIONS, 129. A solenoid with inductance L = 7 mH and, resistance R 44 is first connected to a, source of direct voltage Vo and then to a, source of sinusoidal voltage with effective, value V = Vo. At what frequency (in KHz) of, the oscillator will the power consumed by the, solenoid be 5.0 times less than in the, former case?, 130. An LCR circuit has L = 10 mH, R = 3 and, C = 1F connected in series to a source of, , 15cos t volt. Compute the average energy, dissipated per cycle (in 10-4) at a frequency, that is 10% lower than the resonance, frequency. Give the answer in nearest integer., 138, , JEE-ADV PHYSICS- VOL- IV, 131. A series LCR circuit with R = 20 , L = 1.5 H, and C = 35 F is connected to a variablefrequency 200 V ac supply. When the, frequency of the supply equals the natural, frequency of the circuit, what is the average, power in Kw transferred to the circuit in one, complete cycle ?, 132. A sinusoidal voltage of peak value 283 V and, frequency 50 Hz is applied to a series LCR, circuit in which R 3 , L = 25.48 mH, and, C = 796 F . Find the impedance of the circuit., 133. Two resistors are connected in series across, 5V rms source of alternating potential. The, potential difference across 6 resistor is, 3Vm . If R is replaced by a pure inductor L of, such magnitude that current remains same,, then the potential difference across L is, L, , R, , 6, , 5V, , 134. In LCR circuit current resonant frequency is, 600 Hz and half power points are at 650 and, 550 Hz. The quality factor is, 135. An ac ammeter is used to measure current in, a circuit. When a given direct current passes, through the circuit, the ac ammeter reads 3 A., When another alternating current passes, through the circuit, the ac ammeter reads 4 A., Then the reading of this ammeter, if dc and ac, flow through the circuit simultaneously is, 136. In a series LCR circuit the voltage across the, resistance, capacitance and inductance is 10, V each. If the capacitance is short circuited,, the voltage across the inductance will be, , 10 x, , 2, what is value of x, 137. An ideal choke takes a current of 10 A when, connected to an ac supply of 125 V and 50 Hz., A pure resistor under the same conditions take, a current of 12.5 A. If the two are connected, to an ac supply of 100 V and 40 Hz, then the, current in series combination of above resistor, and inductor is, , 10 x, 2, , what is the value of x, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 138. In a region of uniform magnetic induction, B 10 2 tesla, a circular coil of radius 30 cm, and resistance 2 ohm is rotated about an axis, which is perpendicular to the direction of B, and which forms a diameter of the coil. If the, coil rotates at 200 rpm the amplitude of the, alternating current induced in the coil is___mA, , SUBJECTIVE TYPE QUESTIONS, 139. A series LCR circuit with L = 0.12H,C=480, nF, R = 23 is connected to a 230 V variable, frequency supply., (a) The source frequency 663 Hz, current amplitude, is maximum and this maximum value is 14.1 A., (b) At the source frequency 663 Hz average power, absorbed by the circuit is maximum and the value, of this maximum power is 2300 W., (c) At the frequencies 648 Hz, 678 Hz of the, source, the power transferred to the circuit is half, the power at resonant frequency. The current, amplitude at these frequencies is 10 A, (d) The Q-factor of the given circuit is 21.7, 140. An ac source of angular frequency is fed, across a resistor R and a capacitor C in series., The current registered is I. If now the, frequency of source is changed to / 3 ( but, maintaining the same voltage), the current in, the circuit is found to be halved. Calculate the, ratio of reactance to resistance at the original, frequency ., , 3, 5, 3, 4, B., C., D., 5, 3, 4, 3, 141. A circuit has a coil of resistance 60 ohm and, inductance 3 henry. It is connected in series, with a capacitor of 4 F and A.C. supply, voltage of 200 V and 50 cycle/sec. Calculate, A. the impedance of the coil is 943, B. the impedance of the coil is 843, C. the p.d. across the inductor coil is 1110V, D. the p.d. across the capacitor is 924 V, 142. A circuit consists of a noninductive resistor of, 50 , a coil of inductance 0.3 H and resistance, 2 , and a capacitor of 40F in series and is, supplied with 200 volt rms at 50 cycles/sec., Then, A. the current lag or lead by an angle150 51, B. the power in the circuit is 710.4 W, C. the power in the circuit is 640 W, D. the current lag or lead by an angle12051, A., , NARAYANAGROUP, , ALTERNATING CURRENTS, 143. A coil of resistance 300 and inductance, 1.0 henry is connected across an voltage, source of frequency 300 / 2 Hz. The phase, difference between the voltage and current in, the circuit is, , , , , A., B., C., D., 2, 4, 3, 6, 144. A circuit draws a power of 550 watt from a, source of 220 volt, 50 Hz. The power factor of, the circuit is 0.8 and the current lags in phase, behind the potential difference. To make the, power factor of the circuit as 1.0, The, capacitance should be connected in series with, it is, A. 75 F B. 60 F C. 50 F D. 65 F, , X L 2 a capacitor, X C 8 and a resitance 8 is connected, , 145. An inductor, , in series with an ac source. The voltage output, of A.C source is given by V=10cos 100 t ., A. The impedance of the circuit 10, B. The instant at which voltage between A and, B is equal to half of the voltage output from, 1, , 1, , 17 , , the source at that instant is 100 Tan 24 , , C. The impedance of the circuit is 8, D. The instant at which voltage between A and, B is equal to half of the voltage output from, 1, , 1, , 24 , , the source at that instant is 50 Tan 7 , , 146. In the figure shown V1 ,V2 ,V3 are AC, voltmeters and A is AC ammeter. The readings, of V1 ,V2 ,V3 and A are 10 V, 20 V, 20 V, 2A, respectively. Then, , A, , R, , C, , L, , V1, , V2, , V3, , , 50Hz, , A. the value of R is 5, 1, 103 F, , 1, H, C. the value of L is, 10, D. the value of source voltage is 10 V, , B. the value of C is, , 139
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 147. In the figure shown V1 ,V2 ,V3 are AC, voltmeters and A is AC ammeter. The readings, of V1 ,V2 ,V3 and A are 10 V, 20 V, 20 V, 2A, respectively. If the inductor is short circuited,, then, , A, , R, , C, , L, , V1, , V2, , V3, , , 50Hz, A. the reading of V1 is 2 5V, B. the reading of V2 is 4 5V, C. the reading of V2 is 2 5V, D. the value of A is, , 2, A, 5, , 148. In a series LCR circuit with an ac source, E0 50V , I rns 0.1 amp, R 300 ,, frequency, , PREVIOUS IIT QUESTIONS, 150. An inductor of inductance 2.0 mH is connected, across a charged capacitor of capacitance 5.0, F and the resulting L-C circuit is set, oscillating at its natural frequency. Let Q, denote the instantaneous charge on the, capacitor and I the current in the circuit. It is, found that the maximum value of Q is 200 C., (1998), (a) When Q = 100 C, what is the value of, |dI/dt|?, (b) When Q = 200 C, what is the value of ?, (c) Find the maximum value of I., (d) when I is equal to one-half its maximum value,, what is the value of |Q| ?, 151. When an AC source of emf e = E0 sin (100 t) is, connected across a circuit, the phase, difference between emf e and the current i in, , , as shown in, 4, the diagram. If the current consists possibly, only of R–C or R–L or L–C in series, find the, relationship between the two elements. (2003), , the circuit is observed to be, , i, , e, , 50, v, Hz . The average electric, , , field energy stored in the capacitor and, average magnetic energy stored in the coil are, 25 mJ and 5 mJ. Then, A. capacitance C of capacitor is 20F, B. inductance L of inductor is 2H, C. peak voltage of source is 50 V, D. the sum of rms voltage across the three elements, is 35.4V, 149. An LCR series circuit with 100 resistance, is connected to an AC source of 200V and, angular frequency 300 radians per second., When only the capacitance is removed, the, current lags behind the voltage by 600 . When, only the inductance is removed, the current, leads the voltage by 600 . The average power, dissipated in LCR circuit, A. 200 W B. 100 W C. 50 W D. 400 W, , 140, , (a) R 1k, C 10F (b) R 1k, C 1F, (c) R 1k, L 10H (d) R 1k, L 1H, 152. In an L-R series circuit, a sinusoidal voltage, V = V0 sin t is applied. It is given that L = 35, mH, R = 11 , Vrms = 220 V, /2 = 50 Hz, and = 22/7. Find the amplitude of current in, the steady state and obtain the phase, difference between the current and the, voltage. Also plot the variation of current for, one cycle on the given graph. (2004), V, , V=V0 somt, , T/4, , T/2, , 3T/2, , 2T, , t, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, 153. You are given many resistances, capacitors, and inductors. These are connected to a, variable DC voltage source (the first two, circuits) or an AC voltage source of 50Hz, frequency (the next three circuits) in different, ways as shown in Column II. When a current, I (steady state for DC or rms for AC) flows, through the circuit, the corrresponding voltage, V1 and V2 . (indicated in circuits) are related, as shown in Column I . match them (IIT-JEE 2010), , Column I, (A) i 0,V1 is proportional to i, (C) V1 =0,V2 =V, , (B) i 0,V2 V1, , (D) i 0, V2 is proportional to i, Column II, V1, , V2, , 6mH, , 3F, , (p), V, , V1, 6mH, , V2, 2, , (q), V, , V1, , V2, , 6mH, , (r), , 2, , V, , , V1, , V2, , 6mH, , 3F, , (s), V, , , V1, , V2, , 1k, , 3F, , (t), V, , , , NARAYANAGROUP, , ALTERNATING CURRENTS, 154. An AC voltage source of variable angular, frequency and fixed amplitude V connected, in series with a capacitance C and an electric, bulb of resistance R (inductance zero). When, is increased :, (A) the bulb glows dimmer, (B) the bulb glows brighter, (C) total impedence of the circuit is unchanged, (D) total impedence of the circuit increases, , LEVEL-VI - KEY, 1) B 2) B 3) D 4) B 5) A 6) C, 7) C 8) A 9) B 10) C 11) A 12) B, 13) B 14) C 15) B 16) C 17) C 18) D, 19) D 20) D 21) C 22) C 23) A, 24) (A),(B),(D), 25) (A),(C),(D), 26) A, B, C, 27) (A),(C),(D), 28) (B),(D), 29) (A), (D), 30) (A), 31) A 32) (A), (D), 33) (B),(C),(D), 34) (A),(B),(C), 35) (A),(B), 36) (A, C,D), 37) (B,C,D), 38) A,D, 39) B,C,D, 40) (A, B,C, D), 41) D 42) B 43) B 44) A 45) C 46) D, 47) A, B,C, D 48) B, C, 49) (A)50) (C), 51) (C)52)(A) 53) (C)54) (A)55) (B) 56) (A), 57) (A)58) (A)59) (A)60) (C)61) A 62) B, 63) B 64) A 65) C 66) D 67) A 68) B, 69) A 70) A 71) A 72) A 73) A 74) D, 75) C 76) B 77) A 78) C 79) C 80) D, 81) B 82) B 83) A 84) B 85) D 86) B, 87) C 88) B 89) A 90) B 91) A 92) C, 93) C 94) B 95) C 96) D 97) B 98) D, 99) A 100) B 101) D 102) B, 103) A-p,s,t B- p,s,tC-p, r D - p, q, 104) A- R , B - P C - S, D - Q, S, 105) A- R , B - S , C - P, D - Q, 106) A- q,, B - s, C- q, r, D - p, 107) A-S,, B-P, R, S, C- Q,S,T, D-Q, 108) A-s;, B-r; C-p, D-q, 109) A-r;, B-s; C-q; D-p, 110) A-p;, B-r; C-s; D-q, 111) A- p, s B - p, t C - q, t D - r, s, 112) A – s;, B – r; C – q; D – p, 113) A-q,r;, B-p,s; C-p,r; D-q,s, 141
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, 114) A-p,q,r; B-p,q,s; C-q; D-p,q,r,s, 115) A-p, s; B-q; C-p, r; D-q, 116) A-p,q,r,s; B-p,r; C-p,q,r,s;, D-r, 117) A-q, r; B-q, s; C-q, r; D-q, s, 118) A-Does not match; B-q, r; C-q,r,s; D-p, 119) C 120) D 121) B 122) A 123) A 124) A, 125) C 126) C 127) A 128) C 129) 2 130) 5, 131) 2 132) 5 133) 4 134) 6 135) 5 136) 1, 137) 1 138) 6 139) (A), (B), (C), (D)140) A, 141) A, C, D 142) A, B, 143) B 144) A, 145) A, B, 146) A, B, C, D, 147) A, B, D, , 148) A, B, C, D 149) D, , 150) (A) 104 A/s, (B) zero, (C) 2)0 A, (D) 1)732 × 10–4 C, , 151) (A), 152) 20 A,, 4, 153) A - r,s,t; B-q,r,s,t; C - p,q; D-q,r,s,t, 154) B, , 8., , Z R 2 2L2, , cos , , R, 1 100, , Z 200 , 2, 2, 2, , R2 2L2 Z 2, , 2, , , , 2L2 4 104 100 3 104 L 100 3, , , , 2 50 L 100 3 L , , 9., , The current leads in phase by X C X L , , i , , 10cos 100 t 37 0 , , For an ideal tranformer (100% efficient), Pin put Poutput V1I1 V2 I 2, 40 12.5 , VI, I1 2 2 , 2.5 A, V1, 40 5, , 3., , Frquency of the current remains same, only, magnitudes of current changes in a transformer., , 4., , I s Ps / Vs I s , , 5., , 6., , Vrms, R 2 L , , 2, , 4.4 103, 0.4 A, 11103, , 5.9 amp, , VL I L 148.2 volt., Resultant voltage = 200 volt, Since v1 and v3 out of phase, the resultant voltage, is equal to v2, , 7., 142, , fo , , 1, 2 LC, , , , v2 200 volt, 1, 1 1, 2 , , 10 6, 4, , XC X L 3, , R, 4, , 0, , = 37, , I m X C X L cos 100 t 37 0 90 0 , , The instantaneous potential difference across AB, is half of source voltage, 6 cos 100 t 530 5 cos100 t, , Solving we get, , n1 V1, 5 V1 , n V 1 40 , 2, 2, , , I, , tan , , The instantaneous potential difference across AB, is, , I dt, dt, , AC across L-R, L-C, L-C-R of circuit, 2., , cos 100 37 0 , , R=8, , 2, , I, , Z, , XC=XL=6, , RMS & Average value of AC, 1., , 100 3, 3, , 2 50, , , 37 0, , LEVEL-VI - HINTS, a b sin t ; I rms , , 2L2 Z 2 R2, , 1000Hz, , cos100 t , , 1, 1 7 / 24 , , 2, , , , 24, 25, , Instantaneous potential difference, 24 24, , = 5, volts, 25 5, 11. Since the circuit is at resonance so current in the, circuit is in the phase with applied voltage. Voltage, across inductor leads the current by / 2 and, across a capacitor lags by / 2 . So the voltage, across resistance is lagging by 900 than the voltage, across capacitor., 12. Voltage across the capacitor and inductor are same, i.e., VL VC IX L IX C, , X L X C (as they are in series), The circuit is in resonance. z = R, Hence, the voltmeter reading across resistor will, be 200 V., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, R, E, irms rms ; ER Ri, Z, Z, 24. It is given that in series circuit, instantaneous current, is zero when instantaneous voltage is maximum., , V 200, , 2A, R 100, V, 200, I', , 2A, X L X C 100, , 13. I R , , 2, R, , I, , I I, , '2, , 23. power factor cos , , 90o R 0, circuit may be of pure capacitor, pure inductor, , 2 2 A, , 14. Let: Vs Vs sin t, , I1 I 01 sin t / 2 ; I 2 I 02 sin t , , tan , , X, tan C ;, R, , , 1, , XC , , So phase difference = 2 ; tan R 2, , , V, X, I, , tan L, 2, 15., R, R2 X , L, , 16. Since Z L Z C . Current will be same,, VL 1 2 30 , , So, VZ L VZC, , 1, 60 volt, , , VR 80 1 80 volt, 2, L, , 20 10 , , 80 60, , 100 2, , 50 , , 2, , 2, , 2, , 2, , 30 30 X L 40, , 30 X L 10 , , X L 10 , , 2, , 100 volt, , X L X L X C2 , , 2, , 2, , 2, , , , E, R (maximum)., , So current is maximum and in phase with applied, voltage., 27. V 200 sin 100t , Comparing the equation with V Vo sin t we, have peak current, Vo 200volt Vrms , , 200, volt 100 2 volt, 2, , 2, , 2, , 28. Pav = VrmsIrms cos , If cos 1 then maximum average power =, If cos 1then average power will be less than, 1000 wattt., So correct option is b, d, , 2, , 29. Here, , 80 20; X L 10 40; X L 50, , 2, , Z R 2 X L X C 60 2 100 20 100, Power factor, cos R / Z 60 /100 0.6, , , 1 , 19. | Erms Ri j L i, , C , , Erms Ri j VL VC , Erms RI, , V 220 V, I 220 /100 2.2 amp, 2, Erms, cos some impedance should be added, Z, , in series such that resonance is to be obtained to, become power factor unity., NARAYANAGROUP, , R 2 XL XC , , 0, , VrmsIrms = 1000 watt., , 18. Total resistance R 20 40 60, The impedance of LCR circuit is given by, , 22. Pave , , I, , XL XC, 0, R, E, , 100 2f 100 f 50 Hz, 2, , 2, R, , V V V , 17. V I, , or combination of inductor and capacitor., 25. At resonance XL XC and Z = R (minimum), , X L L 2fL 2 50 , , 0.4, 40, , , R = 30 , 2, 2, 2, 2, Z R X L 30 40 50, , Irms , , Vrms 200, , 4A ., Z, 50, , 30. R 11, X L X C 120 , 110, 10 A, 11, VC VL 120 10 1200 V ; VR 110 V, 31. Energy = Average power time, when resistance is removed, cos 0, Z 11, I , , , , V, , ; i 0 sin t , 2, Z, 143
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JEE-ADV PHYSICS- VOL- IV, XL, , Voltage gain , , ALTERNATING CURRENTS, , 2, , R, 48. Brightness is proportional to power dissipated, , P , , E R, Z, , 58. AC ammeters are based on heating effect of, current. Heat produced depends on RMS current., Hence AC measuring instrument measures its rms, value., , VR 2.0V sin 10 t , , R, 100, 3, , 49. i , , 2.0 102 A sin 103 t , , 50. XL L 10 4H 4.0 10 ohm, 51. Amplitude of voltage across inductor, V0 I0 X L, 3, , 3, , 59. (a), , 52. rms value of voltage across the source, Vrms , , 2, , Vrms, , |Z|, , , , Vo2dt , , 0, , 2, , (c) Vrms , , R 2 X L X C , , 2, , 2 2 T, .Vo , T, 2, , T/2, , , , 2Vo, 2, , V 2dt , , 0, , , , 2, , 60. Vav , , 1, , , 1000 2 , 6 , 1000 1 10 , , , 2, , 2 Vo, , Area of the graph for time t 0 to T, Time int erval, , 61. P irms Vrms cos ; Vrms 260 volt, irms , , Since the curernt will be same every where in the, circuit, therefore, P.D. across resistor, , Vrms 260, R 10, , 10 ; P.F. = cos , Z, 26, z 26, , 2, , , 1 , , 2, z R L , , , 26, c , , , , , , , VR i rms R 0.0707 1000 70.7 volts, , 53. P.D. across inductor, , 10 , P 10 260 1000 w.., 26 , , VL i rms XL 0.0707 1000 2 141.4 volt, , 54. P.D. across capalitor VC i rms X C, 62. Resonance frequency =, , 1, 0.0707 , 70.7 volts, 1 1000 106, VS, 1 , , R 2 L , C , , , 55., Vout, , Vs, , 3, , 1 T, 2Vo 0, VT V, 2, 2, Vav , o o, T, 2T, 2, , 0.0707 amp, , Vout , , 2Vo, , (d) Vrms Vo So correct option is (a)., , 100, 2, , 0, , (b) Vrms T, , Vrms, , 1 , , R 2 L , , C , , , 1000 , , V 2dt, , , , T /2, , 100 Volt, , Vrms, , , , 2, T, , T /2, , Vrms Vo, , From question, 1000 rad / s, i rms , , 2, T, , Vrms , , 2, Vrms, , , 2.0 102 A 4.0 103 ohm = 80 volts, 100 2, , Vout, 1, VS, , ,, , 57. As, , 2, rms, 2, , V out, CR, VS, , 56. As 0,, , 2, , 1 , , R L , , C , , , NARAYANAGROUP, , R 2 2L2, , 2, , 10 4 10 , 2, , ;, 70. i0 , , R2 2L2, 2, , 1, , 2, , , , R2 2L2, 1 , , R 2 L , , , C, , , 4, , , , 1, 2 LC, , 1, 100 250, , 2 2 103, , 4, , , , i0, E, E, , 2, and 2, 2, R X L XC , R, , 2, , Solving 3R , , 103, R 30, 6, 145
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , 72. Q , , N, , 1, 103, ;L , mH 0.05mH, LC, 500 4 2, , 2, 71. 0 , , 0 L 314, , 10.4, R, 30.01, , 73. RB 3; Z B 5, Z B RB2 X B2 4; p. f , , 3, 0.6, 5, , 100, 1, 2400 2000, Hence N 2 V V2 , , 97. V1i1 V2i2 (or ) P1 P2, 99. Transformer does not work with DC input, L, 1, 100. Tan1 R ; Tan2 CR, 1, 2, 1 2, , 2, rms, , 74. P i RC, 75. RA 6, RB 3, RC 8, , 101., , Z A 7, Z B 5, ZC 10, R(total) = 6+3+8 = 17, , i1 , , 200 2, R12 L , , and i2 , , X A 7 2 62 3.6, , X B 4; X C 6, X(total) = 3.6+4+6=13.6 , Total impedance = R 2 X 2 21.8 m, Power factor =, , R, 17, , 0.78, Z 21.8, , 87. BA cos t, , irms , , 0.1 103 63, , , , 2, , 1, 2, , , 2 63 0.1 , 10 , 63 0.01 , , , , 2, , Z R2 XL XC R, , B 2 A2 2 R , 88. E Vrms irms cos t , 50, 2 Z Z , 2, , 0.1 2 103 632 10 50, 2, , 1, 5, 2, , , 2 63 0.1 , 10 8.12 10 J, , , , 63, 0.01, , , , , 89. , N, , 1, , LC, N, , 1, , 0.1 0.01, , 31.6 rad / s, , 1, 2, 96. V V (we can see from the table), 1, 2, , 146, , then, , I Io sino t, , where, , Eo, R, , Frequency of current is saem as frequency of, , alternating voltage source., tan , , XL XC, 0 0, R, , current is in phase with applied voltage., (B) Only resistor in an ac circuit, Frequency of current is same as frequency of, applied voltage., Reactance of the circuit is zero., , 4.0 10 5 A, , , , , , sin 100t , 2, 6, , 1 , R22 , , C , , 102. At any time total current i i1 i2, 103. (A) R-LC circuit at resonance, At resonance, XL XC Net reactance = 0., , Io , , vrms, BA / 2, , 2, Z, 1 , , 2, L , R, , C, , , , 2, , , , sin 100t , 3, , , 200 2, , If E Eo sin o t, , d, e, BA sin t, dt, , 120, , 1, , tan , , X, 0 0, R, , So current is in phase with applied voltage., (C) Only inductor in ac circuit, Let applied voltage E Eo sin t, , , , , Current in the circuit, I Io sin t 2 , where, , , Io , , , , Eo, XL, , Frequency of current is same as applied voltage., Current lags behind applied voltage., (D) Only capacitor in an a.c. circuit, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , , , , , , Current I Io sin t , 2, , , Frequency of current is same as applied voltage., Current leads the applied voltage by, , , ., 2, , 104. (A) Square wave having peak value Vo., Vav , 2, Vrms, , 2, T, , T/2, , 2, , T, , Vdt Vo, , , 0, , V 2dt Vrms Vo, , 0, , Vo Vrms Vav, , (B) Sinusoidal wave: V = Vo sin t, Vav , , (C) If 2r then XL XC . So voltage will lead, the current., When r , XL XC ., If 2r then, X 'L 2r L 2r L 2XL 2XC ., , T/2, , , , (B) If r and XL XC , Z = R., The circuit will be resistive in nature, so the current, will be in phase with voltage. B S , , 2Vo, V, , Vrms o, , 2, , So if 2r then new inductive reactance will be, two times the capacitive reactance in magnitude., (D) If r then XL XC . The circuit will be, capacitive in nature., So the current will lead the applied voltage., T, , Vo Vrms Vav, , (C) In pure capacitor circuit current leads the, , voltage by ., 2, , (D) Wattless current flows due to reactance of the, circuit., (i.e. XL, XC or both), 1 , , 2, 105. Z R L , C , , , 2, , (a) At resonance, XL XC . So Z = R = 40, (b) At resonance Z = R = 40 ;, 200, 5 A Io 5 2 A, 40, 1, 1, 50, (c) o LC , 5 80 10 6, Irms , , VL Irms XL 250 5 1250 volt, , (d) At resonance, VR IR 5 40 200 volt, , XL, , 106., , XC, r, , , , (A) If r then XL XC , the circuit will be, inductive in nature. So voltage will lead the current., NARAYANAGROUP, , T/2, , 2, T, , In half cycle, Iav , , Idt, 0, , T, , 1, , In full cycle, Iav T Idt, 0, Now calculate for different wave form., (A) Irms , , Io, 2, , 2Io, , I, , 0, In full cycle, av, ., , In half cycle, Iav , , (B) I , 2, Irms, , , XL oL 50 5 250 , , XL, , 1 2, I dt, T 0, , 107. Irms , , Irms , , 4Io, t, T, , 0t, , T /4, , 4, T, , , , I2dt , , 0, , Io, 3, , 4, T, , T, 4, , T /4, , , 0, , 4Io, T, , , 2, , I, , t dt o, 3, , , In half cycle,, , 1 T, I, area 2 2 o Io, Iav , , , T, time, 2, 2, , In full cycle Iav 0, (C) I Io, 2, Irms, , , 2, T, , 0 < t < T/2, , T/2, 2, , 2, , 2, o, , I dt T .I, 0, , , , T 2, Io ;, 2, , Irms Io, , 147
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , In half cycle,, , Ia v, , T, Io, a rea, , 2, Io, T, tim e, 2, , In full cycle Iav 0, (D) I = Io 0 < t < T/2, I=0, T/2 < t < T, 2, rms, , I, , Io, , In full cycle, Iav = Io/2, 110. Impedance Z R 2 X 2, , 37, , V3 5V current is lagging, , 0, , 4, 4, , -370, (B), , 5, , 2, , I dt, Irms , , 0, T, , dt, T, , If I Io sin t then, Irms , , Irms , , 2, I dt, T 0, , ;, , T, I, 1 2, Io sin tdt o, , T0, 2, , So both statements are true but statement-2 is not, correct explanation of statement-1., 122. In a series R-L-C circuit resonance can takes, place, because at a particular frequency of ac, source, the impedance can be minimum., , Reactance X X L X C, 3, 5, 111. (A), , T, , 0, , In half cycle, Iav = Io., , 2, , E0, X XC, ; Tan L, Z, R, , 121. Definition of rms current,, , T, T/2, T, 1 T I2, 1, 1, I2 dt Io2 dt 0dt Io2 o, T0, T 0, T/2, T 2 2, , Irms , , 115. i0 , , V3 5V Voltage is lagging, , 3, 3, , Z R2 XL XC , , 2, , Resonance means current is maximum , Impedance is minimum XL XC ., 123. In series R-L-C circuit Z R2 XL XC 2, At resonance I should be maximum, Impedance Z should be minimum, XL XC and the minimum impedance is R (at, resonance), 124. Z R2 XL XC 2, , V3 1V Voltage is lagging, , (C), , o, , 4, 4, (D), , , , The, , EC X C i ; EL X L i, The voltage of AC source means RMS voltage, 2, , 148, , 2, , ; , , 1, 1, fo , 2foC, 2 LC, , 129. In case of direct current, PDC , , 112. ER Ri and i 5 A, , 1 , , 2, 114. Z R L , , C , , 2foL , , Resonance frequency depends on L and C., , 3V 7V Current is lagging, 3, , E ER2 EL EC , , 1, , At resonance XL XC oL C, , 1, LC, , impedance, , of, , VM2, R, , LR, , circuit, , Z R2 XL2 R2 2L2, , VM, VM, I, , , M, , Z, R 2 2L2, 2, PAC IMR , , VM2R, R2 2L2, , According to the problem, PDC PAC, , NARAYANAGROUP, , is
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JEE-ADV PHYSICS- VOL- IV, VM2, , R, , or, , or , , VM2R, L 2 , R2 1 , , R , , or, , 1, , R, , ALTERNATING CURRENTS, , L 2 , R 1 , , R , , R 2, R 2, 1 2v , 1, L, L, , R, 2 1 2kHz, 2L, 1, 130. 1 , 10 4 rad / s, o, , 2, , 6, LC, 10 10, v, , w wo , , Hence, , 2 3, 1, , ; i A, , R, , 4, , R 6, 2, When R is replaced by L, i does not change, 1, 5, , ; X L 8, 2, 2, 4 X L2, , 133. Initially i , , 10, o 9 103 rad/s, 100, , 1, VL iX L 8 4V, 2, Resonant frequency, f0, , 134. Q , f 2 f1, Band width, 135. At an instant current in the circuit is, , i 3 4 2 sin t, , XL L 9 103 102 90 , , 1, 1, XC , , 111.11, 3, C 9 10 106, X XC XL 111.11 90 21.11, , T, , Reading of ammeter is i , , 1 2, i dt, T 0, , 136. The source voltage is, 2, , V 10 2 10 10 10V, , Impedance Z R2 X2 3 2 21.112, , After capacitor is short circuited, let V1 be voltage, across each of inductor, resistor, 10, 10 V12 V12 2V1 ; V1 , 2, , Z 21.32 , , Io , , =, , 15, i2, 2, 0.704 A ; Pav irms, R oR, 21.32, 2, , 1, 2, 0.704 3 0.74 watt, 2, 3, , f, , W 9 10, , cycle / s So energy per cycle, 2, 2, , 0.74, 5.16 10 4 5 10 4 Joules / cycle, 9 10 3 , = , , 2 , , 131. When frequency of supply is equal to natural, frequency,, , 125, 12.5, ; L , 10, 100, 125, 10, and R , 12.5, In the series combination of L and R, 100, 100, 10, i, , , 2, 2, 2, 2, R 2 40 L , 2, 10 10, , 137. 2 50 L , , NBA, R, , then XL XC . Z R2 XL XC 2 R, , 138. i0 , , V, V 2 200 200, 1, , R, R, 20, Pav 2000 watt 2kW, , 139. (a) Amplitude of current is maximum at resonance, , Pav VIcos V., , 132. To find the impedance of the circuit, we first, calculate X L and X C ., X L 2fL, , 2 3.14 50 25.48 103 8, 1, 1, XC , , 4, 2 fC, 2 3.14 50 796 106, , Therefore,, Z R 2 (X L X C ) 2 32 (8 4) 2 5, NARAYANAGROUP, , frequency., fr , , 1, 2 LC, , Irms , , , , 1, 2 3.14 0.12 480 10 9, , Vrms 230, , 10A, R, 23, , 663 Hz, , Io Irms . 2 14.14 A, , (b) Resonance frequency, power absorbed by, circuit is maximum., So f = 663 Hz., 2, , 2, Pav Irms, R 10 23 2300 watt, , (c) Q factor =, , r, L, R, r , 2 R, 2L, 149
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, , , 23, 23, , 95.833, 2 0.12 0.24, , tan , , 2 f ~ fo 95.833 f ~ fo 15 Hz, , f fo 15 663 15 648 Hz and 678 Hz, I, 1, 1, Pmax Io2R I o 10 Amp, 2, 2, 2, , (d) Q factor, =, , 1 L, 1, 0.12, , 21.7, R C 23 480 109, , I' , , , , 2, , R X, , V, R 2 3XC2, , V, , I, , 140. Z R 2 XC2 ,, , (220) 2 0.8, 70.4 ohm, 550, R, Now, power factor cos or R = Z cos , Z, R = 70.4 × 0.8 = 56.32 ohm, Further,, , So, Z , , 2, C, , R 2 X2C, I', , I, R2 9XC2, , , , 1 R 2 X 2C, , ; R2 9XC2 4R 2 4X2C, 4 R2 9XC2, , 5X2C 3R2 ;, , Z 2 R 2 (L) 2 or (L) (Z2 R 2 ) Or, , XC, 3, , R, 5, , L (70.4)2 (56.32) 2 42.2 ohm, When the capacitor is connected in the circuit,, , 141. Z R2 XL XC 2 ; Irms , , V, 2, , 2, 2 , 1 , Z R L , and, C , , , , Impedance of coil = Z1 R2 XL2, Potential drop across the coil, , cos , , = Irms Z1 Irms R2 XL2, Potential drop across capacitor = VC Irms XC, 142. Total resistance R 50 2 52 , L = 0.3 H, C 40 106 F, , R, 2, 2 , 1 , R L , , C , , , , when cos 1, L , , 2, , Z R2 XL XC ;, , , , XL L 2 50 0.3 30 , , 1, 1, , 250 , C 2 50 40 106, XC XL so current leads the applied voltage., , , , XC , , C, , Irms , , 2, , 30 250 , , 143. XL 2fL 2 , , 1, 1, , (L) 2f (L), , 145., 2, , Vrms, 2, , XC=8, , A, , 8, B, , , , Pav ErmsIrms cos Erms, , 150, , 52 , , 1, C, , 1, 6, ., (2 3.14 50) (42.2) 75 10 F 75F, XL=2 , , 2, , Z R2 XL XC , , (lagging), , 144. As the current lags behind the potential difference,, the circuit contains resistance and inductance., Power, P = vrms × irms × cos ;, Here,, V, i rms rms , where Z [(R 2 (L) 2 )], Z, , 2, 2, Vrms cos , v rms, cos , P, or Z , ;, Z, P, , Let f be half power frequency., , I2R , , , XL, 1 , R, 4, , 2, Erms R Erms, R, , 2, 2, 2, , 300, 1 300 ; R 300 , 2, , a) Impedance of circuit =, , R2 X C X L , , 2, , 2, , Z 82 8 2 10, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- IV, b) The current leads in phase by X C X L , , 37 0, i , , 10cos 100 t 37 0 , Z, , XC=XL=6, , tan , , = 37, , cos 100 37 0 , , XC X L 3, , R, 4, , 0, , R=8, , The instantaneous potential difference across AB, is, I m X C X L cos 100 t 37 0 900 , 6 cos 100 t 530 , The instantaneous potential difference across AB, is half of source voltage., 6 cos 100 t 530 5 cos100 t, , 1, 17 , 7, t , Tan 1 , 100, 24, 24 , 146. Let I r be the rms current through the circuit then, , Solving, Tan 100t , , I, I r 2 A, r 20V , I rC 20 V and I r R 10V, C, Solving we get, 1, 1, R 5, C 103 F and L =, H, , 10, Vs source voltage =, Ir , , , 1 , , R2 L , , C , , , Ir R , , 2, , Ir R , , 2, , I , , I r L r , C , , , 2, , 2, , 102 20 20 10 volts, After the inductor is shorted, Vs, 10, 2, Ir , A, , , 1, 25 10, 5, 2, R 2 2, C, Ir, 4 5 volts, v1 I r R 2 5 volts; v2 , c, 148. Av.Electric field energy =, 1, 2 , 3, CVrms 25 10 J, 2, , 1, 2, C I rms X C 25 103 J, 2, 1, 1, C I 02, 25 10 3 J, 2 2 2, 2, 2 v c, C 20 F, 1 2 , Average magnetic energy LI rms L 2H, 2, , , L , , 2 5 103, , .10 , , 2, , L =1 heriry, , VR I rms R VC 50V = VL I rms L, , .10 .300, , .10 , , 1, 50 , 2 .20 106, t, , 50, I , 2, VR 30V VC 50V VL 10V, 50, rsm voltage of source Erms , 2, Erms 35.36V, 150. (a) 104 A/s (b) zero (c) 2.0 A, (d) 1.732 × 10–4 C, This is a problem of L – C oscillations, Charge stored in the capacitor oscillates simple, harmonically as Q = Q0 sin ( t ± ), Here, Q0 = maximum value of, Q = 200 C = 2 × 10–4 C, .10 2, , 2, , 2, , 102 20 20 10 volts, 147. Let I r be the rms current through the circuit then, I, I r 2 A, r 20V , I rC 20 V and I r R 10V, C, Solving we get, 1, 1, R 5, C 103 F and L =, H, , 10, Vs source voltage =, 1 , , Ir R L , , C , , NARAYANAGROUP, , , , 2, , I , , Ir L r , C , , , 2, , ALTERNATING CURRENTS, , 2, , 1, , 1, , = (2 10–3 H )(5.0 106 F ) = 104 s–1, LC, Let at t = 0, Q = Q0 then, … (1), , =, , 151
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JEE-ADV PHYSICS- VOL- IV, , ALTERNATING CURRENTS, Q(t) = Q0 cos t, dQ, I=, = – Q0 sin t and, …(2), dt, dI, = –Q0 2 cos ( t) …(3), dt, Q, 1, (a) When Q = 100 C or 0 , cos t , 2, 2, , From Eq. (3) :, 1, dI, –4, 4 –1 2 , =, (2.0, ×, 10, C)(10, s, ), 2, dt, dI, 4, dt = 10 A/s, , (b) When Q = 200 C or Q0 then cos ( t) = 1, i.e., t = 0, 2 …., or I = 0, (c) Imax = Q0 = (2.0 × 10–4 C)(104 s–1), Imax = 2.0 A, (b) From energy conservation., 1 Q2, 1 2, 1, LI max = LI2 +, or Q =, 2 C, 2, 2, I, I = max = 1.0 A, 2, , 2, LC ( I max, I2), , Q = (20 103 )(5.0 106 )(22 I 2 ), Q = 3 × 10–4 C or Q = 1.732 × 10–4 C, 151. (a), Current leads voltage in RC cobination., XC, 1, , , 1 , R CR, 4, , 152. 20 A,, 4, tan , , Inductive reactance, XL = L = (50)(2 )(35 × 10–3) » 11, 1, Impedance Z= R 2 X L2 = (11) 2 (11)2 = 11 2 , Given vrms = 220 volt, Hence, amplitude of voltage, v0 = 2 vrms = 220 2 volt, Amplitude of current, i0 =, , 220 2, v0, =, = 20 A, Z, 11 2, , Phase difference, –1, , f = tan, , XL , , , R , , 11 , = tan 11 , –1 , , =, , , 4, , In L – R circuit voltage leads the current., Hence, instantaneous current in the circuit is,, i = (20A) sin ( t – /4), Corresponding i – t graph is shown in figure., , 152, , v, t, , V 220 2 stv , , t 20 sin (t /4), , 20, 0, , T, , 9T/8, , t, , T/8 T/4 T/2 5T/2, 10 2, , 153. In the Circuit (P), as i is the steady state, current, V1 0, i 0 . Hence V2 V ., So (P) C,, In the Circuit (Q), in the steady state, V1 0 as, di, 0 . So V2 V iR, dt, (OR) V2 i and V2 V1, So (Q) B,C,D., In the Circuit (R), the inductive reactance, X L L 6 101 and resistance, R = 2 ., So V1 iX L and V2 iR . Hence V2 V1 ., So (R) A,B,D., In the circuit (S), V1 iX L and V2 iX c , Where, , 10 4, Xc , . So V2 V1 and V1 i also, V2 i ., 3, So (S) A,B,D, In the circuit (T), V1 iR , where R 1000 ,, 10 4, 3, So V2 V1 and V1 i also, V2 i ., So (T) A,B,D, So (A) - r,s,t; (B) - q,r,s,t;, (C) - p,q; (D) - q,r,s,t, di q, Note: For circuit (p), V L 0, dt C, di, d 2i dq, OR CV CL q 0 CL 2 , OR, dt, dt, dt, 1, , d 2i, 1 dq, t ., , So, i io sin , 2, dt, CL dt, CL, , As per given conditions, there will be no steady, state in circuit .p.. So it should not be, considered in options of .c.., Vrms, i , 154. rms, 1 when increases, irms, R2 2 2, c, increases so the bulb glows brighter, and V2 i X c with X c , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , ELECTROMAGNETIC WAVES, SYNOPSIS, , , , , According to Maxwell, an accelerated charge, produces a sinusoidal time-varying magnetic field,, which in turn produces a sinusoidal time varying, electric field. The two fields so produced are, mutually perpendicular. They constitute electro, magnetic waves which can propagate through, empty space., , Displacement current:- According to, , , , C2, , , , Ampere’s circuital Law, the magnetic field B is, related to steady current I as, , , , B . dl , , 0, , I ....(i), , where I is the current travelling through the surface, bounded by closed loop., In 1864, Maxwell showed that relation (i) is logically, inconsistent. He accounted for this inconsistency, as follows: Consider a parallel plate capacitor, having plates P and Q being charged with battery, B., Q, , P, E, , C/S area = A, , B, , B, , , , r, , C1, , , , , , ic, , C2, , During charging, a current IC flows through the, connecting wires which changes with time. This, current will produce magnetic field around the wires, which can be detected using a magnetic compass, needle. Consider two loops c1 and c2 parallel to, the plates P and Q of the capacitor. c1 is enclosing, only the connecting wire attached to the plate P of, the capacitor and c2 lies in the region between the, two plates of capacitor. For the loop c1 , a current, I is flowing through it, hence Ampere’s circuital law, , B.d l 0 I, .... ii , for loop c1 gives , , C2, , Thus, relation (iv) is in contradiction with relations, (ii) and (iii). This led Maxwell to point out that, Ampere’s circuital law as given by (i) is logically, inconsistent., , d, , , , The relations (ii) and (iii) continue to be true even if, two loops c1 and c2 are infinitesimally close to the, plate P of the capacitor. In the other hand, as the, loops c1 & c2 are infinitesimally close, it is expected, , , B.dl B.d l ... iv , , , that, C1, , R, ic, , Since the loop c2 lies in the region between the, plates of the capacitor, no current flows in this, region. Hence Ampere’s circuital law for loop c2, , B.dl 0 .... iii , gives , , , , Idea of Displacement Current : Maxwell, predicted that not only a current flowing in a, conductor produces magnetic field but also a timevarying electric field (i.e., changing electric field) in, a vacuum/free space (or in a dielectric) produces a, magnetic field. It means a changing electric field, gives rise to a current which flows through a region, so long as the electric field is changing there., Maxwell also predicted that this current produces, the same magnetic field as a conduction current, can produce. This current is known as, ‘displacement current’., Thus, displacement current is that current which, comes into play in the region in which the electric, field and hence the electric flux is changing with, time., Maxwell defined this displacement current in space, where electric field is changing with time as, ID 0, , dE, dt, , ... v , , where E is the electric flux., , C1, , NARAYANAGROUP, , 1
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , , i i0e t / , , Maxwell also found that conduction current I , and displacement current I D together have the, , , , property of continuity, although, individually, they, may not be continuous., This idea led Maxwell to modify Ampere’s circuital, law in order to make the same logically consistent., He states Ampere circuital law in the form,, , B.dl 0 I I D , , R, , as, , C, , , , d , , 0 I 0 E , dt , , It is now called as Ampere-Maxwell’s law., This means that out side the capacitor plates, we, , i, , i0 ', i .(say), e, , (Where e is the base of natural logarithm)., If Q be the charge at the mentioned instant then,, the electric field between the plates is, , have only conduction current ic i and no, E, , displacement current id 0 . On the other hand,, inside the capacitor, there is no conduction current, , E ' E(R / 4)2 , , d, E', dt, , Thus, id , corresponding to r will be, , , , , 1 dQ 1(i'), i, 0, , , 160 dt 160 16e0, , Maxwell, in 1862, gave the basic Laws of, electricity and magnetism in the form of four, fundamental equations which are known as, Maxwell’s equations. In the absence of any, dielectric and magnetic material may be stated in, the integral form as below., , 1. Gauss’s Law for electrostatics :This Law gives the total electric flux in terms of, charge enclosed by the closed surface., q, in, In the usual notations E .dS , 0, , This Law states that electric lines of force start from, positive charge and end at negative charge i.e.,, electric lines force do not form closed paths., , 2. Gauss’s Law for magnetism : , Mathematically B.dS 0, , , 2, , , Q, , 160, , MAXWELL’S EQUATIONS, , 2, , (ii) Eq. Reflects that the magnetic field induction B, varies linearly with r: so that it is zero at the axis (r, = 0) and maximum at the periphery of the cylindrical, volume enclosing the plates (i.e., r = R), W.E-1: A circular parallel plate capacitor with, plate radius R is charged by means of a cell,, at time t=0. The initial conduction current is, i0 . Consider a circular area of radius R / 4, coplanar with the capacitor plates and located, symmetrically between them. Find the time, rate of electric flux change through this area, after one time constant., Sol:, The conduction current at the end of one time, constant can be obtained by substituting t in the, expression, , Q R2, , 0 (R)2 16, , Rate of electric flux change is, , i, of the constant current density j given by j = c 2, R, , , , , Q, , 0 0 (R)2, , The electric flux through the specified area is, , ic 0 and there is only displacement current id i, Note: (i) Between the capacitor plates the, displacement current can be treated as the output, , r, id j r 2 ic , R, , R/4, , This Law shows that the no. of magnetic lines of, force entering a closed surface is equal to no. of, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , , , 3., , , , , 4., , , , , , 5., , magnetic lines of force leaving that closed surface., This law tells that the magnetic lines of force form a, continuous closed path., This Law also predicts that the isolated magnetic, monopoles does not exist., , Faraday’s Law of electro magnetic, induction : d, B, induced emf., Mathematic cally E.dl , dt, This law gives a relation between electric field and, changing magnetic flux., This law tells that changing magnetic field is a source, of electric field., , ELECTROMAGNETIC WAVES, , , , Characteristics of EM waves :, 1), 2), 3), , Amper’s-Maxwell’s Law :-, , , Mathematically B.dl 0 ic id , , d , , 0 ic 0 E , dt , , This law states that magnetic field can be produced, by a conduction current as well as by displacement, current., At any instant in a circuit, conduction current is, equal to displacement current., Lorentz Force :- Force acting on a charge ‘q’, moving in a region where electric and magnetic fields, similar to EM waves are existing simultaneously is, , , F q E v B, , , , , , W.E.-2: What is the instantaneous displacement, , 4), 5), , EM waves can be polarised., EM waves are self-sustaining oscillations of electric, and magnetic fields in free space or vaccum. EM, waves travel through vaccum with speed of light, 1, 8, ‘C’ where. C 3 10 m / S ec, 0 0, , charging at rate of 1 0 V / S, , , , 6), , 0 A dV, dV, C, ; I d 106 106 1A, d dt, dt, , The speed of EM waves in any other medium of, permittivity and permeability is, Cmed , , Source of EM waves :, , , , , Accelerated charges radiate energy in the form of, EM waves. So it is source of EM waves, An oscillating charge produces an oscillating electric 7), field inturn which produces an oscillating magnetic, field., The oscillating electric and magnetic fields, NARAYANAGROUP, , B B0 sin kx wt , , In this, E0 & B0 are the amplitudes of the fields, , 6, , dE, d, d V , 0A E 0 A , dt, dt, dt d , , EM waves are transverse in nature whose speed is, same as that of speed of light, , , The two fields E and B have same frequency of, oscillation and they are in phase with each other., Keeping these features in mind, we can assume, that if EM wave is travelling along positive direction, along x-axis, the electric field is oscillating parallel, to the y-axis and that magnetic field is parallel to zaxis, then we can write the electric and magnetic, fields as sinusoidal functions of position ‘x’ and time, ‘t’, , E E0 Sin kx wt ;, , current in space between plates of parallel, plate capacitor of capacitor 1 F which is, , Sol. As I d 0, , regenerate each other and propagate through space, as waves called EM waves., An electric charge oscillating harmonically with, frequency ' ' produces EM waves of same, frequency., , r r , , 1, , , , C0, 1, , r 0 r 0, r r, , C0, n R.I.of medium, C med, , In vaccum, EM waves are of different wavelengths,, but velocity is same., , 3
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , W.E.-3 : Electro magnetic waves travel in a, , 1, B2, 2, u 0 E dV , dV, 2, 20, , medium with speed of 2 108 m / sec . The, relative permeability of the medium is 1 find, relative permittivity., , , Sol. Give C 2 108 m / sec, r 1, Speed of EM waves in medium, C med , , 1, , r 0 r 0, , C0, r r, , ;, , r , , 2, 0, , 3 10 8 , , 2, , c, , 2.25, 2, c2r, 2 10 8 1, , W.E-4: Suppose that the electric field amplitude, of an EM wave is E0 120 N / C and that its, , Energy density of EM wave is U , , B02, 1, 2, 2, , E, sin, kx, , wt, , sin 2 kx t , , , =, 0 0, 2, 20, If we take average over a long time, the sin 2 terms, have an average value of, , frequency 50HZ . Determine (a) B0 , , , and K (b) Find expressions for E and B, E0, Sol. a) i) Using C B we get, 0, B0 , , 120, E0, , 4 107 T = 400nT, C 3 108, , ii) 2 v 2 50 106 3.14 108 rad / Sec, iii) c , iv) K , , C 3 108, , 6m, 50 106, , 2 2 2 3.14, , , 1.05m 1, , 6, 6, , , b) E E0 sin kx t , , 1, C2, 2, , uE , , B0, 1, 1, 1, 1, 0 E 02 0 C 2 B 02 0, B 02 =, 4, 4, 4, 0 0, 4, , uB, 0, , Hence is an EM wave, average energy density of, electric field is equal to average energy density of, magnetic field., The units of uE & u B are Jm 3, average energy density of EM wave, , Energy density of EM waves:, Consider a plane electro magnetic wave propagating, along x-axis. The electric and magnetic fields in a, plane EM wave can be given by, , total EM wave energy, i.e. Intensity I Surface area time, , 400 10 sin 1.05 x 3.14 10 t , 8, , E E0 sin kx t and B B0 sin kx t , , 4, , Now E 0 CB0 and 0 0 , , Intensity of EM wave is defined as the energy, crossing per second per unit area of a source, perpendicular to the direction of propagation of the, wave. It is denoted by I., , 9, , , , 1, B2, 0 E02 0, 4, 4 0, , Intensity of electro magnetic wave:, , B B0 sin kx t , , , , Thus u aV , , 1, 2, , 1, B2, u u E u B 2u E 2u B 0 E02 0, 2, 2 0, , = 120 sin 1.05 x 3.14 108 t , , , , 1, B2, 0E 2 , 2, 2 0, , In any small volume ‘dV’, the energy of electric, , uaV V uaV Act, , At, At, , 1, 2, field is U E 0 E dV and energy of the magnetic, 2, , 1, 2, Interms of electric fields I 0 E0 C ---------(1), 2, , B2, U, , dV, field in volume ‘dV’ is B, 2 0, , B02, C ---------(2), Interms of magnetic field I , 20, , Thus t otal energy of, , Either eq (1) or (2) may be used to find intensity of, EM waves, , EM wave is, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , ii) The intensity of EM radiation from an isotropic, P, where P, point source at a distance r is I , 4 r 2, is power of source, Note: The rate of flow of energy crossing a unit, area in an EM wave is described by the vector ‘S’, called poynting vector which is described by the, expression., 1 , S, EB, 0, , , Since E and B are mutually perpendicular, , E B EB, , , , , , Thus magnitude of Poynting vector, S, , SI unit of S is J sec 1 m 2 (or ) Wm 2, This relation shows that the value of electric vector, at any instant in the EM wave is about 377 times, the value of magnetic vector. It is because of this, reason, optical properties of light is due to electric, field. Average of Poynting vector is given by, E0 B0 1 2, CB02, , E0 C , 20 2, 20, , W.E.-5 : The electric field of an electro magnetic, x, wave is given by E 50 sin t N / C ., c, Find energy contained in a cylinder of crosssection 10cm 2 and length 50cm along x-axis, , Sol: Average volume of energy density uaV, , iii), , 1, 0 E02, 2, , Total volume of cylinder V Al, Total energy of contained in cylinder, 1, , U U aV V 0 E02 Al 5.5 10 12 J, 2, , , U, (total absorption), C, , E, E, , 2, E mC mC C mC C P C , , , When radiation incident on a surface is entirely, reflected back along its original path, magnitude of, , momentum delivered to the surface is p , , 2U, C, , where ‘C’ is velocity of light., iv) When the radiation incident on a surface (Perfect, absorber) radiaton pressure, Pr , , EB E 2, , 0 0C, , I SaV, , surface is p , , F 1 dp 1 d U , 1 dU S, , , , , A A dt A dt C AC dt C, , dU / dt, is called average value of Poynting vector.., A, If the surface is perfect reflector radiation pressure, Pr , , 2S, C, , W.E-6: Light with an energy flux of 18W / cm 2, falls on a non reflecting surface at normal, incidence. If the surface has an area of 20cm 2, then find average force exerted on the surface, during a 30 minute time span., Sol. Total energy falling on the surface is, U 18 10 4 20 10 4 30 60 = 6.48 105 J, , Total momentum delivered (complete absorption), , U 6.48 105, , 2.16 10 3 kg m / sec, 8, C, 3 10, average force exerted, p, , F, , p 2.16 103, , 1.2 106 N, t, 30 60, , W.E.-7: The rms value of electric field of light, coming from sun is 720 N / C . Find average, Electro magnetic waves have linear momentum as, Momentum and Radiation Pressure, , i), , ii), , well as energy. When EM waves strike a surface,, pressure is exerted on it, called radiation pressure., When EM waves are incident on a surface and the, total energy transferred to the surface in a time t is, U then magnitude of momentum transferred to, NARAYANAGROUP, , energy density of em wave., , Sol: Total average energy density =, , 1, 2, 0 E02 0 Ems, 2, , E0 , , 2, 12, E rms , 8.85 10 720 =, 2, , , , 4.58 106 Jm3, 5
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, 2), , Electro magnetic spectrum :, , Wavelength, , The array obtained on arranging all the, electromagnetic waves in an order on the basis of, their wavelength is called the electromagnetic, spectrum, In the order of increasing wavelength, these waves, are (i) Gamma rays, (ii) X-rays, (iii) Ultraviolet rays,, (iv) Visible light, (v) infrared waves, (vi), Microwaves and (vii) Radio waves., Frequency (Hz), , , , Gamma-rays, , to 1018 m . X-rays are produced when highly, energetic cathode rays are stopped by a metal, target of high melting point. They affect the, photographic plate and can penetrate through the, transparent materials. They are mainly used in, detecting the fracture of bones, hidden bullet,, needle, costly material, etc., inside the body and, also used in the study of crystal structure., , 3) Ultraviolet rays : They were discovered by, Ritter in 1801. Their wavelength is of the order, 10 9 m to 4 107 m . In the radiations received, from sun, major part is that of the ultraviolet, radiation. Its other sources are the electric discharge, tube, carbon arc etc. These radiations are mainly, used in excitation of photoelectric effect and to kill, the bacteria of many diseases., , 0.1 A, , 19, , 10, , 1A, 0.1 nm, , 18, , 10, , 400 nm, , X-rays, , 1 nm, , 1017, 10 nm, 1016, , 500 nm, , Ultra violet, 100 nm, , 15, , 10, , Visible, Near IR, , 14, , 10, , 1000 nm, 1 m, , 4) Visible light : This was first studied in 1666 by, Newton. The radiations in the range of wavelength, from 4 107 m to 7 107 m fall in the visible, region. The wavelength of the light of violet colour, is the shortest and that of red colour is the longest., Visible light is obtained from the glowing bodies,, while they are white hot. The light obtained from, the electric bulbs, sodium lamp, fluorescent tube is, the visible light., , 600 nm, , 10 m, , Infra-red, 13, , 10, , Thermal IR, , 100 m, , 12, , 10, 1000 MHz, , For IR, 11, , 10, UHF, , 300 MHz, , 700 nm, , 1000 m, 1 mm, 1 cm, , 1010, , Radar, 10 cm, , 5) Thermal or infrared waves: They were, , 109, VHF, 7-13, , 100 MHz, 30 MHz, , FM, VHF, 2-0, , 1m, 108, , discovered by Herchell in 1800. Their wavelength, is of the order of 7 107 m to 10 3 m . A body on, being heated, emits out the infrared waves. These, radiations have the maximum heating effect. The, glass absorbs these radiations, therefore for the, study of these radiations rock salt prism is used, instead of a glass prism. These waves are mainly, used for therapeutic purpose by the doctors because, of their heating effect., , Radio, TV, 10 m, , 107, 100 m, 106, , AM, 1000 m, Long-waves, , 1), , 6, , X-rays : They were discovered by Roentgen in, 1895. Their wavelength is of the order of 1012 m, , The figure illustrates the general spectrum of the, electromagnetic radiations, in which the wavelength, is expressed in metre., Gamma rays : They were discovered by, Becquerel and Curie in 1896. Their wavelength is, of the order of 10 14 to 1010 m . The main sources, are the natural and artificial radioactive substances., These rays affect the photographic plate. These, rays are mainly used in the treatment of cancer, disease., , 6), , Microwaves: They were discovered by Hertz in, 1888. Their wavelength is in the range of nearly, 10 4 m to 1m. These waves are produced by the, spark discharge or magnetron valve. They are, detected by the crystal or semiconductor detector., These waves are used mainly in radar and long, distance communication., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 7), , Radiowaves: They were first discovered in 1895, by Marconi. Their wavelength is in the range of, 0.1m to 105 m . They can be obtained by the flow, of high frequency alternating current in an electric, conductor. These waves are detected by the tank, circuit in a radio receiver or transmitter., , Application of EM waves, 1), , 2), , 3), , 4), , 5), , Radio and microwave radiations are used in radio, and TV communication system. Microwave, radiations are mainly used in radar and TV, communication., Infrared radiations are used, i) in green houses to keep the plants warm, ii) in revealing the secret writings on the ancient, walls, iii) for looking through haze, fog and mist during, war time, as these radiations can pass through them., Ultraviolet radiations are used, i) in preserving the food stuffs., ii) in the detection of invisible writing, forged, documents, finger prints in forensic laboratory., iii) Ultraviolet radiations are also used for knowing, the structure of the molecules and arrangement of, electrons in the external shells., X-rays many applications these rays provide us, valuable information, i) about the structure of atomic nuclei, ii) in the study of crystal structure, iii) in the fracture of bones etc., rays were used, i) in treatment of cancer and tumours, ii) to produce nuclear reactions., , ADVANCED MAIN POINTS, Radiation Force and Pressure, (i), , Electromagnetic waves (radiation) carry energy and, momentum and exert force on the surface of a body, when they get absorbed or reflected by that surface, (ii) Accoding to Einstein’s mass energy equivalence, , E mc , 2, , linear momentum associated with, , E, ( E mc c and mc is linear, c, momentum of a photon of energy E or linear, momentum of a portion of the wave carrying an, energy E), (iii) According to qunatum theory of radiation, linear, , energy E is P , , NARAYANAGROUP, , ELECTROMAGNETIC WAVES, moment um associated with a photon is, E h h C h, , , where =wave, C C C , length, frequency,,, C= velocity of light, (iv) Intensity of eletromagnetic waves (or photons)I is, the flow of energy through unit area, (perpendiculer o the flow) in unit time., P, , ie I , , 1 dE , , , A dt , , Unit : W m 2, , Case (i):, (i) Consider a beam of eletro magnetic radiation of, intensity I, and of cross sectional area A which falls, on a surface of a body normally, (ii) If the surface absorbs the radiaction falling on it, completely, force excerted by the radiation on the, surface =Rate of change of linear momentum, ie F , , dp dE 1 1 dE IA, , dt C dt C dt C, , !, (iii) Pressure excerted on the surface P , , (i), , Case (ii):, If the surface reflects the radiation completly (falling, on it normally), force excerted on the surface, F, , dp, dE 1 2 dE 2 IA, 2, , , dt, C dt C dt C, , !, pressure on the surface P , , (i), , F I, , A C, , and, , F 2I, , A C, , Case (iii):, If the radiation falls normally and the surface is, partially reflecting and absorbing the remaining with, reflection and absorption coefficients r and a, respectively, then force on the surface, dE 1, dE 1, F r 2 a , C dt, C dt, 1 dE , IA, IA, 2r a 2r 1 r 1 r , C dt , C, C, , a r 1, F I, 1 r , A C, (ii) In this case if surface is partially transmitting wtih, !, Pressure P , , 7
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , Elect romagnetic Spectrum, Name, , S.N, , Freque ncy range, (Hz), , Wavele ngth range (m), , 1., , Gamma rays, , 5 10 22 to 5 1018, , 0.6 10 14 to 10 10, , 2., , X-rays, , 3 10 21 to 1 1016, , 10 13 to 3 108, , 3., , Ult raviolet rays (UV), , 8 1014 to 8 1016, , 4 10 9 to 4 10 7, , 4., , Visible light, , 4 1014 to 8 10 14, , 4 10 7 to 8 107, , 5., , 3 1011 to 4 1014, , 8 10 9 to 3 103, , 6., , Thermal of infrared, rays (IR), Microwaves, , 3 10 8 to 3 1011, , 10 3 to 1, , 7., , Radiowaves, , 3 10 3 to 3 1011, , 10 3 to 10 5, , reflection ,absorption and transmission coefficients, r,a and t respectively, then force on the surface, dE 1, dE, F r2, a, , C dt, C, 1 dE , IA, , 2 r, , 2 r a , C dt , C, and r a t 1, , Bombardment of high, Z target by elec trons, Exc it ation of atoms, and spark, Excitation of atoms,, spar k and arc flame, Exc it ation of atoms, and molec ules, Klystron value or, magnetron value, Oscillating circuits, , Force on the surface parallel to the surface is, , 1, , dt, a, , F Fn Ft , , A, , 2, , 2, , IA, at an angle with normal, C, , to the surface, normal force, Fn, IAcos I 2, , , cos , area, A , A C, , C, , cos , cos , (ii) In this case if radiation is completely reflected at, the same angle, then force on the surface, Pr essure , , 2IA, dE cos 1 2 dE , F 2, cos, cos , C, C dt C dt , and force parallel to the surface =0 ( no change, in linear momentum parallel to the surface), , , A, cos , , If the radiation is absorbed by the surface, completely, force on the surface normal to it is, , IA, dE cos 1 1 dE , Fn , , cos cos , C, C, dt C dt , 8, , Nuclear Origin, , IA, dE sin 1 1 dE , Ft , sin sin , C, C dt C dt , Resultant, force, on, the, surface, , Case (iv) :, Let a parallel beam of radiation falls on a plane, surface at an angle with normal to thesurface and, A be the cross-sectional area of the beam, , (i), , Production, , pressure, , P! , , 2 I cos 2 , F, , C, A , , , cos , , (iii) In this case if the surface partially reflects (at same, angle) and absorbs the remaining with reflection, and absorption coefficients r and a respectively, (r+a=1), then force on surface normal to it due to, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, the reflected and absorbed parts of the radiation is, Fn r , , , ELECTROMAGNETIC WAVES, 3., , 2dE cos 1, dE cos 1, a, , C, dt, C dt, , 1 dE , IA cos, cos 2r a , 2r 1 r , C dt , C, , 4., , IA cos , 1 r , C, Force on the surface parallel to it ( this is due to the, absorbed portion of the radiation only) is, , , 5., dE sin 1 a dE , Ft a, , sin , C, dt C dt , IA, sin 1 r , C, Resultant force on th surface is, IA, 2, 2, F Fn2 Ft 2 , 1 r cos2 1 r sin2 , C, 1 Ft , This force ats at an angle tan F with, n, normal to the surface, , ie, , , 1 r , tan 1 tan , , 1 r , , P! , , Pressure, , , , F, normal force, n, A, area, cos , , IA cos 1 r I, cos 2 1 r , A, C, C, cos , , C.U.Q, 1., , 2., , , , If E and B are the electric and magnetic field, vectors of electromagnetic waves then the, direction of propagation of electromagnetic, waves is along the direction of , , 1. E, 2. B, , , 3. E B, 4. B E, The electromagnetic waves do not transport1. energy, 2. charge, 3. momentum, 4. information, NARAYANAGROUP, , 6., , A capacitor is connected in an electric circuit, with a open key, immediately after pressing, the key, the current in the circuit is 1. zero, 2. maximum, 3. any transient value, 4. depends on capacitor used, Displacement current is continuous1. when electric field is changing in the circuit, 2. when magnetic field is changing in the circuit, 3. in both types of fields., 4. through wire and resistance only, The conduction current is the same as, displacement current when the source is, 1. A.C. only, 2. D.C. only, 3. both A.C and D.C., 4. neither for A.C. nor for D.C., The Maxwells four equations are written as, q0, , E, (ii) B.dS 0, (i) .dS , 0, , d , E, .dl dt B.dS, , d , (iv) B.dl 0 0 E .dS, dt, , The equations which have sources of E and, , B are, 1. (i), (ii), (iii), 2. (i), (ii), 3. (i) and (iii) only, 4. (i) and (iv) only, 7. Out of the above four equations, the equations, which do not contain source field are 1. (i) and (ii)2. (ii) only 3. all of four 4. (iii) only, 8. Out of the four Maxwell’s equations above,, which one shows non-existence of monopoles?, 1. (i) and (iv), 2. (ii) only, 3. (iii) only, 4. none of these, 9. Which of the above Maxwell’s equations shows, that elelctric field lines do not form closed, loops?, 1. (i) only 2. (ii) only 3. (iii) only 4. (iv) only, 10. In an electromagnetic wave the average, energy density is associated with 1. electric field only, 2. magnetic field only, 3. equally with electric and magnetic fields., 4. average energy density is zero., , (iii), , 9
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, 11. The displacement current flows in the dielectric, of a capacitor when the potential difference, across its plates1. becomes zero, 2. has assumed a constant value, 3. is increasing with time, 4. decreasing with time, 12. Select wrong statement from the followingElectromagnetic waves, 1. are transverse, 2. travel with same speed in all media, 3. travel with the speed of light, 4. are produced by accelerating charge., 13. The waves related to telecommuni-cation are1. infra red, 2. visible light, 3. microwaves, 4. ultraviolet rays, 14. The nature of electromagnetic wave is1. longitudinal, 2.longitudinal stationary, 3. transverse, 4. transverse stationary, 15. The frequencies of x-rays , -rays and, ultraviolet rays are respectively a,b and c., Then:, 1. a b, b c, 2. a b, b c, 16., , 17., , 18., , 19., , 10, , 3. a b, b c, 4. a b, b c, Electromagnetic radiation of frequency, 3×105 MHz can be produced by which of the, following wave, 1. Radiowave, 2. X-rays, 3. Ultraviolet, 4. Microwave, Maxwell’s equations describe the fundamental, laws of, 1. electricity only, 2. magnetism only, 3. mechanics only, 4. both 1 and 2, Which of the following statements is not, correct?, 1. photographic plates are sensitive to infrared rays, 2. photographic plates are sensitive to ultraviolet, rays, 3. Infra-red rays are invisible but can cast shadows, like visible light, 4. infrared photons have more energy than photons, of visible light, Radio waves and visible light in vacuum have, 1. same velocity but different wavelength, 2. continuous emission spectrum, 3. band absorption spectrum, 4. line emission spectrum, , 20. Energy stored in electromagnetic oscillations, is in the form of, 1. electrical energy, 2. magnetic energy, 3. both 1 and 2, 4. neither of the above, 21. Whcih wave is not electromagnetic in nature?, 1. micro, 2. radio, 3. X-ray 4. audio, 22. Total energy of EM waves in free space is, given by, E2, B2, , 1., 2 0 20, , 23., , 24., , 25., , 26., , 2., , 0 E 2 0 B 2, , 2, 2, , 0 E 2 B2, E 2 B2, , 3., 4., 2, 20, C, Which of the following waves have the, maximum wavelength?, 1. Ultraviolet rays, 2. I.R. rays, 3. UV rays, 4. radio waves, Electromagnetic waves are transverse in, nature is evident by, 1. polarization, 2. interference, 3. reflection, 4. diffraction, Which of the following are not electromagnetic, waves?, 1. secondary cosmic rays 2. gamma rays, 3. rays, 4. X-rays, , , Let E , B and C represent the electric field,, magnetic field and velocity of an, electromagnetic wave respectively. Their, directions, at any instant point along the unit, vectors given below in order. Which of the, following cannot be true?, 1) kˆ, iˆ, ˆj 2) kˆ, ˆj , iˆ 3) iˆ, ˆj , kˆ 4) ˆj , kˆ , iˆ, , 27. A radiation of energy E falls normally on a, perfectly reflecting surface. The momentum, transferred to the surface, 1) E/C, 2) 2E/C, 3) EC, 4) E / C2, 28.An electromagnetic wave passing through, vacuum is described by the equation;, , E E0 sin kx t and B B0 sin kx t ;, then, 1) E0 B0, , 2) E0 B0 k, , 3) E0 B0 k, , 4) E0 k B0, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , 29. The frequency of visible light is of the order, of, 1. 1015 Hz 2. 1010 Hz 3. 106 Hz 4. 104 Hz, 30. Which of the following wavelength falls in Xray region?, 0, , 0, , -2, , 0, , -3, , 0, , 1. 1A, 2. 10 A, 3. 10 A 4. 10 A, 31. An electro magnetic wave is vaccume has the, , , electric and magnetic field E and B which, are always perpendicular to each other. The, , direction of polarisation is given by X and, , that of wave propagation by K then, (AIE:2012), , , , 1) X || B and K || B E 2) X || E and K || E B, , , , 3) X || E and K || E B 4) X || E and K || B E, Note : Directions q.no. 32 to 46, 1. Both Assertion and reason are true and the, reason is correct explanation of the Assertion., 2. Both Assertion and reason are true, but, reason is not correct explanation of Assertion., 3. Assertion is true, reason is false, 4. Assertion is false, reason is true, 32. Assertion: Displacement current arises on, account of change in electric flux., Reason: I d 0, , dE, dt, , 33. Assertion (A): In an e.m. wave, magnitude of, , magnetic field vector B is much smaller than, , the magnitude of vector E, Reason(R): This is because in an e.m. wave, E/B c 3 108 m / s., 34. Assertion(A): Microwaves have more energy, than the radio waves, hc, , 35. Assertion: Displacement current decreases, with the increase in frequency of a.c. supplied, to a capacitor, Reason: Reactance due to capacitance is, directly proportional to the frequency of a.c., 36. Assertion: The electrostatic field lines cannot, form a closed path., , Reason: E.dl 0, , Reason(R): E h , , NARAYANAGROUP, , 37. Assertion: The magnetic flux through a closed, surface is zero, Reason: Gauss’s law applies in the case of, electric flux only, 38. Assertion: A changing electric-field produces, a magnetic field., Reason:A changing magnetic field produces, an electric field., 39. Statement I: Sound waves are not, electromagnetic waves., Statemenet II: Sound waves require a material, medium for propagation., 40. Statement I: Electromagnetic waves are, transverse in nature., Statement II: The electric and magnetic fields, of an e.m. wave are perpendicular to each, other and also perpendicular to the direction, of wave propagation., 41. Statement I: Electromagnetic waves exert, pressure, called radiation pressure., Statement II: This is because they carry, energy., 42. Statment I: in an electric circuit a capacitor of, reactance 100 is connected across a 220 V, source. The displacement current is 2.2 A., Statement II: The data is insufficient., 43. Statement I: An e.m. radiation of energy 14.4, keV belongs to X-ray region., Statement II: E hv hc / , 44. Statement I : The velcoity of all, electromagnetic waves in vacuum is different., Statement II: The different electromagnetic, waves are of different frequency, 45. Column I, (A) Aveage energy density of electric field in, electromagnetic wave, (B) Average energy density of magnetic field in, electromagnetic wave, (C) Total average energy density of electromagnetic, wave, (D) Intensity of electromagnetic wave, 11
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , MAGNETIC FIELD PRODUCED, BETWEEN PLATES OF PARALLEL, PLATE CAPACITOR, , Column II, (P), , 1, 0 E02 c, 2, , (Q), , 1, 0 E02, 2, , 3., , 2, 0, , 1B, 1, 0 E02, (S), 4 0, 4, 1. A-P, B-Q, C-R, D-S 2. A-Q, B-R, C-S, D-P, 3. A-R, B-S, C-Q, D-P 4. A-Q, B-R, C-S, D-P, 46. Column I, (A) Radiowaves, (B) Infrared radiations, (C)Ultraviolet rays, (D)Gamma rays, Column II, (P) Vibrations of atoms and molecules, (Q) Arc lamp, (R) Nuclear origin, (S) Oscillating circuit, 1. A-P, B-Q, C-R, D-S 2. A-Q, B-R, C-S, D-P, 3. A-R, B-S, C-P, D-Q 4. A-S, B-P, C-Q, D-R, (R), , 2) 2, 9) 1, 16) 4, 23) 4, 30) 1, 37) 3, 44) 4, , 3) 2, 10) 3, 17) 4, 24) 1, 31) 2, 38) 2, 45) 3, , 4) 1, 11) 3, 18) 4, 25) 3, 32) 1, 39) 1, 46) 4, , 5) 2, 12) 2, 19) 1, 26) 4, 33) 1, 40) 1, , 4., , 5., , 6., 6) 4, 13) 3, 20) 3, 27) 2, 34) 2, 41) 2, , 7) 2, 14) 3, 21) 4, 28) 4, 35) 3, 42) 3, 7., , DISPLACEMENT CURRENT, , 2., , A Parallel plate condenser of capacity 100 pF, is connected to 230 V of AC supply of 300rad/, sec frequency. The rms value of displacement, current., 1) 6.9 A 2) 2.3 A 3) 9.2 A 4) 4.6 A, A parallel plate capacitor of plate separation, 2mm is connected in an electric circuit having, source voltage 400V. If the plate area is, 60 cm 2 , then the value of displacement current, for time period of 106 sec will be, , 12, , R, and R from the axis of, 2, condenser (R is the radius of plate) while, charging is, 1) 1:1, 2) 2 :1, 3) 1: 2, 4) 1: 4, The magnetic field between the circular plates, of radius 12 cm separated by distance of 4mm, of a parallel plate capacitor of capacitance 100, pF. along the axis of plates having conduction, current of 0.15 A is, 1) zero, 2) 1.5 T, 3) 15 T, 4) 0.15 T, , distance of, , LEVEL-I (C.W), , 1., , µ 0 ID r, µI, µ I, 2. 0 D, 3. 0 D, 4. zero, 2, 2πR, 2πR, 2πr, A condenser is charged using a constant, current. The ratio of the magnetic fields at a, , 1., , WAVE EQUATION, , C.U.Q - KEY, 1) 3, 8) 2, 15) 4, 22) 4, 29) 1, 36) 1, 43) 2, , The magnetic field between the plates of a, capacitor when r > R is given by -, , 1) 1.062 amp, , 2) 1.062 102 amp, , 3) 1.062 103 amp, , 4) 1.062 10 4 amp, , 8., , 9., , The wave function (in S.I unit) for an, electromagnetic wave is given as -, , x, t 103 sin 3 106 x 9 1014 t , The speed of the wave is, 1. 9 1014 m/s, 2. 3 108 m/s, 3. 3 106 m/s, 4. 3 107 m/s, The velocity of all radiowaves in free space is, 3×108 m/s , the frequency of a wave of, wavelength 150 m is, 1. 45 MHz 2. 2 MHz 3. 2 KHz 4. 20 KHz, The relative permeability of glass is 3/8 and, the dielectric constant of glass is 8. The, refractive index of glass is, 1) 1.5, 2) 1.1414 3) 1.732 4) 1.6, An electromagnetic wave of frequency 3 MHz, passes from Vacuum into a dielectric medium, with permittiviy 4.0 Then (AIE : 2004), 1) Wave length doubled and frequency remains, unchanged, 2) wave length is doubled and frequency becomes, half, 3) wave length is halved and frequency remains, unchanged, 4) wave length and frequency both remain, unchanged, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , RELATION BETWEEN B & E, 10. In a plane electromagnetic wave, the electric, field oscillates sinusoidally at a frequency of, 2 1010 Hz and amplitude 48V/m . The, amplitude of oscillating magnetic field will be, 1, -8, 2, 2. 16×10-8 Wb/m2, 1. ×10 Wb/m, 16, 1, -7, 2, 3. 12×10-7 Wb/m2, 4. ×10 Wb/m, 12, 11. In a plane electromagnetic wave, the electric, field oscillates sinusoidally at a frequency of, 2 1010 Hz and amplitude 48V/m , the, wavelength of the wave will be 1. 1.5 m, 2. 66.6 m 3. 1.5 cm 4. 66.6 cm, 12. In an apparatus the electric field was found to, oscillate with an amplitude of 18Vm 1 . The rms, of the oscillating magnetic field is, 1) 6 108 T, 2) 4.23 108 T, 3) 9 108 T, 4) 7.0 108 T, 13. The amplitude of the sinusoidally oscillating, electric field of a plane wave is 60V/m. Then, the amplitude of the magnetic field is, 1) 12 107 T, 2) 6 107 T, 4) 2 107 T, 3) 6 107 T, , MOMENTUM AND FORCE, 14. Light with energy flux of 18W / cm2 falls on a, non reflecting surface of area 20cm 2 at normal, incidence the momentum delivered in 30, minutes is, 2) 2.16 103 kgms 1, 1) 1.2 106 kgms 1, 3) 1.18 10 3 kgms 1, , 4) 3.2 103 kgms 1, 15. Light with energy flux of 24Wm 2 is incident, on a well polished disc of radius 3.5cm for one, hour. The momentum transferred to the disc, is, 1) 1.1 kg ms 1, 2) 2.2 kg ms 1, 3) 3.3 kg ms 1, , 4) 4.4 kg ms 1, , ELECTROMAGNETIC WAVES, 17. The rms value of the electric field of a plane, electromagnetic wave is 314V/m. The average, energy density of electric field and the, average energy density are, 1) 4.3 107 Jm 3 ; 2.15 107 Jm 3, 2) 4.3 107 Jm 3 ;8.6 10 7 Jm 3, 3) 2.15 107 Jm 3 ; 4.3 107 Jm 3, 4) 8.6 10 7 Jm 3 ; 4.3 10 7 Jm 3, 18. The magnitudes of Poynting vector and, electric field vector are respectively S and E,, then, , , 2, 0, 1. S E , 0, , 2, 3. S E, , NARAYANAGROUP, , 0, 0, , 2, 4. S , , E2, 0, , INTENSITY, 19. If a source of power 4 kW produces 1020, photons/second, the radiation belong to a part, of the spectrum called [AIE 2010], 1. X - rays, 2. Ultraviolet rays, 3. Microwaves, 4. rays, 20. The intensity of electromagnetic wave at a, distance of 1 Km from a source of power 12.56, kw. is, 1) 103Wm 2, 2) 4 103Wm 2, 3) 12.56 103Wm 2 4) 1.256 103Wm 2, 21. The sun delivers 103W / m 2 of electromagnetic, flux incident on a roof of dimensions 8m 20m ,, will be, 1) 6.4 103W, 2) 3.4 104 W, 3) 1.6 105 W, , 4) 3.2 105 W, , LEVEL-I (C.W) - KEY, 1) 1 2) 2 3) 3 4) 3 5) 1 6) 2 7) 2, 8) 3 9) 3 10) 2 11) 3 12) 2 13) 4 14) 2, 15) 2 16) 3 17) 2 18) 1 19) 1 20) 1 21) 3, , LEVEL-I (H.W) - HINTS, , ENERGY DENSITY, 16. The maximum electric field of a plane, electromagnetic wave is 88 V/m. The average, energy density is, 1) 3.4 108 Jm 3, 2) 13.7 108 Jm 3, 3) 6.8 108 Jm 3, 4) 1.7 108 Jm 3, , 2. S E 2 0 0, , V, V, , VC, Z XC, , 1., , Id , , 2., , Id 0, , VA, Td, 13
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , 7., , 0 I d, 2 r, Ir, I, r RB 0 02 ; r R B 0 d, 2 R, 2 R, I r, B 0 d2 ;r 0, 2 R, , C, K, C n, , 8., , RI of glass =, , 9., , C n, E, B, C, C, , , E, C, B, E, C 0, B0, , 3., 4., 5., 6., , 10., 11., 12., 13., , rR B, , 1, 2, 2, 0 EMax, ; U av 0 Emax, 2, nE, hc, 19. P , ; E, t, , P, 20. I , 4 r 2, 21. PA, , 3., , 14, , The voltage between the plates of a parallel, plate condenser of capacity 2.0 F is, charging at a rate of 10Vs 1 . The displacement, current, 2) 2 A, 3) 20 A 4) 2A, 1) 2 mA, , 2), B, , B, , distance, 3), , distance, 4), , B, , B, , distance, , 4., , distance, , The electrical field in the gap of a condenser, charges as 1012Vm 1 s 1 . If the radius of each, plate of the condenser is 3cm, the magnetic, field at the edge of plate in the gap is, 1) 1.67 mT, 2) 0.167 T, 3) 0.5 T, , 4) 5 T, , WAVE EQUATION, 5., , LEVEL-I (H.W), 1., , The graph representing the variation of, induced magnetic field in the gap of the, condenser plates during its charging with the, distance from the axis of the gap is, , 1), , U av E , , DISPLACEMENT CURRENT, , A parallel plate condenser of capacity 10 F, is charged with a constant charging current of, 0.16A. The displacement current is, 1) 0.16 A, 2) 0.16A, 3) 0.96A, 4) 9.6A, , MAGNETIC FIELD PRODUCED, BETWEEN PLATES OF PARALLEL, PLATE CAPACITOR, , , C0, , r r, 0 0, C, , U IAt, 14. p , C, C, 2U 2 IAt, , 15. P , C, C, 2, 16. U av 0 EMax, , 17., , 2., , 6., , An LC circuit contains inductance L 1 H, and capacitance C 0.01 H . The wavelength, of electromagnetic wave generated is nearly, 1. 0.5 m, 2. 5 m, 3. 188 m 4. 30 m, The wave length of the Green light of mercury, is 550nm. If the refractive index of the glass, is 1.5, the time period of the electrical vector, in glass nearly C0 3 108 mS 1 , 1) 1.8 10 9 S, 3) 9 1015 S, , 2) 3.6 1015 S, 4) 2.751015 S, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 7., , ELECTROMAGNETIC WAVES, , The all India Radio, station at Vijayawada, transmits its signals at 840 K C/s. The length, of the radio wave is, 1) 35.7m, 2) 357m, 3) 35.7 km, 4) 3.57 m, , RELATION BETWEEN B & E, 8., , A point source of electromagnetic radiation, has an average power output of 800W, The, maximum value of electric field at a distance, 3.5 m from the source will be 62.6 V/m, the, maximum value of magnetic field will be 1. 2.09 105 T, , 9., , 2. 2.09 106 T, , 3. 2.09 107 T, 4. 2.09 108 T, A plane E.M. wave of frequency 40 MHz, travels along X-axis. At same point at same, instant, the electric field E has maximum value, of 750 N/C in Y-direction. The magnitude and, direction of magnetic field is, 1) 2.5T along X-axis, , 13. In an electromagnetic wave, the amplitude of, electric field is 1V/m. The frequency of wave, is 5 1014 Hz . The wave is propagating along, z-axis. The average energy density of electric, field, in Joule / m3 , will be, 1) 1.1 1011, , 2) 2.2 1012, 4) 4.4 1014, , 3) 3.3 1013, , INTENSITY, 14. About 5% of the power of a 100 W light bulb, is converted to visible radiation. The average, intensity of visible raditaion at a distance of 1, m from the bulb:, 1. 0.4W/m 2 2. 0.5W/m 2 3. 0.6W/m 2 4. 0.8W/m 2, 15. The sun radiates electromagnetic energy at, the rate of 3.9 1026 W . Its radius is, 6.96 108 m . The intensity of sun light at the, solar surface will be (in W / m 2 ), , 2) 2.5T along Y-axis, , 1) 1.4 104 2) 2.8 105 3) 64 106 4) 5.6 107, 16. The intensity of TV broad cast station of, , 3) 2.5T aloing Z-axis, , E 800 Sin 109 t k x V/M is...... and the, , wave length in meter is ......, , 4) 5T along Z-axis, 10. A plane electromagnetic wave of frequency 25, MHz travels in free space along the xdirection. At a particular point in space and, time E 6.3 j . The magnetic field B at this, , 1) 850 wm2 ;0.6, , 2) 425 wm 2 ; 0.6, , 3) 850 wm 2 ; 0.3, , 4) 425 wm 2 ; 0.3, , LEVEL-I (H.W) - KEY, , , 1) 4.2 108 kT, , , 2) 2.1 108 kT, , 1) 3 2) 2 3) 1 4) 2 5) 3 6) 4 7) 2, 8) 3 9) 3 10) 2 11) 1 12) 2 13) 2 14) 1, 15) 3 16) 1, , , 3) 18.9 108 kT, , , 4) 2.1 108 kT, , LEVEL-I (H.W) - HINTS, , point is, , MOMENTUM AND FORCE, 11. Light with energy flux 36w / cm 2 is incident, on a well polished metal square plate of side, 2cm. The force experienced by it is, 1) 0.96 N 2) 0.24 N 3) 0.12 N 4) 0.36 N, 12. The rms value of the electric field of the light, coming from the sun is 720 N/C. The average, total energy density of the electromagnetic, wave is, 1) 3.3 103 J / m3, 2) 4.58 106 J / m3, 3) 6.37 10 J / m, NARAYANAGROUP, , 3, , 4) 81.35 10, , Id C, , 2., , IC I d, , 3., , ENERGY DENSITY, , 9, , 1., , 12, , J /m, , 4., 5., , dv, dt, , 0 i, r 2i0, i, , and, 2 r, R2, , d, B.dl 0 0 dtE ;, B, , n, , 1, 2 LC, , , , , B 2 r , , 1 dE 2, r, C2 dt, , C, n, , 3, , 15
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , MAGNETIC FIELD PRODUCED, BETWEEN PLATES OF PARALLEL, PLATE CAPACITOR, , 6., , , , C0, , ,T , C, C, , 7., , , , C, n, , 8., , Bm , , 9., , B, , E, C, , 10. B , , E, C, , 11. F , , 2U 2 IAt, 2IA, , ;, Ct, Ct, C, , 3., , Em, C, , 4., , 1, 2, 12. U av 0 Emax, 2, , 13., , Vav E , , 14. I , , P, 4 r 2, , 15. I , , P, 4 r 2, , WAVE EQUATION, 5., , 1, 2, 0 Emax, 4, , E0 B0, E02, c, 2 , I, , , , , , c, 16., ;, n, 20 20 C, , , 6., , LEVEL-II (C.W), DISPLACEMENT CURRENT, 1., , 2., , A parallel plate condenser consists of two, circular plates each of radius 2cm separated, by a distance of 0.1mm. A time varying, potential difference of 5 1013V / s is applied, across the plates of the condenser. The, displacement current is, 1) 5.50A, 2) 5.56 102 A, 3) 5.56 103 A, 4) 2.28 10 4 A, A parallel plate condenser has conducting, plates of radius 12cm separated by a distance, of 5mm. It is charged with a constant charging, current of 0.16 A, the rate at which the, potential difference between the plate change, is, 2) 2 1010 Vs 1, 1) 1 109 Vs 1, 12, , 3) 3 10 Vs, 16, , 1, , A condenser has two conducting plates of, radius 10cm separated by a distance of 5mm., It is charged with a constant current of 0.15A., The magnetic field at a point 2cm from the axis, in the gap is, 2) 3 108 T, 1) 1.5 10 6 T, 3) 6 108 T, 4) 3 106 T, An AC rms voltage of 2V having a frequency, of 50 KHz is applied to a condenser of capacity, of 10 F . The maximum value of the magnetic, field between the plates of the condenser if, the radius of plate is 10cm is, 1) 0.4 p 2) 4 T 3) 2 T 4) 40 T, , 9, , 4) 2 10 Vs, , 1, , The wave emitted by any atom or molecule, must have some finitetotal length which is, known as the coherence length. For sodium, light, this length is 2.4cm. The number of, oscillations in this length will be Given, 5900 A0, 2) 4.068 104 Hz, 1) 4.068 105 Hz, 3) 4.068 106 Hz, 4) 4.068 108 Hz, A wave is propagating in a medium of dielectric, constant 2 and relative permeability 50. The, wave impedance is, 2) 376.6 3) 3776 4) 1883, 1) 5 , , RELATION BETWEEN B & E, 7., , The magnetic field in travelling EM wave has, a peak value of 20nT. The peak value of, electric field strength is (AIE : 2013), 1) 6 V/m 2) 9 V/m 3) 12 V/m 4) 3 V/m, , MOMENTUM AND FORCE, 8., , A plane electromagnetic wave of wave, intensity 6W/m 2 stikes a small mirror of area, 40 cm 2 , held prependicular to the approaching, wave. The momentum transfered by the wave, to the mirror each second will be, , 1. 6.4×10-7 kg-m/s, , 2. 4.8×10-8 kg-m/s, , 3. 3.2×10-9 kg-m/s, , 4. 1.6×10-10 kg-m/s, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , 3. 1.73 107, , 4. 1.73 108, , (a), , P0, 1) E0 2 cr 2, 0, , 2) E0 , , P0, 2 0cr 2, , 0.9, , f 0.8, (×1015Hz), , f, (×1015Hz), , medium, , glass, , water, , (b), 1.8, 1.6, 1.2, , 28. A lamp radiates power P0 uniformly in all, directions, the amplitude of electric field, strength E0 at a distance r from it is, , medium, , 1.2, , 11. An electromagnetic radiation has an energy, 14.4 KeV. To which region of electromagnetic, spectrum does it belong?, 1. Infra red region, 2. Visible region, 3. X-rays region, 4. γ- ray region, , 0.9, 0.8, , f, (×1015Hz), , medium, , glass, , 2. 1.73 106, , f, (×1015Hz), , 1.2, , water, , 1. 1.73 105, , 1.2, , vacuum, , 10. A point source of electromagnetic radiation, has an average power output of 800W. The, maximum value of electric field at a distance, 3.5 m from the source will be 62.6 V/m, the, energy density at a distance 3.5 m from the, source will be - (in joule/m3), , vacuum, , ENERGY DENSITY, , glass, , 4. 1.6 10-10 N, , glass, , 3. 3.2 10-9 N, , water, , 2. 4.8 10-8 N, , water, , 1. 6.4 10-7 N, , 15. Electromagnetic waves of frequency, 1.2 1015 Hz enters into water and, subsequently into glass from v a c u u m ., Which of the following graphs correctly, represent the variation of frequency f with, medium? (Giventhat indices of refraction for, water and glass are 4/3 and 3/2 respectively)., , vacuum, , In the above question the radiation force on, the mirror will be, , vacuum, , 9., , ELECTROMAGNETIC WAVES, , medium, , LEVEL-II (C.W.) - KEY, 1) 3 2) 4, 8) 4 9) 4, 15) 1, , 3) 3 4) 3 5) 2 6) 4 7) 1, 10) 4 11) 4 12) 2 13) 3 14) 2, , LEVEL-II (C.W.) - HINTS, 3) E0 , , P0, 4 0cr 2, , P0, 4) E0 8 cr, 0, , Id , , 2., , dv dI d, , dt 0 A, , 3., , B, , 0ir, 2 R 2, , 4., , B, , 0C dv , , 2 R dt , , 5., , No.of oscillations in this length =, , 6., , Wave impedence Z , , INTENSITY, 13. A laser beam can be focussed on an area equal, to the square of its wavelength. A He-Ne laser, radiates energy at the rate of 1mW and its, wavelength is 600 nm. The intensity of, focussed beam will be, 1) 3.2 109 W / m 2, , 2) 2.8 1013W / m 2, , 3) 2.7 109 W / m 2, 4) 3.2 1013 W / m 2, 14. The intensity of solar radiation at the earths, surface is 1KW m 2 . The power entering the, pupil of an eye of diameter 0.5 cm is, 1. 39.2 mw, 2) 19.6 mw, 3) 9.8 mw, 4) 4.9 mw, NARAYANAGROUP, , 0 A dv, , d, dt, , 1., , l, , , E CB, , , , H H, , , 17
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, 7., 8., 9., , Bm , , MAGNETIC FIELD PRODUCED, BETWEEN PLATES OF PARALLEL, PLATE CAPACITOR, , Em, C, , 2U 2 av A, , C, C, Momentum per sec is force, , 4., , P, , 1, 2, 10. Energy density U 0 Em, 2, , 11. , , 12, 400, hc, , 3, 14.4 10, E, , 13. I , , P P, , A 2, , 5., , 6., , 14. P IA I R 2, 15. Frequency remains unchanged as electromagnetic, waves pass from one medium to another., , LEVEL-II (H.W), , The capacity of a parallel plate condenser is, 50 pF. A magnetic field of 4 107 T is produced, at a distance of 10cm from the axis of the gap., The charging current is, 1) 0.1A, 2) 0.2 A, 3) 0.3 A, 4) 0.15 A, The diameter of the condenser plate is 4cm., It is charged by an external current of 0.2A., The maximum magnetic field induced in the, gap, 1) 2 T, 2) 4 T, 3) 6T, 4) 8T, A condenser of capacity 50 p F is connected to, an AC supply of 220 V 50 Hz. The rms value, of magnetic field at a distance of 5cm from, the axis is, 1) 22 1014 T, 2) 22 1012 T, 3) 44 1013 T, , DISPLACEMENT CURRENT, 1., , The area of each plate of a parallel plated, condenser is 144 cm 2 . The electrical field in, the gap between the plates changes at the rate, of 1012 V m1 s 1 . The displacement current is, 4, 0.4, 40, 1, A, A 3), A, A, 2), 4), , , , 10, A condenser having circular plates having, radius 2cm and separated by a distance of, 3mm. It is charged with a current of 0.1 A. The, rate at which the potential difference between, the plates change is, , 3., , 1) 9 1010 V / S, 2) 1.8 1010 V / S, 3) 2.7 106 V / S, 4) 2.7 1010 V / S, An AC source having a frequency of 50 Hz, and voltage supply of 300v is applied directly, to the condenser of capacity 100 F . The peak, and rms values of displacement current are, 1) 9.42 A;, , 9.42, A, 2, , 3) 9.42 2 A;9.42 A, , 18, , 2), , 9.42, A;9.42 2 A, 2, , 4) 9.42 A;9.42 A, , 11, 10 12 T, 5, , WAVE EQUATION, 7., , The velocity of an electromagnetic wave in a, medium is 2 108 mS 1 . If the relative, permeability is 1 the relative permittivity of, 8, 1, the medium is C0 3 10 mS , , 1), 2., , 4), , 1) 2.25, , 2) 1.5, , 3) 4/9, , 4) 2/3, , RELATION BETWEEN B & E, 8., , In a plane electromagnetic wave, the electric, field oscillates sinusoidally at a frequency of, 2 1010 Hz and amplitude 48 V/m. The, amplitude of oscillating magnetic field will be, 1), , 9., , 1, 108 Wb / m 2, 16, , 2) 16 108Wb / m 2, , 1, 107 Wb / m 2, 12, In an apparatus, the electric field was found, to oscillate with amplitude of 18 V/m. The, amplitude of the oscillating magnetic field will, be, 1) 4 106 T, 2) 6 108 T, , 3) 12 107 Wb / m 2, , 4), , 3) 9 109 T, , 4) 11 10 11 T, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , MOMENTUM AND FORCE, , LEVEL-II (H.W) - HINTS, , 10. Light with energy flux 36Wm 2 is incident on, a circular part of radius 1.4 m of a perfectly 1., black body. The force experienced by the body, and the momentum delivered in 10 minutes are, 3., 1) 2.2 N ; 7.2 kgms 1, , Id 0 A, , 3) 0.74 N ; 444 kgms 1, 4) 7.4 N ; 2.2 kgms 1, 11. Light with energy flux 18 wcm 2 is incident on, , dv dI d, 2. dt A, 0, , I rms , , Erms, I, i, I rms 0 4&5. B 0, ;, XC, 2, 2 r, , irms , , Erms, i, , Brms 0 rms 7. RI , XC, 2 r, , 2) 3.5 N ;7.4 kgms 1, 6., , dE, dt, , 8&9. B0 , , 1, C, 0, r r C, , E0, C, , U IAt, IA, a mirror of size 2cm 2cm normally. The force, 10,11. Momentum P , ; Force F , experienced by it and momentumdelivered in, C, C, C, one minute are, IA, F, 12., Force, 1, 1) 0.48 N ; 28.8 kgms, C, , 2) 48 N ; 2.88 kgms 1, 3) 28.8 N ; 4.8 kgms 1, , B0 E0, B02C, I, , 13. Poynting vector = , 14., 2 0, 0, , 4) 0.24 N ; 28.8 kgms 1, 12. Electromagnetic radiation with energy flux, , LEVEL-III, , 50W cm 2 is incident on a totally absorbing 1., surface normally for 1 hour. If the surface has, an area of 0.05m 2 , then the average force due, to the radiation pressure, on it is;, 1) 8.3 10 7 N, 3) 1.2 107 N, , 60 cm 2 , then the value of displacement current, , for 106 sec. will be-, , 2) 8.3 105 N, 4) 1.2 105 N, , 1. 1.062 amp., , ENERGY DENSITY, 13. In an electromagnetic wave in vacuum. The 2., electrical and magnetic fields are 40 V / m, and 0.4 107 T . The Poynting vector, 1) 4.4Wm 1 2) 0.44Wm 1 3) 5.65Wm 1 4) 4.0Wm 1, , INTENSITY, 14. The amplitude of magnetic field at a region, carried by an electromagnetic wave is 0.1T ., The intensity of wave is, 1) 4 Wm2 2) 1.2Wm 2 3) 4Wm 2 4) 1.2 Wm 2, , LEVEL-II (H.W) - KEY, 1) 2, 8) 2, , 2) 4, 9) 2, , 3) 3 4) 2 5) 1 6) 2 7) 1, 10) 3 11) 1 12) 2 13) 4 14) 2, , NARAYANAGROUP, , A parallel plate capacitor of plate seperation, 2 mm is connected in an electric circuit having, source voltage 400V. If the plate area is, , 3., , 2. 1.062 10 2 amp, , 4. 1.062 10 4 amp, 3. 1.062 10 3 amp, A long straight wire of resistance R, radius ‘a’, and length ‘l’ carries a constant current ‘I’. The, poynting vector for the wire will beIR, IR 2, I2 R, I2 R, 1., 2., 3., 4., 2πal, al, al, 2 al, To establish an instantaneous displacement, current of 2A in the space between two parallel, plates of 1 F capacitor, the potential, difference across the capacitor plates will have, to be changed at the rate of, 1) 4 10 4 V / s, 2) 4 106 V / s, , 3) 2 10 4 V / s, , 4) 2 106 V / s, , 19
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, 4., , The sun delivers 103 W/m 2 of electromagnetic 10. A plane electromagnetic wave propagating in, the x-direction has a wavelength of 60 mm., flux to the earth’s surface. The total power that, The electric field is in the y-direction and its, is incident on a roof of dimensions, maximum magnitude is 33Vm -1 . The equation, 8m 20m, will be for the electric field as function of x and t is, 1. 6.4 103 W, 2. 3.4 10 4 W, 2. 33sin 1010 t x / c , 1. 11sin t x / c , 3. 1.6 105 W, 4. none of these, , 5., , 4. 11sin 1010 t x / c , 3. 33sin t x / c , The sun delivers 103 W/m 2 of electromagnetic, flux to the earth’s surface. The total power that, LEVEL-III - KEY, is incident on a roof of dimensions, 1) 2 2) 4 3) 4 4) 3 5) 2 6) 4 7)1 8) 2 9) 3 10) 2, 8m 20m is 1.6 105 W , the radiation force, LEVEL-III - HINTS, on the roof will be , EA 0 A v, =, 1. I D = ε 0 E = ε 0, 1. 3.33 105 N, 2. 5.33 104 N, t, t, t d, 3. 7.33 103 N, 4. 9.33 102 N, 8.85 1012 400 60 104, An electric field of 300V/m is confined to a, ID =, 2 103 106, circular area 10 cm in diameter. If the field is, , 6., , 7., , 8., , increasing at the rate of 20V/m-s, the, magnitude of magnetic field at a point 15cm, from the centre of the circle will be2., 1. 1.85 10-15T, 2. 1.85 10-16 T, , =1.062×10 -2 amp, Hence the corrent answer will be (4)., If V is the potential difference across the wire, then, , 3. 1.85 10-17 T, 4. 1.85 10-18T, A lamp emits monochromatic green light, uniformly in all directions. The lamp is 3%, efficient in converting electrical power to, electromagnetic waves and consumes 100W, of power. The amplitude of the electric field, associated with the electromagnetic radiation, at a distance of 10m from tha lamp will be 1. 1.34 V/m2. 2.68 V/m3. 5.36V/m 4.9.37 V/m, A flood light is covered with a filter that 3., transmits red light. The electric field of the, emerging beam is represented by a sinusoidal, plane wave, , E=, , V IR, =, and magnetic field at the surface of wire, , , B=, , µ 0I, . Hence poynting vector, directed radially, 2πa, , E x = 36sin 1.20×10 7 z-3.6×1015 t V/m, , The average intensity of the light will be1. 0.86W/m 2, 2. 1.72W/m 2, 9., , 20, , 4., 3. 3.44W/m 2, 4. 6.88W/m 2, A plane electromagnetic wave of frequency 40, MHz travels in free space in the X-direction. 5., At some point and at some instant, the electric, , field E has its maximum value of 750 N/C in, Y-direction.The wavelength of the wave is6., 1. 3.5 m, 2. 5.5 m, 3. 7.5 m, 4. 9.5 m, , EB, , IR µ I, , I2R, , 0, inwards, is gives by S= μ = μ 2πa = 2πal, 0, 0, , Hence the correct answer will be (4)., ID =ε 0, , d E, d, d V, =ε 0 EA =ε 0 A , dt, dt, dt d , , or I D =, , ε 0 A dV, dV, =C, d dt, dt, , dV I D, 2, = = -6 =2×106 V/s, dt C 10, Hence the correct answer will be (4)., , , P = 103 160 1.6 105W, , SA 1.6 105, , 5.33 104 N, 8, c, 3 10, (or) F = total power / velocity of light, F PA , , μ 0ε 0 πd 2 dE, B=, , , 2πR 4 dt, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , 2 107 8.85 1012 3.14 0.01 20, 4 0.15, 18, 1.85 10 T, , 4., , , , P, 1, = ε 0 cE02, 2, 4πR, 2, , 3, 1.34V/m, 2 3.14 100 8.85 10 12 3 108, , , , c 0 E 02 3 10 8 8.85 10 12 36 2, , 2, 2, , 8. I av, , , , 9., , , , 1.72W/m 2, , C 3 108, , 7.5m, f 4 107, , 10. 2 v , , 2 c 2 3 108, , 1010 rad/s, 3, , 60 10, , a) 22 109 Jm3, , x, E y E0 sin t 33sin 1010 t x / c , c, , LEVEL - IV, COMPREHENSION TYPE QUESTIONS, COMPREHENSION-1, 1. The figure shown the variation of electric field, E due to electromagnetic wave passing a point, with time. Which part of the electromagnetic, spectrum does it represent?, , 2., , 3., , 66.6, , 58.3, , 41.62, 50, , 25, 33.3, , 16.65, , 8.32, , E(V/m), , –50, , Time, (femto second), , i0, , 0.368i0, 100, , 6., , 50, , b) 15 10 9 Jm3, , c) 11 10 9 Jm3 d) 5 109 Jm3, COMPREHENSION - II, 5. A circular parallel plate capacitor having plate, radius R is charged by means of a cell at time, t = 0. The variation, of the current, through the connecting wires with time is shown, in figure. Consider a plane circular area of, radius R/2 parallel to the plates and situated, symmetrically between them., The, displacement current through this area at, time t = 200 s is:, Conduction current i, , Sav =, , 7., , P, E0 , 2πR2ε0c, , With reference to the problem 3.2 the average, energy density due to electric field is:, , a) 0.135 i0, , page 21, fig. 2, b) 0.092 i0, , c) 0.068 i0, , d) 0.034 i0, , Time(s), , In the above problem no.3.6. assuming i0 =5A,, what will be the time rate of electric flux change, through the area at t 100s ?, a) 2.08 1011 N m2 / C s, b) 4.16 1011 N m2 / C s, , a) -rays, b) Visible light, c) 6.24 1011 N m2 / C s, c) Infrared rays d) Radio waves., d) 1.66 1012 N m2 / C s, In the above problem the rms value of the, COMPREHENSION - III, corresponding magnetic field is:, 7. A charged circular plate capacitor is discharged, a) 1.65 107 T, b) 1.17 10 7 T, by connecting its two plates by a copper wire, d) 1.17 10 9 T, c) 1.65 109 T, at time t = 0. The displacement current, In the above problem the frequency with which, through a plane surface parallel and midway, the energy density at that point oscillates is:, between the two plates, with an area half, a) 33.3 10 15 s1, b) 66.6 10 15 s 1, that of either plate, is shown to vary with time, in the figure. The time constant of the circuit, c) 3 10 13 s1, d) 6 1013 s 1, is:, NARAYANAGROUP, , 21
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , LEVEL - V, Conduction current i, , i0, , SINGLE ANSWER TYPE, 1., , 0.5i0, , 50, , 8., , Time(s), , a) 50s b) 100s, c) 75s d) 144s, In the above problem no. (39) the conduction, current at time t = 100 s is:, a) 0.5i0 b) 0.37i0, c) 0.25i0 d) 0.125i0, , 2., , LEVEL - IV - KEY, 1) C 2) B 3) C 4) B 5) D 6) A 7) C, , 8) A, , LEVEL - IV - HINTS, 1., , From the graph, the time period is T 33.3 10 15 s ., , , c, cT (3 108 ms1 )(33.3 10 15 s), , , 1 10 5 m 0.01mm, , 2., , 3., , 5., , 22, , 3., , E0, B, and Brms 0, c, 2, , Since ua E2 or B2 and E sin(t kx), so, ua sin2 (t kx), Thus, the frequency is double that of the electric or, magnetic field., The total displacement current initially = i0., The displacement current through the circular area, initially (being proportional to the fraction of area), , , 8., , MULTIPLE ANSWER QUESTIONS, , From the graph, E0 50V /m, B0 , , A photon of light enters a block of glass after, travelling through vacuum. The energy of the, photon on entering the glass block, (A) increases because its associated wavelength, decreases, (B) Decreases because the speed of the radiation, decreases, (C) Stays the same because the speed of the, radiation and the associated wavelength do not, change, (D) Stays the same because the frequency of the, radiation does not change, Radiation pressure on any surface :, (A) is dependent on wavelength of the light used, (B) is dependent on nature of surface and intensity, of light used, (C) is dependent on frequency and nature of, surface, (D) depends on the nature of source from which, light is coming and on nature of surface on which it, is falling., , i0, 4, , A parallel beam of radiation of intensity 10W, and of area of cross section 1cm2 is falling on, a plane surface at an angle 600 with normal to, the surface. The surface is partially reflecting, with reflection coefficient 0.5 and absorbing the, remaining.Choose the correct option(s) of the, following:, A) Force on the surface normal to it is 2.5 10 12 N, B) Force on the surface parallel to it is, , 2.5, 1012 N, 3, , 5, 1012 N, 3, D) Net force on the suface acts at an angle 300, with normal to the suface, , C) Net force on the surface , , INTEGER TYPE QUESTIONS, , The conduction current at time t 200s is 4., obviously (0.368)2i0., Total displacement current (or the conduction, current) at time t 100s is twice that of the current, through the area (given in the graph)., , A parallel beam of monochromatic light of, wavelength 663nm is incident on a totally, reflecting plane mirror. The angle of incidence, is 60 and the number of photons striking the, mirror per second is 1.0 1019 . Calculate the, force exerted by light beam on the mirror, , Reqd. Current = 2[(0.5)2 i0 ] 0.5i0 ., , in 10, , 8, , N, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, , MULTIPLE ANSWER QUESTIONS, , LEVEL - V- KEY, 1. D 2. B 3. ABCD, , 4.1, , 2., , LEVEL - V- HINTS, 1., , Frequency of light does not change with medium., I, 2, The radiation pressure = (1 ) Cos , C, where = coefficient of refection, Force on the surface normal to it is, , 2., 3., , IA cos , Fn , (1 r ) , C, , A sphere of radius R is exposed to a parallel, beam of radiation of intensity I as shown in, figure.Choose the correct option(s) of the, following., , R, , 1, 10 104 (1.5), 2, 2.5 1012 N, 3 108, , force on the surface parallel to it is, , Ft , , IAsin, 1 r , C, , 3, 0.5 2.5 12, 2, 10 N, 3108, 3, , 10104 , , Net, , force on t he, 5, 2, 2, 1012 N, F Fn Ft , 3, , surface, , Ft, , , , Tan , , (A) If the surface of the sphere is completely, 2I R 2, reflecting, radiation force in the sphere is, c, (B) If surface of the sphere is completely absorbing,, I R2, radiation force on the sphere is, c, (C) If surface of the sphere is completely reflecting,, I R2, radiation force on the sphere is, c, (D) If surface of the sphere is partially reflecting, with reflection coefficient 0.3 and absorbtion, coefficient 0.7, the radiation force in the sphere is, 1.7I R 2, c, , Ft, 1, , Fn, 3, , 300, , 4., , Fn, dp [mv cos ( mv cos )]dn, F, , dt, dt, dn, , dn h dn , 2mv cos 60 mv 108 N, dt , dt dt , , F, , INTEGER TYPE QUESTIONS, 3., , LEVEL - VI, SINGLE ANSWER QUESTIONS, 1., , The radiation force experienced by body, exposed to radiation of intensity I, assuming, surface of body to be perfectly absorbing is:, , I= Intensity, of radiation, , H, , (A), , R 2 , c, , (B), , RH, c, , RH, , A point source of radiation power P is placed, on the axis of an ideal plane mirror . The, distance between the source and the mirror is, n times the radius of the mirror . The force, P, that light exerts on the mirror is xc n 2 y , , find x +y?, 4., , A point source of radiation of power P is placed, on the axis of completely absorbing disc. The, distance between the source and the disc is 2, times the radius of the disc.The force that light, , (C) 2 c, , exerts on the disc is, , R, NARAYANAGROUP, , RH, , (D) c, , Px, . Then the value of x, 40c, , is, 23
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JEE-ADV PHYSICS- VOL- VI, , ELECTROMAGNETIC WAVES, 5., , Assuming a particle to have the form of a, sphere and to absorb all incident light,the radius, (in mm) of a particle for which its gravitational, attraction to the Sun is counter balanced by, the force that light exerts on it is x 101 .Find, x? The power of light radiated by the Sun, equals P = 4 × 1026 W and the density of the, particle is r = 1.0 g/cm3. Use G=, Nm2/kg2, =, , , , 4 IR 2 2, , cos 3 ( sin d ), , c 0, , 0, , 4 IR2 cos4 2, IR2, I R2, cos 4 , , , , , c 4 0, c, c, 2, , When surface is co mplet ly absorbing, I (dA cos ), dF , c, , 20, 10 11, 3, , 25, and mass of the Sun =2×1030, 8, , , 2, , Net force F dF , 0, , kg, , LEVEL - VI - KEY, 1.D, , 2.B,C 3. 3, , 4. 2, , 5. 6, , =, , LEVEL - VI - HINTS, SINGLE ANSWER QUESTIONS, 1., , 2., , dAcos, r, , dFsin, d, dF, , dF, , Consider a circular strip of radius R sin and of, width Rd ., Amount of energy falling on the strip per sec, 3., IdA cos , Change in momentum due t o reflect ion, 2I, dp , (dA cos ) cos , c, Force on the strip, Net force on the sphere, , 2, , , 2, , 4 IR 2, cos3 sin d, F dF cos , c, 0, 0, 24, , I R2, I R 2 I R2, 0.7, , c, c, c, Note: In all the above cases radiation force =, r adiat io n pr essur e( due t o abso r bt io n), × effective area perpendicular to the flow of, I, 2, energy = R, c, 0.3, , dFcos, , , 2I, 2I, dA cos 2 2 R sin Rd cos 2 , c, c, , , , 2, 2(0.3I )(dA cos ) cos , (0.7 I )( dA cos ), F, cos , c, c, 0, 0, , dFsin, , dF , , 0, , I R cos 2 , c 2 0, , , 2, , dA, , , , , sin 2 d , , when surface is partially reflecting with reflection, coefficient 0.3 and absorbtion coefficient 0.7 net, force on the sphere is, , MULTIPLE ANSWER QUESTIONS, , , , 2, , 2, , 0, I R2, I R 2, I R 2, cos, 2, , , , 2, , , , = 2c, , 2c, c, 2, , I, F effective area, c, , , , I R, c, , , 2 2, , I 2 R sin ( Rd ) cos , c, , INTEGER QUESTIONS, Consider a ring of radius x and of width d x., Power indicent on the ring., Pxdx.nr, , P, , dP= 4 (nr )2 x 2 2xdx.cos =, , , , 2 (nr )2 x 2, , , , 3/2, , No. of photons falling (per unit time) on the area =, h, dP . , momentum given by one photon = 2 cos ., hC, , , force on the ring d F (in the downward direction), , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, =, , ELECTROMAGNETIC WAVES, now dA = dA cos and dA = 2 xdx, , 2d P cos, d P . 2h, ., cos =, hC, C, , , dA = ( 2 xdx ) cos , , r, , F=, , , , P x dx n r, , 2., , , , 0, , 2, , 2 (n r ) x, r, , =, , P n2 r 2, , C, 0, , 2, , , , 3/ 2, , . cos , , A photon will exert force F as shown, only the, Fcos component will remain and F sin will, , C, , Nh, , cancel out as we integrate on the ring F = , , x dx, , [as N photons strike per second and this leads to a, , (n 2 r 2 x 2 )2, , loss of, , 1, P, Pn r, = 2C . n 2 r 2 (n2 1) = 2C(n2 1), 2 2, , momentum per second.], , as only Fcos component of force remains, dA , , h, , dF = h × cos , , , , , P, , , R, , , , , , Nh, , , R, , , , P, , , , (2dx cos ) h, × cos , F= dF = 4(4R2 x 2 ) ×, , , , hC , 0, 0, , , , nr, , , , , , P, , [ r = 4R 2 x 2 ]. Solving we get F = 20c, , SUN, r, , 5., 4 . Number of photons striking per second N =, , A, h, , Area A here is the area perpendicular to the direction, of intensity or direction of energy flow., Consider a ring of radius x and width dx on the, disc. Intensity I on the ring due to source is I =, P, 4r 2, , ., , , , r, , dA, , , F, Fcos, , NARAYANAGROUP, , =d, , s, co, , , , Fsin, , x, dx, , Particle, energy incident/sec on the particle, , P, PR 2, 2, , , R, =, =, 4r 2, 4r 2, , , dp, , Change in, F due to striking of photon F1 , dt, lin. mom. of 1 photon in collision × No. of photons, , P, 2R, , R, , r, , , , dA, , A, , h, PR 2, PR 2, , striking per sec =, = 2, C 4r 2 h, 4r C, Gravitational force on the particle, , GM s m GM s 4, F2 , 2 R 3 ;, 2, 3, r, r, , , , F1 F2, , 3P, PR 2 GM s 4R 3, , R=, , 4GM s C = 0.6, 4r 2C, 3r 2, 25
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , DUAL NATURE OF MATTER AND RADIATION, , , SYNOPSIS, Electron Emission :, , , , Metals have free electrons and these normally, cannot escape out of the metal surface., The free electron is held inside the metal surface, , by the attractive forces of the ions., A certain minimum amount of energy is requried to, be given to an electron to pull it out from the surface, of the metal and this energy is known as “Work, Function”., Work function ( . ) = 5.65 eV, highest (for platinum) , , . =1.88 eV, lowest (for ceasium), Metal Work function Metal Work function, 0 (eV), 0 (eV), , Cs, , 2.14, , Al, , 4.28, , K, , 2.30, , Hg, , 4.49, , Na, , 2.75, , Cu, , 4.65, , Ca, , 3.20, , Ag, , 4.70, , MO, , 4.17, , Ni, , 5.15, , Pb, , 4.25, , Pt, , 5.56, , In 1888 Hallwachs under took the study further., He connected zinc plate to an electroscope. He, found that when zinc plate is illuminated with ultra, violet light it became positively charged. A positively, charged zinc plate became more positively charged, when it is further illuminated with ultra violet light ., From these observations he concluded that, negatively charged particles were emitted by the, zinc plate under the action of ultra violet light. After, the discovery of electron these particles were called, as photo electrons., The emission of electrons from a metal plate when, illuminated by electromagnetic radiation of suitable, wavelength is called Photoelectric effect., , Lenard’s Experimental Study of, Photoelectric effect :, light of same wavelength, , This minimum energy required for the electron, emission can be supplied by any one of the, , following processes., a) Thermionic emission : “Sufficient thermal, energy can be imparted to free electrons” by, suitably heating, b) Field emission: “By applying a very strong , electric field ( 108V / m )”., c) Photo electric emission: “By irradiating the, metal surface with suitable E.M radiaton”., , , Photo electric effect :, , , , Photo electric effect was discovered by Hertz in, 1887. In his experiments, Hertz observed that high, voltage spark passes across the metal electrodes, more easily when cathode is illuminated with ultra, violet rays from an arc lamp., NARAYANAGROUP, , A, Micro, Anmmeter, , V, , K, , (), , , , e, e, e, , C, , -, , +, , The apparatus used for experimental study of, photoelectric effect. A metal plate C called cathode, (emitter) and a metal cup A called anode (collector), are sealed in a vacuum chamber., A beam of monochromatic light enters the window, of a vacuum chamber and falls on cathode C. The, photoelectrons emitted are collected by the anode, A., When key K is open and monochromatic light is, made incident on the cathode, then current is, measured by the ammeter. i.e., even though applied, voltage is zero current flows in the circuit., These photoelectrons emitted from the cathode C, moves towards anode A. But less energetic, electrons comes to rest before reaching the anode., 1
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , , -x, , (-V0), stopping, potential, , , , y, , Saturation, photoelectric, current, , I, O, , (i), , , , intensity, of light, , Observations :, It is observed that saturation photocurrent (i) is, proportional to the intensity (I) of incident light at, a given frequency, , Variation of saturation photo current, with stopping potential at constant, intensity : Keeping the frequency of incident light, and nature of the cathode constant, for different, intensities of incident light photo current is measured., When a graph is plotted with photocurrent on yaxis and applied voltage on x-axis. It is as shown, in figure., , Observations :, The value of stopping potential is independent of, the intensity of incident light, if frequency is constant., (ii) The magnitude of saturation current depends on, the intensity of light. Higher the intensity, larger the, saturation current., Variation of frequency of incident light on, (i), , stopping potential :, , o, , Anode, potential (V), , x, , Variation of Photo current with intensity, of incident light : Keeping the frequency of, incident light and nature of the cathode constant,, for different intensities of incident light saturation, photo current is measured., When a graph is plotted with saturation, photocurrent on y-axis and intensity of incident light, on x-axis, it is as shown in figure., , 2, , i, , Keeping the intensity of incident light and nature of, the cathode constant, for different frequencies of, incident light, photo current is measured., When a graph is plotted with photocurrent on yaxis and applied voltage on x-axis. It is as shown, in figure., photo, current, , photoelectric, current, , When anode is given positive potential w.r.t the, cathode, electrons in the space charge are attracted, towards the anode so photocurrent increases. If, potential of the anode is increased gradually the, effect of space charge becomes negligible at some, potential and then every electron that is emitted from, the cathode will be able to reach the anode. The, current then becomes constant even though voltage, is increased and this current is called saturation, photocurrent., When anode is given negative potential w.r.t the, cathode, the photo electrons will be repelled by, the anode and some electrons will go back to, cathode so current decreases. At some negative, potential anode current becomes zero.This, potential is called stopping potential., The minimum negative potential(V0) given to the, collector with respect to the emitter for which, ‘photocurrent’ becomes zero is called ‘stopping, potential’., Stopping potential is related to maximum kinetic, energy of photoelectrons, because at this potential, even the most energetic electron just fails to reach, the anode., So work done by the stopping potential is equal to, the maximum kinetic energy of the electrons., 1 2, 1, (e)(V0 ) mvmax, 0 ; e V 0 , m v m2 a x, 2, 2, A graph is plotted with current on y-axis and applied, voltage on x-axis. It is as shown in below graph, , v3v2v1, v3 v, 2 v, , 1, , saturation, current, , -V03 -V02 -V01 O, V, anode, retarding, potential, potential, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , Observations :, Larger the frequency of incident radiation, larger is, the stopping potential., So The maximum kinetic energy of the emitted, electrons depends on the frequency of incident light, and nature of the metal plate. Maximum kinetic, energy of photo electrons is independent of the, intensity of incident light., (ii) The saturation photo current is independent of the, frequency of incident radiation., Variation of Stopping potential with, (i), , frequency of incident light :, , stopping, potential (vo), , When a graph is plotted with stopping potential, on y-axis and frequency of incident radiation on xaxis, keeping the metal constant, then it is as shown, in figure., Cs, , O, , vo, , v1, , K Na A1 Cu, , v2, , v3, , v4, , frequency of incident, , (v), , Observations :, (i) Threshold frequency ( n0 )is a characteristic of the, metal plate and at this frequency, kinetic energy of, the photo electrons is zero., (ii) Above threshold frequency, kinetic energy of, photo electrons range from zero to a maximum, value., (iii) Maximum kinetic energy and Stopping potential, increases linearly with increasing frequency as, shown in the above figure., , LAWS OF PHOTOELECTRIC EFFECT:, , , , , , , , , If the frequency of incident radiation is less than a, certain value called threshold frequency, electrons, are not emitted from a given metal surface,, whatever be the intensity of the incident radiation., The maximum kinetic energy of photoelectrons, depends on the frequency of the incident radiation,, but it is independent of the intensity of the radiation., The maximum kinetic energy of photoelectrons is, a linear function of the frequency of the incident, radiation., The saturation photocurrent increases with intensity, of incident radiation, but it is independent of the, frequency of incident radiation., There is no time lag between the incidence of the, incident radiation and the emission of photo, electrons., NARAYANAGROUP, , QUANTUM THEORY OF LIGHT:, , , Since wave theory of light can not explain observed, facts about photo electric effect, Einstein thought, that light has got particle-like behaviour during its, interaction with matter. According to this theory,, light consists of particles associated with definite, amount of energy and momentum. These particles, were latter called as photons., A photon is a packet of energy, given by E hn, where h = 6.62x10-34 Js, (where Planck’s constant, h = 6.62x10-34 Js, = 4.14x10-15eVs), n = frequency of the wave associated with photon, then C n l, C= 3x108 ms-1 = velocity of light, l = wavelength of the wave associated with, hC, photon E hn , l, Conclusions :, (i) Energy of a photon is normally expressed in electron, volt (eV), (ii) Electron volt is the change in the energy of an, electron when it passes through a potential difference, of one volt. 1eV = 1.6x10-19 J, (iii) Charge of a photon is zero, so it can not be, deflected by electric and magnetic fields., (iv) Rest mass of a photon is zero. It is not a material, particle, it is a bundle of energy., E, hn, h, 2 2, 2, C, C, nl, (vi) A photon possesses momentum given by, h hn E, p , , l, C, C, , (v) Effective mass of a photon is m , , INTENSITY, , , Intensity (I) of radiation at a given point is the energy, transmitted through unit area perpendicular to that, area in unit time. Intensity(I) of radiation at a, distance ‘r’from a monochromatic source of power, P, , . If ‘N’ photons are emitted in, ‘P’ is, I , 4p r 2, time ‘t’ by a monochromatic source of power, ‘P’then P N h n N h C . So increase in, t, lt, intensity of monochromatic radiation means, increase in the number of photons incident on unit, area in unit time., 3
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , W.E-1: While working with light and X-rays, there, is a useful relation between the energy of a, photon in electron volts (eV) and the, wavelength of the photon in angstom (A0)., Suppose the wavelength of aphoton is A0 ., Then energy of the photon is, hc, Sol: E hv , , Here wavelength =, , , , , , 1010 m; h 6.62 1034 Js, c 3 108 ms 1, , , , 6.62 10 3 10 , E , , , , 6.62 10 3 10 eV 12400 eV, , , 10 1.6 10 , , , , 34, , 8, , 1010, , 34, , 8, , 10, , 19, , 12400, eV, , Note : ( is taken in A0 and 12400 in A0 eV), E , , 4000 A 400nm then the energy of the, photon is, 0, , hC 12400eVA0 1240eVnm, , , 3.1eV, Sol: E , , 4000 A0, 400nm, W.E-3: A monochromatic source of light operating, at 200 W emits 4 1020 photons per second., Find the wagelength of the light., N, Sol: Power = P h, t, Energy of photon =, , P, 200, , 5 10 19, 20, N 4 10, , t , , 6.62 10 3 10 m 3.972 A, , 34, , 5 10, , 8, , 0, , 19, , EINSTEIN’S PHOTO ELECTRIC, EQUATION:, , , , 4, , For explaining photoelectric effect, Einstein, postulated that light consists of particles called, photons. Energy of a photon of frequency n is, hn ., According to this theory the emission of a, , i), , W hn 0, , Work function of a metal depends on nature of the, metal, it will not depend on frequency and intensity, of the radiation., When a photon of energy hn is absorbed by an, electron, an amount of energy at least equal to work, function W (provided hn >W ) is used up in, liberating the electron from the surface and the, difference ( hn - W) is equal to the maximum, kinetic energy.of that electron., , , , W.E-2: If wavelength of radiation is, , E, , photoelectron was the result of the interaction of a, single photon with an electron, in which the photon, is completely absorbed by the electron., The minimum amount of energy required to eject, an electron from a metal surface is called work, function (W) of that metal. It is also called threshold, energy., The minimum frequency of radiation required to, eject an electron from a metal surface is called, threshold frequency( n0 ) for that met al., , 1 2, mVmax hn W, 2, 1, , (1), , 2, hn W mVmax, (2), 2, 1, 2, hn hn 0 mVmax, (3), 2, The above relation is called the Einstein’s, Photoelectric equation. Here ‘m’ is the mass of, the electron and Vmax is the maximum velocity of, the photoelectrons. Infact, most of the electrons, possess kinetic energy less than the maximum, value, as they lose a part of their kinetic energy, due to collisions before escaping from the metal., Thus from the above discussion the laws of, photoelectric effect from Einstein’s, Photoelectric equations are deduced., From equation (1) maximum kinetic energy of, photoelectrons is KEmax hn hn0 ., For photoelectric emission to take place kinetic, energy of electrons must be positive. It follows that, hn hn 0 n n0 ., It proves that for photoelectric emission to take, place, from a given metal the frequency of the, incident radiation must be greater than threshold, frequency for that metal., If frequency of the incident radiation is less than, threshold frequency then no photoelectric emission, will take place, whatever be the intensity of the, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, incident radiation, or how long it falls on the metal, surface., ii) From equation (1) it follows that maximum kinetic, energy of photoelectrons depends linearly on the, frequency. It proves that the maximum kinetic, energy of photoelectrons increases as frequency, of incident radiation increases., Since Einstein’s equation does not involve a factor, representing intensity, it proves that the maximum, kinetic energy of emitted electrons is, independent of the intensity of incident, radiation., iii) According to Einstein, the photoelectric effect, arises , when a single photon is absorbed by a, single electron. So number of photoelectrons, ejected will be large if intense radiation is incident., This is because intensity of radiation is proportional, to number of photons per unit area per unit time., Hence if intensity of incident radiation is larger, then, number photons incident is larger and number of, electrons ejected is larger., It proves that number of photoelectrons ejected, from a metal surface depends on intensity of, incident radiation. Further, there is no effect of, frequency of incident radiation on number of, photoelectrons emitted. It is because one photon, is capable of ejecting only one electron, provided ,, n n0, iv) According to Einstein, the basic process in, photoelectric emission is absorption of a photon of, light by an electron. So as the photon is absorbed,, emission of electron takes place instantaneously, irrespective of intensity., Conclusions :, (a) Alkali metals can cause photoelectric effect with, visible light., (b) Work function of Alkali metals is around 2eV., (Cs=2.14eV, K= 2.3eV, Na=2.75eV, Al=4.28eV,, Cu=4.65eV, Ag=4.7eV, Ni=5.15eV, Pt=5.65eV), Among all metals work function is least for Cesium(, 2.14eV), (c) Work function W hn 0 , , hC, l0, , where n0 =threshold frequency,,, , l0 =threshold wavelength, (d) Einstein’s equation can be written as follows:, KEmax E W (or) KEmax hn hn0, hC, , 1, 1, 2, m V m2ax E W (or) mVmax, hn hn 0, 2, 2, , NARAYANAGROUP, , hC, , 1, , hC, , 2, (or) 2 mVmax l l, 0, , (f), , eV0 E W (or) eV0 hn h n 0, (or), , eV 0 , , hC, hC, , l, l0, , W.E-4: The work function of a metal is 3.0eV. It, is illuminated by a light of wave length, 3 x 107m. Calculate i) threshold frequency, ii), the maximum energy of photoelectrons,, iii) the stopping potential. (h =6.63 x 10–34 Js, and c = 3 x 108ms–1)., Sol. i) W = h n 0 = 3.0eV = 3 x 1.6 x 10–19J, Threshold frequency, n0 , , W, 31.6 1019, , 0.72 1015 Hz ., h, 6.631034, , ii) Maximum kinetic energy (Kmax) = h ( n – n0 ), c, 3108, 11015 Hz, l 3 107 m, v l , 3107, , Kmax = h ( n – n0 ) = 6.63 x 10–34 (1– 0.72) x 1015, J = 1.86 x 10–19 J., iii) Kmax = e V0 where V0 is stopping potential in, volt and e is the charge of electron, V0 , , K max, e, , . Here Kmax = 1.86 x 10–19 J and, , e = 1.6 x 10–19 C; V0 , , 1.86 1019 J, 1.6 1019 C, , 1.16V, , W.E-5: The work function of a photosensitive, element is 2eV. Calculate the velocity of a, photoelectron when the element is exposed to, 0, , a light of wavelength 4103 A ., Sol. Einstein's photoelectric equation is, 1 2 hc, mv W0, 2, l, , 1 2, 6.62 3, mv , 1026 2 1.61019, 3, 10, 2, 4 10 10, v2 , , 1.765 2, 1012, 9.1, , v, , 1.765 2, 106 = 6.228, 9.1, , hC, , (or) K E m ax l l, 0, (e), , DUAL NATURE, , x 105 ms–1, , 5
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , W.E-6: A metal of work function 4eV is exposed, to a radiation of wavelength 140 x 10–9 m., Find the stopping potential., hc, 6.62 1034 3108, E, eV, l, 140 109 1.6 1019, , Sol. E , , = 8.86eV, , work function W0 = 4eV, eV0 = E –W0 = 8.86 –4 = 4.86eV, Stopping potential V 0 4.86V, , W.E-7: Radiations of wavelength 200nm, propagating in the form of a parallel beam,, fall normally on a plane metallic surface. The, intensity of the beam is 5mW and its cross, sectional area 1.0mm 2 . Find the pressure, exerted by the radiation on the metallic, surface, if the radiation is completely, reflected.[Roorkee 2001], Sol. E , , 12400 12400, , 6.2eV 1018 J, l, 200, , P 5109, , 5109 . momentum of each, E, 1018, , photon, , E, p 3.31027 J / s, C, , ., , Change in, , momentum after each strike = 2p = 6.6 x 10–27 J/s, Total momentum change per second is, F, , dp n 2 p, , 5109 6.6 1027 331018 N, dt, t, , F, , 12, , pressure A 3310, , N / m2, , W.E-8: In a photocell bi chromatic light of wave, length 2480 A0 and 6000A0are incident on a, cathode whose workfunction is 4.8eV. If a, uniform magnetic field of 3x10-5 T exists, parallel to the plate, find the radius of the, circular path described by the photoelectron., (mass of electron is 9 x1031 kg ), Sol. E1 , , 12400, 12400, , 5 eV, l1, 2480, , ;, , E2 , , 12400 12400, , 2.06 eV, l2, 6000, , As E2 < W0 and E1 > W0, photo electric emission, is possible only with 1 ., Maximum K.E of emitted photo electrons, K = E1 – W0 = 0.2 eV., Photo electrons experience magnetic force and, move along a circular path of radius, r, , 6, , mn, 2 mK, , ., Bq, Bq r 5cm, , is incident on an isolated metalic sphere of, radius a. The threshold wavelength is l 0, which is larger than l . Find the number of, photoelectrons emitted before the emission of, photo electrons stops., Sol. As the metallic sphere is isolated, it becomes, positively charged when electrons are ejected from, it. There is an extra attractive force on the, photoelectrons. If the potential of the sphere is, raised to V, the electron should have a minimum, energy W + eV to be able to come out. Thus,, emission of photoelectrons will stop when, hc, W eV, l, , hc 1 1 , ., l l0 , , hc, , or, V e, , = l eV, 0, , The charge on the sphere needed to take its potential, , Number of photons passing a point per second is, n, , W.E-9: A monochromatic light of wavelength l, , to V is Q 4pe0 a V, The number of electrons emitted is, therefore,, n, , Q 4pe0 aV 4pe0 ahc 1 1 , , , e 2 l l0 , e, e, , W.E-10: A small metal plate (work function W) is, kept at a distance d from a singly ionized, fixed, ion. A monochromatic light beam is incident, on the metal plate and photoelectrons are, emitted. Find the maximum wavelength of the, light beam so that some of the photoelectrons, may go round the ion along a circle., Sol. Electron is moving around the ion in a Circle of, 1, , e2, , mV 2, , radius ‘d’. 4pe 2 d ,, 0 d, K .E , , 1 e2, 4pe0 d, , 1 e2, ----------(1), 8pe0 d, , But K .Emax , , l, , mV 2 , , hc, W ----------(2), l, , hc 8pe0 d, e2 8pe0 dW, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , Photo electric cells :, , , , , , , , , , , , , , , , , , , A device which converts light energy into electrical, energy is known as a photo electric cell or photo, cell., It is a technological application of the photo electric, effect., It is a device whose electrical properties are, affected by light:, It consists of a semi-cylindrical photosensitive metal, plate (emitter) and a wire loop (collector)., when light of suitable wavelength falls on the emitter, ,photoelectrons are emitted and enters into the, circuit giving a few amperes of current., It converts a change in intensity of illumination into, a change in photocurrent., These are the devices used to convert light energy, into electrical energy., In the outer photoelectric effect photo electrons, liberated are emitted out of the metal surface., In the inner photoelectric effect photoelectrons, liberated are in the metal surface., There are three types of photo cells., 1) Photo emissive cells 2) Photo voltaic cells, 3) Photo conductive cells., Photo emissive cells depend upon outer, photoelectric effect whereas photo voltaic and, photo conductive cells depend upon inner, photoelectric effect., Photo emissive cells are of two types., 1. Vacuum type and 2. Gas filled type, In the Vacuum type of photo emissive cell, an, evacuated glass tube has its inner surface coated, with an alkali metal., To study stellar spectra potassium hydride, photosensitive surface is used., In the vacuum type, current is directly proportional, to the intensity of incident radiation., In a gas filled emissive cell, an inert gas such as He,, Ar or Ne at a low pressure, some tenths of mm of, Hg, is filled. Gas filled cells produce much more, intense photoelectric current due to ionization by, collision in the gas. But there is no proportionality, between the current and intensity of incident, radiation., Photovoltaic cell is a true cell as it generates e.m.f., without the application of any external potential., When light is incident on a semi conductor coated, on a metallic plate covered by a semi transparent, film electrons are emitted and travel in the direction, opposite to the light rays., NARAYANAGROUP, , , , , , , , , , , The semi conductors used are cuprous oxide,, selenium., The metal plates used are copper plate, iron plate., The semi transparent films used are silver, gold,, platinum., For small load resistances the current is nearly, proportional to the intensity of incident radiation., In a photo conductive cell, conductivity of a semi, conductor is increased when light falls on it. But, the response is slow. The current is not proportional, to the intensity of light. Photo electric cells are used, in exposure meters, to compare intensities of illuminations of two light, sources., in recording and reproduction of sound in films, in video cameras, to study stellar spectra, in electronic relay circuits such as Burglar’s alarm,, counting devices, switching on and off street lights,, etc., , DUAL NATURE OF MATTER (de-BROGLIE HYPOTHESIS) :, , , Photoelectric effect and Compton effect proves that, radiation behaves like particles (photons), where, as Interference and Diffraction proves that radiation, behaves like waves., So ‘radiation has dual nature’ i.e., radiation behaves, like particles when interacting with matter and, radiation behaves like waves when propagating in, a medium., , de Broglie Hypothesis, 1), 2), 3), , , The universe consists of matter and radiation only., Nature loves symmetry, If radiation has dual nature then matter also should, have dual nature., According to de Broglie particles like electron,, proton and neutron, also have both wave and, particle properties. The waves associated with, moving particle are called matter waves and the, wavelength is called the de Broglie wavelength, of a particle., hC, For a photon Energy, E , mC 2, , where m = effective mass then wavelength, h, h, , , mC p, where p = momentum of the photon, 7
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, de Broglie extended the same for particles also., So if a particle of mass ‘m’ is moving with velocity, ‘v’ then its momentum p = mv, hence de Broglie, wave length of the matter wave associated with is, , h, h, , p mv, Davisson and Germer studied the scattering of, electrons by a nikel target. The wavelength of, diffracted electrons was determined by Davisson, and Germer. The experimental values of wavelength, were found to agree with the theoretical, , given by , , (ix) The de-Broglie wavelength of a particle is, independent of nature of the particle and these, waves are not electromagnetic. Diffraction effects, have been obtained with streams of electrons,, protons, neutrons and alpha particles., (x) de-Broglie explains Bohr’s criterion to select the, allowed orbits in which angular momentum of the, h, electron is an integral multiple of, . According, 2, to his hypothesis, an electron revolving round the, nucleus is associated with certain wavelength ‘ ’, , h, value , m v, , Hence it is concluded that electrons behaves like, waves and undergo diffraction., For definite sized objects like a car the, corresponding wavelength is very small to detect, the wave properties. But the de-Broglie wavelength, of the electron is large enough to be observed., Because of their small mass, electrons have a small, momentum and hence large wavelength h / p ., Conclusions :, h, h, (i) deBroglie wavelength p m v, Where momentum p = mv ; m=mass, v= velocity, h, (ii) deBroglie wavelength , 2mK, 2, p, p 2mK, where kinetic energy, K , 2m, (iii) If a particle having charge q starting from rest is, accelerated through a potential difference V then, gain in kinetic energy, K=qV, h, so, deBroglie wavelength 2 m qV, (iv) For electron , (v) For proton , , 12.27 o, 150 o, A, A, V, V, , 0.286 o, 0.082 o, A, A, V, V, 0.202, , Å, E, 0.101, Å, (vii) For particle , V, 0 .2 86, Å, (viii) for neutron , E, (vi) For dueteron , , which depends on its momentum mv. It is given by, , , h, h, , mv p, , In an allowed orbit, an electron can have an integral, multiple of this wavelength., That is the nth orbit consists of n complete deBroglie wavelengths 2rn n n , where n is, , the principle quantum number., where rn is the radius of nth orbit and n is, the wavelength of electron in nth orbit, , n , , 2rn, 2, (or) n (0.53 n2 ) Å ., n, n, , ln 2p 0.53nA, (a), , (b), , Figure (a) shows the waves on a string have a, wavelength related to the length of the string, allowing them to interfere constructively as shown, If we imagine the string bent into a closed circle we, get an idea of how electrons in circular orbits can, interfere constructively as shown in figure(b). If, the wavelength does not fit in to the circumference, the electron interferes destructively, electron can, not exist in such an orbit., , where E = kinetic energy in electron volts, 8, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , W.E-11: A particle of mass ‘m’ projected horizontally, , W.E-16: Photons of energies 4.25eV and 4.7eV, , with velocity u. If it makes an angle q , with the, horizontal after some time, then at that instant,, its de Broglie wavelength is, Sol: For a projectile horizontal component of velocity, is constant., u, , , are incident on two metal surfaces A and B, respectively. The maximum KE of emitted, electrons are respectively TA eV and TB =( TA 1.5)eV. The ratio de-Broglie wavelengths of, photo electrons from them is A : B =1:2, then, find the work function of Aand B, Sol. Debroglie wavelength, , v x u x ; V cos q u, , v, , gt, , de Broglie wavelength, l , , l, , h, h cos q, , mv, mu, , 2, , W.E-12: Electrons are accelerated through a, potential difference of 150V. Calculate the de, Broglie wavelength., Sol. V = 150V; h=6.62 x 10–34Js, m=9.1 x 10–31kg,, e = 1.6 x 10–19C, h, , l , , 2Vem, , 6.62 1034, , , , of molecules of hydrogen and helium which, are at tem peratures 27°C and 127°C, respectively, , , h, , , , mHe THe, 8, , mH TH, 3, , 6.6 1034, 10, , 5000 10, , position and some uncertainity p in the, specification of momentum. The product of x, and p is of the order of h (with , , 1450 m / s, , , , W.E-15: If 10,000V applied across an X-ray tube,, what will be the ratio of deBroglie wavelength, of the incident electrons to the shortest, wavelength of X-ray produced (e/m of electron, is 1.7 x1011C / Kg ), Sol. Debroglie wave length of incident electron is, l1 , , h, 2meV, , ........ 1, , Shortest wavelength of x ray photon is l2 , , , l1 1, , l2 c, , V , , e 0.1, m , 2 , , NARAYANAGROUP, , The matter-wave picture elegantly incorporated the, Heisenberg’s uncertainty principle. According to, the principle, it is not possible to measure both the, position and momentum of an electron (or any other, particle) at the same time exactly. There is always, some uncertainity x in the specification of, , 0, , 9.110, , TA, TA 2eV, TA 1.5, , HEISEN-BERG UNCERTAINITY, PRINCIPLE, , photon with a wavelength of 5000 A, ( h 6.6 x1034 Js , me 9.1x1031 Kg ), 31, , T, , WB 4.7 TB 4.7 0.5 4.2 eV, , so that its momentum is equal to that of a, , h, l, , l, , B, A, (k = k.E = T); l T, k, A, B, , TB TA 1.5 2 1.5 0.5eV, , W.E-14: With what velocity must an electron travel, , Sol. mv v , , 1, , WA 4.25 TA 2.25 eV, , 1A, , W.E-13: Find the ratio of de Broglie wavelength, , h, , 2km, , l , , 0, , 2 9.11031 1.6 1019 150, , H, ; , Sol. Since, m , 3mkT, He, , h, , hc, Ve, , ..... 2, , , , h, ) i.e.,, 2, , xp h ., Equation allows the possibility that x is zero, but, then p must be infinite in order that the product, is nonzero. Similarly, if p is zero, x must be, infinite. Ordinarily, both x and p are nonzero, such that their product is of the order of h ., Now, if an electron has a definite momentum p ,, (i.e., p =0), by the de Broglie relation, it has a, definite wavelength . A wave of definite (single), wavelength extends all over space. By Born’s, probability interpretation this mens that the electron, is not localized in any finite region of space. That, , is, its position uncertainity is infinite x ,, which is consistent with the uncertainity principle., 9
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , , , , , , In general, the matter wave associated with the, electron is not extended all over space. It is a wve, packet extending over some finite region of space., In that case x is not infinite but has some finite, value depending on the extension of the wave, packet. Also, you must appreciate that a wave, packet of finite extension does not have a single, wavelength. It is built up wavelengths spread, around some central wavelength., By de Broglie’s relation, then, the momentum of, the electron will also have a spread - an uncertainity, p . This is as expected from the uncertainity, principle. It can be shown that the wave packet, description together with de Broglie relation and, Born’s probability interpretation reproduce the, Heisenberg’s uncertainity principle exactly., The de Broglie relation will be seen to justify bohr’s, postulate on quantisation of angular momentum of, electron in an atom., Figure shows a schematic diagram of (a) a localised, wave packet, and (b) an extended wave with fixed, wavelength., , (a), , , (b), , Figure (a) the wave packet description of an, electron. The wave packet corresponds to a spread, of wavelength around some central wavelength (and, hence by de Broglie relation, a spread in, momentum). Consequently, it is associated with, , proton is 6 108 m , then the m inim um, uncertainity in its speed is, Sol: p mv , or v , , , x, , , 1.034 1034, 1, , mx 1.67 10 27 6 108 1 ms, , W.E-18: The correctness of velocity of an electron, movign with velocity 50 ms-1 is 0.005%. The, , accuracy with which its position can be, measured will be, Sol: Here, v , , 0.005 50, 0.0025ms 1, 100, , 1.034 1034, , x , , mv 9.11031 0.0025, 4634 105 m, , DAVISSON AND GERMER’S ELECTRON, DIFFRACTION EXPERIMENT, i), , The first experimental evidence of matter wave was, given by two American physicists, Davisson and, Germer in 1927. They also succeeded in measuring, the de - Broglie wave length associated with slow, electrons., ii) A beam of electron emitted by electron gun is made, to fall on nikel crystal cut along cubical axis at a, particular angle., iii) Ni crystal behaves like a three dimensional diffraction, grating and it diffracts the electron beam obtained, from electron gun., iv) The diffracted beam of electrons received by the, detector which can be positioned at any angle by, rotating about the point of incidence., , an uncertainity in position x and an uncertainity, in momentum p . (b) the matter wave, corresponiding to a definite momentum of an, electron extends all over space. In this case,, p 0 and x ., W.E-17: If the uncertainity in the position of, 10, , v) The energy of the incident beam of electron can, also be varied by changing the applied voltage to, the electron gun., vi) According to classical physics, the intensity of, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , scattered beam of electrons was not the same but, different at different angles of scattering. It is, maximum for diffracting angle 500 at 54 volt P.D., vii) It is seen that a bump begins to appear in the curve, for 44 volt electrons. With increasing potential, the, bump moves upwards and becomes most, prominent in the curve for 54 volt electrons at, 500 . At higher potential the bump gradually, disappears., , angle ‘ ’., 90 , , , , and d D cos D sin, 2, 2, , using sin cos, , , 2, , , f, f, l 2 d sin q 2 d sin cos d sin f, , 2, 2, , Incident beam, , l d sin f, , ADDITIONAL, 44V, , 48V, , 54V, , 60V, , 500, , viii) If the de Broglie waves are associated with electron,, then these should be diffracted like x - rays. using, the Bragg’s formula 2d sin n , we can, determine the wavelength of these waves., Where ‘d’ is the distance between the diffracting, 180 , = glancing angle for incident, 2 , , planes. , , , beam = Bragg’s angle., ix) The distance between diffracting planes in Ni crystal for this experiment is d = 0.91A0 and for, n = 1; 2 0.91 10 10 sin 65 = 1.65A0, Now debroglie wave length can also be determined, using the formula ; , 0, , =65, , 12.27, V, , , , 12.27, 54, , 1.67 A0, , =50, , D, , , d, , Atomic, planes, , Thus the deBroglie hypothesis is verified., x) The Bragg’s formula can be rewritten in the form, containing inter atomic distance D and scattering, NARAYANAGROUP, , X-RAYS :, Roentgen discovered the X-rays., i) Most commonly x-rays are produced by the, deceleration of high energy electrons bombarding, a hard metal target., ii) The target should have, a) high atomic weight, b) high melting point, c) high thermal conductivity, iii) They are electromagnetic waves of very short, wavelength. i.e., order of wavelength 0.1A° to, 100A° , order of frequency 1016Hz to 1019 Hz,, order of energy 124eV to 124keV, iv) Most of the kinetic energy of electrons is converted, into heat and only a fraction is used in producing xrays (less than 1% x - rays and more than 99%, heat)., v) Intensity of x-rays depends on the number of, electrons striking the target which inturn depends, on filament current., vi) Quality of x - rays (hard /soft) depends on P.D, applied to x - rays tube., vii) high frequency x-rays are called hard x-rays, viii) low frequency x-rays are called soft x-rays, ix) Penetrating power of x-rays is a function of, potential difference between cathode and target., x) Interatomic distance in crystals is of the order of, the wavelength of x-rays hence crystals diffract xrays., xi) Production of x-rays is converse of photoelectric, effect., , X-RAY SPECTRUM, 11
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , Continuous X-ray spectrum:, , a), , It is produced when high speed electrons are, suddenly stopped by a metal target., It contains all wave lengths above a minimum, wavelength m . ( continuous spectrum ) For a, given accelerating potential, m is called cut off wave, length., Properties of continuous x - rays spectra are, independent of nature of target metal and they, depend only on accelerating potential., , b), , c), , 12, 10, 8, 6, 4, 2, , e), , 12, 10, 8, 6, 4, 2, , a), 40KV, , Intensity, , d), , ii) Characterstic X-ray spectrum:, , 50KV, , b), , 30KV, 20KV, , c), , o -1 -2 -3 -4 -5 -6 -7 -8 -9 -10, 0, Wave length in A, , d), , min , , min, o -1 -2 -3 -4 -5 -6 -7 -8 -9 -10, Wave length in A0, , Produced due to transition of electrons from higher, energy level to lower energy level in target atoms, Wavelengths of these x-rays depend only on atomic, number of the target element and independent of, target potential., Characteristic x-rays of an element consists of K,, L,M and N series., K-series of lines are obtained when transition takes, place from higher levels to k shell, , hc 12400 0, , A, eV, V, , K, , g), , h), , i), j), , With the increase in target potential, min and, wavelength corresponding to maximum intensity, 0 shifts towards minimum wavelength side., At a given potential the range of wave length of, continous x - rays produced is min to ., Efficiency of x - ray tube, , , out put power, 100, input power, , L, M, , M, , +, K, L, M, N, O, , max , , f), , L, , K, , 1, min it is Duane and Hunt’s law, V, Maximum frequency of emitted x - ray photon is, , ev, h, In this spectrum intensity first increases, reaches a, maximum value I, and then decreases., max, Every spectrum starts with certian minimum wave, length called limiting wave length or cut off wave, length min ., , K K, , 35KV, , Intensity, , i), , e), , This spectrum is useful in identifying the elements, by which they are produced., , f), , Relation among the energies Ek Ek Ek ,, Ek EL, , g), , Intensity of x - rays Ik Ik Ik, , h), , Relation among frequences k , k and L is, , nkb nka nLa , , h), , EK E L hn K a , , 1, l Kb, , , , 1, 1, , lKa lLa, , hc, lK a, , input power P = VI. Where V is P.D applied to x ray tube I = anode current, 12, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , EK EM hn K b , EL EM hn La , , DUAL NATURE, iii), , he, lK b, , iv) The intercept on ‘Z’ axis gives the screening constant, ‘b’ and it is constant for all spectral lines in given, series but varies with the series., , he, lK a, , b = 1 for k series ( k , k , k ), , iii) Intensity and wavelength graph, v), , , vi) Ratio of wavelengths k and k lines from a given, min, , 0, , k 32, target is 27 ., k, , k k, , , As target potential V is increased, , a), , 0, , b), , Wavelength of k remains constant., , c), , diffrence between , , m in decreases, , and increases, min, diffrence between k line and k line remains, k, , , , constant., e), , vii) Significance :, a) The elements must be arranged in the periodic table, as per their atomic numbers but not on their atomic, weights., b) Helped to discover new elements like masurium, (43) and illinium (61) etc., c) Decided the positions and atomic numbers of rare, earth metals., , Difference between k 0 increases., , C. U. Q, PHOTO ELECTRIC EFFECT, , MOSELEY’S LAW, i), , “The square root of frequency ( v ) of the spectral, line of the characteristic x-rays spectrum is directly, proportional to the atomic number(z) of the target, element., Z or =a(Z-b), , , 0, , b=1 Z, , The slope(a) of -Z curve varies from series to, series and also from line to line of a given series., For K series, , , , 1., , n1 Z1 1 , , , n 2 Z 2 1, , l2 Z1 1 , , , l1 Z 2 1, , NARAYANAGROUP, , The rest mass of a photon is, 1) zero, , 2., , , , v, , ii), , b = 7.4 for L series, The wavelength of characteristic X-rays is given, 1 1, 1, 2 2 2, by =R(Z-b) n1 n2 , , , , , I, , d), , ak ak ak, , 3., , 2) 1.6 10 19 kg, , 3) 3.1 10 30 kg, 4) 9.1 10 31 kg, The mass of a photon in motion is (given its, frequency = x ), hx, hx 3, 3, 1) 2, 2) hx, 3) 2 4) zero, c, c, Photoelectric effect supports the quantum, nature of light because, 1) There is minimum frequency of light above which, no photo electrons are emitted, 2) The maximum kinetic energy of photo electrons, depends on both frequency and intensity of light, 3) Even when a metal surface is faintly illuminated,, the photoelectrons do not leave the surface, immediately, 4) The maximum K.E. of photo electrons depends, only on the frequency of light and not on intensity, 13
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 4., , Which of the following statement is wrong?, 1) Einstein explained photo electric effect with the, help of quantum theory, 2) Millikan determined the value of planck’s, constant depending upon the property of photo, electric effect, 3) The maximum K.E. of the photo electrons, depends upon the intensity of incident radiation, 4) As the frequency of incident photon increases, the corresponding stopping potential also increases, 5. In photoelectric emission, the energy of the, emitted electron is, 1) larger than that of the incident photons, 2) smaller than that of the incident photons, 3) same at that of the incident photons, 4) proportional to the intensity of the incident light, 6. A laser beam of output power ‘P’ consists only, of wavelength . If Planck’s constant is h and, the speed of light is c, then the number of, photons emitted per second is, 1) P /hc 2) P /h 3) hc/P , 4) hc/P, 7. In photoelectric effect, which of the following, property of incident light will not affect the, stopping potential, 1) Frequency, 2) Wavelength, 3) Energy, 4) Intensity, 8. The best suitable metal for photo electric effect, is, 1) Iron 2) Steel 3) Aluminium, 4) Cesium, 9. Photo electric effect can be explained only by, assuming that light, 1) is a form of transverse waves, 2) is a form of longitudinal waves, 3) can be polarized 4) consists of quanta, 10. When green light is incident on a metal, photo, electrons are emitted by it but no photo, electrons are obtained by yellow light. If red, light is incident on that metal then, 1) No electron will be emitted, 2) Less electrons will be emitted, 3) More electrons will be emitted, 4) we can not predict, 11. If the energy and momentum of a photon are, E and P respectively, then the velocity of, photon will be, 1) E/P 2) (E/P)2, 3) EP 4) 3x107 m/s, 12. The photo electric effect proves that light, consists of, 1) Photons, 2) Electrons, 3) Electromagnetic waves 4) Mechanical waves, 14, , 13. Intensity of light incident on a photo sensitive, surface is doubled. Then, 1) the number of emitted electrons is tripuled, 2) the number of emitted electrons is doubled, 3) the K.E of emitted electrons is doubled, 4) the momentum of emitted electrons is doubled, 14. A point source of light is used in a photoelectric, effect. If the source is moved farther from the, emitting metal, the stopping potential, 1) will increase, 2) will decrease, 3) will remain constant, 4) will either increase or decrease, 15. If the frequency of light in a photoelectric, experiment is doubled, the stopping potenital, will, 1) be doubled, 2) be halved, 3) become more than double, 4) become less than double, 16. With the decrease in the wave length of the, incident radiation the velocity of the, photoelectrons emitted from a given metal, 1) remains same, 2) increases, 3) decreases, 4) increases first and then decreases, 17. Sodium surface is illuminated with ultraviolet, light and visible radiation successively and the, stopping potentials are determined. Then the, potential, 1) is equal in both the cases, 2) greater for ultraviolet light, 3) more for visible light, 4) varies randomly, 18. In photo electric effect, the slope of the, straight line graph between stopping potential, and frequency of the incident light gives the, ratio of Planck’s constant to, 1) charge of electron 2) work function, 3) photo electric current 4) K.E. of electron, 19. From the graph shown, the value of Work, function if the stopping potential (V), and, frequency of the incident light, v , are on y and, x- axes respectively is, V, 3, 2, 1, -1, -2, -3, , 1) 1eV, , v, , 2) 2eV, , 3) 4eV, , 4) 3eV, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , 20. A graph is drawn between frequency of the, incident radiation (on X- axis) and stopping, potential (on Y-axis). Then the slope of the, straight line indicates, 1) h.e 2) h/e, 3) e/h, 4) (e-h), 21. In an experiment of photo electric emission, for incident light of 4000 A0, the stopping, potential is 2V. If the wavelength of incident, light is made 3000 A0 , then the stopping, potential will be, 1) Less than 2 volt, 2) More than 2 volt, 3) 2 volt, 4) Zero, , 27. The correct curve between the stopping, potential (Vo ) and intensity of incident light, (I) is, , 22. Light of wavelength falls on a metal having, , 28. The photo electrons emitted from the surface, of sodium metal are, 1) Of speeds from 0 to a certain maximum, 2) Of same de Broglie wavelength, 3) Of same kinetic energy, 4) Of same frequency, 29. The necessary condition for photo electric, emission is, , work function hc / o Photoelectric effect will, take place only if, 1) 0, , 2) 2 0, , 3) 0, , 4 ) 0 / 2, , 23. Emission of electrons in photoelectric effect, is possible, if, 1) metal surface is highly polished, 2) the incident light is of sufficiently high intensity, 3) the light is incident at right angles to the surface, 4) the incident light is of sufficiently low wavelength, 24. The work function of a metal, 1) is different for different metals, 2) is the same for all the metals, 3) depends on the frequency of the light, 4) depends on the intensity of the incident light, 25. The process of photo electric emission depends, on, 1) Temperature of incident light, 2) Nature of surface, 3) Speed of emitted photo electrons, 4) Speed of the incident light, 26. The threshold wavelength of lithium is 8000, A0. When light of wavelength 9000 A0 is made, to be incident on it, then the photo electrons, 1) Will not be emitted, 2) Will be emitted, 3) Will sometimes be emitted and sometimes not, 4) Data insufficient, NARAYANAGROUP, , 1) h h0, 30., , 31., , 32., , 33., , 2) h h0, , 3) Ek > h0, 4) Ek < h0, At stopping potential, the photo electric, current becomes, 1) Minimum, 2)Maximum, 3) Zero, 4) Infinity, Stopping potential depends on, 1) Frequency of incident light, 2) Intensity of incident light, 3) Number of emitted electrons, 4) Number of incident photons, Work function is the energy required, 1) to excite an atom, 2) to produce X-rays, 3) to eject an electron just out of the surface, 4) to explode the atom, Threshold wavelength depends on, 1) frequency of incident radiation, 2) work function of the substance, 3) velocity of electrons, 4) energy of electrons, , 34. If the work function of a metal is 0 , then its, threshold wavelength will be, 1) hc 0, , 2), , c 0, h, , 3), , h 0, c, , hc, 4) , 0, 15
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 35. The work function of a metal is X eV When, light of energy 2X eV is made to be incident, on it then the maximum kinetic energy of, emitted photo electron will be, 1) 2 eV, 2) 2X eV, 3) X eV, 4) 3X eV, 36. If the distance of 100W lamp is increased from, a photocell, the saturation current i in the, photo cell varies with distance d as, , 37., , 38., , 39., , 40., , 41., , 42., , 16, , 1, 1, 4) i 2, 1) i d2 2) i d, 3) i , d, d, A source of light is placed at a distance 4m, from a photocell and the stopping potential is, then 7.7 volt. If the distance is halved, the, stopping potential now will be, 1) 7.7 volt2) 15.4 volt, 3) 3.85 volt, 4) 1.925 volt, A milliammeter in the circuit of a photocell, measures, 1) number of electrons released per second, 2) energy of photon, 3) velocity of photoelectrons, 4) momentum of the photo electrons, The Einstein’s photoelectric equation is based, upon the conservation of, 1) Mass, 2)momentum, 3) angular momentum 4) energy, The stopping potential of the photocell is, independent of, 1) wavelength of incident light, 2) nature of the metal of photo cathode, 3) time for which light is incident, 4) frequency of incident light, The maximum energy of emitted photo, electrons is measured by, 1) the current they produce, 2) the potential difference they produce, 3) the largest potential difference they can, transverse, 4) the speed with which they emerge, Three metals have work functions in the ratio, 2:3:4. Graphs are drawn for all between the, stopping potential and the incident frequency., The graphs have slopes in the ratio, 1) 2: 3: 4, 2) 4: 3: 2, 3)6: 4: 3, 4)1: 1: 1, , 43. The curve between current (I) and potential, difference (v) for a photo cell will be, , 1) I, , 2) I, V, , V, 3) I, , 4) I, V, , V, , 44. Which conservation law is obeyed in Einstein’s, photo electric equation?, 1) Charge 2) Energy 3) Momentum 4) Mass, 45. In photo electric effect, the photo electric, current, 1) increases when the frequency of incident photon, increases, 2) decreases when the frequency of incident photon, decreases, 3) does not depend upon the photon frequency, but depends on the intensity of incident beam, 4) depends both on the intensity and frequency of, the incident beam., 46. The photoelectric current can be increased by, 1) increasing frequency, 2) increasing intensity, 3) decreasing intensity, 4)decreasing wavelength, 47. The threshold wavelength for sodium is, 5 x 10–7 m. Photoemission occurs for light of, 1) Wavelength of 6 x 10-7 m and above, 2) Wavelength of 5 x 10–7 m and below, 3) Any wavelength, 4) All frequencies below 5 x 1014 Hz, 48. If Planck’s constant is denoted by h and, electronic charge by e, then photoelectric, effect allows determination of:, 1) Only h, 2) Only e, 3) Both h and e, 4) Only h/e, 49. The electron behaves as waves because they, can, 1) be diffracted by a crystal, 2) ionise a gas, 3) be deflected by magnetic fields, 4) be deflected by electric fields, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 50. A nonmonochromic light is used in an, experiment on photoelectric effect. The, stopping potential, 1) is related to the mean wavelength, 2) is related to the longest wavelength, 3) is related to the shortest wavelength, 4) is not related to the wavelength, 51. The incident photon involved in the, photoelectriceffect experiment, 1) completely disappears, 2) comes out with increased frequency, 3) comes out with a decreased frequency, 4) comes out with out change in frequency, 52. In a photoelectric experiment, the maximum, velocity of photoelectrons emitted, 1) depends on intensity of incident radiation, 2) does not depend on cathode material, 3) depends on frequency of incident radiation, 4) does not depend on wavelength of incident, radiation, 53. The number of electrons emitted by a surface, exposed to light is directly proportional to, 1) Frequency of light 2) Work function, 3) Thereshold wavelength 4) Intensity of light, 54. Emission of electrons in photo electric effect, is possible, if, 1) metal surface is highly polished, 2) the incident light is of sufficiently high intensity, 3) the light is incident at right angles to the surface, 4) the incident light is of sufficiently low wavelength, 55. When orange light falls on a photo sensitive, surface the photocurrent begins to flow. The, velocity of emitted electrons will be more when, surface is hit by, 1) red light, 2) violet light, 3) thermal radiations 4) radio waves, 56. When the amplitude of the light wave incident, on a photometal sheet is increased then, 1) the photoelectric current increases, 2) the photoelectric current remains unchanged, 3) the stopping potential increases, 4) the stopping potential decreases, 57. Which of the following is dependent on the, intensity of incident radiation in a photoelectric, experiment, 1) work function of the surface, 2) amount of photoelectric current, 3) stopping potential, 4) maximum kinetic energy, NARAYANAGROUP, , DUAL NATURE, 58. Maximum kinetic energy (E k) of a, photoelectron varies with the frequency ( v ), of the incident radiation as, , a), , Ek, , b), , Ek, , V, , c), , Ek, , d), , V, , 59., , 60., , 61., , 62., , V, , Ek, , V, , 1) a, 2) b, 3) c, 4) d, Which one of the following is true in, photoelectric emission, 1) photoelectric current is directly proportional to, the amplitude of light of given frequency, 2) photoelectric current is directly proportional, to the intensity of light of given frequency at, moderate intensities, 3) above the threshold frequency the maximum, kinetic energy of photoelectrons is inversely, proportional to the frequency of incident light, 4) the threshold frequency depends on the intensity, of incident light, If the work function of the metal is W and the, frequency of the incident light is , then there, is no emission of photoelectrons if, 1) v < W/h, 2) v > W/h, 3) v W/h, 4) v W/h, Kinetic energy with which the electrons are, emitted from a metal surface due to, photoelectric effect is, 1) Dependent of the intensity of illumination, 2) Dependent on the frequency of light, 3) Inversely proportional to the intensity of, illumination, 4) Directly proportional to the intensity of, illumination, When ultraviolet radiation is incident on a, surface, no photoelectrons are emitted. If a, second beam causes emission of, photoelectrons, it may consist of :, 1) radio waves, 2) infrared rays, 3) visible light rays, 4) X-rays, 17
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , number of phenomena observed with light, it, is necessary to retain the wave-nature of light, to explain the phenomena of:, 1) photoelectric effect 2) diffraction, 3) compton effect, 4) black body radiation, , 63. The maximum kinetic energy (Ek) of emitted, photoelectrons against frequency v of incident, radiation is plotted as shown in fig. The slope, of the graph is equal to, K, , 70. In the following diagram if V2 V1 then, , Ek, , f, , Photo electric, current, , V, , 64., , 65., , 66., , 67., , 68., , 69., 18, , 1) charge on electron, 2) work function of emitter, 3) Planck’s constant, 4) ratio of Planck’s constant and chargeon electron, Einstein’s photoelectric equation states that, Ek = h v - W, In this equation Ek refers to :, 1) kinetic energy of all ejected electrons, 2) mean kinetic energy of emitted electrons, 3) minimum kinetic energy of emitted electrons, 4) maximum kinetic energy of emitted electrons, The function of photoelecrtic cell is, 1) to convert electrical energy into light energy., 2) to convert light energy into electrical energy, 3) to convert mechanical energy into electrical, energy, 4) to convert DC into AC., Photoelectric effect can be explained only by, assuming that light:, 1) is a form of transverse waves, 2) is a form of longitudinal waves, 3) can be polarised 4) consists of quanta, When light falls on a photosensitive surface,, electrons are emitted from the surface .The, kineticenergy of these electrons does not, depend on the:, 1) Wave length of light, 2) thickness of the surface layer, 3) type of material used for the layer, 4) intensity of light., Photoelectric effect is described as the ejection, of electrons form the surface of a metal when:, 1) it is heated to a high temparature, 2) light of a suitable wave lenght is incident on it, 3) electrons of a suitable velocity impinge on it, 4) it is placed in a strong electric field, Though quantum theory of light can explain a, , 2, V1, , 71., , 72., , 73., 74., , 75., , 76., , 1, V2, , potential difference, , 1) 1 2 2) 1 2 3) 1 2 4) 1 2, When an X-ray photon collides with an, electron and bounces off, its new frequency, 1) is lower than its original frequency, 2) is same as its original frequency, 3) is higher than its original frequency, 4) depends upon the electron’s frequency, A point source of light is used in a photoelectric, effect. If the source is removed farther from, the emitting metal, the stopping potential, 1) will increase, 2) will decrease, 3) will remain constant, 4) will either increase or decrease, De-Broglie wavelength depends on, 1) mass of the particle 2) size of the particle, 3) material of the particle 4) shape of the particle, The deBroglie wavelength associated with a, particle of mass m, moving with a velocity v, and energy E is given by, 1) h/mv2, 2) mv/h2, 3)h/ 2mE, 4) 2mE / h, Choose the correct statement, 1) Any charged particle in rest is accompanied by, matter waves, 2) Any uncharged particle in rest is accompanied, by matter waves, 3) The matter waves are waves of zero amplitude, 4) The matter waves are waves of probability, amplitude, An electron of charge e and mass m is, accelerated from rest by a potential difference, V. The de Broglie wavelength is, 1) Directly proportional to the square root of, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , potential difference., 2) Inversely proportional to the square root of, potential difference, 3) Directly proportional to the square root of, electron mass, 4) Inversely proportional to the cube root of, electron mass, 77. Which of the following particles - neutron,, proton, electron and deuteron has the lowest, energy if all have the same de Broglie, wavelength, 1) neutron2) proton 3)electron 4) deuteron, 78. The momentum of a proton is p. The, corresponding wavelength is, , 1) More 2) Less, 3) Same, 4) More for lighter particles and less for heavy, particles, 84. The wavelength of matter waves does not depend, on, 1) Momentum 2) Velocity 3) Mass 4) Charge, 85. The wave nature of matter is not observed in, daily life because their wave length is, 1) Less, 2) More, 3) In infrared region 4) In ultraviolet region, 86. The ratio of the wavelengths of a photon and, that of an electron of same energy E will be, [m is mass of electron], , 1) h/p, 2) h p, 3) p/h, 4) hp, 79. A wave is associated with matter when it is, 1) stationary, 2) in motion with a velocity, 3) in motion with speed of light, 4) in motion with speed greater than that of light, 80. An electron of mass 9.1 x 10-31kg and charge, 1.6 x 10-19 C is accelerated through a potential, difference of V volt. The de Broglie wavelength, ( ) associated with the electron is, , 2m, E, 2m, EC, 2), 3) C, 4), E, 2m, E, 2m, 87. One of the following figures respesents the, variation of particle momentum with, associated de Broglie wavelength, , 1), , 12.27, V, , A0, , 2), , 12.27 0, A, V, , 1), , h, 2mKT, , 3) h 2mKT, , 4), , 2), 4), , h, 2mKT, , b)P, , , , , , c) P, , d)P, , , , , , 1) a, 2) b, 3) c, 4) d, 88. A point source causes photoelectric effect, from a small metal plate. Which of the following, curves may represent the saturation, photocurrent as a function of the distance, between the source and the metal?, , 1, , h 2mKT, 82. The wavelengths of a proton and a photon are, same. Then, 1) Their velocities are same, 2) Their momenta are equal, 3) Their energies are same, 4) Their speeds are same, 83. If the value of Planck’s constant is more than, its present value, then the de Broglie, wavelength associated with a material particle, will be, , NARAYANAGROUP, , a) P, , 1, , A0, 12.27 V, 81. The de Broglie wavelength of a molecule of, thermal energy KT (K is Boltzmann constant, and T is absolute temperature) is given by, 3) 12.27 V A0, , 1), , a, b, c, , current, d, , distance, , 1) a, 2) b, 3) c, 4) d, 89. Matter waves are:, 1) electromagnetic waves, 2) mechanical waves, 3) either mechanical or electromagnetic waves, 19
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 4) neither mechanical nor electromagnetic waves, 90. The incorrect statement is, 1) Material wave (de-Broglie wave) can travel in, vacuum, 2) Electromagnetic wave can travel through vacuum, 3) The velocity of photon is the same as light passes, through any medium, 4) Wavelength of de-Broglie wave depends upon, velocity, 91. The magnitude of the de-Broglie wavelength, ( ) of an electron (e),proton(p),neutron (n), and - particle ( ) all having the same, energy of MeV, in the increasing order will, follow the sequence:, 1) e , p , n , , 2) , n , p , e, 3) e , n , p , , 4) p , e , , n, 92. Moving with the same velocity ,one of the, following has the longest deBroglie, wavelength, 1) -particle, 2) -particle, 3) proton, 4) neutron, 93. Debroglie wavelength of a particle at rest, position is, 1) zero, 2) finite, 3) infinity, 4) cannot be calculated, 94. Debroglie wavelength of protons accelerated, by an electric field at a potential difference v, is, , 0.108, 0.202, 0.286, 0.101, 2), 3), 4), V, V, V, V, 95. Debroglie wavelength of uncharged particles, depends on, 1) mass of particle, 2) kinetic energy of particle, 3) nature of particle 4) All above, 96. Debroglie wavelength of a moving gas, molecule is, 1) proportional to temperature, 2) inversely proportional to temperature, 3) independent of temperature, 4) inversely proportional to square root of, temperature, 97. The particles that can be accelerated by an, electric field is, 1) proton, 2) electron, 3) alpha particle, 4) all above, 1), , 20, , 98. If a proton and an electron are confined to the, same region, then uncertainity in momentum, 1) for proton is more, as compared to the electron, 2) for electron is more, as compared to the proton, 3) same for both the particles, 4) directly proportional to their masses, 99. Which phenomenon best supports the theory, that matter has a wave nature ?, 1) electron momentum2) electron diffraction, 3) photon momentum 4) photon diffraction, 100. The wavelength of de-Broglie wave associated, with a thermal neutron of mass m at absolute, temperature T is given by (Here, k is the, Boltzmann constant), 1), , h, 2), 2mkT, , h, 3), mkT, , h, h, 4), 3mkT, 2 mkT, , ASSERTION & RESONING, In each of the following questions, a statement, is given and a corresponding statement or, reason is given just below it. In the statements,, mark the correct answer as, 1) If both Assertion and Reason are true and, Reason is correct explanation of Assertion., 2) If both Assertion and Reason are true but, Reason is not the correct explanation of, Assertion., 3) If Assertion is true but Reason is false., 4) If both Assertion and Reason are false., 101. Assertion (A) : For a fixed incident photon energy,, photoelectrons have a wide range of energies, ranging from zero to the maximum value K max, Reason (R) : Initially, the electrons in the metal, are at different energy level., 102. Consider the following statements A and B,, identify the correct choice in the given, answers., A) Tightly bound electrons of target material, scattered X-ray photon,resulting in the Compton, effect., B) Photoelectric effect takes place with free, electrons., 103. The frequency and intensity of a light source, are both doubled. Consider the following, statements., (A) The saturation photocurrent remains almost the, same., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, (B) The maximum kinetic energy of the, photoelectrons is doubled., 1) Both A and B are true, 2) A is true but B is false, 3) A is false but B is true, 4) Both A and B are false, , STATEMENT TYPE, In each of the following questions, a statement, is given and a corresponding statement or, reason is given just below it. In the statements,, mark the correct answer as, 1) Statement I is true, Statement II is true; statement, II is a correct explanation of statement I., 2) Statement I is true, Statement II is true,, Statement II is NOT a correct explanation for, statement I., 3) Statement I is true, Statement II is false, 4) Statement I is false, Statemenet II is true., 104. A proton and an electron both have energy 50, eV., Statement-I: Both have different wavelengths, Statement-II: Wavelength depends on energy and, not on mass., 105. Statement I: Though light of a single frequency, (monochromic light) is incident ona metal, the, energies of emitted photoelectrons are different., Statement II: The energy of electrons just after, they absorb photons incident on the metal surface, may be lost in collision with other atoms in the metal, before the electron is ejected out of the metal., 106. Statement I: The de Broglie wavelength of a, molecule (in a sample of ideal gas) varies inversely, as the square root of absolute temperature., Statement II:The deBroglie wavelength of a, molecule (in sample of ideal gas) depends on, temperature., 107. Statement-I: A metallic surface is irradiated by a, monochromatic light of frquency v v0 (the, threshold frequency). The maximum kinetic energy, and the stopping potential are K max and v0 are also, doubled., Statement-II: The maximum kinetic energy and, he stopping potential of photoelectrons emitted, from a surface are linearly dependent on the, frquency of incident light. [AIEEE-2011], 108. Statement I: Davisson-Germer experiment, established the wave nature of electrons, Statement II: If electrons have wave nature, they, can interface and show diffraction. [AIEEE-2012], NARAYANAGROUP, , DUAL NATURE, , C.U.Q - KEY, 1) 1, 8) 4, 15) 3, 22) 3, 29) 2, 36) 4, 43) 4, 50) 3, 57) 2, 64) 4, 71) 1, 78) 1, 85) 1, 92) 1, 99) 2, 106) 2, , 2) 1, 9) 4, 16) 2, 23) 4, 30) 3, 37) 1, 44) 2, 51) 1, 58) 4, 65) 2, 72) 3, 79) 2, 86) 3, 93) 3, 100) 3, 107) 3, , 3) 4, 10) 1, 17) 2, 24) 1, 31) 1, 38) 1, 45) 3, 52) 3, 59) 2, 66) 4, 73) 1, 80) 1, 87) 4, 94) 3, 101) 1, 108) 1, , 4) 3, 11) 1, 18) 1, 25) 2, 32) 3, 39) 4, 46) 2, 53) 4, 60) 1, 67) 4, 74) 3, 81) 1, 88) 4, 95) 4, 102) 4, , 5) 2, 12) 1, 19) 4, 26) 1, 33) 2, 40) 3, 47) 2, 54) 4, 61) 2, 68) 2, 75) 4, 82) 2, 89) 4, 96) 4, 103) 2, , 6) 1, 13) 2, 20) 2, 27) 2, 34) 4, 41) 3, 48) 4, 55) 2, 62) 4, 69) 2, 76) 2, 83) 1, 90) 3, 97) 4, 104) 3, , 7) 4, 14) 3, 21) 2, 28) 1, 35) 3, 42) 4, 49) 1, 56) 1, 63) 3, 70) 4, 77) 4, 84) 4, 91) 2, 98) 3, 105)1, , LEVEL-I (C.W), PHOTO ELECTRIC EFFECT, 1., , 2., , 3., , 4., , 5., , The frequency of a photon associated with an, energy of 3.31 eV is (given h = 6.62 x 10-34 Js), 1) 0.8 x 1015 Hz, 2) 1.6 x 1015 Hz, 15, 3) 3.2 x 10 Hz, 4) 8.0 x 1015 Hz, A radiation of wave length 2500 A0 is incident, on a metal plate whose work function is 3.5, eV. Then the potential required to stop the, fastest photo electrons emitted by the surface, is (h = 6.63×10-34Js & c= 3×108 m/s), 1) 1.86V 2) 3.00 V 3) 1.46V 4) 2.15 V, The work function of a metal is 2.5 eV. The, maximum kinetic energy of the photoelectrons, emitted if a radiation of wavelength 3000 A0, falls on it is, (h = 6.63×10-34Js and c= 3 ×108 m/s), 1) 1.12 ×10-19J, 2) 4.8 ×10-19J, 3) 3.2 ×10-19J, 4) 2.61×10-19J, The work function of a substance is 4.0 eV., The longest wavelength of light that can cause, photoelectric emission from this substance is, approximately, 1) 220 nm 2) 310 nm 3) 540 nm 4) 400 nm, A laser used to weld detached retains emits, light with a wavelength 652 nm in pulses that, are of 20ms duration. The average power, during each pulse is 0.6W. The energy in each, pulse and in a single photon are, 1) 7.5 1015 eV , 2.7eV 2) 6.5 1016 eV , 2.9eV, 3) 6.5 1016 eV , 2.7eV 4) 7.5 1016 eV ,1.9eV, 21
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 6., , 7., , 8., , Electrons ejected from the surface of a metal,, when light of certain frequency is incident on, it, are stopped fully by a retarding potential of, 3 volts. Photo electric effect in this metallic, surface begins at a frequency 6 x 1014s-1. The, frequency of the incident light in s-1 is [h=6 x, 10-34J-sec;charge on the electron=1.6x10-19C], 1) 7.5 x 1013, 2)13.5 x 1013, 14, 3) 14 x 10, 4) 7.5 x 1015, The threshold wavelength for emission of, photoelectrons from a metal surface is, 6×10-7m. The work function of the material of, the metal surface is ., 1) 3.3×10-19J, 2) 6.67×10-19J, -19, 3) 1.23×10 J, 4) 2.37 ×10-19J, The maximum velocity of an electron emitted, by light of wavelength incident on the, surface of a metal of workfunction is where, h = Planck’s constant, m = mass of electron, and c=speed of light, , 2(hc ) , 1) , , m, , , , 1/ 2, , 2), , 9., , M, m, , 2), , M, m, , 3), , m, M, , 4), , m, M, , LEVEL-I (C.W) - KEY, , 1/ 2, , 2( hc ) , 2(hc ) , 3) , 4) , , , m, m, , , , , The work function of nickle is 5eV. When light, of wavelength 2000A 0 falls on it, emits, photoelectrons in the circuit. Then the, potential difference necessary to stop the, fastest electrons emitted is (given h=6.67×1034, Js), 1) 1.0V 2) 1.75V, 3) 1.2V 4) 0.75V, , MATTER WAVES, 10. If an electron and a proton have the same KE,, the ratio of the de Broglie wavelengths of, proton and electron would approximately be, 1) 1 : 1837 2) 43 : 1 3) 1837 : 1 4) 1 : 43, 11. If electron is having a wavelength of 100 Ao,, then momentum is (gm cm s-1) units, 1) 6.6 x 10-32, 2) 6.6 x 10-29, -25, 3) 6.6x 10, 4) 6.6 x 10-21, 12. The de-broglie wavelength of an electron and, the wavelength of a photon are same. The ratio, between the energy of the photon and the, momentum of the electron is [M 2006], 1) h, 2) c, 3) 1/h, 4) 1/c, 13. A proton and an alpha particle are accelerated, through the same potential difference. The, ratio of wavelengths associated with proton, and alpha particle respectively is, 1) 1: 2 2 2) 2:1 3) 2 2 :1 4) 4:1, 22, , 1), , 1) 1 2) 3 3) 4 4) 2 5) 4 6) 3 7) 1, 8) 3 9) 3 10) 4 11) 4 12) 2 13) 3 14) 1, 15) 1 16) 2, , 2(hc ), m, , 1/ 2, , 14. Ratio of debroglie wavelengths of uncharged, particle of mass m at 270 C to 1270 C is nearly, 1) 1.16 2) 0.16 3) 1.33, 4)0.8, 15. A particle is projeted horizontally with a, velocity 10m/s. What will be the ratio of deBroglie wvelengths of the particle, when the, velocity vector makes an angle 300 and 600, with the horizontal, 1) 3 :1 2) 1: 3 2) 2 : 3, 4) 3 : 2, 16. A positron and a proton are accelerated by the, same accelerating potential. Then the ratio, of the associated wavelengths of the positron, and the proton will be [ M = mass of proton,, m = mass of positron], , LEVEL-I (C.W) - HINTS, 1., 3., 4., 5., 6., , 7., , E = hv, , 12400, , 2. V0e in A0 0 eV, , , , 12400, , K .E. , 0 eV, 0, in A, , 12400, in A0 = E in eV, hc, , K.E. = V0e and V0e = h[ 0 ], E pt ; E in each photon , , where V0 is the stopping potential and 0 is the, Threshold frequency, hc, 1, E 0, 0 , E mv 2 9. V0 , 8., 0, 2, e, , 10. , , h, h, 11. P , 2mE, , , hc, E, E, , 12., e ph ; ph ph ph C, h, meV h , , e, e , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 8., , p, q m, 1, 13. q m 14. , T, , p, p, 15. , , h, , P m v Take v component, P, , 16. , , h, h, , 2Vmq, p, , 9., , LEVEL-I (H.W), PHOTO ELECTRIC EFFECT, 1., , 2., , 3., , 4., , 5., , 6., , 7., , The threshold wavelength for a surface having, a, threshold, frequency, of, 15, 8, 0.6 x 10 Hz is (given c = 3 x 10 m/s), 1) 4000 Ao, 2) 6000 Ao, o, 3) 5000A, 4) 3500Ao, Two photons of energies twice and thrice the, work function of a metal are incident on the, metal surface. Then the ratio of maximum, velocities of the photoelectrons emitted in the, two cases respectively, is, 1) 2 :1 2) 3 : 1 3) 3 : 2 4) 1 : 2, The photo electric work function for a metal, surface is 4.125 eV. The cut-off wavelength for, this surface, 1) 4125 Ao, 2) 2062.5 Ao, 3) 3006.06 Ao, 4) 6000Ao, The energy of emitted photoelectrons from a, metal is 0.9 eV. The work function of the metal, is 2.2 eV. Then the energy of the incident, photon is, 1) 0.9 eV 2) 2. 2 eV 3) 4. 4 eV 4) 3.1 eV, A photoelectron is moving with a maximum, velocity of 106 m/s. Given e=1.6x10-19 c, and, m = 9.1x 10-31 kg, the stopping potential is, 1) 2.5 V, 2) 2.8 V 3) 2.0 V 4) 1.4 V, A metal of work function 4eV is exposed to a, radiation of wavelength 140×10-9m.Then the, stopping potential developed by it, (h = 6.63×10-34Js and c= 3×108 m/s), 1) 6.42 V 2) 2.94 V 3) 4.86V 4) 3.2 V, Threshold wavelength for a metal having work, function wo is .Then the threshold, wavelength for the metal having work function, 2 w o is, 1) 4 , , 2) 2 , , NARAYANAGROUP, , 3) /2, , 4) /4, , The work function of metals A and B are in, the ratio 1:2. If light of frequencies f and 2f, are incident on metal surfaces A and B, respectively, the ratio of the maximum kinetic, energies of the photo electrons emitted is, (2000 M), 1) 1:1, 2) 1:2, 3) 1:3, 4) 1:4, The threshold wave length for photo electric, emission from a material is 5,200A0, photo, electrons will be emitted when this material is, illuminated with mnochromaic radiation from, a, 1) 50 watt infrared lamp 2) 1 watt infrared lamp, 3) 1 watt ultraviolet lamp, 4) 50 watt sodium vapour lamp, , MATTER WAVES, 10. A particle having a de Broglie wavelength of, 1.0 Ao is associated with a momentum of (given, h = 6.6 x 10-34 Js), 1) 6.6 x 10-26 kg m/s 2) 6.6 x 10-25 kg m/s, 3) 6.6 x 10-24 kg m/s 4) 6.6 x 10-22 kg m/s, 11. The de Broglie wavelength of an electron, having 80 eV of energy is nearly, (1 eV 1.6 x10 19 J , Mass of electron =, , 91031 kg,, , Planck’s, , constant, , 6.61034 Js) (nearly) (2001 E), 1) 140 Ao 2) 0.14 Ao 3) 14 Ao 4) 1.4 Ao, 12. Electrons are accelerated through a p.d. of, 150V. Given m = 9.1x10-31kg,e =1.6x10-19 c,, h = 6.62x10-34 Js, the de Broglie wavelength, associated with it is, 1) 1.5 Ao 2) 1.0 Ao 3) 3.0 Ao, 4) 0.5 Ao, 13. If accelerating potential of an alpha particle, is doubled than its new debrolgie wavelength, becomes, , 1, times of initial 2) 2 times of initial, 2, 3) 1/2 times of initial, 4) 2 times of initial, 14. The ratio of the deBroglie wavelenths of, proton, deuteron and alpha particle, accelerated through the same potential, difference 100V is, 1) 2 : 2 :1, 2) 1: 2 : 2 2, 1), , 3) 1: 2 : 2 2, , 4) 2 2 : 2 :1, 23
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , LEVEL-I (H.W) - KEY, 1) 3, 8) 2, , 2) 4, 9) 3, , 3., , 3) 3 4) 4 5) 2 6) 3 7) 3, 10) 3 11) 4 12) 2 13) 1 14) 4, , LEVEL-I (H.W) - HINTS, , , 3., , in A0 =, , 4., , E K .E 0, , 6., , 8., 9., , E1 0, E2 0, , V1, 2. V , 2, , c, v, , 1., , 12400, in eV, E, , 4., , 1 2, 5. mv eV0, 2, , 12400, , hc, in A0 0 , w0 , , , 7., V0 , 0, e, KE1 E1 W1, hv1 w1, hf w, 1, , , , =, KE2 E2 W2 hv2 w2 2hf 2 w 2, P, , 11. , , NE, hc, ,E , t, , h, , 2mE, 1, , 13., V, , 10. P , 12. , , 5., , h, , , 150, V, , h P2, V q, 14. ,, P 2m, , 6., , LEVEL-II (C.W), PHOTO ELECTRIC EFFECT, 1., , A photometal is illuminated by lights of, wavelengths 1 and 2 respectively. The, maximum kinetic enegies of electrons emitted, in the two cases are E1 and E2 respectively., The work function of metal is., 1), , E2 1 E12, 1, , 2), , E11 E2 2, 1 2, , Ultraviolet light of wavelength 300 nm and, intensity 1.0 W/m2 falls on the surface of a, photoelectric material. If one percent of the, incident photons produce photoelectrons, then, the number of photoelectrons emitted from an, area of 1.0 cm2 of the surface is nearly (in per, second), 1) 9.61 x 1014, 2) 4.12 x 1013, 3) 1.51 x 1012, 4) 2.13 x 1011, Light rays of wavelengths 6000 A and of, photon intensity 39.6 watts/m2 is incident on a, metal surface. If only one percent of photons, incident on the surface emit photo electrons,, then the number of electrons emitted per, second per unit area from the surface will be, [ Planck constant = 6.64 x 10-34 J - S; Velocity, of light = 3 x 108 ms-1], 1) 12 x 1018 2) 10 x 1018 3) 12 x 10174) 12 x 1015, Light of wavelength 4000 Ao is incident on a, metal surface of work function 2.5 eV. Given, h=6.62 x 10-34 Js, c = 3 x 108 m/s, the maximum, KE of photoelectrons emitted and the, corresponding stopping potential are, respectively, 1) 0. 6 eV, 0.6 V, 2) 2.5 eV, 2.5 V, 3) 3.1 eV, 3.1 V, 4) 0.6 eV, 0.3 V, The K.E of the electron is E when the incident, wavelength is . To increase the K.E of the, electron to 2E, the incident wavelength must, be, , hc, 2hc , 1) 2 , 2), 3), 4), , 8., , A photon of energy 15 eV collides with H-atom., Due to this collision, H-atom gets ionized .The, maximum kinetic energy of emitted electron, is :, 1)1.4 eV 2) 5 eV 3)15eV, 4) 13.6eV, The anode voltage of a photocell is kept fixed., The wavelength of the light falling on the, cathode in gradually changed. The plate, current I of the photocell varies as follows :, I, , E11 E22, 3) , 1, 2, , 2., , 1), , Light of wavelength strikes a photo sensitive, surface and electrons are ejected with kinetic, energy E. If the kinetic energy is to be, increased to 2E, the wavelength must be, changed to where, , , 1) , 2, 24, , E2 2 E11, 4) , 1, 2, , I, 2), , O, , , , I, 3), , , 2) 2 3) 4) , 2, , E hc, , E hc, , 2, , 7., , O, , , , I, 4), , O, , O, , , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 9., , DUAL NATURE, , According to Einstein’s photoelectric equation,, the graph between the kinetic energy of, photoelectrons ejected and the frequency of, incident radiation is :, Kinetic energy, , Kinetic energy, 1), , 2), Frequency, , Frequency, , Kinetic energy, , Kinetic energy, 3), , 4), , 14. If the momentum of an electron is changed by, pm , then the de Broglie wavelength associated, with it changes by 0.5%. The initial momentum, of electron will be, 1) pm/200 2) pm/100 3) 200pm 4) 100pm, 15. When the mass of an electron becomes equal, to thrice its rest mass, its speed is, 2, 1, 1, 2 2, c 2) c 3) c, 4) c, 3, 3, 4, 3, 16. Which of the following figures represents the, variation of particle momentum with the, associated de Broglie wave-length ?, , 1), , P, Frequency, , Frequency, , 1), , 10. The graph shown in figure show the variation, of photoelectric current (i) and the applied, voltage (V) for two different materials and for, two different intensities of the incident, radiation., , P, , 3), , 4), , , 17. The de Broglie wave present in fifth Bohr orbit, is :, 4, V, , Identify the pairs of curves that correspond, to (a) different material (b) same intensity of, incident radiations., 1) Curve 1 and 3, Curve 2 and 4, 2) Curve 1 and 2, Curve 3 and 4, 3) Curve 1 and 4, Curve 2 and 3, 4) Curve 1 only, Curve 2 and 4, , MATTER WAVES, 11. A proton when accelerated through a p.d. of V, volt has a wavelength associated with it., An - particle in order to have the same, wavelength must be accelerated through a, p.d. of, 1) V/8 volt 2) V/4 volt 3) V volt 4) 2V volt, 12. An electron of mass m and charge e initially, at rest gets accelerated by a constant electric, field E . The rate of change of de-Broglie, wavelength of this electron at time t ignoring, relativistic effects is, h, eEt, mh, h, 1), 2), 3), 4), 2, 2, eEt, E, eEt, e.E, 13. If the velocity of a particle is increased three, times, then the percentage decrease in its de, Broglie wavelength will be, 1) 33.3% 2) 66.6% 3) 99.9% 4) 22.2%, NARAYANAGROUP, , , , P, , , , 3, 2, , 2), , , i, , 1, , P, , 1), 2), , 3), , 4), , HEISEN-BERG UNCERTAINITY, PRINCIPLE AND DAVISSONGERMER EXPERIMENT, 18. The correctness of velocity of an electron, moving with velocity 50 ms-1 is 0.005%. The, accuracy with which its position can be, measured will be, 1) 4634 103 m, 2) 4634 105 m, 3) 4634 106 m, 4) 4634 108 m, 19. If the uncertainity in the position of an electron, is 10-10 m, then the value of uncertainity in its, momentum (in kg-ms-1) will be, 1) 3.33 x 10-24, 2) 1.03 x 10-24, -24, 3) 6.6 x 10, 4) 6.6 x 10-20, 25
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, 20. a) Name the experiment for which the, adjacent graph, showing the variation of, intensity of scattered electrons with the angle, of scattering was obtained., b) Also name the important hypothesis that, was confirmed by this experiment., 1) A) Davission and Germer, experiment, , I, , B) de Broglie hypothesis, 2) A) Photo electric effect, B) de Broglie hypothesis, 50, , 3) A) Thermionic emission, B) de Broglie hypothesis, , 0, , , , 13. , , h, , 26, , 2., , 3., , h, , 11. p 2mvq ( = constant); v1q1m1 v2 q2 m2, eE, 12. Here u 0, a ' v ?.t t, m, eE, v u at 0, t ; de-Broglie wavelength,, m, h, h, h, , , , mv m eEt / m eEt, Rate of change of de-Broglie wavelength, d, h 1 h, , , dt eE t 2 eEt 2, , m0, 1, , v2, c2, , PHOTO ELECTRIC EFFECT, 1., , hc, hc, W E1; W E2, 1, 2, hc hc, 2hc hc, use and , 1, 1, , n hc , 3&4. P , t , 12400, , 1 E, , , , K, ., E, , 0 6., 5., in A0, 2E, , , 7. Energy of photon ionization energy + K .Emax =15ev, -13.6ev=1.4ev, 10. (a) Curves 1 and 3 (different materials due to, different stopping potentials), (b) Curves 2 and 4 (same intensity due to same, current), , 15. m , , LEVEL-II (H.W), , LEVEL-II (C.W) - HINTS, , 2., , P, , , P, , 0.005 50, 0.0025ms 1, 100, h, 1.034 1034, x , , 4634 10 5 m, 31, mv 9.1 10 0.0025, h 1.034 10 34, , 19. p , 1.034 1024 kg ms 1, 10, x, 10, 20. A) Davission and Germer experiment, B) de Broglie hypothesis, , LEVEL-II (C.W) - KEY, , 1., , 14., , 18. Here, v , , 4) A) Photocell, , 1) 4 2) 3 3) 3 4) 3 5) 1 6) 3 7) 1, 8) 3 9) 4 10) 1 11) 1 12) 1 13) 2 14) 3, 15) 1 16) 4 17) 4 18) 2 19) 2 20) 2, , 1, V, , 4., , 5., , When a metal surface is illuminated by a, monochromatic light of wave - length , then, the potential difference required to stop the, ejection of electrons is 3V. When the same, surface is illuminated by the light of, wavelength 2 , then the potential difference, required to stop the ejection of electrons is V., Then for photoelectric effect, the threshold, wavelength for the metal surface will be, 1) 6 2) 4 / 3, 3) 4 , 4) 8 , o, If U.V. Light of wavelengths 800 A and 700 Ao can, liberate electrons with kinetic energies of 1.8eV and, 4 eV respectively from hydrogen atom in ground, state, then the value of planck’s constant is, 1) 6.57 x 10-34 Js, 2) 6.63 x 10-34 Js, -34, 3) 6.66 x 10 Js, 4) 6.77 x10-34 Js, In a photoelectric effect experiment, photons, of energy 5 eV are incident on a metal surface., They liberate photoelectrons which are just, stopped by an electrode at a potential of -3.5, V with respect to the metal. The work function, of the metal is, 1) 1.5 eV 2) 3.5 eV 3) 5. 0 eV 4) 8.5 eV, The number of photons emitted per second by, a 62W source of monochromatic light of, wavelength 4800 Ao is, 1) 1.5 x 1019, 2) 1.5 x 1020, 20, 3) 2. 5x 10, 4) 4 x 1020, Photons of frequencies 2.2 x 1015 Hz and, 4.6 x 1015 Hz are incident on a metal surface., The corresponding stopping potentials were, found to be 6.6 V and 16.5 V respectively., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , -19, , 6., , 7., , Given e = 1.6 x 10 c, the value of universal, planck’s constant is, 1) 6.6 x 10-34 Js, 2) 6.7 x 10-34 Js, -34, 3) 6.5 x 10 Js, 4) 6.8 x 10-34 Js, If stopping potentials corresponding to, wavelengths 4000A0 and 4500A0 are 1.3V and, 0.9V respectively, then the work function of, the metal is, 1) 0.3eV 2) 1.3eV 3) 1.8eV 4) 5eV, In a photoelectric experiment anode potential, is plotted against plate current, I, , 12. If the energy of a particle is reduced to one, fourth, then the percentage increase in its de, Broglie wavelength will be, 1) 41% 2) 141% 3) 100%, 4) 71%, 13. The de Broglie wavelength associated with an, electron of velocity 0.3 c and rest mass 9.1 x, 10-31kg is, 2) 7.68 x 10-12 m, 1) 7.68 x 10-10 m, -12, 3) 5.7 x 10 m, 4) 9.1 x 10-12 m, 14. The two lines A and B shown in figure are the, graphs of the de Broglie wavelength as a, , 1, (, is the accelerating, V V, potential) for two particles having the same, charge., function of, , C, B, , , , A, , B, , V, , 1) A and B will have same intensities while B and, C will have different frequencies, 2) B and C will have different intensities while A, and B will have different frequencies., 3) A and B will have different intensities while B, and C will have equal frequencies., 4) B and C will have equal intensities while A and, B will have same frequencies., , MATTER WAVES, 8., , An electron moves with a speed of, , 3, 2, , c. Then, , its mass becomes....times its rest mass., 1) 2, 2) 3, 3) 3/2, 4) 4, 9. Photons of energy 2.0 eV fall on a metal plate, and release photoelectrons with a maximum, velocity V. By decreasing by 25% the, maximum velocity of photoelectrons is, doubled. The work function of the metal of the, material plate in eV is nearly, 1) 2.22 2) 1.985 3) 2.35, 4) 1.80, 10. A proton when accelerated through a p.d of V, volt has wavelength associated with it .An, electron to have the same must be, accelerated through a p.d of, V, 1), volt 2) 4V volt 3) 2V volt 4) 1838V volt, 8, 11. The momentum aphoton of electromagnetic, radiation is 3.3x10–29 kgms–1. The frequency, of these waves is:, 1) 3.0x103 Hz, 2) 6.0 x103 Hz, 3)7.5 x1012 Hz, 4)1.5 1013 Hz, NARAYANAGROUP, , A, , 1/ V, , Which of the two represents the particle of, heavier mass ?, 1) A, 2) B, 3) Both A and B, 4) Data insufficient, , HEISEN-BERG UNCERTAINITYPRINCIPLEAND, DAVISSON-GERMEREXPERIMENT, 15. The uncertainity in the position of a particle is, equal to the de-Broglie wavelength. The, uncertainity in its momentum will be, h, 2h, , 3, 1), 2), 3), 4), , 3, h, 2h, 16. If the uncertainity in the position of proton is, 6 108 m , then the minimum uncertainity in its, speed will be, 1) 1 cms 1 2) 1 ms 1 3) 1 mms 1 4) 100 ms 1, 17. From Davisson-Germer experiment an , particle and a proton are accelerated through, the same pd V . Find the ratio of the de, Broglie wavelengths associated with them, 1) 1: 2 2 2) 2 2 :1 3) 1: 2, 4) 2 :1, , LEVEL-II (H.W) - KEY, 1) 3 2) 1 3) 1 4) 2 5) 1 6) 3 7) 4, 8) 1 9) 4 10) 4 11) 4 12) 3 13) 2 14) 1, 15) 1 16) 2 17) 1, 27
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , LEVEL-II (H.W) - HINTS, 1., , 4., , A source of light is placed above a sphere of, radius 10cm. How many photoelectrons must, be emitted by the sphere before emission of, photoelectrons stops? The energy of incident, photon is 4.2 eV and the work function of the, metal is 1.5 eV., 1) 2.08 x 1018, 2) 1.875 x 108, 18, 3) 2.88 x 10, 4) 4 x 1019, Figure shows the variation of the stopping, potential V0 with the frequency v of the, incident radiations for two different, photosensitive material M 1 and M 2 .What are, , 1 1 , hc hc, , eV0 or hc eV0, 0, 0 , , hc, n hc , w eV0 3. E w eV0 4. P , t , , 5. hv w eV0 6. h = w+K.E h = w+ V0 e, , 2., , m0, , 8. m , , 1, , 1, hc, 2, 9. mv max h w & h , 2, , , v2, c2, , 5., , 10. V1 q1 m1 V2 q2 m2, 11. , , 14., , pc, h, , 12. , , v , , h, or, x, , h, 1.034 1034, , 1 ms 1, mx 1.67 1027 6 108, , h, h, 17. p , 2mqV, , LEVEL-III, 1., , 2., , 3., , 28, , the values of work functions for M 1 and M 2, respectively, vo, M1, , M2, , 1) hv01 , hv02 2) hv02 , hv01, , h h, , 15. p , x , , 1, Slope m, , 16. p mv , , 1, E, , v2, h 1 2, c, 13. , m0V, , When a surface 1 cm thick is illuminated with, light of wave lenght the stopping potential, is V0 ,but when the same surface is illuminated, by light of wavelength 3 , the stopping, V0, . The threshold wavelength, potential is, 6, for metallic surface is:, 1) 4, 2) 5, 3) 3, 4) 2, A photon of energy 2.5 eV and wavelength , falls on a metal surface and the ejected, electrons have velocity ‘v’. If the of the, incident light is decreased by 20%, the, maximum velocity of the emitted electrons is, doubled. The work function of the metal is, 1) 2.6 eV 2) 2.23 eV 3) 2.5 eV 4) 2.29 eV, When a metal surface is illuminated by light, of wavelengths 400 nm and 250 nm, the, maximum velocities of the photoelectrons, ejected are V and 2V respectively. The work, function of the metal is, 1) 2hc x 106 J, 2) 1.5hc x 106 J, 6, 3) hc x 10 J, 4) 0.5hc x 106 J, , O, , 6., , v01, , 3) hv01 , hv01 4) hv02 , hv02, , v, , v02, , From the above figure the values of stopping, potentials for M 1 and M 2 for a frequency, , v3 v02 of the incident radiatioins are V1 and, V2 respectively. Then the slope of the line is, equal to, V2 V1, V1 V2, V2, V1, 1) v v 2) v v 3) v v 4) v v, 02, 01, 02, 01, 02, 01, 02, 01, 7. Photoelectric effect experiments are, performed using three different metal plates, p, q and r having work functions p 2.0eV ,, , q 2.5eV and r 3.0eV respectively. A, light beam containing wavelengths of 550 nm,, 450 nm and 350 nm with equal intensities, illuminates each of the plates. The correct IV graph for the experiment is : [Take, hc 1240 eV nm], I, , I, , p, q, , 1), , r, , 2), , p, , q, r, , V, , I, , V, , I, , r, q, , 3), , p, V, , 4), , r, , q, p, , V, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, , 8., , An electron accelerated under a p.d. of V volt, has a certain wavelength . Mass of the, proton is 2000 times the mass of an electron., If the proton has to have the same wavelength, , then it will have to be accelerated under, p.d. of (volts), 1) 100 2) 2000 3) V/2000 4) 2000, , 9., , The graph between the stopiing potential V0 , and 1/ is shown in figure, 1 , 2 and 3, are work functions. Which of the following is, correct, , 13. The kinetic energy of -particles at a distance, 5 10 14 m from the uranium nucleurs will be, (in joules). Which is moving in a field of 1mega, volt potential difference, 1) 6.4 10 13 2) 4.3 1013 3) 2.1 1013 4) 3.4 1014, 14. The stopping potential for the photoelectric, emitted from a metal surface of work function, 1.7 eV is 10.4 V. Find the wavelength of the, radiation used. Also identify the energy levels, in hydrogen atom, which will emit this, wavelength, 1) 1024 A0 , n 3 to n 1, , V0, Metal 1 Metal 2, 1, , 2, , Metal 3, 3, , , 0.001 0.002, , 0.004, , 1/ nm 1 , , 1) 1 : 2 : 3 1: 2 : 3 2) 1 : 2 : 3 4 : 2 :1, 3) tan is directly proportional to hc / e , where, h is Planck’s constant and c is the speed of light, 4) ultraviolet light can be used to emit, photoelectrons from metal 2 and metal 3 only., 10. For certain photosensitive material, a stopping, potential of 3.0 V is required for light of, wavelength 300 nm, 2.0 V for 400 nm and 1.0V, for 600nm. The work function of the material, is (nearly), 1) 2.5 ev 2) 1.5 ev 3) 2.0 ev, 4)1.0 ev, 11. An electron (mass m ) with an initial velocity, v v0iˆ v0 0 is in an electric field E E0iˆ, ( E0 = constant > 0). Its de Broglie wavelength, at time t is given by, 0, eE0t , , , 1) 1 eE0t 2) 0 1 mv 3) 0 4) 0t, 0 , , mv0 , 12. An electron (mass m ) with an initial velocity, v v0iˆ is in an electric field E E0 ˆj . If, h, mv0 , its de Broglie wavelength at time t, is given by, , 0 , , e 2 E 2t 2, , 0, 2) 0 1 m 2 v 2, 0, , 1) 0, 0, 2, , 2 2, 0, 2 2, 0, , 3) 1 e E t, mv, , NARAYANAGROUP, , 2) 1024 A0 , n 2 to n 1, 3) 2044 A0 , n 2 to n 1, , 0, 2 2 2, , , 4) 1 e E2 0 t2 , m v0 , , , 4) 2044 A0 , n 3 to n 1, 15. A graph regarding photoeletric effect is shown, between the maximum kinetic energy of, electrons and the frequency of the incident, light. On the basis of data as shown in the, graph, calculate the work fucntion, Kmax (eV), 8, 6, 4, 2, 0, -2, -4, , 1) 2 eV, A, D, 10 20 30, C, , f 1014 Hz , , 2) 4 eV, , 3) 4.2 eV 4) 2.5 eV, , 16. Light of wavelength 180 nm ejects, photoelectrons from a plate of metal whose, work fucntion is 2 eV. If a unifrom magnetic of, 5, 5 10 T be applied parallel to the plate, what, would be the radius of the path followed by, electrons ejected normally from the plate with, maximum energy, 1) 0.148 m 2) 0.2 m 3) 0.25 m, 4) 0.3 m, 17. Light described at a place by the equation, E , , 100 V / M , , sin 5 1015 s 1 t sin 8 1015 s 1 t , , falls on a metal surface having work fucntion, 2.0 eV. Calculate the maximum kinetic energy, of the photoelectrons, 1) 3.27 eV 2) 5 eV 3) 1.27 eV, 4) 2.5 eV, 18. The electric field associated with a light wave, is given by E E0 sin 1.57 107 m1 x ct ., Find the stopping potential when this light is, used in an experiment on photoelectric effect, with the similar having work function 1.9 eV, 1) 1.2 V 2) 1.1 V 3) 2 V, 4) 2.1 V, 29
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JEE-ADV PHYSICS- VOL- V, , 2., , 3., , 4., , 5., , 6., , 3) even when the metal surface is faintly illuminated,, the photoelectrons leave the surface immediately, 4) electric charge of the photoelectrons is quantized., If the wavelength of light in an experiment on, photoelectric effect is doubled :, 1) the photoelectric emission will not take place, 2) the photoelectric emission may or may not take, place, 3) the stopping potential will increase, 4) the stopping potential will decrease, The frequency and intensity of light source are, both doubled. Which of the following, statement (statements) is (are) true ?, 1) The saturation photocurrent gets doubled., 2) the saturation photocurrent remains almost the, same, 3) the maximum KE of the photoelectrons is more, than doubled., 4) the maximum KE of the photoelectrons get, doubled., In which of the following situations, the heavier, of the two particles has smaller de Broglie, wavelength ? The two particles :, 1) move with same speed, 2) move with the same linear momentum, 3) move with the same kinetic energy, 4) have fallen through the same height, When a monochromatic point source of light, is at a distance of 0.2m from a photoelectric, cell, the cut-off voltage and the saturation, current are respectively 0.6V and 18.0 mA. If, the same source is placed 0.6m away from the, photoelectric cell, then :, 1) the stopping potential will be 0.2 V, 2) the stopping potential will be 0.6 V, 3) the saturation current will be 6.0 mA, 4) the saturation current will be 2.0 mA, In a photoelectric experiment the wavelength, of the incident light is decreased from 6000A0, to 4000A0 while the intensity of radiation, remains the same. Choose the correct, statement(s), 1) the cut-off potential will increase, 2) the cut-off potential will decrease, 3) the photoelectric current will increase, 4) the kinetic energy of the emitted photoelectrons, will increase, NARAYANAGROUP, , DUAL NATURE, , COMPREHENSION TYPE, Passage I:, o, , Photoelectric threhold of silver is 3800 A ., o, , ultraviolet light of 2600 A is incident on, silver surface. (Mass of the electron, 9.111031 kg ), 7. Calculate the value of work function in eV., 1) 1.77 2) 3.27, 3) 5.69, 4) 2.32, 8. Calculate the maximum kinetic energy, (in eV) of the emitted photoelectrons., 1) 1.51 2) 2.36, 3) 3.85, 4) 4.27, 9. Calculate the maximum velocity of the, photoelectrons., 1) 72.89 108, 2) 57.89 108, 3) 42.93 108, 4) 68.26 108, Passage II:, A 100 W point source emits monochromatic, 0, , light of wavelength 6000 A ., 10. Calculate the total number of photons emitted, by the source per second., 1) 5 1020 2) 8 1020 3) 6 1021 4) 3 1020, 11. Calculate the photon flux (in SI unit) at a, distance of 5 m from the source. Given, 8, 1, h 6.6 1034 Js and c 3 10 ms ., 3) 1020, 4) 1022, 1) 1015 2) 1018, 0, , 12. 1.5 mW of 4000 A light is directed at a, photoelectric cell. if 0.10 per cent of the, incident photons produce photoelectrons, find, current in the cell. [Given h 6.6 1034 ms 1, and e 1.6 1019 C ], 1) 0.59 A 2) 1.16 A 3) 0.48 A 4) 0.79 A, Passage III:, When a particle is restricted to move along xaxis between x=0 and x=a, where a is of, nanometer dimension, its energy can take only, certain specific values. The allowed energies, of the particle moving in such a restricted, region, corrospond to the formation of standing, waves with nodes at its ends x = 0 abd x = a., The wavelength of this standing wave is, related to the linear momentum p of the, particle according to the de Broglie relation., The energy of the particle of mass m is related, to its linear momentum as E p 2 / 2m . Thus,, 31
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JEE-ADV PHYSICS- VOL- V, , DUAL NATURE, the energy of the particle can be denoted by a, quantum number ‘n’ taking values, 1,2,3,..........(n=1, called the ground state), corrosponding to the number of loops in the, standing wave. Use the model described above, to answer the following three questions for a, particle moving along the line from x=0 to x=a., 13. The allowed energy for the particle for a, particular value of n is proportional to, 1) a 2, 2) a 3/ 2, 3) a 1, 4) a 2, , 18. In a permoting photoelectric experiment to, study photoelectric effect, intensity of, radiation I , frequency of radiation v ,, work function 0 of the photosensitive, emitter, distance d between emitter and, collector are changed or kept constant. Match, the changes given in Column - I to their effect, given in Column - II., Column - I, A) 0 is decreased, keeping v and I constant, B) d is increased, keeping I , v, 0 constant, C) v is increased, keeping I , 0 , d constant, D) I is increased, keeping v , 0 , d constant, Column - II, p) Saturation photoelectric current increases, q) stopping potential V0 increases, r) Maximum KE K max of photoelectrons, increases, s) Stopping potential remains the same, , 14. If the mass of the particle is m 1.0 10 30 kg, and a=6.6 nm, the energy of the particle in its, ground state is closest to, 1) 0.8 meV 2) 8 meV 3) 80 meV 4) 800 meV, 15. The speed of the particle that can take discrete, values is proportional to, 3) n1/ 2, 4) n, 1) n 3/ 2 2) n 1, 16. Statement-I: When ultraviolet light is incident, on a photo cell, its stopping potential is v0 and, the maximum kinetic energy of the, photoelectrons is K max increase., Statement-II: Photoelectrons are emitted with, speeds ranging from zero to a maximum value, because of the range of frequencies present, in the incident light. [AIEEE-2010], 1) Statement I is true, Statement II is true; statement, II is a correct explanation of statement I., 2) Statement I is true, Statement II is true,, Statement II is NOT a correct explanation for, statement I., 3) Statement I is false, Statement II is true, 4) Statement I is true, Statemenet II is false, 17. Wavelengths associated with different, particles are given in Column - I. Match these, wavelengths with their values given in ColumnII., Column-I, A) Wavelength associated with an electron, accelerated through a pd of 1V, B) Wavelength associated with an -particle, accelerated through a pd of 1V, C) Wavelength associated with a proton, accelerated through a pd of 1V, D) Wavelength associated with a photon of energy, 124.2 eV, Column - II, p) 10nm q) 0.10 A0 r) 0.286 A0 s) 12.27 A0, 32, , LEVEL-IV - KEY, 1) 1,2,3, 6) 1, 4, 12) 3, , 2) 2,4 3) 1,3 4) 1,3,4 5) 2, 4, 7) 2 8) 1 9) 1 10) 4 11) 2, 13) 1 14) 2 15) 4 16) 4, , 17) A-s, B-q, C-r, D-p 18) A-q,r, B-s, C-q,r, D-p,s, , LEVEL-IV HINTS, 5., , (b) Stopping potential remains the same as it, depends on the frequency of incident radiation., (D) Saturation current intensity of incident, 1, radiation 2 . Since r becomes three times, r, 0.6m , , ,, 0.2m , 18.0mA, , 3, 6., , 2, , saturation, , current, , becomes, , 2.0mA ., hc, 0 , when decreases, V0, , increase., , As K max eV0 , and K max, , 7., , E hv ;, , 9., , K, , hc, , , 8. E W .E. K .E., , 1 2, mv, 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , ATOMIC PHYSICS, J.J.Thomson’s, experimental, determination of specific charge of an, electron:, , SYNOPSIS, , , , , , , , , , , , , , , , , , , , At ordinary pressures and ordinary voltages gases, are bad conductors of electricity., The p.d. required to start a spark through a gas is, known as sparking potential. It is much greater, than the p.d. required to maintain the spark once, started., At ordinary pressures for moderately great spark, lengths, the sparking potential is nearly proportional, to the spark length, being about 30,000 V per cm, in air between spherical electrodes of 1 cm diameter, and less in case of pointed electrodes., According to Paschen’s law, the sparking potential, is directly proportional to the pressure (P) of the, gas between the electrodes and the distance (d), between them., At ordinary pressures, the current is due to the, presence of positive and negative ions as in the, case of electrolytic conductivity of a solution., At low pressures, discharge (the passage of current, through gases) occurs at much lower voltages than, at high pressures., At lower pressures negative ion throws off its, attendant atom or molecule and the resultant, negatively charged particle, the electron travels free, and faster than the positive ions., At low pressures in the discharge tube, as the, pressure is reduced different phenomena such as, sparking, positive column, glow of cathode,, Faraday’s dark space, Crooke’s dark space and, the striations of positive column are gradually, observed., Finally at a pressure of about 0.01mm of Hg, pressure, Crooke’s dark space completely fills the, discharge tube with the walls of the tube glowing, with light i.e., producing fluorescence on the walls, of the tube. This is due to some radiation emitted, from the cathode surface to which the name, cathode rays has been given., Cathode rays were first observed by Plucker. J, J Thomson after studying their properties called, them “Streams of negative corpuscles”, while, Johnson stoney who having found from the, electrolysis that electricity was atomic in nature,, suggested the name “electron”., NARAYANAGROUP, , , , Cathode rays can move undeflected through, mutually perpendicular electric and magnetic fields, acting in same region., , + B, , –, C, , +, A, , x, E1, , v, , E2, , 1, , , , , , , , , , v, , S, , A, , –, , , , S1, , S2, F, , If plate E1 is connected to the positive terminal and, E2 to the negative terminal of a battery, vertically, downward electric field (E) is produced, perpendicular to the direction of motion of electrons., Then electrons will be deflected upwards striking, the screen at S1., If the magnetic field (B) alone is applied by using, clockwise current carrying circular coil the, direction of magnetic field is directed into the plane, of the paper. Then by Fleming’s left hand rule, electrons will be deflected downwards striking the, screen at S2., Then both electric and magnetic fields are applied, simultaneously and their values are adjusted so that, the electron beam can move undeflected and strikes, the screen at S., When the electron beam enters the region of, crossed electric and magnetic fields with velocity, v horizontally,, If only electric field is switched on,, Electric force on the electron which is vertically, upwards is, Fe eE 1, , V1, d, 1, V = potential difference between the plates E1 and E2, , Where Electric field E , , d = distance between the plates E1 and E2., If only magnetic field is switched on ,, Magnetic force acting on the electron, which is, vertically downwards is, Fm Bev 2, Where B = magnetic induction field, e = charge of electron, v = velocity of electron, 33
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , , If both electric and magnetic fields are applied, simultaneously and the electrons moves undeflected,, then magnetic and electric forces acting on the, electrons are equal and opposite., Bev = eE, Fm = Fe, E, Velocity of electron v 3, B, As the electron is accelerated from cathode to, anode, the workdone on the electron by the, accelerating voltage is equal to the increase in kinetic, energy of the electron, K.E of electron at anode =, work done on the electron, e, v2, 1 2, 5, mv eV0 4 ; , m 2V0, 2, Where V0 = potential difference between cathode, and Anode ., m = mass of electron, e = charge of electron, v = velocity of electron., substituting equation (3) in (5) we get, , e, E2, , 6, m 2V0 B 2, Since we know E, B and V0, specific charge of, electron can be calculated., The value of specific charge of electron as measured, by J.J.Thomson is, 1.7588 x 1011 C kg-1, Note :, (a) Specific charge of electron is found to be, independent of nature of the gas., (b) Specific charge of electron is found to be, independent of nature of the electrodes., (c) Specific charge of electron decreases with increase, in velocity of electron, because charge is, independent of velocity and mass increases with, increase in velocity, since, , m0, , m, , 1, , , , , , , , F e( E v B) 0, , v2, c2, , 34, , , , , E vB, , , , 2.5 × 10 7 ms-1 enters into the magnetic field, 4 ×10 -3 Wb/m2 directed perpendicular to its, direction of motion. Find the intensity of the, electric field required so that the electron, moves undeflected., Sol. When electron goes undeflected, force due to, electric field is equal and opposite to force due to, magnetic field i.e.,, Ee = Bev, B = 4 ×10-3 Wb/m2 and v = 2.5×107 ms-1, E = Bv = 4 ×10-3 × 2.5 ×107 = 105 V/m, , W.E-2: A mono energetic electron beam with a, speed of 5.2 × 10 6 ms-1 enters into a magnetic, field of induction 3 × 10 -4 T, directed normal, to the beam. Find the radius of the circle traced, by the beam (Take e/m=1.76×1011Ckg -1), 2, Sol. From the equation, Bev = mv, , r, , mv, v, Radius of circle r = Bq B e / m , r , , 5.2 10 6, , , , 3 10 4 1.76 1011, , , , 0.1m, , W.E-3 :The deflecting plates in a Thomson’s setup, are x meters long. Intensity of electric field, applied between the plates is E. The plates, are maintained at a p.d of V volts. Electrons, accelerated through a p.d of V volts enter, from one edge of the plates midway in a, direction parallel to the plates. Find the, deflection at the other edge of the plates., Sol. Motion of a charged particle projected, perpendicular to a uniform electric field is a, parabola. Horizontal distance travelled by the, x, v, Deflection of the electron in y - direction is, , electron in time t is x = v t t , , Where m0 = rest mass of electron,, m = Relativistic mass of electron and, v = Velocity of electron; c = Velocity of light, Note : If cathode rays are moving along +ve xaxis and electric field is applied along +ve y-axis, then magnetic field should be along +ve z-axis so, that cathode rays moves undeflected. This is, because net force on the electron, , , W.E-1: An electron beam moving with a speed of, , , , y, , Ee, 1 Ee x 2, 1 2, at , F = ma = Ee; a , ; y 2, m, 2 m n, 2, , (since uy = 0), We know that,, , y, , Ee, x2, 1, , 2, 4 mn , 2, , , 1, mn 2 eV, 2, , From (1) and (2) y , , ...... (1), , ...... (2), , E 2, x, 4V, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , W.E-4: The specific charge of a proton is 9.6 x107 C/, , Determination of charge of an electron :, Millikan’s oil drop method.[Additional], , Kg. The specific charge of an alpha particle is, q, Sol: Specific change = ;, m, , , , S q m p, , S p q p m, , S, 2 1 1, . ; S 4.8 107 C / kg, 7, 9.6 10 1 4 2, , Millikan’s method to find charge of an oil drop is, based on the measurement of terminal velocity of, the charged oil drop under the action of gravity and, under the combined action of gravity and electric, field., , W.E-5: A beam of electrons enter at right angles, to an uniform electric field with a velocity 3 x, 107 m/s, E= 1800 v/m while travelling through, a distance of 10cm, the beam deflected by 2mm, then specific charge of electron is, Sol: t , , x, 0.1, 1, 1, q, , 108 s ; y at 2 a E, 7, u 3 10, 3, 2, m, , 2 10 3 , , 1 , q 1, 8 , 1800, . 10 , 2 , m 3, , , 2, , q, , ;m, , w2, , , , 2 10 11 C / kg, , W.E-6:In Thomson’s experiment for determining, e/m the potential difference between the, cathode and the anode (in accelerating, column) is same as that between the dflecting, plates (in the region of crossed fields). If the, p.d is doubled, by what factor should the, magnetic field be increased to ensure that the, electron beam remains undeflected, Sol:, , , , 2, 2, 1, m 2 V .q ; 1 m E V .q ; 1 m V2 2 V .q, 2 B, 2 d .B, 2, , q, V, 2 2 ; B 2 V B, m 2d B, B1 , , V ; B1, , 2V, , 2B, , W.E-7: In a cathode ray tube, a p.d of 3000V is, maintained between the deflector plates whose, seperation is 2cm. A magnetic field of 2.5 x, 10-3 T at right angles to the electric field gives, no deflection of electron beam, which recieved, an initial acceleration by a p.d of 10,000V. The, e/m of electron is, Sol: V1 is p.d between the deflector plates and V2 be, the potential diference through which electrons are, accelerated, q, V2, 1, 1, V2, m 2 V2 q m. 21 2 V2 q ; 2 1 2, m 2d B V2, 2, 2 d B, q, 9 106, 11, , m 2 4 10 4 6.25 106 104 1.8 10 c / kg, NARAYANAGROUP, , , , Electric field can be produced between the plates, A and B. Small droplets of non volatile oil (clock, oil or apiezon oil) are sprayed using an atomizier, above the hole in plate A. Some of the droplets, acquire charge due to friction. The chamber is, illuminated by sending light horizontally through it., The drops can be seen by using a telescope placed, perpendicular to the light beam. A drop looks like, a bright star moving upwards or downwards., An electric field is produced between the plates A, and B by connecting upper plate to the positive, terminal and lower plate to the negative terminal of, a battery. The field is adjusted so that an oil drop, slowly moves upwards. The drop experiences, electric force and buoyancy in the upward direction, where as weight and viscous force in the downward, direction., The uniform velocity attained by the drop Ve is, measured by measuring the distance travelled and, time taken for it. Now electric field is switched off, and the drop is allowed to fall freely. The oil drop, now experiences viscous force and buoyancy, directed upwards and weight downwards. The, drop soon moves w ith terminal velocity V g, , downwards. Vg can be measured by measuring, distance travelled by the drop and time taken for it., Charge on the drop can be calculated using 1/ 2, 6ph Vg Ve 9hVg , , , q, , the formula, 2 d d 1 g , E, , , where h = Coefficient of viscosity of air, d = density of oil; d1 = density of air, E = Intensity of electric field between the plates, g = acceleration due to gravity, Ve = terminal velocity of the oil drop in electric field, Vg=terminal velocity of the oil drop in gravitational field, 35
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , , , , After large number of observations Millikan found, that the value of ‘q’ is an integral multiple of a, common value ‘e’ such that q = ne. Where n is an, integer and e is the charge of an electron. It is found, that charge of electron, e = 1.6x10-19 C., When the electric field is applied between the plates, A and B, suppose a negatively charged drop is, moving upwards with constant velocity Ve., upward forces are electric force (Fe) and Buoyancy, (Fb) while downward forces are viscous force (Fv), and weight (Fg)., Net force on the drop is zero., Fe, Fb, ve, , Fe Fb Fg Fv, , Fg, Fv, , qE , , Case-1 : In Millikan’s oil drop method, if the drop, is falling under gravity in the absence of electric field, then, upward forces are Buoyancy (Fb) and viscous, force (Fv) while downward force is weight (Fg)., Net force on the drop is zero., Fv, Fb, , Fb Fv Fg, Fg, , 4p 3 1, 4p 3, R d g 6ph RV , R dg, 3, 3, Case-2: In Millikan’s oil drop method if a, positively charged drop is at rest in the presence of, electric field directed upwards then, upward forces, are Buoyancy (Fb) and electric force (Fe) while, downward force is weight (Fg)., Net force on the drop is zero., , 4p 3 1, 4p 3, R d g, R dg 6ph RVe 1, 3, 3, , R - Radius of the drop, If electric field is switched off, suppose the drop, falls downwards with terminal velocity Vg ., upward forces are Buoyancy and viscous force, while downward force is weight., Net force on the drop is zero., Fv, Fb, vg, , Fb Fv Fg, Fg, , Fe, Fb, v=0, , Fe Fb Fg, Fg, , 4p, 4p, R 3d 1g , R 3d g, 3, 3, Case-3: In Millikan’s oil drop method, if a, positively charged drop is moving upwards with, constant velocity V in the presence of electric field, directed upwards then, upward forces are, buoyancy (Fb ) and electric force (Fe) while, downward forces are weight (Fg) and viscous force, (Fv). Net force on the drop is zero., qE , , 4p 3 1, 4p 3, R d g 6ph RVg , R dg 2, 3, 3, subtracting (2) from (1), qE 6ph RV g 6ph RVe, 6ph R Ve V g , q, 3, E, , 4p 3, R d d 1 g 6ph RVg, from equation (2), 3, 9 hVg, 9hVg, R2 , R, , 4, 2 d d 1 g, 2 d d1 g ;, , , , q, 36, , , 6ph Vg Ve , , E, , 1/ 2, , 5, , 1, 2 d d g , , 9hVg, , v, , Fe, Fb, v, , Fe Fb Fg Fv, , Fg, Fv, , 4p 3 1, 4p 3, qE , Rd g , R dg 6ph RV, 3, 3, Case-4: In Millikan’s oil drop method if a positively, charged drop is moving downwards with constant, velocity V in the presence of electric field directed, upwards then, upward forces are Buoyancy (Fb),, electric force (Fe) and viscous force (Fv) while, downward force is weight (Fg) ., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, Case-6 : In the Millikan’s oil drop experiment, the oil drop is subjected to a horizontal electric, field E and the drop moves with a constant, velocity making an angle q to the horizontal., If the weight of the drop is W, neglecting, W cot q, buoyancy, the charge on the drop is, E, , Net force on the drop is zero., , Fe Fb Fv Fg, , 4p 3 1, 4p 3, R d g 6ph RV , R dg, 3, 3, Case-5 : A charged oil drop having charge q1, is falling freely under gravity in the absence, of electric field with a velocity ‘V1’. It is held, stationary in an electric field. As it acquires a, charge it moves up with a velocity V2. Then, i) The new charge on the drop is q2 = (k + 1)q1, ii) Additional charge aquired is q kq1, qE , , Where k V2, , V1, , In the absence of electric field, drop is falling, downwords with velocity V 1., , Fg Fb Fv, , FV cosq qE, FV sin q Fg W, W cot q, q, E, , W.E-8: In Millikan’s oil drop experiment an oil, drop of radius r and charge Q is held in, equilibrium between the plates of a charged, parallel plate capacitor when the potential, difference is V. To keep another drop of same, oil whose radius is 2r and carrying charge 2Q, in equilibrium between the plates, find the, new potential difference required., , Fg Fb Fv, , QE, , Fg Fb 6ph RV1 (1), , V=0, , Sol. Since drop is at rest QE = mg, In the presence of electric field when the drop is at, rest., , Fg Fb Fe, Fg Fb q1E, , Fe, , V 4p 3, Q , r rg, d, 3, , v=0, , V2, 1, 8, 2, V, , Fb, , (2), , ;, , r3, V, Q, , 3, V2 r2 Q1, ; V r Q, 1 2, 1, , ; V2 4V, , W.E-9: A charged oil drop of charge q is falling, Fg, , In the electric field, after acquiring final charge q2, drop is moving upwards with velocity V2.Then, Fe, , Fg Fv Fb Fe, Fg Fb q2E6phRV2 but,V2 kV1, Fg Fb q2Ek6phRV1 (3), , mg, , Fb, , under gravity with terminal velocity v in the, absence of electric field. A electric field can, keep the oil drop stationary. If the drop, acquires an additional charge, it moves up, with velocity 3v in that field. Find the new, charge on the drop., Sol. In the absence of electric field, , v2, , 6RV, , Fv, Fg, , q1 E q2 E kq1 E, So,the final charge on the drop is q2 = (k + 1)q1, Additional charge aquired is q kq1, NARAYANAGROUP, , mg 6ph RV , , V, (1), , mg, 37
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, when electric field is applied, , W.E-12: A charged oil drop is of charge q is falling, freely under gravity in the absence of electric, field with a velocity ‘v’. It is held stationary, in an electric field, as it acquires a charge it, moves up with a velocity ‘3v. Now the charge, on the drop is, Sol: mg 6 r , Eq = mg, , qE, , V=0, , mg qE (2), , mg, , If q1 = new charge and drop is moving up then, , Eq1 mg 6 r (3 ) , Eq1 = Eq + 3Eq, Eq1 = 4Eq q1 = 4q, , q1E, , W.E-13: A charged oil drop falls with a terminal, q E mg 36ph RV (3), 1, , 3V, mg, 6R3V, , q1E 4mg 4qE, , q1 4q, , velocity V in the absence of electric field. An, electric field E keeps the oil drop stationary, in it. When the drop acquires a charge ‘q’ it, moves up with same velocity. Find the initial, charge on the drop., Sol: mg 6 r ,Eq = mg, , W.E-10: In Millikan’s method of determining the, charge of an electron, the terminal velocities, of oil drop in the presence and in the absence, of an electric field are xcm/s upwards and y, cm/s downwards respectively. Find the ratio, of electric force to gravitational force on the, oil drop. (Neglect Buoyancy), Sol. In Gravitational field, weight = viscous force, , E(Q+q) = mg + 6 r ,E( Q+q) = 2EQ, Q + q = 2Q q = Q, , W.E-14: Two oil drops in Millikan’s experiment, are falling with terminal velocities in the ratio, 1:4. The ratio of their de-Broglie wave length, is, Sol: T r 2, , r VT, , W 6phry........(1), , r1, V, 1, 1 , r2, V2 2, , In electric field,, Electric force = weight + viscous force, Eq W 6phrx ...........(2), , , , Substitute (1) in (2) then Eq 6phr( y x) ...........(3), Electric force, xy, , Gravitational force, y, , 1 m2 2, , ., 2 m1 1, , m1 1, , m2 8, 8 4, . = 32 : 1, 1 1, , Rutherford’s -particle Scattering, Experiment:, , W.E-11: In a Millikan’s experiment an oil drop of, radius 1.5 x 10-6 m and density 890 kg/m3 is, held stationary between two condenser plates, 1.2 cm apart and kept at a p.d of 2.3 kV. If, upthrust due to air is ignored, then the, number of excess electrons carries by the drop, will be, Sol: Eq = mg;, , V, 4, 4, d, .ne r 3 . g ; n r 3 g, d, 3, 3, Ve, , 4 3.375 1018 890 9.8 1.2 102, n , 3, 2.3 103 1.6 1019, , n4, 38, , -particles, , -particles, detector, , , Radio Lead cavity, active, Lead Plate, source, with slit, , Thin Gold Foil, , Experimental Observations:, a), , b), , Most of the - particles were found to pass through, the gold- foil without being deviated from their, paths., Some - particles were found to be deflected, through small angles 90 ., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, Few - particles were found to be scattered at fairly, large angles from their initial path 90, , Incident beam of -particles, , c), , ATOMIC PHYSICS, , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, , 1, K m v2 , 2, Because the positive charge on the nucleus is Ze, and that on the -particle 2e, hence the, electrostatic potential energy of the - particle, when, it is at a distance r0 from the centre of the nucleus, is, 1 (2e)( Ze), ., given by U , 40, r0, Because at r = r0 kinetic energy of the - particle, appears as its potential energy, hence, K=U, 1 (2e)(Ze) 1, ., m v2, 4 0, r0, 2, 1, 4Ze 2, r0 , 4 0 m v 2, Note:, , , , Nucleus, (+Ze), , , , Space, occupied, by Electrons, , , , , , , , , , r0, , d), , , , , A very small number of - particles about 1 in, 8000 practically retracted their paths or suffered, deflections of nearly 180° ., The observation (a) indicates that most of the, portion of the atom is hollow inside., Because - particle is positively charged, from the, observations (b), (c) and (d) atom also have, positive charge and the whole positive charge of, the atom must be concentrated in small space which, is at the centre of the atom is called nucleus. The, remaining part of the atom and electrons are, revolving around the nucleus in circular objects of, all possible radii. The positive charge present in the, nuclei of different metals is different . Higher the, positive charge in the nucleus, larger will be the, angle of scattering of - particle., , , , When a mono energitic beam of particles is, projected towards a thin metal foil, some of the, particles are found to deviate from their original path., This phenomenon is called ray scattering, It is caused by coulomb repulsive force between , particles and positive charges in atom., The number of -particles scattered at an angle , is given by, , , , , , An - particle which moves straight towards the, nucleus in head on direction reaches the nucleus, i.e, it moves close to a distance r0 as shown the, figure., As the - particle approaches the nucleus, the, electrostatic repulsive force due to the nucleus, increases and kinetic energy of the alpha particle, goes on converting into the electrostatic potential, energy. When whole of the kinetic energy is, converted into electrostatic potential energy, the, - particle cannot further move towards the, nucleus but returns back on its initial path i.e, - particle is scattered through an angle of 180°., The distance of - particle from the nucleus in this, stage is called as the distance of closest approach, and is represented by r0., Let m and v be the mass and velocity of the, - particle directed towards the centre of the, nucleus. Then kinetic energy of the - particle, NARAYANAGROUP, , Q n t Z2 e 4, θ, (8πε 0 ) 2 r 2 E 2 sin 4 , 2, , where, Q Total number of particles striking the foil, n number of atoms per unit volume of the foil, r distance of screen from the foil, t thickness of the foil, Z Atomic number of atoms of the metal foil, angle of scattering, E kinetic energy of particles, , Distance of Closest Approach :, , , N=, , N t;, , N Z2, , ; N, , 1, sin4, , , 2, , 1, 1, 2 or N , E, 4, where v is the velocity of particles falling on the, foil., Impact Parameter(b):The perpendicular distance, of the initial velocity vector of the - particle from, N, , , , centre of the nucleus is called “impact parameter”., , , Ze 2 cot , 2, b, 1, 4 0 mv 2, 2, 39
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , Bohr’s model of Hydrogen like atoms :, , , nh, .................. (2), 2mr, After solving the equations, radius of the orbit, n2 h2, r 2, 4 kZme 2, n2 , h2, th, r, , ., n, For n orbit, , ............(3), 4 2 ke 2 mZ , For hydrogen atom Z = 1, radius of the first orbit, (n = 1) is given by r1 0.529 10 10 m 0.53Å, This value is called as Bohr’s radius and the orbit, is called Bohr’s orbit. In general, the radius of the, nth orbit of a hydrogen like atom is given by, n2 , rn 0.53 Å where n 1,2,3,....... (4), Z, , (or) V , , Electron can revolve round the nucleus only in, certain allowed orbits called stationary orbits and, the Coulomb’s force of attraction between electron, and the positively charged nucleus provides, necessary centripetal force., –e, , k(Ze)e mV2, 1, , ; where k , 2, r, 40, r, , , , , , , , +Ze, Nucleus, , Suppose m is the mass of electron, V is the velocity, and ‘r’ is the radius of the orbit, then in stationary, orbits the angular momentum of the electron is an, h, integral multiple of, , where h is the Planck’ss, 2, constant., The, angular, momentum, h, L I mVr n, where n is called, 2, principal quantum number., An electron in a stationary orbit has a definite, amount of energy. It posses kinetic energy because, of its motion and potential energy on account of, the attraction of the nucleus. Each allowed orbit is, therefore associated with a certain quantity of, energy called the energy of the orbit, which equals, the total energy of the electron in it. In these allowed, orbits electrons revolve without radiating energy ., Energy is radiated or absorbed when an electron, jumps from one stationary orbit to another, stationary orbit. This energy is equal to the energy, difference between these two orbits and emitted, or absorbed as one quantum of radiation of, , Velocity of the Electron in the orbit :, The velocit y of an electron in n th orbit, nh, Vn , 2 mrn hence, n2h2 , 2ke 2 Z r , Vn , . n, ..(5), 4 2 kZme2 , h n , i.e the velocity of electron in any orbit is, independent of the mass of electron. The above, equation can also be written as, c Z, Vn , . m / s .......... (6), 137 n, Where ‘c’ is the speed of light in vacuum., , Time period of electron in the orbit :, Angular velocity of electron in nth orbit, V, Z2, 8 3 k 2 e 4 m, n n 0 3 where 0 , ......(7), rn, n, h3, is the angular velocity of electron in first Bohr’s, orbit. The time period of rotation of electron in nth, , 2, n3, n3, T, , , T, , orbit, .........., (8), i.e, ., n 2 0 Z 2, Z2, The time period of rotation increases as n, increases and is independent on the mass of the, electron., , frequency v given by Planck’s equation, , hc, . This is called Bohr ’ss, , frequency condition., Radius of Bohr’s orbit : When mass of the, nucleus is large compared to revolving electron,, then electron revolves around the nucleus in circular, orbit. According to first postulate, k(Ze)e mV 2 , 1 , where, k, , , , r, 4 0 ......(1), r2, According to second postulate, h, mVr n, where n = 1,2,3,4............., 2, E2 E1 hv , , , , 40, , Energy of the electron in the orbit :, , , The kinetic energy of the electron revolving around, the nucleus in nth orbit is given by, 2, 1, 1 2 ke 2 Z , 2, K n mV m , . , 2, 2 h, n, 2 2 k 2 e 4 mZ 2 , mZ 2, Kn , ., K, , .......(9);, 2 , n, h2, n , n2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , If the reference level (zero potential energy level), is at infinity then the electrostatic potential energy, is given by, , with atomic number Z for normal and excited states, as shown in Figure., , 2, 2, k(Ze)e, 2 4 kmZe , kZe , Un , , 2 2, rn, n h, , , Un , , 2, , Z ev, Z22ev, Z ev, , 4 2 k 2 e 4 mZ 2 , 2 ....... (10), h2, n , , Z2ev, 2, , Total energy of the electron in nth orbit, , Z ev, , 2 2 k 2mZ 2e 4, n2 h2, 2 2 k 2 e 4 mZ 2 , En , 2 ........ (11), h2, n , E n K n Un , , The expression of total energy for hydrogen like, atom may be simplified as, Z2, E n 13.6 2 eV , n = 1,2,3..... (12), n, Where -13.6 eV is the total energy of the electron, in the ground state of an hydrogen atom., From the equations (9),(10)&(11) it is clear that, PE : K.E : T.E = -2 : 1 : -1, i.e, , , PE KE TE, , , 2, 1, 1, , n=5, n=4, n=3, , nd, , –0.544eV, –0.850eV, –1.511eV, , sr, , n=2, n=1, , 1 excited state, , ground state, , –3.4eV, –13.6eV, , Energy level diagram of hydrogen atom (Z = 1), for normal and excited states as shown the figure., The energy level diagram of hydrogen like atom, NARAYANAGROUP, , , , When an electron jumps from higher energy level, n2 to a lower energy level n1 in stationary atom,, the difference in energy is radiated as a photon, whose frequency v is given by Planck’s formula., E n 2 E n1 hn, 1, , 0eV, , 2 excited state, , Emission of radiation:, , 1, , 2, (or) hv E 2 E1 13.6Z n 2 n 2 e.V, , The state n = 1 is called ground state and n > 1, states are called excited states. When electron go, from lower orbit to higher orbit speed and hence, kinetic energy decrease, but both potential energy, 1, and total energy increases. E 2 tells us that, n, the energy gap between the two successive levels, decreases as the value of n increases. At infinity, level the total energy of the atom becomes zero., n=, , , , The total energy of the electron is negative implies, the atomic electron is bound to the nucleus. To, remove the electron from its orbit beyond the, attraction of the nucleus , energy must be required., The minimum energy required to remove an, electron from the ground state of an atom is called, its ionization energy and it is 13.6 Z2 eV., In hydrogen atom the ground state energy of, electron is –13.6 eV, so 13.6 eV is the ionization, energy of the Hydrogen atom., , 1, 2, , , 13.6Z 2, e.V since 1eV 1.6 10 19 J, E n , n2, , , , c, , , 1 1, 2 J, 2, n, 1 n2 , , hence h (21.8 1018 ) Z 2 , (or) wave number, , 1, 1, 1 , v R Z 2 . 2 2 m 1, l, n 1 n 2 , , where R is called for “Rydberg constant”, when, the nucleus is infinitely massive as compared to the, revolving electron. In other words the nucleus is, considered to be stationary. The numerical value, of R is 1.097×107 m-1., , Emission Spectrum of Hydrogen atom :, Electron in hydrogen atom, can be in excited state, for very small time of the order of 10-8 second., This is because in the presence of conservative, force, particles of a system always try to occupy, stable equilibrium position and hence minimum, potential energy, which is least in ground state., Because of instability, when an electron in excited, state makes a transition to lower energy state, a, 41
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, photon is emitted. Collection of such emitted, photon frequencies is called an emission spectrum., This is as shown in figure., e–, , Transition, , e, N, , –, , photon, +, , N, , Emission spectrum, , n=, n=5, n=4, n=3, , L, L L , , n=2, k, , P fund series, (I.R. region), , bluegreen, , k, k, , Bracket series, (I.R. region), Paschen series, (I.R. region), , Balmer series, (Visible region), , n=1, Lyman series, (u.v. region), , The Spectral Series of Hydrogen Atom as shown, in figure, are explained below., (a) Lyman Series : Lines corresponding to transition, from outer energy levels n2 2,3, 4,.......... to, first orbit (n1= 1) constitute Lyman series. The, wave numbers of different lines are given by,, v, , , , 1, 1, 1 , R 2 2 , , 1 n 2 , , 1, , 1 1, R 2 2 , max 6568Å, max, 2 3 , 1, 1 , 1, R 2 2 , min 3636Å, min, , 2, Balmer series lies in the visible region of, electromagnetic spectrum. The wavelength of L, line is 656.8 nm (red). The wavelength of L line, is 486 nm (blue green). The wavelength of L line, is 434 nm (violet). The remaining lines of Balmer, series closest to violet light wavelength. The, speciality of these lines is that in going from one, end to other, the brightness and the separation, between them decreases regularly., This series is obtained only in emission spectrum., Absorption lines corresponding to Balmer series, do not exist, except extremely weakly, because very, few electrons are normally in the state n = 2 and, only a very few atoms are capable of having an, electron knocked from the state n = 2 to higher, states. Hence photons that correspond to these, energies will not be strongly absorbed. In highly, excited hydrogen gas there is possibility for, detecting absorption at Balmer-line wavelengths., (c) Paschen Series: Lines corresponding to n2 =, 4,5,6,....... to n1 = 3 constitute Paschen series., , Line corresponding to transition from n2 = 2 to n1, = 1 is first line; its wavelength is maximum., 1, 1 , 1, 1 1 , R 2 2 1.1 10 7 , max, 2 , 1, 1 4 , , max 1212Å, , Similarly transition from n2 to n1 1 gives, line of minimum wavelength., 1, m in, , 1 , 1, R 2 2 1.1 10 7, , 1, , The wave number of different lines are given by, , , , min 912Å, , , , Lyman series lies in ultraviolet region of electro, magnetic spectrum., Lyman series is obtained in emission as well as in, absorption spectrum., (b) Balmer Series: Lines corresponding to, n 2 3, 4,5,........ to n1 = 2 constitute Balmer, series. The wave numbers of different lines are, 1, 1, 1, given by, v R 2 2 , , n2 , 2, Line corresponding to transition n2=3 to n1=2 is, first line, wavelength corresponding to this transition, is maximum. Line corresponding to transition, n2 to n1 = 2 is last line; wavelength of last, line is minimum., 42, , 1, 1, v R 2 2 , n2 , 3, Line corresponding to transition n2 = 4 to, n1 = 3 is first line, having maximum wavelength., Line corresponding to transition n2 to n1 = 3, is last line, having minimum wavelength, 1, max, , 1, 1, R 2 2 , max 18747Å, 3 4 , , 1, 1 1, 1, , R 2 1.1 107 0 , min, , 9, 3, , , , , min 8202Å, Paschen series lies in the infrared region of, electromagnetic spectrum., This series is obtained only in the emission spectrum., (d) Bracket Series: The series corresponds to, transitions from n2 = 5,6,7......., to n 1 = 4., , , The, , wave, , number, , are, , given, , by,, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 1, 1, 1, R 2 2, , n2 , 4, Line corresponding to transition from n2=5 to n1=4, has maximum wavelength and n2 to, n1 = 4 has minimum wavelength., 1, 1 1, R 2 2 , max 40477Å, max, 5 , 4, 1, 1 , 1, R 2 2 14572Å, min, min, , 4, This series lies in the infrared region of, electromagnetic spectrum., (e) Pfund Series: This series corresponds to, transitions from n 2 6,7,8,...., to n1 5 . The, , Limitation of Bohr’s model :, , v, , wave numbers are given by, , , , , v, , , , , , , , , Name of Final, S.No the, series State, (n1), , Initial State, (n2), , Formula, 1 1 , 1, R 2 2 , l, 1 n 2 , , Series, limit, , UV, , 4, R, , l, , 36, 5R, , Visible, , l, , 9, R, , l, , 144, 7R, , Near IR, , l, , 16, R, , l, , 400, 9R, , Middle IR, , 2. Balmer, , n1 = 2 3, 4,5...., , 1, 1, 1 , R 2 2 , l, 2, n 2 , , l, , 3. Paschen, , n1 = 3 4, 5, 6...., , 1, 1, 1, R 2 2 , l, 3, n 2 , , 5, 6, 7...., , 1, 1, 1 , R 2 2 , l, 4, n 2 , , 6 , 7 , 8 ...., , 1, 1, 1 , R 2 2 , l, n 2 , 5, , 4. Brackett n1 = 4, Pfund, , n1 = 5, , Region, , 4, 3R, , n1 = 1, , 2,3, 4,...., , Maximum, wavelength, , l, , 1, 911A, R, , 1. Lyman, , 5., , (n2 n1 )(n2 n1 1), 2, , l, , l, , 25, R, , 9000, l, 11R, , Far IR, , W.E-15: The electron in a hydrogen atom makes, n2, n1, , Note 1: for n2 4 , and n1 = 1, the number of, possible lines are 6., Note 2 : If E is the energy difference between, two given energy states, then due to transition, between these two states wavelength of emitted, , NARAYANAGROUP, , , , 1, 1, 1 , , R, , , 52 n 2 , , 2 , , Line corresponding to transition from n2 = 6 to n1, = 5 has maximum wavelength and n2 to n1 =, 4 has minimum wavelength., 1, 1 1, R 2 2 , max, 5 6 , 1, 1 , 1, R 2 2 , max 74563Å, min, 5 , min 22768Å, This series lies in infrared region of electromagnetic, spectrum., Note : In an atom emission transition may start, from any higher energy level and end at any energy, level below of it. Hence in emission spectrum the, total possible number of emission lines from some, excited state n2 to another energy state n1 ( n 2 ) is, , photon is ( Å ) , , , , Despite its considerable achievements, the Bohr’s, model has certain short coming., It could not interpret the details of optical spectra, of atoms containing more than one electron., It involves the concept of orbit which could not be, checked experimentally, It could be successfully applied only to singleelectron atoms (e.g., H, He+ , Li2+, etc.), Bohr’s model could not explain the binding of, atoms into molecules., No justification was given for the “principle of, quantization of angular momentum”., Bohr’s model could not explain the reason why, atoms should combine to form chemical bonds, and why do the molecules become more stable on, such combinations., Bohr had assumed that an electron in the atom is, located at definite distance from the nucleus and is, revolving with a definite velocity around it. This is, against the Heisenberg uncertainty principle. With, the advancements in quantum mechanics, it be came, clear that there are no well defined orbits; rather, there are clouds of negative charge., , a transition n1 n 2 where n1 and n2 are the, principal quantum numbers of the two states., Assume the Bohr model to be valid. The time, period of the electron in the initial state is, eight times that in the final state. What are, the possible values of n1 and n2 ?, 3, , T1 n13, n , 3, T, , n, Sol. Since,, ; T n3 , As T1 = 8T2, n 1 8, 2, 2, 2, , (or) n1 = 2n2. Thus the possible values of n1 and, n2 are n1 2, n 2 1, n1 4, n 2 2, n1 6, n 2 3; and, so on., , 12400, E (eV ), 43
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , W.E-16: Find the kinetic energy, potential energy, and total energy in first and second orbit of, hydrogen atom if potential energy in first orbit, is taken to be zero., Sol. E1 = -13.60eV; K1 = -E1 = 13.60eV, U1 = 2E1 = -27.20eV E2 3.40eV K2 = 3.40eV, and U2 = -6.80eV, Now, U1 = 0, i.e., potential energy has been, increased by 27.20eV. So, we will increase U and, E in all energy states by 27.20eV while kinetic, energy will remain unchanged., Hence K(eV), U(eV),E(eV), First orbit are, 13.6,, 0,, 13.6, in Second orbit 3.40,, 20.40, 23.80, , W.E-17: A small particle of mass m moves in such, a way that the potential energy U = ar2 where, a is a constant and r is the distance of the, particle from the origin. Assuming Bohr’s, model of quantization of angular momentum, and circular orbits, find the radius of nth, allowed orbit., Sol. The force at a distance r is, F , , dU, 2ar, dr, , Suppose r be the radius of nth orbit. Then the, necessary centripetal force is provided by the, above force. Thus,, , mv2, 2ar, r, , -------- (ii), , Solving Eqs. (i) and (ii) for r, we get, , n2 h2 , r, , 8am 2 , , 1/ 4, , W.E-18:Consider a hydrogen-like atom whose, energy in n th excited state is given, by En , , 13.6Z 2, n2, , when this excited atom makes, , transition from excited state to ground state, most energetic photons have energy, Emax 52.224eV and least energetic photons, have energy Emin 1.224eV . Find the atomic, number of atom and the state of excitation., 44, , Hence,, E1, , E1, n2, , E1 52.224eV ... (1), , E1, , and n 2 (n 1)2 1.224 eV ...... (2), Solving above equations simultaneously, we get, E1 54.4eV and n = 5 Now E1 , , 13.6Z 2, 12, , 54.4eV ., , Hence, Z = 2 i.e, gas is helium originally excited to, n = 5 energy state., , W.E-19: A hydrogen-like atom (atomic number Z), is in a higher excited state of quantum, number n. This excited atom can make a, transition to the first excited state by, successively emitting two photons of energies, 10.20 eV and 17.00 eV respectively., Alternatively the atom from the same excited, state can make a transition to the second, excited state by successively emitting two, photons of energies 4.25 eV and 5.95 eV, respectively. Determine the values of n and Z, (ionization energy of hydrogen atom = 13.6, eV), Sol. The electronic transitions in a hydrogen-like atom, from a state n2 to a lower state n1 are given by, 1, 1, E 13.6Z 2 2 2 . For the transition from, n1 n 2 , , -------- (i), , Further, the quantization of angular momentum, nh, gives, mvr , 2, , Sol. Maximum energy is liberated for transition En 1, and minimum energy for En En1, , a higher state n to the first excited state n1 = 2,, the total energy released is (10.2 + 17.0) eV or, 27.2eV. Thus E = 27.2 eV,,, n1 = 2 and n2 = n., 2, , 1, , 1 , , We have 27.2 13.6Z 4 2 ...... (1), n , , For the eventual transition to the second excited, state n1 = 3, the total energy released is (4.25 +, 5.95) eV or 10.2eV., 2, , 1, , 1, , Thus 10.2 13.6Z 9 2 .....(2), n , , Dividing the Eq. (1) by Eq. (2) we get, , 27.2 9n 2 36, , ., 10.2 4n2 36, , Solving we get n2 = 36 or n = 6, Substituting n = 6 in any one of the above equations,, we obtain Z2 = 9 (or) Z = 3, Thus n=6 and Z=3., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , W.E-20: A doubly ionized lithium atom is hydrogen, , W.E-22: A hydrogen atom initially in the ground, , like with atomic number Z = 3. Find the, wavelength of the radiation required to excite, the electron in Li2+ from the first to the third, Bohr orbit. Given the ionization energy of, hydrogen atom as 13.6 eV., Sol. The energy of nth orbit of a hydrogen-like atom is, , level absorbs a photon, which excites it to the, n = 4 level. Determine the wavelength and, frequency of photon. To find the wavelength, and frequency of photon use the relation of, energy of electron in hydrogen atom is, , given as E n , , 13.6Z 2, n2, , Thus for Li2+ atom, as Z =, , 3, the electron energies of the first and third Bohr, orbits are For n = 1,E1 = –122.4eV, for n = 3, E3, = –13.6eV. Thus the energy required to transfer an, electron from E1 level to E3 level is, E E3 E1, 13.6 ( 122.4) 108.8eV . Therefore, t he, radiation needed to cause this transition should have, photons of this energy. hv = 108.8 eV. The, wavelength of this radiation is or , , hc, , 108.8eV, , 114.25 Å, , W.E-21: A hydrogen atom in a state of binding, energy 0.85 eV makes a transition to a state of, excitation energy of 10.2 eV., (i) What is the initial state of hydrogen atom?, (ii)What is the final state of hydrogen atom ?, (iii) What is the wavelength of the photon, emitted ?, Sol. (i) Let n1 be initial state of electron. Then, E1 , , 13.6, n12, , 0.85 , , eV, , 13.6, n12, , Here E1 = - 0.85 eV, therefore, or n1 = 4, , (ii) Let n2 be the final excitation state of the electron., Since excitation energy is always measured with, respect to the ground state, therefore, , 1, E 13.6 1 2 here E 10.2eV, therefore,, n 2 , , 1, 10.2 = 13.6 1 n2 or, , 2, , , 13.6, eV ., n2, Sol. For ground state n1 = 1 to n2 = 4., Energy absorbed by photon, E = E2 E1, , En= , , 1 1 , = 13.6 2 2 1.6 1019 J, n1 n 2 , 1 1 , 19, = 13.6 2 1.6 10, 1, 4, , , 15 , = 20.4 10-19., 16 , or E = h = 20.4 10-19, , = 13.6 1.6 10, , 20.4 1019, 20.4 1019, Frequency =, =, 6.63 1034, h, = 3.076 1015 = 3.1 1015 Hz., 3 108, c, Wavelength of photon λ = =, =, 3.076 1015, , 9.74 10-8 m. Thus, the wavelength is 9.7 10-8 m, and frequency is 3.1 1015 Hz., , W.E-23: (a) Using the Bohr’s model calculate the, speed of the electron in a hydrogen atom in, the n = 1, 2 and 3 levels., (b) Calculate the orbital period in each of, these levels., Sol. (a) Speed of the electron in Bohr’s nth orbit, , =, , c, 2πKe 2, α where, α =, v, ch, , α = 0.0073, n2 = 2 Thus, the electron, , jumps from n1 = 4 to n2 = 2., (iii) The wavelength of the photon emitted for a, transition between n1 = 4 to n2 = 2, is given by, 1, 1, 7 1, 1, 1, 1, R 2 2 (or) 1.09 10 2 2 =4860Å ., 2 4 , , n2 n1 , NARAYANAGROUP, , 19, , v=, , c, 0.0073, n, , c, 0.0073, 1, = 3 108 0.0073 = 2.19 106 m/s, , For n = 1,, , v1 =, , For n = 2, , v2 =, , c, 0.0073, 2, 45
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , W.E-24: The radius of the innermost electron orbit, , 3108 0.0073, =, = 1.095 106 m/s, 2, v3 =, , For n = 3, , =, , c, 0.0073, 3, , 3 108 0.0073, = 7.3 105 m/s., 3, , (b) Orbital period of electron is given by T =, 2, , 2πr, v, , 2, , n h 4 0, Radius of nth orbit rn , , 2, m 2 e, , (1) 2 (6.6310 34 )2, r1 = 4 9.87 (9 10)9 9 10 31 6.6 1019 ), = 0.53 10-10 m., 2πr1, T1 = v, 1, , For n = 1, , 2 3.14 0.53 1010, = 1.52 10-16 s, =, 2.19 106, For n = 2, radius rn = n2r1, r2 = 22 .r1 = 4 0.53 10-10 m and, velocity vn=, , v1, n, , 2 3.14 4 0.531010 2, Time period T2 =, 2.19 106, = 1.216 10-15 s., For n = 3, radius r3 = 32, r1 = 9 r1 = 9 0.53 10-10 m, v1, 2.19 106, m/s, =, 3, 3, , 2πr3, Time period T3 = v, 3, , 2 3.14 9 0.5310-10 3, = 4.1 10-15 s., =, 2.19 106, , 46, , W.E-25: A 12.5 eV electron beam is used to, bombard gaseous hydrogen at room, temperature. What series of wavelength will, be emitted?, Sol. Energy of electron beam E=12.5 eV, = 12.5 1.6 10-10 J, Planck’s constant h=6.63 10-34 J-s, Velocity of light c=3 108 m/s, 8, 34, hc 6.62 10 3 10, Using the relation E=, =, 12.5 1.6 10 19, λ, = 0.993 10-7 m = 993 10-10 m = 993 A0, This wavelength falls in the range of Lyman series, (912 A0 to 1216A0), thus, we conclude that Lyman series of wavelength, 993 A0 is emitted., , W.E-26:In accordance with the Bohr’s model, find, , v1, 2.19 106, =, v2 =, 2, 2, , and velocity v3 =, , of a hydrogen atom is 5.3 10-11 m. What are, the radii of the n = 2 and n = 3 orbits?, Sol. Given, the radius of the innermost electron orbit of, a hydrogen r1 = 5.3 10-11 m., As we know that, rn = n2r1 For n = 2, radius r2 = 22, r1 = 4 5.3 10-11 = 2.12 10-10 m., For n = 3, radius r3 = 32, r1 = 9 5.3 10-11 = 4.77 10-10 m., , the quantum number that characterises the, earth’s revolution around the sun in an orbit, of radius 1.5 1011 m with orbital speed 3 104, m/s. (Mass of earth = 6.0 1024 kg)., A. Given, radius of orbit r = 1.5 1011 m, Orbital speed v=3 104m/s;, Mass of earth M=6 1024kg, nh, 2πvrm, Angulalr momentum, mvr =, or n =, 2π, h, [where, n is the quantum number of the orbit], 2 3.14 3 104 1.5 1011 6 1024, =, 6.631034, = 2.57 1074 or n = 2.6 1074 . Thus, the, quantum number is 2.6 1074 which is too large., The electron would jump from n=1 to n=3, 13.6, = 1.5 eV.., 32, So, they belong to Lyman series., , E3 =, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , Effect of finite mass of nucleus on, Bohr’s model of an atom, (i), , In the atomic spectra of hydrogen and hydrogen, like atoms a very small deviation with Bohr’s model, results, , (ii) This is in the assuption that the nucleus is infinitely, massive when compared to mass of eletron so that, it remains stationary during the rotation of eletron, around it, (iii) Infact the nucleus is not infinitely massive and hence, both the nucleus and eletron revolve around their, centre of mass with same angular velocity , , ATOMIC PHYSICS, , m, , M 2r 2, Mm 2 r 2, nh, , , , 2, 2, ( M m), ( M m), 2, , nh, mMr 2, nh, 2, ( M m) , ie r , --- (2), 2, 2, ( M m), 2, (vii) A system of this type is equalent a single particle of, mass revolving around the position of the heavier, particle(nucleus) in an orbit of radius r., , 0 n 2 h2, From (1) and (2) r , z e2, (viii) Radius of orbit of such a particle in a quantum state, , cm, , M, r2, , (iv), , m, r1, , n is rn , , 0 n2 h2, n2, , r, , n, z e2, z, , Electron, Nucleus, , (iv) let m be the mass of eletron, M be the mass of the, nucleus , Z be its atomic number and r be the, separation between them. If r1 , r2 are distances of, centre of mass from electron and nucleus, respectively then r1 r2 r and, Mr, mr, , r2 , r1 , M m, M m, , (v) For both the eletron and the nucleus the necessary, centripetal force to revolve in circular orbits is, provided by the electrostaic force between them., 1 Ze 2, Mr, 1 Ze 2, 2, , i.e., mr1 , ;m, 4 0 r 2, M m, 4 0 r 2, , Ze2, and, 4 0 r, the, syst em, , (ix) Potential energy of the system PE , kinetic, KE , , , , energy, , of, , 1, 1, 1, I1 2 I 2 2 I1 I 2 2, 2, 2, 2, , 1, mr12 Mr22 2, , 2, , 1, M 2r 2, m2r 2 2, m, M, , 2, 2 M m 2, M m , , 2, , 3 2, i.e., r , , where , , Ze 2, ----- (1), 4 0, , Mm, called reduced mass, M m, , (vi) From Bohr’s theory of quantization of angular, momentum, total angular momentum of the system, L I1 I 2 , , NARAYANAGROUP, , nh, nh, 2, 2, ; mr1 Mr2 , 2, 2, , , , 1 mMr 2 2, 2 M m 2, , M m, , 1 mM 2 2, , r , 2 M m, , 1 2 2 1 Ze2, KE, , r , ie, 2, 2 4 0 r, , , Ze2 , 3 2, , , r, , , , , 40 , , , Total energy of the system (or equivalent, particle of mass ) E PE KE, , E, , Ze 2, Ze 2 Z e 2, Z 2 e4, , 2 2 2 2 2, 8 0 r, 8 0 n h 0, 8 0 n h, 47
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, ie Energy of nth quantum state, , En =, , -Z2 e4, mZ 2 e4, me4 Z 2 , , , , , 8 02 n2h2 8 02n2 h2 m 8 02h 2 n2m , , z2 , Z2 M, En 13.6 2 eV 13.6 2 , eV, n m, n M m, , , , 13.6 Z 2 , eV, m n 2 , 1, M, , The formulae for rn and En can be obtained simply, by replacing m by in the formulae for stationary, nucleus, If is wave length of photon emitted due, to transition from a quantum state n2 to a quantum, hc me4 z 2, 1, 2 2, state n1 , then 8 0 h 1 m, M, 4 2, , 1 me z, 1, 2 3 , m, 8 0 h c , 1 , M, 2, , 1, R0 Z, , m, , 1 , M, , 1 1, , n12 n22 , , , , 1 1 , , n12 n22 , , , , 1 1 , , n12 n22 where R is Rydberg’ss, 0, , , , constant when the nucleus is stationary, , Excitation by collision, (i), , When an atom is bombarded by particles like, electron, proton, neutron, particle etc, the loss, in KE of the system during collision may be used, in excitation of the atom, , (ii) If loss in KE of the system during collision (during, deformation phase)is less than the energy required, to excite the eletron to next higher energy state,, eletron can’t be excited and the loss of KE of the, system during deformation phase again converts, 48, , into KE of the system and totally there will be no, loss of KE of of the system and hence the collision, is elastic, (iii) If loss in KE of the system during deformation, phase is more than or equal to the energy required, to excite the electron to next higher state, excitation, of the electron may take place and hence kinetic, energy of the system may not be conserved hence, the collision may be inelastic or even perfectly, inelastic., (iv) If loss in KE is sufficient even ionization may take, place.Even though the possible loss in KE is greater, than or equal to excitation energy of electron,, excitation may not take place necessarily and hence, collision may be elastic., (v), , Consider a particle of mass m moving with velocity, u which strikes a stationary hydrogen like atom of, mass M which is in ground state., , (vi) Loss in KE will be maximum in perfectly inelastic, collision. In this case if V is common velocity after, collision, from conservation of linear momentum, mu+0=(M+m)V V , , mu, M m, , Maximum possible loss in KE is, , K , , 1, 1, mu 2 M m V 2, 2, 2, , 1 Mm 2, i.e K , u, 2 M m, , (vii) If E is minimum excitation energy (ex: n=1 to, n=2 in ground state) and if K E eletron can’t, be excited hence there will be no loss of KE of the, system hence the collision is elastic., (viii) If K E the electron may get excited and the, collision may be perfectly in elastic., (ix) If K E the electron may get excited to higher, energy states or even removed from the atom and, may have some kinetic energy. In this case the, collision may be inelastic or may be perfectly, inelastic as there is loss in KE of the system, or, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , even elastic if excitation dose not take place., (x) If KEmin is the minimum KE that should be, pocessed by the colliding particle to excite the, electron, K E for excitation, 1 mM 2, 1, M m, u E mu 2 E , , 2 M m, 2, M , , The emitted wavelengths range from a minimum, value all the way to infinity. Thus the name, continuous spectrum., (iv) The other possibility is that the accelerated electron, might knock out the inner electron of the target, atom, whereby a vacancy is created in the innter, orbit., Electrons from the higher orbit jump in to fill this, vacancy, while releasing the energy difference as, electromagnetic radiations., , m, M m, , KEmin E , E 1 , M , M, , These wavelenghts are characteristic of the material, from which they are emitted. Hence the name, Characteristic Spectra., , In the same way we can calculate KEmin in other, cases where atom is also moving using the, conservation of linear momentum., , Production of x-rays, , (i), , It is also known as inverse photoelectric effect, as, energetic electrons produce electro-magnetic, radiations., , Intensity, , , , P2, , (ii) Their wavelength is of the order 1A0., , P1, I, , , , (iii) They are producted with an apparatus called, Coolidge tube. In a Coolidge tube, electrons are, emitted by thermionic emission, accelerated across, a very high potential difference V and made to hit a, target., , Wavelength, , Incident, electron, , Hv, K X-ray, , K-electron, X-rays are produced and emerge oust of a window., Water is circulated in the target to keep it cool., v), , Continuous and characteristic x-rays, (i), , When an accelerated electron hits the target, the, electron loses its energy in two processes. One, process gives rise to continuous X-rays and the, other process gives rise to characteristic X-rays., , To discuss energy transitions in X-rays, we take, the ground state atom with all electrons intact as, the zero-level. The atom with a vacancy in the Kshell is the one with the highest energy, as a lot of, energy is required to dismiss a K-shell electron., Therefore we have the following energy diagram., , (ii) When the electron loses its kinetic energy in several, collisions with the atoms, a fraction can range from, zero to one., (iii) Corresponding to the maximum kinetic energy lost, by an electron, we have the largest frequency of, the X-ray or the shortest wavelength., hc, hc, K max eV min , min, eV, NARAYANAGROUP, , 49
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , C.U.Q., , 9., , THOMSON’S METHOD, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 50, , If ‘C‘ denotes the velocity of light, velocity of, cathode rays is, 1) equal to C, 2) greater than C, 3) less than C, 4) either greater or less than C, An electron has no initial velocity in a direction, different from that of an electric field. The path, of the electron is, 1) a straight line, 2) a circle, 3) a parabola, 4) an ellipse, Cathode rays enter an electric field normal to, the field. Then their path in the electric field, is, 1) a parabola, 2) a circle, 3) a straight line, 4) an ellipse, When the electron in the discharge tube is, accelerated to high speeds (i.e. comparable, with speed of light), 1) The charge on the electron will decrease, 2) The specific charge will decrease, 3) The charge of the electron will increase, 4) The value of e/m will increase, The discovery of electron was a consequence, of study of the, 1) discharge of electricity through atmosphere., 2) discharge of electricity through rarefied gases., 3) photoelectric effect, 4) nuclear fission., , If a proton and an electron are accelerated, through the same potential difference, 1) both the proton and electron have same K.E, 2) both the proton and electron have same, momentum, 3) both the proton and electron have same velocity, 4) both the proton and electron have same, temperature, A cathode ray tube has a potential difference, of V between the cathode and anode. The, speed of the cathode rays is given by, 1) v V 2) v V 1 3) v V 4) v V2, In a region of space, cathode rays move along, positive z-axis and a uniform magnetic field is, applied along x-axis. If cathode rays pass, undeviated, the direction of electric field will, be along, 1) Negative x-axis, 2) Positive y-axis, 3) Negative y-axis, 4) Positive z-axis, , A cathode ray particle is accelerated from rest, through a potential difference of V volt. The, speed of the particle is, 1), , 2eV, m, , m, 2) eV, , 3), , meV, , 4), , 4 eV, m, , 10. In J.J.Thomson’s method, electric field of, intensity E, magnetic field of induction B and, velocity V of the electrons were in mutually, perpendicular directions. The condition for, velocity is, 1) V = E / B, 2) V = B / E, 11., , 12., , 13., , 14., , 15., , 3) V = B E, 4) V = B / E, A positive ion of mass M kg and charge e, coulomb travels from rest through a p.d. of V, volt. The final K.E. is, 1) e V joule, 2) MeV joule, 3) Me/V joule, 4) eV/M joule, The increasing order of specific charge for, electron(e), proton(p), neutron(n) and alpha, particle( ) is, 1) e, p, n, , 2) n, , p, e, 3) n, p, e, , 4) n, p, , e, A uniform electric field and a uniform magnetic, field are produced pointing in the same, direction. An electron is projected with its, velocity pointed in the same direction., 1) the electron will turn to its right, 2) the electron will turn to its left, 3) the electron velocity will increase in magnitude, 4) the electron velocity will decrease in magnitude, When a positively charged particle enters a, uniform magnetic field with uniform velocity, its trajectory can be, a) straight line, b) a circle, c) a helix, 1) a only, 2) a or b, 3) a or c, 4) any one of a,b and c, The ratio e/m of electron is independent of, , a) Nature of cathode, anode, b) Nature of gas in discharge tube, c) Applied voltage, d) Size of discharge tube, 1) a ,b,c are only correct, 2) a,b,d are only correct, 3) b,c,d are only correct 4) a,b,c,d are correct, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 16. An electron is not deflected on passing through, a certain region, because, 1) there is magnetic field in that region and the, electron enters into it in any direction., 2) there may be magnetic field but the velocityofelectron, may be parallel to the direction of magnetic field, 3) electron is a chargeless particle, 4) there is electric field and the electron enters into, it in any direction., 17. In Thomson’s experiment, when the electron, strikes the undeflected spot, then it moves with, 1) constant acceleration 2)non uniform velocity, 3) constant velocity, 4)constant retardation, 18. An electron enters perpendicular to a uniform, magnetic field with a speed of 108 cm/s. The, particle experiences a force due to the, magnetic field and the speed of the electron, 1) will decrease, 2) will increase, 3) will remain constant 4) may increase or decrease, 19. A neutron, a proton, an electron and an alpha, particle enter a region of constant magnetic, field with equal velocities. The magnetic field, is along the inward normal to the plane of the, paper. The tracks of the particles are labled, in the figure. The electron follows ____ track, and alpha particle follows track ______, , 1) C,D 2) B, A, 3) A,C, 4) A,D, 20. When a charged particle moves through a, magnetic field, the quantity which is not, affected in the magnetic field is, 1) particle velocity, 2)particle acceleration, 3) linear momentum of the particle, 4) Kinetic energy of the particle, 21. A negatively charged electroscope with zinc, disc discharges when irradiated by an, ultraviolet lamp. What caused this?, 1) - particles from the source combine with, electrons of the disc, 2) electrons escape from the disc when ultraviolet, radiation falls on it, 3) ultraviolet rays ionize the air surrounding the, electroscope, 4) the disc becomes hot and thermionic emission, takes place, NARAYANAGROUP, , ATOMIC PHYSICS, 22. The force felt by an electron on entering into, a magnetic field is independent of its, 1) Charge 2) Strength of the field, 3) Mass, 4) Direction of its velocity, 23. When an electron moves through a magnetic, field, its speed will, 1) decrease, 2) increase, 3) remain the same, 4) increase first and then decrease, 24. The direction of deflection of a cathode ray, particle passing through a magnetic field can, be found by, 1) Fleming’s left hand rule, 2) Laplace’s law, 3) Maxwell’s cork screw rule 4) Ampere’s rule, 25. Which of the following particles cannot be, deflected by magnetic field, 1) electrons, 2) neutrons, 3) - particles, 4) protons, 26. A - particle enters a magnetic field making, an angle of 450 with the field lines. The path of, the particle is, 1) circular, 2) elliptical, 3) spiral, 4) a straight line, 27. An electron and a proton are injected into a, uniform magnetic field at right angle to its, direction with the same momentum. Then, 1) electron’s path is less curved than proton’s path, 2) proton’s path will be less curved than electron’s path, 3) the paths of both will be equally curved, 4) both the trajectories will be straight, 28. A proton and an electron simultaneously enter, into a region in which a uniform magnetic field, acts normal to the motion of both the particles., The frequency of revolution of, 1) the proton is greater than that of the electron, 2) the electron is greater than that of the proton, 3) the proton is equal to that of the electron, 4) both are having same frequency, but revolve in, opposite direction, 29. You are sitting in a room in which uniform, magnetic field is present in vertically, downward direction. When an electron is, projected in horizontal direction, it will be, moving in circular path with constant speed, 1) clockwise in vertical plane, 2) clockwise in horizontal plane, 3) anticlockwise in vertical plane, 4) anticlockwise in horizontal plane, 51
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 30. A charged particle of charge Q and mass m, moves with velocity v in a circular path due to, transverse magnetic field, B, then its frequency, is, 1), , QB, 2m, , 2), , QvB, 2m, , 3), , QmvB, 2, , 4), , vB, 2Qm, , 31. A proton and an electron simultaneously enter, a region in which a uniform magnetic field acts, normal to the motion of both the particles. The, frequency of revolution of, 1) proton is greater than that of electron, 2) electron is greater than that of proton, 3) proton is equal to that of electron, 4) proton depends on its velocity, , 32. Imagine that you are sitting in a room with your, back to one wall and that an electron beam, traveling horizontally from the back wall, towards the front one is deflected towards the, right. What is the direction of the magnetic, induction field that exists in the room?, 1) vertically upwards, 2) vertically downwards, 3) horizontal and perpendicular to the direction of, motion of the electron beam, 4) horizontal and parallel to the direction of motion, of the electron beam, 33. Cathode rays are made to pass between the, poles of a magnet as shown in figure. The effect, of magnetic field is, cathode, rays, , N, S, , 1) to deflect them towards the south pole, 2) to deflect them perpendicular to the plane of, the paper and towards the observer, 3) to deflect them towards the north pole, 4) to increase the velocity of the rays, 34. In which of the following fields cathode rays, show minimum deflection, 1) Electric field, 2) Magnetic field, 3) Plasma field, 4) Gravitational field, 52, , MILLIKAN’S OIL DROP EXPERIMENT, 35. An oil drop of mass m and charge +q is, balanced in vacuum by a uniform electric field, of intensity E. The direction of this field should, be, 1) vertically up, 2) vertically down, 3) horizontal 4) inclined at 45 0 to the horizontal, 36. An oil drop of mass m falls through a medium, that offers a viscous drag force F. If the, velocity of the drop is constant it means that, 1) F > mg, 2) F < mg, 3) F > mg, 4) F = mg, 37. An oil drop of mass m falls through a viscous, medium. The viscous drag force (F) is, proportional to the velocity of the drop. At the, instant it begins to fall the net force that acts, on the oil drop is (neglect buoyancy), 1) mg, 2) mg-F, 3) F-mg, 4) F, 38. An oil drop of mass m falls through air with a, terminal velocity in the presence of upward, electric field of intensity E. The drop carries a, charge +q. R is the viscous drag force and F is, the buoyancy force. Then for the motion of, the drop., 1) there is a net force directed upwards, 2) there is a net force directed downwards., 3) mg = Eq + F + R, 4) mg + Eq = F + R, 39. If g E and g M are the accelerations due to, gravity on the surfaces of the earth and the, moon respectively and if Millikan’s oil drop, experiment could be performed on the two, surfaces, one will find the ratio, electronic charge on the moon, to be, electronic charge on the earth, , 1) g E / g M, , 2) 1, , 3) 0, , 4) g M / g E, , 40. In Millikan's oil drop method, electron charge, is accurately measured by conducting an, experiment in which, 1) Apparent weight of drop is balanced by electric, force on it, 2) Apparent weight of drop is balanced by magnetic, force, 3) Electric force of drop is equal to magnetic force, 4) Apparent weight of drop is balanced by force, due to surface tension, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 41. In Millikan‘s experiment, the oil drops acquire, charge by, 1) induction, 2) Friction, 3) electric field, 4) magnetic field, 42. If V is velocity of oil drop, then according to, Stoke‘s law, the viscous drag force on an oil, drop is proportional to, 1, , 1) V, 2) V2, 3) V, 4) V, 43. An oil drop of radius r carrying a charge q, remains stationary in the presence of electric, field of intensity E. If the density of oil is ,, then, 4 3, 1) E r gq, 3, , 4 3, 2) E r g, 3, , 4 3, 4 3, 3, 4) E r / g, 3) E r g / q, 3, 3, 44. The important conclusion given by Millikan’s, experiment about the charge is, 1) Charge is never quantised, 2) Charge has no definite value, 3) Charge is quantised, 4) Charge on an oil drop always increases, , ALPHA RAY SCATTERING, 45. ‘Coulomb’s law correctly describes the electric, force is that (pick the wrong statement), 1) binds the electrons and neutrons in the nucleus, of an atom., 2) binds electrons to nucleus, 3)binds atoms together to form molecules, 4)binds atoms and molecules to from solids, 46. Is the probability of backward scattering, (i.e.scattering of - particals at angles greater, than 900) predicted by Thomson’s model much, less, about the same, or much greater than, that predicted by Ruthorford’s model?, 1) much greater, 2) much less, 3) same, 4) slightly greater, 47. An electron with KE 6 eV is incident on a, hydrogen atom in its ground state.The, collision., 1) must be elastic 2) may be partially elastic, 3) must be completly inelastic, 4) may be partially inelastic., NARAYANAGROUP, , 48. The angular momentum of the -particles, which are scattered through large angles by, the heavier nuclei,is conserved because of the, 1) nature of repulsive forces, 2) conservation of kinetic energy, 3) conservation of potential energy, 4) there is no external torque, 49. The Incorrect statement regarding, Rutherford’s atomic model is, 1) Atom contains nucleus, 2) Size of nucleus is very small in comparison, to that of atom, 3) Nucleus contains about 90 % mass of the atom, 4) Electrons revolve round the nucleus with uniform, speed, 50. -particles are, 1) helium nuclei, 2) sodium nuclei, 3) ionised nuclei, 4) hydrogen nuclei, 51. In scattering experiment,the force that scatters, particles is, 1) nuclear force, 2) coulomb force, 3) Both(1)and(2), 4) gravitational force, , BOHR’S THEORY, 52. Bohr’s atom model assumes, 1) the nucleus is of infinite mass and is at rest, 2)electron in a quantized orbit will not radiate, energy., 3) mass of the electron remains constant., 4) 1,2,3 are correct, 53. The ratio of magnetic dipole moment to angular, momentum in a hydrogen like atom is, e, e, e, 2e, 2), 3), 4), m, 2m, 3m, m, 54. The radius of hydrogen atom,when it is in its, second excited state, becomes:, 1) half, 2) double, 3) four times, 4) nine times, 55. For electron moving in nth orbit of the atom,, the angular velocity is proportional to:, , 1), , 1, 1, 3) n3, 4) 3, n, n, 56. The order of size of nucleus and Bohr’s radius, of an atom respectively are:, , 1) n, , 2), , 1) 1014 m,1010 m, , 2)1010 m,108 m, , 3) 1020 m,1016 m, , 4) 10 8 m,10 6 m, 53
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 57. In the lowest energy level of hydrogen, atom,electron has the angular momentum:, , h, h, 2, 2), 3), 4), 1) h, h, , 2, 58. Atomic hydrogen is excited to the nth energy, level.The maximum number of spectral lines, which it can emit while returning to the ground, state ,is:, , 1, 1, 1) n(n 1) 2) n(n 1) 3) n(n 1) 4) n(n 1), 2, 2, 59. As the orbit number increases ,the distance, between two consecutive orbits in an atom or, ion having single electron:, 1) increases, 2)decreases, 3) remains the same, 4) first increases and then becomes constant, 60. Ionization energy of a hydrogen -like ion A is, greater than that of another hydrogen like, ion B. Let r , u , E and L represents the radius, of the orbit, speed of the electron, energy of, the atom and orbital angular momentum of the, electron respectively in ground state, then, , 1) rA rB 2) u A uB 3) EA EB 4) LA LB, 61. The classification of discrete energy levels in, atoms was first given experimentally by, 1) Thomson’s experiment, 2) Millikan’s oil drop experiment., 3) Frank - Hertz experiment, 4) Leonard experiment, 62. An atomic nucleus contains, 1) only electrons, 2) only protons, 3) only neutrons, 4) both protons and neutrons, 63. On decreasing principal quantum number n,, the values of r and v will, 1)decrease, 2) increase, 3) r will increase but v will decrease., 4) r will decrease but v will increase, 64. The possible values of principal quantum, number can be, 1) 1, 2, 3...8., 2) 0, 1, 2...8., 3) Only zero, 4) only odd numbers, 65. According to Sommerfeld, an electron, revolves round a nucleus in, 1) Circular orbits., 2) Elliptical orbits., 3) Hyperbolic orbits, 4) parabolic, 54, , 66. The main defect of Bohr atom model is, 1) mixing of classical and quantum theories, 2) exclusion of nuclear motion, 3) failed to explain the fine structure of spectral, lines, 4) failed to explain other atoms, 67. According to Bohr hypothesis, discrete, quantity is, 1) Momentum, 2) Angular velocity, 3) Potential energy, 4) Angular momentum, 68. If En and J n are the magnitude of total energy, and angular momentum of electron in the nth, Bohr orbit respectively, then, 1) En Jn2, , 1, 2) E n J 2, n, , 3) En Jn, , 1, 4) En J, , n, , 69. If the radius of first Bohr orbit is r, then radius, of second orbit will be, 1) 2r, , 2), , r, 2, , 3) 4r, , 4), , 2r, , ATOMIC SPECTRA, 70. According to quantum mechanics, one of the, following is wrong about spin of electron, 1) it is related to intrinsic angular momentum, 2) spin is rotation of electron about its own axis, 3) value of spin quantum number must not be 1, 4) +1/2 value of spin quantum number represents, up spin, 71. According to classical theory, the path of, electron in Rutherford atom nuclei is, 1) straight line, 2) spiral, 3) circular, 4) parabolic, 72. One of the following radiations are not emitted, by electron transition in atoms, choose the, option, 1) ultra violet, 2) infrared radiations, 3) visible rays, 4) -rays, 73. The energy emitted by a source is in the form, of, 1)Photons, 2)electrons, 3)protons, 4)neutrons, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 74. The unit of Planck’s constant is equivalent to, that of, 1) energy, 2) angular momentum, 3) velocity, 4) force, 75. Hydrogen atom will be in its ground state,if, its electron is in, 1) any energy level, 2) the lowest energy state, 3) the highest energy state, 4) the intermediate state, 76. The wavelengths involved in the spectrum of, deuterium ( 1D2 ) are slightly different from that, of hydrogen spectrum, because, 1) the size of the nuclei are different, 2) the nucelar forces are different in two cases, 3) the masses of the two nuclei are different, 4) the attraction between electron and the nucelus, is different in the two cases, , 77. When an electron jumps from n1 th orbit to, n2 th orbit then the formula for energy radiated, out is, 1) E1 E2 h, 78., , 79., , 80., , 81., , 2) E2 E1 h, , 4) E2 2 E1 h, 3) E2 E1 h, Which of the following parameters are the, same for all hydrogen -like atoms and ions in, their ground states?, 1) radius of the orbit, 2) speed of the electron, 3) energy of the atom, 4) orbital angular momentum of the electron, The fine structure of hydrogen spectrum can, be explained by, 1) the presence of neutrons in the nucleus, 2) the finite size of nucleus., 3) the orbital angular momentum of electrons, 4) the spin angular momentum of electrons, The visible region of hydrogen spectrum was, first studied by, 1) Lyman, 2) Balmer 3) Pfund 4) Brackett, With the increase in quantum number the, energy difference between consecutive, energy levels, 1) remains constant, 2) decreases., 3) increases, 4)sometimes increases sometimes decreases, , NARAYANAGROUP, , ATOMIC PHYSICS, 82. Hydrogen atom does not emit X-rays because:, 1) its energy levels are too close to each other, 2) its energy levels are too far apart, 3) it has a very small mass, 4) it has a single electron, , 83. An electron makes transition from n= 3 , n=1, state in a hydrogen atom. The maximum, possible number of photons emitted will be, 1) 1, 2) 2, 3) 3, 4) 6, , ASSERTION & REASONING, Choose the correct answer :, 1) A and R are true and R is the correct, explanation of A, 2) A and R are true and R is not the correct, explanation of A, 3) A is true and R is false., 4) A is false but R is true, 84. Assertion (A) : Rydberg’s constant varies with, mass number of a given element., Reason (R): The reduced mass of the electron is, dependent on the mass of the nucleus only, 85. Assertion (A) : Study of discharge of electricity, through gases at low pressures resulted in the, discovery of cathode rays., Reason (R): Cathode rays are deflected by both, magnetic and electric fields and the direction of, deflection shows that they are negatively charged., 86. Assertion (A) : If electrons travel undeflected by, the electric field E and magnetic field B then the, velocity of electrons is given by V =, , E, B, , ., , Reason (R) : When both electric and magnetic, fields are applied simultaneously on electron beam., If force due to electric field is equal and opposite, to force due to magnetic field then they travel, undeflected., 87. Assertion (A) : In the Thomsons e/m experiment, of electrons, the specific charge of electrons is, independent of nature of the gas in the discharge, tube., , Reason (R): Charge of a body is quantized., 88. Assertion (A): Between any two given energy, levels, the number of absorption transition is always, less than number of emission transition., Reason (R): Absorption transitions start from the, lowest energy level only and may end at any higher, energy level. But emission transitions may start from, any higher energy level and end at any energy level, below it., 55
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ATOMIC PHYSICS, , MATCH THE FOLLOWING, 89. In each situation of Column I, a physical, quantity related to orbiting electron in, hydrogen-like atom is given.The terms ‘Z’and, ‘n’given in Column II have usual meaning in, Bohr’s theory. Match the quantities in Column, I with the terms they depend on it Column II, Column-I, a. Frequency of orbiting electron, b. Angular momentum of electron, c. Magnetic moment of electron, d.The average current due to orbiting of electron, Column-II, p. is directly proportional to n, q. is inversely proportional to n, r. is inversely proportional to n3, s.is independent of Z, 1)a -p,r, b -q,s, c -q,s, d -q,r, 2)a -q,s, b -p,r, d -q,r, c-q,s, 3) a-q,r, b-q,s, c-q,s, a-p,r, 4) a-p,r, b-q,r, c-q,s, d-q,s, 90. Column-I, a. Radius of orbit is related with atomic number(Z), b. Current associated due to orbital motion of, electron with atomic number(Z), c. Magnetic field at the center due to orbital motion, of electron related with Z, d. Velocity of an electron related with atomic, number(Z), Column-II, p. is proportional to Z, q. is inversely proportion to Z, r. is proportional to Z2, s. is proportional to Z3, 1) a-q b-r c-r d-p, 2) a-q b-s c-r d-p, 3)a-p b-r c-s d-q, 4) a-p b-s c-s d-q, 91. The spectral lines of hydrogen-like atom fall, within the wavelength range from 950A0 to, 1350A0 Then, match the following., Column-I, a. If it is atomic hydrogen atom and energy, , E 0.85 eV, b., c., d., 56, , If it is atomic hydrogen atom and energy atom and, energy E 3.4 eV, If it is double ionized lithium atom, then, If it is singly ionized helium atom, then, , JEE-ADV PHYSICS- VOL- V, Column-II, p. =1212A0 and it corresponds to transition from, 2 to 1, q. =134A0 and it corresponds to transition from 2, to 1, r., =303A0 and it corresponds to transition from 2, to 1, s. =970A0 and it corresponds to transition from 4, to 1, 1) a-s b-p c-q d-r, 2) a-s b-q c-p d-r, 3)a-p b-q c-r d-s, 4) a-q b-r c-s d-p, 92. Excitation energy of hydrogen atom is 13.6 eV., Match the following., Column-I, Column-II, a. Energy of second, p. 3.4 eV, excited state of hydrogen, b. Energy of fourth state, q 13.6eV of He+, c. Energy of first excited r. 1.5 eV state of Li+2, d. Energy of third excited s.None, state of Be+3, 1) a-r b-p c-s d-q, 2)a-r b-s c-p d-q, 3)a-s b-p c-q d-r, 4)a-q b-p c-s d-r, 93. Match the column–I with column–II, Column – I, Column – II, (a) Radius of orbit depends (p) Increase, on principal quantum, number as, (b) Due to orbital motion, (q) Decrease, of electrons, magnetic, field arises at the centre, of nucleus is proportional, to principal quantum, number as, (c) If electron is going from (r) Is proportional, lower energy level to higher, , to, , 1, n2, , energy level than velocity, of electron will, (d) If electron is going from (s) Is proportional, lower energy level to, to n2, higher energy level than, total energy of electron will, (t) Is proportional, to, 1) a-s b-t c-q d-p, 3)a-p b-q c-t d-s, , 1, n5, , 2) a-t b-s c-p d-q, 4) a-p b-q c-s d-t, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 94. Match the column–I with column–II, A certain amount of He+ ions are subjected to, excitation and subsequently the emission, spectrum for the same is observed. Column –, I shows four different series to which the, spectral lines belong and column – II shows, five different energies of the photons emitted, during the process, Column – I, Column – II, (a) Lyman series, (p) 1.33 eV, (b) Balmer series, (q) 12.09 eV, (c) Paschen series, (r) 51 eV, (d) pfund sereis, (s) 5.5 eV, (t) 3.87 eV, 1) a-r b-q c-s,t d-p, 2) a-q b-r c-p,t d-s, 3)a-r b-p c-q d-t,s, 4) a-s b-q c-p d-t, , 4., , 5., , 6., , C.U.Q - KEY, 1) 3, 8) 3, 15) 4, 22) 3, 29) 2, 36) 2, 43) 3, 50) 1, 57) 3, 64) 1, 71) 2, 78) 4, 85) 2, 92) 1, , 2) 1, 9) 1, 16) 2, 23) 3, 30) 1, 37) 1, 44) 3, 51) 2, 58) 1, 65) 2, 72) 4, 79) 4, 86) 1, 93) 1, , 3) 1, 10) 1, 17) 3, 24) 1, 31) 2, 38) 3, 45) 1, 52) 4, 59) 1, 66) 1, 73) 1, 80) 2, 87) 2, 94) 1, , 4) 2, 11) 1, 18) 3, 25) 2, 32) 2, 39) 2, 46) 2, 53) 2, 60) 2, 67) 4, 74) 2, 81) 2, 88) 1, , 5) 2, 12) 2, 19) 3, 26) 3, 33) 2, 40) 1, 47) 1, 54) 4, 61) 3, 68) 2, 75) 2, 82) 1, 89) 1, , 6) 1, 13) 4, 20) 4, 27) 3, 34) 4, 41) 2, 48) 4, 55) 4, 62) 4, 69) 3, 76) 3, 83) 3, 90) 1, , 7) 3, 14) 4, 21) 2, 28) 2, 35) 1, 42) 4, 49) 3, 56) 1, 63) 4, 70) 2, 77) 1, 84) 4, 91) 1, , MILLIKAN OIL DROP, EXPERIMENT, 7., , A water drop of mass 3.2 x 10-18 kg and, , 8., , carrying a charge of 1.6 1019 C is suspended, stationary between two plates of an electric, field. Given g=10 m/s2, the intensity of the, electric field required is, 1) 2 V/m 2) 200 V/m 3) 20 V/m 4) 2000 V/m, In a Millikan’s oil drop experiment, an oil drop, of mass 0.64 10-14 kg, carrying a charge, 1.6 10-19 C remains stationary between two, plates seperated by a distance of 5 mm. Given, g=9.8 m/s2; the voltage that must be applied, between the plates being, 1) 980 V, 2) 1960 V 3) 3920V 4) 2880V, , LEVEL-I (C.W), J.J.THOMSONS METHOD, 1., , 2., , 3., , An electron passes undeflected through, perpendicular electric and magnetic fields of, intensity 3.4 x 103 V/m and 2 x 10-3 Wb/m2, respectively. Then its velocity in m/s is, 1) 1.7 x 106 2) 6.8 x 106 3) 6.8 4) 1.7 x 108 m/s, The ratio of specific charges of an electron to, that of a hydrogen ion is, 1) 2:1, 2) 1:1, 3) 1: 1840 4) 1840:1, An - particle and a proton are subjected to, the same electric field, then the ratio of the, forces acting on them is, 1) 2: 3, 2) 1: 2, 3) 3 :2, 4) 2 :1, NARAYANAGROUP, , An electron is accelerated in an electric field, of 40 V cm-1. If e/m of electron is 1.76 X 1011, Ckg-1, then its acceleration is, 1) 14.0 x 1014 ms-2, 2) 14.0 x 1010ms-2, 3) 7.0 x 1010ms-2, 4) 7.04 x 1014ms-2, An electron beam moving with a speed of, 2.5×107 ms-1 enters into the magnetic field, directed perpendicular to its direction of, motion. The magnetic induction of the field is, 4×10-3 wb/m2. The intensity of the electric field, applied so that the electron is undeflected due, to the magnetic field is., 1) 104N/C 2) 105N/C 3) 107N/C 4) 103N/C, A particle carrying a charge moves, perpendicular to a uniform magnetic field of, induction B with a momentum p, then the, radius of the circular path is, 1) Be/p, 2) pe/B, 3) p/Be, 4) Bep, , ALPHA RAY SCATTERING, 9., , particles are projected towards the nuclei, , of the different metals, with the same kinetic, energy. The distance of closest approach is, minimum for, 1) Cu(Z=29), 2) Ag(Z=47), 3) Au(Z=79), 4) Pd(Z=46), 10. In Rutherford experments on ray, scattering the number of particles scattered, at 900 be 28 per minute. Then the number of, particles scattered per minute by the same foil, but at 600 are, 1) 56, 2) 112, 3) 60, 4) 120, 57
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ATOMIC PHYSICS, 11. For a given impact parameter (b), if the energy, increase then the scattering angle, 1) Decrease, 2) increase, 3) become zero, 4) become, , will, , BOHR’S MODEL OF ATOM, 12. Find the frequency of revolution of the electron, in the first stationary orbit of H-atom, 1) 6 1014 Hz, 2) 6.6 1010 Hz, 4) 6.6 1015 Hz, 3) 6.6 10 10 Hz, 13. Let the potential energy of a hydrogen atom, in the ground state be zero. Then its energy in, the first excited state will be, 1) 10.2eV 2) 13.6eV 3) 23.8eV 4) 27.2eV, 14. According to bohr model, the diameter of first, orbit of hydrogen atom will be, 1) 1.A0 2) 0.529A0 3) 2.25A0 4) 0.725A0, 15. The angular momentum of electron is J. If, e=charge fo electron, m = mass of electron,, then its magnetic moment will be, mJ, eJ, 2m, emJ, 2), 3), 4), 2e, 2m, eJ, 2, 16. The radius of shortest orbit in one electron, system is 18 pm.It may be., , 1), , 2) 12 H, , 1) 11H, , 3) He , , 4) Li , , 17. In the Bohr model of a hydrogen atom, the, centripetal force is furnished by the coulomb, attraction between the proton and the electron., If a0 is the radius of the ground state orbit, m, is the mass, e is the charge on the elctron and, 0 is the vacuum permittivity, the speed of the, electron is: [CBSE 1998], e, , 1) Zero, , 2), , 0 a0 m 3), , e, 40 a0 m 4), , 4 0 a0 m, e, , 18. The energy necessary to remove the electron, from n=10 state in hydrogen atom will be, 1)1.36 eV 2)0.0136 eV 3)13.6 eV 4) 0.136 eV, 19. The ratio of energies of first two excited states, hydrogen atom is, 1) 3/1, 2) 1/4, 3) 4/9, 4) 9/4, , ATOMIC SPECTRA, 20. The number of different wavelengths may be, observed in the spectrum from a hydrogen, sample if the atoms are excited to third excited, state is, 1) 3, 2)4, 3)5, 4)6, 58, , JEE-ADV PHYSICS- VOL- V, 21. The ratio of the frequencies of the long, wavelength limits of the Balmer and Lyman, series of hydrogen is, 1)27:5, 2)5:27, 3)4:1, 4)1:4, 22. When an electron jumps from higher orbit to, the second orbit in hydrogen ,the radiation, emitted out will be in ( R 1.09 107 m 1 ) :, 1)ultraviolet, 2) visible region, 3)infrared region, 4)X-ray region, 23. The energy required to seperate a hydrogen, atom into a proton and an electron is 13.6eV., Then the velocity of electron in a hydrogen, atom is, 1) 2.2 104 m / s, , 2) 2.2 102 m / s, , 3) 2.2 106 m / s, , 4) 2.2 1010 m / s, , LEVEL-I (C.W) - KEY, 1) 1, 8) 2, 15) 2, 22) 2, , 2) 4 3) 4 4) 4 5) 2 6) 3 7) 2, 9) 1 10) 2 11) 1 12) 4 13) 3 14) 1, 16) 4 17) 3 18) 4 19) 4 20) 4 21) 1, 23) 3, , LEVEL-I (C.W) - HINTS, E, B, , 1., , V, , 2., , specific charge of electron, q m, 1 2, specific charge of hydrogen ion q 2 m1, , 3., , F = Ee, , 4., , a, , Ee, m, , 5., , V, , E, B, , 6., , P mv, r , , 7., , mg = qE, , 8., , V, Eq mg , d, , 9., , r, , mv, Be, , , q mg, , , 1 q1q2, r q2, 4 0 ( KE ), , N, , 10. No. of particles scattered at angle is, , 1, , sin4 , 2, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 12. T , , 2 r, 3 108, 10, where V , ;, V, 137 r 0.53 10 m, , 13. P.E2 = , , 27.2, 27.2 =20.4 eV, 4, , ATOMIC PHYSICS, 3., , 4., , K .E1 13.6 eV , K .E2 3.4 eV, T .e2 20.4 3.4, , 23.8 eV, , 14. Diameter 2r0 =2 x 0.529 =1.058 A0, , 5., , magneticmoment, e, 15. angular momentum 2m, , n 2 rB, 0.53 A, , 0.18 A Z 3, 16. rn , Z, Z, 17., , 6., , 1 e 2 mv 2, , a0, 4 0 a02, , 18. Energy requried = E E10 , , 13.6, 0.136eV, 102, , 19. En 13.6 Z 2 / n 2, 20. N=, , 22., , MILLIKAN OIL DROP, EXPERIMENT, , n( n 1), : where n=4, 2, , 21. B , , 36, 4, : 2 , ,, 5R, 3R, , 7., , B 27, , :, c, 5, , For max :, n1 2 & n2 3;, , 23. Velocity of revolving electron is given as, , mv 2, 1 e2, , ,, r, 4o r 2, , e, v , 4o mr, , 8., , LEVEL-I (H.W), J.J.THOMSONS METHOD, 1., , 2., , A cathode emits 1.8 x 1017 electrons per, second and all the electrons reach the anode, when it is given a positive potential of 400 V., Given e=1.6x10-19C, the maximum anode, current is, 1) 2.88 mA 2) 28.8 mA 3) 7.2 mA 4) 6.4 mA, An electron of mass 9 x 10-31 kg moves with a, speed of 107 m/s. It acquires a K.E. of (in eV), 1) 562.50 2) 1125, 3) 1250, 4) 281.25, NARAYANAGROUP, , Two electron beams having velocities in the, ratio 1:2 are subjected to the same transverse, magnetic field. The ratio of the deflections is, 1) 1:2, 2) 2:1, 3) 4:1, 4) 1:4, The velocity of electrons accelerated by, potential difference of 1×104 V (The charge, of the electron is 1.6×10-19 C and mass is, 9.11×10-31kg) is, 1) 5.93×107 ms-1, 2) 2.94×107 ms-1, 3) 6.87×107 ms-1, 4) 3.98×107 ms-1, Cathode ray tube is operating at 5 KV. Then, the K.E. acquired by the electrons is, 1) 5 eV, 2) 5 MeV 3) 5 KeV 4) 5 V, A steam of similar negatively charged particals, enters an electrical field normal to the electric, lines of force with a velocity of 3 107 m / s ., The electric intensity is 1800 V/m. While, through a distance of 100 m, the electron beam, is deflected by 2mm. Then the specific charge, value of in C Kg-1 is, 1) 2 1010 2) 2 107 3) 2 1011 4) 2 104, , An oil drop having a mass 4.8×10-10g. and, charge 2.4×10-18C stands still between two, charged horizantal plates seperated by a, distance of 1cm. If now the polarity of the plates, is changed the instantaneous acceleration of, the drop is (in ms-2), (g=10ms-2), 1) 5, 2)10, 3) 20, 4) 40, In millkan’s oil drop experiment a charged, drop of mass 1.8 X 10-14 kg is stationary, between the plates. The distance between the, plates is 0.9cm and potential difference is 2000, V. The number of electrons in the drop is (g =, 10ms-2), 1) 2, 2) 4, 3) 5, 4) 1, , ALPHA RAY SCATTERING, 9., , An particle accelerated through V volt is, fired towards a nucleus.Its distance of closest, approach is r.If a proton is accelerated through, the same potential and fired towards the same, nucleus,the distance of closest approach of, proton will be:, 1) r, 2) 2r, 3) r/2, 4) r/4, 59
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 10. In ray scattering, the scattering angle for, impact parameter (b) to become zero is, 1) 0o, 2) 90o, 3) 180o, 4) 45o, 11. The impact parameter at which the scattering, o, , angle is 90 , z 79 and initial energy 10 MeV, is, 1) 1.137 1014 m, , 2) 1.137 1016 m, , 3) 2.24 1017 m, , 4) 2.24 1018 m, , BOHR’S MODEL OF ATOM, 12. When a hydrogen atom emits a photon of, energy 12.1 eV,its orbital angular momentum, changes by, 1) 1.051034 J s, , 2) 2.111034 J s, , 3) 3.161034 J s, 4) 4.221034 J s, 13. An electron is in an excited state in a hydrogen, like atom. It has a total energy 3.4 eV. The, kinetic energy of the electron is E and its de, broglie wavelength is ., , 18. If the potential energy of a H-atom in the, ground state be zero then its potential energy, in the first excited state will be, 1) 10.2 ev 2)20.4 eV 3) 23.8 eV 4)27.2 eV, , ATOMIC SPECTRA, 19. The value of wavelength radiation emitted due, to transition of electrons from n = 4 to, n = 2 state in hydrogen atom will be, , 3R, 16, 36, 5R, 2), 3), 4), 36, 3R, 5R, 16, 20. The maximum number of photons emitted by, an H-atom,if atom is excited to states with, principal quantum number four is:, [AIEEE-2012], 1)4, 2) 6, 3) 2, 4) 1, 21. For certain atom,there are energy levels, A,B,C corresponding to energy values, 1), , E A EB Ec . Choose the correct option if, 1 , 2 , 3 are the wavelength of radiations, corresponding to the transition from C to B,B, to A and C to A respectively:, , 1) E 6.8eV . 6.6 1010 m, 2) E 3.4eV , 6.6 1010 m, 3) E 3.4eV , 6.6 1011 m, 4) E 6.8eV , 6.6 1011 m, 14. The ionisation potential of hydrogen atom is, 1) 12.97V 2) 10.2V 3) 13.6V 4) 27.2V, 15. The energy of electron in an excited hydrogen, atom is -3.4 e V. Its angular momentum, according to bohr’s theory will be, h, h, 3h, 3, 2), 3), 4), , 2, 2, 2 h, 16. The velocity of an electron in its fifth orbit,if, the velocity of an electron in the second orbit, of sodium atom (atomic number=11) is v , will, be:, , 2) 3 , , 3) 1 2 3 0, , 4) 32 3 22, , 22. If 13.6 eV is the energy required to separate, a hydrogen atom into a proton and an electron, then its orbital radius is, 1) 5.3 1011 m, , 2) 5.3 1012 m, , 3) 7.6 1013 m, , 4) 7.6 10 14 m, , LEVEL - I (H.W) - KEY, , 1), , 22v, 5, 2, 1) v, 2), 3) v, 4) v, 5, 2, 5, 17. The ratio of the kinetic energy and the, potential energy of electron in the hydrogen, atom will be, 1)1:2, 2) -1:2, 3) 2:1, 4) -2:1, 60, , , 1 2, 1 2, , 1) 3 1 2, , 1) 2 2) 4, 3) 2 4) 1 5) 3 6) 1 7) 3, 8) 3 9) 1 10) 3 11) 1 12) 2 13) 2 14) 3, 15) 1 16) 4 17) 2 18) 2 19) 2 20) 2 21) 2, 22) 1, , LEVEL - I (H.W) - HINTS, 1., , 2., , i, , ne, t, , K.E. =, , 1 2, 2 mv , , , e , , , , in ev where e is charge of electron, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , y1 v2, , y2 v1, , 3., , LEVEL-II (C.W), J.J.THOMSONS METHOD, , 2V0 e, Where V0 is the P.D, m, K.E. = Ve Where V is P.D, v, , 4., 5., , 1., , 2, , 1 F 2 1 Eq x , y, t , , 2m, 2 m v , mg = Eq and F = (mg+Eq) ma = 2mg a=2g, , 6., 7., , V , Eq mg ne mg, d, 1 q1q2, KE =PE q1V 4 r, 0, Independent of charge of the particle projected, , 8., 9., , 11. b , , 12., , 2., , ze 2 cot / 2 , 4o E, , 1, 1 , E 13.6eV 2 2 , n, n, 1, 2 , , 13. T .E K .E; , , ;, , L , , h, ( n1 n2 ), 2, , 3., , h, 2mK .E, , 14. Eionisation E E1, 15. . En , , 13.6, nh, 3.4 n 2; L , h /, 2, n, 2, , 4., , 1, 16. n , n, 17. PE = -2 K.E, 18. P.E. 0 27.2 27.2 0, P.E . , , 27.2, 27.2 20.4, 4, , 1, 1 , 2 1, 19. RZ n 2 n 2 :, 1, 2 , 20. For maximum no. of spectral lines = N , , n 1, , 1, 1 1, 21. , 3 ?, 3, 1, 2, 22. Total energy of electron in hydrogan atom,, , E 13.6eV 13.6 1.6 1019 J, , , , e 2, 8o r, , r , , NARAYANAGROUP, , e2, 8o E, , 2, , Two ions having masses in the ratio 1:1 and, charges 1:2 are projected into uniform, magnetic field perpendicular to the field with, speeds in the ratio 2:3 .The ratio of the radii, of circular paths along which the two particles, move is, 1) 4:3, 2) 2:3, 3) 3:1, 4) 1:4, In Thomson’s experiment, a magnetic field of, induction 10-2 wb/m2 is used. For an undeflected, beam of cathode rays, a p.d. of 500 V is applied, between the plates which are 0.5 cm apart., Then the velocity of the cathode ray beam is, ...... m/s., 1) 4 x 107, 2) 2 x 107, 8, 3) 2 x10, 4) 107, A cathode ray beam is bent into an arc of a, circle of radius 0.02 m by a field of magnetic, induction 4.55 milli tesla. The velocity of, electrons is (given e = 1.6x10 -19 C and, m = 9.1 x 10-31 kg), 1) 2 x 107m/s, 2) 3 x 107m/s, 7, 3) 1.6x10 m/s, 4) 3.2 x 107m/s, When two electrons enter into a magnetic field, with different velocities, they deflect in, different circular parts, in such a way that the, radius of one path is double that of the other. 1, × 107 ms-1 is the velocity of electron in smaller, circle of radius 2×10-3 m. The velocity of, electron in the other circular path is :, 1) 4 × 107 ms-1, 2) 4 × 106 ms-1, 3) 2 × 107 ms-1, 4) 2 × 106 ms-1, , MILLIKAN OIL DROP, EXPERIMENT, , n, , 5., , A charged oil drop falls with terminal velocity, V0 in the absence of electric field. An electric, field E keeps it stationary. The drop acquires, additional charge q and starts moving upwards, with velocity V0. The initial charge on the drop, was, 1) 4q, 2) 2q, 3) q, 4) q/2, 61
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 6., , A charged oil drop is moving with a velocity, V1.As it acquires an additional charge it moves, up with the velocity V2 in the same electric, filed. It fall freely with a Velocity ‘V’ in the, absence of electric field . The ratio of the, charges before and after acquiring additional, charge is, V1 V, 1) V v, 2, , 1, , 2), , V2 2V1, V1 V2 V V1, 3) V V 4) 2V V, 2V, 2, 1, , ALPHA SCATTERING, 7., , A proton strikes another proton at rest with, speed Vo . Assume impact parameter to be, zero. Their closest distance of approach is, (mass of proton is m), , 13. The de-Broglie wave length of the electron in, the second Bohr orbit is (given r0=0.53A0), 1) 3.33A o, , ATOMIC SPECTRA, 14. The maximum wavelength of Brackett series, of hydrogen atom will be ______ A0, 1)35,890, 2)14,440 3) 62,160 4)40,477, 15. When a hydrogen atom emits a photon in going, from n = 5 to n = 1, its recoil speed is almost, 2) 2 102 m / s, 1) 104 m / s, 3) 4m / s, 4) 8 102 m / s, 16. An orbital electron in the ground state of, hydrogen has the magnetic moment 1 .This, orbital electron is excited to 3rd excited state, by some energy transfer to the hydrogen, atom. The new magnetic moment of the, , e2, e2, e2, 1), 2), 3), 4) zero, 4 mo 2, 0 mo 2, m0 2, 8., , A closest distance of approach of an particle, travelling with a velocity v towards, , electron is 2 , then, , Al13 nucleus is d. The closest distance of, approach of an alpha particle travelling with, velocity 4V towards Fe 26 nucleus is, 1) d/2, 2) d/4, 3) d/8, 4) d/16, , BOHR’S THEORY, 9., , The energy required to excite an electron from, n=2 to n=3 energy state is 47.2 e V. The charge, number of the nucleus, around which the, electron revolving will be, 1) 5, 2) 10, 3)15, 4) 20, 10. The debroglie wave length of an electron in, the first Bohr orbit is, 1) Equal to the circumference of the first orbit, 2) 1/2 th circumference of the first orbit, 3) 1/4 th circumference of the first orbit, 4) 3/4th circumference of the first orbit, 11. If the radius of first Bohr’s orbit is x ,then deBroglie wavelength of electron in 3rd orbit is, nearly:, 1) 2 x, 2)6 x, 3) 9 x, 4) x /3, 12. The angular momentum of the electron in third, orbit of hydrogen atom , if the angular, momentum in the second orbit of hydrogen, atom is L is, 1) L, 2)3L 3) (3/2) L, 4)2/3 L, 62, , 2) 6.66A o 3) 9.99A o 4) 1.06A o, , 1) 1 2 2, , 2) 21 2, , 3) 161 2, , 4) 41 2, , LEVEL-II (C.W) - KEY, 1) 1 2) 4 3) 3 4) 3 5) 3 6) 3 7) 2, 8) 3 9) 1 10) 1 11) 2 12) 3 13) 2 14) 4, 15) 3 16) 4, , LEVEL-II (C.W) - HINTS, 1., , r1 m1 v1 q2, , , r2 m2 v2 q1, , 2., , V, , E, B, , 3., , v, , rBe, m, , 4., , r, , mv, Bq, , 6 rV0 mg ; Eq mg, , 5., , E q1 q 6 rV0 mg 2mg 2 Eq, , 6., , Eq1 - mg = 6aV1 1, , mg 6aV, , Eq2 mg 6 aV2 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 7., , ATOMIC PHYSICS, 2., , Use:LCLM and conservation of energy., ie: m0 = 2m (1), 2, , 2, , e2, mo, 1 , 2 m o , 2, 2 2 4 r, , 8., , 1, 1, q q, mv 2 , 1 2 ;, 2, 4o, r, , 9., , E E3 E2 , , r, , z, , 2, , v, , 2, , 3., , 13.6 Z 2 13.6 Z 2, , 47.2, 4, 9, , Z ?, , 10. mvr , , nh, h, ; 2 , ; 2 r , 2, mv, , 11. mvr , , r0 n 2 , nh , r , r0 x, Z , 2 , , 2 r n , , 12. L , , 4., , nh, 2, , 2r 2ro n 2, , 2ro n, n, n, , 1, 1 1, 2, 14. RZ 2 n 2 Where n1 4; n2 5, 2 , n1, 15. From law of conservation of momentum, , m mass of proton, wave length of emitted photon, qv, r 2, 16. niA iA , , 2r, n ; n , , 5., , An -Particle of energy 5 MeV is scattered, through 1800 by a fixed uranium nuclens.The, closest distance is in the order of:, 1) 1Ao, 2) 10 10 cm 3) 10 12 cm 4) 10 16 cm, , 6., , An alpha nucleus of energy, , 1, n, , A proton and an -particle enter a magnetic, field in a direction perpendicular to it. If the, force acting on the proton is twice that acting, on the -particle, the ratio of their velocities, is, 1) 4 : 1, 2) 1. : 4, 3) 1 : 2, 4) 2 : 1, NARAYANAGROUP, , 1, mv2 bombards, 2, a heavy nuclear target of charge Ze. Then the, distance of closest approach for the alpha, nucleus will be proportional to (AIEEE-2006), , 1), , LEVEL-II (H.W), J.J.THOMSONS METHOD, 1., , In the Millikan’s experiment, the oil drop is, subjected to a horizontal electric field of 2, N/C and the drop moves with a constant, velocity making an angle of 450 with the, horizontal. If the weight of the drop is W, then, the electric charge, in coulomb, on the drop is, 1) W, 2) W/2, 3) W/4, 4) W/8, , ALPHA SCATTERING, , h, h, m p , , m, , z, n2, v, , &, r, , where, Mvr, n, z, , 3) 1 : 1 : 1, 4) 2 : 1 : 1, An electron starts from rest and travels, 0.9 m in an electric field of 200 V/m. After this,, it enters a magnetic field at right angles to its, direction of motion. If the radius of circular, path of the electron is 9 cm, the magnetic field, induction is (Given e=1.6 x 10-19C,, m=9x10-31kg), 1) 5 x 10-4 wb/m2, 2) 5 x 10-5 wb/m2, 3) 5 x 10-3 wb/m2, 4) 5 x 10-2 wb/m2, , MILLIKAN OIL DROP, EXPERIMENT, , 2 r, 6 x, n, , 13. 2r n ; , , A proton, a deuteron and an -particle are, accelerated through the same p.d. of V volt., The velocities acquired by them are in the ratio, 1) 1 : 1 : 2, 2) 1 : 2 : 1, , 1, v, , 2), , 1, Ze, , 3) v 2, , 4), , 1, m, , BOHR’S THEORY, 7., , In the Bohr model of hydrogen atom, the ratio, of the kinetic energy and total energy of, electron in the nth quantum state will be, 1) 1, 2) -1, 3) 2, 4) -12, 63
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 8., , The number of revolutions done by an, electron ‘e’ in one second in the first orbit of, hydrogen atom is, , LEVEL-II (H.W.) - KEY, 1) 1, 8) 1, , 2) 4, 9) 4, , 3) 1 4) 2 5) 3 6) 4 7) 2, 10) 1 11) 3 12) 1 13) 2 14) 1, , 13, , 1) 6.57x1015 2) 6.57 x 10 3) 1000 4) 6.57 x 1014, , LEVEL-II (H.W.) - HINTS, , 10, , 0.5110 , m is the radius of smallest, If , 4, , , electron orbit in hydrogen like atom ,then this, atom is :, 1) hydrogen atom 2) He 3) Li 2 4) Be3, 10. In Bohr’s orbit of hydrogen atom m kg is mass, of an electron and e coulomb is the charge on, it .The ratio (in SI units) of magnetic dipole, moment to that of the angular momentum of, electron is:, 1) e/2m, 2) e/m, 3) 2e/m 4)2e/3m, 11. In a sample of hydrogen like atoms all of which, are in ground state, a photon beam contain, photons of various energies is passed. In, absorption spectrum, five dark lines are, observed. The number of bright lines in the, emission spectrum will be(assume that all, transitions take place), 1) 21, 2) 10, 3) 15, 4) 12, 9., , ATOMIC SPECTRA, , 1., , F Bqv, , 2., , v, , 3., , Eq, Eq, 2, V 2 U 2 2as ; V 2 m S V 2 m S, r, , first line of the lyman series,and f3 be the, frequency of the series limit of the balmer, series, ., 1) f1 f 2 f 3, 2) f 2 f1 f 3, 1, 3) f3 f1 f 2 , 4) f1 f 2 f 3, 2, 13. Ratio of difference of spacing between the, energy levels with n=3 and n=4 and the spacing, between the energy levels with n=8 and n=9, for a hydrogen like atom or ion is, 1) 0.71 2) 0.41, 3) 2.43 4) 14.82, 14. A stationary hydrogen atom emits photon, corresponding to the first line of Lyman series., If R is the Rydberg’s constant and M is the, mass of the atom, then the velocity acquired, by the atom is (neglect energy absorbed by, the photon), 3Rh, 4M, Rh, 4M, 1), 2), 3), 4), 4M, 3Rh, 4M, Rh, 64, , mV, mV, B, Bq, rq, Eq, 45°, , 4., , Tan 450 =, , Eq, w, , Eq = w, W, , 5., , 1, qq, m o 2 1 2, 2, 4 o r, , 6., , 1, 1 2e ze , mv 2 , 2, 4o, r, , 12. Let f1 be the frequency of the series limit of, the lyman series, f 2 be the frequency of the, , 2eV, e, v, m, m, , 7., , 8., , 13.6 Z 2, K .E , n2, 13.6Z 2, T .E , n2, , K .E T .E, T .E K .E, , 1 (t ) n.2 r1 n ?, , 9. rn , , n2, ro :, Z, , er 2 w, 10. Magnetic dipole moment =, 2, 2, Angular momentum = mr w, 11., , n n 1, 2, , Here n 6, , 12. f relates n 1; n , 1, 1, 2, f 2 relates n1 1; n2 2, f 3 relates n1 2 : n2 , , r4 r3 42 32, , 0.41, 13., r9 r8 92 82, 14., , p mv ;, , h, h 1 1 3R, mv v & , , m 4, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , LEVEL-III, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , Magnetic moment due to the motion of the, electron in nth energy state of hydrogen atom, is proportional to, 1) n, 2) n0, 3) n5, 4) n3, The ratio between total accelertion of the, electron in singly ionized helium atom and, hydrogen atom (both in ground state) is, 1) 1, 2) 8, 3) 4, 4) 16, The shortest wavelength of the Brackett series, of a hydrogen like atom (atomic number =Z), is the same as the shortest wavelength of the, Balmar series of hydrogen atom. The value, of Z is, 1) 2, 2) 3, 3) 4, 4) 6, According to Bohr’s theory of hydrogen atoms,, the product of the binding energy of the, electron in the nth orbit and its radius in the n, the orbit, 1) is proportional to n2, 2) is inversely proportional to n3, 3) has a constant value of 10.2eV-A0, 4) has constant value 7.2 eV-A0, If an electron drops from 4 th orbit to 2nd, orbit in an H-atom,then, 1) it gains 2.55 eV of potential energy, 2) it gains 2.55 eV of total energy, 3) it emits a 2.55 eV electron, 4) it emits a 2.55 eV photon, The energy of an atom or ion in the ground, state is -54.4 eV . It may be:, 1) He 2) Li 2 3) hydrogen 4)deuterium, An atom absorbs 2 eV energy and is excited, to next energy state. The wavelength of light, absorbed will be, 1) 2000 A, 2) 4000 A ., 3) 8000 A, 4) 6206 A, When an electron in the hydrogen atom in, ground state absorbs a photon of energy 12.1, eV, its angular momenturm, 1) decreases by 2 .1 1 1 0 3 4 J s, 2) decreases by 1.055 10 34 Js, 3) increases by 2 .1 1 1 0 3 4 J s, 4) increases by 1.055 10 34 Js, NARAYANAGROUP, , ATOMIC PHYSICS, 9., , Magnetic field at the centre (at nucleus) of, the hydrogen like atoms (atomic number =Z), due to the motion of electron in nth orbit is, proportional to, , n3, n4, Z2, Z3, 1) 5, 2), 3) 3, 4) 5, Z, Z, n, n, 10. A neutron moving with a speed v makes a, head on collision with a hydrogen atom in, ground state kept at rest. The minimum, kinetic energy of neutron for which inelastic, collision will take place is (assume that mass, of proton is nearly equal to the mass of, neutron), 1) 10.2 eV 2) 20.4eV 3) 12.1eV 4) 16.8ev, 11. A charged particle is moving in a uniform, magnetic field in a circular path. The energy, of the particle is doubled. If the initial radius, of the circular path was R, the radius of the, new circular path after the energy is doubled, will be, 3) 2R 4) R/ 2, 1) R/2, 2) 2R, 12. An electron in hydrogen atom after absorbing, an energy photon jumps from energy state, n1to n2. Then it returns to ground state after, emitting six different wavelengths in emission, spectrum. The energy of emitted photons is, either equal to, less than or greater than the, absorbed photons. Then n1 and n2 are, 1) n2 4, n1 3, 2) n2 5, n2 3, 3) n2 4, n1 2, 4) n2 4, n1 1, 13. The photon radiated from hydrogen, corresponding to 2nd line of Lyman series is, absorbed by a hydrogen like atom ‘X’ in 2nd, excited state. As a result, the hydrogen-like, atom ‘X’ makes a transition of nth orbit., 1) X=He+, n=4, 2) X=Li++,n=6, 3) X=He+, n=6, 4) X=Li++,n=9, 14. In a hypothetical system, a particle of mass m, and charge -3q is moving around a very heavy, particle charge q. Assume that Bohr’s model, is applicable to this system. Then velocity of, mass m in first orbit is, 1), , 3q 2, 3q 2, 2) 4 h, 2 0 h, 0, , 3q, 3q, 3) 2 h 4) 4 h, 0, 0, , 65
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ATOMIC PHYSICS, 15. Consider a hydrogen-like atom whose energy, in n th excited state is given by E n , , 16., , 17., , 18., , 19., , 20., , 13.6 Z, n2, , When this excited atom makes a transition, from excited state to ground state, most, energetic photons have energy Emax= 52.224, eV and least energetic photons have energy, Emin= 1.224eV. Find the atomic number of atom, and the state of excitation., 1) Z=2; n=5, 2) Z=2, n=4, 3) Z=3, n=6, 4) Z=4, n =6, 29 electrons are removed from Zn atom, (Z=30) by certain means .The minimum energy, needed to remove the 30th electron ,will be:, 1)12.24 keV 2)408 keV 3)0.45 keV4)765keV, The ionisation energy of Li 2 atom in ground, state is :, 1) 13.6 x 9 eV, 2) 13.6J, 3) 13.6 erg, 4)13.6x10 19 J, A photon of energy 15 eV collides with H-atom., Due to this collision, H-atom gets ionized .The, maximum kinetic energy of emitted electron, is :, 1)1.4 eV, 2) 5 eV, 3)15eV, 4) 13.6eV, Monochromatic radiation of wavelength is, incident on a hydrogen sample in ground, state.Hydrogen atoms absorb a fraction of, light and subsequently emit radiation of six, different wavelengths. Find the value of ., 1) 80 nm, 2) 97.5 nm, 3) 105 nm, 4) 60 nm, The figure shows energy levels of a certain, atom, when the system moves from level 2E, to E, a photon of wavelength is emitted. The, wavelength of photon produced during its, transition from level 4/3 E to E level is:, 2E, , (4/3)E, E, , O, , 1) 3 , 66, , 2, , 2)3/4 , , 3) /4, , 4)2 , , JEE-ADV PHYSICS- VOL- V, 21. When the electron in hydrogen atom jumps, from the second orbit to the first orbit, the, wavelength of the radiation emitted is . When, the electron jumps from the third to the first, orbit, the wavelength of the radiation emitted, is:, 4, , 9, 4, , 27, , 32, , 2) 9 , 3) 32 , 4) 27 , 22. The ratio of the largest to shortest wavelengths, in Balmer series of hydrogen spectra is:, 1) , , 1), , 25, 9, , 2), , 17, 6, , 3), , 9, 5, , 4), , 5, 4, , 23. An electron in a hydrogen atom makes a, transition n1 n2, where n1 and n2 are principal, quantum numbers of the states. Assume the, Bohr’s model to be valid. The time period of, the electron in the initial states is eight times, n1, , to that of final state. What is ratio of n, 2, 1) 8:1, 2) 4:1, 3) 2:1, 4) 1:2, 24. Any radiation in the ultra violet region of, Hydrogen spectrum is able to eject, photoelectrons from a metal. Then the, maximum value of threshold frequency of the, metal is, nearly, 1) 3.3x1015 Hz, 2) 2.5x1015 Hz, 3) 4.6 x 1014 Hz, 4) 8.2x1014 Hz, 25. A hydrogen atom emits a photon corresponding, to an electron transition from n=5 to n=1. The, recoil speed of hydrogen atom in nearly equal, to, 1) 10-4 m/s, 2) 2x10-2 m/s, 3) 4 m/s, 4) 8x10-2 m/s, 26. The wave number of energy emitted when, electron jumps from fourth orbit to second orbit, in hydrogen in 20,497 cm-1. The wave number, of energy for the same transition in He+ is, 1) 5,099 cm-1, 2) 20,497 cm-1, -1, 3) 40,994 cm, 4) 81,988 cm-1, 27. In a Bohr atom the electron is replaced by a, particle of mass 150 times the mass of the, electron and the same charge. If a0 is the radius, of the first Bohr orbit of the orbital atom, then, that of the new atom will be, 1) 150 a0, , 2) 150 a0, , a0, , a0, 150, 150, NARAYANAGROUP, , 3), , 4)
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JEE-ADV PHYSICS- VOL- V, 28. If the wavelength of first member of Balmer, series of hydrogen spectrum is 6564 A0 , the, wavelength of second member of Balmer, series will be:, 1) 1215 A0, 2) 4848 A0, 3) 6050 A0, 4) data given is insufficient to calculate the value, 29. A hydrogen like atom (atomic number Z) is in, a higher excited state of quantam number ‘n’., This excited atom can make a transition to the, first excited state by emitting a photon of, energy 27.2eV. Alternatively the atom from, the same excited state can make a transition, to second excited state by emitting a photon, of energy 10.20 eV. The value of n and z are, given (Ionization energy of hydrogen atom is, 13.6eV), 1) n = 6 and z =3, 2) n = 3 and z = 6, 3) n = 8 and z = 4, 4) n = 4 and z = 8, 30. Photons from n=2 to n=1 in Hydrogen atom is, made to fall on a metal surface with work, function 1.2eV. The maximum velocity of photo, electrons emitted is nearly equal to, 1) 6x105 m/s, 2) 3x105 m/s, 5, 3) 2x10 m/s, 4) 18x105 m/s, 31. Let 1be the frequency of the series limit of, the Lyman series and 2 be the frequency of, the first line of the Lyman series and 3 be, the frequency of the series limit of Balmar, series, then, 1) 1- 2= 3, 2) 2- 1= 3, 3) 2 3= 1+ 2, 4) 1+ 2= 3, 32. a)Find the wavelength of the radiation required, to excite the electron in Li from the first to, the third Bohr orbit,, b) How many spectral lines are observed in, the emission spectrum of the above excited, system?, 1) 108.8 eV, 3, 2) 13.6 eV, 4, 3) 54.4 eV, 2, 4) 10.2 eV, 3, 33. Find the wavelengths in a hydrogen spectrum, between the range 500nm to 700nm., 1) 540 nm 2) 580 nm 3) 654 nm 4) 696 nm, 34. The largest wavelength in the ultraviolet, region of the hydrogen spectrum is 122 nm., The smallest wavelength in the infrared region, of the hydrogen spectrum (to the nearest, integer) is (IITJEE 2007), 1) 802 nm 2) 823 nm 3) 1882 nm 4)1648 nm, NARAYANAGROUP, , ATOMIC PHYSICS, 35. The electric potential between a proton and, an electron is given by V V0l n, , r, , where r0, r0, , is a constant. Assuming Bohr’s model to be, applicable, write variation of rn with n, n being, the principal quantum number, 1) rn n 2) rn 1/ n 3) rn n 2 4) rn 1/ n 2, 36. If elements of quantum number greater than, n were not allowed, the number of possible, elements in nature would be ?, 2, , n n 1 , , 2) , 2 , 1, 1, 3) n n 1 2n 1, 4) n n 1 2n 1, 3, 2, 37. Magnetic field at the centre (at nucleus) of, the hydrogen like atoms (atomic number z ), due to the motion of electron in nth orbit is, praportional to, 1, 1) n n 1, 2, , n3, 1) 5, z, , n4, 2), z, , z2, 3) 3, n, , z3, 4) 5, n, , 38. The recoil speed of a hydrogen atom after it, emits a photon in going from n 5 , state to, n 1 state is (in ms-1), 1) 4.718, 2) 7.418 3) 4.178 4) 7.148, 39. The binding energy of an electron in the ground, state of He atom is E0 24.7eV . The energy, required to remove both the electrons from, the atom is, 1) 24.6eV 2) 79.0eV 3) 54.4eV 4) none of these, 40. In hydrogen atom, the radius of nth Bohr orbit, , rn, log, , is Vn . The graph between, r1, , , and log n, , , will be, 1), , 2), , r , log n , r1 , , r , log n , r1 , , log n, , log n, , 3), , 4), , r , log n , r1 , , log n, , r , log n , r1 , , log n, , 67
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , LEVEL - III - KEY, , 41. In hydrogen atom, the area enclosed by nth, , An , log, and, orbit is An . The graph between, A1 , log n will be, 2), A , log n , A1 , , 1), , A , log n , A1 , , 1) 1, 8) 3, 15) 1, 22) 3, 29) 1, 36) 4, 43) 2, , 2) 2, 9) 4, 16) 1, 23) 3, 30) 4, 37) 4, , 3) 1, 10) 2, 17) 1, 24) 1, 31) 1, 38) 3, , 4) 4, 11) 2, 18) 1, 25) 3, 32) 1, 39) 2, , 5) 4, 12) 3, 19) 2, 26) 4, 33) 3, 40) 1, , 6) 1, 13) 4, 20) 1, 27) 4, 34) 2, 41) 1, , 7) 4, 14) 1, 21) 3, 28) 2, 35) 1, 42) 1, , LEVEL - III - HINTS, O, , O, , log n, , log n, , 1., , 4), A , log n , A1 , , 3), A , log n , A1 , , O, , Magnetic Moment angular momentum, , O, , log n, , Magnetic Moment, e, , Angular Momentum 2 m, , nh , M n L 2 , , , , log n, , 42. An electron in the ground state of hydrogen, atom is revolving in anticlockwise direction, in the cicular orbit of radius R. The atom is, placed in a uniform magnetic induction B such, that the plane normal of electron orbit makes, an angle 300 with B, as shown in figure. The, torque experienced by electron willbe, , 2, , 3, , a1 2 , v2 a z , or a z 3 8, 2. a , a2 1 , 1/ z , r, 3., , 13.6 , Z 13.6, or Z=2, , ; 2, 16 4 Z 4, , 4., , En , , 2, , B, , 1, and rn n 2 ; En rn is independent of n, n2, , Hence, E1r1 13.6eV 0.530A = 7.2eV A, =constant, 0, , n, , v, 30°, , ehB, eh, eB, hB, 2), 3), 4), 8m, 8Bm, 8mh, 8em, 43. If we assume only gravitational attraction, between proton and electron in hydrogen atom, and the Bohr’s quantization rule to be followed,, then the expression for the ground state, energy of the atom will be (the mass of proton, is M and that of electron is m), , 1), , 1), , 3), 68, , G 2M 2m2, h2, 22 GM 2m3, h2, , 2), , 4), , 5., , En , , 6., , E, , 7., 8., , hc, ?, , After absorbing a photon of energy 12.1 eV, electron jumps from ground state (n=1) to second, excited state (n=3).Therefore change in angular, , h, momentum L L3 L1 3 , 2, , , , h2, G 2M 2n 2, , 13.6z 2, n2, , E, , 2 2 G 2 M 2 m3, h2, , 13.6 Z 2, eV ; E E4 E2, n2, , 9., , h, h, , , 2 , , 6.6 10 34, J s 2.11 10 34 J s, 3.14, , 0 I n, In fn , . Bn 2r (or) Bn , rn, rn, n, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, Bn, , v / r , n n , rn, , vn, , rn , , 2, , ;, , , , ATOMIC PHYSICS, , Z / n, , n, , 2, , /Z, , 2, , Z3, 5, n, , 10. Let v=speed of neutron before collision, v1= speed of neutron after collision, v2= speed of proton or hydrogen atom after, collision and E=energy of exitation., From conservation of linear momentum, ....(1), mv mv1 mv2, From conservation of energy, 1 2 1 2 1 2, mv mv1 mv2 E, 2, 2, 2, , ....(2), , 16. After removing 29 electrons,Zn atom will become, Zn 29 E= 13.6 Z 2 eV=12,240 eV =12.24KeV, 17. (E)= Li 2 ; E Z 2 ( E ) H ; E 9 13.6 eV, 18. Energy of photon ionization energy + K .Emax =15ev, -13.6ev=1.4ev, 19. The atom is excited to n = 4 state, , hc, E4 E1 12.75 eV 97.5nm, , hc, hc 4, E, 20. 2E E E 1 ; 3 E E 3 2, 1, 2, , 2 E 3, , 1 E 1, 3, , from 1 and 2 As v1 v2 must be real, v2 4, , , , E, 0, m, , 1 2, mv 2 E 2 10.2 20.4 eV, 2, , 11. r KE, , , , r2, , r1, , KE2, KE1, , 12. From n2 4, six lines are obtained in emission, spectrum. Now: E4 2 Eabsorbed, , 1, 1, 1 , 2, 21. RZ 2 n 2 , 2 , n1, 1, 1, 1 , 2, 22. RZ 2 n 2 Where n1 2; n2 3, 2 , max, n1, 1, 1, 1 , RZ 2 2 2 Where n 2; n , 1, 2, min, n1 n 2 , 23. Time period of revolution of electron in the nth orbit, is T n 3, , E43 Eabsorbed and E41E31 , E21 Eabsorbed, Hence, n1 2 and n2 4, 13. Energy of nth state in hydrogen is same as energy, of 3nth state in Li++. 3 1 transition in H would, give same energy as the 9 3 transition in Li++., 2, , 14., , 2, , mv, 1 3q, , r, 4 o r 2, , ; mvr , , nh, 2, , 3q 2, using above equations, putting n = 1; v , 2o h, 15. Maximum energy is liberated for transition En 1, and minimum energy for En En 1, , Hence, , 1, 1 , E (Rhc) Z 2 2 2 , n1 n2 , , NARAYANAGROUP, , 24. , 25. v , , c, , , h, 1, 1, ;, R 1 , m , 25 , , 1, 1 , 2, , , RZ, , ; Z 2, , 26., 2, 2 , n1 n 2 , 27., , rn, , 1, m, , 1, 1, 1 , 2, 28. RZ 2 n 2 for balmer series, 2 , n1, n1=2 and n2= 3,4,5...., , 1, , 1, , 2, 29. E1 27.2 (13.6) 2 z, 4 n , , 1 1 , E2 10.2 (13.6) 2 z 2, 9 n , 69
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JEE-ADV PHYSICS- VOL- V, , LEVEL-IV, COMPREHENSION TYPE, Passage-I:, A particle of charge equal to that of an, electron, -e, and mass 208 times the mass of, electron (called -meson)moves in a circular, orbit around a nucleus of charge+3e.(Take the, mass of the nucleus to be infinite). Assuming, that Bohr model of the atom is applicable to, this system:, 1. Derive an expression for the radius of the, nth Bohr orbit, , 0n2h2, 1), 208 me e 2, , 0n2h 2, 2), 3 me e 2, , 0n2h2, 3), 624 me e 2, , 0n2 h2, 4), 64 me e 2, , 2., , Find the value of n for which the radius of the, orbit is approximately the same as that of the, first Bohr orbit for the hydrogen atom., 1) 10, 2)15, 3) 25, 4) 30, 3. Find the wavelength of the radiation emitted, when the meson jumps from the third orbit, to the first, 1) 0.4500 A0, 2) 0.5500 A0, 3) 0.6500 A0, 4) None of these, (Rydberg’s constant=1.097x107m-1), Passage-II, Phtoelectrons are emitted when, 4000 A0 radiation is incident on a surface of, work function 1.9eV. These photoelectrons, pass through a region has a-particles to form, H e+ ion, emitting a single photon in this process, He+ ions thus formed are in their fourth, excited state., 4. Energy of the fourth excited state is approx, 1) -4.2eV 2) -2.2eV 3) -3.2eV 4) -1.2eV, 5. Energy released during the combination of He+, ions is., 1) 5.38 eV 2) 3.38eV 3) 2.38eV 4) 1.38eV, 6. Energy of emitted photon in range of 3eV&, 4eV after combination is, 1) 3.86eV, 2) 3.24eV, 3) 3.29eV, 4) 5.24eV, NARAYANAGROUP, , ATOMIC PHYSICS, Passage - III:, A sample of hydrogen gas in its ground state, is irradiated with photons of 10.2eV energies., The radiation from the above sample is used, to irradiate two other samples of excited, ionized He + and excited ionized Li 2+ ,, respectively. Both the ionized samples absorb, the incident radiation., 7. How many spectral lines are obtained in the, spectra of Li2+, 1) 10, 2) 15, 3) 20, 4) 17, 8. Which is the smallest wavelength that will be, observed in spectra of He+ ion?, 1) 24.4nm 2) 28.8nm 3) 22.2nm 4) 30.6nm, 9. How many spectral lines are observed in, spectra of He+ ion?, 1) 2, 2) 4, 3) 6, 4)8, 10. Which is the smallest wavelength observed in, the spectra of Li2+?, 1) 8.6nm, 2) 10.4nm 3) 12.8nm 4)14.6nm, 11. Consider the spectral line resulting from the, transition n = 2 to n = 1 in the atoms and the, ions given below. The shortest wavelength is, produced by, 1) hydrogen atom, 2) deuterium atom, 3) singly ionized helium, 4) doubly ionized lithium, 12. A gas of monoatomic hydrogen is bombared, with a stream of electron that have been, accelerated from rest through a potential, difference of 12.75 volt. In the emission, spectrum, one cannot observe any line of, 1) Lyman series, 2) Balmer series, 3) Paschen series, 4) Pfund series, 13. An electron jumps from the 4th orbit to 2nd orbit, of hydrogen atom. Given the Rydberg’s, constant r = 105 cm–1, the frequency in hertz, of the emitted radiation will be, 3, , 5, , 9, , 15, , 1) 16 10, , 3, , 15, , 2) 16 10, 3, , 3) 16 10, 4) 1015, 4, 14. With increasing quantum number, the energy, difference between adjacent energy levels in, atoms, 1) decreases, 2) increases, 3) remains constant, 4) decreases for low Z and increases for high Z, 71
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 15. Which of the following statements are true, regarding Bohr’s model of hydrogen atom?, (I) orbiting speed of electron decreases as it, shifts to discrete orbits away from the nucleus, (II) radii of allowed orbits of electron are, proportional to the principal quantum number, (III) frequency with which electron orbits around, the nucleus in discrete orbits is inversely, proportional to the principal quantum number., (IV) binding force with which the electrons are, bound to the nucleus increases as it shifts to, outer orbits., Select the correct answer, 1) I and III, 2) II and IV, 3) I, II and III, 4) II, III and IV, 16. A Hydrogen atom and a Li++ ion are both in, the second excited state. If lH and lLi are their, respective electronic angular momenta, and, EH and ELi their respective energies, then, 1) lH > lLi and |EH|>|ELi| 2) lH = lLi and |EH|<|ELi|, 3) lH = lLi and |EH|>|ELi| 4) lH < lLi and |EH|<|ELi|, 17. Consider the spectral line resulting from the, transition n = 2 n = 1 in the atoms and ions, given below, the shortest wavelength is, produced by, 1) Hydrogen atom 2) deuterium atom, 3) Singly ionized helium4) doubly ionized lithium, 18. For hydrogen like system,, (I) Ratio of magnetic moment to angular, momentum is e/2m., (II) energy of the electron is directly proportional, to 1/n2., (III) angular momentum varies inversely with n., (IV) radius of an orbit is inversely related to n., 1) I, II, 2) II, IV, 3) I, II, IV 4) III, IV, 19. As per Bohr model,, (I) minimum energy required to remove an, electron from ground state of doubly ionized, Li atom(z = 3) is 122.4 eV., (II) energy of transition n = 3 to m = 2 is less, than that of m = 2 to n = 1., (III) minimum energy required to remove an, electron from ground state of singly-ionised, He atom (z = 2) is 27.2 eV., (IV) A transition from state n = 3 to n = 2 in a, hydrogen atom results in U-V radiations, 1) II, III, IV, 2) I, II, 3) II, III, 4) I, III, IV, 72, , 20. Assertion: A discharge tube appears dark,, when evacuated to very low pressures., Reasoning: No colour is left at such low, pressures., 1) If both assertion and reason are true and reason, is the correct explanation of assertion., 2) If both assertion and reason are true but reason, is not the correct explanation of assertion., 3) If assertion is true and reason is false., 4) If assertion is false but reason is true, 21. Assertion: An electron in hydrogen atom, passes from n = 4 to n = 1 level. The maximum, number of photons that can be emitted is 6., Reasoning: maximum number of photons, omitted can only be 4., 1) If both assertion and reason are true and reason, is the correct explanation of assertion., 2)If both assertion and reason are true but reason, is not the correct explanation of assertion., 3)If assertion is true and reason is false., 4)If assertion is false but reason is true, , LEVEL - IV - KEY, 1) 3 2) 3 3) 2 4) 3 5) 2 6) 2 7) 2, 8) 1 9)3, 10) 2 11) 4 12) 4 13) 3 14) 1, 15) 1 16) 2 17) 4 18) 1 19) 2 20) 3 21) 3, , LEVEL - IV - HINTS, n 2 h 2 o, ; Put Z 3; m 208 me, me 2 Z, , 1., , rn , , 2., , rn ao , , n 2 h 2 o, o h 2, , n= 25, 624 me e 2 me e 2 , , hc, E, , hc, Z2, 1.18 eV ; E 13.6 eV 2, 4. K max , , n, 3., , 5., 6., 7., 8., , Difference of energy when photoelectron recombines, with particle =1.18-(-2.2)=3.38eV, For de-excitation from n = 5 to n = 3; E= 3.86 eV, Li+2is excited from n 3 to n 6 So, N 15, He+is excited from n = 2 to n = 4, , hc, 24.4 nm, E, 4 x (4 1), N, 6, 9., 2, hc, 10.4 nm, 10. min , E, , min , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, ejecting the photoelectrons, the ratio of the, number of photoelectrons from beam A to that, from B is, , LEVEL-V, SINGLE ANSWER TYPE QUESTIONS, 1., , 2., , A hydrogen atom in ground state is moving, with initial kinetic energy K. It collides head, on with a He+ ion in ground state kept at rest, but free to move. The minimum value of K so, that both the particles can excite to their first, excited states is, (take mass of He+ ion =, 4 times the mass of hydrogen atom), A) 63.75eV B) 51 eV, C) 54.4eV D) 13.05eV, Given X-ray spectrum is for a Coolidge tube, having acceleratin g p oten tial V. If, accelerating potential is decreased to V/4,, then C becomes four times with, change in anode element. If Z is the atomic, number of the original element, then the atomic, number of new element is (neglect screening, effect)., , I, , 2, , 5., , (A), 6., , Ka, 7., , c, 3., , 4., , , , , , a) Z, b) Z/2, c) 2Z, d) Z/3, When stopping potential is applied in an, experiment on photoelectric effect, no, photocurrent is observed. This means that, (A) the emission of photoelectrons is stopped, (B)the photoelectrons are emitted but are, reabsorbed by the emitter metal, (C) the photoelectrons are accumulated near the, collector plate, (D)the photoelectrons are dispersed from the sides, of the apparatus., Two separate monochromatic light beams A, and B of the same intensity (energy per unit, area per unit time) are falling normally on a, unit area of a metallic surface. Their, wavelength are A and B respectively.., Assuming that all the incident light is used in, NARAYANAGROUP, , 2, B , , A , A , , , (A) (B), (C) (D) B , A , A , B , B , When a centimetre thick surface is illuminated, with light of wavelength , the stopping, potential is V. When the same surface is, illuminated by light of wavelength 2 , the, stopping potential is V/3. The threshold, wavelength for the surface is :, , 8., , 4, 3, , (B) 4 , , (C) 6 l, , (D), , 8, 3, , When monochromatic light falls on a, photosensitive material, the number of, photoelectrons emitted per second is n and, their maximum kinetic energy is Kmax. If the, intensity of the incident light is doubled, keeping the frequency same, then :, (A) both n and Kmax are doubled, (B) both n and Kmax are halved, (C) n is doubled but Kmax remains the same, (D) Kmax is doubled but n remains the same, The frequency and intensity of a light source, are both doubled. Consider the following, statements., (i) The saturation photocurrent remains, almost the same., (ii) The maximum kinetic energy of the, photoelectrons is doubled., (A) Both (i) and (ii) are true, (B) (i) is true but (ii) is false, (C) (i) is false but (ii) is true, (D) both (i) and (ii) are false, The work function for aluminium surface is 4.2, eV and that for sodium surface is 2.0 ev. The, two metals were illuminated with appropriate, radiations so as to cause photo emission. Then:, (A) Both aluminium and sodium will have the same, threshold frequency, (B) The threshold frequency of aluminium will be, more than that of sodium, (C) The threshold frequency of aluminium will be, less than that of sodium, (D) The threshold wavelength of aluminium will be, 73
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , distance , D), , current , , C), , distance , , distance , , current , , B), , current , , A), , current , , more than that of sodium, 9. A point source of light is used in a photoelectric, effect. If the source is moved farther from the, emitting metal, the stopping potential, (A) will increase, (B) will decrease, (C) will remain constant, (D) will either increase or decrease, 10. A point source causes photoelectric effect, from a small metal plate. Which of the following, curves may represent the saturation, photocurrent as a function of the distance, between the source and the metal?, , limit of Lyman series, 2 = the wavelength of, the series limit of Balmer series & 3 =the, wavelength of first line of Lyman series is :, (A) 1 2 3, , distance , , (B) 3 1 2, , (C) 2 3 1, (D) none of these, 17. According to Bohr’s theory of the hydrogen, atom, the speed v n of the electron in a, stationary orbit is related to the principal, quantum number n as (C is a constant) :, (A) vn = C/n2, (B) vn = C/n, (C) vn = C × n, (D) vn = C × n2, 18. The orbital speed of the electron in the ground, state of hydrogen is v. What will be its orbital, speed when it is excited to the state, having energy – 3.4 eV ?, (A) 2 v, , 11. The innermost orbit of the hydrogen atom has, a diameter of 1.06 Å. What is the diameter of, the tenth orbit :, (A) 5.3 Å (B) 10.6 Å (C) 53 Å (D) 106 Å, 12. The energy difference between the first two, levels of hydrogen atom is 10.2 eV. What is, the corresponding energy difference for a, singly ionized helium atom ?, (A) 10.2 eV (B) 20.4 eV (C) 40.8 eV (D) 81.6eV, 13. An energy of 24.6 eV is required to remove, one of the electrons from a neutral helium, atom. The energy (in eV) required to remove, both the electrons from a neutral helium atom, is : [ JEE’95,1 ], (A) 38.2 (B) 49.2 (C) 51.8 (D) 79.0, 14. Which energy state of doubly ionized lithium, (Li++) has the same energy as that of the, ground state of hydrogen?Given Z for Li=3:, (A) n = 1 (B) n = 2 (C) n = 3 (D) n = 4, 15. If an orbital electron of the hydrogen atom, jumps from the ground state to a higher energy, state, its orbital speed reduces to half its initial, value. If the radius of the electron orbit in the, ground state is r, then the radius of the new, orbit would be :, (A) 2r, (B) 4r, (C) 8r, (D) 16r, 74, , 16 The relation between 1 =wavelength of series, , (B), , v, 2, , (C), , v, 4, , (D), , v, 8, , 19. In the Bohr model of the hydrogen atom, the, ratio of the kinetic energy to the total energy, of the electron in a quantum state n is, , 20., , 21., , 22., , 23., , (A) – 1, , (B) + 1, , (C) 1n, , 3, (A) 23, , 5, (B) 27, , 7, (C) 29, , 1, , (D) n2, The total energy of the electron in the first, excited state of hydrogen is – 3.4 eV. What is, the kinetic energy of the electron in this state?, (A) +1.7eV (B) +3.4eV(C) +6.8eV (D) –13.4eV, The total energy of the electron in the first, excited state of hydrogen is – 3.4 eV. Then, the potential energy of the electron is, (A) –1.7eV (B) –3.4eV (C) –6.8eV (D) –13.4eV, The highest energy state, that unexcited, hydrogen atoms can reach when they are, bombarded with 12.2eV electron, is, (A) n = 1 (B) n = 2 (C) n = 3 (D) n = 4, The ratio of the wavelengths of the longest, wavelength lines in the Lyman and Balmer, series of hydrogen spectrum is, 9, (D) 31, , 24. The frequency of the first line in Lyman series, in the hydrogen spectrum is n. What is the, frequency of the corresponding line in the, spectrum of doubly ionized Lithium ?, (A) n, (B) 3 n, (C) 9 n, (D) 27 n, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 25. The electron in a hydrogen atom makes a, transition from an excited state to the ground, state. Which following statements is true?, (A) Its kinetic energy increases and its potential, and total energies decrease, (B) Its kinetic energy decreases, potential energy, increases and its total energy remains the same, (C) Its kinetic and total energies decrease and its, potential energy increases, (D) Its kinetic, potential and total energies decrease, 26. Three photons coming from excited atomichydrogen sample are picked up. Their energies, are 12.1eV, 10.2eV and 1.9eV. These photons, must come from, (A) a single atom, (B) two atoms, (C) three atoms, (D) either two atoms or three atoms, 27. The ionization energy of hydrogen atom is 13.6, eV. Hydrogen atoms in the ground state are, excited by electromagnetic radiation of energy, 12.1 eV. How many spectral lines will be, emitted by the hydrogen atoms?, (A) one (B) two, (C) three (D) four, 28. The wavelength of the first line in balmer, series in the hydrogen spectrum is . What is, the wavelength of the second line :, (A), , 20, 27, , (B), , 3, 16, , (C), , 5, 36, , (D), , 3, 4, , 29. An electron with kinetic energy 5 eV is incident, on a hydrogen atom in its ground state. The, collision, (A) must be elastic, (B) may be partially elastic, (C) must be completely inelastic, (D) may be completely inelastic, 30. Consider a photon of continuous X-ray coming, from a Coolidge tube. Its energy comes from, (A) the kinetic energy of the stricking electron, (B) the kinetic energy of the free electrons of the, target, (C) the kinetic energy of the ions of the target, (D) an atomic transition in the target, 31. 50% of the X-rays coming from a Coolidge, tube is able to pass through a 0.1 mm thick, aluminum foil. The potential difference, between the target and the filament is, increased. The thickness of aluminimum foil,, NARAYANAGROUP, , which will allow 50% of the X-ray to pass, through, will be, (A) zero (B) < 0.1 mm (C) 0.1mm (D) 0.1mm, 32. The characteristic X-ray spectrum is emitted, due to transition of, (A) valence electrons of the atom, (B) inner electrons of the atom, (C) nucleus of the atom, (D) both, the inner electrons and the nucleus of the, atom, 33. According to Moseley’s law the ratio of the, slopes of graph between and Z for K and, K is :, (A), , 32, 27, , (B), , 27, 32, , (C), , 33, 22, , (D), , 22, 33, , 34. An X-ray tube is operated at 66 kV. Then, in, the continuous spectrum of the emitted X-rays:, (A) wavelengths 0.01 nm and 0.02 nm will both, be present, (B) wavelengths 0.01 nm and 0.02 nm will both, be absent, (C) wavelengths 0.01 nm will be present but, wavelength 0.02 nm will be absent, (D) wavelength 0.01 nm will be absent but, wavelength 0.02 nm will be present, 35. The energy of a photon of frequency n is E =, hn and the momentum of a photon of, wavelength is p = h / . From this statement, one may conclude that the wave velocity of, light is equal to :, (A) 3 × 108 ms–1 (B), , E, p, , (C) E p (D), , E, , p, , 2, , 36. A particle of mass M at rest decays into two, particles of masses m1 and m2, having non zero, velocities. The ratio of the de Broglie, 1, wavelengths of the particles, is :, 2, m1, , m2, , m2, , (A) m, (B) m, (C) 1 : 1 (D) m, 2, 1, 1, 37. Let p and E denote the linear momentum and, the energy of a photon. If the wavelength is, decreased,, (A) both p and E increase, (B) p increases and E decreases, (C) p decreases and E increases, (D) both p and E decreases, 75
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 38. The wavelength of de Broglie waves 45. The potential difference applied to an X-ray, tube is 5 kV and the current thrrough it is 3.2, associated with an electron (mass m, charge, mA. Then the number of electrons striking, e) accelerated through a potential difference, the target per second is [JEE 2002 (Scr)], of V is given by (h is Planck’s constant) :, (A) = h/mV, (B) = h/2 meV, A) 2 1016 B) 5 1016 C) 1 1017 D) 4 1015, 46. A Hydrogen atom and Li++ ion are both in the, (C) = h/ meV, (D) = h/ 2meV, 39. If a hydrogen atom at rest, emits a photon of, second excited state. If lH and lLi are their, wavelength , the recoil speed of the atom of, respective electronic angular momenta, and, mass m is given by :, EH and ELi their respective energies, then, h, mh, (A) m (B) (C) mh (D) none of these, [JEE 2002 (Scr)], 40. Two particles of masses m and 2m have equal, A) lH lLi and EH ELi, kinetic energies. Their de Broglie wavelengths, B) lH lLi and EH ELi, are in the ratio of :, (A) 1 : 1 (B) 1 : 2, (C) 1 : 2 (D) 2 : 1, C) lH lLi and EH ELi, 41. The wavelength of de Broglie wave associated, with a thermal neutron of mass m at absolute, D) lH lLi and EH ELi, temperature T is given by (here k is the, 47. If the atom 100Fm257 follows Bohr model and, Boltzmann constant) :, the radius of last orbit of 100Fm257 is n times, h, h, h, h, the Bohr radius, then find n [JEE 2003], (A), (B), (C), (D), mkT, 2mkT, 3mkT, 2 mkT, A) 100 B) 200, C) 4, D) 1/4, 42. A proton and an electron are accelerated by 48. The attractive potential for an atom is given, the same potential difference. Let e and p, r, denote the de Broglie wavelengths of the, by V V0 ln r , v0 and r0 are constants, 0, electron and the proton respectively., and r is the radius of the orbit. The raidus r, (A) e p (B) e p (C) e p, of the nth Bohr’s orbit depends upon principal, (D) e and p depends on the accelerating, quantum number n as : [JEE 2002 (Scr)], potential difference., 1, 1, 43. A positronium ‘atom’ is a system that consists, A) r n B) r 2 C) r n 2 d) r , n, n, of a positron and an electron that orbit each, other. The ratio of wavelength of the spectral 49. The wave length of K β X-ray of certain metal, lines of positronium with those of ordinary, is 12.42 pm. It takes 10keV to remove the, hydrogen is, electron from M shell of an atom of that, A) 2 : 1 B) 1 : 1, C) 2 : 3, D) 4 : 1, metal. The minimum accelarating voltage that, 44. The intensity of X-rays from a coolidge tube, should be applied across the X-ray tube, so, is plotted against wavelength as shown in, that a K α X-ray would be produced is (hc =, the figure. The minimum wavelength found is, 1242 eV nm), c and the wavelength of the K line is k ., A) 10kV B) 100kV C) 110kV D) 90kV, As the accelerating voltage is increased, 50., A photon of 10.2 eV energy collides with a, (JEE-2001), hydrogen atom in ground state inelastically., I, After few microseconds one more photon of, energy 15 eV collides with the same hydrogen, atom. Then what can be detected by a, suitable detector. [JEE 2005 (scr)], A) one photon of 10.2 eV and an electron of, energy 1.4 eV, , , , B) 2 photons of energy 10.2 eV, A) k c increases B) k c decreases, C) 2 photons of energy 3.4 eV, D) 1 photon of 3.4 eV and one electron of 1.4 eV, C) k increases, D) k decreases, c, , 76, , k, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 51. The potential energy of a particle of mass m, E0 0 x 1, , is given by V x , and 2, x 1 1, 0, are de-Broglie wave lengths of the particle,, when 0 x 1 and x 1 respectively. If the, , 1, total energy of particle is 2 E0 , find [JEE, 2, 2005], , ATOMIC PHYSICS, 56. In a photoelectric experiment anode potential, is plotted against plate current [JEE 2004, (scr)], I, C, , B, A, , V, 1, A) 2 B), C) 2 :1, D) 4 :1, 2, (A) A and B will have different intensitites while B, 52. The largest wavelength in the ultraviolet region, and C will have differen frequencies., of the hydrogen spectrum is 122 nm. The, (B) B and C will have different intensities while A, smallest wavelength in the infrared region of, and C will have different frequencies., the hydrogen spectrum (to the nearest integer), (C) A and B will have different intensities while A, is [JEE 2007], and C will have equal freuencies., A) 802 nm, B) 823 nm, (D) A and B will have equal intensities while B and, C) 1882 nm, D) 1648 nm, C will have different fequencies., 53. Electrons with de-Broglie wavelength fall, MULTIPLE ANSWER QUESTIONS, on the target in an X-ray tube. The cut-off, th, wavelength of the emitted X-rays is [JEE 57. Let An be the area enclosed by the n orbit in, a hydrogen atom. The graph of ln (An / A1), 2007], against In(n), 2h, 2mc 2, (A)will pass through the origin, A) 0 , B) 0 , mc, h, (B)will be certain points lying on a straight line with, slope 4, 2m2c 2 3, , , C) 0, D) 0 , (C) will be a monotonically increasing nonlinear curve, h2, (D) will be a circle, 54. Which one of the following statements is 58. The potential difference applied to an X-ray, WRONG in the context of X-rays generated, tube is increased. As a result, in the emitted, from a X-ray tube ?, radiation,, A) Wavelength of characteristic X-rays decreases, (A) the intensity increases, when the atomic number of the target increases, (B) the minimum wavelength increases, B) Cut-off wavelengthof the continuous X-rays, (C) the intensity remains unchanged, depends on the atomic number of the target, (D) the minimum wavelength decreases, C) Intensity of the characteristic X-rays depends, on the electrical power given to the X-rays tube. 59. Photoelectric effect supports quantum nature, of light because, D) Cut-off wavelength of the continuous X-rays, (A)there is a minimum frequency below which no, depends on the energy of the electronsin the Xphotoelectrons are emitted, rays tube., (B)the maximum kinetic energy of photoelectrons, 55. In a coolidge tube, the potential difference, depends only on the frequency of light and not on, across the tube is 20 kV and 10 mA current, its intensity, flows through the voltage supply. Only 0.5%, (C) even when the metal surface is faintly illuminated, of the energy carried by the electrons striking, the photoelectrons (if n ³ nth) leave the surface, the target is converted into x-rays. The x-ray, immediately, beam carries a power of, (D) electric charge of the photoelectrons is, A) 0.1 W B) 1W, C) 2 W, D) 10 W, quantized, NARAYANAGROUP, , 77
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 60. Select the correct alternative(s):, When photons of energy 4.25 eV strike the, surface of a metal A, the ejected photo, electrons have maximum kinetic energy TA eV, and de Broglie wave length A . The maximum, kinetic energy of photo electrons liberated, from another metal B by photons of energy, 4.70 eV is TB = (TA - 1.50) eV. If the deBrogleie wave length of these photo electrons, is B = 2 A , then:, [JEE ’94, 2], (A) the work function of A is 2.25 eV, (B) the work function of B is 4.20 eV, (C) TA = 2.00 eV, (D) TB = 2.75 eV, 61. When a hydrogen atom is excited from ground, state to first excited state then, (A) its kinetic energy increases by 10.2 eV., (B) its kinetic energy decreases by 10 .2 eV, (C) its potential energy increases by 20.4 eV, (D) its angular momentum increases by 1.05 × 10–, 34 J-s., 62. Two electrons starting from rest are, accelerated by equal potential difference., (A) they will have same kinetic energy, (B) they will have same linear momentum, (C)they will have same de Broglie wave length, (D) they will produce x-rays of same minimum wave, length when they strike different targets., 63. When an electron moving at a high speed, strikes a metal surface, which of the following, are possible ?, (A)the entire energy of the electron may be, converted into an X-ray photon, (B)any fraction of the energy of the electron may, be converted into an X-ray photon, (C)the entire energy of the electron may get, converted to heat, (D)the electron may under go elastic collision with, the metal surface., 64. The graph between 1/ and stopping potential, (V) of three metals having work functions, 1 , 2 and 3 in an experiment of photoelectric, effect is plotted as shown in the figure. Which, of the following statement(s) is/are correct, (JEE-2006), V, metal 1, , metal 2, , , 0.001 0.002, , 78, , 0.004, , metal 3, , A) Ratio of work functions 1 : 2 :3 1: 2 : 4, B) Ratio of work functions 1 : 2 :3 4 : 2 :1, C) tan is directly proportional to hc/e, where h, is Planck’s constant and c is the speed of light, D) The violet colour light can eject photoelectrons, from metals 2 and 3., 65. Consider a hypothetical atom with single, electron. In this atom, when an electron deexcites from energy level n = x to n = 2,, wavelength ( ) of the radiation emitted is, given by , , Ax 2, x2 4, , (where A is a constant)., , Choose the correct alternatives., A) Least energetic photon emitted during such a, transition will have wavelength 1.8A., B) Most energetic photon emitted in such a, transition will have wavelength A., C) lonization potential of the atom in its ground state, hc, , is 1.8eA, D) lonization potential of the atom in its first excited, hc, , state is eA, 66. If in a hydrogen atom, radius of nth Bohr orbit, is rn, frequency of revolution of electron in nth, orbit is fn, and area enclosed by the nth orbit is, An, then which of the following graphs are, correct, , rn, , a), , b), , r , log n , r1 , , log n, , n, , c), , A , log n , A1 , , d), log n, , f , log n , f1 , , log n, , nm–1, 1/, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 67. The electrons in a hydrogen atom make a, transition n1 n2 are the principal quantum, numbers of two states. Assume the Bohr’s, model to be valid. The time period of the, electron in the initial state is eight times that, in the final state. The possible values of n1, and n2 are :, a) n1 4, n2 2, , b) n1 8, n2 2, , c) n1 8, n2 1, , d) n1 6, n2 3, , 68. When a monochromatic point source of light, is at a distance of 0.2m from a photoelectric, cell, the cut-off voltage and the saturation, current are respectively 0.6V and 18.0 mA. If, the same source is placed 0.6m away from the, photoelectric cell, then :, a) The stopping potential will be 0.2V, b) The stopping potential will be 0.6V, c) The saturation current will be 6.0 mA, d) The saturation current will be 2.0mA, , COMPREHENSION TYPE QUESTIONS, PARAGRAPH - I, In a mixture of H – He+ gas (He+ is singly, ionized He atom), H atoms and He+ ions are, excited to their respective first excited states., Subsequently, H atoms transfer their total, excitation energy to He+ ions (by collisions)., Assume that the Bohr model of atom is exactly, valid. ( IIT 2008), 69. The quantum number n of the state finally, populated in He+ ions is :, (A) 2, (B) 3, (C) 4, (D) 5, 70. The wavelength of light emitted in the visible, region by He+ ions after collisions with H, atoms is, (B) 5.6 × 10–7 m, (A) 6.5 × 10–7 m, (C) 4.8 × 10–7 m, (D) 4.0 × 10–7 m, 71. The ratio of the kinetic energy of the n = 2, electron for the H atom to that of He+ ion is:, (A), , 1, 4, , (B), , 1, 2, , NARAYANAGROUP, , (C) 1, , (D) 2, , PARAGRAPH - II, The co llecter of the photocell (in photoelectric, experiment) is made of tungsten while the, emitter is of Platinum having work function of, 10 eV. Monochromotic radiation of wavelength, 124 Å & power 100 watt is incident on emitter, which emits photo electrons with a quantum, efficiency of 1%. The accelerating voltage, across the photocell is of 10,000 volts, (Use:hc=12400eV Å), radiation, 124A, , Platinum, emitter, , Tungsten, target, , VA = 10,000 V, , 72. What is the power supplied by the accelerating, voltage source., (A) 100 watt, (B) 10 watt, (C) 0.1 watt, (D) 1 watt, 73. The minimum wavelength of radiation coming, from the tungsten target (collector) is, (A) 124 Å, (B) 1.24 Å, (C) 1.23 Å, (D) 12.3 Å, 74. If the source of monochromatic radiation of, wavelength 124 Å has an efficiency of 50%,, and the power of X ray emitted by the tungsten, target is 3W, the overall efficiency of the, apparatus for X-ray production is, (A) 1 % (B) 0.1% (C) 1.5% (D) 0.67%, , PARAGRAPH - III, The key feature of Bohr’s theory of spectrum, of hydrogen atom is the quantization of angular, momentum when an electron is revolving, around a proton. We will extend this to a, general rotational motion to find quantized, rotational energy of a diatomic molecule, assuming it to be rigid. The rule to be applied, is Bohr’s quantization condition. (JEE 2010), 75. A diatomic molecule has moment of inertia I., By Bohr’s quantization condition its rotational, energy in the nth level (n = 0 is not allowed) is, , 1 h2, (A) 2 2, n 8 I, (C) n, , h2, 8 2 I, , 1 h2, (B), n 8 2 I, 2, (D) n, , h2, 8 2 I, 79
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 76. It is found that the excitation frequency from, ground to the first excited state of rotation for, 4, 11, the CO molecule is close to 10 Hz . Then, , the moment of inertia of CO molecule about, its centre of mass is close to (Take h =, 2 1034 Js), A) 2.76 10-46 kg m2 B) 1.87 10-46 kg m2, C) 4. 67 10-47 kg m2 D) 1.17 10-47 kg m2, 77. In a CO molecule, the distance between C, (mass = 12 a.m.u) and O (mass = 16 a.m.u.),, 5, 27, where 1 a.m.u. 10 kg, is close to, 3, A) 2.4 10-10 m, B) 1.9 10-10m, -10, C) 1.3 10 m, D) 4.4 10-10 m, , PARAGRAPH - IV, When a particle is restricted to move along xaxis between x=0 and x=a, where a is of, nanometer dimension, its energy can take only, certain specific values. The allowed energies, of the partcile moving in such a restricted, region, correspond to the formation of standing, waves with nodes at its ends x=0 and x=a. The, wavelength of this standing wave is related to, the linear momentum p of the particle, according to the deBroglie relation. The, energy of the particle of mass m is related to, , p2, . Thus, the, 2m, energy of the particle can be denoted by a, quantum number ‘n’ taking values 1,2,3,........., (n=1, called the ground state) corresponding, to the number of loops in the standing wave., Use the model described above to answer the, following three questions for a particle moving, in the line x=0 to x=a. Take h=6.6x10-34 JS, and e=1.6x10-19C (JEE-2009), 78. The allowed energy for the particle for a, particular value of n is proportional to, A) a 2 B) a 3/ 2 C) a 1, D) a 2, 79. If the mass of the particle is m=1.0x10-30kg, and a=6.6nm, the energy of the particle in its, ground state is closes to, A) 0.8 meV, B) 8 meV, C) 80 meV, D) 800 meV, 80. The speed of the particle, that can take, discrete values, is proportional to, A) n 3/ 2 B) n 1, C) n1/ 2, D) n, its linear momentum as E , , 80, , PARAGRAPH - V, In a photoelectric effect set up, a point source, of light of power 3.2 x 10 –3 W emits, monoenergetic photons of energy 5eV. The, source is located at a distance of 0.8m from, the centre of a stationary metallic sphere of, work function 3eV and radius 8 x 10–3m. The, efficiency of photoelectron emission is one for, every 106 incident photons. Assume that the, sphere is isolated and initially neutral and the, photoelectrons are initially swept away after, emission., 81. Find the number of photons emitted per second, a) 105, , b) 2 x 1015 c) 4 x 1015 d) 6 x 1015, , 82. Find the maximum kinetic energy of, photoelectrons :, a) 8 10 20 J, c) 24 10 20 J, , b) 16 10 20 J, d) 32 10 20 J, , STATEMENT TYPE QUESTIONS, A) Statement - 1 is True, Statement - 2 is, True; Statement - 2 is a correct explanation, for Statement - 1., B) Statement - 1 is True, Statement - 2 is True;, Statement - 2 is NOT a correct explanation, for Statement - 1., C) Statement - 1 is True, Statement - 2 is, False, D) Statement - 1 is False, Statement - 2 is, True., 83. STATEMENT - 1 : If the accelerating potential, in an X-ray tube is increased, the wavelengths of, the characteristic X-rays do not change, STATEMENT - 2 : When an electron beam, strikes the target in an X-ray tube, part of the kinetic, energy is converted into X-ray energy [JEE 2007], 84. STATEMENT - 1 : A metallic surface is irradiated, by a monochromatic light of frequency v v0 (the, threshold frequency). The maximum kinetic energy, and the stopping potential are Kmax and V0, respectively. If the frequency incident on the surface, is doubled, both the Kmax and V0 are also doubled., STATEMENT - 2 : The maximum kinetic energy, and the stopping potential of photoelectrons emitted, from a surface are linearly dependent on the, frequency of incident light., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , MATRIX MATCH TYPE QUESTIONS, 85. Related to photoelectric effect, in Column I,, Some physical quantities change while in, column II effects of changes are given. Match, the entires of Column I with the entries of, Column II., COLUMN - I, A) Intensity of incident light changes, B) Frequency of incident light changes, C) Target material changes, D) Potential difference between the emitter and, collector changes, COLUMN - II, P) K max of emitted photoelectrons changes, Q) Stopping potential changes, R) Saturation current changes, S) Time delay in emission of photoelectrons changes, 86. In each situation of Column I, a physical, quantity related to orbiting electron in, hydrogen-like atom is given. The terms ‘Z’ and, ‘n’ given in Column II have usual meaning in, Bohr’s theory. Match the quantities in Column, I with the terms they depend on it in Colimn II, COLUMN - I, A) Frequency of orbiting electron, B) Angular momentum of orbiting electron, C) Magnetic moment of orbiting electron, D) The average current due to orbiting electron, COLUMN - II, P) Is directly proportional to Z 2, Q) Is directly proportional to n, R) Is inversly proportional to n3, S) Is independent of Z, 87. Match the following:, COLUMN - I, A) The voltage applied to X-ray tube is increased, B) In photoelectric effect, work function of the, target is increased, C) Stopping potential decreases, D) Wavelength of K X-ray increased, COLUMN - II, P) Average KE of the electrons decreases, Q) Average KE of the electrons increases, R) Cut-off wavelength decreased, S) Atomic number of target material decreases, NARAYANAGROUP, , ATOMIC PHYSICS, 88. Match the following:, COLUMN - I, A) K photon of aluminium, B) K photon of aluminium, C) K photon from sodium, D) K photon of beryllium, COLUMN - II, P) Will be most energetic among the four, Q) Will be least energetic among the four, R) Will be more energetic than the K photon of, lithium, S) Constant speed, , INTEGER TYPE QUESTIONS, 89. The radius of an - particle moving in a circle, in a constant magnetic field is half the radius, of an electron moving in circular path in the, same field. The de Broglie wavelength of particle is n times that of the electron. Find n, (an integer)., 90. The de Broglic wavelength of an electron, moving with a velocity of 1.5 108 m / sec is, equal to that of a photon. Find the ratio of the, energy of the photon to that of the kinetic, energy of the electron., 91. ‘S’ is isotropic point source producing, monochromatic radiation with power P. Force, P, . Find the value of ‘n’, nC, (C is speed of light)., 92. A monochromatic sources of light operating, at 200 W emits 4 1020 photons per second., , on hemisphere is, , Find the wavelength of the light in 107 m ., 93. Find the recoil speed (appoximately in m/sec), when a hydrogen atom emits a photon during, the transition from n = 5 to n = 1., 94. an atom of atomic nmuber Z = 11 emits K, wavelength which is . Find the atomic, number for an atom that emits K radiation, with wavelength 4 (an integer), 81
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 95. An electron in the nth excited state in a, hydrogen atom comes down to the first excited, state by emitting 10 different wavelengths., Find the value of n (an integer), 96. In ground state, find the ratio between total, acceleration of an electron in a singly ionized, helium atom and hydrogen atom., 97. The shortest wavelength of the Bracktt series, of a hydrogen like atom (atomic number Z ) is, the same as the shortest wvelength of the, Balmer series of hydrogen atom. Find value, of Z ., 98. A hydrogen like atom (atomic number Z ) is, in a higher excited state of quantum number, 6. The excited atom can make a transition to, the first excited state by successively emitting, two photons of energies 10.2 eV and 17.0 eV, respectively. Determine the value, of Z .(Ionization energy of H-atom is13.6 eV.), 99. The electric potential between a proton and, , STATEMENT QUESTIONS, 83.B 84.D, MATRIX MATCHING QUESTIONS, 85.A-R, B-P,Q, C-P,Q, D-S, 86.A-P,R, B-Q,S, C-Q,S, D-P,R, 87.A-Q,R, B- P,R, C-P,R, D-S, 88.A-R,S; B-P,R,S; C-R,S; D-Q,R,S, INTEGER QUESTIONS, 89.1 90.4 91.4 92.4 93.4 94.6 95.6, 96.8 97.2 98.3 99.1, , LEVEL - V- HINTS, 1., , mu + 0 = (m + M) V V , Max.loss, K , , r, an electron is given by V V0 ln r , Where V0, 0, , is, , 1, 1, 1, M , mu 2 ( M m)V 2 mu 2 , , 2, 2, 2, M m, , K E , , , 1, M , mu 2 , E, 2, M m, , 1, m, M m, , mu 2 E , KEmin E 1 , 2, M , M, , 1, i.e KEmin 51 1 63.75eV, 4, , LEVEL - V- KEY, , 82, , in, , mu, m M, KE, , For both to get excited to their first excited states,, energy required is E 10.2 10.2 4 51eV . , For, both, to, get, excit ed, , and r0 are constants and r is the radius of, the electron orbit around the proton. Assuming, Bohr’s model to be applicable, it is found that, r is proportional to n x , where n is the, principal quantum number. Find the value of, x., SINGLE ANSWER QUESTIONS, 1.A 2.B 3.B 4.A 5.B 6.C 7.B, 8.B 9.C 10.D 11.D 12.C 13.D 14.C, 15.B 16.D 17.B 18.B 19.A 20.B 21.C, 22.C 23.B 24.C 25.A 26.D 27.C 28.A, 29.A 30.A 31.D 32.B 33.A 34.D 35.B, 36.C 37.A 38.D 39.A 40.D 41.C 42.C, 43.A 44.A 45.A 46.B 47.D 48.A 49.C, 50.A 51.A 52.B 53.A 54.B 55.B 56.A, MULTIPLE ANSWER QUESTIONS, 57.A,D 58.A,B,C 59.A,B,C 60.B,C,D, 61.A,C,D 62.A,B,C, 63.A,B,C,D 64.A,C, 65.A,B,D 66.A,B,C 67.A,D 68.B,D, COMPREHENSION QUESTIONS, 69.C 70.C 71.A 72.A 73.C 74.A 75.D, 76.B 77.C 78.A 79.B 80.D 81.C 82.D, , Max. loss in KE of the system takes place in, perfectly inelastic collision. From conservation of, linear momentum, , 2., , 4, , In the first case: 3RZ, , 2, , , , hc, eV, , 1, 1 1 , RZ 2 , , 1 4 , K, , 4, , 4hc, , Z, , In the second case: 4 3RZ 2 eV Z 2 2, 2, , 3., , 4., , Emission of photo electron is independent of, external factor. It depends only on the nature of, the material and wavelength of incident light, The number of photo electron depends on the, Number of photons Number of photon, I, , ·I, , = hc / = hc , Ratio of no. of photo electrons=, , A, B, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 5., , hC, , , = + eV, , ....(i), , hC, 2, , =+, , eV, 3, , ....(ii), , 3 x II – I , , , 7., , 8., , 3, hc, – 1, 2, , , , ATOMIC PHYSICS, energy required to remove both the, electrons 24.6 54.4 79.0eV, 14. En (Li2+) = E1 (H), , = 2 =, , hc, 4, , th = 4, Since frequency of light is same, KEmax remains same, and if intensity of the light is doubled, number of, photons doubles and hence n also doubles., nh, , when frequency is doubled for I to, tA, become doubled n should remain the, same. saturation photo current remains the same, but maximum kinectic energy becomes more than, , – 13.6, 1, 4, , With distance intensity will fall as, , 1, r, , 2, , 10. I , , 16., , r10 =, , 12. En = 13.6, , 13.6(1)2, (1)2, , EHe =, , 13.6(2)2, (1)2, , –, –, , 13.6(1) 2, ( 2) 2, , 13.6(2)2, ( 2) 2, , K, 1, , = E – E1 and, , K, 2, , = E – E2, , K, 3, , = E2 – E1 , , 1, 1, , –, , 18. E , v, , = 10.2eV, , ..........II, , 1, 2, , =, , 1, 3, , e2, , z, , o, , h n, , 1, n2, , z, n, , =, , cz, c, = ( For H -atom z =1), n, n, n22, , 13.6, 3.4, , = n2 n2 = 2, 1, , (z = 1), , v2 =, , v, 2, , ....... (B), , – | T.E | | KE |, , = –(PE), 2, 2, , Where T.E. =, K. E.=, , – me 4, 8 02 n2h2, , me 4, 8 20, , 2 2, , n h, , , , KE, = –1, TE, , 20. In a quantum state of a Bohr‘s atom –T.E = KE, – (–3.4) = KE, 3.4 ev = KE, 21. P.E = 2(T.E.), = 2 (– 3.4 ) = – 6.8 eV, 22., , E = 13.6, , = 40.8 eV, , 13. After removing one electron He + becomes, hydrogen like and energy required to remove the, second, electron, , h, 2, , r` = 2 ·, , 19. According to the Bohr model, , × 1.06 Å = 106 Å., , Z2, n2, , EH =, , v0, 2, , V=2, , 1, square of the distance from the source. i.e i 2, r, 11. r, , m, , 17. Velocity of an electron is given by, , intensity of light which is inversely propotional to, , 102, , n=3, , h, , 1, .Saturation photo current is propotional to, r2, , n2, , , , nh, 2, , th n = 2 mvr =, , . When source, , moves away from the metal intensity of light falling, on the metal decreases but frequency remains the, same. As stopping potential is independent of, intensity and depends on frequency, stopping, potentail remains the same., , n, , 1, 1, , mv0 r = 1. 2 ..........I, , it has higher work function. ( W h 0 ), 9., , = –13.6 ×, , 2, , 15. Since speed reduces to half , KE reduced to, , I, , doubled (from KEmax h w ), The threshold frequency for Al must be greater as, , 32, , 1 , 1, 2 2 , n , 1, , 13.6 – 12.2 =, n2 =, , = 12.2, , 13.6, n2, , 3 . 6, 13.6 – 12 .2, , =, , 13.6, n=3, 1.4, , 13.6 Z 2 , 0, 13.6 4 54.4eV, n2, , , NARAYANAGROUP, , 83
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 32. The characteristic x-rays are obtained due to the, transition of electron from inner orbits., , 1, 1, 1 , 2, 23. RZ 2 2 , n1 n2 , for longest wave lengths in Lyman series n1 1, n2 2, and for Balmer series n1 2, n2 3, , , 2, , , , 33., , K , , = K (Z – ) ×, , K , , = K (z – ) ×, , Z, 24. E = 13.6 2 , n , 2, , EH =, , 13.6(1), (1) 2, , ELi =, , 13.6(3) 2, (1) 2, , 2, , –, , 13.6(1), ( 2) 2, , –, , 13.6(3) 2, (2)2, , 25. Kinetic energy (K.E.) =, Potential (P.E.) =, , = 10.2 eV = h, =91.80eV=h(9), , 13.6 z 2, , 2(13.6) z 2, n2, , Total energy (T.E.) =, , eV, , n2, , eV, , 28., , =R, , 1 1, – , 4 9, , similarly, 2 =, , 1, 2, , 13.6 z 2, n2, , eV, , 16, 3R, , 1 =, 1, , 49, 5R, , 16, 3, , ×, , =, , =, , 1, 9, , 32, 27, , hc, hc, 1240, , 0.0188nm, =, E, eV 66 103, , 34. min =, , 35. C = · =, , h E, E, ·, =, p h, p, , h, and from conservation of linear momentum, p, both particles will have same magnitude of linear, momentum after decay., , 5, 49, , 37. E , , hc, h, and P , , , , =, , h, =, P, , h, =, 2mK, , i.e mv , 40. =, , h, h, v, , m, h, 2mK ., , 2, , =, , 20, 27, , , , h, 2meV, , 39. From conservation of linear momentum, after, emission of a photon magnitude of linear momentum, of the atom = magnitude of linear momentum of, photon, , , , 29. 5eV electron cannot excite a hydrogen atom and, hence there will be no loss of K.E.and hence the, collision is elastic., 30. The continuous X-rays are produced due to gradual, retardation of the striking electrons., 31. With increase of potential difference, x-ray of higher, energy will be produced, To stop them, thicker foil, is required., 84, , K, , 1, , 1 , , = R 4 – 42 , =, , Ratio for, , 8, 9, 3, 4, , 1, , 1, 4, , 36. =, , When an electron in H–atom makes a transition, from an excited state to the ground state, value of, n decreases, hence K.E. increases and its P.E. &, T.E. decrease., 26. 12.1 = E(n = 3) – E (n = 1), 10.2 = E(n = 2) – E (n = 1), 1.9 = E(n = 3) – E (n = 2), At least two atoms must be involved as there can, not be two transitions from same level from same, atom., 27. 12.1 eV radiation will excite a hydrogen atom in, ground state to n = 3 state., n, number of possible transitions = C1 = 3C1 = 3., 1, 1, , K, , 1–, , h, = =, P, , K. = kinetic energy, , m2 k2, , m2, , 2, , m , m1 k 1, 1, 1, h, , 2m KE, , h, , =, , 3, , 2m KT , 2, , , , , h, 3mKT, , Where KE is kinetic energy and K is Boltzman ' s constant, , h, h, 42. 2mKE 2mqV . As q and V are same, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 1, e p, m, 43. (A) Here the two particles have same mass ‘m’., So reduced mass is, for both , , , , m1m2, m2 m, , , m1 m2 2m 2, , m, E, , 1, 2 , En m m 2, , i .e E K 1 0 10 0 1 1 0 e V, , hc, as V increases min decreases but, eV, the peak of characteristic x-rays spectrum remains, unchanged., , 44. As min , , dn , dn , 3, 19, 45. i e 3.2 10 1.6 10 , dt , dt , dn, 2 1016, dt, nh, lH lLi ( l is independent of z), 2, , Z2, ELi EH, n2, , 2 , , 8 18 32 , , 40, , n 1 n 2 n 3 n 4 n 5, n2, 25, rn 0.53 A0 0.53 , z, 100, 1, 1, 0.53 n 0.53 n , 4, 4, , ( Bohr radius = 0.53A0 ), dV, vr, 0 0 and force on the electron, 48. E , dr, r, , F Ee. i.e, , nh, mv2 v0r0e, , and mvr , 2, r, r, , NARAYANAGROUP, , kinetic energy of striking electron = 110eV, eV 110eV V 110V, 50. 10.2 eV photon will be absorbed completely for, transition of electron from n = 1 to n = 2 and it will, be emitted back when electron is deexcited. 15, eV photon will knockout K - shell elecron by, absorbing 13.6 eV and rest 1.4 eV appears as its, KE., , h, 1 p1 p2, 2mK 2, 2 E0, , , , , 2, h, 51. 2, p1, 2mK1, 2 E0 E0, p2, 52. Infrared region Paschen series n 3, , 1 1 , 1, RZ 2 2 2 , , n1 n2 , , 47., 100 , , 1242eV nm, = 105 eV = 100KeV, -3, 12.42×10 nm, , i .e E K = E M E K E K E M E K , , i.e, energy of each level is halved, Their difference will also be halved., Hence n1 2n, , E 13.6, , 49. EKβ =, , For K to be produced, electron from k shell is to, be knocked out, Kinetic energy of the striking electron should, be energy of k shell, , 1, n, , 46. l , , mn 2 h 2, v0 r0e r n, 4 2 m 2 r 2, , and energy of M shell is EM = 10 KeV, , Here m = electron mass En m, , , , , , , , 1, 1 , 2 1, 1.1107 1 2 2 , , 3 , , , , 9, 900, , 109 823nm, 7, 1.1 10, 1.1, 2, , h, 2, , 2, 53. hc KEe pe h, c, 2me, 2m, 2m 2, 2 hcm 2, 2cm 2, c , c , h2, h, 54. C , , hc, eV, 85
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, Independent of atomic no. of target, 55. Power drawn by coolidge tube, 20 103 10 103 200 W, 0.5, 200 1W, 100, 56. A and B have same stopping potential A and B, will have same frequency. B and C have same, saturation photo current B and C will have same, intensity , but different frequencies as their stopping, potentials are different., 57. An = rn2 = (r0 n2)2 = r02 n4 and A1=r02, , n, , = n (n4) = 4 n n., , ln, , An, A1, ln n, , h, , or , , 2Km, , 1, , B, A, , KA, , 13.6, n, , 2, , eV, P.E.=, , 2(13.6 ), n, , 2, , eV, , , , (increase), Angular momentum L =, h, , nh, 2, , h, , L= 2 (2–1)= 2 =1.05×10–34J-s(increase), 62. Kinetic energy of an electron is given by, 1, 2, , mv2 = eV = same for both electron, , Where v = speed of electron, V = Potential difference, Linear momentum of two electron may be different,, in direction but their magnitude i.e. mv = 2mk will, be same. De–Broglie wavelength is given by, =, , h, mv, , = same for both electrons, , Minimum wavelength of produced x–rays is given, by, hc, eV, , min =, , = same for both electrons., , 63. The electron will not undergo elastic collision with, the metal surface, 64. eV , , hc, , , , V , , TA, , = K or 2 = T – 1.5 or TA = 2eV, B, A, , 1 , , , , hc hc 1 , = , e e e e, , Slope = tan , , From equation (i), WA = 4.25 – TA = 2.25 eV, From equations (iii) ,, WB + 1.95 eV = (2.25 + 1.95) eV, or WB = 4.20 eV., TB = 4.70 – WB = 4.70 – 4.20 = 0.50 eV., 61. K.E.=, , 1, , K, , K = KE of elecron, , , 1 , =–10.2eV (decrease), (1)2 , , P.E.= 2(13.6) (2)2 (1)2 =20.4eV, , k=, , 58. As the potential difference is increased, the kinetic, energy is increased. The total energy of x-rays, emitted will also increase hence intensity will, increase.The shorter wavelengths will also, decrease., 59. The photo electric effect can be explained if photon, is considered as particle i.e.quantum nature, 60. Kmax = E – W, Therefore : TA = 4.25 – WA, ......(i), TB = (TA – 1.50) = 4.70 – WB ....(ii), Equation (i) and (ii) gives,, WB – WA = 1.95 eV, ......(iii), de-Broglie wavelength is given by, =, , , , 2, (2), , = (PE)f – (PE)i, , Power of x-ray , , An, A1, , 1, , K.E.= 13.6 , , , , hc, e, , hc, 1 1 1, 1 : 2 : 3 : : 1: 2 : 4, , 1 2 3, , Ax 2, 1 4 1 1 , 2 2, 65. 2, x 4, A2 x , for least energetic photon x 3, , , , 1, , max, , , , 4 1 1, , m ax 1.8 A, A 4 9 , , for Hydrogen z = 1 ; K = Kf – Ki, 86, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , for most energetic phot on, , , , 1, , min, , x, 71. K.E. = – T.E. = 13.6, , 4 1 1 , min A, A4 , , 1 41 1 , , 0.25 A, A 12 2 , hc, hc, V , i.e eV , , e 0.25 A, for ionization potential in its first excited state, , i.e, , eV , , hc, hc, V , , eA, , 66. a) rn n 2 therefore graph between rn , n will be, parabola, , rn, rn, 2, b) rn n r n log r, 1, 1, , , 2log n hence the, , graph is a straight line passing through origin, , =, , 100 .01 124, 12400, , efficiency, , (in Å), , =0.01A; Power = 100W, hc, , 73. Maximum energy of incoming electron = – +, eV, 12400, , , , = 124 10 10,000 eV = 10,090 eV, , , , min =, , 2, , 12400, 10090, , = 1.23 Å, , 74.Power provided by accelerating potential = 100 W, Power consumed by the source of, 100, , 124 Å= (0.5) = 200 W, , An, 4, c) rn n A n n A n, 1, 2, , eV, , n2, , 72. P = V is, where V = accelerating voltage, is = saturation photocurrent, Quantum, is = Power of source of light 12400, , for ionization potential in its ground state, , 1 4 1, 1 , , , A, A 2 2 2 , , Z2, , 4, , Overall, , A , log n 4 log n; i.e the graph is a straight, A1 , line passing through origin., , net power of X ray, , efficiency= net power supplied, , =, , 3W, 100W 200 W = .01 = 1%, 2, , fn , 1, fn, 1, d) f n 3 f n 3 log f 3log n, 1, 1, i.e the graph is a straight line passing through origin, with negative slope., rn, 3, 67. Tn v Tn n, n, , proportional to intensity of light. i.e i I , , 1, r2, , hv kEn 2 kEn 1, 76. I 1.87 1046 kgm 2, , I m1r12 m2 r22 d 1.3 1010 m, 78. E , , p2, de-Broglie’s expression for wave length is, 2m, , 1 , , 69. EHe = 13.6 × 4 4 2 eV , EHe = EH, n , , 70. The wavelength corresponding to transition from n, = 4 to n = 3 in He+ corresponds to visible region., , NARAYANAGROUP, , nh, L2 nh 1, n2h2, KE, , , and, =, , , 2, 2 I 2 2 I, 8 2 I, , m2 d, m1d, 77. r1 m m and r2 m m, 1, 2, 1, 2, , given t n1 8Tn 2 n1 2n 2, 68. Stopping potential remains same as frequency, remains the same. Saturation current is directly, , 1, , 75. L , , λ, =a, 2, , 87
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , , , h, h, h2, p, , i.e E , and a, p , , 2m 2, , E, , 1, a2, , 86. A) f , , h2, , E, , a ) 8meV, 2, 79., (in, ground, state, 2m 4a , 2, , 80. p , , D) If we change the potential difference between, emitter and collector, then time taken for electrons, to eject chages., , h, h, , 2a, v, and n a , , m, 2, n, , B) L , , mZ 2e4, Z2, , f, , 4 02 h3n3, n3, nh, Ln, 2, , C) Magnetic moment:, , , M, , v , , E 1 5eV 5 1.6 10 19 J, , Power of point source 3.2 10 3 J / s, Energy emitted per second E 2 3.2 10 3 J, E2, 15, No. of photons n E 4 10 photons, 1, , 82., , K.E max 5 3 ev 32 1020 J, STATEMENT QUESTIONS, , 83. Characteristic X-ray depends only upon the torget, metar., 84. KE max = eVs = h -W when is doubled, KE max and Vs become more than doubled and, vary linearly with frequency., , MATRIX MATCH QUESTIONS, 85. A) If intensity changes, then number of photons/, time incident on the metal surface change and hence, number of photoelectrons liberated change, so, saturation photocurrent changes. Stopping potential, and KEmax will remain the same., B) From eV0 hf and K max hf , If f changes, then V0 and K max change., C) From eV0 hf and K max hf , If target material changes, then changes, then, V0 and K max change., 88, , e, e, A, r2, 2, , r, T, v, , e, e, e, vr , mvr L, 2, 2m, 2m, , λ= a, , h n, vn, m 2a, 81. Energy of emitted photons, , M IA , , e nh , M n, 2m 2 , , ev, e mZe 2 Ze2, Z2, I, , , , I, , , , D), n3, 2 r 2 n 2 h2 0 2 0 nh, 87. A) KE of electrons striking the target KE = eV., So, if V increases, KE increases., Cut-off wavelength: min , , 12400 o, A, V, , So, if V increases, min decreases., B) If work function of target is increased in, photoelectric effect, the KE of photoelectrons, emitted decreases from: KE hf W0, hc, hc, Now, hf0 W0 W0 0 W, 0, 0, , If work function W0 increases, then cut-off, wavelength 0 decreases., C) eV0 kE , where V0 is stopping potential. If, V0 decreases, then KE decreases. V0 is also, decreased by increasing W0 , hence 0 decreases, as explained above., 3, 2, Rc z 1 . If k increases, then f k, 4, decreases and hence Z decreases., 88. From Moseley’s law:, , D) f k , , f , , 1, 1 , 2 1, R z 2 2 , , n1 n2 , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , For Al, Z is highest and for K , n1 1 and n2 3 ;, and for K , n1 1 and n2 2, Hence, order of frequency:, f k Al f k Al f k Na f k Be f k Li , Speed will be same (c) for all photons of any, frequency., , , , , , , P, , 1, , , , 1 P, , P, , , , 1 P, , , e, , e, 89. r 2 re B q 2 B q 2e 2 e P Pe, , e, , de-Broglie’s wavelength , , h, e, p, , 2, , 1 Z 2 1, , 4 11 1 2 On solving Z 2 6, 95. When the electron jumps from the nth state to the, ground state, kthe number of possible emission lines, is, , n n 1, . Here, the number of possible, 2, , emission lines is, , n 1 n 2 , 2, , Z2, a, , 3, 96. a v / r , 1/ z a Z, 2, , 90. Speed of photon c 3 108 m / sec . Let be, the wavelength of the photon. The de Broglie, wavelength of the electron, , h, h, . Given , , Now, mv, mv, , energy of photon, h, 2hc, , , 2, 1, K .E. of electron, mv 2 mv , 2, , , , 2c, v, , 91. dF , , h, c 2 3 108, , , , , &, , , 4, , =, mv, 1.5 108, , , I, ds , c, P, 4R 2 c, , , , ds cos , , P, P, , R 2 n 4, 2, 4R c, 4c, 92. The energy of each photon is, 200 J / S, hc, 5 1019 J Wavelength = , 20, 4 10 / s, E, 34, , 5 10, , 19, , J, , 4.0 107 m, , 1 , , 93. E photon 13.6 1 eV 13.0eV, 25 , E / c mv (momentum conserved), , E, 13 1.6 1019, v, , 4m / sec, mc 1.67 1027 3 108, 2, , 1 Z 2 1, 2, 94. , Z1 1, 2, , 1, 2, , sin ce, Z 1 , , , , , NARAYANAGROUP, , 2, , 1 1 , 13.6eV 2 2 z 2 10.20 17.00 27.20eV, 6 2 , 27.20, 1 1, 2, Z 2 , or Z 3, 13.6 Z 9, 36 4 , r , , 99. V V0 1n r .Magnitude of force on electron, , , J s 3 108 m / s , , 3, , 13.6 13.6 , , , or Z 2, 16 4 , 98. For the first excited stare, n 2 . hence, , Z , , F dF cos F , , 6.63 10, , , 3, , a1 Z1 2 , a Z 1 8, 2, 2 , 97. The shortest wavelength of Brackett series is, corresponding to the transition of electron between, n1 4 and n2 ., Similarity, the shortest wavelength of Balmer series, is corresponding to the transition of electron, between n1 2 and n2 ., , 0, , , , v re, dV, e 0 0 . If v is the orbital speed of, dr, r, the electron and m is its mass, then, F, , mv 2 V0 r0 e, , mv2 V0 r0e .... (i), r, r, , Bohr’s quantum condition is mvr , v, , nh, 2, , nh, ...... (ii) using (ii) in (i), we get, 2 mr, , , 2, h2, r2 2, n r n, 4 mV0 r0e , 89
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 5., , LEVEL-VI, SINGLE ANSWER QUESTIONS, 1., , Two photons having, (A) equal wavelengths have equal linear momenta, (B) equal energies have equal linear momenta, (C) equal frequencies have equal linear momenta 6., (D) equal linear momenta have equal wavelengths., , 2., , The work function of a certain metal is ., 0, When a monochromatic light of wavelength, 0 is incident such that the plate gains a, total power P. If the efficiency of photoelectric, emission is % and all the emitted, photoelectrons are captured by a hollow, conducting sphere of radius R already charged, to potential V, then neglecting any interaction, between plate and the sphere, expression of, potential of the sphere at time t is :, , hC, , (A) V +, , Pe t, , 100 Pe t, 4 0 RhC, , (B)V– 400 RhC, 0, Pe t, , (C) V, , 3., , (D) 4 RhC, 0, A metal plate is exposed to light with, wavelength . It is observed that electrons, are ejected from the surface of the plate. When 7., a retarding uniform electric field E is imposed,, no electron can move away from the plate, farther than a certain distance d. Then the, threshold wavelength 0 for the material of, plate is (e is the electronic charge, h is Planck's, constant and c is the speed of light), 1, , 1, , 1, , eEd , , (B) 0 = hc , , , , hc, eEd, , eEd, , (C) 0 = 0 –, 4., , hc , , , (A) 0 = , eEd , , 1, , (D) 0 = 0 – hc, In a photoelectric experiment, with light of, wavelength , the fastest electron has speed, v. If the exciting wavelength is changed to, , 3, ,, 4, , the speed of the fastest emitted electron will, become, (A) v, , 3, 4, , (C) less than v, 90, , (B) v, 3, 4, , 4, 3, , (D) greater than v, , 4, 3, , 8., , Two hydrogen atoms in ground state are, moving in opposite directions with the same, speed and collide head on. The minimum, kinetic energy of each hydrogen atom for the, collision to be inelastic so that both the atoms, are excited is, (A) 13.6 eV (B) 10.2 eV (C) 20.4 eV (D) 27.2 eV, Both the frequency and the intensity of a beam, of light falling on the surface of photoelectric, material are increased by a factor of two. This, will :, (A)increase both, the maximum kinetic energy of, the photo-electrons, as well as photoelectric, saturation current by a factor of two., (B)increase the maximum kinetic energy of the, photo-electrons by a factor greater than two and, would increase the photoelectric saturation current, by a factor of two., (C)increase the maximum kinetic energy of the, photoelectrons by a factor greater than two and, will have no effect on the magnitude of the, photoelectric saturation current produced., (D)increase the maximum kinetic energy of the, emitted photo-electrons by a factor of two but will, have no effect on the saturation photoelectric, current., Radiation of frequency 1.5 times the threshold, frequency is incident on a photosensitive, material. If the frequency of incident radiations, is halved and the intensity is doubled, the, number of photoelectrons ejected per second, becomes, (A) zero, (B) half of its initial value, (C) one fourth the initial value, (D) three fourth the initial value, A parallel beam of monochromatic radiation, of cross-section area A( < a2), intensity I and, frequency is incident on a solid conducting, sphere of work function 0, [h > 0 ] and radius 'a'. The sphere is, grounded by a conducting wire . Assume that, for each incident photon one photoelectron is, ejected. Just after this radiation is incident, on initially uncharged sphere, the current, through the conducting wire is:, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, Ae, , (A) h, , I,A, a, , Ae, , (B) 2h, 2Ae, , (C) h, , (D) none of these, 9., , A parallel beam of light of intensity I and cross, section area S is incident on a plate at normal, incidence. The photoelectric emission, efficiency is 100%, the frequency of beam is, and the work function of the plate is (h, > ). Assuming all the electrons are ejected, normal to the plane and with same maximum, possible speed. The net force exerted on the, plate only due to striking of photons and, subsequent emission of electrons is, S 2h, , , , (A) h 2m(h ) , , , 2S h, , , , (B) h 2m(h ) , , , S h, , , , (C) h 2m(h ) , , , 2S h, , , , (D) h m(h ) , , , 10. The electric field of an electromagnetic wave, changes with the time as, , E K 1 cos t cos t, where, = 5 × 1015 s -1 , = 2 × 1016 s1 and K is, constant. This radiation is incident on a sample, of hydrogen atoms initially in ground state., Assume that atoms absorb light as photons., Neglecting recoil of hydrogen nucleus on, ionisation, what will be the energy of ejected, electrons from hydrogen. [The ionisation, energy of hydrogen atom = 13.6 eV and, h = 2 × 6.6 × 10-16 eV-s], (A) 0.7 eV (B) 0.9 eV (C) 1.4 eV (D) 2.9 eV, 11. Electrons in a sample of gas containing, hydrogen like atom (Z=3) are in fourth excited, state. When photons emitted only due to, NARAYANAGROUP, , transition from third excited state to second, excited state are incident on a metal plate, photoelectrons are ejected. The stopping, potential for these photoelectrons is 3.95 V., Now, if only photons emitted due to transition, from fourth excited state to third excited state, are incident on the same metal plate, the, stopping potential for the emitted, photoelectrons will be approximately equal to, (A) 0.85 V (B) 0.75 V (C) 0.65 V (D) None of these, 12. The photon radiated from hydrogen, corresponding to 2nd line of Lyman series is, absorbed by a hydrogen like atom ‘X’ in 2nd, excited state. As a result the hydrogen like, atom ‘X’ makes a transition to nth orbit. Then,, (A) X = He+, n = 4 (B) X = Li++, n = 6, (C) X = He+, n = 6 (D) X = Li++, n = 9, 13. In the hydrogen atom, an electron makes a, transition from n = 2 to n = 1. The magnetic, field produced by the circulating electron at, the nucleus, (A) decreases 16 times (B) increases 4 times, (C) decreases 4 times (D) increases 32 times, 14. An electron in a hydrogen atom makes a, transition from first excited state to ground, state. The equivalent current due to circulating, electron, (A) increases 2 times (B) increases 4 times, (C) increases 8 times (D) remains the same, 15. An particle with a kinetic energy of 2.1 eV, makes a head on collision with a hydrogen, atom moving towards it with a kinetic energy, of 8.4 eV. The collision., (A) must be perfectly elastic, (B) may be perfectly inelastic, (C) may be inelastic, (D) must be perfectly inelastic, 16. A monochromatic radiation of wavelength , is incident on a sample containing He+. As a, result the Helium sample starts radiating. A, part of this radiation is allowed to pass through, a sample of atomic hydrogen gas in ground, state. It is noticed that the hydrogen sample, has started emitting electrons whose maximum, Kinetic Energy is 37.4 eV. (hc = 12400 eV Å), Then is (A) 275 Å (B) 243 Å (C) 656 Å (D) 386 Å, 91
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 17., , The difference between the longest, wavelength line of the Balmer series and, shortest wavelength line of the Lyman series, for a hydrogen like atom (Atomic number Z), equal to . The value of the Rydberg, constant for the given atom is, , 5 1, 31 1, 5 Z2, (A), (C), (D) none, 2 (B), 31 ·Z, 5 ·Z2, 36 , 18. A particle of mass ‘m’ is projected from ground, with velocity ‘u’ making angle ‘ ’ with the, vertical. The de-Broglie’s wave length of the, particle at the highest point is :, (B), , (A) , , h, h, h, (C), (D), mu sin , mu cos , mu, , 19. A particle of charge q0 and of mass m0 is, projected along the y-axis at t = 0 from origin, with a velocity V0. If a uniform electric field, E0 also exists along the x-axis, then the time, at which debroglie wavelength of the particle, becomes half of the initial value is :, m 0v0, m 0v0, (A) q E (B) 2 q E (C), 0, 0, 0, 0, , m 0v0, m 0 v0, 3 q 0 E 0 (D) 3 q 0 E 0, , 20. If we assume that penetrating power of any, radiation/particle is inversely proportional to, its De-broglie wavelength of the particle then:, (A) a proton and an -particle after getting, accelerated through same potential difference will, have equal penetrating power., (B) penetrating power of -particle will be greater, than that of proton which have been accelerated, by same potential difference., (C) proton's penetrating power will be less than, penetrating power of an electron which has been, accelerated by the same potential difference., (D) penetrating powers can not be compared as, all these are particles having no wavelength or wave, nature., 21. The energy ratio of two K photons obtained, in x-ray from two metal targets of atomic, numbers Z1 and Z2 is:, (A), , Z1, Z2, , (B), , Z1 , , , Z2 , , 2, , Z 1, (C) Z 1 1 , 2, , , 2, , ( Z1 1), , (D) ( Z 1), 2, , 22. In an x - ray tube, if the accelerating potential, difference is changed, then:, (A) the frequency of characteristic x - rays of a, material will get changed, 92, , (B) number of electrons emitted will change, (C) the difference between 0 (minimum, wavelength) and k (wavelength of ka x - ray) will, get changed, (D) difference between k and k will get, changed., 23. A beam of electrons striking a copper target, produces X-rays.Its spectrum is as shown., Keeping the voltage same if the copper target, is replaced with a different metal, the cut-off, wavelength and characteristic lines of the new, spectrum will change in comparision with old, as :, relative, intensity, , cutoff, wavelength, , wavelength, , (A) Cut-off wavelength will remain unchanged, while characteristic lines will be different., (B) Both cut-off wavelength and characteristic lines, will remain unchanged., (C) Both cut-off wavelength and characteristic lines, will be different., (D) Cut-off wavelength will be different while, characteristic lines will remain unchanged., 24. The wavelengths of K x-rays of two metals, 4, , 1, , ‘A’ and ‘B’ are 1875 R and 675 R, respectively, where ‘R’ is rydberg constant., The number of elements lying between ‘A’ and, ‘B’ according to their atomic numbers is, (A) 3, (B) 6, (C) 5, (D) 4, 25. A cobalt (atomic no. = 27) target is bombarded, with electrons, and the wavelengths of its, characteristic x-ray spectrum are measured., A second weak characteristic spectrum is also, found, due to an impurity in the target. The, wavelengths of the K lines are 225.0 pm, (cobalt) and 100.0 pm (impurity). Atomic, number of the impurity is (take b = 1), (A) 39, (B) 40, (C) 59, (D) 60, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 26. The wavelength of the first spectral ine in the, Balmer series of hydrogen atom is 6561 A0., The wavelength of the second spectral line in, the Balmer series of singly-ionized helium, atom is (JEE 2011), A) 1215 A0 B) 1640 A0 C) 2430 A0 D) 4687 A0, 27. Two identical non relativistic particles move, at right angles to each other, possessing de, Broglie wavelengths λ1 & λ 2 . Find the de, Broglie wavelength of each particle in the, frame of their centre of mass., λ1λ 2, 2λ1λ 2, λ1λ 2, A) λ1 + λ 2 B) λ 2 + λ 2 C) λ + λ D) λ 2 + λ 2, 1, , 2, , 1, , 2, , 1, , 2, , 28. Which of the following statements about xrays is false?, , , , , , A) E Kα +E Lβ = E Kβ +E Mα = E Kγ, where E is the energy of respective X-rays, B) For the harder X-rays, the intensity is higher, than soft X-rays, C) The continuous and characteristic X-rays differ, only in the method of creation, D) The cut-off wavelength λ min depends only on, the acceleration voltage applied between the target, and the filament., 29. Peak emission from a black body at a certain, temperature occurs at a wavelength . On, increasing its temperature, total radiation, emitted is increased 16 times. At the initial, temperature, when peak radiation from the, black body is incident on a metal surface, it, does not cause photoemission from surface., After the increase of temperature the peak, radiation from black body caused, photoemission. To bring these photoelectrons, to rest, a potential equivalent to the excitation, energy between n = 2 to n = 3 Bohr levels of, hydrogen atom is required. If work function, of metal is 2.24 eV, then value of is, [hc = 12400 ev – Å], A) 3000 Å B) 6000Å C) 9000Å D) 12000Å., 30. An X-ray tube is operating at 50 kV and 20, mA.The target material of the tube has a mass, of 1.0 kg and specific heat 495 Jkg-1C-1. One, percent of the supplied electric power is, converted into X-rays and entire remaining, energy goes into heating the target. Then the, incorrect option is, (A) the average rate of rise of temperature of the, target would be 2oC/sec, NARAYANAGROUP, , (B) the minimum wavelength of the X-rays emitted, is about 0.25 10-10 m, (C) a suitable target material must have a high, melting temperature, (D) a suitable target material must have low thermal, conductivity, , MULTIPLE ANSWER QUESTIONS, 31. Hydrogen atoms absorb radiation of, wavelength 0 and consequently emit, radiations of 6 different wavelengths of which, two wavelengths are shorter than 0 . Then,, (A) The final excited state of the atoms is n 4, (B) the initial state of the atoms may be n 2, (C) The initial state of the atoms may be n 3, (D) there are three transitions belonging to Lyman, series, 32. A neutron moving with a speed v,makes a head, on collision with a hydrogen atom in ground, state which is at rest.Choose the corret options, (s) of the following (assume mass of neutron, is nearly equal to the mass of hydrogen atom)., (A) For the collision to the inelastic, the minimu, value of initial kinetic energy of neutron before, collision is 20.4 eV, (B) For the collision to be inelastic, the minimum, value of initial kinetic energy of the neutron before, collision is 10.2 eV., (C) For any value of intial kinetic energy of neutron, the collision may be elastic., (D) If initial kinetic energy of neutron before, collision is 27.2 eV, the collision may be elastic or, inelastic or prefectly inelastic., 33. When a point light source, of power W, emiting, monochromatic light of wavelength is kept, at a distance a from a small photosensitive, surface of wok function and area S. Then, (A) number of photons strking the surface per unit, time as W S / 4 hca 2, (B) the maximum energy of the emitted, hc , photoelectrons is, , (C) the stoppeing potential needed to stop the, most energetic emitted photelectrons as, , e, , hc , , , (D) photoemission occurs only if lies in the range, 0 hc / ., 93
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 34. In Bohr’s model of the hydrogen atom, let R., V. T and E represent the radius of the orbit,, speed of the electron, time period of, revolution of electron, and the total energy of, the electron, respectively. The quantities, proportional to the quantum number n are, T, V, (D), R, E, 35. For a certain metal, the K absorption edge is, , (A) VR, , (B) RE, , (C), , at 0.172A0. The wavelength of K , K and, K lines of K series are 0.210A0, 0.192A0, and, , 0.182A0, respectively. The energies of K , L, and M orbits are, respectively. Then, , EK , EL and, , (a) EK 13.04keV, , (b) EL 7.52keV, , (c) EM 3.21keV, , (d) EK 13.04keV, , EM ,, , 36. Photons of energy 5 eV are incident on, cathode. Electrons reaching the anode have, kinetic energies varying from 6eV to 8eV., hv, –, , – – –, , 5V, , (A) Work function of the metal is 2 eV, (B) Work function of the metal is 3 eV, (C) Current in the circuit is equal to saturation value., (D) Current in the circuit is less than saturation value., 37. The maximum kinetic energy of, photoelectrons ejected from a photometer, when it is irradiated with radiation of, wavelength 400 nm is 1 eV. If the threshold, energy of the surface is 1.9 eV., (A) The maximum K.E. of photo electrons when it, is irradiated with 500 nm photons will be 0.42 eV., (B) The maximum K.E. in case (a) will be 1.725eV., (C) The longest wavelength which will eject the, photo electrons from the surface is nearly 610 nm., (D) Maximum K.E. will increase if the intensity of, radiation is increased, 94, , 38. A beam of light having frequency is incident, on an initially neutral metal of work function, (h > ) . Then, (A) all emitted photoelectrons have kinetic energy, equal to (h - ) ., (B) all free electrons in the metal, that absorb, photons of energy h completely, may not be, ejected out of the metal., (C) after being emitted out of the metal, the kinetic, energy of photoelectrons decreases continuously, as long as they are at a finite distance from metal., (D) the emitted photo electrons move with constant, speed after escaping from the electric field of metal., 39. A gas containing hydrogen like atoms with, atomic number Z, emits photons in transition, n + 2 n; where n = Z and work function of, metal is 4.2eV . The photons are incident on, the metal emitting electrons of minimum de, Broglie wavelength 5A 0, A) The maximum K.E of photoelectron is 6 eV, B) The maximum K.E of photoelectron is 5 eV, C) The value of Z is 2 D) The value of Z is 3, 40. The radiation emitted when an electron jumps, from n=4 to n=3 in a lithium atom z = 3 falls, on a metal surface to produce photoelectron., When the photoelectrons with maximum kinetic, energy are made to move perpendicular to a, magnetic field of 2x10-4T, it traces a circular, , 9.1, cm . [RhC=13.6eV](mass, 1.6, of electron=9.1x10-31kg), A) The wavelength of radiation falling on metal is, 208 nm (nearly), B) The work function of metal is 3.95 eV, C) The kinetic energy of photoelectrons is 6eV, D) The energy of incident photon is 5.95eV, 41. Which of the following statements about the, photoelectric effect, are true, A) greater the frequency of the incident light, greater, is the stopping potential., B) greater the energy of photons is, the smaller is, the stopping potential., C) greater the intensity of light is, greater is the cut, off frequency., D) greater the frequency of incident light is, greater, is max kinetic energy of ejected electrons., path of radius, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 42. The work function of a metal is 2.5 eV. A, monochromatic light of wavelength 3000 Å, falls on it, A) The maximum kinetic energy of ejected, electron is 1.64 eV approximately, B) The minimum kinetic energy of ejected electron, is zero, C) The stopping potential is 1.64 V approximately, D) Electrons can not be ejected, , COMPREHENSION TYPE QUESTIONS, PARAGRAPH - I, A physicist wishes to eject electrons by, shining light on a metal surface. The light, source emits light of wavelength of 450 nm., The table lists the only available metals and, their work functions., Metal, W0(eV), Barium, –, 2.5, Lithium, –, 2.3, Tantalum, –, 4.2, Tungsten, –, 4.5, 43. Which metal(s) can be used to produce, electrons by the photoelectric effect from, given source of light ?, (A) Barium only, (B) Barium or lithium, (C) Lithium,tantalum or tungsten, (D) Tungsten or tantalum, 44. Which option correctly identifies the metal, that will produce the most energetic electrons, and their energies ?, (A) Lithium, 0.45 eV (B) Tungston, 1.75 eV, (C) Lithium, 2.30 eV (D) Tungston, 2.75 eV, 45. Suppose photoelectric experiment is done, separately with these metals with light of, wavelength 450 nm. The maximum magnitude, of stopping potential amongst all the metals, is (in volts), (A) 2.75 (B) 4.5, (C) 0.45 (D) 0.25, PARAGRAPH -II, The photocurrent is measured with the help, of an ideal ammeter. Two plates of potassium, oxide of area 50 cm2 at separation 0.5 mm, are used in the vacuum tube. Photo current, in the circuit is very small so we can treat, potentiometer circuit an independent circuit., Light, , 1, Violet, , 2, Blue, , 3, Green, , 4, Yellow, , in Å 4000-4500 4500-5000 5000-5500 5500-6000, , NARAYANAGROUP, , 5, Orange, , 6, Red, , 6000-6500, , 6500-7000, , C, , A, , A, , V, , P, , 8W, 100cm, Q, , J, 20V, , 2W, , 46. The number of electrons appeared on the, surface of the cathode plate, when the jockey, is connected at the end ‘P’ of the, potentiometer wire. Assume that no radiation, is falling on the plates., (A) 8.85 × 106, (B) 11.0625 × 109, (C) 8.85 × 109, (D) 0, 47. When radiation falls on the cathode plate a, current of 2 A is recorded in the ammeter.., Assuming that the vacuum tube setup follows, ohm’s law, the equivalent resistance of, vacuum tube operating in this case when, jockey is at end P., (B) 16 × 106 , (A) 8 × 108 , (C) 8 × 106 , (D) 10 × 106 , 48. It is found that ammeter current remains, unchanged (2 A) even when the jockey is, moved from the end ‘P’ to the middle point of, the potentiometer wire. Assuming all the, incident photons eject electron and the power, of the light incident is 4 × 10–6 W. Then the, colour of the incident light is, (A) Green (B) Violet (C) Red (D) Orange, 49. Which of the following colour may not give, photoelectric effect from this cathode, (A) Green (B) Violet (C) Red (D) Orange, 50. When other light falls on the anode plate the, ammeter reading remains zero till, jockey is, moved from the end P to the middle point of, the wire PQ. Thereafter the deflection is, recorded in the ammeter. The maximum, kinetic energy of the emitted electron is, (A) 16 eV (B) 8 eV (C) 4 eV (D) 10 eV, PARAGRAPH -III, Two initially uncharged concentric thin, conducting spherical shells of radius a and, 2a are as shown and the inner shell is, grounded. The work function of outer shell is, 0 .At time t = 0, a continuous parallel beam, of monochromatic radiation of cross-section, area A and intensity I is incident on outer, shell. The energy of each photon is h such, 95
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, that h > 0 .Assuming for each incident, photon one photoelectron is ejected, answer, the following questions., I, A, , a, , 2a, , 51. The time after t = 0, at which charge on outer, sphere becomes constant., (A), (C), , 8 0 ah(h 0 ), Ae, , 2, , 2 0 ah(h 0 ), A e2, , (B), , 4 0 ah(h 0 ), A e2, , (D) None of these, , 52. Net charge flowing from inner shell to ground, during the above time is:, (A), , 16 0 a(h – 0 ), e, , (B), , 8 0 a(h – 0 ), e, , (C), , 4 0 a(h – 0 ), e, , (D) None of these, , 53. The maximum potential difference between, inner and outer shell is :, (A), , h 0, e, , (B), , h 0, 2e, , (C), , 2 h 0, 3, e, , (D) None of these, PARAGRAPH -IV, A monoenergetic beam of particles, each, of them having kinetic energy E is incident on, a sample of singly ionised Helium gas in, ground state. The Helium sample may start, emitting radiation since the kinetic energy of, incident particles is sufficiently high. A partt, of this radiation (if any) is allowed to pass, through atomic Hydrogen sample in ground, state. A detector placed near the Hydrogen, records both radiation and electrons. Assume, all He+ ions to be initially at rest., 54. If maximum kinetic energy of electrons, intercepted by the detector is 27.2 eV, the, minimum kinetic energy of particle must, be (A) 81.6 eV (B) 96.8 eV (C) 102eV (D) 40.8eV, 55. If maximum kinetic energy of electrons, intercepted by the detector is 27.2 eV, the K.E., of particle must be lower than (A) 81.6 eV, (B) 96.8 eV, (C) 102 eV, (D) there is no upper limit, 96, , 56. If no electrons are intercepted by the detector,, the energy of photons detected by it can be (A) 10.2 eV, (B) 40.8 eV +, (C) 12.1 eV, (D) None of these, PARAGRAPH -V, The properties of X-ray as put forward by, Rontgen in his pioneering paper on the topic, are as follows :, 1. X-ray posses a very strong penetrative power., It can penetrate wood upto 3 cm and an, aluminium foil upto 15 mm. If the hand is held, between the discharge tube and the screen the, dark shadow of the bones is visible within the, slightly dark shadow of the hand.”, 2. Photographic plates and films “show, themselves susceptible to x-rays.” Hence,, photography provides a valuable method of, studying the effects of x-rays., 3. X-rays are neither reflected nor refracted (so, far as Rontgen could discover). Hence, “Xrays cannot be concentrated by lenses.”, 4. X-rays discharge electrified bodies, whether, the electrification is positive or negative., 5. X-rays are generated when the cathode rays, of the discharge tube strike any solid body. A, heavier element, such as platinum, however,, is much more efficient as a generator of xrays than a lighter element, such as aluminum., Most of Rontgen’s observations stood the test, of time, though some of them needed to be, modified later., Now today we know if electron are accelerated, through a potential difference V, then, maximum energy of emitted photon could be, hc, , hc, , Emax = eV, = eV, min =, eV, min, min is also called cut off wavelength. Since, electron may loose very small energy in a, given collision, the upper value of will, approach infinity. When X-ray is produced in, an X-ray tube, two types of X--ray spectra are, observed : continuous spectra and line spectra., A continuous spectrum is produced by, bremsstrahlung, the electromagnetic radiation, produced when free electrons are accelerated, during collisions with ions., A line spectrum results when an electron, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , having sufficient energy collides with a heavy, atom, and an electron in an inner energy level, is ejected from the atom. An electron from an, outer energy level then fills the vacant inner, energy level, resulting in emission of an Xray photon., , Intensity, , P2, P1, , (C) Emission of X-ray photons as a result of, electronic transitions in atoms, (D) Both A and C, 58. Rontgen conducted an experiment where, Parallel X-rays are passing through an optical, lens available at his time with focal length 10, cm then the correct ray diagram for the, paraxial rays will be :, , I, , X-ray, , , , wavelength, , The electron knocks out an inner shell electron, of the atom with which it collides. Let us take, a hypothetical case of a target atom whose Kshell electron has been knocked out as shown., -, , e, , Incident, electron, , (A), , X-ray, , 20cm, , Hv, K X-ray, , K-electron, , This will create a vacancy in K-shell. Sensing, this vacancy an electron from a higher energy, state may make a transition to this vacant, state. When such a transition takes place the, difference of energy is converted into photon, of electromagnetic radiation, which is called, characteristic X-rays. Now depending upon, the shell from which an electron makes a, transition to K-shell we may have different, lines in the K series of X-rays e.g. if electron, from L shell jumps to K shell we have K , if, electron from M shell jumps to K shell we have, K X-ray and so on. Moseley conducted, many experiments on characteristic X-rays,, the findings of which played an important role, in developing the concept of atomic number., Moseley’s observations can be expressed as, = a (Z – b), where a and b are constants. Z is the atomic, number of target atom and n is the frequency., 57. In intensity v/s wavelength graph, which of the, following process is responsible for the, intensity of peaks P1 and P2 ?, (A) Bremsstrahlung, (B) Absorption of X-ray photons resulting in, electronic excitations in atom, NARAYANAGROUP, , 10cm, , (B), , X-ray, , 10cm, , (C), , X-ray, , (D), 59. Use the data given in the table shown to, arrange the elements in modern periodic table:, , Element, , Wavelength, , Atomic weigth, , P, , 1.94A, , 55.8, , Q, , 1.79A, , 58.9, , R, , 1.66A, , 58.7, , (A) PQR, , (B) PRQ, , (C) RQP (D) QRP, 97
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ATOMIC PHYSICS, 60. X-rays of high penetrating power are called, hard X-ray. Hard X-rays have energy of the, order of 105 eV. The minimum potential, difference through which the electrons should, be accelerated in an X-ray tube to obtain Xray of energy 105 eV is :, (A) 105 eV (B) 50 kV (C) 50 keV (D) 105 V, PARAGRAPH -VI, A positronium atom consists of an electron, and a positron revolving about their common, centre of mass.Calculate, 61. Separation between the electron and positron, in their first excited state., A) 4.232Å B)5.232Å C) 6.2Å D) 7.Å, 62. Kinetic energy of the electron in ground state, A) 4.5eV B) 3.4eV C) 7.6eV D) 8.0eV, PARAGRAPH -VII, A neutron beam, in which each neutron has, same kinetic energy, is passed through a, sample of hydrogen like gas (but not, hydrogen) in ground state and at rest. Due to, collision of neutrons with the ions of the gas,, ions are excited and then they emit photons., Six spectral lines are obtained in which one of, the lines is of wavelength (6200/51) nm, 63. Which gas is this ?, A) H, B) He , C) Li D) None, 64. What is the minimum possible value of kinetic, energy of the neutrons for this to be possible., The mass of neutron and proton can be, assumed to be nearly same. Use hc = 12400, A) 10.5eV B)63.75 eV C) 6.9eV D) 8.9eV, PARAGRAPH -VIII, If an electron jumps from mth orbit to the nth, orbit (m > n). The energy of the atom changes, from Em to En. This extra energy Em – En is, emitted as a photon whose wavelength is given, by, 1, , 1, 1 , 1, RZ2 2 2 Where R=1.09×107m–, , m , n, , (Rydberg constant) A photon ejected from, the transition of electron from mth excited state, of He+ ion to nth state is allowed to fall on a, photoelectric material with work function, , = 7eV. [Given h = 4.14 × 10–15 eVs and c = 3 ×, 108 ms–1], 65. The photo-electric effect will take place for, the transition of electron from, A) 3rd orbit to 2nd orbit B) 4th orbit to 3rd orbit, C) 5th orbit to 3rd orbit, D) Photoelectron can not be ejected, 98, , JEE-ADV PHYSICS- VOL- V, 66. The stopping potential for the photo-electrons, ejected due to transition from 2nd orbit to, ground state is, A) 30.5 V B) 40.5 V C) 50.5 V D) 60.5 V, PARAGRAPH -IX, A beam of alpha particles is incident on a, target of lead. A particular alpha particle, comes in “head-on” to a particular lead, nucleus and stops 6.5 × 10-14 m away from the, centre of the nucleus. (This point is well, outside the nucleus ). Assume that the lead, nucleus which has 82 protons, remains at rest., The mass of alpha particle is 6.6 × 10-27 kg., 67. The electrostatic potential energy at the, instant that the -particle stops is, A) 3.63 MeV, B) 40 MeV, C) 45 MeV, D) 36.3 MeV, 68. The initial kinetic energy (in Joules) did the, alpha particle have, A) 3.82 × 10-13 J, B) 5.82 × 10-13 J, C) 1.32 × 10-12 J, D) 3.82 × 10-12 J, 69. What was the initial speed of alpha particle?, A) 1.32 × 107 m/s, B) 13.2 × 102 m/s, 7, C) 0.13 × 10 m/s, D) 132 × 102 m/s, , STATEMENT TYPE QUESTIONS, (A) Statement-1 is True, Statement-2 is True;, Statement-2 is a correct explanation for, Statement-1, (B) Statement-1 is True, Statement-2 is True;, Statement-2 is NOT a correct explanation for, Statement-1, (C) Statement-1 is True, Statement-2 is False, (D) Statement-1 is False, Statement-2 is True, 70. STATEMENT-1 : Though light of a single, frequency (monochromatic light) is incident on a, metal, the energies of emitted photoelectrons are, different, STATEMENT-2 : The energy of electrons just, after they absorb photons incident on metal surface, may be lost in collision with other atoms in the, metal before the electron is ejected out of the metal., 71. STATEMENT-1 : When a beam of highly, energetic neutrons is incident on a tungsten target,, no X-rays will be produced., STATEMENT-2 : Neutrons do not exert any, electrostatic force on electrons or nucleus of an, atom., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 72. STATEMENT–1 : In the duration electron jumps, Column-II, from first excited state to ground state in a stationary, (p) is independent of n., (q) is directly proportional to n, isolated hydrogen atom, angular momentum of the, (r) is inversely proportional to n., electron about the nucleus is conserved., (s) is directly proportional to L., STATEMENT–2 : As the electron jumps from, first excited state to ground state, in a hydrogen 78. In the shown experimental setup to study, photoelectric effect, two conducting electrodes, atom, the electrostatic force on electron is always, are enclosed in an evacuated glass-tube as, directed towards the nucleus., shown. A parallel beam of monochromatic, 73. STATEMENT-1 : In process of photoelectric, radiation, falls on photosensitive electrode., emission, all emitted electrons donot have same, Assume that for each photon incident, a, kinetic energy., photoelectron is ejected if its energy is greater, STATEMENT-2 : If radiation falling on, than work function of electrode. Match the, photosensitive surface of a metal consists of, statements in column I with corresponding, different wavelengths, then energy acquired by, graphs in column II., electrons absorbing photons of different, Parallel beam, wavelengths shall be different., of ligth, 74. STATEMENT-1 : The de-Broglie wavelength of, a molecule (in a sample of ideal gas) varies inversely, as the square root of absolute temperature., STATEMENT-2 : The rms velocity of a molecule, (in a sample of ideal gas) depends on temperature., 75. STATEMENT–1 : When applied potential, Column I, difference between cathode and anode is increased, (A) Photocurrent versus intensity of radiation is, in a coolidge tube, cut off wavelength decreases., represented by, STATEMENT–2 : Cut off wavelength for X(B) Maximum kinetic energy of ejected, rays depends on atomic number of target material., photoelectrons versus frequency for electrodes of, 76. STATEMENT – 1 : In photoelectric effect,, different work function is represented by, electron absorbing the photon can not be a free, (C) Photo current versus applied voltage for, electron., different intensity of radiation is represented by, STATEMENT – 2 : A free electron can’t absorb, (D) Photo current versus applied voltage at constant, a photon completely, intensity of radiation for electrodes of different work, function., MATRIX MATCH TYPE QUESTIONS, Column II, 77. The energy, the magnitude of linear, momentum, magnitude of angular momentum, (p), and orbital radius of an electron in a hydrogen, atom corresponding to the quantum number n, are E, p, L and r respectively. Then according, to Bohr's theory of hydrogen atom, match the, expressions in column-I with statement in, (q), column-II., Column-I, (A) Epr, (B), , p, E, , (C) Er, (D) pr, , NARAYANAGROUP, , (r), , (s), , 99
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 79. In the shown experimental setup to study, photoelectric effect, two conducting electrodes, are enclosed in an evacuated glass-tube as, shown. A parallel beam of monochromatic light,, falls on photosensitive electrodes. The emf of, battery shown is high enough such that all, photoelectrons ejected from left electrode will, reach the right electrode. Under initial, conditions photoelectrons are emitted. As, changes are made in each situation of column, I; Match the statements in column I with, results in column II., , Parallel beam, of light, , Column I, A) If frequency of incident light is increased keeping, its intensity constant, B) If frequency of incident light is increased and its, intensity is decreased, C) If work function of photo sensitive Electrode is, increased, D) If intensity of incident light is increased, Keeping its frequency constant, Column II, (p) magnitude of stopping potential will increase, (q) current through circuit may stop, (r) maximum kinetic energy of ejected, photoelectrons will increase, (s) saturation current will increase, , INTEGER TYPE QUESTIONS, 80. In the figure shown electromagnetic radiations, of wavelength 200nm are incident on the, metallic plate A. The photo electrons are, accelerated by a potential difference 10V., These electrons strike another metal plate B, from which electromagnetic radiations are, emitted. The minimum wavelength of the, emitted photons is 100nm. The work function, of the metal ‘ ’ is 38 10 x find the value of, x. use hc = 12400 eVÅ, use Rch = 13.6 eV., 100, , 200nm, A, , B, min=100nm, , 10V, , 81. Consider Bohr's theory for hydrogen atom., The magnitude of angular momentum, orbit, radius and frequency of the electron in nth, energy state in a hydrogen atom are , r & f, respectively. Find out the value of ' x ', if (frl), is directly proportional to nx., 82. A gas of H-atoms in excited state n2 absorbs a, photon of some energy and jump in higher, energy state n1. Then it returns to ground state, after emitting six different wavelengths in, emission spectrum. The energy of emitted, photon is equal, less or greater than the energy, of absorbed photon. If n1 x n2 find x ?, 83. H-atoms in a sample are excited to same, energy level n = 5 and it is found that photons, of all possible wavelengths are released when, the atoms return back to the ground state. The, minimum number of H atoms in the sample is, 84. A proton is fired from very far away towards a, nucleus with charge Q 120e, where e is the, electronic charge. It makes a closest approach, of 10 fm to the nucleus. The de Broglie, wavelength (in units of fm) of the proton at its, start is : (take the proton mass,, 5, h, m p 1027 k g ;, 4.2 1015 J .s / C ;, 3, e, , 1, 9 109 m / F ; 1 fm 1015 m ), 4 0, [JEE-2012], 85. The average life time of an excited state of an, , electron in Hydrogen is of order of 108 s . It, , , , , , 6, makes x 10 revolutions when it is in the, , state n 2 and before it suffers a transition, n 1 state. Find the value of x. Given, h, = 64×1014 , where m is mass of, 2, 2, 4π ma 0, electron and 'a 0 ' is Bohrs radius., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 86. A monochromatic light source of frequency v, illuminates a metallic surface and ejects, photoelectrons. The photo electrons having, maximum energy are just able to ionize the, hydrogen atoms in ground state. When the, whole experiment is repeated with an incident, radiation of frequency, , 5v / 6 ,, , the, , photoelectrons so emitted are able to excite, the hydrogen atom beam which then emits a, radiation of wavelength 1215 A 0 . Then the, frequency v will be equal to k ×1015 Hz . The, value of k is (consider nearest integer), 87. Find the ratio of De-broglie wavelength of a, proton to that of a -particle if they are being, subjected to the same magnetic field so that, the radii of their path are equal to each other, , assuming the field induction vector B is, perpendicular to the velocity vectors of the, proton and the -particle., 88. A small particle of mass m moves in such a, 1, m2 r 2 where, 2, is a constant and r is distance of particle, from origin. Assume Bohr’s model of, quantization of angular momentum and, circular orbits, the radius of nth allowed orbit, , LEVEL - VI - SOLUTIONS, SINGLE ANSWER QUESTIONS, 1., , 2., , 1, P t e , electric potential after t sec V 100hc 4 R, , , 0, 3. The maximum kinetic energy of the electrons, immediately upon ejection is the difference between, the energy of the incident photon and the threshold, hc, , 1 eEd , Therefore, K = Eed and l0 = hc , , , , 4., , 1 2 hc, mv , 2, , , Clearly v' , 5., , 1, , is proportional to n x then find x., , LEVEL - VI - KEY, , NARAYANAGROUP, , hc, , energy. K = , This kinetic energy of, 0, ejected electron is converted to electrostatic, potential energy, DU = eEd, as electrons come to, rest while moving in the direction of electric field., , way that potential energy U , , 1.D 2.B 3.B 4.D 5.B 6.C 7.A 8.A 9.C 10.D, 11.B12.D 13.D14.C 15.C 16.B 17.C 18.B19.C, 20.B 21.C 22.C 23.A 24.D 25.B 26.A 27. D 28.B, 29.B 30.D 31.A,B,D 32.A,C,D 33.A,B,D 34.A,C,D, 35.A,B,C36.A,D 37.A,C 38.B,C 39.A,C 40.A,B,D, 41.A,D 42.A,B,C 43.B 44.A 45.C 46.C 47.C 48.D, 49.C 50.B 51.A 52.C 53.A 54.A 55.B 56.D 57.C, 58.D 59.A 60.D 61.A 62.B 63.B 64.B 65.A 66.B, 67.A 68.B 69.A 70.A 71.D 72.D 73.B 74.B 75.C, 76.A 77.A-r, B-q,s C-p, D-q,s 78.A-r, B-s , C-p,D-q, 79.A-p,r B-p,r C-q D-s 80.1 81.0 82.2 83.6, 84.7 85.8 86.5 87.2 88.2, , Two photons having equal wavelengths, have equal, linear momenta in magnitude not neccessarily in, same direction., P t e, Charge on the sphere after t sec = , 100 hc, , 6., , ;, , 1, , 1, hc, 4hc, mv '2 =, , –f=, (, 3, , /, 4, ), 3, 2, 4, v, 3, , Let K be the initial kinetic energy of each, atom.Maximum loss in KE takes place in perfectly, inelastic collision. From conservation of linear, momentum mu ( mu ) 2mv v 0, where v = common velocity after collision., Maximum possible loss in KE is, 1, 1, 1, K mu 2 2 2m 0 2 2 mu 2 2 K, 2, 2, 2, Minimum energy required to excite both the, electrons is E 10.2 10.2 20.4eV ., For inelastic collision with two electrons excited, K E 2 K 20.4 K min 10.2eV, The maximum K.E. of ejected photoelectron is, (K.E)max = h – 0, If the frequency of photon is doubled, maximum, kinetic energy of photo electron becomes, , max, , K.E., , , K.E.max, , =2h – 0 ; (K.E.), max, , Photo current i , , 0, ), 2 >2, =, h 0, 2(h , , int ensity of beam, h, , 101
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 7., 8., , If intensity and frequency both are doubled, the, photocurrent remains same., Under the given condition energy of photon, becomes less than the work function of the metal., Hence photo-emission stops altogether., Number of photons incident on sphere per, second =, =, , A, h, , A, h, , Ae, h, , dn IS, dn h, I , = number of photon, dt h, dt S, falling per second = number of electrons emitted, per second from the metal ( efficiency of, emission is 100 %), Force on the plate due to the absorption of, dn p dn h IS, , photos is F1 , dt, dt c, c, Force on the plate due to the ejection of, elect r ons, from, it, is, dn, IS, F2 , 2m KE , 2m h , dt, h, force, on, t he, met al, To t al, IS IS, F1 F2 , 2m h , c h, hIS IS, IS h, , 2m h 2m h , h , , h h, 10. The expression for electric field can also be, 1, 2, , written as E=Kcos t+ K cos t +, K cos t. The three terms correspond, , to photons of energies, , , , h, , , h, and, 2, 2, , .The last exceeds the ionisation, , 2, energy by 2.9 eV. That difference equals to the, KE of ejected electrons., 1, , 1, , 11. hv = 13.6 (3) 2 2 2 = 2.75 eV for, 5 , 4, transition from 4th to 3rd excited state, 1, , 1, , hv = (13.6) × (3)2 2 2 = 5.95 eV, 4 , 3, For transition from 3rd to 2nd excited state, 3.95 = 5.95 – , 2 eV, for longer wavelength eVs=2.75–2 =0.75 eV, 102, , 0i, 2r, , B=, , 0 e, 2rT, , current through conducting wire = photo, , 9., , h, , 13. B =, , photo electrons ejected per second, , current ejected from surface of sphere =, , 1, 2, , 1 , 1 1 , 2 1, 12. 13.6 13.6 Z 2 2 this is, 1 9 , 3 n , satisfied for Z 3 and n 9 of the given options, , 14. i =, , q, T, , and, , i=, , e, T, , [r n2 , T n3 ] B , , 3, and T n i , , 1, n5, , 1, n3, , 15., Pi = m .4. E 4mE = 0, For completely inelastic collision both come to, rest aft er collisio n and net energy of, 4E + E = 10.5 eV is lost. But electron in ground, state of H-atom can accept only an energy of, 10.2 eV.Hence the collision may be inelastic but, it can never be perfectly inelastic., 16. Maximum energy of radiation incident on Hsample = KEmax of electron + 13.6 eV = 51eV, this energy corresponds to the transition n = 4, to n = 1 in Helium. For electrons of the He to, 12400, 243 A0, get excited to n = 4; , 51, 1, 1, R z 2 (1 0) , Lym an , 17. , R z2, Lyman, , 1, , 1 1, 5 , R z2 R z2 , Balmer, 4 9, 36 , 36, 31, B alm er 5 R z 2 ; R =, 5z 2, h, h, 18. mv mu sin , 19. a , , E0 q0, h, , v u at , , . For, mv, m, , de-Broglie’s wave length to become half of its, initial value, its speed is to be increased to 2v0, E0 q0, t , v y v0 and v 2 vx2 v 2y, m, mv, i.e 4v02 vx2 v02 vx 3 v0 t 3 0 0, q0 E0, vx 0 , , h, 20. 2mq.V, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , 21. The energy of K x-ray photons is directly, propotional to (Z–1)2.The energy ratio of two K, photons obtained in x-ray from two metal targets, of atomic numbers Z1 and Z2 is, , Z1 1 , , , Z 2 1, , 2, , ., , 22. As the accelerating potential difference is changed, only the minimum wavelength changes., It has no effect on wavelengths of characteristic xrays (whether they are produced or not)., 23. Cut-off wave length depends on accelerating, voltage and characteristic wavelength depends on, nature of the target material., 24. Using, , 1, 1 , 1, = R(z – 1)2 2 2 , , n1 n2 , , 12 22, h, 1, 1, h, i.e v1c v 2 c , , , 2m 12 22, 2m 1 2, de-Broglie’s wavelength of one of the particles in, the center of mass frame of reference is, 21 2, h, , , m v1c, 12 22, 28. Penetrating power of hard x-rays is more than soft, x-rays and penetrating power depends on, frequency but not on intensity., 4, , E1 T1 , T2 2T1, 29. E eA T E T , 16E1 T2 , 4, , from wein’s displacement law, , 1875R, 4, , 1 , , 2 2, 1 1, KEmax eVs 13.6 1.89 eV, 4 9, hc, i.e eVs W 1.89 2.24 4.13 eV, 2, , 3, , 2 , = RZ1 1 4 ; z1 = 26, , , , 3, , 2 , For metal B; 675R = RZ 2 1 4 ; z2 = 31, , , , Therefore, 4 elements lie between A and B., 25. Using Mosely’s law for both cobalt and impurity, , = K (Z – 1), c, co, , c, , = K(Z – 1), , 12400, 3000 A0 6000 A0, 4.130, 30. Power reaching the target P Vi 1000W ., 2 , , c, , The rate of heat production , , = K (Zco – 1)and = K (Zx – 1), x, , 1, 1 , 2 1, RZ 2 2 , n1 n2 , , , 1, 1 1 3R 4, 0, 4R , , 16 1215A, 4 16 , h ˆ, h ˆ, h, 27. , and let v1 m i , v2 m j and, mv, 1, 2, m v1 mv2 v1 v2 h iˆ h ˆj, , 2m1, 2m2, 2m, 2, In centre of mass frame of reference, h ˆ, h ˆ, v1c v1 v c , i, j, and, 2m 1, 2m 2, vc , , v2c v 2 vc , , h ˆ, h ˆ, i, j, 2m 1, 2m 2, , NARAYANAGROUP, , 99, 1000 990W, 100, , ms d, d, hc, 990 , 20 c and min , s, dt, dt, eV, , co, Z 1, x, x, Z co 1 Zx =40, , 1, 1 1 5R, R , 26., 6561 4 9 36, , m T b, , 1 1 2T2 2 , , For K photon ; n1 = 1, n2 = 2, For metal A ;, , 4, , MULTIPLE ANSWER QUESTIONS, 31., , n n 1, 6n 4, 2, n=4, 3, 2, 1, , If the initial state were n 3 , in the emission, spectrum, no wavelengths shorter than 0, would have occurred. This is possible if initial, state were n 2 ., 32. Let m,M be the masses of neutron and hydrogen, atom respectively.Maximum possible loss takes, place in perfectly inelastic collision .From, co nser vat ion o f linear mo ment um, 103
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ATOMIC PHYSICS, , mv 0 M m v ' where v ' common, velocity after collision. Maximum possible, 1 2 1, 2, loss in KE is K mv M m v ', 2, 2, 1, M , 2, i.e K 2 mv M m , , , Least energy required to excite the electron, (from n 1 to n 2) is E 10.2 eV, If excitation takes place, KE of the system is, not conserved hence collison inelastic. i.e to, excite the electron K E, 1, mu 2, 2, , m, 1, M , , 2, , E mu E 1 , , M, , m, 2, , , M, , m, i.e KEmin E 1 10.2 1 1 20.4eV, M, If initial KE of the neutron is less than 20.4eV,, loss in KE will be less than 10.2 eV hence, electron can’t be excited hence atom can’t, absorb any energy internally and hence there will, no loss of kenitic energy of the system., The collision is elastic., Even though the possible loss in KE of the, system is greater than or equal the excition, energy, excitation may not take place necessarily, hence there may not be loss in KE of the system, hence for any value of initial KE of the neutron, the collision may be elastic but for collision to, be inelastic excitation should take place hence, initial KE of the neutron should not be less than, 20.4 eV., If initial KE of neutron is 27.2 eV, maximum, possible, lo ss, in, KE, is, M , 1, 1, K mu 2 , 27.2 13.6 eV, 2, 2, M m , This is enough to excite the electron and if this, takes place as both move together the collison, is perfectly inelastic., The loss may be less t han 13.6eV(say, 10.2eV),In this case excition may take place and, as both can’t move together the collision may, be inelastic.Excitation may not take place hence, there may not be any loss in KE actually, hence, the collision may be elastic., 33. The energy of each photon is hc / so that the, number of photons released per unit time is, , W / hc / . These phtons are spread out in all, , JEE-ADV PHYSICS- VOL- V, of an area S is a fraction S / 4 a 2 of the total, number of photons emitted. The maximum kinetic, energy of the emitted photoelectrons is, hc, 1, KEmax hv hc , , , The stopping potential is given by eVs KEmax ., 1, 1, hence Vs KEmax hc , e, e, For photoemission to be possible, we must have, hc, hc, hv , , , Thus the permitted range of values of is, 0 hc / ., 34. In Bohr’s model of hydrogen atom, 1, 1, R n2 , v n , T n3 and E n2, T, V, VR n , TE n , R n E n, 35. Energy of K absorption edge, 1242eVnm, 3, E K' , 0.0172nm 72.21 10 eV 72.21KeV, hc, Energy of K line Ek e, , 1242eVnm, 59.14 KeV, 0.021nm, , Energy of K shell = K k E K' , , , 59.14 72.21 KeV 13.04keV, Energy of L shell = Ek 72.21keV, , 64.69keV 72.21keV 7.52keV, Energy of M shell =, 1242eVnm, Ek Ek' , 72.21keV, 0.018nm, 69keV 72.21keV 3.21keV, 36. KEmax = 5 eV, when these electrons are accelerated through 5V,, they will reach the anode with maximum energy =, , 5 5 eV 10 8 2eV, Just after emission kinetic energies of electrons, range from 0 to 3 eV.If all the electrons reach the, anode, photo current reaches saturation and in this, situation range of KE s of electrons reaching anode, , directons over an area 4 a 2 so that the share, 104, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , should be 5eV to 8eV . As given KE range is, 6eV ot 8eV, some of the emitted electrons are not, reaching the anode, hence the current is less than, the saturation value., hc, hc, 37. KEmax W 1 1.9 2.9 eV, e, e, hc, 2.9 400 eV nm, e, hc 2.9 400, , 2.32 eV, for 500 nm, E , e500, 500, i.e KEmax 2.32 1.9 0.42 eV, hc 2.9 400, , 610 A0, eW, 1.9, 38. Some electrons in the interior of the metal may, use all the energy absorbed by them in comming, on to the suface hence those can’t come out., Afer being emitted eletrons get retarted due to, positive potential devolped on the metal (due to, loss of electrons), 0 , , 39., , 1, 1 , 13.6z2 2 , , 2 W, n n 2 , , KE max and KEmax , , h2, 2m 2, , 40. Energy of emitted photons, , minimum. W0 for lithium = 2.3 eV, 45. The maximum magnitude of stopping potential, will be for metal of least work function. required, stopping potential is Vs =, , C, , A, , P, , 8, 100cm, Q, , J, , 20V, , 46. Q = CV ; ne =, , 2, , 0 A, d, , mv 2mKEmax, B2q2r 2, , KEmax , 2eV, Bq, Bq, 2m, W E KEmax 3.95eV, , 48., , n=, , hc, , , , , , , , , , 1 hc, , , , Stopping potential = 1.64V, e , , 12400, , 43. E =, E = 2.75 eV, ...........(1), 4 .500 Å, for photoelectric effect E>W0 (work function)., 44. E = W0 + Ek, (Ek) = E – W0, for maximum value of (Ek ), W0 should be, NARAYANAGROUP, , 0.5 10 3 1.6 10 19, , 16 V, V, =, 2 10 6 A, , nhC, P=, , , R=, , 16, , = 8 × 106 , , where n = no. of photons incident per unit time., Also,i=ne, ihc, ihC, , i.e P =, e, eP, , =, , 4.14 1015 3 108, , , , , , 2.5, eV, 10, 3000 10, , , = 1.64 eV ; K .E.min = zero, , V, , 2.85 10 12 10, , n = 8.85 × 109, 47. Equlvalent resistance, , 42. K.E.max , , A, , V, , 1 1, 1 1 , E 13.6 z 2 2 13.6 9 5.95eV, 9 16 , n1 n2 , hc 1240, , 208.4 nm, E 5.95, , 41. KEmax and stoping potantial increase with, increase in frequency and threshold frequency, depends only on nature of the metal., , = 0.45 volt., , PARAGRAPH -II, An experimental setup of verification of, photoelectric effect is shown in the diagram. The, voltage across the electrodes is measured with, the help of an ideal voltmeter, and which can be, varied by moving jockey ‘J’ on the potentiometer, wire. The battery used in potentiometer circuit, is of 20 V and its internal resistance is 2 .The, resistance of 100 cm long potentiometer wire is, 8., , 2, , r, , hv 0, e, , (2 106 )(6.6 1034 )(3 108 ) =6187 Å, (4 106 )(1.6 1019 ), , Which came in the range of orange light., 49. The range of wavelength for red light is beyond, the wavelength of incident light., 50. Stopping potential VS = 8V and, KE = eVS= 8eV, 51. The number of photoelectrons ejected in time t, IAt, is n=, ., hv, The charge q on sphere keeps on increasing till, it acquire potential equal to stopping potential, , 105
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, Vs , , q, ne, hv , =, 4 o ( 2a ) 4 o ( 2a ), e, , Solving we get t =, , r/2, , 8 0 ah(h 0 ), , a, 2a, , 4 0 a(h – 0 ), e, , KE 2 1st excitation energy of He+.=2 × 40.8, = 81.6 eV, 55. The K.E. of particle should not be so large, that it excites electron in He+ to 2nd excited state., i.e KE 2 2nd excitation energy of He+., = 2 × 48.4 = 96.8 eV, 56. Electrons will not be detected if and only if there, is no excitation of the He+ sample. Thus no, photons are emitted by He+ and hence by H, sample, and hence none are detected by the, detector., 57. From the passage, (A) and (C) are correct., 58. From point (3), X-Rays are neither reflected, nor refracted and hence can not be concentrated, by lenses., , 59. Using, = a(Z – b) More the wavelength,, lesser will be the atomic number. Required order, is : PQR., 60. If electrons are accelereted through a potential, difference V, the maximum energy of emitted, photon could be, Emax = eV. i.e 105 eV = eV ; V = 105 V., , 106, , Ke, r2, , = mw2, , r, 2, , 2, , 2, , , , n2 h 2, , hv , e, 54. Maximum energy of photon incident in H sample = 13.6 eV + 27.2 = 40.8 eV, this corresponds to the transition 2 to 1 in He+, sample., For this electrons in He+ sample must be excited, to a maximum of 1st excited state., Since mass of -particle and He+ ion is almost, same, the -particles transfer 1/2 of their, kinetic energy to excitation energy (40.8eV) for, He+ion., , 61. (i), , r r, nh, and m 2 , ×2=, ...(2), , By (1) and (2) r = 22m K e 2 =, , 53. The potential difference is Vs , , 2, , +e, r/2, , A e2, , 52. The charge on inner sphere is q' q, =, , C.M., , –e, , .....(1), , n2 h2 , , , 2, 42m K e 2 × 2 = (0.529 Å) × 2 n, , , , For first excited state n = 2, r = 8 × 0.529 Å ;, r = 4.232 Å, 62. (ii) KE =, , r, 1, m 2 , 2, , , , Ke 2, 4 (0.529 Å 2), , =, , 2, , =, , 1, 2, , 13.6 eV, 4, , ., , m . 2 r 2, Ke 2, =, =, 4, 4r, , = 3.4 eV, , 63. The H-type atom is in the third excited state i.e. n, = 4. Energy corresponding to wave length, nm =, , 12400 51, 62000, , 6200, 51, , = 10.2 eV, , This is the E2 – E1 for H and E4 – E2 for He+., we get z = 2 for 4 to 2 radiation, Hence the atom is Helium ion, 64. Let u be the speed of neutron before collision, u, m, , 4m, , At end of the deformation phase (when the kinetic, energy of (neutron + He+) system is least), ucm, m, , ucm, 4m, , Where ucm is velocity of centre of mass. From, conservation of momentum ucm =, The loss of kinetic energy =, , mu, m 4m, , 1, 1, mu2 – m, 2, 2, , =, , u, 5, 2, , u, –, 5, , 2, , u, 4 1, 1, 2, 4m 5 = 2 mu , 5 , , 2, , , If K is the kinetic energy of nuetron then the, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, , maximum loss in K.E. of system is, , 4K, this excites, 5, , the electron from n 1 to n 4, 4, 5, , K = 51 eV or, , Kmin =, , K=, , 51 5, = 63.75 eV, 4, , 255, = 63.75 eV, 4, , 1, , 1, , 1, , , , 7, 65. 1.09 10 4 = 165 nm, , 4, 9, , , hc, hc, 7eV 0 177 nm, 0, 0 Photoelectric effect will occur, , 66.Stoping potential=, , hc 1 1 hc, , 40.5V, e e e , , , 1 q1q2, 67. U 4 r , q1 2e , q2 82e, 0, , U 9 109., , 2 82 1.6 1019 , 6.5 10 14, , r1 and v2 0( particle stops ) a n d, K 2 0 , U1 0 K1 U 2 5.82 1013 J ., , 2k, 1.32 107 m/s ., m, , STATEMENT TYPE QUESTIONS, 70. Even though all the electrons absorb the same, amount of energy, as different electrons may, come from different depths from the surface they, may lose different amounts of energy in collisions, with other atoms while comming on to the surface, hence the emitted photo electrons have different, kinetic energies., 71. Neutrons will knock out electrons in target leading, to production of characteristic X-ray., Hence statement-1 is false., 72. As electron jumps from n = 2 to n = 1, angular, nh , momentum , does not remain conserved., 2 , 73. Both statement I and II are true; but even it, , 1, , , T, , ., , 75. The cutoff wavelength for x-rays depends on, applied voltage between cathode and anode and, not on atomic number of target. Hence statement2 is false., 76. A free electron can’t absorb a photon completely., , MATRIX MATCH QUESTIONS, 77. E , , 1, n, , 2, , ,, , V, , (A) Epr , (B), , 1, n, , 2, , , , 1, n, 1, n, , and r n2, × n2 or Epr , , p 1 2, n, E n, , 5.82 1013 J, , 5.82 1013, MeV 3.63 MeV, 1.6 10 13, 68. Apply conservation of energy K1 + U1 = K2 + U2, , 1, mv 2 so v , 2, , molecules varies as , , (C) Er , , 2, , i.e U , , 69. K , , radiation of single wavelength is incident on, photosensitive surface, electrons of different KE, will be emitted., 74. de-Broglie wavelength associated with gas, , (D) Pr , , 1, n2, , or, , 1, n, , p, n, E, , × n2 orEr is independent of n, , 1, × n2 or pr n, n, , 78. Consider two equations, eVs , , 1, 2, mv max, hv 0, 2, , …(1), , int ensity, …(2), hv, A) As frequency is increased keeping intensity, 1, 2, constant. | VS | will increase, m(v max ) will, 2, increase and saturation current will decrease., B) As frequency is increased and intensity is, decreased, Vs will increase, KEmax will increase and, saturation current will decrease., C) If work function is increased photo emission, may stop., D) If intensity is increased keeping frequency, constant saturation current will increase., , No of photoelectrons ejected/sec. , , INTEGER QUESTIONS, 12400, , 79. Energy of incident photons = 2000 =6.2 eV, maximum kinetic energy of ejected electrons =, , 6.2 eV, Maximum kinetic energy of electrons striking plate, , NARAYANAGROUP, , 107
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JEE-ADV PHYSICS- VOL- V, , ATOMIC PHYSICS, 83., , B = 6.2 10 eV ., Minimum wavelength of photons emitted from, B 1000 A, , i.e 6.2 10 , 80. f r =, , (7), 1, K(Q)e, K(120e)e, 0 mv 2 , , 15, 2, 10 10, 10 10 15, 9, 1 5, 9 10 120 (1.6 1019 )2, 10 27 V 2 , 2 3, 10 1015, 9 6 109 120 2.56 10 38, V, 50 10 42, , 0, , 12400, 12.4 3.8eV, 1000, , nh, nhv, v, r 2 =, 2r, 4 2, , V 331.776 1013, h, mv, 4.2 1015 1.6 10 19, 4.2 4.8 10 34, , , 0.07 10 13, 21, 5, , , 0, 57.6, 5, 1, 13, 27, 10 331.776 10, 3, 7 1015 7fm, , , nh, 2 , mv 2 k e2 , r r2 , , v=, , nh, , ke 2, , m vr , , fr =, , , , ke 2, 2, nh, ke 2, , n h 2 = 2 , 4 2, , nh, nh, nh, 2, 84. I 2 mr 2 f 2 f 4 2 mr 2, , Ke 2 n0, x0, 2, , r r0 n 2 and N ft 8 106, , 81. As the electron returns to ground state after emitting, six different wavelengths in emission spectrum, there, must be a difference of '2' between the ground and, excited states. Also the ground state should be n =, 2. Hence n1 = 4 , n2 = 2., , Possible transitions, n=5 a, n=4, b, n=3, C, n=2, d, 82., n=1, Starting level n = 5, Possinle transtions, , n=5, n=4, n=3, n=2, , n=5, n=4, n=3, , b, , n=1, , e, f, , n=1, Starting level n = 4, Possinle transtions, , mv p, h, , P Bqr and , Bq Bq, p, , 1 P2 Bq2 r q2 2, , 2, 2 P1 Bq1r q1 1, , 87. F , , dU, mv 2, m 2 r ;, m 2 r, dr, r, , V r, mvr , , …(i), , nh, nh, ; V, 2, 2mr, , …(ii), , a, 1, , Starting level n = 2, Possinle transtions, , All possible transition are shown in the figures, above. It can be seen, to get all possible transitions, we need at least six atoms (a, b, c, d, e, f,)., 108, , 5, h W 10.2 eV 5 1015 H z, 6, , i.e, , n=1, , Starting level n = 3, Possinle transtions, , 12420, 10.2 eV, 1215, , a, , n=2, , n=2, , in second case KEmax , , 86. r , , n=5, n=4, n=3, , a, , 85. h W 13.6 eV and, , 1, nh 2, 2, , r, , n, (i) & (ii) r , , 2, , m, , , , , x2, , * * *, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , NUCLEAR PHYSICS, , , SYNOPSIS, NUCLEUS: The nucleus of an atom is at the, , , , , , , , centre. Most of the mass of an atom is at the, centre. The entire positive charge of an atom lies, in the nucleus., All atomic nuclei are made up of elementary, particles called protons and neutrons. Proton is, the nucleus of the hydrogen atom. It has a, positive charge of 1.6 × 10–19 C having a mass of, 1.6726 × 10-27 kg. This is nearly equal to 1836, times the electron mass. Neutron is electrically, neutral (i.e. neutron carries no charge). Mass of, neutron is slightlygreater thanthat ofthe proton (1.6750, × 10-27 kg). Both the proton and neutron together, constitute the nucleus. They are called nucleons., Generally, atomic number is denoted by Z and mass, number is denoted by A and (A-Z) gives, number of neutrons (N) in the nucleus., N A Z ; A = Z + N, Nucleus is positively charged and its shape is, considered as spherical., , TYPES OF NUCLEI:, , , , , ISOTOPES: Atomic nuclei having same atomic, number but different mass numbers are known as, isotopes. They occupy same position in the periodic, table and possess identical chemical properties., They have same proton number., Ex: 1) 3Li6, 3Li7 2) 1H1,1 H2,1 H3, ISOTONES : Atomic nuclei having same number, of neutrons are called isotones., 37, , 39, , 17, , 18, , Ex.: 1) 17 Cl ,19 K , 2) 7 N ,8 O ,9 F, , , , ISOBARS: Atomic nuclei having same mass, number but different atomic numbers are called, Isobars. They have same number of nucleons., Ex.:1) 18Ar40,20 Ca40 , 2), , , , 19, , 32, , G e 7 6 ,3 4 S e 7 6, , ISOMERS: Atomic nuclei having same mass, number and same atomic number but different, nulear properties are called isomers., Ex:- m 35 Br 80 metastable Bromine and g35 Br 80, ground state Bromine are two isomers with, different half lives, NARAYANAGROUP, , ISODIAPHERS: Nuclei having different Atomic, number (Z) and mass number (A) but with same, excess number of neutrons over protons (A-2Z), are called isodiaphers. Ex:- 11 Na 23 ,, , 13 Al, , 27, , SIZE OF THE NUCLEUS:, , , , Nuclear sizes are very small and are measured in, fermi (or) femtometer. 1 fermi=10-15m, Radius of the nucleus depends on number of, nucleons. R R0 A, , , , , , 1, , 3, , above equation does not apply to heavy nuclei, Value of Ro = 1.4 x 10-15m, Radius of the nucleus is in the order of 10-15 m., Size of an atom is in the order of 10-10m., If an -particle with an initial kinetic energy E, approaches a target of atomic number Z, if the, distance of closest approach is “d” then, 1 2 e 2, E (Where ‘e’ is charge of an, 4 d, electron) If “v” represents the initial velocity of , particle, (m is mass of “ ” particle) then, 1 2 e 2 1 2, mv, 4 d, 2, Note : If a particle of charge q, mass m is projected, towards a nucleus of charge Q with velocity v, from infinity then the distance of closest approach, 1 qQ 1 2, d is give by 4 d 2 mv, , , Note : If R, S and V be the Radius, surface area, and volume of a nucleus with mass number A then, R, , 1, A3, , 1, , 2, , 2, A 3, A 3, R, S, 1 1 ;S R 2 A 3 1 1 , S 2 A 2 , R 2 A 2 , , V R3 A , , V1 A1, , V2 A 2, , Note : If a stationary nucleus splits in to two lighter, nuclei with mass numbers A1 and A2 then according, to law of conservation of linear momentum, the two, lighter nuclei move in opposite directions with equal, momenta hence m1v1 = m2v2, 109
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, Ratio of velocities of the two nuclei, v1 m 2 A 2 R 2 , , , , v2, m1, A1 R 1 , , ii), , 3, , m A R 3 , , Ratio of kinetic energy of the two nuclei, KE1 m 2 A 2 R 2 , , , , KE 2 m1, A1 R 1 , , 3, , 2, , KE p & KE 1 when pis constant, , 2m, m, , , , , , 1.6605651024 gm 1.6605651027 Kg, , MASS - ENERGY EQUIVALENCE :, , DENSITY OF THE NUCLEUS:, , , , , , , , According to Einstein’s mass-energy equivalence, principle, mass is another form of energy. Mass, can be converted into energy & energy can be, converted into mass according to the equation E=, mC2, Here m is the mass that disappears and E is the, energy liberated. C is the velocity of light in vacuum., When 1 amu of mass is converted in to energy, Energy liberated is given by, E = (1.660565 x 10–27) x 9 x 1016 J= 931.5 MeV, hence 1 amu of mass is equivalent to 931.5, , Density of nucleus is independent of mass number, of the atom., Density of the nucleus is 1.45 x 1017 Kgm-3., The density is maximum at the centre and gradually, falls to zero as we move radially outwards., Radius of the nucleus is taken as the distance, between the centre and the point where the density, falls to half of its value at the centre., Density of nucleus is of the order of, 1014 gm / cc 1017 kg / m3, , MeV of energy 1 amu = 931.5 MeV/C 2, The masses of electron, proton and neutron in terms, of various units are :, Mass of the electron = me = 9.1095 × 10-31 kg, = 0.000549 u = 0.511 MeV/C2, Mass of the proton = mp = 1.6726 × 10-27 kg, = 1.007276 u = 938.28 MeV/C2, Mass of the neutron = mn = 1.6750 × 10-27 kg, = 1.008665 u = 939.573 MeV/C2 ., , W.E-1: Compare the radii of the nuclei of mass, numbers 27 and 64., A. The ratio of the radii of the nuclei is, 1, , 1, , R1, A1 3 27 3, 3, ( R = R0A1/3) =, =, , , R2, 4, A2 64 , , W.E-2: The radius of the oxygen nucleus 168O is, , A., , The attractive force which holds the nucleons, together in the nucleus is called nuclear force., , 205, 82, , Properties of nuclear forces :, , Pb ., R0 = 2.8 x 10-15 m, A0 = 16, APb =205 R A1/3, 1/3, , 1), , 1/3, , ATOMIC MASS UNIT (A.M.U):, , 110, , NUCLEAR FORCES :, , 2.8 x 10-15m. Find the radius of lead nucleus, , R0 A0 , 2.8 1015 16 , , , =, , RPb APb , RPb, 205 , RPb = 6.55 x 10-15m., , i), , Now, the mass of 1 gm -mole of carbon is 12 gm, and according to Avogadro’s Hypothesis it has N, (Avogadro’s Number) atoms. Thus, the mass of, one atom of carbon is (12/N) gm. According to, the definition., 1, 1amu 1u mass of one carbon atom, 12, 1, 1 12 1, gm, gm , 12 N, N, 6.0231023, , The masses of atoms, nuclei, sub atomic particles, are very small. Hence, a small unit is used to express, these masses. This unit is called as atomic mass, unit (amu). 1 amu is equal to one twelth part of, the mass of carbon (6C12) isotope., Mass of 6C12 is exactly 12 amu, , 2), 3), 4), 5), 6), 7), 8), , Nuclear forces are strongest forces in nature., Nuclear forces are about 1038 times as strong as, gravitational forces.The relative strengths of the, gravitational, Coulomb's and nuclear forces are, , Fg : Fe : Fn 1:1036 :1038, Nuclear forces are short range forces ., Nuclear forces are basically strong, attractive forces, but contain a sm all, component of repulsive forces., Nuclear forces are saturated forces., Nuclear forces are charge independent., Nuclear forces are spin-dependent., Nuclear forces are exchange forces., Nuclear forces are non-central forces., NARAYANAGROUP
Page 1766 : JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , MASS DEFECT, BINDING ENEGRY,, EINSTEIN’S MASS ENERGY, RELATION, , m ZMp ( A Z)Mn Mnucleus, , Z = Atomic number; Mp= Mass of proton, Mn = Mass of neutron; A=Mass number, M nucleus = Mass of nucleus., BINDING ENERGY: The energy required to, bring the nucleons from infinity to form the nucleus, is called binding energy or it is the energy required, to split a nuclens into nucleons., It is energy equivalent of mass defect BE=, 2, [ m]C, NOTE: BE = mass defect x 931.5 MeV if mass, is expressed in a.m.u., B.E. per nucleon = Binding fraction, Binding Energy, Mass Number, , =, , the average energy needed to separate a nuclei in, to its individual nucleons., Binding energy is not a measure of stability of a, nucleus., , PACKING FRACTION OF A NUCLEUS :, Packing fraction : It is defined as the mass defect, m M A, , A, A, If the packing fraction is negative then the nucleus, is more stable., If the packing fraction is positive then the nucleus, is unstable., , per nucleon. Packing fraction =, , , , Packing fraction is zero for 6 C 12, Packing fraction measures the stability of a nucleus., Smaller the value of packing fraction, large is the, stability of the nucleus., NARAYANAGROUP, , (1), (2), (3), (4), (5), , m x 931 Mev, A, , Averge Binding energy or Binding energy, fraction: It is the Binding energy per nucleon (or), , , Binding energy per nucleon (Mev), , , , When matter is completely annihilated, energy, released is E = mc2, The energy equivalent to 1 amu is 931.5 M eV =, 1.4925 1010 J ., MASS DEFECT: Atomic mass is always less, than the sum of the masses of constituent particles., The difference between the total mass of the, nucleons and mass of the nucleus of an atom, gives mass defect., , (6), , (7), , 40 Fe 56, 8.79ONeCa, C, He, , Region of, Greater stability, 40 A 180, , NF, , 7.6, , Li, 2, H, 1, , 1, 0, , 50, 100, Mass number (A), , 150, , 200, , 210, , The main features of binding energy curve shown, in figure are :, The minimum value of binding energy per nucleon, is in the case of deuteron (1.11MeV)., BE, The maximum value of, is 8.7MeV for the, A, nucleus 28Fe56 (iron) which is the most stable., Binding energy is high in the range 28<A<138. The, binding energy of these nuclei is very close to 8.7, MeV., Further increase in the mass number, binding energy, per nucleon decreases and consequently for the, heavy nuclei like uranium it is 7.6 MeV., In the region of smaller mass numbers, the binding, energy per nucleon curve shows the characteristic, minima and maxima. Minima are associated with, nuclei containing an odd number of protons and, 14, neutrons such as 36 Li,10, 5 B,7 N and the maxima are, associated with nuclei having an even number of, 16, protons and neutrons such as 42 He,12, 6 C,8 O ., Nuclei with A > 220 are distinctly unstable. That, means from A > 220 single heavy nucleus breaks, into two nearly equal nuclei with mass number A <, 150 and so which are most stable. This process, takes at right of the BE curve as shown in figure ., This process explains the nuclear fission., Light nuclei such as hydrogen combine to form, heavy nucleus to form helium for greater stability., This process takes at left of the BE curve as shown, in figure. This process explains the nuclear fusion., Binding energy per nucleon (Mev), , , , VARIATION OF B.E. PER NUCLEON, WITH MASS NUMBER, , +, , E, , Ene, y rgy++, g, r, ne, , Mass number (A), 111
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NUCLEAR PHYSICS, Note : Iron 28 Fe56 whose binding energy per nucleon, stands maximum at 8.7 MeV is most stable and, will undergo neither fission nor fusion., Exo-ergic Reaction : The reaction in which, energy will be released is called exo-ergic, Reaction. A + B C + D + Q, Here A and B are called Reactants, C and D are called Products, Q is the amount of energy released, In an Exo - ergic Reaction, Mass of reactants > Mass of products, , m MR MP MA MB MC M D , Energy Released Q m C2 joule [ m is in kg], m931.5 MeV ( m is in amu), If Binding energies are given then for Exo-ergic, reactions., (B.E) Products > (B.E) Reactants., Energy released Q = (B.E)P – (B.E)R, = [(B.E)C + (B.E)D] – [(B.E)A + (B.E)B], Endo-ergic Reaction : The reaction in, which energy will be absorbed is called Endoergic Reaction. A + B C + D – Q, Here A and B are called Reactants, C and D are called Products, Q is the amount of energy absorbed, In an Endo - ergic Reaction, mass of reactants < Mass of products, , m MP M R MC MD M A MB , Energy absorbed Q m C2 joule [ m is in kg], m931.5 MeV ( m is in amu), If Binding energies are given then for, Endo-ergic reaction., (B.E) Products < (B.E) Reactants., Energy absorbed Q = (B.E)R – (B.E)P, = [(B.E)A + (B.E)B] – [(B.E)C + (B.E)D], Note: A nuclear reaction can occur only if certain, conservation laws are followed. These are :, 1. Conservation of mass number A., 2. Conservation of charge., 3. Conservation of energy, linear momentum and, angular momentum., 112, , JEE-ADV PHYSICS- VOL- V, , W.E-3 : Find the binding energy of, , 56, 26, , Fe . Atomic, mass of Fe is 55.9349u and that of Hydrogen, is 1.00783u and mass of neutron is 1.00876u, Sol. Mass of the hydrogen atom mH = 1.00783u; Mass, of neutron mn = 1.00867 u; Atomic number of iron, Z = 26; mass number of iron A = 56; Mass of iron, atom Ma = 55.9349u, Mass defect m = [ZmH+(A-Z)mn] - Ma, = [ 26 x 1.00783+(56-26)1.00867]-55.93493, u = 0.5287 u., Binding energy = (m)c 2 = (0.52878 ), c2 = (0.52878)(931.5MeV) = 492.55 MeV, , W.E-4 : Find the energy required to split 168O, nucleus into four - particles. The mass of, - particle is 4.002603u and that of oxygen, is 15.994915u., Sol. Mass of -particle = 4.002603 u, Mass of oxygen = 15.994915u, B.E=[Mass of 4 -particles - Mass of oxygen] x, 931.5MeV, B.E = [ 4 x 4.002603 - 15.994915] x 931.5 MeV, = (16.010412-15.994915) x 931.5 MeV, = 0.015497 x 931.5 ; B.E = 14.43 MeV, , W.E-5 : Calculate the binding energy per nucleon, of 4020Ca. Given that mass of 4020Ca nucleus =, 39.962589 u, mass of proton = 1.007825 u., mass of Neutron = 1.008665 u and 1 u is, equivalent to 931 MeV., Sol. A =40, Z = 20, A -Z = 20, m = Zm p A Z mn M n, = { (20 x 1.007825+(20x1.008665)} - 39.962589, = 40.329800 - 39.962589 ; m = 0.367211, 1, Binding energy per nucleon =, m 931 0.367211 931, 8.547 MeV ., =, A, 40, , W.E-6 : The binding energies per nucleon for, deuterium and helium are 1.1 MeV and 7.0, MeV respectively. What energy in joules will, be liberated when 2 deuterons take part in the, reaction., Sol. 12 H 12 H 24 He Q, Binding energy per nucleon of helium ( 24 He ) =7 MeV, Binding energy = 4 x7 = 28 MeV, Binding energy per nucleon of deuterium ( 12 H )=1.1MeV, Binding energy = 2 x 1.1 = 2.2 MeV, Energy liberated (Q) = (28 -(2.2)2] = 23.6 Mev., i.e. Q=23.6x106x1.6x10-19; Q 37.76 1013 J, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , W.E-7: The kinetic energy of -particles emitted, Ra 226, , Rn 222, , in the decay of 88, into 86, is, measured to be 4.78 MeV. What is the total, disintegration energy or the ‘Q-value of this, process’ ?, Sol. The standard relation between the kinetic energy, of the -particle (KE) and the Q-value (or total, disintegration energy) is, A - 4 , KEa = , .Q, A , , A , Q = , .KEa, A - 4 , , 226 , 226, , 4.78MeV , 4.78MeV, , 226 4 , 222, Q 4.865MeV 4.87MeV, , W.E-8: A nucleus X-initially at rest, undergoes, alpha-decay, according to the equation., A, 92, , X 228, Y α, z, , The α -particle in the above process is found, to move in a circular track of radius, 1.1102 m in a uniform magnetic field of, , 3.0 103 T ., The energy (in MeV ) released during the, process and binding energy of the parent, nucleus X, respectively., Given: m y 228.03 amu ; mα 4.003 amu ,, , m 10 n 1.009 amu ; m 10 H 1.008 amu ,, 1amu 1.66 10 27 ; kg 931.5 MeV / c 2, A, 4, Sol: The given equation is 92, X 228, z Y 2 He, , A 228 4 232 ; 92 z 2, , z 90, , rqB, mα vα2, qvα B ; vα , mα, r, , , 1.1 10 2 2 1.6 10 19 3 103, 4.003 1.66 10 27, , 4.0 106 m / s, From conservation of linear momentum, mα vα m y v y, vy , , mα vα 4.003 4.0 10, , my, 228.03, , 6, , , , 4, , 7.0 10 m / s, , There fore, energy released during the process, NARAYANAGROUP, , 1.66 10 27 , , 1, 2, 2, mα vα m y v y , 2 1.6 1013 , 2, 4.003 4.0 106 2 228.03 7.0 104 2 MeV, , , , 0.34 MeV , , 0.34, amu 0.000365amu, 931.5, , 232, Therefore, mass of 92, X m y mα 0.000365, , 232.033365 u, Mass defect Δm 92 1.008 232 92 , , 1.009 232.033365 ., Binding energy 1.962635 931.5 MeV, 1828.2 MeV, NATURAL RADIO ACTIVITY :, Spontaneous decay of naturally occurring, unstable nuclei by emission of certain sub, particles (like , , and radiation) is called, natural radio activity., The emission of these rays takes place because of, the instability of the nucleus. In the process of, emitting these rays a nucleus tries to attain the, stability., In general natural radioactivity takes place in heavy, nuclei beyond lead in the periodic table. There are, also naturally radioactive light nuclei, such as, potassium isotope 19K40 , the carbon isotope 6 C14, and the rubidium isotope 37 Rb87 ., Regarding radioactivity., (i) It is completely uneffected by the physical and, chemical conditions to which the nucleus is, subjected i.e we cannot change the radio activity, by applying high temperature, high pressure and, strong electric field etc., (ii) The nucleus can disintegrate immediately (or) it may, take infinite time., (iii) The energy liberated during the radioactive decay, comes from individual nuclei., , MODES OF DECAY :, The radioactive nucleus before decay is called, a parent nucleus, the nucleus resulting from, its decay by particles (Radiation) emission is, called daughter nuclei., 113
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, This daughter nuclei may be stable (or) unstable., X, , , , , Parent, , Y, , , , R, , , , zX, , A, , Ex:, , z 2 Y A 4 2 He 4 Q, 88 Ra, , radiating b rays, it is said to undergo, b decay.., , Q, , Daughter Radiation Energy, , Here R may be either particle (or) particle, (or) radiation. Q is the energy of the emitted, particles (or radiation)., -DECAY : When a nucleus disintegrates by, radiating -rays, it is said to undergo decay. An - particle is a helium nucleus. Thus a, nucleus emitting an particle losses two protons, and two neutrons, as a result its atomic number Z, decreases by 2, the mass number A decreases by, 4 and the neutron number N decreases by 2., , , , -DECAY : When a nucleus disintegrates by, , 226, , 86 Rn 222 2 He 4 4.87MeV, , Both electric charge and nucleon number are, conserved in the process of a decay.., Application : When a stationary Radio active, nucleus x decays into another nucleus y by emitting, an a -particle. x y a particle Q, Applying LCLM if a particle moves forward with, a momentum ‘P’ then daughter nucleus y recoils, with same momentum ‘P’ so that total momentum, of the system is zero. Hence Py Pa, The energy released ‘Q’ is in the form of K.E of, daughter nucleus ‘y’ and ‘ a ’ particle., , i), , b particles are nothing but electrons. Hence when, , a nucleus emits a b particle, the atomic number, (Z) increases by 1 unit, but the mass number does, not change., The general form of b decay can be written as, Z, , X A Z1 Y A 1 e 0 ., , Ex: 90 Th 234 91 Pa 234 1 e 0, Both electric charge and nucleon number are, conserved in b decay also., -DECAY: When a nucleus disintegrates by, radiating g rays, it is said to undergo, g decay.., Gamma rays are nothing but electromagnetic, radiations of short wavelengths (not exceeding, 10–10m.), The emission of g rays from the nucleus does, not alter either atomic number Z or mass number, A. It just results in the change of the energy state of, a nucleus., When a parent nucleus emits an a or a b particle,, the daughter nucleus may be formed in one of, excited states. Such a nucleus will eventually comes, to the ground state. In this process g radiation, will be emitted., , Q KE y KEa, , Z, , Example:, KE y Ma, Ratio of kinetic energies KE M, a, y, , M, 1, 1 a, KEa, My ;, M y , , KE a Q , M a M y , , KEa KE y, KEa, , , , M y Ma, My, , 87, , 38Sr 87 ., , is isomer of 38Sr ., Note :When a Radio active nucleus emits an a - particle, followed by two b - particles, its isotope is formed., , M, , a, , KE y Q , ;, Ma M y , , 2 b, , a, , A, Z2Y A4 , Z X A4, Z X , , Note :When a Radio active nucleus emits a b - particle, its isobar is formed., ZX, , Notice that KE is very close to (but smallerthan) Q., 114, , 38*Sr, , 38 Sr, , 1, P2, ( KE , and KE a when ‘P’ is same), m, 2m, , KE y, , X A Z X A g Photon ( s ), , A, , b, , Z1Y A, , Note: When a Radio active nucleus emits a g - particle, its isomer is formed, g, , Z, , X A , Z X A, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , Deflection of Radioactive radiations in, electric and magnetic fields :, , , W.E-11: How many and - particles are emitted, when uranium nucleus (92 U 238 ) decay to, , , 82, , , , E, , B, , , , , , , , , , , , , , , , Pb 214 ?, , Sol. Let n be the number of - particles and m be the, number of - particles emitted., U 238 82 Pb214 n2 He4 m 1e0, As mass is conserved, 238 = 214 + 4n + m (0), = 214 + 4n ; 4n = 24; n = 6, As charge is conserved , 92 = 82 + 2n + m (-1), 10 = 2(6) - m n 6 ; m = 2., 92, , Ra, , Ra, Lead block, , 6a - particles and 2 - particles are emitted, X, X, X, , B, , X, , , , , , , , X, X, , X, , X, , X, , X, , X, , , , RADIOACTIVE DECAY LAW:, , X, , Ra, Lead block, , W.E-9:The nucleus 2310Ne decays by emission., Write down the -decay equation and, determine the maximum kinetic energy of the, electrons emitted. Given that :, m (2310Ne) = 22.994466u; m(2311Na)=22.989770 u, Sol., , 23, 10, , Ne , , 23, 11, , _, , _, , Na e v Q, , For - decay, Q = [M(x) - M(y)]C2, = [22.994466-22.989770]931.5, = 0.004696 x 931.5 = 4.37 MeV, particle. Given that mass of proton = 1.0073u,, mass of neutron = 1.0087u. and mass of particle = 4.0015u., Sol. mP = 1.0073u, mN = 1.0087u, M = 4.0015u, , , , 2, , He 4 Z X A, , From eqn (1), , dN, dt .......(2), N, , Integrating eq (2) on both sides, , , , = Zm p ( A Z ) mn M 931.5, 2 1.0073 2 1.0087 4.0015 931.5MeV, = 0.0305 931.5 MeV ; B.E = 28.4 MeV, NARAYANAGROUP, , The proportionality constant is called decay, constant (or) disintegration constant. The negative, sign indicates that as time increases N decreases., , , , B.E = m x 931.5 MeV, , , , Let ‘N’ be the number of radioactive atoms present, at a time ‘t’and N0 is the initial number of radio active, nuclei. Let dN atoms disintegrate in time ‘dt’., According to the law of radioactive decay, , dN , dN , dt N ; dt = – lN ....... (1), , W.E-10: Calculate the binding energy of an -, , N = A -Z =4 -2 =2, , Based on their experimental observations and, analysis of certain radioactive materials Rutherford, and Soddy formulated a theory of radioactive, decay. According to them, After decay of a nucleus the new product (daughter), of nucleus has totally different physical as well as, chemical properties., The rate of radioactive decay (or) the number, of nuclei decaying per unit time at any instant, is directly proportional to the number of, nuclei (N) present at that instant and is, independent of the external physical, conditions like temperature, pressure etc., , , , dN, dt, N, , log e N l t C ..... 3, , Here C is the constant of integration, At t = O, N= N0 Substituting in eqn (3),, we get,, , log e N 0 C, 115
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, log e N t loge N 0 ;, log e N loge N 0 t, , N, N, t, log e t ; N e ; N N elt ... (4), 0, 0, N 0 , This shows that the number of radioactive nuclei, decreases exponentially with time., , Above equation is known as the decay law (or), the law of radio-active decay. It is an, exponential law., taking logarithm on both sides for the above, N, equation. loge N = loge No - lt ; t loge 0, N, , 1 N 0 , ln , l N , , number of nuclei decreases to, , Nx, , N0, , O, , time, (a), , No. of nuclei decayed, , No. of nuclei left, , t, , which have activities R1, R2, ........... and Rn. Then, the resultant activity R = R1 + R2 +............... +, Rn. If nucleus decays simultaneously more, than one process is called parallel decay., The S.I unit of activity is Becquerel (Bq) and, other units are curie (Ci) and Rutherford (Rd)., 1 Bq = 1 decay per second,, 1 Rd = 106 decays per second., 1 Ci = 3.7 × 1010 decays per second., Note : Curie is approximately equal to the activity, of one gram of pure radium., DECAY CONSTANT ( ) : It gives the ability of a, nucleus to decay. The decay constant for a, given radio active sample is defined as the, reciprocal of the time during which the, , original value., , Ny, , N, , N0, , N0, O, , N0, , t, (b), , e, , 0, In(N0), In(N), , 1), , time(t), , 2), 3), , (C), 4), , ACTIVITY (R) :, The number of decays per unit time (or) decay, rate is called activity (R), | R |, , dN d, N 0e t (or) R lN lN 0elt (or), , dt, dt, , , , , , R R 0e t , where R 0 N 0 is the decay rate, at t = 0, called initial activity., , R, R0, , If a nucleus can decay simultaneously by n processes,, 116, , 1, times their, e, , 1, , , t, , Larger value of corresponding to decay in, smaller time and vice versa., = 0 for stable nuclei., Decay constant is the characteristic of the sample, taken and does not vary with time., If a nucleus can decay simultaneously by more than, one process (say n), which have decay constants, , 1, 2 ...........and n , then the effective decay, constant is 1 2 .......... n . This is, called parallel decay., , HALF LIFE (T) : As the name suggests, the half life, of a radioactive sample is defined as “The time, interval during which the activity of a radio, active sample falls to half of its value, (or) The, time interval during which the number of, radio active nuclei of a sample disintegrate to, half of its original number of nuclei” Half lives, vary from isotope to isotope. While T may be as, small as 10-16 s, its largest value may be as big as, 109 years., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, Eg: Half-life of uranium, half-life of krypton, , NUCLEAR PHYSICS, , U is 4.47 x 10 years, 238, 92, 9, , 8936 Kr is 3.16 minutes., , Relation between decay constant ( l ) and, half life period (T)., From Law of Radioactive decay, , when N =, , N, e t, N0, , 1, N0, , t T elT or 2 elT, 2, 2, , taking logirthms on both sides ln 2 T, (or) loge 2 lT T , , 2.303log10 2 0.693, , , , , ln 2 2.303log 2 0.693, , T= l , l, l, The above relation establishes that the half - life (T), depends upon the decay constant of the, radioactive substance. The value of is different, for different radioactive substances., , ii), , Note :, Half life is the characteristic property of the sample, and T cannot be changed by any known method., At any given instant whatever be the amount of, the undecayed sample, it will be reduced to exactly, half its value after a time equal to the half life of the, sample., , iii), , In parallel decay 1 2 ........ n hence, , i), , 1 1 1, 1, ........ , where T is t he, T T1 T2, Tn, equivalent half-life and T1, T2 ...........Tn are the, half-lives in individual decay., Application :, In a radioactive sample the number of nuclides, undecayed after n-half lives (i.e., t = nT) is, N, 1, t = nT , l n 0, N, l, , n ln 2 1 N 0 , , ln , or, l, l N , , , Note: The number of nuclei remain in the sample, after half of half life period (t=1/2T) is given by, 1, , 1 n, 1, 2, N N 0 here n then N N 0 1 , 2 , 2 , 2, N, , N0, 2, , taking N0 = 100, N = 50 2 = 70.7, , 70.7% of nuclei remain and 29.3% of nuclei, decayed., , (N/ N0), 1, T, 0.5, 0.25, 0.125, 0.0625, 0, , T 2T 3T, , 1, N0, ; or N N 0 , 2, N, , NARAYANAGROUP, , time, , 4T, , AVERAGE LIFE (OR) MEAN LIFE :, The phenomenon of radioactivity is random, because we just can’t predict which of the atoms, in a given sample will decay first and when. Hence, radioactivity process totally depends on chance., In decay process some of the atoms of the given, sample may have very short life span, and others, may not decay even after a very large span of time., So to determine the ability of the nucleus to decay, it would be useful to calculate the average life., Hence average life is defined as the total life time, of all the nuclei divided by the total number, of original nuclei., , life span of individual nucleus t, i.e t , Total number of original nuclei, N0, Let N0 be the radio active nuclei that are present at, t = 0 in the radioactive sample., The number of nuclei which decay between t and (t, + dt) is dN i.e the life time of these nuclei is ‘t’., The total life time of these dN nuclei is (t dN), The total life time of all the nuclei present initially, t , , in the sample =, , tdN [ N = 0 at infinity], t 0, , n, , n, or 2 , , Half life, , Average life time =, , tdN, N0, , But, , dN, N, dt, 117
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, dN = Ndt N 0 e t dt ( N N 0 e t ), , , t, 0, , log e N log e N 0 t log N log N 0 , , N 0 e t, 1, dt ; t , l, N0, , , log N , t log N 0, 2.303, , The mean life (or) average life of a radio active, sample is reciprocal to decay constant., , Slope of the graph m tan , , We know that N N 0 elt ; When t t ,, , 2.303 tan , , Half life period T , , 1, xt, e t, , , N N0, 0 0.37N0 37% of N0, e, Hence average life period of a radio active sample, can also be defined as “The time interval during, which 63% of sample decays or sample reduces, to 37% of its original amount”., , RELATION BETWEEN HALF LIFE, PERIOD AND AVERAGE LIFE, PERIOD, We know that T , , t, 2.303, , 0.693, 1, &t , l, l, , , 2.303, , 2.303 log 2, l, , 2.303 log 2, T log 2 cot q, 2.303 tan q, Note : In radioactive sample decay, The probability survival of nucleus after time, N, Ps , e t ., N0, The probability of nucleus to disintegrate in time t, T, , 1), , 2), , is Pd 1 Ps 1 e t ., , W.E-12: A radioactive sample has an activity of, , T, 1.443T, Hence T 0.693t (or) t , 0.693, , From the above equation it is clear that average, life period is 44.3% greater than half life period., , 5.13 x107 Ci. Express its activity in ‘becqueral’, and ‘rutherford’., Sol. Since 1 Ci = 3.7 x 1010 decays per second,, activity = 5.13 x 107 Ci, =5.13 x 107 x 3.7 x 1010 Bq = 1.9 x 1018 Bq, Since , 1 x 106 decay per second = 1Rd, Activity =1.9 x 1018 Bq =, , R, R0, , 1.9 1018, 1106, , Rd = 1.9 x 1012 Rd., , W.E-13:A radioactive substance has 6.0 1018, R0, 2, , R0, e, , 0, , T , , t, , Determination of decay constant l and half, , active nuclei initially. What time is required, for the active nuclei of the same substance to, become 1.01018 if its half-life is 40 s., Sol. The number of active nuclei at any instant of time t,, N0, elt, ;, N, , life period (T) of a radioactive sample, graphically, t , , N , loge 0 lt, N , , N , loge 0 , N , l, , , , N , 2.303 log10 0 , N , l, , In this problem, the initial number of active nuclei,, 18, N0 = 6.0 1018 ; N 1.010 , T 40s,, l, , 118, , If N0 and N be the number of atoms present, undecayed initially and after a time t, then, , t, , We know that N=N0 e t taking log on both sides, , =, , 0.693 0.693, , 1.733 102 s1 ., T, 40, 6.0 1018 , , 2.303log10 , 18 , 1.0 10 , , 1.733102, 2.303 log10 6 , 1.73310, , 2, , , , 2.303 0.7782, 1.733102, , 103.4 s., , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , W.E-14: A radioactive sample can decay by two, , W.E-16: A certain substance decays to 1/32 of its, , different processes. The half-life for the first, process is T1 and that for the second process, is T2. Find the effective half-life T of the, radioactive sample., Sol. Let N be the total number of atoms of the, , initial activity in 25 days. Calculate its half life., , radioactive sample initially. Let, , dN1, dt, , and, , dN 2, dt, , be the initial rates of disintegrations of the, radioactive sample by the two processes, dN, , dN, , 1, Sol. 1g , , , 1, 1, 1, 1, 2, 3, 4, g , g , g , g, , , , 2, 4, 8, 16, , 1, 5, , 32 g n 5, , n , , t, t1/ 2, , t1/ 2 , , t 25, , n 5 ; t1/2 5 days, , respectively. Then dt1 l1 N and dt2 l2 N, , W.E-17: The half -life period of a radioactive, , Where 1 and 2 are the decay constants for the, first and second processes respectively., The initial rate of disintegrations of the radioactive, sample by both the processes, , substance is 20 days. What is the time taken, for 7/8th of its original mass to disintegrate?, Sol. Let the initial mass be one unit., , =, , dN1 dN 2, , l1 N l2 N l1 l2 N ., dt, dt, , If is the effective decay constant of the, radioactive sample , its initial rate of disintegration., dN, lN, dt, , But, , 7 1, , 8 8, , A mass of 1 unit becomes, , 1, unit in 1 half life, 2, , 1, 1, unit becomes unit in 2nd half life, 2, 4, , 1, 1, unit becomes unit in 3rd half life, 4, 8, Time taken = 3 half lifes = 3 x 20 = 60 days, , dN dN1 dN 2, , , dt, dt, dt, , l N l1 l N, , W.E-18: How many disintegrations per second will, , l l l, , Occur in one gram of 238 U , if its half-life, 92, against -decay is 1.42 x 1017s?, , 0.693 0.693 0.693, , , T1, T2, T, , Sol. Given Half -life period (T) =, , 1, 1, 1, TT, ;T 1 2 ., T T1 T2, T1 T2, , W.E-15: Plutonium decays with a half life of, 24,000 years. If plutonium is stored for 72,000, years, what fraction of it remains?, A. T1/2 = 24,000 years, Duration of time (t) = 72,000 years, t, Number of half lifes (n) = T, , 1/2, , , , 72000, 3, 24000, , 1, 1, 1, 1, 2, 3, 1g , g , g , g, 2, 4, 8, Fraction of plutonium remains =, NARAYANAGROUP, , Mass reamaining = 1 , , 1, g, 8, , 0.693, , , , 1.42 1017 s, 0.693, , 4.88 1018, 17, 1.42 10, Avagadro number (N) = 6.023 1023 atoms, 238, n = Number of atoms present in 1 g of 92 U , , , , N, A, , 6.023 10 23, 25.30 10 20, 238, , Number of disintegrations =, , dN, n, dt, , = 4.88 x10-18 x 25.30 x 1020, = 1.2346 x 104 disintegrates/sec, 119
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , W.E-19: One gram of radium is reduced by 2, milligram in 5 years by -decay. Calculate, , W.E-21: Obtain the amount of, , Co necessary to, provide a radioactive source of 8.0 mCi, , the half-life of radium., Sol: Initial mass = 1 g, t = 5 years, , strength. The half -life of, , 2, Reduced mass = 2mg = 2 x 10 g =, g, 1000, -3, , 60, 27, , Co is 5.3 years., , Sol. Half - life of 60 27 Co = 5.3 years, = 5.3x365x24 x 60 x 60 S = 5.3 x 3.15 x 107 s, , 10 3 x, k - mole, 60, , 2, 998, , Remaining mass = 1 1000 1000, , Now `x` gm of 60 27 Co contains, , N 998, , ( Mass Number of atoms), N0 1000, , x 103, 6.025 1026, 60, N = 0.1004x x 1023 = 1.004 x x 1022 atoms, =, , N, e t ; 998 e t, N0, 1000, , Required strength of 60 27 Co = 8 m Ci, = 8 x 10-3 x 3.7 x1010 dis/s, We know that, decay rate (R) = N, , 1000 , 1000, et e5 log e , 5, 998, 998 , , N, , 2.303 (3.0000 -2.9991) = 5 , , , , 2.303 1 0.0009, 5, , (T1/2) =, , 0.693, 0.693 5, , = 1671.7 years, , 2.303 0.0009, , ARTIFICIALTRANSMUTATION OF, ELEMENTS, , is 5000 years . In how many years, its activity, will decay to 0.2 times of its initial value?, Given log105 = 0.6990., Sol. T = 5000 years,, , The conversion of one element into another, by artificial means is called artificial, transmutation of the element. Rutherford, performed number of experiments in which the, atoms of different stable elements, such as nitrogen,, aluminium, phosphorus, etc, were bombarded by, high speed particles from natural radioactive, substances. Finally in 1919, he discovered the, phenomenon of artificial transmutation., , N, 2 1, , =, 0.2, =, N0, 10 5, , , , 0.693 0.693, , T, 5000, , N, e t, N0, 1, 1, t, = t 5 e, 5, e, 5, loge = t, , B, , 2.303 0.6990 5000, 0.693, , t = 11614.6years 1.1615 104 years, , A, R, , B, , S, F, , M, , Fig., , 2.303 0.6990 = t, t, , R 8 3.7 107, , 0.693 / T1/ 2, , 29.6 107, , x5.3x3.15x107 = 713.0909 x 1014, 0.693, 1.004x x 1022 = 713.0909 x 1014, x = 710.2499 x 10-8 kg = 7.1 x 10-6 g, , W.E-20: The half-life of a radioactive substance, , 120, , 60, 27, , i., , The apparatus used by Rutherford is as shown in, Fig., It consists of a chamber A provided with an, adjustable rod, carrying a radio - active substance, R (Radium C)., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, ii., , The side of the glass tube facing ‘R’ is covered by, metal plate with a central hole which is closed by a, thin silver foil ‘F’., iii. A screen ‘S’, coated with a fluorescent material, like zinc sulphide is arranged infront of the silver, foil and the scintillations produced on it can be, observed through the microscope ‘M’., iv. The side tubes B, B were used to fill various gases, in the chamber., v. The source of - particles, Ra was placed on a, small disc at R. Its distance from F was adjustable., vi. The radio-active substance emits -particles, whose range in air was found to be about 7cm., vii. When the glass tube is filled with nitrogen gas,, scintillations are observed, even when ‘R’ is at a, distance of 40cm from the foil., viii. These particles producing scintillations can not be, -particles as they can not have such a long range., ix. Rutherford concluded that nitrogen nucleus hit by, an (2He4 )-particle transmutes into oxygen, nucleus along with a proton (1H1)., x. The nuclear reaction causing artificial transmutation, can be represented as 7N14+ 2He4 8O17 + 1H1, Thus an atom of nitrogen is transformed into an, isotope of oxygen. This process is called, transmutation of elements., * High energy - particles were used in the, discovery of artificial transmutation and neutron, because - particles produce intense ionisation of, the medium through which they pass and can be, stopped after travelling a few mm in air., Significance :, i) It leads to the discovery of proton and neutron., ii) It helps to produce radio isotopes., iii) It helps to produce transuranic elements., , RADIO ISOTOPES AND THEIR USES:, , 1), , Radio isotopes have very short half lives and hence, used for various purposes., Medical applications :, , a), , 0 n1 11 Na 24* 2 He 4, Radio – sodium is used to find out how a given, medicine is circulated in the body. It is also used to, find out circulatory disorders in blood vessels., , b), , 1 H2 15 P32* 1 H1, Radio – phosphorus is used in the treatment of skin, diseases. It is also used for the treatment of blood, disorders., , 13 Al, , 15 P, , 27, , 31, , NARAYANAGROUP, , NUCLEAR PHYSICS, c), , 0 n1 53 I128* rays, Radio – iodine is used in the treatment of thyroid, glands. Radio - iodine ( I131) is used for diagnosis, and treatment of brain tumor and for the study of, pumping condition of heart., , d), , 27 Co, , e), 2), a), b), 3), a), b), c), d), 4), 5), , 6), a), b), c), d), 7), a), , b), , 127, 53 I, , 59, , 0 n1 27 Co60* rays, , Radio – cobalt is used in the detection and, treatment of cancer, Radio-iron is used to detect anemia and treat, anemia., In Geology:, Radio carbon(C14) is used to determine the age, of fossils by radio - carbon dating, Radio isotopes are used to determine the age of, rocks by the ratio of U 238 to Pb 206, In industry :, Radio – isotopes are used to find the wear and, tear of machine parts, Radio isotopes are used to detect flaws in metal, structures, Radio isotopes are used for treatment of alloys such, as quenching , annealing and hardening., Radio isotopes are used in the selection of, appropriate lubricants., In research: Radio - isotopes are used in the, study of nuclear disintegrations of elements., In food preservation: By exposing vegetables, and other food stuffs to radiations from radio - active, isotopes, their shelf life can be increased., In agriculture:, Radio phosphorus (P32) is used to study the uptake, of phosphorus by plants using., Radio sulphur (S34) is used to study the transport, of minerals in plants ., Radio zinc is used to develop new species of plants, by causing genetic mutation., Irradiation by g - radiations of seeds to improve, yields., In Chemistry :, Radio oxygen (O 18 ) is used to study the, mechanisms of photosynthesis and hydrolysis of, ester, Radio isotopes are used in the chemical analysis of, solubility of sparingly soluble salts such as PbSO4, and AgCl and determination of trace amounts of, elements in industrial raw materials and products., 121
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , NEUTRON, , , , It is electrically neutral and its mass is slightly greater, than that of proton. It was discovered by chadwick, Bothe – Becker equation:, 4 Be, , , , , , , , 9, , 4, , 2 He , , 13, , 6C, , 12, , , , 1, , 6 C 0 n, , , , Neutron is unstable outside the nucleus., 0, , , , CHAIN REACTION:, , , n 1 1 H 1 1 e 0 (anti neutrino), , It has high penetrating power and low ionizing, power., Slow moving neutrons are called thermal, neutrons. Fast moving neutrons convert into thermal, neutrons when they pass through a substance called, moderator., Thermal neutrons have an average energy of nearly, 0.025 eV. Fast moving neutrons have an average, energy of 2 MeV., , NUCLEAR FISSION, , , , , , , , , , , , , , , 122, , Nuclear Fission is a nuclear reaction in which a, heavy atomic nucleus like U235 splits into two, approximately equal parts, emitting neutrons and, liberating large amount of energy., Bohr and Wheeler proposed liquid drop model to, explain this fission process., Nucleus of U235 undergoes fission when it is struck, by slow neutrons. This fission is not due to the, impact of neutron., Energy of about 200 MeV is released during one, 235, fission reaction of 92U . The most probable, nuclear fission reaction is, 235, 1, 141, 92, 1, U + n Ba + Kr +3 n +energy, 92, 0, 56, 36, 235 0, There is no guarantee that U always breaks into, Barium and Krypton., On an average,in the fission of U235, 2.5 neutrons, are emitted per fission when fission occurs due to, slow neutrons. U235 undergoes fission with fast, neutrons also. But this probability is minimum., Fission fragments are unstable and emit neutrons, some time after fission reaction which are called, “delayed neutrons”, 99% of neutrons emitted during fission process, are prompt., Delayed neutrons play an important role in chain, reaction, , , , If the mass of fissionable material exceeds a, critical value, chain reaction or self propagating, fission reaction takes place., The rate of reaction increases in geometric, progression during uncontrolled chain reaction., Chain-reaction : The process of continuation of, nuclear fission which when once started continues, spontaneously without the supply of additional, neutrons from outside is defined as chain reaction., Reproduction factor (K): “It is the ratio of number, of neutrons in any particular generation to the, number of neutrons in the preceeding generation., Case(i): K<1 ; Chain react ion is not, maintained.(sub-critical state), Case (ii): K=1 : Chain reaction is maintained at, steady rate. (critical state). In the state electricity is, produced in the reactors at steady rate, Case (iii): K>1 : Chain reaction becomes self, sustained and lead to atomic explosion(super critical, state), Uncontrolled chain reaction takes place in atom, bomb., , NUCLEAR REACTOR OR ATOMIC PILE:, , , , , , , Nuclear reactor is a device in which nuclear fission, is produced by controlled self sustaining chain, reaction. And is used for the production of nuclear, power (energy)., The essential parts of a nuclear reactor are (i) the, fuel, (ii) moderator, (iii) control rods, (iv) coolant,, (v) radiation shields., THE FUEL: The common fuels used are uranium, 235, , U , enriched uranium (U, Pu and T h ., 238, , 236, , ) and plutonium, , 23 2, , MODERATOR:, The function of a moderator is to slow down the, fast moving neutrons to increase the rate of fission., The commonly used moderators in the order of, efficiency are (i) Heavy water (ii) graphite,, (iii) Berillium and Berillium Oxide, Heavy water is a best moderator, A good moderator should have, (1) low atomic mass, (2) poor absorption of neutrons, (3) good scattering property., (4) The size of moderator atom should be nearly, of same size as that of the size of a prompt neutron., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , CONTROL RODS:, , , , , , Note: If ‘x’gm of fuel with mass number ‘A’, completely undergo nuclear fission in time t sec in, a reactor then its power is given by, , The function of a control rod is to absorb, (capture) the neutrons., Cadmium, Boron and steel rods are used as control, rods in a nuclear reactor., Cadmium rods are best control rods, They regulate the net rate of neutron production, and hence they control the intensity of fission, process., , x, A, Number of atoms (nuclei) present in x gm of fuel, , Number of moles in x gm of fuel =, , x, n N A . Where N is Avogadro number, A , A, , COOLANT:, , , The function of a coolant is to keep the reactor, temperature at a low value so that there may not, be any danger of heat damage to the reactor., Air and CO2 are used as gaseous coolants. Water,, Organic liquids, Helium, Liquid Sodium are used, as liquid coolants. Liquid sodium is best coolant., Protective shield: The process of preventing, radioactive effect around nuclear reactor is called, Protective Shield., During the working of a nuclear reactor dangerous, radiations such as high energy neutrons, gamma rays, and thermal radiations are produced. To protect, the persons working there, the reactor is thoroughly, shielded with concrete wall of several feet thick, and lined with metals like lead., , xN A E, nE, P, At, t, Uses of Nuclear Reactors:, 1) To generate electric power., 2) To produce nuclear fuel plutonium -239 and, other radioactive materials which have a wide, variety of applications in the fields of medicine,, industry and research., , power P , , USES OF ATOMIC ENERGY :, 1., , POWER OF A NUCLEAR REACTOR, In the nuclear reactor, large amount of heat will be, generated in the core. These reactors have, elaborate cooling systems that use water. This water, absorbs the heat and produces steam. This steam, in turn is used to run the steam turbines which, ultimately generate electric power. Such reactors, are called power reactors., The power generated by a nuclear reactor is, nE, n, here be the number of fissions per, t, t, second and E be the energy released in each fission, E = 200 MeV 200106 1.61019 J, = 3.2 x 10-11J, Note : Number of fissions per sec in a reactor of, n P, power 1 W is given by , t E, 1, , = 3.125 x 1010 fissions per sec, 3.21011, , 2., , 127, 53 I, , P, , Note : If only x% of energy released in fission is, converted into electrical energy then out put power, x n E , , of reactor is, , 1 0 0 t , NARAYANAGROUP, , Generation of electric power : The coolant in a, nuclear reactor absorbs the heat generated as a, result of the chain reaction and it releases the heat, to the water which is converted into high pressure, steam. This steam is used to drive turbine and, operate the electric generator., Production of radio isotopes : A small amount of, the pure element is placed in an aluminium container, and the container is placed in the reactor for a few, days. The element absorbs neutrons and the, element becomes radioactive isotope., , 3., , 4., , 5., 6., , 0 n1 53 I128* ., Radioiodine obtained in this way can be used to, treat the thyroid gland. These radio isotopes have, a number of applications in the field of medicine ,, agriculture, industry and basic research., Source of neutrons : A large number of neutrons, are produced in a reactor. They are used in research, . The effect of neutrons on biological tissues is, studied. A new branch of physics called Neutron, Physics has come up., Atomic energy is used to create artificial lakes, to, divert the course of a river , to make tunnels for, laying new railway tracks etc., Atomic energy is used for driving automobiles,, submarines and war - planes., Atomic energy is used in war - fare for creating, destructive atom bombs and hydrogen bombs., 123
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , NUCLEAR FUSION:, , , , , , , , , , , , , , The phenomenon in which two lighter nuclei, combine to form a heavier nucleus of mass less, than the total mass of the combining nuclei is called, nuclear fusion. This mass defect appears as energy., At temperatures of about 107K, light nuclei combine, to give heavier nuclei. Hence, fusion reactions are, called thermo nuclear reactions., Nuclear fusion takes place in the sun and other, stars., Energy produced in a single fission of 92U235 is, larger than that in a single fusion of Hydrogen into, Helium., But fusion produces more energy than fission per, nucleon., In fission, 0.09% of mass is converted into energy., In fusion 0.66% of mass is converted into energy., Hydrogen bomb is a fission – fusion bomb., , , , In the sun both proton - proton cycle and, carbon- nitrogen cycles occur with equal, probabilities. In stars, whose interior temperatures, are less than that of the sun, proton -proton cycle, dominates the energy generation. Again in stars,, whose interior temperatures are more than that of, the sun, the energy generation is mainly due to, carbon- nitrogen cycle., The core temperature of heavier stars may be larger, than that of the sun and much larger nuclei may be, formed., NOTE: Due to enormous energy released in Sun, and Stars the atmosphere of them will be in ionised, state which is called Plasma (Which contains fast, moving neutrons and electrons). Nuclear fusion can, not be controlled., , STELLAR AND SOLAR ENERGY:, Stellar and solar energy is due to fusion., The cycles that occur are., Proton - Proton Cycle & Carbon - Nitrogen Cycle, , PROTON - PROTON CYCLE:, The Thermonuclear reactions involved are:, 1, 1, 2, 0, 2( H )+2( H ) 2( H ) + 2( e ) + Q1, 1, , 1, , 21H, , 2, , 2, , 3, , 1, , 1, , H, , 1, , 2, , 3, , +1, , He Q2, 3, , 2, , 4, , 2, , He + He ( He ) + 2( H ) + Q3, 2, 2, 2, 1, On adding up these reactions, we obtain., 1, 4, 0, 4( H ) He + 2 e + Q, 1, , 2, , +1, , Where Q Q1Q2 Q3 is the total energy evolved, in the fusion of 4 hydrogen nuclei (protons) to form, Helium nucleus. The value of Q as calculated from, mass defect comes out to be 26.7MeV, , CARBON - NITROGEN CYCLE:, Proposed by bethe. It consists of following reactions., 13, , 6, , 13, , 0, , C 12 1 H 1 7 N 13 Q1 ; N C +( e )+ Q2, 7, , 13, 1 H 1 7 N 14 Q3 ;, 6C, , 15, , 15, , 6, , 1, , 14, 1 H 1 8 O15 Q4, 7N, , 0, , O N +( e )+ Q5 ;, , 8, , 7, , +1, , 15, 1 H 1 6 C 12 2 He 4 Q6, 7N, On adding all these 6 equations, we get, 1, 4, 0, 4 H He + 2( e )+ Q, , 1, , 2, , +1, , Where Q Q1Q2 Q3 Q4 Q5Q6 The value of Q, as calculated from mass defect is 26.7 Mev., 124, , W.E-22: An explosion of atomic bomb releases an, energy of 7.6 x 1013 J. If 200 MeV energy is, , released on fission of one 235 U atom calculate, (i) the number of uranium atoms undergoing, fission. (ii) the mass of uranium used in the, atom bomb, Sol:E=7.6x103J;Energy released per fission = 200 MeV, = 200 106 1.6 x 10-19 = 3.2 x 10-11 J, Total energy, Number of uranium atoms (n) =, Energy per fission, , 7.6 1013, n, = 2.375 1024 atoms, 11, 3.2 10, Avagadro number (N) = 6.023×1023 atoms, Mass of uranium =, n × 235 2.375 × 1024 × 235, =, = 92.66g, N, 6.023 × 10 23, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , W.E-23: Calculate the energy released by fission, from 2 g of 23592 U in kWh. Given that the, energy released per fission is 200 MeV., Sol. Mass of uranium = 2g, Energy released per fission = 200 MeV, = 200 x 106 x 1.6 x 10-19 = 3.2 x 10-11 J, Number of atoms in 2 gram of uranium is, , n, , 2 6.023 1023, = 5.125 1021 atoms, 235, , NUCLEAR PHYSICS, , W.E-26: How long can an electric lamp of 100W, be kept glowing by fusion of 2.0 kg of, deuterium ? Take the fusion reaction as, 2, H+21H 32He+n+3.27 MeV, 1, Sol. 1 H 2 1 H 2 3 2 He n 3.27 MeV, No. of atoms in 2 kg of 1H2 = 2/2 x 6.023 x 1026, = 6.023 x 1026 atoms, In the above reaction two deuterium nuclei are combined, Power (p) = w x rate of fusion., Number of atoms, = 3.27 MeV x Time exp ended, , Total energy released = No. of atoms x energ, released per fission, = 5.125 x 1021 x 3.2 x 10-11 = 16.4 1010J, 10, Energy in Kwh = 16.4 10 Kwh, 36 105, , = 0.455 105 Kwh = 4.55 104 Kwh, , W.E-24: 200 Mev energy is released when one, nucleus of 235U undergoes fission. Find the, number of fissions per second required for, producing a power of 1 megawatt., Sol. Energy released = 200MeV, = 200 x 106 x 1.6 x 10-19 = 3.2 x 10-11J, P = 1 mega watt = 106 watts., , Total energy, No.of fissions per second (n)= Energy per fission, n=, , 106, 3.125 1016 Fissions, 11, 3.2 10, , W.E-25: How much 235U is consumed in a day in, an atomic power house operating at 400 MW,, provided the whole of mass235U is converted, into energy?, Sol. Power = 400 MW = 400 x 106 W;, time = 1 day = 86, 400 s., Energy produced , E = power time = 400 , 106 86,400 = 3.456 1013 J., As the whole of mass is converted into energy , by, Einstein's mass -energy relation., E = Mc2, , E 3.456 1013, , 3.84 10 4 kg 0.384 g., 2, 8 2, c, (3 10 ), NARAYANAGROUP, , 100 = 3.27 x 106 x 1.6 x 10-19 x, , x , =, , 6.023 1026, 2x, , 3.27 1.6 6.023 1011, = 15.756 x 1011S, 2, , 15.756 1011, 15.756 1011, =, =5x104 years, 365 24 60 60, 3.15 107, , W.E-27: Suppose India had a target of producing, by 2020 AD, 200,000 MW of electric power,, ten percent of which was to be obtained from, nuclear power plants. Suppose we are given, that, on an avedrage, the efficiency of, utilization ( i.e conversion to electric energy), of thermal energy produced in a reactor was, 25%. How much amount of fissionable, uranium would our country need per year by, 2020 ? Take the heat energy per fission of 235U, to be about 200 MeV., Sol. Required power from nuclear plants, = 10% of 2,00,000 Mw = 2 x 1010w, Required electric energy from nuclear plants in one, year = 2 x 1010 x 365 x24 x 60 x 60, = 2 x 1010 x 3.15 x 107 = 6.30 x107 J, Available electric energy per fission = 25% of 200, MeV, = 50 MeV = 8 x 10-12J, Req. no. of fissions per year =, , 6.30 1017, 8 10 12, , =0.7875x1029, , 0.7875 1029, Req. no. of moles of U =, =0.1307 x106, 6.023 1023, Required mass of U238 = 0.1307 x 235 x106 g, = 30.71 x106 gm = 30.71 x106 x 10-3 kg, = 0.03071 x 106 kg = 3.071 x 104 kg, 238, , 125
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , W.E-28: Calculate the energy released by the, fission 1 g of 235 U in joule, given that the, energy released per fission is 200 MeV., ( Avogadro’s number = 6.023 1023 ), Sol. The number of atoms in 1 g of 235 U, =, , Avogadro ' s number, Mass number, , =, , 6.0231023, 2.5631021, 235, , Energy released per fission = 200 MeV, , , , PAIR AND PRODUCTION AND PAIR, A N N I H I L A T I O N : When an energetic γ photon falls on a heavy nucleus, it is absorbed by, the nucleus and a pair of electron and positon is, produced. This phenomenon is called as pair, production and can be represented by the following, equation:, , hv, , γ photon , , , , 1, , β0, , , , Positron , , = 200106 1.61019 3.21011 J ., Energy released by 1 g of 235 U, = Number of atoms energy released per fission, , uranium, the mass lost is 0.92 milligram. The, efficiency of power house run by it is 10%. To, obtain 400 megawatt power from the power, house, how much uranium will be required, per hour? (c = 3 x 108 ms–1), Sol. Power to be obtained from power house = 400, mega watt\ Energy obtained per hour = 400, meagwatt x 1hour = (400 x 106 watt) x 3600 second, = 144 x 1010 joule, Here only 10% of input is utilised. In order to obtain, 144 x 1010 joule of useful energy, the output energy, 10 E, , electron , , photon, Heavy nucleus 10, The rest mass energy of electron or positron is:, , E0 m0c 2 9.1 1031 3 108 , , 2, , 8.2 1014 J 0.51MeV ., Hence for pair production, the minimum energy of, γ -photon must be 2 0.51 1.02MeV . If the, energy of γ -photon is less than this, there may be, Compton’s effect. If energy of γ -photon is greater, than E0 , then extra energy will become kinetic, energy of the particles. If E is the energy of γ photon, then kineric energy of each particle will, , E 2 E0, 2, , from the power house 100 = 144 x 1010 J, , be, K electron K positron , , E = 144 x 1011 joule, Let, this energy is obtained from a mass–loss of, , The inverse process of pair production is called, pair annihilation. According to it when electron, and a positron come close to each other, annihilate, each other and produces minimum two γ -photons., Thus, , m kg., Then, m , , m c 2 144 1011 joule, 1441011, , 310 , 8, , 2, , 16105 kg, , = 0.16 g, , Since 0.92 milli gram (= 0.92 x 10–3 g) mass is lost, in 1 g uranium, hence for a mass loss of 0.16g, the, uranium required is , , 1 0.16, 3, , 0.9210, , 1, , β0, , Positron , , , , 1, , β0, , electron , , , , 2hf, γ photon , , +10, 2 ( photon ), , 174 g, , Thus to run the power house, 174 gm uranium is, required per hour., 126, , β0, , +10, , = 2.5631021 3.21011 J 8.2021010 J, , W.E-29: In the process of nuclear fission of 1 gram, , 1, , 10, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , ADDITIONAL INFORMATION, ELEMENTARY PARTICLES :, We have realized, so far that there are only four, fundamental constituents of matter. We can, describe various physical processes involving, atoms, molecules and nuclei in terms of electrons,, protons, neutrons and photons. The first three are, the building blocks of atoms and hence matter. The, fourth one (i.e photon) is the quantized energy which, is exchanged whenever electronic or nucleonic, transition is involved., Subsequently many more elementary particles and, antiparticles have been discovered, using giant and, modern accelerating machines., The particles which are not constituted by any, other particles are called Elem entary, particles. A brief discussion of important, fundamental particles is as follows., i) Electron : It was discovered in 1897 by Thomson., Its charge is –e and mass is 9.1 X 10–31 kg. Its, symbol is e– (or 1 b 0 ). It is a stable particle having, 1, 2, Proton : It was discovered in 1919 by Rutherford, in artificial nuclear disintegration. It has a positive, charge +e and its mass is 1836 times (1.673 X 10–, 27kg) the mass of electron. In free state, the proton, is a stable particle. Its symbol is P+. It is also written, , spin =, ii), , as 1 H 1 . It is a stable particle having spin = 1/2., iii) Neutron : It was discovered in 1932 by Chadwick., Electrically it is a neutral particle. Its mass is 1839, times (1.675 X 10–27 kg) the mass of electron. In, free state the neutron is unstable. Inside the nucleus, the neutron is stable. Its symbol is n (or) 0 n1 ., iv) Positron : It was discovered by Anderson in 1932., It is the antiparticle of electron, i.e., its charge is +e, and its mass is equal to that of electron. Its symbol, is e+ (or 1 b 0 ), v) Antiproton : It is the antiparticle of proton. It was, discovered in 1955. Its charge is –e and its mass is, equal to that of proton. Its symbol is P–., vi) Antineutron : It was discovered in 1956. It has, no charge and its mass is equal to the mass of, neutron. The only difference between neutron, and antineutron is that their magnetic momenta, will be equal in magnitude and opposite in, direction. The symbol for antineutron is n ., NARAYANAGROUP, , NUCLEAR PHYSICS, vii) Neutrino and antineutrino : The existence of, these particles was predicted in 1930 by Pauli while, explaining the emission of b - particles from, radioactive nuclei, but these particles were actually, observed experimentally in 1956. Their rest mass, and charge are both zero but they have energy and, momentum. These are mutually antiparticles of each, other. They have the symbol n and n, viii) Pi - Mesons : The existence of pi - mesons was, predicted by Yukawa in 1935, but they were, actually discovered in 1947 in cosmic rays. Nuclear, forces are explained by the exchange of pi-mesons, between the nucleons. pi - mesons are of three, types, positive p - mesons p , negative pi, 0, mesons p and neutral p - mesons p ., , Charge on p is e. Whereas mass of p is 274, times the mass of electron. p 0 has mass nearly, 264 times the electronic mass. These are unstable, having half life 10–8 sec and spin = 0, ix) Mu-Mesons : These were discovered in 1936, by Anderson and Neddermeyer. These are found, in abundance in the cosmic rays at the ground level., There are two types of mu-mesons. Positive mumeson m and negative mu-meason m . There, , x), , is no neutral mu-meson. Both the mu-mesons have, the same rest mass 207 times the rest mass of the, electron. These are unstable having half life 10–6, sec and spin = 1/2., Photon : These are bundles of electromagnetic, energy and travel with the speed of light. Energy, and momentum of a photon of frequency n are, , hn, hn and c respectively. They posses no charge., no mass and spin = 1 and are stable., xi) Gravitons : Hypothetical particles that carry, gravitational energy are called Gravitons. They, possess no mass, no charge and spin = 2 as, proposed by Dirac., , Antiparticles :An antiparticle is a form of matter, that has the same mass as the particle but carries, an opposite charge and / or a magnetic moment, that is oriented in an opposite direction relative to, the spin., 127
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , Name of the particle, , Antiparticle, (e+1), Anti proton (P-), , Electron (e-1), , Positron, , Proton (P+), , Anti Neutron n, , Neutron (n), , Anti Neutrino n , , Neutrino n , Positive Pi-Meson p , , THE QUARK MODEL:, According to this model, each baryon and meson, particle is made up of quarks. Quarks are supposed, to be the ultimate building blocks of these particles, and hence matter. They are point particles with no, internal structure. Unlike any other charged, particles, quarks are supposed to have fractional, charges and spin 1/2., , Negative Pi-Meson p , , Positive Mu-Meson m Negative Mu-Mesonm , , Note: A few electrically neutral particles, like the, photon and neutral p meson are their own, antiparticles. A collision between a particle and an, antiparticle results in annihilation of matter., , Name, , Q u arks, Sym bo l, C h a rg e, , Up, , u, , , , D ow n, , d, , , , S tr a n g e, , s, , , , C harm ed, , c, , , , Top, , t, , , , B o tto m, , b, , , , Classification of particles based on spin, 1), 2), , Bosons : These particles have spin in the integral, multiples of unity, Fermions : These particles have spin in the integral, multiples of 1/2., , Classification of particles based on rest mass, 1) Photons : Particles with zero rest mass, 2) Leptons : Lighter particles, 3) Mesons : Particles with intermediate mass, 4) Baryons : Heavier particles, Classification of particles based on interaction, 1) Photons: representing electromagnetic interactions., 2) Leptons: representing weak interactions., 3) Hadrons: representing strong interactions., 4) Gravitons: representing gravitational interactions., , 2, 3, 1, 3, 1, 3, 2, 3, 2, 3, 1, 3, , A n tiq u a r k s, Sym bo l, C h arg e, , e, , u, , , , e, , d, , , , e, , s, , , , e, , c, , , , e, , t, , , , e, , b, , , , 2, 3, 1, 3, 1, 3, 2, 3, 2, 3, 1, 3, , e, e, e, e, e, e, , These quarks combine to form some well known, Mesons and Baryons. A meson consists of a quark, and an antiquark whereas a baryon contains three, quarks., , Elementary Particles, , Bosons, , Mass less Boson, , Particle, , Fermions, , spin, , proton, , (u, u, d), , e, , 1, 2, , neutron, , (u d d), , 0, , 1, 2, , pi-meson, , (ud), , e, , Kaons, , , , , , , , , +, +, , , Leptons, , Baryons, , Tau Neutrinos Hyperons, , ( ), Lambda Sigma, , , , 128, , charge, , Meson, , Etameson Pions, , ( ), Gravition Photons, , Electrons Muon, , , ( ), (e ), , Quarks, , Xi, , , Omega, , , Nucleons, , Proton Neutron, , 0, , Table - 2, , Photons and leptons are considered to be, elementary particles and they are not composed of, quarks., Note : Isolated quark doesn’t exist., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , W.E-30: An electron-positron pair is produced, , W.E-32: Obtain the maximum kinetic energy of, , when a -ray photon of energy 2.36MeV, passes close to a heavy nucleus. Find the, kinetic energy carried by each particle, produced, as well as the total energy with, each., , -particles, and the radiation frequencies of, decays in the decay scheme shown in Fig., 14.6. You are given that, m(198Au) = 197.968233 u; m(198Hg) = 197.966760 u, 198, 79, , Au, , Sol. The reaction is represented by, , , , 1, 2, , 0, , Page 24 Table 2, , , , 1.088Mev, , 0, , ( 1e ) ( 1e ), sothat, , 1, , 3, 0.412Mev, , 2, , 2, , E m0C K .E electron + m0C K .E positron, 2, , 2.36MeV = 2m0.C 2 K.E (electron) + K.E (positron), , 198, 80, , Sol. -rays are electro magnetic radiations having, , = 1.02 MeV + K .E(e ) K .E(e ), 1, 2, , , K.E. of (e ) K .E(e ) (2.36 1.02)MeV ,, , (K.E. carried each) = 0.67 MeV (motional energy), , E, where h = plank`s, h, constant = 6.625 x 10-34 J.S, , energy E = h =, , 1) Frequencies of 1 , 2 and 3 are calculated as follows, , 1 , , Total energy shared by each particle is obviously, m0C 2 K .E 0.51MeV 0.67MeV = 1.18MeV.., , W.E-31: A gamma ray photon of energy 1896 MeV, annihilates to produce a proton-antiproton, pair. If the rest mass of each of the particles, involved be 1.007276 a.m.u approximately,, find how much K.E these will carry?, Sol. Working on the same lines as an electron-positron, pair production, we notice that the reaction. , proton + antiproton, has the energy balance, E = m0, m0, , (proton), , C2, , 2, (antiproton)C, , + K.E (proton) +, , + K.E, , (antiproton), , But m0C2 = energy equivalent of 1.007276 a.m.u, , 0, , Hg, , 2 , , E (1.088 0)MeV 1.088 106 1.6 1019, , , h, 6.625 10 34 J .s, 6.625 1034, = 0.2627 x 1021 = 2.627 x 1020 Hz, E (0.412 0) MeV 0.412 1.6 1019 106, , , h, 6.625 1034 J .s, 6.625 1034, = 0.0995 x1021 = 9.95 x 1019Hz, E (1.088 0.412) 106 1.6 10 19, , h, 6.625 10 34, = 0.1632 x 1021 = 1.632 x 1020 Hz, , 3 , , 2) Now maximum K.E of 1 = [M(19879Au), M(19880Hg)-, , 1.088 2, ]c, 931.5, , 1, U ), 931.5, = [197.968233-197.966760-0.001168]931.5 MeV, = 0.000305 x 931.5 = 0.284 MeV, , ( 1 amu = 931.5 MeV 1MeV =, , 938 MeV. [ 1.007276 x 931 938 MeV], , Maximum K.E of 2 =, , Thus K.E of each particle, , 0.412 2, ]c, 931.5, = [197.968233-197.966760-0.000442]931.5, = 0.001031 x 931.5 =0.9603 MeV, , =, , 1, [1896 MeV - 2 × 938 MeV] = 10MeV.., 2, , NARAYANAGROUP, , [M(19879Au)-M(19880Hg)-, , 129
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NUCLEAR PHYSICS, , C.U.Q, COMPOSITION OF NUCLEUS, ISOTOPES,, ISOBARS, ISOTONES SIZE OF THE, NUCLEUS, 1., , 2., , 3., 4., , The particle A is converted to C via following, reactions then, A B + 2He4, B C + 2-1e0, 1) A and C are isobars 2) A and C are isotopes, 3) A and B are isobars4) A and B are isotopes, The particles which can be added to the nucleus, of an atom without changing its chemical, properties are, 1) electrons 2) protons 3) neutrons 4) positron, The radius of the nuclues is proportional to,, (if A is the atomic mass number), 1) A, 2) A3, 3) A1/3, 4) A2/3, The radius of a nucleus mainly depends on, 1) Proton number, 2) Electron Number, 3) Mass number, 4) Neutron number, , The nuclei 6 C13 & 7 N14 can be described as, 1) isotones 2) isobars 3) isomers 4) isotopes, 6. The graph of n (R/ R 0 ) versus n A, (R =radius ) of nucleus. A is atomic number is, 1) Straight line 2) Parabola 3) Ellipse 4) Circle, 7. The nucleus of 56Ba141 contains, 1) 85 protons, 56 neutrons, 2) 55 protons ,86 neutrons, 3) 56 protons,85 neutrons, 4) 86 protons,55neutrons., 8. The nuclear size is measured in units of, 1) Angstrom 2) Fermi, 3) Bar 4) Light-year, 9. Nuclides which have the same mass number, are called (NCERT), 1) Isotopes 2) Isobars 3) Isotones 4) Isomars, 10. Observe the following statements regarding, isotones i) 39 K19 and 40 Ca20 are isotones, ii) Nuclides having different atomic numbers, (z) and mass number (A) but same number of, neutrons (n) are called Isotones iii) 1 9 F9 and, 23, Na11 are isotones The correct answer is, 1) i,ii and iii are correct 2) only i and ii are correct, 3) only i and iii are correct 4) only ii and iii are correct, 11. A and B are isotopes. B and C are isobars. All, three are radioactive. Which one of the, following is true., 1) A, B and C must belong to the same element, 2) A, B and C may belong to the same element, 3) It is possible that A will change to B through a, radioactive-decay process, 4) It is possible that B will change to C through a, radioactive-decay process, 5., , 130, , JEE-ADV PHYSICS- VOL- V, , MASS DEFECT, BINDING ENEGRY,, PACKING FRACTIONAND MASS ENERGY, RELATION, 12. M,Mn & Mp denotes the masses of a nucleus, of ZXA a neutron, and a proton respectively., If the nucleus is separated in to its individual, protons and neutrons then, 1) M=(A-Z)Mn+ZMp 2) M=ZMn+(A-Z)Mp, 3) M>(A-Z)Mn+ZMp 4) M<(A-Z)Mn+ZMp, 13. The difference between the mass of a nucleus, and the combined mass of its nucleons is, 1) zero, 2) positive, 3) negative 4) zero, positive or negative, 14. The mass number of a nucleus is, 1) Always less than atomic number, 2) Always more than atomic number, 3) Equal to atomic number, 4) Sometimes more or equal to atomic number., 15. If M is atomic weight , A is mass number, then (M-A)/A represent, 1) Mass defect, 2) Packing fraction, 3) Binding Energy, 4) Chain Reaction, 16. The difference between mass of the nucleus, and total mass of its constituents is called, 1) Packing fraction, 2) Mass defect, 3) Binding energy, 4) Binding energy per neucleon, 17. The parameter used to measure the stability, of the nucleus is, 1) Average binding energy2) No of protons, 3) No of neutrons, 4) No of electrons, 18. When the number of nucleons in a nuclues, increases the binding energy per nucleon, 1) Increase continuously with mass number, 2) Decreases continuously with mass number, 3) Remains constant with mass number, 4) First increases and then decreases with, increase in mass number, 19. Maximum value of binding energy per nucleon, for most stable nuclei is, 1) 8MeV 2) 8.8MeV 3) 7.6MeV 4) 1.1MeV, 20. The binding energy per nucleon is maximum, at A=56 and its value is around_Mev/ Nucleon, 1) 8.4, 2) 8.7, 3) 9, 4) 7.8, 21. Average binding energy per nucleon over a, wide range is, 1) 8MeV 2) 8.8MeV 3) 7.6MeV 4) 1.1MeV, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 22. The wrong statement about binding energy is, 1) It is the sum of the rest mass energies of, nucleons minus the rest mass energy of the nucleus., 2) It is the energy released when the nucleons, combine to form a nucleus., 3) It is the energy required to break a given nucleus, into its constituent nucleons., 4) It is the sum of the kinetic energies of all the, nucleons in the nucleus., 23. The binding energies of a deutron and an particle are 1.125, 7.2 MeV/nucleon, respectively. The more stable of the two, is, 1) deutron, 2) -particle, 3) both, 4) sometimes deutron and sometimes -particle, 24. Mass defect of an atom refers to, 1) inaccurate measurement of mass of neutrons, 2) mass annihilated to produce energy to bind the, nucleons, 3) packing fraction, 4) difference in the number of neutrons and, protons in the nucleus, 25. The stability of a nucleus can be measured by, 1) Average binding energy, 2) Packing fraction, 3) Ratio of number of neutrons and protons, 4) All the above., 26. In a nuclear reaction some mass converts into, energy. In this reaction total B.E of reactants, when compared with that of products is, 1) always greater, 2) always les, 3) either greater or less, 4) always equal, RADIOACTIVITY, , , RAYS, HALF, LIFE, MEAN LIFE, DECAY CONSTANT, 27. The age of pottery is determined by, archeologists using a radioisotope of, 1) carbon 2) cobalt 3) iodine 4) phosphorus, 28. During an artificial transmutation the nucleus, emits, 1) -particles, 2) -particles, 3)always neutrons, 4) may emit protons or neutrons, 29. When two deuterium nuclei fuse together to, form a tritium nucleus, we get a, 1) neutron, 2) deuteron, 3) alpha particle, 4) proton, NARAYANAGROUP, , NUCLEAR PHYSICS, 30. Identify the correct statement/statements, a) Radiation causes genetic mutation, b) Restriction in blood circulation can be, detected using radio-iodine, c) Hydrocarbon plastics are used as, moderators in a nuclear reactor, d) The damage caused due to -radiation is, small due to its small penetrating power, 1) a,b,c 2) a,c,d 3) b,c,d 4) a,b,d, 31. Identify the correct ascending order of , , and with reference to their ionizing power, I) -ray II) -ray, III) -ray, 1) II,III,I 2) I,III,II 3) II,I,III 4) I,II,III, 32. Two identical nuclei A and B of the same, radioactive element undergo b decay. A, emits a b particle and changes to A'. B emits, a b particle and then a g –photon, immediately afterwards, and changes to B'., 1) A' and B' have the same atomic number and, mass number, 2) A' and B' have the same atomic number and, different mass numbers, 3) A' and B' have different atomic numbers but the, same mass number, 4) A' and B' are isotopes, 33. Identify the correct ascending order of , , and with reference to their penetrating, power, I) -ray, II) -ray III) -ray, 1) II,III,I, 2) II,I,III, 3) I,II,III, 4) III,I,II, 34. If a beam consisting of , and radiation, is passed through an electric field, perpendicular to the beam, the deflections, suffered by the components, in decreasing, order, are, 1) , , 2) , , 3) , , 4) , , , 35. Decrease in atomic number is observed during, A) -emission, B) -emission, C) Positron emission D) electron capture, 1) B is correct, 2) A and B are correct, 3) A,C and D are correct, 4) Only C, 131
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 36. When 15 P 30 decays to become 14 Si 30 ,the, particle released is, 1) electron, 2) -particle, 3) neutron, 4) positron, 37. During - decay, a neutron inside nucleus, converts into proton, electron and x. Then the, particle x is, 2) neutrons, 1) - meson, 3) anti-neutrino, 4) -meson, 38. If a nucleus emits a gamma-ray, its atomic and, mass number____ but there will be ___in the, energy of the nucleus.Select suitable pair, 1) Remain same, increase, 2) Remain same, decrease, 3) Decrease, increase, 4) increase, decrease, 39. In the following nuclear reaction, Al27+2He4 15P30+X,X will be, 13, 1) Proton, 2) Electron, 3) Neutron, 4) -particle, 9, 40. In nuclear reaction 4Be +2He4-->6C12+X,X will, be, 1) Proton 2) Neutron 3) -particle 4) -particle, 41. In nuclear reaction 2 He 4 z X A z 2 Y A 3 R R, denotes, 1) electron 2) positron 3) proton 4) neutron, 42. A positron is emitted by radioactive nucleus, of proton no 90. The product nucleus will have, proton number, 1) 91, 2) 90, 3) 89, 4) 88, 43. 13Al27 + -particle neutron +’X’ then ‘X’, is, 1) 15p31, 2) 14Si30, 3) 15p30, 4) 15Si30, 44. The penentrating power of beta particle, compared to alpha particle is, 1) Less, 2) More, 3) Equal, 4) Can be more or less, 45. In a nuclear reactor , heavy water is used as a, 1) Controlling material 2) Moderator, 3) Fuel, 4) Heat exchanger, 46. The units of radioactivity is, 1) Fermi 2) Farad, 3) Curie, 4) Hertz, 47. The half-life of a radioactive isotope is 3 hours., The value of its disintegration constant is, 1) 0.3 hour-1, 2) 0.693 hour-1, 3) 0.231 hour-1, 4) 0.231 min-1, 132, , 48., , 49., , 50., , 51., 52., , 53., , 92 U, , 238, , 82Pb 206 8 42 He . The number of , , particles released in this reaction is, 1) 6, 2) 3, 3) 1, 4) 10, The activity in any nucleus is measured in, 1) Curie, 2) Rutherford, 3) Both 1 &2, 4) Newton, The -particles are, 1) high energy electrons, 2) positively charged hydrogen ions, 3) high energy -radiation, 4) doubly positively charged helium nuclei, -particles carries, 1) Mass 1 2) Mass 2 3) Mass 3 4) Mass 4, At a specific instant, the emission of radio, active compound is deflected in a magnetic, field. The compound can emit, i) electrons, ii) Positrons, , 2, iv) neutrons, iii) He, 1) i,ii,iii, 2) i,ii,iii,iv 3) iv, 4) ii,iii, The atomic number (A ) and mass number (M), of the nuclide formed where three alpha , 238, and two particles are emitted from 92, U, , 1) A 87, M 233, 2) A 86, M 226, 3) A =88, M = 227, 4) A = 88, M = 226, 54. Element ZM A emits one (alpha) particle, followed by two (beta) particles. Among the, following the daughter element is, 1), , Z 2, , M A4, , 2), , Z 2, , MA, , A4, 3) Z M A 4, 4) Z2 M, 55. Particles which can be added to the nucleus, of an atom without changing its chemical, properties are called., 1) Neutrons, 2) Electrons, 3) Protons, 4) Alpha Particles, 56. An Electric field can deflect, 1) - particles, 2) X - rays, 3) Neutrons, 4) - rays, 57. On the bombardment of Boron with neutron,, - particle is emitted and product nucleus, formed is ..............., , 1) 6C12, 2) 2Li6, 3) Li 8, 4) 4Be9, 3, 58. The one has maximum activity, 1) Uranium, 2) Plutonium, 3) Radium, 4) Thorium, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , 59. Among the following one is not emitted by a, radioactive substance, 1) Electrons 2) - rays 3) Positron 4) Protons, 60., , 228, 90, , 212, Th 83, Bi . the number of and, , particals given out during the process are, 1) 4 , 7 2) 4 ,1 3) 4, 4) 7 , 61. The reciprocal of radioactive decay constant, is called, 1) Half life period, 2) Whole life period, 3) Average life period, 4) Avagadro number, 62. The missing particle in the reaction, 1, 1, , p , , , , , 0, 1, , e, , 1) deuteron 2) proton 3) neutron 4) - particle, 63. In the Radioactive transformation, , , , R, , X, , Y, , Z ; the neuclii R and, Z are, 1) Isotopes 2) Isobars 3) Isomers 4) Isotones, 64. In the reaction 15 P 30 14 Si 30 , The change, requires the emission of, 1) - particle, 2) -particle, 3) neutron, 4) positron, 65. When a radioactive substance is subjected to, a vacuum, the rate of disintegration per second, 1) increases considerably 2) is not affected, 3) increases only if the products are gases, 4) suffers a slight decrease, 66. A radioactive nuclide can decay, simultaneously by two different processes, which have decay constants 1 and 2 . The, effective decay constant of the nuclide is ., 1) 1 2, , 2) , , 1, 1 2 , 2, , 1 1 1, 3) , 4) 12, 1, 2, 67. A sample of radioactive material is used to, provide desired doses of radiation for medical, purposes. The total time for which the sample, can be used will depend, 1) only on the number of times radiation is drawn, from it, 2) only on the intensity of doses drawn from it, 3) on both (a) and (b), 4) neither on (a) nor on (b), NARAYANAGROUP, , 68. A fraction f1 of a radioactive sample decays, in one mean life, and a fraction f 2 decays in, one half-life., 1) f1 f 2 2) f1 f 2 3) f1 f 2, 4) May be (a), (b) or (c) depending on the values, of the mean life and half-life., , NUCLEAR FORCES, 69. The short range attractive nuclear forces that, are responsible for the binding of nucleons in, a nucleus are supposed to be caused by the, role played by the particles called, 1) Positron, 2) m-Meson, 3)K-Meson, 4) - Meson, 70. The strong interaction exists in, 1) Gravitational forces, 2) Electrostatic force of attraction, 3) Nuclear forces, 4) Magnetic force on a moving charge, 71. Nuclear forces are, 1) Non-central forces 2) saturated, 3) Spin dependent, 4) All the above, 72. Identify the correct statement/statements, a) At greater distances nuclear forces are, negligible, b) Nuclear forces are non central forces, c) Nuclear forces are weakest in nature, d) Nuclear forces are charge dependent forces, 1) a, b 2) b, c 3) c, d, 4) a, d, 73. Which of the following is not correct about, nuclear forces?, 1) They are short range attractive forces, 2) They are independent of charge, 3) They change to repulsion at very close distance, 4) They obey inverse square law, 74. Among the following, short range, charge, independent and spin dependent forces are, 1) Gravitational forces, 2) Nuclear forces, 3) Electromagnetic forces 4) Weak forces, 75. Let Fpp,Fpn and Fnn denote the magnitudes of, the nuclear force by a proton on a proton ,by a, proton on a neutron and by a neutron on a, neutron respectively when the separation is, less than one fermi, then (2008 E), 1) Fpp>Fpn=Fnn, 2) Fpp=Fpn=Fnn, 3) Fpp>Fpn>Fnn, 4) Fpp<Fpn=Fnn, 133
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 76. Two protons are kept at a seperation of, 10nm.Let Fn be the nuclear force and Fe be, electrostatic force between them. Then, 1) Fe=Fn 2) Fe>>Fn 3) Fe<<Fn 4) Fn= 3Fe, 77. Two nucleons are at a separation of, 1x10–15m. The net force between them is F1 if, both are neutrons, F2 if both are protons and, F3 if one is a proton and other is a neutron. In, such a case, 1) F2>F1>F3, 2) F1=F2>F3, 3) F1=F2=F3, 4) F1=F3>F2, 78. Two protons attract each other when, 1) the distance between them is 10-10 m, 2) the distance between them is 10-1 m, 3) the distance between them is 10-15 m, 4) the distance between them is 10-6 m, 79. Among gravitational,electrostatic and nuclear, forces,the two attractive forces between two, neutrons are, 1) Electrostatic and nuclear, 2) Electrostatic and gravitational, 3) Gravitational and nuclear 4) Electrostatic, 80. Among the following interactions one is of, least significant in nuclear physics is, 1) nuclear interaction, 2) gravitational interaction, 3) electrostatic interaction, 4) electromagnetic interaction, 81. The origin of nuclear force between nucleons, is due to the exchange of, 1) Mesons 2) Photons 3) Positrons 4) Electrons, , DISCOVERY OF NEUTRON, 82. Which of the following particles is most, unstable?, 1) Neutron 2) Proton 3) Electron 4) particle, 83. A free neutron decays spontaneously into:, 1) a proton, an electron and an anti-neutrino, 2) a proton, an electron and a neutrino, 3) a proton and electron, 4) a proton, an electron, a neutrino and an anti-neutrino, 84. Neutron was discovered by the experiment of, , , 2) Artificial transmutation of , 1) Artificial transmutation of, , 4 Be, 7N, , , , 11, , by - particles, by -particles, , 3) Rutherfored scattering of alpha particles by heavy, nuclei, 4) Bequerel with radio activity, 134, , 85. The average life of an isolated neutron is, 1) 1500 s 2) 1000 s 3) 1200 s 4) 3 minutes, 86. The energy of thermal neutrons is, 1) < 1 ev 2) > 1 ev 3) = 2 Mev4)= 4 Mev, 87. A nucleus with an excess of neutrons may, decay with the emission of, 1) a neutron, 2) a proton, 3) an electron, 4) a positron, 88. The most penetrating atom smashing particle, is, 1) neutron, 2) proton, 3) alpha particle, 4) deuteron, 89. Which of the followig is formed by decay of a, free neutron?, 1) A number of electrons 2) Two Protons, 3) A proton and an electron 4) An - particle, 90. In neutron discovery experiment Beryllium, target is bombarded by, 1) Protons, 2) Alpha particles, 3) Neutrons, 4) Deutrons, 91. Slow neutrons are sometimes refer to as, thermal neutrons because, 1) they are sort of heat radiations, 2) they are in thermal equilibrium, 3) they are capable of generating heat, 4) their energies are of same order as that of, molecular energies at ambient temperatures., 92. Thermal neutrons are, 1) Prompt neutrons, 2) Slow neutrons, 3) Neutrons which are in the nucleus, 4) Neutrons from the sun, 93. In neutron discovery experiment, Be is, bombarded with, 1) Proton, 2) Deutrons, 4) -particle, 3) -particle, 94. The process of producing a new stable nucleus, from the other stable nucleus is called, 1) Nuclear reaction 2) Artificial transmutation, 3) Nuclear fusion 4) Nuclear fission, NUCLEAR FISSION, 95. At least how many thermal neutrons should, be available to start a fission reaction, 1) 2, 2) 3, 3) 1, 4) 4, 96. Which of the following changes in the artificial, transmutation of elements?, 1) number of neutrons 2) number of electrons, 3) atomic weight, 4) nucleus, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 97. During the fission process of Uranium,the, amount of energy liberated per fission is nearly, 1) 100MeV 2) 200MeV 3) 150MeV 4) 300MeV, 98. The number of neutrons that are released on, an average during the fission of U235nucleus, is, 1) 3, 2) 1, 3) 2.5, 4) 5, 99. Nuclear fisssion can be explained by, 1) Optical model of the nucleus, 2) Shell model of nucleus, 3) Collective model of the nucleus, 4) Liquid drop model of the nucleus, 100. Percentage of mass lost during the fission of, U235 approximately is, 92, 1) 0.01% 2) 0.1%, 2) 0.7%, 4) 0.9%, 101. Most of energy released in the fission is, carried by, 1) neutrons, 2) fission fragments, 3) neutrons and fragments carry equally, 4) positrons, 102. Regarding Prompt neutrons, 1) They are highly energetic 2) They constitute 99%, 3) Cannot initiate chain reaction, 4) 1,2,3 are correct, 103. Nuclear reactions obey the law of, conservation of, 1) Mass and energy, 2) Charge, 3) Momentum, 4) All the above, 104. The critical mass of a fissionable material is, 1) 0.1kg equivalent, 2) The minimum mass needed for chain reaction, 3) The rest mass equivalent to 1020 joule, 4) 0.5kg, 105. For fast chain reaction, the size of U235 block,, as compared to its critical size, must be, 1) greater 2) smaller 3) same 4) anything., 106. The critical mass of fissionable uranium-235, can be reduced by, 1) adding impurities, 2)heating material, 3) surrounding it by a neutron-reflecting material, 4) surrounding it by a neutron-absorbing material, 107. Nuclear energy is released in fission since, binding energy per nucleon is, 1) smaller for fission fragments than for parent nucleus, 2) the same for fission fragments and parent nucleus, 3) larger for fission fragments than for parent nucleus, 4) sometimes larger and sometimes smaller, NARAYANAGROUP, , NUCLEAR PHYSICS, 108. In a critical chain reaction, 1) energy is released at increasing rate, 2) energy is released at steady rate, 3) energy is released at decreasing rate, 4) energy is not released, 109. Among the following one is wrong, 1) The energy of thermal neutrons is about 25 meV, 2) In a nuclear reactor, when neutrons multiplication, factor, K = 1 then the reaction is said to be critical, 3) 92 U 235 undergoes fission by bombardment of high, energy neutron, 4) On average 2.5 neutrons are emitted per fission of, U 235, 110. When 1gm of U235 is completely annihilated, energy liberated is E1 and when 1 gm of U235, completely undergoes fission the energy, liberated is E2 , then, 1) E1 > E2 2) E1 = E2 3) E1 < E2 4) E1 E2, 111. In the process of fission, the binding energy, per nucleon, 1) Increases, 2) Decreases, 3) Remains unchanged, 4) Increases for mass number A < 56 nuclei but, decreases for mass number A> 56, 112. Assertion (A) : Fragments produced in the, fission of U 235 are radioactive., Reason (R) : The fragments have abnormally, high proton to neutron ratio, 1) Both A and R are true and R is correct, explanation of A, 2) Both A and R are true and R is not correct, explanation of A, 3) A is true but R is false 4) A is false but R is true, 113. The products of the fission of U235 by thermal, neutron are, 1) Ba 141 and Kr 92 and 3 neutron always, 2) Xe 140 , Sr 94 and 2 0n1 always, 3) can be different in each fission, 4) should have same mass number, 114. Consider the following statements A and B., Identify the correct in the given answer., A) p-n; p-p; n-n forces between nucleons are, not equal and charge dependent., B) In nuclear reactor the fission reaction will, be in accelerating state if the value of neutron, reproduction factor k>1., 1) Both A and B are correct, 2) Both A and B are wrong, 3) A is wrong B is correct, 4) A is correct B is wrong, 92, , 135
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NUCLEAR PHYSICS, 115. The process of fission is responsible for the, release of energy in, 1) The hydrogen bomb 2) The atom bomb, 3) The sun, 4) The star, 116. The working principle in atom bomb is, 1) under-critical chain reaction, 2) Critical chain reaction, 3) super-critical chain reaction 4) All the above, 117. Heavy water is, 1) Water at 40C, 2)Watercontaining various salts, 3) Compound of heavy oxygen and hydrogen, 4) Compound of oxygen and deuterium., 118. Nuclear reactor is surrounded by concrete walls, to, 1) Strengthen the construction, 2) Control the chain reaction, 3) form a protective shield 4) as moderator, 119. The operation of a nuclear reactor is said to, be critical, if the multiplication factor (K) has, a value, 1) 1, 2) 1.5, 3) 2.1, 4) 2.5, 120. Cadmium and Boron rods are used in a, nuclear reactor to, 1) Slow down the neutrons, 2) Absorb excess number of thermal neutrons, 3) speed up neutrons, 4) absorb fast neutrons, 121. The coolant in the nuclear reactor is, 1) Liquid sodium, 2) cadmium, 3) Deuterium, 4) Liquid hydrogen, 122. Substance used to slow down the fast neutrons, released during nuclear fission is called., 1) Fuel, 2) Moderator, 3) Controlling rods, 4) Reflecting rods, 123. If the neutron reproduction factor K is, a) greater than 1 the fission reaction is accelerated, b) less than 1 the fission reaction retards, c) equal to 1 the fission reaction is at steady state, 1)only a,b are true, 2)only b,c are true, 3)only a,c are true, 4)only a,b,c true, 124. Unstable fission fragments decay by emitting, neutrons and electrons, neutrons so emitted, are called, 1) prompt nuetrons, 2) delayed neutrons, 3) stray neutrons, 4) sustained neutrons, 125. Chain reaction can be initiated by, 1) prompt neutrons, 2) delayed neutrons, 3) slowed prompt neutrons 4) 2 or 3, 136, , JEE-ADV PHYSICS- VOL- V, 126. The man-made element which was made in the, nuclear reactor is, 1) polonium 2) plutonium 3) thorium 4) uranium, 127. In a fast breeder reactor, the main charm is, that the nuclear ash is that it is, 1) more fissile than parent fuel, 2) not dangerous as a potential pollutant, 3) easily disposed off, 4) stable in terms of further decay, 128. If “x” gm of a nuclear fuel of mass number, “A” undergoes fission inside a reactor then, the number of fissions will be (N-Avagadro, number), 1) NA/x, 2) NAx 3) Nx/A, 4) Ax/N, 129. The reactor which produces power due to, fission by fast neutron and at the same time, regenerates more fissionable material than it, consumes is, 1) Thermal reactor, 2) Breeder reactor, 3) Both the above, 4) Neither 1 & 2, 130. The reactor in which the number of fissionable, nuclides produced are more than the used is, called, 1) breeder reactor, 2) Pressurised reactor, 3) Heterogeneous reactor, 4) Homogenerous reactor, 131. A good moderator should, 1) be a gas, 2) have appetite for neutrons, 3) be lighter in mass number, 4) heavier in mass number, 132. Who designed the atomic reactor?, 1) Wilson 2) Fermi 3) Rutherford 4) Teller, 133. From the following that are conserved in, nuclear reactions are, 1) mass number and energy, 2) mass number and charge number, 3) charge number and mass, 4) mass number, charge number and energy, 134. (A) Fission is a thermonuclear process, (B) Fusion is a thermonuclear process, (C) Fusion is exothermic, (D) Fission is exothermic, Choose the correct answer, 1) A and B are correct, 2) B and C are correct, 3) A and C are correct, 4) B,C and D are correct, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 135. Consider the following statements A and B., Identify the correct choice in the given answer., (A) p-p,p-n,n-n forces between nucleons are, not equal and charge dependent, (B) In nuclear reactor the fission reaction will, be in accelerating state if the value of neutron, reproduction factor k >1, 1) Both A and B are correct, 2) Both A and B are wrong, 3) A is wrong and B is correct, 4) A is correct and B is wrong., 136. A chain reaction in fission of Uranium is, possible because, 1) Large amount of energy is released, 2) Two intermediate size nuclear fragments are, formed, 3) More than one neutron is given out in each fission, 4) Fragments in fission are radioactive, 137. A slow neutron can cause fission in, 1) U238, 2) U235, 3) Pb206, 4) Sr90, 138. Heavy water is used as moderator in a nuclear, reactor. The function of the moderator is, 1) To slow down the neutrons to thermal energies, 2) To control the energy released in the reactor, 3) To cool the reactor faster, 4) To absorb neutrons and stop chain reaction, 139. To control fission process of the reactor , the, following material is used_______, 1) Graphite 2) Cadmium 3) Gold 4) Uranium, 140. Nuclear fission is caused by, 1) fast protons, 2) fast neutrons, 3) Slow protons, 4) slow neutrons, 141. The liquid drop model of nucleus was proposed, by, 1) Bohr,Wheeler, 2) Fermi, 3) Rutherford, 4) Chadwick, 142. In nuclear reactor, Cadmium rods are used as, 1) Fuel, 2) Moderator, 3) Control Rods, 4) None, 143. The material used to slow neutrons in a, reactor is called, 1) Controlrod, 2) Moderator, 3) Fuel, 4) Heat exchanger, , NUCLEAR FUSION, 144. Atomic mass of the most useful material for, fusion reaction is, 1) 1, 2) 4, 3) 235, 4) 292., NARAYANAGROUP, , NUCLEAR PHYSICS, 145. Average K.E of thermal neutron is of the order, of (in KeV), 1) 3.0, 2) 0.03, 3) 0.3, 4) 0.003, 146. Inside the sun, 1) Four nuclei of hydrogen combine to form two, nuclei of helium, 2) Four nuclei of hydrogen combine to form four, nuclei of helium, 3) Four nuclei of hydrogen combine to form one, nucleus of helium, 4) Four nuclei of hydrogen is transformed into one, nucleus of helium, 147. As the age of star increases, 1) Helium quantity increases, 2) Helium quantity decreases, 3) Helium quantity does not change, 4) Helium, Hydrogen both quantities increases, 148. In the carbon cycle of nuclear fusion carbon acts, like a, 1) Moderator, 2) Activator, 3) Catalyst, 4) Controller, 149. In a fusion process a proton and neutron, combine to give a deuterium nucleus.If mo and, mp be the mass of neutron and proton, respectively the mass of deuterium nucleus is, 1) equal to mo+mp, 2) more than mo+mp, 3) less than mo+mp, 4) can be less than or more than(mo+mp), 150. The binding energies of the atoms of elements, P and Q are EP and EQ respectively. Three, atoms of elements Q fuse to form one atom of, element P. In this process the energy released, is e. The correct relation between EP, EQ and, e will be, 1) EQ=3EP+e, 2) EQ=3EP-e, 3) EP=3EQ+e, 4) EP=3EQ-e, B.E., for deutron and an - particle are, A, X1 and X2 respectively. The energy released, in the fusion of deuterium into, -particle is, , 151. The, , 1) 4 (X2-X1) 2) 2 ( X2-X1) 3) 4 (X2+X1) 4), , X 2 X1, 4, , 152. If Q1 and Q2 are the energies released in the, fusion of hydrogen in Carbon - nitrogen cycle, and proton - proton cycle respectively then, 1) Q1 > Q2 2) Q1 = Q2 3) Q1 < Q2 4) Q1 > Q2, 137
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NUCLEAR PHYSICS, 153. Fusion reaction is initiated with the help of, 1) low temperature, 2) high temperature, 3) neutrons, 4) any particle, 154. In an exo-ergic reaction the binding energies, of reactants and products are E1, E2, respectively then, 1) E1 <E 2 2) E1 =E 2 3) E1 >E 2 4) E1 E2, 155. In an endo-ergic reaction the binding energies, of reactants and products are E1, E2, respectively, 1) E1 <E 2 2) E1 =E 2 3) E1 >E 2 4) E1 E 2, 156. Among the following reactions which is, impossible, 1) 2He4+4Be9= 0n1+6C12 2) 2He4+7N14= 1H1+8O17, 3) 4(1H1) =2He4+2 (-1e0) 4) 3Li7+1H1 = 4Be8, 157. If the nuclei of masses X and Y are fused, together to form a nucleus of mass m and, some energy is released, then, 1) X+Y=m 2) X+Y<m 3) X+Y> m 4) X-Y=m, 158. Fusion reactions take place at about, 1) 3 102 K, 2) 3 103 K, 3) 3 104 K, 4) 3 106 K, 159. The percentage of mass lost during nuclear, fusion is, 1) 0.1%, 2) 0.4%, 3) 0.5%, 4) 0.65%, 160. Nuclear fusion requires high temperature, because, 1) All nuclear reactions absorb heat, 2) The particles can not come together unless they, are moving rapidly, 3) The binding energy must be supplied from an, external source, 4) The mass defect must be supplied, 161. Among the following true option is, 1) Energy released per nucleon is same in both, fission and fusion reactions, 2) Energy released per nucleon is more in fission, than in fusion reaction, 3) Energy released per nucleon is less in fission, than in fusion reaction, 4) No energy is released in fusion reaction, 162. Fusion reaction take place at high temperature, because, 1) atoms are ionized at high temperature, 2) molecules break up at high temperatures, 3) nuclei break up at high temperature, 4) kinetic energy is high enough to overcome, repulsion between nuclei, 138, , JEE-ADV PHYSICS- VOL- V, 163. In the carbon cycle from which stars hotter, than the sun obtain their energy the 6 C12, isotope, 1) splits up into three alpha particles, 2) fuses with another 6 C12 nucleus to form12 Mg 24, 3) is completely converted into energy, 4) is regenerated at the end of the cycle, 164. Source of solar energy can be said to be due, to natural fusion in which hydrogen gets, converted into helium with carbon serving as, a natural catalyst. This carbon cycle was, proposed by, 1) Bethe 2) Yukawa 3) Fermi 4) Soddy, 165. In Carbon-Nitrogen fusion cycle , protons are, fused to form a helium nucleus, positrons and, release some energy.The number of protons, fused and the number of positrons released, in this process respectively are, 1) 4,4, 2) 4,2, 3) 2,4, 4) 4,6, 166. Nuclear fission and fusion can be explained, on the basis of ____, 1) Einstein theory of relativity, 2) Einstein specific heat equation, 3) Einstein mass-energy relation, 4) Einstein photo electric equation, 167. Energy in the sun is due to, 1) Fossil fuels, 2) Radioactivity, 3) Fission, 4) Fusion, 168. The overall process of carbon nitrogen fusion, cycle results in the fusion of 4 protons to yield, helium nucleus and --1) positron, 2) two electrons, 3) two positrons, 4) An electron., 169. The nucleus finally formed in fusion of protons, in proton-proton cycle is that of, 1) Heavy hydrogen, 2) Carbon, 3) Helium, 4) Lithium, 170. 41H1 2 He4 2e 26MeV . The above, reaction represents ......., 1) Fusion, 2) Fission, 3) b-decay, 4) g-decay, 171. The source of stellar energy is __process, 1) Nuclear fission, 2) Nuclear fusion, 3) Nuclear fission&fusion 4) Nuclear decay, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 172. Fusion reaction takes place at very high, temperature because., 1) atoms are ionised at high temperatures., 2) molecules breakup at high temperature., 3) nuclei break up at high temperature., 4) kinetic energy is high enough to overcome, repulsion between nuclei, 173. In the carbon cycle, from which stars hotter, than the sun obtain their energy the 6 C12, isotope, 1) splits into three alpha particles, 2) fuse with another 6C12 nucleus to form 12 Mg 24, 3) is completely converted into energy., 4) is regenerated at the end of the cycle., , PAIR PRODUCTION & PAIR, ANNIHILATION, 174. The phenomenon of pair production is, 1) The production of an electron and a positron, from radiations, 2) Ejection of an electron from a metal surface, when exposed to ultraviolet light, 3) Ejection of an electron from a nucleus, 4) Ionization of a neutral atom, 175. When the particle and its antiparticle unite,, the result is, 1) a heavier particle, 2) two or more smaller particles 3) photons, 4) partly matter and partly photons., 176. Particles and their antiparticles have, 1) The same masses but opposite spins, 2) The same masses but opposite magnetic moment, 3)The same masses and same magnetic moment, 4) Opposite spins and same magnetic moment, 177. To produce pair production, the minimum, energy of -ray should be (in MeV), 1) 0.15, 2) 1, 3) 1.02, 4) 1.5, 178. The rest mass energy of electron or positron, is (in MeV), 1) 0.51, 2) 1, 3) 1.02, 4) 1.5, 179. A positron and an electron come close, together to give a neutral one called, 1) Electronium, 2) Positronium, 3) -photon, 4) -particle, 180. Positronium is converted into, 1) 2 Photons each of energy 0.51MeV, 2) 1 Photon of energy 1.02 MeV, 3) 2 Photons each of energy 1.02MeV, 4) One Photon of energy 0.51MeV, NARAYANAGROUP, , NUCLEAR PHYSICS, 181. In pair annihilation two -ray photons are, produced it is due to, 1) Law of conservation of energy, 2) Law of conservation of mass, 3) Law of conservation of momentum, 4) Law of conservation of angular momentum, 182. In pair annihilation the least number of - ray, photons produced is, 1)2, 2)3, 3)4, 4)1, 183. The number of protons, electrons and neutrons, in the nucleus of 13Al27 is, 1) 13, 13, 14, 2) 13, 0, 14, 3) 14,14, 13, 4) 14, 0, 13, 39, 184. 19, K and 40, 20 Ca are, 1) Isotopes 2) Isobars 3) Isotones 4) Isodiaphers, 185. K40,Ar40,Ca40 are, 1) Isobars 2) Isotopes 3) Isotones 4) Isogonals, 186. Of the following atoms, 14, 13, 236, , 7N14 , 8 O16 and 86 Rn232 a pair of, 6 C , 7N , 88Ra, isobars is :, 1) 6 C11, 7N13, 2) 7 N13 , 7N14, , 3) 6 C14 , 7N14, 4) 6 C14 , 8 O16, 187. Of the following a pair of isotones is, 1) 6 C11, 7N13, 2) 7 N13 , 7N14, 3) 6 C14 , 7N14, 4) 6 C14 , 8 O16, 188. Of the following a pair of isotopes is, 2) 7 N13 , 7N14, 1) 6 C11, 7N13, 3) 6 C14 , 7N14, 4) 6 C14 , 8 O16, 189. Of the following a pair of isodiaphers is, 1) 88 Ra236 , 86Ra232, 2) 7 N13 , 7N14, 3), , 6, , C14 ,7 N14, , 4), , 6, , C14 , 8 O16, , ASSERTION & REASONS, These Questions consist of two statements, each printed as Assretion and Reason. While, answering these questions you are required, to choose any one of the following four, responses, 1) A and R are true and R is the correct, explanation of A., 2) A and R are true and R is not the correct, explanation of A., 3) A is true, R is false., 4) A is false, R is true., 139
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 190. (A): Free Neutron decays into proton, electron, and antinuetrino, (R): Neutron is unstable outide the nucleus, 191. (A): Nuclear forces arise from strong, coulombic interactions between protons and, neutrons, (R): Nulear forces are independent of charge, of the nucleons, , MATCHING TYPE, 192. Match list I with list II, List I, a) Red dwarfs, , List II, e) Carbon nitrogen, cycle, b) stars having higher temperatures f) Isotope of carbon, c) Blood circulation problems g)proton - proton, cycle, d) Radio carbon dating, h)radio sodium, 1) a- e, b - g, c - h , d - f 2) a - g, b - e, c - f, d- h, 3) a - g, b - e, c - h, d - f 4) a - f, b - e, c - g , d - h, 193. Matching, List – II, List – I, A) K < 1, e) Critical state, B) K = 1, f) Sub critical state, C) K > 1, g) Super critical, 1) A f , B e , C g, 2) A e , B f , C g, 3) A g , B e , C f, 4) A e, B g C f, 194. Matching., List - I, List - II, A) Radio Iodine, e) Lukamia, B) Radio sodium, f) age of ancient, objects, C) Radio phosphorous, g) Restriction in, blood circulation, D) Radio carbon, h) Functioning of, thyroid gland, 1) A h , B g , C e , D f, 2) A g , B h , C f , D e, 3) A h , B e , C g , D f, 4) A h , B g , C f , D e, 195. Match the following., List – I, List – II, A) Moderator, e) Absorbs heat, B) Control rods, f) Prevent neutrons, exposed outside, C) Radiation shielding g) Absorb neutrons, D) Coolant, h) Slow down neutrons, 1) A – h, B – g, C – f, D – e, 2) A – g, B – h, C – f, D – e, 3) A – h, B – g, C – e, D – f, 4) A – h, B – f, C – e, D – g, 140, , 196. This question contains Statement -1 and Statement 2. Of the four choice given after the statements,, choose the one that best describes the two, statements., (2008-AIEEE), Statement - 1:, Energy is released when heavy nuclei undergo, fission or light nuclei undergo fusion. and, Statement - II, For heavy nuclei,binding energy per nucleon, increases with increasing Z while for light nuclei, it decrease with increasing Z., 1) Statement - 1 is false,Statement -2 is true, 2) Statement - 1is true, Statement -2 is true ;, Statement -2 is correct explanation for Statement -1., 3) Statement - 1 is true, Statement -2 is true ;, Statement -2 is not a correct explanation for, statement -1, 4) Statement -1 is true, Statement -2 is False., , C.U.Q - KEY, 1) 2, 8) 2, 15) 2, 22) 4, 29) 4, 36) 4, 43) 3, 50) 4, 57) 4, 64) 4, 71) 4, 78) 3, 85) 2, 92) 2, 99) 4, 106) 3, 113) 3, 120) 2, 127) 1, 134) 4, 141) 1, 148) 3, 155) 3, 162) 4, 169) 3, 176) 2, 183) 2, 190) 1, , 2) 3, 9) 2, 16) 2, 23) 2, 30) 2, 37) 3, 44) 2, 51) 4, 58) 3, 65) 2, 72) 1, 79) 3, 86) 1, 93) 3, 100) 2, 107) 3, 114) 3, 121) 1, 128) 3, 135) 3, 142) 3, 149) 3, 156) 3, 163) 4, 170) 1, 177) 3, 184) 3, 191) 4, , 3) 3, 10) 2, 17) 1, 24) 2, 31) 1, 38) 2, 45) 2, 52) 1, 59) 4, 66) 1, 73) 4, 80) 2, 87) 3, 94) 2, 101) 2, 108) 2, 115) 2, 122) 2, 129) 2, 136) 3, 143) 2, 150) 3, 157) 3, 164) 1, 171) 2, 178) 1, 185) 1, 192) 3, , 4) 3, 11) 4, 18) 4, 25) 4, 32) 1, 39) 3, 46) 3, 53) 4, 60) 2, 67) 4, 74) 2, 81) 1, 88) 1, 95) 3, 102) 4, 109) 3, 116) 3, 123) 4, 130) 1, 137) 2, 144) 1, 151) 1, 158) 4, 165) 2, 172) 4, 179) 2, 186) 3, 193) 1, , 5) 1, 12) 4, 19) 2, 26) 4, 33) 2, 40) 2, 47) 3, 54) 3, 61) 3, 68) 1, 75) 2, 82) 1, 89) 3, 96) 4, 103) 4, 110) 1, 117) 4, 124) 2, 131) 3, 138) 1, 145) 2, 152) 2, 159) 4, 166) 3, 173) 4, 180) 1, 187) 4, 194) 1, , 6) 1, 13) 3, 20) 2, 27) 1, 34) 3, 41) 4, 48) 1, 55) 1, 62) 3, 69) 4, 76) 3, 83) 1, 90) 2, 97) 2, 104) 2, 111) 1, 118) 3, 125) 4, 132) 2, 139) 2, 146) 3, 153) 2, 160) 2, 167) 4, 174) 1, 181) 3, 188) 2, 195) 1, , 7) 3, 14) 4, 21) 1, 28) 4, 35) 3, 42) 3, 49) 3, 56) 1, 63) 1, 70) 3, 77) 4, 84) 1, 91) 4, 98) 3, 105)1, 112) 3, 119) 1, 126)2, 133)4, 140)4, 147)1, 154)1, 161)3, 168)3, 175)3, 182)1, 189)1, 196)4, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 8., , LEVEL-I (C.W), SIZE OF THE NUCLEUS, 1., , The density of a nucleus in which mass of each, nucleon is 1.67 10 27 kg and R0 1.4 10 15 m is, 1) 1.453 1017 kg / m3, , 2) 1.453 1016 kg / m 3, , 3) 1.453 10 21 kg / m 3 4) 1.453 1010 kg / m 3, 2., , r1 and r2 are the radii of atomic nuclei of mass, numbers 64 and 27 respectively. The ratio, , r1 / r2 is, 3., , 1) 64 / 27 2) 27 / 64 3) 4 / 3, 4) 1, The mass number of a nucleus is 216. The, size of an atom without changing its chemical, properties are called, 1) 7.2 10 13 cm, 2) 7.2 1011 cm, 3) 7.2 10 10 cm, , 4) 3.6 1011 cm, , MASS DEFECT, BINDING ENERGY,, PACKING FRACTION AND MASS, ENERGY RELATION, 4., , 5., , 6., , Energy released as mass of 2 amu is, converted into energy is, 1) 1.5 x 10-10 J, 2) 3 x 10-10 J, 3) 1863 J, 4) 931.5 Mev, A 1 MeV positron encounters a 1 MeV, electron travelling in opposite direction. The, total energy released is (in MeV), 1) 2, 2) 3.02, 3) 1.02, 4) 2.04, The binding energies of the nuclei A and B are, Ea and Eb respectively. Three nuclei of the, element B fuse to give one nucleus of element, ‘A’ and an energy ‘Q’ is released. Then, Ea , Eb , Q are related, 1) Ea 3Eb Q, , 7., , 2) 3Eb Ea Q, , 3) Ea 3Eb Q, 4) Eb 3Ea Q, The binding energies per nucleon for, deuterium and helium are 1.1 MeV and 7.0, MeV respectively. The energy in joules will, be liberated when 106 deuterons take part in, the reaction, 1) 18.88103 J, 2) 18.88105 J, 3) 18.88 107 J, NARAYANAGROUP, , 4) 18.881010 J, , 9., , 1 Kg of iron (specific heat 120 Cal kg-1C-1) is, heated by 10000C. The increase in its mass, is, 1) Zero, 2) 5.6 x 10-8 Kg, -16, 3) 5.6 x 10 Kg, 4) 5.6 x 10-12 Kg, In nuclear fission, 0.1% mass is converted into, energy. The energy released in the fission of 1Kg, mass is, 1) 2.5 x 105 KWH, 2) 2.5 x 107 KWH, 9, 3) 2.5 x 10 KWH, 4) 2.5 x 10-7 KWH, , Y, -DECAY,, DECAY,, DECAY, 10. After the emission of one -particle, followed by two -particles from 238, 92 U , the, number of neutrons in the newly formed, nucleus is, 1) 140, 2) 142, 3) 144, 4) 146, 11. A radioactive nucleus undergoes a series of, decays according to the sequence, , , , A , A1 , A2 , A3 . If the mass, , number and atomic number of A3 are 172 and, 69 respectively, then the mass number and, atomic number of A is, 1) 56, 23 2) 180, 72 3) 120, 52 4) 84, 38, 12. The number of and particles emitted, in the conversion of 90Th232 to 82Pb208 are, 1) 6, 4, 2) 4, 6, 3) 8, 6, 4) 6, 8, , RADIOACTIVITY, HALF LIFE, MEAN, LIFE, DECAY CONSTANT, 13. The decay constant of a radio active, element, which disintegrates to 10 gms from, 20 gms in 10 minutes is, 1) 0.693 min 1, 2) 6.93 min 1, 3) 0.693 sec 1, 4) 0.0693 min 1, 14. Half life period of radium is 1600 years., 2 gm of radium undergoes decay and gets, reduced to 0.125 gms in, 1) 3200 years, 2) 25600 years, 3) 8000 years, 4) 6400 years, 15. After a certain lapse of time, the fraction of, radioactive polonium undecayed is found to be, 12.5% of initial quantity. The duration of this, time lapse is (if the half-life of polonium is 138, days), 1) 414days 2) 407 days 3) 421 days 4) 410 days, 141
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 16. Two radioactive substances X and Y initially, contain an equal number of atoms. Their halflives are 1 hour and 2 hours respectively. Then, the ratio of their rates of disintegration after, two hours is, 1) 1:1, 2) 2:1, 3) 1:2, 4) 2:3, 17. 1 g of a radio active substance disintegrates, at the rate of 3.7 1010 disintegrations per, second. The atomic mass of the substance, is226. Calculate its mean life., 1) 1.2 10 5 s, 2) 1.39 1011 s, 3) 2.1 10 5 s, , 4) 7.194 1010 S, , NUCLEAR FISSION-NUCLEAR REACTOR, 18. No. of uranium 235 nuclei required to undergo, fission to give 9 x 1013 joule of energy is, 1) 2.8125 x 1024, 2) 28.125 x 1024, 3) 281.25 x 1024, 4) 28215 x 1024, 19. The energy supplied by a power plant is 40, million kilowatt hour. It is supplied by, annihilation of matter, the mass that is, annihilated is., 1) 1.6 gm 2) 1.6 kg 3) 1.6 mg 4)1.6 amu., , 1/3, , 2., , 4., 5., 6., 7., , B.E., B.E., , E 4, of He , of Deuterium , A, A, , , 8., , mc 2 J MS t , , 9., 10., , 238 , , 92 X, , d, , Am, 3 Am, 3 Am, 3m, , , , 3, 1 3, 4, 4, , R, A, 4, , R03, 3, , , 0, R, 3, 4, , R, A, 0 , 3, , , , 234 He 4 2 e0, 2, 1, , , , , 11. A , A1 , A2 , A3, e0, , 72, , 1, 73 X, X 180 , , 180, , 71, , 2 He, X 176 , 69 X 172, , 4, , 2 He, , 71 X 176, , 4, , 12., 13., , 90, , Th 232 82 Pb 208 62 He 4 41 e0, , 0.693 / T1, 2, t, , dN, N , here N avagadrono /238,, dt, , 0.693 / T1, 2, , 16., , A N, , N, 1, 17. N 2n, 0, , 18. 200MeV 1 fission, 9 1013 J ?, , 9 1013, Number of fissions =, 200 106 1.6 1019, 19. E 40 106 36 105 J ; E mc 2, 40 106 36 105 m 9 1016, , LEVEL - I (C.W) - HINTS, , 142, , 92 U, , E, 0.1 2, mc ; in kwh , 100, 36 105, , no of neutrons=234-92=142, , Binding energy per nucleon of 2 He4 =7 Mev], 1) 8.1 MeV 2) 5.9 MeV 3) 23.6 MeV 4) 2 MeV, , 1., , E, , 15., , 1) 1 2) 3 3) 1 4) 2 5) 2 6) 1 7) 3, 8) 4 9) 2 10) 2 11) 2 12) 1 13) 4 14) 4, 15) 1 16) 3 17) 4 18) 1 19) 1 20) 3 21) 1, , 7.2 1013 cm, , 3Eb Ea Q, , [Binding energy per nucleon of 1H2 = 1.1 Mev, , LEVEL - I (C.W) - KEY, , 4, 3, , E MC 2, Total energy=rest mass energy+KE, , 14. N N 0 e, , 21. The minimum amount of energy released in, annhilation of electron-positron is, 1) 1.02MeV 2) 0.58MeV 3)185MeV 4) 200MeV, , , , 1/3, , of two 1H2 to form a 2 He4 nucleus will be, , PAIR PRODUCTION & PAIR, ANNIHILATION, , 1/3, , 64 , , 27 , , 3. R R0 A1/3 1.2 10 13 216 , , NUCLEAR FUSION, 20. The amount of energy released in the fusion, , r1 A1 , As , r2 A2 , , 20., , 1, , H 2 1 H 2 2 He4 Q, , E = B.E. of products - B.E. of reactants, = 4(7) - 4(1.1) = 23.6 MeV, 21. 0.51 MeV + 0.51 MeV = 1.02 MeV, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 7., , LEVEL-I (H.W.), , X 200 A110 B90 energy The binding, , SIZE OF THE NUCLEUS, 1., , 2., , Consider the nuclear reaction:, , Assume that the nuclear mass is of the order, of 10-27kg and the nuclear radius is of the order, of 10-15m.The nuclear density is of the order, of, 1) 102Kg/m3, 2) 1010kg/m3, 3) 1017Kg/m3, 4) 1031 kg/m3, Highly energetic electrons are bombarded on, a target of an element containing 30 neutrons., The ratio of radii of nucleus, 1) 25, 2) 26, 3) 56, 4) 30, , energy per nucleon for X 200 , A110 and B90 is, respectively 7.4MeV, 8.2MeV and 8.2Mev., The energy released is, 1) 200MeV, 2) 160MeV, 3) 110MeV, 4) 90MeV, , Y, -DECAY,, DECAY,, DECAY, 8., , The isotope, , 238, 92, , U decays successively to form, , 234, 90, , 234, 234, 230, 226, Th, 91, Pa, 92, U , 90, Th and 88, Ra , The, radiations emitted in these five steps are, 1) , , , , , 2) , , , , , , MASS DEFECT, BINDING ENEGRY,, PACKING FRACTION AND MASS, ENERGY RELATION, , 9., , 3., , Sun radiates energy at the rate of 3.6x1026J/, s. The rate of decrease in mass of sun is (Kgs, 1, )., 1) 12 x 1010, 2) 1.3 x 1020, 3) 4 x 109, 4) 3.6 x 1036, , a - particle to form 147 N contains, 1) 8 neutrons, 2) 10 neutrons, 3) 7 neutrons, 4) 6 neutrons., 10. A nucleus X initially at rest, undergoes, alpha decay according to the equation, , 4., , A slow neutron strikes a nucleus of, splitting it into lighter nuclei of, , 141, 56, , 235, 92, , U, , Ba and, , 92, 36, , 5., , Kr along with three neutrons. The energy, released in this reaction is (The masses of, uranium, barium and krypton of this reaction, are 235.043933, 140.917700 and 91.895400 u, respectively. The mass of a neutron is, 1.008665u), 1) 740.69 MeV, 2) 156.9 MeV, 3) 186.9 MeV, 4) 209.8 MeV, The energy required to separate the typical, , middle mass nucleus 120, 50 Sn into its constituent, , 6., , nucleons ( Mass of 120, 50 Sn 119.902199u ;, mass of proton=1.007825 u and mass of, neutron = 1.008665 u), 1) 951 MeV, 2) 805 MeV, 3) 1021MeV, 4)1212 MeV, The mass defect in a nucleus is 3.5 amu. Then, the binding energy of the nucleus is, 1) 32.58MeV, 2) 325.85 MeV, 3) 3260.25MeV, 4) 3.258 MeV, NARAYANAGROUP, , 3) , , , , , 4) , , , , , The nuclide which disintegrates by emitting, , 232, A, z X 90, , Y What fraction of the total, energy released in the decay will be the, kinetic energy of the alpha particle, , 1), , 90, 92, , 2), , 228, 232, , 3), , 228, 232, , 4), , 1, 2, , RADIOACTIVITY, HALF LIFE, MEAN, LIFE, DECAY CONSTANT, 11. A radio active sample contains 600 radio, active atoms. Its half life period is 30, minutes. The no. of radio active atoms, remaining, if the decay occurs for 90, minutes is, 1) 300, 2) 200, 3) 400, 4) 75, 12. Radio active carbon - 14, in a wood sample, decays with a half life of 5700 years. The, fraction of the radio active carbon - 14, that, remains after a decay period of 17,100, years is, 1) 1/4, 2) 3/4, 3) 1/8, 4) 7/8, 143
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 13. The half-life of 238 U for decay is 4.5 109, years. The number of disintegrations per, second occur in 1 g of 238 U is, (Avogadro’s number= 6.023 1023 mol 1 ), 1) 1.532 10 4 s 1, , 5. E m 931.5MeV, 6. Binding energy = m 931.5MeV, 7. Energy released=BE of products-BE of reactants, 9., , 2) 1.325 104 s 1, , 3) 1.412 10 4 s 1, 4) 1.235 104 s 1, 14. A certain substance decays to 1/32 of its initial, activity in 25 days. The half-life is, 1) 1 day 2) 3 days 3) 5 days 4) 7 days, , 10., , 7, , N 14 6 X 14 1 e 0, , K.E A 4, , Where A 232 , K.E, A, n, , N 1, t, ,n , 11. N 0 2 , T1, 2, , NUCLEAR FISSION-NUCLEAR REACTOR, 15. The energy released by the fission of 1 g of, 235, U in joule, given that the energy released, per fission is 200 MeV. (Avogadro’s, number 6.023 10 23 ), 1) 8.202 10, 3) 8.202 1010, , 2) 8.202 10, 4) 8.202 1014, , 12, , 8, , 12., , t N 1, n ;, , T1 N 0 2 , , n, , 2, , 13. N , , N0, 2n, , dN, N here N avagadrono / 226, 1 / , dt, 15. E (avagadrono./235)X-200x106x1.602x10-19J, 16. Energy relesed for nucleon in fussion = 7.28 (energy, , 14., , NUCLEAR FUSION, 16. The ratio of the amounts of energy released, as a result of the fussion of 1 kg hydrogen E1 , , released per nucleon in fission) E1 7.28 E2, , and fission of 1 kg of 92 U, E2 will be, 1) 1.28, 2) 3.28, 3) 5.28, 4) 7.28, 236, , LEVEL-II (C.W), SIZE OF THE NUCLEUS, , LEVEL - I (H.W.) - KEY, , 1., , 1) 3 2) 2 3) 3 4) 4 5) 3 6) 3 7) 2, 8) 3 9) 1 10) 2 11) 4 12) 3 13) 4 14) 3, 15) 3 16) 4, 2., , LEVEL - I (H.W.) - HINTS, 1., , d, , 3M n, 4 R 3, , R radius, 1/3, , 2. As R R0 A1/3, 14 , , A, , 4, , 3., , A 56 (or) Z A N, , Z 56 30 26, 3., , dE dm 2, , c, dt, dt, , 4., , BE mc 2 m 931.5MeV, , 144, , MASS DEFECT, BINDING ENEGRY,, PACKING FRACTION AND MASS, ENERGY RELATION, , 1/3, , R A , R A, 1 1 , , R2 A2 , RHe 4 , 1/3, , 1/3, , A nucleus x 235 splits into two nuclei having, the mass numbers in the ratio 2:1. The ratio, of the radii of those two nuclei is, 1) 2:1, 2) 1:2, 3) 21/3:1, 4) 1:21/3, A match box of 5 cm x 5cm x 1 cm dimensions, is filled with nuclear matter. Its weight is in the, order of, 1) 10g, 2) 108 g, 3) 1012g, 4) 1015g, , 4., , If the speed of light were 2/3 of its present, value, the energy released in a given atomic, explosion will be decreased by a fraction of, 1) 2/3, 2) 4/9, 3)4/3, 4) 5/9, The binding energy per nucleon of C12 is 7.68, MeV and that of C13 is 7.47 Mev. The energy, required to remove one neutron from C13 is, 1) 495 MeV, 2) 49.5 MeV, 3) 4.95 MeV, 4) 0.495 MeV, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 5., , NUCLEAR PHYSICS, , The binding energy per each nucleon in the, neighbourhood of medium nuclei is 8.5 MeV, and the binding energy per each nucleon is, about 7.6 MeV in the neighbourhood of, Uranium. The energy released in the fission, of U 236 is, 1) 212 eV, 2) 212 MeV, 3) 2.12 MeV, 4) 0.9 MeV, , 12. A nuclear reactor generated power at 50%, efficiency by fission of, fragment of, , RADIOACTIVITY, HALF LIFE, MEAN, LIFE, DECAY CONSTANT, 7., , 8., , 9., , The mass of one curie of U 234 is, 1) 3.7 1010 g, 2) 3.7 10 10 g, 4) 1.438 10 11 g, 3) 6.25 10 34 g, A radio active isotope having a half life of 3, days was received after 9 days. It was found, that there was only 4 gms of the isotope in the, container. The initial weight of the isotope, when packed was, 1) 8 g, 2) 64 g, 3) 48 g, 4)32 g, Half life of a radio active element is 5 min., 10 sec. Time taken for 90% of it to, disintegrate is nearly, 1)100 min 2) 1000sec 3) 104 sec 4)104 min, , 10. The half life of, , 238, 92 U, , NUCLEAR FUSION, 13. In nuclear fusion,One gram hydrogen is, converted into 0.993gm.If the efficiency of the, generator be 5%,then the energy obtained in, KWH is, 1) 8.75 103 2) 4.75 103 3) 5.75 103 4) 3.73 103, , PAIR PRODUCTION & PAIR, ANNIHILATION, 14. A photon of energy 1.12 Mev splits into, electron positron pair. The velocity of electron, is (Neglect relativistic correction), 1) 3 x 108 ms-1, 2) 1.33 x 108 ms-1, 8, -1, 3) 6 x 10 ms, 4) 9 x 108ms-1, , LEVEL - II (C.W.) - KEY, 1) 3, 8) 4, , 2) 4, 9) 2, , 3) 2, 10) 1, , 1., , M, 3. E mc 2, V, Energy required =13 7.47-12 7.68 = 4.95 MeV, Q BE product BEreac tan ts, , R A1/ 3 2. D=, , Bq, , 2), , 2.4 10 Bq, , 3) 1.82 106 Bq, , 4), , 4.02 108 Bq, , 6., , 22, 10, , 7., , N 234 gm, , 1) 1.23 10, , 5, , NUCLEAR FISSION-NUCLEAR, REACTOR, , NARAYANAGROUP, , Ne 22 He4 6 c12, , 3.7 1010 xgm, x , , 11. In a thermo nuclear reaction 103 Kg of, hydrogen is converted into 0.99 103 kg of, helium. If the efficiency of the generator is 50%,, the electrical energy generated in KWH is, 1) 105, 2) 1.5 105 3) 1.25 105 4) 1.3 105, , 4) 3 5) 2 6) 1 7) 4, 11) 3 12) 1 13) 1 14) 2, , LEVEL - II (C.W.) - HINTS, , 4., 5., , 4, , with the emission of two, , respectively. The amount of U 235 consumed, per hour to produce 1600Mw power is, 1) 128 gm 2)1.4 kg 3) 140.5 gm 4)281 gm, , undergoing decay, , is 4.5 109 years. Its activity of 1 g sample, is, , into equal, , U 235 and Pd116 is 7.2 MeV and 8.2 MeV, , 22, , Ne nucleus, after absorbing energy, decays, into two - particals and an unknown nucleus., The unknown nucleus is, 1) Carbon 2) Nitrogen 3) Boron 4) oxygen, , 235, , -rays of 5.2 MeV each and three neutrons., The average B.E. per particle of, , Y, - DECAY,, DECAY,, DECAY, 6., , 116, 46 Pd, , 92 U, , 8., , n, , 3.7 1010 234, N, , t 9, W, 3; W n0, 2, t1 3, 2, , W0 2n , W 23 4 32 gm, 145
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , RADIOACTIVITY, HALF LIFE, MEAN, LIFE, DECAY CONSTANT, , t1, , 9., , N , t 2 log 0 , log 2, N , , 10. Activity or rate of decay is R N W here, 0.693, , T1/2 and N No.of atom is 1 gm of, 1 6.025 10, 238, 92 U , 238, , 6., , 23, , 25.3 10, , 20, , atoms ; R N, , output, E mc 2, , 11., ; effieiency =, input, t, t, 12., , 5., , 7., , BE product BEreac tan t , 50P, 1600 MW ; P , 100, t, , output, 13. Efficency =, 14. E , mc 2, , 8., , 4., , 146, , A nucleus with mass number 220 initially at, rest emits an particle. If the Q value of, the reaction is 5.5MeV then the kinetic energy, of the particle is (in MeV), 1) 4.4, 2) 5.4, 3) 5.6, 4) 6.5, , 3) 1/4, , 4) 3/4, , When, , 92 U, , 235, , undergoes fission. About 0.1%, , of the original mass is conveted into energy., Then the amount of, , 92 U, , 235, , should undergo, , fission per day in a nuclear reactor so that it, provides energy of 200 mega watt electric, power is, , MASS DEFECT, BINDING ENEGRY,, PACKING FRACTION AND MASS, ENERGY RELATION, , Y, - DECAY,, DECAY,, DECAY, , 2) 1/2, , NUCLEAR FISSION-NUCLEAR, REACTOR, , A nucleus splits into two nuclear parts having, radii in the ratio 1:2. Their velocities are in, the ratio, 1) 8:1, 2) 6:1, 3) 4:1, 4) 2:1, , 2. The atomic mass of 7N15 is 15.000108 amu and, that of 8O16 is 15.994915 amu. The minimum, energy required to remove the least tightly, bound proton is (mass of proton is 1.007825, amu), 1) 0.01 3018 amu, 2) 12.13 MeV, 3) 13.018 meV, 4) 12.13 eV, 3. Assume that a neutron breaks into a proton, and electron. The energy released during this, process is (in MeV) (Mass of neutron=, 1.6725 10-27Kg=mass of proton=1.6725 1027, Kg, mass of electron=9 10-31 Kg)[AIEE-12], 1) 0.73, 2) 7.10, 3) 6.30, 4) 5.4, , undergoes radiorctive decay with a half, , 1) 1, , LEVEL-II (H.W), 1., , 221, 87 Ra, , life of 4 days. The probability that a Ra, nucleus will disintegrate in 8 days is, , 1, , mv2 2 2E0, 2, , , SIZE OF THE NUCLEUS, , If the activity of 108Ag is 3 micro curie, the, number of atoms present in it are( =0.005, sec-1 ), 1) 2.2x107 2) 2.2x106 3) 2.2x105 4) 2.2x104, The half life period of Pb210 is 22 years. If, 2g of Pb210 is taken, then after 11 years the, amount of Pb210 will be present is, 1) 0.1414g 2) 1.414g 3) 2.828g 4) 0.707g, , 1) 9.6 10 2 kg, , 2) 4.8 102 kg, , 3) 19.2102 kg, , 4) 1.2 102 kg, , PAIR PRODUCTION & PAIR, ANNIHILATION, 9., , A -ray photon creates an electron, positron, pair. The rest mass of electron is 0.5 MeV., KE of the electron - positron pair system is, 0.78 MeV. Then the energy of -ray photon, is (in MeV), 1) 1.78, 2) 0.28, 3) 1.28, 4) 0.14, , LEVEL-II(H.W) - KEY, 1) 1, 8) 3, , 2) 2, 9) 1, , 3) 1, , 4) 2, , 5) 1, , 6) 2, , 7) 4, , LEVEL-II(H.W) - HINTS, 3, , 1., , m1 R1 V2, m1v1 m2v2 ; m R V, 2, 2, 1, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 2., , NUCLEAR PHYSICS, 6., , m m p m nitrogen moxygen, , energy required = m 931.5 Mev, 3. m m p m e m n ;Energy released= m×c2, 4., , 216, A4, K.E , 5.5 5.4MeV, Q , 220, A , , 5., , dN , , , dt , , N, , 7., , n, , t, t1/2, , , , t1, , 6., , t, , N , log 0 , log 2, N , , No, 4, , Dacayed atoms N o N No , Probability of decay , , 8., , N o 3N o, , 4, 4, , No N 3, , No, 4, , E=MC 2 9. E 2 EK E0 , , LEVEL-III, 1., , 2., , 3., , 4., , 5., , 50% of a radio active substance decays in 5, hours. The time required for the 87.5%, decay is, 1)10hours 2) 15hours 3)12.5hours 4)17.5hours, 4 grams of radioactive substance A left, 1/2 gm after some time. 1 gram of another, radioactive substance B left 1/4 gm in the, same period. If half life of B is 2 hours,, the half life of A is (in hours), 1) 3/4, 2) 4/3, 3) 1/4, 4) 1/2, One mole of a emitter of half life equal, to 2days was placed in a sealed tube for 4, days at S.T.P. volume of helium collected is, 1) 22.4 lit, 2) 16.8 lit 3) 11.2 lit 4) 5.6 lit, 3 rutherfords of a radio active isotope of, half-life equal to 3 days was received after, 12 days. Initial isotope packed was, 1) 48 rutherfords, 2) 12 rutherfords, 3) 25 rutherfords, 4) 36 rutherfords, The half life of a radio active substance is, 6 hours. The amount of the substance, undergone disintegration when 36 gms of it, undergoes decay for 18 hours is, 1) 31.5 gm 2) 4.5 gm 3) 18 gm 4) 9 gm, NARAYANAGROUP, , 8., , 2, , N, 8, 2 ; N o 2n N ; o 22 4, 4, N, , Remaining atoms = N , , 7., , 9., , The radio active nuclides A and B have half, lives t and 2t respectively. If we start an, experiment with one mole of each of them,, the mole ratio after time interval of 6t will, be, 1) 1 : 2, 2) 1 : 8, 3) 1 : 6, 4) 1 : 1, 20% of a radio active element disintegrates, in 1hr. The percentage of the radio active, element disintegrated in 2hrs will be, 1) 36%, 2) 64%, 3) 60%, 4) 40%, The C14 to C12 ratio in a certain piece of, wood is 25% of that in atmosphere. The, half life period of C14 is 5,580 years. The, age of wood piece is (in years), 1) 5,580 2) 2790, 3) 1395, 4) 11,160, A radioactive sample can decay by two, different processes. The half-life for the, first process is T1 and that for the second, process is T2 . The effective half-life T of the, radioactive sample is, 1 1 1, 1) T T1 T2, 2) T T T, 1, 2, , T1 T2, T1 T2, 3) T T T, 4) T T T, 1 2, 1 2, 10. The age of the wood if only 1/16 part of, original C14 is present in its piece is (in, years) (T of C14 is 5,580 years), 1) 5580, 2) 11,160 3) 22320 4) 16740, , C14, C12, ratio to be 0.5 times of that in a living plant, The number of years back the plant died, will be (T of C14 = 5,580 years), 1) 2,790 years, 2) 5,580 years, 3) 11,160 years, 4) 27,900 years, 12. A piece of wood collected from cro-Magnon, caves gave 4 disintegrations / min. A, freshly cut wood of the same weight gives, 16 d.p.m. The cro-magnon man lived about, (Half life of C14 is 5760 years. Assume the, activity is due to C14 only), 1) 5700 years ago, 2) 2900 years ago, 3) 11520 years ago, 4) 1400 years ago, 13. The number of U 238 nuclei in a rock sample, equal to the number of Pb 206 atoms. The half, life of U 238 is 4.5 109 years.The age of the, rock is, 1) 4.5 109 y, 2) 9 109 y, 3) 13.5 109 y, 4) 18 109 y, 11. A piece of wood is found to have the, , 147
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 14. Equal masses of two samples A and B of, charcoal are burnt and the activity of resulting, carbon-di-oxide from two samples is, measured. The gas from sample A gives104, counts per month and that from sample B gives, 2.5 103 counts per month. The age difference, of two sample is ( half life of C14 is 5730 years), 1) 5730y 2) 11460y 3) 17190y 4) 22920y, 15. The half life of a radioactive substance is 20, minutes.The approximate time interval, 2, t2 t1 between the time t2 , when 3 of it has, 1, decayed and time t1 and of it had decayed, 3, is:, (AIEEE-2011), 1) 14 min 2) 20 min 3) 28 min 4) 7 min, 16. A charged capactor of capacitance C is, discharged through a resistance R. A radio, active sample decays with an average life t., Find R interms of C and t in order that the, ratio of the electrostatic energy stored in the, capacitor to the activity of the radio active, sample remains constant with time, 1), , 2t, C, , 2), , C, 2t, , 3) 2t C, , 2), , m, M, , 3), , 4) 1010 yr, 3) 7.5 109 yr, 21. The half-life of a radioactive sample is T. If, the activities of the sample at time t1 adn t2 ( t1, < t2) are R1 and R2 respec tively, then the, number of atoms disintegrated in time t2 t1 is, proportional to, 1) R1 R2 T, , 2) R1 R2 T, , R1 R2, R1 R2, 3) R R T, 4), T, 1, 2, 22. Considering a hypothetical annihilation of a, stationary electron with a stationary positron,, the wavelengthof resulting radiation? (with, usual notations), , 4) t C, , 17. Uranium-238 decays to thorium-234 with halflife 5 109 yr .The resulting nucleus is in the, excited state and hence further emits rays, to come to the ground state. It emits, 20 rays per second. the emission rate will, drop to 5 rays per second in, 1) 1.25 109 yr, 2) 1010 yr, 3) 10 8 yr, 4) 1.25 109 s, 18. A sample of radioactive material has mass m,, decay constant and molecular weight M., Avagadro constant = NA. The initial activity, of the sample is, 1) m, , 20. In moon rock sample the ratio of the number, of stable argon-40 atoms present to the, number of radioactive potassium-40 atoms is, 7:1. Assume that all the argon atoms were, produced by the decay of potassium atoms,, with a half-life of 2.5 109 yr. The age of the, rock is, 1) 2.5 109 yr, 2) 5.0 109 yr, , 1), , h, 2h, 2), 2m0C, m0C, , 3), , h, m0C, , 4), , h 2, m0C, , 23. A radioactive nucleus can decay by two, different processes. The half life for the first, process is 2t and that for the second process, is t. The effective disintegration constant of, nucleus is, 1), , 3, 3ln 2, ln 2, 3ln 2, 2), 3), 4), 2t, 3t, t, 2tln2, , 24. A proton with kinetic energy, K strikes another, proton at rest. If the collision is head - on, find, the correct graph between K and the distance, of closest approach, r., , mN A, 4) mN Ae, M, , 19. A sample of radioactive material has mass m,, decay constant and molecular weight M., Avagadro constant = NA . The activity of the, sample after time t will be, , mN A t, e, M , mN A t, 3) , e, M , , 1) , , 148, , mN A t, e, M , , 2) , 4), , m, 1 e t, , , , , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , 1), , 1, 2, , 2), , 1, 2 1, , 3), , 2 1, 2, , 4), , Binding energy per nucleon Mev, , 25. The fraction of a radiactive sample will decay, during half of its half-life period is, 1, 2, , 26. A small quantity of a solution containing N24, radio - nuclide of half - life T and activity R0 is, , T , R0 , 1, n, , , 1), ln(2) VR1 , 3), , T VR1 , 1 n, , ln(2) R0 , , T VR0 , 1 n, , 2), ln(2) , R1 , 4), , T R1 , 1 n, , ln(2) VR0 , , 27. A nucleus with mass number 220 initially at rest, emits an - particle. If the Q value of the, reaction is 5.5 MeV, calculate the kinetic, energy of -particle., 1) 4.4 MeV 2) 5.4 MeV 3) 5.6MeV 4) 6.5 MeV, 28. Some amount of radioactive substance (halflife=10 days) is spread inside a room and, consequently the level of radiation becomes, 50times the permissible level for normal, occupancy of the room. The room be safe for, occupation after, 1) 20days 2) 34.8 days3) 56.4 days4) 62.9 days, 29. In the options given below, let E denote the, rest mass energy of a nucleus and n a, neutron. The correct option is, 97, 1) E 92236 U E 137, 53 I E 39 Y 2E n , 97, 2) E 92236 U E 137, 53 I E 39 Y 2E n , , X, W, , 5.0, , 0, , 3, , injected into blood of a person. 1cm of sample, of blood taken from the blood of the person, shows activity R1. If the total volume of the, blood in the body of the person is V, find the, timer after which sample is taken., , Y, , 8.5, 8.0, 7.5, , Z, , 30, 60 90 120, Mass Number of nuclei, , 1) Y 2Z, 2) W X Z, 3) X Y Z, 4) W 2Y, 32. When 3 Li 7 nuclei are bombarded by protons,, and the resultant nuclei are 4 Be8 , the emitted, particles will be., (AIEEE 2006), 1) alpha particles, 2) beta particles, 3) gamma photons, 4) neutrons, 33. A sample of uranium is a mixture of three, isotopes 92 U 234 , 92 U 235 and 92 U 238 present in, the ratio 0.006% , 0.71% and 99.284%, respectively. The half lives of then isotopes, are 2.5 105 years, 7.1 108 years and, , 4.5 109 years respectively. The contribution, to activity (in %) of each isotope in the sample, respectively, 1) 51.41%, 2.13%, 46.46%, 2) 51.41%, 46.46%, 2.13%, 3) 2.13%,51.41%, 46.46%, 4) 46.46%, 2.13%, 51.41%, 34. The table that follows shows some, measurements of the decay rate of a sample, of 128 I , a radio nuclide often used medically, as a tracer to measure the rate at which iodine, is absorbed by the thyroid gland., , 94, 3) E 92236 U E 140, 56 Ba E 36 Kr 2E n , 94, 4) E 92236 U E 140, 56 Ba E 36 Kr 2E n , 30. Four different radioactive elements are kept, in separate containers. In the begining the, container A has 200 g-atom with half-life of 2, days , B has 20 g-atom with half-life of 20 days,, C has 2g-atom with half-life 200 days and D, has 100g-atoms with half-life of 10 days. In, the begining the maximum activity exhibited, by the container is, 1) A, 2) B, 3) C, 4) D, 31. Binding energy per nucleon versas mass, number curve for nuclei is shown in the fig., W,X,Y and Z are four nuclei indicated on the, curve. The process that would release energy, is, , NARAYANAGROUP, , ln A, 6.2, 6, 4, 2, 0, , 225, , t(min), , 50 100 150 200, , The half life t1/ 2 for this radio nuclide., 1) 25 min 2) 50 min 3) 2.5 min 4) 5 min, 149
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 35. The fraction f of radioactive element decayed, change with respect to time (t). The curve, representing the correct variation is, , LEVEL - III - HINTS, 1., , W , , W0, 2n, , n, , t, t 1 ; t1/ 2 5hrs, 2, , 1) f, , 2) f, , O, , 3), , O, , t, , t, , W, , 3., , W0 n t, ;, t1/ 2 . initially value of Helium 22.4, 2n, lits no of half lives =2, , 4., , , 4, W n, W n0 ,, t1, 3, 2, 2, , f, 4), , f, , O, , t, , 36. The rate of decay (R) of nuclei in a radiocative, sample is plotted against time (t). Which of, the following best represents the resulting, curve?, , 1) R, , O, , t, , 7., , 8., , N N 0 et, , t, 9., , 3), , R, 4), , R, , O, , t, , 10., 11., 12., , O, , t, , 37. The probability of survival of a radioactive, nucleus for one mean life is, 1), , 1, e, , 2) 1 , , 1, ln 2, ln 2, 3), 4) 1 , e, e, e, , 38. A radioactive isotope is being produced at a, constant rate A. The isotope has a half -life T, initially there are no nuclei, after a time, t >>T, the number of nuclei becomes constant., The value of this constant is, AT, A, 1) AT 2) ln 2 3) AT ln 2 4) ln 2 , T, , LEVEL - III - KEY, 1) 2, 8) 4, 15) 2, 22) 3, 29) 1, 36) 1, 150, , 2) 2, 9) 2, 16) 1, 23) 2, 30) 1, 37) 1, , 3) 2, 10) 3, 17) 2, 24) 3, 31) 4, 38) 4, , 4) 1, 11) 2, 18) 3, 25) 3, 32) 3, , 5) 1, 12) 3, 19) 2, 26) 1, 33) 1, , 6) 2, 13) 1, 20) 3, 27) 2, 34) 1, , 7) 1, 14) 2, 21) 1, 28) 3, 35) 3, , 12, , W0, ;n 3, 2n, W, W n0 . Intial amount W0 , after one hour amount, 2, 80, W0, of remaining sample =80%W0 ; W1 , 100, , 5., , 2) R, , O, , W, , t, , O, , t, , t, W0, n, , ;no, fo, half, lives, t1/2, 2n, , 2., , 13., , W, , 1 2, dN, N ; 1, dt, T, n = 4 half lifes, t = 1 half life, t = 2 half lifes, No N 1, ; N, Given N , 2, 2, o, that means t = 1 half life, , 14. t , , N0, N, log 2, , t1/ 2 log, , 2, 1, t1 / 20, 3 2, t2-t1= 20 mins., , 15. At t1, , At t2, , 1, 1, t2 / 20, 3 2, , 1, 1, 2, CVo, CV 2, Uc, constant; 2, 16., 2, A, No, N, Where V Voe t / CR , N N oe t and , , 1, t, , 17. t = 2 half lifes; t nT 2 5 109 year, , 1010 year, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 26. Total volume of blood,, , N Am, M, N Am, where No , M, , 18. Activity = N where N , 19. activity = N o e t, , V, , 20. Let the number of radioactive Potassium atoms, present initially (t=0) is N 0 and the number of stable, argon atoms at t=0 is zero. After time t the number, of stable argon atoms is m and the radioactive, potassium atoms is N 0 m given that, N0 m 1, 7, 1, , m N0 and N 0 m N 0, m, 7, 8, 8, , 2 half-lives N 0 / 4 and after 3 half-lives it reduces, to N 0 / 8, , t nT 3 2.5 109 years, Thus,, , 7.5 109 year, 21. Activity R N so that, 0.6931, N1 N2 , T, , R1 R2 T, 0.6931, , N1 N2 R1 R2 T ; t2 t1 R1 R2 T, 22., , hc hc, , m0C 2 m0C 2, , , 23. 1 2, 24. At closest approach, both protons willbe moving, with same velocity., 2, , 1 2, 1 v0 , e2, K .E mv0 2. m , 2, 2 2, 4 0 r, 1 2, e2, K, mv0 , , 4, 4 0 r 2, e2, Kr , constant ,, 2 o, n, , 25., , 1/ 2, , 2K1 (216m), , 1/ 2, , 2 K 2 (4m) , , ; K 2 54 K1, , From the above K 2 5.4 MeV, 28. Since the intial activity is 50 times the activity for, safe occupancy, therefore , R0 50 R where, R N since,, , 29. (a): Rest mass energy of U will be greater than the, reat mass energy of nuclei into which it breaks. The, consistituent nuclei and neutrons will have kinetic, energy also, as a result of conservation of linear, momentum., 30. The activity of radioactive element is given by, R N , , 0.6931N, N, Thus, R . The value of, T, T, , N, is largest for the element in the container A., T, 31. Energy of products must be more then the reactants, to release energy., 32. 3 Li 7 1 H 1 4 Be8 , 33: Let m is the total mass of the uranium mixture. The, 235, U 234 , 92U and, 0.006, 238, m, , m, in, the, mixture, are, ;, U, 1, 92, 100, 0.71, 99.284, m2 , m , and m3 , m., 100, 100, If N A is the Avogadro number, then number of, , masses of the isotopes, , K, , 1, r, , t, , T, t 1, N, 1, 1 T, Here, t or , N0 2 , 2 T 2, 2, 1, , N 1 2, 1 N0 N 1 1 2 1, , ;, N0, 2, 2, N0 2 , 2, NARAYANAGROUP, , VR , R , ln(2)t, T, ln 0 , ln 1 t , ;t, ln(2) VR1 , T, R0 , 27. K1 K 2 5.5 MeV, , RN, n, t /T, R, N 1 1, , , R0 N 0 2 2 or, t /10, 1, 1, , 2, 50, , , R1 N1 and R2 N2, , N1 N2 , , VR1, R0e t, e t, , (or), R0, R1, , From conservation of linear momentum, , since after one half-life N 0 reduces to N 0 / 2 after, , R1 R2 N1 N2 , , , , Total activity R, Activity per cm3 R1 , , 92, , atoms of threeisotopes are; N1 , , m1 N A, ,, M1, 151
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , N2 , , m2 N A, m3 N A, , and N 3 , M2, M3, , Comprehension-2:, Atomic nucleus is central core of every atom, in which the whole of positive charge and, almost entire mass of atom is concentrated. It, is a tiny sphere of radius R is given by, , Activity of radioactive sample A λN ., As λ , , 0.693, ,, t1/ 2 , , A, , 0.693, N, t1/ 2, , R Ro A1/ 3 , where Ro 1.4 1015 m , a, constant and A is the mass number of the, nucleus, , If t1 , t 2 and t3 be the half lives, then, , A1 : A2 : A3 , , A1 : A2 : A3 , , or, , , N1 N 2 N 3, :, :, t1 t2 t3, , 4., , m1, m, m, : 2 : 3, M 1t1 M 2t2 M 3t3, , 0.006, 0.71, 99.284, :, :, 5, 8, 234 2.5 10 235 7.5 10 238 4.5 10 9 , , 51.41% : 2.13% : 46.46%, 34. A A0e λt ln A ln A0 λt, λ , T1/ 2, , ln A0 ln A 6.2 0, , 0.0275min 1, t, 225, 0.693, , 25min, λ, , 5., , 6., , N N 0e λt, , e λt, N0, N0, , e, 38., , 1, λ, , , , The questions given below are based on the, following paragraph (AIEEE-2010), A nucleus of mass M + m is at rest and, decays into two daughter nuclei of equal mass, , 1, ;, e, , A N ; N , , A AT, , In2, , 7., , LEVEL-IV, Comprehension-1:, , 1., 2., 3., , 152, , The count rate meter is used to measure the, activity of a given amount of a radio active, element. At one instant, the meter shows 475, counts/minute. Exactly 5 minutes later, is, shown 270 counts/minute then, The decay constant is, 1) 0.82/ minute, 2) 0.113/ minute, 3) 0.166/ minute, 4) 0.182/ minute, Mean life of the sample is (in minutes), 1) 6.35, 2) 7.45, 3) 8.85, 4) 9.95, Half life of the sample is (in minutes), 1) 6.13, 2) 8.42, 3) 8.85, 4) 9.92, , 4) 7 1015 m, , Comprehension-3:, , So, in one mean life, required probability is, λ, , 1) Is a parabola, 2) Is a straight line passing through origin, 3) Is a straight line have an intercept, 4) Is an ellipse, On increasing the value of ‘A’ the density of, the nucleus, 1) Increases, 2) Decreases, 3) Remains constant, 4) None, The radius of the nucleus of mass number 125, is, 1) 175 1015 m, 2) 35 10 15 m, 3) 70 1015 m, , 37. Probability of survival for any nucleus at time t is, , P, , R, , and ‘log A’’, Ro , , A graph between ‘ log , , 8., , M, each. speed of light is c., 2, The binding energy per nucleon for the parent, nucleus is E1 and that for the daughter nuclei, is E2 then, 1) E2 = 2E1 2) E1 > E2 3) E2 > E1 4) E1= 2E2, The speed of daughter nuclei is, m, 2m, m, m, 2) c, 3) c, 4) c, M m, M, M, M m, Choose correct statement from the following., A)Large mass number nuclei undergo fission, B)Low mass number nuclei undergo fusion, C)For heavy nuclei the decrease in binding energy, per nucleon shows the contribution of the increasing, coulomb repulsion., 1) A, B are correct, 2) A, B, C are correct, 3) B , C are correct, 4) A , C are correct, , 1) c, 9., , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 10. Which of the following are not fundamental, particles i)Electron, ii) Photon, iii) a - Particle iv) Deutron, 1) Only i & ii are true, 2) Only ii & iii are true, 3) Only i &iii are true, 4) Only iii & iv are true, 11. When a nucleus with atomic number Z and, mass number A undergoes a radioactive decay, process, a) Both Z and A will decrease, if the process is decay, b) Z will decrease but A will not change, if the process, is + decay, c) Z will increase but A will not change, if the process, is – decay, d) Z and A will remain unchanged, if the process is , decay, 1) a & b are true, 2) b & d are true, 3) a, b & c are true, 4) a, b, c, d are true, 12. A nuclide A undergoes a decay and another, nuclide B undergoes b – decay, a) All the –particle emitted by A will have almost, the same speed, b) The –particle emitted by A may have widely, different speeds, c) All the –particle emitted by B will have almost, the same speed, d) The –particle emitted by B may have widely, different speeds, 1) a, b are true, 2) b, c are true, 3) b, d are true, 4) a, d are true, 235, 13. In the fission of U, i) Slow neutron is absorbed by U235, ii) The products in the process are not same always,, their atomic number varies from 34 to 58, iii) About 200 Mev energy is released per fission, iv) The product are always Ba and Kr, 1)Only i, ii & iii are true 2)Only ii & iii are true, 3)All are true, 4)Only i , ii & iv are true, 14. Which of the following statements are correct, i) Positron is predicted by Dirac and, discovered by Anderson, ii) Liquid drop model of nucleus is developed, by Bohr and Wheeler, iii) Carbon cycle was proposed by Bethe, iv) Fission reaction is first observed by, OttoHahn and Strassman, 1) All are true, 2) Only i, ii & iv are true, 3) Only i, iii & iv are true 4) Only iii & iv are true, NARAYANAGROUP, , NUCLEAR PHYSICS, 15. Consider the following two statement A and B, and identify the correct answer given below :, A) Nuclear density is same for all nuclei, , B) Radius of the nucleus (R) and its mass, number (A) are related as A R1 / 6, 1) A and B are true, 2) A and B are false, 3) A is true but B is false 4) A is false but B is true, 16. Consider the following statements A, B and, identify the correct choice in the given answers, A: Density of a nucleus is independent of its, mass number, B: Beryllium is used as moderator in nuclear, reactors, 1) A and B are correct 2) A and B are wrong, 3) A is correct, B is wrong4) A is wrong, B is correct, 17. Consider the following statements (A) and (B), and identify the correct answer given below., Statement (A) : Positive values of packing, fraction implies a large value of binding energy., Statement (B) : The difference between the, mass of the nucleus and the mass number of, the nucleus is called packing fraction, 1) (A) and (B) are correct, 2) (A) and (B) are false, 3) (A) is true (B) is false, 4) A is false, B is true, , LEVEL - IV - KEY, 1) 2 2) 3 3)1, 4) 2 5) 3 6) 4 7) 3, 8) 2 9) 2 10) 4 11) 4 12) 4 13) 1 14) 1, 15) 3 16) 1 17) 2, , LEVEL - IV - HINTS, 1., , A A0e t ; 270 475e t, On solving we get 0.113 / min, , 2., , , , 3., , T=0.693, , 7., , binding energy per nucleon of daughter nuclei is, more than that of the parent nuclei., According to conservation of momentum, , 8., 0, , 1, , 6., , R R0 A1/ 3, , M, M, 1 M 2 1 M 2, V1 , V 2 ; V1 V 2 ; mc 2 . V1 . V 2, 2, 2, 2 2, 2 2, , *****, 153
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , ADVANCED MAIN POINTS, , , Q-Value of Energy of a Reaction :, Consider the nuclear reaction a x y b ;, where a - bombarding particle x - Target nucleus, y - daughter nucleus b - emitted particle. Let, m1 , m 2 , m3 and m4 be the masses of a, x, y & b, respectively and k1 , k 2 , k 3 and k 4 be their KEs., From principle of conservation of energy, , (iii) In decay process, the energy Q is shared by, the anti-neutirinos and the beta particle. The Kinetic, Energy of the beta particle can be anything between, zero and a maximum value Q., Endoergic Collision / Reaction:, (i) If Q is negative, the reaction is endoergic., (ii) Some minimum energy called threshold energy is, requied to intitiate the nuclear reaction., , m, , E th Q 1 1 . Where m is the mass of the, 1, m2 , , m1c 2 k1 m 2 c 2 k 2 m 3c 2 k 3 m 4 c 2 k 4 or, , bombarding particle and m2 is the mass of the target, nucleus. The threshold energy is somewhat, greaterthan Q because the outgoing particles must, have some kinetic energy to conserve momentum., , 2, m1 m 2 m 3 m 4 c k 3 k 4 k 1 k 2 , The Q - value is, , k 3 k 4 k1 k 2 m1 m 2 m3 m 4 c 2, , , Q-Values of various decays:-, , a), , For decay ; ZA X ZA42Y 42 He, , A , 2, A4 , 4, , m, X, , m, Y, , m, He, , , , c, , , , , 2, Q, Z2 , Z , , b), , For decay;, , A, Z, , X, , A, Z 1, , Y 01 e , , For decay;, , A, Z, , X, , A, Z 1, , Y 01 e , , Q m Z X A m Z1 Y A 2m e C 2, d), , For K - capture; ZA X 01 e , , A, Z 1, , Y , , Q m Z X A m Z1 Y A c 2, , , Types of nuclear collisions:a) Exoergic reaction / collision :, i), , If Q - value is positive, rest mass energy is, converted into kinetic mass energy, radiation or, both., (ii) In -emision, kinetic Energy of the emitted, A4, particle A Q, , , Where A is the mass number of parent nucleus and, Q is the Q- value of the reaction., 154, , Radioactivity law for different types of, disintegration., , a., , Only disintegration:-, , b., , dN, N N N 0e t, dt, Disintegration with continueous production:formation rate, decay, , A , q, B stable , , dN, 1, N q N q N 0 q e t , dt, , , Q m z XA m z 1 YA c2, c), , , , Successive Disintegration, A parent nucleus may decays into a daughter nucleus,, which may decay into another daughter nucleu and, so no. Such decay is called successive, disintegration or series decay. The chain stops only, when the end product is stable., c. For successive disintegration of the products:, decay, decay, A , B , C stable , 1, 2, , For A, , , , dNA, 1 NA NA N0 e1t, dt, , For B, dN B, N0 1, 1t, 2 t, + dt 1 N A 2 N B N B e e , 2, 1, , 2 e t, 1.e t , dN C, N, , N, 1, , , 2 NB . c, , 0 , For C, dt, 2 1 2 1 , 1, , 2, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , If ‘C’ is the final stable product, then the decay, and recovery curves for the substances A,B and C, are as shown, , N, no.of atoms, , A, , B, , N0, , C, , B, A, , time, , O, , At peak of B, the rate of formation of B = rate of, disintegration of B. Hence 1 N A 2 N B . If ‘C’, is also an unstable product, then the decay and, recovery curves for the substances A,B and C are, as shown, , t, , O, , Transient Equlibrium, If the parent is long-lifed compared to daughter, but half-life of the parent not very large i.e TA TB, Then 1 2 , but 1 0., , N, no.of atoms, , hence N B 2 N 0 1, i.e Thus B is in permanent of secular equlibrium, with A (parent), , N, , N0, , N0, , For 2 be large e 2t 0, , A, , B, , NB, 1, For N , A, 2, 1, , C, , After sufficiently long time e 2t becomes negligible, conpared to e 2 t ., , time, , O, , Also, N 0 N A N B N C ; at t=0, N B 0 ;, at 0 t ; N B increases, 1 2 , at t ln ; NB max imum, 2 1 1, , at t ; N, , B, , 1, NB, So that N cons tan t, A, 2, 1, After sufficient time, the ratio of parent atom to, daughter atom become constant and both eventually, decay with same half life. This is known as transient, equilibrium., , N, , decreases to zero, , Permanent or Secular Equlibrium, If half lives TA , TB of the species A and B are such, that TA TB i.e parent nuclei has longer half life,, then their decay constants obey 1 2 ., Let us choose, TA and TB 0, , 1 0 and so 2 1 2, 1 N 0, 2t, then N B 1 e , 2, NARAYANAGROUP, , e 2t , 1 e 1t , , , , N0, 1, N0, 2 1, , O, , A, , B, , t, , Note : It is to be noted that if the parent has shorter, half-life than that of the daughter 1 2 , no state, of equilibrium is attained, , 155
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, d., , Simultaneous disintegration of parent nuclei:-, , B, 1, , 1, , Vx, , Vx, , Va, , a, , x, , x, , a, , Before, , A, , 1, , Va, , After, , By conservatyion pf momentum, , 1), , 2, , Pf Pf, , C, eff 1 2 or, , Px1 Pa1 Px Pa, , The K>E before and after collision are, 2, , 1, 1 1, TT, Teff 1 2, Teff T1 T2, T1 T2, , Where T1 and T2 are the respective half lifes, , e., , Radioactive equilibrium:- After a period of, time, successive daughter nucleus decays at the, same rate as it is formed. The situation is called, radioactive equilibrium., A N A B N B C N C =-----, , PHOTO NUCLEAR REACTION, The reaction in which a photon causes the target, mucleus to change its form are called photo nuclear, reaction. In this type of reaction, striking particles, are ray photons of high energy and they are, completely absorbed by the target nucleus., , After collision, the internal nuclear energy U of the, particles may be different. Hence by conservation, of energy., , K f U f Ki U i or K f Ki U i U f, , or Q U i U f Q K f K i or Q U, Case (i) : If U 0 and hence Q 0 , there is no, change in KE of system. The collision is said to be, elastic., eg: 2 He 4 79 Au197 79 Au197 2 He 4, Case (ii) : IF U f U i , hence Q 0, i.e K f K i, the collision is inelastic and exoergic, , X Y b Q, , eg: 1 H 1 3 Li 7 1 H 1 , , hvmin m y mb mx c 2 Q, , Case (iii) : If U f U i ,, , hv min B .E Q, , hence Q 0 , i.e K f K i ,, , eg : 1 H 3 1 H 2 0 n1 Q, , collision is inelasitc and endoergic., , SCATTERING REACTION, Basically here no reaction is involved, there is only, an exachange of energies between target and, stirking particle. Hence the nature of the target, nucleus as well as the product nucleus is the same., , 156, , Px2, Pa2, Px1, Pa1, Ki , , and K f , , 2mx 2ma, 2mx 2ma, , Note : In scattering and photo-electic reactuins,, the minimum energy required to the projectile in, endoergic reaction when target is at rest is, Emin , , ma M x, Q , where Q is B.E, Mx, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 5., , LEVEL-V, , is A1 at time t1 and A 2 at time t 2 t 2 t1 ., Its mean life is T then, A1 A 2, B) t t cons tan t, A) A1t1 A2 t 2, 2, 1, , SINGLE ANSWER TYPE QUESTIONS, 1., , The dependence of nuclear force on distance, between nucleons is not known precisely, but, approximate variation is shown graphically., From graph which of following statements, cannot be concluded?, , t1 t 2 , , 6., , F(Nuclear force), 0.5 fermi, , 10 fermi, R(Separation, between sncleon), , a) Nuclear force is repulsive for separation less than, 0.5 fermi, b) Nuclear force is attractive for separation less than, 0.3 fermi, c) Nuclear force is attractive for separation more than, 0,5 fermi, d) Nuclear force is negligible when separation between, nucleon is more than 10 fermi, 2. Consider fission reaction, :92 U 236 X117 Y117 20 n1 . Two nuclei of, same mass number 117 are formed along with, two neutrons. The binding energy per nucleon, of X is 8.5 MeV, where as that of U 236 is 7.6, MeV the total energy liberated will be about, A) 2 MeV B) 20 MeV C) 195.4 MeV D) 2000 MeV, 3. The half lives of a radioactive sample are 30, years and 60 years for two decay processes., If the sample decay by both the processes, simaltaneouly. The time after which, only onefourth of the sample will remain is, A) 10 years B) 20 years C) 40 years D) 60 years, 4. Two radioactive samples of different elements, (half lives t1 and t 2 respectively) have same, number of nucleii at t = 0. The time after which, their activites are same is, t1t 2, t2, t 1t 2, t2, A) 0.693 t t ln t B) 0.693 ln t, 2 1 1, 1, , t 1t 2, C) 0.693 t t, 1, , 2, , NARAYANAGROUP, , , , ln, , t2, t1, , D) None of these, , The activity of sample of radioactive materials, , 7., , 8., , C) A A e T, D) A2 A1e T , 2, 1, Find the decay constant of 55 Co radio nuclide, if its activity is known to decrase 4% per hour.., The decay product is non-radioactive., A) 1.1105 s 1, B) 2.2 105 s 1, t1 / t 2, , D) 4.4 105 s 1, C) 3.3105 s 1, A radioactive isotope is being produced at a, constantrate X. Half -life of the radioactive, substance is Y. After some time, the number, of radioactive nuclei become constant. The, value of this constant is., XY, A) ln 2 B) XY C) XY ln 2 D) X / Y, Nuclei of radioactive element A are produced, at rate ' t 2 ' (where t is time) at any time t. The, element A has decay constant . Let N be the, number of nuclei of element A at any time t., At time t t 0 , dN / dt is minimum. The the, number of nuclei of element A at time t t 0 is, A), , 2t 0 t 02, 2, , 2t 0 t 02, C), , 9., , B), , t 0 t 02, 2, , 2t 0 t 02, D), , , The count rate for 100cm3 of a radio-active, liquid is C. Some of this liquid is now discarded., The count rate of the remaining liquid is found, to be C/10 after three half-lives. The volume, of the remaining liquid in cm 3 is, , a) 20, b) 40, c) 60, d) 80, 10. The count rate observed from a radioactive, source at ‘t’ second was N 0 and at 4t second, 11 , N0, . The count rate observed, at t, 16, 2, second will be, N, N, N, A) 0, B) 0 C) 0 D) None of these, 128, 64, 32, , it was, , 157
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 11. The energy released by the fission of a single, uranium nucleus is 200 MeV. The number of, fissions of uranium nucleus per second, required to produce 16 MW of power is, (Assume efficiency of the reactor is 50%), A) 2 106 B) 2.5 106 C) 5 106 D) 1 1018, 12. A star initially has 1040 deuterons. It produces, , 16. For a certain radioactive substance, it is, observed that after 4 hours, only 6.25% of the, original sample is left undecayed. It follows, that, A)the half life of the sample is 1 hour, , energy via the processes 12 H 12 H 13 H p, , C) the decay constant of the sample is ln2 hour 1, D) after a further 4 hours, the amount of the, substance left over would by only 0.39% of the, original amount, 17. Choose the correct options, A) By gamma radiations atomic number is not changed, B) By gamma radiations mass number is not changed, C) By the emission of one and two particles, isotopes are produced, D) By the emission of one and four particles isobars, are produced, 18. A nuclide A undergoes decay and another, , and 12 H 13 H 24 He n . Where the masses, of the nuclei are:, m 2 H 2.014 amu, m p 1.007 amu,, , m n 1.008 and m 4 He 4.001 amu. If, the average power radiated by the star is, 1016 W, the deuteron supply of the star is, exhausted in a time of order of, A) 106 s, B)108 s, C) 1012 s D)1016 s, , MULTIPLE ANSWER QUESTIONS, 13. Regarding the nuclear forces, choose the, correct options., A) They are short range forces, B) They are charge independent forces, C) They are not electromagnetic forces, D) They are exchange forces, 14. Regarding a nucleus choose the correct, options., A) Density of a nucleus is directly proportional to, mass number, A, B) Density of all the nuclei is almost constant, of, the order of 1017 kg / m3, C) Nucleus radius is of the order of 10 15 m, D) Nucleus radius is directly proportional to mass, number, A, 15. Choose the correct alternative, A) K wavelength emitted by an atom of atomic, number Z = 21 is then K wavelength emitted, 4, by an atom of atomic number Z = 31 is, 9, B) Half life of radioactive substance is 5 years,, Probability that a nucleus decays in 10 years is 3/4, C) Mass number of a nucleus is always greater, than its atomic number, D) Gamma rays are emitted due to nuclear, deexcitation, 158, , B) the mean life of the sample is 1 ln 2 hour, , nuclide B undergoes decay, a) All the particles emitted byA will have almost, the same speed, b) The particles emitted by A may have widely, different speed., c) All the particles emitted by B will have almost, thee same speed, d) The particles emitted by B may have widely, different speeds, 19. Assume that the nuclear binding energy per, nucleon (B/A) versus mass number (A) is as, shown in Fig. Use this plot to choose the, correct choice (s) given below :, , B/A, 8, 6, 4, 2, 0, 100, , 200, , A, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, a) Fusion of two nuclei with mass number lying in, the range of 1 < A < 50 will release energy, b) Fusion of two nuclei with mass number lying in, the range of 51 < A < 100 will release energy, c) Fission of a nucleus lying in the mass range of, 100 < A < 200 will release energy when broken, into equal fragments., d) Fission of a nucleus lying in the mass range of, 200 < A < 260 will release energy when broken, into equal fragments., 20. From the following equations pick out the, possible nuclear fusion reactions :, a) 6 C 13 1 H 1 6 C 14 4.3MeV, , NUCLEAR PHYSICS, 22. If TH is the half-life and TM is the mean life., Which of the following statements is correct, A) TM TH, B) TM TH, C) Both are directly proportional to square of the, decay constant, D) TM 0, 23. ‘n’ number of particles per second are, being emitted by N atoms of a radioactive, element. The half-life of element will be, A), , n, s, N, , B), , N, s, n, , C), , 0.693N, s, n, , D), , 0.693n, s, N, , b) 6 C12 1 H 1 7 N 13 2MeV, 14, , 1, , 15, , N 1 H 8 O 7.3MeV, , c), , 7, , d), , 92, , 36, , Sr 94 0 n1 0 n1 200 MeV, , U 235 0 n1 54 Xe140 , , COMPREHENSION QUESTIONS, Comprehension : 1, Many unstable nuclei can decay, spontaneously to a nucleus of lower mass but, different combination of nucleons. The process, of spontaneous emission of radiation is called, radioactivity. Three types of radiations are, emitted by radioactive substance., Radioactive decay is a statistical process., Radioactivity is indepndent of all external, conditions., The number of decays per unit time or decay, rate is called activity. Activitiy exponentially, decreases with time. Mean lifetime is always, greater than half-life time., 21. Choose the correct statement about, radioactivity, A) Radioactivity is a statistical process, B) Radioactivity is independent of high temperature, and high pressure, C) When a nucleus undergoes or decay,,, its atomic number changes, D) All of these, NARAYANAGROUP, , Comprehension : 2, We have two radioactive nuclei A and B. Both, convert into a stable nucleus C. Nucleus A, converts into C after emitting two -particles, and three -particles. Nucleus B converts, into C after emitting one, -particle anf five -particles. A time t =0,, , nuclei of A are 4N0 and that of B are N0. Halflife of A (into the conversion of C) is 1 min and, that of B is 2min. Initially number of nuclei of, C are zero., 24. If atomic numbers and mass numbers of A and, B are Z1, Z2 ,A1 and A2 respectively. Then, A) Z1 Z 2 4, , B) A1 – A2 = 4, , C) both (a) and, (b) are correct, D) both (a) and, (b) are wrong., 25. What are number of nuclei of C, when number, of nuclei of A and B are equal ?, A) 2N0, B) 3N0, C), , 9N0, 2, , D), , 5N 0, 2, , 26. At what time rate of disintegrations of A and, B are equal., A) 4 min, B) 6 min, C) 8 min, D) 2 min, 159
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , Comprehension : 3, A, , The mass of nucleus z X is less than the sum, of the masess of A Z number of neutrons, and Z number of protons in the nucleus. The, energy equivalent to the corresponding mass, difference is known as the binding energy of, the nucleus. A heavy nucleus of mass M can, , C) 1.21023, , D) 1.21030, , Comprehension : 5, , massws m 3 and m 4 can undergo complete, fusion and form a heavy nucleus of mass M, “only if m3 m 4 M ". The masses of some, , N 0 , N 0 , zeroo, , netural atoms are given in the table below., , N1 , N 2 , N3 and N 4 are the number of nuclei, of A,B,X and Y at any instant t., , only if m1 m2 M. Also two light nuclei of, , H, , 1.007825u, , 2, 1, , H, H, , 4.002603u, , Li, , 7.016004u, , H, , 3.016050u, , 4, 2, , 6, 3, , Li, , 6.015123u, , 7, 3, , 70, 30, , Zn, , 69.925325u, , 82, 34, , 152, 64, , Gd, , 151.91980u, , 206, 82, , Pb, , 208.980388u, , 210, 84, , Po 209.982876u, , Bi, , Se, , and, , zero, , respectively., , 2.014102u, , 3, 1, , 209, 83, , C) 2.51014, D) 2.51019, 31. Find the number of decays in the next 6.15 y., A) 1.21014, B) 1.21018, , Consider radioactive decay of A to B which, further decays either to X or Y, 1 , 2 and 3, are decay constants for A to B decay, B to X, decay and Bto Y decay respectively. At t 0 ,, the num,ber of nuclei of A,B,X and Y are, , break into two light nuclei of mass m1 and m 2, , 1, 1, , 30. Find the number of decays in the next 10 hours, A) 2.5104, B) 2.5107, , A, , 1, , 27. The correct statement is, , B, 3, , 81.916709u, 205.97445u, , Y, 32. The net rate of accumulation of B at any instant, is, A) N1 1 N 2 2 N 3 3, , a. The nucleus 63 Li can emit an alpha particle, , B) N1 1 N3 2 N 4 3, , 210, b. The nucleus 84, Po can emit a proton, c. Deuteron and alpha particle can undergo, complete fusion., , C) N1 1 N 2 2 N 2 3, , 70, d. The nuclei 30, can usdergo, Zn and 82, 34Se, complete fusion., 28. The kinetic energy (in keV) of the alpha, particle, when the nucleus at rest undergo, alpha decay, is, A. 5319, B. 5422 C. 5707 D. 5818, , Comprehension : 4, , X, , 2, , D) N1 1 N 2 2 N 2 3, 33. The number of nuclei of B will first increase, and then after a maximum value, it decreases, for, A) 1 2 3, B) 1 2 3, C) 1 2 3, D) For any values of 1 , 2 and 3, 34. At t , which of following is incorrect?, , 410 tritium atoms are contained in a vessel., The half-life of decay of tritium nuclei is 12.3 y., 29. Find the activity of the sample, A) 71014 s 1, B) 71018 s1, 23, , C) 7 1024 s1, 160, , D) 7104 s 1, , A) N 2 0, , N0 2, B) N 3 , , 2N 0 3, C) N 4 , 2, 3, , D) N1 N 2 N 3 N 4 2N 0, , 2, , 3, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , Comprehension : 6, Nuclei of a radioactive element X are being, produced at a constant rate q and this element, decays to a stable nuclues Y with decay, constant and half-life T1/ 2 . At time t = 0, theree, are N 0 nuclei of the element X., 35. The number NX of nuclei of x at time t T1/ 2, is, q N 0, A), 2, , 1, B) 2 N 0 q , , , q 1, , C) N 0 , 2 , , , q 1, , D) N 0 , 2 , , , 36. The number of NY of nuclei of Y at time t is, q N 0, A) qt , , , , t q N 0, e , , , , q N 0, B) qt , , , , t, e, , , q N 0, C) qt , , , , t q N 0, e , , , , q N 0 t, D) qt , e, , , , , NUCLEAR PHYSICS, 38. Assuming 200 MeV of thermal energy is, liberated from each fission event on an, average, find the number of events that should, take place every day., A) 3.241024, B) 3.241021, C) 3.24 10 22, D) 3.241020, 39. Assuming the fission to take place largely, through 235 U , at what rate will the amount of, 235, , U decrease ? Express your answer in kg/, , day., A) 3.264 B) 2.264 C) 4.264 D) 1.264, 40. Assuming that uranium enriched to 3% in 235 U, will be used, how much uranium is needed per, month (30 days) ?, A) 631.5 kg, B) 1263 kg, C) 2263 kg, D) 3263 kg, , Comprehension : 8, The nuclear charge (Ze) is non-uniformly, distributed within a nucleus of radius R. The, charge density r (charge per unit volume), is dependent only on the radial distance r from, the centre of the nucleus, as shown. The, electric field is only along radial direction, , 37. The number NY of nuclei of Y at t T1/ 2 is, A) q, , ln 2 3 q N 0 , , , , , 2, , , B) q, , ln 2 3 q N 0 , , , , 2, , , , C) q, , ln 2 1 q N 0 , , , , 2, , , , D) q, , ln 2 1 q N 0 , , , , 2, , , , Comprehension : 7, A town has a population of 1 million. The, average electric power needed per person is, 300 W. A reactor is to be designed to supply, power to this town. The efficiency with thermal, power is converted into electric power is aimed, at 25%., NARAYANAGROUP, , (r), , d, , a, , r, , R, , 41. The electric field at r = R is :, a) independent of a, b) directly proportional to a, c) directly proportional to a2, d) inversely proportional to a, 42. For a = 0, the value of d(maximum value of , as shown in the figure) is :, a), , 3Ze, 4 R 3, , b), , 3Ze, R3, , c), , 4Ze, 3 R 3, , d), , Ze, 3 R 3, 161
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , MATRIX MATCHING QUESTIONS, 43. Match Column - I with Column -II, COLUMN - I, COLUMN-II, a) -decay, p) Neutrino, q) Tunnel effect, b) -decay, c) -decay, r)Atomic number decreases by 1, d)K-capture s) No change in atomic number, 44. Match Column - I with Column -II, COLUMN - I, COLUMN-II, a) Energy of thermal neutrons, p) 0.025eV, b) Energy of X-rays, q) 8eV, c) Binding energy per nucleon, r) 10eV, d)Photoelectric threshold of a metal s) 3eV, 45. Some laws/processes are given in column-I., Match these with physical phenomena given, in column, COLUMN - I, a) Nuclear fusion b) Nuclear fission, d) Exothermic nuclear reaction, c) -decay, COLUMN-II, p) converts some matter into energy, q) Generally possible for nuclei with low atomic, number, r) Generally possible for nuclei with higher atomic, number, s) Essentially proceeds by weak nuclear forces, 46. Match Column - I with Column -II, COLUMN - I, COLUMN-II, a) -decay, p) Neutrino, b) -decay, q) Antinutreno, c) Electrom capture, r) Positron, d) Pauli’s exclusion principle s) X-ray, , INTEGER TYPE QUESTIONS, 47. Assume that the mass of nucleus is given by, M Amp where A is the mass number and mp, =1.00727u. The density of the matter in the, nucleus is about y 1017 kg / m3 . Then y is___, (Take R0 1.5Fm ), 48. If, , 92, , U 238 changes to, , 85, , At 210 by series of, , and decays, Find the number of decay, undergone., 162, , 49. At a given instant, there are 25% undecayed, radio-active nuclei in a sample. After 8 sec,, the number of undecayed nuclei reduced to 12.5, %. The time after which the number of, undecayed nuclei will further reduce to 6.25%, of the reduced number is ___sec, 50. There are two radio-active nuclei A and B. A, is an alfa-emitter while B is a beta-emitter;, Their disintegration constant are in the ratio, of 1 : 4. The ratio of number of nuclei of A and, B at any time ‘t’ such that probabilities of, getting number of alfa and beta particles are, same at the instant is y : 1. Then y is___, 51. A small quantity of solution containing Na24, radionuclide (with a half life of 15 hours) of, activity 1.0 curie is injected into the blood, of a person. A 1 cm3 blood sample shows 296, disintegrations per minute after 5 hours., Assuming uniform spreading of the, radionuclide in the person’s blood, estimate, the total volume of his blood.(in litres), 52. The radioactivity of a sample R1 at a time T1, and R2 at a time T2. If the half-life of the, specimen is T, the number of atoms that have, disintegrated in the time (T2 – T1) is equal to, n R1 R2 T, . Here n is some integral, ln 4, number. What is the value of n?, 53. In a nuclear reactor an element X decays to a, radio active element Y at a constant rate 1015, atoms per sec. Each decay releases 100 MeV, energy. Half life of Y equals T and decays to a, stable product Z. Each decay of Y releases 50, MeV. All energy released inside the reactor, is used to produce electricity at an efficiency, of 25%. Calculate the electrical power in kw, generated in the reactor in steady state., 54. A charged capacitor of capacity C is, discharged through a resistance R. A, radioactive sample decays with an average life, J. If the ratio of electrostatic field energy, stored in the capacitor to the activity of the, radioactive sample remains constant with time, , then R xJ / C . Where x is ______, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , 55. A radio nuclide with half-life T days emits , particles of average kinetic energy E J. The, radionuclide is used as a source in a machine, which generates electric eneergy with, efficiency 25%. The number of moles of the, nuclide required to generate electrial energy, at an initial rate P is, , n , , yT P, E N ln 2 , , where ‘y’, , is., 56. The average lifetime for the n = 3 excited state, of a hydrogen like atom is 4.8x10–8 sec and, that for the n=2 state is 12.8 x 10–8sec. The, ratio of the average number of revolutions, made in the n = 2 state to the average number, of revolutions made in the n = 3 state before, any transitions can take place from these, states is., , 61. The amount of heat generated by 1.0 mg of a, Po 210 preparation during the one mean lifetime, period of there nuclei, if the emitted a-particles, are known to possess the kinetic energy 5.3, MeV is n(0.2 MJ). Assume all daughter nuclei, are practically formed in ground state. What, is the value of n?, , LEVEL - V - KEY, SINGLE ANSWER TYPE, 2.C, 9.C, , 1.B, , 8.A, , 13. A,B,C,D, 17.A,B,C, , 7.A, , 14. B,C 15.A,B,D, 18. A,D 19.B,D, , 16.A,B,C,D, 20. B,D, , COMPREHENSION, 21.D, , 57. In an decay, the kinetic energy of, particle is 48 MeV and Q-value of the, reaction is 50 MeV. The mass number of the, , 35.A, 42.B, , 900, where x is [Assume, x, the daughter nucleus is in ground state), 58. Consider a nuclear reaction A B C . A, nucleus A moving with kinetic energy of 5, MeV collides with a nucleus B moving with, kinetic energy of 3 MeV and forms a nuclues, C in excited state. Find the kinetic energy of, nucleus C just after its formation if it is formed, in a state with excitation energy 10.65 MeV., Takes masses of nuclei of A, B and C as 25.0,, 10.0, 34.995 amu 1 amu = 930 MeV/c2., , 43., 44., 45., 46., , 59. Q value of the reaction, N4 O17 p, , 1., 2., , [the masses of N14 ,He4 ,p,O17 are respectively, 14.00374u, 4.0026u, 1.00783u and 16.99913u, is x/5 MeV. The value of x is nearly [1u = 931.5, Me V], , 6.A, , MULTIPLE ANSWER, , 28 A, , mother nucleus is, , 3. C 4. A 5.C, 10. B 11. D 12.C, , 22.A, 29.A, 36.C, , 23.C, 30.C, 37.C, , 24.C, 31.A, 38.A, , 25.C, 32.C, 39.D, , 26.B, 33.A, 40.B, , 27. C, , 34.B, 41.A, , MATRIX MATCHING, A (q), B (p,r), C (s), D (p,r), A (p), B (r), C (q), D (s), A (p,q), B (p,r), C (p,s), D (p,q,r), A (p,r), B (q), C (s), D (p,q), , INTEGER ANSWER, 47.1 48.7, 54.2 55.4, 61. 8, , 49.8, 56.9, , 50.4, 57.9, , 51. 6 52.2, 58.2 59.3, , 53. 6, 60.1, , LEVEL - V - SOLUTIONS, SINGLE ANSWER QUESTIONS, , 3., , Characterstics of nuclear forces, energy released =, ( 2 117 8.5) – (236 7.6) MeV, 1 1 1, 1 2 or t t t , where t is T1/ 2, 1, 2, , T1/ 2 = 20 years, , 60. In a nuclear reactor 0.96 grams of 92 U 235, is consumed in one day. If 0.1% of mass of, , 4. N1 N i e 1t , A1 N11 & N 2 N i e 2t , A2 N 2 2, , U 235 is available as energy find the power, 92, , It is given that after time ‘t’, N 2 = N1 . Solving the, above equations, we get the time ‘t’., , of reactor (in MW), NARAYANAGROUP, , 163
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , MULTIPLE ANSWER, , 5., , A1 A0 e t1 and A2 A0 e t2 ., , 6., , 1, Dividing both, we get and mean life is, , t, t, Activity A N 0 e A 0 e, Since activity decreases at 4.0% per hour, activity, of 55 Co radio nuclide at t 1hr., , 7., , A A 0 A 0 where 0.04, The number of radiocative nuclei becomes constant, when the rate of production is equal to the rate of, decay X N, 2, , 8., , dN 2, d N, dN, t N 2 2t , dt, dt, dt, , Intial count rate of 1cm3 of liquid , , C, 100, , 1 C, After 3 half-lives, count rate of the liquid , 8 100, Let V be the volume of the remaining liquid. The, VC C VC, , V 80, CR of this liquid , ;, 800 10 800, 10. Let initially substance have Ni nuclei then N= Ni, dN, t, e t ; dt N i e, dN, N i e t =N0, it is given that, At t=t,, dt, N, dN, N i e 4 t 0, And at t= 4t,, dt, 16, 3, , t, Dividing both, we get e 16 ; At t=11t/2,, 11, 8, 3, t, t, t, dN, N, N i e 2 N i e 2 e 2 0, dt, 64, 11. No of fissions = total energy produced /(energy, released per fission efficiency), 12. mass defect m= 0.026 amu, Energy released per reaction is, = 0.026 931.5 Mev= 24.2 Mev, , = 3.87 10 12 J, Energy released by consuming 1040 deuterons is, , 1040, = 12.9 1028 J, 3, Power = energy/ time, , = 3.87 10 12 , , 164, , 1 E2 ( Z 2 1)2, , , 15., 2 E1 ( Z1 1)2, N0, , that means, t = 4 T1/ 2, 16, 17. basic definitions of isobars and isotopes, , 16. 6.25% means, N , , 18. The total kinetic energy in the process will be taken, by the particles while in decay the total, energy is shared by anti-neutrino and the particle., , dN, d2 N, For, to be min , 2 0, dt, dt, 9., , 13. from the property of nuclear forces, it is answered., 14. density of all nuclei is almost equal and radius is of, the order of fermi., , 19. In fusion two or more lighter nuclei combined to, make a comparatively heavier nucleus. In fission a, heavy nucleus breaks into two or more lighter nuclei., Further energy will be released in a nucler process, if total binding energy increases. Hence B,D arre, correct, 20. Carbon-nitrogen cycle, , COMPREHENSION TYPE, Comprehension : 1, 21. Radio activity defnistion, 1, 0.693, 22. TM and TH , , , dN, 23. A N, and n N, dt, Comprehension : 2, 24. For, , A C ; AAZ, , 1, , 1, , CAZ1138 22 He4 31 e0, , For B C ; A AZ22 CAZ2234 2 He 4 51 e0, Hence A1 8 A 2 4 and z1 3 z 2 3, 25. The no. of nuclei of A and B are equal after t = 4, min and the number of nuclei of A and B are, N0, each . Hence number of nuclei of C formed, 4, N 15 N0, from A 4N0 0 , and number of nuclei, 4, 4, N 0 3N 0, , . Hence, of C formed from B N 0 , 4, 4, 18N 0 9N 0, , total number of nuclei of C formed, 4, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , dN , ln 2 , A t, 26. dt A A 1 4N0 e -----(1), , A, , , , dN , ln 2 , B t, dt B NB 2 N0e -------(2), , B, , , Given, ln 2 , t, 1 , , , dN , dN , , dt dt 4N 0e, , A , B, , ln 2 , t, 2 , , N , 0e, 2, , t ln 2, 2, , 8 e t 6 min, Comprehension :3, Q, m M reac tan ts M products, 27., C2, Q, If 2 0, decay is possible, C, 210, 206, 28. 84 Po 42 He 82, Pb, , Q 209.982876 4.002603 205.97455 C2, 5.422 Me V, From conservation of momentum., K1 5.319 MeV 5319 Ke V, Comprehension : 4, , 29., , ln 2 4 1023, A 0 N 0, 7 1014 / s, 12.3 365 24 60 60, , 30. Number of decay N 0 1 e t where, ln 2, / hrs, 12.3 365 24, 31. Similar to the above problem, Comprehension :5, 32. Rate of accumulation = rate of formation -rate of, disintegration, 33. The initial rate of formation and disintegration of B, , , are respectively N 0 1 and N 0 2 3 ., B will first increase if, , N 0 1 N 0 2 3 or 1 2 3, 34. At t is 2 N 0 ., Also, the total number of nuclei at any instant, remains the same, N1 N 2 N3 N 4 2N 0, The ratio of X and Y formed are in the ratio 2 : 3, and the total number of nuclei of X and Y at t is, 2 N0 ., NARAYANAGROUP, , Comprehension : 6, 1, t, dNX, q NX N X q (q N 0 e , , dt, Also,, dN Y, 1, t, 36. dt N X q (q N0 e , , 35., , q N 0 t q N 0, N Y qt , e , , , , 37. In the above answer sub t = T1/ 2, Comprehension : 7, n, 6, 38. P E 300 10, t, , n, 25, , , 6, 19, , 200 10 1.6 10 , 100, 24 60 60 , n 3.24 1024, 39. 6.023 1023 Nuclei weigh 235 g for 3.24 1024, 235, 3.24 1024 g, nuclei requires, 6.023 1023, 1.264 kg / day, 40. Similar to above problem., Comprehension : 8, 1 Q, 41. E 4 R 2 ;Q total ch arg e Ze, 0, , E, , Ze, 4 0 R 2, , E at r = R is independent of a, 42. versus r graph for a = 0 will be as shown, (r), d, , , R, r, r, , dR r, d, , , , for r R, R Rr, R, R, , Q Ze 4x 2 dx , 0, , dR 3, 3Ze, or d , 3, R 3, 165
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , MATRIX MATCH TYPE, 43. disintegration, 45. Nuclear reactions, , 44. Order of energy, 46. Disintegration, , INTEGER ANSWER, m A 1.000727u, , y 1017, 3, 4, V, R 0 A1/3 , 3, Change in mass number = 238-210 = 28, 28, 7, No.of particles emitted =, 4, To reduced by half number the time reuired = one, half life = 8 sec, N, , 4, N11 N 2 2 A 2 , N B 1 1, 3, Consider a 1 cm sample of the blood. If its initial, activity was Ao, then the activity A after time t will, be A A0 e t . In terms of half life T1/2, the, d, , 47., 48., , 49., 50., 51., , 1, expression can be written as A A0 , 2, 5 hr /15 hr, , t / T1/ 2, , ,, , 21 / 3, , 296, 1, 296 , A0 , i.e.,, or A0 , =6.22s-1., 60 s, 2, , 60s , Since the total initial activity is given to be, A0' 1.0 10 6 curie = 1.0 x 10-6 x (3.7 x 1010, disintegration/second) = 3.7 x 104 s-1,, The total volume of the blood is V =, Total activity, 3.7 10 4 s 1, =, activity per cm 3, 6.22 s 1 / cm3, 3, 3, = 5.95 x 10 cm = 5.95 litres., 52. R1 N1 ; R 2 N 2 No. of atoms decayed in, R R2, T1 T2 N1 N 2 1, , R R 2 T 2 R1 T2 T n 2, 1, log 2, log 4, 53. At steady st ate energy released per sec, r E1 E2 ; 25% ; r 105, E1 100 10 6 1.6 10 19 1.6 10 11 J, E2 50 106 1.6 10 19 0.8 10 11 J, 2, 1, Q 0 e t / RC , , E 2C, 54., For this ratio to be constant,, , A, A 0 et, 2t, 2 1, 2, t or , ; , with time;, RC, CR J CR, , 166, , 55. Let ‘n’ be the number of moles of the ratio nuclide, ln 2 , number of nuclei in the nuclide = nN; , ;, T, Nn ln 2 , Rate of decay A nN , T, Rate of release of energy = AE, Rate of generate of electrical energy P = AE, 4TP, 25 nNE ln 2 , n , , NE ln 2 , 100, T, 56. No. of revolutions before transition = frequency x, 1, time Also frequency 3, n, 3, , 1, 8, 12.8 10, 2, , 9, So required ratio 1 3, 8, 4.8 10, 3, A4, 57. KE , Q A 100, A , 58. m A C 2 K A m B C 2 k B m C C 2 k c , excitation energy., mA mB mC c2 k A k B k C excitation, , energy k C 2MeV, 59. Q 1.00783 16.99913 14.00307 4.0026 931.5 MeV, 3, = MeV.., 5, energy, E, mc 2, 60. Power =, ; P ; P, time, T, t, 0.1 2, M C, , 100 , , ; M = 0.96 grams = 96 x 10-5, P, t, kg ; t = 1day = 86400 seconds, 2, , 5, 8, 0.1 96 10 3 10 , = 1 x 106 W = 1 MW, P, , 100, 86400, , 61. Number of nuclei initially present =, 103, 6.023 1023 2.88 1018, 210, , The no. of decay in one mean lifetime, 1, 1 2.88 1018, e, 1, , , 18 , 13, Energy released= 2.88 10 1 5.3 1.6 10 J, e, , , , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 6., , LEVEL-VI, SINGLE ANSWER QUESTIONS, 1., , Two radioactive materials X1 and X 2 have, decay constants 11 and respectively. If, initially they have same number of nuclei, then, ratio of number of nuclei of X1 to X 2 will be, , 7., , 1, after a time, e2, , 2., , 1, 1, 1, 1, A) 5, B) 11, C) 10 , D) 9 , Radon 220 decays to Bismuth 212 by the, following series of decay, 86, 84, , Rn, Po, , 220, , 216, , 84 Po, 82 Pb, , 216, , 212, , 4, , 2 He ; T1/ 2 0.016s, , Pb 212 83 Bi 212 1 e 0 ; T1/ 2 10.6s, If certain mass of radon is allowed to decay in, certain container, after five minutes element, with greatest and least mass will respectively, be, A) Radon, bismuth, B) Polonium, lead, C) Lead, bismuth, D) Bismuth, lead, 238, If a U preparation of mass 1gram amits, 1.24 104 alpha particles per second, the halflife of the preparation is, A) 4.5 109 years, B) 1.2 109 years, 82, , 3., , 4., , 5., , D) 9.1109 years, C) 2.2 109 years, If the specific activity of C14 nucleid in some, ancient wooden items amounts to 3/5 of that, in lately felled trees, the age of that ancient, items is [ T1/ 2 for C14 is 5570 years], A) 4112 years, B) 2092 years, C) 5570 years, D) 2785 years, In the uranium ore, the ratio of U 238 nuclei to, Pb 206 nuclei is 2.8. If it is assumed that all the, lead Pb 206 to be a final decay product of the, uranuium series, th age of the ore is [ T1/ 2 for, U 238 is 4.5 109 years], B) 2.0 109 years, A) 4.5 109 years, C) 3.2 109 years, D) 6.4 109 years, NARAYANAGROUP, , , , 8., , , , R, 1 e t, D)None of these., , Two identical samples (same material and, same amount) P and Q of a radioactive, substances having mean life T are observed, , C), , 4, , 2 He ;T1/ 2 55s, , The specific activity of a preparation consisting, of radioactive Co58 and non-radioactive Co59, is equal to 2.2 1012 disintegration per sec per, gram. Thehalf-life of Co58 os 71.3 days. The, ratio of the mass of radioactive cobalt in that, preparation to the total mass of the, preparation in percentge is, A) 2.1 % B) 99% C) 0.1818% D) 97.9 %, An unstable element is produced in nuclear, reaction at a constnat rate R. Its disintegration, constant is . Find number of nuclei after time, ‘t’ if initialy it was Zero, R t, R, t, A) e, B) 1 e, , , , , , , , to have activities A P and A Q respectively at, the time of observation. If P is older than Q,, then the difference in their age is., AP , AP , AQ , AQ , A) T ln A B) T ln A C) T A D) T A , P, P, Q, Q, 9. A radionuclide is produced at constant rate ‘q’, & having half life T. Find time after which, activity of nuclei will be A, if initially number, of nuclei were zero., , A), , TA, q ln(2), , B), , T, A, ln, ln(2) q, , A, T, ln 1 , D)None of these, ln(2) q , 10. A fertiliser solution containing 1 mole of a, radioactive material having half life of, 14.3days was injected into the roots of a plant., The plant was allowed 70 hours to settle down, and then activity was measured in its fruit. If, the activity ameasured was 1 micro curie, the, percentage of activity transmitted from the, root to the fruit in steady state is [1 curie =, 3.7 1010 sec1 ], A) 1.26 1011 %, B) 2.52 1011 %, C) 0.63 1011 %, D) 1.78 1011 %, C), , 167
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , and B decays to C with decay constant 2 ., Initially at t = 0 number of nuclei of A and B, are 2 No and N0 respectively. At t = to, no. of, 3N 0, nuclei of B is, and nuclei of B stop, 2, changing. Find t0 ?, A), , 1 21 , ln , , 1 3 2 B), , 1 81 , ln , , 1 3 2 , , 1 71 , 1 41 , ln , ln , D), , 1 3 2 , 1 3 2 , 13. A stationary nucleus of mass 24 amu emits a, gamma photon. The energy of the emitted, photon is 7 MeV. The recoil energy of the, nucleus is, A) 2.2 keV B) 1.1 keV C) 3.1 keV D) 22 keV, 14. A nucleus with mass number 220 initially at, rest emits an particle. If the Q-value of, the reaction is 5.5 MeV, the kinetic energy of, the particle is (in MeV), A) 4.4, B) 5.4, C) 5.6, D) 6.5, , 17. At t = 0, number of radioactive nuclei of a, radiaoctive substance are x and its, radioactivity is y. Half-life of radioactive, substance is T. Then, , x, A) is constant throughout, y, B), , x, T, y, , C) Value of xy remains half after one half-life, D) Value of xy becomes one fourth after one halflife, 18. A nucleus A (parent) decays into B (Daughter), with half life T1 and B decays into C with half, life T2. Graph is drawn between number of, atoms/activity versus time. Select the correct, graph(s), , Parent, , C), , A) When T1 >> T2, , 168, , Daughter, Time, , ., , B)When T1 >> T2, , No. of atoms, , N0, , MULTIPLE ANSWER QUESTIONS, , Parent, Daughter, Time, , C)When T1 < T2, , Activity, , Parent, Daughter, Time, N0, , ., , D)When T1 < T2, , No. of atoms, , 15. Two radioactive substances have half lives T, and 2T. Initially they have equal number of, nuclei. After time, t = 4T, the ratio of their number of nuclei is x, and the ratio of their activity is y. Then, A) x = 1/8 B) x = 1/4 C) y = 1/2 D) y = 1/4, 16. Two radioactive nuclei A and B are initially in, the ratio 1 : 4. Also initial activities of the nuclei, are in the ratio 1 : 8. Given that, half life of A, is 2 years, choose the correct alternatives, A) Half life of nuclei B is 1 year, B) At t = 4 years, activities of A and B are equal, C) At t = 6 years, ratio of number of nuclei of A to, that of B is 2 : 1, D) Fraction of nuclei decayed in one mean life for, A and B are f1 and f 2 respectively, f1 f 2, , Activity, , 11. To investigate the beta-decay of Mg 23, radionuclide, a counter was activated at the, moment t=0. It registered N1 beta-particles, by a moment t1 2sec , and by a moment t2 3t1, the number of registered beta-particles was, 2.66 times greater. the mean life time of the, given nuclei is, A) 3 sec B) 7 sec C) 16 sec D) 14 sec, 12. Nucleus A decays to B with decay constant 1, , Parent, Daughter, Time, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , COMPREHENSION QUESTIONS, , Sta, ble, n/p, rat, ios, , No. of neutrons (n), , Comprehension : 1, Various thumb rules have been proposed by the, scientific community to explain the mode of, radioactive decay by various radio-isotopes. One, of the major rules is called the n/p ratio. If all the, known isotopes of the elements are plotted on a, graph of number of neutrons (n) versus number of, protons (p), it is observed that all isotopes lying, outside of a ‘stable’ n/p ratio region are radioactive, as shown in figure. The graph exhibits straight line, behaviour with unit slope up to p = 25. Above p =, 25, those isotopes with an n/p ratio lying below, stable region usually undergo electron capture while, those with n/p ratios lying above the stable region, usually undergo beta decay. Very heavy isotopes, (p > 83) are unstable because of their relatively, large nuclei and they undergo alpha decay. Gamma, ray emission does not involve the release of a, particle. It represents a change in an nucleus where, the energy of the nucleus changes from a higher, energy level to a lower energy level., -decay, , -decay, , Electron capture, , 25 No. of protons (p), , 19. The radioisotope of magnesium with atomic, amss 27 undergo radioactive decay by, A) Electron capture, B) alpha decay, C) Beat decay, D) Gamma ray emission, 20. Th-230 undergeos a series of radioactive, decay processes resulting in Bi-214 being the, final product. What was the sequence of the, processes that occurred?, A) , , , , , B) , , , , , C) , , , , D) , , , , , 21. Which of the following represents the relative, penetrating power of the three types of, radioactive emission in decreasing order?, A) , B) , C) , D) , Comprehension : 2, The radionuclide 56 Mn is being produced in a, cyclotron at a constant rate P by bombarding a, manganese target with deuterons. 56 Mn has a halflife of 2.5h and the target contains large number of, NARAYANAGROUP, , NUCLEAR PHYSICS, only the stable manganese isotopes 56 Mn . The, reaction that produces 56 Mn is 56 Mn d 56 Mn p, After being bombarded for a long time, the activity, of 56 Mn becomes const ant, equal to, 13.86 1010 s 1 (Use ln 2 = 0.693; Avagardo, number 6 1023 ; atomic weight of, 56, , Mn 56 g mol1 ), 22. At what constant rate P, 56 Mn nuclei are being, produced in the cyclotron during the, bombardment?, A) 2 1011 nuclei / s B)13.86x1010 nuclei / s, C) 9.6 1010 nuclei / s D) 6.93 1010 nuclei / s, 23. After the activity of 56 Mn becomes constant,, number of 56 Mn nuclei present in the target is, equal to, A) 5 1011 B) 20 1011 C) 1.2 1014 D) 1.8 1015, 24. After a long time bombardment, number of, 56, Mn nuclei present in the target depends, upon i) The number of 56 Mn nuclei present, the start of the process ii) Half-lie of the 56 Mn, iii) The constant rae of production P, A) All (i), (ii) and (iii) are correct, B) Only (i) and (ii) are correct, C) Only (ii) and (iii) are correct, D) Only (i) and (iii) are correct, Comprehension : 3, When subatomic particles undergo reactions,, energy is conserved, but mass is not necessarily, conserved. However, a particle’s mass, ‘contributes’ to its total energy, in accordance with, Einstein’s famous equation, E mc 2, In this equation, E denotes the equivalent energy, when a particle of mass m is converted into energy., The particle can also have additional energy due to, its motion and its interactions with other particles., Consider a neutron at rest, and well separated from, other particles. It decays into a proton, an electron,, and an undetected third particle:, Neutron proton+ electron + third particle, The table below summarizes some data from a single, nuetron decay. Column 2 shows the rest mass of, the particle times the speed of light squared., Particle, Neutron, Proton, Electron, , Mass x c2, ( MeV), 940.97, 939.66, 0.51, , Kinetic Energy, ( MeV), 0, 0.02, 0.42, 169
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, 25. Assuming the table contains no major errors,, what can we conclude about the (mass x c2) of, the undetected third particle?, A) It is 0.90 MeV, B) It is 0.36 MeV, C) It is less than or equal to 0.90 MeV; but we, cannot be more precise, D) It is less than or equal to 0.36 MeV; but we, cannot be more precise, 26. From the given table, which properties of the, undetected third particle can we calculate?, A) Total energy, but not kinetic energy, B) Kinetic energy, but not total energy, C) Both total energy and kinetic energy, D) Neither total energy nor kinetic energy, 27. Consider an ensemble of particles, all of which, have the same positive kinetic energy but, different masses. For this ensemble, which, graph best represents the relationship between, the particle’s mass and its total energy?, B), Energy, , Energy, , A), , Mass, , Mass, , Energy, , Mass, , Mass, , 28. Could this reaction occur?, Proton neutron + other particles, A) Yes, if the other particles have much more kinetic, energy than mass energy, B) Yes, but only if the proton has potential energy, (due to interactons with other particles), C) No, because a neutron is more massive than a, proton, D) No, because a proton is positively charged while, a neutron is electrically neutral, Comprehension : 4, A beam of alpha particles is incident on a target of, lead. A particular alpha particle comes in ‘headon’ to a particular lead nucleus and stops, 6.50 10 14 m away from the center of the nucleus., (This point is well outside the nucleus) Assume that, the lead nucleus, which has 82 protons, remains at, rest. The mass of alpha particle is 6.64 1027 kg., 29. The electrostatic potential energy at the, instant when the alpha particle stops is, A) 36.3 MeV, B) 45.0 MeV, C) 3.63 MeV, D) 40.0 MeV, 170, , D) 0.13 107 ms 1, C) 13.2 102 ms 1, Comprehension : 5, A deuterium reaction that occurs in an experimental, fusion reator is in two stages;, i), , ii), , Two deuterium 1 D 2 nuclei fuse together to form, a tritium nucleus with a proton as by product written, as D (D, p) T., A tritium nucleus fuses with another deuterium, nucleus to form a helium 4 He 2 nucleus with nuetron, as a by product written s T (d,n) 4 He 2 . Given, 2, , D1 2.014102 amu, , 3, , T1 3.016049 amu; 2 He 4 4.002603amu, , H1 1.007825amu; o n1 1.00665 amu, 32. Total energy released in of the two processes, is, A) 4.03 MeV, B) 17.58 MeV, C) 20.61 MeV, D) 21.61 MeV, 33. The energy released in the combined reaction, per deuterium is, A) 1.34 MeV, B) 5.86 MeV, C) 6.87 MeV, D) 7.21 MeV, 34. Percentage of mass of initial deuterium that, is released a energy is, A) 0.5R% B) 0.7% C) 0.38% D) 0.18%, Comprehension : 6, 1, , D), Energy, , C), , 30. Initial kinetic energy (in MeV) of the alpha, particle is, A) 36.3, B) 0.36, C) 3.63 D) 2.63, 31. The initial speed of the alpha particle is, A) 132 10 2 ms 1, B) 1.32 10 7 ms 1, , Suppose a nucleus X undergoes -decay, 225, . The emitted -particle is found, 92 X Y , to move along a helical path in a uniform magnetic, field B = 5 T. Radius and pitch traced by the particle are R = 5 cm and P = 7.5 cm respectively., [Given m(Y) = 221.003 U, m() = 4.003 U, m(n) =, 1.009 U, m (P) = 1.008 U, 1 U = 931 MeV/c2, = 1.6 × 10-27 kg]., , 35. Speed of the particle after the decay is, A) 1.2 107 m / s, , B) 9 106 m / s, , C) 1.5 107 m / s, D) 8 105 m / s, 36. Total energy released during decay is, approximately (in MeV), A) 2.5, B) 4.7, C) 9.9, D) 8, 37. Binding energy per nucleon of nucleus X is, approximately (in MeV), A) 2.3, B) 4.7, C) 6, D) 7.8, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , which further decays to, , 234, 91 Pa, , as A of decay constant 1 and, , A), , 238, . Let 92, U be called, 234, 90, , Th is called as B, , 234, of decay constant 2 and stable nuclei 91, Pa be, called as C. Here A is called parent nucleus and B, is called daughter nucleus of A. Any two adjacent, nuclei may be consider parent or daughter nuclei., , Suppose N1 , N 2 and N 3 are number of nuclei A,, B and C respectively at time ‘t’., 1, 2, Then we can write A , B , C, dN1, 1 N1, Rate of disintegration of A , dt, dN 2, 1 N1 2 N 2, dt, Rate of formation of nuclei C is equal to, , If at t = 0, there are N 0 number of nuclei ofA where, as nuclei B and C are absent in the sample Answer, the following questions, 38. Number of nuclei of B at any time t is, N 0 1 1t 2 t, A) e e , 2, 1, , N 0 1, , B) e, 2, 1, , 1 t, , e 2t , , N 0 1 1t 2 t, N 0 2 1t, 2 t, C) e e D) e e , 2, 1, 2, 1, , 39. Number of nuclei of nuclei C at time t is, , , 1, 2, 2 t, 1t , A) N 0 1 e e , , 2, 1, 2, 1, , 1, 2, 2 t, 1t , B) N 0 e e , 2, 1, 2, 1, , 1, 2 t, 1t, C) N 0 e e , 2, 1, , 1, , 2, 2 t, 1t, D) N 0 e e 1, 2, 1, 2, 1, , NARAYANAGROUP, , N2, N3, N1, , N3, N2, N1, , Time, , C), , Time, , D), N3, N1, N2, , N3, N2, N1, , Time, , Time, , Comprehension : 8, A radionuclide with decay constant is being, produced in a nuclear reactor at a rate a 0 t per, second, where a 0 is a positive constant and t is, , Rate of distintegration of B , , dN 3, 2 N2, dt, , B), , No of nuclei, , 238, Thorium 92, Th , which is again an unstable nucleus, , No of nuclei, , U is an unstable nucleus. It decays to, , 40. The graph between N1 , N 2 and N 3 with time, can be best represent by, No of nuclei, , Uranium, , 238, 92, , No of nuclei, , Comprehension : 7, , the time. During each decay, E0 energy is released., The production of radionuclide starts at time t = 0., 41. Which differential equation correctly, represents the above process?, dN, dN, a 0 t N, N a 0 t, A), B), dt, dt, dN, dN, N a 0 t, a 0 t N, C), D), dt, dt, 42. Instantaneous power developed at time ‘t’ due, to the decay of the radionuclide is, a, , a, , a0, , a0, , , 0, 0 t , A) a 0 t e E 0, , , , , t, , a, , a, , a, , a, , , 0, 0 t , B) a 0 t e E 0, , , , , , , , t, , , , 0, 0, C) a 0 t e E 0 D) a 0 t e E 0, , , , , 43. Average power developed in time ‘t’ due to, decay of the radionuclide is, a, a0t a0 a0, , 2 20 e t E 0, A) , t t, 2, , , a, a 0t a 0 a 0, , , 2 20 e t E 0, t t, 2, , , B) , , a t, , a, , a, , a, , a t, , a, , a, , a, , , , t, 0, 0, 0, 0, C) 2 2 t 2 t e E 0, , , , 0, t , 0, 0, 0, D) 2 2 t 2 t e E 0, , , 171
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, Comprehension - 9, A nucleus kept at rest in free space, brakes up into, smaller nuclei of masses ‘m’ and ‘2m’. Total energy, generated in this fission is E. The bigger part is, radioactive, emits five gamma ray photons in the, direction opposite to its velocity and finally comes, torest.[Given h 6.6 1034 Js , m 1 1026 Kg ;, E 3.63 108 mc2 ; C 3 108 m / s ), 44. Fractional loss of mass in the fission is, A) 1.21 108, B) 2.56 108, D) 3.52 108, C) 1.73 108, 45. Velocity of small daughter nucleus is, A) 5.6 104 m / s, B) 6.6 104 m / s, C) 7.6 104 m / s, D) 8.6 10 4 m / s, 46. The wave length of the gamma ray is, A) 0.02 A 0 B) 0.03A 0 C) 0.15A 0 D) 0.05A 0, , MATRIX MATCHING QUESTIONS, 47. At t = 0, ‘x’ nuclei of a radioactive substance, emit ‘y’nuclei per second., Match Column - I with Column -II, COLUMN - I, COLUMN-II, p) ln 2 x / y , , A) Decay constant , B) Half life, C) Activity after time t , , q) x / y, 1, , , r) y / e, , 1, s) None, , 48. Match Column - I with Column -II, COLUMN - I, A) After emission of one and one , B) After emission of two and one , C) After emission of one and two , D)After emission of two and two , COLUMN-II, p) atomic number will decrease by 3, q) atomic number will decrease by 2, r) atomic number will decrease by 8, s) atomic number will decrease by 4, 49. Match Column - I with Column -II, COLUMN - I COLUMN-II, A) X-rays, p) Stream of electrons, , B) -decay q) Stream of He nuclei, C) -decay r) Maximum penetrating power, D) -decay s) Stream of positrons, t) E.M. waves, , D)Number of nuclei after time t , , 172, , 50. Consider two radioactive nuclei A and B . Both, convert into a stable nucleus C. Nucleus A, converts into C after emitting two - particles, and three -particles . Nucleus B converts, into C after emitting one -praticle and five, - particles. At time t =0 , no.of nuclei of A, are 4N0 and that of B are N0. In the conversion, of A into C half life A is 1 minute and that B in, conversion of B into C is 2 minute. Initially, no.of nuclei of C are zero then Match the, following table:, COLUMN - I, COLUMN-II, A) The difference between atomic, p) 6, no.of A and that of B is, B)The difference between mass number, q) 4, of A and that of B is, C) If at an instant number of nuclei of A, r)18, are equal to number of nuclei of B then, at the instant the ratio between no.of, nuclei of C and no.of nuclei of B is, D) The time t at which rate of disintegrations s) 2, of A are equal to that of B in minutes is, t) 0.25, 51. Match the following table :, COLUMN - I, COLUMN-II, A) Stability of nucleus p) Depends on mass number, B) Density of nucleus q) Packing fraction, C) mass defect, r) binding energy per nucleon, D) Dimensionless parameter, s) Independent of mass number, 52. In column I consider each process just before, and just after it occurs. Initial system is isolated, from all other bodies . Consider all product, particles including photons, neutrinos. Match, the system in Column I with results they, produce in Column II, Match the following table :, COLUMN - I, A) Spontaneous radio active decay of, B) Fusion reaction of 2 hydrogen nuclei, C) Fission of U 235 nuclear is initiated, D) decay ( negative beta decay), COLUMN-II, p) Number of protons is increased a uranium, nucleus at rest as given by the reaction, 238, , 234, , 4, , 92 U 90 Th 2 He ......, q)Momentum is conserved as given by the reaction, 1, , 1, , 2, , 1 H 1 H 1 H ......, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , r) Mass in converted to energy vice versa by a, thermal neutron as given by the reaction, 1, , 235, , 144, , STATEMENT QUESTIONS, , 89, , 0 n 92 U 56 Ba 36 Kr 3 01n ..., s) Charge is conserved, , INTEGER TYPE QUESTIONS, 53. The ratio of the radii of the 79 Au197 nucleus to, that of the radius of its innermost Bohr orbit, is nearly 16x. Find the value of x. (Given, R0=1.2x10-15m)., 54. The isotope 92 U 238 successively undergoes, eight -decays and six -decays. The, resulting isotope is found to be, , 92 5y, , X 2384x ., , 64., , Find the value of ratio x / y ., 55. The alpha activity of a 10kg sample of 92 U 235, that is used in a nuclear reactor is found to be, x 10 8 Bq. If the half life of uranium-235 for, , 56., , 57., , 58., , 59., , emitting alpha particles is 7.04 108 year, then, find the value of ‘x’. (Given 1year = 3.15x107s), The mean lives of a radioactive substance are, 1620 year and 405 year for alpha emission and, beta emission respectively. The time during, which three-fourth of a sample will decay if it, is decaying by both emissions simultaneously, is nearly 50 p year. Find the value of ‘p’., Energy is released from stars due to nuclear, fusion taking place in two cycles proton proton cycle and corbon nitrogen cycle. In, these cycles four hydrogen nuclei combine to, form helium nucleus. In this fusion reaction in, addition to energy how many number of, nutrinos are released ?, Mean free path ( ) in a fission reaction is the, average distance covered by a neutron, between two fissions. Generation of nearly, 200 neutrons are needed to fission all the, nuclei in 10kg of 92 U 235 . If 4cm and the, fastest moving neutrons generated in a fission, reaction have an energy of 2MeV, then the, time for 200 neutron generation is found to be, p 107 seconds. Find ‘p’., There is a stream of neutrons with kinetic, energy 0.0327 eV. If the half life of neutron is, 700 sec, then fraction of neutrons that decay, before they travel a distance of 10m is found, to be px10-6 (nearly). Find the value of ‘p’., , NARAYANAGROUP, , 65., , 66., , 67., , This section contains 2 questions. Each question, contains STATEMENT–1 (Assertion) and, STATEMENT–2 (Reason). Each question has 4, choices (A), (B), (C) and (D) out of which ONLY, ONE is correct., (A) Statement-1 is True, Statement-2 is True;, Statement-2 is a correct explanation for, Statement-1, (B) Statement-1 is True, Statement-2 is True;, Statement–2 NOT a correct explanation for, Statement-1., (C) Statement–1 is True, Statement–2 is False, (D) Statement –1 is False, Statement–2 is True., Half-life for certain radioactive element is 5min. four, nuclei of that element are obseved at a certain, instant of time. After 5 minutes, STATEMENT-I: It can be definitely said that two, nuclei will be left undecayed., STATEMENT-II: After half-life that is 5 minutes,, half of total nuclei will disintegrate. So, only two, nuclei will be left undecyed., STATEMENT-I: It is easy to remove a proton, from 4020 Ca nucleus as compared to a neutron., STATEMENT-II: Inside nucles neutrons are, acted on only by attractive forces but protons are, also acted on by repulsive forces., STATEMENT-I: It is possible for a thermal, neutron to be absorbed by a nucleus where as, proton or an alpha particle would need a much, larger amount of energy for being absorbed by the, same nucleus., STATEMENT-II: Neutron is electrically neutral, but proton and alpha particle are positively charged., STATEMENT-I:Consider the following nuclear, reaction of an unstable 146 C nucleus initially at rest., The decay 146 C 147 N 01e . In a nuclear, reaction total energy and momentum is conserved, experiments show that the electrons are emitted, with a continous range of kinetic energies upto, some maximum value., STATEMENT-II: Remaining energy is released, as thermal energy., , LEVEL - VI - KEY, SINGLE ANSWER, 1.A, 8.B, , 2.C, 9.C, , 3.A, 4.A, 10. A 11.C, , 5.B, 12.D, , 6.C, 13.B, , 7.B, 14.B, , MULTIPLE ANSWER, 15. B,C, , 16.A,C, , 17. A,B,D 18. A,D, 173
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , COMPREHENSION, 19.C, 26.A, 33.C, 40.B, , 20.B, 27.C, 34.D, 41.C, , 21.D, 28.B, 35.C, 42.C, , 22.B, 29.C, 36.B, 43.A, , 23.D, 30.C, 37.D, 55.A, , 24.C, 31.B, 38.C, 45. B, , 25.D, 32.A, 39.A, 46.D, , MATRIX MATCHING, 47., 48., 49., 50., 51., 52., , A (s), B (p), C (r), D (s), A (s), B (p,r), C (s), D (q,r), ,t), A (q,r), B (p), C (s), D (r,t), A q, B q, C r,, D p, A p,q,r B s, C p,q,r D q, A q,r,s B q,r,s C q,r,s,D p,q,r,s, ,s, , INTEGER TYPE, 53.6, , 54.4, , 55.8, , 56.9, , 57.2, , 58.4, , 59.4, , STATEMENT TYPE, 60. D 61. A 62. A 63. C, , LEVEL - VI - SOLUTIONS, SINGLE ANSWER QUESTIONS, 1., , N1= N0 e11t and N2= N0 e t It is given that, N1, 1, 2 , solving the above two equations, we, N2, e, 1, 5, small T1/2 indicates that more such reactions take, place., A0 N 0 1.24 10 4 sec1 ;, , get t , 2., 3., , N0 , , 4., , 1, Avogaodronumber ?, 238, , ; T1/ 2 , , 3, A0 A0 .e t, 5, , U 238 Pb 206, , (radio nuclide) (stable nuclide), number of Pb 206 nuclei = number of decayed, urnium nuclei = N 0 1 e t In present sample, N U, , 238, , N Pb, , 2.8 N .e 28, N 1 e 10, , t, , 0, , 8., , dN, dN, R N , dt , solving t his, dt, R N, equation with initial conditions, we get the desired, answer, , dN , ( t t ), Ap , N i e, and, dt, , p, , dN , (t ), Aq , N i e, dt q, Dividing both and solving, we get t ., dN, dN, R N , dt , solving t his, 9., dt, R N, equation with initial conditions, we get the desired, answer, 10. R0 N 0 After 70 hours, R R0 e t, percentage of activit y tansmitted =, , 11., , activity bound in, 100%, R, Mg 23 given one -particle in one disintegration., , number of -particle = number if decayed nuclei, = number of disintegrations = N 0 1 e t , N 0 1 e t1 N1 & N 0 1 e t2 N 2 2.66 N1 ; t 2 3t1, , after simplification e 2 0.882; , 12., , 1, 16sec, , , dN B, 1 N A 2 N B, dt, , h E, ., c, Apply law of conservation of linear momentum and, , 13. momentum of a photon, P=, , get the recoil energy of nucleus and E , , t, , 206, , 0, , solve for time, t., 6. Cobalt prepartion = Co58 Co59, Here Co58 is radioactive Co59 is non-radioactive, Activity means, it is only for Co58 Specific activity, measured for total preparation. specific activity of, 174, , 7., , 1 4 , 3N , 0 1 (2N 0 )e t 0 2 0 ; t 0 ln 1 , 2 , 1 3 2 , , solve for time t=?, 5., , mass of Co58, above result, , 100%, total mass of cobalt preparation, 1gm, , ln 2, ?, , , Activity of ancient wood = 3/5 x activity of lately, felled trees, A, , cobalt preparation = 2.21012 dis.sec 1 .gm 1 = N , Mass of N nuceli of Co58 = N 58 1.6 1024 gm, , p2, ., 2m, , A 4, 14. KE , Q, A , , MULTIPLE ANSWER TYPE, N1 1, A1 1 N1 1, 15. x N 4 and y A N 2, 2, 2, 2 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , N1 1, A1 1 N1 1, 1 1, 16. N 4 & A N 8 2, 2, 2, 2 2, 2, , x 1, T, 17. x= N and y= N ; , y 0.693, xy N 2 , so after one half- life, it’s value becomes, 1/4th of its value., 18. When T1 >> T2, the no of nuclei ofA (parent) almost, remains same., when T1 < T2, the no of nuclei ofA (parent) decrease, exponentially., , COMPREHENSION TYPE, 19. n/p ratio for magnesium is 27/12>1., 230, 214, 20. 90, Th 83, Bi x 1e 0 y 2 He 4 z, Hence, 90 83 x 2y and 230 = 214 + 4y, Solving, we get y = 4 and x = 1. The order of the, rections is irrelevant, 21. Depends on ionization power., 22. In equilibrium, Rate of decay = Rate of production, 23. As Rate of decay = Rate of production, P Pt, P N N 1/ 2 1.8 1015, ln 2, Pt1/ 2, 24. As N ln 2 . It is dependent upon P and t1/ 2 ., In equilibrium, rate of production = rate of decay., Large initial number will only make equlibrium come, sonner. Hence, Initial number of 56 Mn nuclei will, make no difference, 25. From the passage, sub-atomic reactions do not, conserve mass. So, we cannot find the third, particle’s mass by setting m neutron equal to, m proton m electron third particle . By constrast, the total, energy in this case, the sum of ‘mass energy’ and, kinetic energy, is conserved. If E denotes total, energy, then, E neutron E proton E electron E third particle, The neutron has energy 940.97 meV. The proton, has energy 939.66 MeV + 0.02 MeV = 939.68, MeV. The electron has energy 0.51 MeV + 0.42, MeV = 0.93 MeV. Therefore, the third particle, has energy, E third particle E neutron E proton E electron, We just found the third particle’s total energy, the, sum of mass energy and kinetic energy. Without, more information, we cannot figure out how much, of that energy is mass energy., NARAYANAGROUP, , NUCLEAR PHYSICS, 26. Look to the solution of the above problem., 27. But E varies linearly with m. Since the particles in, the ensemble all have positive kinetic energy, the, total energy is positive even when the mass is zero, [relativistic physics permits this, because kinetic, energy no longer equals (1/2) mv2). So, the graph, does not begin at the origin., 28. A proton with only mass-energy can never decay, into a neutron as the mass of the neutro is more, than that of the proton. However, If the proton, has some additional energy in the form of PE such, that its energy is more than rest mass- energy of, the neutron, it can decay into neutron and release, some energy., 29. If the particles are treated as point charges,, 1 q1q 2, U, 4 0 r where Q1 2e , q 2 82e , and, r 6.5 1014 m , 9, , U 910 x, , 282 1.6021019 , 6.501014 m, , 2, , 5.821013 J, , =3.63MeV, 30. Apply conservation of energy; K1 U1 K 2 U 2 ., , r1 and U1 0 . and K 2 0 ., From PCE K1 U 2 5.82 10 13 3.63MeV, 31. K , , 1, mv 2 , 2, , 2K, V, , m, , 2 5.82 10 13 J , , 32 to 34., 35. R , , 6.64 10, , 27, , kg, , 1.32 107 ms 1, , Mass defect concept, mvsin , vsin 1.2 107 m / s (Charge of qB, , particle = 3.2 10-19 C), 2m, PqB, P, v cos v cos , 9 106 m / s, qB, 2m, V , , vsin 2 v cos 2 1.5 107 m / s, , m y v y m v v y 2.715 105 m / s, TE released during an -decay of the nucleus X, is,, E, 0.005 U, 37. Mass lost during -decay, m , 931, , 36., , mass of nucleus X,, m x m y m m 225.038 U, mass defect in nucleus X,, 175
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JEE-ADV PHYSICS- VOL- V, , NUCLEAR PHYSICS, , , , , , m d 92m p 225 92 m n m x U 1.895 U, , BE per nucleon in nucleus X =, md 931, 7.84 MeV, 225, , 38. Successive disintegration, 39. Successive disintegration, 40. Successive disintegration, dN, dN, a 0 t N;, N a 0 t, 41., dt, dt, a0 t a0 a0 t, 42. N 2 2 e , , a a, , , Pinst NE0 a 0 t 0 0 et E0, , , , t, , a0, , , , a0, , a t e, , t, , 0, , 43., , Pav , , 0, , , E 0 dt, , As N , , 44. m 3.63 108, m, Fractional loss = m 1.21x108, m, , 45. mv1 2mv 2 v1 2v 2, From PCE :, , -----(1), , 1, 1, mv12 2m v 22 E -----(2), 2, 2, , From (1) and (2), , 1, 2E, 4E, mv12 , v1 , 2, 3, 3m, , E, . Momentum of the proton, 3, , 5hc, Momentum of the larger nucleus., , , 5hc, 4Em, 5hc 3, 2mv2 , , , 3, 2mc 3.63, , MATRIX MATCHING TYPE, , 176, , N , N m A = 2.56 1025, M , , Radio active decay, Disintegration, Fundemental particles, Radio activity, , ln(2)N, 16, T1/ 2 = 2.52 10 decays / year, , 56. Total decay constant, , , m C2 E m 3.63 108 m, , 47., 48., 49., 50., , 0.529 1010, r, 6.69 1013 m, Z, r, 96 16x x 6, R, 8 6, 206, 54. 92 U 238 , ., 82 Pb, 55. No. of radioactive atoms is, , = 8 108 Bq, , a, a, a t a, , 0 0 20 20 e t E 0, t t, 2, , , Hence, , 53. R R 0 A1/ 3 6.98 1015 m, , t, , 0, , , , INTEGER TYPE, , R N , , dt, , 2, 46. mv 2 , , 51. Properties of nucleus, 52. Nuclear reactions, , 1, per year.., 324, , N0, N, t, t, we have 0 N 0e e 4, 4, 4, , 2, log e 2 450 50 9 year.., , 57. In nuclear fusion 2 positrons and 2 nuetrinos are, released, , 2E, 9, 2 107 m/s, t 2 10 s, 58. v , m, v, For 200 neutrons generation: T=200t, t, , 4 10 7 s, 2E, 2500m / s, m, Time taken to cover 10m distance,, 10, t , 4 103 s, 2500, , 59. v , , Fraction of neutrons decayed in time t, N 0.693, 6, is N T t 4 10, 1/ 2, , STATEMENT TYPE, 60. Half-life is the time in which the probability of, remaing matter is half of the intial amout, 61. Nuclear forces, 62. Nuclear forces, 63. Binding energy, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , SEMI CONDUCTOR DEVICES, , , SYNOPSIS, Classification of solids: Solids can be, , , classified as metals, insulators and semiconductors, , basing on conductivity (or) band theory, Classification according to conductivity is as follows, , , , Metals have high conductivity 10 8 to10 2 Sm 1 , , The valance band may be completely filled and the, conduction band partially filled with an extremely, small energy gap between them E g 0, Eg: Sodium (Na), The valence band is completely filled and, conduction band is empty but the two overlap each, other. Eg: Zn, , Overlapping, conduction band, , (or) high resistivity 10 to10 m , 11, , , , 19, , Semiconductors have conductivity 10 to 10 Sm, 5, , 6, , 1, , , , (or) resistivity 10 5 to10 6 m , intermediate to, metals and insulators, , Energy Band theory in solids: An isolated, , , , , , , , , , , , , , atom has well defined energy levels and energy of, an electron depends on its orbit (Principal quantum, number), But in solids atoms are so close such that outer, orbits are very close (or) overlaped to form energy, band., Inside the crystal each electron has a unique position, and no two electrons see exactly same pattern of, surrounding charges and each electron has different, energy level., Different energy level with continuous energy, variation form energy bands (According to Pauli’s, principle), The energy band formed by a series of energy bands, containing valance electrons is valance band., At 0 K, electrons start filling energy level in valance, band starting from the lowest one., The highest energy level, occupied by an electron in, the valance band at 0K is called Fermi level., , The lowest unfilled energy band formed just above, valance band is called conduction band., Depending on the forbidden energy gap between, valance band and conduction band, the solids are, , classified as conductors, insulators and, semiconductors., 1) Conductors : The energy band structure in, conductors have two possibilities, NARAYANAGROUP, , Electron energies, , Insulators have low conductivity 1019 to1011 Sm 1 , , (Eg 0), Ev, EC, , Valence, band, conductors, Conduction, band, , EC, Ev, , Valence, bond, conductors, , 2) Insulators: In insulators forbidden energy gap, is quite large. E g 3ev, Eg. Energy gap for diamond is 5.5 eV., , Electron energies, , , , Electron energies, , (or) low resistivity, , Empty, conduction, band, , EC, Eg > 3eV, Ev, , Valence, band, Insulators, , 3) Semi conductors: Semi conductors are the, basic materials used in the present solid state, devices like diode, transistor, Ic’s., The energy band structure of the semiconductors, is similar to insulators but in their case, the size of, forbidden energy gap is much smaller than that of, insulators. Eg 0.2eV to 3eV, 177
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JEE-ADV PHYSICS- VOL- V, , Electron energies, , SEMI CONDUCTOR DEVICES, , EC, Eg < 3eV, , , Ev, , Semiconductors, , , , , , , , , , , , Eg: Forbidden energy gap for Ge is 0.67eV, for Si, is 1.1 eV and for GaAs is 1.41eV, The vacancy in the covalent band is called hole., The electrons and holes move in opposite directions. , The difference in the resistivity of C, Si and Ge, depends upon the energy gap between their , conduction and valence bands. For C (diamond),, Si and Ge, the energy gaps are 5.4 eV, 1.1 eV and, 0.7 eV, respectively. Sn also is a group IV element , but it is a metal because the energy gap in its case is, 0 eV., Intrinsic semi conductor: Semiconductors , in their purest form having equal number of electrons, in CB and holes in VB is called as intrinsic semi, conductors., These electrons (free from covalent bonds) and, holes are called intrinsic carriers and such a, semiconductor is called intrinsic semiconductor., Ex:- Germanium (Ge) and Silicon (Si) are the purest, , form of intrinsic semi conductors., In an intrinsic semiconductor if ne denotes the, electron number density in conduction band , n the , h, , hole number density in valence band and ni the, number density (or) concentration of intrinsic, carriers, then ne nh ni, , Doping: The process of adding impurity to a pure, , , , , , 178, , semi conductor (in 1 : 106 ratio also called 1ppm), so as to improve its conductivity is called doping., Pentavalent impurity atoms are called donor, impurity atoms., Trivalent impurity atoms are called acceptor impurity, atoms., Extrinsic semi conductor: The semi, conductor formed by adding a small quantity of, impurity (tri or penta valent) to pure semi conductor, is called as extrinsic semi conductor., The extrinsic semi conductor formed by adding , pentanalent impurity atoms is called n-type semi, conductor., , The extrinsic semi conductor formed by adding, trivalent impurity atoms is called p-type semi, conductor., Fermi level in n-type semi conductors (also known, as donor energy level) lies in forbidden energy gap, and is very close to conduction band ( 0.01eV, below conduction band), Fermi level in p-type semi conductor (also known, as acceptor energy level) lies in forbidden energy, gap and is very close to valence band ( 0.01 to, 0.05 eV above valence band), When ever a covalent bond breaks an electron hole, pair is formed., When valency electron combines with a hole, a, covalent bond formed and at the starting point of, the electron a hole is formed., If a conduction electron combines with a hole a, covalent bond is formed without the formation of a, hole any where., In a doped (or) extrinsic semi conductor number, density of electrons in conduction band (ne), number, density of holes in the valence band nh and, number density of electrons in conduction band (or), holes in valence band in a pure semi conductor ni , then they are related as ne nh ni2 ., The intrinsic concentration ni varies with T as, ni2 AoT 3e Eg / KT (Ao is constant), The fraction of electrons of valence band present, in conduction band is given by, , , Eg, , f e KT, Where K is Boltzman’s constant, Eg is forbidden, energy gap and T is temperature, The Energy Gap: Experimentally it has been, found that the forbidden energy region Eg depends, on temperature., For silicon Eg T 1.21 3.60 104 T, For germanium Eg T 0.785 2.23 104 T, At room temperature, , 300K for, , silicon, , E g 1.1ev and for germanium E g 0.72ev, Adding equal concentration of donor and acceptor, or atoms to P type and N type semiconductor, respectively results in an intrinsic semiconductor., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , , , , , , SEMI CONDUCTOR DEVICES, If is the resistivity of the material of the, semiconductor, then the resistance offered by the, semiconductor to the flow of current is given by, , When concentration of donor atoms exceeds the, acceptor concentration in P type semiconductor, it, changes from P type to N type semiconductor., In semi conductors the total current I is the sum of, electron current Ie and holes current Ih, I = Ie + Ih, Electrical conductivity e h, , , .........(5), A, Since V=RI, from the equations (4) and (5),, we have, , ne e e nh h e, , V RI , , R, , For intrinsic semi conductor ne nh ni so that, , or V e ne ve nh vh .............(6), , ni e e h , , If E is the electric field set up across the, , Electrical conductivity of Semiconductors:, (Formulae derivation), , , V, ........(7), , From the equations (6) and (7) , we have, , semiconductor, then E , , Consider a block of semiconductor of length l , area, , of cross-section A and having number density of, , E e ne ve nh vh , , electrons and holes as ne and nh respectively ., , , By that on applying a potential difference, say V, a, current I flows through it as shown in figure., , , , The electron current I e and the hole current I h , , v , 1, v, e ne e nh h ......... 8 , E, , E, On applying electric field, the drift velocity acquired, by the electrons (or holes) per unit strength of, electric field is called mobility of electrons ( or holes), , or, , constitute the total current I flowing through the, semiconductor i.e ., I I e I h ............1, , – +, +, , ve, v, and h h, E, E, Therefore, the equation (8) becomes, , e , , –, –, , +, , –, +, , 1, e ne e nh h ......... 9 , , , l, , 1, is called the conductivity of, , the material of semiconductor., , Also, , , I, , e ne e nh h .......(10), , V, , , If ne is the density of conduction band electrons in, , , , , the semiconductor and Ve is the drift velocity of, electrons, then electron current is given by, , , , I e ene Ave ........(2), , , , Also, the hole current I h enh Avh ......(3), Using the equation (2) and (3) , the equation (1), becomes, I ene Ave enh Avh or I eA ne ve nh vh ........ 4 , NARAYANAGROUP, , , eA ne ve nh vh , A, , , , Electrons mobility is greater than the hole mobility, Mobility is a property of the semiconductor itself., It does not depends on the doping concentration., The mobility of an electron or hole generally, decreases with increase temperature., Resistance of semi conductors decrease with the, increase in temperature so semi conductors are, insulators at low temperature but becomes slightly, conducting at room tempeture., P- type (or) n- type semi conductor material is, elctrically neutral., 179
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-1:The number of silicon atoms per m3 is, 5 1028 . This is doped simultaneosuly with, 5 1022 atoms per m3 of Arsenic and 5 1020, , per m3 atoms of Indium. Calculate the, number of electrons and holes. Given that, nt 1.5 1016 m 3 .Is the material - n type or ptype?, Sol. Arsenic is donor impurity No. of donor atoms, , W.E-3:An N-type silicon sample of width 4 10 3 m, thickness 0.25mm and length 6 102 m carries, a current of 4.8mA when the voltage is applied, across the length of the sample. What is the, current density? If the free electron density is, 1022 m 3 , then find how much time it takes for, the electrons to travel the full length of the, sample ., Sol. The current density J is given by, I, 4.8 103, 4.8 10 3, , , A 4 103 25 10 5 , 106, , J, , added, N D 5 1022 m 3 ,, Indium is acceptor impurity, no.of accepter atoms, , The drift velocity vd given by, , added N A 5 1020 m 3, , vd , , Therefore, no.of free electrons creat ed, , The time taken ‘t’ is given by, , ne N D 5 10, , 22, , t, , Now, ne nh , therefore, net no.of free electrons, created,, ne1 ne nh 5 1022 5 1020 4.95 10 22 m 3, , Also net no. of holes created, 2, , 16, n12 1.5 10 , n 1 , 4.55 109 m 3, 22, ne, 4.95 10, 1, h, , As n1e n1h , the resulting material is n-type, semiconductor., , W.E-2: A semiconductor has an electron, concentration of 0.45 1012 m 3 and a hole, concentration of 5.0 1020 m 3 . Calculate its, conductivity . Given electron m obility, 0.135m2V 1 s 1; hole mobility 0.048m2V 1 s 1 ,, Sol. The conductivity of a semicon ductor is the sum of, the conductivities due to electrons and holes and is, given by, , e h ne ee nheh e ne e nh h , As per given date , ne is negligible as compared to, nh , so that we can write, , 1.6 10 19 C 5.0 10 20 m 3 0.048m 2V 1s 1 , 1, , 3.84 m 3.84 Sm, 180, , L 6 102, , 0.02sec, vd, 3, , W.E-4:The energy gap of pure Si is 1.1 eV. The, mobilities of electrons and holes are, respectively 0.135m 2V 1s 1 and 0.048m 2V 1s 1, and can be taken as independent of, tem perature. The intrinsic carrier, concentration is given by ni n0 e Eg / 2 kT ., Where n0 is a constant, Eg The gap width and, k The Boltzmann’s constant whose value is, 1.38 10 23 JK 1. The ratio of the electrical, conductivities of Si at 600K and 300K is., Sol. The total electrical conductivity of a semiconductor, is given by e ne e nh h , For an intrinsic semiconductor, ne nh ni, We can thus write for t he conductivity, e e h ni or e e h n0e E / 2 kT, g, , As the mobilities e , h are independent of, temperature, they can be regarded as constant. The, ratio of the conductivities at 600 K and 300 K is, E / 2 k 600, , 600 e e h n0e g, Eg /1200 k, then, e n e Eg / 2 k300 e, e h 0, 300, , As per given data E g 1.1eV, 1.38 1023 , k 1.38 1023 JK 1or , eVK 1, 19 , 1.6 10 , , enh h, , 1, , J, 4800, , 3m / s, ne 1022 1.6 1019, , 1, , k 8.625 10 5 eVK 1, , Solving we get the ratio of electrical conductivities, is 4 104, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-5: In a p-n junction diode, the current I can, eV, , W.E-6: The energy of a photon of sodium light, , , , expressed as I I 0 exp 2k T 1 where I 0 is, B, , called the reverse saturation current,V is the, voltage across the diode and is positive for, forward bias andnegative for reverse bias,, and I is the current through the diode, K B is, , 589nm equal to the band gap of a, semiconducting material (a) Find the, minimum energy E required to create a holeelectron pair (b) Find the value of E/kT at a, temperature of 300K, Sol. (a) The energy of the photon in, , the Boltzmann constant 8.6 10 5 eV / K and, T is the absolute temperature. If for a given, diode I o 5 1012 A and T= 300K, then, (a) What will be the forward current at a, forward voltage of 0.6V?, (b) What will be the increase in the current if, the voltage across the diode is increased to, 0.7V?, (c) What is the dynamic resistance?, (d) What will be current if reverse bias voltage, changes from 1Vto 2V?, , eV , , (Wavelength of photon =589nm=5890 A0 ), Thus the band gap is 2.1ev . This is also the minimum, energy E required to push an electron from the, valence band into the conduction band. Hence the, minimum energy required to create a hole-electron, pair is 2.1eV, (b) At T=300K,kT 8.62 105 eV / K 300 K , 25.86 10 3 eV Thus ,, , 12, 5, 1, Sol. I o 5 10 A, k 8.6 10 eVk, 5, , 8.6 10 1.6 10, , , , a) I I0 e, , ev, , 2 kT, , 19, , Jk, , E, 2.1eV, , 81, kT 25.86 103 eV, , 1, , The availale thermal energy is nearly 81 times less, than that of the required energy to create electron, hole pair. So it is difficult for the thermal energy to, create the hole-electron pair but a photon of light, can do it easily, , , , 1 ,, , For V=0.6V,, 19, , I 5 10, , 12, , 1.610 0.6, , , 28.610 51.610 19 300, e, 1, , , , , , 5 10 12 e 23.52 1, 5 10 12 1.256 1010 1 0.0628.A, , b) For v=0.7v, we have, , , , 19, , I 5 10, , 12, , 1.610 0.7, , , 28.6105 1.61019 300, e, 1, , , , , , , , I 5 1012 e 27.32 1, 5 1012 6.054 1011 1 3.0271 A, , I 3.271 0.0628 2.9643 A, c) I 2.9643, v 0.7 0.6 0.1V, dynamic resistance , , v, 0.1, , 0.0337, I 2.9643, , d) For change in voltage from 1 to 2v, the current, will remain equal to I 0 5 1012 A . It shows that, the diode possesses practically infinite resistance, in reverse biasing, NARAYANAGROUP, , 12400 12400, , 2.1eV, , 5890, , , , , , P-n junction: In a single Ge (or) Si crystal by, doping one part with trivalent impurity and the other, part with pentavalent impurity , a p - n junction is, formed, At p-n junction migration of majority charges i.e., holes form p side to n side and electrons from n, side to p - side due to concentration difference is, called diffussion, At p-n junction , the donor impurity atoms, become +ve ions and the acceptor impurity atoms, become -ve ions, so that the ions act as two, electrodes and the junction is called p-n junction, diode., The ions at junction develop a potential difference, called potential barrier and this barrier prevents, further diffusion of charges across junction, Around junction electron combine with holes to, form a small region, free of charge carriers is called, depletion region., The large electric field of intensity (E) is directed, from n-type semi conductor to p-type at the, junction., 181
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , , If potential barrier is Vb, width of depletion layer, Vb, ., d, Only inside depletion layer the +ve and -ve, immobile ions exist whereas outside this layer the, material is neutral, Potential barrier depends on nature of semi, conductor crystal, temperature and amount of, doping, The symbol of p-n junction diode is given below, n, P, , is d then E , , , , , , , , , , Direction of diffusion current is from p- type to, n-type, The current flows through the diode in the below, circuit is, I, R, , VB, , E, , I, , E VB, R rf, , Where R= external resistance,, rf = resistance of diode in forward bias,, , Semi conductor diode :, 1) Forward bias :, , VB =barrier potential., , v, , i) Power developed across the diode VB I, ii) Power developed across the resis, W, p, , , , , , , , –, –, –, –, –, , +, +, +, +, +, , n, , When an external voltage V is applied across a, semiconductor diode such that p-type is connected, to +ve terminal and n-type to -ve terminal of, battery ( in general p - type to high voltage and, n-type to low voltage), the diode is said to be, forward biased., External voltage V is greater and opposite to, barrier potential Vb. So width of depletion layer, and resistance decrease., Effective barrier voltage under forward bias is VbV., , Vb, , 1, 2, 3, , , , , , tor E VB I ,, The external voltage beyond which diode current, start increasing rapidly is called knee voltage, (Vknee), The below diagram show forward bias of junction, diode., a) 5v, , 2v, , –5v, , b) –3v, , –5v, c), , 2v, d), , 2) Reverse bias :, , , , , 182, , Fig. p-n junction diode under forward bias. Barrier, potential (1) without battery, (2) low battery, voltage, and (3) high voltage battery., Resistance of ideal diode in forward bias is zero, If external voltage (V) is greater than barrier, voltage then majority charge carriers diffuse across, the junction and constitute diffusion current (I =, Ie + Ih ), , v, , W, p, , –, –, –, –, –, , –, –, –, –, –, , –, –, –, –, –, , +, +, +, +, +, , +, +, +, +, +, , +, +, +, +, +, , n, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , , , , , , SEMI CONDUCTOR DEVICES, , When an external voltage V is applied across a, semiconductor diode such that p-type is connected, to -ve terminal and n-type to +ve terminal of, battery ( in general p - type to low voltage and, n-type to high voltage), the diode is said to be, reverse biased., External voltage V is greater and in same direction, to barrier potential Vb, so width of depletion layer, and resistance increase., Effective barrier voltage under reverse bias is, , I(mA), , 100, 80, 60, 20, 100 80 60 40 20, Vbr, , 2, 1, , Vb, , , , , , , , , 30, I(A), , Fig : Avalanche breakdown, Note:The potential barrier existing across an unbiased, i), , Fig : Diode under reverse bias, Barrier potential, under reverse bias., Resistance of ideal diode in reverse bias is infinity, upto a large external voltage (V)., At lower external voltage few covalent bonds are, broken to liberate electrons and holes and these, constitute reverse saturation current, At very high reverse bias voltage all the covalent, bonds are broken to liberate large number of, electrons called as breakdown voltage., The below diagram show reverse bias of p n, junction diode., , 0.2 0.4 0.6 0.8 1.0 V(V), , 10, 20, , Reverse bias, , Vb V ., , Forwardbias, , 40, , ii), , p-n junction is VB volt, The minimum kinetic energy required by a hole to, diffuse from the p-side to the n-side is ' eVB ', If the junction is forward biased at V volt, then the, minimum kinetic energy required by a hole to diffuse, , from the p-side to the n-side is e VB V , iii) If the junction is reverse biased at V volt, then the, minimum kinetic energy required by a hole to diffuse, from the p-side to the n-side is e VB V , , W.E-7:The V-I characteristic of a silicon diode is, shown in the Fig. Calculate the resistance of, the diode at a) I D 15mA and b VD 10V, I(mA), 30, Silicon, , a) 2v, , 5v, , –3v, , b) –5v, , 20, 15, 10, –10V, 0, , 2v, c), , –5v, d), , , Thus there is an unexpected release (avalanche), of large number of electrons and holes there by, sharp increase in current takes place at a voltage, called avalanche breakdown voltage., NARAYANAGROUP, , 0.5, , 0.8, , V(V), , –1A, , Sol. Considering the diode characteristics as a straight, line between I=10mA to I=20mA passing through, the origin, we can calculate the resistance using, Ohm’s law, a) From the curve at I=20mA,V=0.8V, I=10mA,V=0.7V, rfb V / I 0.1V / 10mA 10, b) From the curve at, V 10V , I 1 A, Therefore, rrb 10V / 1 A 1.0 10 7 , 183
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-8: Two junction diodes, one of germanium, (Ge) and other of silicon (Si) are connected, as shown in fig to a battery of 12V and a load, resistance 10k . The germanium diode, conducts at 0.3V and silicon diode at 0.7V., When current flows in the circuit, the, potential of terminal Y will be, , W.E-10: Find the voltage VA in the circuit shown, in figure . The potential barrier for Ge is, 0.3 V and for Si is 0.7 V, 24V, , Si, , Ge, , Ge, , Si, , VA, , 10k, , 12V, , Sol. The Ge diode conducts for a p.d of 0.3V, therefore, the current passes through it and the Si diode do, not conduct. Hence the potential of terminal, Y 12 0.3 11.7V, , 3k, , Sol. In the situation given, germanium diode will turn on, first because potential barrier for germanium is, smaller. The silicon diode will not get the opportunity, to flow the current and so remains in open circuit., The equivalent circuit is as in figure, 24V, , W.E-9:Find maximum voltage across AB in the, circuit shown in Fig. Assume that diode is, ideal, , 0.3V, , +, 30V, , 5k, A, , VA, , 10k, , 15k, 3k, , B, , Sol. As the diode is treated ideal, its forward resistance, R f zero . It acts as short circuit. So 10 k is in, , parallel with 15 k and the effective resistance, across AB is, RAB, , 10 15 10 15, , , 6k , 10 15, 25, , 6k is in series with 5k, total resistance, = RT 6k 5k 11k ,, , VA 24 0.3 23.7V, , W.E-11:Considering the circuit and data given in, the diagram, calculate the currents flowing, in the diodes D1 and D2 Forward resistance, of D1 and D2 is 20, 100, +, 1V –, , D1, , D2, , V=30V. Current drawn from the battery is, I, , V, 30V, , 2.72mA, RT 11k , , VAB IRAB 2.72mA 6k 16.32V, 184, , Sol. Since the positive terminal of battery is connected, to P-type of both diodes D1 and D2 , they are, forward biased. These diodes are replaced by with, their forward resis tance as shwon is Fig:, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-13: A potential barrier of 0.50V exists across, , 100 I, I2, , I1, +, 20, , 1V, , 20, , The resistance of 20 and 20 in parallel,, 1, 1, 1, 20, , or R , 10, R 20 20, 2, Therefore, total current I in the circuit, I, , a p-n junction, a) If the depletion region is 5.0 10 7 m, wide,what is the intensity of the electric field, in this region?, b) An electron with a speed of 5.0 105 m / s, approaches the p-n junction from the n-side., With what speed will it enter the p-side ?, Sol. a) The electric field is E=V/d, , , 1, 1, , amp, 100 10 110, , 0.50V, 1.0 106 V / m, 7, 5.0 10 m, –, , 1 1, 1, , amp, and I1 I 2 , 2 110 220, , b), , +, , p, , n, , v2, , v1, d, , W.E-12:In the circuit shown, the potential drop, , Let the electron has a speed v1 when it enters the, , across each capacitor is ( assuming the two, diodes are ideal ), C1 = 4F, , depletion layer and v2 when it comes out of it. As, the potential energy increases by e 0.50 V . From, the principle of conservation of energy, , D2, , 1, 1, m12 e 0.50 m22, 2, 2, , +, D1, , E = 24V, , C2 = 8F, , 2, 1, 9.1 1031 5 105 , , 2, , Sol. The diode D1 is reverse biased (open circuit), but, , 1, 1.6 1019 0.5 9.1 10 31 22, 2, , the diode D 2 is forward (short circuit)., , 4F, , +, , Solving this. 2 2.7 105 m / s, , Application of junction diode as a, Rectifier: The process of conversion of ac to, , 8F, , 24V –, the potential of the battery divides across the, two capacitors in the inverse ratio of their capacities., , i.e,, , V1 C2 8 2, , , V2 C1 4 1, , V1 , , 2, 2, E 24 16V, 3, 3, , 1, 1, V2 E 24 8V, 3, 3, NARAYANAGROUP, , dc is called rectification, tha arrangement is called, rectfier. They are, , , Half wave rectifier, P, , n, , AC, , ~, , RL, , O/P, , Half Wave Rectifier, , VO, , Vi, , t, I/P, , O/P, , t, , 185
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JEE-ADV PHYSICS- VOL- V, , , SEMI CONDUCTOR DEVICES, , In full wave rectifier the efficiency is, , 0.812, , RL, rf RL . Its maximum value is 81.2%, , (where rf forward resistance of diode and, , , RL load resistance.), Maximum current, Im , , , , I rms , , , , , P, 2 I m / RL 0.812RL 0.812, d .c. , r, rf RL, Pa.c. I m 2, 1 f, , rf RL , RL, 2, , Vm, Vm is max imum voltage , RL rf, , , , , , 2, , I , Pa.c m rf RL , 2, Full-wave rectification efficiency,,, 2, , 2I, Mean dc current I dc m, , , , , , For a full-wave rectified cycle, I r .m.s I m / 2, , Im, , 2, Value of ripple factor = 0.482, The ripple frequency is twice to the frequency of, applied emf., The value of dc component in out put voltage is, more than the ac., 2, Input ac power = I rms RL rf, , The efficiency will be maximum if rf is zero, maximum efficiency = 81.2%, The efficiency of a full-wave rectifier is double the, efficiency of a half-wave rectifier, Full-wave bridge rectifier: In a centre tap, full-wave rectifier, it is difficult to locate the centre, tap on the secondary winding, which can be, overcome in bridge rectifier. The circuit is shown, in Figure Four diodes D1 , D2 , D3 and D4 are used, in the circuit., , , , P, D4, , D1, , Out put dc power = I dc2 RL, , AC, supply, , –, RL, , Efficiency of full-wave rectifier:, (Formula derivation), , Vout, , D3, , D2, , +, , Q, , Efficiency of full-wave rectifier is given by, , Vin, , Secondary voltage, , V, +, , , , –, , , d, , Pd .c, Pa .c, , t, –, , Vout, , , D1 D3, , , Instantaneous current is i r R, f, L, Average d.c. current is I d .c , , D2 D4, , D1 D3, , D2 D4, , t, , Ripple factor: The ratio of r.m.s value of a.c, component to d.c. component in the rectifier output, is called ripple factor., , 2Im, , , 2, , 2, , 2I m , d.c . power output I d .c RL , RL, , 2, , 2, a.c. input power is given by Pd .c I r .m.s rf RL , , NARAYANAGROUP, , +, , I a .c., I r . m. s , Ripple factor I , 1, d . c., I d .c. , , Ripple factor decides the effectiveness of a rectifier., The smaller value of ripple factor shows lesser a.c., component; hence more effectiveness of rectifier., 187
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, i) For half-wave rectification,, , Im, 2, , ;, , I d .c. , , 1.21, , Im, Im , , as I r .m.s 2 and I d .c , , , ii) For Full-wave rectification:, I r . m. s ., , , I d .c. 2 2, , Im, 2Im , , as I r .m. s 2 and I d .c , , , , The role of capacitor in filtering:, , , , , , 188, , output with, capacitor, input filter, , Form factor: It is ratio of I r .m.s . and I d .c., I, , F r .m.s 1.57, I d .c . 2, , , , dc, , RL, , a) Full wave rectifier with capacitor filter, , 2Im, , , i) For half-wave rectification:, , , , –, , IL, , (a), , Im / 2 , 1 0.48, Ripple factor , 2Im / , , F, , C, , Y, , 2, , , , +, , ac input, , I r . m. s . , , Rectifier, , ac, , 2, , Im / 2 , Ripple factor , 1, I, /, , m, , ii) For full-wave rectification, In full-wave rectification ,, , dc component, , x, , I, I, I r .m.s . m ; I d .c. m, 2, , , When the voltage across the capacitor is rising, it, gets charged. If there is no external load,it remains, charged to the peak voltage of the rectified output., When there is a load,it gets discharged through the, load and the voltage across it begins to fall.In the, next half-cycle of rectified output it again gets, charged to the peak value in figure., The rate of fall of the voltage across the capacitor, depends upon the inverse product of capacitor C, and the effective resistance RL used in the circuit, and is called the time cosntant., To make the time constant large value of C should, be large.So capacitor input filters use large, capacitors .The output voltage obtained by using, capacitor input filter is nearer to the peak volatge, of the rectified voltage.This type of filter is most, widely used in power supplies., , t, , (b), , b) Input and out put voltage of rectifier in (a), , W.E-14:A p-n diode is used in a half wave rectifier, with a load resistance of 1000 . If the, forward resistance, , r , f, , of diode is, , 10 ,calculate the efficiency of this half wave, rectifier., Sol. Load resistance RL 1000, Forward resistance of the diode rf 10, Efficiency of half wave rectifier, 0.406 RL 0.406 1000, 0.4019, , , 1010, rf RL , The percentage efficiency of the half wave, rectifier 40.19%, , W.E-15:Afull wave rectifier uses two diodes with, a load resistance of 100 . Each diode is, having negligible forward resistance . F i n d, the efficiency of this full wave rectifier., Sol. Forward resistance of the diode rf 0 ,, ; Load resistance, RL 100; ?, efficiency of full wave rectifier, 0.812 100, 0.812, 100, The percentage efficiency of the full wave rectifier, = 81.2%, , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , W.E-16: If a p-n junction diode, a square input, , SEMI CONDUCTOR DEVICES, , , Symbol of zener diode is, , , , Zener diode is used as voltage regulator its circuit, diagram is, , signal of 10V is applied as shwon, +5v, , RL, –5v, , Then the out put signal across RL will be, 1), , 2), –5V, , –10V, 4) 10V, , 3) 5V, , 1) I I Z I L, , 2) VO / P VZ I L RL, , 3) VI / P IR VZ, , VI /P IR VZ VO / P VI / P I Z I L R, As VI / P changes current through zener diode, Sol. The junction diode will conduct when it is forward, biased. Therefore, the output voltage will be, obtainded during positive half cycle only . So option, is (3), , p-n junction diodes are specially used, as :, , , , change so that VO/ P (or) VZ remain constant., W.E-17:For the circuit shown in figure,Find, 1)the output voltage;, 2)the voltage drop across series resis, tance;, 3)the current through Zener diode., , 1) Zener diode, 2) Opto electronic junction devices, , +, , 1) Zener diode:A heavily doped p-n junction, , 120V, , diode used to operate in reverse bias is called, zener diode., In Zener diode even at low reverse voltage, direct, breakage of covalent bonds due to very strong, electric field releases large number of electron hole pairs. This voltage where current increases, sharply is called as zener break down voltage., I(mA), , R = 5k, , IL, IZ, 50V, , , , Sol. From the figure R 5k 5 103 ;, input voltage Vin 120V ; zener voltage ,Vz 50V, 1) Output voltage Vz 50V, 2) Voltage drop across series tesistance R=, Vin Vz 120 50 70V, V, , 80, 60, Forwardbias, , 40, 20, , 0.2 0.4 0.6 0.8 1.0 V(V), , 20, , NARAYANAGROUP, , Current through R= i , , , 100 80 60 40 20, , Reverse bias, , 50, , 3, z, 3) Load current I L R 10 103 5 10 A, L, , 100, , 10, , RL, 10k, , 30, I(A), , Vin Vz, R, , 70, 14 10 3 A, 5 103, , According to Kirchoff’s first law I I L I z, Zener current, I Z I I L 14 103 5 103 9 103 9mA, 189
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 2) Opto electronic junction devices, Photodiode, , Light Emitting Light (LED), , hv, n, , n, , p, A, , p-side, , , , , n-side, , Fig : An illuminated photodiode under reverse bias, Photocurrent is proportional to incident light, intensity., Photodiode can be used as a photodetector to, detect optical signals., , , , , , , mA, , Light-emitting diode (LED) is a forward-biased, p-n junction diode which emits visible light when, energised ., The energy of radiation emitted by LED is equal, to or less than the band gap of the semiconductor., The band width of emitted light is 100 A0 to 500 A0, or in other words it is nearly (but not exactly), monochromatic., , Reverse bias, , R, , I1, I2, I3, I3 > I2 > I1, , G, , B, , I, , A, , Fig : I-V characteristics of a photodiode for, , O, , different illumination intensity I 3 I 2 I1 ., , Solar Cell:, , to be more mA than the current in the, reverse bias A . What is the reason then, to operate the photodiodes in reverse bias ?, Sol. Consider the case of an n-type semiconductor. The, majority carrier density (n) is considerably larger, than the minority hole density p(n>>p) On, illumination, let the excess electrons and holes, generated be n and p , respectively, p1 p p, , n1 n n ;, 1, , Here n1 and p are the electron and hole, concentrations at any particular illumination and n, and p are carriers concentration when there is no, illumination. Remember n p and n p ., Hence, the fractional change in the majority carriers, (i.e., n / n )would be much less than that in the, minority carrier dominated reverse bias current is, more easily measurable than the fractional change, in the forward bias current. Hence,photodiodes are, preferably used in the reverse bias condition for, measuring light intensity, , V, , The I-V characterstics of L.E.D:, , W.E-18: The current in the forward bias is known, , 190, , Y, , volts, , IL, , p, , n, , Depletion, region, , , , , , , Fig : A typical illuminated p-n junction solar cell, Unlike a photodiode, a solar cell is not given any, biasing. It supplies emf like an ordinary cell., Sunlight is not always required for a solar cell., Semiconductors with band gap close to 1.5ev. are, ideal materials for solar cell fabrication, Si and GaAs are preferred material for solar cells., I, , VOC (open circuit voltage), , ISC, , Short circuit current, , Fig : I-V characteristics of a solar cell, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , , , Selection of Solar material:, i) band gap ( 1.0 to 1.8eV), ii) high optical absorption (10 4 cm 1 ), iii) good electrical conductivity, iv) availability of the raw material and, v) cost., , , , , , , , , , , , , Transistors: Transfer + resistor = Transistor, , Transistor Configurations: In electronic, , Transistors are current operated solid-state devices., Silicon is the element from which most of the, transistors and other semiconductor components, are made today., Transistor has three regions known as the emitter, (E), base (B) and collector (C), 1) Emitter is heavily doped and medium in size., 2) Base of a transistor is lightly doped and very, thin., 3) Collector of a transistor is moderately doped, and large in size., Transistor has two junctions, 1) emitter-base junction, 2) Collector-base junction, In a transistor the emitter base junction is forward, biased and the collector base junction is reverse, biased., , circuits transistors are connected in three ways., They are, 1) Common base configuration ,, 2) Common emitter configuration,, 3) Common collector configuration, Common Base Configuration: In this, configuration base is common to both input and, output., Base terminal is earthed and input is given across, base - emitter and output is taken across base collector as shown in figure, , This mode is called grounded base configuration., , In an n-p-n Transistor:, Emitter, , Collector, , E, , , , , E, Input, , , , p n, B, , p, , C, Output, , C, , E, Input, , B, , Output, , Current amplification factor of common base, , , , , IC , configuration for ac I , E constant V, Current amplification factor of common base, I, c, configuration for dc, IE, Values of range from 0.95 to 0.99., Input resistance of transistor in CB configuration, , , , VBE , is Rin I , , B V, Output resistance of transistor in CB configuration, , CB, , , , B, , Base, , , , , , , C, , N P N, , , The conventional current is from emitter to base., The emitter junction is forward biased and the, collector junction is reverse biased., Here also IE = IC + IB, IC is 97 to 98% of IE and IB is 2 to 3% of IE., Thus the collector current is less than the emitter, current ( IC < IE), , The current is due to electrons inside and outside, the n-p-n transistor and they are the majority charge, carriers, The conventional current flows from base to emitter., The emitter junction is forward biased and the, collector junction is reverse biased., The emitter current ( IE) is the sum of base current, ( IB) and collector current ( IC ), i.e.,IE = IB + IC, IC is 97 to 98% of IE and IB is 2 to 3% of IE., , CB, , In a p-n-p Transistor:, , VCB , is Rout I , , C I, Voltage gain = current gain x resistance gain., R, Av out, Rin, Power gain = Voltage gain x current gain., E, , Emitter, , P, , Collector, , C, , N P, Base, , , , E, , , , B, , The current is due to holes inside as they are the, majority charge carriers and due to electons outside, the p-n-p transistor., NARAYANAGROUP, , , , Ap Av 2 , , Rout, Rin, 191
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , Common Emitter configuration: In this, , , , configuration emitter is common to both input and, output., The emitter is earthed and input is given across base, - emitter and output is taken across collecter emitter as shown in fig, This mode is called grounded emitter configuration., C, B, Input, , VCE, , Current amplification factor of common emitter, I, , , , , , , , Power gain = Voltage gain x current gain., Rout, Rin, , , , configuration collector is common to both input and, output., The collector is earthed and input is given across, base - collector and output is taken across emitter, -collector as shown in figure, This mode is called grounded collector, configuration., E, B, Input, , Relation between & :I C, , ac current gain in C.B , = I, E, ac current gain in C.E , = I, B, I C, I, I C, I E, , C , , , I B I E IC 1 I C 1 , I E, , , (or) 1 , 1 , , , , I E, 1, , , 1, I B 1 , , Comparative study of CB, CE, CC, configurations, , 1., , Input, resistance, , Minimum, (50–20 ), , Medium, (1–2K ), , Maximum, (150–180K ), , 2., , Output, resistance, , Maximum, (1–2M ), , More, (50K ), , Minimum, (1K ), , 3., , Current, gain, , Minimum, = 0.95 – 0.99, , More, = 20 – 500, , Maximum, = 20 – 500, , 4., , Voltage, gain, , Medium, , Maximum, , Minimum, , 5., , Power, gain, , Medium, (20–30), , Maximum, (30–40), , Minimum, (10), , 6., , Useful in, application, of, , Current, , Power, , Impedance, matching, , 7., , Phase, difference, , 0, , rad (or) 180°, , 0, , E, , p, Output, , n, p, , B, C, , 192, , B, , CB, CE, CC, S.No Parameter configuration configuration configuration, , Common collector configuration: In this, , , VCE , is Rout I , , E I, , Relation between , and :, , R out, Rin, , A p Av 2 , , , , Output resistance of transistor in CCconfiguration, , , , , , IB, , Voltage gain = current gain x resistance gain., Av , , , , CE, , Output resistance of transistor in CE configuration, is R out, , Input resistance of transistor in CC configuration, , I C, , Values of range from 20 to 500., Input resistance of transistor in CE configuration, , VCE, , IC, , IE, IB, , CE, , , , Current amplification factor of common emitter, , V, , , , configuration for dc, V, , E, , BE, is Rin I , , B V, , Current amplification factor of common collector, , BE, is Rin I , , B V, , c, configuration for dc I, B, , , , , , , Input, , IC , configuration for ac I , , B constant, , , , CE, , Output, , B, E, , , , I , , Output, , n, p, , Current amplification factor of common collector, E, configuration for ac I , B constant V, , , , C, , p, , , , Output, , Input, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , Characterestics of a transistor:, , , , n–p–n Transistor (C.E), , collector current I c is many times greater than, , IC, , B, +A–, , R2, VBB, , 1), , base current I b . A small change in input current, , – +, mA, C, , IB, , Ib , , R1, , +, , E VCE –, IE, , VBE, , VCC, , Input characterestics: Input characterstics are, drawn between VBE verses IB at constant VCE ., , , , , The output voltage Vce is fixed (say at zero volts)., , RB, , VCE = 0, , VCE = 10V, , Vi, , , , IB, , , VBE (volt), , 2) Output characterestics:Out put characterstics, are drawn between VCE verses IC at constant IB, Active region, IB = 60A, 50A, , Saturation, region, , 5, 4, , 30A, , 2, , 20A, 10A, IB = , , 1, , C, , RC, , E, IE, , VBB, , +, V0, –, , VCC, , Appling KVL to the input and output sides of the, circuit, We get, Vi I B RB VBE and VO VCC I C RC, In the case of Si transistor, as long as input Vi is, less than 0.6 V, the transistor will be in cut off state, and current IC will be zero. Hence Vo = VCC When, Vi becomes greater than 0.6 V the transistor is in, active state ,so Vo decreases linearly till its value, becomes less than about 1.0 V., Graph:, Cutt off, region Active, V0, region, , 40A, , 3, , B, , IB, , R2, , values (say 10V, 20V etc.) of Vce ., , 6, , current I c . This happens when emitter junction, is forward biased and collector junctions is reverse, biased., The transistor works as an amplifier when operated, in the active region., When the transistor is used in the cut off (or), saturation state it acts as a switch, , IC, , upto 1 volt and the corresponding base current I b, is noted down. This process is repeated for different, , IC(mA), , produces a large change in the output, , Transistor as a switch, , The input voltage Vbe is changed in steps (say 0.1V), , (A), , Active (or) Linear region: In this region, , Saturation, region, , AV, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9 VCE(V), , Cutt-off region, , , , , , Saturation region :- In this region the collector, current becomes almost independent of the base, current. This happens when both junctions are, forward biased., Cut off region:- In this region the collector curent, is almost zero. This happens when both the junctions, are reverse biased., NARAYANAGROUP, , Vi, , , , If we plot the V0 vs Vi curve [also called the, transfer characterstics of the base biased transistor, in figure,we see that between cut off state and active, state and also between active state and saturation, state there are regions of non-linearty showing that, the transition from cut off state to active state and, from active state to saturation state are not sharply, defined., 193
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , Transistor as an Amplifier, , W.E-19:Current amplification factor of a common, , (CE configuration), , base configuration is 0.88. Find the value of base, current when the emitter current is 1 m A., Sol. In a common -base arragement, the current, , The process of raising the strength of a weak input, signal to a strong output signal is called, ‘amplification’., Amplifier : It is a device which increases the weak, input signal into strong output signal, Amplifier has wide applications in industries, T.V,, radio and communication systems., Amplifiers are of two types, 1) Power amplifiers 2) Voltage amplifiers, 1) Power amplifier: Amplifier which is used to, raise the power level is known as “Power amplifier.”, 2) Voltage amplifier: The amplifier which is used, to raise voltage level is known as voltage amplifier., IB, , IC, , IB, , Vi, ac input, ~, , VBE, , VCE, , V0, RL ac output, , I C , IC, , , , , amplification factor, I E VCB I E, , Given 0.88, I E 1mA, Collector current, I C I E 0.88 1 0.88mA, Now since I E I B I C, , I B I E I C 1 0.88 0.12mA, , W.E-20: In a transistor, the emitter circuit, resistance is 100 and the collector resistance, is 100 . The power gain, if the emitter and, collector currents are as sumed to be equal,, will be, Sol. If I C I B 1, R, AP 2 L, Ri, , Fig:N-P-N TRANSISTOR AS AMPLIFIER, , i), , Voltage gain: It is defined as the ratio of change, in output voltage to the change in input voltage., , V, R (I ), R, Av CE L c L, VBE, Ri (Ib ), Ri, Negative sign indicates input and output voltages, are in opposite phase., ii) Power gain: It is defined as the ratio of output, power to the input power., , Note-1:, In common base amplifier, the phase difference, between the input and output signals is zero, Note-2:, In common emitter amplifier, the phase difference, between input and output signals is , 194, , 100 103, 1000, , 100, , , the transistor when VCE is 10V and I c 4.0mA, Sol. Consider any two characteristics for two values of, I B which are above and below the given value of, I C , Here I C 4.0mA., (Choose characteristics for I B 30 and 20 A ), At VCE 10V we read the two values of I C from, the graph Then, , Collector current (I C) in mA, , Rout, Rin, , RL, , Ri, , WE:21:From the output characteristics shown in, Fig .Calculate the values of ac and dc of, , output power I outVout, power gain = input power I V, in in, Power gain= current gain × voltage gain, Ap Av 2 , , Base current, , Base current, , 10, , 60A, 8, 50A, 6, , 40A, 30A, , 4, , 20A, 2, , 10A, 0, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , Collector to emitter voltage (VCE) in volts, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , I B 30 20 A, I C 4.5 3.0 mA 1.5mA, Therefore, i , ac c 1.5mA / 10 A 150, I B VCE, , For determining dc calculate the two values of dc, for the two characteristics chosen and find their, mean . Therefore for, I C 4.5mA and I B 30 A, , The value of input voltage below which the transistor, remains cutoff is given by VIL 0.6V ,VIH 2.8V ., Between 0.0V and 0.6V, the transistor will be in, the ‘switched off state. Between 2.8V and 5.0V,it, will be in ‘switched on’ state, Note that the transistor is in active state when I B, varies from 0.0mA to 0.20 mA., In this range, I C I B is valid. In the, In this range, I C I B is valid., In the saturation range, I C I B, , W.E-23:Two amplifiers are connected one after the, , I, dc C 4.5mA / 30 A 150 and for, IB, , other in series (cascaded). The first amplifier has, a voltage gain of 10 and the second has a voltage, gain of 20. If the input signal is 0.01 volt,, calculate the out put ac signal., Sol. When the amplifiers are Connected in series, the, net voltage gain is equal to the product of the gains, of the individual amplifiers., Av Av1 Av11 , hear Av1 10 , and Av11 20, , I C 3.0mA and I B 20 A, , dc 3.0mA / 20 A 150, Hence dc 150 150 / 2 150, , W.E-22:In Figure the VBB supply can be varied, form 0V to 5.0V. The Si transistor has, , dc 250 and Rc 1K , VCC 5.0V . Assume, that when the transistor is saturated, VCE 0V, , Voutput, also Av V, input, Voutput, Vinput, , we can write, , Av1 Av11 : Vin 0.01V, , and VBE 0.8V . Calculate (a) the minumum, base current, for which the transistor will, reach saturation. Hence, (b) determine V1, when the transis tor is switched on(c) find the, , W.E-24: In a single state transistor amplifier,, , ranges of V1 for which the transistor is, switched off and switched on., , 1mA. If collector load RC 2k and, , RB, , B, IB, , IC, , C, E, , RC, , V0, , Vout Vin Av1 Av11 0.01 10 20 2V, , when the signal changes by 0.02V, the base, current by 10 A and collector current by, RL 10k , Calculate: (i) Current Gain, , (ii)Input impedance,(iii) Effective AC load,, (iv)Voltage gain and (v) Power gain., , IE, VBB, , VCC, , V1, , Sol. Given at saturation VCE 0V , VBE 0.8V, , VCE VCC IC RC I C VCC / RC, 5.0V /1.0 K 5.0mA, Therefore, I B IC / 5.0mA / 250 20 A, , The input voltage at which the transistor will go, into saturation is given by, VIH VBB I B RB VBC, 20 A 100 K 0.8V 2.8V, NARAYANAGROUP, , i, , 1mA, , c, Sol. i) Current Gain i 10 A 100, b, ii) Input impedance, , Ri , , VBE, 0.02, , 2000 2k , ib, 10 A, , iii) Effective (a,c) load, 2 10, 1.66k , 2 10, RAC 100 1.66, 83, iv) Voltage gain AV R , 2, in, , RAC RC || RL, , RAC , , v) Power gain, Ap Current gain Voltage gain, 100 83 8300, 195
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , Fig. illustrates the principles of feedback., , W.E-25: An n-p-n transistor in a common emitter, mode is used as a simple voltage amplifier, with a collector current of 4mA. The terminal, of a 8V battery is connected to the collector, , , es, , , e0, , , , es, , through a load resistance RL and to the base, through a resistance RB . The collector-emitter, , , , The gain of the amplifier without feedback is A. If, a signal eS is applied at the input terminals of the, amplifier, then let the output voltage be e0., , , , If a fraction ( , , VBE 0.6V and base current amplification, and RB, IB, , IC, , RB, , +, , RL, , C, , 8V, , B, E, , Sol., , , , VCE, VCE, , VBE, , VBE, , 8V VCE, , 8V 4V 4V, , Now I C RL 4V, , VCC I B RB VBE, or I B RB = Potential difference across RB, , , , , , , VCC VBE 8. 0.6 7.4V, Again , I B , , I C 4 103, , 4 10 5 A, , 100, , 196, , , , , 7.4, 1.85 105 185k , 4 105, , Concept of feedback: When a part of the, output voltage (or current) of an amplifier is injected, back into the input circuit, feedback is said to exist., If the voltage feedback is in phase with the applied, voltage, the feedback is said to be positive or, regenerative;, If the voltage feedback is in opposite phase to the, incoming signal, the feedback is said to be negative, or degenerative., , e0, A, , e s 1 A, The positive feedback thus, increases the gain of, the amplifier. If too much positive feedback is, applied so that 1 A = 0, the gain of the amplifier, becomes infinite., For stable oscillation A 1 (Barkhausen’ss, criteria), In this case the amplifier becomes unstable and, the output can be obtained with no external input, signal, i.e., the amplifier becomes an oscillator., In the case of negative feedback ei es e f, In the case of negative feedback the voltage, feedback e0 is in opposite phase to the applied, voltage eS, so that gain with negative feedback, Af , , 4, RL , 103 1k , 4 103, Further for base emitter equation, , , , ) of this output voltage is, e0, fedback into the input in phase with the applied, signal, then the actual input voltage of the amplifier,, ei es e f, ei = es + e0 (for positive feed back), This total input voltage multiplied by the gain A of, the amplifier must be equal to the output voltage, , e0 A e 0 Aes e0 1 A Aes f r o m, which the gain Af with feedback is,, , Potential difference across RL, , , , ef, , i.e. e0 Aei e0 A es e0 Aes Ae0, E, , RB , , e0, , , , voltage VCE 4V , base-emitter voltage, factor d .c 100 .Calculate the values of RL, , A, , e0, A, becomes, A f e 1 A, s, The negative feed back also reduces noise &, distortion in an amplifier., Note:, i), , When 1 A 1, Af | A , feed back is negative, , ii), , When 1 A 1, Af | A , feed back is positive, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , Digital Eletronics, , Transistor as an oscillator:The simplest, eletrical oscillating system consists of an inductance, L and capacitor C connected in parallel., , L, , 1) Binary number system: It is a two-valued, , C, , , , , Once an eletrical energy is given to the circuit, this, energy oscillates between capacitance (in the form, of eletrical energy) and inductance ( in the form of, magnetic energy) with a frequency , , , , , 1, , 2 LC, The amplitude of oscillations is damped due the, presence of inherent resistance in the circuit, In order to obtain oscillations of constant amplitide,, an arrangement of regenerative or positive, feedback from an output circuit to the input circuit, is made so that the circuit losses may be, compensated., , W.E-27:Convert the decimal number 23 into, binary number., 2 23, 2111, 2 5 1, , Sol., , Af , , A, 100, , 20, 1 A 1 (1/ 25) 100, , 1, 25, iii) Out put voltage, , ii) , , , , , , of 23 is found to be 10111. i.e., 23(10) 10111(2), In binary representation of any number, the first bit, is the ‘most significant bit’ (MSB) and the last bit is, the ‘least significant bit’ (LSB)., Binary number can be converted into decimal, number by multiplying each bit with 2n and adding, them as mention below. where n is positon of the, bit from right side (LSB) to leftside (MSB)., , 4 3 210, , Sol. 10111(2) 1 24 0 23 1 22 1 21 1 20, 16 0 4 2 1 23(10), , 1, 1 0.04 volt, 25, , 10111(2) 23(10), 1, i, , v) New increased input voltage V Vi 1 A, , NARAYANAGROUP, , Arranging the remainders in the reverse order(from, bottom to top shown below) the binary equivalent, , decimal number., , iv) Feedback voltage, , 1, , , 50 1 100 250mV, 25, , , 2 0 1, , W.E-28: Convert the binary number 10111 into, , V0 ' A f Vi 20 50mV 1volt, , V0 ' , , 2 2 1, 2 1 0, , W.E-26:In a negative feedback amplifier, the gain, without feedback in 100, feed back ratio is 1/, 25 and input voltage is 50m V. Calculate, (i)gain with feedback (ii) feedback factor, (iii) output voltage (iv) feedback voltage, (v) new input voltage so that output voltage, with feedback equals the output voltage, without feedback, Sol. i) Gain with feedback, , system developed by George Boole. Only two, digits 0 and 1, called bits, are used in binary system., But binary addition (in this 1+1=10) is different from, addition in Boolean algebra (in this 1+1=1 and, remaining will be same i.e., 0+0=0, 1+0=1,, 0+1=1)., A number in a decimal system can be converted, into binary by the successive division of 2 until the, quotient is zero. The remainders obtained in the, successive divisions, taken in the reverse order, (from bottom to top shown below) give the binary, representation of that number., , 2) Boolean Algebra: Only two states or values of, a varible are allowed in Boolean algebra. In logic, these two states correspond to ‘on’ and ‘off’ and, ‘saturation’ of electronic devices., 197
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , , , , , , , , , , , , The two allowed states in Boolean alegbra are, represented by the digits 0 and 1. 0 can also, represent Off, low, false,No, 0 V 1 can also, represent ON, high, true, Yes, 5 V, The variables of Boolean algebra are subjected to, three operations., The OR addition indicated by a plus (+) sign., The AND multiplication indicated by a cross (X), or a dot (.) sign., The NOT operation indicated by a bar over a, variable., OR Addition: The + sign in Boolean algebra, represents OR addition. The equation Y = A+B is, read as ‘Y equals A OR B’. The, AND multiplication: In Boolean algebra the, equation Y=AXB or Y=A.B or Y=AB is read as, ‘Y equals A AND B’, NOT operation: The NOT operation on a, , Theorem 1: The complement of the sum of two, (or more) variables is equal to the product of, complements of the variables i.e. A B A B, , , complements of the variables i.e A B A B ., , , Logic gates: A digital circuit having a certain, logical relationship between the input and the output, voltages is called a logic gate., There are three basic logic gates, (1) OR gate (2) AND gate (3) NOT gate, , , , , , , , AND, , NOT, , 0+0=0, , 0.0=0, , 0=1, , 0+1=1, , 0.1=0, , 1=0, , 1+0=1, , 1.0=0, , 1+1=1, , 1.1=1, , A D1, A, 5V, , B, , D2, , Symbol OR gate, , OR gate, , , , , , Associative laws:, , , Basic OR and AND Relations:, OR, A+0=A, A+1=1, A+A=A, , B, , Y, , 0V, , Some useful laws of Boolean algebra, Commulative laws: A+B=B+A; A.B=B.A, A+(B+C) = (A+B) +C ; A.(B.C) = (A.B).C, Distributive laws:A.(B+C) = A.B +A.C, , The OR gate, 1) It is a logic gate which has two or more inputs, and one output., , The equation Y A is read as ‘Y equals NOT A’., Rule for OR, AND and NOT functions in Boolean, algebra, OR, , , , A.B A B, , variable A is represented by A., , , , , A B A .B, Theorem 2 : The complement of the product of, two (or more) variables is equal to the sum of, , AND, A.0=0, A.1=A, A.A=A, , If its one or more inputs are high, then iis output, will be high. Therefore it has a logic of OR., Boolean expression for OR gate:A + B = Y, which reads as A OR B is equal to Y., Truth table for OR gate A, B, Y = A +B, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, The AND gate: It is logic gate which has two or, more inputs and one output, A, , D1, Y, , 5V, , B, , D2, , A, , Y, , B, , 0V, , A+ A =1, , A. A =0, , , , Double complement function: A A, , , , De Morgan’s theorems: DeMorgan’s, theorems (or rules) are very useful in simplifying, complicated logical expressions and can be stated, as under :, , 198, , Circuit of AND gate, , Symbol of, AND gate, , , , If its all inputs are high, then it is output will be high, (1). Therefore it has a logic of AND., , , , Boolean expression for AND gate:, A. B = Y reads as A AND B is equal to Y., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , , SEMI CONDUCTOR DEVICES, , Truth table for AND gate, A, B, Y = A. B, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, The NOT gate: It is the logic gate which has one, input and one output, , , , , , , VCC=+5V, RL, RB, , Y, , Y, , 1, 5V, 0, , , , Circuit of NOT gate, Symbol of NOT gate, If its input is high (1), then its output will be low(0)., Therefore it has a logic of NOT., , , , Boolean expression for NOT gate: A Y, , reads as ‘A NOT is equal to Y’., Truth table for NOT gate, A, YA, 0, 1, 1, 0, Combination of gates: In complicated digital, circuits used in calculators, computers etc. the, different types of combination of three basic logic, gates are used., NOR gate:This logic gate is the combination of, OR gate and NOT gate. OR + NOT = NOR, In this logic gate the output of OR gate is given to, the input of NOT gate as shown in the below figure., A, A, Y, Y, , , , B, OR, , , , , , B, NOT, , NOR gate, , Boolean expression for NOR gate:, Y A B . Which reads as A OR B negated., Truth table for NOR gate, AB, A+ B, Y A B, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, NAND gate:This logic gate is the combination of, AND gate and NOT gate., AND+NOT NAND, NARAYANAGROUP, , , , In this logic gate the output of AND gate is given to, the input of NOT gate as shown below, A, A, Y, Y, , B, B, AND, NOT, NAND gate, Boolean expression for NAND gate Y A.B, Truth table for NAND gate, AB, A.B, A.B, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, Uses of NOR gate and NAND gate: The, NAND gate and NOR gates are the building blocks, of digital circuits. All the basic gates (OR, AND, and NOT ) can be obtained by the repeated use of, NAND or NOR gates., , NOT gate from NAND gate:, 1) Diagram, B, , Y, , 2) Truth table, AB, Y A.A A, 0, 0, 1, 1, 1, 0, , AND gate from NAND gate, 1) Diagram, A, , B, 2) Truth table, A, B, 0, 0, 0, 1, 1, 0, 1, 1, , Y', , Y, , Y, 0, 0, 0, 1, , OR gate from NAND gate, 1) Diagram, , A, , A, Y, , B, B, 199
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 1) two input XOR gate, , 2) Truth table, A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , A, , Y A B, , B, , 1, 0, 1, 0, , 1, 1, 0, 0, , A, , A, , 0, 1, 1, 1, , B, A, , 2) circuit symbol, , 1) Diagram, , A, , A, , Y=A, , Y, , 2) Truth table, A, B, 0, 0, 1, 1, , B, , Y A.A A, 1, 0, , 3) truth table, , AND gate from NOR gate, , A, 0, 0, 1, 1, , 1) Diagram, , A, , A, , B, , B, , 2) Truth table, AB, Y, 0, 0, 0, 1, 1, 0, 1, 1, , Y=A.B, , , , , 0, 0, 0, 1, , , , 1) Diagram, A, A+B, , , 200, , Y, 0, 1, 1, 0, , OR, AND and NOT gates., It is also called exlusive NOR gate., The output of a two input XNOR gate is 1 only, when both the inputs are same., The Boolean equation is Y A.B A.B, XNOR gate is inverse of XOR gate., 1) Diagram, A, B, Y, , Y= A+B = A+B, Y, 0, 1, 1, 1, , XOR GATE: XOR gate is obtained by using OR,, , , B, 0, 1, 0, 1, , XNOR GATE: XNOR gate is obtained by using, , OR gate from NOR gate, , B, 2) Truth table, A, B, 0, 0, 0, 1, 1, 0, 1, 1, , Y, , Y, , B, , B, , NOT gate from NOR gate, , Y, , AND and NOT gates. It is also called exlusive OR, gate., The output of a two input XOR gate is 1 onl, y when the two inputs are different., The Boolean equation is Y A.B B. A, , 2) circuit symbol, , 3) truth table, A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 1, 0, 0, 1, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-29:The Boolean expression of the output Y, of the inputs A and B for the circuit shown in, the fig:, A, , A, , 2, , B, , then, A X A B. A B. A, , A. 1 B A.B A A.B, , 1, Y, , B, , A X A B A B. A A A A 1, , So X A A.B , which is AND gate with, inputs as A and B, , W.E-32:You are given two circuits as shown in, figure which consist of NAND gates. Identify, the logic operation carried out by the two, circuits., , Sol. The output of AND gate 1 is AB, The output of AND gate 2 is AB, the output of OR gate is Y AB AB, , A, , A, Y, , W.E-30:The diagram of a logic circuit is given, below. The output of the circuit is represented, by, w, X, F, w, Y, , Sol. F W X .W Y , , W .W W .Y X .W X .Y, W W .Y X .W X .Y, W 1 Y X .W X .Y W XW X .Y, W 1 X X .Y W X .Y, , W.E-31:The logic circuit and its truth table are, given, what is the gate X in the diagram, , Y, , B, , B, Sol. From fig(a). The output of NAND gate is, connected to NOT gate ( obtained from NAND, gate) Let Y 1 be the output of NAND gate and the, final output of the combination of two gates is Y., The output of a NAND gate is O only when both, the inputs are zero, while in NOT gate, the input, gets inverted. Truth table for the arrangement, , A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Y', 1, 1, 1, 0, , Y, 0, 0, 0, 1, , A, 0, = 1, 0, 1, , B, 0, 0, 1, 1, , Y, 0, 0, 0, 1, , It is the truth tables of AND gate. Therefore the, given circuit acts as AND gate., b) The output of two NOT gates are connected to, NAND gate Let Y1 and Y2 be the outputs of the, two NOT gates and the final output of the, combination of three gates be Y. In a NOT gate., the input gets inverted, while the output of a NAND, gate is ‘O’ only when both the inputs are zero., Truth table for given arrangement., , A, , Y, X, B, , Sol. From the truth table we note that Y A B, i.e., it is for OR gate (or), NARAYANAGROUP, , A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Y, 0, 1, 1, 1, 201
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , W.E-33:Justify the output waveform (Y) of the OR, , A B and A B as shown. These inputes when, applied to AND gate give the required output., , gate for the following inputs A and B given is, Fig, t1, , t2, , t3, t4, , t5, , Y A B . A B , , t6, , A+B, , A, , A, (input), , B, , B, , A+B, , INTEGRATED CIRCUITS: An entire, , Y, (0 input), , Sol. Note the following:, At t t1 ;, , A 0, B 0; HenceY 0, , For t1 tot2 ; A 1, B 0;, , HenceY 1, , For t2 tot3 ; A 1, B 1;, , HenceY 1, , For t3 tot4 ; A 0, B 1;, , HenceY 1, , For t4 tot5 ; A 0, B 0;, , HenceY 0, , For t5 tot6 ; A 1, B 0;, , HenceY 1, , For t t6 ; A 0, B 1;, , HenceY 1, , Therefore the wave form Y will be as shown in the, Fig, , W.E-34:Draw logic diagrams for the Boolean, expressions given below., i) A B A B Y ii) A B A B Y, Sol. (i) The required logic diagram for the given Boolean, expression is given in figure. Here the input B before, applying to first AND gate and input A before, applying to second AND gate have been inverted., The output of these gates are, therefore, AB and, AB respectively. These outputs are fed to OR gate, which gives Y AB AB as shown in figure., A, B, , , , , , , , , circuit ( consisting of many passive components, like R and C active devices like diode and, transistor) on a small single block (or chip) of a, semicon ductor is known as Integrated Circuit (IC)., The most widely used technology is the Monolithic, Integrated Circuit., The chip dimensions are as small as1mm 1mm or, it could even be smaller., Depending on nature of input signals, IC’s can be, grouped in two categories:, (a) linear or analogue IC’s and, (b) digital IC’s. The linear IC’s process analogue, signals varies linearly with the input., One of the most useful linear IC’s is the operational, amplifier., The digital IC’s process signals that have only two, value., They contain circuits such as logic gates. Depending, upon the level of integration (i.e,. the number of, circuit components or logic gates), the IC’s are, termed as, i) Small Scale Integration, SSI ( logic gates 10), ii) Medium Scale Integration, MSI, ( logic gates < 100), iii) Large Scale Integration, LSI, ( logic gates <1000), iv) very large scale integration VLSI, ( logic gates > 1000)., , AB, , BA, , Plastic, case, , ‘chip’, Connection, From‘chip, to pin, , Y =AB + BA, , ii) The required logic diagram for the given Boolean, expression is shown in figure. The input B to first, OR gate and input A to second OR gate have been, inverted. The output of these gates are, therefore,, 202, , Y =(A+B). (A+B), , metal, pin, , notch, Small, dot, , 0.1 inch (2.5mm), pin 1, , The casing and connection of a ‘chip’, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , C.U.Q, , SEMI CONDUCTOR DEVICES, 8., , INTRINSIC AND EXTRINSIC, SEMICONDUCTORS, 1., , 2., , 3., , 4., , 5., , 6., , 7., , Which property makes the crystalline solids, to have sharp melting point?, 1) Equal strength of all the interatomic bonds, 2) anisotropic, 3) long range order, 4) short range order, Which of the following is not a property of, crystalline substance, 1) sharp melting point 2) bounded by flat surface, 3) isotropic, 4) long range order, A strip of copper and another of germanium, are cooled from room temperature to 80K. The, resistance of:, 1) each of these decreases, 2) copper strip decreases and that of germanium, decreases, 3) copper strip decreases and that of germanium, increases, 4) each of these increases, The difference in the variation of resistance, with temperature in a metal and a, semiconductor arises essentially due to the, difference in the :, 1) crystal structure, 2) variation of the number of charge carriers with, temperature, 3) type of bonding, 4) variation of scattering mechanism with, temperature, The manifestation of band structure in solids, is due to :, 1) Heisenberg’s uncertainty principle, 2) Pauli’s exclusive principle, 3) Bohr’s correspondence principle, 4) Boltzman’s law, A solid which is not transparent to visible light, and whose conductivity increases with, temperature is formed by:, 1) ionic bonding, 2) covalent bonding, 3) vander Wall’s bonding 4) metallic bonding, To a germanium sample, traces of gallium are, added as an impurity. The resultant sample, would behave like :, 1) A conductor, 2) A p-type semiconductor, 3) A n-type semiconductor 4) An insulator, NARAYANAGROUP, , 9., , 10., , 11., , 12., , 13., , 14., , In semiconductors at a room temperature, 1) The valence band is partially empty and the, conduction band is partially filled., 2) The valence band is completely filled and the, conduction band is partially filled, 3) The valence band is completely filled, 4) The conduction band is completely empty, Identify the property which is not characteristic, for a semiconductor?, 1) at a very, low temperatures, it behaves like an insulator, 2) at higher temperatures two types of charge, carriers will cause conductivity, 3) the charge carriers are electrons and holes in the, valence band at higher temperatures, 4) the semiconductor is electrically neutral., Carbon, silicon, and germanium have four, valence electrons each. At room, temperature, which one of the following, statements is most appropriate?, 1) The number of free conduction electrons is, negligibly small in all the three., 2) The number of free electrons for conduction is, significant in all the three., 3) The number of free electrons for conduction is, significant only in Si and Ge but small in C., 4) The number of free conduction electrons is, significant in C but small in Si and Ge, The relation between number of free electrons, (n) in a semiconductor and temperature (T) is, given by, 1) n T 2) n T 2 3) n T 4) n T 3/ 2, An electrically neutral semi conductor has, 1) equal amounts of negative and positive charges, 2) no minority charge carriers, 3) no majority charge carriers 4) no free charges, There is no hole current in conductors because, they have, 1) high conductivity, 2) high electron density, 3) no valence band, 4) overlapping of valence and conduction bands., In the insulators, 1) the valence band is partially filled with electrons, 2) the conduction band is partially filled with, electrons, 3) the conduction band is partially filled with, electrons and valence band is empty, 4) the conduction band is empty and the valence, band is filled with electrons., 203
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SEMI CONDUCTOR DEVICES, 15. In semiconductors the forbidden energy gap, between V.B. and C.B. is of the order of, 1) 1eV, 2) 5eV 3) 1 keV 4) 1MeV, 16. The level formed due to impurity atom, in the, forbidden energy gap, very near to the valence, band in a p-type semicoductor is called, 1) acceptor level, 2) donar level, 3) conduction level, 4) forbidden level, 17. The bond in semiconductors is, 1) covalent 2) ionic 3) metallic 4) hydrogen, 18. On increasing temperature, the conductivity, of pure semiconductors, 1) decreases, 2) increases, 3) remains unchanged 4) becomes zero, 19. The mobility of free electrons is greater than, that of free holes because, 1) they carry negative charge, 2) they are light, 3) their mutual collisions are less, 4) they require low energy to continue their motion, 20. A semiconductor at 0 K behaves as, 1) conductor, 2) insulator, 3) super conductor 4) extrinsic semiconductor, 21. Carbon, Silicon, and Germanium have four, valence electrons each. These are, characterised by valence and conduction bands, separated by energy band gap respectively, equal to Eg C , Eg S i and Eg Ge .Which of the, following statement is true?, 1) Eg S i Eg Ge Eg C, 2) Eg C Eg Ge Eg S i, 3) Eg C Eg S i Eg Ge, 4) Eg C Eg S i Eg Ge, 22. The valency of impurity element for making, p-type semiconductor is, 1) 5, 2) 4, 3) 3, 4) 7, 23. In n-type semiconductors the electron, concentration is equal to, 1) density of donor atoms, 2) density of acceptor atoms, 3) density of both type of atoms, 4) neither density of acceptor atoms nor density of, donor atoms, 204, , JEE-ADV PHYSICS- VOL- V, 24. Which of the following statements is not true?, 1) the resistance of intrinsic semiconductors, decreases with increase of temperature., 2) doping pure Si with trivalent impurities give ptype semiconductors, 3) t he majority charge carries in n-type, semiconductors are holes, 4) a p-n junction can act as a semiconductor diode, 25. p-type semi conductor is, 1) negatively charged 2) positively charged, 3) neutral, 4) may be positive or negative, 26. n-type semi conductor is, 1) negatively charged 2) positively charged, 3) neutral, 4) may be positive or negative, 27. An electric field is applied across a semi, conductor. Let n be the number of charge, carriers. As temperature increases, n will, 1) increase, 2) decrease, 3) does not change 4) may increase or decrease, 28. In a n - type semiconductor , the fermi energy, level lies, 1) in the forbidden energy gap nearer to the, conduction band., 2) in the forbidden energy gap nearer to the valence, band., 3) in the middle of forbidden energy gap, 4) outside the forbidden energy gap, 29. An n-type and p-type silicon can be obtained, by doping pure silicon with, 1) Arsenic and phosphrous 2) Indium and aluminium, 3) Phosphorous and indium 4) aluminium and boron, 30. The width of forbidden gap in silicon crystal is, 1.1 ev. When the crystal is converted into a ntype semiconductor the distance of fermi level, from conduction band is, 1) Greater than 0.55eV 2) Equal to 0.55 eV, 3) lesser than 0.55eV 4) Equal to 1.1 eV, 31. In extrinsic semiconductors, 1) the conduction band and valence band overlap, 2) the gap between conduction band and valence, band is near about 16eV, 3) the gap between conduction band and valence, band is near about 1eV, 4) The gap between conduction band and valence, band will be 100eV and more, 32. The element that can be used as acceptor, impurity to dope silicon is, 1) Antimony, 2) Arsenic, 3) Boron, 4) phosphorous, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 33. Among the following, the wrong statement in, the case of semiconductor is, 1) Resistivity is in between that of a conductor and, insulator, 2) Temperature coefficient of resistance is negative, 3) Doping increases conductivity, 4) At absolute zero temperature it behaves like, a condutor, 34. The value indicated by fermi energy level in, an intrinsic semiconductor is, 1) the average energy of electrons and holes, 2) the energy of electrons in conduction band, 3) the energy of holes in valence band, 4) the energy of forbidden region, 35. The conduction band and valency band of a, good conductors are, 1) well separated, 2) just touch, 3) very close, 4) overlap, 36. Two pieces one of germinium and the other of, aluminium are cooled from T1K to T2K. The, resistance of, 1) aluminium increases and that of germanium, decrease, 2) each of them decreases, 3) aluminium decreases and that of germanium, increases, 4) each of them increases, 37. In intrinsic semiconductor at room, temperature the no. of electrons and holes are, 1) equal 2) zero, 3) unequal 4) infinite, 38. Band gap in insulator is of the order, 1) 6 eV, 2) 0.60 eV 3) –6 eV 4) 0eV, 39. In p-type semiconductor conduction is due to, 1) greater number of holes and less number of, electrons, 2) only electrons, 3) only holes, 4) greater number of electrons and less number, of holes, 40. In an intrinsic semiconductor, the fermi nergy, level is, 1) nearer to valence band than conduction band, 2) equidistant from conduction band and valence, band, 3) nearer to conduction band than valence band, 4) bisecting the conduction band, 41. With increase in temperature in an intrinsic, semiconducter the ratio of conduction, electrons and holes is, 1) 1:1, 2) 1: 2, 3) 2 :1, 4) 1 : 3, NARAYANAGROUP, , SEMI CONDUCTOR DEVICES, 42. To obtain n-type extrinsic semiconductor, the, impurity element to be added to germanium, should be of valency, 1) 2, 2) 5, 3) 4, 4) 3, 43. The majority carriers in a p-type semiconductor are........, 1) Electrons 2) Holes 3) Both, 4) Impurities, 44. The objective of adding impurities in the, extrinsic semiconductor is, 1) to increase the conductivity of the semiconductor, 2) to increase the density of total current carries, 3) to increase the density of either holes or electrons, but not both, 4) to eliminate the electron -hole pairs produced in, intrinsic semiconductor., 45. In intrinsic semiconductor conductivity is, 1) low, 2) average 3) high, 4) very low, 46. In intrinsic semiconductor conductivity is due, to, 1) doping, 2) breaking of covalent bonds, 3) free electrons 4) holes, 47. When the conductivity of a semi conductor is, only due to breaking of covalent bonds, the, semi conductor is called, 1) n-type 2) p-type 3) intrinsic 4) extrinsic, 48. Assertion (A) : C,Si and Ge have 4 Valancy, each but C is an insulator where as Si and Ge, are semi conductors, Reason (R) : Energy gap is least for Ge, less, for Si compared to C, So that free electrons, for conduction in Ge and Si are significant but, negligible small for C., 1) A, R are true and R explains A correctly, 2) A, R are true and R do not explain A correctly, 3) A is true, but R is false 4) R is true, but A is false, , JUNCTION DIODE, 49. The potential barrier at PN junction is due to, 1) fixed acceptor and donor ions on either side of, the junction, 2) minority carriers on either side of the junction, 3) majority carriers on either side of the junction, 4) both majority and minority carriers on either side, of junction, 50. A PN junction diode cannot be used, 1) as rectifier, 2) for converting light energy to electric energy, 3) for getting light radiation, 4) for increasing the amplitude of an ac signal, 205
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SEMI CONDUCTOR DEVICES, , JEE-ADV PHYSICS- VOL- V, , 51. A full wave rectifier along with the output is 59. The main cause of avalenche breakdown is, shown in fig. the contributions from the diode, 1) collision by ionisation, (2)are, 2) high doping, 3) recombination of electrons and holes, D, 4) low doping, AC, O/P, C, , 60. The main cause of Zener breakdown is, 1) the base semiconductor being germanium, 2) production of electron - hole pairs due to thermal, D, exitation, Full Wave Rectifier, 3) low doping, 4) high doping, 61. When p-n junction is foward biased, the current, across the junction is mainly due to, Vi, 1) diffusion of charges 2) drifting of charges, 3) both diffusion and drifting of charges, I/P, 4) holes only, 62. The current through any p-n junction is due to, A, B, C, D, a) drift of charge carriers, V, b) diffusion of charge carriers, c) different concentrations of same type of, O/P, charge carriers in different regions, 1) C 2) A,C 3) B, D 4) A,B,C,D, d) same concentrations of same type of charge, 52. A full-wave rectifier is used to convert ‘n’, carriers in different regions, Hz a.c into d.c,then the number of pulses per, 1) a,b and c, 2) a and b only, second present in the rectified voltage is., 3) only d, 4) a,b,c,d, 1) n, 2) n/2, 3)2n, 4)4n, 63. The thickness of depletion layer is, 53. If the input frequency of half-wave rectifier, approximately, is n Hz ac, then its output is, 1) 1 m 2) 1mm 3) 1cm, 4) 1m, 1) a constant dc, 2) n/2 Hz pulsating dc, 64. The depletion region is, 3) n Hz pulsating dc, 4) 2n Hz pulsating dc, 1) region of opposite charges, 54. p-n junction diode acts as, 2) neutral region, 1) ohmic resistance, 2) non-ohmic resistance, 3) region of infinite energy, 3) both 1 and 2, 4) amplifier, 4) region of free current carriers, 55. The process of converting alternating current, 65., The diffusion current in a p-n junction is greater, into direct current is known as, than the drift current when the junction is, 1) modulation, 2) amplification, 1) forward biased, 3) detection, 4) rectification, 2) reverse biased, 56. On increasing reverse voltage in a p-n junction, 3) un biased, diode the value of reverse current will, 4) both forward and reverse biased, 1) gradually increases, 66., Germanium diode, 2) first remains constant and then suddenly increase, 1) may be used as rectifier because it offers a, 3) remains constant, 4) gradually decrease, relatively low resistance for forward bias and very, 57. In forward bias the depletion layer behaves, high resistance for reverse bias., like, 1) an insulator, 2) a conductor, 2) may be used as a rectifier because it offers a, 3) a semiconductor, 4) capacitor, relatively high resistance for forward bias and very, 58. p-n junction in reverse bias behaves like, low resistance for reverse bias., 1) an inductor, 2) a condenser, 3) cannot be used as a rectifier, 3) amplifier, 4) an off switch, 4) may be used as an amplifier, 1, , 2, , 0, , 206, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 67. The output of the given circuit in Fig., R, vm sin t , , 68., , 69., , 70., , 71., , 1) would be zero at all times, 2) would be like a half-wave rectifier with positive, cycles in output, 3) would be like a half-wave rectifier with negative, cycles in output, 4) would be like that of a full-wave rectifier, Diode is forward biased and the applied voltage, is greater than the potential barrier then, I) resistance of the junction in the forward bias, decreases, II) potential barrier remains same, III) width barrier remains decreases, IV)p-type is at higher potential than the n-type, 1) all are true, 2)all are false, 3) I,III,IV are true, 4)I,II, III are true, When a junction diode is reverse biased, the, current called drift current is due to, 1) majority charge carriers of both n&p sides, 2) minority charge carriers of both n&p sides, 3) holes of both n &p sides, 4) conduction band electrons of n-side only, Among the following one statement is not, correct when a junction diode is in forward bias, 1) the width of depletion region decreases, 2) free electron on n- side will move towards the, junction, 3) holes on p -side move towards the junction, 4) electron on n- side and holes on p-side will move, away from junction, Consider a.p-n junction as a capacitor, formed, with p and n material acting as thin metal, electrodes and depletion layer width acting as, seperation between them. Basing on this, assume that a n-p-n transistor is working as, an amplifier in CE configuration. If C1 and C2, are base-emiter and collector emitter junction, capacitances, then :, 1) C1 C2, , 2) C1 C2, , 3) C1 C2, , 4) C1 C2 0, , NARAYANAGROUP, , SEMI CONDUCTOR DEVICES, 72. Consider the following statements A and B and, identify the correct answer, 1) : Germanium is preferred over silicon in the, construction of zener diode., 2) : Germanium has high thermal stability than, silicon in the construction of Zener diode, 1) Both 1 & 2 are true 2) Both 1 & 2 are false, 3) 1 is true but 2 is false 4) 1 is false but 2 is true, 73. A Zener diode when used as a voltage regulator, is connected, (a) in forward bias, (b) in reverse bias, (c) in parallel to the load, (d) in series to the load, 1) (a) and (b) are correct 2) (b) and (c) are correct, 3) (a) only is correct, 4) (d) only is correct, 74. Consider the following statements A and B and, identify the correct answer, (A) A Zener diode is always connected in reverse, bias to use it as voltage regulator, (B) The potential barrier of a p - n junction lies, between 0.1 to 0.3V, approximately, 1) A and B are correct 2) A and B are wrong, 3) A is correct but B is wrong, 4) A is wrong but B is correct, 75. Consider the following statements A and B, identify the correct of the give answer., A) The width of the depletion layer in a p-n, junction diode increases in forward bias., B) In an intrinsic semiconductor the fermi, energy level is exactly in the middle of the, forbidden gap, 1) A is true and B is false 2) Both A and B are false, 3) A is false and B is true 4) Both A and B are true, 76. The potential in the depletion layer due to, 1) Electrons, 2) Holes, 3)Ions, 4) Forbidden band, 77. Pickout the incorrect statement regarding, reverse saturation current in the p-n junction, diode, 1) this current doubles for every 1000C rise of, temperature, 2) this current is due to minority carriers, 3) the current carriers are produced by thermal, agitation, 4) reverse saturation current is also known as, leakage current, 78. When the p-n junction diode is reverse biased,, the thickness of the depletion layer, 1) increases, 2) decreases, 3) becomes zero, 4) remains constant, 207
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 79. p-n junction diode can be used as, 1) amplifer 2) detector 3) oscillator 4) capacitor, 80. A p-n junction diode is reverse biased.Then, 1) more current flows, 2) the barrier potential decreases, 3) the barrier potential increases, 4) resistance offered is low, 81. In the middle of the depletion layer of reverse, biased p-n junction, the :, 1) electric field is zero 2) potential is maximum, 3) electric field is maximum 4) potential is zero, 82. When p-n junction diode is forward biased then, 1) the depletion region is reduced and barrier height, is increased, 2) the depletion region is widened and barrier height, is reduced, 3) both the depletion region and barrier height are, reduced, 4) both the depletion region and barrier height are, increased, 83. In a p-n junction photo cell, the value of the, photo electromotive force produced by, monochromatic light is proportional to, 1) The barrier voltage at the p-n junction., 2) The intensity of the light falling on the cell, 3) The frequency of the light falling on the cell., 4) The voltage applied at the p-n junction, 84. In Fig. V0 is the potential barrier across a, p-n junction, when no battery is connected, across the junction:, , 86. Application of a forward bias to a p-n junction:, 1) increases the number of donors on the n-side, 2) increases the electric field in the depletion zone, 3) increases the potential difference across the, depletion zone, 4) widens the depletion zone, 87. In a p-n junction diode, not connected to any, circuit, 1) the potential is the same everywhere, 2) the p-type side is at a higher potential than the ntype side., 3) there is an electric field at the junction directed, from the n-type side to the p-type side., 4) there is an electric field at the junction directed, from the p-type side to the n-type side., 88. Select the correct statement from the following:, 1) A diode can be used as a rectifier, 2) A triode cannot be used as a rectifier, 3) The current in a diode is always proportional to, the applied voltage, 4) The linear portion of the I -V characteristic of a, triode is used for amplification without distortion, 89. When a p-n junction diode is forward-biased,, energy is released at the junction due to the, recombination of electrons and holes. This, energy is in, 1) Visible region, 2) Infrared region, 3) UV region, 4) X-ray region, 90. In Fig. assuming the diodes to be ideal:, , 1, , D1, , 2, 3, , -10V, , A, , R, , V0, , D2, B, , 1) 1 and 3 both correspond to forward bias of, junction, 2) 3 corresponds to forward bias of junction and 1, corresponds to reverse bias of junction, 3) 1 corresponds to forward bias and 3, corresponds to reverse bias of junction, 4) 3 and 1 both correspond to reverse bias and 3, corresponds to reverse bias of junction, 85. There is a sudden increase in current in zener, diode is, 1) Due to rupture of bonds, 2) Resistance of deplection layer becomes less, 3) Due to high doping 4) Due to less doping, 208, , 1) D1 is forward biased and D2 is reverse biased, and hence current flows from A to B, 2) D2 is forward biased and D1 is reverse biased, and hence no current flows from B to A and vice, versa, 3) D1 and D2 are both forward biased and hence, current flows from A to B, 4) D1 and D2 are both reverse biased and hence, no current flows from A to B and vice-versa, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 91. A Si and a Ge diode has identical physical 95. Assertion (A) : Putting p-type semiconductor, slabs directly in physical contact can not form, dimensions.The band gap in Si is larger than, p-n junction, that in Ge.An identical reverse bias is applied, Reason (R) : The roughness at contact will be, across the diodes., much more than inter atomic crystal spacing, 1)The reverse current in Ge is larger than that in Si, 0, 0, 2)The reverse current in Si is larger than that in Ge, 2A, to 3A and continuous flow of charge, , 3)The reverse current is identical in the two diodes, , , 4)The relative magnitude of the reverse currents, carriers is not possible, cannot be determined from the given data only, 96. Assertion (A) : Eventhough forward bias is, 92. The correct curve between potential(V) and, known to be more ( mA) than the current in, distance(d) near p-n junction is, the reverse bias ( A) , photodiodes are still, 1), V, V, used in reverse bias., n, n, P, P, Reason (R) : The fractional change in majority, charge carriers would be much, d, d, 2), 1), less than that in minority charge carriers due, to photo-effects so that minority carrier, reverse bias current is more easily to measure, V, V, with photo diodes than majority carrier forward, current., n, n, P, P, 97. Match List-I with List-2, LIST-1, LIST-2, 3), 4), d, d, A) Intrinsic, E) Donor level, semiconductor, B) N-type, F) Acceptor level, 93. The graph given below represents the I-V, semiconductor, characterstics of a zener diode.Which part of, C) p-type, G) Immobile ions, the characterstics curve is most relavant for, semiconductor, its operation as a voltage regulator., D) Depletion layer H) Fermi energy, I(mA), levels is at the, centre of forbidden, Forward bias, Reverse bias, energy gap, a, V, I) Fermi energy level, is near the conduction, V(V), c, d, b, band, e, Current, A B C D, A B C D, 1) F E,G H I, 2) H E,I F G, I(A), 3) E F G H,I 4) I, H G E,F, 98. Match the following, 1)ab, 2)bc, 3)cd, 4)de, LIST - I, LIST - II, a., Antimony, e. P - type impurity, 1) A, R are true and R explains A correctly, b. Indium, f. N - type impurity, 2) A, R are true and R do not explain A correctly, c. Carbon, g. not semiconductor, 3) A is true, but R is false, d., Silicon, h. semiconductor, 4) R is true, but A is false, 1)a-f,, b-e,, c-g,, d-h, 2)a- e, b- g, c- h, d- f, 94. Assertion (A) : Si and GaAs are preferred, 3) a-g, b-h, c-f, d-e, 4) a-h, b-f, c-e, d- g, materials for solar cells, Reason (R) : Energy gap of Si is 1.1eV and, TRANSISTORS, that of GaAs is 1.53eV which give maximum 99. The correct relation between current gains , irradiance where as other materials like CdS, and is, or CdSe E g 2.4 eV and PbS E g 0.4 eV , , , 1, 1) , 2) , 3) 1 4) , 1, , , 1, , , , give minimum irradiance., z, , NARAYANAGROUP, , 209
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SEMI CONDUCTOR DEVICES, 100. Transistors are made of, 1) insulators, 2) conductors, 3) alloys, 4)doped semi-condcutors, 101. In n-p-n transistor the arrow head on emitter, represents that the conventional current flows, from, 1) base to emitter, 2) emitter to base, 3) emitter to collector 4) base to collector, 102. In a junction transistor the emitter, base and, collector are made of, 1) extrinsic semi conductors, 2) intrinsic semi conductors, 3) both 1 and 2, 4) metal, 103. In a transistor, 1) both emitter and the collector are equally doped, 2) base is more heavily doped than collector, 3) collector is more heavily doped than the emitter, 4) the base is made very thin and is lightly doped, 104. In a transistor, 1) length of emitter is greater than that of collector, 2) length of collector is greater than that of emitter, 3) both emitter and collector have same length, 4) any one of emitter and collector can have greater, length, 105. In transistor the emitter current is, 1) slightly more than the collector current, 2) slightly less than the collector current, 3) equal to the collector current, 4) equal to the base current, 106. In the use of transistor as an amplifier, 1) the emitter - base junction is reverse biased and, the collector base junction is also reverse biased, 2) the emitter - base junction is forward biased and, the collector - base junction is reverse biased, 3) both the junctions are forward biased, 4) any of the two junctions may be forward biased., 107. One way in which the operation of an npn, transistor differ from that of a pnp transistor, is that, 1) the emitter junction is reverse biased in npn, 2) the emitter junction injects minority carriers into, the base region of the pnp., 3) the emitter injects holes into the base of the pnp, and electrons into the base region of npn, 4) the emitter injects holes into the base of npn, 108. npn transistors are preferred to pnp transistors, because they have, 1) low cost, 2) low dissipation energy, 3) capable of handling large power, 4) electrons have high mobility than holes and hence, high mobility of energy, 210, , JEE-ADV PHYSICS- VOL- V, 109. A CE transistor amplifies a weak current, signal because collector current is, 1) times Ib, 2) times IC, 3) times Ib, 4) times IC, 110. When a positive voltage signal is applied to, the base of a common emitter npn amplifier, 1) The emitter current decreases, 2) The collector voltage becomes more positive, 3) The collector voltage becomes less positive, 4) The collector current decreases, 111. In case of common emitter p-n-p transistor, input characteristic is a graph drawn, 1) With IC on y-axis and VCE on x-axis keeping IB, constant, 2) With IB on y-axis and VBE on x-axis keeping, VCE constant, 3) With IC on y-axis and IB on x-axis keeping VCE, constant, 4) With VBE on y-axis and VCE x-axis keeping IB, constant, 112. The output characterstics of an n-p-n transistor, represent,[ I C Collector current, VCE , potential difference between collector and, emitter, I B Base current, VBB voltage given, base; VBE the potential difference between, base and emitter], 1)change in I C as I B and VBB are changed, 2)Changes in I C with changes in VCE ( I B cosntant), 3)changes in I B with changes in VCE, 4)Change in I C as VBE is changed, 113. In a transistor the base is made very thin and, is lightly doped with an impurity, because, 1) to enable the collector to collect about 95% of, the holes or electrons coming from the emitter side, 2) to enable the emitter to emit small number of, holes or electrons, 3) to save the transistors from high current effects, 4) to enable the base to collect about 95% of holes, or electrons coming from the emitter side, 114. A p-n-p transistor is said to be in active region, of operation, When, 1) Both emitter junction and collector junction are, forward biased, 2) Both emitter junction and collector junction are, reverse biased, 3) Emitter junction is forward biased and collector, junction is reverse biased, 4) Emitter junction is reverse biased and collector, junction is forward biased, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 115. An n-p-n transistor power amplifier in C-E, configuration gives, 1) Voltage amplification only, 2) Current amplification only, 3) Both current and voltage amplification, 4) Only power gain of unity, 116. When n-p-n transistor is used as an amplifier:, 1) electrons move from base to collector, 2) holes moves from emitter to base, 3) holes move from collector to base, 4) holes move from base to emitter, 117. The part of a trasistor which is heavily doped, to produce large number of majority carriers, is :, 1) emitter, 2) base, 3) collector, 4) can be any of the above three, 118. When n-p-n transistor is used as an amplifier, 1) electrons move from collector to base, 2) holes move from collector to base, 3) electrons move from base to emitter, 4) holes move from base to emitter, 119. A n-p-n transistor conducts when, 1) both collector and emitter are positive with, respect to the base, 2) collector is positive and emitter is negative with, respect to the base, 3) collector is positive and emitter is at same, potential as the base, 4) both collector and emitter are negative with, respect to the base, 120. In a common -base amplifier, the phase, difference between the input signal voltage and, output voltage is :, , SEMI CONDUCTOR DEVICES, 124. A three terminal device with one terminal, common to both the output and input is called, 1) rectifier 2) transistor 3) diode 4) triode, 125. Input and output signal of an amplifier in CE, configuration are always, 1) Equal, 2) Inphase, 3) Having a phase difference 4) Out of phase, 126. Ge transistor can be operated at a, temperature, 1) upto 900C, 2) upto 400C, 0, 3) upto 100 C, 4) upto 200C, 127. Transistor amplifier circuit with a feed back, circuit is called, 1) oscillator, 2) detector, 3) modulator, 4) all, 128. A pulsating voltage is a mixture of an a.c, componet and a d.c component. the circuit used, to seperate a.c and d.c component is called, 1) an oscillator, 2) an amplifier, 3) a rectifier, 4) a filter, 129. The and of a transistor are always, , 1) > 1 , <1, 2) < 1 , >1, 3) = , 4) 1, 130. In case of NPN transistor, emitter current is, always greater than collector current,, because:, 1) Collector side is revese biased and emitter side, is forward biased, 2) Collector being reverse biased attracts more, electrons, 3) Some electrons are lost in base, 4) Collector side is forward biased and emitter side, is reverse biased., , , 131., When a transistor amplifier having current, 2) , 3) zero, 4), 1), 4, 2, gain of 75 is given an input signal,, 121. In a common emitter amplifier, the phase, VI 2sin(157t / 2) , the output signal is, difference between the input signal voltage and, found to be VO 200sin(157t 3 / 2) . The, output voltage is, transistor is connected as:, , , 1) A common collector amplifier, 1), 2) 0, 3) , 4), 2, 4, 2) A common base amplifier, 122. When n-P-n transistor is used as an amplifer, 3) A common emitter amplifier 4) An oscillator, 1) Electrons move from emitter to collector, 132. An oscillator is an amplifier with, 2) Holes move from emitter to base, 1) A large gain, 2) Negative feedback, 3) Electrons move form collector to base, 3) Positive feedback 4) No feedback, 4) Holes move from base to collector, 133. In which of the transistor configurations, the, 123. In a PNP transistor the base is the N-region., voltage gain is highest?, Its width relative to the P-region is, 1) Common-base, 2) Common-emitter, 1) smaller 2) larger 3) same, 4) not related, 3) Common-collector 4) Same in all threee, NARAYANAGROUP, , 211
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 134. A working transistor with its three legs marked, P, Q, and R is tested using a multimeter to R, and the other (positive) terminal to P or Q., Some resistance is seen on the multimeter., Which of the following is true for the, transistor?, 1) It is a pnp transistor with R as emitter, 2) It is an npn transistor with R as collector, 3) It is an npn transistor with R as base, 4) It is pnp transistor with R as collector, 135. Match the following Column I with Column II, Column I, , Column II, , n-p-n transistor, , A, , p-n-p transistor, , B, , Light emmitting, diode, , E, C, , N, , C, B, , c. Power gain, d. Resistance gain, 1) a-e, b-f, c-g, d-h, 3) a-g, b-h, c-e, d-f, , I c , , g. , I b , , 2, , RL, Ri, , RL, h. R, i, 2) a- f, b- g, c- h, d-e, 4) a- h, b- e, c- f, d- g, , 1) If both A and R are correct and R is the correct, explanation of A, 2) If both A and R are correct and R is not the, correct explanation of A, 3) If ‘A’ is correct and ‘R’is incorrect, 4) If ‘A’ is incorrect and ‘R’is correct, 138. Assertion (A) :- Common emiter mode of a, transistor is widely used, Reason (R) :-Current gain, voltage gain, and, power gain are maximum in C.E mode of a, transistors., 139. Assertion (A) :- Transistor in C.E. mode can, be used as amplifier, Reason (R), :A, small, change in base current produces a relatively, large change in collector current, , LOGIC GATES, Zener diode, , E, D, , C, , 140. In the Binary number system the number 100, represents :, B, 1) one, 2) three, 3) four 4) hundred, 141., Among, the, following, is, not, the function of NOT, 1) I-A,II-B,III-C, IV-D 2) I-D, II-A, III-B, IV-C, gate is, 3) I-C, II-D, III-B, IV-A 4) I-B, II-A, III-C, IV-D, 1) stop a signal, 2) invert an input signal, 136. Match List-I with List-2, 3), complement, a, signal, List-1, List-2, 4) change the logic in a digital circuit, A) Emitter, E) Transistor, 142. Digital circuit can be made by repetitive use, B) Base, F) Moderately doped, of, C) Collector, G) lightly doped, 1) OR gates, 2) AND gates, D) Transfer of, H) Heavily doped, 3) NOT gates, 4) NAND gates, resistance, 143. Among the following one gives output 1 in the, I) Largest physical size, AND gate, A B C D, A B C D, 1) A = 0, B = 0, 2) A = 1, B = 1, 1) F E H,I G 2) G F,I E H, 3) A = 1, B = 0, 4) A = 0, B = 1., 3) H G F,I E 4) E H,I G F, 144. Person who use Boolean algebra for describing, the operation of logic gates first was, 137. Match the following for C E transistor amplifier, 1) Boole 2) Shannon 3) Schottky 4) Zener, LIST - I, LIST - II, 145. NAND and NOR gates are called universal, I c, gates because they, a. Current gain, e. I, 1) are universally available, b, 2) can be combined to produce OR, AND and, I c RL, NOT gates, b. Voltage gain, f. I R, 3) are widely used in the Integrated circuits, b, i, 4) can be easily manufactured., 212, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 146. In positive logic, the logic state 1 corresponds, to, 1) positive voltage, 2) zero voltage, 3) lower voltage level 4) higher voltage level., 147. In Boolean algebra A + B = Y implies that :, 1) sum of A and B is Y, 2) Y exists when A exists or B exists or both A and, B exist, 3) Y exists only when A and B both exist, 4) Y exists when A or B exist but not when both A, and B exist., 148. In the Boolean algebra, the following one is, wrong, 1) 1+0 = 1 2) 0+1=1 3) 1+1=1 4) 0+0 =1, 149. In Boolean algebra A.B = Y implies that:, 1) product of A and B is Y, 2) Y exists when A exists or B exists, 3) Y exists when both A and B exist but not when, only A or B exists, 4) Y exists when A or B exists but not both A and B, exist., 150. In the Boolean algebra, the following one is, wrong, 1) 1.0 = 0 2) 0.1 = 0 3) 1.1 = 0 4) 1.1 = 1, 151. The following truth table is for, A, B, Y, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1) NAND gate, 2) AND gate, 3) XOR gate, 4) NOT gate, 152. The output of a 2-input OR gate is zero only, when its, 1) both inputs are 0, 2) either input is 1, 3) both inputs are 1, 4) either input is zero., 153. Boolean algebra is essentially based on, 1) symbols 2) logic, 3) truth, 4) numbers, 154. The value of A A in the Boolean algebra is, 1) A, 2) A, 3) 0, 4) 1, 155. The value of A.A in Boolean algebra is, 1) 0, 2) 1, 3) A, 4) A, 156. The following is NOT equal to 0 in the Boolean, algebra is, 1) A.0, 2) A.A, 3) A.0, 4) A A, 157. An AND gate is followed by a NOT gate in, series. With two inputs A & B, the Boolean, expression for the out put Y will be :, 1) A.B, 2) A + B 3) A+B 4) A.B, NARAYANAGROUP, , SEMI CONDUCTOR DEVICES, 158. NOR gate is the series combination of, 1) NOT gate followed by OR gate, 2) OR gate followed by NOT gate, 3) AND gate followed by OR gate, 4) OR gate followed by AND gate, 159. The gate that has only one input terminal, 1) NOT, 2) NOR 3) NAND 4) XOR, 160. AND gate:, 1) It has no equivalence to switching circuit., 2) It is equivalent to series switching circuit., 3) It is equivalent to parallel switching circuit, 4) It is a mixture of series and parallel switching, circuit, 161. The gate that can act as a building block for, the digital circuits is, 1) OR, 2) NOT 3) AND 4) NAND, 1) Statement -1 is false, statement -2 is true, 2) statement 1-is true statement -2 is true, statement -2 is correct explanation of, statement-1., 3) statement 1-is true statement -2 is true, statement -2 is not correct explanation of, statement-1., 4) statement 1-is true statement -2 is false, 162. Statement -1: NAND or NOR gates are called, digital building blocks., Statement-2: The repeated use of NAND (or, NOR) gates can produce all the basis or, complicated gates., 163. Statement-1: NOT gate is also called invertor, circuit, Statement-2: NOT gate inverts the input order., 164. Match the quantities of column I with the, column II when A=1 and B=0, Column - I, Column - II, a) A.B A, , (p) A, , , c) A B . A, d) A B .B, , (q) B, , b) A.B A, , 1) a-q,s ;, 2) a-p,s ;, 3) a-q,r ;, 4) a-r,s ;, , b-p,r ;, b-p,s ;, b-p,s ;, b-p,r ;, , (r) A + B, (s) A.B, c-q,s ;, c-q,r ;, c-q,r ;, c-q,s ;, , d-q,s, d-r,q, d-s,p, d-p,s, 213
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , C.U.Q - KEY, 01)1 02)3 03)3 04)2 05)2 06)2 07)2, 08)1 09)3 10)3 11)4 12)1 13)4 14)4, 15)1 16)1 17)1 18)2 19)4 20)2 21)3, 22)3 23)1 24)3 25)3 26)3 27)1 28)1, 29)3 30)3 31)3 32)3 33)4 34)1 35)4, 36)3 37)1 38)1 39)1 40)2 41)1 42)2, 43)2 44)3 45)1 46)2 47)3 48)1 49)1, 50)4 51)3 52)3 53)3 54)2 55)4 56)2, 57)2 58)4 59)1 60)4 61)1 62)1 63)1, 64)4 65)1 66)1 67)3 68)3 69)2 70)4, 71)1 72)2 73)2 74)3 75)3 76)3 77)1, 78)1 79)2 80)3 81)3 82)3 83)2 84)2, 85)1 86)1 87)3 88)1 89)2 90)2 91)3, 92)1 93)4 94)1 95)1 96)1 97)2 98)1, 99)1 100)4 101)1 102)1 103)4 104)2 105)1, 106)2 107)3 108)4 109)1 110)3 111)2 112)2, 113)1 114)3 115)3 116)1 117)1 118)4 119)2, 120)3 121)3 122)1 123)1 124)2 125)4 126)1, 127)1 128)4 129)2 130)3 131)3 132)3 133)2, 134)3 135)3 136)3 137)1 138)1 139)1 140)3, 141)1 142)4 143)2 144)2 145)2 146)4 147)2, 148)4 149)3 150)3 151)1 152)1 153)2 154)4, 155)1 156)1 157)4 158)2 159)1 160)2 161)4, 162)2 163)2 164)1, , 3., , 4., , 5., , 6., , 7., , In a p-n junction the depletion region is 400nm, wide and electric field of 5 105 Vm 1 exists in, it. The minimum energy of a conduction, electron, which can diffuse from n-side to the, p-side is, 1) 4eV, 2) 5eV, 3) 0.4eV 4) 0.2eV, The reverse bias in a junction diode is changed, from 5V to 15V then the value of current, changes from 38 A to 88 A. The resistance, of junction diode will be, 1) 4 x 105 2) 3 x105 3) 2 x 105 4) 106 , A diode made of silicon has a barrier potential, of 0.7V and a currrent of 20mA passes through, the diode when a battery of emf 3V and a, resistor is connected to it. The wattage of the, resistor and diode are, 1) 0.046W, 0.014W 2) 4.6W, 0.14W, 3) 0.46W, 0.14W, 4) 46W, 14W, In a half wave rectifier output is taken across, a 90 ohm load resistor. If the resistance of diode, in forward biased condition is 10ohm, the, efficiency of rectification of ac power into dc, power is, 1) 40.6% 2) 81.2% 3) 73.08 % 4) 36.54%, In a full wave rectifier output is taken across, a load resistor of 800 ohm. If the resistance of, diode in forward biased condition is 200 ohm,, the efficiency of rectification of ac power into, dc power is, 1) 64.96% 2) 40.6% 3) 81.2% 4) 80%, , TRANSISTORS, 8., , In a P-N-P transistor, the collector current is, 10 mA. If 90% of the holes reach the collector,, then emitter current will be:, 1) 13 mA 2) 12 mA 3) 11 mA 4) 10 mA, INTRINSIC, EXTRINSIC, 9., A transistor has a base current of 1mA and, SEMICONDUCTORS AND DIODES, emitter current 100mA. The current transfer, 1. The electrical conductivity of a semiconductor, ratio will be, increases when electomagnetic radiation of, 1) 0.9, 2) 0.99 3) 1.1, 4) 10.1, wavelength shorter than 2480nm is incident on, it. The band gap in (eV) for the semiconductor 10. When the base -emitter voltage of a transistor, connected in the common -emitter mode is, is, changed by 20 mV the collector current is, 1) 0.7eV 2) 0.5eV 3) 2.5eV 4) 1.2eV, changed by 25 mA. Find the transconductance., 2., Pure Si at 300 K has equal electron, 1) 1.25 1 2) 2.501 3) 0.5 1 4) 5.5 1, ni concentrations of 1.5 1016 m 3 . Doping by 11. In a transistor circuit the base current changes, from 30 A to 90 A . If the current gain of, indium increases nh 4.5 1022 m3 . ne in the, the transistor is 30, the change in the collector, doped Si is, current is, 1) 5 109 2) 7 109 3) 9 109 4) 8 109, 1) 4 mA, 2) 2 m A 3) 3.6 mA 4) 1.8 mA, , LEVEL-I (C.W), , 214, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 12. The current gain of transistor in a common, emitter circuit is 40. The ratio of emitter, current to base current is, 1., 1) 40, 2) 41, 3) 42, 4) 43, 13. In a common base configuration the emitter 2., current changes by 5mA when emitter voltage, is changed by 200 mV at a fixed collector to, base voltage. The input resistance is, 2) 1000 3) 2.5 4) 4 , 1) 40 , 14. For a common base amplifier, the values of, resistance gain and voltage gain are 3000 and, 2800 respectively. The current gain will be, 3., 1) 0.93, 2) 0.83 3) 0.73, 4) 0.63, , 3) B, , 2) 1 0 .1 0, , 3) 1 0 .1 0, , 4) 1 1.1 0, , LEVEL-I (C.W) - KEY, 1) 2 2) 1 3) 4 4) 3 5) 1 6) 4 7) 1, 8) 3 9) 2 10) 1 11) 4 12) 2 13) 1 14) 1, 15) 3 16) 4 17) 2 18) 3 19) 4 20) 3 21) 4, NARAYANAGROUP, , , , , , 0.5eV, , We know that ofr a doped semiconductor in thermal, equilibrium, we have ne nh ni2, As, , per, , given, , data,, , ni 1.5 1016 m 3, , nh 4.5 10 22 m 3, 2, , 16, 6, ni2 1.5 10 m, , , 5.0 109 m 3, n, Thus e, 22 3, 4.5 10 m, nh, , W Vq =Edq, , 5 105 400 109 1.6 1019 J, eV, 1.6 1019, = 2000 X 10-4 eV = 0.2 eV, 4., , 4) A .B, 7., 20. In the Boolean algebra, the following one, which is not equal to A is, 8., 1) A.A, 2) A + A 3) A.A, 4) A A, 21. The logic expression which is NOT true, 9., in Boolean algebra is, 1) 1 1 .1 0, , hc, 6.6 1034 3 10 8, , 2480 109 1.6 1019, , , , 16. Decimal number 15 is equivalent to the binary, number:, 1) 110001 2) 000101 3) 101101 4) 001111, 17. Binary number 1001001 is equivalent to the 5., decimal number:, 1) 37, 2) 73, 3) 41, 4) 32, 18. In the Binary number system 1+1=, 1)2, 2)1, 3)10, 4)100, 19. If A = B = 1, then in terms of Boolean algebra, 6., the value of A.B + A is not equal to, 1) B.A+B 2) B+A, , Eg , , 5 105 400 10 9 1.6 10 19 J, , 15. In a transistor amplifier =62,RL= 5000 , and internal resistance of the transistor, is 500 . Its power amplification will be, 1) 25580 2) 33760 3) 38440 4)55280, , LOGIC GATES, , LEVEL-I (C.W) - HINTS, , R, , , , V, 15 5, 10, , , 6, I 88 38 10, 50 10 6, , 106 10, 105 2 105 , 5, 5, , P = VI (for resistor), 3 0.7 20 10 3 2.3 20 10 3, , 46 10 3 0.046W, for diode, P 0.7 20 10 3 0.014W, , , , 0.406 RL, rf RL, , , , 0.812 RL, rf RL, , IC , , 90, 10, I E I E IC, 100, 9, , I, c , Ie Ib Ic, Ie, , I c, 25 103, , 1.25 1, 10. g m , 3, VBE 20 10, I, , C, 11. IB I E IC and I, B, , IC, I E IC I B IC, 1 1, 12. I and I I, IB, B, B, B, , 13. VBE = I e Ri, 215
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 6., , 14. Av AR , R, , 2, L, 15. AP R, i, , 2 15, , 7., , 271, , 16., , 2 3 1, 2 1 1, 0 1, , 1510 1111 2, , 65 432 1 0, , 17. 1001001 2 1 26 0 25 0 24 1 23, 0 22 0 21 1 20 73, 18. 1+1=10 in binary number system, 19. If A=B=1; then A.B+A=1.1+1=1, then A.B 1.1 0.1 0, 20. A A 0, 21. logic, OR and AND operation, , TRANSISTORS, 8., , 9., , 10., , LEVEL-I (H.W), INTRINSIC, EXTRINSIC, SEMICONDUCTORS AND DIODES, 1., , 2., , 3., , 4., , 5., , The electrical conductivity of a semi, conductor increases when electromagnetic, radiation of wavelength shorter than 1240, nm is incident on it. The forbidden band, energy for the semi conductor is (in eV), 1) 0.5, 2) 0.97 3) 0.7, 4) 1.1, A semiconductor is known to have an electron, concentration of 5 1013 / cm 3 and hole, concentration of 8 1012 / cm3 . The, semiconductor is, 1) n-type 2) p-type 3) intrinsic 4) insulator, A potential barrier of 0.5V exists across a p-n, junction. If the width of depletion layer is 106, m, then intensity of electric field in this region, will be, 1) 1 x 106 V/m, 2) 5 x 105 V/m, 4, 3) 4 x 10 V/m, 4) 2 x 106 V/m, A p - n junction diode has breakdown voltage, of 28V. If applied external voltage in reverse, bias is 40V the current through it is, 1) Zero, 2) infinite 3) 10 A 4) 15A, The value of current in the following diagrams, is (diode assumed to be ideal one), 3, -4V, -1V, 1) 0.1 amp 2) 0.01amp 3) 1 amp 4) zero, , 216, , A half -wave rectifier is used to convert, 50Hz A.C. to D.C. voltage. The number, of pulses per secon d in the rectified, voltage are, 1) 50, 2) 25, 3) 100 4) 75, If a full wave rectifier circuit is operating, from 5 0H z mai n s, t h e f u n d a men t al, frequency in the ripple will be, 1) 25 Hz 2) 50 Hz 3) 70.7 Hz 4) 100 Hz, , 11., , 12., , 13., , 14., , 15., , 16., , In an npn transistor the base and collector, currents are 100 A and 9mA respectively.., Then the emitter current will be, 1) 9.1mA 2) 18.2mA 3) 3.91 A 4)18.2 A, A change of 8mA in the emitter current brings, a change of 7.9mA in the collector current. The, change in base current required to have the, same change in the collector current is, 1) 0.01mA 2) 1A, 3) 10mA 4) 0.1mA, For a p-n-p transistor in CB configuration, the, emitter current IE is 1mA and = 0.95. The, base current and collector current are, 1) 0.95 mA, 0.05mA 2) 0.05mA, 0.95mA, 3) 9.5mA, 0.5mA, 4) 0.5mA, 9.5mA, If a change of 100 A in the base current of, an n-p-n transistor in CE causes a change of, 10mA in the collector current, the ac current, gain of the transistor is, 1) 10, 2) 100, 3) 1000 4)10000, For a common emitter amplifier, current gain, is 70. If the emitter current is 8.4mA, then the, base current is, 1) 0.236mA 2) 0.118mA 3) 0.59mA 4) 8.3mA, The base current of a transistor is 105 A and, the collector current is 2.05mA. Then of the, transistor is, 1) 1.952, 2)19.52 3) 195.2 4) 1952, For a transistor the value of is 0.9. Its , value is, 1) 9, 2) 0.9, 3) 0.09, 4) 90, For a transistor the current amplification factor, is 0.8. The transistor is connected in common, emitter configuration, the change in collector, current when the base current changes by 6mA, is, 1) 6mA, 2) 4.8mA 3) 24mA 4) 8mA, A change of 400mV in base-emitter voltage, causes a change of 200 A in the base current., The input resistance of the transistor is, 2) 6K 3) 2K 4) 8K, 1) 1K, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 17. In a common base circuit, if the collector base, voltage is changed by 0.6V, collector current, changes by 0.02mA. The output resistance will, be, 1) 104 2) 2 x 104 3) 3 x 104 4) 4 x 104 , 18. A common emitter transistor amplifier has a, current gain of 50. If the load resistance is, 4k , and input resistance is 500 , the, voltage gain of amplifier is, 1) 100, 2) 200 3) 300 4) 400, , LOGIC GATES, 19. Equivalent of decimal number 8 in the binary, number is, 1)10, 2)101, 3)1000 4)1011, 20. The equivalent of 110in the decimal number is, 1)2, 2)4, 3)8, 4)6, 21. If A = 1, B = 0 then the value of A B in terms, of Boolean algebra is, 1) A, 2) B, 3) B + A 4) A.B, 22. In the Boolean algebra : A + B =, 1) A B 2) A.B, 3) A B 4) A B, 23. The following one represents logic addition is, 1) 1 + 1 = 2, 2) 1 + 1 = 10, 3) 1 + 1 = 1, 4) 1 + 1 = 11, 24. In the Boolean algebra : A.B = ......, 1) A B 2) A.B, 3) A B 4) A + B, 25. In the Boolean algebra, which gate is, expressed as Y A B, 1) OR, 2) NAND 3) AND 4) NOR, 26. The truth table for NOT gate is, 1 0 , 1 0 , 0 1, 1 1 , 1) , 2) , 3) , 4) , , , , , 0 0, 0 1, 1 1, 0 0, , LEVEL-I (H.W) - KEY, 1) 2, 8) 1, 15) 3, 22) 3, , 2) 1, 9) 4, 16) 3, 23) 3, , 3) 2, 10) 2, 17) 3, 24) 1, , 4) 2, 11) 2, 18) 4, 25) 4, , 5) 4 6) 1 7) 4, 12) 2 13) 2 14) 1, 19) 3 20) 4 21) 2, 26) 3, , LEVEL-I (H.W) - HINTS, 1., 2., 3., , E, , hc, , 12000, , , , , , 12400 A, , 0, , e , v, , ne nh, E, , V, d, , ;, , 1024, 106, 1018, , NARAYANAGROUP, , SEMI CONDUCTOR DEVICES, 4., , Since Va Vb, , 5., 6., 7., , diode does not conduct in reverse bias as V p Vn, n Hz, 2n Hz, , 8., , Ie =Ib +Ic, , 9., , IB I E IC, , Ic, 10. I , Ie Ib Ic, e, Ic, 11. I, , b, , Ic, 12. I and I e Ib I c, b, Ic, 13. I, b, , 14. , , , 1 , , 15. , , I c, , and I, 1 , b, , 16. VBE I e Ri, 17. VCB IC Rout, RL, 18. AV R, i, , 28, 2 40, 2 20, 19., 2 1 0, 0 1, , 810 1000 2 , , 21 0, , 20. 110 2 1 22 1 21 0 20 6, 21. If A=1, B=0, then A B 1 0 0 0 0 B, 22. A A and B B, 23. acording to Boolean algebra, 1+1=1, 24. According to Demorgon’s theorem, A B A B, 25. Y A B is the output of NOR gate, 217
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 5., , LEVEL-II (C.W), INTRINSIC,EXTRINSIC, SEMI CONDUCTORS AND DIODES, 1., , 2., , P, , N, , N, , P, , a), , Mobilities of electrons and holes in a sample, of intrinsic germanium at room temperature, , P N, , P N, , N, , N, , are 0.36m 2 / Vs and 0.17m 2 / Vs .The electron, and hole densities are each equal to 2.5, 1019 m 3 .The electrical conductivity of, germanium is, , b), , 1) 0.47 S / m, , 1) circuits a and b, 2) circuits b and c, 3) circuits a and c, 4) all the circuits, If V1 > V2, r is resistance offered by diode in, forward bias then current through the diode is, , P, , 2) 5.18 S / m, , 3) 2.12 S / m, 4) 1.09 S / m, 6., In a p-n junction diode the thickness of, deplection layer is 2 106 m and barrier, potential is 0.3V. The intensity of the electrical, field at the junction is, 1) 0.6 106 Vm 1 from n to p side, , R, , V1, , V2, , V1 V2, V1 V2, 3), 4) None, R r, Rr, A PN junction diode when forward biased has, a drop of 0.5V which is assumed to be, independent of current. The current in excess, of 10mA through the diode produces large, joule heating which damages the diode. If we, want to use a 1.5V battery to forward bias the, diode, the resistor used in series with the diode, so that the maximum curent does not exceed, 5mA is, , 1) 0 2), 7., , 3) 1.5 105 Vm 1 from n to p side, 4) 1.5 105 Vm 1 from p to n side, A potential barrier V volts exists across a P-N, junction. The thickness of the depletion region, is ‘d’. An electron with velocity ' v ' approches, P-N junction from N-side. The velocity of the, electron acrossing the junction is, , P, , 1), , 2Ve, v , m, 2, , 2), , P, , c), , 2) 0.6 106 Vm 1 from p to n side, , 3., , Two similar p-n junctions can be connected in, three different ways as shown in the figures., The two connections across which the potential, difference is same are, , 2Ve, v , m, , N, , 0.5V, , 2, , R, 1.5V, , 4., , 2Ve, 3) v, 4), m, In the following , reverse biased diode is, +10, R, , 1), , -12V, , R, , 2), , +5V, , , , 8., , , , 1) 2 x 102 , 2) 2 x 105 , 3) 2 x 103 , 4) 2 x 104 , The circuit shown in figure (1) Contains two, diodes each with a forward resistance of 50, ohm and with infinite reverse resistance. If the, battery voltage is 6 V, the current through the, 100 ohm resistance is., , -10, , +5V, , D1, , 150, , D2, , 50, , 6V, , 100, , R, , 3), , R, -10V, , 4), 1) 0.01A, , 218, , 2) 0.02A 3) 0.03A 4) 0.04A, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 9., , SEMI CONDUCTOR DEVICES, , 4 ideal diodes are connected as shown in the 14. A p–n junction (D) shown in the figure can act, as a rectifier. An alternating curent source (V), cir cuit the cur r ent t hr ough 50 is, is connected in the circuit, 50, , D, , 5V, , 1) 0.1 A, 2) 0.5 A 3) 0.6 A 4) 1 A, 10. Find the effective resistance between A &, B, , R, , , , V, , I, , 1), t, , I, 2R, , R, , 2), A, , t, , B, , I, 2R, , 3), , R, , t, , I, , 2, 3R, 2R, 2), 3), 4) R, 3R, 2, 3, 4), 11 The equivalent resistance of the circuit across, t, 15. In the figure, the input is across terminals A, AB is given by, and C and the output is across B and D. Then, 2, 4, the output is, , 1), , B, , A, , B, , 5, , A, , 4, , 8, , C, , 1) 4, 2) 13, 3) 4 or 13, 4) 4 zero, 12. The equivalent resistance between A and B is, 36, B, , A, , 18, , 1) 36 if VA VB, , 2) 18 if VA VB, , 3) Zero if VA VB and 54 if VA VB, 4) Zero if VA VB and 54 if VA VB, 13. A junction diode is connected to a 10 V source, and 103 rheostat. The slope of load line on, the characteristic curve of diode by (in A/V), 1) 10-1, 2) 10-2, 3) 10-3, 4) 10-4, NARAYANAGROUP, , D, , 1) Zero, 2) Same as the input, 3) Full wave rectified 4) Half-wave rectified, 16. The I-V characteristic of an LED is, I, B, G, Y, R, , I, , O, , O, O, , V, V, , V, R, , R, Y, G, B, , Y G B, , I, , I, , O, , V, 219
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 17. In the figure shown, the potential drop 23. In a single state transistor amplifier, when the, signal changes by 0.02 V, the base current, across each capacitor is (assume diodes, change by 10A and collector current by 1mA., to be ideal), C =4F, If collector load RC = 2kW and RL 10k ,, Calculate Current Gain, Input impedance,, Effective a.c.load, Voltage gain and Power, , C =8F, E=24V, gain., , 1)50, 2k ,1.66k , 83, 8300, 2) 100, 1k ,1.66k , 83, 8300, 3) 100, 2k ,1.66k , 83, 830, 1) 12V, 12V, 2) 16V, 8V, 4) 100, 2k ,1.66k , 83, 8300, 3) zero, 24V, 4) 8V, zero, 18. From the circuit shown below, the maximum, LOGIC GATES, and minimum values of zener diode current are 24. On subtracting 010101 from 101010, we get:, 1) 001011 2) 001100 3) 010101 4) 011111, 5K, I, 25. The minimum number of gates required to, I, realise this expression Z DABC DABC is, 80-120V, 1) One, 2) Two, 3) Eight 4) Five, 50V, 26. In the circuit below, A and B represent two, inputs and C represents the output, the circuit, 1) 6mA, 5mA, 2) 14mA, 5mA, reperesents, 3) 9 mA, 1mA, 4) 3mA, 2mA, 1, , 2, , L, , 10K, , Z, , A, , TRANSISTORS, , C, 19. In an n-p-n transistor circuit, the collector, current is 10mA. If 90% of the electrons, B, emitted reach the collector., 1) the emitter current , the will be 9mA, 1) NOR gate 2) AND gate 3) gate 4) OR gate, 27., In the following circuit the output Y becomes, 2) the base current will be 1 mA, zero for the inputs, 3) the emitter current will be 11mA, A, 4) both 2 & 3, B, 20. The constant of a transistor is 0.9. What, Y, would be the change in the collector current, corresponding to a change of 4 mA in the base, C, current in a common emitter arrangement ?, 1) A 1, B 0, C 0 2) A 0, B 1, C 1, 1) 30mA 2) 63mA 3)36 mA. 4) 3.6 mA, 3) A 0, B 0, C 0 4) A 1, B 1, C 0, 21. A voltage amplifier operated from a 12 volt, battery has a collector load 6kW . Calculate the 28. The combinatin of gates shown below yields, maximum collector current in the circuit., A, 1) 0.5mA 2) 1 mA 3) 3 mA 4) 2mA, X, 22. In a transistor circuit shown here the base, B, 1) NAND gate, 2) OR gate, curent is 35 A The value of the resistor Rb is, 3), NOT, gate, 4) XOR gate, E, C, 29. The logic gate having an output of 1 is, B, RB, , 0, 1, , RL, , (i), 0, 1, , -, , +, , 9V, , 1) 124k, 3) 352k, 220, , 0, 1, , 2) 257k, 4) None of these, , (ii), 0, 1, , (iii), , 1)(iv), , 2)(i), , (iv), , 3)(ii), , 4)(iii), , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 30. In a given circuit as shown the two input wave 4., forms A and B are applied simultaneously, , 6., , For reverse biasing of an ideal diode, the potential, of n-side should be higher then potential of p-side., Only option (4) is satisfying the criterion for reverse, biasing., In (1) one is forward biased and the other one is, reverse biased.In (2) both are forward biased, In (3) both are reverse biased, total P.D, i, total resistence, , 7., , R, , A, Y, , 5., , B, , The resultant wave form at Y is, 1), , 3), , 2), , 4), , 31. The output of the combination of the gates 8., shown in the figure below is, A, , Y, , B, , 2) A B A B, , 1) A + A. B, , , , 3) A.B A.B, , 4), , A B A.B , , E Vb 1.5 0.5, 1, , , 3, i, 5 10, 5 103, , =0.2´ 103 =2´ 102 W, As per given circuit, diode D1 is forward biased, and offers a resistance of 50 ohm. Diode D2 is, reverse biased and as its corresponding resistance, is infinite, no current flows through it. Thus the, equivalent circuit is as shown in Figure(2). As all, the three resistances are in series, the current, through then is, I, , 6V, 50, , 150, 100 , , , 32. The expression of Y in following circuit is, A, D, , Y1, Y2, , , Y, , 1) ABCD, 2) A+BCD, 3) A +B+C+D, 4) AB+CD, 33. How many NAND gates are used in an OR, gate ?, 1) four, 2) two, 3) three 4) Five, , LEVEL-II (C.W) - KEY, 1) 3, 8) 2, 15) 3, 22) 2, 29) 4, , 2) 3, 9) 1, 16) 4, 23) 4, 30) 1, , 3) 2, 10) 2, 17) 2, 24) 3, 31) 1, , 4) 4, 11) 3, 18) 3, 25) 1, 32) 3, , 5) 2, 12) 4, 19) 4, 26) 4, 33) 3, , 6) 2, 13) 3, 20) 3, 27) 4, , LEVEL-II (C.W) - HINTS, , 2.., 3., , , , 1, e( e ne h nh ), , , v, 0.3, 5, 1, , d 2 106 1.5 10 Vm, Its direction from n to p side., E, , v 2 u 2 2as, NARAYANAGROUP, , 150, , 50, , B, C, , 1., , D1, , 7) 1, 14) 3, 21) 4, 28) 2, , 6, A 0.02 A, 300, 6V, , 9., , i, , 100, , V, 5, , R 50, , 10. current does not flow through the diode., 11 If VA VB both the junction diodes are forward, biased and the given circuit diagram becomes a, balanced Wheatestone bridge The equivalent, resistance in this case becomes 4 ,if VA VB the, diodes are reveres biased in that case 4,5 and, 4 are in series, 12. During FB, diode conducts and during RB, diode, does not conduct., I, 13. Slope =, V, 14 Given figure is halfwave rectifier, 15. The given question is based on Bridge Rectifier., This is the most widely used full-wave rectifier. It, makes use of four diodes D1 , D2 , D3 , D4, connected in the four arms of a bridge. Bridge, rectifier does not require a centre-tapped, transformer., 221
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 17. The diode connected parallel to the battery is 28., X A.B A B A B . It is OR gate, reverse bias. so current will not pass through, it. so total emf divided among C1 and C 2 and 29. Y A.B 1.0 1 NAND, in the inverse ratio of their capacities, 30. Logic of A is 1010, Logic of B is 1001, Logic of Y is 0111, q V1 C2, 31., The input to OR gate is A and A.B. Hence, V ;, C V2 C1, Y = A + BA, VC2, VC1, 32. Y1 A D, Y2 A D B ,, V1 , ,V2 , C1 C2, , C1 C2, , 18. V iR, 90, 10, I E I E I C and, 19. I C , 100, 9, 20. , , Ie =Ib +Ic, , I , , , 1 , , C, But I B , VCE, , Y A D B C, 33. For this purpose we use NAND gate in manner as, shown. The first two NAND gates are operated as, NOT gates adn their output are fed to the third The, resulting circuit is OR gate. Here, there gates are, used, A, , A, , I C B 9 4mA, Y=(A.B), =A+B, , Vcc, 21. I c R, L, , B, B, , 22 V ib Rb, RB , , V, 9, , 257 k , ib 35 106, i c, , 23. (i) Current Gain i, , LEVEL-II (C.W), INTRINSIC,EXTRINSIC, SEMI CONDUCTORS AND DIODES, , b, , V, , 1., , BE, (ii) Input impedance Ri i, b, , R R, , C L, (iii) Effective (ac) load RAC R R, C, L, , Ec Conduction band gap, , R, , Eg Forbidden band gap, , AC, (iv) Voltage gain Av R, in, , Ev Valence band gap, , (v) Power gain, Ap Av , 1, 1, , 1, 1, , 1) All Ec , Eg & Ev decrease, , 1, 1, , 1 01 01 0, , 24., , 2) All Ec , Eg & Ev increase, , 0 1 0 10 1, , 3) Ec and Ev increase, but Eg decrease, , 0 1 0 10 1, , 25. Z = DABC + DA BC, , , , , , = DA BC + BC = DA 1, = DA i.e One AND gate., 26. It is diode-diode logic ckt of OR gate, , 27. Here Y A.B .C A.B C A B C, , Y 0 if A 1, B 1 and C 0, 222, , If the lattice constant of this semiconductor, is decreased, then which of the following is, correct?, , 2., , 4) Ec and Ev decrease but Eg increase, A Ge specimen is doped with Al. The, concentration of acceptor atoms is, 10 21 atoms / m 3 . Given that the intrinsic, concentration of electron -hole pairs is, 1 0 19 / m 3 , the concentration of electrons in, the specimen is, 1) 1017 / m3 2) 1015 / m3 3) 104 / m3 4) 102 / m3, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , The following data are for intrinsic germanium 9., at 300 K. ni 2.4 1019 / m 3 ,, , 3., , e 0.39m2 / Vs, h 0.19m2 / Vs. Calculate the, conductivity of intrinsic germanium., 1) 4.3 Sm1 2) 1.21Sm1 3) 2.22 Sm 1 4) 4.22 Sm 1, The diagram correctly represent the, direction of flow of charge carriers in the, forward bias of p-n junction is, , 4., , 1), , P --, , +, +, +, +, , N, , 3), , P --, , +, +, +, +, , N, , 2), , P --, , +, +, +, +, , N, , 4), , P --, , +, +, +, +, , N, , Two ideal junction diodes D1, D2 are connected, as shown in the figure. A 3V battery is, connected between A and B. The current, supplied by the battery if its positive terminal, is connected to A is, D1, , 20, D2, , 6., , 7., , 8., , 0V, , -5V, , B, , 4, B, , A, , In the figurs shown below, , +10V, , A, , 1) 0.1A, 2) 0.3 A 3) 0.9 A 4) 90 A, 10. Find the effective resistance between A and B, 2, , 5., , 10, , 3, , 6, , -2V, , 1) 5/18 2) 9/5 3) 18/5 4) 5/9 , 11. The peak voltage in the output of a half-wave, Fig. b, Fig. a, diode rectifier fed with a sinusoidal signal, 1) In both Fig a and Fig b the diodes are, without filter is 10V. The d.c. component of the, forward biased, output voltage is, 2) In both Fig, a and Fig b the diodes are, reverse biased, 2) 10 / V, 1) 10 / 2 V, 3) In Fig a the diode is foward biased and in, 3) 10 V, 4) 20 / V, Fig b, the diode is reverse biased, 4) In Fig a the diode is reverse biased and in 12. In the figure shown the potential drop across, the series resistor is, fig b, it is forward biased, A P-N junction diode can withstand currents, 2K, up to 10 mA. Under forward bias, The diode, 120V, 90V, 20K, has a potential drop of 0.5 V across it which is, assumed to be independent of current.The, maximum voltage of the battery used to, forward bias the diode when a resistance of, 1) 30 V, 2) 60 V 3) 90 V 4) 120 V, 13. A 220V ac supply is connected between points, 200 is connected in series with it is, A and B (Fig). What will be the potential, 1) 2.5V, 2) 2.6V 3) 2.7V 4) 2.8V, difference V across the capacitor?, A cell of emf 4.5V is connected to a junction, 1) 220V 2) 110V, diode whose barrier potential is 0.7V. If the, A, external resistance in the circuit is 190 , the 220V ac, C, V, current in the circuit is, 3) 0V, 4) 220 2V, B, 1) 20 mA 2) 2m A 3) 23mA 4) 200mA, 14 In the circuit shown (Fig). if the diode forward, VA and VB denote potentials of A and B, then, voltage drop is 0.3V, the voltage difference, the equivalent resistance between A and B in, between A and B is :, the adjoining circuit is(ideal diode), A, 1) 1.3V, 30, 0.2mA, A, , B, , 5K, , 2) 2.3V, C, , 30, , 1) 15 if VA VB, 3) Both 1 and 2, NARAYANAGROUP, , 2) 30 if VA VB, 4) neither 1 nor 2, , 3) 0, , 5K, B, , 4) 0.5V, 223
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , 25. The tuned collector oscillator circuit used in, the local oscillator of a radio receiver makes, In a n-p-n transistor 10 electron enter the, -6, use of a tuned circuit with L = 60 H and C =, emitter in 10 s. If 2% of the electrons are, 400 pF. Calculate the frequency of oscillations., lost in the base, the current transfer ratio, and the current amplification factor are, 1) 1.03 KHz, 2) 1.03 Hz, 1) 0.98, 49, 2) 0.49, 49, 3) 1.03 GHz, 4)1.03 MHz, 3) 0.98, 98, 4) 0.49, 98, LOGIC GATES, In a common base mode of transistor, collector, current is 5.488 mA for an emitter current of 26. When we add binary numbers 111 and 111 we, get the binary number:, 5.60mA. The value of the base current, 1) 222, 2) 1000 3) 1110, 4) 000, amplification factor () will be:, 1) 48, 2) 49, 3) 50, 4) 51, 27. If A= B=C=1 and X ABC BCA C AB ,, Current amplification factor of a common base, then X=, configuration is 0.88. Find the value of base, 1) 0, 2) 1 3) 100, 4) 110, current when the emitter current is 1 mA., 28. The input of A and B for the, 1) 0.12 mA.2)0.1 mA 3) 0.5 mA 4) 0.2 mA, _______ _______ , For a transistor and IB 25A . Find the, Boolean expression A B . A.B 1 is, value of IE ., , , , 1) 1mA, 2) 1.025mA 3) 2mA 4) 1.2 mA, 1) 0,0, 2) 0, 1, 3) 1,0, 4) 1, 1, IC, IC, 29. Write the name of the follwoing gate that the circuit, In a transistor if I and I , If varies, shown in figure., E, B, , TRANSISTORS, , 15., , 16., , 17., , 18., , 19., , 10, , between, , 20, 100, and, ,then the value of, 21, 101, , lied, , between, 1)1-10 2)0.95-0.99 3)20-100 4)200-300, 20. For a transistor x , , 21., , 22., , 23., , 24., , 224, , 1, 1, & y where & are, , , , +5V, D1, V0, D2, , current gains in common base and common, emitter configuration. Then, 1)AND gate 2)OR gate 3)NOR gate 4)XOR gate, 1) x y 1, 2) x y 1, 30. Consider a two-input AND gate of figure below., 3) 2 x 1 y, 4) x y 0, Out of the four entries for the Truth Table given, A voltage amplifier operated from a 12 volt, here, the correct ones are, battery has a collector load 6kW . Calculate the, Input Output, maximum collector current in the circuit., B, Y, A, A, 1) 0.5mA 2) 1 mA 3) 3 mA 4) 2mA, 1 0, 1, 0, Y, 2 1, 0, 0, A CE amplifier is designed with a transistor, B, 1, 1, 3 1, having 0.99 . Input impedance is 1 k and, 4 0, 0, 1, load is 10 k . Voltage gain will be:, 1) All, 2) 1 and 2 only, 1) 9900, 2) 99000 3) 99, 4) 990, 3) 1, 2 and 3 only, 4) 1, 3 and 4 only, In a common emitter amplifier the load 31. A Truth table is given below. The logic gate, resistance of the output circuit is 792 times the, having following truth table is, resistance of the input circuit. If = 0.99,, A, B, Y, the voltage gain is, 0, 0, 1, 1) 79200 2) 39600 3) 7920 4) 3960, 1, 0, 0, I n a t r ansistor amplifier = 62, RL = 5000 , 0, 1, 0, and internal resistance of the transistor is, 1, 1, 0, 500 . Its power amplification will be, 1)NAND gate, 2)NOR gate, 1) 25580 2) 33760 3) 38440 4) 55760, 3)AND gate, 4)OR gate, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , LEVEL-II (H.W) - HINTS, , 32. For a logic 0101 the waveform is, 1), , 2), , 3), , 33., , 2., , When Ge specimen is doped with Al, then, concentration of acceptor atoms is also called, concentration of holes. Using formula, ni 2 n0 p0, where, n1 concentration of electron hole pair = 1019 m 3, , 4), , For the given combination of gates, if the, logic states of inputs A, B, C are as follows, A=B=C=0 and A=B=1, C=0 then the logic, states of output D are, , n0 concentration of electron, p0 concentration of holes = 1021 atom m 3, 2, , A, , 1019 10 21 n0 n0 1017 m3, , B, , 3., 4., , Y, , C, , 5., 1) 0,0, 2) 0,1, 3) 1, 0, 4) 1,1, 34. Identify the gate represented by the block, diagram as shown in fig., 6., A, Y, B, 7., 1) AND gate, 2) NOT gate, 3) NAND gate, 4) NOR gate, 8., 35. The Boolean expression for the gate, 9., circuit shown below is, , A, , Y, , i ni e(e h ), In forward bias holes move from P to n, side, electrons move from n to P, for forward bias, V p Vn , and for reverse, bias V p Vn, i, , E Vb, R, , V =4.5- 0.7=3.8V ; R 190 , i =, , V, R, , R in the case of reversebias, If positive terminal is connected to A, D1 is in, forward bias and D2 is in reverse bias. So 20W is, ineffective., , V, 3, , 0.3 A, R 10, 10. Use Wheat stones bridge principle, i , , 1) A A 1, 2) A 1 1, V, 10, 3) A A A, 4) A 0 A, 11. Vdc m volt, , , 36. The output Y of the gate circuit shown in the, 12. Vin V VZ, figure below is, 13. Capacitor once gets charged upto maximum, A, Y, potential .After that for any other lesser value, B, of p.d across A and B diode is reverse biased, 1) A.B, 2) A.B, 3) A.B, 4) A B, and it does not allow charge to flow in opposite, direction., , LEVEL-II (H.W) - KEY, 1) 4, 8) 3, 15) 1, 22) 4, 29) 1, 36) 2, , 2) 1, 9) 2, 16) 2, 23) 1, 30) 3, , 3) 3, 10) 3, 17) 1, 24) 3, 31) 2, , NARAYANAGROUP, , 4) 3, 11) 2, 18) 2, 25) 4, 32) 1, , 5) 3, 12) 1, 19) 3, 26) 3, 33) 4, , 6) 1, 13) 4, 20) 2, 27) 1, 34) 4, , 7) 1, 14) 2, 21) 4, 28) 1, 35) 1, , FromVrms , , Vmax, 2, , Vmax Vrms 2 220 2 volts, VA 0.2 103 5 103 0.3, 3, 3, 14. 0.2 10 5 10 VB 0, VA VB 2.3 volt, , 225
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 15., , Ic, 98, , , Ic , Ie , , Ie, 100, 1 , , , , I, , C, 16. I e Ib I c and I, B, 17. In a common-base arrangement, the current, amplifica-tion factor., , IC, IE, , , , 35. The input of OR gate is A & A, Y A A 1, 36. The input of AND gate is A and B Y A.B, , LEVEL-III, 1., , , , that of holes in a semi-conductors is, , and IB IE IC, I, , C, 18. I IC B 40 25 106 and I E I B IC, B, , , 1, I c, I c, 20. I , I, e, b, , the ratio of currents is, , 19. , , Battery voltage, , 2., , 12 volt, , 21. I Collector load , =, 6k, 22. , , R AC, , and Av R, 1 , in, , A RL, , V, Ri, 1 , 2 RL, 24. AP R, i, 23. , , 25. , , 26., , 4) 88.0cm 2 / Vs, , A pure silicon crystal of length l 0.1m and, , e , , and holes h as 0.135m 2 / Vs and, , 0.048m 2 / Vs , respectively, If the voltage, applied across it is 2V and the intrinsic charge, , 2 LC, , concen-tration is ni 1.5 106 m 3 , then the, total current flowing through the crystal is., 1) 8.78 1017 A, 2) 6.25 1017 A, , 1 11, , A B 0 , A.B 0, , 30. Check the truth table, 31. Y A B, B, 1, 0, 1, 0, , Y, 1, 0, 0, 0, , 34. The boolean expression of this gate is, Y=(A+B).(A+B) = (A+B) + (A+B) = (A+B), which is for NOR gate., 226, , 7, , then the ratio of, 4, , 4, 2, area A 10 m has the mobility of electron, , 1, , 1110, 27. Check the value of x by substituing A=B=C=1, 28. A B 1, A.B 1, , 33., , 3., , 7, and, 5, , drift velocities is, 1) 5/8, 2) 4/5, 3) 5/4, 4) 4/7, If the resistivity of copper is 1.7 10 6 cm,, then the mobility of electrons in copper, if each, atom of copper contributes one free electron, for conduction, is [The atomic weight of copper, is 63.54 and its density is 8.96 g/cc]:, 2) 503.03cm 2 / Vs, 1) 23.36cm 2 / Vs, 3) 43.25cm2 / Vs, , 111, 111, , A, 1, 1, 0, 0, , If the ratio of the concentration of electrons to, , IC E 0.88 1 0.88mA., , 4., , 3) 7.89 1017 A, 4) 2.456 1017 A, Find the current produced at room temperature, in a pure germanium plate of area 2 10 4 m2, and of thickness 1.2 103 m when a potential, of 5 V is applied across the faces., Concentration of carriers in germanium at room, temperature is 1.6 106 per cubic metre. The, mobilities of electrons and holes are, 0.4m 2V 1s 1 and 0.2m 2V 1s 1 respectively.., The heat energy generated in the plate in 100, second is, 1) 2.4 1011 J, 2) 3.4 1011 J, 3) 5.4 1011 J, , 4) 6.4 1011 J, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, 5., , SEMI CONDUCTOR DEVICES, , An n-type semiconductor has impurity level 10. For a CE-transistor amplifier, the audio signal, 20meV below the conduction band. In a, voltage across the collector resistance of 2k, thermal collision, transferble energy is KT., is 2V. Suppose the current amplification factor, The value of T for which electrons start to jump, of the transistor is 100. Find the input signal, in conduction band is :, voltage and base current, if the base resistance, 1) 232 K 2) 348 K 3) 400 K 4) 600 K, is 1k ., Assume that the number of hole-electron pair, 1) 0.02V 2) 0.01V 3) 0.03V 4) 0.04V, in an intrinsic semiconductor is proportional 11. In the circuit, transistor has a current, to e E / 2 KT . Here E = energy gap and, () 100 . What should be the base resistor, 5, k 8.62 10 eV / kelvin, R B neglect VBE , so that VCE 5V :, The energy gap for silicon is 1.1eV. The ratio, R =1k, of electron hole pairs at 300K and 400 K is :, RB, 2, , 5.31, , 5, C, 1) e, 2) e, 3) e, 4) e, VCC=10V, VCE, B, E, I n the cir cuit shown in figur e (1), t he V 0,I1, I D1 ,, , 6., , C, , 7., , and I D3 are respectively, , 1) 200 k 2) 1 k 3) 500 k 4) 2 k, 12. An N P N transistor is connected in, l, l, l, l, +, +, v, v, D, D, common - emitter configuration in which, = 10V, Si, Si, 0.7V –, 10V, – 0.7V, collector supply is 8V and the voltage drop, –, –, across the load resistance of 800 connected, (A), (B), in the collector circuit is 0.8V. If current, 1) 0.5V, 25 mA , 15 mA, amplification factor is 25/26 (If the internal, 2)0.7V, 28.18 mA , 14.09 mA, resistance of the transistor is 200 ), the, collector-emitter, voltage, voltage gain and, 3) 0.4V, 15 mA , 20 mA, power gain are respectively., 4) 0.3V, 15.06 mA , 20.18 mA, 1) 5.2V, 1.86, 3, 2) 6.2V, 186, 5.5, 8. For a junction diode , the ratio of forward, 3) 7.2V, 3.86, 3.698 4) 8.2V, 4.91, 3.15, 13 For a CE transistor amplifier, the audio signal, current I f and reverse current is, voltage across the collector resistance of 2 k , [ I e electronic charge,, is 2V. Suppose the current amplification factor, V = voltage applied across junction,, of the transistor is 100. The value of RB in, k = Boltzmann constant, series with VBB supply of 2V, if the DC base, T = temperature in kelvin], eV / kT, current has to be10 times the signal current is, 1 4) eV / kT 1, 1) e v / kT, 2) eV / kT 3) e, 1) 4 k, 2) 14 k 3) 28 k 4) 54 k, 9. In the diagram D an ideal diode and an 14 Figure shows the transfer characteristics of a, alternating voltage of peak value 10V is, base biased CE transistor.Which of the, connected as input V1 . Which of the following, following statements are true ?, diagram represents the correct waveform of, V, output voltage V ?, 0.33 k, , +, , +, , 1, , D1, , 1, , 1, , D2, , 0, , 2, , 0, , 0, , R, D, Vi, , , , 5V, , +5V, , 0, , +, V0, –, , 0, (1), , +10V, +5V, , 0, 0, (2) –5V, , NARAYANAGROUP, , 2V V, , A) At V1 0.4 V transistor is in active state, , +5V, 0, , 0.6V, , 0, , 0, , +5V, 0, , 0, (3) –5V, , (4), , –10V, , B) At V1 1 V it can be used as an amplifier, C) At V1 0.5 V , It can be used as a switch turned offf, D) At V1 2.5 V , it can be used as a switch turned on, 1)A,B,C 2)B,C,D 3)A,C,D 4)A,B,D, 227
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 15. The following configuration of gates is, equivalent to, A, , 19. The logic circuit shown below has the input, waveforms ‘A’ and ‘B’ as shown. Pick out the, correct output waveform, , A, , B, Y, , Y, B, 1) NAND gate, 2) XOR gate, 3) OR gate, 4) NOR gate, 16. The combination of the gates shown below, produces, , Input A, , Input B, , G1, , A, , G3, , B, , G4, , G2, , 1), , 1) AND gate, 2) XOR gate, 3) NOR gate, 4) NAND gate, 17. Which of the following truth tables is true?, 2), , A, Y, B, , 1), , A B Y 2) A, 0 0 0, 0, 1 0 0, 1, 0 1 0, 0, 1 1 0, 1, 3) A B Y 4) A, 0 0 0, 0, 1 0 1, 1, 0 1 1, 0, 1 1 1, 1, 18. Truth table for system of, as shown in figure is :, A, , B Y, 0 0, 0 0, 1 1, 1 1, B Y, 0 0, 0 1, 1 1, 1 0, four NAND gates, , Y, B, , 228, , 1), , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 0, 1, 1, 0, , 3), , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 1, 1, 0, 0, , 2), , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 0, 0, 1, 1, , 4), , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 1, 0, 0, 1, , 3), , 4), , 20. Logic gates X and Y have the truth tables, shwon below, , P, Q, , R, , X, , P, , Y, , R, , P, , Q, , R, , P, , R, , 0, 1, 0, 1, , 0, 0, 1, 1, , 0, 0, 0, 1, , 0, 1, , 1, 0, , When the output of X is connected to the input, of Y, the resulting combination is equivalent, to a single ., 1) NOT gate, 2) OR gate, 3) NOR gate, 4) NAND gate, , LEVEL-III - KEY, 1) 3 2) 3 3)1 4) 4 5)1 6) 1 7) 2, 8) 3 9) 4 10) 2 11)1 12) 3 13) 2 14) 2, 15) 2 16) 4 17)1 18)1 19)1 20) 4, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , LEVEL-III - HINTS, 1., , SEMI CONDUCTOR DEVICES, 5., , I n e Avd, , T , , I e ne vd e, , I h nh vd h, , , , ne 7 I e 7, Here n 5 , I 4, h, h, , 6., , 7 7 vd e, , 4 5 vd h, , , 2., , vd e, vd h, , Mobility of electron , , 7., , , - (1), ne, , 1, , , N0 d, atomicweight, , 8., , AV, I e ne e nh h , l, Putting the values, we get, I 8.78 1017 A, ne n h ni, , ni e e h 1.6 1.6 0.6 1013, Current produced in germanium plate., V , I JA E . A A, d , Heat generated in the plate, H V I t, 13, 11, 5 1.28 10 100 6.4 10 joule, , 1.1, , 28.62105, , VR E VD, , R, R, , , , 10V 0.7V, 0.33k , , =, , I1, 14.09mA, 2, , Current in junction diode , l f l0 eeV / kT 1 In, forward baising, V is positive and in reverse bias V, is negative Then ,, I r I0, If, Ir, , 43.25cm 2 / Vs, , NARAYANAGROUP, , 1 , 1, , , , e, e, 400 300 , Since applied voltage is greater than 0.7 V and, direction of current is same as direction of arrow, head, both the diodes are in “on” state (see above, figure(2)). Now V 0is parallel to D1and D2., Hence V0=0.7 V, , , Hence I D1 I D2 , , 1, , 22, 8.5 10 1.6 1019 1.7 106, , 4., , N1 e E / 2 KT1, , N 2 e E / 2 KT2, , 28.18 mA, Bot he t he silecon diodes have similar, characteristics., , 6.023 1023 8.96, n, 8.5 1022, 63.54, From Eqs. (iii) and (iv) , we get, , 3., , 20 103 eV, = 232 K, 8.62 105 eV / K, , Current I1 , , 1, From Eqs. (i) and (ii) , we get , ne , Here, n = number of free electrons per unit volume, n, , 20 10 3, eV, k, , E 1 1 , , 2 K T2 T1 , , 5 7 5, , 7 4 4, , also resistivity , , KT 20 103 eV, , , , I 0 (eeV / KT 1), , I0, , e eV / kT 1, , 9., , For V1 5V , the diode is forward-biased, output, will be fixed at 5V. For V1 5V , the diode is, reverse biased. The output will follow V, , 10. Here; v0 2V , 100,, RC (collector resistance) = 2 k , RB 1 k , , RC , RB, v0, As Av v R , vi v0 R, 1, B, C, , 1 , , , , 1k 1 , 2V , , 0.01V, 2 k 100 , 229
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, 11. Applying KVL to QCETSRQ, we get, V I R V and calculate I, CC, , C, , C, , CE, , C, , The signal current through the base is therefore ,, given by, , iB , , ic, , , The D.C base current has to be10 iB, RB , , , IC, I, IB C, IB, , , Applying KVL to PUTSRQP, we get, V I R V here V is negligible, CC, , B, , VCC, , B, , BE, , BE, , V, I B RB I B CC, RB, , 12. The circuit arrangement is shown in figure, lC, C, lB B, , VCE, , RB, , E, , + VBE, , 800 RL 0.8V, , -, , +, , VCE= 8V, , collector current IC =, , , Voltage drop across R L, RL, , 0.8, 103 amp., 800, , 26, 103 1.04 103 amp., 25, , R, Voltage gain L, RC, , 2 0.6 14 k , , 0.10, 14. From the given transfer characteristics of a base, biased common emitter transsistor, we note that, ii) When V1 1V ( which is in between 0.6V to, 2.V), the transistor circuit is in active state and it, used as an amplifier, iii) When V1 0.5V there is no collector current ,, The transistor is in cut-off state. The transistor circuit, can be used as a switch turned off, iv) When V1 2.5V the collector current becomes, maximum and transistor is in saturation and can be, used as switch turned on state, , , , 2, , 16. The output of G1 is A G 2 is B . Hence output, of G 3 is AB y = AB i.e., NAND gate., , , , 17. Y A.B . A.B, , , , , , 230, , , , , , , 18. Y A. A.B . A.B.B A. A.B A.B.B , , , , , , , , , A. A A.B A.B B.B, , , , Y A.B A.B ., 19. Truth table:, , A B, 1 1, 1 0, 13 The output AC voltage is 2V. So the AC collector, 0 1, 0 0, 2, 1mA, current ic , 20., Verify, truth, table, 2000, 2, , , , , , , , RL 25 800, , RC 26 200, , 25 , 4 3.698, 26 , , , , 15. Y A B . A.B A B . A B, , Y A. A B A B .B, , 25 800 100, , , 3.846, 26 200 26, , Power gain 2, , , VBE 0.6V, , Y A.B A.B . it is XOR gate, , IC, 25 10 3, , , , Current gain, I B (or) 26, IB, , , , lB, , A. A A.B A.B B.B, , Now VCE 8 0.8 7.2 volt, , IB , , RB , , VBB VBE , , Y, 1, 0, 0, 0, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- V, , SEMI CONDUCTOR DEVICES, , LEVEL-IV, , (R) y A.B, , COMPREHENSION TYPE, , (S) y = A + B, , PASSAGE - I, A block of pure silicon at 300K has a length of, 10cm and an area of 1.0cm2 . A battery of emf, 2V is connected across it. The mobility of, 2 1, , 1) A R, B S , C p, D Q, 2) A S , B R , C P , D Q, 3) A P, B Q, C R, D S, , 1, , electron is 0.14 m v S and their number, density is 1.5 1016 m-3. The mobility of holes, 1., , is 0.05m 2 v 1 S 1, The electron current is, 4, , 2., , 3., , 4) A S , B P , C Q, D R, , LEVEL-IV - KEY, 5, , 1) 6.72 10 A, , 2) 6.72 10 A, , 3) 6.72 106 A, The hole current is, , 4) 6.72 107 A, , 1) 2.0 107 A, , 2) 2.2 107 A, , 1) 4, , 2) 3, , 7) 2, , 8) 1, , 3) 4, , 4) 2, , 5) 4, , 6) 3, , LEVEL-IV - HINTS, 1., , E, , V, 2, , 20V / m, l 0.1, , 3) 2.4 107 A, 4) 2.6, The total current in the block is, , A 1.0cm 2 , ve e E 0.14 20 2.8ms 1, , 1) 2.4 107 A, , I e ne AeVe 6.72 10 7 A, , 2) 6.72 107 A, , 3) 4.32 107 A, 4) 9.12 107 A, PASSAGE:2, The input and output resistances in a common base, amplifier circuits are 400 and 400K , respectively. The emitter current is 2mA and current, gain is 0.98., 4. The collector current is, 1) 1.84mA 2)1.96mA 3)1.2mA 4)2.04mA, 5. The base current is, 1) 0.012mA, 2) 0.022mA, 3)0.032mA, 4) 0.042mA, 6. Voltage gain of transistor is, 1)960, 2)970, 3)980, 4)990, 7. Power gain of transistor is, 1) 950, 2) 960, 3)970, 4)980, 8. The logic circuits are given in column I and, the Boolean expressions in column II., Column - I, Column - II, (P) y A B, , 2., , In a pure semi conductor,, ne nh 1.5 1016 m 3, , Vh h E 0.05 20 1.0m / s, I h nh Aevh 2.4 107 A, , 3., , Total current, , I Ie Ih, 9.12 107 A, 4., , I c I e 0.98 2 1.96mA, , 5., , I b I e I c 2 1.96 0.04mA, , 6., , Av , , 7., , Power amplification,, , R0, 400 103, 0.98 , 980, Ri, 400, , AP Av, 0.98 980 960, , ***, , (Q) y A.B, NARAYANAGROUP, , 231
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , COMMUNICATION SYSTEMS, Ø, , SYNOPSIS, Introduction: Communication is an act of, , Ø, , exchange of information between the sender and, the receiver. Over decades, methods have been, evolved to develop languages, codes, signals etc, to, make, communication, effective., Communication through electrical signals has, made things much simpler because they can be, transmitted over extremely large distances in, extremely short time as their speed is, Ø, 3 ×108 m / s ., Modern communicaton has its roots in the 19t h, and 20th century in the work of scientists like, J.C. Bose, F.B. Morse, G.Macroni and Alexander, Graham Bell. The pace of development seems, to have increased dramatically after the first half, of the 20th century. We can hope to see many, more accomplishments in the coming decades., The aim of this chapter is to introduce the, concepts of communication, namely the mode of, communication, the need for modulation,, production and deduction of amplitude, modulation., , Communication is basically of two types:, a) Point to point :- This takes place between a, transmitter and a receiver. Telephonic, conversion between two persons is a good, example of it., b) Broad cast mode :- Here, a large number of, receivers receive the information from a single, transmitter. Radio and television are good, examples of broadcast mode., , Elements of Communication System, Basic units of a communication system., Transmitted signal, Information, source, , Transmitter, , Message, signal, , Channel, , Noise, , Ø, , Message signal, , Receiver, , User, Information, , Ø, , Received, signal, , a) Transmitter: The part of the communication, system, which sends out the information is called, transmitter., b) Transmission channel: The medium or, the link, which transfers message signal from the Ø, transmitter to the receiver of a communication, system is called channel., c) Receiver: The part of the communication Ø, system, which picks up the information sent out, by the transmitter is called receiver. The receiver Ø, consists of, 26, , Basic Terminology Used in Electronic, Communication System, Some important terms needed to understand the, basic elements of communication, a) Information : It is nothing but, the message, to be conveyed. The message may be a symbol,, code, group of words etc. Amount of information, in message is measured in “bits”, b) Communication Channel : Physical medium, through which signals propagate between, transmitting and receiving stations is called, communication channel., Transmitter: Essential components of, transmitter are as follows., a)Transducer : Converts sound signals into, electric signal. The device which converts a, physical quantity (information) into electrical, signal is known as transducer., b)Modulator : Mixing of audio electric signal, with high frequency radio wave., c)Amplifier: Boosting the power of modulated, signal., d)Antenna : Signal is radiated in the space, with the aid of an antenna., Receiver: Basic componenets of receiver., a) Pickup antenna: To pick the signal, b) Demodulator: To separate out the audio, signal from the modulated signal, c) Amplifier: To boost up the weak audio signal, d) Transducer: To convert back audio signal, in the form of electrical pulses into sound waves., Message Signal: Information converted in, electrical form and suitable for transmission is, called signal., A signal is defined as a single-valued function, of time (that conveys the information) and which,, at every instant of time has a unique value., Types of message signals, a) Analog signal: A signal, which is a, continuous function of time (usually a sinusoidal, function) is called analog signal., b) Digital signal: A discrete signal, (discontinuous function of time) which has only, two levels is called digital signal., Noise : This refers to undesired signals which, disturb the transmission and processing of, signals., Attenuation : It is the loss of strength of a signal, during propagation in a medium., Amplification : It is the process of increasing, the strength of the signal (amplitude) using an, amplifier., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , Ø, Ø, , Range :It is the maximum distance from a source, upto which the signal is received with sufficient, strength., Repeaters : These are the devices used to, increase the range of communication system, Band width of signals (speech, T.V and digital, data) : Band width is the frequency range over, which an equipment operates., (or), It is the portion of the spectrum occupied by the, signal., , (voltage), -0.6, -0.4, -0.2, 0, 0.1, , extending to infinity and, , f 0 = 1/ T0 . The, , ( f0 ) , fundamental ( f0 ) + second, harmonic ( 2 f 0 ) and fundamental, ( f0 ) +, second harmonic ( 2 f 0 ) + third harmonic ( 3 f 0 ) ,, fundamental, , are shown in the same figure to illustrate this, fact. It is clear that to reproduce the rectangular, wave shape exactly we need to superimpose all, the harmonics f 0 , 2 f 0 , 3 f 0 , 4 f 0 ..... which, implies an infinite bandwidth. However, for, practical purposes, the contribution from higher, harmonics can be neglected this limiting the, bandwidth. As a result, received waves are a, distorted version of the transmitted one. If the, bandwidth is large enough to accommodate a few, harmonics, the information is not lost and the, rectangular signal is more or less recovered. This, is so because the higher the harmonic. Less is, its contribution to the wave form., NARAYANAGROUP, , 0.3, , 0.4, , unit, , 0.5, , -0.2, -0.4, -0.6, , a) Rectangular wave, b) Fundamental ( f 0 ), , Bandwidth of signals :, In a communication system, the message signal, may be voice, music, picture or data etc. Each, of these signals has a spread of different range, of frequencies. Hence, the type of communication, system needed depends upon the band of, frequencies involved. Speech signal requires the, band width of 2800 Hz (3100 Hz to 300 Hz)., For music, a bandwidth of about 20KHz is, required (due to high frequency produced by, musical instruments). The audible range of, frequencies extends from 20Hz to 20KHz. Video, signals require band width of 4.2 MHz for, picture transmission. However, a band width of, 6MHz is needed for T.V signals. (as it contains, both voice and picture), Digital signals are in the form of rectangular, waves as shown in Fig. One can show that this, rectangular wave can be decomposed into a, superposition of sinuosidal waves of frequencies, f 0 . 2 f 0 ,3 f 0, 4 f 0....nf 0 where n is an integer, , 0.2, , (time ), , Ø, , COMMUNICATION SYSTEM, , ( f0 ) + second harmonic ( 2 f0 ), d) Fundamental ( f 0 ) Second harmonic, ( 2 f0 ) + third harmonic ( 3 f0 ), c) Fundamental, , Ø, , Bandwidth of transmission medium :, , The most used transmission media are wire, free, space, and fibre optic cable. Different, transmission media offer different band width., Coaxial cable offers a band width of about 750, MHz. Radio wave communication through free, space takes place over a wide range of, frequencies from 100kHz-GHz., Service, Frequency bands Comments, ----------------------------------------------------------Standard, 540-1600 kHz, AM broadcast, --------------------------------------------------FM broadcast 88-108 MHz, ----------------------------------------------------Television, 54-72 MHz, VHF(Very high, frequencies), 76-88 MHz, TV, 174-216 MHz, UHF(ultra high, frequencies), 420-890 MHz, TV, ----------------------------------------------------------Cellular Mobile 896-901 MHz, Mobile to base, station, 840-935 MHz, Base station to, mobile, ----------------------------------------------------Satellite, 5.925-6.425 GHz, Uplink, Communication 3.7 - 4.2 GHz, Downlink, --------------------------------------------------Ø Optical communication using fibres is performed, in the frequency range of 1 THz to 1000 THz, (microwaves to ultraviolet)., Ø An optical fibre can offer a transmission band, width in excess of 100 GHz., Ø COMMUNICATION CHANNELS:, The medium or the link, which transfers message, signal from the transmitter to the, receiver, of a communication system is called, 27
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COMMUNICATION SYSTEM, Communication channel., (i) Space communication, (i) Ground wave propagation, (ii) Space wave propagation. (Tropospheric, wave propagation. Surface wave propagation.), (iii) Sky wave propagation: A new dimension, recently added to space communication is, satellite communication., ·, 2) Line communication, i) Two wire transmission line, ii) Coaxial cable, iii) Optical fibre cable, (1)SPACE COMMUNICATION, ØPropagation of EM waves in the atmosphere, The communication process utilizing the, physical space around the earth is termed as, space communication. Electromagnetic waves, which are used in Radio, Television and other, communication system are radio waves and, microwaves., The velocity of electromagnetic waves of, different frequency in a medium is different. It, is more for red light and less for violet light., Electromagnetic waves are of transverse nautre., Ø Earth’s atmosphere, (i) Earth’s atmosphere is a gaseous envelope, which surrounds the earth., (ii) Earth atmosphere mainly consists of nitrogen, 78% , oxygen 21% along with a little protion of, argon, carbon dioxide, water vapour,, hydrocarbons, sulphur compounds and dust, particles., (iii) The density of the atmospheric air goes on, decreasing as we go up., (iv) The electrical conductivity of the, atmospheric air increases as we go up., (v) The various regions of earth’s atmosphere are:, Ø Troposphere. It extends upto a height of 12km, Ø Stratosphere. It extends from 12km to 50km., There is an ozone layer in stratosphere which, mostly absorbs high energy radiations like, ultraviolet radiations. etc. coming from outer, space., Ø Mesosphere. It extends from 50km to 80km., Ø Ionosphere.(i) It extends from 80 km to 400km., (ii) In this region, the temperature rises to some, extent with height, hence it is called, Thermosphere., (iii) The ionosphere which is composed of, ionised matter (i.e. electrons and positive ions), plays an important role in space communication., (iv) The ionosphere is subdivided into four main, layers as D , E, F1 and F2 ., (v) D- layer is at a virtual height of 80km from, surface of earth and having electron density, ≈ 109 m −3 ., 28, , JEE-ADV PHYSICS- VOL- VI, (vi) The extent of ionisation of D layer depends upon, the altitude of sun. This layer disappears at night. It, reflects very low frequency (VLF) and low, frequency (LF) electromagnetic waves, but absorbs, medium frequency (MF) and high, frquency, (HF) electromagnetic waves to a certain degree,, (vii) E-layer is at a virtual height of 110km, from, the surface of earth, having electron density, ≈ 1011 m −3 . The critical frequency* of this layer, is about 4MHz. This layer helps to MF surfacewave propagation a little but reflects some high, frequency waves in day time. It exists in day as, well as in night time., (viii) F1 − layer is at a virtual height of 180km, from the surface of earth, having electron density, ≈ 5 ×1011 m−3 . The critical frequency for this, layer is 5 MHz. It reflects some of the high frequency, waves but most of the high frequency w a v e s, pass through it and they get reflected froml a y e r, F2 at night time., (ix) F2 layer is at a virtual height of about 300km, in day time and about 350km in night time. The, electron density of this layer is ≈ 8 × 1011 m −3 ., The cirtical frequency of this layer is 8MHz in, day time and 6MHz in night time. It reflects back, the electromagnetic waves of frequency upto 30, MHz but cannot reflect back the electromagnetic, waves of frequency 40MHz or more. It exists in, day as well as night time, 3. The electromagnetic waves of frequency ranging, from a few kilo hertz to a few hundered m e g a, hertz are called radio waves., The various frequency ranges used in radio waves, or micro wave communication system are a, s, follows:, (a) Medium frequency band (M.F) 300 to, 3000kHz., (b) High frequency band (H.F) 3 to 30 MHz, (c) very high frequency band (V.H.F) 30 to, 300 MHz., (d) Ultra high frequency band (U.H.F) 300 to, 3000MHz, (e) Super high frequency band (S.H.F) 3000 to, 30,000 MHz., (f) Extra high frequency band ( E.H.F) 30 to, 300 GHz, The radio waves emitted from a transmitter antenna, can reach the receiver antenna by the following, mode of operation., NARAYANAGROUP
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NARAYANAGROUP, , F M br oad castin g an d, m icr o wa v e lin ks, D u e to cu rvatu r e of th e ear th, th e w av es are b lo ck ed at a, p o int, , S ho r t w av e b ro ad ca stin g, , Ø, , Day and night, , F2 (Thermosphere), , Efficiently reflects HF, waves, particularly at night, , Partially absorbs HF waves, yet allowing them to reach, F2, , Helps surface waves, reflects HF, , Reflects LF, absorbs MF, and HF to some degree, , VHF (up to several GHz), , Ø, i), , 300 km at night 250400 km during day, time, , Day time, merges with, F2 at night, , FI (Parts of Mesophere) 170-190 km, , Day only, Day only, , 65-75 km, , Parts of Ionosphere : D, (part of Stratosphere), , Day and night, , E (Part of Stratosphere) 100 km, , 10 km, , Frequencies most, affected, , KENNELY HEAVISIDE LAYER :, At 110 km above the surface of earth the, concentration of electrons is very large. This layer, is called Kennely Heaviside layer., ii) The thickness of this layers is about of few km., iii) Beyond this layer the electron concentration, decreases upto 250 km, iv) From 250 km to 400 km, a layer of large, concentration of electrons called Apple ton layer, exists., v) Above appleton layer, ie above Ionosphere the, temperature is 927.60 C ., , Troposphere, , Approximate height, , Exists during, , L in e o f sigh t co m m u n ication, T he r ad io w a v es travel fr om, tra ns mitting anten n a to, receiv ing ant en na alon g a, stra igh t line, G r eater th an 4 0 M H z, , L ay er s of atm o sp h er e, D ue to r ef lection of r ad io, w a ves fr o m th e lay er s, h av in g h ig h er elec tr on, d ens ity, 3 M H z to 3 0 M H z, , D ep en d s o n th e ang le of, incid en ce on th e, ion o sp h er e. 1 5 0 k m to, 3 0 0 0 km, , S pa ce W av e Pro p ag at io n, , S ky W av e Pr o pag a tio n, , Name of the layer, , A tten u atio n in cr ea ses, w ith fr eq u en cy, , A tten ua tio n, , R ang e, , U s es, , D ep en d s o n p o w er and, frequ en cy L es s th an 2, MH z, In med iu m w a ve B r oad, ca stin g, D ep en d s o n h eig ht of th e, A nt en na and C ur vatur e, of ea rth, , F req u en cy, , C ha nn el, M eth o d, , G rou n d W av e, Pro p ag a tio n, G ro un d, W av e g lid es o v er th e, sur face of ea rth, dif fractio n ef fect, , JEE-ADV PHYSICS- VOL- VI, COMMUNICATION SYSTEM, , Ground wave propagation : In this method, the, radio waves are guided along the surface. The wave, induces charges on the earth. These charges travel, with the wave and this forms a current. Now the, earth behaves like a leaky capacitor in carrying the, induced current. The wave loses some energy, as, energy is spent due to flow of, , 29
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COMMUNICATION SYSTEM, , Ø, , charge through the earth’s resistance. The wave also, looses energy due to diffraction as it glides along, the ground. The loss of energy increases as the, frequency increases. Thus ground propagation is, suitable upto 2MHz. As they loose energy they, cannot go to long distances on the ground., Maximum range of the ground wave can be, increased by increasing the power of the transmitter., Sky wave propagation : Above 2MHz and upto, 30MHz, long distance communication takes place, through ionosphere. The ionosphere reflects the, radio waves back to the earth. This method is, called sky wave propagation. It is used for, shortwave broad casting services. Ionosphere, is a thick blanket of 65 km to 400 km above the, earth’s surface. UV rays and other higher energy, radiation coming from space results in the, ionization of air molecules. The ionosphere is, further divided into serval layers as shown in, table below. It should be understood that degree, of ionization changes with height. This is because, the density of atmosphere decreases with height., At great heights, the radiation is intense, but the, molecules available are few. On the other hand,, near the earth’s surface the molecular, concentration is high but the intensity of radiation, is low and thus again the ionization is low., Logically, the peak of ionization density occurs, at some intermediate heights. The ionosphere, acts as a mirror (reflector) for frequencies of 330 MHz. Electromagnetic waves of frequences, greater than 30 MHz pass through the atmosphere, and skip., The process of bending of EM waves is similar, to total internal reflection in optics. The bending, of waves can be easily explained on the basis of, variation of refractive index of the ionosphere, with change in electron density. Suppose that a, radio wave enters the ionosphere from the, underlying unionized medium. Since the refractive, index of ionosphere decreases from D layer to F2, layer, consequently, the incident ray will move away, from the normal drawn at the point of incidence Ø, following the ordinary laws of refraction, , 30, , JEE-ADV PHYSICS- VOL- VI, , Ionosphere, , Unionized medium, Radio wave, , During the propagation in ionosphere the angle of, refraction gradually increases and the ray goes on, bending more and more till at some point, the angle, of refraction becomes 900 and the wave travels, parallel to the earth surface. This point is called point, of reflection. Then the ray tends to move in the down, ward direction and comes back to earth because, of symmetry. Super high frequency (SHF) waves, propagate as sky waves taking reflection at satellite., Ionosphere, , The sky wave propagation can cover a very long, distance and so round the globe communication, is possible., (c) The sky waves being electromagnetic in nature,, changes the dielectric constant and refractive index, of the ionosphere. The effective refractive index, of the ionosphere is, , Ne2 , neff = n 0 1 −, 2, ε 0mω , , 1/ 2, , 1/2, , 80 .5 N , = n0 1 −, , f2 , , , Where n0 = refractive index of free space, N =, electron density of ionosphere, ε 0 = permittivity, of free space, e = charge on electron, m = mass, of electron w = angular frequency of EM wave., (d) As we go deep into the ionosphere, N, increases so ne f f decreases. The refractions or, bending of the beam will continue and finally it, reflects back., (e) The highest frequency of radio wave, which, gets reflected to earth by the ionosphere after, having been sent straight to it is, Critical frequency (fc), If maximum electron density of the ionosphere, is Nmax per m3 , then fc ≈ 9(N max )1 / 2 . Above f c, a, wave will penetrate the ionosphere and is not, reflected by it., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , (f): The highest frequency of radio waves which, when sent at some angle of incidence, towards, the ionosphere, get reflected and return to the, earth is Maximum usable, Ø, , Ø, , MUF =, , frequency (MUF), , hR = height of the receiving antenna., If the Population density around the tower is, given, the number of persons covered by the, transmitting tower, = (Area covered by the tower) × Population, density., ∴ No. of persons = π d 2 × covered population, density (Here d = radius of the area covered by, single transmitting tower of height hT ), Television broadcast, microwave and satellite, communications are a few examples of, communication systems that use space wave, propagation. The figure below illustrates the, various modes of wave propagation., , fc, cos θ, , (g) The smallest distance from a transmitter along, the earth’s surface at which a sky wave of a fixed, frequency but more than f c is sent, back to the earth is Skip distance., (h) The fluctuation in the strength of a signal at, a receiver due to interference of two waves is, fading. Fading is more at high frequencies. It, results into errors in data transmission and, retrieval., Wave propagation : This method is used for, line-of-sight [LOS] communication and also for, satellite communication. At frequencies above, 40MHz, communication is mainly by LOS, method. At such frequencies, relatively smaller, antenna can be erected above the ground., Because of LOS propagation, the direct waves, get blocked, at some point due to the curvature, of the earth as shown in the figure., Ø, , dm, dT, hT, , dR, hR, , Range of TV transmission :, As the frequency range of TV signals is 100200 MHz, such signal transmission via ground, waves is not possible. In such situations, we use, line of sight transmission., P, , For the signal to be received beyond the horizon,, the receiving antenna must be high enough to, intercept the LOS waves. If the transmitting, antenna is at a height hT then it can be shown, , A, , that the distance to the horizon d T is given by, , d T = 2 RhT where ‘R’ is the radius of earth., Similarly if the receiving antenna is at a height, hR , the distance to the horizon d R is, d R = 2 RhR, ∴ The maximum distance d M between the two, , antennas is d M = 2 RhT + 2 RH R where, R = Radius of the earth., hT = height of the transmitting antenna and, NARAYANAGROUP, , O, , Let CP be the TV tower on the earth’s surface. It’s, antenna is at P. Let PC = h. When TV broadcast is, made, the signal can reach the earth upto A to B., There will be no reception of the signal beyond A, and B. Arc length CA and CB is the range of TV, transmission. If O is the centre of the earth, OA =, OB = R is the radius of the earth, from right angled, triangle OAP, OP 2 = OA2 + PA2, 31
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , ( h + R), , 2, , basic principle of satellite communication is shown, in figure. A communication satellite is a spacecraft, placed in an orbit around the earth. The frequencies, used in satellite communication lie in UHF/, microwave regions. These waves can cross the, ionosphere and reach the satellite., , = R 2 + PA2, , PA = PB = d, , ( h + R), , 2, , = R2 + d 2, , h 2 + R2 + 2Rh = R2 + d 2, As h << R we can ignore h 2, , Ø, , Satellite-1, , d 2 = 2 Rh and d = 2 Rh, Range of TV transmission depends upon the, height of the transmission antenna. Broadcasts, are made from tall transmitting antenna., A repeater is a combination of a receiver and a, transmitter. A repeater, picks up the signal from Ø, the transmitter, amplifies and retransmits it to, the receiver sometimes with a change in carrier, frequency. Repeaters are used to extend the, range of a communication system as shown in Ø, figure. A communication satellite is essentially, a repeater station in space. Use of repeater, station to increase the range of communication, , Ø, Mountain, , Ø, , These problems are solved by using, geostationary satellite as a communication, satellite., Ø, Satellite Communication: Long distance, communication beyond 10 to 20 MHz was not, possible before 1960 because all the three modes, of communication discussed above failed Ø, (ground waves due to conduction losses, space, wave due to limited line of sight and sky wave, due to the penetration of the ionosphere by the, high frequencies beyond f c )., Ionosphere behaves as a rarer medium by which, carrier wave is reflected back if its frequency, , f ( ≤ f c ) where, , f c is called a “critical, 1, , frequency” and is given by f c = f 0 ( Nmax ) 2, [ N max = maximum electron density.], Satellite communication made this possible. The, 32, , Geostationary, orbit, , Earth, , Satellite-3, , Satellite-2, , A geostationary satellite has the same time period, of revolution of earth. It locates at the height of, 36000 km above the earth’s surface (well above, the ionosphere)., A communication satellite is a spacecraft placed, in an orbit around the earth which carriers a, transmitting and receiving equipment called, radio transponder. It amplifies the microwave, signals emitted by the transmitter from the surface, of earth and send to the receiving station on earth., The transmitted signal is UP-LINKED and, received by the satellite station which DOWNLINKS it with the ground station through its, transmitter., The up-link and down-link frequencies are kept, different (both frequencies being in the regions, of UHF/microwave)., At least three geo-stationary satellites are, required which are 120° apart from each other, to have the communication link over the entire, globe of earth., Satellite technology is very useful in collecting, information about various factors of the, atmosphere which governs the weather and, climatic conditions., The satellite communication can be used for, establishing mobile communication with great, use the communication satellites are now being, used in Global Positioning System (GPS). The, ordinary users can find their positions within, accuracy of 100m., There are two types of satellites used for long, distance transmission., , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , Ø, , Ø, , Ø, , Ø, , Ø, , Ø, , Ø, , (i) Passive satellite: It acts as reflector only for the, signals transmitted from earth. Moon the natural, satellite of earth is a passive satellite., (ii) Active satellite: It carries all the equipment, used for receiving signals sent from the earth,, processing them and then re-transmitting them to, the earth. Now a days active satellites are in use., The Indian communication satellites INSAT-2B and, INSAT-2C are positioned in such away in the outer, space that they are accessible from any place in, India., REMOTE SENSING AND APPLICATION, OF SATELLITE COMMUNICATION., Remote sensing is the technique to obtain, information about an object by observing it from, a distance and without coming to actual contact, with it., There are two types of remote sensing, instruments: active and passive. Active, instruments provide their own energy to, illuminate the object of interest, as radar does., They send an energy pulse to the object and then, receive and process the pulse reflected from the, object. Passive instruments sense only radiations, emitted by the object or solar radiation reflected, from the object., The remote sensing is done through a satellite., The satellite used in remote sensing should move, in an orbit around the earth in such a way that it, always passes over the particular area of the, earth at the same local time., The orbit of such a satellite is known as sunsynchronous orbit. A remote sensing orbit can, be circular polar orbit or in highly inclined, elliptical orbit., A remote sensing satellite takes, photographs, of a particular region which nearly the same, illumination every time it passes through that, region., The most useful remote sensing technology is, that it makes possible the repetitive surveys of, vast areas in a very short time, even if the areas, are inaccessible., Space based remote sensing is a new technology., It has high potential for applications in nearly, all aspects of resource management., The Indian remote sensing satellites are, IRS-1A, IRS-1B, and IRS-1C., NARAYANAGROUP, , COMMUNICATION SYSTEM, Ø, , Ø, , Ø, , Remote sensing is applied in (i) Meteorology, (ii) Climatology (iii) Oceanography, (iv) Archaeology, geological surveys. (v) Water, resource surveys, (vi) Urban land use surveys. (vii), Agriculture and forestry and natural disaster., (viii) To detect movements of enemy army. (ix)To, locate the place where underground nuclear, explosion has carried out., 2.LINE COMMUNICATION, Line communication means interconnection of two, points with the help of wires for exchange of, information. There are three principal types, (i)Two Wire Transmission Line, (ii) Coaxial wire lines (coaxial cables), (iii) optical fibers, , Two Wire Transmission Line, , The most commonly used two wire lines are:, Parallel wire, twisted pair wires and co-axial cable., (1) Parallel wire line: In a two wire transmission line,, two metallic wires (may be hard or flexible) are, arranged parallel to each other inside a protective, insulation coating . Commonly used to connect an, antenna with TV receiver. Such wires, can suffer from interferences and losses., (2) Twisted pair wire: It consists of two insulated, copper wires twisted around each other at regular, intervals to minimize electrical interference (to, connect telephone systems). Used to connect, telephone systems. It works well up to small, distances. They cannot transmit signals over very, large distances. They transmit both, the analog and, digital signals. They are easy to install and cost, effective., Ø Coaxial wire lines: It consists of a central, copper wire (which transmits surrounded by a PVC, insulation over which a sleeve of copper mesh (outer, conductor) is placed. The outer conductor is, normally connected to ground and, thus, it, provides an electrical shield to the signals, carried by the central conductor. The outer, conductor is externally covered with a polymer, jacket for protection., , 33
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , (a) Co-axial line wires can be used for microwaves, Ø TELEPHONE LINKS, and, ultra-high, frequency, waves. (1) A telephone (the most common means of, (b) The communication through co-axial lines is more, communication) link can be established with the help, efficient than through a twisted pair wire lines., of co-axial cables, ground waves, sky waves,, (c) Co-axial cables can be gas filled also. To reduce, microwaves or optical fiber cables., flash over between the conductor handling high, (2) Simultaneous transmission of a number of messages, power, N 2 -gas is used in the cable., over a single channel without their interfering with, Ø Impedance of Line, one another is called multiplexing., (1)Each portion of the transmission line can be, (3) Twisted pair wire lines provide a band width of 2, considered as a small inductor, resistor and, MHz, while co-axial cable provides a band width, capacitor. As a result each length of transmission, of 20 MHz. For further increase in band width, we, line has characteristic impedance., use (i) microwave link (ii) communication satellite, (2) In case of co-axial cable, the dielectric can be r, link., epresented by a shunt resistance G., Ø OPTICAL COMMUNICATION, (3) When co-axial cable is used to transmit a radio, The use of optical carrier waves for transmission, frequency signal, X L and X C are large as, of information from one place to another is called, compared to R and G respectively. Hence R and, optical communication., G can be neglected., The information carrying capacity ∝ bandwidth, (4) In co-axial cable, R is zero, so no loss of energy, ∝ frequency of carrier wave. Because of high, and hence no attenuation of frequency signal, frequency (10 12 Hz to 10 16 Hz) optical, occurs when transmitted along it. That’s why cocommunication is better than others. (radio and, axial cables are specially used in cable TV, microwave frequencies,106 Hz – 1011 Hz)., network., Basic optical communication link is a point to, Ø Characteristic impedance (Z 0 ) : It is defined, point link having transmitter at one end, receiver, as the impedance measured at the input of a line, at the other end and consists of three components, of infinite length., namely, (1) Optical source and modulator, 276, 2s, a)For parallel line Z0 = k log d, (2) Optical signal detector or photodetector, (3) Optical fibre cable through which optical, d = Diameter of each wire, signal is transmitted., s = Separation between the two wires, k = Dielectric constant of the insulating medium Ø Optical sources for communication, b) For co-axial line wire, , Z0 =, , 138, k, , log, , D, d, , d = Diameter of inner conductor, D = Diameter of outer conductor, c), , Ø, , L, At radio frequency Z 0 =, , C, , The usual range of characteristic impedance for, parallel wire lines is 150W to 600 W and for, co-axial wire it is 40 W to 150 W., Velocity factor of a line (v. f.) : It is the ratio, of reduction of speed of light in the dielectric of, Ø, the cable, v. f . =, , v Speed of light in medium, 1, =, =, c Speed of light in vacuum, K, , For a line v.f. is generally of the order of 0.6 to 0.9., 34, , links, Light emitting diodes (LED) and diode lasers, are preferred for optical source. LEDs are used, for small distance transmission while diode laser, is used for very large distance transmission., For optical communication, light is to be, modulated with the information signal. The, frequency and intensity of light is sensitive to, temparature changes,which is to be avoided. So, suitable arrangement isrequired to obtain, thermal stability., , Optical signal, photodetector:, , detector, , or, , The optical signal reaching the receiving end has to, be detected by a detector which converts light, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , Ø, , Ø, , intoelectrical signals, So that the transmitted, information may be decoded., The optical detector should have, (i) size compatible with the fibre, (ii) High sensitivity at the desired optical wavelength, (iii) High response for fast speed data transmission/, reception., Semiconductor based photo-electors are used, because they fullfill the above criteria, MODULATION AND ITS NECESSITY :, Message signals are also called base band signals., Which essentially designate the band of frequencies, representing the original signal, as delivered by the, source of information. No signal, is a single, frequency sinusoid, but it spreads over a range of, frequencies called the signal bandwidth to transmit, an electric signal frequency less than 20 kHz) over, a long distance directly. It is clear that low frequency, waves, can not travel long distances. Hence, to, transmit low frequency wave over long distance,, we take the help of high frequency waves called, carrier wave. The low frequency wave is, superposed over high frequency carrier wave. This, process is called the modulation. The low frequency, wave is called the modulating wave and the high, frequency wave is called the carrier wave, and, the resultant wave is called modulated wave. In, this section we will discuss in detail about, modulation. What is it? What is the need of, modulation or how modulation is done etc., No signal in general, is a single frequency but it, spreads over a range of frequencies called the, signal bandwidth. Suppose we with to transmit, an electronic signal in the audio-frequency, (20Hz-20kHz) range over a long distance. Can, we do it? No it cannot because of the following, problems., Size of antenna : For transmitting a signal we, need an antenna. This antenna should have a size, comparable to the wavelength of the signal. For, an electromagnetic wave of frequency 20kHz,, wave length is 15km. Obviously such a long, antenna is not possible and hence direct, transmission of such signal is not practical., The linear size of the antenna must be the order, of the wave length and for effective transmission, λ, its length must be h =, 4, NARAYANAGROUP, , COMMUNICATION SYSTEM, so that antenna properly senses the time variation, of the signal., Example1: For an electromagnetic wave of, f = 20 kHz, λ = 15 km Obviously, such a long, antenna is not possible to construct and operate., Hence direct transmission of such baseband signals, is not practical., Example2: If f = 1 MHz, then λ = 300 m, h = 75 m, Therefore, there is a need of translating the, information contained in our original low, frequency baseband signal into high or radio, frequencies before transmission., Ø Additional Information:, (a) The distance between transmitting antenna, and the horizon, Dt = 2Rht ., Where ht = height of transmitting antenna, R = Radius of the earth, (b) The distance between receiving antenna and, the horizon, Dr = 2Rhr ., Where hr = height of receiving antenna, (c) The maximum distance between the, transmitting antenna and receiving antenna Dm ., Dm = Dr + Dt, , Dm =, , 2 R hr + 2 R h t ., , Where R is the radius of earth., hr > h t so then the receiving antenna intercepts, , Ø, , the line of sight waves., Single antenna, (d) The radius "d" of the area covered by a single, transmitting tower of height h is given by d =, , 2R eh . Where Re is the radius of the Earth., , Ø, , (e) If the Population density around the tower is, given, the number of persons covered by the, tower is, = (Area covered by the tower) x Population, density No. of persons covered = π d2 x, Population- density., Effective power radiated by an antenna, Power radiated by an antenna is proportional to, l , 2 . Where l is length of the antenna ., λ , 35
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , Ø, , Ø, , For a good transmissmission,high powers are Ø, required, hence low wavelength i.e high, frequency transmissions are needed., Mixing up of signals from different, transmitters, Suppose many people are talking at the same time Ø, or many transmitters are transmitting baseband, information signals simultaneously. All these signals, will get mixed up and there is no simple way to, distinguish between them. This points out towards, a possible solution by using communication at high, frequencies and alloting a band of frequencies to, each mesage signal for its transmission., In doing so, we take the help of a high frequency, signal, known as the carrier wave, and a process, known as modulation which attaches information, to it., Modulation: The process of superimposing, information contained in the low frequency ,, message signal on a high frequency carrier wave,, near transmitter is known as modulation., , According to the type of modulation, For sinusodial continuous carrier waves, i) Amplitude Modulation (AM), ii) Frequency Modulation (FM), iii) Phase Modulation, For pulsed carrier waves, i) Pulse Amplitude Modulation (PAM), ii) Pulse Time Modulation (PTM), a) Pulse Position Modulation (PPM), b) Pulse Width Modulation (PWM) or Pulse, Duration Modulation (PDM), iii) Pulse Code Modulation (PCM), MODULATION, , Continuous wave, modulation, , Amplitude, modulation, , Frequency, modulation, , Pulse wave, modulation, , Phase, modulation, , Pulse, Amplidude, , Source of, Original Information, signal, , Modulated, signal, , Oscillator, , Pulse width, modulation, , The carrier wave may be continuous (sinusoidal), or in the form of pulses as shown in figure(2)., Time period T, , ω=, , 2π, T, , Amplitude, time, , Eqution representing sinusoidal carrier wave can, , c ( t ) = Ac sin ( ωct + φ ) ----(1), where c(t) is the signal strength (voltage or, current),, Ac is the amplitude, , (ω t + φ ), c, , (a), , is called argument of Phase angle of, , the carrier wave, , Pulse fall, Pulse, Amplitude, , (b), , Therefore depending upon the specific, characteristicof carrier wave which is being, varied in accordance with the message signal,, modulation can basically be differentiated as, (i) continuous wave modulation; and, (ii) pulse wave modulation., , 36, , Pulse, position, modulation, row, , I) Continuous Wave Modulation, , TYPES OF MODULATION:, , Pulse, Pulse, rise duration, , Pulse code, modulation, , Modulator, Baseband, signal, Carrier, wave, , Ø, , Pulse time, modulation, , ωc ( = 2π f c ) is the angular frequency, φ is the initial phase of the carrier wave. During, the process of modulation, any of the two, parameters, viz amplitude or phase angle, of the, carrier wave can be controlled by the message, or information signal. This results in two types, of modulations:, i) Amplitude modulation (AM), ii) Angle modulation, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , Angle modulation again can be of two types. They, are, i) Frequency modulation (FM), ii) Phase modulation (PM), As shown in figure., , c ( t ) = Ec sin ωc t represent carrier wave. Here, ωc = 2π f c is the angular frequncy of the carrier, signal, , 1, c(t) 0, -1, , (a), 0, , 0.5, , 1, , 1.5, , 2, , 2.5, , 3, , 1, m(t) 0, -1, , (b), 0, , 2, c(t), for AM 0, m, -2, , 0.5, , 1, , 1.5, , 2, , 2.5, , 3, (c), , 0, , 0.5, , 1, , 1.5, , 2, , 2.5, , Relative Voltage, Current, or Power, , +, , Wave, , One, Cycle, , One, Cycle, Peak, to peak, Amplitude, , 0, , _, , Time, Starting point, , Time Axis, , 3, , 1, c(t), for FM 0, m, -1, 0, , 0.5, , 1, , 1.5, , 2, , 2.5, , 0, , 0.5, , 1, , 1.5, , 2, , 2.5 time 3, , (d), , The modulated signal cm ( t ) can be written as, , (e), , cm ( t ) = ( Ec + Em sin ωmt ) sin ωct ----(1), , 3, , 1, c(t), for PM 0, m, -1, , II) Pulse Wave Modulation., The significant characteristics of a pulse are: i), Pulse amplitude, ii) Pulse duration or pulse width, iii) pulse position (denoting the time of rise or, fall of the pulse amplitude) as shown in figure, (3)., Types of pulse modulation:, a) pulse amplitude modulation (PAM),, b) pulse duration modulation (PDM) or pulse, width modulation (PWM), c) pulse position modulation (PPM)., , I Continuous Wave Modulation:, 1) Amplitude Modulation:, The method in which the amplitude of carrier is, varied in accordance with the modulating signal, keeping the frequency and phase of carrier wave, constant is called amplitude modulation (AM)., Here we explain amplitude modulation process, using a sinusoidal signal as the modulating, signal., Let m ( t ) = Em sinωmt represent the message or, the modulating or base band signal. Here, ω m = 2π f m is the angular frequency of the, message signal., , E, , cm ( t ) = Ac 1+ m sinωmt sinωct -----(2), Ec, , , cm ( t ) = Ec sin ωct + µ Ec sin ω mt sin ω ct --(3), Using the trignometric relation, sin A sin B =, , 1, ( cos ( A − B ) − cos ( A + B ) ) , we, 2, , can write cm ( t ) of equation (3) as, cm ( t ) = Ac sin ωct +, , -----(4), , µAc, 2, , cos (ωc − ωm ) t −, , µ Ac, 2, , cos (ωc + ωm ) t, , Here, , ωc − ωm = 2π ( f c − f m ) = Lower side band, frequency (LSB), ωc + ωm = 2π ( f c + f m ) =Upper, frequency (USB), , side, , band, , Here (m or µ = Em / Ec ) is the modulation index;, (or) modulating factor., In practice, µ is kept ≤ 1 to avoid distorition., Am, , Depth of modulation = A × 100 = µ × 100, C, Depth of modulation interms of Emax and Emin, mE C, , +, , EC, 0, _, A-F Signal, , A.M. Wave, NARAYANAGROUP, , 37
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , Emax = Ec + Em = E c (1 + m), , Amplitudechangeincarrier wave 2A − A, =, = 100%, AmplitudeofCW, A, , Emin = Ec − E m = E c (1 − m), , Case-III: When the amplitude of the signal is half of, that of CW., , Emax Ec + Em Ec (1 + m), =, =, Emin Ec − Em Ec (1 − m), , , , , A, , Amplitude of CW changes from A to A+ 2 =1.5A, A, +, , , , 1.5A, , 0.5A, =, , The Band width of AM wave is " 2 fm ", Carrier, Signal, A.M. Wave, The modulated signal now consists of the carrier, 0.5A, wave of frequency ωc plus two sinusoidal, = 0.5, Modulation factor =, A, waves each with a frequency slightly different, = 5 0 %, fromωc , known as side bands. The frequency Case-IV: When the amplitude of signal is 1.5 times, spectrum of the amplitude modulated signal is, that of the CW., shown in, Amplitude of the modulated wave changes from, 2.5 A to A, AG, µAG, Amplitude, 2, , A, , Carrier, , 2.5A, , 1.5A, =, , +, , Signal, , A.M. Wave, , 2.5A − A, = 1.5 = 150 %., A, In this case the quality of signal is lost, As long as the broadcast frequencies (carrier Note:A carrier wave is modulated by a number of, waves) are sufficiently spaced out so that, sine waves with modulation indices m1 , m2 and, sidebands do not overlap, different stations can, m3 . The total modulation index of the wave is, operate without interfering with each other., m = m12 + m22 + m32, Special cases of Amplitude modulation:, CaseI: In the absence of signal., Ø Power out put in AM wave, ( ωC -ω m) ωC ( ωC+ωm) ω in radians, , figure (5) A plot or amplitude versus ω for, an amplitude modulated signal, , A, A, , A, , + No Signal =, Carrier, , Signal, , O, x100 = 0%, A, Case-II: When the signal amplitude is equal to CW, wave., Amplitude varies from 2A to zero., , Carrier, , 38, , A, , +, , A, , Signal, , 2A, =, , A.M. Wave, , Pt = Pc + Ps, where Pt is power transmitted, , A.M. Wave, , Modulation factor ma =, , A, , Modulation factor ma =, , Pc is power of carrier wave, Ps is total power of side bands, The equation of a carrier wave Yc = Ac sin ( wc t + φ ), 2, , Power of carrier wave P = [ Arms ], c, R, , 2, , Ac , 2, , , A, 2, =, = c, R, 2R, , The power of side bands = The power of lower, side band + the power of upper side band, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, Here the modulating signal Am sin ωm t is added to, , 2, ( µ Ac ), A, , 2 / 2 µ c / 2 , + 2, , Ps = , R, R, 2, , =, , the carrier signal Ac sinωct to produce signal, , x ( t ) = Am sin ωm t + Ac sinω ct, , µ Ac, µ, =, Pc, 4R, 2, 2, , 2, , 2, , This signal is passed through a square law device, which is a non linear device that can give tha output, , y ( t ) = Bx ( t ) + Cx 2 ( t ) where B and C are, , µ2, Pc, Pt = Pc + Ps = Pc +, 2, , constants, , µ , = Pc 1 +, , 2 , , , y ( t ) = BAm sin ωm t + BAc sin ωc t, , Thus,, , 2, , +C Am2 sin 2 ωmt + Ac2 sin 2 ωct + 2 Am Ac sin ω mt sin ωct -(6), , µ2 , ∴ Pt = Pc 1+, − − − −(1), 2 , , ⇒, , = BAm sinω mt + BAc sinωct +, , CAm2, CA2, CA2, + Ac2 − m cos2ω mt − c cos2ω ct, 2, 2, 2, , + CAm Ac cos ( ωc − ωm ) t − CAm Ac cos ( ωc + ω m ) t -(7), , Pt, µ, = 1+, − − − −(2), Pc, 2, 2, , Where, , 2, , i , µ2, ⇒ t =1 +, − − − − (3) Q P α i 2 , 2, ic , , the, , trignometric, , relations, , sin 2 A = (1 − cos2 A) / 2 and the relation for, sinA sinB mentioned earilier are used., In equation, there is a dc term c / 2 ( Am2 + Ac2 ), and sinusoids of frequencies, , Example:, , If the modulation factor is 1 ie 100 %, , modulation then the useful power is, , 1, of the, 3, , total power radiated. The remaining 2/3 power is, contained by carrier wave, , Pc, 2, 2, Ps, ma, 1, =, =, =, =, 2, PT 2 + ma 2 3 and PT 2 + ma 3, 2, , m2, Transmission Efficiency η =, 2 + m2, , PRODUCTION OF AMPLITUDE, MODULATED WAVE, Amplitude modulation can be produced by a variety, of methods. A conceptually simple method is shown, in the block diagram of Fig., , ω m , 2ωm , ωc , 2ωc , ωc − ω m and ωc + ωm . As, shown in figure this signal is passed through a, band pass filter which rejects dc and the, sinusoids of frequencies ω m , 2ω m and 2ωc and, retains the frequencies ω c , ω c − ωm and ωc + ωm ., The output of the band pass filter therefore is of, the same form as equation(4) and is therefore an, AM wave., It is to be mentioned that the modulated signal, cannot be transmitted as such. The modulator is, to be followed by a power amplifier which, provides the necessary power and then the, modulated signal is fed to an antenna of, appropriate size for radiation as shown in, figure(7)., Transmitting, Antenna, m(t), Message, signal, , Amplitude, Modulator, , Power, Amplifier, , Carrier, , Figure(7), NARAYANAGROUP, , 39
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, Ø, , A-M Transmitter., In the block diagram of the AM transmitter the, r-f section consists of an oscillator feeding a, buffer, which in turn feeds a system of frequency, multipliers and/or intermediate power, amplifiers. If frequency multiplication is, unneccessary, the buffer feeds directly into the, intermediate power amplifiers which, in turn, drive, the final power amplifier. The input to the antenna, is taken from the final power amplifier., , 3) The AM wave input is shown in figure. It appears, at the output of the diode across PQ as a rectified, wave (since a diode conducts only in the positive, half cycle). This rectified wave after passing through, the RC network does not contain the radio, frequency carrier component. Instead, it has only, the envelope of the modulated wave., P, , AM, Antenna, , CSC, , Buffer, , Int, Power, Amp, , R, , Output, , Power, Amp, , AM Wave, Audio, Amp, , C, , Wave Input, , A-M, Mod, , (a), , Mike, , Envelope m(t) Output, Detector, , Rectifier, (b), , (c), , Amplitude-Modulated Transmitter, , Ø, , Detection of amplitude modulated wave, The transmitted message gets attenuated in, propagating through the channel. The receiving, antenna is therefore to be followed by an, amplifier and a detector. In addition, to facilitate, further processing, the carrier frequency is, usually changed to a lower frequency by what is, called an intermediate frequency (IF) stage, preceding the detection. The detected signal, may not be strong enough to be made use of and, hence is required to be amplified. A block, diagram of a typical receiver is shown in figure., Receiving, Antenna, , Received, Signal, , Ø, , 40, , Amplifier, , If, stage, , Detector, , Amplifier, , time, AM input wave, , time, , time, , Rectified, wave, , Output (without RF, component), , The capacitor connected in parallel with resistance, R provides very low impedance at the carrier, frequency and a much higher impedance at the, modulating frequency. As a result capacitor, effectively shorts or filters out the carrier, there by, leaving the original modulating signal, In the actual circuit the value of RC (The time, 1, , constant, t=RC ) is chosen such that f << RC ;, c, where f C = frequency of carrier signal., Distortion in diode detectors: There are two types, of distortions in diode detectors. Namely, a) Negative peak clipping, , Output, , Simple demodulator circuit :, 1) A diode can be used to detect or demodulate, an amplitude modulated (AM) wave., 2) A diode basically acts as a rectifier i.e. it, reduces the modulated carrier wave into positive, envelope only., , time, , Figure shows the negative peak missing in the, output, message, We, know, that, where, E, I, Modulation Index(m)= m = m, E C Ic, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, Im =, , COMMUNICATION SYSTEM, , W.E-1: The load on an Am diode detector consists, of a resistance of 50 K Ω in parallel with a, , Em, E, Ic = c, and, Zm, Rc, , capacitor of 0.001 µ F . Determine the, maximum modulation index that the detector, can handle without distortion when, modulation frequency is (i) 1 kHz (ii) 5 kHz., , Here z m is audio load resistance of diode and Rc is, dc diode resistance while z m < Rc, , hence I m > Ic . This makes the modulation index, 1, Zm = Rc P C =, in the demodulated wave higher than it was in, 1, 1, modulated wave applied at the detector. In turn Sol:, 2 +, 2, ( Rc ) ( X c ), there is an increase in the chance of over, modulation for modulation index nearer to 100%, Due to this over modulation there is a negative, Z, M max = m, clipping of the detector wave., Rc, b) Diagonal clipping:, 1, 1. Diode ac load may no longer be purely resistive,, =, 1, 1, it can contain reactive component., Rc, 2 +, 2, 2. At high modulation depths current will be, ( Rc ) ( X c ), changing so fast that the time constant of the load, 1, may be too slow to follow the changes. As a, =, 2, result current will decay exponentially. Hence, 1 + ( 2π fCRc ), output voltage follows the discharge law of the, CR circuit, For f =1 kHz, Voltage, , Too quick, , (a), , time, , Mmax=, , Voltage, , Too slow, , Ø, (b), , 3., , 1, = 0.945, 1 + 0.098696, for f = 5 kHz, Mmax =, , reduced, sensitivity, , time, , =, , 1, 1+ 2.4674, , = 0.537, , Limitation of amplitude modulation, (i) Noisy reception, (ii) Low efficiency, (iii) Small operating range, (iv) Poor audio quality, , Condition necessary for avoiding distortion of, 2) Angle modulation:, this type is as follows., Im =, , Em mEc, E, =, Ic = c, and, Zm, Zm, Rc, , md =, , I m mEc Z m mRc, =, =, Ic, Ec Rc, Zm, , Maximum value of mdMaximum = 1, So maximum permissible transmitted modulation, Zm, Zm, index will be M max imum = md R = 1 × R, c, c, NARAYANAGROUP, , The angle of the carrier wave is varied according to, the base band signal while the amplitude is maintained, constant. This method provides better discrimination, against NOISE and INTERFERENCE then the, Amplitude modulation, , There are two ways of varying the angle of the, carrier., , θ i ( t ) is the angle of moduilated sinusoidal, carrier assumed to be a function of the message, signal., 41
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , e) Carrier swing (CS): The total variation in, frequency from the lowest to the highest is called, the carrier swing i.e. CS = 2 δ f, f) Frequency modulation index (mf -) :, The ratio of maximum frequency deviation to the, modulating frequency is called modulation index., , Equation of the angle modulated wave is, , S ( t ) = Ac cos (θi ( t ) ), where Ac is amplitude of the carrier., Instantaneous frequency of the angle modulated, 1 dθ i ( t ), 2π dt, , mf =, , dθ ( t ), here i, is the angular velocity of the PHASOR, dt, of length Ac, a) By varying the frequencyωc , Frequency, Modulation., b) By varying the phaseφc , Phase Modulation., FM Modulator, m(t), , FM, Modulator, , Integrator, , PM, Modulator, , The number of side bands depends on the, modulation index mf ., Band width = 2n × fm ; where n = number of, significant side band pairs, h) Deviation ratio: The ratio of maximum, permitted frequency deviation to the maximum, permitted audio frequency is, , PM, Modulator, , m(t) Different-iator, , (∆f ), , max, known as deviation ratio. ( = ( f ), ), m max, i) Percent modulation: The ratio of actual, frequency deviation to the maximum followed, frequency deviation, , FM, Modulator, , A comparison of FM and PM modulators., , i)Frequency modulation (FM), , δ = ( fmax − fc ) = fc − fmin = k f ., , Em, 2π, , k f = Constant of proportionality., It determines the maximum variation in frequency, of the modulated wave for a given modulating, signal., 42, , (∆f ) actual, (∆f ) max, , j) Frequency spectra of fm waves under, various conditions., Amplitude, , fm=1kHz, ∆f=0.5, fm, , ∆f=5kHz, ∆f=0.5, fm, Frequency, , fm=1kHz, ∆f=1.0, fm, , ∆f=5kHz, ∆f=1.0, fm, , Amplitude, , Frequency, , Amplitude, , Frequency, , Amplitude, , The method in which the frequency of carrier is, varied in accordance to the modulating signal,, keeping the amplitude and phase of the carrier, the same is called Frequency modulation (FM), a) Audio quality of AM transmission is poor., There is need to eliminate amplitude sensitive, noise. This is possible if we eliminate amplitude, variation., b) In FM the overall amplitude of FM wave, remains constant at all times., c) In FM, the total transmitted power remains, constant., d) Frequency deviation: The maximum, change in frequency from mean value (f c) is, known as frequency deviation. This is also the, change or shift either above or below the, frequency f c and is called as frequency deviation., , m=, , Amplitude, , m(t), , ( fc ± fm), ( fc ± 2 fm), ( fc ± 3 fm) ......., , Amplitude, , m(t), , g) Frequency spectrum: FM side band, modulated signal consist of infinite number of, side bands whose frequencies are, , Frequency, , fm=1kHz, ∆f=0.5, fm, , Amplitude, , FM Modulator, , k f Em, δ, f, −f, f −f, = max c = c min =, fm, fm, fm, fm, , fm=5kHz, ∆f=5.0, fm, , Frequency, , Frequency, , fm=1kHz, ∆f=10.0, fm, , fm=5kHz, ∆f=10.0, fm, , Amplitude, , wave is f i ( t ) =, , Frequency, , Frequency, , (A) Constant Modulating, Frequency, ∆f Frequency, Deviation, , (B) Constant Frequency, Deviation, , fm Modulation, Frequency, , ∆f Modulation, fm, Index, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , W.E-2: The carrier frequency is 500 kHz. The ii)Pulse time modulation (PTM):, modulating frequency is 15 kilohertz and the, deviation frequency is 75 kilohertz. Find, a) modulation index, b) Number of side bands, c) Band width, ∆f, Sol: MI = f = 5, m, , 620, , 605, , 590, , 560, , 575, , 545, 530, , 500 kHz, , 515, , 485, , Transmitter Deviation, 75 kHz, , 75 kHz, Bandwidth, , To determine total bandwidth for this case,, we use: bW = f m × ( number of sidebands ), bW = 15 x 16=240 kHz, ii) Phase Modulation (PM), phase of carrier is varied, in accordance with modulating signal keeping, namplitude and frequency constant. We use the, term phase shift to characterize such changes. If, phase changes after cycle k, the next sinusoidal, wave will start slightly later than the time at, which cycle k completes., time, , II) pulse wave modulation., Here the carrier wave is in the form of pulses., Pulse modulation is an analog process as the, modulating signal is analog. No. of quantization, levels = 2n, bitrate = sampling rate × no. of bits per sample, The common pulse modulating systems are:, , i) Pulse amplitude modulation(PAM):, The amplitude of the pulse varies in accordance, with the modulating signal., NARAYANAGROUP, , a), , Pulse width modulation (PWM):, The pulse duration varies in accordance with, the modulating signal, or the width of the, unmodulated signal is constant., b) Pulse position modulation (PPM):, In PPM, the position of the pulses of the carrier, wave train is varied in accordance with the, instantaneous value of the modulating signal., , iii) Pulse code modulation (PCM):, , 470, , 440, , 425, , 410, , 395, , 380, , 455, , We can have that there are 16 significant, sidebands for a modulation index of 5., , Or, Pulse duration modulation(PDM), , The pulse amplitude, pulse width and pulse, position modulations are not completely digital., A completely digital modulation is obtained by, pulse code modulation (PCM) by following, three operations., (a) Sampling: It is the process of generating pulses, of zero width and of amplitude equal to the, instantaneous amplitude of the analog signal. The, number of samples taken per second is called, sampling rate., (b) Quantization: The process of dividing the, maximum amplitude of the analog voltage signal, into a fixed number of levels is called, quantization., (c) Coding: The process of digitizing the quantised, pulses according to some code is called coding., Ø Digital Communication And Quantisation, , Of Message Signal, (Data Transmission and Retrieval), , Data means facts, concepts or instructions, suitable for communication, interpretation or, processing by human beings or by automatic, means. Data in most cases consists of pulse type, of signals., In digital communication the modulating signals, are discrete and are coded as represen tation of, message signals to be transimitted. There are a, number of encoding steps in digital communication,, which makes its circutary complicated. Digital, communication, is error free and noise, free., The source encoder converts the informat i o n, into binery code. Encoder first digitise the analog, waveform. Some times an additional encoding, 43
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, called channel encoding is carried out. In the Ø, final step, before transmission,the channel codes, modulate a continuous wave form., The pulse code modulated (PCM) signal is a series, of 1’s and 0’s. The following three modulation, techniques are used to transmit a PCM signal., (1) Amplitude shift keying (ASK):, Two different amplitudes of the carrier represent, the two binary values of the PCM signal. This, method is also known as on-off keying (OOK), 1: Presence of carrier of same constant amplitude., 0: Carrier of zero amplitude., (2) Frequency shift keying (FSK): The binary, values of the PCM signal are represented by two, frequencies., 1: Increase in frequency, 0: Frequency unaffected, (3) Phase shift keying (PSK): The phase of the, carrier wave is changed in accordance with, modulating data signal., 1: Phase changed by p, 0: Phase remains unchanged., The analog signal is sampled by the sampler., The sampled pulses are then quantized. The, encoder codes the quantized pulses according, to the binary codes. After modulating the PCM, signal (by ASK, FSK or PSK method)the, modulated signal is, then transmitted into free, space in the form of bits., , (A) Carrier Wave, 0, , 0, , 1, , 1, , 0, , 1, , 0, , 0, , 0, , (B) Modulating source code, , (C) ASK Modulated wave, , (D) FSK Modulated wave, , (E) PSK Modulated wave, , 44, , 1, , 0, , MODEM AND FAX, (1) Modem: Modems are used to interface two, digital sources/receivers., i) Word modem implies, MODulator and DEModulator. Both the, functions, (modulation and demodulation) are, included in a signal unit., Modem-1, Business, machine, or, computer, or data, code, generator, , Modulator, , Section, Demodu-lator, , Section, Transmitter side, , Modem-2, , Communication, Channel, , Computer, Data, display, machine, , Receiver side, , ii) Modems are placed at both ends of the, communication circuit., iii) The modem at the transmitting station changes, the digital output from a computer to an analog, signal, which can be easily sent via, communication channel. While the receiving, modem reverses the process., iv)There are three modes of operation of a, modem., (a) Simplex mode: Data is transmitted in only, one direction., (b) Half duplex: Data is transmitted between the, transmitter and the receiver in both direction,, but only in one direction at a time., (c) Full duplex: Data are transmitted between, the transmitter and receiver in both directions at, the same time., (2) Fax (Facsimile transmission): The electronic, reproduction of a document at a distance place, is known as facsimile transmission (FAX)., The original written document is converted into, transmittable codes and is converted in to, electrical signals, which are then modulated and, transmitted on to the receiving end., Ø The Internet : It is a system with billions of, users worldwide. It permits communication and, sharing of all types of information between any, two or more computers connected through a large, and complex network. It was started in 1960’s, and opened for public use in 1990’s. With the, passage of time it has witnessed tremendous, growth and it is still expanding its reach. Its, applications include., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, Ø, , Ø, , Ø, , Ø, , Ø, , Ø, , Email: It permits exchange of text/ graphic material, using email software. We can write a letter and send, it to the recipient through ISP’s (internet Service, Providers) who work like the dispatching and, receiving post offices., File transfer: A FTP (File Transfer Programmes), allows transfer of files/software from one, computer to another connected to the internet., World Wide Web (WWW): Computers that, store specific information for sharing with others, provide websites either directly or through web, service providers. Government departments,, companies, NGO’s (Non-Government, Organizations) and individuals can post, information about their activities for resticted, or free use on their websites. This information, becomes accessible to the users. Several search, engines like Google, Yahoo!etc. help us in finding, information by listing the related websites., Hypertext is a powerful feature of the web that, automatically links relevant information from, one page on the web to another using HTML, (hypertext markup language), E-commerce: Use of the internet to promote, business using electronic means such as using, credit cards is called E-commerce. Customers, view images and receive all the information, about various products or services of companies, through their websites. They can do on-line, shopping from home/office. Goods are, dispatched or services are provided by the, company through mail/courier., Chat:Real time conversation among people with, common interests through typed messages is, called chat. Everyone belonging to the chat, group gets the message instantaneously and can, respond rapidly., D.Mobile telephony : The concept of mobile, telephony was developed first in 1970’s and it, was fully implemented in the following decade., The central concept of this sytem is to divide, the service area into a suitable number of cells, centered on an office called MTSO (Mobile, Telephone Switching Office). Each cell contains, a low-power transmitter called a base station, and caters to a large number of mobile receivers, (popularly called cell phones). Each cell could, have a service area of a few square kilometers, NARAYANAGROUP, , COMMUNICATION SYSTEM, or even less depending upon the number of, customers. When a mobile receiver crosses the, coverage area of one base station, it is necessary, for the mobile user to be transferred to another, base staion. This process is called handover or, handoff. This process is carried out very rapidly,, to the extent that the consumer does not even, notice it. Mobile telephones operate typically, in the UHF range of frequencies (about 800-950, MHz), , EXAMPLE PROBLEMS, W.E-3:How many AM broadcast stations can be, accommodated in a 100 kHZ bandwidth if the, highest modulating frequency of carrier is 5, kHZ?, Sol. Any station being modulated by a 5 kHz singal, will produce an upper side frequency 5 kHz, above its carrier and a lower side frequency 5, kHz below its carrier, thereby requiring a, bandwidth of 10 kHz. Thus, Number of stations, accommodated, Total bandwidth, 100, =, =10, Bandwidth per station 10, W.E-4: How many 500 kHz waves can be on a, 10km transmission line simultaneously?, Sol: Let λ be the wavelength of 500 kHz signal. Then,, , c 3.0 ×108, λ= =, m = 600m, f 5.0 ×105, The number of waves on the line can be found, from,, d 10 ×103, =, = 16.67, λ, 600, W.E-5: A two wire transmission line has a, capacitance of 20 pF/m and a characteristic, impedance of 50Ω, a) What is the inductance per metre of this, cable?, b) Determine the impedance of an infinitely, long section of such cable., Sol: a) The characteristic impedance. Z = L / C, n=, , L = ( Z 2 ) ( C ) = ( 50 ) ( 20 ×10−12 ) H = 0.05H / m, 2, , b) The characteristic impedance of a, transmission line is the impedance that an infinite, length of line would present to a power supply, at the input end of the line. Thus, Z∞ = Z 0 = 50Ω, 45
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COMMUNICATION SYSTEM, , JEE-ADV PHYSICS- VOL- VI, , W.E-6: T.V. transmission tower at a particular, , i) f c = 100,000Hz = 100kHz, station has a height of 160m., ii) f c − fs = 100,000 − 2000 = 98kHz, a) What is the coverage range?, b) How much population is covered by, iii) f c + fs = 100kHz + 2kHz = 102kHz, transmission, if the average population, Therefore, frequency spectrum of modulated wave, density around the tower is 1200 per km 2 ?, extends from 98kHz to 102 kHz is called band, c) What should be the height of tower to, width., double the coverage range, W.E-8: The antenna current of an AM transmitter, Sol: a) Coverage range d = 2 Rh, is 8A when only the carrier is sent but it, 3, increases to 8.93A when the carrier is, = 2 × 6400 × 10 ×160m, modulated. Find percent modulation., = 45.254km, Sol. The modulated or total power carried by AM, b) Population covered, , = ( populationdensity ) × ( area covered ), = (1200 ) × (π d 2 ), , m2 , P, =, P, , wave T, C 1+, 2 . If R is load resistance., , , = ( 2400π Rh ) = 2400 × 3.14 × 6.4 ×103 × 0.16, , I m is the current when carrier is modulated and, , = 77.17 lac, , I c the current when unmodulated, then, , c) Coverage range ∝ h, Therefore coverage range can be doubled by, making height of the tower four times to 640m., So, height of the tower should be increased by, 480 m., W.E-7: An audio signal given by, , es = 15sin2π ( 200t ) modulates a carrier wave, , PT I m2 R, =, ;, PC I c2 R, , ∴I +, , m 2 I m2 R, = 2, 2, Ic R, , Given I m = 8.93 A, Ic = 8 A, 8.93 2 , 2, ∴ m = 2 , − 1 ∴ m = 0.7, 8.0 , , , Therefore, percentage modulation = 70%, W.E-9: A sinusoidal carrier voltage of 80 volts, a) Percent modulation, amplitude and 1 MHz frequency is amplitude, b) Frequency spectrum of the modulated wave., modulated by a sinusoidal voltage of, Sol: a) Signal Amplitude, B = 15, frequency 5kHz producing 50% modulation., Carrier amplitude, A = 60, Calculate the amplitude and frequency of, lower and upper side bands., B 15, m= =, = 0.25, Sol: Amplitude of both LSB and USB are equal and, A 60, given by, ∴ Percentage modulation = 0.25 ×100 = 25%, b) By comparing the given equations of signal, mEc 0.5 × 80, =, =, = 20volts, and carrier with their standard form, 2, 2, es = Es sin ωs t = Es sin2π f s t and, Now frequency of LSB = f − f, given by es = 60sin2π (100,000t ) . If calculate, , ec = Ec sin ωc t = Ec sin2π f c t, we have signal frequency f s = 2000Hz and, carrier frequency f c = 100,000Hz, The frequencies present in modulated wave, 46, , c, , s, , = (1000 −5) kHz = 995kHz, Frequency of USB = f c + fs, , = (1000 +5 ) kHz =1005kHz, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , W.E-10: The load current in the transmitting W.E-12: A transmitting antenna at the top of a, antenna of an unmodulated AM transmitter, is 6 Amp. What will be the antenna current, when modulation is 60%., Sol: Total power carried by AM wave, , m2 , PT = PC 1 +, ......(1), 2 , , , tower has a height 32m and the height of the, receiving antenna is 50m. What is the, maximum distance between them for, satisfactory communication in LOS mode?, Given radius of earth 6.4 ×106 m ., Sol: d m = 2 × 64 × 105 × 32 + 2 × 64 ×105 × 50m, , Where Pc is the power of carrier component and, = 64 ×10 2 × 10 + 8 ×103 × 10m, m is the modulation factor. If R is the resistance,, = 144× 102 × 10m = 45.5km, I m the antenna load current when modulation is W.E-13: A message signal of frequency 10 kHz, and peak voltage of 10 volts is used to, 60% and I c is the antenna load current when un, modulate a carrier of frequency 1 MHz and, modulated, then, peak voltage of 20 volts. Determine, PT I m2 R, m2 Im2, a) modulation index, = 2 ,∴1 +, = 2, using (1), b) the side bands produced., PC I c R, 2, Ic, Sol: a) modulation index = 10/20 = 0.5, 2, b) The side bands are at, m , I, =, I, 1, +, , , c, , , or m, 2 , (1000 + 10 ) kHz = 1010, , , kHz & (1000 − 10kHz ) = 990kHz . are, , Given I c = 6 Amp, m = 0.6, 1/2, , ( 0.6 )2 , I m = 6 1 +, = 6 [1.086] = 6.52 Amp, 2 , , W.E-11: A carrier wave of 1000 W is subjected to 1., 100% modulation. Calculate (i) Power of, modulated wave, (ii) power is USB, (iii) power, is LSB, Sol: i) Total power of modulated wave, , m2 , 12 , PT = PC 1 +, =, 1000, , 1 + = 1500watt, 2 ;, 2, , , , 2., , 1, PSB, 2, Where power carried by side bands is given by, amplitude modulation and detection, 3., , ii) Power in USB =, , m2 , 12 , PSB = PC , =, 1000, ;, = 500watt, 2 , 2, 1, 1, PSB = × 500 = 250watt, 2, 2, iii) Since power in LSB = Power in USB, PUSB =, , PLSB = PUSB = 250watt, NARAYANAGROUP, , 4., , frequency., , C. U. Q, A transducer used at the transmitting end,, serves the purpose of converting, 1) electrical signal to sound form, 2) sound signal to electrical form, 3) electrical signal to magnetic form, 4) sound signal to magnetic form, In a communication system, noise is most likely, to affect the signal, 1) at the transmitter, 2) in the medium of transmission, 3) information source signal, 4) at the destination, Device that converts one form of energy into, another is called, 1) transmitter, 2) transducer, 3) receiver, 4) channel, The part of communication system that, extracts the signal at the output of the channel, is, 1) transducer, 2) transmitter, 3) receiver, 4) receiver or transmitter, 47
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COMMUNICATION SYSTEM, 5., , The attenuation of a signal is compensated by, 1) rectifier, 2) oscillator, 3) modulator, 4) amplifier, , BAND WIDTH OF SIGNALS & BAND, WIDTH OF TRANSMISSION, 6., , 7., , 8., , 9., , 10., , 11, , 12., , 13., , 48, , JEE-ADV PHYSICS- VOL- VI, 14. The short wave Radio broadcasting band is, 1) 7 MHz to 22 MHz, 2) 88 MHz to 108 MHz, 3) 30 KHz to 300 KHz, 4) 3 GHz to 30 GHz, 15. The FM Radio broad casting band is, 1) 5 MHz to 30 MHz, 2) 88 MHz to 108 MHz, 3) 30 KHz to 300 KHz, 4) 3 GHz to 30 GHz, 16. The TV broad casting bands are, 1) MF and HF bands, 2) VHF and UHF bands, 3) UHF and SHF bands, 4) SHF and EHF band, 17. A: Satellite communication uses different, frequency bands for uplink and downlink, B: Bandwidth of video signals is 4.2 MHz, 1) A is true but B is false, 2) A is false but B is true, 3) A and B are false, 4) A and B are true, 18. A: The frequency band of VHF is greater than, UHF of TV transmission, B: Optical fiber transmission has frequency, band of 1 THz to 1000 THz, 1) A is true but B is false, 2) A is false but B is true, 3) A and B are false 4) A and B are true, 19. The higher frequency TV broad casting bands, range is, 1) 54 - 72 MHz and 76 to 88 MHz, 2) 174 - 216 MHz and 420 to 890 MHz, 3) 896 to 901 MHz and 840 to 935 MHz, 4) 5.925 to 6.425 GHz and 3.7 to 4.2 GHz, 20. Frequency ranges for micro waves are :, 1) 3 ×109 to 3 ×10 4 Hz, 2) 3 ×1013 to 3 ×109 Hz, , Modern communication systems use, 1) analog circuits, 2) digital circuits, 3) combination of analog & digital circuits, 4) radio circuits, The audio signal, 1) can be sent directly over the air for large, distance, 2) can not be sent directly over the air for large, distance, 3) possesses very high frequency, 4) possesses very low frequency, A digital signal possess, 1) continuously varying values, 2) only two discrete values, 3) only four discrete values, 4) constant values, Digital signals, 1) provide continuous set of values, 2) represent values as randomly, 3) Utilise Decimal code system, 4) Utilise binary code system, Digital signals, i) do not provide a continuous set of values., ii) represents values as descrete steps., iii) can utilize binary system, iv) can utilize decimal as well as binary system., The true option is., 1) (i) & (ii) only, 2) (ii) & (iii) only, 3) (i), (ii) & (iii) only 4) (i),(ii),(iii)& (iv), A digital signal, 1) is less reliable than analog signal, 2) is more reliable than analog signal, 3) 3 ×1014 to 3 ×109 Hz, 3) is equally reliable as the analog signal, 4) Not at all reliable, 4) 3 ×1011 to 3 ×109 Hz, The band width required for transmiting video 21. The frequency band used for radar relay, signal is, systems & T.V is, 1) 50 KHz 2) 1 MHz 3) 4.2 MHz 4) 6 MHz, 1) UHF, 2) VLF, 3) VHF, 4) EHF, Band width of an optical fiber is, 22. For TV transmission the frequency range, 1) more than 100 GHz 2) few kHz, employed, (Karnataka 1990, 1989), 3) less than 1MHz, 4) less than 1GHz, 1) 30 - 300 MHz, 2) 30 - 300 GHz, 3) 30 - 300 KHz, 4) 30 - 300 Hz, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 23. The frequency which is not part of AM, broadcast, 1) 100 kHz, 2) 700 kHz, 3) 600 kHz, 4) 1500 kHz, 24. Cellular Mobile works in the frequency range, of, 1) 840 to 935 MHz, 2) 3.7 to 4.2 GHz, 3) 420 to 890 MHz, 4) 30 to 300 GHz, 25. Frequency range used in down linking in, satellite communication is, 1) 0.896 to 0.901 GHz, 2) 0.420 to 0.890 GHz, 3) 5.925 to 6.425 GHz, 4) 3.7 to 4.2 GHz, 26. In the satellite communication, the uplinking, frequency range is, 1) 0.896 to 0.901 GHz, 2) 0.420 to 0.890 GHz, 3) 5.925 to 6.425 GHz, 4) 3.7 to 4.2 GHz, 27. In a communication system, noise is most likely, to affect the signal, 1) At the transmitter, 2) In the channel or in the transmission line, 3) In the information source, 4) At the receiver, 28. The frequency of a FM transmitter without, signal input is called, 1) Lower side band frequency, 2) Upper side band frequency, 3) Resting frequency, 4) None of these, 29. Indicate which one of the following system is, digital, 1) Pulse position modulation, 2) Pulse code modulation, 3) Pulse width modulation, 4) Pulse amplitude modulation, 30. Television signals on earth cannot be received, at distances greater than 100km from the, transmission station. The reason behind this, is that, 1) The receiver antenna is unable to detect the, signal at a distance greater than 100 km, 2) The TV programme consists of both audio, and video signals, 3) The TV signals are less powerful than radio, signals, 4) The surface of earth is curved like a sphere, NARAYANAGROUP, , COMMUNICATION SYSTEM, 31. Audio signal cannot be transmitted because, 1) The signal has more noise, 2) The signal cannot be amplified for distance, communication, 3) The transmitting antenna length is very small to, design, 4) The transmitting antenna length is very large, and impracticable, , PROPAGATION OF EM WAVES IN, THE ATMOSPHERE, 32. A signal emitted by an antenna from a certain, point can be received at another point of the, surface in the form of, 1) sky wave, 2) ground wave, 3) sea wave, 4) both 1 and 2, 33. An antenna is a device, 1) That converts electromagnetic energy into, radio frequency signal, 2) That converts radio frequency signal into, electromagnetic energy, 3) That converts guided electromagnetic waves, into free space electromagnetic waves and viceversa, 4) None of these, 34. An antenna, 1) Converts AF wave to RF wave, 2) RF signal into electromagnetic energy, 3) Converts the guided EM waves into free space, EM waves and vice versa, 4) Super imposes AF wave on RF wave, 35. An antenna behaves as a resonant circuit only, when the length is, 1) equal to ?/4, 2) equal to ?/2, 3) equal to the integral multiples of ?/2, 4) equal to 3?/4, 36. If audio signal is transmitted directly into, space, the length of the transmitting antenna, required will be, 1) extremely small, 2)extremely large, 3) infinitely large, 4) moderate, 37. The height of the antenna, a) limits the population covered by the, transmission, b) limits the ground wave propagation, c) effectively used in line of sight, communication, 1) a & b are true, 2) b & c are true, 3) c & a are true, 4) a, b,c are true, 49
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COMMUNICATION SYSTEM, 38. Statement A: If the antenna is vertical the, vertically polarised EM wave is radiated, Statement B: The vertically polarised EM, wave has electrical variations in the vertical, plane, 1) A is true but B is false, 2) A is false but B is true, 3) A and B are false 4) A and B are true, 39. A: It is necessary for transmitting antenna, must be at same height as that of receiving, antenna for line of sight communication., B: EM waves of frequency beyond 40 MHz,, propagate as space waves., 1) both A and B are correct, 2) both A and B are wrong, 3) only A is correct, 4) only B is correct, 40. Broadcasting antennas are generally, 1) Omnidirectional type, 2) Vertical type, 3) Horizontal type, 4) None of these, , MODULATION, 41. The process of superimposing signal frequency, (i.e. audio wave) on the carrier wave is known, as, 1)Transmission, 2) Reception, 3)Modulation, 4) Detection, 42. In frequency modulation, 1) The amplitude of modulated wave varies as, frequency of carrier wave, 2) The frequency of modulated wave varies as, amplitude of modulating wave, 3) The amplitude of modulated wave varies as, amplitude of carrier wave, 4) The frequency of modulated wave varies as, frequency of modulating wave, 43. For transmitting audio signal properly, 1) it is first superimposed on high frequency, carrier wave, 2) it is first superimposed on low frequency, carrier wave, 3) It is sent directly without superimposing on, any wave, 4) it is superposed with carrier wave of high, velocity, 50, , JEE-ADV PHYSICS- VOL- VI, 44. The process of recovering the audio signal, from the modulated wave is known as, 1) amplification, 2) rectification, 3) modulation, 4) demodulation, 45. The most commonly employed analog, modulation technique in satellite, communication is the, 1) amplitude modulation, 2) frequency modulation, 3) phase modulation, 4) amplitude & phase modulation, 46. The need for doing modulation is, 1) to increase the intensity of audio signal, 2) to decrease the intensity of audio signal, 3) to transmit audio signal to large distances, 4) to increase the frequency of the audio signal, 47. The type of modulation is employed in India, for radio transmission is, 1) pulse modulation, 2) frequency modulation, 3) amplitude modulation, 4) phase modulation, 48. Modulation is used to, 1) reduce the bandwidth, 2) to seperate the transmission of different users, 3) to ensure that intelligence may be, transmitted to long distances, 4) to allow the uses of practical antenna, 49. The process of translating the information, contained by the low base band signal to high, frequencies is called, 1) Detection, 2) Modulation, 3) Amplification, 4) Demodulation, 50. During the process of modulation the RF wave, is called, 1) Modulating wave 2) Modulated wave, 3) Carrier wave, 4) Audio wave, 51. Modulation is required to, a) distinguish different transmissions, b) ensure that the information may be trans, mitted over long distances, c) allow the information accessible for, different people, 1) a & b are true, 2) b & c are true, 3) c & a are true, 4) a, b & c are true, 52. The physical quantities of the wave used for, modulation, 1) Amplitude only, 2) Amplitude and frequency, 3) Amplitude, frequency and phase, 4) Only frequency, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 53. In Amplitude modulation, 1) The amplitude of the carrier wave varies in, accordance with the amplitude of the, modulating signal, 2) The amplitude of carrier wave remains constant,, frequency changes in accordance with the, modulating signal, 3) The amplitude of carrier wave varies in, accordance with the frequency of the, modulating signal, 4) The amplitude changes in accordance with, the wave length of the modulating signal, 54. Amplitude modulation is used for broad, casting because, 1) it is more noise immune, 2) it requires less transmitting power, 3) it has simple circuit, 4) it has high fidelity (faithful reproduction), 55. In amplitude modulation, carrier wave, frequencies are ............... than that compared, to those in frequency modulation, 1) lower, 2) higher, 3) same, 4) lower or higher, 56. AM is used for broad casting because,, 1) it is more noise immune than other modulating, systems, 2) it requires less transmitting power compared, with other systems, 3) its use avoids receiver complexity, 4) no other modulation system can provide the, necessary bandwidth, faithful transmission., 57. Draw backs of Amplitude modulation, 1) During transmission extreneous noise creeps, in., 2) Most of the transmitting power is wasted, as, it does not contain useful information., 3) The reception is not clear in the case of weak, signals due to noise, 4) The receiver set is complex, 58. In amplitude modulation, 1) only amplitude is changed but frequency, remains same, 2) both amplitude & frequency changes equally, 3) both amplitude & frequency changes, unequally, 4) only frequency changes but amplitude remains, constant., NARAYANAGROUP, , COMMUNICATION SYSTEM, 59. The limitation of amplitude modulation is, 1) clear reception, 2) high efficiency, 3) small operating range, 4) good audio quality, 60. In frequency modulation, 1) Frequency of CW remains constant but, amplitude changes in accordance with, modulating wave frequency, 2) Frequency of CW changes in accordance with, the modulating wave frequency but the amplitude, also changes., 3) Frequency of CW changes in accordance with, the frequency of modulating wave frequency but, the amplitude remains constant., 4) Frequency of CW changes in accordance, with the amplitude of modulating wave, amplitude, 61. In T.V. broadcasting both picture and sound, are transmitted simultaneously. In this, 1) audio signal is frequency modulated and video, signal is amplitude modulated, 2) both audio and video signals are frequency, modulated, 3) audio signal is amplitude modulated and video, signal is frequency modulated, 4) both audio and video signals are amplitude, modulated, 62. Effective power radiated by an antenna is, 1) Proportional to the square at the length of the, antenna, 2) inversely proportional to the wavelength, 3) inversely proportional to the square of the, wavelength, 4) proportional to the wavelength, 63. The concepts of communication are, a) mode of communication, b) need for modulation, c) types of modulation, d) detection of modulated wave, 1) a, b, c are true, 2) b, c, d are true, 3) c, d, a are true, 4) a, b, c & d are true, 64. The difference between phase and frequency, modulation, 1) practically they are same but theoretically, they differ, 2) lies in the poorer audio response of phase, modulation, 3) lies in the poorer audio response of frequency, modulation, 4) lies in the definitions of modulation and their, modulation index, 51
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, 65. Basically, the product modulator is, 1) An amplifier, 2) A mixer, 3) A frequency separator, 4) A phase separator, 66. Which of the following is the disadvantage of, FM over AM, 1) Larger band width requirement, 2) Larger noise, 3) Higher modulation power, 4) Low efficiency, 67. Audio signal cannot be transmitted as such, because, 1) the signal has more noise, 2) the signal cannot be amplified for distance, communication, 3) the transmitting antenna length is very small, to design, 4) the transmiting antenna length is very large, and impracticable, 68. The examples of broadcast are, A) radio, B) television, C) telephony, D) internet, 1) A & B, 2) A, B & D, 3) A,B & C 4) B & D, 69. The waves relavent to telecommunications, are, 1) visible light, 2) infrared, 3) ultraviolet, 4) microwave, 70. While tuning in a certain broad cast station, with a receiver, we are actually, 1) varying the local oscillator, 2) varying the resonant frequency of the circuit, for the radio signal to be picked up, 3) tuning the antenna, 4) varying the current of receiver set, 71. Long distance short-wave radio broadcasting, uses, 1) Ground wave, 2) Ionospheric wave, 3) Direct wave, 4) Sky wave, 72. Radio waves of constant amplitude can be, generated with, 1) filter, 2) rectifier 3) FET 4) oscillator, 73. Advantage of HF transmission is, A) that the length of antenna is small, B) that the antenna can be mounted at larger, heights, C) that the power radiated is more for a given, 52, , 74., , 75., , 76., , 77., , 78., , 79., , 80., , 81., , length of antenna, 1)a & b, 2) b& c 3) a & c 4) a,b & c, The intensity of the ground waves decrease, with increase of distance due to, 1) Interference 2) Diffraction, 3) Polarization 4) Due to unknown reason, High frequency waves are, 1) absorbed by F layer, 2) reflected by the E layer, 3) capable of use for long distance transmission, 4) affected by the solar cycle, As the e.m. waves travel in free space, 1) absorption takes place, 2) attennuation takes place, 3) refraction takes place, 4) reflection takes place, The electromagnetic waves of frequency 80, MHz and 200 MHz, 1) can be reflected by troposphere, 2) can be reflected by ionosphere, 3) can be reflected by mesosphere, 4) cannot be reflected by any layer of earth’s, atmosphere, The better propagation mode to propagate, television frequency and radar signals is, 1) satellite communication, 2) ground propagation, 3) polarized communication, 4) skywave communication, Micro wave link repeaters are typically 50, km apart, 1) because of atmospheric attenuation, 2) because of the earths curvature, 3) to ensure that signal voltage may not harm the, repeater, 4) to reduce the interference of microwaves, Attenuation of ground waves is due to, 1) Diffraction effect, 2) Radio waves induce currents in the ground, because of the polarisation, 1) a & b are true, 2) Only a is true, 3) Only b is true, 4) Both a & b false., The ground wave eventually disappears, as, one moves away from the transmitter, because, of, 1) interference from the sky wave, 2) loss of line of signal condition, 3) maximum single - hop distance limitation, 4) diffraction effect causing tilting of the wave, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 82. The range of ground wave transmission can, be increased by, 1) increasing the power of transmitter with the use, of HF, 2) increasing the power of transmitter with the, use of VLF, 3) decreasing the power and increasing the, frequency of radio waves, 4) decreasing both power and frequency of radio, waves, 83. Space wave propagation is used in, a) microwave communication, b) satellite communication, c) TV transmission, 1) Only a, 2) Both a & b, 3) Both b & c, 4) a ,b & c, 84. Frequencies in the UHF range normally, propagate by means of:, 1) Ground waves, 2) Sky waves., 3) Surface waves, 4) Space waves., 85. When a sky wave is reflected onto the ground, 1) frequency of the reflected wave is different, to that of incident wave, 2) there is a phase difference introduced to the, reflected wave, 3) the reflected wave is out of phase with incident, wave and reach the receving antenna along with, the direct wave from transmitting antenna causing, interference., 4) the waves are not reflected by the ground., 86. The electromagnetic waves of frequency 2, MHz to 30 MHz are, (1)In ground wave propagation, (2)In sky wave propagation, (3)In microwave propagation, (4)In satellite communication, 87. Among the following frequencies one will be, suitable, for, beyond-the, horizon, communication using sky waves is, 1) 10 kHz 2) 10 MHz 3) 1GHz 4)1000GHz, 88. Among the following, the waves which can, penetrate the ionosphere are, 1)10GHz 2)10MHz 3)20MHz 4) 25 MHz, 89. A: At great heights from surface of earth, and close to earth ionisation of air molecules, is low., B: EM waves of frequencies beyond 30 MHz, penetrate ionosphere and escape., 1) both A and B are correct, 2) both A and B are wrong, 3) only A is correct, 4) only B is correct, NARAYANAGROUP, , COMMUNICATION SYSTEM, 90. Through which mode of propagation, the radio, waves can be sent from one place to another, 1) Ground wave propagation, 2) Sky wave propagation, 3) Space wave propagation, 4) All of them, 91. The frequency at which communication will not, be reliable is, 1) 100 KHz, 2) 1MHz, 3) 10GHz, 4) 100 GHz, 92. The frequency above which radiation of, electrical energy is practical is, 1) 0.2 kHz, 2) 2 kHz, 3) 20 kHz, 4) 2Hz, 93. The radio waves of frequency 300 MHz to, 3000 MHz belong to, 1) High frequency band, 2) Very high frequency band, 3) Ultra high frequency band, 4) Super high frequency band, 94. Coaxial cable is an example of, 1) Optical fibre, 2) Free space, 3) Wire medium, 4) Sea medium, 95. Optical fibre communication is generally, preferred over general communication system, because, 1) it is more efficient 2) of signal security, 3) both (1) & (2), 4) it is easily available, 96. The attenuation in optical fibre is mainly due, to, 1)Absorption, 2)Scattering, 3)Neither absorption nor scattering, 4)Both 1 and 2, 97. Consider telecommunication through optical, fibres. Which of the following statements is, not true [AIEEE 2003], (1)Optical fibres may have homogeneous core, with a suitable cladding, (2)Optical fibres can be of graded refractive, index, (3)Optical fibres are subject to electromagnetic, interference from outside, (4)Optical fibres have extremely low, transmission loss, 98. The phenomenon by which light travels in an, optical fibres is, 1)Reflection 2)Refraction, 3)Total internal reflection 4) Transmission, 53
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, 99. A laser is a coherent source because it contains 4., 1) Many wavelengths, 2) Uncoordinated wave of a particular wavelength, 3) Coordinated wave of many wavelengths, 4) Coordinated waves of a particular wavelength, 100. In which of the following remote sensing 5., technique is not used, 1)Forest density, 2)Pollution, 3)Wetland mapping, 4)Medical treatment, , C. U. Q - KEY, 1) 2, 8) 2, 15) 2, 22) 1, 29) 2, 36) 2, 43) 1, 50) 3, 57) 2, 64) 1, 71) 3, 78) 1, 85) 3, 92) 3, 99) 4, , 2) 2 3) 2, 9) 4 10) 3, 16) 2 17) 4, 23) 1 24) 1, 30) 4 31) 4, 37) 4 38) 4, 44) 4 45) 2, 51) 4 52) 3, 58) 1 59) 3, 65) 2 66) 1, 72) 4 73) 4, 79) 2 80) 1, 86) 2 87) 2, 93) 3 94) 3, 100) 4, , 4) 3, 11) 2, 18) 2, 25) 4, 32) 4, 39) 4, 46) 3, 53) 1, 60) 3, 67) 4, 74) 2, 81) 4, 88) 1, 95) 3, , 5) 4, 12) 3, 19) 2, 26) 3, 33) 3, 40) 2, 47) 3, 54) 3, 61) 1, 68) 2, 75) 2, 82) 2, 89) 1, 96) 4, , 6., 6) 2, 13) 1, 20) 2, 27) 2, 34) 3, 41) 3, 48) 1, 55) 1, 62) 3, 69) 4, 76) 2, 83) 4, 90) 4, 97) 3, , 7) 2, 14) 1, 21) 1, 28) 3, 35) 1, 42) 2, 49) 2, 56) 3, 63) 4, 70) 3, 77) 4, 84) 4, 91) 1, 98) 3, , LEVEL - I (C.W), 1., , 2., , 3., , An optical communication system is operating, at a wavelength of 800 nm, it’s optical source, frequency is, 1) 3.8×1014 Hz, 2) 3.8×1012 Hz, , 7., , 8., , 9., , A T V transmitting antenna is 80 m tall. If the, receiving antenna is on the ground the, service area is, 1) 12 sq km, 2) 3215 sq km, 3) 144sq km, 4) 32 sq km, The maximum distance upto which TV, transmission from a TV tower of height ‘h’, can be received is proportional to, 1) h1/2, 2) h, 3) h 3/2, 4) h2, In short wave communication waves of which, of the following frequencies will be reflected, back by the ionospheric layer, having electron, density 1011 per m3, 1)2 MHz, 2)10 MHz, 3)12 MHz, 4)18 MHz, In an amplitude modulated wave for audio, frequency of 500 cycle/second, the, appropriate carrier frequency will be, 1)50 cycles/sec, 2)100 cycles/sec, 3)500 cycles/sec, 4)50,000 cycles/sec, The modulation index of an FM carrier having, a carrier swing of 200 kHz and a modulating, signal 10 kHz is, 1)5, 2)10, 3)20, 4)25, If a number of sine waves with modulation, indices n 1 , n2 , n3 ........ Modulate a carrier, wave, then total modulation index (n) of the, wave is, 1)n1 + n2 .... + 2(n1 + n2 .....), 2) n1 − n2 + n3 .........., , 3) n12 + n22 + n32 ........., 4)n1 + n2 ...., 3) 3.8×1010 Hz, 4) 3.8×109 Hz, 10. A sky wave with a frequency 55 MHz is, incident on D-region of earth’s atmosphere, A carrier wave of 1000 kHz is used to carry, at 45o . The angle of refraction is (electron, the signal, the length of the transmitting, density for D-region is 400 electron/cm3 ), antenna will be equal to, 1)60°, 2)45°, 3) 30°, 4)15°, 1) 3 m, 2) 30 m, 11. What should be the maximum acceptance, 3) 300 m, 4) 3000 m, angle at the aircore interface of an optical, A transmitting antenna is at a height of 40 m, fibre if n1 and n2 are the refractive indices of, and the receiving antenna is at a height of 60, the core and the cladding, respectively, m. The maximum distance between them for, satisfactory communication is nearly, 1) sin− 1(n2 / n1 ), 2) sin− 1 n12 − n22, 1) 22.5 km, 2) 27.5 km, −1 n 2 , −1 n1 , 3) 50 km, 4) 25 km, 3) tan n1 , 4) tan n2 , , , 54, , , , , , , , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , 12. The characteristic impedance of a coaxial, cable is of the order of, 1)50 Ω, 2)200 Ω, 3) 130 Ω, 4)None of these, 13. The velocity factor of a transmission line x. If, dielectric constant of the medium is 2.6, the, value of x is, 1)0.26, 2)0.62, 3)2.6, 4)6.2, 14. For television broadcasting, the frequency, employed is normally, 1)30-300 MHz, 2)30-300 GHz, 3)30-300 KHz, 4)30-300 Hz, 15. Because of tilting which waves finally, disappear, 1) Microwaves, 2)Surface waves, 3)Sky waves, 4)Space waves, 16. A radar has a power of 1kW and is operating, at a frequency of 10 GHz. It is located on a, mountain top of height 500m. The maximum, distance upto which it can detect object, located on the surface of the earth (Radius of, earth 6.4 ×106 m ) is, 1) 80 km 2) 16 km 3) 40 km, 4) 64km, , Also, , c = n? ⇒ n = c/?, length of antenna ≈ ?, , 13., , area covered, A = pd 2 where d =, , 5., 6., 7., 8., , d = 2Rh ⇒ d a h, By using fc ≈ 9(N max )1 / 2 ⇒ fc ≈ 2 MHz, Carrier frequency > audio frequency, CS = 2 ´ ∆ f or ∆ f = CS/2, ∆f =, , 10., , 200, = 100 kHz, 2, , n eff = n 0, , ; Now, , mf =, , 1., , 2., , 2Rh, , ∆ f 100, =, = 10, fm, 10, , 80 . 5 N , 80 . 5 × ( 400 × 10 6 ), 1−, = 1 1−, ≈ 1, 2, ν, ( 55 × 10 6 ) 2, , , , NARAYANAGROUP, , r = i = 45°, , = sin− 1 n12 − n22, , 1, k, , =, , 1, 2 .6, , = 0. 62, , ( R + h ) 2 − R2, , = 2 Rh + h 2, , LEVEL - I ( H. W ), , 4., , 4., , ⇒, , 1, = 2Rh = 2 × 6400 × km = 80km, 2, , 3., , 2Rh T + 2Rh R, , v. f . =, , Range =, , C, 3 × 108, =, = 300 m, ν 1000 ×10 3, distance between transmitter and receiver, is s =, , sin r = sin i, , 14. VHF (Very High Frequency) band having frequency, range 30 MHz to 300 MHz is typically used for TV, and radar transmission., 15. conceptual, 16. Range of radar on earth surface (optical distance,, for space wave, i.e., line of view), , Here l = λ =, 3., , ⇒, , 12. conceptual, , LEVEL - I ( C. W ) - HINTS, 1., 2., , sin i, sin r, , 11. Core of acceptance angle θ, , LEVEL - I ( C. W ) - KEY, 1) 1 2) 3 3) 3 4) 2 5) 1 6) 1 7) 4, 8) 2 9) 3 10) 2 11) 2 12) 3 13) 2 14) 1, 15) 2 16) 1, , neff =, , 5., , An optical communication system is operating, at a wavelength of 600 nm, it’s optical source, frequency is, 1) 3.8×1014 Hz, 2) 3.8×1012 Hz, 3) 3.8×1010 Hz, 4) 5×1014 Hz, A carrier wave of 2000 kHz is used to carry, the signal, the length of the transmitting, antenna will be equal to, 1) 3 m, 2) 30 m 3) 150 m, 4) 3000 m, A transmitting antenna is at a height of 25 m, and the receiving antenna is at a height of 64, m. The maximum distance between them for, satisfactory communication is nearly, 1) 22.5 km, 2) 46.5 km, 3) 50 km, 4) 25 km, A T V transmitting antenna is 100 m tall. If, the receiving antenna is on the ground the, service area is, 1) 12 sq km, 2) 4000sq km, 3) 144sq km, 4) 32 sq km, The maximum distance upto which TV, transmission from a TV tower of height ‘h’, can be received is proportional to (R earth, radius), 1, , 1) ( Rh ) 2, , 2) h, , 3) h 3/2, , 4) h2, 55
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, 6., , In short wave communication waves of which, of the following frequencies will be reflected, back by the ionospheric layer, having electron, density 1012 per m3, 1)2 MHz, 2) 9 MHz, 3)12 MHz, 4)18 MHz, 7. In an amplitude modulated wave for audio, frequency of 1000 cycle/second, the, appropriate carrier frequency will be, 1)50 cycles/sec, 2)100 cycles/sec, 3)500 cycles/sec, 4) 40,000 cycles/sec, 8. The modulation index of an FM carrier having, a carrier swing of 300 kHz and a modulating, signal 10 kHz is, 1) 15, 2)10, 3)20, 4)25, 9. A sky wave with a frequency 64.4 MHz is, incident on D-region of earth’s atmosphere, at 45o . The angle of refraction is (electron, density for D-region is 400 electron/cm3 ), 1)60°, 2) 90°, 3) 45°, 4)15°, 10. The velocity factor of a transmission line x. If, dielectric constant of the medium is 1.8, the, value of x is, 1)0.26, 2)0.62, 3) 0.74, 4)6.2, 11. A radar has a power of 1kW and is operating, at a frequency of 10 GHz. It is located on a, mountain top of height 700m. The maximum, distance upto which it can detect object, located on the surface of the earth (Radius of, earth 6.4 ×106 m ) is, 1) 80 km 2) 16 km 3) 40 km, 4) 92 km, , 4., , = 46.5 km, Area covered A = π d 2 ; where d = 2 Rh, , 5., , d = 2 Rh ; d ∝ Rh ; d ∝ ( Rh ) 2, , 6., , By using f c ≈ 9 ( N max ) 2 ; f c ≈ 9 (1012 ) 2, , 7., 8., , f c ≈ 9 MHz, Carrier frequency > audio frequency, CS = 2 ∆f, , 1, , 9., , 2) 3, 8) 1, , 3) 2, 9) 3, , 2., , C, 3 × 108, =, = 150m, ν 2000 ×103, Distance between transmitter and receiver is, , 3., , d = 2 RhT + 2RhR, = 2 × 6400 ×103 × 25 + 2 × 6400 ×103 × 64, 56, , ∆f 150KHz, =, = 15, f m 10KHz, , neff = n0 1 −, , 80.5 N, ν2, , 80.5 × 400 ×106, , ( 64.4 ×10 ), , 6 2, , ; ≈1, , sin i, ⇒ sin i = sin r ; ⇒ i = r = 450, sin r, , 1, 1, = 0.74, ; =, k, 1.8, 11. Range of Radar on earth surface (optical, distance, for space wave, i.e line of view), Range =, , ( R + h )2 − R 2, , = 2Rh + h2, , = 2Rh = 2 × 6400 × 0.7, , [ 700m = 0.7km] = 92 km, LEVEL - II (C. W), , C = nλ ; n =, , here l = λ =, , mf =, , 10. V . f =, , 4) 2 5) 1 6) 2, 10) 3 11) 4, , C, λ, Length of antenna ≈ λ ,ν = 2000KHz, , CS 300 KHz, =, = 150KHz, 2, 2, , Also neff =, , LEVEL - II ( H. W ) -HINTS, 1., , ∆f =, , = 1 1−, , LEVEL - II (H. W ) - KEY, 1) 4, 7) 4, , 1, , 1, , 1., , The maximum distance between the, transmitting and receving TV towers is 72 km, . If the ratio of the heights of the TV, transmitting tower to receiving tower is, 16, : 25, the heights of the transmitting and, receiving towers are, 1) 51.2 m ; 80 m, 2) 40 m ; 80 m, 3) 80 m ; 125 m, 4) 25 m ; 75 m, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 2., , 3., , 4., , 5., , 6., , 7., , COMMUNICATION SYSTEM, , The maximum distance between the 9., transmitting and receiving TV towers is D. If, the heights of both transmitting and receiving, towers are doubled then the maximum, distance between them becomes, 10., 1) 2D, 2) 2 D 3) 4D, 4) D/2, A T.V. tower has a height of 10 m is in a region, of average population density 100 p /km2 ., The number of people that can recieve the, transmission is nearly, 1) 1,28,000 2) 64,000 3) 2,56,000 4) 32,000, At certain distance from a transmitting tower 11., a receiver tower of height 20 m is used to, receive direct signal. Another tower is, installed beyond the first along the same line, of sight to receive the signals from the same, transmitter. Its height is 44% more than the, first receiving tower. Then the separation, between the two receiving towers is, 12., 1) 6.4 km 2) 3.2 km 3) 1.6 km 4) 0.8 km, The TV signals have a band width of 6MHz., The number of TV channels that can be, 13., accommodated in a band width 12 GHz is, 1) 2, 2) 20, 3) 200, 4) 2000, A carrier wave of peak voltage 12 V is used, to transmit a message signal. The peak, voltage of the modulating signal in order to, 14., have a modulation index of 75% is, 1) 6 V, 2) 9 V, 3) 4 V, 4) 15 V, A modulating signal is a square wave as shown, in the figure, , 1, m(t), in Volts, , The, , 1, , carrier, , wave, , 2, , is, , t(s), , given, , by, , c ( t ) = 2sin ( 8π t ) volts ., The modulation index is, 1) 2, 2) 0.75 3) 0.5, 8., , 4) 1.5, , An AM wave is given by, V = 1500 [1 + 0.5sin12560t ] sin (5.26 ×105 t ) ., , The modulating frequency is, 1) 2.0 kHz, 2) 1.0 kHz, 3) 12.5 kHz, 4) 50 kHz, The amplitude modulated current is given by, i = 125 [1 + 0.6sin2900t ] sin ( 5.50 ×105 t ) ., , The RMS value of carrier current will be, 125, 100, 75, 50, A 2), A 3), A 4), A, 1), 2, 2, 2, 2, An audio signal 25sin2π (1400t ) amplitude, , modulates 80sin2π (105 t ) . The two side band, frequencies are, 1) 98.6 kHz, 101.4 kHz, 2) 92.5 kHz, 105.5 kHz, 3) 94 kHz, 102.5 kHz, 4) 96 kHz, 106 kHz, If f c and f m are the frequencies of carrier, wave and signal, then the band width is, 1) f m, 2) 2 f m 3) f c, 4) 2 f c, The number of AM broadcast stations that, can be accommodated in a 300kHz band width, for the highest modulating frequency 15 kHz, will be, 1) 10, 2) 5, 3) 7, 4) 12, A 1000 kHz carrier is simultaneously, modulated with 300Hz, 800 Hz and 2 kHz, audio waves. The frequencies present in the, output are, 1) 999.7 kHz, 1000.3kHz, 999.2kHz, 2) 1000.8kHz, 998 kHz, 1002kHz, 3) 1002.8 kHz, 996kHz, 1106 kHz, 4) Both (1) and (2), 15. Depth of modulation in terms of Emax and Emin, is, Emax + Emin, E − Emin, m a = max, 1) m a =, 2), Emin, Emax, Emax − Emin, 3) m a = E + E, max, min, , Emax + Emin, 4) m a = E − E, max, min, , 16. If Vc is amplitude of carrier wave in AM, transmitter where modulation factor is m then, amplitude of side bands can be, 5, modulates 60sin2π (10 ) t . The depth of, V, m, V, Vc 3) mVc, 1) c, 2), 4) c, modulation is, 2m, 2, 2, 1) 25%, 2) 20% 3) 50%, 4) 40%, A audio signal 15sin2π (1500t ) amplitude, , NARAYANAGROUP, , 57
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COMMUNICATION SYSTEM, , JEE-ADV PHYSICS- VOL- VI, , 17. If a carrier wave of amplitude 10mV is 26. In an FM system a 7 kHz signal modulates, 108MHz carrier so that frequency deviation, modulated by an audio signal of amplitude, is 50 kHz. The frequency modulation index is, 2mV, then amplitude of LSF or USF is given, 1) 7.143, 2) 8, 3) 0.71, 4) 350, by, 27., Maximum, usable, frequency, (MUF), in F1) 0.2 mV 2) 0.5 mV 3) 1 mV, 4) 2 mV, region layer is x, when the critical frequency, 18. The amplitude modulated current is given by, is 60 MHz and the angle of incidence is 70°., Then x is, i = 125 [1 + 0.6sin2900t ] sin106 t . The depth, (1)150 MHz, 2)170 MHz, of modulation is, (3)175 MHz, 4)190 MHz, 1) 60%, 2) 6%, 3) 36%, 4) 66%, 28. Suppose that the modulating signal is, 19. If minimum voltage in an AM wave was found, m ( t ) = 2cos ( 2π f mt ) and the carrier signal is, to be 2V and maximum voltage 10V. The %, modulation index, xc ( t ) = Ac cos ( 2π f c t ) which one of the, 1) 80%, 2) 66.67%, following is a conventional AM signal without, 3) 64.25%, 4) 76.25%, over modulation, 20. Sinusoidal carrier voltage of frequency 1.5, 1) x ( t ) = Ac m( t ) cos ( 2π f c t ), MHz and amplitude 50 V is amplitude, modulated by sinusoidal voltage of frequency, 10 kHz producing 50% modulation. The lower, and upper side-band frequencies in kHz are, 1) 1490, 1510, 2)1510, 1490, 3), , 1, 1, ,, 1490 1510, , 4), , 1, 1, ,, 1510 1490, , 2) x ( t ) = Ac (1 + m ( t ) ) cos ( 2π f c t ), 3) x ( t ) = Ac cos ( 2π f c t ) +, Ac, m ( t ) cos ( 2π f ct ), 4, , 4) xc ( t ) = Ac cos ( 2π f mt ) +, 21. A transmitter supplies 9 kW to the aerial when, unmodulated. The power radiated when, cos ( 2π f c t ) + Ac sin ( 2π f mt ) sin ( 2π fct ), modulated to 40% is, 29. C ( t ) and m ( t ) are used to generate an AM, 1) 5 kW, 2) 9.72 kW, signal. The modulation indes of generated AM, 3) 10 kW, 4) 12 kW, 22. The total power content of an AM wave is, PTotal SB, signal, is, 0.5., Then, the, quantity, =, 1500 W. For 100% modulation, the power, PCarrier, transmitted by the carrier is, 1) 1/8, 2) 1/4, 3) 2/3, 4) 9/8, 1) 500 W, 2) 700 W, 30. Which of one of the following modulated, 3) 750 W, 4) 1000 W, signals are recovered up to a scaling factor, 23. The bit rate for a signal, which has a sampling, using envelope detedtor, rate of 8 kHz and where 16 quantization levels, 1) 20cos200π t + 30m ( t ) cos200π t, have been used, is, 2) 20cos200π t + 16 m ( t ) cos200π t, 1) 32000 bits/sec, 2) 16000 bits/sec, 3) 64000 bits/sec, 4) 72000 bits/sec, 3) 10m ( t ) cos400π t, 4) None, 24. The antenna current of an AM transmitter is, 8 A when only carrier is sent but increases to 31. A message signal m ( t ) = 4cos2000π t, 8.96 A when the carrier is modulated, modulates the carrier C ( t ) = cos2π f ct where, sinusoidally. The percentage modulation is, 1)50%, 2)60% 3)65%, 4)71%, f c = 1MHz to produce an AM signal. For, 25. In a diode AM-detector, the output circuit, demodulation using envelope detector the, consist of R = 1kW and C = 10 pF. A carrier, time constant RC should satisfy, signal of 100 kHz is to be detected. Is it good, 1) 0.5ms < RC < 1ms 2) 1µ s = RC = 0.5ms, 1) Yes, 2) No, 3) RC ? 1µ s, 4) RC ? 1µ s, 3) Information is not sufficient 4) None of these, 58, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , 32. A 1 MHz carrier signal is modulated by a, symmetrical sinusoidal wave for period of, 100 µ s in a nonlinear (square law device)., Which of the following frequencies will not be, present in the modulated signal, 1) 990 kHz, 2) 1010 kHz, 3) 1020 kHz, 4) 1030 kHz, 33. A given AM transmitter develops an, unmodulated power out put of 1kW across, 50 Ω resistance. When a message signal of, amplitude 5 V is applied on it then the side, bands carry 40% of power of carrier., Amplitude of the carrier signal used is, 1) 505.952 V, 2) 126.488, 3) 252.976, 4) 316.22, m ( t ) = cos ( 4π ×10 t ), 3, , 34. Let, , and, , C ( t ) = 5cos ( 2π × 106 t ) are the message and, , y = 2Rh T + 2R(1.44h1 ), find y -x, 5., , 12 ×109, = 2000, Number of channels =, 6 ×106, , 6., , µ=, , Am, 3 A, ⇒ = m, Ac, 4 12, , 7., , µ=, , Am 1, = = 0.5, Ac 2, , 8., , µ=, , Am, 15, × 100 =, ×100 = 25%, Ac, 60, , 9., , C m ( t ) = A c (1 + µ sin ωm t ) sin ωc t, fm =, , ωm 12560, =, = 2khz, 2π, 2π, , 10. I rms =, , I peak, 2, , =, , 125, 2, , carrier signals modulation index is 0.5. What 11. LSB = f − f = 105 − 1400 = 98.6khz, c, m, is the efficiency achieved?, USB = f c + fm = 10 5 + 1400 = 101.4khz, 1) 8.33% 2) 11.11% 3) 20 %, 4) 25%, 12. band width = ( f c + f m ) − ( f c −f m ) = 2fm, LEVEL– II (C. W ) - KEY, 1) 3, 8) 1, 15) 3, 22) 4, 29) 1, , 2) 2, 9) 1, 16) 2, 23) 1, 30) 2, , 3) 1, 10) 1, 17) 3, 24) 4, 31) 2, , 4) 2, 11) 1, 18) 1, 25) 2, 32) 3, , 5) 4, 12) 2, 19) 2, 26) 1, 33) 4, , 6) 2, 13) 1, 20) 1, 27) 3, 34) 2, , 7) 3, 14) 4, 21) 2, 28) 3, , LEVEL - II ( C. W ) -HINTS, 1., , ht 16, =, hr 25 ; d =, 72 × 10 3 =, , 2R, , 2 Rht +, , 16, hr +, 25, , 2 Rhr, , 2 Rhr, , ⇒ h r = 125 m & ht = 80 m, , 2Rh T + 2Rh R when both hT and, , 2., , s=, , 3., , hR are doubled then s becomes 2 times., Number of people that can receive, = area × population density, , 4., , bandwidth, , 300, , 13. n = signalbandwidth = 2 × 15 = 10, 14. frequencies in output are f c − f m & f c + f m, 15. E max = Ac + Am, E min = Ac − A m ⇒ µ =, , Am E max − E min, =, Ac E max + E min, , 16. Amplitude of LSB = USB = µ, , Ac, 2, , A, , A, , A, , A, , c, m, c, m, 17. Amplitude of LSB = USB = µ 2 = A × 2 = 2, c, , 18. µ = 0.6 ×100 = 60%, 10 − 2, 19. µ = 10 + 2 × 100, 20. (1) Here, fc = 1. 5 MHz = 1500kHz,, \ Low side band frequency, , fm == 10 kHz, , = fc − fm = 1500kHz − 10 kHz = 1490kHz, , = pd 2 × population density, Upper side band frequency, distance between transmitter and first receiver, = fc + fm = 1500kHz + 10 kHz = 1510 kHz, is x and between transmitter and second receiver, m2 , (0 .4 )2 , is y., 21. (2) Pt = Pc 1 + 2 = 9 1 + 2 , , , , , x = 2Rh T + 2Rh1, NARAYANAGROUP, , 59
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , 0 .16 , = 9 1 +, , 2 , , , 30. For option 1 m =, , (Q m = 40% = 0.4), , = 9 (1.08) = 9.72 kW, 22. (4), \, , Pt, m2, = 1+, Pc, 2, , or, , 2 , Pc = 1500 , , 2 +1, , 30, = 1.5, 20, , m>1, , 16, = 0.8, m<1, 20, For option 3 signal contains only side bands so, message cannot be detected., , For option 2 m =, , 2 , Pc = Pt , 2, 2 + m , , ∴ m = 100% = 1 = 1000 W, , 31. f c = 1 MHz and f m = 2000 Hz, 23. (1) If n is the number of bits per sample, t h e n, 1, 1, number of quantisation level = 2n Since the, = RC =, number of quantisation level is 16, fc, fm, ⇒ 2n = 16 Þ n = 4, 32. The frequency components of modulation signal, bit rate = sampling rate × no. of bits per sample, are from nonlinear device are, fc, ,, = 8000 × 4 = 32,000 bits/sec., 2, f c ± f m , f c ± 3 f m ...., I , m2, 24. (4) We know that I t = 1 + 2, So the frequencies present in the AM are 100, c, kHz, 990 kHz, 1010 kHz, 970 kHz, 1030 kHz, Here, I t = 8. 96A and I c = 8 A, 2, , m2, 8 .96 , , = 1+, 2, 8 , , or, , m2, = 0. 254, 2, , or, , or, , AC 2, Ac 2, = 1kW ⇒, = 103, 33., 2R, 2 × 50, , m2, 1. 254 = 1 +, 2, , m2, 34. η =, 2 + m2, , m2 = 0. 508, , or m = 0.71 = 71%, 25. (2)For demodulation, , 1, << RC, fc, , LEVEL - II (H. W), , 1, 1, =, = 10 − 5 s, f c 100 × 10 3, RC = 10 3 × 10 × 10 − 12 s =, , 1., 10–8 s, , 1, , We see that f here is not less than RC as, c, required by the above condition. Hence, this, is not good., 26. (1) Frequency modulation index, =, , Frequency deviation, 50, =, = 7.143, Modulating frequency 7, , 27. (3), , 2., , f, 60, MUF = c =, = 175 MHz, cos θ cos 70°, , 28. Convention AM signal withour over, modulation means m<1, , x ( t ) = Ac (1 + m ( t ) ) cos ( 2π f c t ), For, option, , 1, , 1, x ( t ) = Ac 1 + m ( t ) cos ( 2π f c t ) ; ∴ m =, 4, , , 4, , 29., 60, , PTotal SB, =, PCarrier, , ( 2)= 1, , 2, Pc m, , Pc, , C, , 3., , The maximum distance between the, transmitting and receving TV towers is 64 km, . If the ratio of the heights of the TV, transmitting tower to receiving tower is 4 : 9 ,, the heights of the transmitting and receiving, towers are, 1) 51.2 m ; 80 m, 2) 1280 m ; 2880 m, 3) 80 m ; 125 m, 4) 25 m ; 75 m, The maximum distance between the, transmitting and receiving TV towers is D. If, the heights of both transmitting and receiving, towers are halved then the maximum distance, between them becomes, 1) 2D, 2) D/ 2 3) 4D, 4) D/2, A T.V. tower has a height of 5 m is in a region, of average population density 100 p /km2 ., The number of people that can recieve the, transmission is nearly, 1) 1,28,000, 2) 64000, 3) 2,56,000, 4) 32,000, , 8, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 4., , 5., , 6., , 7., , COMMUNICATION SYSTEM, , At certain distance from a transmitting tower, a receiver tower of height 180 m is used to, receive direct signal. Another tower is, installed beyond the first along the same line, of sight to receive the signals from the same, transmitter. Its height is 21% more than the, first receiving tower. Then the separation, between the two receiving towers is, 1) 6.4 km 2) 3.2 km 3) 1.6 km 4) 4.8 km, The TV signals have a band width of 4MHz., The number of TV channels that can be, accommodated in a band width 16 GHz is, 1) 2, 2) 20, 3) 200, 4) 4000, A carrier wave of peak voltage 10 V is used, to transmit a message signal. The peak, voltage of the modulating signal in order to, have a modulation index of 80% is, 1) 6 V, 2) 9 V, 3) 8 V, 4) 15 V, A modulating signal is a square wave as shown, in the figure, , 1, m(t), in Volts, , 1, , 2, , 11. An audio signal 5sin2π (1600t ) amplitude, modulates 20sin2π ( 2 ×105 t ) . The two side, , 12., , 13., , 14., , t(s), , 15., The, , carrier, , wave, , is, , c ( t ) = 1.33sin ( 8π t ) volts, The modulation index is, 1) 2, 2) 0.75 3) 0.5, 8., , given, , by, , ., , 4) 1.5, , A audio signal 20sin2π (1500t ) amplitude, , 16., , band frequencies are, 1) 98.4 kHz, 101.6 kHz, 2) 92.5 kHz, 105.5 kHz, 3) 32.1 kHz 31.59 kHz, 4) 96 kHz, 106 kHz, If 2 f c and 3 f m are the frequencies of carrier, wave and signal, then the band width is, 1) f m, 2) 6 f m 3) f c, 4) 2 f c, The number of AM broadcast stations that, can be accommodated in a 140kHz band width, for the highest modulating frequency 10 kHz, will be, 1) 10, 2) 5, 3) 7, 4) 12, A 2000 kHz carrier is simultaneously, modulated with 600Hz, 800 Hz and 4 kHz, audio waves. The frequencies present in the, output are, 1) 2000.6 kHz,2000.8 kHz,1996 kHz, 2) 1999.4 kHz, 2004 kHz, 1999.2kHz, 3) 1002.8 kHz, 996kHz, 1106 kHz, 4) Both (1) and (2), If a carrier wave of amplitude 20mV is, modulated by an audio signal of amplitude, 4mV, then amplitude of LSF or USF is given, by, 1) 0.2 mV, 2) 0.5 mV, 3) 1 mV, 4) 2 mV, The amplitude modulated current is given by, , i = 70 [1 + 0.06sin2900t ] sin106 t . The depth, modulates 40sin2π (10 ) t . The depth of, of modulation is, modulation is, 1) 60%, 2) 6%, 3) 36%, 4) 66%, 1) 25%, 2) 20% 3) 50%, 4) 40%, 17. If minimum voltage in an AM wave was found, 9. An AM wave is given by, to be 1.1 V and maximum voltage 10V. The, % modulation index, V = 1500[1+ 0.5sin62800t] sin (5.26 ×105 t) ., 1) 80.2% 2) 66.67% 3) 64.25% 4) 76.25%, The modulating frequency is, 18. SinusoidalcarriervoltageoffrequencyMHz, 3, 1) 20 kHz 2) 10 kHz 3) 12.5 kHz 4) 50 kHz, and amplitude 50 V is amplitude modulated, 10. The amplitude modulated current is given by, by sinusoidal voltage of frequency 10 kHz, 5, producing 50% modulation. The lower and, i = 75[1 + 0.3sin2003t ] sin ( 7 ×10 t ) ., upper side-band frequencies in kHz are, The RMS value of carrier current will be, 1) 1490, 1510, 2) 3010, 2990, 125, 100, 75, 50, 1, 1, 1, 1, A 2), A 3), A 4), A, 3) 1490 , 1510, 4) 1510 , 1490, 1), 2, 2, 2, 2, 5, , NARAYANAGROUP, , 61
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, 19. A transmitter supplies 6 kW to the aerial when, unmodulated. The power radiated when, modulated to 60% is, 1) 5 kW, 2) 9.72 kW, 3) 7.08 kW, 4) 12 kW, 20. The total power content of an AM wave is, 6000 W. For 100% modulation, the power, transmitted by the carrier is, 1) 500 W, 2) 700 W, 3) 750 W, 4) 4000 W, 21. The bit rate for a signal, which has a sampling, rate of 8 kHz and where 32 quantization levels, have been used, is, 1) 32000 bits/sec, 2) 40000 bits/sec, 3) 64000 bits/sec, 4) 72000 bits/sec, 22. The antenna current of an AM transmitter is, 10 A when only carrier is sent but increases, to 10.1 A when the carrier is modulated, sinusoidally. The percentage modulation is, 1)50%, 2)60% 3) 200%, 4)71%, 23. In a diode AM-detector, the output circuit, consists of R = 1k Ω and C = 10 pF. A carrier, signal of 1000 MHz is to be detected. Is it, good, 1) Yes, 2) No, 3) Information is not sufficient 4) None of these, 24. In an FM system a 9 kHz signal modulates, 108MHz carrier so that frequency deviation, is 60 kHz. The frequency modulation index is, 1) 7.143, 2) 8, 3) 6.67, 4) 350, 25. Maximum usable frequency (MUF) in Fregion layer is x, when the critical frequency, is 60 MHz and the angle of incidence is 60°., Then x is, 1)150 MHz, 2)170 MHz, 3)120 MHz, 4)190 MHz, , LEVEL -II ( H. W ) - KEY, 1) 2, 8) 3, 15) 4, 22) 3, , 2) 2, 9) 2, 16) 2, 23) 1, , 3) 2, 10) 3, 17) 1, 24) 3, , 4) 4 5) 4 6) 3 7) 2, 11) 1 12) 2 13) 3 14) 4, 18) 2 19) 3 20) 4 21) 2, 25) 3, , LEVEL - II ( H. W ) -HINTS, 1., 2., 62, , hT 4, d = 2 RhT + 2RhR ; h = 9 ; d = 64 km, R, , D = 2 RhT + 2RhR, , 3., , A = π D2 = π [ 2 Rh2 ], n = ( Populationdensity ) A, , 4., 5., 6., , d = 2 Rh2 − 2 Rh1 ; h2 = h1 − 21% of h1, 16GHz, 4MHz, Modulation index, n=, , =, , Amplitudeof modulatingsignal, Amplitudeof carrierwave, am, ac, , 7., , m=, , 8., , am, depth = % µ = a ×100, c, fm =, , wm, 2π, , 10. irms =, , i0, 2, , 9., , 11. LSB = f c − f m = 10 5 − 1600 = 98.4KHz, USB = fc + f m = 105 + 1600 = 101.6 KHz, , 12. Frequencies in out put are 2 f c − 3 f m and, 2 fc + 3 f m, bandwidth, 140, 13. n = signalbandwith = 2 ×10 = 7, , 14. Frequencies in out put are f c − f m and f c + f m, 15. Amplitude, of, L. S.B = U .S .B = µ, , Ac Am Ac Am, =, . =, 2, Ac 2, 2, , 16. Depth of modulation = 0.06 ×100 = 6%, 17. µ =, , 10 − 1.1, ×100 = 80.2%, 10 + 1.1, , 18. f c − f m and f c + f m, , m2 , P, =, P, 1, +, c , = (6 K W, 19. t, 2 , , , 2, , 0.6 ) , (, ) 1 +, , 2 , , , , 2 , 20. Pc = Pt , 2, 2 +m , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 21. No. of quantization levels = 2 n, bit rate = sampling power x no. of bits per sample, , COMMUNICATION SYSTEM, 6., , Vm = 2sin12π × 103 t . The band width and LSB, if carrier wave has a frequency, , 2, , It , m2, 22. = 1 +, 2, Ic , 1, 23. For demodulation f << RC, c, 24. Frequency modulation index, =, , 7., , frequencydeviation, Modulating frequency, , 25. MUF =, , fc, cos θ, , LEVEL - III, 1., , 2., , 3., , 4., , 5., , The audio signal voltage is given by, , The tuned circuit of an oscillator in a simple, AM transimitter employs a 250 micro henry, coil and 1nF condenser. If the oscillator output 8., is modulated by audio frequency upto 10KHz,, the frequency range occupied by the side bands, in KHz is, 1) 210 to 230, 2) 258 to 278, 3) 308 to 328, 4) 118 to 128, A TV tower has a height of 70m. If the, average population density around the tower, is 1000km −2 , the population covered by the, TV tower, 9., 1) 2.816 ×106, 2) 2.86 ×109, , 3.14 ×106 rad / s, 1) 12 KHz; 494 KHz, 2) 6 KHz; 313 KHz, 3) 6 KHz; 494 KHz, 4) 18 KHz; 494 KHz, A TV transmission tower at a particular, station has a height of 160m. Radius of earth, is 6400km, i. The range it covers is 45255 m, ii. The population that it covers is77.42 lakhs., When population density is 1200km −2, iii. The height of antenna should be increased, by 480 m to double the coverage range, 1) i and ii are true, 2) ii and iii are true, 3) i and iii are true, 4) i, ii and iii are true, The tuned circuit of the oscillator in a, simple AM transmitter employs a, 40 µ H coil and 12 nanofarad(nF) capaci, tor. If the oscillator output is modulated, by audio frequency of 5 kHz,Which of the, following frequencies doesn’t appear in, the out put AM?, , 1) f USB = 225 KHz, , 2) fUSB = 235 kHz, , 3) fc=230 kHz, 4) f c = 235 kHz, A 400 watt carrier is modulated to a, depth of 80%. Calculate the total power, 3) 2.816 ×103, 4) 2.816 ×1012, in the modulated wave., 1) 528 W 2) 128 W 3) 256 W 4) 400 W, A carrier wave is modulated by a number of, 10., Calculate modulation index if carrier waves is, sine waves with modulation indices 0.1, 0.2,, modulated by three signals with modulation, 0.3. The total modulation index (m) of the, indices as 0.6, 0.3 and 0.4, wave is, 1) 1.0, 2) 0.70 3) 0.78, 4) 1.3, 1) 0.6, 2) 0.2, 3) 0.14 4) 0.07, 11. A 1000 KHz carrier is simultaneously, The maximum peak-to-peak voltage of an AM, modulated with fm1 = 300 Hz, fm2 = 800 Hz and, wave is 16 mV and the minimum peak-to-peak, fm3 = 1 KHz audio sine waves. What will be, voltage is 4mV. The modulation factor is equal, the frequencies present in the output?, to, f LSB1 = 999.7 KHz, f LSB2 = 999.2 KHz, 1) 0.6, 2) 0.3, 3) 0.8, 4) 0.25, a), b), An AM wave is expressed as, f USB1 = 1000.3KHz, f USB 2 = 1000.8 KHz, , e = 10 (1 + 0.6cos2000π t ) cos2 ×108 π t v o l t s ,, the minimum and maximum values of, modulated carrier wave are, 1) 10 V, 20 V, 2) 4 V, 8 V, 3) 16 V, 4 V, 4) 8 V, 20 V, NARAYANAGROUP, , f LSB3 = 999 KHz, c) f, USB3 = 1001KHz, 1) a only, 3) a, b and c, , f LSB3 = 990KHz, d) f, USB3 = 1010KHz, 2) b, c and d, 4) a, c only, 63
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , LEVEL-III - KEY, , The antenna current of an AM transmitter is, 8A when only the carrier is sent, but it, 1) 3, 2) 1 3) 3 4) 1 5) 3 6) 1, increases to 8.93A when the carrier is, 7) 4, 8) 2 9) 1 10) 3 11) 3, modulated by a single wave. The percentage, LEVEL-III - HINTS, modulation is, 1, 1) 60.1% 2) 70.1% 3) 80.1%, 4) 50.1%, = 318khz,f m = 10khz, 1. f c =, 4. The antenna current is 8A only when only, 2π LC, carrier wave is transmitted and 9.6A when, LSB = f c − fm = 308, AM wave is transmitted. If carrier power is, USB = f c + fm = 328, 10kw, the modulating power is, 2. Population = Population density × Area, 1) 14.4 kW 2) 1.2 kW 3) 2 kW, 4) 4.4 kW, = Population density × π× 2Rh, Passage - I, A message signal of frequency 10kHz and, 3. m = m12 + m22 + m32, peak voltage 10V is used to modulate a carrier, (16 − 4) / 2, wave of frequency 1MHz and peak voltage, 4. µ = (16 + 4) / 2, 20volts., 5. The modulation index is, E max = (1 + µ ) E c, 5. E = 1 − µ E, 1) 0.2, 2) 0.3, 3) 0.5, 4) 0.8, ( ) c, min, 6. Among the following are USB and LSB are, 6. band width = 2 f m, 1) 1010kHz, 990 kHz 2) 990 kHz, 1010kHz, LSB= f c − f m, 3) 950 kHz, 1050kHz 4) 1050kHz, 990 kHz, 7. The bandwidth of signal is, 7. d = 2Rh, 1) 10kHz, 2) 20 kHz, Population = Population density× πd 2, 3) 30 kHz, 4) 40 kHz, dα h, Passage - II, 2, 8. When only carrier is transmitted antenna, m , P, =, P, 1, +, 9., Total, C, current observed is 8A when it is modulated, 2 = 528 watts, , with 500 Hz sine wave antenna current, becomes 9.6A. The percentage of modulation, 10. m t = m12 + m22 + m32 = 0.78, is, 1) 80%, 2) 20% 3) 94.26% 4) 83.76%, LEVEL - IV, 9. 1% of 1012 Hz of a satellite link was used for, telephony. The number of channels or, 1. Consider telecommunication through optical, subscribers if each channel is of 8 kHz are, fibres. Among the following statements one, 1) 2.5 ×107, 2) 1.25 ×10 6, is not true, 1)optical fibers can be of graded refractive index, 3) 2.5 ×108, 4) 1.25 ×108, 2)optical fibers are subjected to electromagnetic 10. A carrier is modulated simultaneously by 3, interference from out side, sine waves of modulation indices 0.3, 0.4 and, 3)optical fibers have extremely low transmission, 0.45 respectively. The net modulation index, loss, is, 4)optical fibers have homogeneous core with a, 1) 1.15, 2) 0.67, 3) 0.57, 4) 0.84, suitable cladding, 11., A, 1, kW, carrier, is, modulated, to, a, depth of, 2. A 600 W carrier is modulated to a depth of, 80%. The total power in the modulated wave, 75% by a 400Hz sine wave. The total antenna, is, power is, 1) 1.32, 2) 1.56, 3) 1.84, 4) 1.96, 1) 769 W, 2) 796 W 3) 679 W 4)637.5 W, 64, , 3., , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , Passage - III, The expression for an AM voltage is, 12., 13., 14., , 15., , e = 5 (1 + 0.6cos1000π t ) cos5 ×106 π t, The peak value of carrier wave is, 1) 30 V, 2) 6 V, 3) 3 V, 4) 5 V, The percentage of modulation is, 1) 0.6% 2) 6%, 3) 60%, 4) 66%, Frequency of audio signal is, 1) 1000 Hz 2) 5000 Hz, 3) 500Hz, 4) 250 Hz, Frequency of carrier signal is, 1) 5 ×10 6 Hz, 2) 1000Hz, , 23., , 24., , 3) 2.50 ×103 Hz, 4) 2.50 ×106 Hz, 16. Frequencies of sideband is, 1) 2.5005 × 106 Hz, 2.4995 × 106 Hz, 2) 2.505 × 106 Hz, 2.495 × 106 Hz, 25., 6, 6, 3) 2.505 × 10 kHz, 2.495 × 10 kHz, 4) 2.505 MHz, 2.495kHz, 26., Passage - IV, An amplitude modulated voltage is expressed, as e = 10 (1 + 0.8cos2000π t ) cos3 ×106 π tvolt, 17. The peak value of carrier wave is, 1) 5 V, 2) 8 V, 3) 10 V, 4) 100 V, 18. The depth of modulation is, 1) 8, 2) 0.8, 3) 800, 4) 80, 19. The frequency of carrier wave is, 1) 3 ×106 Hz, 2) 1.5 ×103 Hz, , 27., , 3) 3 ×103 Hz, 4) 1.5 ×106 Hz, 20. The minimum and maximum value of, 28., modulated carrier wave is, 1) 2V, 18V 2) 4V, 16V 3) 2V, 8V 4) 4V, 12V, 21. The frequency of modulating signal is, 1) 1000 Hz, 2) 100Hz, 3) 10MHz, 4) 100 MHz, 22. The bandwidth of modulated signal is, 29., 1) 1000 Hz, 2) 2000 Hz, 3) 1500 Hz, 4) 2500 Hz, , ASSERTION AND REASON, Instructions for the following questions:, In each of the following questions a statement, of Assertion (A) is given followed by 30., corresponding statement of Reason (R) just, below it. Of the statements mark the correct, answers., NARAYANAGROUP, , 1) A and R are true and R is the correct, explanation of A, 2) A and R are true but R is not correct, explanation of A, 3) A is true but R is false, 4) A is false but R is true., A: It is not necessary for a transmitting antenna, to be at the same height as that of receiving, antenna for line-of sight communication, R: If the signal is to be received beyond the, horizon then the receiving antenna must be high, enough to intercept the line-of sight waves, A: Low frequency audio signals cannot be, transmitted directly over long distances, R: To transmit low frequency audio signals over, long distances, the audio signals are, superimposed on a high frequency carrier signal, A: The microphone acts as a transducer., R: In radio communication, microphone is used, to convert sound signal into electrical signal., A: A FM signal is less susceptible to noise than an, AM signal., R: In FM transmission, the message signal is in, the form of frequency variations of carrier, waves. During modulation process, the noise, gets amplitude modulated., A: Artifical satellites are not necessary for long, distance TV transmission., R: The TV signals are not reflected by the, ionosphere. They are reflected back to the earth, from an earth station by making use of artifical, satellites., A: Greater the height of a TV transmitting, antenna, greater is its coverage., R: The height of transmitting antenna is directly, proportional to the square of the distance of TV, coverage, A: Long distance radio broadcasts use short, wave bands., R: The short waves are absorbed by the earth, but reflected by F layer in ionosphere. After, reflection, they reach the surface of earth back, only at a large distance from the transmitter., A: TV signals cannot be transmitted via sky, wave propagation., R: Ionosphere is unable to reflect radio waves of, frequencies greater than 40MHz., 65
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , MATRIX MATCHING, , 35. Match List-1 with List-2, List-1, List-2, 31. Match the frequency band with the type of use, Communication mode Example, Frequency Band, Type of use, a) Point to point, a) LF, e) Radio Broad casting, communication, e) RADAR, b) HF, f) Marine and, b) broadcast, navigational aid, communication, f) AM Radio, c) VHF, g) Satellite, c) Line of sight, g) FM Radio, communication, communication, d) SHF, h) TV Broad casting, d) Satellite, h) Traditional telephony, 1) a - e; b - f; c - h; d - g, communication, 2) a - f; b - e; c- g; d- h, i) Mobile telephony, 3) a - f; b - e; c - h; d - g, j) TV, 4) a - e; b - g; c - f; d - h, 1) a-h; b-f,g,j; c-e;d-i,j, 2) a-e; b-i,j; c-h;d-g,f, 32. Match layer of ionosphere and height of the, 3) a-f ; b-h,i,j; c-g;d-e, layer from surface of earth for atmosphere, 4) a-g; b-f,g,j; c-e;d-i,j, a) D, e) 250 - 400 km, 36. Match List-1 with List-2, b) E, f) 100 km, List-1, List-2, c) F, g) 65 - 75 km, Name of device, use, d) F2, h) 170 - 190 km, a) Antenna, e) sends out information, 1) a-g; b-f; c-h; d-e, 2) a-h; b-g; c-e; d-f, b) Transmitter, f) picksup information, 3) a-g; b-h; c-f; d-e, 4) a-g; b-h; c-e; d-f, c) Receiver, g) converts energy in, 33. Match layer of ionosphere and frequencies, one form to another, most affected, form, a) D, e) helps surface waves and reflects, d) Transducer, h) radiates signal, HF, i) recieves signal, b) E, f) efficiently reflects HF waves, 1) a-e; b-g,h;c-f;d-h 2) a-h,i;b-e;c-f;d-g, c) F, g) partially absorbs HF, 3) a-f; b-e;c-f,i;d-h 4) a-h;b-g,i;c-f;d-e, d) F2, h) reflects LF and absorbs MF, 37. Match List-1 with List-2, 1) a-g; b-f; c-h; d-e, 2) a-h; b-e; c-g; d-f, List-1, List-2, Name, Frequencies most, 3) a-h; b-e; c-f;d -g, 4) a-e; b-h; c-g; d-f, affected, 34. Match the frequency band with the type of, a) Part of, e) VHF(upto, use service frequency band, stratosphere(D), several GHz), a) AM broadcast, e) 88 - 108 MHz, b), Part, of, f) reflects LF, b) satellite, f) 896 - 935 MHz, mesophere ( F1 ), communication, c) FM broadcast, d) Cellular mobile, 1) a-g; b-h; c-f; d-e, 3) a-f; b-h; c-e; d-g, , 66, , g) 540 - 1600 kHz, h) 3.7 - 6.5 GHz, 2) a-h; b-g; c-e; d-f, 4) a-g; b-h; c-e; d-f, , c) Part of, , g) Partially absorbs HF, , thermosphere( F2 ), d) Troposphere, 1) a-e;b-f,g,h;c-i;d-h, 3) a-f,g,h;b-g;c-i;d-e, , h) Efficiently reflects HF, 2) a-f;b-f,g,h;c-f;d-e, 4) a-g;b-f,g,h;c-f;d-e, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , COMMUNICATION SYSTEM, , 38. Match List-1 with List-2, List-1, List-2, Type of propagation Frequencies, a) sky waves, e) 1.5 MHz, b) space wave, f ) 20 MHz, c) ground wave, g) 30MHz, d) micro wave, h) 50MHz, i) 3 GHz, 1) a-e,i;b-e;c-h;d-i, 2) a-f,g;b-h;c-e;d-i, 3) a-i,g;b-e;c-h;d-f, 4) a-g;f-h;c-i,b;d-e, , LEVEL - IV - KEY, 1) 2, 7) 2, 13) 3, 19) 4, 25) 1, 31) 3, 37) 3, , 2) 1, 8) 3, 14) 3, 20) 1, 26) 1, 32) 1, 38) 2, , 3) 2, 9) 2, 15) 4, 21) 1, 27) 4, 33) 2, , 4) 4, 10) 2, 16) 1, 22) 2, 28)1, 34) 4, , 5) 3, 11) 1, 17) 3, 23) 1, 29) 1, 35) 1, , LEVEL - IV - HINTS, 2., , 6) 1, 12) 4, 18) 2, 24) 1, 30) 1, 36) 2, , 7., , bandwidth of signal = 2fm, , 8., , it, µ2, = 1+, ic, 2, , 9., , 1% of 1012 Hz = 1010 Hz, Number of channels =, , 1010, = 1.25 ×10 6, 3, 8 ×10, , 10. m = m12 + m22 + m32, , µ2 , P, =, P, 11. t, c 1 +, 2 , , 12. eqn of AM voltage is given by, e = Ac (1 + µ sin ωm t ) sin ωc t, ∴ Ac = 5 v, , 13. µ = 0.6 ×100 = 60%, ω, 1000π, 14. f m = 2mπ = 2π = 500, , µ2 , Pt = Pc 1 +, , 2 , , ( 0.75) 2 , = 600 1 +, , 2 = 768.5W, , , 15., , fc =, , ωc 5 × 106 π, =, 2π, 2π, , 16. USB = f c + f m, LSB = f c − f m, , 2, , 3., , it , µ2, µ2, 8.93 , ⇒, =, 1, +, =1 +, , 2, 2, 8 , ic , 2, , 17. eqn of AM voltage is given by, e = Ac (1 + µ sin ωm t ) sin ωc t, , ⇒ µ = 0.701 = 70.1%, 2, , 4., , Pt it , P 9.6 , = ⇒ t =, = 14.4kw, Pc ic , 10 8 , , ∴ Pm = 14.4 −10 = 4.4 kW, 5., 6., , ∴ A c = 10 V, , 10, µ=, 20, , USB = f m + f c, LSB = f m − fc, NARAYANAGROUP, , 2, , 18., , µ = 0.8, , 19. f c =, , ωc 3 × 106 π, =, 2π, 2π, , 20., , E min = (1 − µ ) A c ,, , 21., , fm =, , E max = (1 + µ ) A c, , ωm 2000π, =, 2π, 2π, , 22. bandwidth =2 f m, , 67
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EXPERIMENTAL PHYSICS, , JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, SYNOPSIS, Familiarity with basic approach and observations of the experiments and activities:, 1., , Vernier Callipers, It is used to measure internal and external diameter and depth of a vessel, , 2., , Screw Gauge, It is used to determine thickness of a thin wire;, diameter of a thin wire and thickness of a sheet., , 3. Simple Pendulum, Dissipation of energy by plotting a graph between square of amplitude and time., 4., , Meter Scale, Mass of a given object by ‘ principle of moments’, , 5., , Young’s modulus, Determination of Young’s modulus of elasticity, of the material of a metallic wire., , 6., , Surface tension, Determination of Surface Tension of water by, capillary rise and effect of detergents., , 7., , Viscosity, Determination of coefficient of viscosity of a, given viscous liquid by measuring terminal velocity of a given spherical body., , 8., , Cooling a curve, Plotting a cooling curve for the relationship between the temperature of a hot body and time., , 9., , Speed of sound, Determination of speed of sound in air at room, temperature using a resonance tube., , 10. specific heat, Determination of specific heat of a given, (i) Solid and (ii) liquid by the method of mixtures., 11. Resistivity, Determination of resistivity the material of a, given wire using meter bridge, 12. Ohm’s Law, Resistance of a given wire using Ohm’s Law., , 68, , 13. Potentiometer, Comparison of emf of two primary cells, Determination of internal resistance of a cell, 14. Figure of merit, Resistance and figure of merit of a galvanometer by half - deflection method., 15. Determination of focal length of, (i) Convex mirror, (ii) Concave mirror and, (iii) Convex lens., 16. Prism, Using parallax method - plot of angle of, deviation versus angle of incidence of a, triangular prism, 17. Refractive index of glass, Refractive index of a glass slab using a travelling microscope., 18. p - n junction diode, Characteristic curves of a p - n junction diode, in forward and reverse bias., 19. Zener diode, Characteristic curves of a Zener diode and, finding reverse breakdown voltage., 20. Transistor, Characteristic curves of a transistor and finding, current gain and voltage gain., 21. Identification of devices, Identification of (i) diode, (ii) LED, (ii) transistor (iv) IC (v) resistor and, (vi) capacitor from mixed collection of such, items., 22. Multimeter, Using multimeter to, (i) identify base of a transistor, (ii) distinguish between n-p-n and p-n-p type, transistor, (iii) See the unidirectional flow of current in the, case of a diode and an LED, (iv) Check the correctness or otherwise, of a, given electronic component such as diode,, transistor or IC., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENT- 1, VERNIER CALLIPERS, , , , , , , , , , , , , , , , , , It is used to measure length of objects, diameter, of cylinders (both internal and external) and, depth of cylindrical vessels, upto an accuracy, of 0.1mm., It consists of a main scale M graduated in, centimeter and a vernier scale V which can slide, along the length of main scale., In general 10 vernier scale divisions will have, a length of 9mm on the main scale., Main scale has two fixed jaws A and C (upper, and lower) whereas vernier scale has two jaws, B and D ( upper and lower) movable by nature., Jaws A and B are used to determine external, diameter of a cylindrical vessel whereas C and, D are used to determine internal diameter., In an error free vernier callipers, the zero of, vernier coincides with the zero of main scale, when the jaws A and B are made to touch each, other. If it is not so, the instrument has zero, error which may be positive or negative., If zero of vernier scale lies to the right of zero, of main scale, when the jaws A and B are made, in contact with each other, the error is said to be, positive and the correction will be negative., If zero of vernier scale lies to the left of zero of, main scale, then the error is said to be negative, and the correction will be positive., Least count or, Vernier constant 1MSD 1VSD, = value of one main scale division - value of, one vernier scale division, When 10 vernier scale divisions = 9 main scale, divisions then, , 1MSD 1mm ; 1VSD , , 9, mm, 10, , 9, 1, mm , 10 10, 0.1mm 0.01cm, , LC 1 , , , , , After keeping the object between the jaws A and, B tightly, then the value of main scale coinciding with the zero of vernier scale is noted as, main scale reading (MSR)., The division of vernier scale that coincides exNARAYANAGROUP, , EXPERIMENTAL PHYSICS, actly with any other main scale division is noted as, vernier coincidence (VC), , , Then total reading MSR VC LC , , , , where LC least count 0.01cm, A metallic strip connected to the back of vernier, scale is used to measure the depth of a, cylindrical vessel., Fixed, Upperjaw, , C, , D, , Movable, upperjaw, Screw, Mainscale, , 1 2 3 4 5 6 7 8 9 1011 121314, Protecting, rod, Vernier Scale, , A, Fixed, lower jaw, , B Movable, lowerjaw, , EXERCISE - 1, 1., , 2., , 3., , 4., , The zero error in a Vernier callipers is said to, be positive when, 1) zero of vernier scale is towards the right of, the zero of main scale, 2) zero of vernier scale coincides with the zero, of main scale, 3) zero of vernier scale is towards the left of the, zero of main scale, 4) the vernier scale is not visible clearly., Depth of a cylindrical vessel can be, measured by using Vernier Callipers by the help, of, 1) thin metallic strip projecting at the back, 2) the lower pair of jaws, 3) the upper pair of jaws, 4) both lower and upper pairs of jaws, Using upper pair of jaws of Vernier, Callipers, we can measure, 1) the depth of a cylindrical vessel, 2) the outer diameter of a vessel, 3) the inner diameter of a vessel, 4) the thickness of a thin wire, The Vernier constant of a Vernier Callipers, is, 1) the difference between one main scale, division and one Vernier scale division, 2) the sum of one main scale division and one, Vernier scale division, 3) the ratio of one main scale division to one, Vernier scale division, 4) the product of one main scale division and, one vernier scale division, 69
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EXPERIMENTAL PHYSICS, , JEE-ADV PHYSICS- VOL- VI, , vernier scale coincides with the main scale., The zero error in a Vernier Callipers is said to, Mass of the cube is 2.736g. The density of, be negative when, the cube is approximate appropriate, 1) zero of Vernier Scale is towards right of the, significant figures is, zero of main scale, 2) zero of Vernier scale coincides with the zero, 2) 2.66gcm 3, 1) 1.33gcm 3, of main scale., 3) 2.667 gcm 3, 4) 2.5 gcm 3, 3) zero of Vernier scale is towards left of the, zero of main scale, 12. A) Vernier callipers with 20 divisions on, sliding scale, coinciding with 19 main scale di4) the Vernier scale is not visible clearly., visions, 6. The vernier constants of two vernier, calB) A screw gauge of pitch 1mm and 100, lipers A and B are 0.01cm and 0.01 mm redivisions on the circular scale, spectively. The one which can measure the, C) An optical instrument that can measure, length of a small cylinder more accurately is, length to within a wavelength of light, 1) A, 2) B, Out of A, B and C the most precise device for, 3) Both A and B with same accuracy, measuring length is, 4) accuracy does not depend on vernier, 1) A only, 2) B only, constant, 3) C only, 4) All are equally accurate, 7. Each division on the main scale of a vernier, th, callipers is 1mm. Out of the following, the 13. The n division of main scale coincides with, vernier scale which has vernier constant, th, n 1 division of vernier scale. Given one, 0.01mm is, main scale division is equal to ' a ' units. The, 1) 99 mm divided into 100 divisions, 2) 9 mm divided into 10 divisions, least count of vernier is, 3) 90 mm divided into 100 divisions, n, a, a, 4) 9 mm divided into 100 divisions, 1), 2), 3) an, 4), a 1, n 1, n, 8. A vernier callipers has 20 divisions on the vernier scale which coincides with 19mm on the 14. The vernier scale of a travelling microscope, has 50 divisions which coincide with 49 main, main scale. Its least count is, scale divisions. If each main scale division is, 1) 0.5mm, 2) 1mm, 0.5mm, the minimum inaccuracy in the, 1, measurement of distance is, 3) 0.05mm, 4) mm, 4, 1) 0.1mm, 2) 0.001 mm, 9. Least count of a vernier callipers is 0.01cm., 3) 0.01mm, 4) 1mm, Using this, the diameter of a sphere is 15. The vernier constant of a vernier callipers is, measured as 1.95cm. Radius of the sphere to, 0.1mm and it has a positive zero error of, the correct significant figure will be, 0.04cm. While measuring diameter of a rod,, 1) 0.98cm, 2) 0.975 cm, the main scale reading is 1.2 cm and 5th ver-3) 1.0 cm, 4) 1 cm, nier, division is coinciding with any scale, 10. The main scale of vernier callipers is divided, division. The correct diameter of the rod is, into 0.5mm and its least count is 0.005cm., 1) 1.21cm, 2) 1.21 mm, Then the number of divisions on vernier scale, 3) 1.29mm, 4) 1.29 cm, is, 16. When the two jaws of a vernier callipers are, 1) 10, 2) 20, 3) 30, 4) 40, in touch, zero of vernier scale lies to the right, 11. The side of a cube is measured by a vernier, of zero of main scale and coinciding with vercalliper (10 divisions of vernier scale, nier division 3. If vernier constant is 0.1mm,, coincide with 9 divisions of main scale, where, the zero correction is, 1 division of main scale is 1mm). The main, 1) 0.03cm, 2) 0.03cm, scale reads 10mm and first division of, 3) 0.03mm, 4) 0.03mm, 5., , 70, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 17. You are given two different vernier, calipers A and B haiving 10 divisions on, vernier scale that coincide with 9 divisions, on the main scale each. If 1 cm of main, scale A is divided into 10 parts and that of, B in 20 parts, then least count of A and B, are, 1) 0.001 cm and 0.005 cm, 2) 0.01 cm and 0.05cm, 3) 0.01 cm and 0.005cm, 4)0.01 cm and 0.001cm, 18. For measuring depth of a beaker using, vernier callipers. observed readings are, given as, S NO, MSR (cm), VSD, 1, 0.5, 8, 2, 0.5, 4, 3, 0.5, 6, If zero error is - 0.03cm, then mean, corrected depth is, 1) 0.56cm, 2) 0.59cm, 3) 0.53cm, 4) 0.52cm, 19. The main scale of a vernier callipers reads, in millimetre and its vernier is divided into, 10 divisions which coincide with 9 divisions, of the main scale. When there is nothing, between the jaws of the vernier callipers,, the 7th division of vernier scale coincides, with a division of main scale and in this case, zero of vernier scale is lying on right side, of the zero of main scale. When a cylinder, is tightly placed along its length between, the jaws, the zero of the vernier scale is, slightly left to be 3.1cm and 4th VSD, coincides with a scale division. The length, of the cylinder is, 1) 3.2cm 2) 3.07cm 3) 3.21cm 4) 2.99cm, 20. In a vernier callipers, smallest division on, the main scale is 1mm, while the vernier, scale have 20 divisions. When fixed jaw, touches with movable jaw, zero of vernier, scale lies on the right of the zero of the, main scale and 15th division of vernier, scale coincides with any division of main, scale. What is the type of zero error and, its value?, 1) positive, 0.75mm 2) negative, 0.75mm, 3) positive, 0.15mm 4) negative, 0.15mm, NARAYANAGROUP, , EXPERIMENTAL PHYSICS, , Paragraph:, A vernier callipers used by student has 20 divisions in 1cm on main scale. 10 vernier divisions coincide with 9 main scale divisions., When jaws are closed, zero of main scale is, on left of zero of vernier scale and 6th division of vernier scale coincides, with any mains scale divisions. He placed, a wooden cylinder in between the jaws and, measure the length. The zero of vernier, scale is on right of 3.2 cm mark of main, division. When he measures diameter of, cylinder he finds that zero of vernier scale, lies on right of 1.5 cm mark of main scale, and 6th division of vernier scale coincides, with, any main, scale, division., 21) Least count and zero error of vernier, callipers are, 1) 0.05cm, +0.3cm, 2) 0.05mm, -0.3mm, 3) 0.005cm, +0.03cm 4) 0.05cm, -0.3cm, 22) Correct values of measured length and, diameter are, 1) 3.27cm, 1.5cm, 2) 3.21cm, 1.5cm, 3) 3.27cm, 1.93cm, 4) 3.27cm, 1.56cm, 23) In a vernier callipers, N divisions of the main, scale coincide with N+m divisions of the vernier scale. What is the value of m for which, the instrument has minimum least count?, N, N, 4), 10, 2, 24) 1cm on the main scale of a vernier callipers, divided into 10 equal parts. If 10 divisions of, vernier coincide with 8 small divisions of main, scale. then the least count of the vernier calliper is?, 1) 0.01 cm2)0.02cm 3) 0.05cm 4) 0.005cm, 25) The vernier constant of a travelling microscope is 0.001 cm. If 49 main scale divisions, coin cide with 50 vernier scale divisions. then, the value of 1 main scale division is, 1) 0.1mm 2) 0.5mm 3) 0.4mm 4) 1mm, 26) 1 cm of main scale of a vernier calliper is divided into 10 divisions. The least count, of, the callipers is 0.05 cm, then the varnier scale, must have, 1) 10 divisions, 2) 20 divisions, 3) 25 divisions, 4) 50 divisions, , 1) 1, , 2) N, , 3), , 71
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 27) A spectometer gives the following reading when, used to measure the angle of a prism., main scale reading :58.5 degree vernier scale, reading : 09divisions Given that 1 division on, main scale corresponds to 0.5 degree. Total, divisions on the vernier scale is 30 and match, with 29 divisions of the main scale. The angle, of the prism from the above data., 1) 58.590 2) 58.77 0 3) 58.650 4) 590, 28) A student measured the length of a rod and, wrote it as 3.50 cm. which instrument did the, use to measure it?, 1) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm., 2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm., 3) A meter scale, 4) A vernier calliper where the 10 divisions in, vernier scale matches with 9 divisions in main, scale and main scale has 10 divisions in 1 cm., 29) A varnier callipers has 1 mm marks on the main, scale. It has 20 equal divisions on the vernier, scale, which match with 16 main scale division. For this vernier callipers, the least count, is, 1) 0.02 mm, 2) 0.05 mm, 3) 0.1 mm, 4) 0.2 mm, 30) The least count of vernier callipers is 0.1mm., The main scale reading before the zero of the, vernier scale is 10 and the ze roth division of, the vernier scale coincides with the main scale, division. Given that each main scale division, is 1 mm, the measured value should be expressed, 1) 1 cm, 2) 2 cm, 3) 0.5 cm, 4) 0.1 cm, 31) An experiment is performed to find the refrac, tive index of glass using a travelling microscope. In this experiment distance are measured by, 1) a vernier scale provided on the microscope, 2) a standard laboratory scale, 3) a meter scale provided on the microscope, 4) a screw gauge provided on the microscope, , 72, , EXERCISE - 1 - KEY, 1) 1, 8) 3, 15) 1, 22) 2, 29) 4, , 2) 1,2,3 3) 34) 1, 9) 1 10) 1 11) 2, 16) 1 17) 3 18) 2, 23) 1 24) 2 25) 2, 30) 1 31) 1, , 5) 3, 12) 3, 19) 2, 26) 2, , 6) 2, 13) 2, 20) 1, 27) 3, , 7) 1, 14) 3, 21) 3, 28) 4, , EXERCISE - 1 - HINTS, 8), 9), , 19, 20, daimeter d=2r=1.95, LC 1 , , 1.95, 0.975, 2, = 0.98 cm (after rounding the figure), , r, , 1main scale division, 10) LC number of vernier scale divisions, , 11) LC 1MSD 1VSD 0.1mm, side of the cube, , 10mm 1 0.1 10.1mm 1.01cm, , , , mass, 2.736, , 2.66 g / cm3, 3, volume 1.01, , 12) LC of ‘A’ = 0.005 cm, LC of ‘B’0.001 cm, Wavelength of light =, , 105 cm 0.00001cm, 13), , N 1VSD N MSD, N , 1VSD , MSD, N 1 , LC = 1 MSD - 1 VSD, N , = 1 MSD - , MSD, N 1, N 1 N , MSD , , N 1 , , a, (MSD = a), N 1, 14) minimum inaccuracy = Vernier constant, , =, , , , 1, 0.5 mm, 50, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 15) diameter 1.2 5 0.01 0.04 , , 1.21cm, 16) Error is positive, zero correction is negative, 17) A., Least count = 1-, , 9, 10, , 1, mm, 10, = 0.1 mm, = 0.01 cm, , , = 1 MSD -, , 1, MSD, 50, 1 MSD = 50 (LC), = 50(0.001)cm, = 0.05cm, =0.5mm., 26) Let Vernier scale has N divisions. Then:, (N) VSD = (10) MSD, , or, , LC =, , 10 , 1 VSD = MSD, N, LC = 1 MSD - 1 VSD, , B., Least count 1 , , 10, 20, , 10 , = 1 MSD, N, , 10, mm, 20, =0.5 mm, =0.05cm, 23) ( N m)VSD ( N ) MSD, , , or, , 49, MSD, 50, LC = 1 MSD - 1 VSD, , NARAYANAGROUP, , 29, main scale divi60, , sion, , , 1, , , , MSD, 1 N / m , For minimum value of least count m should be, minimum or 1., 24) 10 VSD = 8 MSD, 1 VSD = 0.8 MSD, Now LC = 1 MSD - 1 VSD, =1 MSD - 0.8 MSD, , 1 MSD=, , , , 10 1cm , , , N 10 , , 27) 1 vernier scale division =, , N , 1 MSD , MSD, N m, , = 0.02cm., 25) 50 VSD = 49 MSD, , 0.025cm 1 , , Solving we get,, N = 20., , N , 1VSD , MSD, N m, LC = 1 MSD -1 VSD, , = 0.2 MSD = 0.2 , , 49, MSD, 50, , 1, cm, 10, , 1VSD , , 29, 29 , 0.50 , 30, 60 , , 0, , The least count = 1MSD - 1 1 VSD, 0, , 0, , 0, , 1 29 1 , , 2 60 60 , so, reading=main scale reading + vernier scale, reading, = MSR + N X LC, 1 , 58.50 9 , 60 , , 0, , 58.650 ., , 28) Least count of vernier calliper is, , 1, mm, 10, , = 0.1mm., = 0.01 cm., 29) Least count of vernier callipers, LC = 1 MSD = - 1 VSD, 73
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , Smallestdivisiononmainscale, , Numberofdivisionsonvernierscale, , , , , 20 divisions of vernier scale = 16 divisions of main, scale, 16, mm 0.8mm, 20, LC = 1 MSD - 1 VSD, = 1mm - 0.8 mm, =0.2mm., 30) Given, least count (LC) = 0.1mm, main scale reading = 0, vernier scale reading = 10, Length measured with the vernier callipers = reading before the zero of vernier scale + number of, varnier divisions coinciding with any main division, X least count, =10mm X 0 X 0.1mm, =10mm, =1 cm, 31) Vernier scale is provided in the microscope, 1VSD , , , , , , , , , , , , EXPERIMENT - 2, SCREW GAUGE, , , , , , , , It is used to measure thickness / diameter of a thin, sheet or a thin wire., It works on the principle of a micrometer screw, It consists of a pitch scale marked in millimeter and, a circular scale or head scale on which 100 divisions are marked., When the head scale completes one rotation (i.e.,, by 100 divisions) then the screw moves by 1mm., Least count, (LC)=, , , , Pitch of the screw (PS), , correctedheadscalereading LC, d= PSR corrected HSR LC , Object, Reference, line, , Fixed, stud, A, , B, , 60, 50, 40, 30, 20, , O, , Movable, stud, , Circular, scale, , Pitch, scale, , Ratchet, screw, , distance travelled by the screw, number of rotations made, , Least count LC , , 0.001cm, 74, , = pitch scale reading +, , pitch of thescrew, number of divisions on the circular scale, , (PS) , , , , , , Screw gauge is more accurate than vernier, callipers., When the zero of circular scale coincides with, the reference line perfectly when the screw is, completely rotated, then the the screw gauge is, said to have no error (error free instrument), When the zero of circular scale advances, beyond the reference line the instrument is said, to have negative error and the correction is, positive, When the zero of circular scale is left behind, the reference line, then the instrument is said to, have positive error and the correction is negative, Some times the screw does not move forward, or backward for a little distance even though the, head is rotated. This error is called backlash, error., To avoid backlash error in a screw gauge, the, screw is well greased and is always turned in, the same direction while taking an observation., After inserting a thin wire (or) a metal sheet, between the jaws, rotating the screw - until it is, held tightly without any pressure. Then pitch, scale reading and head scale reading are to be, noted., Thickness (or) diameter, , 1mm, 0.01mm or, 100, U-shaped frame, of stainless steel, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , ERROR CORRECTION, , 0, , 8), , Circular scale, , Circular scale, , 5, , 0, , 0, , 45, , Reference line, , Reference line, , (A) Positive zero error, , (B) Negative zero error, , EXERCISE - 2, 1), , 2), , 3), , 4), , 5), , 6), , 7), , 9), , The least count of a screw gauge is 0.005mm, and it has 100 equal divisions on its head scale., Then the distance between two, consecutive threads on its screw is, 1) 0.5mm 2) 0.05mm 3) 0.01mm 4) 0.1mm, The diameter of a wire is measured by, using a screw gauge having least count, 0.01mm. If the diameter is found to be, 0.20mm, then the error in the cross-section, of the wire will be, 1) 5%, 2) 10%, 3) 1%, 4) 2.5%, , In a screw gauge, the main scale has divisions, in millimeter and circular scale has 50 divi- 10) The least count of a screw gauge is 1 mm, 100, sions. The least count of screw gauge is, 1) 2 microns, 2) 5 microns, and the pitch of the screw is 1mm . The maxi3) 20 microns, 4) 50 microns, mum percentage error of the instrument is, The diameter of a wire is measured with a, 1) 5%, 2) 2%, 3) 1%, 4) 10%, screw gauge having least count 0.01mm. Out, 11), The, radius, of, a, ball, bearing, measured, by a, of the following the one which correctly, screw gauge is 3.75mm. The pitch of the, expresses its diameter is, screw is 1mm and it has 100 division on its, 1) 2.00 mm 2) 0.2 mm 3) 0.02 cm 4) 0.002mm, head scale. The percentage error in the volA screw gauge has 1.0mm pitch and 200, divisions on the circular scale. The least count, ume of the ball bearing which is perfectly, of the instrument is, spherical by shape is, 1) 2%, 2) 1.5%, 3) 0.8% 4) 1%, 1) 5 103 mm, 2) 5 104 mm, 12), The, length,, breadth, and, thickness of a small, 3) 5 102 mm, 4) 5 105 mm, uniform rectangular glass strip are 4.25cm,, In a screw gauge, keeping pitch of the screw, 6.25mm and 2.75mm. Its length is measured, constant, if we increase the number of head, by vernier callipers of least count 0.01cm and, scale divisions, then its accuracy of, breadth and thickness were measured by, measurement, screw gauge having least count 0.01mm. The, 1) increases, 2) decreases, 3) does not change, 4) cannot be predicted, percentage error in the measurement of volOut of the following three devices, the one, ume of the strip is, which is more accurate to measure length is, 1) 0.76% 2) 1.36% 3) 2.13% 4) 1.76%, (i) a meter rod, 13) Length of a thin cylinder as measured by ver(ii) a vernier callipers with least count 0.01cm, nier callipers having least count 0.01cm is, (iii) a screw gauge with a pitch 0.5mm, 3.25cm and its radius of cross-section is meahaving number of divisions on the circular, sured by a screw gauge having least count, scale as 100, 0.01mm as 2.75mm. The percentage error in, 1) (i), 2) (ii), 3) (iii), the measurement of volume of the cylinder, 4) all the three are equally accurate., Pitch of the screw gauge is 0.5mm. Its head, will be, scale contains 50 divisions. The least count, 1) 2%, 2) 3%, 3) 1%, 4) 1.5, of it is, 14) When circular scale of a screw gauge carry1) 0.01mm 2) 0.1mm 3) 0.25mm 4) 0.02mm, ing 100 divisions is given four complete rotaWithout changing the number of divisions on, tions, the head of the screw moves through, the circular scale, if the pitch of the screw, 2mm. The pitch and least count of screw gauge, gauge is halved, then its accuracy of, are respectively., measurement, 1)1mm and 0.005 mm, 1) decreases, 2) increases, 2) 0.05 mm and 0.001 mm, 3) remains unaffected, 3) 0.5mm and 0.005 mm, 4) increases or decreases depending on the, weight., 4) 0.005mm and 0.005 mm, NARAYANAGROUP, , 75
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EXPERIMENTAL PHYSICS, , JEE-ADV PHYSICS- VOL- VI, , 15) A student measured the diameter of a wire 21) While measuring diameter of a wire using a, screw gauge the main scale reading is 7mm, using a screw gauge with least count 0.001cm, and zero of circular scale is 35, divisions, and listed the measurement. The correct, above, the, reference, line., If, the, screw, gauge, measurement is, 1) 5.3 cm 2) 5.32 cm 3) 5.320 cm 4) 5.3200 cm, has a zero error of 0.003cm , the correct, 16) A screw gauge having 100 equal divisions and, diameter of the wire is (given least count =, a pitch of length 1mm is used to, measure, 0.001cm), the diameter of a wire of length 5.6cm. The, 1) 0.735cm, 2) 0.732cm, main scale reading is 1mm and 47th circular, 3) 0.738 cm, 4) 7.38 cm, division coincides with the main scale. The 22) When a screw gauge is completely closed,, curved surface area of the wire to the apzero of circular scale is 6 divisions below the, proximate significant figures is, reference line of graduation. If least count, 1) 2.6cm2, 2) 2.587cm2, of screw gauge is 0.001cm, the zero correction is, 3) 2.58cm2, 4) 2.5872cm2, 1) 0.006cm, 2) 0.006cm, 17) Two full turns of the circular scale of a screw, gauge cover a distance of 1mm on its main, 3) 0.006mm, 4) 0.006mm, scale. The total number of divisions on the 23) For the given figure, calculate zero correccircular scale is 50. Further it is found that, tion., the screw gauge has a zero error of, 5, 0, 0.03mm . While measuring the diameter of, a thin wire, a student notes the main scale, 1) - 0.02mm 2) + 0.02mm, reading of 3mm and the number of circular, scale divisions in line with the main scale as, 3) - 0.03mm 4) +0.03mm, 35. The diameter of the wire is, 1) 3.32mm 2) 3.73 mm 3) 3.67mm 4) 3.38 mm 24) The pitch of a screw guage is 0.5 mm and there, are 50 divisions on circular scale.When there, 18) The density of a solid ball is to be determined, is nothing between the two ends (studs) of, in an experiment. The diameter of the ball is, screw guage, 45th division of circular scale is, measured with a screw gauge, whose pitch is, 0.5mm and there are 50 divisions on the circoincide with screw guage, and in this situacular scale. The reading on the main scale is, tion zero of main scale is not visible. When a, 2.5mm and that on the circular scale is 20 diwire is placed between the studs, the linear, visions. If the measured mass of the ball has, scale reads 2 divisions and 20th division of cira relative error of 2%, the relative, percular scale coincides with reference line. For, centage error in the density is, this situation mark the correct statement(s)., 1) 0.9% 2) 2.4%, 3) 3.1% 4) 4.2%, 1) Least count of the instrument is 0.01mm, 19) a screw gauge gives the following reading, 2) Zero correction for the instrument is, when used to measure the diameter of a wire., +0.45mm, Main scale reading 0mm, circular scale, 3) Thickness of wire is 1.65mm, reading = 52 divisions. Given that 1mm on, 4) All of the above, main scale corresponds to 100 divisions of the 25) In a screw guage, the value of one division, circular scale. The diameter of the wire from, on the linear scale is 1mm, while the, the above data is, circular scale have 100 divisions. Without, 1) 0.052cm 2) 0.026cm 3) 0.005cm 4) 0.52cm, any object for measurement, while the, 20) The circular scale of a screw gauge has 200, screw touches the stud, the zero on, divisions. When it is given 4 complete, rocircular scale advances 27 divisions beyond, tations, it moves through 2mm. The least, the reference line. What is the type and, count of the screw gauge is, amount of zero error?, 1) 0.25 102 cm, 2) 0.25 103 cm, 1) positive, 0.27mm 2) negative, 0.27mm, 3) positive, 0.027mm 4) negative, 0.027mm, 3) 0.001cm, 4) 0.001mm, 76, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 26) When a screw guage is completely closed, zero 8., of circular scale is 7 division above the reference line of graduation. If LC of s c r e w 10., guage is 103 cm, the zero error is, 16., 1) 7 103 cm, 2) 7 103 cm, 3) 0.007mm, 4) 0.007mm, 27) A screw gauge gives the following reading, when used to measure the diameter of a wire., Main scale reading: 0 mm Circular scale reading: 52 divisions Given that 1 mm on main, scale corresponds to 100 divisions of the circu, lar scale. The diameter of wire from the above, 17., data is:, 1) 0.052 cm, 2) 0.026 cm, 3) 0.005 cm, 4) 0.52 cm, 28) If in a screw gauge, zero mark of the circular, scale remains on right of reference line and, does not cross it and 2nd division on circular, scale comes on reference line. then zero correction is, 18., 1) +0.02 mm, 2) -0.02 mm, 3) +0.002 mm, 4) -0.002 mm, 29) On measuring diameter of a wire with help of, screw gauge, main scale reading is 1 mm and, 6th division of circular scale lying over reference line. On measuring zero error, it is, found that zero of circular scale has advanced, from reference line by 3 divisions on circular, scale, then corrected diameter is, 1) 1.09 mm, 2) 1.06 mm, 19., 3) 1.03 mm, 4) 1.60 mm, , EXERCISE - 2 - KEY, 1) 3, 6) 1, 11) 3, 16) 1, 21) 3, 26) 1, , 2) 1, 7) 2, 12) 1, 17) 4, 22) 1, 27) 1, , 3) 1, 8) 1, 13) 3, 18) 3, 23) 3, 28) 2, , 4) 1, 9) 2, 14) 3, 19) 1, 24) 4, 29) 1, , 5) 3, 10) 3, 15) 3, 20) 2, 25) 3, , EXERCISE - 2 -HINTS, 1., 3., 6., , 1, = 0.02mm = 20 microns, 50, 1, 1, 102 5 10 3 mm, LC , ;, 200, 2, 0.5, 5, LC , ;, = 0.01 mm, 50, 500, , LC , , NARAYANAGROUP, , d = 0.005 X 100 = 0.05 mm, Percentag error , , 1, 100 = 1 %, 100, , Value of 1MSD 1mm 0.1cm, Value of 1CSD , , 1, mm 0.001cm, 100, , diameter d 1.47 mm 0.147cm, and l 5.6cm, curved surface area A 2 r l, , pitch 0.5mm, least count 0.01mm, , zeroerror 0.03mm, , correction 0.03mm, diameter 3.35 0.03 3.38mm, Least count of screw gauge , , 0.5, 0.01mm, 50, , diameter 2.5 20 0.01 2.7mm, Relative error in density, , M, D , , 3, , M, D , , relative error, , , 0.01 , 100 3.11%, 100 2% 3 , , 2.7 , Diameter, , MSR CSR LC 0 , , 1, 52 , 100, , 0.52 mm 0.052cm, 20. LC , , , , pitch of the screw, number of circular scale divisions, , 2, 0.25 103 cm, 4 200 , , 21. diameter 0.7 0.035 0.003 0.738cm, 22. Here zero error is positve 0.006cm, , zero correction is 0.006cm, 26. As zero of circular scale is above the reference, 77
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, line of graduation, zero correction is positive, and zero error is negative., Zero error = 7 103 cm, 27 Diameter of wire,, d MSR CSR LC, 0 52 , , 1, 0.52mm, 100, , =0.052cm, 28) Zero error = +0.02mm, zero correction = -0.02 mm, 29) Observed reading = 1 mm + 6 X 0.01mm, = 1.06mm, Zero error = -0.03mm, Zero correction, = + 0.03mm, Corrected reading= 1.06+0.03=1.09mm, , 1, , E kr02 e t, 2, , , PROCEDURE, , , , , , , , EXPERIMENT - 3, , , , To study dissipation of energy of a simple , pendulum by plotting a graph between square of, amplitude and time., If the acceleration of a body in motion is di- , rectly proportional to its displacement in opposite, direction then the motion of the body is, said to be simple harmonic motion, , , , F x and F k x, mg, where k is force constant , x, , , , , , The maximum displacement of a vibrating, particle in simple harmonic motion is called, amplitude, Energy stored in an oscillating simple pendulum having maximum displacement ' r0 ' is, , 1 2, k r0 ., 2, , , , , , , 78, , , , Find the mass of the bob of a simple pendulum, and calculate force constant (k), Suspend the pendulum with the help of a thread, of length 2m from a stand, keeping it at the edge, of a table., A meter scale marked 50cm 0 50cm is to, be placed horizontal on the table just below the, bob such that it is above the zero mark on the, scale., Oscillating the bob simple harmonically the, amplitude of oscillation is marked on a paper, strip placed on the meter scale after every two, oscillations. The oscillation time is measured, using a stop clock., Taking 10 to 15 readings a graph is plotted taking, time on x-axis and square of the amplitude on, y-axis,, Energy of an oscillating pendulum is directly, proportional to square of amplitude. Since, amplitude decays with the passage of time,, energy of oscillating pendulum dissipates with, time., While performing the experiment, the spinning, motion of the bob should be avoided., , 50 40 30 20 100 10 20 30 40 50, , r1, , r2, , r3, r5, , Paper, strip, Meter, scale, , r4, , Decay of, amplitude, , r6, , If the angular displacement of an oscillating, simple pendulum decreases gradually with time, EXERCISE - 3, then the oscillations of the pendulum are called, 1) The necessary and sufficient condition for, damped oscillations., simple harmonic motion is, Since the damping force is directly proportional, 1) Constant period, to the velocity of the bob, the decay of energy, 2) Constant acceleration, E E0 e t . where ' ' is decay constant., 3) Proportionality between restoring force and, displacement from equilibrium position, The dissipation of energy of a simple pendulum, 4) Acceleration is inversely proportional to, is due to damping, displacement from equilibrium position, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 2), , 3), , 4), , 5), , The work done by the string of a simple 8), pendulum during one complete oscillation is, 1) equal to the total energy of the pendulum, 2) equal to the kinetic energy of the pendulum, 3) equal to the potential energy of the pendulum, 4) zero, In the case of forced vibrations, resonance 9), will become very sharp when, 1) damping force is small, 2) restoring force is small, 3) applied periodic force is small, 4) quality factor is small, The one which is not the cause of damping of, an oscillating simple pendulum, 1) Friction due to two halves of split cork used, to suspend the simple pendulum, 2) air currents due to the use of a fan, 3) Opening of doors and windows of the room, 4) Closing of doors and windows of the room, The time period of an oscillating simple, pendulum is 1s when its amplitude of vibration, is 4cm. Its time period when its amplitude is, 6cm is, 1), , 6), , EXPERIMENTAL PHYSICS, , 3, s, 2, , 2), , 2, s, 3, , 3) 4 s, , 4) 1s, , A simple pendulum suspended from the ceiling, of a train has a time period ‘T’ when the train, is at rest. If the train is accelerating uniformly, at ‘a’ then its time period, 1) increases, 2) decreases, 3) unaltered, 4) becomes infinity, The graph between time period (T) and length, , l of a simple pendulum is, 1), , 2), T, , O, , T, , l, , l, , O, 4), , 3), T, , O, , T, , O, , l, , l, , The graph between square of amplitude and, time elapsed is, 10) The time period of a simple pendulum inside, a stationary lift is ‘T’. If the lift accelerates, upwards uniformly with, , g, , then its time, 4, , period would be, 1) 2 5 T 2), , 2T, 5, , 3) 2T, , 4), , T, 2, , 11) The time period of an oscillating simple, pendulum is given as T 2, , 7), , A simple pendulum is oscillating in a stationary, lift. When the lift falls freely, the frequency, of oscillations of the pendulum is, 1) zero, 2) infinity, 3) unaltered, 4) negative, NARAYANAGROUP, , l, where l is, g, , its length and is about 1m having 1mm, accuracy. Its time period is 2s . The time for, 100 oscillations is measured by a stopwatch, having least count 0.1s . The percentage, error in the measurement of ' g ' is, 1) 0.4% 2) 0.1%, , 3) 0.3%, , 4) 0.2%, 79
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 12) The bob of a simple pendulum is hollow and is, and is filled with sand. Making a small hole, at the bottom of the bob and when it is made, to oscillate simple harmonically, then its time, period of oscillation, 1) first increases and then decreases to its, original value, 2) first decreases and then increases to its, original value, 3) remains unchanged, 4) increases gradually to infinity., 13) A particle executes simple harmonic motion., Then the graph of velocity as a function of its, displacement is, 1) a straight line, 2) a circle, 3) an ellipse, 4) a hyperbola, 14) The graph plotted between acceleration and, displacement of a particle in SHM is, 1) a straight line, 2) a circle, 3) an ellipse, 4) a hyperbola, 15) The tension in the string of an oscillating, simple pendulum is, 1) a constant, 2) maximum at extreme position, 3) maximum while crossing the mean position, of rest, 4) zero at mean position, 16) The time period of a simple pendulum, oscillating in a freely falling lift is, 1) infinity, 2) zero, 3) negative, 4) 2 sec., 17) When the displacement is half of the, amplitude of a simple pendulum then the, fraction of total energy which is kinetic is, , 1, 1), 4, , 2, 2), 3, , 4, 3), 5, , 3, 4), 4, , 19) If the time period of a pendulum is 1 sec, then, what is the length of the pendulum at point of, intersection of l T and l T 2 graph., , l–T graph, l–T2 graph, , 1) 25 cm 2) 21 cm, , 3) 22 cm, , 4) 27 cm, , EXERCISE - 3 - KEY, 1) 3 2) 4 3) 1 4) 4 5) 4 6) 2 7) 1, 8) 2 9) 2 10) 2 11) 4 12) 1 13) 3 14) 1, 15) 3 16) 1 17) 4 18) 2 19) 1, , EXERCISE - 3 - HINTS, 5., , Time period is independant of amplitude, , 10., , T1, , T2, , 11., , g, l, , T , 100 2 , 100 , g, T , l, , , ga, g, , 17. KE , , 1, a, m 2 a 2 y 2 and y , 2, 2, , 18. F , , du, 8sin 2 x, dx, , , , , , For small oscillation, sin 2 x 2 x, , F 16 x k 16 Nm1, Time period T 2, , 1 , s, 16 2, , 18) The potential energy of a particle of mass 1kg 19. T 2 l ; T 2 42 l, g, g, in motion along x-axis is given by, , U 4 1 cos 2 x joule, where ' x ' is in, meter. The period of small oscillations in, second is, 1) , , 80, , 2), , , 2, , 3), , 3, 2, , 4) 2, , l, , g, 9.8, , 0.25 25 cm, 2, 4, 4 9.8, , EXPERIMENT - 4, To find the mass of object of any shape and size, using a uniform meter scale applying the, principle of moments., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , , , , , EXPERIMENTAL PHYSICS, , The turning effect of a force acting on a body about , an axis is called ‘ moment of the force ‘ or ‘ torque’, Torque is a vector quantity and is equal to the product of the magnitude of the applied force and the , perpendicular distance to the point of application, of the force from the axis of rotation or fulcrum., If ‘O’ is the fulcrum and a force ‘F’ is applied at ‘A’, to produce torque, then, , , O, , A, , x, , , , F, , While performing the experiment, a uniform meter, scale is placed horizontally at 50 cm mark on a, wedge of hard wood or plastic., Hanging the body of which the mass is to be, determined on one side at a certain distance,, known weights are hung on the other side at, different distances, the system is to be made in, equilibrium, To bring the system in equilibrium, the distances, to the known weights are to be changed., When the system is in equilibrium, then the sum, of clockwise moments is equal to the sum of anti, clockwise moments., , moment of force or torque = (force) (perpendicular distance), F OA F x , , , , , , , , , , , , , , , , SI unit of torque is newton meter (Nm) and its, dimensional formula is M L2T 2 which is same, as the dimensional formula of work done., After the application of several coplanar forces, on a rigid body, if the body is in equilibrium, then, (i) the algebraic sum of forces acting on the body , is zero and, (ii) the alegraic sum of moments of forces about, the fulcrum in that plane is zero [ principle of, moments], If the force has a tendency to rotate the body in, clockwise direction, then the corresponding moment is said to be clockwise moment. If the force, has a tendency to rotate the body in anti, clockwise direction, then the corresponding 1), moment is said to be anti clockwise moment., Principle of moments can also be stated as ‘ the, sum of clockwise moments on a body is equal to, the sum of anti clockwise moments’ when the, body is in equilibrium, The point at which the total weight of a body is 2), supposed to act is known as centre of gravity of, the body. For a body to be in stable equilibrium, its centre of gravity must be at its lowest, position., When a body is balanced with respect to its centre of gravity, then its moment about the axis, passing through its centre of gravity is zero., A body is said to be in equilibrium if it remains in its, state of rest or of uniform motion with the time., NARAYANAGROUP, , Applying the principle of moments, anti clockwise moment = sum of clock-wise moments, , Mg OA m1 g OB m2 g OC , m OB m2 OC , M 1, gram, OA, , , , , , EXERCISE - 4, A meter scale is balanced on a knife edge at, its centre. When two coins each of 5g are put, one on the other at 12cm mark, the scale is, found to be balanced at 45cm mark. The mass, of the meter scale is, 1) 100 g 2) 33 g, 3) 66 g, 4) 99 g, A cubical block of side ‘L’ rests on a rough, horizontal surface with coefficient of friction, ' ' . A horizontal force F is applied on the, block as shown. If the coefficient of friction, is sufficiently high so that the block does not, slide before toppling, the minimum force required to topple the block is, , 81
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 6), , F, , L, , A balance is made of a rigid rod free to, rotate about a point not at the centre of the, rod. When an unknown mass ‘m’ is placed in, the left pan, it is balanced by a mass ' m1 ', , mg, 4) mg 1 , 2, 3) A wheel of radius ' r ' and mass ' m ' stands in, front of a step of height ' h ' . The least hori1) mg, , 2), , mg, 4, , placed in the right pan and similarly when the, mass ‘m’ is placed in the right pan, it is bal-, , 3), , anced by a mass ' m2 ' in the left pan. Neglecting the masses of the pans, ‘m’ is, , zontal force which should be applied to the, axle of the wheel to allow it to raise onto the, step is, , 1), , m1 m2, 2, , m12 m22, 2, , 3), F, , r, , r–h, , 7), , x, mg, , 1), , 3), 4), , mgh 2r h , rh, mg h 2r h , rh, , 2) mgh r h , , 4), , Calculate the force ' F ' that is applied, horizontally at the axle of the wheel which is 8), necessary to raise the wheel over the, obstacle of height 0.4m. Radius of the wheel is, , , , 2, 1m and its mass is 10 kg g 10ms, , 9), , 1), 82, , L, 3, , 2), , L, 2, , 3), , 2L, 3, , 4), , 3L, 4, , 2), , xy, , x2 y2, 3), 2, , 4), , x y, 2, , x2 y 2, 2, , A weightless ladder 6m long rests against a, frictionless wall at an angle of 600 from the, , 1), , 5), , 4), , m12 m22, 2, , horizontal. A 60kg manstanding on it is 2m, from the top of the ladder. A horizontal force, is applied at the lower end to keep it from, slipping. The magnitude of the force is, , , , 1) 100 N 2) 66 N, 3) 167 N 4) 133.3 N, Two men ' A ' and ' B ' are carrying a uniform, bar of length ' L ' on their shoulders. The bar, is held horizontally such that A gets one fourth load. If ‘A’ is at one end of the bar, the, distance of ‘B’ from that end is, , m1m2, , A false balance has equal arms. An object, weighs ' x ' when placed in one pan and ' y ', when placed in the other pan. The true weight, of the object is, 1), , mgh, r, , 2), , 10, 20, N 2), N 3) 3 N, 3, 3, 20, , 4) 20 3 N, , An uniform metre scale of weight 50 g is balanced at 60 cm mark, when a weight of 15 g, is suspended at 10 cm mark. Where must a, weight 100 g be suspended to balance metre, scale., 1) 72.5 cm 2) 70 cm 3) 71.5 cm 4) 74.5 cm, , EXERCISE - 4 - KEY, 1) 3, 6) 2, , 2) 3, 7) 2, , 3) 3, 8) 2, , 4) 4, 9) 1, , 5) 3, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXERCISE - 4 - HINTS, 1), , 2), , Applying law of moments,, , Let the weights of the pans be W1 and W2 and, true weight be W., , l0 g 45 12 mg 50 45 , , Then W W1 l x W2 l .......(1), , m 66 g, , and W W2 l y W1 l .......(2), , applying the conditon of rotational equilibrium, , L, F L mg ;, 2, 3), , 7), , A, , Applying the condition of rotatinal equilirbium,, , F r h mgx, 2, , 2m, , 2, , 2, , but r x r h x , , C, , h 2r h , , 4), , 6m, , 60°C, D, , 8), , F , , x y, 2, , W , , mg, F, 2, , W, , mg h 2r h , , 60°, O, , B, , r h, , F, , applying principle of moments, , Taking moments about ‘A’, , F 0.6 100 0.8 , , F OA W CD , , F 133.3N, , F 6sin 600 60 2cos 600 , , x, A , , L, , 2, , B, , F , , 5), W, 4, , W, , 3W, 4, , 10, , Taking moments about ‘A’, , x, , 50, , 60, , x, , 100, , 0, 10, , 9), , 3W, L, x W ;, 4, 2, , 20, N, 3, , 15gf, , 50gf, , 1, , x–60, 100gf, , Let x be the point where 100 g of weight is suspended Anti clock wise moment, 15 g 50 50 g 10 1250 g cm, Clock, wise, moment, , 2L, 3, , 100 g x 60 100 x g cm 6000 g cm, , 6), , by the law of moments clock wise moments =, anti clock wise moments, 100 x g cm 6000 g cm 1250 g cm, , Let ‘L’ be the length of the rod and knife edge is, placed at a distance ' x ' from the left pan, About the knife edge, , mg x m1g L x ............(1), , 100 x g cm 7250 g cm ; x , , m2 g x mg L x .............(2), , EXPERIMENT - 5, , dividing (1) by (2), , m m1, , m m1m2, m2 m, NARAYANAGROUP, , 7250, 72.50 cm ., 100, , , , Determination of Young’s modulus of the, material of a given wire by Searles’s method, In Searle’s apparatus two wires A and B of the, same material, length and area of cross - section, suspended from a rigid support carry at their ends, 83
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , , , , , , , , , , , , two rectangular metal frames of same mass and, dimensions as shown., One of the wires ‘A’ is called reference wire, and the other wire ‘B’ is called experimental, wire., The metal frame which is attached to the, reference wire ‘A’ carries a constant weight to, keep the wire stretched without any kinks., The metal frame which is attached to the, experimental wire ‘B’ carries a variable weight, hanger over which allotted weights can be, slipped as required., A spirit level (SL) is placed with one end to the, frame attached to the reference wire and rests, horizontally on the tip of a micrometer screw, which can be worked in the frame attached to, the experimental wire along a vertical scale, marked in millimeter., While doing the experiment, a suitable load, (equal to the fixed weight) is kept on the hanger, so that the experimental wire is straight without, kinks., The micrometer screw is adjusted so that the air, bubble in the spirit level (SL) comes to the, centre marked with a circle and the reading of, , , , tion r of the experimental wire can be determined by taking 7 to 8 readings., , , , , A load of 500 gram is added to the weight hanger, and after adjusting the screw so that the air, bubble is again at the middle of the spirit level, and the reading R1 of the, micrometer, screw is noted., Loads are increased in steps of 500 gram and in, each case, the micromet er readings, , , , , , , , , , 84, , From the graph, slope tan , , BC, e, , AC Mg, , B, extension, (e), O, , , , , , A, , , , C, , Load, (Mg), , Young’s modulus of the wire can also be, determined from the graph, using the formula, , Y, , are t he , corresponding elongations in the experimental, wire for the loads 500g; 1000g ; 1500g......., The graph plotted between the load and, extension will be a straight line and gives the, elongation e for a load Mg , , FL Mg L , , the Young’s modulus of, Ae r 2 e , , the material of the wire can be calculated., , When the maximum load (far below the, breaking point of the wire) is reached, the load, is decreased in the same steps and the reading, are noted again., From, t he, above, observations,, , R1 R0 ; R2 R0 ; R3 R0 ...., , mg, , in the, Substituting the values of ' r ' , L and, e, formula, for, Young’s, modulus, , Y, , R2 , R3........ are noted., , , Using a meter scale, the actual length L of the, experimental wire can be measured., , the micrometer screw R0 is noted., , , By using a screw gauge, the radius of cross - sec-, , L, r tan , 2, , The following are the possible sources of errors in this experiment,, (a) The support may yield when the load is, attached at the lower end of the experimental, wire and thereby the measured value of, elongation (increased length) of the wire may, not be correct., (b) While the experiment is carried out, external temperature may change which causes some, increase in the length of experimental wire, thereby the measured value of increase in length, becomes incorrect., The above said errors are minimised by using, the reference wire. The yield of support or the, change of temperature affects both experimental, and reference wires. The relative increase in, length of the experimental wire with respect to, the reference wire will give correct increase in, length., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 5), , Rigid, support, , 6), Reference, wire, , A, , B, , Experimental, wire, , Metal, frame, , 2X, 1), Y, 7), , SL, , Spirit, level, , Young’s modulus for a perfect rigid body is, 1) zero, 2) 1, 3) infinite, 4) negative, When an elastic material with Young’s modulus ‘Y’ is subjected to a stretching stress ‘X’,, the elastic energy stored per unit volume is, , 1), 8), Variable, weighthanger, , EXERCISE - 5, 1), , 2), , A load of 3kg produces an extension of 1.5 mm, in a wire of length 3m and diameter 2mm., Young’s modulus of the material of the wire, is, 1) 1.87 1010 Nm 2, 2) 3.5 1010 Nm2, 3) 15 1010 Nm2, 4) 2.5 1010 Nm2, 9), The force required to stretch a steel wire, 1cm 2 in cross - section to increase its length, by 1% if its Young’s modulus is 2 1011 Nm2, is, 1) 105 N, 2) 3 105 N, , 3), , 4), , X 2Y, 3), 2, , X2, 4), 2Y, , A wire can support a load ‘W’ without, breaking. It is cut into two equal parts. The, maximum load that each part can support is, , Micrometer, screw, , Constant, weight, (Fixedload), , Y2, 2), 2X, , W, 4, , 2), , W, 2, , 3) W, , 4) 2W, , A metal ring of initial radius ' r ' and cross sectional area ' A ' is fitted onto a wooden disc, of radius R r . If Young’s modulus of the, metal is ‘Y’ then the tension in the ring is, 1), , AYR, r, , 2), , AY R r , r, , 3), , Y Rr, A r , , 4), , Yr, AR, , A metal beam supported at the two ends is, loaded at the centre. If ‘Y’ is Young’s, modulus, then the depression at the centre is, proportional to, 1), , 1, Y, , 2) Y, , 3), , 1, Y2, , 4) Y 2, , 10) Hooke’s law is applicable when, 3) 2 105 N, 4) 4 105 N, intermolecular distance is, A wire is made of a material of density, 1) much smaller than the distance of equilibrium, 10 g / cm3 and breaking stress 5 109 Nm2 ., 2) approximately equal to the distance of, equilibrium, If g 10ms 2 the length of the wire that will, 3) much larger than the distance of equilirbium, break under its own weight when suspended, 4) zero, vertically is, 11) The material which practically does not, 1) 5 102 m, 2) 5 103 m, exhibit elastic after effect is, 1) rubber 2) quartz 3) copper 4) steel, 3) 5 104 m, 4) 5 105 m, 12) A force ‘F’ is needed to break a copper wire, Two steel wires of lengths 1m and 2m have, of diameter ‘d’. The force needed to break, diameters 1mm and 2mm respectively. If they, another copper wire of diameter ‘2d’ is, are stretched by forces of 40N and 80N reF, spectively, the ratio of their elongations is, 1), F, 2), 2F, 3), 4F, 4), 1) 2:1, 2) 2:3, 3) 3:4, 4) 1:1, 4, NARAYANAGROUP, , 85
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 13) When temperature of a material increases, its 18) The stress - strain graphs for materials A and, Young’s modulus, B are as shown. Choose the correct alterna1) increases, 2) decreases, tive, 3) remains same, 4) becomes infinity, 14) The load versus extension graph for four wires, of same material is shown. The thinnest wire, stress, stress, A, is represented by the line, B, strain, , strain, , 1) OA, , 2) OB, , 1) material A is stronger than material B, 2) material B is stronger than material A, 3) OC, 4) OD, 3) Young’s modulus of A is greater than that of B, 4) Young’s modulus of B is greater than that of A, 15) The graph shows the change ' l ' in the length, of a thin uniform wire used by the application 19) A steel rod has a radius 10mm and a length of, 1m. A force stretches it along its length and, of force ‘F’ at different temperatures T1 and, produces a strain of 0.32%. Young’s modulus, T2 . The variation suggests that, 11, 2 , the magnitude of, of steel is, , 2 10 Nm, , T2, , force stretching the rod is, T1, , 1) T1 T2 2) T1 T2, , F, , 3) T1 T2 4) T1 T2, , O, , L, , 1) 100.5 kN, , 2) 201 kN, , 3) 78 kN, , 4) 150 kN, , 20) A square lead slab of side 50cm and thickness, 10cm is subjected to a shearing force of, , 16) When the stress is increased beyond the elas9 104 N . The lower edge of the slab is fixed, tic limit, the length of the wire starts increasing without increasing the force. This point is, to the floor. The upper edge of the slab is, called, displaced by 0.16mm. Young’s modulus of lead, 1) yield point, 2) inverse point, is ( y 3 ), 3) breaking point, 4) triple point, 1) 1.9 109 Nm 2, 2) 1.7 1010 Nm 2, 17) In the experiment to determine Young’s modulus of the material of a wire under tension, 3) 1.1 1010 Nm 2, 4) 5.6 109 Nm 2, used in the arrangement as shown. The percentage error in the measurement of length 21) Two wires of the same material have equal, lengths but A is thicker than other B. Which, is ‘a’, in the measurement of the radius of the, of the two has greater value of Young’s moduwire is ‘b’ and in the measurement of the, lus ?, change in length of the wire is ‘c’. Percent1) A, 2) B, age error in the measurement of Young’s, 3), same, for, A, and, B, 4) cant perdict, modulus for a given load is, 22) Maximum permissible load of given wire. If, area of cross section= r 2 and breaking, stress is F is, 1) a 2b c 2) a 2b c, 1, 1, 2, 2, 1) F r, 2) F r, 3, 2, 3) a 2b c 4) a 2b, 3) F r 2, 86, , 4) 3F r 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 23) Statement-I : In Searle’s experiment, extension versus load curve is drawn as shown. In, the plot the first two readings are not lying, on the straight line., Extension, , Load, , Statement-II : Experiment is perfomed, incorrectly, 1) Statement-I is true, Statement-II is true, Statement-II is a correct explanation for Statement-I, 2) Statement-I is false, Statement-II is true, Statement-II is not a correct explanation forStatementI, 3) Statement-I is true, Statement-II is false, 4) Statement-I is false, Statement-II is true, 24) Assertion (A):In Searle’s experiment, it is better to start the experiment with some initial load, on the hanger., Reason(R): In Searle’s experiment, it is desirable that wire is striaght and no kinks are there., 1) If both assertion and reason are true and the, reason is the correct explanation of the assertion., 2) If both assertion and reason are true but reason is not the correct explanation of the assertion., 3) If assertion is true but reason is false., 4) If assertion is false but reason is true., 25) A student performs an experiment to, determine the Young’s modulus of a wire,, exactly 2m long, by Searle’s method. In a, particular reading, the student measures, the extension in the length of the wire to, be 0.8 mm with an uncertainity of the wire, to be 0.8mm with an uncertainity of, 0.05 mm at a load of exactly 1.0kg. The, student also measures the diameter of the, wire to be 0.4mm with an uncertainity of, Take, g 9.8ms 2 (exact)., 0.01 mm., Young’s modulus obtained from the, reading, is, , 26) Which of the following is wrong regarding, Searle’s apparatus method in finding Young’s, modulus of a given wire?, 1) Average elongation of wire will be determined with a particular load while increasing, the load and decreasing the load., 2) Reference wire will be just taut and experimental wire will undergo for elongation., 3) Air bubble in the spirit level will be disturbed from the central position due to relative, displacement between the wires due to elongation., 4) Average elongation of the wires is to determined by increasing the load attached to both, the wires., 27) Two wires A and B have same lengths and, made of the same material but A is thicker, than B. Both are subjected to the same extending load. Which will extend more?, (a) A, (b) B, (c) Same extension, (d) Can’t predict, , EXERCISE - 5 - KEY, 1) 1, 8) 2, 15) 2, 21) 3, , 1., , 4) 2.0 0.05 10 Nm, NARAYANAGROUP, , 2, , Y, , 4) 4 5) 3, 11) 2 12) 3, 18) 2&3, 24) 2 25) 2, , 6) 4, 13) 2, 19) 2, 26) 4, , 7) 3, 14) 1, 20) 2, 27) 2, , , , mgL, r2 e, , , , Y A e , L, , 2., , F, , 3., , stress , L, , F L A g , , L g, A, A, , stress, g, , 4., , e1 F1 L1 r22, , e2 F2 L2 r12, , 6., , Energy , , 3) 2.0 0.1 1011 Nm 2, 11, , 3) 3, 10) 2, 17) 3, 23) 3, , EXERCISE - 5 - HINTS, , 1) 2.0 0.3 1011 Nm 2, 2) 2.0 0.2 1011 Nm 2, , 2) 3, 9) 1, 16) 1, 22) 3, , 1, stress strain, 2, 87
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 8., , 17., , Y, , T 2 r , A 2 R r , , EXPERIMENT - 6, , l, Y, r l , 100 , 2, , 100, Y, l , r , l, , , , 18. Since the elastic limit of ‘B’ is greater than that, of ‘A’, ‘B’ is stronger than ‘A’, Since for a given stress, strain in ‘A’ is smaller , than that of ‘B’. since Y , , 1, , Young’ss, strain, , modulus of ‘A’ is greater than that of ‘B’, , , , 2, 19) F Y r, , Ll, , , , F L, , A L, and Y 3, , 20) , , , , Determination of Surface Tension of water by Capillary rise method and the effect of, detergents, When a capillary tube open at both ends is dipped, vertically into a liquid, then the liquid, either rises or falls inside the tube due to surface, tension. This phenomenon is called capillarity., When a glass capillary tube of inner radius ' r ', open at both ends is inserted vertically into a, liquid of density ' ' and surface tension ‘T’ such, that only its lower end dips in the liquid, then, the liquid rises up into the capillary tube by a, height ‘h’ above the free surface of the liquid., In this case, the surface tensional force ‘T’ acts, along the tangent to the liquid meniscus., The total force acting upwards on the meniscus, is, , 2 r T cos, , where ' ' is angle, contact. This force acting upward about, circle of contact supports the weight of, liquid column below the meniscus inside, capillary tube., , 21) Young’s modulus does not depend on length or, cross-section area of the wire but it depends on, property of the material., MgL, 4MgL, , 2, L D 2 L, substituting the given values, we get, , of, the, the, the, , 25) Young’s modulus Y , , Y, , 4 1 9.8 2, 2, , 0.4 103 0.8 103, , 2 1011 N m2, Now,, , 2D L , , , , D, L, , 2 0.01 0.05 , 11, 2, , , 2 10 N m, 0.4, 0.8, , , , = 0.2 1011 N m 2, Hence, the Young’s modulus obtained from the, reading is 2 0.2 1011 N m 2, 27) As, Young’s moduls,, Y, , l A AB, F .l, F .l, 1, l , ; l ; l A, Al, YA, A, B, A, , As,, , lB l A, 88, , , , AA AB, , Weight of the liquid column inside the capillary, tube, , r, , W mg r 2 h g, 3, , , , Equating the upward force with weight of the, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , liquid column, surface tension of the liquid i.e, Fupward Fdownward, , cal cross - wire, the reading ' d1 ' is noted. Similarly, by coinciding the right edge of the circle with, the vertical cross-wire, the reading ' d 2 ' is noted., The difference between these two readings gives, the internal diameter of the capillary bore, from, , r, , 2 r T cos r 2 h g, 3, , , r, , , , rh g, r, , 3, 2 T cos r h g ;, T , 3, , 2cos , , , , if r h then, , T , , , r, can be neglected, 3, , , , , , , , , , , , , , , equation T , , EFFECT OF DETERGENTS, , For glass capillary tube dipped in water, 00, , , , rh g, 2, , Arrange the adjustable stand on a table such that, its base is perfectly horizontal., Water is filled in the flat bottom dish placing it, , on the stand., Three capillary tubes of different radii are fixed, on the stand along with a needle in increasing, order of their radii., Capillary tubes are dipped into water in the dish , such that the pin just touches the surface of, water., The least count of travelling microscope for both, horizontal and vertical scales is to be determined. , Raising the microscope to a suitable height, pointing towards the first capillary tube, the meniscus of water is coincided with the horizontal, cross-wire and the reading h1 on the vertical, scale is noted., By coinciding the tip of the pin with the, horizontal cross-wire, the reading h2 on the, vertical scale is noted. The difference between, these two readings gives the height of water column h in the capillary tube., Repeating the same experiment with the other, capillary tubes, heights of water column in them, are measured., By placing the first capillary tube perfectly, horizontal and by focussing the travelling, microscope, we can observe a circular bore in, it., Coinciding the left edge of the circle with the vertiNARAYANAGROUP, , rh g, , surface tension of water, 2, , can be determined., , rh g, 2cos , , and cos 1 ; T , , , , , which the radius of cross section r can be determined., Substituting the values of ' h ' and ' r ' in the, , When some impurity is dissolved in water, its, density and surface tension undergo a change., The detergents added to water significantly, reduce its surface tension and hence improve its, wetting property. That is why a detergent, solution in water is able to provide cleaning, action., The effect of detergent on the surface tension of, water can be expressed as h , , 2T, h T, r g, , As the effect of detergent is to decrease surface, tension of water by their addition, it follows that, a solution of a detergent will rise to lesser height, than water., The cleansing effect of a detergent is inversely, proportional to the height. The detergent, solution which rises to a minimum height is most, effective., capillary, tubes, , Pin, , Stand, Water, , Adjustable, stand, , 89
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXERCISE - 6, 1), , 2), , 3), , If a liquid used in capillary rise method for, measuring surface tension whose surface, tension is more than that of water, then rise, in capillary tube, 1) increases, 2) decreases, 3) remains same, 4) cant say, What will happen if the double mass of, detergent is added in same volume of, water, the rise in capillary tube? 9), 1) decreases, 2) increases, 3) remains same, 4) cant say, In experiment for measuring surface tension, by capillary rise method, reading or positions, A,B, C and D for internal diameter of capillary tube are given as under, A(cm) = 1.006, B(cm) = 1.009, C(cm) = 1.004, D(cm) = 1.009, , 4), , 5), , 8), , A, C, , EXERCISE - 6 - KEY, D, , 1) 1, 6) 1, , B, , Mean intenal radius of capillary is, 1) 0.002cm 2) 0.003cm 3) 0.004cm 4) 0.005cm, 5), Consider two capillers A and B with radii, r, ri r , r2 . In which case liquid rise more?, 2, 1) A 2) B 3) same level 4) data insufficient, While measuring surface tension of water using capillary rise method, height of the lower, meniscus from free surface of water is 3cm, while inner radius of capillary tube is found, to, be, 0.5cm., Then, compute, surface tension of water using this data. (Take, contact angle between glass and, water as, and, g 9.81m s 1 ), 00, , 1) 0.72Nm 1 2) 0.77Nm 1 3) 1.67Nm 1 4) 8.67Nm 1, 6) In previous question if we add some detergent, to water, then, 1) liquid level in capillary tube is less than 3 cm, 2) liquid level in capillary tube is greater than 3 cm , 3) liquid level in capillary tube is equal to 3 cm, 4) anything may happen, 7) While measuring surface tension of water using capillary rise method the necessary precaution to be taken is/are, 1) capillary tube should be clean while water, should have some grease, 2) both capillary tube and water should be clean, , 3) no need to take care of temperature of water, 4) none of these, 90, , A capillary tube A is dipped in water., Another identical tube B is dipped in, detergent-water solution. Which of the, following shows the correct nature of, meniscus of the liquid column in the two, capillary tubes ?, 1) Both shows convex meniscus, 2) Meniscus in both the cases are concave, 3) In A meniscus is concave while in B it is convex, 4) In A meniscus is convex while in B it is concave, Find the depression of the miniscus in the capillary tube of diametre 0.4 mm dipped in a beaker containing mercury (density of mercury, = 13.6 103 kg / m3 and surface tension of the, mercury is 0.49 N / m and angle of contact, 1350 )., 1)0.025cm 2) 0.021 cm 3) 0.020cm 4) 0.027cm, 2) 1, 7) 2, , 3) 1, 8) 2, , 4) 2, 9) 1, , 5) 2, , EXERCISE - 6 - HINTS, T, , hrdg, 2 cos , , hrdg, 9) T 2 cos , , EXPERIMENT - 7, Determination of coefficient of viscosity of a, given viscous liquid by measuring terminal velocity of a given spherical body., , glycerine, 50cm, , P, , Q, R, , Rubberband, , Rubberband, Rubberband, , If a small spherical body falls under gravity in a liquid, a resistance is offered to it by the liquid. According to stoke’s law this opposing force F is given, F 6 r, by, where r radius of the spherical body, coefficient of viscosity of the fluid, terminal velocity of the spherical body, In the steady state, the force F is equal to the net, downward force acting on the spherical body, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , F weight of the sphere - upthrust of the liquid, , 4, 4, 4, r 3 g r 3 g, r3 g, 3, 3, 3, , where density of the solid and is density, of the liquid., , , , , , , , , 4, 6 r r 3 g, 3, , 2, 2 r g, , , 9, , When , and are const ant , then, , , constant ., r2, When a graph is plotted between terminal, velocity taken along y-axis and square of the , , , , 2, radius r of the spherical bodies taken along x- , , axis, then the slope of the graph gives, from the graph , , , . slope, r2, , MN , , LN r 2, , , , , , +, , , +, O, , , , , , , , , , N, , , , 2, , r, , St oke’s law holds good only when, (i) the size of the sphere is greater than the space, between the molecules of the liquid, (ii) the liquid is of infinite extent, (iii) the velocity of the sphere is less than the, critical, velocity, in, t he, liquid. , The velocity of free fall is affected by the, nearness of the walls of the vessel. Therefore the, sphere should travel in the middle of the vessel., If ' ' is the velocity in a vessel of radius ‘R’ and, '1 ' is the velocity in a vessel of infinite radius, then, r, , 1 1 2.4 .This is called Ladenburg correction., R, , , Applying, Ladenburg, 2, g r2, 9, r, , 1 2.4 , R, , NARAYANAGROUP, , the average time t 2, Average distance travelled by t he ball, d , , +, , L +, , Fill the clean glass jar with viscous liquid, glycerine., Take 5 steel balls according to the increasing radii, and find the diameters of the spherical balls using, screw gauge and find their corresponding radii, Drop the first ball into the jar. The ball initially, accelerates and then falls with terminal velocity., Start the first stop watch when the ball is just crossing, ‘P’., Mark three levels P, Q and R using three rubber, bands wrapping them around the glass jar such that, PQ Q R ., When the ball just reaches the level ' Q ' stop the, first stop watch and immediately start the second, stop watch., When the ball just reaches the level ‘R’ stop the, second stop watch., If ' t1 ' and ' t2 ' are the times taken by the ball to, move from P to Q and Q to R respectively, then, t1 t2, , M +, , , , PROCEDURE, , correction,, , PQ QR, 2, , d, ., t, Repeat the above steps with the remaining four balls, also., , Then the terminal velocity , , Then coefficient of viscosity of the fluid can, be det ermined using t he formula, 2, g r2, 9, r, , v 1 2.4 , R, , Plotting a graph between terminal velocity , , , , 2, along y-axis and square of the radius r along, , x-axis, we get a straight line of which the slope is, , , r2, , EXERCISE - 7, 1), , Terminal velocity is, 1) the velocity of flow upto which the flow is, streamlined, 2) the velocity at the bottom of the container, 3) the constant velocity of fall of a body through a, viscous liquid, 4) equal to escape velocity of a satellite, 91
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 2), , Viscosity is a transport phenomenon explained, using the concept of transfer of, 1) mass, 2) kinetic energy, 3) potential energy, 4) momentum, 3) If temperature increases viscosity of liquids, and gases respectively, 1) increases & decreases 2) decreases & increases, 3) increases & increases4) decreases & decreases, 4) The viscous drag on a liquid layer does not, depend on, 1) nature of the liquid 2) velocity gradient, 3) velocity, 4) area, 5) Stoke’s theorem is applicable only to, 1) non-viscous liquids 2) viscous liquids, 3) solutions, 4) pure metals, 6) A steel ball falls slowly through water than, through air because, 1) there is no surface tension 2) density of air is low, 3) upthrust of air is low 4) viscosity of air is low, 7) A liquid is flowing uniformly. The net, external force causing the liquid to flow is, 1) equal to viscous force 2) more than viscous force, 3) less than viscous force, 4) not related to viscous force, 8) Two steel balls of radii R1 and R2 R1 R2 , are dropped through a tube full of glycerine., Their terminal velocities are v1 and v2 in the, experiment, then, 1) v1 v2 2) v1 v2 3) v1 v2, 4) v1 and v2 are independent of R1 and R2, 9) A liquid flows through a horizontal tube of, variable diameter. Then the pressure is lowest, where, 1) velocity is lowest, 2) velocity is highest, 3) diameter is largest, 4) both velocity and diameter are largest, 10) In laminar flow of a fluid, the velocity of the, fluid in contact with the walls of the tube is, 1) maximum, 2) between ‘0’ and maximum, 3) equal to critical velocity 4) zero, 11) A lead shot of diameter 1mm falls through a, long column of glycerine. The variation of the, velocity ‘v’ with distance covered (s) is, represented by, 1), , 2), , v, , 12) A metallic sphere of mass M falls through glycerine with a terminal velocity ‘v’. When another sphere of the same metal having mass, ‘8M’ is dropped into the same column of, glycerine, the terminal velocity of the ball will, be, 1) 2V, 2) 4V, 3) 6V, 4) 8V, 13) A lead sphere is dropped into a medium. As, the sphere falls, the velocity of lead sphere, 1) remains constant throughout, 2) decreases and finally becomes zero, 3) decreases for some time and then becomes constant, 4) increases for some time and then decreases, 14) Coefficient of viscosity of a gas vary with, temperature (T) as, , 1, 1, , 2 4), T, T, 15) A spherical metal ball of mass ' m ' and, radius r is falling through a viscous, 1) T 2) T 2 3) , , medium. The value of its terminal velocity is, proportional to, 1), , 1, r, , 2), , m, r, , Density of oil is 1.5 103 Kg / m3 , density of, copper is 8.9 103 Kg / m3 ., 1) 9.9 101 kg m 1 s 1 2) 9.0 101 kg m 1 s 1, 3) 8.0 101 kg m 1 s 1 4) 8.5 101 kg m 1 s 1, , EXERCISE - 7 - KEY, 1) 3 2) 4 3) 2 4) 1 5) 2 6) 4 7) 1, 8) 2 9) 2 10) 4 11) 3 12) 2 13) 3 14) 1, 15) 2 16) 1, , EXERCISE - 7 - HINTS, 12) V r 2, , v, , 3), , S, , 2 2 10, , 9, , 4), , v, , O, , 92, , O, , S, , v, , S, , 4) m, , 16) The terminal velocity of a copper ball of radius, 2 mm falling through a tank of oil at 200 C is, 6.5 cm/s. Find the viscosity of the oil at 200 C ., , 16) Coefficient of viscosity , O, , m, r, , 3), , O, , S, , , , 3 2, , , , 2, 2 r g , 9, t, , 9.8 8.9 103 1.5 103 , 6.5 102, , 2 10 6 9.8 7.4 103, , 9.9 10 1 kg m 1 s 1, 2, 9, 6.5 10, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXPERIMENT - 8, ms, , To plot a cooling curve for the relationship, between the temperature of a hot body and time., Newton’s law of cooling says that whenever a hot, body looses heat by radiation, the rate of loss of, heat is directly proportional to the excess, temperature of the body over that of its, surroundings provided, the difference of, temperature is not too large (not more than 300 C ), , , , Thermometers, T2, , (or), , d, k, K 0 where K , dt, ms, , (or), , K dt, , d, , 0, , (or) log 0 Kt C, , , where C integration constant, The above equation shows that the graph, between log 0 and time (t) will be a straight, line with slope ‘K’ and intercept ‘C’, , T1 stirrer, lid, , d, d, 0 ms, k 0 , dt, dt, , +, , calorimeter, , , log( – 0), , +, +, , +, O Time(t) , , Doublewalled, vessel, , PROCEDURE, , , , , , If, , dQ, is the rate of loss of heat, ' ' the, dt, , , , temperature of the body at any instant and ' 0 ', the temperature of the surroundings, then, , dQ, dQ, 0 (or), k 0 , dt, dt, , , , Hence negative sign indicates that as the time, increases, 0 decreases., , , , , If m mass of the body, s=specific heat and, , d, the rate of fall of temperature, then the rate, dt, of cooling, , dQ, d, ms, dt, dt, , NARAYANAGROUP, , , , Check the two thermometers T1 and T2 by, measuring the temperature of the same sample of, water contained in the beaker. Let ' x ' and ' y ', be their readings., Taking thermometer T1 t o be correct,, correction is applied to the thermometer T2, Take out the copper calorimeter from the double walled vessel and fill about 2/3 of its volume with, warm water having temperature of about 800C. Put, it back into the enclosure and cover it with a lid., Insert the thermometer T1 and stirrer in the, calorimeter and thermometer T2 in water, contained in the double - walled vessel., Note the temperature of thermometer T2 when the, difference of readings on t he t wo, thermometers is about 300C., Note the reading on thermometer T1 first after every, 1 minute for 10 minutes, then after every 2 minutes, and finally after every 5 minute intervals by, continuously stirring the water gently with the stirrer., This should be continued until the temperature of, hot water comes to about 50C higher than that of, the enclosure., 93
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , Note the final temperature of the enclosure after 5), the final reading on T1 is taken, similarly note the, final temperature of the enclosure after the final, reading on T2 is taken., , , , Find the excess of temperature 0 , , , , , Plot a graph between log 0 and time (t), The temperature falls quickly in the beginning and, then slowly, as the difference of temperature goes, on decreasing. This verifies Newton’s law of, cooling, , EXERCISE - 8 - KEY, 1) 4, , EXERCISE - 8, 1), , axis and temperature along Y-axis, it, appears as, , , , t, , , , t, , , , , , t, , 2), , 3), , 4), , , , t, , As difference of temperature of body and , surroundings increases, rate of cooling, 1) increases, 2) decreases, , 3) does not depend on it4) increases or decreases, In the experiment to study relationship , between temperature of the body and time if a, graph is plotted between log 0 taking it, along y-axis and time (t) taking along x-along,, , it is, 1) exponential, 2) straight line, , 3) parabola, 4) hyperbola, In the above experiment, it is given that, at time t=5min, temperature of water, , 0, C 61 and temperature of water in, enclosure 0 0 C 30 . At t=8 min, if, , 0 0 C 30 ,, , then, , 0 C , , will, , be, , 1) 600 C 2) 610 C 3) 610 C 4) cant predict, 94, , 2) 1, , 3) 2, , 4) 2, , 5) 1, , EXPERIMENT - 9, , If a cooling curve is drawn taking t(s) along X-, , , , While drawing cooling curve between the, temperature of hot water and time we should, stir the water uniformly, this has been done to, ensure that, 1) temperature of water in the calorimeter is same, at all placed, 2) cooling will occur fast to save the time of, experiments, 3) We can stir water non-uniformly also, 4) none of these, , Find the speed of sound at room temperature using, a resonance tube, The apparatus consists of a vertical uniform, cylindrical metal tube of length 1m and about 4 to, 5cm diameter., A uniform transparent plastic pipe is connected to, the metal tube parallel to it., The lower end of the metal tube is drawn into a, nozzle and it is connected to a water reservoir by, means of a uniform rubber pipe., The whole apparatus is fixed to a vertical wooden, stand and a meter scale is fixed adjacent to the metal, tube parallel to it., The water reservoir can be fixed at any desired, position on a vertical rod., The metal tube and plastic tube are filled with water, and the reservoir is partiallty filled with water., Level of water in the metal tube and plastic tube is, equal., A vibrating tuning fork of known frequency (f) is, held at the mouth of ‘A’ when the length of air, column in ‘A’ is practically zero. No sound is heard, from the tube., Now lower the reservoir to increase the length of, air column., As the length of air column increases, intensity of, sound heard from the tube increases gradually and, reaches a maximum then dies., The length of the air column corresponding to, maximum intensity of sound is measured as ' l1 ' ., The loud sound is heard because the natural, frequency of the air column of length ' l1 ' matches, with the frequency of the tuning fork and hence, resonance occurs. This is the first position of, resonance., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , Transparent, plasticpipe, A, , B, , EXERCISE - 9, 1), , Water, reservoir, , Meter scale, , metal, tube, , Rubber, tube, , 2), , Wooden, stand, , , , In this position, one node and one antinode are 3), formed in the tube considering end correction,, AN, , l1 e , , , ...... 1, 4, , , 4, , l1, , 4), , N, , , , When we lower the reservoir further, at another 5), length of air column again intensed sound is heard., This is the position of second resonance., Considering end correction,, AN, , 6), N, l, 3, l2 e , ...... 2 , 4, , 2, , 3, 4, , AN, , N, , 7), , , From (1) and (2), l2 l1 2 l2 l1 , 2, We know, velocity of sound V f , V 2 f l2 l1 , , , End correction in this case can be determined using, the formula e , , , , l2 3l1, 2, , 8), , Speed of sound at 00C in air can be determined by, , , , , the formula V0 Vt 1 , , t , 546 , , Where t=room temperature in 0C, NARAYANAGROUP, , In a resonance tube, we get, 1) stationary longitudinal wave, 2) stationary transverse wave, 3) progressive longitudinal wave, 4) progressive transverse wave, A tuning fork of frequency 500 Hz is sounded, and resonance is obtained at 17cm and 52cm, of air column respectively in a resonating air, column apparatus. The velocity of sound in, air is, 1) 650ms 1 2) 350ms 1 3) 700ms 1 4) 190ms 1, In a resonance apparatus, the first and, second resonating lengths of air column are, 15cm and 48cm respectively. The end, correction for this appratus is, 1) 6cm, 2) 3 cm, 3) 1.5 cm 4) 2 cm, When a stationary wave is formed, its, frequency is, 1) same as that of the individual waves, 2) twice that of individual wave, 3) half as that of an individual wave, 4) four times as that of an individual wave, The apparatus used to find the speed of sound, in a gas is, 1) Melde’s apparatus, 2)Quinke’s tube apparatus, 3) Kundt’s tube apparatus, 4) Newton’s apparatus, The change in speed of sound in a gas is, independent of change of, 1) temperature of the gas, 2) pressure of the gas, 3) density of the gas, 4) both pressure and temperature, The correct statement out of the following is, 1) both sound and light waves in air are, transverse, 2) both sound and light waves in air are, longitudinal, 3) sound waves in air are transverse and light waves, in air are longitudinal, 4) sound waves in air are longitudinal and light waves, in air are transverse, A tuning fork is used as a source of, standard frequency, because, 1) it is U - shaped body, 2) it is made of a metal, 3) it has two symmetrical prongs, 4) it retains its frequency despite small changes in, temperature, 95
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EXPERIMENTAL PHYSICS, , JEE-ADV PHYSICS- VOL- VI, , If air is replaced by hydrogen in an organ pipe 14) An open pipe is in resonance in its fundamental, mode. Air, hydrogen and ethane are filled in, then, succession in this pipe. The speed of sound is, 1) the fundamental frequency decreases, different in these three media on account of, 2) the fundamental frequency increases, which, 3) the fundamental frequency remains same, 1) only the wavelength changes, 4) stationary waves are not formed, 2) both frequency and wavelength change, 10) With increase in temperature the frequency, 3) only the frequency changes, of sound in an organ pipe, 4) neither frequency nor wavelength changes, 1) decreases, 2) increases, 15) An air column in a pipe closed at one end is in, 3) remains same, 4) becomes zero, resonance with a vibrating tuning fork of, frequency 264Hz. If speed of sound in air is, 11) When a closed organ pipe of length ' l ' if the, 330ms-1, length of air column will be (in cms), velocity of sound is ‘v’ then the, 1) 31.25 2) 62.5, 3) 93.75 4) 125, fundamental frequency will be, 16) Two closed organ pipes give 10 beats between, V, the fundamentals when sounded together. If, and only even harmonics are present, 1), the length of the shorter pipe is 1m then the, 4l, length of the longer pipe will be (speed of sound, V, in air is 340ms 1 ), 2), and only odd harmonics are present, 1) 2.87 m 2) 0.87 m 3) 1.13 m 4) 2.13 m, 2l, 17) Stationary waves are setup in an air column., V, Velocity of sound in air is 330ms -1 and, and even as well as odd harmonics are present, 3), frequency is 165 Hz. The distance between, 2l, two successive nodes is, V, 1) 2m, 2) 1m, 3) 0.5 m 4) 4m, 4), and only odd harmonics are present, 18) An open pipe of length ' l ' vibrates in, 4l, fundamental mode. The pressure variation is, 12) There are two organ pipes of exactly the same, maximum at, length and material but of different radii. The, frequencies of their fundamental notes are, l, 1) from the ends 2) the middle of the pipe, such that, 4, 1) wider pipe has lower frequency, l, 2) narrow pipe has lower frequency, 3) the end of the pipe 4) from its ends, 8, 3) both will have the same frequency, 19), A, glass, tube, 1.5m, long, and, open at both ends,, 4) both will have infinite frequency, is, immersed, vertically, in, a water tank, 13) In an open end organ pipe of length ' l ' if the, completely. A tuning fork of frequency 660 Hz, speed of sound is ‘V” then the fundamental, is vibrated. It is kept at the upper end of the, frequency will be, tube and the tube is gradually raised out of, water. The total number of resonances heard, V, 1), and both odd as well as even harmonics, before the tube totally comes out of water is, 2l, (velocity of sound in air 330ms 1 ), are present, 1) 12, 2) 6, 3) 8, 4) 4, V, 20) A vertical tube is made to stand in water so, 2), & both odd as well as even harmonics are present, that the water level can be adjusted. Sound, 4l, waves of frequently 320Hz are sent into the, top of the tube. If standing waves are produced, V, 3), and only even harmonics are present, at two successive water levels of 20 cm and 73, 2l, cm, what is the speed of sound waves in the, air in the tube (in m/s), V, 4), and only even harmonics are present, 1) 339, 2) 332, 3) 334, 4) 336, 9), , 4l, , 96, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXERCISE - 9 - KEY, 1) 1 2) 2, 8) 4 9) 2, 15) 1&3, , 3) 3 4) 1 5) 3 6) 2 7) 4, 10) 2 11) 4 12) 1 13) 1 14) 3, 16) 3 17) 2 18) 2 19) 2 20) 1, , EXPERIMENTAL PHYSICS, , , , , The principle of method of mixtures is that the heat, lost by the hot body is equal to the heat gained by, the cold body if there is no loss of heat to the, surroundings., When a hot solid is dropped into the calorimeter, containing water, then heat lost by solid is equal to, the heat gained by calorimeter and water., A clean, dry and empty calorimeter with stirrer is, , , , taken and its mass W1 is determined with a common, balance., It is filled with water to one - third of its volume and, , , , EXERCISE - 9 - HINTS, 2., , V 2 f l2 l1 , , 3., , e, , l2 3l1, 2, , V, V, l , 15) n1 , ;, 4l, 4n1, 16), , n2 3n1, , , v1 1 , 10 ; Here l1 1m ; v 340ms 1, 4 l1 l2 , , v, 17) distance , 2 2n, v, ; l2 3l1 ; l3 5l1......., 19) l1 , 4n, , contents t10C is noted with a thermometer.., , , , , , , , V 2 320 73 20 ; V = 339 m/s., , EXPERIMENT - 10 ‘A’ & ‘B’, , , , , , , , To determine the specific heat of the given solid, and a liquid by method of mixtures, Calorimeter is used to determine specific heat, capacity of a solid. Calorimeter (C) is a, cylindrical copper vessel provided with a copper, stirrer (S) and is kept in a wooden box containing, cotton to prevent the loss or gain of heat., , To prevent heat loss due to radiation, the, calorimeter is brightly polished., A thermometer (T) is inserted in the calorimeter to, measure the temperature of its contents., When bodies at different temperatures are in, thermal contact with each other, then heat flows, from hot body to the cold body until they attain a, common resultant temperature., NARAYANAGROUP, , The hot solid is now dropped into t he, calorimeter quickly and by stirring the system, the, find resultant temperature t30C is noted., , Length of water level l 20 cm, Velocity of sound in air V 2n l2 l1 , , The solid of which the specific heat capacity is to, be determined is taken in the form of small pieces, and is heated in a steam heater to a steady, temperature t20C, , 20) Frequency n 320 Hz, Second length of water level l2 73 cm, , its mass W2 is determined., Keeping the calorimeter in the wooden case and, initial temperature of the calorimeter and its, , The mass of calorimeter with water and solid W3, is determined using common balance., According to the law of method of mixtures, heat, lost by the solid = heat gained by calorimeter +, heat gained by water, , W3 W2 S t2 t3 W1s t3 t1 , W2 W1 t3 t1 S2, specific heat of the solid, W1s W2 W1 S2 t3 t1 , S, cal / g 0C, W3 W2 t2 t3 , , Specifc heat capacity of a liquid also can be, determined in the same method replacing water by, the liquid. Then, Heat lost by the solid = Heat gained by, calorimeter + Heat gained by the liquid, , W3 W2 S t2 t3 , W1 s t3 t1 W2 W1 S1 t3 t1 , 97
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, where S1 specific heat capacity of the liquid, S1 , , W3 W2 S t2 t3 W1s t3 t1 cal / g / 0 C, W2 W1 t3 t1 , Thermometer, Stirrer, , 6), , 7), , 8), , Wooden, lid, felt, wateror, liquid, , 9), , If temperature scale is changed from 0C to 0F,, the numerical value of specific heat, 1) decreases, 2) increases, 3) remains constant, 4) becomes infinity, Which of the following has maximum specific, heat ?, 1) water 2) alcohol 3) glycerine 4) oil, 100g ice at 0C is mixed with 100g water at, 1000C. The resultant temperature of the, mixture is, 1) 100C 2) 200C, 3) 300C 4) 400C, A liquid of mass ‘M’ and specific heat ‘S’ is at, a temperature ‘2T’. Another liquid of thermal, capacity 1.5 times the first liquid at a, , calorimeter, , temperature, , T, is added to it. The resultant, 3, , temperature of the mixture will be, , EXERCISE - 10 - A & B, 1), , 2), , 3), , 4), , 5), , 98, , 1), , 2T, 3, , 2), , T, 2, , 3) T, , 4), , 4T, 3, , Heat capacity of a body depends on, 10) A liquid ‘A’ of specific heat 0.5 at 600 C is, 1) the quantity of heat energy supplied to it, 2) rise in temperature of the body, mixed with another liquid ‘B’ of specific heat, 3) the mass of the body, 0.3 at 200 C . After mixing, the resultant, 4) the material of the body, temperature of the mixture is 300 C . The ratio, If heat energy is supplied to a body, its, of masses of A and B respectively is, temperature, 1) 1:2, 2) 1:3, 3) 1:5, 4) 2:3, 1) must increase, 11) An aluminium vessel of mass 0.5kg contains, 2) may increase, 3) may remain constant, 0.2kg of water at 200 C . A block of iron of, 4) must decrease, mass 0.2kg at 1000 C is gently put into the, Calorimeter usually is made of copper, water. The equilibrium temperature of the, because, mixture is found to be 250 C . The specific heat, 1) it is cheaper and easily available, capacity of iron would be, 2) it does not get rusted, 3) its emissive power is more, Saluminium 910JKg 1K 1 , Swater 4200JKg 1K 1 ., 4) its thermal conductivity is high, The temperature of the furnace would be, The direction of flow of heat between two, 1) 1000 C 2) 3540 C 3) 953.60 C 4) 893.60 C, bodies is determined by, 1) kinetic energy, 12) An electric heater of power 150 W is immersed, 2) internal energy, in 0.75 Kg of ice at 00C in a lagged container, 3) total energy, of negligible heat capacity. The temperature, 4) the temperature difference between the, remains constant for 27.5 minutes and then, bodies, rises to 40.00C in further 14 minutes. Calculate, Specific heat of a gas in an isothermal process, the specific heat capacity of water., is, 1) 4.2 103 J Kg 1 K 1 2) 4.7 103 J Kg 1 K 1, 1) infinite, 2) zero, 3) 4.0 103 J Kg 1 K 1 4) 3.0 103 J Kg 1 K 1, 3) negative 4) either zero or one, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , PROCEDURE, , EXERCISE - 10-KEY, 1) 3&4 2) 2&3, 6) 1 7) 1 8) 1 9) 3, , 3) 4, 10)3, , 4) 4 5) 1, 11) 2 12) 1, , , , , EXERCISE - 10 - HINTS, 8&9) Heat lost = Heat gained, , , , 10) m1 s1 t1 m2 s2 t2, , 11) Apply the principle of method of mixture, 12) Let ‘S’ be the specific heat capacity of water heat, taken by water at 00C to rise it to 400C., ms t 0.75 S 40 0 30 s J kg, , Heat given by the heater in 14 minutes p t, , , , 150 14 60 126000 J acc to principle, method of mixtures heat taken by water = heat , supplied by the water 30 s 12600, , s, , Arrange the apparatus according to the circuit as, shown., Connect the resistance wire whose resistance is to, be determined in the right gap of the meter bridge., Connect a known resistance (resistance box) in the, left gap of the meter bridge., Taking out some resistance (say 2 ) in the, resistance box making use of the jockey, the, balancing length ' l1 ' is to be determined, when the, deflection in the galvanometer is zero, For different values of resistances in the, resistance box the corresponding balancing lengths, are to be noted., In each case, the unknown resistance (X) can be, determined by using the formula, , 126000, 4200 J Kg 1 K 1, 30, , , , or 4.2 103 J Kg 1 K 1 ., , Calculating the length ' l ' and radius of crosssection (r) of the wire of unknown resistance, the, resistivity of its material can be determined by using, , EXPERIMENT - 11, To find the resistance of given wire using meter, bridge, (), Key, , –, , +, , the formula , , , Resistance Box, , RB, , 1), , X, , J, A, , l1, , l2, , B, , HR G, , , , , The practical form of Wheatstone bridge is meter, bridge., When the meter bridge is balanced, then the, galvanometer shows zero deflection. Then, R, l, l1, 1 , X, l 2 100 l1, , X r 2 , l, , Using meter bridge, the laws of resistances when, connected in series and in parallel can also be, verified, , 2), , In a meter bridge apparatus, the bridge wire, should posses, 1) high resistivety and low temperature coefficient, 2) high resistivety and high temperature coefficient, 3) low resistivety and high temperature coefficient, 4) low resistivety and low temperature coefficient, In post office box, the graph of galvonameter, deflection versus resistance R (pulled out of, resistance box) for the ration 100 : 1 is given, as shown (due to unsuitable values of R,, galvonometer shows deflection). The two, consecutive values of R are shown in the figure., , .Where X is unknown resistance, , R is known resistance in the resistance box and l1, is balancing length on the meter bridge wire., , Deflection, , 5, 326, 320, –3, , NARAYANAGROUP, , ohm m, , EXERCISE - 11, , Unknown resistance, , R, , R l1, , X l2, , R(), , 99
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, The value of unknown resistance would be, 7), 1) 3.2 2) 3.24 3) 3.206 4) 3.2375, 3) Consider the meter bridge shown in the figure, below, 9, , x, , G, 10cm, , The resistance X has temperature coefficient, , 1 and form resistance box 9 shown has, 2 . For shown situation balance point is at, 10cm from left end, if temperature of system, increases by T due to joule heating, then the, shift in the balance point is (Assume that only, the resistance of X and resistance box changes, due to change in temperature and there is no, other effect)., 1) 9 1 2 T, , 2) 9 1 2 T, , 1, 1, 1 2 T, 4) 1 2 T, 9, 9, In meter brdge experiment the observation, table and circut diagram are shown in figure, , 3), 4), , S NO, , R , , l(cm), , 1, 2, 3, 4, , 1000, 100, 10, 1, , 60 cm, 13 cm, 1.5 cm, 1.0 cm, , Two resistances are connected in two gaps, of slide wire bridge. The balance point is, at 40cm from left end. A resistance X is, connected in series with smaller resistance, R and balance point shifts to 40 cm from, right end. What is the value of X if R, is 4 ?, 1) 4, 2) 5, 3) 6, 4), 7, 8) In meter bridge for measurement of, resistance, the known and the resistances are, interchanges. The error so removed is, 1) end correction, 2) end correction, 3) index error 4) due to temperature effect, 9) Statement-I : In a meter bridge experi ment ,, null point for an unknown resistance is measured., Now the unknown resistance is put inside an, enclosure maintained at a higher temperature. The, null point can be obtained at the same point as, before by decreasing the value of the standard, resistance., Statement-II : Resistance of metal increases with, increase in temperature., 1) Statement-I is true, Statement-II is false, 2) Statement-I is true, Statement-II is true,, Statement-II is a correct explanation for statementI, 3) Statement-I is false, Statement-II is true,, Statement-II is not a correct explanation for, Statement-I, 4) Statement-I is false, Statement-II is true, 10) If each of the resistances in the network in, figure R, the equivalent resistance between, terminals A and B is, Q, , l, R, R, , , , R, , R O, A, , 5), , 6), , 100, , B, Which is the readings is not taken correctly ?, P, S, R, 1) 1, 2) 2, 3) 3, 4) 4, In above question the value of unknown, 1) 5 R, 2) 2 R, 3) 4 R, 4) R, resistance is, 11) A, B, C and D are four resistnaces of 2 ,, 1) 664 2) 100 3) 348 4) 864, 2 , 2 and 3 respectively. They are used, In the measurement of resistance by a meter, to form a wheatstone bridge. The resistnace, bridge, the known and the unknown, ‘D’ is short circuited with a resistnace ‘R’ in, resistances are interchanged to eliminate, order to get the bridge balanced. The value of, 1) end error, 2) index error, ‘R’ will be, 3) error due to thermo-electric effect, 4) random error, (1) 4 (2) 6 , (3) 8 , (4) 3, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXERCISE - 11-KEY, 1) 4, 6) 1, 11) 2, , 2) 2, 7) 2, , 3) 1, 8) 1, , 4) 4, 9) 4, , , , We require a battery, ammeter, voltmeter,, rheostat, one-way key, resistance wire (R) and, connecting wires to do this experiment., , , , Ohm’s law states that the electric current i , , 5) 1, 10) 4, , EXERCISE - 11 - HINTS, 7), , through a conductor is directly proportional to the, potential difference (V) across the two ends of the, conductor provided the physical, conditions, like temperature, pressure etc., across the, conductor remain constant., , If R and S (R<S) are two resistances in two, gaps and balance point is on 40cm from, left end, R, l, 40 2, , , , .......(i), S 100 l 60 3, Now X is in series with R, l | 60cm, , R X, l|, 60 3, , , , |, S, 100 l 40 2, Divide (i) by (ii), we get, , 1, i.e., i V or i V, R, , V iR or R , , ------(ii), , Here ‘R’ is electric resistance of the, conductor., ‘R’ depends on the nature of the conductor, its, dimensions and physical conditions. ‘R’ is, independent of ‘V’ and ' i ' ., , R, 2 2, , R X 3 3, , or, , 4, 4, , 4 X 9, , PROCEDURE, , , , R 4 given , , or 16 4 X 36 or X 5, 10) When cell is connected to points A and B no current , will flow in the resistance of arm PQ. Effective, resistance of arm APS will be in parallel to the total, resistnace of arm AQS., , 2R 2R, R, Equivalent resistnace , 2 R 2R, , 11) The bridge will be balanced when the shunted, , 3s, resistance of value 2 i.e. 2 , 3 s, s6 , , , EXPERIMENT - 12, To determine the resistance of a given wire using , Ohm’s Law., , A, , V, i, , –, , +, , Rh, R, , key, (), , , , Connect the circuit as shown., Clean the ends of the connecting wires with sand, paper and make the connections tight., See that the positive terminals of voltmeter and, ammeter are joined to the positive terminal of the, battery., Determine the least count of ammeter and, voltmeter and note the zero error if any., Adjust the sliding contact of rheostat such that small, current passes through the resistance wire., Change the rheostat contact slowly, so that both, ammeter and voltmeter show full division, readings and not in fraction., Note the readings of ammeter and voltmeter in each, case., Taking a minimum of six sets of readings plot a graph, between potential difference ‘V’ (taken along xaxis) and current ' i ' (taken along y-axis). It is a, straight line passing through the origin., We can calculate the resistance (R) of the given, wire by finding the slope and also using the, formula R , , V, ., i, , V, , NARAYANAGROUP, , 101
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , i 1, Slope of V i graph is G where G is, V R, , 7), , electric conductance of the wire., , B, , i, O, , , 8), , A, , , C, V, , The rheostat used in this experiment should be of, low resistance., , 9), , 10), , EXERCISE - 12, 1), , 2), , 3), , 4), , 5), , 6), , 102, , Out of the following, the one which is not the, active material in a ‘lead storage cell’ is, 1) lead peroxide, 2) lead sulphate, 3) sponge lead, 4) sulphuric acid, The current capacity of the charged secondary, cell does not depend on, 1) rate of discharge, 2) temperature, 3) amount of active material 4) rate of charging, The correct method of connecting circuit in, Ohm’s law experiment is, 1) voltmeter in series and ammeter in parallel, 2) both voltmeter and ammeter are in series, 3) both voltmeter and ammeter are in parallel, 4) voltmeter in parallel and ammeter in series, Two identical batteries are connected such that, their positive terminals are together and, negative terminals are together. Then, 1) the emf of the combination is zero, 2) potential difference across each cell is zero, 3) current in the circuit is zero, 4) resistance in the circuit is zero, The material of wire chosen for rheostat is, 1) copper, 2) aluminium, 3) constantan, 4) lead, A rheostat is used in an electric circuit, 1) to change the resistance of the circuit, 2) to change the potential difference, 3) to change emf, 4) to change the current through a particular, instrument, , 11), , 12), , 13), , 14), , 15), , Resistance of conductor depends on, 1) applied potential difference across the, conductor and electric current passing through the, conductor, 2) length and area of cross-section of the, conductor, 3) temperature of the conductor, 4) both 2 and 3, 1 kg piece of copper is drawn into a wire 1 mm, thick and another piece into a wire 2 mm thick., Compare the resistances of these wires, 1) 2 : 1, 2) 4 : 1, 3) 8 : 1, 4) 16 : 1, The resistance of dry human body is, measured by, 1) Ohm’s law apparatus 2) slide wiere bridge, 3) AVO meter, 4) meter bridge, Inside a resistance box various resistance coils, are connected, 1) in series, 2) in parallel, 3) some in series and some in parallel, 4) all in parallel except one., By increasing the temperature, specific, resistance of a conductor, 1) increases, 2) decreases, 3) remains same, 4) becomes zero, If the positions of voltmeter and ammeter are, interchanged in Ohm’s law circuit, then, 1) both the instruments will be damaged for flow, of large current, 2) No effect on the readings of both the instruments, 3) both will show the reading out of scale, 4) no instrument will be harmed due to the flow of, very small current, By increasing the temperature, specific, resistance of a semiconductor, 1) increases, 2) decreases, 3) remains same, 4) becomes zero, If the resistivity of an aloy is ' and that of, constituent metals is ' ' then, 1) ' 2) ' , 3) ' , 4) There is no relation between and ', V i graphs of anichrome wire at three, different temepratures t1 , t2 and t3 are shown., from the graph,, t, 1) t1 t2 t3, 3, , 2) t1 t2 t3, , t2, t1, , 3) t1 t2 t3, 4) t1 t2 t3, , O, , V, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 16) The resistance of an incadescent lamp is, 22) A sser tion (A ): In Ohm’s law experiment, 1) greater when switched off, potential drop across a resistance wire was, 2) greater when switched on, measured as 5.0 V and current was measured, 3) zero when switched on, as 2.00A. The maximum permissible error in, 4) infinity when suitched off, resistance is 25%, 17) The figure shows the variation of ‘V’ with ‘i’ at, Reason(R) : In experiment of assertion the least, count of ammeter is 0.01A., temperatures T1 and T2 . Then T1 T2 is, 1) If both assertion and reason are true and t the, proportional to, reason is the correct explanation of the assertion., T, 2) If both assertion and reason are true but reason, is not the correct explanation of the assertion., 3) If assertion is true but reason is false., T, 2, 4) If assertion is false but reason is true., V, , 23) In Measuring the values of resistance,, 2, different students connect ammeter and, O, voltmeter as shown in following figures. Which, of these will be correct ?, 1) tan 2 2) tan 3) sin , 4) cos 2 , 1, , 2, , 18) The resistance of a wire of length ' l ' and, diameter ' d ' is ' R ' . The wire is stretched to, double its length. the resistance of the wire, will now be, , R, 1), 2, , K, , (), , v, , A, K, , R, , (), , v, 2) 2R, , 3) 4R, , NARAYANAGROUP, , R, , 4) 16R, , 19) What is immaterial for an electric fuse?, 1) its specific resistance, 2) its radius, 3) its length, 4) current flowing through it, 20) The conductivity of a super conductor is, 1) infinity 2) large 3) very small 4) zero, 21) Assertion (A): In Ohm’s law experiment, the, reading of voltmenter and ammeter are 13.5V and, 0.40A respectively, then the measured computed, value of R is 33.75 upto correct SD’ss, Reason(R) : The reliability in computed value, can not be more than the reliability in measured, values., 8), 1) If both assertion and reason are true and the, reason is the correct explanation of the assertion., 2) If both assertion and reason are true but, reason is not the correct explanation of the, assertion., 3) If assertion is true but reason is false., 4) If assertion is false but reason is true., , A, , 1) Both figures are correct, 2) Both are wrong, 3) Only figure(i) is correct, 4) Only figure (ii) is correct, , EXERCISE - 12 - KEY, 1) 2, 6) 4, 11) 1, 16) 2, 21) 4, , 2) 2, 7) 4, 12) 4, 17) 2, 22) 2, , 3) 4, 8) 4, 13) 2, 18) 3, 23) 1, , 4) 1&3 5) 3, 9) 3 10) 1, 14), 15) 2, 19) 3 20) 1, , EXERCISE - 12 - HINTS, R1 l1 D22, , R2 l2 D12, 1, , 10 3 , 4, , 2, , l1d , , 2 10 3 , 4, , 2, , l2 d, , R1, 2, l1 4 l2 ; R 4 2 16, 2, 103
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 14) The resistivity of a metal increases when it is , converted into an alloy., 15) Slope =, , i 1, = conductance (G), V R, , 18) When a wire is stretched ‘n’ times its resistance, becomes n2 times., , EXPERIMENT - 13 A, , To compare the emfs of two given primary cells , using potentiometer, Rh, , (), , +, A, –, , Q, , 400cm, 300cm, 200cm, , Key, , –, , P 0cm, +, , E, , , , , , J, , E1, + –, E2, + –, , , , 100cm, G, , RB, , , , We require two cells (Leclanche cell, Daniel cell),, an ammeter, a galvanometer, a battery, a , rheostat of low resistance, Resistance box, a twoway key, a jockey, a potentiometer and connecting, wires to perform this experiment., , In a potentiometer, the potential gradient across the, , E, , constant, wire remains constant. ie.,, ., , l, , , , To compare emfs of two given cells, we use the, , E1 l1, , formula, E 2 l2, , Taking out a 2000 resistance in the resistance, box, the wire is pressed with the jockey at zero, end P then the galvanometer deflects in one, direction. Similarly when pressed at ' Q ' the other, end of the potentiometer wire,, galvanometer, deflects to the other side. If opposite deflections, in galvanometer are observed, the connections, are perfect and correct., Connecting the cell of emf E1 with two - way key,,, move the jockey on the potentiometer wire tapping, it so as to obtain a balancing point where, galvanomter shows zero deflection. This, balancing length ' l1 ' is to be noted., Connecting the two way key to the cell of emf E2 ,, repeating the same experiment, the, balancing, length ' l2 ' is to be noted., The emfs of both the cells are compared using the, , E1 l1, , E 2 l2, , formula, , In this experiment emf of the battery in the, primary circuit should be more than the emf of either, of the two cells., The ammeter reading should remain constant for a, particular set of observation., Balancing length should be obtained carefully by, tapping the wire with jockey without sliding., , EXPERIMENT - 13 B, To determine the internal resistance of a given, primary cell using potentiometer., , Where E1 and E2 are emfs of both the cells. l1, , , , , , , Connect the circuit as shown. In the primary circuit, a battery of known emf E, a plug key, ammeter and, a rheostat are connected., The primary circuit gives the required potential, difference to the potentiometer., , In the secondary circuit, two cells of emfs E1 and, , E2 are connected along - with a two - way key,,, galvanometer and a jockey., 104, , P 0cm, , 300cm, 200cm, , K1, , –, , J, , 100cm, , +, E, , E1, + –, , RB G, ( ), , PROCEDURE, , 400cm, , Q, , (), , and l2 are the corresponding balancing lengths of, potentiometer wire., , Rh, +, A, –, , R, , K2, , To perform this experiment, we require a, potentiometer, a battery, two one - way keys,, rheostat of low resistance, a galvanometer, a, resistance box, an ammeter, a Leclanche cell, a, jockey and connecting wires., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , , EXPERIMENTAL PHYSICS, , The internal resistance of the given primary cell can, be determined by the formula, 1), , l l , r 1 2 R, l2 , where l1 is balancing length without shunt, resistance in the secondary circuit., , 2), , l2 is balancing length with the shunt resistance in, the secondary circuit, , R is shunt resistance connected in parallel to the, primary cell, , , the internal resitance is expressed interms of E and, 3), V, i, , E, Rr, , V iR , , V, E, V, R, E, r, , , ;, ; 1, R Rr, E Rr V, R, , 4), , E V , E V r, ; r R, , V, R, V , , EXERCISE - 13 - A & B, Potentiometer is superior to voltmeter, because, 1) potentiometer has high resistance, 2) potentiometer has low resistance, 3) potentiometer does not draw any current from, the unknown source of emf to be measured, 4) potentiometer has greater size than voltmeter, Potentiometer wire is made of manganin, because it has, 1) high conductivity, 2) negligible melting point, 3) high temperature coefficient of resistance, 4) negligibly small temperature coeficient of, resistance., In a potentiometer experiment the balancing, with a cell is at length 240 cm on shunting the, cell with a resistance of 2 , the balancing, length becomes 120 cm. The internal, resistance of the cell is, 2) 4 , 3) 0.5 4) 1 , 1) 2 , In a potentiometer whose wire resistance is, 10 . The potential fall per cm is V volts is, , , , Repeat the experiment for different values of ‘R’, , V, volt / cm . The resistance that, 4, must be connected in series with the, potentiometer wire is, 1) 40 2) 30 , 3) 20 4) 10 , Sensitivity of a potentiometer, 1) increases with the increase of length of the wire, 2) decreases with the increase of length of the wire, 3) increases with the decrease of length of the wire, 4) does not depend on the length of the wire, Potentiometer is an ideal instrument, because, 1) no current is drawn from the source of, unknown emf, 2) current is drawn from the source of unknown, emf, 3) it gives deflection even at null point, 4) it has variable potential gradient, For a potentiometer to funciton, the emf of the, cell (E) in the primary circuit compared to the, , , , Determine the internal resistance of the primary, , 1, emf of the cell E in the secondary circuit, , l1 l2 , cell using the formula r l R, 2 , , should have a relation, 1) E E1 2) E E1 3) E E1 4) E E1, , reduced to, , E Emf of the cell, V Potential differance across the wire, , PROCEDURE, , , Connecting the circuit as above and opposite, deflections are observed in the galvanometer to, confirm the perfectness of the circuit., , , , Putting on the key K1 and keeping the key K 2, open, applying a resistance 2000 in the, , 5), , 6), , resistance box the balancing point ' l1 ' is obtained, by tapping the jockey at different points on the, potentiometer wire., , , By keeping both the keys K1 and K 2 closed, again, balancing length is found. It is observed that l2 l1, , NARAYANAGROUP, , 7), , , , 105
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 8), , 9), , 10), , 11), , 12), , In a potentiometer, balance point is obtained,, when, 1) the emf of the battery in the primary circuit is, equal to the emf of the experimental cell in the, secondary circuit., 2) the potential difference of the wire between the, +ve end and jockey becomes equal to the emf of, the experimental cell., 3) the potential difference of the wire between the, +ve end and jockey becomes equal to the emf of, the battery connected in the primary circuit., 4) the potential difference across the, potentiometer wire becomes equal to the emf of, the battery., The sensitiveness of a potentiometer wire can, be increased by, 1) decreasing the length of the wire, 2) increasing the emf of the battery in the primary, cirucit, 3) decreasing the potential gradient on its wire, 4) increasing the potential gradient on its weire, If the current in the primary circuit is, decreased, then the balancing length is, obtained at, 1) lower length, 2) higher length, 3) same length, 4) at zero length, In potentiometer experiment, the unknown, potential difference is compared with, 1) unknown resistance 2) known resistance, 3) known standard resistance, 4) internal resistance of the cell, On increasing the resistance of primary, circuit of potentiometer, its potential, gradient, 1) becomes more, 2) becomes less, 3) does not change, 4) becomes inifnite, The specific resistance and area of, crosssection of the potentiometer wire are ' ' and, ' A ' respectively. If a current ' i ' passes, through the wire, its potential gradient will, be, , 15) In a potentiometer of ten wires each of 1m,, the balance point is obtained on the sixth wire., To shift the balance point to eighth wire, we, should, 1) increase the resistance in the primary circuit, 2) decrease the resistance in the secondary circuit, 3) decrease the resistance in series with the cell, whose emf is to be determined, 4) increase the resistance in series with the cell, whose emf is to be dtermined., 16) The pontential gradients on the potentiometer, wire are V1 and V2 with an ideal cell and a, real cell of same emf in the primary circuit,, then, 1) V1 V2 2) V1 V2 3) V1 V2 4) V1 V2, 17) The balancing lengths on a potentiometer wire, are 800cm and 600cm when two cells of emfs, E1 and E2 are connected in the secondary, cirucit first to help each other and next to, oppose each other, then, , E1, , E2, , 1) 1/11, 2) 14/11, 3) 7/1, 4) 4/3, 18) Six identical cells of no internal resistance are, connected in series in the secondary circuit of, a potentiometer and the balancing length is ' l ' ., If two of them are wrongly connected, the, balancing length becomes, 1), , l, 4, , 2), , l, 3, , 3) l, , 4), , 2l, 3, , 19) The balancing length on a potentiometer wire, with a cell of emf 2V and internal resistance, 1 connected in the secondary circuit with no, load is 200cm. If a resistor of 19 is, 13), connected parallel to the cell, the balancing, length, 1) decreases by 10% 2) decreases by 20%, 3) decreases by 5%, 4) increases by 10%, 20) A potentiometer wire of length 10m and, i, iA, resistance 20 is connected in series with, i, 1), 2), 3), 4) iA, another resistance of 80 and a battery of, A, , A, emf 4V. The potential gradient on the wire, 14) At the moment when the potentiometer is, will be (in mV/Cm), balanced,, 1) 0.8, 2) 0.16, 3) 0.2, 4) 0.4, 1) current flows only in the primary circuit, 2) current flows both in primary and secondary, EXERCISE - 13 A & B - KEY, circuits, 1)3, 2)3, 3)1, 4)1, 5)1, 6)1, 7)1, 3) current flows only in the secondary circuit, 8)2, 9)3, 10)2, 11)3, 12), 2, 13)1, 14)1, 4) current does not flow in any circuit (primary or, 15)1 16)2 17)3 18)2 19)3 20)1, secondary), 106, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXERCISE - 13 A & B - HINTS, 3), , PROOF: effective resistance R , , l1 240 cm , l2 120 cm , R 2 , r = ?, , R G S GS, , l l , 240 120 , r 1 2 R , 2 2, 120 , l2 , V R1, , V 10, 4, , 4), , RG GS RS, , G, , RS, RS, , E, , E1 E2 l1, E, l l, , 1 1 2, E1 E2 l2, E2 l1 l2, , 18), , E1 6 E l, , , E2 2 E l2, , , , , , l1 l2 , r, , 19), , R, l2 , , , , 20) V1 : V2 R1 : R2 1: 4 V1 , , Potential gradient , , 1, 4 0.8V, 5, , 0.8 1, Vm 0.08Vm 1, 10, , To determine the resistance of a galvanometer by , half - deflection method, +, , –, , (), , The figure of merit K R G , , , where E = emf of the cell, and deflection prodced with resistance R, The maximum current that passes through the, galvanometer is ig nK , where n total number, of divisions on the galvanometer scale on either side, of zero, The required shunt resistance for conservaiton, G, G, ig G S 1 , ; S1 , 1, i, n 1, S , 1, i ig ;, ig, i, where n i ,, , EXPERIMENT - 14, , K1, , 1 , , ig n i , , , g, The length of the shunt wire required for, , r S, l, 2, , conservation is, , 1, , , , where r radius of cross-section of wire, specific resistance of the material of the wire, , R, , G, , PROCEDURE, , (), , S, , +, , –, , (), , K2, , K, , R, , , , , G, , , , RG RS GS, G R S RS, , R1 40 , , 17), , , , GS, GS, , To perform this experiment, we require a , galvanometer, a voltmeter, a battery of two cells,, two resistance boxes, two one - way keys and, connecting wires, If R = resistance connected in series to, galvanometer and S = shunt resistance, then the , resistance of the galvanometer as found by half, defleciton method is G , NARAYANAGROUP, , RS, RS, , , , Make the connections according to the circuit, shown, Take out the plug marked 5000 from the, resistance box and insert the key K1 only.., Adjust the value of ‘R’ so that the deflection is, maximum even in number and within the scale., Note the defleciton ' ' ., Inserting the key K 2 , without changing the value, of ‘R’, adjust the value of ‘S’ such that the, deflection in the galvanometer reduces to exactly, half the value of deflection obtained in the, previous step. Note the value of resistance ‘S’., Repeat the experiment taking out different values, of ‘R’ and adjusting the value of ‘S’ every time., Using the formula G , , RS, , the resistance of, RS, , galvanometer can be determined., 107
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EXPERIMENTAL PHYSICS, , , , , , , 1), , 2), , 3), , 4), , 5), , 108, , JEE-ADV PHYSICS- VOL- VI, , In this experiment, the emf of the cell or battery 6), should be constant., When a graph is plotted between resistance (R), , The value of shunt to be used to pass 1/10th of, the total current through a galvanometer of, resistance 100 is, 1) 41.11 2) 31.11 3) 21.11 4) 11.11, 1, 7), A galvanometer has a sensitivity of 10, and , we get a straight line as shown., divisions per milli ampere. Its resistance is, , 50 . The shunt required to change its, K, sensitivity, to 2 divisions per milli ampere is, Slope of the graph , where K= figure of merit, E, 1) 42.5 2) 32.5 3) 22.5 4) 12.5, 8), The ratio of the shunt resistance and the, KG, resistance of a galvanometer is 1:499. If the, OA, , Intercept, E, full scale deflection current of the, galvanometer is 2mA, the range of the, ammeter is, 1) 4A, 2) 3A, 3) 2A, 4) 1A, 9) An ammeter reads upto 1 ampere. Its internal, resistance is 0.81 ohm. To increase the range, to 10A, the value of shunt resistance required, is, 1) 0.03 2) 0.3, 3) 0.9 4) 0.09, EXERCISE - 14, 10) The relation between voltage sensitivity ' V ', The figure of merit of a galvanometer is, 1) the voltage required to produce a deflection of, and current sensitivity ' i ' of moving coil, one division, galvanometer if its resistance is ‘G’ is, 2) the current required to produce a diflection of, i, 10 divisions, 2) V , 1) V G i, G, 3) the voltage required to produce a deflection of, 2, 2, 10 divisions, 3) V i G, 4) V i G, 4) the current required to produce a deflection of 11) The sensitivity of a galvanometer of resistance, one division, 990 is increased by 10 times. The shunt, Galvonometer shows deflection when current, used is, passes through it. This is because,, 1) 100 2) 120 3) 110 4) 50 , 1) it is based on heating effect of current, 12) When a high resistance ‘R’ is connected in, 2) it is based on magnitude effect of current, series with a voltmeter of resistance ‘G’, the, 3) it is based on induced emf, range of the voltmeter increases 5 times. Then, 4) it is based on electrolysis, G : R will be, The sensitivity of a galvanomter depends on, 2) 1: 2, 3) 8 :1, 4) 1: 4, 1) 4 :1, 1) the cylindrical magnetic field used in the, EXERCISE - 14 - KEY, galvanometer, 1) 1 2) 2 3) 4 4) 2 5) 2 6) 4 7) 4, 2) area of the coil, 3) torsion constant of the spring, 8) 4 9) 4 10) 2 11) 3 12) 4, 4) all the above, EXERCISE - 14-HINTS, In an ammeter 5% of the total current is, passing through the galvanometer of, ig, ig, S , S, S, resistance ‘G’. the resistance of shunt (S), i, , , 6), 7) ig , 4), required will be, i GS, i GS, G S , 1) 19G, 2) G/19, 3) 20G, 4) G/20, S, 1, S , i given that, , The sensitivity of a galvanometer does not 8) ig , , G S , G 499, depend upon, ig, 1) the very strong magnetic field in the permanent, S, G, 990, , , 110 , 9), 11) S , magnet, n 1 10 1, i GS, 2) the current it measures, G 1, ., 3) a very thin and weak suspension, 12) R G n 1 G 5 1 4G ;, R 4, 4) large number of turns in the coil., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, Note this position and measure its distance from, , EXPERIMENT - 15 A, To determine the focal length of a concave mirror., , , the pole of the mirror as v ., Repeat the same experiment for different values, of object distance, , Concave, mirror, , Image, , F, , C O, , Pole, , object, needle, v, , , , , , Focal length of the mirror can be calculated by, using the formula., , f , , u, , Image, needle, , , , To perform this experiment, we require optical, bench, upright stands, optical needles and a concave mirror., When an object needle is placed at a distance, ' u ' from the pole of a concave mirror, by , removing the parallax between the image needle, and the inverted image of the object needle, formed by the concave mirror the position of the, image is located. The distance to the image, needle from the pole of the mirror is ' v ' . If the, , uv, in each case and finally take the avuv, , erage., When a graph is plotted between object distance, , u along negative x-axis and, along negative y-axis,, , image distance, a rectangular, , hyperbola is obtained., The angle bisector OC drawn at an angle 450, with x-axis, the lines OP and OQ are perpendicular to X and Y axes respectively, u, , X, , P, , O, 45, , focal length of the concave mirror is ' f ' then, , f , , and calculate the, , corresponding image distance v , , , I, , u , , 0, , C, , uv, uv, , PROCEDURE, , , , , , , , , Q, v, , Find the rough focal length of the concave, , mirror by focussing the image of distant object, on a wall., Take an optical needle as an object needle (O)., The tip of the object needle and the pole of the, concave mirror should be on the same line., , Place the object needle at a distance u , 1.5, times the roughly calculated focal length of the, mirror. Place another image needle behind the, object needle on the same side of the mirror., Adjust the position of the image needle and, coincide it with the image formed by the, , concave mirror without parallox., In this position, the real image formed by the, mirror and the image needle will move together, when our eye is moved towards left or right., NARAYANAGROUP, , Y, , From the graph OP OQ 2 f, , f , , OP OQ, 4, , When a graph is plotted between, , 1, taken along, u, , 1, taken along negative, v, y-axis, we get a straight line which cuts both x, and y axes at P and Q, respectively.., negative x-axis and, , From the graph,, , OP OQ , , 1, f, 109
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , 2, f , OP OQ, , X, , P, , 1, u, , , , O, 1, v, , Q, , , Y, , EXPERIMENT - 15 B, To find the focal length of a convex mirror using , a convex lens, L, , , o, , P, , C, , , L, M, I, , O, , , , , , , , To perform this experiment, we require, an , optical bench, convex lens, convex mirror, a lens, holder, a mirror holder, two optical needles and, a meter scale., , If ‘R’ is radius of curvature of a convex mirror,, then its focal length f , , R, 2, , , , PROCEDURE, , , , , 110, , Clamp the holder with the convex lens in a fixed, upright such that the lens surface is vertical and , perpendicular to the length of the optical bench., Take a thin optical needle as an object (O) and, mount it in a laterally movable upright which is, kept at zero on the optical bench., , Keep the object upright at a certain distance say, 40cm from the lens. Adjust the height of the object needle to make its tip lie on the horizontal, line through the optic centre of the lens., Clamp the holder with convex mirror near the, lens upright keeping the reflecting surface of, the mirror towards the lens and adjust the height, of the mirror to make its pole lie on the horizontal line passing through the optic centre of the, lens, making the surface of the mirror vertical, and perpendicular to the length of the optical, bench., By adjusting the positions of convex lens and, convex mirror, see the inverted image of the, object needle formed by reflection from the, convex mirror., Adjust the height of the needle so that the tips, are seen in line with your right eye open. When, you move your eye towards right, the tips get, separated. It means the tips have parallax., Move the convex mirror backwards until, tip-to-tip parallax is removed and note this, position., Remove the convex mirror keeping upright in, its position. See with your right eye open from, the other end of the optical bench. You can see, an inverted and enlarged image of the object, needle., Mount the second optical needle (image needle), in the fourth upright near the other end of the, optical bench., Adjust the height of the image needle so that its, tip is seen in line with the tip of the image. By, removing parallax, again note the position., By changing the position of object needle repeat, the experiment 5 to 6 times and note the, observations., The distance between the image needle and convex lens in each case gives radius of, curvature (R) of the convex mirror when the mirror, is placed very close to the convex lens., The focal length of the convex mirror can be, obtained by the formula f , , R, 2, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , EXPERIMENT - 15 C, To find the focal length of a convex LENS, Image, C, , Object, , , , , , u, , O, Image, needle, , , , Plot a graph between ' u ' (along negative x-axis), and ' v ' (along positive y-axis). It will be a, rectangular hyperbola. When an angle bisector, is drawn, it cuts the curve of a point ‘C’., From the graph, CP CQ 2 f . Therefore the, coordinates of C are 2 f ,2 f, , , , v, , For performing the experiment, we require an, optical bench, three uprights, a convex lens, a, lens holder, two optical needles and a meter, scale., If ' u ' is distance to the object from the convex, lens ‘V’ is distance to the image from the, convex lens and ' f ' is focal length of the, convex lens, then, , Focal length of convex lens from u graph, is, , f , , CP CQ, 4, Y, , v, C, , 1 1 1, by convention , f v u, Find rough focal length of the given convex lens, by focusing the image of a distant object on the, wall. In this case, distance to the image from, the convex lens given us the approximate value, , , , of its focal length f ., , , , , Place the optical needle (object needle) on one, side of the lens at a certain distance (lies be, tween ' f ' and '2 f ' of the convex lens) and the, image needle on the other side., The image pin is moved such that the image of, the object pin coincides with the image pin. By, eliminating parallax error, the distance to the, object pin from the lens is noted as object, , u, , Plot a graph between, , u ,, , From the graph, OP OQ , , uv, u v, , NARAYANAGROUP, , 1, therefore, the, f, , focal length of convex lens from, is f , , 1 1, graph, u v, , 2, OP OQ, Y, Q, , find t he, , corresponding values of (v) and determine the, focal length of given convex lens using the, formula f , , 1, (along negative x-axis), u, , 1, (along positive y-axis). It will be a, v, straight line cutting x-axis and, y-axis at ' P ', and ' Q ' respectively.., , convex lens and image pin as image distance, (V), For different values of, , O, , P, , and, , distance u and the distance between the, , , , Q, , 450, X, , PROCEDURE, , , , , 1, v, , X, , P, , 1, u, , O, , 111
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , EXERCISE - 15, 1), , 2), , 3), , If a lens in made of 3 largers of different refractive indices, then number of images of an, object formed by the lens is equal to, 1) 1, 2) 2, 3) 3, 4) 4, A concave mirror has focal length 20cm. Distance between the two positions of the object for which the image size is double that of, the size of the object is, 1) 20 cm 2) 40 cm 3) 30 cm 4) 60 cm, A concave mirror of focal length ' f ', produces an image ' n ' times the size of an, object. If the image formed is real, then the, distance of object from the mirror is, , 8), , An object is placed at a distance of f/2 from a, convex lens where f focal length of the lens., The image will be, 1) at, , 3f, , real and inverted, 2, , 2) at 2 f , virtual and erect, , 9), , 3) at 2 f , real and inverted, 4) at one of the focii, virtual, erect and double, the size, The focal length of a convex lens is 30cm and, the size of the real image is, , 1, th of the ob4, , ject. Then the object distance is, 1) 150 cm, 2) 90 cm, 1) n 1 f, 2), n, 3) 60 cm, 4) 30 cm, 10) A convex lens is dipped in a liquid, whose ren, fractive index is equal to the refractive index, 3) n 1 f, 4) n 1 f, , of the lens. Then its focal length, 1) becomes zero, A convex lens of focal length ' f ' produces a, 2) becomes infinite, real image ' n ' times the size of object. The, 3) remains unchanged, distance of object from the lens is, 4) becomes reduced but not zero, , n 1 f, , 4), , 1) nf, , 3), 5), , 6), , 7), , 112, , 2), , n 1 f, n, , f, n, , 4) n 1 f, , 11) If fV and f R are the focal lengths of a, convex lens for violet and red light respectively. Then, 1) fV f R, , 2) fV f R, , 3) fV f R, 4) fV f R, When a convex lens of refractive index 3/2, and focal length 20 cm is dropped into water 12) An object of height 5m is placed 10cm from a, convex mirror of radius of curvature 30 cm., of refractive index 4/3. Its focal length in waThe nature and size of the image formed is, ter is, 1) virtual, erect, behind the mirror, 10 cm in size, 1) 20, 2) 40 cm 3) 80 cm 4) 10 cm, 2) virtual, erect, behind the mirror, 3cm in size, The minimum distance between an object and, 3) real, inverted, infront of the mirror, 10 cm in, its real image formed by a convex lens of fosize, cal length ' f ' is, 4) real, inverted, infront of the mirror, 3cm in, 1) f, 2) 2 f, 3) 3 f 4) 4 f, size, A convex lens is making full image of an ob- 13) Two objects A and B when placed in turn in, front of a concave mirror of radius of, ject. If half of the lens is covered by an, curvature 15cm, give images of equal size. If, opaque object, then, ‘A’ is three times the size of ‘B’ and is placed, 1) half the image is not seen, 30 cm from the mirror, the distance of ‘B’ from, 2) full image is seen with same intensity, the mirror is, 3) full image is seen with decreased intensity, 1) 20cm 2) 15 cm 3) 12.5 cm 4) 9.5 cm, 4) no image is observed, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 14) A concave mirror forms a real image 5 times 20) One surface of a lens is convex and the other, the size of the object placed at a distance of, is concave, if the radii of curvature are r1 and, 20cm from it. The radius of curvature of the, r2 respectively, the lens will be convex if, mirror is, 1), , 100, 50, cm 2), cm 3) 100cm 4) 50cm, 3, 3, , 15) A mirror produces on a screen an image of, the sun 2cm in diameter. If the sun’s disc subtends an angle 0.1 radian on the surface of, the earth, then the radius of curvature of the, mirror is, 1) 20 cm 2) 40 cm 3) 200 cm 4) 400 cm, 16) A small strip of plane mirror ‘A’ is set with its, plane normal to the principal axis of a convex, mirror ‘B’ and placed 10 cm in front of ‘B’, which it partly covers. An object is placed, 20cm from ‘A’ and the two virtual images, formed by reflection in ‘A’ and ‘B’ coincide, without parallax. The radius of curvature of, B is, 1) 20cm 2) 22.5 cm 3) 27.5cm 4) 30 cm, 17) An object is placed in front of a mirror at a, distance of 60cm. If its two times diminished, image is formed on the screen, the focal length, of the mirror is, 1) 20cm 2) 45 cm 3) 15cm, 4) 90 cm, 18) An object is placed on the principal axis of a, convex mirror. Distance of object from the, mirror is 40cm. A plane mirror is placed between the object and the convex mirror, covering lower half below principal axis of the, mirror. Distance between the object and the, plane mirror is 30cm. If there is no parallax, between the two images formed by plane mirror and convex mirror, the focal length of the, convex mirror is, 1) 20 cm 2) 40 cm 3) 60 cm 4) 80 cm, 19) 'U' shaped wire is placed in front of a concave mirror of radius of curvature 20cm as, shown., The total length of the image of the wire, ABCD is nearly, A, , D, , 1) r1 r2, , 2) r1 r2, , 3) r1 r2, , 1, 4) r1 r, 2, , EXERCISE - 15 - KEY, 1) 3, 7) 3, 13) 2, 19) 2, , 2) 1 3) 2 4) 3 5) 3 6) 4, 8) 4 9) 1 10) 2 11) 1 12) 2, 14) 1 15) 2 16) 4 17) 1 18) 2, 20) 3, , EXERCISE - 15 - HINTS, 1), , Number of images formed by the lens is equal to, number of different media., , 2), , v, 2, u, , 3), , v, n, u, , 4), , v, n, u, f liq, , 5), , f air, , , , g 1 l, g l , , 8), , 1 1 1, v, and m , u v f, u, , 9), , 1 1, 1, v 1, u 4v and , u v, f, u 4, , 12), , sizeof theimage v, 1 1 1, , , and, size of the object u, u v f, , 13), , M A 1 f uB, f, , , MB 3, f, f uA, , v, 5, u, u 20cm; v 100cm ,, , 14) M , , 5cm, B, , C, , 10cm, , 1) 2.5 cm, 3) 12.5 cm, NARAYANAGROUP, , 30cm, , 2) 6 cm, 4) 15 cm, , 1 1 1, , f u v, 113
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 15) Since the image is real, it is a concave mirror, , , image, , 40, v, 1, Magnification of AB 1 3 , u1 4o 3, , , f, sun, , , , 1, 1, 1, , , v2 15cm, 10 v2 30, , arc, 2, 0.1 rad, radius f, , A1 B1 , , f 20cm and R 2 f 40cm, 16) u 30cm v 10cm, , AB 5, cm, 3, 3, , Magnification of CD , , 1 1 1, and R 2 f, f v u, , C1 D1 , , v2 15 1, , cm, u2 30 2, , CD 5, cm, 2, 2, , Total length of the image of the wire, , B, , 5 5 5, A1B1 B1C1 C1D1 ; 3 3 2 6cm, , IA, IB, A, , Object, , Image, , 20cm, , 1 1, 1, 20) f 1 r r , 1 2, , 10cm, , v 1, u, , v 30cm, 17) M , u, 2, 2, , 1 1, 6, r1 r2, , r1 r2, , EXPERIMENT - 16, To determine angle of minimum deviation for a, triangular prism by plotting a graph between, angle of deviation and angle of incidence., , 1 1 1, f 20cm, f v u, 18) u 40cm ; v 20cm, , 1 1 1, , f v u f 40cm, , M, , P, , I, , O, i2, D2, , 20cm, , 10cm, 40cm, , 19) Focal, , lengt h, , of, , concave, , mirror, , , , To perform this experiment, we require, triangular glass prism, drawing board,, chart paper, drawing pins, pencil, scale and a, protractor., , , , If ‘A’ is angle of the prism, m is angle of, minimum deviation and refractive index of the, mat erial of t he prism is ' ' then, , R, f 10cm, 2, for the left arm (AB) of the U - wire, u 40cm, , 1 1 1, 40, v1 , cm, f v u, 3, for the right arm (CD), u2 30cm, 114, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , A m , sin , , 2 , , A, sin , 2, , , , m, , PROCEDURE, , , , , Fix a white chart paper on a drawing board and, take 8, sketches of triangular based glass prism,, Draw normals to each and every sketch of the , triangle on one side., Using protractor keep angles, , 300 ;330 ;360 ;390....., , , , , , , , , from each normal, , respectively and draw lines showing the inci- , dent ray., Fix two pins P and Q vertically on the line, representing incident ray 2cm apart., Place the glass prism in the sketch and by, viewing the two pins from the other side of the, prism, fix two more pins R and S such that all, the four pins are on the same line without parallax., , Removing the glass prism from the sketch, and, removing the two pins R and S join the two, positions of R and S using a scale and a pencila., This line represents the ray of emergence, By extending ray of incidence and ray of, emergence, the angle between the incident ray, and the emergent ray called as angle of deviation is measured using a protractor.., , , , The experiment is repeated in the remaining , sketches at different angles of incidence, , 330 ,360 ,390....... Every, , , t he, , corresponding angle of deviation is measured. 1), When graph is plotted taking angle of incidence, , i , , , time,, , on x-axis and the angle of deviation , , on y-axis, we get a parabola which is known as 2), i curve., From i curve, it is observed that as ' i ', increases, the value of ' ' decreases gradually, to a minimum value, beyond which ' ' increases, with the increased values of ' i ' ., NARAYANAGROUP, , 0, , i, , i=e, i, , e, , For every angle of deviation, there will be two, corresponding values of angles of incidence,, such as angle of incidence i and angle of, emergence e ., When the angle of deviation in the prism is, minimum, then angle of incidence i and angle, of emergence e are equal. Therefore angle of, refraction r1 at the first face of the glass prism, is equal to angle of refraction r2 at the second face of the prism., From i graph, the angle of minimum, deviation can be determined, and using the, , A m , sin , , 2 , , , formula, the value of, A, Sin, 2, refractive index of the material of the prism also, can be calculated., For an equilateral triangular glass prism, angle, of the prism A 600 ., , EXERCISE - 16, When light falls on a prism, the resultant can, be, 1) inversion, 2) magnification, 3) elongation, 4) deviation, The angular dispersion produced by a prism, 1) increases if the average refractive index, increases, 2) increases if the average refractive index, decreases, 3) remains constant without depending on, refractive index, 4) no relation with average refractive index, 115
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 3), , 4), , When a glass prism is placed in water, the angle 10) If angle of incidence, emergence and, of minimum deviation, deviation are 450 ,550 and 400 then the, 1) increases, 2) decreases, angle of the prism is, 3) remains same, 4) cannot be predicted, If the angle of incidence and angle of, re1) 550, 2) 400, 3) 450, 4) 600, 0, fraction at the first refracting surface are 45 11) A given ray of light suffers minimum deviation in an equileteral prism ‘P’. Additional, and 300 respectively, then the refractive inprisms Q and R of identical shape and of the, dex of the material of the prism is, same material as ‘P’ are now added. The ray, 1, 1, will now suffer, 1) 2, 2), 3), 4), , 2, , 5), , 6), , 7), , 8), , The angle of minimum deviation of a prism of, refractive index 3 is equal to its refracting, angle. Then the refracting angle of the prism, is, 2) 600, 3) 750, 4) 900, 1) 450, , P, , R, , sin r, , 2 1 , , 2) sin , 2 , 1, , , , , , 1, 1, 3) 2cos , 4) cos , 2, 2, The refracting angle of a prism is ‘A’ and the, refractive index of the material of the prism, , 5., , A, 6., . The angle of minimum deviation, 2, , A m , sin , , 2 here A , , m, A, sin, 2, , , , is cot , , of the prism is, 1) 2A, 3), 116, , Q, , 1) greater deviation, 2) no deviation, 3) same deviation as before, A prism of refractive index, 2 has, 4) total internal reflection, refracting angle 600 . At what angle must a 12) The maximum refractive index of a prism, ray of light be incident on it so that it underwhich permits the passage of light through it,, goes minimum deviation ?, when the refracting angle of the prism is 900, 1) 300, 2) 450, 3) 600, 4) 900, is, 0, A certain prism of refracting angle 60 and, 3, 3, of refractive index 2 is immersed in a liquid, 1) 3, 2) 2, 3), 4), 2, 2, of refractive index 2 . Then the angle of, EXERCISE - 16-KEY, minimum deviation will be, 1) 4, 2) 1 3) 2 4) 2 5) 2 6) 2, 2) 450, 3) 600, 4) 750, 1) 300, 7) 1, 8) 3 9) 2 10) 4 11) 3 12) 2, A prism of refractive index ' ' and angle, ‘A’ is placed in the minimum deviation, EXERCISE - 16-HINTS, position. If the angle of minimum deviation, sin i, is ‘A’, then the value of ‘A’ in terms of ' ' is, 4. , 1 , 1) sin , 2, , 9), , 2, , 2, , , A, 2, , 2) 2A, 4), , , A, 2, , 7., , g, , l, , sin i, A, sin, 2, A m , sin , , 2 , A, sin, 2, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , 2A, 2, , A, sin, 2, sin, , 8., , , , A m, 2 cot A, A, 2, sin, 2, , sin, , , , 9., , , , , , 10. i e A , 3, , 1, , 11., , 2, , 4, , , , No deviation occur on interfaces 2 and 3 as there, is no change in medium. How ever deviation at, interface ‘4’ is same as on interface 2 with only, prism P., A, 12. cos ,, 2, 2,, , 90 , cos , 2, , =, , , , , EXPERIMENT - 17, To find refractive index of a glass slab using, , travelling microscope, Main, Scale, , Main, Scale, , R2, , R1, , Vernier, scale, , P1, , Main, Scale, , , , , V, , V, , , , , , P1, , NARAYANAGROUP, , S, , real thickness of glass slab, apparent thickness of the glassslab, , Put the microscope on the table near a window, for getting sufficient light, By adjusting the horizontal screws on the base,, make the base of the microscope perfectly, horizontal, Adjust the eye-piece of the microscope and its, position, so that the cross wires are clearly, visible, Determine the vernier constant of the vernier, scale of the microscope, Make a cross marked as ‘P’ on the base of the, microscope., Adjust and focus the microscope on the cross, marked as ‘P’ and avoid parallax between the, cross-wires and image of the cross ‘P., Note down the main scale reading and vernier, , , , scale reading R1 on the vertical scale., Keep a glass slab of thickness ‘t’ on the cross, ‘P’. Now due to refraction of light the cross, , , , marked as ‘P’ looks as if it is shifted up to ' P1 ', a little., Move the microscope further upwards and, focus the image of P1 , and note the reading on, , V, R3, , realdepth, apparent depth, , PROCEDURE, , max 2 ., , Compound, microscope, , To perform this experiment, we require a travelling microscope, lycopodium powder and three, glass slabs of same material but different thicknesses, A travelling microscope is a compound, microscope fitted on a vertical scale and can move, up and down carrying a vernier scale moving along, the main scale., The reading of travelling microscope is the result of main scale and vernier scale measurements., When we look into a denser medium standing, in a rarer medium then the refractive index of, the denser medium with respect to the rarer medium, , the vertical scale of the microscope as ' R2 ' ., , 117
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , , , , 1), , 2), , 3), , 4), , 5), , 6), , Now sprinkle few particles of lycopodium 7), powder on the surface of the slab., By moving the microscope further up, focus the lycopodium particles with cross wires and note the, reading on the vertical scale as R3 ., , In a travelling microscope, focal length of objective is, 1) greater than focal length of eye - piece, 2) smaller than focal length of eye - piece, 3) equal to the focal length of eye - piece, 4) zero, Now the refractive index ' ' of the slab can be, 8) In a travelling microscope, the final image is, determined by the formula, formed at, R R, 3 1, 1) infinity, R3 R2, 2) at the focus of eye - piece, 3) at the focus of objective, EXERCISE - 17, 4) at least distance of distinct vision from the, As light enters from air into glass slab, its, eye - piece, wavelength, 9) In a travelling microscope 49 main scale divi1) increases, 2) decreases, sions coincide with 50 vernier scale, divi3) remains constant, sions. The least count of microscope when, 4) may increase or decrease, one main scale division equals 0.5mm is, Travelling microscope is used for, 1) 0.01 cm, 2) 0.001 cm, 1) determination of focal length of convex mir3) 0.002 cm, 4) 0.02 cm, ror, 2) determination of focal length of convex lens 10) If ‘t’ is the real thickness, ' ' is refractive, 3) determination of refractive index of a prism, index of a glass slab then the shift of the im4) determination of refractive index of glass slab, age with reference to the object is given by, Refractive index of a medium depends on, 1, 1) wavelength, t, 1), t, 1, , , 2), , 1 , 2) frequency, , , 3) surrounding medium, 4) size of the medium, 1 , The absolute refractive index of a medium, 3) t 1, 4) t 1, other than air is always, , 1) less than unity, 11) A ray of light passes normally through a slab, 2) equal to unity, 1.5 of thickness ‘t’. If the speed of light, 3) more than unity, in vaccum be ‘c’, then time taken by the ray, 4) may be more or less than unity, The refractive index of a material will be less, to go across the slab will be, than unity when, t, 3t, 2t, 4t, 1) material is placed in optically rarer medium, 1), 2), 3), 4), c, 2c, 3c, 9c, 2) material is placed in optically denser medium, 3) material is placed in vaccum, 12) A ray of light incident on a transparent block, 4) material is placed in air, at an angle of incident 600 . If the refractive, If V1 and V2 are velocities of light in two difindex of the block is 1.732, the angle of deferent media, then the ratio of wavelengths, viation of the refracted ray is, 1, 1) 150, 2) 250, 3) 300, 4) 450, , of light in the same media, , 2, , V1, 1), V2, 118, , V2, 2), V1, , 3), , V1V2, , EXERCISE - 17 - KEY, , 4), , V1, V2, , 1) 2, 7) 2, , 2) 4, 8) 4, , 3) 1&3 4) 3 5) 2 6) 1, 9) 2 10) 2 11) 2 12) 3, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXERCISE - 17 - HINTS, 9), , least count =, , 11) t ' , , valueof 1MSD, number of VSDs, , x 3 x 3t, , , c, 2c 2c, , sin i, , sin r, , 12) i r, , sin 60, 3, sin r ., , EXPERIMENTAL PHYSICS, , , , , , , , 60 30 300 ., , r 300, , EXPERIMENT - 18, To draw characteristic curves of a p - n junction , diode in forward and reverse bias, , Potential, devider, , P, , , , A, , n , , , V, , , , , , , (), , , , forward biased p - n junction diode, , Potential, devider, , , , , P, , , , A, , n , , , V, , , , , PROCEDURE, (), , , , In forward bias, electrons move towards, p, - region and holes move towards n- region. Thus, the junction diode conducts electricity in forward, bias., When the applied voltage is greater than the, barrier voltage, then the diode conducts, electricity, Germanium diode needs 0.3V and silicon diode, needs 0.7V nearly to just conduct electricity, In reverse bias p - end of p - n junction diode is, connected to negative terminal of the battery and, n - end is connected to positive terminal of the, battery., In reverse bias, majority carriers which are, electrons in the n - region flow into the positive, of the battery which is the high potential area,, leaving a positive charge near the deflection, layer. The electrons from the negative of the, battery flow into p - region leaving negative, charge near the depletion layer. Thus the depletion layer, increases and the potential barrier, increases., In reverse bias, the minority carriers contribute, some flow of current as they are forward biased. This current is called leakage current., As temperature increases, leakage current, increases and it is in the order of few, microampere in germanium and few, nanoammpere in silicon diodes., When the applied voltage increases gradually,, at a particular voltage, breakdown occurs and, suddenly the current increases due to formation, of more number of electron - hole pairs. This, voltage in the reverse bias is called break down, voltage. At this stage, the junction is damaged., , (i) Forward - biased p - n junction diode, , , reverse biased p - n junction diode, In forward bias, p - end of p - n junction diode , is connected to positive terminal of the battery, and n - end is connected to negative terminal of , the battery. In forward bias, the electrons which, are the majority charge carriers in n - region are, at negative potential and the holes which are the , majority carriers in p - region are at, positive potential., NARAYANAGROUP, , Make the circuit as shown for forward bias of p, - n junction diode., Find the least count and zero error (if any) of, voltmeter and milliammeter., Bring contact of potential divide (rheostat) near, negative end and insert the key (k). Initially, voltmeter and milliammeter give zero reading., Move the contact a little towards the positive, end to apply a forward - biased voltage of 0.1V., By increasing this voltage gradually, the, corresponding currents are to be noted., 119
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , (ii) Reverse - biased p - n junction diode, , , , , , , 5), Make the circuit as shown for reverse bias of, p - n junction diode., Bring contact of potential divider (rheostat) near, positive end and insert the key (k). Voltmeter, and microammeter give zero reading., Move the contact towards negative end to, apply a reverse - biased voltage of 0.5V then a, feebly reverse current starts flowing. By, increasing this voltage the corresponding currents are to be noted until the current increases, suddenly representing reverse break down stage. 6), Note the current and pluckout the key., If voltage is taken along x - axis and the, corresponding current along y - axis and plot a, graph, it is called v - i characteristic curve of, p - n junction diode., 7), If, , Vr, , 0, , Vf, , 8), Ir, , , , In the forward bias, the voltage where the, current starts increasing rapidly is called knee, voltage., , EXERCISE - 18, , 1), , 2), , 3), , 4), , 120, , In a junction diode, the direction of diffusion, current is, 1) from n - region to p - region, 2) from p - region to n - region, 3) from n - region to p - region if the junction is, forward biased and vice - versa if it is reverse, biased, 4) from p - region to n - region if the junction is, reverse biased and vice - versa if it is reverse, biased., Resistivity of a semiconductor depends on, 1) shape of semiconductor, 2) atomic nature of semiconductor, 3) length of semiconductor, 4) shape and atomic nature of semiconductor, In V - i characteristics of p - n junction in, reverse biasing results in, 1) leakage current, 2) the current which cannot be neglected, 3) no flow of current 4) large current, In the middle of the depletion layer of a, reverse biased p - n junction diode, 1) the potential is zero, 2) the electric field is zero, 3) the electric field is maximum, 4) the potential is maximum, In a p-n junction diode having depletion layer, of thickness 106 m , the potential across it is, 0.1V. The electric field produced is, 1) 107 Vm1, 2) 106 Vm1, , 9), If a full wave rectifier circuit is operating from, 50 Hz mains, the fundamental frequency in the, ripple will be, 1) 25 Hz 2) 50 Hz 3) 70.7 Hz 4) 100 Hz, Avalanche breakdown in a semi conductor di3) 105Vm1, 4) 105Vm1, ode occurs when, 10) In an unbiased p - n junction diode electric, 1) the potential barrier becomes zero, field at the junction is of the order of, 2) the forward current exceeds a certain value, 3) forward bias exceeds a certain value, 2) 106 Vm 1, 1) 10Vm 1, 4) reverse bias exceeds a certain value, On increasing the reverse voltage in a p - n, 3) 106 Vm 1, 4) 0.1Vm 1, junction diode the value of reverse current, 11) The resistance of an ideal p - n junction, will be, diode in forward biased condition is, 1) increased gradually 2) increased suddenly, 1) zero, 2) infinite, 3) constant, 4) decreased gradually, 3) negative, 4) finite, Change in temperature, 12) The value of current if diode is ideal is, 1) increases forward resistance, 2) decreases forward resistance, 3, 1V, 3) affects v - i characteristics of p - n junction, 4V, diode, 1) 0, 2) 1 amp, 4) does not affect v- i characteristics of p - n, 3) 1.66 amp, 4) 15 amp, junction diode., NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 13) The resistance of reverse biased p - n 21) The resistance of the diode in forward bias, condition is 20 and infinity in the reverse, junction diode is about, biased condition. The current in the circuit is, 1) 1 2) 102 3) 103 4) 106 , 14) In reverse biased condition, the width of depletion layer, 1) increases 2) decreases 3) remains same, 4) first increases and then decreases, 15) In a p - n junction diode, the barrier, potential opposes diffusion of, 1) free electrons from n - region, 2) holes from p - region, 3) majority charge carriers from both the regions, 4) minority charge carriers from both the regions, 16) The potential barrier in the depletion layer is, due to, 1) ions, 2) electrons, 3) holes, 4) forbidden band, 17) Conductivity of a semiconductor increases 9), when a radiation of wavelength is less than, 2480nm is incident on it. The forbidden gap, 12), is, 1) 0.5 J 2) 0.5 eV 3) 1 eV, 4) 2 eV, 18) Which of the following diodes is forward biased ?, 17), , 1), , 9V, , 5V, , 2), , V, , V, , 80, , 6V, , 4), 2V, , 5V, , V, , 3V, , V, , 3), , 2), , 4), V, , 3V, , V, , 1) 4, 7) 1, 13) 4, 19) 1, , NARAYANAGROUP, , 2) 4, 8) 2, 14) 1, 20) 3, , V, , V, , 0V, 2) 0.02 A, 4) 0.01 A, , 4) zero, , 3) 2 4) 3 5) 2 6) 2, 9) 3 10) 3 11) 1 12) 2, 15) 3 16) 1 17) 2 18) 3, 21) 3, , EXERCISE - 18 - HINTS, E, i, , V, d, , V 4 1, , 1 amp, R, 3, , 12400 , energy , eV, 0, , in, A, , , V, R, , V, , 20) In the figure shown, current passing through , the diode is, 30, A, B, , 3V, 1) 0.1 A, 3) zero, , 3) 0.04 A, , EXERCISE - 18- KEY, , EXPERIMENT - 19, Draw the characteristic curve of Zener diode, and to determinate its reverse breakdown, voltage., , Rin, , V, , 2V, , 20) diode is reversely biased, , 19) Which of the following diodes is reverse biased ?, , 1), , , , , 1) 0.08A 2) 0.1A, , 21) i , , 3), , , , , , , , , V Vin, , , , IL, A, , Iin, , Iz, RL, V, , Vout, , To perform this experiment, we require a Zener, diode VZ 6V , a 10V batt ery, a high, resistance rheostat, voltmeter, milliammeter, a, 20 resistance, one-way key and connecting, wires., 121
Page 1983 :
JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , , , Zener diode is a very highly doped p - n junction, , diode. This heavy doping results in a low value, of reverse breakdown voltage., The reverse breakdown voltage of a Zener diode, is called Zener voltage VZ . The reverse, current that results after the breakdown is called, , Plot a graph between input voltage Vin and, output voltage Vout along x-axis and y-axis, respectively. The shape of the graph is as shown., , Zener current iZ ., , , Vout, , From the circuit iin iL iZ, where iin input current (total current in the, circuit), iZ current through Zener diode, , , iL current through load resistance RL , , , Output voltage or Zener voltage, , Vin, , 0, , When a graph is plotted between voltage taken, on x-axis and current taken along y-axis for a, Zener diode, we get a graph as shown., , Vout or VZ Vin Riniin, , Forward, bias, , or Vout RLiL, , , i, Vz, , Initially as Vin increases, iin increases a little, , 0, , V, , then Vout increases at breakdown, , Vout Vin Rin iin, , , This constant value of Vout which is reverse, breakdown voltage is also called Zener Voltage., , PROCEDURE, , , , , Connect the circuit as shown, Adjust the potential divider (Rheostat) near , negative end., Move the contact a little towards positive end, , , , to apply some reverse biased voltage Vin . 1), Initially milliammeter reads zero., As Vin is further increased, iin starts flowing., , , , , Then Vout becomes less than Vin . Note the, values of Vin ; iin and Vout, , , 2), , Go on increasing Vin in small steps of 0.5V and, note down the corresponding values of iin and, , Vout ., , , At one stage, as Vin is increased further, iin, increases by large amount and Vout does not, increase. This is reverse breakdown situation., , 122, , 3), , Here VZ represents output voltage or Zener, voltage., In forward bias, Zener diode works just similar, to an ordinary p - n junction diode., In reverse bias, Zener diode works very, effectively., , EXERCISE - 19, Zener diode can be used as, 1) half - wave rectifier, 2) oscillator, 3) voltage regulator, 4) transformer, Zener break down will occur if, 1) impurity level is low, 2) impurity level is high, 3) impurity is less in n-side, 4) impurity is less in p-side, Zener diode is used for, 1) rectification, 2) stabilization, 3) amplification, 4) producing oscillations in an oscillator, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , 6), , 7), , 8), , 9), , 5k, , 80 120V, 50v, , , 10k, , 3) 2, 9) 3, , 4) 2 5) 2, 10) 1, , 6) 3, , EXERCISE - 19 - HINTS, V, 50, , 5 103 A 5 mA, 3, RL 10 10, , IL , , for 80 V VS 30 V, , IS , , VS, 30, , 6 mA, RS 5 103, , (minimum current), , I Z 1 mA, for 120 V VS 70 V, IS , , VS, 70, , 14 mA, RS 5 103, , I Z 14 5 9 mA (maximum current), VS 120 90 30 V, , EXPERIMENT - 20, To draw the charact eristic curves of a, Transistor and finding current gain and voltage, gain., ib , A, ib, A, , , K1, , , V Vbe, Rh1 , , , , C, b, , e, , , V Vce, , , np n, transistor, , , K2, , , , 120V, 90v, , 3) 90 V, , 20k, , 4) 120 V, , To perform this experiment, we require one, n - p - n transistor, a 3V battery, a 30V battery,, two high resistance rheostats, one voltmeter, one ammeter, , 0 500mA ,, , one, , voltmeter 0 30V , two one - way keys and, , , 2k, , NARAYANAGROUP, , 2) 2, 8) 2, , 0 3V ,, , 1) 6 mA, 5 mA, 2) 14 mA, 5 mA, 3) 9 mA, 1 mA, 4) 3 mA, 2 mA, 10) In the figure shown the potential drop across, the series resistor is, , 2) 60 V, , 1) 3, 7) 2, , IL, IZ, , 1) 30 V, , EXERCISE - 19 - KEY, , (), , 5), , A Zener diode when used as a voltage regulator, is connected in, i) forward bias, ii) reverse bias, iii) parallel with the load resistance, iv) series with the load resistance, 1) i and ii only are correct, 9), 2) ii and iii only are correct, 3) only i is correct, 4) only iv is correct, Avalanche breakdown in a semiconductor, diode occurs when, 1) forward current exceeds certain value, 2) reverse bias exceeds a certain value, 3) forward bias exceeds a certain value, 4) the potential barrier is reduced to zero, The sharp range of breakdown voltage in, Zener diode is, 1) 0.1 to 10 V, 2) 1 to 20V, 3) 0.05 to 0.1 V, 4) 20 to 200 V, Zener diode will function more effectively in, 10), 1) forward bias, 2) reverse bias, 3) both forward and reverse bias, 4) neither forward nor reverse bias, In the breakdown region, Zener diode behaves, as a, 1) constant current source, 2) constant voltage source, 3) constant resistance source, 4) constant power source, The maximum and minimum values of zener, diode current are, , (), , 4), , EXPERIMENTAL PHYSICS, , connecting wires, An n - p - n transistor has one p - type wafer in, between two n - type wafers., In common emitter circuit of transistor,, emitter - base makes input section and collector, - emitter makes output section., 123
Page 1985 :
JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , , The emitter -base circuit is forward biased and, collector - base circuit will be reverse biased., , Resistance offered by base junction is called, input resistance Ri and it has low value due, , , , to forward biasing., Resistance offered by collector junction is called, output resistance R0 and it has high value due, , PROCEDURE, Make the connections as shown in the circuit and, bring the movable contact of rheostat to be minimum, so that voltmeters show zero reading, , INPUT CHARACTERISTICS, , , Apply forward - biased voltage at the emitter base junction. Note the, base - emitter, , to reverse biasing., Due to high output resistance a high resistance, , voltage Vbe and the base current ib keeping, , is used as load resistance RL , , collector - emitter voltage Vce constant at, , , , The ratio of output resistance to input resistance, is called as resistance gain., , 0.2V. The voltage Vbe should be increased by, , , , The ratio of change in collector current ic , , steps of 0.2V and corresponding values of ib, are noted., , , , to the change in base current ib gives the, , Vce=0.2V Vce=0.4V, , current gain in common emitter configuration., , ib, , i, c, ib, , , , , The product of current gain and resistance gain, is known as voltage gain of a transistor. It is, about 50 times the resistance gain., Input resistance, , , V , Ri be , ib Vce constant, , , , Output resistance, , V , R0 ce , ic ib constant, , , , 0, , Vbe, , OUTPUT CHARACTERISTICS, Keep the collector voltage zero. Make all, readings zero., Make the base current ib 25 A by adjusting, base voltage. You will be able to read even, same collector current although the collector, voltage is zero., Increase the collector voltage by steps of 0.2V, and note the corresponding collector currents, Ib= 75A, , Resistance gain , , R0, Ri, , Ib= 50A, Ic, , ic , , , , Current gain, ib Vce constant, Voltage gain =, (current gain) (resistance gain), , R , 0, Ri , , 124, , 0, , Ib= 25A, , Vce, , TRANSFER CHARACTERISTICS, , , Make the collector - emitter voltage Vce 2V, by moving the sliding contact of Rheostat Rh2 , , , , Keep the base current ib at 100 A , by changing, NARAYANAGROUP
Page 1986 :
JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , Vbe record the collector current ic in the collector 7) In an n - p - n transistor the collector current is, , , , 24 mA. If 80% of electrons reach the, collector, its base current is, 1) 35 mA, 2) 25 mA, 200 A,300 A.... keeping Vce constant at 2V.., 3) 15 mA, 4) 6 mA, 8) A transistor is used in common - emitter, V = 2V, configuration. Given its 0.9 . The change, in collector current when the base current, i, changes by 2mA is, 1) 0.9mA, 2) 18 mA, 3) 20 mA, 4) 0.1 mA, i, 9) Transistor acts like, 1) oscillator, EXERCISE - 20, 2) amplifier, 3) both as oscillator and amplifier, The number of depletion layers in a transistor, 4) a rectifier, is, 10) In common emitter transistor, the input, 1) 1, 2) 2, 3) 3, 4) 4, An n - p - n transistor conducts when, resistance is 200 and bad resistance is, 1) collector is positive and emitter is negative, 40 k . If current gain is 80 then voltage gain, with respect to base, is, 2) collector is positive and emitter is at same, 1) 16, 2) 160, 3) 1600, 4) 16000, potential as the base, 3) both collector and emitter are negative with 11) Transistor means transfer of, 1) current, 2) voltage, respect to the base, 3) resistance, 4) all of these, 4) both collector and emitter are positive with, 12) In a transistor the region which is heavily, respect to the base, A transistor has a base current of 1 mA and, doped is, emitter current 100 mA. The current transfer, 1) emitter, 2) base, ratio will be, 3) collector, 4) all the three regions, 1) 0.9, 2) 0.99, 3) 1.1, 4) 10.1, 13) Transistor amplifier circuit with a feed back, The relation between and of a transistor, circuit is called, 1) oscillator, 2) detector, is, 3) modulator, 4) rectifier, 1) 1, 2) 1, circuit., Repeat the above steps with base currents, , ce, , c, , b, , 1), , 2), , 3), , 4), , 5), , 6), , 14) The value of current gain in common base, 3) (1 ), 4) ( 1), configuration is, The voltage gain of a transistor is higher in, 1) 1, 2) 1, 3) 1, 4) 1, the configuration of, 1) common emitter, 15) The value of amplification factor in, 2) common base, common emitter configuration is, 3) common collector, 1) 1, 2) 1, 3) 1, 4) 1, 4) all the three (1, 2, and 3), At the base emitter junction of a transistor, EXERCISE - 20 - KEY, one finds, 1) forward bias, 1) 2, 2) 3 3) 2 4) 3 5) 3 6) 4, 2) narrow depletion layer, 7) 4, 8) 2 9) 3 10) 4 11) 3 12) 1, 3) low resistance, 13) 1, 14) 2 15) 4, 4) all the three (1, 2 and 3), NARAYANAGROUP, , 125
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , , EXERCISE - 20 - HINTS, IC, I e , I e Ib I c, , 3), , , , 7), , 80 , ic 24mA , ie, 100 , , 8), , , , ic, ic, 0.9 , ie, ic ib, , 10) voltage gain =, (current gain) (resistance gain), , EXPERIMENT - 21, Identification of Diode; LED; Transistor : IC:, Resistor : Capacitor from mixed collection of, 1), such items., , p q, , Electrolytic, capacitor, , r, , s, , 2), , Siver, Ring, n, , P, , Carbon resistor, , integrated Circuit, , , , Junction Diode, , , 3), Black, dot, , , Mica capacitor, , p, , n, LED, , , , , , , , , 126, , e, , 4), , c, b, Transistor, , We require fixed collection of items like LED, 5), transistor, capacitor, resistor, p - n junction diode, etc., to perform this experiment., The identification of different items from a mixed 6), collection depends on their appearance and, working., An integrated chip (IC) is in the form of a chip, with flat back and has multiple terminals, say 8, or more. Therefore it can be easily identified., However there are ICs UM-66 and 77 series 1), which have only three pins., A transistor is a three terminal device that can, be identified by just appearance., , A resistor, a capacitor, a diode and an LED are, two terminal devices. For their identification, the following facts are used., i) A transistor is a two terminal device. It, conducts both with DC and AC voltage. Further, a resistor offers same resistance to the current, even when battery terminals are reversed., ii) A capacitor is a two terminal device which, does not conduct with DC voltage applied either, way. However it conducts with AC voltage., iii) A diode is a two terminal device that, conducts only when it is forward biased., iv) An LED is also a two terminal device which, conducts and emits light only when it is forward, biased., , PROCEDURE, INTEGRATED CIRCUIT (IC): From t he, collection of given items, pick the one having, more than three legs (pins). It is an IC which, consists of more than three i.e., 8, 10, 14 and 16, legs. It has a component of a cylindrical shape, with metal casing or made of flat back chip., TRANSISTOR : Pick the item which has three, legs. It is a transistor., All the other remaining items are two - terminal, devices. These items can be distinguished from, each other by using a multimeter as an ohmmeter., Set the multimeter in continuity test mode or, resistance measurement mode., RESISTOR: A resistor will show same value, of resistance on the multimeter even when current, is reversed., CAPACITOR: A capacitor does not show any, continuity as it blocks DC even when the current, through it is reversed. However a damaged or, short - circuited capacitor will show almost zero, resistance in most of the cases, DIODE: A diode will show a small value of, resistance when it is forward biased and very, high resistance when it is reverse biased., LED: When its p - lead is connected to positive, terminal and n - lead to common terminal gives, out faint light and when its connections are, reversed, a very high resistance is shown by the, multimeter and there is no emission of light., , EXERCISE - 21, A multimeter is a device which cannot be used, as, 1) an ammeter, 2) a voltmeter, 3) an ohmmeter, 4) a magnetometer, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 2), , An LED operates under the condition of, 1) reverse bias, 2) forward bias, 3) both in forward and reverse bias, 4) no biasing., 3) Which of the following is not a two legged, device ?, 1) resistor, 2) capacitor, 3) p - n junction diode 4) integrated circuit, 4) Which of the following is not used in making, LEDs ?, 1) Ga As 2) Cd S, 3) Ga P, 4) Ga As P, 5) Silicon and Germanium p - n junction diodes, are not used for making LEDs, 1) as their energy band gap is very large, 2) as their energy band gap is very small, 3) as greater percentage of energy is in the form, of heat, 4) as there is no energy gap in them, 6) The main precaution while connecting LED, is, 1) it should always be reverse biased, 2) it should always be forward biased, 3) it should never be reverse biased, 4) none of these, 7) Which of the following is an active component, ?, 1) resistor, 2) capacitor, 3) transistor, 4) inductor, 8) In case of an IC, the pin number is counted, with respect to a tap provided on it, 1) clockwise, 2) anticlockwise, 3) alternately, 4) diagonally, 9) Multimeter used for AC measurement of, 1) bridge rectifier, 2) p - n -p transistor, 3) n - p - n transistor 4) LDR, 10) Which of the following devices work on AC, as well as DC ?, 1) LED, 2) resistor, 3) diode, 4) capacitor, , , , EXPERIMENTAL PHYSICS, , IDENTIFICATION OF BASE OF A, TRANSISTOR:, The collector lead of a transistor is identified, from the fact that it lies on the outer side and is, far from the other two leads (emitter and base), which are situated close to each other., In order to identify the base of the transistor touch, the two probes of the multimeter to the extreme, two legs of the transistor. Note whether the, resistance between the two legs is low or high., Now interchange the probes and again note, whether the resistance is low or high., If the resistance is high in both the cases, the, central leg of the transistor is base and the two, ext reme legs are emitter and collector, respectively. It is because emitter - collector, junction - offers high resistance in both the, directions., If the resistance is low in one direction and high, in the other direction, then one of the two extreme, legs of the transistor is base., ii) DISTINGUISH BETWEEN n - p - n AND p n - p TRANSISTORS, After identifying the base of the transistor, connect negative terminal of multimeter to base, and positive terminal of multimeter to any other, terminal out of the emitter and collector. If it, gives reading in the multimeter it is p - n - p, transistor., If no reading is observed, then it is n - p - n, transistor., iii) TO SEE UNIDIRECTIONAL FLOW OF, CURRENT IN THE CASE OF A DIODE AND, LED:, A p - n junction diode and an LED (light emitting, diode) conduct only when forward biased and, do not conduct when reverse biased., The LED when forward biased not only conducts, but also emits light which helps us to distinguish, between junction diode and LED., Connect the two probes of a multimeter to the, EXERCISE - 21-KEY, two end terminals of the junction diode and note, whether resistance is low or high. Then, 1) 4, 2) 1 3) 4 4) 2 5) 3 6) 3, interchange the two probes and again note the, 7) 3, 8) 2 9) 1 10) 2, resistance whether it is high or low., EXPERIMENT - 22, If the resistance of the diode is high in the first, case, then it will be low in the second case or, Using multimeter to (i) identify base of transistor, vice-versa. It shows the unidirectional flow of, (ii) distinguish between n - p - n and p - n - p, current in junction diodes., transistors (iii) see the unidirectional flow of, , For LED also repeat the above two steps. The, current in case of a diode and an LED and (iv), LED glows by emitting light when its resistance, check the corrections or otherwise of a given, is low and it does not glow when its resistance, electronic component (diode, transistor or IC), is high. This shows unidirectional flow of, To perform this experiment, we require diode,, current through LED., transistor, LED, IC and a multimeter., , NARAYANAGROUP, , i), , 127
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, iv) CHECKING A DIODE OR TRANSISTOR, OF IC IN WORKING MODE:, A forward biased diode offers low resistance., Most of the multimeters now- a-days have a, diode testing option, marked by the diode, symbol., To check a p - n junction diode connect the, positive lead to p - connection and negative lead, to n-connection, a low resistance must be, observed. Reversing the terminals, high, resistance must be observed. If these are not, observed, then the p-n junction diode is a, damaged one., To check a transistor first identify the three pins., Connect the multimeter leads to emitter end base, pins. Note the state of conduction., Note t he state of conduction even by, interchanging the leads. If there is lot of, difference in the resistance, then connect the, multimeter leads to base and collector pins., If the resistance of base - emitter junction and, base - collector junction are low in one direction, and high in the other direction, then the given, transistor is in working order., In case if it shows low resistance in both the, directions, then the transistor is a damaged one., IDENTIFICATION OF TERMINALS OF AN IC, The terminals of an IC are commonly known as, pins., The pin configuration is according to, manufacturer’s choice., However the pin number can be identified by, using the following procedure., a) Some of the ICs such as voltage regulator ICs, (IC-7805) and tone generator ICs (UM-66) have, only three pins., b) The multiple ICs like IC-555 come in two, packages. There is a cut, dot or a tab from which, the pin number is to be counted in anticlockwise, direction., , 1), , 128, , 2), , 3), , The arrow in a given transistor indicates, 1) direction of flow of electrons, 2) direction of flow of holes, 3) both the directions of flow of holes and, electrons, 4) neither the direction of flow of holes nor of, electrons, The correct relation between current gain,, resistance gain and power gain is, 1) power gain = (current gain) (resistance gain), 2) power gain = (current gain)2 (resistance gain), 3) power gain =, , current gain , , 2, , resistance gain , , 4) power gain = (current gain) (resistance gain)2, 4) Digital multimeters use the following, component for display, 1) transistor, 2) p - n junction diode, 3) LED, 4) all of these, 5) Multimeter is used as a voltmeter when, 1) low resistance is connected in parallel to a, galvanometer, 2) high resistance is connected in parallel to a, galvanometer, 3) low resistance is connected in series to a, galvanometer, 4) high resistance is connected in series to a, galvanometer., 6) In electronic industry, now-a-days Carbon, resistors gaining popularity have percentage, accuracy as, 1) 20%, 2) 10%, 3) 5%, 4) 2%, 7) Carbon resistors commonly used in electronic, circuits are made of, 1) copper and carbon 2)magnesium and carbon, 3) carbon and clay, 4) carbon and constantan, 8) When a multimeter is connected to a, transistor, it conducts when, 1) base - emitter is forward biased, 2) base - collector is forward biased, 3) base - emitter is reverse biased, 4) base - collector is reverse biased, 9) A digital multimeter when used as a continuity, tester, 1) it is put in the lowest resistance range, 2) it is put in the highest resistance range, 3) it gives a beep, 4) it will be put off, EXERCISE - 22, 10) A digital multimeter consists of the following, Which is not the function of analog, component, 1) an amplifier 2) analog to digital converter, multimeter?, 3) numeric / a - numeric digital display, 1) to identify the base of a transistor, 4) all the above, 2) to identify the terminals of an IC, 3) to measure the temperature gradient, EXERCISE - 22 - KEY, 4) to check whether the given electronic, 1) 3, 2) 2 3) 2 4) 3 5) 4 6) 4, component is in working order or not, 7) 3, 8) 1 9) 3 10) 4, NARAYANAGROUP
Page 1990 :
JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 6., , ADVANCED, MAIN POINTS, 1., , Least count of a screw guage =, pitch of screw, number of divisions on circular scale, , 2., 3., , 4., , Least count of Vernier calliper = 1 main scale, division - 1 vernier scale division, In both screw gauge and vernier calliper actual, reading = observed reading - zero error, = observed reading + zero correction, Determining the value of ‘g’ using a simple, pendulum, Theoretically T=2, , 5., , L, L, g 4 2 2, g, T, , where L = length of the thread + radius of the bob, Determining Young’s Modulus of a given wire, by “Searle’s Method”:, To determine Young’s Modulus, we can perform, an ordinary experiment. Lets hang a weight ‘m’, from a wire, from Hook’s law:, x, mg, l, Y x 02, A, r Y, l0 , , , mg, , , Determining specific heat capacity of an, unknown liquid using calorimeter:, Figure shows the Regnault’s apparatus to, determine the specific heat capacity of a unknown, liquid. A solid sphere of known specific heat, capacity S1 having mass m1 and initial temperature, , 1 , is mixed with the unknown liquid filled in a, calorimeter. Let masses of liquid and calorimeter, are m2 and m3 respectively, specific heat capacities, are S 2 and S3 and initially they were at room, temperature 2 . When the hot sphere is dropped, in it, the sphere looses heat and the liquid, calorimeter system takes heat. This process, continues till the temperature of all the elements, becomes same (say )., Heat lost by hot sphere = m1s1 1 , Heat taken by liquid and calorimeter =, , m2 s2 2 m3 s3 2 , If there were no external heat loss, Heat given by sphere = Heat taken by liquid Calorimeter system, , m1s1 1 m2 s2 2 m3 s3 2 , m1s1 1 m3 s3, s, , , 2, Get, m2 2 , m2, m, , If we change the weight, the elongation of wire will, increase proportionally. If we plot elongation v/s, mg, we will get a straight line, x, , tan , , By measuring the final (steady state) temperature, of the mixture, we can estimate s2 : specific heat, capacity of the unknow liquid. To give initial, temperature 1 to the sphere, we keep it in steam, chamber "O" , hanged by thread. Within some, time (say 15 min) it achieves a constant temperature, 1 ., , Steam, , l0, r 2 Y, , Steam, Chamber, ‘O’, , mg, , l0 , ,, r 2Y , , By measuring its slope and equating it to , we can estimate Y ., NARAYANAGROUP, , Disk D, 129
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 8., Water, Heat, , C, , Cabrimeter, , Now the calorimeter, filled with water (part C) is, taken below the steam chamber, the wooden, removable disc D is removed, and the thread is, cut. The sphere drops in the water calorimeter, system and the mixing starts., If sp.heat capacity of liquid s2 were known and, that of the solid ball s1 is unknown then we can, find s1 , 7., , m 2 s 2 m 3 s 3 2 , m 1 1 , , Determining speed of sound using resonance, tube:Resonance tube is a kind of closed organ, pipe. So its natural frequencies are, V 3V 5V, ,, ,, ......, 4leq 4leq 4leq, If the air column resonates with a tuning fork of, frequency f 0 then, , 2 n 1, , V, f0, 4 leq, , leq 2 n 1 , , V, 4 f0, , leq l e where l is length of the air column in, tube and e is end correction., , l1 e , , V, 3V, l2 e , V 2 f 0 l2 l1 , ,, 4 f0, 4 f0 so, , 9., , Verification of Ohm’s law using voltmeter and, ammeter: Ohm’s law states that the electric current, I flowing through a conductor is directly, proportional to the P.D. (V) across its ends, provided that the physical conditions of the, conductor (such as temperature, dimensions, etc.), are kept constant. Mathematically., V I or V IR, Here R is a constant known as resistance of the, conductor and depends on the nature and, dimensions of the conductor., METER BRIDGE: Meter bridge is a simple, case of Wheatstone’s-Bridge and is used to find, the unknown resistance. The unknown resistance, is placed in place of R, and in place of S , a known, resistance is used, using R.B. (Resistance Box)., There is a 1m long resistance wire between A and, C. The jockey is moved along the wire. When, , R 100 l S l then the Bridge will be, balanced, and the galvanometer will gives zero, deflection. " l " can be measured by the meter, scale., The unknown resistance is, l, ............(1), 100 l, If length of unknown wire is L and diameter of the, wire is d, then specific resistance of the wire, RS, , d2 , R, d 2 l , , , , 4, , S, from eq.(1), 4 L 100 l , , L, , Un known D, , Max Permissible Error in speed of sound due, , R, , to error in f 0 , l1 , l2 :, for Resonance tube experiment, , V 2 f 0 l2 l1 , ln V ln 2 ln f0 ln l2 l1 , max. permissible error in speed of sound =, , S, C, , G, , A, , Q, , P, R, , f, l l1, dV , 0 2, V , f0, l2 l1 , , max, 130, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , Unknown, resistance, , S, RB, , R, , P, , EXPERIMENTAL PHYSICS, , Known resistance, in resistance box, , O Knitting Needle, x, , P, , Q, , A, , C, l, +, , (100–l), , –, , K, , End Corrections: In meter Bridge circuit, some, extra length (is found under metallic strips) at end, points A and C. So some additional length, , A T, , TB, , and should be included at ends for accurate, results. Hence in place of l we use l and in, place of 100 l , we use 100 l (where , , A, y, B, Determination of index correction, , and are called end correction). To estimate , and , we use known resistance R1 and R2 at, the place of R and S in meter Bridge. Suppose we, get null point at l1 distance then, , Index error = Observed distance - Actual distance, (Just like zero error in screw gauge, it is the excess, reading)., , R1, l1 , , R2 100 l1 ...................(i), , To determine index error, mirror and object needle, and placed at arbitrary position. For measuring, acual distance, a knitting needle is just fitted between, the pole of mirror and object needle “O”. The, length of knitting needle will give the actual object, distance while the seperation between indices A, and B at that instant is the observed distance., , Now we interchange the position of R1 and R2, and get null point at l2 distance then, R2, l2 , , R1 100 l2 .................(ii), , Solving, , , , equation, , (i), , and, , (ii), , get, , R2l1 R1l2, R l R2l2, and 1 1, 100, R1 R2, R1 R2, , These end corrections and are used to, , So index error is e = Observed distance - Actual distance, =Seperation between indices A and B - Length of, knitting needle, , modify the observations, , once we get e, in every observation, we get, , INDEX ERROR: In u - v method, we require, , Actual distance = Observed distance (seperation, between the indices) - Excess reading (e), , the distance between object or image from the pole, (vertex) of the mirror (actual distance). But, practically we measure the distance between the, indices A and B. (Observed distance), which need, not exactly coincides with object and pole, there, can be a slight mismatch called index error, which, will be constant for every observation., NARAYANAGROUP, , *There is an another term, Index corraction which, is invert of index error., Index correction = - index error, , 131
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , LEVEL-V, 10, , 0, , SINGLE ANSWER QUESTIONS, 1., , The length of a rectengular plate is measured, by a meter scale and is found to be 10.0 cm., Its width is measured by vermier callipers as, 1.00 cm. The least count of the meter scale, and varmier callipers are 0.1cm and 0.01cm, respectively (Obviously). Maximum permissible, error in area measurement is, , (mm), 60, , (A) 11mm (B) 11.65mm (C) 11.650mm (D) 11.6mm, , 8., , (A) 0.2 cm 2 (B) 0.1 cm 2 (C) 0.3 cm2 (D)Zero, 2., , 3., , In the previous question, minmum possible error in area measurement can be, (A) 0.02 cm 2, , (B) 0.01 cm 2, , (C) 0.03 cm2, , (D)Zero, , , 2 L , To estimate ‘g’ from g 4, , error in, T2 , , measurement of L is 2% and error in measurement of T is 3% . The error in estimated, ‘g’ will be, , (A) 8%, 4., , (B)10 %, , 132, , (D) 20 %, , (B) 44.0 kg, (D) 44.00 kg, , In ohm’s law experiment, potential drop across, a resistance was measured as v 5.0 volt and, current was measured as i 2.00 amp. Find, the maximum permissible error in resistance., (A) 1.5% (B) 2.5%, , 7., , (C)15 %, , (C) 1%, , 9., , In a complete rotation, splindle of a screw, 1, gauge advances by, mm. There are 50, 2, divisions on circular scale. The main scale has, 1, , , 1, mm marks is graduated to mm , 2, 2, , , If a wire is put between the jaws, 3 main scale, divisions ar e clear ly visible, and 20th division, of circular scale coincides with the reference, line. Find diameter of wire in correct S.F., (A)1.7mm (B)1.70mm (C)3.40mm (D)3.20mm, Read the vernier, 0 mm, , 10, , 20, , 30, , 40, , 50, , 30, , 40, , 50, , 0.1 mm, , (D) 5%, , The mass of a ball is 1.76 kg. The mass of 25, such balls is, (A) 0.44 103 kg, (C) 44 kg, , 6., , (C) 3%, , The dimensions of a rectangular block measured with a vernier callipers having least, count of 0.1 mm is 5 mm 10 mm 5 mm . The, maximum percentage error in measurement, of volume of the block is, (A) 5 %, , 5., , (B) 6%, , 70, , 5, , 0 mm, , 10, , 20, , 0.1 mm, , (A) 15.6 mm, (B) 15.60 mm, (C) 14.6 mm, (D) 14.4 mm, 10. Read the special type of vernier, 0 mm, , 10, , 20, , 30, , 40, , 50, , 0 mm, , 10, , 20, , 30, , 40, , 50, , (D) 5%, , Read the normal screwgauge main scale has, only mm marks. Circular scale has 100, division. In complete rotation, the screw, advances by 1 mm., , (A) 13.6 mm, (C) 12.6 mm, , (B) 13.60 mm, (D) 12.60 mm, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 11. Find the thickness of the object using the, defected vernier callipers, 0 mm, , 10, , 20, , 0, , 1, , 0.1 mm, , 30, , 40, , 14. For the third resonance, which option shows, correct mode shape for displcement variation, and pressure variation., , 50, , (A), 0 mm, , 10, , 20, , 30, , 40, , 50, , (B), Displ., , Pressure, , Displ., , Pressure, , 0.1 mm, , (A) 14.0 mm, (C) 15.2 mm, , (B) 14.6 mm, (D) 15.20 mm, , (C), , 12., , (D), , Pressure, , 0 mm, , 10, , 20, , 30, , 40, , 50, , 0.1 mm, , 0 mm, , 10, , 20, , 30, , 40, , 50, , 0.1 mm, , (A) 11.8 mm, (C) 12.2 mm, , (B) 12.0 mm, (D) 11.2 mm, , 13. The main scale of a vernier callipers reads 10, mm in 10 divisions. 10 divisions of Vernier, scale coincide with 9 divisions of the main, scale. When the two jaws of the callipers touch, each other, the fifth division of the vernier, coincides with 9 main scale divisions and the, zero of the vernier is to the right of zero of, main scale. When a cylinder is tightly placed, beyween the two jaws, the zero vernier scale, lies slightly to the left of 3.2 cm and the fourth, vernier division coincides with a main scale, division. the diameter of the cylinder is:, (A)3.09 cm, (C)3.04 cm, NARAYANAGROUP, , (B)3.14 cm, (D) none of these, , Displ., , Pressure Displ., , 15. A cube has a side of length 1.2 10 2 m ., Calculate is volume: [IIT 2003], (B) 1.73 106 m3, (A) 1.7 106 m3, (D) 1.732 106 m3, (C) 1.70 106 m3, 16. A wire of length l 6 0.06 cm and radius, r 0.5 0.005 cm and m 0.3 0.003 g ., Maximum percentage error in density is:, [IIT 2004], (A) 4, (B) 2, (C) 1, (D) 68, 17. A screw gauge having 100 equal divisions and, pitch of length 1mm is used to measure the, diameter of a wire of length 5.6 cm. The main, scale reading is 1mm and 47th circular division, coincides with the main scale. the surface area, of wire cm2 to appropriate significant figure, is equal to (use 22 / 7 ):[IIT-2004], (A) 2.1cm2 (B) 2.6cm2 (C) 5.2cm2 (D)1.3cm2, 18. The number of circular division on the shown, screw gauge is 50. It moves 0.5 mm on main, scale for one complete rotation and main scale, has 1/2 mm marks. The diameter of the ball, is:, [JEE 2006], 0, , 5, , (A) 2.25 mm, (C) 1.20 mm, , 2 25, , (B) 2.20 mm, (D) 1.25 mm, 133
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 19. If nth division of main scale coincides with, (n+1)th divisions of vernier scale. Given one, main scale division is equal to ‘a’ units. The, least count of the vernier is equal to:, [IIT 2003], a, n 1, n, (A), (B), (C), (D) n, n 1, a, a, 20. The side of a the cube is measured by vernier, callipers (10 divisions of a vernier scale, coincide with 9 divisions of mainscale, where, 1 division of main scale is 1mm). The main, scale reads 10 mm and first division of vernier, scale coincides with the main scale. Mass sof, the cube is 2.736 g. The density of the cube in, appropriate significant figures is equal to:[IIT, 2005], (A) 1.33 gm / cm 3, (B) 0.66 gm / cm3, (C) 2.66 gm / cm 3, (D) 4.88 gm / cm3, 21. Student I, II and III perform an experiment, for measuring the acceleration due to gravity, (g) using a simple pendulum. They use different, lengths of the pendulum and/or record time for, different number of oscillations. The, observations are shown in the table., Least count for length = 0.1 cm, Least count for time = 0.1 s, Student Length of the pendulum (cm), I, 64.0, II, 64.0, III, 20.1, Number of, Total time for, Time period, oscillations(n) oscillations(n), (s), 8, 128.0, 16.0, 4, 64.0, 16.0, 4, 36.0, 9.0, If EI , EII and EIII are the percentage errors, in g, i.e.,, , g, , 100 , , g, , , for students, I, II and III,, , respectively,, (A) EI 0, , (B) EI is minimum, , (C) EI EII, (D) EII is maximum, 22. In the given circuit, no current is passing, through the galvanometer. If the crosssectional diameter of the wire AB is doubled,, then for null point of galvanometer, the value, 134, , of AC would be:, (A) 2 X, , (B) X, , G, , A, , X, , C, , X, (C), B, 2, , (D) 3X, , 23. The pitch of a screw gaguage is 0.5 mm, and there are 100 divisions on its circular, scale. The instrument reads +2 divisions when, nothing is put in-between its jaws. In measuring, the diameter of a wire, there are 8divisions on, the main scale and 83rd divison coincides withe, the reference line. Then the diameter of the, wire is, (A)4.05 mm, (B)4.405 mm, (C)3.5 mm, (D)1.25 mm, 24. The pitch of a screw gauge having 50 divisions, on its circular scale is 1 mm.When the two jaws, of the screw gauge are in contact with each, other, the zero of the circlar scale lies 6, divisions below the line of graduation. When, a wire is placed between the jaws, 3 linear scale, divisions are clearly visible while 31st divisions, on the circular scale coincides with the, reference line. The diameter of the wire is:, (A) 3.62 mm, (B) 3.50 mm, (C) 3.5 mm, (D) 3.74 mm, 25. The smallest division on the main scale of, vernier callipers is 1 mm, and 10 vernier, divisions coincide with 9 main scale divisions., While measuring the diameter of a sphere, the, zero mark of the vernier scale lies between 2.0, and 2.1 cm and the fifth division of the vernier, scale coincides with a main scale division. Then, diameter of the sphere is, (A) 2.05 cm, (B) 3.05 cm, (C) 2.50 cm, (D)None of these, , MULTIPLE ANSWER QUESTIONS, 26. While findling specific heat capacity using, calorimeter, error might occur due to :, A) Absence of water equivalent, B) Absence of heat loss reducing covers., C) Presence of stirrer, D) No loss comes, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 27. In Searle’s apparatus :, A) One wire is reference wire., B) Load cannot be increased beyond limit of, elasticity, C) Spirit level should be adjusted for each reading, D) No Vernier scale is used, 28. A student uses a simple pendulum of exactly, 1m length to determine g , the acceleration, due to gravity. He uses a stop watch with the, least count of 1 sec for this and records 40, seconds for 20 oscillations. For this, observation, whih of the following statement(s), is (are) true ? [IIT-2010], (A) Error T in measuring T, the time period, is, 0.05 seconds, (B) Error T in measuring T, the time period, is 1, second, (C) Percentage error in the determination of g is, 5%, (D) Percentage error in the determination of g is, 2.5 %, , PASSAGE TYPE QUESTIONS, Passage-1, In u - v method, we require the distance between, object or image from the pole (vertex) of the mirror, (actual distance). But practically we measure the, distance between the indices A and B. (Observed, distance), which need not exactly co-inside with, object and pole, there can be a slight mismatch, called index error, which will be constant for every, observation., O Knitting Needle, x, , P, , the pole of mirror and object needle “O”. The, length of knitting needle will give the actual object, distance while the seperation between indices A, and B at that instant is the observed distance., So index error is e = Observed distance - Actual distance, =Saperation between indices A and B - Length of, knitting needle once we get e, in, every observation, we get, Actual distance = Observed distance (seperation, between the indices) - Excess reading (e), There is an another term, Index corraction which, is invert of index error., Index corraction = - index error, 29. To find index error for u , when a knitting, needle of length 20.0 cm is adjusted between, pole and object needle, the saperation between, the indices of object needle and mirror was, observed to be20.2 cm. Index corraction for u, is, (A)-0.2 cm (B)0.2 cm (C)-0.1 cm (D)0.1 cm, 30. To find index error for v, when the same, knitting needle is adjusted between the pole, and the image needle, the saperation between, the indices of image needle and mirror was, found to be 19.9cm. Index error for v is, (A) 0.1cm (B)-0.1 cm (C) 0.2 cm (D)-0.2cm, 31. In some observation, the observed object, distance (Saperation between indices of object, needle and mirror) is 30.2 cm, and the observed, image distance is 19.9 cm. Using index, corraction from previous two equations,, estimate the foccus distance of the concave, mirror!, (A) 36 cm (B) 20 cm (C) 12 cm (D) 8 cm, , Passage-2, , A T, , TB, , A, y, B, Determination of index correction, , Index error = Observed distance - Actual distance, (Just like zero error in screw guage, it is the excess, reading)., To determine index error, mirror and object needle, and placed at arbitery possition. For measuring, acual distance, a knitting needle is just fitted between, NARAYANAGROUP, , Figure shows an electrical calorimeter to determine, specific heat capacity of an unknown liquid, First, of all, the mass of empty calorimeter ( a copper, container) is measured and suppose it is ‘ m1 ’., Then the unknown liquid is poured in it. Now the, combined mass of calorimeter + liquid system is, measured and let it be ‘ m2 ’. So the mass of liquid, , m2 m1 . Initially, temperature 0 ., , is, , both were at room, , Now a heater is immersed in it for time interval ‘t’., The voltage drop across the heater is ‘V’ and, current passing through it is ‘ I ‘. Due to heat, 135
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, supplied, the temperature of both the liquid and, calorimeter will rise simultaneously. After t sec;, heater was switched off, and final temperature is, r . If there is no heat loss to surroundings., Heat supplied by the heater = Heat absorbed by, the liquid + heat absorbed by the calorimeter, , VI t m2 m1 Sl f, , 0 m1SC f 0 , , VI t m S, 1 C, f 0, The specific heat of the liquid Sl , m2 m1 , A, V, , Mass of empty calorimeter was 1.0 kg and the, combined mass of calorimeter, +liquid is 3.0 kg. The specific heat capacity of, the calorimeter 3.0 10 3 J / kg 0 C .The, falls in temperature 350 second after switching, off the heater was 7.00 C .Find the specific heat, capacity of the unknown liquid in proper, significant figures., (A) 3.5 103 J / kg 0 C, , (B) 3.50 103 J / kg 0C, , (C) 4.0 103 J / kg 0C (D) 3.500 103 J / kg 0C, 33. If mass and specific heat capacity of, calorimeter is negligible, what would be, maximum permissible error in Sl . Use the data, mentioned below., , Heater, Stirrer, Unknown liquid, , m1 0, Sl 0, m2 1.00kg ,, V 10.0V , I 10.0 A,, t 1.00 102 sec, 0 150 C ,, , Calorimeter, , Corrected 1 650 C, , Temperature , , (A) 4%, (B) 5%, (C) 8%, (D)12%, 34. If the system were loosing heat according to, Newton’s cooling law, the temperature of the, mixture would change with time according to, (while heater was on), , , , , , , , time (t), , Radiation correction: There can be heat loss to, environment. To compensate this loss, a correction, is introduced., Let the heater was on for t sec, and then it is, switched off. Now the temperature of the mixture, falls due to heat loss to environment. The, temperature of the mixture is measured t/2 sec. after, switching off. Let the fall in temperature during this, time is , Now the corrected final temperature is taken as, 'f f , 32. In this experiment voltage across the heater, is 100.0 V and current is 10.0A, and heater, was switched on for t=700.0 sec. Initially all, elements were at room temperature, o 10.00 C and final temperature was, measure as f 73.00 C ., , 136, , (B), , (A), t, , , , t, , , , (C), , (D), t, , t, , MATRIX MATCHING QUESTIONS, 35. Column I, A) Back - lash error B) Zero error, C) Vernier callipers, D) Error in screw gauge, Column II, (P) Always subtracted, (Q) Least count = 1 MSD - 1 VSD, (R) May be -ve or +ve, (S) Due to loose fittings, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , ASSERSON AND REASON QUESTIONS, (A) Assertion is True, Reason is True; Reason, is correct explanation for Assertion., (B) Assertion is True, Reason is True; Reason, is not correct explanation for Assertion., (C) Assertion is True, Reason is False., (D) Assertion is False, Reason is True., 36. Assertion (A): Least count of all screw based, instruments is same., Reason (R): Least count for all screw based, instruments are found using the ratio pitch per, division of circular scale., 37. Assertion (A): Backlash error can be minimised, by turning the screw in one direction only when, fine adjustment is done., Reason (R): Backlash error is caused due to wear, and tear or loose-fittings in screws., , LEVEL-V - HINTS, 1., , dA , 0.01 0.01 ; A 0.02, max, dA max 0.02 10.0cm2 0.2cm 2, 2., , dA , , 0 dA min 0, A min, for minimum error use absolute sign., , 3., , dg, dL, dT, , 2, Answer 8%, g, L, T , , 4., , 0.1 0.1 0.1, dv , v 5 10 5 0.05, max, , SUBJECTIVE TYPE QUESTIONS, 38. A screw gauge having 100 equal divisions and a, pitch of length 1mm is used to measure the, diameter of wire of length 5.6cm. The main scale, reading is 1mm and 47th circular division coincides, with the main scale. Find the curved surface area, of the wire in cm2 to appropriate significant figures., (Use 22 / 7 ) (IIT JEE -2004), 39. In Searle’s experiment , the diameter of the wire, as measured by a screw gauge of least count, 0.01 cm is 0.050cm. The length , measured by a, scale of least count 0.1cm, is 110.0cm.When a, weight of 50N is suspended from the wire, the, extension is measured to be 0.125cm by a, micrometer of least count 0.01 cm. Find the, maximum error in the measurement of Young’s, modulus of the material of the wire from these, data.., (IIT -JEE 2004), 40. The side of a cube is measured by vernier callipers, (10 divisions of the vernier scale coincide with 9, divisions of the main scale , where 1 division of, main scale is 1mm) .The main scale reads 10mm, and first division of vernier scale coincides with, the main scale. Mass of the cube is 2.736g. Find, the density of the cube in appropriate significant, figures., , (A) ; A l b, dA, dl db, 0.1 0.01, , , , , A, l, b, 10.0 1.00, for max error use positive sign., , 5., , 6., , dv , So v 100 5, max, Answer should be in 3 significant figures. As 25 is, exact number of balls., R, , v i, dR , v, , , v i 1 , , i, R max v, i, , v 5.0 volt, , , , v 0.1 volt, , i 2.00 amp i 0.01 amp, , dR , 0.1 0.01 , %, , , , 100% 2.5%, R max 5.0 2.00 , value of R from the observation, v 5.0, R , 2.5, i 2.00, So we can write R 2.5 2.5% , , LEVEL-V - KEY, 01) A 02) D 03) A 04) A, 08) B 09) A 10) B 11) A, 14) B 15) A 16) A 17) B, 21) B 22) B 23) B 24) B, 27)A,B,C 28) A,C, 32) A 33) C 34) C, 35) A-S ; B-P,R; C-Q;, 36) D 37)A 38) 2.6, NARAYANAGROUP, , 05) B, 12) C, 18) C, 25) A, 29) B, , 06) B 07) B, 13) A, 19) A 20) C, 26)A,B, 30) B 31) C, , 7., 8., , 1, , 1/ 2mm , 3 mm 2.0 , , 2, , 50 , 1.5 0.20 1.70 mm, , 9., D-R,S, 39) 4.89 40) 2.66, , 1 mm , , Object thickness=11 mm 65 100 = 11.65 mm, , , dia.of e is, , Thickness of the object = (main scale, reading)+(vermier scale Reading) (least count), where least count = (Main scale division- vernier, Scale division), 137
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 10., , 11., , 12., , 13., , 15., , = 1 mm-0.9 mm (from fig 1) = 0.1 mm, So thickness of the object, =15 mm + (6) (0.1 mm) = 15.6 mm, Thickness of the object = (main scale, reading)+(vermier scale Reading) (least count), where least count = (Main scale division- vernier, Scale division), = 1 mm - 19/20 mm (from fig 1) =0.05 mm, So thickness of the object, =13 mm + (12) (0.05 mm) = 13.60 mm, From first figure, Excess reading (zero error)=0.6, mm. If an object is placed, vernier gives 14.6 mm, in which there is 0.6 mm excess reading, which has, to be subtracted. So actual thickness=14.60.6=14.0 mm we can also di it using the formula, Actual reading = observed-excess reading, reading (zero error), =14.6-0.6 =14.0 mm, Zero error=main scale reading + (vernier scale, reading) (least count), = -1 mm+ 6(0.1 mm) = -0.4 mm, observed reading = 11.8 mm, So actual thickness=11.8-(-0.4)=12.2 mm, Zero error=0.5 mm=0.5 m, diseved reading of cylin, 3der=3.1 cm +(4) (0.01cm), =3.14 cm, actual thickness of cylinder =(3.14)-(0.06), =3.09 cm, Number of significant figures should be two., , 16. , , m, d dm dl, dr, , , 2, 2, m, l, r, , l r, , 21., 23., , 24., , 25., , 26., , 27., , 17. s l d, 28., , 22, 5.6cm 0.147 cm, 7, 2, , 2.5872 cm 2.6 cm, , 2, , 0.5mm, 0.01mm, 18. L.C , 50, Zero error = 5 L.C 0.05mm, MSR 2 0.5mm 1mm, CSR 25 0.01mm 0.25 mm, So diameter = MSR + CSR - Zero error, 1.20mm, n, a, a, 19. L.C 1 MSR 1 VSR a , n 1, n 1, , 138, , m, v, More is number of oscillations less will be error in, measurement of time., 0.5, L.C , mm 0.005mm, 100, Zero error = 2 0.005 mm 0.01mm, So zero correction is 0.01mm, M.S.R = 4 mm, C.S.R = 83 L.C 0.415mm, So diameter = 4mm 0.415 mm - 0.01 nm, = 0.405 mm, 1, L.C , mm 0.02mm, 50, M.S.R = 3 mm, C.S.R = 31 0.02 mm 0.62mm, Zero error correction = -0.12 mm, So diameter = 3.50 mm, L.C = 0.1 mm = 0.01 cm, M.S.R = 2.0 cm, V.S.R = 0.05 cm, So diameter = 2.05 cm., Water equivalent is important since the calorimeter, also will lose or gain heat. Heat loss has to, reduced (by Radiation and / or Convection)., So, Choices (a) and (b) are correct., Use of stirrer, maintains uniform temperature So,, choices (c) and (d) are wrong., These are facts used in setting up and use teh, Searle’s apparatus., Choie (d) is wrong, , 20. , , g 2T, T, t, , 0.05, , T 0.05 s ;, g, T, T, t, , % error =, , g, 100 5%, g, , 29.. Index error (excess reading)=Observed readingActual reading, =20.2-20.0 =0.2 cm, 30 e = 19.9 cm -20.0cm = -0.1 cm, 31. u = 30.2-0.2 (excess reading), =30.0 cm, v=19.9-(-0.1) (excess reading), =20.0 cm., 1 1 1, f 12.0 cm, f v u, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , 32. S l , , 100.0 10.0 700.0 , , 80.0 10.0, 3.0 1.0, , 1.0 3.0 10 3 , , 3.5 103 J / kg 0C, According to addn and multiplication of S.F, 33. If m1 0, Sl 0 ;, , EXPERIMENTAL PHYSICS, , 22, 1.47 56mm 2, 7, 2.58724cm 2 2.6cm 2, 39. Maximum percentage error in Y is given by, Dl , , y, , VIt, Sl , m2 f 0 , , dSl V I t m2 r 0, , , , , Sl, V, I, t, m2, r o, 0.1 0.1 0.01 10 2 0.01 1 1, , , , , 8%, 10.0 10.0 1.00 102 1.00 50, 34. As the temperature increases, heat loss to, surrounding increasses. After some time the rate at, which heat is lost becomes equal to rate at which, heat is supplied and an equilibrium or steady state, is achieved. Hence temperature becomes constant, after some time., C is correct, 35: Back - lash error is caused by loose fittings, wear, and tear etc., in the screw mechanisms, , Y , D x L, 2, , , , , L, Y max, D x, 0.001 0.001 0.1 , 2, , , 0.0489, 0.05 0.125 110 , , , , So maximum percentage error = 4.89%, 40. Least count of vernier callipers =, 1 division of main scale, Number of divisions in vernier scale, 1, 0.1mm, 10, The side of cube = 10mm+1×0.1mm=1.01cm, Nowdensity, , , So, A S ., Zero error may be +ve or -ve but will always be, , , , subtracted So, B P and R ., In vernier callipers, least count is the difference, , Mass, 2.736 g, , 2.66 g cm 3, 3, 3, Volume 1.01 cm, , (to correct number of significant figures), , LEVEL-VI, , between 1 MSD 1VSD ., So, C Q ., Error is screw gauge may be +ve or -ve and may, be due to loose fitting of the circular scale., , So, D R and S , 36. Assertion is false since itdepends on the scale used., Reason is true by concept, 37. Both are true reason explains assertion.Since, running over the defect reapeatedly will cause large, deviation., 1mm, 0.01mm, 100, Diameter = MSR+CSR, = 1mm+47 (0.01)mm=1.47mm, Surface area, , 38: Least count , , W, L, , 2, x, D, 4, , SINGLE ANSWER QUESTIONS, 1., , In, , resonance, , tube, , exp. we find, l1 25.0cm and l2 75.0 cm . If there is no, error in frequency. What will be max, permissible error in speed of sound (take, , 2., , f 0 325 Hz )., (A) 0.2m/s (B) 0.65m/s (C) 1.3m/s (D) 2.6m/s, In Searle’s exp to find Young’s modulus, the, diameter of wire is measured as D=0.05 cm, length of wire is L=125 cm , and when a, weight, m=20.0 kg is put, extension in wire, was found to be 0.100 cm . Find maximum, permissible percentage error in young’s, modulus Y ., (A) 2.1% (B) 3.2%, , NARAYANAGROUP, , (C) 4.3% (D) 5.4%, 139
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, 3., , To find the value of 'g' using simple, pendulum. T=2.00 sec; l=1.00 m, was, measured. Estimate maximum permissible, 2, error in 'g' . use 10 , , (A) 0.1 m / s 2, 4., , (B) 0.2 m / s 2, , 8., , (C) 0.3 m / s 2, (D) 0.4 m / s 2, A student performs an experiment for, , 4 2l , g, , 2 , l 1m, and he, determination of, T , commits an error of l . For T he takes the, time of n oscillation with the stop watch of least, count t and he commits a human error of 0.1, sec. For which of the follwing data, the, measurement of g will be most accurate?, , 9., , temperature comes out to be 300 c , then total, heat loss to surrounding during the, experiment, is (Use the specific heat capacity, of the liquid from previous question)., (A)20 kcal (B)15 kcal (C)10 kcal (D)8 kcal, If a tuning fork (340 Hz 1%) is used in the, resonance tube method, and the first and, second resonance lengths are 24.0 cm and, 74.0 cm respectively. Find max. permissible, error in speed of sound., (A) 5.03 m/s, (B) 0.503 m/s, (C) 2.51 m/s, (D) 0.251 m/s, If emf of battery is 100 V ,then what was the, resistance of Rheostat adjusted at reading, ( i=2A,V=20V), , A L 0.5, t 0.1, n 20, B L 0.5, t 0.1, n 50, C L 0.5, t 0.01, n 20, D L 0.1, t 0.05, n 50, 5., , 4A, 3A, 2A, 1A, 0, , In the experiment, the curve between, X and W is shown as dotted line (1). If, we use an another wire of same material, but, with double length and double radius. Which, of the curve is expected., , 10 20 30 40, , +, , R, , 1, 3, , +, w, , 7., , 140, , (), Rh, , 2, , 6., , K, , –, , A, , x, , V, , (A)1, (B)2, (C) 3, (D)4, If we use very thin and long wire, output X , , (A) Sensivity , of experiment is, input W , increase, (B) Young’s modulus will remain unchange, (C) Wire may break or yield during loading., (D) All of the above, If accidently the calorimeter remained open, to atmosphere was for some time during the, experiment, due to which the steady state state, , V, , –, , (A)10 (B) 20 (C) 30 (D) 40, 10. If three wires of same material but different, dimension were used in place of unknown, resistance, we get these I-V curve, , Match the column according to correct curve:, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , Wire dimension, Corresponding curve, (p) l 1m, radius 1mm, (i) Curve(1), (q) l 1m, radius 2mm, (ii)Curve(2), (r) l , , 1, 1, m, radius mm, 2, 2, , 13. If by mistake, voltmeter is connected in series, with the resistance then i-v curve expected is, (Here i = reading of ammter ,v=reading of, voltmeter), , , (iii)Curve(3), , +, , –, , (A) ( p ) (ii ); q iii ; r i , , Rh, , (B) ( p ) (iii ); q ii ; r i , (C) ( p ) (i ); q ii ; r iii , (D)None of these, 11. i v/s V curve for a non-ohmic resistance is, shown. The dynamic resistance is maximum, at point, , V, , (A), , (B), v, , (C), (A) a, (B) b, (C) c (D)same for all, 12. If by mistake Ammeter is connected parallel, to the resistance then i-v curve expected is, (Here i = reading of ammeter, v=reading of, voltmeter), +, , –, Rh, , V, , A, , A, , v, , (D), v, , v, , 14. In the experment of ohm’s law. When potential, difference 10.0 V is applied, current measured, is 100A. If length of wire of found to be 10.0, cm, and diameter of wire is 2.50 mm, then the, maximum permissible, (A)1.8% (B)10.2% (C) 3.8% (D) 5.75%, 15. From some instruments, current measured is, 1=10.0Amp., potential different measured is, V 100.0 V, length of wire is 31.4 cm, and, diameter of wire is 2.00 mm (all in corract, significant figure). The resistivity of wire(in, correct significant figure) will be (use 3.14), (A)1.00 104 m, (B)1.0 10 4 m, , (A), v, , (C), , v, , (D), v, , NARAYANAGROUP, , (C) 1 104 m, , (B), , v, , (D)1.000 104 m, 16. In the previous question. maximum, permissible error is resistivity and resistance, measurement will be (respecitvely), (A) 2.14%, 1.5%, (B)1.5%, 2.45%, (C) 2.41%, 1.1%, (D)None of these, 141
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, Q.No.17,18,19. Consider the meter bridge circuit., , G1, R, , S, RB, , R2, P, , Q, , A, , G2, , Rr, C, , R1, , (A), , l, +, , –, , (100–l), , K, , V, , 17. If resistance S in RB 300 , then the, balanced length is found to be 25.0 cm from, end A.The diameter of unknown wire is 1mm, and length of the unknown wire is 31.4 cm. The, specific resistivity of the wire should be, (A) 2.5 104 m, (C) 4.5 104 m, , (B) 3.5 104 m, (D)None of these, , G1, R1, G2, , Rr, , R2, , (B), , 18. In the previus question. If R and S are, interchanged, ballanced point is shifted by, (A)30 cm, (C)50 cm, , V, , (B)40 cm, (D)None of these, , 19. In a meter bridge, null point is at l 33.7cm ,, when the resistance S is shunted by 12, resistance the null point is found to be shifted, by a distance of 18.2 cm. The value of, unknown resistance R should be, , R1, G1, G2, , Rr, , (C), , R2, , (A) 13.5 (B) 68.8 (C) 3.42 (D) None, 20. In u-v method to find focus distance of a, concave mirror, if object distance was found, to be 10.0 cm and image distance was also, found to be 10.0 cm then find max, permissible error in f, due to error is u and v, measurement., (A) 5 0.002 cm, (C) 5 0.02 cm, , (B) 5 0.01 cm, (D) 5 0.05 cm, , 21. To verify ohm’s law, a student is provided with, a test resistor RT , a high resistance R1 , a small, resistance R2 , two identical galvanometers G1, and G2 , and a variable voltage source V. The, correct circuit to, carry out the experiment is [IIT - 2010], , 142, , V, R2, G1, , Rr, , (D), , G2, R1, , V, , 22. A vernier calipers has 1mm mark on the main, scale. It has 20 equal divisions on the Vernier, scale which match with 16 main scale divisions., For this Vernier calipers, the least count is, [IIT - 2010], (A) 0.02mm, , (B) 0.05 mm, , (C) 0.1mm, , (D) 0.2 mm, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , 23. A meter bridge is set up as shown in figure, to, determine an unknown resistane ‘X’ using a, standard 10 ohm resistor. The galvanometer, shows null point when tapping - key is at 52, cm mark. The end-corrections are 1 cm and 2, cm respectively for the ends A and B. The, determined value of ‘X’ is, [IIT - 2011], , X, , 10, , 26. A student is performing the experiment of, resonance column. The diameter of the, column tube is 4 cm. The frequency of the, tuning fork is 512 Hz. The air temperature is, 380 C in which the speed of sound 336 m/s., The zero of the meter scale coincides with the, top end of the Resonance column tube. When, the first resonance occurs, the reading of the, water level in the column is, [IIT - 2012], (A) 14.0 cm, (B) 15.2 cm, (C) 16.4 cm, (D) 17.6 cm, , MULTIPLE ANSWER QUESTIONS, A, , B, , (A) 10.2 ohm, (B) 10.6 ohm, (C) 10.8 ohm, (D) 11.1 ohm, 24. The density of a solid ball is to be determined, in an experiment. The diameter of the ball is, measured with a screw gause, whose pitch is, 0.5 mm and there are 50 divisions on the, circular scale. The reading on the main scale, is 2.5 mm and that on the circular scale is 20, divisions. If the measured mass of the ball, has a relative error of 2%, the relative, percentage error in the density is [ IIT - 2011], (A) 0.9% (B) 2.4 % (C) 3.1 % (D) 4.2 %, 25. In the determination of Young’s modulus, 4MLg , , Y , by using Searle’s method, a wiree, ld 2 , , of length L = 2m and diameter d = 0.5 mm is, used. For a load M = 2.5 kg, an extension, l 0.25mm in the length of the wire is, observed. Quantities d and l are measured, using a screw gauge and a micrometer,, respectively. They have the same pitch of 0.5, mm. The number of divisions on their circular, scale is 100. The contributions to the maximum, probable error of the Y measurement [IIT-2012], (A) due to the errors in the measurements of d and l, are the same., (B) due to the error in the measurement of d is twice, that due to the error in the measurement of l, (C) due to the error in the measurement of l is twice, that due to the error in the measuremenet of d., (D) due to the error in the measurement of d is four, times that due to the error in the measurement of l ., NARAYANAGROUP, , 27. If S and V are one main scale and one vernier, scale and n -1 divisions on the main scale, are equivalent to n divisions of the vernier ,, then, A) Least count is S/n B) Vernier constant is S/n, C) The same vernier constant can be used for, circular verniers also, D) The same vernier constant cannot be used for, circular verniers., 28. In a resonance tube apparatus , the first and, second resonance lengths are l1 and, l2 respectively. If v is the velocity of wave., Then, V, , A) Frequency is , u 2 l l , 2, 1, B) End correction, is e , , l 2 3l1, 2, , C) End correction is, e , , l2 3l1, 2, , v, D) Frequency is , u 4 l l , 2, 1, , 29. The pitch of a screw-gauge having 50, divisions on its circular scale is 1 mm. When, the two jaws of the screw gauge are in contact, with each other, 47th division of circular scale, is on the refference line. When a wire is, placed between the jaws. 3 linear scale, divisions are clearly visible while 31st division, on the circular scale coincides with the, refference line. Then, (A) Zero error in the screw gause is -0.94, (B) Zero error in the screw gause is - 0.06 mm, (C) diameter of the wire is 3.68 mm, (D) diameter of the wire is 3.56 mm, 143
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, , COMPREHENSION TYPE QUESTIONS, Passage-I, Consider the meter bridge circuit without neglecting, S, RB, , P, , Q, , A, , C, l, +, , –, , (100–l), , K, , 30. If we used 100 and 200 resistance is place, of R and S, we get null deflection at, l1 33.0cm . If we intercharge the resistance,, the null difflection was found to be at, l2 67.0 cm .The end corraction and , should be:, (A) 1cm, 1cm B) 2cm, 1cm, (C) 1cm, 2cm (D)None of these, 31. Now we start taking obsevation. At the, possition of R, unknown resistance is used,, and at possition of S, 300 resistence is used., If the ballanced length was found to be l=26cm,, estimate the unknow n resistance., (A)108 (B) 105.4 (C) 100 (D)110, 32. If the unknown Resistance calculated without, using the end corraction, is R1 and using the, end corrections is R2 then, (A) R1 R2 when balanced point is in first half, (B) R1 R2 when balanced point is in first half, (C) R1 R2 when balanced point is in second half, (D) R1 R2 always, , Passage-II, In the, Ohm’s law experiment to find resistance of, unknown resistor R, folloeing two arrangement(a), and (b) are possible., A, , V, i, , V=voltage reading of voltmeter, i=currwnt Reading, of ammeter., But unfortunatly the ammeters and voltmeter used, are not ideal, but having resistance, RA and RV respectively.., 33. For arrangement (s), the measured resistance, is, , and corrections , , R, , Rmeasured , , A, , (A) R RV, , (B) R RA, , RRV, (C) R R, , RRV, (D) R R RA, , V, , V, , 34. For arrangement (b), the measured resistance, is, (A) R RV, , (B) R RA, , RRV, (C) R R, , RRV, (D) R R RA, , V, , V, , 35. You are given two resistor X and Y . Whose, resistance is to determined, using an ammeter, of RA 0.5 and a voltmeter of RV 20 K ., It is known that X is in range of a few ohm’s, and Y is in range of several keloohm’s. Which, of the circuit is preferable to measured X and, Y-Resitor Circute, x, (a), y, (b), (A) x (a ), y b (B) x (b), y a , (C) x (a), y a (D) x (b), y b , , MATRIX MATCHING QUESTIONS, 36. Column I, A) Velocity of a wave, B) Least count, C) For a sphere of radius R 2.1 0.5 cm, D) Circular vernier scale, Column II, (p) Pitch / Number of circular scale divisions, (q) Error in surface area is 26.4 cm2, , 1, V, (a), , V, (b), , The resistance measured is given by, 144, , (r), , m m, , (s) 1 MSD - 1 VSD, NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 37. Column I, A) Vernier constant, B) Micrometer screw, C) Travelling, microscope, D) Cathetometer, , Column II, (P) 0.001 cm, (Q) 1 MSD - 1 VSD, (R) Pitch / Number of, circular scale divisions, (S) 0.01 cm, , EXPERIMENTAL PHYSICS, , LEVEL-VI - HINTS, 1., , dv max max of [2 f0 l2 l2 , 2 f 0 l2 l1 , l1 25.0 cm, l1 0.1 cm place value of last number , , ASSERSON REASON QUESTIONS, (A) Assertion is True, Reason is True; Reason, is correct explanation for Assertion., (B) Assertion is True, Reason is True; Reason, is not correct explanation for Assertion., (C) Assertion is True, Reason is False., (D) Assertion is False, Reason is True., 38. Assertion(A):Screw gauge with a pitch of 0.5mm, is more accurate than 1mm for same number of, circular scale divisions., Reason(R):Higher pitch can make the device more, accurate., 39. Assertion(A):Time period of a hollow ball will be, more than that of a solid ball of same radius., Reason(R): Time period is independence of mass, or distribution but on l ,where l is the distance, between the point of suspension and the centre of, the bob., , l1 75.0 cm, l2 0.1 cm place value of last number , , So max permissible error i speed of sound, , dv max 2 325Hz 0.1 cm 0.1 cm 1.3 m / s, V 2 f 0 l2 l1 , 2 325 Hz 75.0 cm 25.0 cm 325m / s, so V 325 1.3 m / s, 2., , m 20.0 kg, l 125 m, , 2) C, 9) D, 16) C, 23) B, , 27)A,B,C, , 3) B, 10) A, 17) A, 24) C, , 4) D, 11) C, 18) C, 25) A, , 28)A,B, , 3., , x 0.001 cm, , T 2, , l, g, , g, , 4 2 l, T2, , 0.01 0.01 , , 2, 100% 2%, 1.00 2.00 , , 4 2l 4 10 1.00, g, , , 10.0 m / s 2, 2, 2, value of, T, 2.00 , dg , dg max, 2 /100 so, 10.0, g max, 2, , so dg max 0.2 max error in ' g ', 100, , 29) B,C, , so ' g ' 10.0 0.2 m / s 2, , 36) A R, B P,S, C Q, D P, , NARAYANAGROUP, , , , dg , l T, 2, T, g max l, , 30) A 31) A 32) A 33) B 34) C 35) B, , 38)C 3S9)D40) 0.5, , l 1 cm, , , , 0.1kg, 1cm 0.001cm 0.001cm , dY , Y 20.0kg 125cm 0.05cm 0.100cm 100% 4.3%, max , , , 5) C 6) D 7) B, 12) C 13) D 14) C, 19) B 20) D 21) C, 26) B, , 37) A P,Q,S, B R, C P,, D P,Q, , m 0.1 kg, , , , x 0.100 cm, , LEVEL-VI - KEY, 1) C, 8) A, 15) A, 22) D, , mg, mgl, x, Y Y , 2, d /4, / 4 d 2 x, l, , m l, d x, dY , Y m l 2 d x, , max, , SUBJECTIVE TYPE QUESTIONS, 40. In an experiment of simple pendulum, time period, measured was 50s for 25 vibrations, when the length of the simple pendulum was taken, 100cm. If the least count of stop watch is 0.1s, and that of meter scale is 0.1cm, calculate the, maximum possible, percentage error in the, measurement of value of g., , V 2 f 0 l2 l1 ; dv 2 f 0 dl2 dl1 , , 4., , Here T , , total time, t, dt, so dT , total oscillation n, n, 145
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, g L, T, , 2, g, L, T, , i, , In option (D) error in L L is minimum and, number of repetition of measurement are maximum, so dT will be less. So in this case, the error in g is, minimum., 5., , Emf, 100, 2, Rrh 40, R Rrh, 10 Rrh, , 10., R, , pl, pl, 2, A r, , for cases ( p) R , , X, lo, initially slope , , 2, W r0 y0 , , 6., , 2l0 , 2, , (2r0 ) y0, , , , l0, 1, 2 r02 y0, , (1), (2) 2, , for case(r ) R , , 1/ 2 , 2, 1/ 2 , , , , 7., , 8., , Heat given by the sphere, =(1000) (1/2) (80-30)=25,000 cal, Heat absorted by the water calorimeter system, =(900) (1) (40-30) + (200) (1/2) (40-30), =10,000 cal, So heat loss to surrounding =15,000 cal, , l1 20.0 cm l1 0.1 cm, , q line 3 , p line 2, dv, , 14., 15., , from the curve slope , , ; R 10, for second reading, , 146, , 1, , 1, , dp , 0.01 0.1 0.1 0.01 , 2, , , , , 2.50 10.0 10.0 1.00 3.8%, p max, , p, , D 2 V (3.14)(2.00 103 )2 100.0 , , , , 4L, 4 0.314 , I, 10.0 , , and answer should be in three S.F, f0, 1, 1% , f0, 100, , f 0 l2 l1, dv , , , v, f0, l2 l1, max, , 9., , r line 1, , 11. Dynamic rasistance R di di / dv slope, at Pointt. c, slope is min , So R is max, 12. As ammeter has very low resistance most of current, will pass through the ammeter so reading of, ammeter (i) will be very large.Voltmeter has very, high resistance so reading of voltmeter will be very, low., 13. Due to high resistance of voltmeter, reading of, ammeter will be very low, , l2 74.0 cm l2 0.1 cm, f 0 340 HZ 1% , , so Rr R p Rq, , i 1, , so sloper slope p slopeq, v R, , Y, , y , l0 r x, 2 , , , r, x, y max l0, , 2, , and alope of i-v curve, , so slope will be valued, Ans. will be (C), Maximum permissible error in ‘Y’ due to error in, measuring l2 , r , x (If there is no tolerance in mass):, l0, mg, r2x, If there is no tolarence in mass max permissible, error in Y is, , 1, , for case(q) R , , In second case:, ( slope)1 , , (1), , so p 1.00 104 m, dR, , i, , v, , , , 16. , v, R max i, 0.1, 0.1, , 1.1%, 10.0 100.0, , 1 1 1, , v R 10, , 17., , dR , , 2.41%, R max, , R, 25, , R 100;, 300 75, R d 2, p, 2.5 104 m, 4L, , NARAYANAGROUP
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JEE-ADV PHYSICS- VOL- VI, 18. If R and S wave interchanged,, l 75, 100 l 25, Balance point will be shifited by 5-25=50 cm, , 33.7 18.2, R, , 125/12 5 100 33.718.2, , R, 33.7, 19. S 100 33.7, 1, , 1, , 1, , df max, , , , 1, , , , 1, , 10 10 , , , , 1, " f 5 cm, f, , u v , 2 2 f 2, v , u, , 0.1 0.1, max 2 2 52 0.05 cm, 10 10 , so, f 5 0.05 cm, , df, , 21. G1 should acts as a voltmeter and G2 should act, as an ameter., 22. L.C. 1 M.S.D 1 V.S.D, 16 , 4, 1 M.S.D 1 1mm 0.2mm, 20 , 5, , 23. X 48 2 10 52 1, X, , 530, 10.6, 50, , 24. diameter 2.5 , 25. L.C , , Choice (d) is wrong and choice (c) is correct,, since for all vernier scales similar approach can, be used., 28. For end correction e, l1 e , , 0.5, 20, 50, , 0.5, 0.005mm, 100, , v, v , ,, 2 l2 l1 , 2 2v, So, choice (a) is correct and choice (d) is wrong, Put 4 l1 e and l2 e , , l2 3l1, 2, So, choice (b) is correct and choice (c) is wrong, 29. As zero on the circular scale is above refference, line, zero error is negative. Therefore zero, correction is positive., , l2 3l1 2e , e , , Passage-1, , 200 33 100 67 1 cm, 2 1, 1 2, 30 R R , 100 200, 1, 2, R l Rl, , R1l1 R2l2, 100 200 67 100 1cm, 100 , R1 R2, 100 200, , , , l, , R, , R, 26 1, 27, m, 31. 100 l 300 300 100 26 1 75, eq, , R, , 300 27, 75, , 108, , RP 10 0.1 0.2 0.3 , , , , RP, 7 2, 5, 7 , , d 2 0.005 10 3 1, 2, , , d, 0.5 10 3, 50, , , , , , L e, 4, 4, , 16.4 12 15.2cm, 27. Least count = 1 MSD - 1VSD = S-V, S, n 1 , S , , nV n 1 S , S , n , n , It is also called vernier constantSo choices (a), and (b) are correct., NARAYANAGROUP, , 3, 4, , Then l2 e 3 l1 e , , 3, Y l 2 d l 0.005 10 1, , , ; l, 0.25 10 3 50, Y, l, d, , 26. L e , , , 3, and l2 e , 4, 4, , l2 l1 , , solving get R 68.8, 20. u v f, , EXPERIMENTAL PHYSICS, , 10, 0.05 0.04 0.04, 7, ,, 10, 1.3, 0.13 , 0.18 0.2, 7, 7, , RP 1.4 0.2 , 32. If ballance point is firsthalf say l 40, If ballance point is secondhalf say l 70, maximum permissible error in p: the specific resistivity of wire, from meter bridge is, , p, , D2 S l, 4 L 100 l, 147
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JEE-ADV PHYSICS- VOL- VI, , EXPERIMENTAL PHYSICS, Assume that know resistance in RB(S),and total, length of wire is presicely known, then lets find, maximum permissible error in p due to error in p, due to error in measurement of l (Ballance length), and D(diameter of wire)., S , In p In , 2 In D In l In 100 l , 4L , , dp , will be least when l 100 l , p max, if max imum , i.e. l 50 cm, , Passage-2, Solution:Let total current is I 0 . Then in circuit a , i, , I0, , A, , vernier scale. So, A P and Q and S ., Micro - meter screw is similar to screw gauge, with perpendicular distance being moved for, rotation. So, B R ., For travelling microscopes, the least count is 0.001, cm. It can be found using LC= 1MSD - 1 VSd, So, C P and Q ., Cathetometer used for the finding vertical height, uses a ; So, D P and Q ., 38. Least, Pitch, , count= No.of circular scale divisions, So, less the pitch more will be the accuracy., So,Assertion is true and reason is wrong., 39. Assertion is false,since effective length is not altered., , V, , I0, , l, , Reason is true since T 2 g, So,choice (d) is correct., 40: The time period of a simple pendulum is given by, , V, So R RA Rmeasured, I, in circuit (B), i, , I0, , A, , V, , RRV, RRV, i ; So Rmeasured , R RV, R RV, To measure x , circuit (b) shsould be used, as, RRV, Rmeasured , R as R R, A, R RA, 148, , r, ,, r, , S= 26.4 cm2., 37. Vernier constant is the least count of devices with, , l 100, dp , l, l, , due to error in l only is , p, l, 100, , l, l, 100 l , max, , V i ' RV , , So, B P and (S), and D P ., , The error is surface area S 2, , dp , D l, l, , 2, D, l 100 l, p max, , R RA R V, , (E.M) ; So, A R ., Least count is defined as in (P) and (S), , r = 2.1 and r 0.5, , dD dl, dl, 2, , D, l 100 l, , RV, , m m refer to velocity of wave, , for a sphere of radius 2.1 0.5 cm,, , dp, dD dl d 100 l , 2, , p, D, l, 100 l , , i, , 1, 36: Dimensionally, , 2, l, 42l, 2 4 l, T 2 orT , or g 2, g, g, T, As 4 and are constants, maximum permissible, g l 2T, , error in g is given by, g, l, T, Here, l 0.1cm,l 1m 100cm, T 0.1s,T 50 s, , , , g 0.1, 0.1 0.1 0.1 , , 2, , , , g 100, 50 100 25 , , or, , g, 0.1 0.1 , 100 , , 100 0.1 0.4 0.5%, g, 100 25 , NARAYANAGROUP
