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The Most Accepted, , CRASH COURSE, PROGRAMME, , JEE Main in, , 40 DAYS, PHYSICS
Page 4 : Arihant Prakashan (Series), Meerut, All Rights Reserved, , © PUBLISHERS, No part of this publication may be re-produced, stored in a retrieval system or distributed, in any form or by any means, electronic, mechanical, photocopying, recording, scanning,, web or otherwise without the written permission of the publisher. Arihant has obtained all, the information in this book from the sources believed to be reliable and true. However,, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute, accuracy of any information published and the damages or loss suffered there upon., All disputes subject to Meerut (UP) jurisdiction only., , ADMINISTRATIVE & PRODUCTION OFFICES, Regd. Office, ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002, Tele: 011- 47630600, 43518550; Fax: 011- 23280316, , Head Office, Kalindi, TP Nagar, Meerut (UP) - 250002, Tele: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648, , SALES & SUPPORT OFFICES, Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune, , ISBN : 978-93-13199-30-4, , Published by Arihant Publications (India) Ltd., For further information about the books published by Arihant, log on to www.arihantbooks.com or email to
[email protected], , /arihantpub, , /@arihantpub, , Arihant Publications, , /arihantpub
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PREFACE, It is a fact that nearly 10 lacs students would be in the race with you in JEE Main, the gateway, to some of the prestigious engineering and technology institutions in the country, requires, that you take it seriously and head-on. A slight underestimation or wrong guidance will ruin, all your prospects. You have to earmark the topics in the syllabus and have to master them in, concept-driven-problem-solving ways, considering the thrust of the questions being asked in, JEE Main., The book 40 Days JEE Main Physics serves the above cited purpose in perfect manner. At, whatever level of preparation you are before the exam, this book gives you an accelerated way, to master the whole JEE Main Physics Syllabus. It has been conceived keeping in mind the, latest trend of questions, and the level of different types of students., The whole syllabus of Physics has been divided into day-wise-learning modules with clear, groundings into concepts and sufficient practice with solved and unsolved questions on that, day. After every few days you get a Unit Test based upon the topics covered before that day., On last three days you get three full-length Mock Tests, making you ready to face the test. It is, not necessary that you start working with this book in 40 days just before the exam. You may, start and finish your preparation of JEE Main much in advance before the exam date. This will, only keep you in good frame of mind and relaxed, vital for success at this level., , Salient Features, Ÿ Concepts discussed clearly and directly without being superfluous. Only the required, Ÿ, Ÿ, Ÿ, Ÿ, Ÿ, , material for JEE Main being described comprehensively to keep the students focussed., Exercises for each day give you the collection of only the Best Questions of the concept,, giving you the perfect practice in less time., Each day has two Exercises; Foundation Questions Exercise having Topically Arranged, Questions & Progressive Question Exercise having higher Difficulty Level Questions., All types of Objective Questions included in Daily Exercises (Single Option Correct,, Assertion & Reason, etc)., Along with Daywise Exercises, there above also the Unit Tests & Full Length Mock Tests., At the end, there are all Online Solved Papers of JEE Main 2019; January & April attempts., , We are sure that 40 Days Physics for JEE Main will give you a fast way to prepare for Physics, without any other support or guidance., , Publisher
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CONTENTS, Preparing JEE Main 2020 Physics in 40 Days !, Day 1., , Units and Measurement, , Day 2., , Kinematics, , Day 3., , 1-11, , Day 21. Magnetic Effect of Current, , 241-253, , 12-24, , Day 22. Magnetism, , 254-263, , Scalar and Vector, , 25-36, , Day 23. Electromagnetic Induction, , 264-274, , Day 4., , Laws of Motion, , 37-53, , Day 24. Alternating Current, , 275-287, , Day 5., , Circular Motion, , 54-62, , Day 25. Electromagnetic Wave, , 288-295, , Day 6., , Work, Energy and Power, , 63-77, , Day 7., , System of Particles and Rigid Body 78-87, , Day 26. Unit Test 5 (Magnetostatics,, EMI & AC, EM Wave), , 296-301, , Day 8., , Rotational Motion, , Day 27. Ray Optics, , 302-315, , Day 9., , Gravitation, , Day 28. Optical Instruments, , 316-322, , Day 29. Wave Optics, , 323-333, , Day 30. Unit Test 6 (Optics), , 334-339, , Day 31. Dual Nature of Matter, , 340-349, , Day 32. Atoms, , 350-358, , Day 33. Nuclei, , 359-366, , 88-97, 98-109, , Day 10. Unit Test 1 (Mechanics), , 110-116, , Day 11. Oscillations, , 117-130, , Day 12. Waves, , 131-143, , Day 13. Unit Test 2, (Waves and Oscillations), , 144-150, , Day 14. Properties of Matter, , 151-166, , Day 34. Electronic Devices, , 367-377, , Day 15. Heat and Thermodynamics, , 167-184, , Day 35. Gate Circuit, , 378-385, , Day 16. Transfer of Heat, , 185-194, , Day 36. Communication Systems, , 386-394, , Day 37. Unit Test 7 (Modern Physics), , 395-401, , Day 38. Mock Test 1, , 402-407, , Day 39. Mock Test 2, , 408-413, , Day 40. Mock Test 3, , 414-420, , Day 17. Unit Test 3, (General Properties of Matter), , 195-201, , Day 18. Electrostatics, , 202-218, , Day 19. Current Electricity, , 219-233, , Day 20. Unit Test 4 (Electrostatics &, Current Electricity), , 234-240, , Online JEE Main Solved Papers 2019, , 1-32
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SYLLABUS, PHYSICS, NOTE The syllabus contains two Sections - A & B. Section A pertains to the Theory Part, having 80%, weightage, while Section B contains Practical Component (Experimental Skills) having 20% weightage., , SECTION- A, UNIT 1 Physics and Measurement, , UNIT 4 Work, Energy and Power, , Physics, technology and society, SI units, Fundamental, and derived units. Least count, accuracy and precision, of measuring instruments, Errors in measurement,, Significant figures. Dimensions of Physical quantities,, dimensional analysis and its applications., , Work done by a constant force and a variable force;, kinetic and potential energies, work-energy theorem,, power., Potential energy of a spring, conservation of, mechanical energy, conservative and nonconservative, forces; Elastic and inelastic collisions in one and two, dimensions., , UNIT 2 Kinematics, Frame of reference. Motion in a straight line: Positiontime graph, speed and velocity. Uniform and nonuniform motion, average speed and instantaneous, velocity., , UNIT 5 Rotational Motion, , Scalars and Vectors, Vector addition and Subtraction,, Zero Vector, Scalar and Vector products, Unit Vector,, Resolution of a Vector. Relative Velocity, Motion in a, plane, Projectile Motion, Uniform Circular Motion., , Centre of mass of a two-particle system, Centre of, mass of a rigid body; Basic concepts of rotational, motion; moment of a force, torque, angular, momentum, conservation of angular momentum, and its applications; moment of inertia, radius of, gyration. Values of moments of inertia for simple, geometrical objects, parallel and perpendicular axes, theorems and their applications., Rigid body rotation, equations of rotational motion., , UNIT 3 Laws of Motion, , UNIT 6 Gravitation, , Force and Inertia, Newton's First Law of motion;, Momentum, Newton's Second Law of motion; Impulse;, Newton's Third Law of motion. Law of conservation, of linear momentum and its applications, Equilibrium, of concurrent forces., Static & Kinetic friction, laws of friction, rolling friction., Dynamics of uniform circular motion: Centripetal, force and its applications., , The universal law of gravitation., Acceleration due to gravity and its variation with, altitude and depth., Kepler's laws of planetary motion., Gravitational potential energy; gravitational potential., Escape velocity. Orbital velocity of a satellite., Geo-stationary satellites., , Uniformly accelerated motion, velocity-time, position, time graphs, relations for uniformly accelerated, motion.
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UNIT 7 Properties of Solids & Liquids, , UNIT 11 Electrostatics, , Elastic behaviour, Stress-strain relationship, Hooke's., Law, Young's modulus, bulk modulus, modulus of, rigidity., Pressure due to a fluid column; Pascal's law and its, applications., Viscosity, Stokes' law, terminal velocity, streamline and, turbulent flow, Reynolds number. Bernoulli's, principle and its applications., Surface energy and surface tension, angle of contact,, application of surface tension - drops, bubbles and, capillary rise., Heat, temperature, thermal expansion; specific heat, capacity, calorimetry; change of state, latent heat., Heat transfer-conduction, convection and radiation,, Newton's law of cooling., , Electric charges Conservation of charge, Coulomb's, law-forces between two point charges, forces, between multiple charges; superposition principle, and continuous charge distribution., Electric field Electric field due to a point charge,, Electric field lines, Electric dipole, Electric field due, to a dipole, Torque on a dipole in a uniform electric, field., Electric flux, Gauss's law and its applications to find, field due to infinitely long, uniformly charged, straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell., Electric potential and its calculation for a point, charge, electric dipole and system of charges;, Equipotential surfaces, Electrical potential energy of, a system of two point charges in an electrostatic, field., Conductors and insulators, Dielectrics and electric, polarization, capacitor, combination of capacitors in, series and in parallel, capacitance of a parallel plate, capacitor with and without dielectric medium, between the plates, Energy stored in a capacitor., , UNIT 8 Thermodynamics, Thermal equilibrium, zeroth law of thermo-dynamics,, concept of temperature. Heat, work and internal, energy. First law of thermodynamics., Second law of thermodynamics: reversible and, irreversible processes. Camot engine and its efficiency., , UNIT 9 Kinetic Theory of Gases, Equation of state of a perfect gas, work done on, compressing a gas., Kinetic theory of gases - assumptions, concept of, pressure. Kinetic energy and temperature: rms speed, of gas molecules; Degrees of freedom, Law of, equipartition of energy, applications to specific heat, capacities of gases; Mean free path, Avogadro's, number., , UNIT 10 Oscillations and Waves, Periodic motion - period, frequency, displacement as, a function of time. Periodic functions. Simple harmonic, motion (S.H.M.) and its equation; phase; oscillations of, a spring - restoring force and force constant; energy in, S.H.M. - kinetic and potential energies; Simple, pendulum - derivation of expression for its time, period; Free, forced and damped oscillations,, resonance., Wave motion Longitudinal and transverse waves,, speed of a wave. Displacement relation for a, progressive wave. Principle of superposition of, waves, reflection of waves, Standing waves in strings, and organ pipes, fundamental mode and harmonics,, Beats, Doppler effect in sound., , UNIT 12 Current Electricity, Electric current, Drift velocity, Ohm's law, Electrical, resistance, Resistances of different materials, V-I, characteristics of Ohmic and nonohmic conductors,, Electrical energy and power, Electrical resistivity,, Colour code for resistors; Series and parallel, combinations of resistors; Temperature, dependence of resistance., Electric Cell and its Internal resistance, potential, difference and emf of a cell, combination of cells in, series and in parallel., Kirchhoff's laws and their applications., Wheatstone bridge, Metre bridge., Potentiometer - principle and its applications., , UNIT 13 Magnetic Effects of Current, and Magnetism, Biot-Savart law and its application to current, carrying circular loop. Ampere's law and its, applications to infinitely long current carrying, straight wire and solenoid. Force on a moving, charge in uniform magnetic and electric fields, Cyclotron., Force on a current-carrying conductor in a uniform, magnetic field. Force between two parallel current-
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carrying conductors-definition of ampere. Torque, experienced by a current loop in uniform magnetic field,, Moving coil galvanometer, its current sensitivity and, conversion to ammeter and voltmeter., Current loop as a magnetic dipole and its magnetic, dipole moment. Bar magnet as an equivalent solenoid,, magnetic field lines; Earth's magnetic field and magnetic, elements. Para, dia and ferro-magnetic substances, Magnetic susceptibility and permeability, Hysteresis,, Electromagnets and permanent magnets., , UNIT 14 Electromagnetic Induction, and Alternating Currents, Electromagnetic induction; Faraday's law, induced emf, and current; Lenz's Law, Eddy currents. Self and mutual, inductance., Alternating currents, peak and rms value of alternating, current/ voltage; reactance and impedance; LCR series, circuit, resonance; Quality factor, power in AC circuits,, wattless current. AC generator and transformer., , UNIT 15 Electromagnetic Waves, Electromagnetic waves and their characteristics., Transverse nature of electromagnetic waves., Electromagnetic spectrum (radio waves, microwaves,, infrared, visible, ultraviolet, X-rays, gamma rays)., Applications of e.m. waves., , UNIT 16 Optics, Reflection and refraction of light at plane and spherical, surfaces, mirror formula, Total internal reflection and, its applications, Deviation and Dispersion of light by a, prism, Lens Formula, Magnification, Power of a Lens,, Combination of thin lenses in contact, Microscope and, Astronomical Telescope (reflecting and refracting) and, their magnifying powers., Wave optics wave front and Huygens' principle, Laws of, reflection and refraction using Huygen's principle., Interference, Young's double slit experiment and, expression for fringe width, coherent sources and, sustained interference of light. Diffraction due to a single, slit, width of central maximum. Resolving power of, , microscopes and astronomical telescopes,, Polarisation, plane polarized light; Brewster's law,, uses of plane polarized light and Polaroids., , UNIT 17, Dual Nature of Matter and Radiation, Dual nature of radiation. Photoelectric effect,, Hertz and Lenard's observations; Einstein's, photoelectric equation; particle nature of light., Matter waves-wave nature of particle, de Broglie, relation. Davisson-Germer experiment., , UNIT 18 Atoms and Nuclei, Alpha-particle scattering experiment;, Rutherford's model of atom; Bohr model, energy, levels, hydrogen spectrum., Composition and size of nucleus, atomic masses,, isotopes, isobars; isotones. Radioactivity-alpha,, beta and gamma particles/rays and their, properties; radioactive decay law. Mass-energy, relation, mass defect; binding energy per nucleon, and its variation with mass number, nuclear, fission and fusion., , UNIT 19 Electronic Devices, Semiconductors; semiconductor diode: I-V, characteristics in forward and reverse bias; diode, as a rectifier; I-V characteristics of LED,, photodiode, solar cell, and Zener diode; Zener, diode as a voltage regulator. Junction transistor,, transistor action, characteristics of a transistor, transistor as an amplifier (common emitter, configuration) and oscillator. Logic gates (OR,, AND, NOT, NAND & NOR). Transistor as a switch., , UNIT 20 Communication Systems, Propagation of electromagnetic waves in the, atmosphere; Sky and space wave propagation,, Need for modulation, Amplitude and Frequency, Modulation, Bandwidth of signals, Bandwidth of, Transmission medium, Basic Elements of a, Communication System (Block Diagram only), , SECTION- B, UNIT 21 Experimental Skills, Familiarity with the basic approach and observations, of the experiments and activities, 1. Vernier callipers - its use to measure internal and, external diameter and depth of a vessel., , 2. Screw gauge - its use to determine thickness/, diameter of thin sheet/wire., 3. Simple Pendulum - dissipation of energy by, plotting a graph between square of amplitude, and time.
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5. Young's modulus of elasticity of the material of, a metallic wire., , 15. Focal length of, (i) Convex mirror, (ii) Concave mirror, (iii) Convex lens, , 6. Surface tension of water by capillary rise and, effect of detergents., , 16. Using parallax method. Plot of angle of deviation, vs angle of incidence for a triangular prism., , 7. Coefficient of Viscosity of a given viscous liquid, by measuring terminal velocity of a given, spherical body., , 17. Refractive index of a glass slab using a travelling, microscope., , 4. Metre Scale - mass of a given object by, principle of moments., , 8. Plotting a cooling curve for the relationship, between the temperature of a hot body and, time., 9. Speed of sound in air at room temperature, using a resonance tube., , 18. Characteristic curves of a p-n junction diode in, forward and reverse bias., 19. Characteristic curves of a Zener diode and finding, reverse break down voltage., 20. Characteristic curves of a transistor and finding, current gain and voltage gain., , 10. Specific heat capacity of a given (i) solid and (ii), liquid by method of mixtures., , 21. Identification of Diode, LED, Transistor, IC, Resistor,, Capacitor from mixed collection of such items., , 11. Resistivity of the material of a given wire using, metre bridge., , 22. Using multimeter to, (i) Identify base of a transistor., (ii) Distinguish between npn and pnp type, transistor., (iii) See the unidirectional flow of current in case of, a diode and an LED., (iv) Check the correctness or otherwise of a given, electronic component (diode, transistor or IC)., , 12. Resistance of a given wire using Ohm's law., 13. Potentiometer, (i) Comparison of emf of two primary cells., (ii) Internal resistance of a cell., 14. Resistance and figure of merit of a, galvanometer by half deflection method.
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HOW THIS BOOK IS, , USEFUL FOR YOU ?, As the name suggest, this is the perfect book for your recapitulation of the whole syllabus, as it, provides you a capsule course on the subject covering the syllabi of JEE Main, with the smartest, possible tactics as outlined below:, , 1., , REVISION PLAN, The book provides you with a practical and sound revision plan., The chapters of the book have been designed day-wise to guide the students in a planned, manner through day-by-day, during those precious 35-40 days. Every day you complete a, chapter/a topic, also take an exercise on the chapter. So that you can check & correct your, mistakes, answers with hints & solutions also have been provided. By 37th day from the date, you start using this book, entire syllabus gets revisited., Again, as per your convenience/preparation strategy, you can also divide the available 30-35, days into two time frames, first time slot of 3 weeks and last slot of 1 & 1/2 week. Utilize first, time slot for studies and last one for revising the formulas and important points. Now fill the, time slots with subjects/topics and set key milestones. Keep all the formulas, key points on a, couple of A4 size sheets as ready-reckner on your table and go over them time and again. If you, are done with notes, prepare more detailed inside notes and go over them once again. Study, all the 3 subjects every day. Concentrate on the topics that have more weightage in the exam, that you are targeting., , 2., , MOCK TESTS, Once you finish your revision on 37th day, the book provides you with full length mock tests, for day 38th, 39th, & 40th, thereby ensures your total & full proof preparation for the final, show., The importance of solving previous years' papers and 10-15 mock tests cannot be, overemphasized. Identify your weaknesses and strengths. Work towards your strengths i.e.,, devote more time to your strengths to be 100% sure and confident. In the last time frame of 1, & 1/2 week, don't take-up anything new, just revise what you have studied before. Be examready with quality mock tests in between to implement your winning strategy., , 3., , FOCUS TOPICS, Based on past years question paper trends, there are few topics in each subject which have, more questions in exam than other. So far Physics is concerned it may be summed up as, below:, Electricity, Magnetism, Modern Physics, Mechanics, Radioactivity, Wave Options and Heat, Transfer. More than 80% of questions are normally asked from these topics., However, be prepared to find a completely changed pattern for the exam than noted above as, examiners keep trying to weed out 'learn by rot practice'. One should not panic by witnessing a, new pattern , rather should be tension free as no one will have any upper hand in the exam.
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4., , IMPROVES STRIKE RATE AND ACCURACY, The book even helps to improve your strike rate & accuracy. When solving practice tests or, mock tests, try to analyze where you are making mistakes-where are you wasting your time;, which section you are doing best. Whatever mistakes you make in the first mock test, try to, improve that in second. In this way, you can make the optimum use of the book for giving, perfection to your preparation., What most students do is that they revise whole of the syllabus but never attempt a mock and, thus they always make mistake in main exam and lose the track., , 5., , LOG OF LESSONS, During your preparations, make a log of Lesson's Learnt. It is specific to each individual as to, where the person is being most efficient and least efficient. Three things are important - what, is working, what's not working and how would you like to do in your next mock test., , 6., , TIME MANAGEMENT, Most candidates who don't make it to good medical colleges are not good in one area- Time, Management. And, probably here lies the most important value addition that's the book, provides in an aspirant's preparation. Once the students go through the content of the book, precisely as given/directed, he/she learns the tactics of time management in the exam., Realization and strengthening of what you are good at is very helpful, rather than what one, doesn't know. Your greatest motto in the exam should be, how to maximize your scoring with, the given level of preparation. You have to get about 200 plus marks out of a total of about 400, marks for admission to a good NIT (though for a good branch one needs to do much better, than that). Remember that one would be doomed if s/he tries to score 400 in about 3 hours., , 7., , ART OF PROBLEM SOLVING, The book also let you to master the art of problem solving. The key to problem solving does, not lie in understanding the solution to the problem but to find out what clues in the problem, leads you to the right solution. And, that's the reason Hints & Solutions are provided with the, exercises after each chapter of the book. Try to find out the reason by analyzing the level of, problem & practice similar kind of problems so that you can master the tricks involved., Remember that directly going though the solutions is not going to help you at all., , 8., , POSITIVE PERCEPTION, The book put forth for its readers a 'Simple and Straightforward' concept of studies, which is, the best possible, time-tested perception for 11th hour revision / preparation., The content of the book has been presented in such a lucid way so that you can enjoy what, you are reading, keeping a note of your already stressed mind & time span., Cracking JEE Main is not a matter of life and death. Do not allow panic and pressure to create, confusion. Do some yoga and prayers. Enjoy this time with studies as it will never come back.
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DAY ONE, , Units and, Measurement, Learning & Revision for the Day, u, , u, , u, , Physics, Units, Significant Figures, , u, , Accuracy and Precision, , u, , Errors in Measurement, , u, , Dimensions of Physical, Quantities, , Physics, Physics is the study of matter and its motion, as well as space and time using concepts, such as energy, force, mass and charge. It is an experimental science, creating theories, that are tested against observation., , Scope and Excitement, Scope of Physics is very vast, as it deals with a wide variety of disciplines such as, mechanics, heat, light, etc., It also deals with very large magnitude of astronomical phenomenon as well as very, small magnitude involving electrons, protons, etc., , Nature of Physical Laws, Physics is the study of nature and natural phenomena. All observations and experiments, in physics lead to certain facts. These facts can be explained on the basis of certain laws., , PREP, MIRROR, , Physics, Technology and Society, , Your Personal Preparation Indicator, , Connection between physics, technology and society can be seen in many examples like, working of heat engines gave rise to thermodynamics. Wireless communication, technology arose from basic laws of electricity and magnetism. Lately discovery of, silicon chip triggered the computer revolution., , Units, Measurement of any physical quantity involves comparison with a certain basic, widely, accepted reference standard called unit., , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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02, , DAY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , Fundamental and Derived Units, , Base, Quantity, , Fundamental units are the units which can neither be, derived from one another, nor they can be further resolved, into more simpler units., These are the units of fundamental quantity. However,, derived units are the units of measurement of all physical, quantities which can be obtained from fundamental units., , Basic Units, Name and, Symbol, , Thermodyn Kelvin (K), -amic, temperature, , The mole is the amount of substance, of a system, which contains as many, elementary entities as there are, atoms in 0.012 kg of carbon-12., , Luminous, intensity, , candela (cd), , The candela is the luminous, intensity in a given direction of a, source emitting monochromatic, radiation of frequency 540 × 1012 Hz, , Supplementary Units, , 3. MKS System In this system also length, mass and time, have been taken as fundamental quantities and, corresponding fundamental units are metre, kilogram, and second., 4. International System (SI) of Units It is an extended, version of the MKS (Metre, Kilogram, Second) system. It, has seven base units and two supplementary units., Seven base quantities and two supplementary, quantities, their units along with definitions are, tabulated below., , Length, , Basic Units, , metre (m), , Mass, , kilogram (kg), , It is the mass of the international, prototype of the kilogram (a, platinum iridium alloy cylinder), kept at International Bureau of, Weights and Measures, at Sevres, (France)., , Time, , second (s), , The second is the duration of, 9,192, 631,770 periods of the, radiation corresponding to the, transition between the two hyperfine, levels of the ground state of cesium133 atom., , Electric, current, , Ampere (A), , The ampere is that constant current,, which if maintained in two straight,, parallel conductors of infinite length, placed 1 m apart in vacuum would, produce a force equal to 2 × 10−7, Nm −1 on either conductor., , Supplementary, Quantity, , Name and, Symbol, , Plane angle, , radian, (rad), , It is angle subtended at the centre by, an arc of a circle having a length, equal to the radius of the circle., , Solid angle, , steradian, (sr), , It is the solid angle which is having, its vertex at the centre of the sphere,, it cuts-off an area of the surface of, sphere equal to that of a square with, the length of each side equal to the, radius of the sphere., , NOTE, , Definition, , • Angle subtended by a closed curve at an inside points, is 2π rad., , Definintion, The metre is the length of path, travelled by light in vacuum during, a time interval of 1/299,792,458 part, of a second., , the thermodynamic temperature of, the triple point of water., , and having a radiant intensity of, 1, W sr −1 in that direction., 683, , 2. FPS System (Foot, Pound, Second) It is also called the, British Unit System. This unit measures, Length in foot, (foot), Mass in gram (pound), Time in second (s)., , Name and, Symbol, , 1, th fraction of, 273.16, , mole (mol), , 1. CGS System (Centimetre, Gram, Second) are often used, in scientific work. This system measures, Length in, centimetre (cm), Mass in gram (g), Time in second (s)., , Base, Quantity, , The kelvin is, , Amount of, substance, , System of Units, A complete set of these units, both fundamental and derived, unit is known as the system of units. The common systems, are given below:, , Definintion, , • Solid angle subtended by a closed surface at an inside, point is 4 π steradian., , Significant Figures, In the measured value of a physical quantity, the digits about, the correction of which we are sure, plus the last digits which, is doubtful, are called the significant figures., Larger the number of significant figures obtained in a, measurement, greater is the accuracy of the measurement., , Rules for Counting Significant Figures, l, , l, , l, , All the non-zero digits are significant. In 2.738 the number, of significant figures is 4., All the zeros between two non-zero digits are significant,, no matter where the decimal point is. As examples 209 and, 3.002 have 3 and 4 significant figures, respectively., If the measured number is less than 1, the zero (s) on the, right of decimal point and to the left of the first non-zero, digit are non-significant. In 0.00807, first three underlined, zeros are non-significant and the number of significant, figures is only 3.
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UNITS AND MEASUREMENT, , DAY ONE, , l, , l, , l, , l, , The terminal or trailing zero (s) in a number without a, decimal point are not significant. Thus, 12.3 m = 1230 cm, = 12300 mm has only 3 significant figures., , l, , LC =, , In addition or subtraction, the final results should retain as, many decimal places as there are in the number with the, least decimal place. As an example sum of 423.5 g, 164.92 g, and 24.381 g is 612.801 g, but it should be expressed as, 612.8 g only because the least precise measurement (423.5, g) is correct to only one decimal place., , Errors in Measurement, The difference in the true value (mean value) and measured, value of a quantity is called error of measurement. Different, types of error are given below:, (i) Absolute error,, a + a2 + a3 + ... an 1 i = n, amean = a0 = 1, = ∑ ai, n, n i= 1, ∆a1 = mean value − observed value, ∆a1 = a0 − a1, ∆a2 = a0 − a2, :., :., :., , ∆an = a0 − an, , In multiplication or division, the final result should, retain as many significant figures as there are in the, original number with the least significant figures., For example Suppose an expression is performed like, , (ii) Mean absolute error,, , l, , l, , The preceding digit is raised by 1 if the insignificant, digit to be dropped is more than 5 and is left unchanged, if the latter is less than 5. e.g. 18.764 will be rounded off, to 18.8 and 18.74 to 18.7., , n, , [| ∆a1| + | ∆a2| + | ∆a3| + .... + | ∆an|], =, n, ∆amean, (iii) Relative or fractional error =, amean, ∆amean =, , (243, . × 1243) / (44 . 65) = 676 . 481522, Rounding the above result upto three significant figures, result would become 676., , Rules for Rounding off the, Uncertain Digits, , Value of 1 main scale division, Total number of vernier scale division, , (ii) Least count of screw gauge, Value of 1 pitch scale reading, LC =, Total number of head scale division, , A choice of change of units does not change the number, of significant digits or figures in a measurement., , Rules for Arithmetic Operations, with Significant Figures, l, , (i) Least count of vernier callipers, , The trailing zero(s) in number with a decimal point are, significant. Thus, 3.800 kg has 4 significant figures., , To remove ambiguities in determining number of, significant figures, a measurement is usually expressed, as ‘a × 10b ’, where 1 ≤ a ≤ 10 and b is the order of, magnitude., , 03, , ∑| ∆ai|, , i= 1, , n, , (v) Percentage error,, δ a = Relative error × 100 % =, , ∆amean, × 100%, amean, , Combination of Errors, (i) If X = A + B, then (∆X ) = ± (∆A + ∆B), ∆X , ∆A ∆B ∆C , (ii) If X = ABC, then , =±, +, +, , X max, B, C , A, , If the insignificant figure is 5 and the preceding digit is, even, then the insignificant digit is simply dropped., However, if the preceding digit is odd, then it is raised, by one so as to make it even. e.g. 17.845 will be rounded, off to 17.84 and 17.875 to 17.88., , ∆B, ∆C , ∆X , ∆A, (iii) If X = A k B l C n , then , +l, +n, = ± k, X , B, C , A, , Dimensions of Physical Quantities, Accuracy and Precision, The accuracy of a measurement is a measure of how close, the measured value is to the true value of the quantity., However, precision tells us to what resolution or limit, the, quantity is measured by a measuring instrument., , Least Count, The least count of a measuring instrument is the least, value, that can be measured using the instrument. It is, denoted as LC., , The dimensions of a physical quantity are the powers to which, the fundamental (base) quantities are raised, to represent that, quantity., Consider the physical quantity force., ‘Force = mass × acceleration = mass × length × (time)−2 ’, Thus, the dimension of force are 1 in mass [M], 1 in length [L] and −2 in time [ T−2 ], that is [MLT−2 ]., l, , Dimensions of a physical quantity do not depend on its, magnitude or the units in which it is measured.
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04, , DAY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , Principle of Homogeneity of, Dimensions and Applications, According to this principle, a correct dimensional equation, must be homogeneous, i.e. dimensions of all the terms in a, physical expression must be same., LHS (dimension) = RHS (dimension), , (i) To check the correctness of a given physical equation., (ii) Derivation of formula., (iii) Dimensional formula is useful to convert the value of a, physical quantity from one system to the other. Physical, quantity is expressed as a product of numerical value, and unit. In any system of measurement, this product, remains constant., Let dimensional formula of a given physical quantity be, [Ma Lb T c ].If in a system having base units [M1L1T1] the, numerical value of given quantity be n1 and numerical, value n2 in another unit system having the base units, [ M2 , L2 , T2 ], then, Q = n1u1 = n2u2, L b1, , T c1], , =, , n2 [Mb2, , L b2, , T c2], , a, , b, , M L T , n2 = n 1 1 1 1 , M2 L2 T2 , , SI Unit, Watt (W), , [ML2 T −3 ], , Pascal (Pa), or Nm −2, , [ML−1 T −2 ], , Frequency, angular frequency, , Hz or s −1, , [T −1 ], , Angular momentum, , kg m2 s −1, , [ML2 T −1], , Pressure, stress, coefficient of, elasticity (ρ, σ, η), , Torque, Gravitational constant (G), , kg m2, , [ ML2 ], , Acceleration, acceleration due, to gravity, , ms −2, , [LT −2 ], , Force, thrust, tension, weight, , Newton (N), , [MLT −2 ], , kg ms −1 or Ns, , [MLT −1 ], , Joule (J), , [ML2 T −2 ], , Surface area, area of, cross-section, , m2, , [L2 ], , Electric conductivity, , Sm −1, , [M −1 L−3 T3 A2 ], , Pa, , [ML−1 T −2 ], , m2 N −1, , [M −1 LT2 ], , Wb, , [ML2 T −2 A −1 ], , Wb / m2, , [MT −2 A −1 ], , Intensity of a wave, , Wm −2, , [MT −3 ], , Photon flux density, , m −2 s −1, , [L−2 T −1 ], , Luminous energy, , Lm s, , [ML2 T −2 ], , Luminance, , Lux, , [MT −3 ], , Jkg −1 K −1, , [L2 T −2 K −1 ], , Jkg −1, , [L2 T −2 ], , Wm −1 K −1, , [MLT −3 K −1 ], , Electric voltage, , JC −1, , [ML2 T −3 A −1 ], , Magnetisation, , Am −1, , [L−1 A], , Magnetic induction, , T, , [MT −2 A −1 ], , Planck’s constant, , J-s, , [ML2 T −1 ], , 2, , Nm, 2, , N m kg, , −2, , [ML T ], −2, , Radioactive decay constant, , Bq, , [T −1 ], , MeV, , [ML2 T −2 ], , Work, energy, KE, PE, thermal, energy, internal energy, etc., , Young’s modulus,, Bulk modulus, Compressibility, Magnetic Flux, Magnetic Flux density (σ ), , Specific heat capacity, , Dimensional, Formula, , Power, , Moment of inertia, , c, , Dimensions of Important Physical Quantities, Physical Quantity, , Dimensional, Formula, , Linear momentum, impulse, , Uses of Dimensions, , n1[M a1, , SI Unit, , Physical Quantity, , −1 3, , −2, , [M L T ], , Latent heat of vaporisation, Coefficient of Thermal, conductivity, , Binding energy
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UNITS AND MEASUREMENT05, , DAY ONE, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 In which of the following systems of units, a Weber is the, (a) CGS, , (b) MKS, , (c) SI, , (d) FPS, , 2 In an experiment, the angles are required to be, measured using an instrument. 29 divisions of the main, scale exactly coincide with the 30 divisions of the vernier, scale. If the smallest division of the main scale is, half-a-degree (0.5°), then the least count of the, j AIEEE 2009, instrument is, (a) one minute, (c) one degree, , 3.50 cm.Which instrument did he use to measure it?, j JEE Main 2014, (a)A meter scale, (b) A vernier calliper where the 10 divisions in vernier scale, matches with 9 divisions in main scale and main scale, has 10 divisions in 1 cm, (c) A screw gauge having 100 divisions in the circular scale, and pitch as 1 mm, (d) A screw gauge having 50 divisions in the circular scale, and pitch as 1 mm, , 4 N division on main scale of a vernier callipers coincide, with (N + 1) division of the vernier scale if each division on, main scale is of “a” units, least count of instrument is, N+ 1, a, , (b), , a, N+ 1, , (c), , N −1, a, , (d), , a, N −1, , 5 One 8 centimetre on the main scale of a vernier calliper is, divided into 10 equal parts. If 10 of the divisions of the, vernier coincide with small divisions on the main scale,, the least count of the callipers is, (a) 0.005 cm, (c) 0.01 cm, , (b) 0.02 cm, (d) 0.05 cm, , 6 The respective number of significant figures for the, numbers 23.023, 0.0003 and 2.1 × 10−3 are, (a) 5, 1, 2, , (b) 5, 1, 5, , (c) 5, 5, 2, , (d) 4, 4, 2, , 7 A bee of mass 0.000087 kg sits on a flower of mass, 0.0123 kg. What is the total mass supported by the stem, of the flower up to appropriate significant figures?, (a) 0.012387 kg, (c) 0.0124 kg, , (b) 0.01239 kg, (d) 0.012 kg, , 8 The radius of a uniform wire is r = 0.021 cm. The value of π, is given to be 3.142. What is the area of cross-section of, the wire up to appropriate significant figures?, (a) 0.0014 cm2, (c) 0.001386 cm2, , (a) 9.71 ms −1, (c) 9.7087 ms −1, , (b) 9.708 ms −1, (d) 9.70874 ms −1, , 10 If the length of rod A is 3. 25 ± 0.01 cm and that of B is, 419, . ± 0.01 cm, then the rod B is longer than rod A by, (a) (0.94 ± 0.00) cm, (c) (0.94 ± 0.02) cm, , (b) (0.94 ± 0.01) cm, (d) (0.094 ± 0.005) cm, , 11 You measure two quantities as A = 1. 0 m ± 0.2 m,, B = 2.0 m ± 0.2 m. We should report correct value for, AB as, , (b) half minute, (d) half degree, , 3 A student measured the length of a rod and wrote it as, , (a), , 9 A man runs 100.5 m in 10.3 s. Find his average speed up, to appropriate significant figures., , unit of magnetic flux?, , (b) 0.00139 cm2, (d) 0.0013856 cm2, , (a) 1.4 m ± 0.4 m, (c) 1.4 m ± 0.3 m, , (b) 1.41m ± 0.15 m, (d) 1.4 m ± 0.2 m, , 12 A student measured the length of the pendulum 1.21 m, using a metre scale and time for 25 vibrations as 2 min, 20 sec using his wrist watch, absolute error in g is, (a) 0.11 ms − 2, (c) 0.44 ms − 2, , (b) 0.88 ms − 2, (d) 0. 22 ms − 2, , 13 The absolute error in density of a sphere of radius, 10.01 cm and mass 4.692 kg is, (b) 4.692 kgm − 3, (d) 1.12 kgm − 3, , (a) 3.59 kgm − 3, (c) 0, , 14 A sphere has a mass of 12.2 kg ± 0.1 kg and radius, 10 cm ± 0.1 cm, the maximum % error in density is, (a) 10%, , (b) 2.4%, , (c) 3.83%, , (d) 4.2%, , 15 If error in measurement of radius of sphere is 1%, what, will be the error in measurement of volume?, (a) 1%, , (b), , 1, %, 3, , (c) 3%, , (d) 10%, , 16 What is the percentage error in the measurement of the, time period T of a pendulum, if the maximum errors in the, measurements of l and g are 2% and 4%, respectively?, (a) 6%, , (b) 4%, , (c) 3%, , (d) 5%, , 17 The density of a material in the shape of a cube is, determined by measuring three sides of the cube and its, mass. If the relative errors in measuring the mass and, length are respectively 1.5% and 1% , the maximum error, j, in determining the density is, JEE Main 2018, (a) 2.5%, (c) 4.5%, , (b) 3.5%, (d) 6%, , 18 Which one of the following represents the correct, dimensions of the coefficient of viscosity?, (a) [ML−1T−2 ], −1 −1, , (c) [ML T ], , (b) [MLT−1 ], (d) [ML−2 T−2 ]
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06, , DAY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , 19 Which of the following sets share different dimensions?, (a) Pressure, Young’s modulus, stress, (b) Emf, potential difference, electric potential, (c) Heat, work done, energy, (d) Dipole moment, electric flux, electric field, , (a) [M2LT−3 ], , (a) ms −1 and ms−3, (c) m2 s and ms2, , (a) Angular momentum and Planck’s constant, (b) Impulse and momentum, (c) Moment of inertia and moment of a force, (d) Work and torque, , 31 In the following dimensionally consistent equation, we, have, F =, , electronic charge, electric permittivity, Planck’s constant, and velocity of light in vacuum respectively, are, (c) [M0LT0 ], , (d) [M0L0 T], , X, + Y , where F = force., Linear density, , The dimensional formula for X and Y are, (a) [M2L0 T−2 ]; [MLT−2 ], −2, , dimensions of capacitance and magnetic induction, respectively, dimensions ofY are, , 32 The dimensions of self-inductance are, , using the codes given below the lists., , (a) [MLT C ] (b) [MT C, , ], , −1 −1, , (c) [MT C ], , −2, , (b) [L2 T2 ], , (c) [L2 T−2 ], , (d) [LT−1 ], , 25 If the acceleration due to gravity is 10 ms –2 and units of, length and time are changed to kilometre and hours, respectively, the numerical value of acceleration is, (a) 360000, , (b) 72000, , (c) 36000, , (d) 129600, , 26 If E = energy, G = gravitational constant, I = impulse and, M = mass, then dimensions of, (a) time, , (b) mass, , GIM 2, are same as that of, E2, , (c) length, , (d) force, , 27 Let [ ε 0 ] denotes the dimensional formula of the, , permittivity of vacuum. If M = mass, L = length, T = time, j, JEE Main 2013, and A = electric current, then, (a) [ε0 ] = [ M −1 L−3 T 2 A], (c) [ε0 ] = [ M −2 L2 T −1 A −2 ], , A., , Boltzmann constant, , p., , [ML2T −1 ], , B., , Coefficient of viscosity, , q., , [ML−1T −1 ], , C., , Planck constant, , r., , [MLT −3K −1 ], , D., , Thermal conductivity, , s., , [ML2T −2K −1 ], , (d) [MT C ], , meaning, are, , (b) [ε0 ] = [ M −1 L−3 T 4 A 2 ], (d) [ε0 ] = [ M −1 L2 T −1 A 2 ], , 28 With the usual notations, the following equation, 1, s = u + a ( 2 t − 1) is, 2, (a) only numerically correct, (b) only dimensionally correct, (c) Both numerically and dimensionally correct, (d) Neither numerically nor dimensionally correct, , 2013 Main, , Column II, , −1, , 1, 24 Dimensions of, , where symbols have their usual, µ 0ε 0, (a) [L−1T], , j, , Column I, , (coulomb) is given as, −2, , (b) [ML2 T−2 A −2 ], (d) [ML−2 T−2 A −1 ], , 33 Match List I with List II and select the correct answer, , (b) [M−3 T 4L−2Q4 ], (d) [ML2 T −2 A −2 ], , 2, , (d) [M0L0 T0 ]; [ML0 T0 ], , (c) [MLT ]; [ML T ], , 23 The dimensions of magnetic field in M,L,T and C, −1 −1, , (b) [M2L−2 T−2 ]; [MLT−2 ], , 2 −2, , (a) [ML−2 T−2 A −2 ], (c) [ML2 T−2 A −1 ], , 22 In the relation X = 3YZ , X and Z represent the, 2, , (a) [MT −1Q−1], (c) [M−3 T −1L−1Q4 ], , (d) [LT−3 ], , (b) ms−2 and ms, (d) ms and ms−1, , –1, , e2, , where e, ε 0, h and c are the, 4πε 0hc, , (b) [ML0 T0 ], , (c) [ML3 T−2 ], , the velocity is measured in ms −1, then units of a and c are, , identical dimensions?, , (a) [M0L0 T0 ], , (b) [MT−2 ], , 30 The velocity of a particle is given as v = a + b t + c t 2. If, , 20 Out of the following pairs, which one does not have, , 21 The dimensions of, , a, a − t2, in the equation p =, where, p, b, bx, is pressure, x is distance and t is time, are, , 29 The dimensions of, , Codes, A B C D, (a) p q r, (c) s, , s, , r p q, , A B, , C D, , (b) s q, , p, , r, , (d) r, , q, , p, , s, , Direction (Q. Nos. 34-37) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 34 Statement I The order of accuracy of measurement, depends on the least count of the measuring instrument., , Statement II The smaller the least count, the greater is, the number of significant figures in the measured, value.
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UNITS AND MEASUREMENT, , DAY ONE, 35 Statement I The dimensional method cannot be used to, obtain the dependence of the work done by a force F on, the angle θ between force F and displacement x., , Statement II Angle can be measured in radians but it, has no dimensions., 36 Statement I The mass of an object is 13.2 kg in the, measurement there are 3 significant figures., , 07, , Statement II The same mass when expressed in, grams as 13200 g has five significant figures., 37 Statement I Method of dimensions cannot be used for, deriving formula containing trigonometrical ratios., , Statement II This is because trigonometrical ratios, have no dimensions., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The dimensions of angular momentum, latent heat and, capacitance are, respectively., (a), (b), (c), (d), , j, , JEE Main (Online) 2013, , [ML2 T1A 2 ], [L2 T−2 ], [M−1L−2 T2 ], [ML2 T−2 ], [L2 T2 ], [M−1 L−2 T4 A 2 ], [ML2 T−1 ], [L2 T−2 ], [ML2 TA 2 ], [ML2 T−1 ], [L2 T−2 ], [M−1L−2 T4 A 2 ], , (b), , ρ, σλ, , (c), , λ, σρ, , (d) ρλσ, , 3 Dimensions of resistance in an electrical circuit, in terms, of dimensions of mass M, length L, time T and current I,, would be, (b) [ML2 T−2 ], (d) [ML2 T−3I−2 ], , (a) [ML2 T−3I−1 ], (c) [ML2 T−1I−1 ], , 4 In the relation p =, , α, e, β, , −, , αz, kθ, , , p is pressure, z is distance, k, , is Boltzmann constant and θ is the temperature. The, dimensional formula of β will be, 0 2, , 0, , 2, , (a) [M L T ], (c) [ML0 T –1 ], , (b) [M L T], (d) [M0L2 T−1 ], , 5 The dimensions of σb 4( σ = Stefan’s constant and, b = Wien’s constant) are, (a) [M0L0 T0 ], , (b) [ML4 T−3 ], , (c) [ML−2 T], , (d) [ML6 T−3 ], , 6 If Planck’s constant (h) and speed of light in vacuum (c), are taken as two fundamental quantities, which one of the, following can, in addition,be taken to express length,, mass and time in terms of the three chosen fundamental, quantities?, , (i) Mass of electron (m e ), (ii) Universal gravitational constant (G), (iii) Charge of electron ( e), (iv) Mass of proton (m p ), (a) (i),(ii) and (iii), (c) (i), (ii) and (iv), , (b) 163.62 ± 2.6 cm2, (d) 163.62 ± 3 cm2, , 8 Resistance of a given wire is obtained by measuring the, , on surface tension ( σ ), density (ρ) and wavelength (λ )., The square of speed (v ) is proportional to, σ, ρλ, , cm and 10 .1 cm , respectively. The area of the sheet in, appropriate significant figures and error is, (a) 164 ± 3 cm2, (c) 163.6 ± 2.6 cm2, , 2 The speed (v ) of ripples on the surface of water depends, , (a), , 7 The length and breadth of a rectangular sheet are 16.2, , (b) (i) and (iii), (d) (i) only, , current flowing in it and the voltage difference applied, across it. If the percentage errors in the measurement of, the current and the voltage difference are 3% each, then, error in the value of resistance of the wire is, (a) 6%, , (b) zero, , (c) 1%, , (d) 3%, , 9 A screw gauge gives the following reading when used to, measure the diameter of a wire., Main scale reading : 0 mm, Circular scale reading : 52 divisions, Given that 1 mm on main scale corresponds to, 100 divisions of the circular scale., The diameter of wire from the above data is, (a) 0.052 cm, (c) 0.005 cm, , (b) 0.026 cm, (d) 0.52 cm, , 10 A screw gauge with a pitch of 0.5 mm and a circular, scale with 50 divisions is used to measure the thickness, of a thin sheet of aluminium. Before starting the, measurement, it is found that when the two jaws of the, screw gauge are brought in contact, the 45th division, coincides with the main scale line and that the zero of the, main scale is barely visible. What is the thickness of the, sheet, if the main scale reading is 0.5 mm and the 25th, division coincides with the main scale line?, (a) 0.75 mm, (c) 0.70 mm, , (b) 0.80 mm, (d) 0.50 mm, , 11 The following observations were taken for determining, surface tensionT of water by capillary method. Diameter, of capillary, d = 1.25 × 10−2 m rise of water,, h = 1.45 × 10−2m. Using g = 9.80 m/s 2 and the simplified, dhg, relation T =, × 103N/m, the possible error in surface, 4, j, tension is closest to, JEE Main 2017 (Offline), (a) 1.5%, , (b) 2.4%, , (c) 10%, , (d) 0.15%
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08, , DAY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , 12 A student measures the time period of 100 oscillations of, a simple pendulum four times. The data set is 90s, 91s,, 92s and 95s. If the minimum division in the measuring, clock is 1s, then the reported mean time should be, j, , JEE Main 2016 (Offline), , Direction (Q. Nos. 15-16) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , (a) (92 ± 2) s (b) (92 ± 5 ) s (c) (92 ± 18, . ) s (d) (92 ± 3) s, , 13 The period of oscillation of a simple pendulum is, T = 2π L / g . Measured value of L is 20.0 cm known to, 1mm accuracy and time for 100 oscillations of the, pendulum is found to be 90 s using a wrist watch of, resolution.The accuracy in the determination of g is, j, , (a) 2%, , (b) 3%, , (c) 1%, , 15 Statement I The value of velocity of light is 3 × 108 ms −1, , JEE Main 2015, , (d) 5%, , 14 The current voltage relation of diode is given by, − 1) mA, where the applied voltage V is in volt, I = (e, and the temperature T is in kelvin. If a student makes an, error measuring ± 0.01V while measuring the current of, 5 mA at 300K, what will be the error in the value of, j JEE Main 2013, current in mA?, 1000V /T, , (a) 0.2 mA, , (b) 0.02 mA, , (c) 0.5 mA, , (d) 0.05 mA, , and acceleration due to gravity is, 10 ms −2 and the mass of proton is 1.67 × 10−27 kg., , Statement II The value of time in such a system is, 3 × 107 s., 16 Statement I The distance covered by a body is given by, s =u +, , 1a, , where the symbols have usual meaning., 2 t, , Statement II We can add, substract or equate, quantities which have same dimensions., , ANSWERS, SESSION 1, , SESSION 2, , 1 (c), , 2 (a), , 3 (b), , 4 (b), , 5 (b), , 6 (a), , 7 (d), , 8 (a), , 9 (a), , 10 (c), , 11 (d), , 12 (d), , 13 (a), , 14 (c), , 15 (c), , 16 (c), , 17 (c), , 18 (c), , 19 (d), , 20 (c), , 21 (a), , 22 (b), , 23 (c), , 24 (c), , 25 (d), , 26 (a), , 27 (b), , 28 (d), , 29 (b), , 30 (a), , 31 (a), , 32 (b), , 33 (b), , 34 (b), , 35 (a), , 36 (c), , 37 (a), , 1 (d), 11 (a), , 2 (a), 12 (a), , 3 (d), 13 (b), , 4 (a), 14 (a), , 5 (b), 15 (b), , 6 (a), 16 (d), , 7 (a), , 8 (a), , 9 (a), , 10 (b)
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UNITS AND MEASUREMENT, , DAY ONE, , 9, , Hints and Explanations, SESSION 1, , 9 Average speed = 100.5 m, , 1 A weber is the unit of magnetic flux in, , Value of main scale division, =, Number of divisions on, vernier scale, =, , 1, 1, 1° 1°, MSD =, ×, =, = 1 min, 30, 30, 2, 60, , 3 If student measures 3.50cm, it means, that there is an uncertainly of, , 1, cm., 100, , The distance has four significant figures, but the time has only three. Hence, the, result must be rounded off to three, significant figure, to 9.71 ms −1 ., , 10 As, A = 3.25 ± 0.01 cm, Y =B − A, = 4.19 − 3.25 = 0.94 cm, , For vernier scale with 1 MSD = 1 mm, and 9 MSD = 10 VSD, , and ∆Y = ∆B + ∆A, , LC of vernier calliper = 1 MSD – 1VSD, 1, 9, 1, = 1 −, cm, =, 10 , 10 100, , ∴, , 4 (N + 1) VSD = N MSD, ∴, , 1 VSD =, , N, MSD, N +1, , Least count, = (1 MSD − 1VSD) (value of MSD), , N , a, = 1 −, × a=, N + 1, N +1, , , 5 1 MSD = 1 cm = 0.1 cm,, , 10, 10 VSD = 8 MSD, Hence, we get, 8, 8, 1 VSD =, MSD =, × (0.1) = 0.08 cm, 10, 10, Thus, the least count = 1 MSD –, 1VSD, = 01, . − 0.08 = 0.02 cm, , 6 Number of significant figures in, , 23.023 = 5, Number of significant figures in, 0.0003 = 1, Number of significant figures in, 2.1 × 10−3 = 2, , 7 The mass of the bee has 2 significant, figures in kg, whereas the mass of the, flower has three significant figures., Hence, the sum (0.012387) must be, rounded off to the third decimal place., Therefore, the correct significant figure, is 0.012., , 8 A = πr 2 = 3.142 × (0.021) 2, = 0.00138562 cm ., 2, , Now, there are only two significant, figures in 0.021 cm. Hence, the result, must be rounded off to two significant, figures as A = 0.0014 cm2 ., , 01, ., 01, . , = , + 3×, × 100, 12.2, 10 , = 3 . 83 %, , 15 As, V = 4 πr 3, , 3, ∆V, ∆r, Hence,, × 100 = 3, × 100, V, r, = 3 × 1% = 3%, , and B = 4.19 ± 0.01 cm, ∴, , M, ,, 4 3, πr, 3, dρ, ∆M 3∆r , , × 100 = , +, × 100, M, ρ, r , V, , ms −1, = 975728, ., , SI system., , 2 Least count, , 14 ∴ Density, ρ = M =, , 10.3 s, , 16 Since, the time period,, , = 0.01 cm + 0.01 cm = 0.02 cm, , T =, , Y = (0.94 ± 0.02) cm, , B = 2.0 m ± 0.2 m, x = AB = (1. 0)(2. 0) = 1.414 m, , 1, 1, = ± × 2% + × 4% = ± 3%, 2, , 2, , Rounding off to two significant digits,, x = AB = 1.4 m, , 17 Q Density, ρ = Mass = M3 or ρ = M3, , ∆x 1 ∆ A ∆ B , = , +, 2 A, x, B , 1 0.2, 0.2 , 0.6, = , +, =, 2 1.0 2 . 0 2 × 2.0, 0.6 x, ∆x=, = 0 .15 × 1.414, 2 × 2.0, = 0.2121, Rounding off to one significant digit,, Now,, , Volume L, L, ∆ρ ∆M, 3∆L, =, +, ρ, M, L, , ⇒ Error in density, , So, maximum % error in measurement, of ρ is, ∆ρ, ∆M, 3∆L, × 100 =, × 100 +, × 100, ρ, M, L, or % error in density = 1.5 + 3 × 1, % error = 4.5%, F, 18 By Newton’s formula η =, A ( ∆V / ∆Z ), , ∆ x = 0.2 m, AB = x ± ∆x = (1. 4 ± 0. 2) m, , 2, 12 As, T = 2π l or g = 4 π2 l, , g, , ∴ Dimensions of η, Dimensions of force, =, (Dimensions of area × Dimensions, of velocity gradient), [MLT −2 ], = 2, = [ML−1 T −1 ], [L ][T −1 ], , T, , ∆g, ∆L 2∆T , ⇒ , =, +, , L, T , g, So, absolute error in g is, ∆L, 2∆T , ∆g = , +, g, L, T , 0.01 2 × 1 , = , +, × 9.8, 1.21, 140 , , 19 Dipole moment = charge × distance, Electric flux = electric field × area, , = (0.0227 × 9.8) = 0.22 ms, , 13 ∴ ρ =, , M, , =, , −2, , 20 I = mr 2, , 4.692 × 3, , 4 3 4 × 3.14 × (10.01) × 10, πr, 3, ρ = 1.12 × 103 kg - m −3, ∆ρ ∆M, 3∆r, =, +, ρ, M, r, 3, , 0.001 3 × 0.01 , 3, ∴ ∆ρ = , +, × 1.12 × 10, 4.692, 10.01 , = 3.59 kgm −3, , l, g, , Thus, for calculating the error, we get, 1 ∆l, ∆T, 1 ∆g , =± , +, , T, l, 2, 2 g , , , 11 Here, A = 1.0 m ± 0.2 m,, , Hence,, , 1, 2π, , −6, , ∴ [I ] = [ML2 ], τ moment of force = r × F, ∴ [τ] = [L][MLT −2 ] = [ML2 T −2 ], 2, , 21 e, , , , 4πε0hc , , =, , [AT]2, −1 −3, , [M L T A ] ⋅ [ML2 T −1 ] ⋅ [LT −1 ], , = [M 0L0T 0], , 4, , 2
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10, , DAY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , , , 22 X = [C ] = [M −1 L−2 T2Q 2 ],, −1, , −1, , Z = [B ] = [MT Q, Y =, , X, , =, , Z2, , −1, , ], , 2 −2, , 2, , [M T L Q ], , = [M −3 T 4 L−2Q 4 ], , [MT −1Q −1 ]2, , 23 From the relation F = qvB, [MLT −2 ] = [C][LT −1 ][B ], [B ] = [MC −1 T −1 ], , ⇒, ⇒, , 24 As we know that, formula of velocity is, v =, , 1, 1, = [LT −1 ]2 ∴, ⇒ v2 =, µ 0ε0, µ 0ε0, , 1, = [L2 T −2 ], µ 0ε0, , , T1 , T , 2, , 25 n2 = n1 L1 , L2 , , −2, , metre sec , = 10 , km h , , −2, , m, sec , n2 = 10 3 , 10 m 3600sec , = 129600, , −2, , −1 3 −2, −1, 2, , ] × [M]2, 26 GIM2 = [M L T ] × 2[MLT, −2 2, , E, , , , [ML T ], , So, the dimensions of Y are the same as, that of F, i.e. [Y ] = [F ] = [MLT −2 ], X , Now, [MLT −2 ] = , ML− 1 , ⇒, X = [M 2 L0T − 2 ], dI, and is, the current varies at a rate, dt, dI, given by e = − L , where e is the, dt, electromotive force (emf) induced in the, coil., Now, the dimensions of emf are the same, as that of the potential difference,, i.e. [ML2 T −3 I −1 ], −e, Now, L =, ., dI / dt, Hence, the dimensions of L are, dimensions of e, [ L] =, dimensions of I / dimensions of t, [ML2 T −3 I −1 ], =, [I / T], = [ML2 T −2 I −2 ], , GIM, , 2, , are same as that, , E2, , 33 (A) U = 1 kT, , 2, ⇒ [ML2 T −2 ] = [k ] K, , of time., , ⇒, , 27 Electrostatic force between two charges,, 1 q1 q2, F =, 4 πε0 R2, q q, ε0 = 1 2 2, ⇒, 4 πFR, Substituting the units., Hence, ε0 =, , C2, 2, , N-m, , =, , (B) F = ηA, ⇒ [η] =, , (D), , 28 s = distance travelled, u = velocity., So, dimensionally it is not a correct, equation., , ⇒, , 2, , −1 −1, , [L LT L ], , = [ML−1 T −1 ], , [h] = [ML2 T −1 ], , k A∆θ, dQ, =, dt, l, [ML2 T −3 L], ⇒ [k] =, = [MLT −3 K −1 ], [L2 K], , Greater is the number of significant, figures obtained in a measurement,, greater is its precision and for this the, least count of the measuring instrument, should be smaller., , [T2 ], [T2 ], =, −1 −2, [ p][ x] [ML T ][L], = [M −1 T 4 ], , or [b] =, , 35 Work done is W = Fx cos θ. Since, θ is, , a, [T2 ], ∴ =, = [MT −2 ], b [M −1 T 4 ], , 30 Unit of a = unit of v = m/s = ms, and unit of c = unit of, s2, , [MLT −2 ], , the least distance (resolution/accuracy),, that can be measured using the device., , ⇒ [ pbx] = [a] = [T2 ], , =, , dv, dx, , 34 The least count of a measuring device is, , a − t2, bx, pbx = a − t 2, , m /s, , ⇒, , [MLT −2 ] [L2 ], , = [M −1 L−3 T 4 A2 ], , 29 p =, , [K] = [ML2 T −2 K −1 ], , (C) E = hν ⇒ [ML2 T2 ] = [h] [T −1 ], , [AT ]2, , −1, , v, , t2, = m / s 3 = ms −3, , 37 It is true that trigonometrical ratios do, not have dimensions. Therefore, method, of finding dimensions cannot be utilized, for deriving formula involving, trigonometrical ratio., , SESSION 2, 1 Angular momentum = r × P = [LM LT −1 ], = [ML2 T −1 ], , 32 The self-inductance L of a coil in which, , = [M 0L0T], So, dimensions of, , , , X, 31 [F ] = , + [Y ]., Linear, density , , , dimensionless, the dependence of W, on θ cannot be determined by the, dimensional method., , 36 The degree of accuracy (and hence the, number of significant figures) of a, measurement cannot be increased by, changing the unit., , Q, = [L2 T −2 ], M, q ( AT )2, Capacitance, C = =, V, W, , W, as, V =, , q , , 2 2, −1 −2 +2, = [A T M L T ], = [M −1 L−2 T 4 A2 ], Latent heat, L =, , 2 Let v ∝ σ aρb λc . Equating dimensions on, both sides, [M 0 L1 T −1 ] = k [MT −2 ] a [ML−3 ] b [L] c, [M 0LT −1 ] = k [M] a +b [L] −3b +c [T] − 2 a, Equating the powers of M, L, T on both, sides, we get, a + b = 0 and −3b + c = 1; − 2a = − 1, 1, 1, 1, Solving, we get a = , b = − , c = −, 2, 2, 2, σ, ∴ v ∝ σ 1 /2ρ−1 /2 λ−1 /2, ∴ v2 ∝, ρλ, Potential difference, 3 Resistance, R =, Current, V W, =, =, I, qI, (Q Potential difference is equal to work, done per unit charge), So, dimensions of R, [Dimensions of work], =, [Dimensions of charge], [Dimensions of current], [ML2 T −2 ], =, = [ML2 T −3 I −2 ], [IT] [I], , 4 In the given equation,, dimensionless, kθ, ∴ [α] = , z , ⇒ [α] =, , αz, should be, kθ, , [ML2 T −2 K −1 × K], = [MLT – 2 ], [L], , α , and [ p] = , β , α [MLT −2 ], ⇒ [ β] = =, = [M 0L2 T 0], –1 –2, p, [ML, T, ], , , 5 λ mT = b or b 4 = λ4mT 4, and, or, , energy, = σT 4, area × time, σ =, , energy, (area × time) T 4
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UNITS AND MEASUREMENT, , DAY ONE, energy 4, σb 4 = , λm, area × time , or [σb 4 ] =, , [ML2 T −2 ], [L2 ][T], , [L4 ] = [ML4 T −3 ]., , 6 h = [ML2 T −1 ]; c = [LT −1 ], me = [M ],, G = [ M −1 L3 T −2 ], e = AT; m p = M ,, 1, , 2, , −1, , −1, , hc [M L T ] [LT ], =, = [ M 2], G, [M −1 L3 T −2 ], ⇒, , M =, h, =, c, , hc, G, [ML2 T −1 ], [LT −1 ], , = [ML], , h, h G, Gh, ⇒, L =, =, = 3 /2, cM c hc, c, From c = [ LT −1 ],, T =, , L, Gh, Gh, = 3 /2 = 5/2, c, c c, c, , Hence, out of (i), (ii) and (iii) any one, can be taken to express L, M, T in terms, of three chosen fundamental quantities., , 7 Here, l = (16.2 ± 0.1 ) cm ;, b = (10.1 ± 0.1) cm, A = l × b = 16.2 × 10.1 = 163.62, Rounding off to three significant digits,, A = 164 cm2, 0 .1, 0 .1, ∆ A ∆ l ∆b , +, =, +, =, l, A, b 16.2 10.1, 1.01 +1.62, =, = 2.63 cm2, 16.2 × 10.1, Rounding off to one significant figure,, ∆ A = 3cm2, ∴, , A = (164 ± 3) cm2, , 8 From Ohm’s law, R = V, I, , ⇒ ln R = ln V − ln I, ∆R ∆V, ∆I, ⇒, =, +, = 3% + 3% = 6%, R, V, I, , 9 Diameter of wire, d = MSR + CSR × LC, 1, = 0.52 mm = 0.052 cm, 100, pitch, 10 Least count =, (number of division on, circular scale), = 0 + 52 ×, , =, , 0.5, mm, 50, , LC = 0.01, ∴, Negative zero error = −5 × LC = −0.005 mm, Measured value = main scale reading, + screw guage reading − zero error, = 0.5 mm + {25 × 0.01 − (−0.05)} mm, = 0.80 mm, , 11 By given formula, we have surface, tension,, dhg, N, Q r = d , × 103, T =, , , , 4, m, 2, ∆T ∆d ∆h, [given, g is constant], =, +, ⇒, T, d, h, So, percentage error is, ∆T, =, × 100, T, ∆d ∆h , = , +, × 100, d, h , 0.01 × 10−2 0.01 × 10−2 , =, +, × 100, −2, 1.45 × 10−2 , 1.25 × 10, = 1.5%, ∆T, ∴, × 100 = 1.5%, T, , ∆g, ∆g, ∆L 2∆T, or, =, +, × 100%, g, L, T, g, ∆L , 2∆T × 100%, = , × 100% + , , L , T , 1, 1, = , × 100 % + 2 ×, × 100%, 200, , 90, = 2.72% = 3%, Thus, accuracy in the determination of g, is approx 3 %., , ∴, , 14 Given, I = (e 1000V / T − 1) mA, dV = ± 0.01V, T = 300 K, I = 5mA,, I = e 1000V / T − 1, I + 1 = e 1000V / T, Taking log on both sides, we get, 1000V, T, d (I + 1) 1000, =, dV, I +1, T, log (I + 1) =, , ⇒, , dI, 1000, =, dV, I +1, T, ⇒, , dI =, , 1000, × (I +1) dV, T, , dI =, , 1000, × (5 + 1) × 0.01, 300, , 12 Arithmetic mean time of a oscillating, Σ xi, N, 90 + 91 + 92 + 95, =, = 92 s, 4, Mean error is, Σ | x − xi | 2 + 1 + 3 + 0, =, =, = 1.5, N, 4, Given, minimum division in the, measuring clock, is 1 s. Thus, the, reported mean time of a oscillating, simple pendulum = (92 ± 2) s., simple pendulum =, , 13 Given, time period, T = 2π L, g, , Thus,changes can be expressed as, 2T, ∆L ∆g, =, =±, ±, T, L, g, According to the question, we can write, ∆L, 0.1cm, 1, =, =, L, 20.0cm 200, 90, Again time period T =, s, 100, ∆T, 1, 1, and ∆T =, =, s ⇒, T, 90, 200, L, Now, T = 2π, g, , 11, , = 0.2 mA, So, error in the value of current is, 0.2 mA., , 15 [ c ] = [LT −1 ] = 3 × 108 ms −1, and [g ] = [LT −2 ] = 10 ms −2, So,, , c, [LT −1 ], =, =T, g [LT −2 ], , ∴, , T=, , 3 × 108, = 3 × 107 s, 10, , 16 The physical quantities can be equated,, added or subtracted only when they, have same dimensions. The distance, 1a, covered by a body is s = u +, 2t, [LT −2 ], −1, [L] = [LT ] +, [T], [L] = [LT −1 ] + [LT −3 ], As every term of equation is not is not, having same dimensions, so it is a, wrong expression for distance.
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DAY TWO, , Kinematics, Learning & Revision for the Day, u, , u, , u, , u, , Frame of Reference, Motion in a Straight Line, Uniform and Non-uniform Motion, Uniformly Accelerated Motion, , u, , u, , Elementary Concept of Differentiation and, Integration for Describing Motion, Graphs, , Frame of Reference, The frame of reference is a suitable coordinate system involving space and time used as a, reference to study the motion of different bodies. The most common reference frame is, the cartesian frame of reference involving (x, y, z and t)., (i) Inertial Frame of Reference A frame of reference which is either at rest or moving, with constant velocity is known as inertial frame of reference. Inertial frame of, reference is one in which Newton’s first law of motion holds good., (ii) Non-Inertial Frame of Reference A frame of reference moving with some, acceleration is known as non-inertial frame of reference. Non-inertial frame of, reference in one which Newton’s law of motion does not hold good., , Motion in a Straight Line, The motion of a point object in a straight line is one dimensional motion. During such a, motion the point object occupies definite position on the path at each instant of time., Different terms used to described motion are defined below:, , Distance and Displacement, l, , l, , Displacement is shortest distance between initial and final, positions of a moving object. It is a vector quantity and its SI, unit is metre., , Y, , r1, , ∆r = r2 – r1, , l, , l, , Displacement of motion may be zero or negative, but path length or distance can never be negative., , Displacement, Distance, A, , From the given figure, mathematically it is expressed as,, l, , PREP, MIRROR, , Distance is the total length of the path travelled by a particle in a given interval of, time. It is a scalar quantity and its SI unit is metre (m)., , ∆r, r2, , B, , O, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , X, , For motion between two points displacement is single valued while distance depends, on actual path and so can have many values., Magnitude of displacement can never be greater than distance. However, it can be, equal, if the motion is along a straight line without any change in direction., , Your Personal Preparation Indicator, , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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KINEMATICS, , DAY TWO, , Speed and Velocity, l, , l, , Speed is defined as the total path length (or actual distance, covered) by time taken by object., Distance, Speed =, Time taken, It is scalar quantity. Its SI unit is m/s., Total distance travelled, Average Speed, vav =, ∆t, , Average and Instantaneous Acceleration If velocity of a, particle at instant t is v1 and at instant t 2 is v2 , then, v2 − v1 ∆v, Average acceleration, aav =, =, t2 − t 1, ∆t, l, , l, , l, , l, , l, , When a body travels equal distance with speeds v1 and v2 ,, the average speed (v) is the harmonic mean of the two, speeds., 2 1, 1, +, =, v v1 v2, , Total displacement x2 − x 1 ∆x, =, =, Total time taken, t2 − t 1, ∆t, Here, x2 and x 1 are the positions of a particle at the time t 2, and t 1 respectively, with respect to a given frame of, reference., , l, , l, , l, , and, , l, , The instantaneous speed is average speed for infinitesimal, small time interval (i.e. ∆t → 0), ∆s ds, i.e. Instantaneous speed, v = lim, =, ∆t → 0 ∆ t, dt, The instantaneous velocity (or simply velocity) v of a, ∆x dx, moving particle is v = lim, =, ∆t → 0 ∆ t, dt, , l, , l, , An object is said to be in non-uniform motion if its, undergoes equal displacement in unequal intervals of, time., howsoever small these intervals may be., , Acceleration, Acceleration of an object is defined as rate of change, of velocity. It is a vector quantity having unit m/s2 or ms −2 ., It can be positive, zero or negative., , …(ii), …(iii), , Equation of uniformaly accelerated motion under gravity, are, 1, (i) v = u − gt, (ii) h = ut − gt 2 (iii) v2 = u2 − 2 gh, 2, , Elementary Concept of Differentiation, and Integration for Describing Motion, , For a moving body speed can never be negative or zero, while velocity can be negative and zero., , An object is said to be in uniform motion if its velocity is, uniform i.e. it undergoes equal displacement in equal may, be intervals of time, howsoever small these interval., , v2 = u2 + 2 as, , …(i), , where, u = initial velocity, v = velocity at time t, and s = displacement of particle at time t ., , It (at a particular time) can be calculated as the slope (at, that particular time) of the graph of x versus t ., , l, , For uniformly accelerated motion are given below, 1, s = ut + at 2, 2, , Average velocity =, , Uniform and Non-uniform Motion, , A motion, in which change in velocity in each unit of time, is constant, is called an uniformly accelerated motion. So,, for an uniformly accelerated motion, acceleration is constant., Equations of motion, v = u + at, , l, l, , ∆v dv, =, ∆t dt, , Uniformly Accelerated Motion, , When a body travels for equal time with speeds v1 and v2 ,, the average speed v is the arithmetic mean of the two speeds., v + v2, vav = 1, 2, Velocity is defined as ratio of displacement and, corresponding time interval taken by an object., Displacement, i.e. velocity =, time interval, , Instantaneous acceleration, a = lim, , ∆t→ 0, , l, l, , 13, , At an instant t , the body is at point P( x, y, z)., dx, Thus, velocity along X-axis, vx =, dt, dvx, Acceleration along X -axis is ax =, dt, dy, Velocity along Y-axis is v y =, dt, dvy, Acceleration along Y-axis is a y =, dt, dz, dv, Similarly,, and az = z, vz =, dt, dt, For a accelerating body, (i) If ax variable, x = ∫ vx dt ,, , ∫ dvx = ∫ ax dt, (ii) If a y is variable, y = ∫ v y dt , ∫ dv y = ∫ a y dt, (iii) If a z is variable, z = ∫ vz dt , ∫ dvz = ∫ az dt, Also, distance travelled by a particle is s = ∫ | v| dt, (i) x-component of displacement is ∆x = ∫ vx dt, (ii) y-component of displacement is ∆y = ∫ v y dt, (iii) z-component of displacement is ∆z = ∫ vz dt
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14, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , Graphs, , Velocity-Time Graph, , During motion of the particle, its parameters of kinematical, analysis changes with time. These can be represented on the, graph, which are given as follows:, , Position-Time Graph, (i) Position-time graph gives, displacement at any instant., , instantaneous, , value, , of, , (iii) The s-t graph cannot make sharp turns., , Different Cases of Position-Time Graph, s- t Graph, , Slope = constant,, v = constant, a=0, , s, , s =νt, t, , Uniformly, accelerated, motion with, u = 0, s = 0 at, t =0, , s, , Uniformly, accelerated, motion with, u ≠ 0 but s = 0, at t = 0, , s, , Uniformly, retarded motion, , Different Cases in Velocity-Time Graph, Different Cases, Uniform motion, , v- t Graph, , 1, s = at 2, 2, , u = 0, i.e., Slope of s-t graph, at t = 0, should be zero., , v = constant, , 1, s =ut + at 2, 2, , v, , Uniformly, accelerated, motion with, u ≠ 0 but s = 0, at t = 0, , v, , Uniformly, decelerated, motion, , v, , v = at, , u, , Positive constant, acceleration because θ is, constant and <90º but the, initial velocity of the particle, is positive, , v = u + at, , t, Slope of v-t graphs = – a, (retardation), , u, v = u – at, t0, , Non-uniformly, accelerated, motion, , θ is decreasing, so, v is decreasing, a is, negative, , t0, , So slope of v-t graph is, constant u = 0 i.e., so, a = constant u = 0, i.e. v = 0 at t = 0, , t, , t, Slope of v-t graph increases, with time., θ is increasing, so,, acceleration is increasing, , v, , t, , s, , (i) θ = 0º, (ii) v = constant, (iii) Slope of v-t graph = a = 0, , t, Uniformly, accelerated, motion with, u = 0 and s = 0, at t = 0, , t, Slope of s-t graph gradually, goes on increasing, , The main Features of Graph, , v, , Slope = v = 0, , t, Uniform motion, , (iii) Area under v-t graph with time axis gives the value of, displacement covered in given time., , The main Features of Graph, , s, , At rest, , (ii) The slope of tangent drawn on graph gives instantaneous, acceleration., , (iv) The v-t curve cannot take sharp turns., , (ii) The slope of tangent drawn to the graph at any instant of, time gives the instantaneous velocity at that instant., , Different Cases, , (i) Velocity-time graph gives the instantaneous value of, velocity at any instant., , t, Non-uniformly, decelerating, motion, , θ is decreasing, so, acceleration decreasing, , v, , t, t
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KINEMATICS, , DAY TWO, , 15, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 An aeroplane flies 400 m from North and then flies 300 m, South and then flies 1200 m upwards, then net, displacement is, (a) 1200 m, , (b) 1300 m, , (c) 1400 m, , (d) 1500 m, , 2 The correct statement from the following is, (a) A body having zero velocity will not necessarily have, zero acceleration, (b) A body having zero velocity will necessarily have zero, acceleration, (c) A body having uniform speed can have only uniform, acceleration, (d) A body having non-uniform velocity will have zero, acceleration, , 3 A vehicle travels half the distance L with speed v1 and, the other half with speed v 2, then its average speed is, (a), , v1 + v 2, 2, , (b), , 2v1 + v 2, v1 + v 2, , (c), , 2v1v 2, v1 + v 2, , (d), , 2 (v1 + v 2 ), v1v 2, , 4 A particle moves along the sides AB, BC, CD of a square, of side 25 m with a velocity of 15 m/s. Its average, velocity is, B, , C, , 7 A body is thrown vertically upwards in air, when air, resistance is taken into account, the time of ascent is t1, and time of descent is t 2, then which of the following is, true?, (a) t1 = t 2, , (b) t1 < t 2, , (c) t1 > t 2, , (d) t1 ≥ t 2, , 8 A stone falls freely from rest and the total distance covered, by it in the last second of its motion equals the distance, covered by it in the first three seconds of its motion. The, stone remains in the air for, (a) 6 s, , (c) 7 s, , (d) 4 s, , 9 The motor of an electric train can give it an acceleration, of 1 ms – 2 and brakes can give a negative acceleration of, 3 ms – 2. The shortest time in which the train can make a, trip between the two stations 1215 m apart is, (a) 113.6 s, (c) 60 s, , (b) 56.9 s, (d) 55 s, , 10 A train is moving along a straight path with a uniform, acceleration. Its engine passes a pole with a velocity of, 60 kmh −1 and the end (guard’s van) passes across the, same pole with a velocity of 80 kmh −1. The middle point, of the train will pass the same pole with a velocity, (a) 70 kmh −1, (c) 65 kmh −1, , v, , (b) 5 s, , (b) 70.7 kmh −1, (d) 75 kmh −1, , 11 The acceleration experienced by a moving boat after its, A, , (a) 5 m/s, , D, , (b) 7.5 m/s, , (c) 10 m/s, , (d) 15 m/s, , 5 A body sliding down on a smooth inclined plane slides, down 1/4th of plane’s length in 2 s. It will slide down the, complete plane in, (a) 4 s, , (b) 5 s, , (c) 2 s, , (d) 3 s, , engine is cut-off, is given by a = −kv 3, where k is a, constant. If v 0 is the magnitude of velocity at cut-off,, then the magnitude of the velocity at time t after the, cut-off is, (a), (c), , 6 Three particles P , Q and R are situated at corners of an, equilateral triangle of side length (d). At t = 0, they, started to move such that P is moving towards Q, Q is, moving towards R and R is moving towards P at every, instant. After how much time (in second) will they meet, each other?, Q, , d, u, , (b), , 2d, 3u, , (d), , v0, 1 + 2 kt v 02, v0, 1 + 2 kt v 02, , 12 A body moving with an uniform acceleration describes, 12 m in the 3rd second of its motion and 20 m in the, 5th second. Find the velocity after the 10th second., (a) 40 ms–1, (c) 52 ms–1, , (b) 42 ms–1, (d) 4 ms–1, , maximum speed of 40 ms –1 in 20 s. It travels at this, speed for 20 s and is brought to rest with an uniform, retardation in the next 40 s. What is the average velocity, during this period?, , u, , (a), , 1 − 2 kv 02, , (b), , 13 A train accelerating uniformly from rest attains a, u, , P, , v0, 2 kt v 02, v0, , u, , (c), , R, , 2d, (3) u, , (d), , d, (3) u, , 80, ms–1, 3, (c) 40 ms–1, (a), , (b) 25 ms–1, (d) 30 ms–1
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16, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 A frictionless wire AB is fixed on a sphere of radius R. A, very small spherical ball slips on this wire. The time taken, by this ball to slip from A to B is, , 21 Look at the graphs (i) to (iv) in figure carefully and, choose, which of these can possibly represent, one-dimensional motion of particle?, x, , x, , A, q, B, , O, , t, , t, , R, C, , 2 gR, (a), g cosθ, , cosθ, (b) 2 gR, g, , (c) 2, , R, g, , (d), , (i), , gR, g cosθ, , (ii), , x, , x, , 15 A balloon is going upwards with velocity12 ms –1. It, releases a packet when it is at a height of 65 m from the, ground. How much time the packet will take to reach the, ground if g = 10 ms –2 ?, (a) 5 s, , (b) 6 s, , (c) 7 s, , (d) 8 s, , 16 A ball is dropped from the top of a building. The ball, takes 0.5 s to fall past the 3 m length of window some, distance below from the top of building. With what speed, does the ball pass the top of window?, (a) 6 ms–1, , (b) 12 ms–1, , t, , (c) 7 ms–1, , (d) 3.5 ms–1, , 17 A body starts from the origin and moves along the axis, , (iii), , (b) 22 ms – 2, , (c) 12 ms –2, , 22 Figure shows the time-displacement curve of the particles, P and Q . Which of the following statement is correct?, , acceleration varies with time as a = ( 5t + 6) ms –2. If it, starts from the origin, the distance covered by it in 2 s is, (a) 18.66 m, , (b) 14.33 m, , (c) 12.18 m, , (d) 6.66 m, , 19 A rod of length l leans by its upper end against a smooth, vertical wall, while its other end leans against the floor., The end that leans against the wall moves uniformly, downwards. Then,, (a) the other end also move uniformly, (b) the speed of other end goes on increasing, (c) the speed of other end goes on decreasing, (d) the speed of other end first decreases and then increases, , 20 Which of the following distance-time graphs is not possible?, x, , x, , t, , t, (a), , (b), , x, , Q, , t, , O, , (a) Both P and Q move with uniform equal speed, (b) P is accelerated and Q moves with uniform speed but, the speed of P is more than the speed of Q, (c) Both P and Q moves with uniform speeds but the speed, of P is more than the speed of Q, (d) Both P and Q moves with uniform speeds but the speed, of Q is more than the speed of P, , 23 The velocity versus time curve of a moving point is shown, in the figure below. The maximum acceleration is, , 80, 70, 60, 50, 40, 30, 20, 10, , x, , Velocity (ms–1), D, , B, , E, , C, , F, A, O 10 20 30 40 50 60 70 80 90 Time (s), , t, (c), , P, , x, , (d) 10 ms –2, , 18 A point initially at rest moves along the x-axis. Its, , (iv), , (a) Both (i) and (ii), (b) Only (iv), (c) Only (iii), (d) Both (iii) and (iv), , such that the velocity at any instant is given by, v = 4 t 3 − 2 t where, t is in second and the velocity in, ms −1. Find the acceleration of the particle when it is at, a distance of 2 m from the origin., (a) 28 ms –2, , t, , (d), , t, , (a) 1 ms−2, (c) 2 ms−2, , (b) 6 ms −2, (d) 1.5 ms−2
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KINEMATICS, , DAY TWO, 24 The velocity-time graph of a body in a straight line is as, v (ms–1), 2, 1, 1, , 2, , 3, , 4 5, , t(s), , –1, –2, , The displacement of the body in five seconds is, (b) 3 m, (c) 4 m, (d) 5 m, (a) 2 m, , 25 All the graphs below are intended to represent the same, motion. One of them does it incorrectly. Pick it up., ª JEE Main 2018, Distance, , Velocity, (a), , Position, , 27 When two bodies move uniformly towards each other the, distance between them decreases by 8 ms –1. If both, bodies move in the same direction with different speeds,, the distance between them increases by 2 ms −1. The, speeds of two bodies will be, , shown in figure., , 0, , 17, , Time, , (b), , (a) 4 ms−1 and 3 ms−1, (b) 4 ms−1 and 2 ms−1, (c) 5 ms−1 and 3 ms−1, (d) 7 ms−1 and 3 ms−1, , Direction (Q. Nos. 28-30) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below., (a) Statement I is true, Statement II is true, Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true, Statement II is, not the correct explanation for Statement I, (c) Statement I is true, Statement II is false, , Position, , (d) Statement I is false, Statement II is true, , Velocity, , 28 Statement I A particle moving with a constant velocity,, (c), , Time, , Time, , (d), , changes its direction uniformly., Statement II In a uniform motion, the acceleration is zero., , 26 A body is thrown vertically upwards. Which one of the, following graphs correctly represent the velocity versus, time?, ª JEE Main 2017 (Offline), v, , (a), , v, t, , (b), , t, , 29 Statement I Two objects moving with velocities v1 and v 2, in the opposite directions, have their relative velocity, along the direction of the one with a larger velocity., Statement II The relative velocity between two bodies, moving with velocity v1 and v 2 in same direction is given, by v = v1 − v 2, , 30 Statement I Acceleration of a moving particle can be, (c), , v, t, , (d), , change without changing direction of velocity., , v, , Statement II If the direction of velocity changes, so the, direction of acceleration also changes., t
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18, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The velocity-time plot for a particle moving on a straight, , 10, v (ms–1), , 10, , 20, , (a) t 2, , 30, , (a) the particle has a constant acceleration, (b) the particle has never turned around, (c) the average speed in the interval 0 to 10 s is the same, as the average speed in the interval 10 s to 20 s, (d) Both (a) and (c) are correct, , 2 A body is at rest at x = 0. At t = 0, it starts moving in the, positive x-direction with a constant acceleration. At the, same instant another body passes through x = 0 moving, in the positive x-direction with a constant speed. The, position of the first body is given by x1(t ) after time t and, that of the second body by x 2(t ) after the same time, interval. Which of the following graphs correctly, describes ( x1 − x 2 ) as a function of time t?, (x1 – x2), , (b), O, , t, , t, , (d), , (c), O, , t, , O, , t, , 3 The velocity of a particle is v = v 0 + gt + ft 2. If its position, is x = 0 at t = 0, then its displacement after unit time, (t = 1) is, (a) v 0 + 2g + 3f, g, f, (b) v 0 + +, 2 3, (c) v 0 + g + f, g, (d) v 0 +, +f, 2, , (d) t 3, , (b) 3 s, (d) None of these, , 6 A point moves with a uniform acceleration and v1, v 2, v 3, denote the average velocities in three successive, intervals of time t1, t 2, t 3. Which of the following relations is, correct?, (a) (v1 − v 2 ) : (v 2, (b) (v1 − v 2 ) : (v 2, (c) (v1 − v 2 ) : (v 2, (d) (v1 − v 2 ) : (v 2, , − v 3 ) = (t1 − t 2 ) : (t 2 + t 3 ), − v 3 ) = (t1 + t 2 ) : (t 2 + t 3 ), − v 3 ) = (t1 − t 2 ) : (t1 − t 3 ), − v 3 ) = (t1 − t 2 ) : (t 2 − t 3 ), , 7 From the top of a tower of height 50 m, a ball is thrown, vertically upwards with a certain velocity. It hits the, ground 10 s after it is thrown up. How much time does it, take to cover a distance AB where A and B are two points, 20 m and 40 m below the edge of the tower?, ( take,g = 10 ms – 2), (b) 1 s, , (c) 0.5 s, , (d) 0.4 s, , 8 Car A is moving with a speed of 36 kmh −1 on a two lane, , (x1 – x2), , (x1 – x2), , (a) 2 s, (c) 4 s, , (a) 2.0 s, O, , (c) t 1/ 2, , ground in 5 s. If the stone is stopped after 3 s of its fall, and then allowed to fall again, then the time taken by the, stone to reach the ground after covering the remaining, distance is, , –10, , (a), , (b) t, , 5 A stone is dropped from a certain height and reaches the, , t (s), , (x1 – x2), , 4 A particle located at x = 0 at time t = 0, starts moving, along the positive x-direction with a velocity v that varies, as v = α x . The displacement of the particle varies, with time as, , line is as shown in figure, then, , road. Two cars B and C, each moving with a speed of, 54 kmh −1 in opposite directions on the other lane are, approaching car A. At certain instant of time, when the, distance AB = AC = 1 km, the driver of car B decides to, overtake A before C does. What must be the minimum, acceleration of car B, so as to avoid an accident ?, (a) 1 ms−2, , (b) 4 ms−2, , (c) 2 ms−2, , (d) 3 ms−2, , 9 The displacement x of a particle varies with time,, according to the relation x =, , a, (1 − e −bt ). Then, b, , (a) the particle can not reach a point at a distance x from its, starting position, if x > a / b, (b) at t = 1/ b, the displacement of the particle is nearly, (2 / 3) (a /b), (c) the velocity and acceleration of the particle at t = 0 are a, and − ab respectively, (d) the particle will come back its starting point as t → ∞
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KINEMATICS, , DAY TWO, 10 From the top of a tower, a stone is thrown up which, , 16 From a tower of height H, a particle is thrown vertically, , reaches the ground in time t1. A second stone thrown, down, with the same speed, reaches the ground in time, t 2. A third stone released from rest, from the same, location, reaches the ground in a time t 3. Then,, , (a) 2gH = n 2u 2, (c) 2gH = nu 2 (n − 2), , (b) t 32 = t12 − t 22, , (c) 8, , (d) 10, , a = 3 t 2 + 2 t + 2 where, t is time. If the particle starts out, with a velocity v = 2 ms −1 at t = 0, then the velocity at the, end of 2 s is, (b) 18 ms−1, , (c) 12 ms−1, , (a) 2 s, , h, m from the ground, 9, 8h, (c), m from the ground, 9, , 1 2, ft, 6, 1, (d) s = f t 2, 4, (b) s =, , 1 2, ft, 72, , 14 The displacement of a particle is given by x = (t − 2)2, , (b) 8 m, , (c) 12 m, , 7h, m from the ground, 9, 17h, (d), m from the ground, 18, , 19 Two stones are thrown up simultaneously from the edge, , of a cliff 240 m high with initial speed at 10 ms −1 and, 40 ms −1, respectively. Which of the following graph best, represents the time variation of relative position of the, second stone with respect to the first? (Assume stones, do not rebound after hitting the ground and neglect air, resistance, take g = 10 ms −2. The figures are schematic, and not drawn to scale, ª JEE Main 2015, 240, , where, x is in metres and t in seconds. The distance, covered by the particle in first 4 seconds is, (a) 4 m, , (d) 1 s, , (b), , (a), , (d) 27 ms−1, , a distance s, then continues at constant speed for time t, f, and then decelerates at the rate to come to rest. If the, 2, total distance travelled is 15 s, then, , (c) s =, , (c) 8 s, , h metre. It takes T second to reach the ground. What is, T, the position of the ball in s?, 3, , 13 A car, starting from rest, accelerates at the rate f through, , (a) s = f t, , (b) 4 s, , 18 A ball is released from the top of a tower of height, , 12 The acceleration in ms − 2 of a particle is given by,, , (a) 36 ms−1, , dv, = − 2.5 v , where, v is, dt, the instantaneous speed. The time taken by the object, to, come to rest, would be, , decelerated at a rate given by, , penetrate two plancks of equal thickness. The number of, such plancks penetrated by the same bullet, when the, velocity is doubled, will be, (b) 6, , (b) gH = (n − 2)2 u 2, (d) gH = (n − 2)2 u 2, , 17 An object, moving with a speed of 6.25 ms −1, is, , (d) t 3 = t1 t 2, , 11 A bullet moving with a velocity of 100 ms −1 can just, , (a) 4, , upwards with a speed u. The time taken by the particle to, hit the ground, is n times that taken by it to reach the, highest point of its path. The relation between H, u and n, is, ª 2014 JEE Main, , (y2 – y1)m, , 240, , 1, 1 1, =, −, t 3 t 2 t1, t + t2, (c) t 3 = 1, 2, (a), , 19, , (a), , (d) 16 m, , (y2 – y1)m, , (b), , 15 A metro train starts from rest and in five seconds, , (a) 12.2 s, (c) 9 s, , (b) 15.3 s, (d) 17.2 s, , t, , 8, , 12, , t(s), , 12, , (y2 – y1)m, , 240, , 240, , achieves 108 kmh −1. After that it moves with constant, velocity and comes to rest after travelling 45 m with, uniform retardation. If total distance travelled is 395 m,, find total time of travelling., , (c), , (y2 – y1)m, , (d), 8, , 12, , t(s), , 8, , 12, , ANSWERS, SESSION 1, , SESSION 2, , t(s), , 1 (a), , 2 (a), , 3 (c), , 4 (a), , 5 (a), , 6 (b), , 7 (b), , 8 (b), , 9 (b), , 10 (b), , 11 (d), , 12 (b), , 13 (b), , 14 (c), , 15 (a), , 16 (d), , 17 (b), , 18 (a), , 19 (c), , 20 (c), , 21 (d), , 22 (c), , 23 (b), , 24 (b), , 25 (b), , 26 (b), , 27 (c), , 28 (d), , 29 (d), , 30 (c), , 1 (d), 11 (c), , 2 (b), 12 (b), , 3 (b), 13 (c), , 4 (a), 14 (b), , 5 (c), 15 (d), , 6 (b), 16 (c), , 7 (d), 17 (a), , 8 (a), 18 (c), , 9 (b), 19 (c), , 10 (d), , t(s)
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20, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , 6 The person at P will travel a distance, , 1 Displacement along North, , = 400 − 300 = 100 m, Upward displacement = 1200 m, ∴ Net displacement, , PO, with velocity along PO = u cos 30°, Here,, Q, u, , = (100)2 + (1200)2, . m ≈ 1200 m, = 120415, upwards, at the highest point of its, motion, the velocity of the body, becomes zero but acceleration is not, zero., , 3 Time taken to travel first half distance,, t1 =, , L /2, L, =, v1, 2v 1, , Time taken to travel second half, distance,, , u 30°, , u, B, PO = PB sec 30°, d, 2, d, = ×, =, 2, 3, 3, ∴ Time of meeting,, d/ 3, distance, t =, =, velocity u cos 30°, , =, , L, L, +, 2v 1 2v 2, , =, , =, , Total distance, Total time, L, L1, 1, +, 2 v 1 v 2 , 2v 1 v 2, v1 + v2, , 4 Since, average velocity, total displacement, total time taken, 25, =, 75, , 15, 25 × 15, =, = 5m/s, 75, =, , 5 As, u = 0 and a is a constant, l, 1, …(i), = a(2) 2, 4 2, 1, …(ii), l = at 2, 2, On dividing Eq. (ii) by Eq. (i), we get, l, t2, =, ,, l / 4 (2) 2, t = 4s, , ⇒, ⇒, , goes on decreasing as air resistance is, small. When a body falls down, its, velocity goes on increasing as air, resistance is large, t 2 increases., g, 8 As, sn = u + (2n − 1), 2, g, = 0 + (2n − 1), 2, Distance travelled in the first three, second, 1, From s = ut + at 2, 2, 1, 9, s3 = 0 × 3 + × g × 32 = g, 2, 2, As,, S n = s3, 9, g, 2, 9, 5s, , 9 Let s1 be the distance travelled by the, train moving with acceleration 1ms – 2, for time t 1 and s2 be the distance, travelled by the train moving with, retardation 3 ms – 2 for time t 2 . If v is the, velocity of the train after time t 1 , then, v = 1 × t1, 1, t2, s1 = × 1 × t 12 = 1, 2, 2, Also,, , v = 3t 2, , t1, = 56.9 s, 3, , 10 From v 2 − u2 = 2as, , d/ 3, , g, (2n − 1) =, 2, 2n − 1 =, n=, , s2 = vt 2 −, , Total time = t 1 + t 2 = t 1 +, , 7 When a body is thrown up, its velocity, , L1, 1, +, , 2 v 1 v 2 , , ∴ Average speed =, , R, , u 3 /2, 2d, s, =, 3u, , Total time = t 1 + t 2, , =, , O, , 30°, , P, , L /2, L, t2 =, =, v2, 2v 2, , =, , A, , C, , 2 When a body is projected vertically, , 1, × 3 × t 22, 2, 3, = t 1 t 2 − t 22, 2, From Eqs. (i) and (iii), we get, t, t 1 = 3t 2 or t 2 = 1, 3, t 12, t 1 3 t 12 2 2, s1 + s2 =, + t1 ×, − ×, = t1, 2, 3 2, 9 3, 2 2, 1215 = t 1, 3, 3 × 1215, ⇒, t1 =, 2, = 42.69 s, and, , …(i), …(ii), …(iii), , 802 − 602, =s, 2a, s=, =, , 6400 − 3600, 2a, 1400, a, , The middle point of the train has to, cover a distance, s 700, =, 2, a, From v 2 − u2 = 2as, 700, = 1400, a, 2, v = 1400 + 3600, , v 2 − 602 = 2a ×, , v =, , 5000, , = 707, . kmh −1, , 11 Given, acceleration a = − kv 3, Initial velocity at cut-off, v 1 = v 0, Initial time of cut-off, t = 0 and final time, after cut-off, t 2 = t, dv, dv, Again, a =, = − kv 3 or 3 = − kdt, dt, v, Integrating both sides, with in the, condition of motion., v dv, t, ∫v 0 v 3 = − ∫0 k dt, − 1 = −[kt ] t, or, 0, 2v 2 , 1, 1, or, −, = kt, 2v 2 2v 20, v0, or, v =, 1 + 2 kt v 20
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12 Using, sn = u + a (2n − 1), , 2, a, 12 = u + (2 × 3 − 1) …(i), 2, a, 20 = u + (2 × 5 − 1) …(ii), 2, On subtracting Eq. (i) from Eq. (ii),, we get, a, 8 = (10 − 6) = 2a, 2, a = 4 ms – 2, 4, From Eq. (i), 12 = u + × 5, 2, u = 2 ms −1, From v = u + at = 2 + 4 × 10, = 42 ms –1, v = u + at 1, , 13 As,, , …(i), , 40 = 0 + a × 20, , s = 65m, t = ?, 1, As, s = ut + at 2, 2, ⇒ 65 = − 12t + 5 t 2, 5t − 12t − 65 = 0, , Subtract Eq. (i) from, Eq. (ii), we get, , 2, , s1 = 400 m, s2 = v × t 2, , 1, 5, 3 = 5 + t = + 5 t, 4, 4, , a = − 1 ms, , …(ii), , From, , –2, , s3 = 800 m, , = 400 + 800 + 800, = 2000 m, , Let, , ∴Total distance travelled, = s1 + s2 + s3, , and total time taken = 20 + 20 + 40, = 80 s, 2000, ∴ Average velocity =, = 25ms −1, 80, , 14 Acceleration of the body down the, , plane = g cos θ, Distance travelled by ball in time t, second is, 1, ...(i), AB = (g cos θ) t 2, 2, From ∆ ABC,, ...(ii), AB = 2R cos θ, From Eqs. (i) and (ii), we get, 1, 2R cos θ = g cos θ t 2, 2, 4R, 2, t =, g, R, g, , 21 In one dimensional motion, there is a, , ∴, Let, , single value of displacement at one, particular time., , 22 As x-t graph is a straight line in either, , v = 4t 3 − 2t, dx, = 4t 3 − 2t, dt, On integration, we get,, x = 2 = t 4 − t2, , (x, 0), Differentiating it w.r.t. t, we get, dy, dx, + 2y, =0, 2x, dt, dt, dy / dt, dx, =−y, dt, x, y, and, vx = − v y, x, As, y decreases, x increases, so v x, decreases., v x becomes zero when y is zero., , 20 The distance travelled can never be, , 3m, , v = u + at ,, , 17, , (y, 0), , negative in one dimensional motion., , 7, v = 0 + 10 ×, = 3.5 ms − 1, 20, , Also, v 2 − u2 = 2as, 02 − 402 = 2(−1) s3, , x, , 5, 3 − = 5t, 4, 7, 7, s, = 5 t or t =, 4, 20, , = 40 × 20 = 800 m, …(iii), v = u + at, 0 = 40 + a × 40,, , t =2, , the end points of the rod at a given, location, then x2 + y 2 = l 2, , This gives,, 12 ± 144 + 1300 12 ± 38, t =, =, = 5s, 10, 10, 16 From s = ut + 1 at 2 ,, 2, 1, x = 0 + × 10t 2 = 5 t 2 …(i), 2, 1, Also, x + 3 = 0 + × 10(t + 0.5) 2, 2, 1, …(ii), = 5 t 2 + + t , , , 4, , 402 − 0 = 2 × 2 × s1, , or, , 19 If ( x, 0) and ( y , 0) are the coordinates of, , 2, , Now, v − u = 2as, , and, , At t = 2s,, 5, x = × 8 + 3 × 4 = 18.66 m, 6, , 15 a = + g = 10 ms – 2 ,, , a = 2 ms –2, 2, , 21, , KINEMATICS, , DAY TWO, , …(i), , case, velocity of both is uniform. As the, slope of x - t graph for P is greater,, therefore, velocity of P is greater than, that of Q., , 23 Maximum acceleration is represented by, the maximum slope of the velocity-time, graph. Thus, it is the portion CD of the, 80 − 20, graph, which has a slope =, 40 − 30, = 6 ms −2 ., , t2 = α, 2 = α2 − α, , …(ii), , t =α, 2, , α2 − α − 2 = 0, (α − 2) (α + 1) = 0, ∴, α = 2, α = − 1,, which is not possible, t 2 = α = 2 or t = 2,, Differentiating Eq. (i) w.r.t. t,, dv, = 12t 2 − 2, dt, , 24 Displacement is the algebraic sum of area, under velocity-time graph., As, displacement = area of triangles, + area of rectangle, v (ms–1), A, , 2, , E, O, 1, , a = 12 × 2 − 2 = 22 ms – 2, , 18 Acceleration, a = dv = 5t + 6, dt, On integrating, we get, 5, dx, v = t2 + 6 t =, 2, dt, Integrating again,, 5, 6, x = t3 + t2, 6, 2, , 2, B, , C, 3, , F, H, 4, , 5, G, , t (s), , D, , ∆ OAB + ∆ ABC + ∆CDH + HEFG, =, , 1, 1, 1, ×2×2+ ×1×2+, 2, 2, 2, × 1 × (− 2) + 1 × 1, , =2+ 1−1+ 1=3m
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22, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , 25 If velocity versus time graph is a straight, line with negative slope, then, acceleration is constant and negative., With a negative slope distance-time, 1, graph will be parabolic s = ut − at 2 ., , , 2, So, option (b) will be incorrect., , 4 Given, v = α x, , t, dx, = ∫ α dt, 0, x, [Q at t = 0, x = 0 and let at any time t,, particle is at x], x, , ∫0, , a constant rate, then it increases in, negative direction with same rate., , 27 Case I, Case II, Relative velocity is v 1 − v 2 = 2, On solving,, v 1 = 5 ms −1 , v 2 = 3 ms −1, , 28 When a particle moves with constant, velocity, then acceleration of particle is, zero and hence particle is not able to, change the direction. Hence, statement I, is false while statement II is true., Hence, correct answer is (d)., , 29 When two objects moving in opposite, direction, then their relative velocity, becomes (v 1 + v 2 ), hence statement I is, false. When moves in same direction,, then relative velocity v = (v 1 − v 2 ),, hence statement II is true. Hence,, correct answer is (d)., , 30 Without changing direction of velocity,, it is possible to change the acceleration, of a moving particle, hence statement I, is true, while statement II is false., Hence, correct answer is (c)., , SESSION 2, 1 The slope of velocity-time graph gives, acceleration. Since, the given graph is a, straight line and slope of graph is, constant. Hence acceleration is, constant. Thus, (a) is correct. The area, of v-t graph between 0 to 10 s is same as, between 10 s to 20 s., 2 As, x1 ( t ) = 1 at 2 and x2 ( t ) = vt, 2, 1, ∴ x1 − x2 = at 2 − vt (parabola), 2, Clearly, graph (b) represents it correctly., , 3 As, v = v 0 + gt + ft 2 or, dx, = v 0 + gt + ft 2, dt, ⇒ dx = (v 0 + gt + ft 2 ) dt, So,, ⇒, , x, , 1, , ∫ 0 dx = ∫ 0(v 0 +, x = v0 +, , gt + ft 2 ) dt, , g, f, +, 2, 3, , Q v = dx , , , , dt , , dx, =α x, dt, dx, or, = α dt, x, On integration,, , or, , 26 Initially velocity keeps on decreasing at, , Relative velocity is v 1 + v 2 = 8, , 7 Given, v = − u, a = g = 10 ms –2 ,, , ⇒, or, or, ∴, , x1 /2, 1/2, , x, , = αt, α, t, 2, 2, α, x=, × t2, 4, x ∝ t2, , x1 /2 =, , 5 From s = ut + 1 at 2 ,, , 2, 1, s = 0 + × 10 × 52 = 125m, 2, Distance covered in 3 s,, 1, = 0 + × 10 × 32 = 45m, 2, Distance to be covered = 125 − 45, = 80 m, 1, From s = ut + at 2, 2, 1, 80 = 0 + × 10t 2, 2, 80, = 16, ⇒, t2 =, 5, ∴, t = 4s, , 6 Suppose velocity at O = zero, As average velocity in interval t 1, is v 1 ,, ∴ Velocity at A = v 1, As average velocity in interval t 2, is v 2 ,, ∴ Velocity at B = (v 2 − v 1 ), As average velocity in interval t 3, is v 3 ,, Velocity at C = (v 3 − v 2 + v 1 ), Using, v = u + at, …(i), v 1 = 0 + at 1, …(ii), (v 2 − v 1 ) = 0 + a(t 1 + t 2 ), (v 3 − v 2 + v 1 ) = 0 + a(t 1 + t 2 + t 3 ), …(iii), Subtract Eq. (i) from Eq. (iii), we get, (v 3 − v 2 ) = a (t 2 + t 3 ), (v 2 − v 1 ) a(t 1 + t 2 ), =, (v 3 − v 2 ) a(t 2 + t 3 ), (v 1 − v 2 ) t 1 + t 2, =, (v 2 − v 3 ) t 2 + t 3, , If t 1 and t 2 are the timings taken by the, ball to reach the points A and B, respectively, then, 1, × 10 × t 12, 2, 1, 40 = − 45 t 2 + × 10 × t 22, 2, On solving, we get, t 1 = 9.4 s and t 2 = 9.8 s, Time taken to cover the distance AB,, = (t 2 − t 1 ), = 9.8 − 9.4 = 0.4 s, 20 = − 45 t 1 +, , 0, , Divide Eq. (ii) by Eq. (iv), we get, , s = 50 m, t = 10 s, 1, As, s = ut + at 2 ,, 2, 1, ⇒ 50 = − u × 10 + × 10 × 102, 2, On solving,, u = 45ms −1, , …(iv), , 8 Let us suppose that the cars A and B are, moving in the positive x-direction. Then,, car C is moving in the negative, x-direction., Therefore,, v A = 36 kmh −1 = 10 ms −1, v B = 54 kmh −1 = 15 ms −1, and v C = − 54 kmh −1 = − 15 ms −1, Thus, the relative speed of B with respect, to A is,, v BA = v B − v A, = 15 − 10 = 5 ms −1, and the relative speed of C with respect, to A is,, v CA = v C − v A = − 15 − 10, = − 25ms −1, At time t = 0, the distance between A, and B = distance between A and, C = 1 km = 1000 m., The car C covers a distance AC = 1000 m, and reaches car A at a time t given by, AC, t =, |v CA|, =, , 1000 m, 25 ms −1, , = 40 s, , Car B will overtake car A just before car, C does and the accident can be avoided if, it acquires a minimum acceleration a, such that it covers a distance,, s = AB = 1000 m in time t = 40 s, travelling with a relative speed of, u = v BA = 5ms −1 ., This gives, from, s = ut +, , 1 2, at , a = 1 ms −2, 2
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KINEMATICS, , DAY TWO, We know that the relation for the, stopping distance s is, , 9 Velocity of the particle is given by, dx d a, −bt , −bt, v =, =, (1 − e ) = ae, dt dt b, , , Thus, choice (b) is correct. At t = 0, the, value v and α are v = ae −0 = a, and α = − ab e −0 = ab, The displacement x is maximum, when, t → ∞,, a, a, i.e., x max = (1 − e − ∞ ) =, b, b, 10 We know that, h = ut + 1 gt 2, 2, 1 2, h = − ut + gt, ⇒, 2, 2u, and, …(i), t =, = t1 − t2, g, , 2, , 2, , Hence,, , 100 , s1 u1 , 1, = = , =, 200 , s2 u2 , 4, , Thus,, , s2 = 4 s1 = 8 x, , or, , Hence, the number of plancks, s, = n2 = 2 = 8, x, dv, 2, 12 Given, a =, = 3t + 2t + 2, dt, ⇒, dv = (3t 2 + 2t + 2) dt, 2, , v, , 2, , 2, , =, , v, , 2, , ⇒, , 3t 3, , 2t 2, v −u= , +, + 2t , 3, 2, , 0, , ⇒, , v = u + [ t 3 + t 2 + 2t ] 20, , O, , v = 2 + [ 2 + 2 + 2 × 2], 2, , = 2 + 16, = 18 ms −1, , 13 The velocity-time graph for the given, , C, , situation can be drawn as below., Magnitudes of slope of OA = f, , B, A, , 1 t , 1 2, f =, ft, 2 6, 72, , 14 Here, x = ( t − 2)2, , + 2t + 2) dt, , 3, , t, 6, , From Eq. (i), we get, 1, s = f (t 1 ) 2, 2, , On integrating, this gives, , ∫u dv = ∫0(3t, , t1 =, , 4 ms-1, , Q e −1 ≈ 1 , , , , 3, , ( f t 1 )t, 12s, =, 1, s, ( f t 1 )t 1, 2, , s ∝ u2, , As,, , s = 1 f t2 , , 1, , , 2, , …(ii), or, ( f t 1 ) = 12s, From Eqs. (i) and (ii), we have, , Since,, v = 0,, So,, 2as = − u2, , – 4 ms-1, , At t = 1/ b , the displacement of the, particle is, a, a, 1, 2 a, x = (1 − e −1 ) ≈ 1 − =, b, b , 3 3 b, , h, , s + ( f t 1 ) t + 2 s = 15 s, , v 2 = u2 + 2as, , Acceleration of the particle is given by, dv, d, α =, =, (ae −bt ) = − abe −bt, dt, dt, , A, , or, , 23, , Y, , C, , B, 2s, , D, 4s, , t, , A, , dx, = 2 ( t − 2) ms −1, dt, dv, Acceleration, a =, = 2 ms −2, dt, (i.e. uniform), Velocity, v =, , When t = 0, v = −4 ms −1 ,, 1 2, gt 1, 2, 1, Case II h = + ut 2 + gt 22, 2, 1 2, Case III h = gt 3, 2, This gives,, 2h 2u, =, t 2 + t 22, g, g, Case I, , h = − ut 1 +, , –1, , …(ii), , v (ms ), , O, , …(iv), , …(v), , t3 = t1 t2, , 11 Given that the initial velocity of the, , bullet in the first case is u1 = 100 ms −1 ., , Initial number of plancks, n1 = 2, Initial stopping distance, = s1 = n1 x = 2 x,, with x as the thickness of one planck., Similarly, Initial velocity of the bullet in, second case,, u2 = 2 × 100 = 200 ms −1, , t = 2 s, v = 0, t = 4s, v = 4 ms −1, , B, , …(iii), , C, t1, , t2, , t, D, , and slope of, , Solving these, give us, t 32 = (t 1 − t 2 ) t 2 + t 22, ⇒, , A, , t (s), , Velocity (v ) - time (t) graph of this motion, is as shown in figure., X, , E, , BC =, , f, 2, , v = f t1 =, , f, t2, 2, , t 2 = 2t 1, In graph area of ∆ OAD gives distance,, 1, …(i), s = f t 12, 2, Area of rectangle ABED gives distance, travelled in time t, s2 = ( f t 1 )t, Distance travelled in time t 2 ,, 1f, s3 =, (2t 1 ) 2, 22, Thus,, ⇒, , Distance travelled, = Area AOB + Area BCD, 4×2 4×2, =, +, = 8m, 2, 2, , 15 Given, v = 108 kmh −1 = 30 ms −1, From first equation of motion, v = u + at, ∴, 30 = 0 + a × 5, (Q u = 0), or, a = 6 ms − 2, So, distance travelled by metro train in 5, s, 1, 1, s1 = at 2 = × (6) × (5)2 = 75 m, 2, 2, Distance travelled before coming to rest, = 45m, So, from third equation of motion, , s1 + s2 + s3 = 15 s, s + ( f t 1 ) t + f t 12 = 15 s, , or, , 02 = (30)2 − 2a′ × 45, 30 × 30, = 10 ms −2, a′ =, 2 × 45
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24, , DAY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , Time taken in travelling 45 m is, 30, = 3s, 10, Now, total distance = 395 m, i.e., 75 + s ′ + 45 = 395 m, or s′ = 395 − (75 + 45) = 275 m, 275, ∴, t2 =, = 9.2 s, 30, Hence, total time taken in whole, journey = t 1 + t 2 + t 3, = 5 + 9 . 2 + 3 = 17. 2 s, , or, ⇒, , 1, s = ut + gT 2, 2, 1, h = 0 + gT 2, 2, T =, , s, , u, g, , O t=T, , H, , At,, , t2, , If t 2 is the time taken to hit the ground,, 1 2, gt 2, 2, But, [Given], t 2 = nt 1, nu 1 n2u2, So, − H = u, − g, g, 2, g2, nu2 1 n2u2, −, g, 2 g, , 2gH = n2u2 − 2nu2, , ∫ 6.25, , −, , 0, , v 1 /2 ] 0625, ., , = 2(− 6.25) = 2 × 2.5, ⇒, , 10 ms-1, , t, , = [2, , 2h , , g , , h, m, 9, Hence, the position of ball from the, h 8h, ground = h − =, m, 9, 9, , 40 ms-1, , v −1 /2 dv = − 2.5 ∫ dt, 2.5 [ t ] t0, , , ∴ T =, , , s=, , t = 2s, , 240 m, , ⇒, , dt, dv, = − 2.5dt, v, , 0, , 1, T2, g⋅, 2, 9, g, 2h, ×, s=, 18, g, s=, , can be applied to predict the nature of, motion of one particle with respect to, the other., , 17 Given, dv = − 2.5 v, , ⇒, , ⇒, , 2, , 19 Central idea concept of relative motion, , 2gH = nu2 (n − 2), , ⇒, , or, , 1 T , g , 2 3, , Cliff, , t 2 − 2 t − 48 = 0, , or, , t 2 − 8t + 6t − 48 = 0, t = 8, − 6, , [As, t = − 6s is not possible], i.e.,, t = 8s, Thus, distance covered by second, particle with respect to first particle in 8, s is, s 12 = (v 21 ) t = (40 − 10) (8s), , T, s,, 3, , s= 0+, , or, , 1 n2u2, nu2, n2u2 − 2nu2, −, =, 2 g, g, 2g, , t =, , or, , or, , Ground, , i.e. − H = ut 2 −, , H =, , t=0, u=0, T, t=, 3, , h, , t1 u, , −H =, , (Q u = 0), , 2h , , g , , 16 Time taken to reach the maximum, height, t 1 =, , Consider the stones thrown up, simultaneously as shown in the diagram, below. As motion of the second particle, with respect to the first we have relative, acceleration, |a21 |=|a2 − a1 |= g − g = 0., Thus, motion of first particle is straight, line with respect to second particle till, the first particle strikes ground at a time, is given by, 1, −240 = 10t − × 10 × t 2, 2, , 18 From equation law of motion gives,, , t3 =, , = 30 × 8, = 240 m, Similarly, time taken by second particle, to strike the ground is given by, 1, −240 = 40 t − × 10 × t, 2, or, − 240 = 40 t − 5t 2, or, , 5 t 2 − 40 t − 240 = 0, , or, , t 2 − 8 t − 48 = 0, t 2 − 12 t + 4 t − 48 = 0, , or, or, , t (t − 12) + 4 (t − 12) = 0, t = 12, − 4, , (As, t = − 4 s is not possible), i.e., t = 12 s, Thus, after 8 s, magnitude of relative, velocity will increase upto 12 s when, second particle strikes the ground., Hence, graph (c) is the correct, description.
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DAY THREE, , Scalar, and Vector, Learning & Revision for the Day, u, , u, , u, , Scalar and Vector Quantities, Laws of Vector Addition, Subtraction of Vectors, , u, , u, , u, , Multiplication or Division of, a Vector by a Scalar, Product of Vectors, Resolution of a vector, , u, , u, , u, , Relative velocity, Motion in a Plane, Projectile Motion, , Scalar and Vector Quantities, A scalar quantity is one whose specification is completed with its magnitude only., e.g. mass, distance, speed, energy, etc., A vector quantity is a quantity that has magnitude as well as direction. e.g. Velocity,, displacement, force, etc., , Position and Displacement Vectors, A vector which gives position of an object with reference to the origin of a coordinate, system is called position vector., The vector which tells how much and in which direction on object has changed its, position in a given interval of time is called displacement vector., , General Vectors and Notation, l, , l, , PREP, MIRROR, , Zero Vector The vector having zero magnitude is called zero vector or null vector. It, is written as 0. The initial and final points of a zero vector overlap, so its direction is, arbitrary (not known to us)., , Your Personal Preparation Indicator, , Unit Vector A vector of unit magnitude is known as an unit vector. Unit vector for A, is A$ (read as A cap)., A=AA, , Direction, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , Magnitude, l, , Orthogonal Unit Vectors The unit vectors along X -axis,s, Y-axis, $ $j and k$ . These are the orthogonal unit, and Z-axis are denoted by i,, vectors., $i = x , $j = y , k$ = z, x, y, z, , Y, j, i, X, Z, , k, , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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26, , l, , l, , l, , l, , l, , DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , Parallel Vector Two vectors are said to be parallel, if they, have same direction but their magnitudes may or may not, be equal., Antiparallel Vector Two vectors are said to be anti-parallel, when, (i) both have opposite direction, (ii) one vectors is scalar non zero negative multiple of, another vector., Collinear Vector Collinear vector are those which act along, same line., Coplanar Vector Vector which lies on the same plane are, called coplanar vector., Equal Vectors Two vectors A and B are equal, if they have, the same magnitude and the same direction., , Laws of Vector Addition, 1. Triangle Law, If two non-zero vectors are represented by the two sides of a, triangle taken in same order than the resultant is given by the, closing side of triangle in opposite order, i.e., R=A+B, The resultant R can be calculated as, | A + B| = R =, , R=, , A2 + B2 + 2 AB cos θ, B, A+, , B sin θ, , B, θ, , α, A, , If resultant R makes an angle α with vector A, then, B sin θ, tan α =, A + B cos θ, , –B, , If θ be the angle between A, and B,, then | A − B| =, , A2 + B2 − 2 AB cos θ, , If the vectors form a closed n sided polygon with all the sides, in the same order, then the resultant is zero., , Multiplication or Division, of a Vector by a Scalar, The multiplication or division of a vector by a scalar gives a, vector. For example, if vector A is multiplied by the scalar, number 3, the result, written as 3A, is a vector with a magnitude, three times that of A, pointing in the same direction as A. If we, multiply vector A by the scalar − 3, the result is − 3 A , a vector, with a magnitude three times that of A, pointing in the direction, opposite to A (because of the negative sign)., , Products of Vectors, The two types of products of vectors are given below, , The scalar product of two vectors A and B is defined as the, product of magnitudes of A and B multiplied by the cosine of, smaller angle between them. i.e. A ⋅ B = AB cos θ, , Properties of Dot Product, l, , B, , C, According to parallelogram law of, vector addition, if two vector acting, R, on a particle are represented in Q, β, magnitude and direction by two, α, adjacent side of a parallelogram, then, P, A, the diagonal of the parallelogram, represents the magnitude and direction of the resultant of the, two vector acting as the particle., , i.e., R = P+Q, Magnitude of the resultant R is given by, , l, , l, , l, , | R | = P2 + Q2 + 2 PQ cos θ, Q sin θ, P sin θ, ⇒ tan β =, P + Q cos θ, Q + P cos θ, , Subtraction of Vectors, Vector subtraction makes use of the definition of the negative, of a vector. We define the operation A − B as vector − B added, to vector A. A − B = A + ( − B), , Dot product or scalar product of two, vectors gives the scalar two vectors, given the scalar quantity., , B, , It is commutative in nature., i.e. A ⋅ B = B ⋅ A ., Dot product is distributive over the, addition of vectors., i.e. A ⋅ (B + C) = A ⋅ B + A ⋅ C, $i ⋅ i$ = $j ⋅ $j = k$ ⋅ k$ = 1, because angle, , θ, O, , A, B cos θ, Component of, vector B along A, , between two equal vectors is zero., l, , tan α =, , B, A, , Scalar or Dot Product, , B cos θ, , 2. Parallelogram Law, , Thus, vector subtraction is, really a special case of vector, addition., The, geometric, construction for subtracting, two vectors is shown in the A – B, above figure., , If two vectors A and B are perpendicular vectors, then, A ⋅ B = AB cos 90 ° = 0 and $i ⋅ $j = $j ⋅ k$ = k$ ⋅ $i = 0, , The Vector Product, The vector product of A and B, written as A × B, produces a, third vector C whose magnitude is C = AB sin θ. where, θ is the, smaller of the two angles between A and B., Because of the notation, A × B is also known as the cross, product, and it is spelled as ‘A cross B’.
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DAY THREE, , SCALAR AND VECTOR, , Properties of Cross Product, l, , l, , Relative Velocity, , Vector or cross product of two vectors gives the vector, quantity., , The time rate of change of relative position of one object with, respect to another is called relative velocity., , Cross product of two vectors does not obey the, commutative law. i.e. A × B ≠ B × A ;, , Different Cases, , Here, A × B = − B × A, l, , l, , l, , l, , Case I If both objects A and B move along parallel straight, lines in the opposite direction, then relative velocity of B, w.r.t. A is given as,, , Cross product of two vectors is distributive over the, addition of vectors., A × (B + C) = A × B + A × C, , vBA = vB − (− v A ) = vB + v A, , Cross product of two equal vectors is given by A × A = 0, $i × $i = (1 × 1 × sin 0 ° ) n$ = 0, Similarly,, $j × $j = (1 × 1 × sin 0 ° ) n$ = 0, k$ × k$ = (1 × 1 × sin 0 ° ) n$ = 0, , If both objects A and B move along parallel staight lines in the, same direction, then, v AB = v B − v A, Case II Crossing the River To cross the river over shortest, distance, i.e. to cross the river straight, the man should swim, upstream making an angle θ with OB such that, OB gives the, direction of resultant velocity ( vmR ) of velocity of swimmer v M, and velocity of river water v R as shown in figure. Let us, consider, , Cross product of two perpendicular vectors is given as, A × B = ( AB sin 90 ° ) n$ = ( AB) n$, For two vectors A = a $i + a $j + a k$, x, , y, , and B = b x $i + b y $j + b z k$ ., $i, $j, A × B = ax, bx, l, , 27, , ay, by, , z, , B, , vm, , Cross product of vectors $i , $j and k$ are following cyclic rules, $ $j × k$ = $i and k$ × $i = $j, as follows $i × $j = k,, j, , i, , vR, , A, , k$, az, bz, , ⊕, , vmR, θ, O, , v, In ∆OAB, sin θ = R and vmR = v2m − v2R, vm, The time taken to cross the river given by, d, d, t1 =, =, 2, vmR, v − v2, m, , k, Cyclic representation for unit vectors $i , $j and k$, NOTE, , • Vector triple product is given by, , A × (B × C) = B (A ⋅ C) − C (A ⋅ B), , Resolution of a Vector, Y, The process of splitting of a single, vector into two or more vectors in, different direction is called resolution, of a vector. Consider a vector A in the, A, X -Y plane making an angle θ with the y, X -axis. The X and Y components of, A are Ax and A y respectively., O, Thus A = A = ( A cos θ)i$, x, , xi, , A y = A yj, , along X -direction, = ( A sin θ)$j along Y-direction, , From triangle law of vector addition, |A|= | A xi + A yj| =, and, , tan θ =, , Ay, Ax, , Ax2 + A2y, , Ay , = θ = tan −1 , , Ax , , R, , Case III To cross the river in possible shortest time The, man should go along OA. Now, the swimmer will be going, along OB, which is the direction of resultant velocity v mR of, vm and vR ., AB vR, In ∆OAB, tan θ =, =, OA vm, and, , vmR = v2m + v2R, vR, x, , A, , B, , A, d, , θ, Ax, , –vmR, , vm, , v, , θ, X, , upstream, , O, , downstream, , Time of crossing the river,, t =, , d, OB, =, =, vm vmR, , x2 + d 2, v2m + v2R, , The boat will be reaching the point B instead of point A. If, AB = x,, v, x, dv, then,, tan θ = R =, ⇒ x= R, vm d, vm
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DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , Motion in a Plane, , l, , B, , Suppose O be the origin for, measuring time and position of, the object (see figure)., , r2, , r1, O, , l, , l, , A, , l, , x1, , x2, , l, , Displacement of an object, form position A to B is, AB = r = r2 − r1 = ( x2 − x1 ) i$ − ( y2 − y1 ) $j, , l, , l, , 0, , 0, , 0, , x = x0 + vx t , y = y0 + v y t, l, , A particle moving in xy-plane (with uniform acceleration),, then its equation of motion for X and Y-axes are, vx = ux + ax t , v y = u y + a yt, 1, 1, x = x 0 + ux t + ax t 2 , y = y0 + u yt + a yt 2, 2, 2, a = a $i + a $i, x, , Projectile Motion, uy = u sin θ, , Y, A, , B, , A, , u cos θ, , h, , B, , X, , Resolving u in two components, we have, ux = u cos θ, u y = u sin θ, ax = 0, a y = − g., g, x2, 2u2 cos2 θ, , Vertical height covered, h =, , l, , u sin 2θ, Horizontal range, R = OB = ux T , R =, g, , α g, t, =, 0, O, , Let a particle be projected in horizontal direction with speed u, from height h., , 2u sin θ, g sin α, , u2 sin2 θ, 2 g cos α, , l, , Horizontal range, R =, , 2u2 sin θ cos (θ + α ), g, cos2 α, , l, , Maximum range occurs when θ =, , l, , Rmax =, , u2, when projectile is thrown upwards., g(1 + sin α ), , l, , Rmax =, , u2, when projectile is thrown downwards., g (1 − sin α ), , π α, −, 2 2, , Projectile Motion Down, an Inclined Plane, A projectile is projected down the plane from the point O with, an initial velocity u at an angle θ with horizontal. The angle of, inclination of plane with horizontal α. Then,, x, , u, , θ, O, , Projectile Motion in Horizontal, Direction From Height ( h), , X, , ay = g cos α, , θ, , A, , α, , s, co, g, α, , (9, , –, 0°, , =, α), , α, , 90, , 2, , NOTE Maximum range occurs when θ = 45°, , –, , Maximum height, h =, , u2 sin2 θ, 2g, , l, , a x=, , P, t=T, , l, , θ, , O ux = u cos θ D, R, , u, , in α, gs, , Time of flight on an inclined plane, T =, , u, , Equation of trajectory, y = x tan θ −, , v, , vy, , Velocity of projectile at any, time, v = u2 + g2t 2, , Y, , Let a particle is projected at, an angle θ from the ground with initial velocity u., , l, , vx, , β, , l, , y, , Projectile is an object which, once projected in a given, direction with given velocity, and is then free to move, under gravity alone. The, path described by the, projectile is called its, trajectory., , 2h, g, , Horizontal range, R = u, , P, , x, , h, , Let a particle be projected, up with speed u from an Y, inclined plane which, makes an angle α with the, horizontal and velocity of, projection makes an angle, θ with the inclined plane., , A particle moving in X -Y plane (with uniform velocity), then, its equation of motion for X and Y axes are, v = v $i + v $j, r = x $i + y $j and r = xi$ + y$j, y, , X, y, , 2h, g, , Time of flight, T =, , u, , O, , Projectile Motion Up an Inclined Plane, , r −r, Velocity, v = 2 1, t2 − t1, , x, , gx2, 2u2, , α, , Let the object be at position A y, 2, and B at timing t 1 and t 2 , where, y, 1, OA = r1 , and OB = r2, , Equation of trajectory, y =, , °–, , 28, , A, , l, , Time of flight down an inclined plane, T =, , l, , Horizontal range, R =, , g, , sin, , α, , g sin (90°– α), = g cos α, , 2u sin(θ + α ), g cos α, , u2, [sin (2θ + α ) + sin α ], g cos2 α
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SCALAR AND VECTOR, , DAY THREE, , 29, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 10 A ball is thrown from the ground with a velocity of, , 1 If A and B are two non-zero vectors having equal, magnitude, the angle between the vectors A and A − B is, (a) 0°, (b) 90°, (c) 180°, (d) dependent on the orientation of A and B, , (a) 2 s, , 2 A vector having magnitude of 30 unit, makes equal, angles with each of the X , Y and Z -axes. The, components of the vector along each of X , Y and Z -axes, are, (a) 10 3 unit (b), , 10, unit, 3, , (c) 15 3 unit, , 20 3 ms −1 making an angle of 60° with the horizontal., The ball will be at a height of 40 m from the ground, after a time t equal to (take, g = 10 ms –2 ), , (d) 10 unit, , 3 A particle has an initial velocity 3 $i + 4$j and an, , (b) 3 s, , (b) 7 2 units (c) 7 units, , shown in the figure. At the same time, another body is, projected vertically upwards from B with velocity v 2 . The, point B lies vertically below the highest point. For both, v, the bodies to collide, 2 should be, v1, v1, v2, , (d) 8.5 units, , 4 Unit vector perpendicular to vector A = 3 $i + $j and, , A, , 30°, , B = 2 $i − $j − 5 k$ both, is, 3 $j − 2 k$, 11, − $j + 2 k$, (c) ±, 13, , (a) 2, , ($i − 3 $j + k$ ), 11, $i + 3 $j − k$, (d) ±, 13, , (a) ±, , (b) ±, , which displaces it from its origin to the point r = (2 $i − $j ) m., The work done on the particle (in joule) is, (b) + 7, , (c) + 10, , (d) + 13, , (a) π, , π, (c), 2, , velocity of u ms . If the steps are h metre high and b, metre wide, the ball will hit the edge of the nth step,, where n is, 2 hu 2, (c), gb, , (a) 200 ms, , (b) 400 ms, , (c) 600 ms, , −1, , (d) 700 ms, , −1, , 9 A projectile is fired at an angle of 30° with the horizontal, such that the vertical component of its initial velocity is, T, 80 ms − 1. Its time of flight is T . Its velocity at t = has a, 4, magnitude of nearly (take, g =10 ms – 2 ), (a) 180 ms − 1 (b) 155 ms − 1 (c) 145 ms – 1, , (b) 20 cm, , (c) 50 cm, , (d) 100 cm, , (b) R = 3 h1h2, (d) R = h1h2, , angle of 60° with the horizontal direction with a velocity of, 147 ms −1. Then, the time after which its inclination with, the horizontal is 45°, is, , hu 2, (d), gb 2, , of 200 m. A bullet pierces A and B. The hole in B is 40 cm, below the hole in A. If the bullet is travelling horizontally, at the time of hitting A, then the velocity of the bullet at, A is, −1, , (a) 10 cm, , 14 A projectile is thrown in the upward direction making an, , 8 Two paper screens A and B are separated by a distance, , −1, , (d) 1, , horizontal distance of 100 m. If the gun can induce a, speed of 500 ms −1 to the bullet, at what height above the, bird must he aim his gun in order to hit it?, (take, g = 10 ms – 2 ), , (a) R = (h1h2 )1 / 4, (c) R = 4 h1h2, , −1, , 2hu 2, (b), gb 2, , 3, 2, , plane for two angles of projection. If h1 and h2 are the, greatest heights in the two paths for which this is, possible, then, , π, (d), 4, , 7 A ball rolls off the top of a stair way with a horizontal, , 2hu, (a), gb 2, , (c), , 13 A cannon ball has the same range R on a horizontal, , 6 If A × B = B × A, then the angle between A and B is, π, (b), 3, , (b) 0.5, , B, , 12 A person aims a gun at a bird from a point, at a, , 5 A force F = ( 5 $i + 3 $j + 2 k$ ) N is applied over a particle, , (a) − 7, , (d) 3 s, , 11 A body is projected with a velocity v1 from the point A as, , acceleration of 0.4 $i + 0.3 $j. Its speed after 10 s is, (a) 10 units, , (c) 2 s, , (d) 140 ms − 1, , (a) 15 s, , (b) 10.98 s, , (c) 5.49 s, , (d) 2.745 s, , 15 A projectile projected with a velocity u at an angle θ, passes through a given height h two times at t1 and t 2 ., Then,, (a) t1 + t 2 = T (time of flight) (b) t1 + t 2 =, (c) t1 + t 2 = 2T, , T, 2, , (d) t1t 2 = T, , 16. A particle is projected at angle of 60° with the horizontal, having a kinetic energy K. The kinetic energy at the, highest point is, (a) K, (c) K / 4, , (b) zero, (d) K / 2
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30, , DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 17 A boy playing on the roof of a 10 m high building throws, a ball with a speed of 10 ms −1 at an angle of 30° with, the horizontal. How far from the throwing point, will the, ball be at the height of 10 m from the ground?, (take, g = 10 ms – 2 , sin 30° =, (a) 5.20 m, , (b) 4.33 m, , 1, 3, ), , cos 30° =, 2, 2, (c) 2.60 m, , (d) 8.66 m, , 18 A ball projected from ground at an angle of 45° just, clears a wall in front. If point of projection is 4 m from, the foot of wall and ball strikes the ground at a distance, of 6 m on the other side of the wall, the height of the, j JEE Main (Online) 2013, wall is, (a) 4.4 m, , (b) 2.4 m, , (c) 3.6 m, , (d) 1.6 m, , 19 Neglecting the air resistance, the time of flight of a, projectile is determined by, (a) U vertical, , (b) U horizontal, , 2, 2, (c) U = U vertical, + U horizontal, , 2, 2, (d) U = (U vertical, + U horizontal, )1 / 2, , 20 The horizontal range of a projectile is 4 3 times its, maximum height. Its angle of projection will be, (a) 45°, , (b) 60°, , (c) 90°, , express train applied brakes to avoid collision. If the, retardation due to brakes is 4 ms −2 , the time in which the, accident is avoided after the application of brakes is, (a) 4.25 s, , (b) 5.25 s, , If the velocity of water is 4 kmh −1, the time taken for going, upstream of 8 km and coming back is, (a) 2 h, (b) 2 h and 40 min, (c) 1 h and 20 min, (d) Cannot be estimated from the given information, , 27 A car is travelling with a velocity of 10 kmh −1 on a straight, road. The driver of the car throws a parcel with a velocity, of 10 2 kmh −1 when the car is passing by a man, standing on the side of the road. If the parcel is to reach, the man, the direction of throw makes the following angle, with the direction of the car., (a) 135°, , (c) tan−1 ( 2 ), , (b) 45°, , circular path as shown in the figure., Y, P (x, y), , (b) 900 m, (d) 750 m, , in time t1. The time taken to cover the same distance up, and down the stream is t 2 . Then, the time the swimmer, would take to swim across a distance 2 d in still water is, (b), , t 22, t1, , (c) t1 t 2, , (d) (t1 + t 2 ), , 24 A man standing on a road has to hold his umbrella at 30°, with the vertical to keep the rain away. He throws the, umbrella and starts running at 10 kmh −1. He finds that the, raindrops are hitting his head vertically. The actual, speed of raindrops is, (a) 20 kmh −1, (c) 20 3 kmh −1, , O, , X, , The movement of p is such that it sweeps out a length, s = t 3 + 5, where, s is in metre and t is in second.The, radius of the path is 20 m. The acceleration of P when, t = 2s is nearly, (a) 13 ms −2, , (b) 12 ms −2, , (c) 7.2 ms−2, , (d) 14 ms −2, , 29 For a particle in uniform circular motion the acceleration, , 23 A swimmer crosses a flowing stream of width d to and fro, , t12, t2, , 20, , (b) tan−1 ( 3 / 2), (d) 45°, , (a) 800 m, (c) 400 m, , (a), , m, , Elevation angle of the projectile at its highest point as, seen from the point of projection is, , water.How long does he take to cross a river 1km wide, if, the river flows steadily 3 kmh−1 and he makes his strokes, normal to the river current.How far down the river does, he go, when he reaches the other bank?, , 1, (d) tan , 2, , 28 A point P moves in counter- clockwise direction on a, , (d) 30°, , 22 A man can swim with a speed of 4 kmh−1 in still, , (d) 7.25 s, , 26 A boat takes 2 h to travel 8 km and back in a still water lake., , 21 A projectile is fired at an angle of 45° with the horizontal., , (a) 60°, (c) tan−1 (1 / 2), , (c) 6.25 s, , (b) 10 3 kmh −1, (d) 10 kmh −1, , 25 A passenger train is moving at 5 ms −1 . An express train, , is travelling at 30 ms −1, on the same track and rear side of, the passenger train at some distance. The driver in, , a at a point P (R , θ ) on the circle of radius R is (here, θ is, measured from the X -axis), v2, v2, v2, v2, cos θ $i +, sin θ $j (b) −, sin θ $i +, cos θ $j, R, R, R, R, v2 $ v2 $, v2, v2, (c) −, cos θ $i −, sin θ $j (d), i +, j, R, R, R, R, (a) −, , Direction (Q. Nos. 30-34) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below :, (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true
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SCALAR AND VECTOR, , DAY THREE, 30 Statement I Rain is falling vertically downwards with a, velocity of 3 kmh −1. A man walks with a velocity of 4 kmh −1., Relative velocity of rain w.r.t. man is 5 kmh −1., , 33 Statement I The resultant of three vectors OA, OB and, OC as shown in the figure is R (1 +, the circle., C, , 31 Statement I For the projection angle tan−1( 4) , the, , O, , horizontal and maximum height of a projectile are equal., , 32 Statement I In order to hit a target, a man should point, his rifle in the same direction as the target., Statement II The horizontal range of bullet is dependent, on the angle of projection with the horizontal., , 2 ). R is the radius of, B, , Statement II Relative velocity of rain w.r.t. man is given, by, v rm = v r − v m, , Statement II The maximum range of a projectile is, directly proportional to the square of velocity and, inversely proportional to the acceleration due to gravity., , 31, , 45°, 45°, , A, , Statement II OA + OC is acting along OB and, (OA + OC ) + OB is acting along OB., , 34 Statement I Angle between $i + $j and $i is 45°., Statement II $i + $j is equally include to both $i and $j and, the angle between $i and $j is 90°., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 If F1 and F2 are two vectors of equal magnitude F such, that | F1 ⋅ F2 | = | F1 × F2 |, then | F1 + F 2 | is equal to, , 4 A projectile is given an initial velocity of (i + 2 j) ms −1, where, i is along the ground and j is along the vertical.If, g = 10 ms −2 , the equation of its trajectory is, (a)Y = X − 5 X, (c) 4Y = 2 X − 5 X 2, 2, , h, θ, , 2) F, , (c) F 2, , (b) 2F, (d) None of these, , 2 If a stone is to hit at a point which is at a distance d away, and at a height h above the point from where the stone, starts, then what is the value of initial speed u if stone is, launched at an angle θ?, (a), , g, d, cosθ 2 (d tanθ − h), gd 2, h cos2 θ, , (c), , (b), , d, g, cosθ 2 (d tanθ − h), gd 2, (d − h), , (d), , 3 A projectile can have the same range R for two angles of, projection. If T1 and T2 be the time of flight in the two, cases, then the product of the two time of flight is directly, proportional to, (a) R, (c), , 1, R2, , (b)Y = 2 X − 5 X, (d) 4Y = 2 X − 25 X 2, 2, , 5 A particle of mass m is projected with a velocity v making, d, , (a) (2 +, , JEE Main 2013, , j, , u, , (b), , 1, R, , (d) R 2, , an angle of 30° with the horizontal.The magnitude of, angular momentum of the projectile about the point of, projection when the particle is at its maximum height h is, 3 mv 2, 2g, mv 3, (c), 2g, , (a), , (b) zero, (d), , 3 mv 3, 16 g, , 6 The coordinates of a moving particle at any time t are, given by x = αt 3 and y = βt 3 . The speed of the particle at, time t is given by, (a) t 2 α 2 + β 2, , (b) α 2 + β 2, , (c) 3 t α 2 + β 2, , (d) 3 t 2 α 2 + β 2, , 7 A ball whose kinetic energy is E , is projected at an angle, of 45° with respect to the horizontal. The kinetic energy of, the ball at the highest point of its flight will be, (a) E, (c), , E, 2, , (b), , E, 2, , (d) zero
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32, , DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , (a) π, , −1, , 5 m - min . A man on the South bank of the river, is, capable of swimming at 10 m - min−1 in still water, he, wants to swim across the river in the shortest time. He, should swim in a direction, (a) due to North, (c) 30° West of North, , (b) 30° East of North, (d) 60° East of North, , 9 A ship A is moving Westwards with a speed of 10 kmh −1, , and 60° with the vertical are as shown in the figure. Two, blocks A and B are placed on the two planes. What is the, relative vertical acceleration of A with respect to B?, A, , B, , (b) 5h, 50 2 km, (d) 10 2 h, 50 2 km, , 60°, , 10 A particle is moving Eastwards with a velocity of 5 ms −1., In 10s , the velocity changes to 5 ms −1 Northwards. The, average acceleration in this time is, 1, ms −2 towards North-East, 2, 1, (b) ms −2 towards North, 2, (c) zero, 1, ms −2 towards North-West, (d), 2, , (a), (b), (c), (d), , (a), , 30°, , 4.9 ms −2 in horizontal direction, 9.8 ms −2 in vertical direction, zero, 4.9 ms −2 in vertical direction, , 14 A particle is moving with velocity v = k (Y i+ X j), where k, is a constant.The general equation for its path is, (a)Y = X 2 + constant, (c) XY = constant, , 11 A boy can throw a stone upto a maximum height of 10 m., The maximum horizontal distance that the boy can throw, the same stone upto will be, (a) 20 2 m, (c) 10 2 m, , (b), , 13 Two fixed frictionless inclined plane making an angle 30°, , and a ship B, 100 km South of A is moving Northwards, with a speed of 10 kmh −1. The time after which the, distance between them is shortest and the the shortest, distance between them are, (a) 0 h, 100 km, (c) 5 2 h, 50 km, , π v4, 2 g2, v2, (d) π, g, , v4, g2, v2, (c) π 2, g, , 8 A river is flowing from West to East with a speed of, , (b) 10 m, (d) 20 m, , 12 A water fountain on the ground sprinkles water all around, , (b)Y 2 = X + constant, (d)Y 2 = X 2 + constant, , 15 The maximum range of a bullet fired from a toy pistol,, mounted on a car at rest is R 0 = 40 m. What will be the, acute angle of inclination of the pistol for maximum range, when the car is moving in the direction of firing with, uniform velocity v = 20 ms −1, on a horizontal surface?, j, JEE Main (Online) 2013, (take,g = 10 ms−2 ), (a) 30°, (c) 75°, , it. If the speed of water coming out of the fountain is v,, the total area around the fountain that gets wet is, , (b) 60°, (d) 45°, , ANSWERS, SESSION 1, , SESSION 2, , 1 (d), , 2 (a), , 3 (b), , 4 (b), , 5 (b), , 6 (a), , 7 (b), , 8 (d), , 9 (c), , 10 (c), , 11 (b), , 12 (b), , 13 (c), , 14 (c), , 15 (a), , 16 (c), , 17 (d), , 18 (c), , 19 (a), , 20 (d), , 25 (c), , 26 (b), , 27 (b), , 28 (d), , 29 (c), , 30 (a), , 5 (d), 15 (b), , 6 (d), , 7 (c), , 8 (a), , 9 (b), , 10 (d), , 21 (c), , 22 (d), , 23 (a), , 24 (a), , 31 (b), , 32 (d), , 33 (a), , 34 (a), , 1 (a), 11 (d), , 2 (b), 12 (a), , 3 (a), 13 (d), , 4 (b), 14 (d)
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SCALAR AND VECTOR, , DAY THREE, , 33, , Hints and Explanations, SESSION 1, , 5 Work done in displacing the particle, W = F⋅ r = (5i$ + 3 $j + 2 k$ ) ⋅ (2 $i − $j ), , 1 Suppose angle between two vectors A, and B of equal magnitude is θ. Then,, angle between A and A − B will be, 180° − θ, θ, or 90° − ., 2, 2, Hence, this angle will depend on the, angle between A and B or θ., B, , A –B, , 6 ( A × B) = (B × A ) (given), ⇒, or, A, , a, q, , = 5 × 2 + 3 × (−1) + 2 × 0 = 10 − 3, = 7J, , ∴, , or, ⇒, or, , A– B, , ⇒, , 2 Ax = Ay = Az, Now, A =, , ( A × B) − (B × A ) = 0, ( A × B) + (A × B) = 0, [Q(B × A ) = −( A × B)], 2( A × B) = 0, 2 AB sinθ = 0, sinθ = 0, [Q|A| = A ≠ 0, |B| = B ≠ 0], θ = 0 or π, , A, 30, Ax =, =, = 10 3, 3, 3, , Similarly A y = A x = 10 3 unit, , 3 u = 3 i + 4 j; a = 0.4 $i + 03, . $j, v = u+a t, = 3 $i + 4 $j + (0.4 $i + 03, . $j ) 10, = 3 $i + 4 $j + 4 $i + 3 $j, = 7 $i + 7 $j, , second. Vertical distance travelled by, 1, the ball = nh = gt 2, ...(i), 2, Horizontal distance travelled by the ball, nb, = nb = ut or t =, u, , A, , B, , = − 5$i + 15$j − 5k$, |A | = 32 + 12 = 10, |B| = (2)2 + (−1)2 + (−5)2 = 30, , =, , A ⋅B, |A ||B|, 5, 1, −, 10 3 2 3, 11, 2 3, −5i$ + 15$j − 5k$, , ∴ sin θ = 1 − cos 2 θ =, ∴, , n$ = ±, , =±, , or, , 11, 10 ⋅ 30 ⋅, 2 3, ($i − 3 $j + k$ ), 11, , v 12 sin2 30° v 22, =, 2g, 2g, v2, 1, = sin 30° = = 0.5, v1, 2, , 12 Let the gun be fired with a velocity, u from the point O on the bird at B,, making an angle θ with the horizontal, direction. Therefore, the height of the, aim of the person be at height BA (h ), above the bird., A, u, q, , 200, u, 2, 40, 1, 200 , , = × 9.8 , , , , 100 2, u, , On solving,, , u = 700 ms −1, , 9 Vertical component of the initial, velocity,, u y = u sin30°, uy, 80, or u =, =, = 160 ms −1, sin30° 1 / 2, t =, , 2u sin 30° 2 × 80, T, =, =, = 4s, 4, 4× g, 4 × 10, , v x = u cos 30° = 160 ×, = 80 3 ms, ∴ v =, , =, or, , 200 = ut or t =, Also,, , B, , Here, horizontal range, , 200 m, , 3, 2, , −1, , v 2x + v 2y = u2x + (u y − gt ) 2, , 1, × 10 × t 2, 2, , highest point if both cover the same, vertical height in the same time. So,, , 40 cm, , The unit vector in the normal direction, is, A×B, n$ = ±, |A| |B| sin θ, $i $j, k$, , −, , 11 The two bodies will collide at the, , O, , u, , 4 Given A = 3 $i + $j, B = 2 $i − $j − 5k$, , 1, 0, −1 − 5, , 2, , or 5 t 2 − 30t + 40 = 0, or t 2 − 6t + 8 = 0 or t = 2 or 4, The minimum time t = 2 s, , 8 Refering to the figure,, , 2, , Here, A × B = 3, 2, , 3t, , So, 40 = 20 3 ×, , So from Eq (i), we get, 2, 2u2 h, 1 nb , nh = g , or n =, 2 u , gb 2, , Speed = 7 + 7 = 7 2 units, , cos θ =, , 2, , 7 Let the ball strike the, nth step after t, , A2x + A2y + A2z = 3 A x, , 2, , 10 As, s = u sinθt − 1 g t 2, , or, , u2 sin 2θ, = 100, g, , (500) 2 sin 2θ, = 100, 10, 100 × 10, 1, sin 2θ =, =, = sin 14 ′, (500) 2, 250, , 2θ = 14 ′ or θ = 7 ′, 7, π, rad, =, ×, 60 180, arc, As, angle =, radius, AB, ∴, θ=, OB, or, , or, , AB = θ × OB, 7, π, =, ×, × (100 × 100) cm, 60 180, = 20 cm, , 13 The cannon ball will have the same, horizontal range for the angle of, projection θ and (90° − θ) . So,, u2 sin2 θ, u2 cos 2 θ, and h2 =, h1 =, 2g, 2g, 2, , = (80 3 )2 + (80 − 10 × 4) 2, = 144.3 ms −1, = 145ms −1, , h1 h2 =, , 1 u2 sin θ cos θ , 1 R2, , = ×, 4, g, 4, 4, , , or R = 4 h1 h2
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34, , DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 Horizontal component of the velocity at, an angle 60°, = Horizontal component of the velocity, at an angle 45°., i.e. u cos 60° = v cos 45°, 1, 1, 147 × = v ×, 2, 2, 147, ms −1, or, v =, 2, Vertical component of u = u sin 60°, 147 3, m, =, 2, Vertical component of v = v sin 45°, , 18 As, range = 10 =, ⇒, , and t 2 =, ∴, , 4 u2 sin2 θ − 8 gh, 4 u2 sin2 θ − 8 gh, 4 u2 sin2 θ − 8 gh, , 2g, 2 u sinθ, t1 + t2 =, =T, g, 1, mv 2 cos 2 θ = K cos 2 θ, 2, K, = K (cos 60° ) 2 =, 4, 10 ms–1, , 17, , R = 4 3H, , u sin 2θ, u2 sin2 θ, =4 3, g, 2g, 2, , ∴ 2sin θ cos θ = 2 3 sin2 θ, cos θ, or, = 3, sin θ, or, cot θ = 3 = cot 30°, , 30°, , P, , If t is the time taken by the swimmer to, swim a distance 2d in still water, then, 2d, t =, u, 2ud, 4d 2, 2d, t2 × t = 2, ×, = 2, 2, u −v, u, u − v2, t2, From above, t 2 × t = t 12 or t = 1, t2, ground, the rain comes to him at an, angle 30° with the vertical. This is the, direction of the velocity of raindrops, w.r.t. to the ground., , 21 Maximum height,, u2 sin2 45°, u2, =, = AC, 2g, 4g, , Horizontal range,, u2 sin 2 × 45° u2, R=, =, g, g, , X, , v be the velocity of the river flow. Then, 2d, t1 =, 2, u − v2, 2 ud, d, d, and t 2 =, +, =, u−v, u + v u2 − v 2, , 24 When the man is at rest w.r.t. to the, , θ = 30°, H =, , Ground, u2 sin 2θ, OP = R =, g, , Here, v rg = velocity of the rain w.r.t. the, ground., vm g, , A, , u, , 30°, , =, , 10 × sin(2 × 30° ), 10, , =, , 10 3, = 5 3 = 8. 66 m, 2, , 2, , 45°, a, , O, , ∴, , ∴, , B, , 23 Let u be the velocity of the swimmer and, , 10 m, , O, , 2, × U vertical, g, , angle θ with the horizontal., Then horizontal range,, u2 sin 2θ, R=, g, u2 sin2 θ, and maximum height H =, 2g, , 16 Kinetic energy at highest point,, ( KE )H =, , 2u sinθ, g, , 20 Let u be initial velocity of projection at, , ∴, , 2g, 2 u sin θ −, , g × 16, 1, 2 2 v 20 cos 2 45°, , T = 2t =, , vr, , Speed of river, v r = 3 kmh −1, Width of the river d = 1 km, Time taken by the man to cross the, river,, Width of the river, 1km, t =, =, Speed of the man, 4 kmh −1, 1, 1, = h = × 60 = 15 min, 4, 4, Distance travelled along the river, = vr × t, 1 3, 3000, = 3 × = km =, = 750 m, 4 4, 4, , 10 × 16, 1, 2 2 × 10 × 10 × 1, 2, = 4 − 0.8 = 3.2 ≈ 3.6 m, , Given,, , β, O, , = 4×1−, , =, , 2g, 2 u sin θ +, , gx2, 1, 2, 2 2v 0 cos 2 θ, , = 4 tan 45° −, , 15 For a projectile fired with a velocity u, , ⇒ t1 =, , vm, , (as, g = 10ms −2 ), , Y = x tan θ −, , ∴, , v, , 1 km, , 6m, 10 m, , 19 Time of flight (T ) is 2t, , inclined at an angle θ with the, horizontal., 1, h = u sinθ t − g t 2, 2, ⇒ g t 2 − 2u sinθ t + 2h = 0, , D, , Wall, , ∴ u = 10ms −1, , But, , ∴, , 2u sin θ ±, , A, , 45º, 4m, , 147, 1, ×, 2, 2, 147, m, =, 2, , ∴ t =, , 22 Given, speed of man v m = 4 km/h, , v, , =, , v y = u y + at, 147 147 3, =, − 9. 8 t, 2, 2, 147, or 9. 8 t =, ( 3 − 1), 2, ∴, t = 5.49 s, , u2 = 10 g, , u2 sin 2θ, g, , R/2, , H, C, R, , OC = R/2 = u2 / (2 g ), AC u2 /4 g 1, tan α =, = 2, =, OC, u /2 g, 2, α = tan −1 (1 /2), , B, , vr m, , vr g, , v rm = v rg + v gm = v rg, − v mg = v rg + (− v mg ), v mg = velocity of the man w.r.t the, ground,, and v r m = velocity of the rain w.r.t., the man.
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SCALAR AND VECTOR, , DAY THREE, Here, v rg sin 30° = v mg = 10 kmh −1, 10, ⇒ v rg =, = 20 kmh −1, sin 30°, , 29 For a particle in uniform circular motion, y, , 25 Initial relative velocity of the express, train w.r.t. the passenger train,, uep = ue − u p = 30 − 5 = 25ms −1, Final relative velocity of the express, train w.r.t. the passenger train, v ep = 0, (because express train comes to rest, relative to passenger train), From first equation of motion,, v ep = uep − at ⇒ 0 = 25 − 4t, ⇒ 4t = 25, ⇒ t = 6.25 s, , 26 Total distance travelled by the boat in 2, , h in still water = 8 + 8 = 16 km, Therefore, speed of boat in still water,, 16, vb =, = 8 km h −1, 2, Effective velocity when boat moves, upstream = v b − v w = 8 − 4, = 4 km h −1 ,, Therefore, time taken to travel from one, 8, end to other = = 2 h, 4, Effective velocity when boat moves, downstream = v b + v w = 8 + 4, = 12 km h −1, The time taken to travel 8 km distance, 8 2, =, = h = 40 min, 12 3, Total time taken = 2 h + 40 min, = 2 h and 40 min, , 27 Let v 1 be the velocity of the car and v 2, be the velocity of the parcel. The, parcel is thrown at an angle θ from Q, it, reaches the man at M., , ac, , P (R, q), x, , O, , vr = − 3 $j (vertically downward), Velocity of man, vm = 4 $i, ∴ Relative velocity of rain w.r.t. man, v = v − v = −3 $j − 4 i$, m, , ∴ |vrm | = (−3)2 + (−4) 2, = 5kmh −1, , 31 Horizontal range of the projectile,, u2 sin 2θ, R=, g, R max =, , u2, g, , M, , (θ = 45° ), , u2 sin2 θ, 2g, , θ, , Q, , v1, , u2 sin2 θ u2 sin 2θ, =, g, 2g, A, , v, 10, 1, =, = cos 45°, ∴ cos θ = 1 =, v 2 10 2, 2, So, θ = 45°, , 28 Given, s = t 3 + 5, ds, = 6t, dt, and rate of change of speed, dv, at =, = 6t, dt, ∴ Tangential acceleration at t = 2s, ∴ Speed v =, , a t = 6 × 2 = 12 ms −2, and at t = 2s, v = 3(2)2 = 12 ms −1, ∴ Centripetal acceleration,, 2, , v, 144, ms −2, ac =, =, R, 20, Net acceleration = a2c + a2t ≈ 14 ms −2, , So,, ∴, , or tanθ = 1 or θ = 45°, ∴|F1 + F2 |, = F 2 + F 2 + 2 F F cos 45°, = (2 +, , 2) F, , 2 Vertical distance covered by a projectile, is given by, , sin2 θ, = 2sin θ cos θ, 2, tanθ = 4, θ = tan −1 (4), , …(i), , d = u cos θ × t, d, or, t =, u cos θ, From Eq. (i), we get, d, 1, d2, h = u sin θ ×, − g 2, u cos θ 2 u cos 2 θ, g, d, or h =, cos θ 2(d tan θ − h ), , 3 We know that the range is, 2u sin θ, g, , According to the question, the range of, the projectile is the same for, complementary angles,, 2u sin θ, So, T1 =, g, 2u sin (90° − θ), ⇒, T2 =, g, , =, , 2u2 sin 2θ 2R, =, g2, g, , ∴ T1 T2 ∝ R, , 32 To hit a target, the man should aim his, rifle at a point higher than the target as, the bullet suffers a vertical deflection, 1, ( y = gt 2 ) due to acceleration due to, 2, gravity., , 33 OA = OC, , 4 Initial velocity, u = i + 2 j ms −1, Magnitude of velocity, u = (1)2 + (2)2 = 5ms −1, Equation of trajectory of projectile, Y = X tan θ −, , OA + OC is along OB (bisector) and its, magnitude is, 2R cos 45° = R 2, (OA + OC ) + OB is along OB and its, magnitude is, R 2 + R = R(1 +, , 1 2, gt, 2, , Again, the range of the projectile is, u2 sin 2θ, R=, g, 2u sin θ 2u sin (90° − θ), ∴ T1 T2 =, ×, g, g, , If H = R, then, v2, , SESSION 2, , T =, , The maximum height attained by the, projectile, H =, , So, θ = 45°, Hence, angle between $i and $j is 90°., , h = (u sinθ) t −, , 30 Velocity of rain, , r, , ( i$ + $j ) ⋅ $i, 1, =, = cos 45°, $, $, $, 2, |( i + j )| ⋅ |i|, , 1 F F cos θ = F F sin θ, , v2, towards centre of circle, a =, R, 2, v, (− cos θ $i − sin θ $j ), ∴ a =, R, v2, v2, or a = − cos θ $i −, sin θ $j, R, R, , rm, , 34 cos θ =, , 35, , 2), , ∴ Y = X ×2−, =2X −, , g X2, (1 + tan2 θ), 2u2, tanθ Y = 2 = 2, , X 1, , , 10( X )2, [1 + (2)2 ], 2( 5)2, , 10( X 2 ), (1 + 4) = 2 X − 5X 2, 2× 5
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36, , DAY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 5 Angular momentum of the projectile,, L = mv h r ⊥ = m(v cos θ)h, , where, h is the maximum height, v 2 sin2 θ , = m (v cos θ) , ;, 2g , 3 mv 3, mv 3 sin2 θ cos θ, [Qθ = 30°], L =, =, 2g, 16 g, , 6 Since, x = αt and y = βt, 3, , ∴, Now,, , 3, , r = x $i + y $j = αt 3 $i + βt 3 $j, dr, v=, = α t 2 × 3 i$ + βt 2 × 3 $j, dt, , Thus,|v| = (3 α t ) + (3 βt ), 2 2, , =, , 9 Let the ships A and B be at positions as, shown in figure when the distance, between them is shortest., , = 3t, , α +β, 2, , N, , W, , v1=10, , 2, , 1, 1 , mu2x = m , 2, 2 , 1 u2 , E, = m =, 2 2, 2, , E′ =, , A, , 10 v1 = + 5$i , v2 = + 5 $j, , ∴, a = g sinθ, where, a is along the inclined plane., ∴ Vertical component of acceleration is, g sin2 θ., ∴ Relative vertical acceleration of A, with respect to B is, g, g (sin2 60° − sin2 30° ) = = 4.9 ms −2, 2, (in vertical direction), , 14 Given, velocity v = kY i + kX j, dX, dY, = kY ,, = kX, dt, dt, dY, dY dt, kX, =, ×, =, dX, dt dX, kY, , |∆ v |= 5 2, |∆ v | 5 2, 1, ms −2, =, =, t, 10, 2, N, , Dv, , YdY = XdX ;, , y, v2, , 8 Let the swimmer starts swimming with, , x, W, , v1, , – v1, , Y2 = X 2 + C, where, c = constant., 2, 15 According to question, u = 40, , g, , E, ∴, , d, , q, , A, t = d / v cos θ, , vr, , As component of AB will be v cos θ, This, time will be minimum, when, cos θ = max = 1, i.e. θ = 0°., So, the swimmer should swim in North, direction., , u = 20 ms −1 and T =, , –v2, = 2 × 20 ×, , S, For direction,, , v, , v2, g, , ∴Total area around fountain,, v4, A = πR2m = π 2, g, , ∆v = v2 − v1 = 5$j − 5$i, ∴ a=, , velocity v along AC in a direction, making an angle θ with AB as shown in, the figure. If d is the width of the river,, time taken by the swimmer to cross the, river will be, B, C, , Rm =, , 13 mg sinθ = ma, , S, 1, = 100 ×, = 50 2 km, 2, d, 50 2, Shortest time, t =, =, = 5h, vr, 10 2, , 2, , Q 1 mv 2 = E , , , 2, , , E, , – v1=10 kmh-1, , B, , u , , 2, , D, , d 90°, 45° C, , v2=10 kmh-1, , component of velocity is zero and only, horizontal component is left which is, u x = u cos θ, Given,, θ = 45°, u, u x = u cos 45° =, ∴, 2, Hence, at the highest point kinetic, energy, , the fountain,, , The shortest distance between A and C, is d given by, d = AC = AB sin 45°, , 2 2, , 7 At the highest point of its flight, vertical, , 12 Maximum range of water coming out of, , = 10 2 kmh −1 along BC, , kmh-1, , u2 (10 2 )2, =, = 20 m, g, 10, , R max =, , Thus,, , Relative velocity of B w.r.t. A is, v r = v 12 + v 22 = 102 + 102, , 9 α2t 4 + 9 β2t 4, 2, , Range is maximum when projectile is, thrown at an angle of 45°., , 5, = −1, 5, 1, Average acceleration is, ms −2 towards, 2, North-West., , =, , tan α = −, , = 10 2 ms ., , 4, =2 2, 2, , v = 20 ms −1 , then, (v cos θ + 20) × t = 40, (20cos θ + 20) × 2 2 = 40, , can throw stone is, , −1, , 1, 1, ×, 2 10, , When car is moving with speed,, , 11 Maximum speed with which the boy, u = 2gh = 2 × 10 × 10, , 2u sin 45°, g, , ⇒, , cos 2, , θ, 1, =, ≈ 60°, 2 2 2
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DAY FOUR, , Laws, of Motion, Learning & Revision for the Day, u, , u, , u, , Concept of Forces, Inertia, Newton’s Laws of Motion, , u, , u, , Principle of Conservation of, Linear Momentum, Free Body Diagram, , u, , u, , u, , Connected Motion, Equilibrium of concurrent Forces, Friction, , Concept of Forces, A push or a pull exerted on any object, is defined to be a force. It is a vector quantity., Force can be grouped into two types:, l, , l, , Contact forces are the forces that act between two bodies in contact, e.g. tension,, normal reaction, friction etc., Non-contact forces are the forces that act between two bodies separated by a distance, without any actual contact. e.g. gravational force between two bodies and electrostatic, form between two charges etc., , Inertia, The inability of a body to change by itself its state of rest or state of uniform motion along, a straight line is called inertia of the body., As inertia of a body is measured by the mass of the body. Heavier the body, greater the, force required to change its state and hence greater is its inertia. There are three type of, inertia (i) inertia of rest (ii) inertia of motion (iii) inertia of direction., , Newton’s Laws of Motion, First Law of Motion (Law of Inertia), It states that a body continues to be in a state of rest or of uniform motion along a straight, line, unless it is acted upon by some external force the change the state. This is also, called law of inertia., If F = 0, ⇒ v = constant ⇒ a = 0, This law defines force., The body opposes any external change in its state of rest or of uniform motion., It is also known as the law of inertia given by Galileo., l, l, l, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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38, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , Linear Momentum, It is defined as the total amount of motion of a body and is, measured as the product of the mass of the body and its, velocity. The momentum of a body of mass m moving with a, velocity v is given by p = mv., Its unit is kg-ms–1 and dimensional formula is [ML T –1], , Second Law of Motion, The rate of change of momentum of a body is directly, proportional to the applied force and takes place in the, direction in which the force acts., dp, dp, or F = k, According to second law, F ∝, dt, dt, where, k is constant., dp d, m dv, as,, =, (mv) = ma or, = ma, dt dt, dt, i. e., second law can be written as, dp, F =, = ma, dt, The SI unit of force is newton (N) and in CGS system is dyne., 1 N = 10 5 dyne, , Applications of Conservation, of Linear Momentum, The propulsion of rockets and jet planes is based on the, principle of conservation of linear momentum., u dm, Upward thrust on the rocket, F = −, − mg and if effect, dt, u dm, ., of gravity is neglected, then F = −, dt, Instantaneous upward velocity of the rocket, m , v = u ln 0 − gt, m, l, , l, , and neglecting the effect of gravity, m , m , v = u ln 0 = 2.303 u log 10 0 , m, m, , l, , Impulse, Impulse received during an impact is defined as the product, of the average force and the time for which the force acts., Impulse,, I = Fav t, Impulse is also equal to the total change in momentum of the, body during the impact., Impulse,, , I = p2 − p1, , Impulse = Change in momentum, , m , m , vb = u loge 0 = 3.303 u log 10 0 , mr , mr , , Apparent Weight of a Boy in a Lift, Actual weight of the body is mg. Here we consider the, apparent weight of a man standing in a moving lift., l, , Third Law of Motion, , l, , To every action, there is an equal and opposite reaction., l, , F12 = − F21, Action and reaction are mutually opposite and act on two, different bodies., , l, , The force acting on a body is known as action., , l, , When a forc acts on a body, then the reaction acts normally, to the surface of the body., , Principle of Conservation, of Linear Momentum, It states that if no external force is acting on a system, the, momentum of the system remains constant., dp, According to second law of motion, F =, dt, If no force is acting, then F = 0, dp, = 0 ⇒ p = constant, ∴, dt, or, , m1v 1 = m2 v2 = constant, , where, m0 = initial mass of the rocket including that of the, fuel,, u = initial velocity of the rocket at any time t ,, m = mass of the rocket left,, v = velocity acquired by the rocket,, dm, = rate of combination of fuel., dt, Burnt out speed of the rocket is the speed attained by the, rocket when the whole of fuel of the rocket has been burnt., Burnt out speed of the rocket, , l, , l, , If lift is accelerating upward with acceleration a, then, apparent weight of the body is R = m(g + a)., If lift is accelerating downward at the rate of acceleration a,, then apparent weight of the body is R = m(g − a), If lift is moving upward or downward with constant, velocity, then apparent weight of the body is equal to, actual weight., If the lift is falling freely under the effect of gravity, then it, is called weightlessness condition., , Free Body Diagram, A free body diagram (FBD) consists of a diagramatize, representation of a single body or sub-system of bodies, isolated from its surroundings showing all forces acting on it., While sketching a free body diagram the following points, should be kept in mind., l, , l, , Normal reaction (N) always acts, normal to the surface on which the, body is kept., , A, , B, , When two objects A and B are connected by a string, the, tension for object A is towards B and for object B, it is, towards A.
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40, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , Net acceleration,, Net accelerating force, (m1 − m2 )g, a=, =, Total mass, (m1 + m2 + M ), Tension, T1 = m1(g − a) =, , (M + 2 m1)m2 g, (M + m1 + m2 ), , a, , in, gs, , m2, , θ, , θ, m 2g, , Whenever an object actually slides or rolls over the surface of, another body or tends to do so, a force opposing the relative, motion starts acting between these two surfaces in contact. It, is known as friction or the force due to friction. Force of, friction acts in a tangential direction to the surfaces in contact., , a, , m1, , Net acceleration,, (m − m2 sin θ)g, , if m1g > m2 g sin θ, a= 1, (m1 + m2 ), and, , a=, , Types of Friction, The four types of friction are given below, , (m2 sin θ − m1)g, , if m1g < m2 g sin θ, m1 + m2, , 1. Static Friction It is the opposing force that comes into, play when one body is at rest and a force acts to move it, over the surface of another body., , and tension in the string, m m (1 + sin θ), T = m1(g − a) = 1 2, (m1 + m2 ), , It is a self adjusting force and is always equal and opposite, to the applied force., , For a pulley and block system on a smooth double inclined, plane as shown in the figure, we have, N, , a, , T, , N, , m2, , θ1, sin, , g, m2, , Net acceleration, a =, , m2g cos θ2, , θ2, , θ2, , m1g cos θ1, , θ1, , (m2 sin θ2 − m1 sin θ 1)g, ,, (m1 + m2 ), , for, θ2 > θ 1, m2 > m1, and tension in the string,, T =, , Applied, force F, , a, , sin, , g, m1, , 2. Limiting Friction It is the limiting (maximum) value of, static friction when a body is just on the verge of starting, its motion over the surface of another body., N, , T, , m1, , P, Q, R, =, =, sin α sin β sin γ, , Friction, T, , m2 g cos θ, , α, , Q, , T, , m2, , θ, , R, , β, γ, , For the system of block and pulley, with a smooth inclined, plane as shown in the figure, we have, N, , l, , P, , (M + 2 m2 )m1g, (M + m1 + m2 ), , and Tension, T2 = m2 (g + a) =, l, , Lami Theorem For three concurrent forces in equilibrium, position., , m1m2 (sin θ 1 + sin θ2 )g, (m1 + m2 ), , Equilibrium of Concurrent Forces, If a number of forces act at the same point, they are called, concurrent forces., The necessary condition for the equilibrium of a body under, the action of concurrent forces is that the vector sum of all the, forces acting on the body must be zero., Mathematically for equilibrium,, Σ Fnet = 0 or ΣFx = 0, ΣF y = 0 and ΣFz = 0, , f = µN, , The force of limiting friction f l between the surfaces of, two bodies is directly proportional to the normal reaction, at the point of contact. Mathematically,, fl ∝ N or fl = µ l N, f, ⇒, µl = l, N, where, µ l is the coefficient of limiting friction for the, given surfaces in contact., 3. Kinetic Friction It is the opposing force that comes into, play when one body is actually slides over the surface of, another body. Force of kinetic friction fk is directly, f, proportional to the normal reaction N and the ratio k is, N, called coefficient of kinetic friction µ k , value of µ k is, slightly less than µe (µ k < µ l )., Whenever limiting friction is converted into kinetic, friction, body started motion with a lurch.
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LAWS OF MOTION, , DAY FOUR, 4. Rolling Friction It is the opposing force that comes into, play when a body of symmetric shape (wheel or cylinder, or disc, etc.) rolls over the surface of another body. Force, of rolling friction fr is directly proportional to the normal, reaction N and inversely proportional to the radius (r ) of, the wheel., N, N, Thus,, or fr = µ r, fr ∝, r, r, The constant µ r is known as the coefficient of rolling, friction µ r has the unit and dimensions of length., Magnitudewise µ r << µ k or µ l ., l, , l, , In limiting condition, f1 = mg sin φ, ⇒, , N = mg cos φ, f1, = tan φ, N, f1, = µ s = tan φ, N, , ∴, , Acceleration of a Block on, Applying a Force on a Rough Surface, l, , The value of rolling friction is much smaller than the, value of sliding friction., , Acceleration of a block on a horizontal surface is as shown, in the figure., N, , Ball bearings are used to reduce the wear and tear and, energy loss against friction., , Angle of friction is defined as the angle θ which the resultant R, of the force of limiting friction fl and normal reaction N,, subtends with the normal reaction., , mg, , F − f F − µmg, =, m, m, F, or, a=, − µg, m, where, µ = coefficient of kinetic friction between the two, surfaces in contact., a=, , The tangent of the angle of friction is equal to the coefficient, of friction. i.e. µ = tan θ, N, , θ, , fl, , F, , f = µN, , Angle of Friction, , R, , 41, , l, , Applied force F, , Acceleration of block sliding down a rough inclined plane, as shown in the figure is given by, a = g(sin α − µ cos α ), , Angle of Repose, , N, f=, , Angle of repose is the least angle of the inclined plane, (of given surface) with the horizontal such that the given body, placed over the plane, just begins to slide down, without, getting accelerated., , µN, , ma, mg, , N, , sin, , α, , α, , mg, , α, , mg cos α, , fl, l, , mg, , nφ, , φ, , si, , φ, , mg, , mg cos φ, , Retardation of a block sliding up a rough inclined plane as, shown in the figure is a = g(sin α + µ cos α ), N, F, , The tangent of the angle of repose is equal to the coefficient of, friction., Hence, we conclude that angle of friction (θ) is equal to the, angle of repose (φ)., , ma, α, α, sin, µN mg, mg cos α, mg, α
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42, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 Five forces inclined at an angle of 72° w.r.t. each other, act on a particle of mass m placed at the origin. Four, forces are of magnitude F1 and one has a magnitude F2., Find the resultant acceleration of the particle., F2 − F1, m, F2 + F1, (c), m, (a), , F2 − 4F1, m, , F1, , applying a force by hand. If the hand moves by 0.2 m, while applying the force and the ball goes upto 2 m, height further, find the magnitude of the force., (take, g = 10 ms −2), (b) 16 N, , (c) 20 N, , F2, , F2, , 2 A ball of mass 0.2 kg is thrown vertically upwards by, , (a) 4 N, , F3, , F3, , (b) Zero, (d), , (b) F1, , (a) F1, , (d) 22 N, , 3 A player catches a cricket ball of mass 150 g, moving at, , (c), , F3, , F2, , (d) F1, , F3, , F2, , 8 A string of negligible mass, going over a clamped pulley, of mass m supports a block of mass M as shown in the, figure. The force on the pulley by the clamp is given by, , a rate of 20 ms −1. If the catching process is completed in, 0.1 s, the force of the blow exerted by the ball on the, hand of the player is equal to, (a) 150 N, , (b) 3 N, , (c) 30 N, , m, , (d) 300 N, , 4 A ball of mass m is thrown vertically upwards with a, velocity v. If air exerts an average resisting force F , the, velocity with which the ball returns to the thrower is, (a) v, , mg, mg + F, , mg − F, (c) v, mg + F, , (b) v, , F, mg + F, , mg + F, (d) v, mg, , 5 A block of mass 200 g is moving with a velocity of 5 ms −1, along the positive x-direction. At time t = 0, when the, body is at x = 0, a constant force 0.4 N is directed along, the negative x-direction , is applied on the body for 10 s., What is the position x of the body at t = 2.5 s?, (a) x = 6.75 m, (c) x = 6 m, , (b) x = 6.25 m, (d) x = 6.50 m, , 6 Two trains A and B are running in the same direction on, parallel tracks such that A is faster than B. If packets of, equal weight are exchanged between the two, then, (a) A will be retarded but B will be accelerated, (b) A will be accelerated but B will be retarded, (c) there will not be any change in the velocity of A but B, will be accelerated, (d) there will not be any change in the velocity of B, but A, will be accelerated, , 7 Which of the four arrangements in the figure correctly, shows the vector addition of two forces F1 and F2 to yield, the third force F3?, , M, , (a), , 2Mg, , (c) 2Mg, , (b), , [(M + m)2 + m 2 ] g, , (d), , [(M + m) + m]2 g, , 9 A bullet is fired from a gun. The force on the bullet is, given by F = ( 600 − 2 × 10 5t ), where F is in newton and t, is in second. The force on the bullet becomes zero as, soon as it leaves the barrel. What is the average impulse, imparted to the bullet?, (a) 9 N-s, , (b) Zero, , (c) 0.9 N-s, , (d) 1.8 N-s, , 10 Two wooden blocks are moving on a smooth horizontal, surface, such that the mass m remains stationary with, respect to the block of mass M as shown in the figure., The magnitude of force P is, , m, , P, M, , b, , (a) g tan β, (c) (M + m) cosec β, , (b) mg cos β, (d) (M + m) g tan β
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LAWS OF MOTION, , DAY FOUR, , 18 A satellite in a force-free space sweeps out stationary, , 11 The figure below shows a uniform rod of length 30 cm, , dM, = αv , where M is the, dt, mass and v is the velocity of the satellites and α is a, constant. The deceleration of the satellite is, , having a mass of 3.0 kg. The strings as shown in the, figure are pulled by constant forces of 20 N and 32 N., Find the force exerted by the 20 cm part of the rod on, the 10 cm part. All the surfaces are smooth and the, strings are light, , 20 N, , (a) 36 N, , F, , F, , 10 cm, , 20 cm, , (b) 12 N, , interplanetary dust at a rate, , (a), , (c) 64 N, , (d) 24 N, , αv 2, M, , (c) −, , (a) g, g, , m1, , (a) 24 N, m2, , pulley, connects two blocks of masses m1 and m 2, (vertically). If the acceleration of the system is ( g/8), then, the ratio of masses is, (c) 4 : 3, , (c) (g − a), g, , (b) a, a, , (b) 74 N, , (c) 15 N, , angle θ and length l, kept inside an elevator going up, with uniform velocity v. Time taken by the block to slide, down the length of the incline, if released from rest is, , v, , (d) 50%, , 15 A man slides down a light rope, whose breaking strength, is η times his weight. What should be his maximum, acceleration, so that the rope does not break?, (a) g(1 − η), , (b) ηg, , (c), , g, 1+ η, , (d), , 480 N, , are connected by a metal rod of mass 8 kg. The, system is pulled vertically up by applying a force, of 480 N as shown in figure. The tension at the, mid-point of the rod is, , M1, , (c) 240 N, , 2 mg, 3A, , (b), , 4 gm, 3A, , (c), , gm, A, , (d), , (b), , 2l, g, , (c), , 2l, g sin θ, , (d), , 2l, sinθ, , body of mass m rests on it. If the coefficient of friction is, µ, then the minimum force that has to be applied parallel, to the inclined plane, so as to make the body to just, move up the inclined plane, is, , T, , 10 kg, , 100 N, , 40 kg, , M, , 3 mg, 4A, , (b) µ mg cosθ, (d) µ mg cosθ + mg sinθ, , rests on the top of the slab. The static coefficients of, friction between the block and the slab is 0.60, while the, kinetic coefficient is 0.40., The 10 kg block is acted upon by a horizontal force of, 100 N. If g = 9.8 ms −2 , the resultant acceleration of the, slab will be, , 17 Two blocks of masses m and M are, , (a), , 2l, (g + a) sin θ, , 23 A 40 kg slab rests on a frictionless floor. A 10 kg block, , M2, , (d) 190 N, , connected by means of a metal wire of, cross-sectional area A passing over a, frictionless fixed pulley as shown in the, T, figure. The system is then released. If, m, M = 2 m , then the stress produced in the, ª JEE Main (Online) 2013, wire is, , (a), , (a) mg sinθ, (c) µ mg cosθ − mg sinθ, , ª JEE Main (Online) 2013, , (b) 96 N, , q, , 22 A plane is inclined at an angle θ with the horizontal. A, , g, 1− η, , 16 Two blocks of mass M1 = 20kg and M 2 = 12 kg, , (a) 144 N, , (d) 49 N, , (d) 5 : 3, , inextensible string passing over a smooth frictionless, pulley. The amount of mass that should be transferred, from one to another, so that both the masses move 50 m, in 5 s is, (c) 70%, , (d) a, g, , 21 A block A is able to slide on the frictionless incline of, , 14 Two bodies of equal masses are connected by a light, , (b) 40%, , (d) −αv 2, , hangs his bag on the string and the balance reads 49 N,, when the lift is stationary. If the lift moves downwards, with an acceleration of 5 ms −2 , the reading of the spring, balance would be, , 13 A light string passing over a smooth light, , (a) 30%, , αv 2, 2M, , 20 A spring balance is attached to the ceiling of a lift. A man, , (b) 9.8 ms −2, (d) 4.8 ms −2, , (b) 9 : 7, , (b) −, , lift drops a ball inside the lift. The acceleration of the ball, as observed by the man in the lift and a man standing, stationary on the ground are respectively, , 32 N, , to a string, are hanging over a light, frictionless pulley. What is the acceleration of, the masses produced when system is free to, move? (take, g = 9.8 ms −2 ), , (a) 8 : 1, , 2 αv 2, M, , 19 A lift is moving down with an acceleration a. A man in the, , 12 Two masses m1 = 5 kg and m2 = 4.8 kg, tied, , (a) 0.2 ms −2, (c) 5 ms −2, , 43, , (a) 0.98 ms −2, (c) 1.52 ms −2, , (b) 1.47 ms −2, (d) 6.1 ms −2
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44, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , f, , 24 A horizontal force of 10 N is necessary to, just hold a block stationary against a wall., The coefficient of friction between the, block and the wall is 0.2. The weight of, the block is, (a) 20 N, (c) 100 N, , 10 N, , coefficient of friction between the blocks is 0.1 and, between block B and the wall is 0.15, the frictional force, applied by the wall in block B is, ª JEE Main 2015, F, , A, , (b) 50 N, (d) 2 N, , B, , 20 N 100 N, , 25 What is the maximum value of the force F , such that the, block as shown in the arrangement, does not move?, F, , m = Ö3 kg, m=, , 60°, , (a) 20 N, , (b) 10 N, , (a) 100 N, , 1, 2Ö3, , (c) 12 N, , (d) 15 N, , by pressing it with a finger. If the coefficient of friction, between the block and the wall is µ and the acceleration, due to gravity is g, then minimum force required to be, applied by the finger to hold the block against the wall?, Mg, 2µ, , (b), , Mg, µ, , (c), , 2M, µg, , (c) 120 N, , (d) 150 N, , 31 A point particle of mass m, moves along the uniformly, , 26 A block of mass M is held against a rough vertical wall, , (a), , (b) 80 N, , (d), , 2 Mg, µ, , rough track PQR as shown in the figure. The coefficient, of friction between the particle and the rough track, equals µ . The particle is released, from rest , from the, point P and it comes to rest at a point R. The energies,, lost by the ball, over the parts, PQ and QR , of the track,, are equal to each other, and no energy is lost when, particle changes direction from PQ to QR. The values of, the coefficient of friction µ and the distance x(= QR), are, respectively close to, P, , 27 A wooden block of mass M resting on a rough horizontal, surface, is pulled with a force F at an angle with the, horizontal. If µ is the coefficient of kinetic friction between, block and the surface, then acceleration of the block is, F, (cos φ + µ sin φ) − µg, M, (c) µ F cos φ, (a), , F, sin φ, M, (d) µ F sinφ, , h=2m, 30º, O, Horizontal, surface, , (b), , 28 A block of mass 10 kg is placed at a distance of 5 m, from the rear end of a long trolley as shown in the figure., The coefficient of friction between the block and the, surface below is 0.2. Starting from rest, the trolley is, given an uniform acceleration of 3 ms −2. At what, distance from the starting point will the block fall off the, trolley? (take, g = 10 ms −2), , R, Q, , ª JEE Main 2016 (Offline), (b) 0.2 and 3.5 m, (d) 0.29 and 6.5 m, , (a) 0.2 and 6.5 m, (c) 0.29 and 3.5 m, , 32 Two masses m1 = 5 kg and m2 = 10 kg connected by an, inextensible string over a frictionless pulley, are moving, as shown in the figure. The coefficient of friction of, horizontal surface is 0.15. The minimum weight m that, should be put on top of m2 to stop the motion is, ª JEE Main 2018, , 5m, m, , F, , f, a=3 ms–2, , Rear, end, , T, , m2, , Trolley, T, , (a) 15 m, , (b) 25 m, , (c) 20 m, , m1, , (d) 10 m, , 29 A body starts from rest on a long inclined plane of slope, 45°. The coefficient of friction between the body and the, plane varies as µ = 0.3 x , where x is distance travelled, down the plane. The body will have maximum speed (for, g = 10 m s −2) when x is equal to ª JEE Main (Online) 2013, (a) 9.8 m, , (b) 27 m, , (c) 12 m, , (d) 3.33 m, , 30 Given in the figure are two blocks A and B of weight 20 N, and 100 N, respectively. These are being pressed, against a wall by a force F as shown in figure.If the, , m1g, , (a) 18.3 kg, , (b) 27.3 kg, , (c) 43.3 kg, , (d) 10.3 kg, , 33 The minimum force required to start pushing a body up a, rough (frictional coefficient µ) inclined plane is F1 while, the minimum force needed to prevent it from sliding, down is F2. If the inclined plane makes an angle θ from, F, the horizontal such that tan θ = 2 µ, then the ratio 1 is, F2, (a) 4, , (b) 1, , (c) 2, , (d) 3
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LAWS OF MOTION, , DAY FOUR, 34 A box of mass 80 kg kept on a horizontal weighing, machine of negligible mass, attached to a massless, platform P that slides down at 37° incline. The weighing, machine read 72 kg. Box is always at rest w.r.t. weighing, machine. Then, coefficient of friction between the, platform and incline is, P, , 45, , choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true, Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true, Statement II is, not the correct explanation for Statement I, (c) Statement I is true, Statement II is false, (d) Statement I is false, Statement II is true, , 36 Statement I When the car accelerates horizontally along, a straight road, the accelerating force is given by the, push of the rear axle on the wheels., , 37°, , (a), , 12, 13, , (b), , 13, 24, , (c), , 3, 5, , (d), , 11, 24, , 35 A block of mass m is placed against a vertical surface by, a spring of unstretched length l. If the coefficient of, friction between the block and the surface is µ, then, choose the correct statement., 2mg, ,, µd, block will not be in equilibrium., (b) Minimum spring constant kmin to, keep the block of mass m in, mg, equilibrium is, ., µd, 2mg, ,, (c) If spring constant is k =, µd, mg, the normal reaction is, ., µ, (a) If spring constant k =, , Statement II When the car accelerates, the rear axle, rotates with a greater frequency., , 37 Statement I It is easier to pull a heavy object than to, push it on a level ground., Statement II The magnitude of frictional force depends, on the nature of the two surfaces in contact., , 38 Statement I A cloth covers a table. Some dishes are, kept on it. The cloth can be pulled out without dislodging, the dishes from the table., , d, m, , Statement II For every action there is an equal and, opposite reaction., , m, , 39 Statement I A bullet is fired from a rifle. If the rifle recoils, l, , freely, the kinetic energy of the rifle is less than that of the, bullet., Statement II In the case of a rifle-bullet system, the law, of conservation of momentum is violated., , 40 Statement I Newton’s second law is applicable on a, , (d) In the part (c), force of friction is 2mg., , Direction (Q. Nos. 36-40) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, , body with respect to an inertial frame of reference., Statement I In order to apply Newton’s second law on a, body observed from a non-inertial frame of reference. We, apply line pseudo force an imaginary force., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A flexible uniform chain of mass m and length l, A, , suspended vertically, so that its lower end just touches, the surface of a table. When the upper end of the chain, is released, it falls with each link coming to rest the, instant it strikes the table. The force exerted by the chain, on the table at the moment when y part of the chain has, already rested on the table is, (a), , 3myg, l, , (b), , 3mg, l, , (c), , 2 mg, 3 l, , (d), , B, , 1 mg, 3 l, , 2 Two fixed frictionless inclined plane making the angles, 30° and 60° with the vertical are shown in the figure. Two, blocks A and B are placed on the two planes. What is the, relative vertical acceleration of A with respect to B?, , 60°, , (a), (b), (c), (d), , −2, , 4.9 ms in horizontal direction, 9.8 ms−2 in vertical direction, zero, 4.9 ms−2 in vertical direction, , 30°
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46, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , 3 If M is the mass of a rocket, r is the rate of ejection of, gases with respect to the rocket, then acceleration of the, dv, rocket,, is equal to, dt, (a), , ru, (M − r t ), , (b), , (M − r t ), ru, , (c), , ru, (M + r t ), , (d), , ru, M, , vertical walls by pressing one wall A by his hands and, feet and B with his back. The coefficient of friction is 0.8, between his body and the wall. The force with which the, person pushes the wall is, (b) 50 N, , (c) 150 N, , g cos2 θ , (b) , 2 , 1+ cos θ , g sin2 θ , (d) , , 1+ sinθ , , 9 A carriage of mass M and length l is joined to the end of, , 4 A person 40 kg is managing to be at rest between two, , (a) 100 N, , g sin2 θ , (a) , 2 , 1+ sin θ , g tan2 θ , (c) , 2 , 1+ tan θ , , a slope as shown below. A block of mass m is released, from the slope from height h. It slides till end of the, carriage. The coefficient of friction between block and, carriage is µ (the friction between other surfaces are, negligible). Then, minimum height h is, m, , (d) 200 N, , h, , 5 An insect crawls up a hemispherical surface very slowly., The coefficient of friction between the insect and the, surface is 1/3. If the line joining the centre of the, hemispherical surface to the insect makes an angle α, with the vertical, the maximum possible value of α is, given by, a, , M, Smooth, , M, (a) µ 1 + l, , m, m, , (c) µ 1 + l, , M, , m, (b) µ 2 + l, , M, m, (d) 2 µ 1 + l, , M, , 10 System as shown in figure is in equilibrium, (a) cotα = 3, (c) cosec α = 3, , (b) sec α = 3, (d) None of these, , 6 A given object takes n times more time to slide down a, 45° rough inclined plane in comparison to slides down a, perfectly smooth 45° incline. The coefficient of kinetic, friction between the object and the incline is, (a), , 1, 1− n, , 2, , (b) 1 −, , 1, n, , (c), , 2, , 1−, , 1, n, , 2, , (d), , 1, 1 − n2, , 7 When a body slides down from rest along a smooth, inclined plane making an angle of 45° with the, horizontal,it takes timeT . When the same body slides, down from rest along a rough inclined plane making the, same angle and through the same distance,it is seen to, take time pT , where p is some number greater than 1., The coefficient of friction between the body and the, rough plane will be, (a) 1 − p 2, , (b) 1 −, , 1, , (c) p 2 − 1, , p2, , (d) p 2, , 8 Block A of mass m is placed over a wedge of the same, mass m. Both the block and wedge are placed on a fixed, inclined plane. Assuming all surfaces to be smooth., Then, displacement of the block A in ground frame in, 1 s is, A, B, , Fixed, incline, , and at rest. The spring and string are, massless, now the string is cut. The, acceleration of the masses 2 m and m just, after the string is cut, will be, , 2m, , g, upwards, g downwards, 2, g, (b) g upwards, downwards, 2, (c) g upwards, 2g downwards, (d) 2g upwards, g downwards, (a), , m, , 11 A block of mass m is at rest under the, action of a force F , acting against a wall,, as shown in the figure. Which of the, following statement is incorrect?, , F, , (a) f = mg (where, f is the frictional force), (b) F = N (where, N is the normal force), (c) F will not produce torque, (d) N will not produce torque, , 12 The upper-half of an inclined plane with an inclination φ, is, perfectly smooth, while the lower half is rough. A body, starting from rest at the top will again come to rest at, the bottom, if the coefficient of friction for the lower half is, given by, (a) 2 sin φ, , (b) 2 cos φ, , (c) 2 tan φ, , (d) tan φ, , 13 A block of mass m is placed on a surface with a vertical, cross-section given by y = x 3/ 6. If the coefficient of, friction is 0.5, the maximum height above the ground at, which the block can be placed without slipping is, ª JEE Main 2014, , q, , (a) 1 / 6 m, , (b) 2 / 3 m, , (c) 1 / 3 m, , (d) 1 / 2 m
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LAWS OF MOTION, , DAY FOUR, 14 A particle of mass m is at rest at the origin at time t = 0. It, , (a)T, (b)T, (c)T, (d)T, , is subjected to a force F( t ) = F0 e −bt in the x -direction. Its, speed v ( t ) is depicted by which of the following curves?, , ª JEE Main 2013, F0, mb, , F0, mb, , (a), , = 1125 N, = 2250N, = 375 N, = 750N, , 17 In the system shown in figure, masses of the blocks are, such that when system is released, acceleration of pulley, P1 is a upwards and acceleration of block 1 is a1,, upwards. It is found that acceleration of block 3 is same, as that of 1 both in magnitude and direction., , (b), v (t), , = 2250N, F, = 1125 N, F, = 1125 N, F, = 1125 N, F, , 47, , v (t), t, , t, , F0, mb, , F0, mb, , a, (c), , P1, , P2, , (d), , a1 1, , v (t), , v (t), t, , 3, , t, , 4, , 2, , 15 The figure shows the position-time ( x -t ) graph of, , a, Given that, a1 > a > 1., 2, , one-dimensional motion of a body of mass 0.4 kg. The, magnitude of each impulse is, , Match the following., 2, , Column I, , x (m), , 0, , 2, , 4, , 6 8 10 12 14 16, t (s), , (b) 0.8 N - s, (d) 0.2 N - s, , (a) 0.4N - s, (c) 1.6N - s, , A., , Acceleration of 2, , 1., , 2 a + a1, , B., , Acceleration of 4, , 2., , 2 a − a1, , C., , Acceleration of 2 w.r.t.3, , 3., , Upwards, , D., , Acceleration of 2 w.r.t. 4, , 4., , Downwards, , Codes, A, , 16 A worker is raising himself and the crate, on which he stands with an acceleration, of 5 ms −2 by a massless rope and pulley, arrangement. Mass of the worker is 100, kg and that of the crate is 50 kg. If T is, the tension in the rope and F be the force, of contact between the worker and the, floor and if g = 10 ms −2, then, , Column II, , B, , C, , (a) 2,3, , 1,4, , 4, , 3, , (b) 2,4, , 1, , 4, , 1,3, , 1,3, , 2, , 1,2, , (c), , 4, , D, , (d) No above matching is correct, , ANSWERS, SESSION 1, , SESSION 2, , 1 (a), , 2 (d), , 3 (c), , 4 (c), , 5 (b), , 6 (a), , 7 (a,c), , 8 (b), , 9 (c), , 10 (d), , 11 (d), , 12 (a), , 13 (b), , 14 (b), , 15 (a), , 16 (d), , 17 (b), , 18 (b), , 19 (c), , 20 (a), , 21 (c), , 22 (d), , 23 (a), , 24 (d), , 25 (a), , 26 (b), , 27 (a), , 28 (a), , 29 (d), , 30 (c), , 31 (c), , 32 (b), , 33 (d), , 34 (b), , 35 (b), , 36 (a), , 37 (b), , 38 (b), , 39 (c), , 40 (a), , 1 (a), 11 (d), , 2 (d), 12 (c), , 3 (a), 13 (a), , 4 (d), 14 (c), , 5 (a), 15 (b), , 6 (b), 16 (c), , 7 (b), 17 (a), , 8 (a), , 9 (c), , 10 (a)
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48, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, 1 According to polygon law, resultant of, four forces, each of magnitude F1 acting, at an angle of 72°, is along the fifth side, of the polygon taken an in opposite, order. As F2 is acting along this side of, polygon, therefore the net force on the, particle = F2 − F1, F − F1, Acceleration (a) = 2, m, , 2 The situation is as shown in the figure., At an initial time, the ball is at P, then, under the action of a force (exerted by, hand) from P to A and then from A to B,, let acceleration of ball during the, motion from P to A is ams −2 [assumed, to be constant] in an upward direction, and velocity of ball at A is v ms −1 ., , ⇒, F × 01, . = |p f − pi |, As the ball will stop after catching,, p i = mv i = 015, . × 20 = 3 and p f = 0, ⇒, F × 01, . =3, ⇒, F = 30 N, , 4 For an upward motion, Retarding force = mg + F, mg + F, Retardation (a) =, m, v2, v2m, Distance, s =, … (i), =, 2a 2(mg + F ), For the downward motion, net force, = mg − F, mg − F, ∴ Acceleration (a ′ ) =, m, v ′2, v ′2 m, As,, Distance (s ′ ) =, =, 2a ′ 2(mg − F ), s = s′, ∴, , B, , 2m, , v′= v, , mg − F, mg + F, , 5 Given, u = 5ms −1 ,, A, 0.2 m, P, , Then for PA, v 2 = 02 + 2a × 0.2, 0 = v2 − 2 × g × 2, , For AB,, ⇒, , v = 2g × 2, 2, , From above equations,, a = 10 g = 100 ms −2, F, , a=10g, , mg, , Then for PA, from FBD of ball is, F − mg = ma, [F is the force exerted by hand on the, ball], ⇒ F = m(g + a) = 0.2(11 g ) = 22 N, , 3 This is the question based on, impulse-momentum theorem., |F∆t | = | Change in momentum|, , along positive x-direction, F = − 0.4 N,, along negative x-direction, M = 200g = 0.2 kg, Thus, the acceleration, F, 0.4, a=, =−, = − 2 ms −2, M, 0.2, The negative sign showing the, retardation., The position of the object at time t is, given by, 1, x = x 0 + ut + at 2, 2, At t = 0, the body is at x = 0, therefore, x 0 = 0., 1, Hence,, x = ut + at 2, 2, Since, the force acts during the time, interval from t = 0 to t = 10 s, the, motion is accelerated only within this, time interval. The position of the body, at t = 2.5 s is given by, 1, x = 5 × 2.5 + × (−2) × (2.5)2, 2, = 6.25m, , 6 Initially, the momentum of the packet, in train A is more than in train B. When, packets are changed, the packet, , reaching train A being of lower, momentum will retard the train A but, packet reaching train B, being of higher, momentum will accelerate B., , 7 Parallelogram law of vector addition, F3 = F1 + F2, So, option (a) and (c) both can be, satisfied., , 8 Force on the pulley, by the clamp, = Resultant of forces (M + m ) g acting, along horizontally and mg acting, vertically downwards, = (Mg + mg )2 + (mg )2, = [(M + m )2 + m 2 ] g, , 9 As, F = 600 − 2 × 105t, At, t = 0, F = 600 N, According to question,, F = 0, on leaving the barrel,, ⇒, 0 = 600 − 2 × 105t, 600, ∴, t =, = 3 × 10−3 s, 2 × 105, This is the time spent by the bullet in the, barrel, 600 + 0, Average force =, = 300 N, 2, Average impulse imparted = F × t, = 300 × 3 × 10−3 = 0.9 N-s, , 10 Different forces involved are as shown in, the figure., R, , f co, , sb, , b, f, P, , mg, , sin, , b, , b, M, mg cos b, , mg, , b, , Observing the figure, we have, acceleration of the system,, P, a=, M + m, Pm, M + m, If f is pseudo force on m in the direction, opposite to force P, then, Pm, f =, M + m, , Force on block of mass (m ) =
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LAWS OF MOTION, , DAY FOUR, As it is clear from the figure,, f cos β = mg sin β, Pm, cos β = mg sin β, (M + m ), P = g (M + m ), , sin β, cos β, , P = (M + m ) g tanβ, , or,, , 11 Net force on the rod,, , f = 32 − 20 = 12 N, Acceleration of the rod, f, 12, =, =, = 4 ms −2, 3, m, Equation of motion of the 10 cm part is, F − 20 = m × a = 1 × 4 ,, F = 4 + 20 = 24 N, Similarly, equation of motion of 20 cm, part is, 32 − F = m × a = 2 × 4,, F = 32 − 8 = 24 N, , 12 On releasing, the motion of the system, will be according to the figure, , a, a, , or 8 m1 − 8 m2 = m1 + m2, or, 7m1 = 9 m2, m1, 9, =, m2, 7, , Now, for the metal rod, tension at both of, its end are dissimilar and, …(ii), T1 − T2 = 80, (Q g = 10 ms −2 ), Now, from Eqs. (i) and (ii), we get, T1 = 230 N and T2 = 150 N, ∴ Tension at mid-point, = T1 − 4 g = 190 N, , 14 As, s = ut + 1 at 2, 2, , 1, × a × (5)2, 2, , ⇒ 50 = 0 × 5 +, , 100, = 4 ms −2, 25, Let, mass of one become m1 and that of, other m2 , where m1 > m2 . As m1 moves, downwards with acceleration, a = 4 ms −2, , ∴ a=, , m − m2 , a= 1, g, m1 + m2 , So,, , m − m2 , 4= 1, 10, m1 + m2 , m1 − m2 , 2, a, 4, =, , = =, m1 + m2 g 10 5, , ∴ Percentage of mass transferred, m − m2 , = 1, × 100, m1 + m2 , 2, = × 100 = 40%, 5, , 15 As, mg − R = ma, , m2 g, , The equations of motion of blocks, are,, … (i), m1 g − T = m1 a, and, … (ii), T − m2 g = m2 a, On solving,, m − m2 , … (iii), a= 1, g, m1 + m2 , Here, m1 = 5 kg,, m2 = 4.8 kg, g = 9.8 ms −2, ∴, , 5 − 4.8 , a = , × 9.8, 5 + 4.8 , 0.2, =, × 9.8, 9.8, = 0.2 ms −2, , 13 As, a =, , (m1 − m2 ) g, g, = ,, 8, m1 + m2, m1 − m2, 1, =, m1 + m2, 8, , 2m × 2m , =, g, m + 2m , (where, m1 = m and m2 = 2m), 4, = mg, 3, Force ( Tension), ∴ Stress =, Area, 4, mg, 4 mg, = 3, =, A, 3A, , 18 It is known that the thrust, , dM , = − v , = − v (αv ), dt , , Hence, the retardation produced, thrust, αv 2, =, =−, mass, M, ball is g as will be observed by a man, standing stationary on the ground. The, man inside the lift is having its own, downward acceleration, a. Therefore,, relative acceleration of the ball as, observed by the man in the lift will must, be = (g − a)., , Man, , m2, , 2m1 m2 , g, m1 + m2 , , 17 Tension, T = , , 19 When ball dropped, acceleration of the, , m1, , m 1g, , mg − ηmg = ma, mg (1 − η ) = ma ⇒ a = g(1 − η ), , 20 When the lift is stationary, then, R = mg, 49 = m × 9.8, 49, kg = 5kg, m=, 9.8, If a is the downward acceleration of the, lift then,, R = m(g − a), = 5(9.8 − 5) = 24 N, , 16 For block of mass M1 ,, , 480 − T1 − 20 g, =a, 20, 480 N, , T2, , M1, , M2, , 21 The situation is given in figure below, T1, , 49, , M2, in, , Also, for block of mass M2 ,, T2 − 12 g, =a, 12, Since, a is common for all the, individuals of the system, 480 − T1 − 20 g T2 − 12 g, ⇒, =, 20, 12, After taking g = 10 ms −2 this gives, 5T2 + 3T1 = 1440, , q, , gs, , q, , q, g cos q, g, , From equation of motion,, 1, g sinθ t 2 = l, 2, ⇒, …(i), , t =, , 2l, g sinθ
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50, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , 22 To move the body up the inclined, , The distance covered by the trolley in, this time,, 1, 1, s ′ = ut + at 2 = 0 + × 3 × 10 = 15m, 2, 2, , f, , plane, the force required,, = mg sinθ + µR, = mg sin θ + µmg cos θ, , F, , N, , 29 From Newton’s second law,, , 23 Limiting force of friction of block on, , slab µ m1 g = 0.6 × 10 × 9.8 = 58.8 N, Since, the applied force = 100 N on, block, which is greater than the force of, limiting friction, the block will, accelerate on the slab, due to which, the, force acting on the slab will be that due, to the kinetic friction (µ k m1 g )., Hence, acceleration of the slab,, µ m g, 0.4 × 10 × 9.8, = 0.98 ms–2, a= k 1 =, 40, m2, , 24 Let, R be the normal contact force by, wall on the block., , Mg, , …(i), f = Mg, and, …(ii), F =N, But force of friction ( f ) = µN, = µ F …(iii), [using Eq. (ii)], From Eqs. (i) and (iii), we get, Mg, µF = Mg or F =, µ, , 27 Here, R = Mg − F sin φ, ∴, , f, , f = µR = µ(Mg − F sin φ), F sin f, , R, , 10 N, , f, , mg sin θ − µ mg cos θ , ⇒ v = 2, x, , , m, = 2 gx sin θ − 0.6 x2 g cos θ, v should be maximum when, , d 2gx sin θ − 0.6 x2 g cos θ ), , ⇒, , =0, dx, By differentiating, we get x = 3 .33 m, , 30 In vertical direction, weights are, , F cos f, , fA, , w, , R = 10N, f L = w and f = µR, µR = w or w = 0.2 × 10 = 2 N, , ∴, , 25 Free body diagram (FBD) of the block, (shown by a dot) is as shown in the, figure, N, f, , F cos 60°, , Vertical, , Horizontal, , mg + F sin 60°, , For vertical equilibrium of the block, N = mg + F sin 60°, F, … (i), = 3g + 3, 2, For no motion, force of friction, f ≥ F cos 60°, or, µN ≥ F cos 60°, 3F F, 1 , or, 3g +, ≥, 2 2, 2 3, F, or F ≤ 2 g or 20 N, 2, Therefore, maximum value of F is 20 N., or, , g≥, , 26 Given, mass of the block = M, Coefficient of friction between the block, and the wall = µ, Let, a force F be applied on the block to, hold the block against the wall. The, normal reaction of mass be N and force, of friction acting upward be f . In, equilibrium, vertical and horizontal, forces should be balanced separately., , dv, =0, dx, , balanced by frictional forces., Consider FBD of block A and B as shown, in diagram below., , F, , R, , mg sin θ − µ mg cos θ, =a, m, Now, distance covered by the particle,, v 2 = u2 + 2as, , mg, , Net force, F cos φ − f = Ma, 1, ∴ a=, [F cos φ − f ], M, 1, ⇒ a=, [F cos φ − µ (Mg − F sin φ)], M, F, µF, =, cos φ − µg +, sin φ, M, M, F, =, (cos φ + µ sin φ) − µg, M, , 28 Given, acceleration of the trolley, (a) = 3 ms −2 ., , Therefore, the force acting on the block, is F = ma = 10 × 3 = 30 N., The weight mg of the block is balanced, by the normal reaction R. The force of, limiting friction is given by, f, f, µ =, =, R mg, f = µmg = 0.2 × 10 × 10 = 20 N, The net force on the block is towards, right and is given by, F ′ = F − f = 30 − 20 = 10 N, F ′ 10, So,, a′ =, =, = 1 ms −2 ., m 10, Let, t be the time taken for the block to, fall off from the rear end for the trolley., Then, the block has to travel a distance, s′ = 5m to fall off. Now, since the, trolley starts from rest. So, u = 0 and, 1, using s = ut + at 2 , we can determine, 2, t as 10 s., , fB, , F, , N, , 20 N, , fA, , 100 N, , As the blocks are in equilibrium, balance, forces are in horizontal and vertical, direction., The system of blocks (A + B ), F =N, For block A, f A = 20 N, and for block B,, f B = f A + 100 = 20 + 100 = 120 N, , 31 Energy lost over path PQ = µ mg cos θ × 4, P, h=2m, , 4m, , 30º, O 23m Q, , x, , R, , Energy lost over path QR = µ mg x, i.e. µ mg cos 30° × 4 = µ mg x, (Q θ = 30° ), x = 2 3 = 3.45 m, From Q to R energy loss is half of the, total energy loss., 1, i.e. µ mg x = × mgh ⇒ µ = 0.29, 2, The values of the coefficient of friction µ, and the distance x(= QR ) are 0.29 and 3.5.
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LAWS OF MOTION, , DAY FOUR, 32 Motion stops when pull due to, , m1 ≤ force of friction between m and m2, and surface., ⇒, m1 g ≤ µ(m2 + m )g, ⇒ 5 × 10 ≤ 01510, . ( + m ) × 10, ⇒, m ≥ 2333, . kg, Here, nearest value is 27.3 kg, So, m min = 27.3 kg, , 33 F1 = mg (sin θ + µ cos θ) [as body just in, position to move up, friction force, downward], F2 = mg (sin θ − µ cos θ) [as body just in, position to slide down, friction upward], F2, , m, , F1, , 35 Free body diagram of the block is, , N, , ∴, , mg, Here, N = F s = kd, and mg = f ≤ µ N = µ kd, mg, or k ≥, µd, Hence, (b) is the correct option., , 36 When a car accelerates, the engine, rotates the rear axle which exerts a push, on the wheels to move., Statement II does not explain correctly,, Statement I., , F1 sin θ + µ cos θ, =, F2, sin θ − µ cos θ, tan θ + µ 2µ + µ, =, =, =3, tan θ − µ, 2µ − µ, , 34 Here, (80 − 72)g = ma y, , a y = 1 m/s2, , or, , 5, ∴asin37° = 1 m/s or a = m/s2, 3, R, 2, , Correct explanation is there is increase, in normal reaction when the object is, pushed and there is decrease in normal, reaction when the object is pulled (but, strictly, not horizontally)., , 38 The cloth can be pulled out without, dislodging the dishes from the table due, to law of inertia, which is Newton’s first, law. While, Statement II is true, but it is, Newton’s third law., , 39 If the bullet is fired from the rifle, the, , m mg, mg, 37°, (a), , ax= a cos 37°, 37°, a, ay=a sin 37°, (b), Now, we apply Newton’s second law of, motion on the box in the direction of, acceleration,, 5, mg sin 37° − µ mg cos 37° = m ×, 3, 13, or µ =, 24, Hence, (b) is the correct option., , the weight of the chain on the table, + momentum of the chain transmitted, on the table, ⇒, , 37 Both Statements are correct. But, q, , SESSION 2, 1 Suppose, F = force on the table due to, , f, Fs, , 51, , momentum of bullet-rifle system is, conserved., It means, Mb v b = M r v r, … (i), 1, 2, M v, E k(b ) 2 b b, and, =, 1, E k( r ), M r v 2r, 2, M, = r, Mb, As, M r > Mb (mass of rifle is greater, than the mass of bullet)., Hence, E k(b ) > E k( r ). So, the kinetic, energy of bullet is greater than the, kinetic energy of rifle., , 40 In order to apply Newton’s second law, on a body observed from a non-inertial, frame of reference pseudo force is, considered in a direction opposite to, real acceleration., , F = F1 + F2 (Let), m, Now, F1 =, yg ,, l, m, dp = dmv =, dy 2gy, l, dp, m dy, = F2 =, 2gy, l dt, dt, dy, m, =, (2gy ) , = 2gy , dt, , l, 2 myg, m, ∴, F =, yg +, l, l, 3 myg, =, l, , dy, , y, , 2 Force applying on the block, F = mg sinθ, mg sinθ = ma, ∴, a = g sinθ, where, a is along the inclined plane., , or, , ∴ Vertical component of acceleration is, g sin2 θ., ∴ Relative vertical acceleration of A with, respect to B is, g (sin2 60° − sin2 30° ], =, , g, = 4.9 ms −2, 2, , [in vertical direction], , 3 Here, initial mass of the rocket = M, dm, =r, dt, Relative velocity of gases w.r.t. rocket, =v, then, acceleration of the rocket, u(dm/dt ), F, a=, =, dm, m , × t , M −, , , dt, ur, =, (M − r t ), , 4 Balanced horizontal force, N 1 = N 2, Balanced vertical force, 2µN 1 = mg, mg, 40 × 10, µN 1 =, =, 2, 2, = 200 N
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52, , DAY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , ∴Displacement of block A in 1 s is, 1 2g sin2 θ , 2, s = 01, . + , × (1), 2 1 + sin2 θ , , ∴ Man pushes the wall with 200 N, , a, mN1, , g sin2 θ , s= , 2 , 1 + sin θ , Hence, (a) is the correct option., or, , θ = 45°, , mg, , Here, θ = 45°, N1, , ∴, , N2, , R a, , mg sin a, , mg, , F = mg sinα, R = mg cos α, F, = tanα, ⇒, R, 1, i.e., µ = tan α =, 3, ⇒ cotα = 3, , a = g (sin θ − µ cos θ), , acceleration of body,, a = g sinθ, If s be the distance travelled, then, 1, …(i), s = g sinθt 12, 2, On rough inclined plane, the, acceleration is given by, a = g sin θ − µg cos θ, 1, …(ii), ∴ s = (g sin θ − µg cos θ)t 22, 2, From Eqs. (i) and (ii), we get, sin θ, t 22, =, g sin θ − g cos θ, t 12, , ∴, , n2 =, , or, , t 22, t 12, , = n2, , =, , n2 − 1, n2, n −1, 2, , n2, , µ = 1−, , ×, , N, , or s =, , g (1 − µ )p2T 2, , 2 2, From Eqs. (i) and (ii), we get, gT 2, , =, , or, , 1−µ =, , …(ii), , g (1 − µ )p2T 2, , 2 2, or (1 − µ )p2 = 1, 1, p2, , 2 2, , Velocity of block, just before reaching, the carriage is, v = 2gh, Now, acceleration of block,, − µmg, a1 =, = −µ g, m, Acceleration of carriage,, µ mg, a2 =, M, At t = 0, motion of block as seen from the, carriage is, u rel = v = 2gh, and, , , 1 , or µ = 1 − 2 , p , , , 8 Free body diagram for A and B is, N¢, , N, , ⇒, , ⇒, , , sin 45° n2 − 1 , =, ×, , cos 45°, n2, , , 1, n2, , 1, n2, , 7 On smooth inclined plane Acceleration, of a body sliding down a smooth, inclined plane, a = g sinθ, , mg, mg, For A, …(i), mg − N = m (asin θ), [as block has only vertically downward, acceleration], For B, …(ii), (N + mg )sin θ = ma, On solving Eqs. (i) and (ii), we get, 2g sin θ , a= , , 2, 1 + sin θ , , The acceleration of block A is, 2g sin2 θ, a A = asin θ =, 1 + sin2 θ, , m, arel = a1 − a2 = − µg 1 + , , M, , Now, relative velocity of block when, block moves through distance x with, respect to carriage,, v 2rel = u2rel + 2 arel x, , A, , 1 , , 2, 1 , , 2, , mmg, a2, mg, , 2, , B, , sin θ, sin θ − µ cos θ, , = 1−, , mg, , g(1 − µ ), , As,sin 45° = cos 45° = 1 , , , , 2, Again using equation of motion, 1, s = ut + at 2 , we get, 2, 1 g (1 − µ ), s = 0( pT ) +, ( pT )2, 2, 2, , N, , Solving, we get, n2 − 1 sin θ, µ =, ×, cos θ, n2, =, , a1, , mmg, , mmg, , = g(sin 45°− µ cos 45° ) =, , 6 On smooth inclined plane, the, , t 1 = nt 1, , N, , of the body, , and, , But, , 9, , g, , 2, Let the travelled distance be s., Using equation of motion,, 1, s = ut + at 2 , we get, 2, gT 2, 1 g 2, …(i), s = 0.t +, T or s =, 2 2, 2 2, On rough inclined plane Acceleration, , 5 As, it is clear from the figure, F, , a = g sin 45° =, , ⇒, , m, 02 = 2gh − 2µg 1 + ⋅ l, , M, (when x = l , then v rel = 0), m, 2 gh = 2µg 1 + ⋅ l, , M, m, h = µ 1 + ⋅ l, , M, , Hence, (c) is the correct option., , 10 Initially under the equilibrium of mass, m, T = mg, Now, the string is cut. Therefore, T = mg, force is decreased on the mass m, upwards and downwards on the mass, 2m., mg, (downwards), ∴, am =, =g, m, mg, g, and, (upwards), =, a2m =, 2m, 2
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LAWS OF MOTION, , DAY FOUR, 11 This is the equilibrium of coplanar, forces., Hence, ΣF x = 0, ∴, F =N, ∴, ΣF y = 0, f = mg , Στc = 0, ∴, τN + τ f = 0, Since,, τf ≠ 0, τN ≠ 0, ∴, Thus, N will produce torque., , 12 According to the work-energy theorem,, ∑ W = ∆K = 0, ⇒ Work done by friction + Work, done by gravity = 0, l, ⇒ − (µ mg cos φ) + mgl sin φ = 0, 2, µ, or cos φ = sin φ or µ = 2 tan φ, 2, , 13 A block of mass m is placed on a surface, with a vertical cross-section, then, y, m, y, θ, , x, , x3 , d , 6, dy, x2, tanθ =, =, =, 2, dx, dx, At limiting equilibrium, we get, µ = tan θ, x2, 0.5=, ⇒ x2 = 1 ⇒ x = ±1, 2, Now, putting the value of x in, x3, y=, , we get, 6, When x = 1, y =, , (1)3, 1, =, 6, 6, , increasing curve with decreasing slope, with time., F, F, dv, a=, = 0 e −bt =, m, m, dt, v, tF, ⇒ ∫ dv = ∫ 0 e −bt, 0, 0m, ⇒, , (−1)3, −1, =, 6, 6, , So, the maximum height above the, ground at which the block can be, placed without slipping is 1/6 m., , 14 As the force is exponentially decreasing,, so it’s acceleration, i.e. rate of increase, of velocity will decrease with time., Thus, the graph of velocity will be an, , F, = 0 e −bt, mb, 0, , t, , F0 0, (e − e −bt ), mb, F, = 0 (1 − e −bt ), mb, F, with v max = 0 [ a + t = ∞ ], mb, =, , motion is uniform because of impulse, direction of velocity changes as can be, seen from the slope of the graph., 2, Initial velocity, v 1 = = 1 ms −1, 2, Final velocity, v 2 = −2 / 2 = −1 ms −1, p i = mv 1 = 0.4 N - s, and, p f = mv 2 = −0.4 N - s, Now, impulse,, J = p f − p i = − 0.4 −0.4, = − 0.8 N - s, ⇒, , |J|= 0.8 N - s, , 16 Free body diagram for crate, T, , T, a=5 ms–2, , F, , F, MC g, mC g, , So, for vertical equilibrium of the crate., T − F − Mc g = Mc a, [Q Mc = mass of crate = 50 kg], ⇒, , T − F − 500 = 250, , ⇒, , T − F = 750, , Free body diagram for worker, , F, a=5 m/s–2, , T, , mW g, T + F − mwg = mwa, [Q m w = mass of worker], ⇒ T + F − 1000 = 500, …(ii), ⇒, T + F = 1500, Solving Eqs. (i) and (ii), we get, T = 1125N and F = 375N, , 17 Let the accelerations of various blocks are, , 15 From the graph, it is a straight line so,, , When x = − 1, y =, , 0, , t, , F 1 −bt, v = 0, e, m −b , , 53, , …(i), , as shown in figure. Pulley P2 will have, downward acceleration a., , a, , P1, , P2, a, , a1 1, a2, , 3, a4, 2, , 4, , Analysing the diagram,, a + a2, a= 1, ⇒ a2 = 2a − a1 > 0, 2, So, acceleration of block 2 is upward, Hence, (A)→ (2,3), a + a4, and, a= 1, 2, ⇒, a4 = 2a + a1 > 0, So, acceleration of block 4 is downward., Hence, (B)→ (1,4)., This is downwards., Hence, (c)→ (4), Acceleration of block 2 with respect to, block 4., a2 /4 = a2 − (− a4 ) = 4 a > 0, This is upward., Hence (D)→(3) (C)→ (4), Hence, A → (2,3), B → (1,4), C → (4),, D → (3)
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DAY FIVE, , Circular, Motion, Learning & Revision for the Day, u, , u, , Concept of Circular Motion, Dynamics of Uniform Circular Motion, , u, , u, , Forces in Circular Motion, Applications of Centripetal and Centrifugal Forces, , Concept of Circular Motion, Circular motion is a two dimension motion. To bring circular motion in a body it must be, given some initial velocity and a force. Circular motion can be classified into two typesUniform circular motion and Non uniform circular motion., When an object moves in a circular path at a constant speed then the motion is said to, be a uniform circular motion., When an object moves in a circular path with variable speed, then the motion is said to, be non-uniform circular motion., , Terms Related to Circular Motion, 1. Angular Displacement, It is defined as the angle turned by the particle from some reference line. Angular, displacement ∆θ is usually measured in radians., Finite angular displacement ∆θ is a scalar but an infinitesimally small displacement is, a vector., , 2. Angular Velocity, It is defined as the rate of change of the angular, displacement of the body., From figure a particle moving on circular track of radius r is, showing angular displacement ∆θ in ∆t time and in this time, period, it covers a distance ∆s along the circular track, then, dθ, ∆θ , ∴ Angular velocity, ω = lim =, ∆t→ 0 ∆t , dt, It is an axial vector whose direction is given by the right, hand rule. Its unit is rad/s., , v2, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , B, r2, ∆θ, , O, , v1, , r1, A, , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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CIRCULAR MOTION, , DAY FIVE, 3. Angular Acceleration, It is the rate of change of angular velocity., Thus,, , α=, , dω d2θ, =, dt dt 2, , 55, , Applications of Centripetal and, Centrifugal Forces, Some of the most important applications of centripetal and, centrifugal forces are given below, , Its unit is rad/s2 ., , Motion of a Vehicle on a Level Circular Road, , Dynamics of Uniform, Circular Motion, , When a vehicle negotiates a circular path, it requires a, centripetal force., , If a particle, is performing circular motion with a uniform, speed, then motion of the particle is called uniform circular, motion. In such a case,, v2, dv, [Q v = rω], = 0 and ac = ω2 r =, r, dt, Thus, if a particle moves in a circle of radius r with a uniform, v2, speed v, then its acceleration is , towards the centre. This, r, acceleration is termed as centripetal acceleration., NOTE, , • In non-uniform circular motion Resultant acceleration of, the body is a =, , ar2 + aT2 =, , v4, r2, , + r2 α2, , In such cases the lateral force of friction may provide the, required centripetal force. Thus, for maintaining its circular, path required centripetal force., mv2 , , ≤ frictional force (µ mg), r , Maximum speed vmax = µrg, where, µ = coefficient of friction between road and vehicle, tyres and r = radius of circular path., , Bending of a Cyclist, When a cyclist goes round turns in a circular track, then angle, made by cyclist with vertical level is given by, , aT, a, , tan θ =, , Tangential, acceleration, , v2 , = tan −1 , rg , , O ar, Radial, acceleration, , Forces in Circular Motion, , v2, ≈θ, rg, , Banking of a Curved Road, For the safe journey of a vehicle on a curved (circular) road,, without any risk of skidding, the road is slightly raised, towards its outer end., , In circular motion of an object two kinds of forces occur, which are described below, , Let the road be banked at an angle θ from the horizontal, as, shown in the figure., , Centripetal Force, , If b is width of the road and h is height of the outer edge of, the road as compared to the inner edge, then, , The centripetal force is required to move a body along a, circular path with a constant speed. The direction of the, centripetal force is along the radius, acting towards the centre, of the circle, on which the given body is moving., , tan θ =, , Centripetal force,, mv2, 4π 2, F =, = mrω2 = mr 4π 2 ν2 = mr 2, [Q v = rω], r, T, Work done by centripetal force is always zero as it is, perpendicular to velocity and hence instantaneous, displacement., , R R cos θ, R sin θ, , O, , Centrifugal Force, ‘Certrifugal force can be defined as the radially directed, outward force acting on a body in circular motion, as observed, by a person moving with the body.’, Mathematically, centrifugal force =, , F, θ, , θ, , A, mg, B, , Outer, edge, raised, X, , In case of friction is present between road and tyre, then, Maximum speed,, vmax =, , 2, , mv, = mrω2 [Q v = rω], r, , v2 h, =, rg b, , where, , rg (µs + tan θ), ,, 1 − µs tan θ, , µs = coefficient of static friction.
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56, , DAY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Motion of a Cyclist in a Death Well, , l, , For equilibrium of cyclist in a death, well, the normal reaction N provides, the centripetal force needed and the, f, N, force of friction balances his weight mg., Cyclist, 2, mv, Thus,, N =, mg, r, and, f = µN = mg, rg, ⇒, vmax =, µ, , v2 = v2L − 2 gr (1 − cos θ), , l, , l, , T = mg cos θ +, , and, l, , TH = 0, TL = 6 mg, vL = 5 rg, vH = rg, , and, In general,, , B, mg, , NOTE, , TH, , TL, , TL − TH = 6 mg, , • When a vehicle is moving over a convex bridge, the, maximum velocityv =, , v, , O, , At the lowest point A, the tension TL, and the weight mg are in mutually, opposite directions and their, resultant provides the necessary, centripetal force,, mv2L, mv2L, i.e. TL − mg =, or TL = mg +, r, r, , θ, , C, , rg , where r is the radius of the road., , • When the vehicle is at the maximum height, the reaction of, the road, is N1 = mg −, , mg, , mv 2, r, , N1, , vL, , A, , mv2, r, , In the critical condition of just looping the vertical loop,, (i.e. when the tension just becomes zero at the highest point, B), we obtain the following results, , r, , Motion along a Vertical Circle, In non-uniform circular motion speed of vH, object decreases due to effect of gravity, as the object goes from its lowest, position A to highest position B., , In general, if the revolving particle, at any instant of time,, is at position C, inclined at an angle θ from the vertical, then, , mg, , A, N2, , At the highest point B, tension TH and the weight mg are in, the same direction and hence,, mv2H, mv2H, or TH =, TH + mg =, − mg, r, r, Moreover, vL and vH are correlated as v2H = v2L − 2 gr ., , B, , • When the vehicle is moving in a dip B,, , then, , mv 2, N2 = mg +, r, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A particle is moving along a circular path of radius 5 m,, , moving with a uniform speed of 5 m s −1. What will be the, average acceleration, when the particle completes half, revolution?, −2, , (b) 10/ π ms, (d) None of these, , (a) zero, (c) 10 ms −2, , 2 A cyclist goes round a circular path of circumference, 34.3 m in 22 s, the angle made by him with the vertical, will be, (a) 45 °, , (b) 40°, , (c) 42°, , (d) 48°, , 3 A particle undergoes a uniform circular motion. About, which point on the plane of the circle, will the angular, momentum of the particle, remain conserved?, (a), (b), (c), (d), , About centre of the circle, On the circumference of the circle, Inside the circle, Outside the circle, , 4 A wheel is rotating at 900 rpm about its axis. When the, power is cut-off, it comes to rest in 1 min. The angular, retardation in rads – 2 is, (a) π / 2, , (b) π / 4, , (c) π / 6, , 5 A roller is made by joining together, , (d) π / 8, B, , D, , two corners at their vertices O. It is, kept on two rails AB and CD which, O, are placed a symmetrically (see the, figure), with its axis perpendicular, to CD and its centre O at the centre, A, C, of line joining AB and CD (see the, figure). It is given a light path, so that it starts rolling with, its centre O moving parallel to CD in the direction shown., As it moves, the roller will tend to ª JEE Main 2016 (Offline), (a) turn left, (b) turn right, (c) go straight, (d) turn left and right alternately
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CIRCULAR MOTION, , DAY FIVE, 6 A particle moves in a circular path with decreasing, , 13 A cyclist starts from centre O of a circular, , speed. Choose the correct statement., (a) Angular momentum remains constant, (b) Acceleration a is acting towards the centre, (c) Particle moves in a spiral path with decreasing radius, (d) The direction of angular momentum remains constant, , (a) 0.1 ms−2 along RO, , 7 A mass of 2 kg is whirled in a horizontal circle with the, help of a string, at an initial speed of 5 rev/min. Keeping, the radius constant, the tension in the string is doubled., The new speed is nearly, (a) 14 rpm, (c) 2.25 rpm, , (b) 10 rpm, (d) 7 rpm, , 8 A cyclist is riding with a speed of 27 kmh−1. As he, approaches a circular turn on the road of radius 80 m, he, applies brakes and reduces his speed at the constant, rate of 0.5 ms −1 every second. What is the magnitude, and direction of the net acceleration of the cyclist on the, circular turn?, (a) 0.86 ms−2 at 54° to the velocity, (b) 0.6 ms−2 at 54° to the velocity, (c) 0.3 ms−2 at 75° to the velocity, (d) 0.7 ms−2 at 68° to the velocity, , (c) 1ms, , −2, , (d) 0.1 rad s−2 along RO, , along RO, , m with a constant speed of 10 ms −1. A plump bob is, suspended from the roof of the car by a light rigid rod of, length 1.00 m. The angle made by the rod with the track, is, (a) 60°, , (b) 30°, , (c) 45°, , (d) zero, , 15 A frictionless track ABCD ends in a semi-circular loop of, radius R. A body slides down the track from point A which, is at a height h = 5 cm. Maximum value of R for the body, to successfully complete the loop is, , D, , h, , speed ν 0. If the coefficient of friction between the tyres, and the road is µ, the shortest distance in which the car, can be stopped is, , v , (c) 0 , µg , , (d), , v0, µ, , 2, , 12 A racing car travel on a track, (without banking) ABCDEFA. ABC is, a circular arc of radius 2R . CD and, FA are straight paths of length R and, DEF is a circular arc of radius, R = 100 m. The coefficient of friction, B, on the road is µ = 0 .1. The maximum, speed of the car is 50 ms −1. The, minimum time for completing one round is, (a) 89.5 s, , (b) 86.3 s, , (c) 91.2 s, , A, , 15, cm, 4, , (c), , 10, cm, 3, , (d) 2 cm, , (b) 16 rad s −1, (d) 2 rad s −1, C, , 17 A bob of mass m suspended by a light, string of length L is whirled into a, vertical circle as shown in figure. What, will be the trajectory of the particle, if, the string is cut at B., (a), (b), (c), (d), , F, , R, E, 90° R, O 2R, , (b), , (a) 4 rad s −1, (c) 21 rad s −1, , 11 A car is moving along a straight horizontal road with a, , v0, µg, , B, , stone of mass 500 g is tied to it and revolved in a circular, path of radius 4 m in a vertical plane. If g = 10 ms −2 , then, the maximum angular velocity of the stone will be, , (c) 3.92 ms −1 (d) 5 ms −1, , (b), , C, , 16 A weightless thread can bear a tension upto 3.7 kg-wt. A, , length 1.96 m is moving in a horizontal circle. The string, will break if the tension is more than 25 N. What is the, maximum speed with which the ball can be moved?, , v 02, 2 µg, , 2R, , (a) 5 cm, , 10 A ball of mass 0.25 kg attached to the end of a string of, , (a), , P, , A, , (b) m1 : m2, (d) 1 : 1, , (b) 3 ms −1, , O, , (b) 0.01ms−2 along OR, , radii r1 and r2, respectively. Their speeds are such that, they make complete circles in the same time t. The ratio, of their centripetal acceleration is, ª AIEEE 2012, , (a) 14 ms −1, , R, , 14 A car is moving in a circular horizontal track of radius 10, , 9 Two cars of masses m1 and m2 are moving in circles of, , (a) m1r1 : m2r2, (c) r1 : r2, , Q, , park of radius 1 km and moves along the, path OPRQO as shown in figure. If he, maintains constant speed of10 ms −1, what, is his acceleration at point R in magnitude, and direction?, , 57, , B, L, , Vertically upward, Vertically downward, Horizontally towards left, Horizontally towards right, , 18 A particle is moving in a vertical circle.The tensions in the, D, C, , string when passing through two positions at angles 30°, and 60° from vertical (lowest positions) areT1 and T2 ,, respectively.Then, (a) T1 = T2, (b) T2 > T1, (c) T1 > T2, , (d) 41.3 s, , A, , (d) tension in the string always remains the same
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58, , DAY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , 19 A small body of mass m slides down from the top of a, , Statement II Components of the frictional force are, providing the necessary tangential and centripetal, acceleration, in the above situation., , hemisphere of radius R. The surface of block and, hemisphere are frictionless. The height at which the, body lose contact with the surface of the sphere is, 3, (a) R, 2, , 2, (b) R, 3, , 1, (c) R, 2, , 21 Statement I A particle moving in a vertical circle, has a, , 1, (d) R, 3, , maximum kinetic energy at the highest point of its, motion., Statement II The magnitude of the velocity remains, constant for a particle moving in a horizontal plane., , Direction (Q. Nos. 20-24) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below :, , 22 Statement I The centripetal force and the centrifugal, force never cancel out., Statement II They do not act at the same time., , (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 23 Statement I Improper banking of roads causes wear and, tear of tyres., Statement II The necessary centripetal force in that event, is provided by friction between the tyres and roads., , 24 Statement I When a particle moves in a circle with a, uniform speed, there is a change in both its velocity and, acceleration., , 20 Statement I A car is moving in a horizontal circular plane, with varying speed, then the frictional force is neither, pointing towards the radial direction nor along the, tangential direction., , Statement II The centripetal acceleration in circular, motion is dependent on the angular velocity of the body., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A heavy sphere of mass m is, suspended by a string of length l., The sphere is made to revolve, about a vertical line passing, through the point of suspension,, in a horizontal circle such that the, string always remains inclined to, the vertical making an angle θ., What is the period of revolution?, l, g, l sinθ, (c)T = 2 π, g, (a)T = 2 π, , 3 Two small spherical balls are free to move on the inner, surface of the rotating spherical chamber of radius, R = 0.2 m. If the balls reach a steady state at angular, position θ = 45°, the angular speed ω of device is, , q, T cosq, , T, , w, , q, , 3m, mg, , l cosθ, g, l tanθ, (d)T = 2 π, g, , q, , (b)T = 2 π, , (a) 8 rad s−1, (c) 3.64 rad s−1, , 2 Two wires AC and BC are tied at C of, , 4 The skate board negotiates the circular, , A, , B, , 30°, , small sphere of mass 5 kg, which, revolves at a constant speed v in the, horizontal circle of radius 1.6 m. The, minimum value of v is, , 45°, , −1, , (a) 8.01 ms, (b) 1.6 ms−1, (c) 0, (d) 3.96 ms−1, , (b) 2 rad s−1, (d) 9.34 rad s−1, , C, 1.6 m, , surface of radius 4.5 m. At θ = 45°, its, q, speed of centre of mass is 6 ms −1. The, combined mass of skate board and the, v, person is 70 kg and his centre of mass is, 0.75 m from the surface. The normal reaction between the, surface and the skate board, wheel is, (a) 500 N, (c) 1157 N, , (b) 2040 N, (d) zero
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CIRCULAR MOTION, , DAY FIVE, , 59, , 5 A particle is moving with a uniform speed in a circular, R/4, , orbit of radius R in a central force inversely proportional, to the nth power of R. If the period of rotation of the, particle is T , then, ª JEE Main 2018, n, , +1, , O, , 6 A bob of mass M is suspended by a massless string of, length L. The horizontal velocity v at position A is just, sufficient to make it reach the point B. The angle θ at, which the speed of the bob is half of that at A, satisfies, B, , (c), , A, , π, π, < θ<, 4, 2, 3π, (d), < θ< π, 4, , ‘weightlessness’ as they go round the top of a hill whose, radius of curvature is 20 m. The speed of the car at the, top of the hill is between, (b) 15 ms −1 and 16 ms −1, (d) 13 ms −1 and 14 ms −1, , 8 Two particles revolve concentrically in a horizontal plane, in the same direction. The time required to complete one, revolution for particle A is 3 min, while for particle B is, 1 min. The time required for A to complete one revolution, relative to B is, (a) 2 min, , (b) 1.5 min, , 5 , (d) cos−1 , , 2 3, , (c) 1 min, , (b) 1. 2 m/s 2, (d) 2.0 m/s 2, m, , 11 A circular tube of mass M is, , 7 A roller coaster is designed such that riders experience, , (a) 14 ms −1 and 15 ms −1, (c) 16 ms −1 and 17 ms −1, , 5, (c) cos−1 , 6, , (a) 0.8 m/s 2, (c) 1.6 m/s 2, , v, , (b), , π, 3π, < θ<, 2, 4, , 5 , (b) cos−1 , , 3, , k = 0.5 m/s 2. Then total acceleration of the point at the, moment when it has covered the nth fraction of the circle, 1, after the beginning of motion, where n =, 10, , L θ, , π, 4, , 2, (a) cos−1 , 3, , 10 A point moves along a circle with a speedV = kt , where, , O, , (a) θ =, , R, , q, , (b)T ∝ R 2, (d)T ∝ R n / 2, , (a)T ∝ R 3 / 2 for any n, (c)T ∝ R (n + 1)/ 2, , (d) 1.25 min, , 9 A skier plans to ski on smooth fixed hemisphere of radius, , m, , placed vertically on a horizontal, surface as shown in the figure., q, Two small spheres, each of, mass m, just fit in the tube, are, released from the top. If θ gives, the angle between radius vector, of either ball with the verticle,, M, then for what value of the ratio , tube breaks its contact, m, with ground when θ = 60º. (Neglect any friction)., (a), , 1, 2, , (b), , 2, 3, , (c), , 3, 2, , (d) None of these, , 12 A particle is moving in a circle of radius R in such a way, that at any instant the normal and tangential component, of its acceleration are equal. if its speed at t = 0 is vo . The, time taken to complete the first revolution is, , R. He starts from rest from a curved smooth surface of, R , height . The angle θ at which he leaves the, 4, , R, v0, R, (c), (1 + e −2 π ), v0, , R −2 π, e, v0, R, (d), (1 − e −2 π ), v0, , (a), , hemisphere is, , (b), , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (a), , 3 (a), , 4 (a), , 5 (a), , 6 (c), , 7 (d), , 8 (a), , 9 (c), , 10 (a), , 15 (d), , 16 (a), , 17 (b), , 18 (c), , 19 (b), , 20 (a), , 5 (c), , 6 (d), , 7 (a), , 8 (b), , 9 (c), , 10 (a), , 11 (a), , 12 (b), , 13 (a), , 14 (c), , 21 (d), , 22 (c), , 23 (a), , 24 (b), , 1 (b), 11 (a), , 2 (d), 12 (d), , 3 (c), , 4 (c)
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60, , DAY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , 8 Speed of the cyclist (v ) = 27 kmh −1, 5, = 27 ×, 18, Q1 kmh −1 = 5 ms −1 , , , , , 18, 15, −1, −1, =, ms = 7.5 ms, 2, , 1 The change in velocity, when the, particle completes half the revolution is, given by, ∆v = 5ms −1 − (−5ms −1 ) = 10ms −1, Now, the time taken to complete half, the revolution is given by, πr, π ×1, t =, =, = πs, v, 1, So, the average acceleration, ∆v 10, ms −2, =, =, π, t, , 2 Here, 2πr = 34.3 ⇒ r = 34.3, 2π, 2 πr, 2 πr, and, v =, =, T, 22, Angle of banking,, , v2 , θ = tan −1 = 45°, rg , , 3 In uniform circular motion, the only, force acting on the particle is centripetal, (towards centre). Torque of this force, about the centre is zero. Hence, angular, momentum about centre remains, conserved., , 4 Use, , ω = ω0 + α t, , ω = 0 and t = 60 s, Then, Eq. (i) gives, π, α = − rad s −2, 2, , 5 As, the wheel rolls forward the radius of, the wheel decreases along AB, hence for, the same number of rotations it moves, less distance along AB, hence it turns, left., , 6 A particle moves in a spiral path with, decreasing radius., , 7 The tension in the string will provide, necessary centripetal force., , ∴, ⇒, ∴, , T = mrω, = mr 4 π2 n2, T ∝ n2, 2, , T1 n1 , = , T2 n2 , T = 5, , , 2T n2 , , a q, O, , 2, , 2, , n22 = 25 × 2, n2 = 5 2 ≈ 7 rpm, , ac, , ∴ Centripetal acceleration acting on the, cyclist, v 2 (15 / 2)2, 225, ac =, =, =, ms −2, r, 80, 4 × 80, = 0.70 ms −2, Tangential acceleration applied by, brakes, a T = 0.5ms −2, Centripetal acceleration and tangential, acceleration act perpendicular to each, other., ∴ Resultant acceleration,, a=, , …(i), , Here, ω 0 = 900 rpm, = (2 π × 900)/60 rad s − 1, , ∴, , Radius of the circular turn (r ) = 80 m, aT, , a2c + a2T, , = (0.7)2 + (0.5)2, = 0.49 + 0.25, = 0.74 = 0.86 ms −2, If resultant acceleration makes an angle, θ with the direction of velocity, then, a, 0.7, tanθ = c =, = 1.4 = tan 54° 28 ′, a T 0.5, θ = 54° 28 ′, , 9 As their period of revolution is same, so,, their angular speed is same centripetal, acceleration is a = ω 2 r ., a1 ω r1, r, =, = 1, a2 ω2 r2, r2, 2, , Thus,, , 12 Balancing frictional force for centripetal, force,, mv 2, = f = µN = µmg, r, where, N is normal reaction, ∴, v = µrg, (where, r is radius of the circular track), For path ABC, Path length, 3, = (2 π 2R ) = 3 πR = 3 π × 100, 4, = 300 π m, v 1 = µ 2 Rg, =, ∴t 1 =, , 01, . × 2 × 100 × 10 = 1414, . ms −1, 300 π, = 66.6 s, 14.14, , For path DEF, Path length, 1, π × 100, = (2 πR ) =, = 50 π, 4, 2, v 2 = µRg = 01, . × 100 × 10, = 10 ms −1, 50 π, = 5 πs = 157, . s, t2 =, 10, For paths CD and FA,, Path length = R + R = 2R = 200 m, 200, t3 =, = 4. 0 s, 50, ∴ Total time for completing one round, t = t1 + t2 + t3, . + 4. 0 = 86.3 s, = 66. 6 + 157, , 13 Acceleration of the cyclist at point R, = centripetal, acceleration (a c ), v2, (10)2, 100, ac =, =, =, r, 1000 1000, = 01, . ms −2 , along RO, , 14 Centrifugal force on the rod,, F =, , mv 2, along BF ., r, O, , 10 As, T = mv 2 /r, , q, , Hence, v = Tr /m, = 25 × 1.96 / 0.25 = 14 ms −1, , 11 Retarding force,, F = ma = µR = µ mg, a=µg, Now, from equation of motion,, v 2 = u2 − 2as, ∴, , 0 = u − 2as, u2, u2, v2, or s =, =, = 0, 2a 2µg 2µg, 2, , C, , B, , F, , A, mg, , Let θ be the angle, which the rod makes, with the vertical., Forces parallel to the rod,, mv 2, mg cos θ +, sin θ = T, r
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CIRCULAR MOTION, , DAY FIVE, , = 1 = tan 45°, θ = 45°, , Here, , 15 Velocity at the bottom is 2gh, , 3.7 × 10 = 0.5 × 4ω + 0.5 × 10, 2, , 32, = 16 or ω = 4 rad s − 1, 2, , 17 When bob is whirled into a vertical, circle, the required centripetal force is, obtained from the tension in the string., When string is cut,tension in string, becomes zero and centripetal force is, not provided, hence bob start to move, in a straight line path along the, direction of its velocity., At point B, the velocity of B is vertically, downward therefore, when string is cut, at B, bob moves vertically downward., 2, , 18 T = mv + mg cos θ, r, , ∴, , T ∝ cos θ, T1, cos 30°, 3 /2, 3, =, =, =, T2 cos 60° 1 / 2, 1, , or, , T1 > T2 ., , 19 Suppose body slips at point B, A, , B, , h, , q, , g, , m, , O, , mg cos θ =, , g cos θ =, , component of frictional force is, responsible for changing the speed of, car while component along the radial, direction is providing necessary, centripetal force, hence net friction, force is neither towards radial nor along, tangential direction., , 21 As the kinetic energy at the highest, , 23 If the roads are not properly banked, the, , 16 As, T max = mrω2 + mg, ω2 =, , 20 In the present case, the tangential, , centrifugal forces act at the same time, on two different bodies. Thus, they, never cancel out., , = (2 × 5)/5 = 2 cm, , sq, co mg, , mv 2, R, mg sinq, , R, , mv 2, R, [Q v = 2g (R − h )], , 2g (R − h ), R, , ⇒, , force of friction between tyres and road, provides the necessary centripetal force,, which causes the wear and tear of tyres., , T = 2π, , T1 sin 30° + T2 sin 45°, , r, , P, q, , mg, , From the figure,, T1 cos 30° + T2 cos 45° = mg, mv 2, T1 sin 30° + T2 sin 45° =, r, mv 2, mg −, r, T1 =, ⇒, ( 3 − 1), 2, But T1 ≥ 0, O, θ, T, , shown feature. The velocity of particle, in circular motion, … (i), v = rω e t, y, er, q, , l cos θ, g, , T1 cos 30° + T2 cos 45°, , 2, , 24 A particle in a circular motion has the, , et, , [Qr = l sinθ], , 4 π2, l cos θ, g, , T2 =, , ⇒, , 22 We know that centripetal and, , 5gR, , Hence, R = 2h/5, , or, , ∴, , cos θ = h , , R , , point is zero., , For completing the loop,, 2gh =, , sin θ, 4 π2, =, l sin θ, cos θ gT 2, , B, , C, , F, , A, , mg, mv 2, r ≥0, 3−1, 2, , mg −, , x, x, , Thus, we see that velocity of the, particle is r ω along e t or in tangent, direction. So, it changes as the particle, rotates the circle. Acceleration of the, particle, dv, …(ii), a = − ω2 r e r +, et, dt, Thus, acceleration of a particle moving, in a circle has two components one, along e t (along tangent) and the other, along e r (or towards centre). Of these, the first one is called the tangential, a t and other is called centripetal a r ., From Eq. (ii), it is obvious that, acceleration depends on angular, velocity (ω ) of the body., , SESSION 2, 2, , 1 Here, mv = T sinθ and mg = T cos θ, r, Dividing these two, we get, r ω2 4 π2 r, v2, tan θ =, =, =, rg, g, gT 2, , ⇒, , mg ≥, , ∴, , v max =, , mv 2, ⇒ v≤, r, , rg, , rg, , = 1.6 × 9.8 = 3.96 ms −1, , 3 Given, R = 0.2 m, From the figure,, r = 3 R + R sin 45°, w, N sin 45°, N, °, , (10)2, v2, This gives tan θ =, =, rg, 10 × 10, , 2 (R − h ), R, h 2 (R − h ), =, R, R, 2, h= R, 3, , cos θ =, , mrw2, , 45, , Force perpendicular to the rod, mv 2, mg sin θ −, cos θ, r, The rod would be balanced if, mv 2, mg sin θ −, cos θ = 0, r, mv 2, mg sin θ =, cos θ, r, , 61, , N cos 45°, R sin 45°, 3R, mg, r, , In the frame of rotating spherical, chamber,, N cos 45° = mrω2, N sin 45° = mg
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62, , DAY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , mg, , ⇒ tan 45° =, g, =, r, , ⇒ω=, , g, , =, , mrω2, , g, R, 2, , 3R +, , = 3.64 rad s, , −1, , N, , 2π, ω1 − ω2, 2π, 2π, where, ω1 =, and ω2 =, T1, T2, 2π, T1T2, 3×1, or t =, =, ∴t =, T2 − T1 3 − 1, 2π 2π , −, , , T1 T2 , or t = 1.5 min, Hence, (b) is the correct option., , 8 Here, (ω1 − ω2 )t = 2 π or t =, , θ, , C, θ, mg cos θ, mg, , From the diagram,, mv 2, r, mv 2, N =, + mg cos θ, ∴, r, From the figure,, θ = 45° , m = 70 kg, g = 9.8 ms −2 ,, N − mg cos θ =, , v = 6 ms −2 , r = CO = 4.5 − 0.75= 3.75 m, Putting these values in above equation,, we get, N = 1157 N, , 5 Q Force = Mass × Acceleration = mω2 R, and given, F ∝, So, we have, k, Rn, ⇒, , 1, , k, , ⇒ F =, , Rn, , Rn, , 2, , 2π , = m , R, T , 4 π2 m n, ⋅R, k, , T2 =, , +1, , n +1, , ⇒T ∝ R, , 2, , 6 Velocity of the bob at the point A, v =, , …(i), , 5gL, , From conservation of energy,, i.e. KE + PE = constant, 2, v = v 2 − 2gh, …(ii), ⇒, , 2, …(iii), h = L (1 − cos θ), Solving Eqs. (i), (ii) and (iii), we get, 7, cos θ = −, 8, 7, θ = cos −1 − = 151 °, ⇒, 8, i.e., 3π / 4 < θ < π, , 7 Balancing the force, we get, Mg − N = M, , 2, , v, R, , N, n, Mg, , 9 At the time of leaving contact, normal, reaction must be zero., v, q, , O, mg, N =0, m(2gh ), mv 2, ∴mg cos θ =, =, R, R, 2h, or, cos θ =, R, 2 R, or, cos θ = + R(1 − cos θ), , R 4, 5, 5, or cos θ =, or 3cos θ =, 2, 6, 5, ∴, θ = cos −1 , 6, Hence, (c) is the correct option., 10 Q v = ds = kt, dt, s, t, 1, ∴ ∫ ds = k ∫ t dt ⇒ s = kt 2, 2, 0, 0, , i.e., , ∴, , Mv, = Mg, R, , or v =, , Rg, , or, , = 0.50 1 + 16 × (314, . )2 × (01, . )2, = 0.8 m / s 2, Hence, (a) is the correct option., , dv, v2, dv, 1, or 2 = dt, =, dt, R, R, v, t, v, 1, −2, ∫ v dv = R ∫ dt, 0, v, 0, , 1, 1 t, v − v = R, 0, , R − v 0t, 1, 1, t, or, =, − =, v, v0 R, Rv 0, R. v 0, or, v =, R − v 0t, dx, As, v =, dt, dx, Rv 0, ∴, =, dt, R − v 0t, or, , …(i), , at + an = [k 2 + 16π2 n2 k 2 ], , = k 1 + 16 π2 n2, For weightlessness, N = 0, , or, , Now, tangential acceleration,, dv, d, aT =, = (kt ) = k, dt, dt, and normal acceleration,, v2, k 2t 2, an =, =, r, r, k 2 4 πrn, or, an =, ×, = 4 πnk, r, κ, 2, , If N is normal reaction, then, mv 2, N + mg cos θ =, R, m, or N + mg cos θ =, × 2gR(1 − cos θ), R, or N + mg cos θ = 2 mg (1 − cos θ), or N = 2m − 3mg cos θ, …(i), The tube will breaks its contact with, ground when,, 2N cos θ ≥ Mg, where, we put the value of N from Eq. (i), is above relation, then we get, 4 mg cos θ − 6mg cos 2 θ = Mg, , 12 Given, at = ar, , After completion of n th fraction of circle, …(ii), s = 2πrn, From Eqs. (i) and (ii), we get, 4 πrn , t2 = , k , , 2, , [After applying conservation of energy], where, h = R(1 − cos θ), v = 2gR (1 − cos θ), ∴, , Put θ = 60º (given), ∴ 4 mg cos 60º −6 mg cos 2 60º = Mg, 3mg, M 1, or 2mg −, = Mg ⇒, =, 2, m 2, Hence, (a) is the correct option., , q, , ∴a =, , 2, , v = 2gh, , v = 20 × 10 = 14.14 ms −1, , So,, , Thus, the speed of the car at the top of, the hill is between 14 ms−1 and 15 ms−1 ., , O, , 4, , 11 Speed of each particle at angle θ is, , Putting the values,, R = 20 m , g = 10 ms −2, , rω2, , or, or, , x, , t, , 0, , 0, , Rv, , ∫ dx = ∫ R − v00 t dt, −1 , t, x = Rv 0 [ln(R − v 0 t )] 0, v0, , v t, x = − R ln 1 − 0 , , R , 1, −, v, t, x, 0 , or, ln , =−, R , R, v t, or, 1 − 0 = e −x /R, R, v t, or, 1 − e −x /R = 0, R, R, or, t =, (1 − e − x / R )., v0, After completing one revolution, x = 2πR, R, (1 − e −2 π )., ∴ t =, v0, or, , Hence, (d) is the correct option.
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DAY SIX, , Work, Energy, and Power, Learning & Revision for the Day, u, , u, , Work, Conservative and, Non-conservative Force, , u, , u, , u, , Energy, Work-Energy Theorem, Law of Conservation of Energy, , u, , u, , Power, Collision, , Work, Work is said to be done, when a body a displaced through some distance in the direction, of applied force. The SI unit work is joule (J) and in CGS it is erg., 1 joule (J) = 107 erg, The work done by the force F in displacing the body through a distance s is, W = (F cos θ)s = Fs cos θ = F ⋅ s, where, F cos θ is the component of the force, acting along the direction of the, displacement produced. SI unit of work is joule (J)., 1 J = 1 N-m, Work is a scalar quantity. Work can be of three types, (i) Positive work, (ii) Negative work and, (iii) Zero work., l, , l, , l, , Positive work If value of the angle θ between the directions of F and s is either zero or, an acute angle., , PREP, MIRROR, , Negative work If value of angle θ between the directions of F and s is either 180° or, an obtuse angle., , Your Personal Preparation Indicator, , As work done W = F ⋅ s = F s cos θ, hence work done can be zero, if, (i) No force is being applied on the body, i.e. F = 0., (ii) Although the force is being applied on a body but it is unable to cause any, displacement in the body, i.e. F ≠ 0 but s = 0., (iii) Both F and s are finite but the angle θ between the directions of force and, displacement is 90°. In such a case, W = F ⋅ s = F s cos θ = F s cos 90 ° = 0, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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64, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , Work Done by Variable Force, , l, , Work done by a variable force is given by W = ∫ F ⋅ ds, B, Force, , l, , where, v is velocity and t is time., , C, Displacement, , It is equal to the area under the force-displacement graph,, along with proper sign., Work done = Area of ABCDA, , Conservative and, Non-conservative Force, A force is said to be conservative if work done by or against, the force in moving a body depends only on the initial and, final positions of the body and not on the nature of path, followed between the initial and the final position., Gravitational force, force of gravity, electrostatic force are, some examples of conservative forces (fields)., A force is said to be non-conservative if work done by or, against the force in moving a body from one positions to, another, depends on the path followed between these two, positions. Force of friction and viscous force are the examples, of non-conservative forces., , Potential Energy, Potential energy is the energy stored in a body or a system by, virtue of its position in a field of force or due to its, configuration. Potential energy is also called mutual energy or, energy of the configuration., Value of the potential energy in a given position can be, defined only by assigning some arbitrary value to the, reference point. Generally, reference point is taken at infinity, and potential energy at infinity is taken as zero. In that case,, r, , U = − W = − ∫ F ⋅ dr, ∞, , Potential energy is a scalar quantity. It may be positive as well, as negative., Different types of potential energy are given below., , Gravitational Potential Energy, It is the energy associated with the state of separation between, two bodies which interact via the gravitational force., l, , Energy, Energy is defined as the capacity or ability of a body to do, work. Energy is scalar and its units and dimensions are the, same as that of work. Thus, SI unit of energy is J., , l, , Some other commonly used units of energy are, 1 erg = 10 −7 J,, 1 cal = 4.186 J ~, = 4.2 J,, 1 kcal = 4186 J,, 1 kWh = 3.6 × 10 6 J,, , Kinetic Energy, , l, , Kinetic energy is the capacity of a body to do work by, virtue of its motion. A body of mass m, moving with a, 1, velocity v, has a kinetic energy, K = mv2 ., 2, Kinetic energy of a body is always positive irrespective of, the sign of velocity v. Negative kinetic energy is impossible., Kinetic energy is correlated with momentum as,, K =, , p2, 2m, , or, , p = 2 mK, , The gravitational potential energy of two particles of, masses m1 and m2 separated by a distance r is, − Gm 1 m2, ., U=, r, If a body of mass m is raised to a height h from the surface, of the earth, the change in potential energy of the system, (earth+body) comes out to be, mgh, ∆U =, h, , 1 + , , R, or, ∆U ≈ mgh if h << R, Thus, the potential energy of a body at height h, i.e. mgh is, really the change in potential energy of the system for, h << R., , and 1 electron volt (1 eV) = 1.60 × 10 −19 J, , l, , Relation between kinetic energy and force is, KE, v ×t, =, Force, 2, , A, , D, , l, , Kinetic energy for a system of particle will be, 1, K = Σ mi v2i, 2 i, , For the gravitational potential energy, the zero of the, potential energy is chosen on the ground., , Electric Potential Energy, The electric potential energy of two point charges q1 and q2, separated by a distance r in vacuum is given by, 1, q1q2, U=, 4 π ε0 r, 1, N-m2, where,, = 9.1 × 10 9, = constant, 4 π ε0, C2
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WORK, ENERGY AND POWER, , DAY SIX, Potential Energy of a Spring, Whenever an elastic body (say a spring) is either stretched or, compressed, work is being done against the elastic spring, 1, force. The work done is W = kx2 ,, 2, where, k is spring constant and x is the displacement., 1, And elastic potential energy, U = k x2, 2, If spring is stretched from initial position x1 to final position, x2 , then, work done = Increment in elastic potential energy, 1, = k ( x22 − x12 )., 2, , Work-Energy Theorem, Accordingly, work done by all the forces (conservative or, non-conservative, external or internal) acting on a particle or, an object is equal to the change in its kinetic energy of the, particle. Thus, we can write, W = ∆K = K f − K i, We can also write, K f = K i + W, Kinetic energy after , Which says that , , the net work is done, , l, , l, , The mechanical energy E of a system is the sum of its kinetic, energy K and its potential energy U., E = K +U, When the forces acting on the system are conservative in, nature, the mechanical energy of the system remains constant,, K + U = constant ⇒ ∆K + ∆U = 0, There are physical situations, where one or more nonconservative force act on the system but net work done by, them is zero, then too the mechanical energy of the system, remains constant., If, Σ Wnet = 0, Mechanical energy, E = constant., , Instantaneous power,, dW F ⋅ ds, Pinst =, =, = F⋅v, dt, dt, Some other commonly used units of power are, 1 kW = 103 W,, 1 MW = 10 6 W,, 1 HP = 746 W, , Collision, The physical interaction of two or more bodies in which each, equal and opposite forces act upon each other causing the, exchange of energy and momentum is called collision., Collisions are classified as, (i) elastic collisions and, (ii) inelastic collisions., , Elastic Collision in One Dimension, In a perfectly elastic collision, total energy and total linear, momentum of colliding particles remains conserved., Moreover, the forces involved in interaction are conservative, in nature and the total kinetic energy before and after the, collision, remains unchanged., m1, , Kinetic energy before The net , =, +, , the net work done work done, , Law of Conservation of Energy, , u1, , m2, , v1, , u2, m1, , l, , l, , It is a quantity that measures the rate at which work is, done or energy is transformed., W, Average power (P)av =, t, The shorter is the time taken by a person or a machine in, performing a particular task, the larger is the power of that, person or machine., Power is a scalar quantity and its SI unit is watt, where,, 1 W = 1 J/s, , v2, m2, , B, A, After collision, , A, B, Before collision, , In above figure, two bodies A and B of masses m1 and m2 and, having initial velocities u1 and u2 in one dimension, collide, elastically and after collision move with velocities v1 and v2 ,, then we find that, l, , l, , Relative velocity of approach = Relative velocity of, separation, i.e. u1 − u2 = v2 − v1, m − m2 , 2 m2 , v1 = 1, u1 + , u2, m1 + m2 , m1 + m2 , m − m1 , 2 m1 , and v2 = , u2, u1 + 2, m, +, m, m1 + m2 , 1, 2, , Elastic Collision in Two Dimensions, l, , In this type of collision, the two particles or objects moving, along different directions collide with each other., , Power, l, , 65, , v1, m1, A, A, m1, Before collision, , θ, , B, u1, m2, , u1, , φ, B, m2, v2, After collision
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WORK, ENERGY AND POWER, , DAY SIX, , 67, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, simultaneously by two forces, one of 4N and other of 3N,, acting at right angles to each other. The work done by, the body after 20 s is, (a) 500 J, , (b) 1250 J, , (c) 2500 J, , length 10 m and height 5 m and then released to slide, down to the bottom. The coefficient of friction between, the body and the plane is 0.1. What is the amount of, work done in the round trip?, (b) 15 J, , (c) 5 3 J, , (c), , t, , (d), t, , (d) 5000 J, , 2 A body of mass 500 g is taken up an inclined plane of, , (a) 5 J, , KE, , KE, , 1 A body of mass 2 kg, initially at rest, is acted upon, , 5, (d), J, 3, , 8 A particle is moving in a circular path of radius a under, the action of an attractive potentialU = −, , k, . Its total, 2r 2, , energy is, (a) −, , k, 4a 2, , (b), , k, 2a 2, , (c) zero, , (d) −, , 3 k, 2 a2, , 9 The potential energy of a 1 kg particle free to move along, 3 A block of mass 5 kg is initially at rest on a horizontal, frictionless surface. A horizontal force F = ( 9 − x ) $i, Newton acts on it, when the block is at x = 0. The, maximum work done by the block between x = 0 and, x = 3 m in joule is, 2, , (a) 18 J, , (b) 15 J, , (c) 20 J, , (d) 24 J, , 4 An object is displaced from point A (2m, 3m, 4m) to a, point B (1m, 2m, 3m) under a constant force, F = ( 2$i + 3 $j + 4 k$ ) N, then the work done by this force in, this process is, (a) 9 J, , (b) −9 J, , (c) 18 J, , x 4 x 2, the x-axis is given by V ( x ) = , J., −, 2, 4, The total mechanical energy of the particle is 2 J. Then,, the maximum speed (in ms −1) is, (a), , 3, 2, , (b) 2, , (c), , 1, 2, , (d) 2, , 10 A particle is placed at the origin and a force F = kx acts, on it (where, k is a positive constant). If U( 0) = 0, the, graph of U( x ) versus x will be (where, U is the potential, energy function), , (d) −18 J, , U(x), , U(x), , 5 An open water tight railway wagon of mass 5 × 10 kg, 3, , coasts with an initial velocity of 1.2 ms −1 on a railway, track without friction. Rain falls vertically downwards on, the wagon., , (a), , What change occurs in the kinetic energy of the wagon,, after it has collected 103 kg of water?, (a) 900 J, , (b) 300 J, , (c) 600 J, , (d) 1200 J, , x, , (b), , U(x), , (c), , x, , U(x), , x, , (d), , x, , 6 A particle moves in a straight line with retardation, proportional to its displacement. The loss in kinetic, energy of the particle, for any displacement x, is, proportional to, (a) x 2, , (b) e x, , (c) x, , (d) loge x, , 7 Which of the diagrams as shown in figure most closely, shows the variation in kinetic energy of the earth as it, moves once around the sun in its elliptical orbit?, KE, , KE, , (a), , (b), t, , t, , 11 A bullet fired into a fixed target losses half of its velocity, after penetrating distance of 3 cm. How much further it, will penetrate before coming to rest, assuming that it, faces constant resistance to its motion?, (a) 3.0 cm, , (b) 2.0 cm, , (c) 1.5 cm, , (d) 1.0 cm, , 12 A 1.5 kg block is initially at rest on a horizontal frictionless, surface, when a horizontal force in the positive direction, of x-axis, is applied to the block. The force is given by, F = ( 4 − x 2 ) i Newton, where x is in metre and the initial, position of the block is at x = 0. The maximum kinetic, energy of the block between x = 0 and x = 2. 0 m is, (a) 6.67 J, , (b) 5.33 J, , (c) 8.67 J, , (d) 2.44 J
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68, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 13 A block of mass M moving on a, , 18 Two balls of masses m1 and m2 are separated from each, , M, , frictionless horizontal surface,, collides with a spring of spring constant k and, compresses it by length L. The maximum momentum of, the block, after collision is, (a) L Mk, , (b), , kL2, 2M, , (c) zero, , (d), , ML2, k, , 14 A cyclist rides up a hill with a constant velocity., Determine the power developed by the cyclist, if the, length of the connecting rod of the pedal is r = 25 cm, the, time of revolution of the rod is t = 2 s and the mean force, exerted by his foot on the pedal is F = 15 kgf., (a) 115.6 W, , (b) 215.6 W, , (c) 15.6 W, , (d) 11.56 W, , 15 Power supplied to a particle of mass 2 kg varies with time, as P = 3 t 2/ 2 W, where t is in second. If velocity of the, particle at t = 0 is v = 0, the velocity of the particle at t = 2 s,, will be, (c) 2 ms −1, , (b) 4 ms −1, , (a) 1 ms −1, , (d) 2 2 ms −1, , 16 A body is moving unidirectionally under the influence of a, source of constant power supplying energy. Which of the, diagrams as shown in figure correctly shows the, displacement-time curve for its motion?, d, , other and a charge is placed between them. The whole, system is at rest on the ground. Suddenly, the charge, explodes and the masses are pushed apart. The mass m1, travels a distance S1 and then it stops. If the coefficient of, friction between the balls and the ground are the same,, mass m2 stops after covering the distance, m1, S1, m2, m2, (c) S2 = 1 2 S1, m2, , (b) S2 =, , 19 A shell is fired from a cannon with a velocity v ms −1 at an, angle θ with the horizontal direction. At the highest point in, its path, it explodes into 2 pieces of equal masses. One of, the pieces retraces its path to the cannon. The speed in, ms −1 of the other piece, immediately after the explosion is, (a) 3 v cosθ, , (b) 2 v cosθ, , (d), , 3, v cosθ, 2, , elastic one dimensional collision with a stationary particle, of mass m. They are in contact for a very short interval of, time T ., F0, , (b), , F, , t, , T/2, Time, , t, , d, , (c) v cosθ, , 20 A particle of mass m moving with a velocity u makes an, , d, , (a), , m2, S1, m1, m 2, (d) S2 = 22 S1, m1, , (a) S2 =, , d, , T, , T, and, 2, T, then decreases linearly to zero in further time interval ., 2, The magnitude of F0 is, The force of interaction increases from zero to F0 in, , (c), , (d), , t, , t, , 17 Two identical ball bearings in contact with each other, and resting on a frictionless table are hit head-on by, another ball bearing of the same mass moving initially, with a speed v as shown in, 1, , 2, , 3, , v, , 2, , 3, , (a), , 1, , 2, , 3, , (b), v/2, , v=0, 1, , 2, , v=0, , 3, , (c), , v, , 1, , 2, , 3, , v/1, , v/2, , v/3, , (d), v/3, , mu, T, , (b), , 2mu, T, , (c), , mu, 2T, , (d) None of these, , 21 A block of mass 0.50 kg is moving with a speed of, , 2.00 ms −1 on a smooth surface. It strikes another mass of, 1.00 kg and then they move together as a single body., The energy loss during the collision is, (a) 0.16 J, , If the collision is elastic, which of the following is a, possible result after collision?, 1, , (a), , (b) 1.00 J, , (c) 0.67 J, , (d) 0.34 J, , 22 A ball hits the floor and rebounds after an inelastic, collision. In this case, (a) the momentum of the ball just after the collision is the, same as that just before the collision, (b) the mechanical energy of the ball remains the same in, the collision, (c) the total momentum of the ball and the earth is, conserved, (d) total mechanical energy of the ball and the earth is, conserved
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WORK, ENERGY AND POWER, , DAY SIX, 23 Two particles of masses m1 and m2 in projectile motion, , have velocities v1 and v2 respectively at time t = 0. They, collide at time t 0. Their velocities become v1′ and v′2 at, time 2t 0 while still moving in air. The value of, | (m1 v′1 + m2 v′2 ) − (m1 v1 + m2 v2 )| is, (b) (m1 + m2 )gt 0, 1, (m1 + m2 ) gt 0, 2, , (a) zero, (c) 2 (m1 + m2 )gt 0, , (d), , 24 A ball is dropped on the ground from a height of 2m. If, the coefficient of restitution is 0.6, the height to which the, ball will rebound is, (a) 0.72 m, , (b) 1.72 m, , (c) 2.72 m, , (d) 1 m, , 25 A body of mass m is accelerated uniformly from rest to a, speed v in a time interval T . The instantaneous power, delivered to the body as a function of time (t ), is given by, (a), , mv, , 2, , T2, , t, , (b), , mv, , 2, , T2, , t2, , 2, , (c), , 1 mv, t, 2 T2, , 2, , (d), , 1 mv 2, t, 2 T2, , 26 A person trying to lose weight by burning fat lifts a mass, of 10 kg upto a height of 1 m, 1000 times. Assume that, the potential energy lost each time he lowers the mass is, dissipated. How much fat will he use up considering the, work done only when the weight is lifted up? Fat supplies, 3.8 × 107 J of energy per kg which is converted into, mechanical energy with a 20% of efficiency rate., ª JEE Main 2016 (Offline), (Take, g = 9.8 ms −2), (a) 2.45 × 10−3 kg, (c) 9.89 × 10−3 kg, , (b) 6.45 × 10−3 kg, (d) 12.89 × 10−3 kg, , 27 A time dependent force F = 6t acts on a particle of mass, 1 kg. If the particle starts from rest, the work done by the, force during the first 1 s will be ª JEE Main 2017 (Offline), (a) 22 J, (c) 18 J, , (b) 9 J, (d) 4.5 J, , their formula given in column II and select the correct, option from the choices given below., Column I, Spring energy, , 1., , B., , Kinetic energy, , 2., , C., , Potential energy, , 3., , A, 3, 3, 1, 2, , B, 2, 1, 2, 3, , C, 1, 2, 3, 1, , Direction (Q. Nos. 29-32) These question consists of two, statements each printed as Statement I and Statement II., While answering these questions you are required to choose, any one of the following five responses., (a) If both Statement I and Statement II are correct and, Statement II is the correct explanation of Statement I., (b) If both Statement I and Statement II are correct but, Statement II is not correct explanation of Statement I., (c) If Statement I is correct but Statement II is incorrect., (d) If Statement I is incorrect but Statement II is correct., (e) If both Statement I and Statement II are incorrect, , 29 Statement I A body cannot have energy without having, momentum but it can have momentum without having, energy., Statement II Momentum and energy have same, dimensions., , 30 Statement I When a machine-gun fires n bullets per, second with kinetic energy K, then the power of the, machine-gun is P = nK ., Work nk, Statement II Power =, =, Time, 1, , 31 Statement I A quick collision between two bodies is more, violent than a slow collision; even when the initial and the, final velocities are identical., Statement II The momentum is greater in the first case., , 32 Statement I A point particle of mass m moving with speed v, , 28 Match the categories of energy given in column I with, , A., , (a), (b), (c), (d), , 69, , Column II, 1, mv 2, 2, mgh, 1 2, kx, 2, , collides with stationary point particle of mass M. If the, 1, , maximum energy loss possible is given as f mv 2 , then, 2, , M , f =, ., M + m, Statement II Maximum energy loss occurs when the, particles get stuck together as a result of the collision., ª JEE Main 2013
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70, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A block of mass 0.5 kg has an initial velocity of 10 ms −1, while moving down an inclined plane of angle 30°, the, coefficient of friction between the block and the inclined, surface is 0.2. The velocity of the block, after it covers a, distance of 10 m, is, (a) 17 ms −1, (c) 24 ms −1, , (b) 13 ms −1, (d) 8 ms −1, , A, II, h, , C, , Which of the following statement is correct ?, (a) Both the stones reach the bottom at the same time but, not with the same speed, (b) Both the stones reach the bottom with the same speed, and stone I reaches the bottom earlier than stone II, (c) Both the stones reach the bottom with the same speed, and stone II reaches the bottom earlier than stone I, (d) Both the stones reach the bottom at different times and, with different speeds, , 3 Suppose the average mass of raindrops is 3 × 10−5 kg and, their average terminal velocity 9 ms −1. Then the energy, transferred by rain to each square metre of the surface at, a place which receives 100 cm of rain in a year is, (a) 302 × 102 J, (c) 4 . 05 × 104 J, , (d) ka 2, , a length of 60 cm hangs freely from the edge of the table., The total mass of the chain is 4 kg. What is the work done, in pulling the entire chain on the table?, (a) 7.2 J, , (b) 3.6 J, , (c) 120 J, , (d) 1200 J, , 7 A 70 kg man leaps vertically into the air from a crouching, , q2, , q1, , (c) −ka 2, , (b) 2 ka 2, , 6 A uniform chain of length 2 m is kept on a table such that, , other steep meet at A from where two stones are allowed, to slide down from rest, one on each track as shown in, figure., , B, , xy-plane. Starting from the origin, the particle is taken, along the positive x-axis to the point (a , 0) and is then, taken parallel to the y-axis to the point (a , a ). The total, work done by the force is, (a) −2 ka 2, , 2 Two inclined frictionless tracks, one gradual and the, , I, , 5 A force F = − k ( y i + x j), acts on a particle moving in the, , (b) 102 × 105 J, (d) 9. 2 × 103 J, , 4 A ball of mass m is dropped from a height h on a, massless platform fixed at the top of a vertical spring as, shown below. The platform is depressed by a distance x., What will be the value of the spring constant?, m, , h, , position. To take the leap the man pushes the ground, with a constant force F to raise himself., The centre of gravity rises by 0.5 m before he leaps. After, the leap the centre of gravity rises by another 1 m. The, maximum power delivered by the muscles is (take,, ª JEE Main (Online) 2013, g = 10 ms −2 )., (a) 6. 26 × 103 W at the start (b) 6. 26 × 103 W at take off, (c) 6. 26 × 104 W at the start (d) 6. 26 × 104 W at take off, , 8 If two springs S1 and S 2 of force constants k1 and k 2,, respectively are stretched by the same force, it is found, that more work is done on spring S1 than on spring S 2., The correct option is, (a) w1 = w 2, (c) k1 > k2, , (b) k1 < k2, (d) None of these, , 9 At time t = 0 s particle starts moving along the x-axis. If its, kinetic energy increases uniformly with time t, the net, force acting on it must be proportional to, (a) t, , (b) constant, , (c) t, , (d), , 1, t, , 10 It is found that, if a neutron suffers an elastic collinear, collision with deuterium at rest, fractional loss of its, energy is Pd ; while for its similar collision with carbon, nucleus at rest, fractional loss of energy is Pc . The values, of Pd and Pc are respectively, (a) (.89, .28), (c) (0, 0), , (b) (.28, .89), (d) (0, 1), , 11 A particle of mass m moving in the x-direction with speed, , (b) 2mgh/x, , (a) 2mg/x, (c) 2 mg (h + x) / x, , 2, , (d) 2 mg (h) + 2mghx/x 2, , 2v is hit by another particle of mass 2m moving in the, y-direction with speed v. If the collision is perfectly, inelastic, the percentage loss in the energy during the, ª JEE Main 2015, collision is close to, (a) 44%, , (b) 50%, , (c) 56%, , (d) 62%
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WORK, ENERGY AND POWER, , DAY SIX, 12 Consider a rubber ball freely falling from a height, , 17 A mass m moves with a velocity v and collides, , h = 4.9 m onto a horizontal elastic plate. Assume that the, duration of collision is negligible and the collision with the, plate is totally elastic. Then, the velocity as a function of, time the height as function of time will be, v, , v, v1, (b), , t, , O, , v, , v, , O, , t1, , (d), , 3t1, , 2t1, , 4t1, , 3v, , (c), , +v1, , t, , O, , –v1, , 2, v, 3, , v, 3, , (d), , 18 The potential energy function for the force between two, atoms in a diatomic molecule is approximately given by, a, b, U( x ) = 12 − 6 , where a and b are constants and x is the, x, x, distance between the atoms. If the dissociation energy of, the molecule is D = [U( x = ∞ ) − Uat equilibrium ], D is, , –v1, , +v1, , (b), , +v1, , t, , O, , (c), , inelastically with another identical mass. After collision,, v, in a direction, the 1st mass moves with velocity, 3, perpendicular to the initial direction of motion. Find the, speed of the second mass after collision., (a) v, , (a), , 71, , 3t1, t1, , 2t1, , 4t1, , t, , (a), , –v1, , b2, 2a, , (b), , b2, 12a, , (c), , b2, 4a, , (d), , b2, 6a, , 19 Under the action of force, 3 kg body moves such that its, 13 A body of mass m = 10−2 kg is moving in a medium and, experiences a frictional force F = −kv . Its initial speed is, 1, v 0 = 10 ms −1. If after 10 s, its energy is mv 02, the value of, 8, ª JEE Main 2017 (Offline), k will be, 2, , (a) 103 kgs−1, (c) 10−1 kgm−1 s−1, , (b) 10−4 kgm−1, (d) 10−3 kgm−1, , its ends A and B which are close together. At a given, instant end B is released. The tension at A when B has, l, fallen a distance x < is, 2, 3 − 3x , , 4 , , (a), , w 3x, − 2, , 2 l, , (b), , w, 2, , (c), , w , 3x , ⋅ 1+, 2 , l , , (d), , w 3x, + 4, , 2 l, , 15 A body is moved along a straight line by a machine, delivering a constant power. The distance moved by the, body in time t is proportional to, (b) t 3 / 2, , 1/ 4, , (d) t 1/ 2, , (c) t, , strikes a stationary particle of the same mass. If the final, total kinetic energy is 50% greater than the original, kinetic energy, the magnitude of the relative velocity, between the two particles after collision, is, ª JEE Main 2018, , (b) 2 v 0, (d), , v0, 2, , (b) 16 J, , (c) 24 J, , (d) 30 J, , exerts a restoring force of magnitude F = ax + bx 2, where,, a and b are constants. The work done in stretching the, unstretched rubber band by L is, ª JEE Main 2014, 1 2, 3, (b) (aL + bL ), 2, 1 aL2 bL3 , (d) , +, , 2 2, 3 , , (a) aL + bL, 2, , (c), , 3, , aL2 bL3, +, 2, 3, , 21 A spring gun having a spring of spring constant k is, placed at a height h. A ball of mass m is placed in its, barrel and compressed by a distance x. Where shall we, place a box on the ground, so that the ball lands in the, box?, (a), , 16 In a collinear collision, a particle with an initial speed v 0, , v, (a) 0, 4, v, (c) 0, 2, , (a) 8 J, , 20 When a rubber band is stretched by a distance x, it, , 14 A uniform chain of length l and weight w is hanging from, , (a) t 3 / 4, , t3, , where, 3, x in meter and t in second. Then, work done by the force, in the first 2 s is, position x as a function of time t is given by x =, , kh, x, mg, , (b), , 2kh, x, mg, , (c), , kh, x, 2mg, , (d), , kh, 2mg, , 22 A particle of mass m is projected from the ground with an, initial speed u 0 at an angle α with the horizontal. At the, highest point of its trajectory, it makes a completely, inelastic collision with another identical particle, which, was thrown vertically upward from the ground with the, same initial speed u 0. The angle that the composite, system makes with the horizontal immediately after the, collision is, (a), , π, 4, , (b), , π, +α, 4, , (c), , π, −α, 4, , (d), , π, 2
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72, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 23 Two small particles of equal masses start moving in, opposite directions from a point A in a horizontal circular, orbit. Their tangential velocities are v and 2 v, respectively, as shown in the figure. Between collisions,, the particles move with constant speeds. After making, how many elastic collisions, other than that at A, these, two particles will again reach the point A?, v, , (a) Statement I is correct, Statement II is correct; Statement, II is the correct explanation for Statement I, (b) Statement I is correct, Statement II is correct; Statement, II is not the correct explanation for Statement I, (c) Statement I is correct; Statement II is incorrect, (d) Statement I is incorrect; Statement II is correct, , 24 Statement I Two bodies of different masses have the, same momentum, and their kinetic energies are in the, inverse ratio of their masses., , A, 2v, , Statement II Kinetic energy of body is given by the, relation., 1, KE = mv 2, 2, (a) 4, , (b) 3, , (c) 2, , 25 Statement I An object is displaced from point, , (d) 1, , A ( 2 m , 3 m , 4 m ) to a point B (1 m , 2 m , 3 m ) under a, , Direction (Q. Nos. 24-25) Each of these questions, contains two statements : Statement I (Assertion) and, Statement II (Reason). Each of these questions also has four, alternative choices, only one of which is the correct answer., You have to select one of the codes (a), (b), (c), (d) given, below, , constant force F = ( 2 i + 3 j + 4 k ) N. The work done by the, force in this process is −9 J., Statement II Work done by a force, an object can be, given by the relation,, W =∫, , r2, r1, , F⋅d r, , or W = F⋅s, , ANSWERS, SESSION 1, , SESSION 2, , 1 (c), , 2 (c), , 3 (a), , 4 (b), , 5 (c), , 6 (a), , 7 (d), , 8 (c), , 9 (a), , 10 (a), , 11 (d), , 12 (b), , 13 (a), , 14 (a), , 15 (c), , 16 (b), , 17 (b), , 18 (c), , 19 (a), , 20 (b), , 23 (c), , 24 (a), , 25 (a), , 26 (d), , 27 (d), , 28 (b), , 29 (e), , 30 (a), , 3 (c), 13 (b), 23 (c), , 4 (c), 14 (c), 24 (a), , 5 (c), 15 (b), 25 (a), , 6 (b), 16 (b), , 7 (b), 17 (c), , 8 (b), 18 (c), , 9 (d), 19 (c), , 10 (a), 20 (c), , 21 (c), , 22 (c), , 31 (a), , 32 (a), , 1 (b), 11 (c), 21 (b), , 2 (c), 12 (c), 22 (a)
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WORK, ENERGY AND POWER, , DAY SIX, , 73, , Hints and Explanations, SESSION 1, 1 Resultant force,, F = 32 + 42 = 5N, F, 5, a=, = = 2.5 m/s2, m 2, u = 0, t = 20 s, 1, ∴ s = ut + at 2, 2, 1, = 0 × 20 + × 2.5 × (20)2 = 500 m, 2, Hence, work done, W = F ⋅ s = 5 × 500 = 2500 J, , 2 Work done in the round trip = total, work done against friction while, moving up and down the plane, = 2( µ mg cos θ) × s, 3, = 2 ( 01, . × 0.5 × 10 ×, × 10), 2, , 5 3, 3, = 5 3 J Q cos θ =, =, , 10, 2, , , A, 10, , O, , 5, , θ, B, , 5√3, , 3 Work done among horizontal is, dW = F ⋅ dx, dW = (9 − x2 ) ⋅ dx, , or, , For total work done, integrate both sides, with proper limit, W, , ∫0, , dW =, , 3, , ∫0, , (9 − x2 )dx, 3, , or, , , x3 , W = 9 x −, , 3 0, , 27, = 27 − − (0 − 0) = 18 J, , 3, , ∴ Maximum work done is 18 J., , 4 Since, F = constant,, we can also use, W = F ⋅s, Here, s = r f − ri = ($i + 2 $j + 3 k$ ), −(2$i + 3 $j + 4 k$ ), ∴, , = (− $i − $j − k$ ), W = (2i$ + 3 $j + 4 k$ ) ⋅ (− $i − $j − k$ ), = −2 − 3 − 4 = −9 J, , 5 If v ′ is the final velocity of the wagon,, then applying the principle of, conservation of linear momentum, we, get, 5 × 103 × 1.2 = (5 × 103 + 103 ) × v ′, v ′ = 1 ms–1, , Change in KE, 1, 1, = (6 × 103 ) × 12 − (5 × 103 )(1.2)2, 2, 2, [minus sign for the loss, = − 600 J, in kinetic energy], vdv, 6 a = − kx or, = − kx, dx, ⇒, v dv = − kxdx, Let the velocity change from v 0 to v., ⇒, , v, , ∫v, , 0, , ⇒, ⇒, ⇒, , x, , dv = − ∫ k xdx, 0, , v 2 − v 20, k x2, =−, 2, 2, v 2 − v 20 , mk x2, m, =−, 2, 2, , , ∆k ∝ x2, [∆k is loss in kinetic energy], , 7 As, the earth moves ones around the, sun in its elliptical orbit, its KE, is, maximum when it is closest to the sun, and minimum when it is farthest from, the sun. Also, KE of the earth is never, zero during its motion choice (d) is, correct., 8 ∴ Force = − dU, dr, d −k , k, ⇒, F =−, , =− 3, dr 2r 2 , r, As particle is on circular path, this force, must be centripetal force., mv 2, ⇒, |F |=, r, k, mv 2, 1, k, So,, =, ⇒, mv 2 = 2, r, 2, r3, 2r, ∴ Total energy of particle = KE + PE, k, k, = 2 − 2, 2r, 2r, =0, Total energy = 0, 2, 4, 9 V ( x) = x − x J, 4, 2, , , , For minimum value of V ,, , or, , 10 From F = − dU, dx, , U(x ), , x, , x, , 0, , 0, , ∫0, , dU = − ∫ F dx = − ∫ (kx )dx, , ∴, , U ( x) = −, , kx2, as U(0) = 0, 2, , 11 According to the work-energy theorem,, W = ∆K, Case I, 2, , −F ×3=, , 1 v0, 1, m − mv 20, 2 2, 2, , where, F is the resistive force and v 0 is, the initial speed., Case II Let, the further distance, travelled by the bullet before coming to, rest is s., 1, ∴ − F (3 + s ) = K f − K i = − mv 20, 2, 1, 1, 2, − mv 0(3 + s ) = − mv 20, ⇒, 8, 2, 1, or, (3 + s ) = 1, 4, 3, s, or, + =1, 4 4, or, s = 1 cm, , 12 From the work-energy theorem, kinetic, energy of the block at a distance x is, K =, , x, , x, , ∫ 0 F dx = ∫ 0 (4 −, , dV, =0, dx, , x2 )dx = 4 x −, , x3, 3, , For kinetic energy to be maximum,, dK, =0, dx, 3, d , x , 4x −, =0, dx , 3 , 4 − x2 = 0 or x = ± 2 m, At x = + 2m ,, , 4 x3 2 x, −, =0, 4, 2, ⇒, x = 0, x = ± 1, 1 1 −1, So, V min ( x = ± 1) = − =, J, 4 2, 4, Now, K max + V min, = total mechanical energy, 1, ⇒ K max = + 2, 4, 9, or K max =, 4, ⇒, , mv 2, 9, =, 2, 4, 3, v =, ms −1, 2, , or, , d2K, , = negative, dx2, i.e. Kinetic energy is maximum., x3, K max = 4 x −, 3, 23, = 4 (2) −, = 5.33 J, 3, , 13 According to the conservation of energy,, , ⇒, or, ⇒, , 1 2 1, kL = Mv 2, 2, 2, (Mv )2, 2, kL =, M, [Q p = Mv ], MkL2 = p2, p = L Mk
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74, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 v = rω = r 2 π = 1 × 2 π = π ms −1, t, , 4, , 2, , P = F × v = (15 × 9.8) ×, , π, = 115.6 W, 4, , 15 From the work-energy theorem,, Kf, , ∆KE = W net, − K i = ∫ P dt, , 2 3, 1, mv 2 − 0 = ∫ t 2 dt, 0 2 , 2, 1, 3 2, (2)v 2 = ∫ t 2dt, 2, 2 0, 2, , 3 t 3 , = × = 4, 2 3 0, v = 2 ms −1, , ⇒, , 16 Here, P = [ ML2 T −3 ] = constant, As mass M of body is fixed, [L 2 T −3 ] = constant, [L2 ], , ⇒, , [T3 ], , = constant, , ⇒, , [L] ∝ [T3 /2 ], , ⇒, , Displacement ∝t 3 /2, , 17 As the ball bearings are identical, their, masses are equal. In elastic collision,, their velocities are interchanged. In, collision between 1 and 2; velocity of 1, becomes zero and velocity of 2 becomes, v. In collision between 2 and 3, velocity, of 2 becomes zero and velocity of 3, becomes v., , 18 From the conservation of momentum,, , ⇒, , …(i), m1 v 1 = m2 v 2, 1, m1 v 12 = f1 S 1 = µ m1 g S 1 …(ii), 2, 1, …(iii), and, m2 v 22 = f2 S 2 = µm2 gS 2, 2, where, µ = coefficient of friction, On dividing Eq. (ii) by Eq. (iii), we get, m1 v 12, mS, …(iv), = 1 1, m2S 2, m2 v 22, also,, , Using Eqs. (i) and (iv), we get, v1, mS, = 1 1, v2, m2 S 2, m2, mS, m2, = 1 1 ⇒ S 2 = 12 S 1, m1, m2S 2, m2, , 19 Velocity at the highest point = v cos θ, Applying the principle of conservation, of linear momentum, we get, 2m (v cos θ) = m(− v cos θ) + mv ′, v ′ = 3 v cos θ, , ⇒, , 21 m1 u1 + m2 u2 = (m1 + m2 )v, ⇒, v = 2/3 ms −1, Energy loss, 2, 1, 1, 2, = (0.5) × (2)2 − (1.5) × , 3, 2, 2, = 0.67 J, , 22 In an inelastic collision only, momentum of the system may remain, conserved. Some energy can be lost in, the form of heat, sound, etc., , 23 |(m1 v1′ + m2 v2′ ) − (m1 v1 + m2 v2) |, = |change in momentum of the two, particles |, = |External force on the system |, × time interval, = (m1 + m2 ) g (2t 0 ) = 2 (m1 + m2 ) gt 0, , 24 Let v is the final velocity of ball on, reaching the ground, then, v = 2 gh = 2 × g × 2, v =2 g, , or, , p=, , 1, , ∫0 6t dt, , m, = 3 kg , s , , Also, change in kinetic energy, ∆p2, 32, ∆k =, =, = 4.5, 2m, 2×1, From work-energy theorem,, Work done = change in kinetic energy, So, work done = ∆k = 4.5 J, 28 Kinetic energy = 1 mv 2, 2, Potential energy = mgh, 1, Spring energy = kx2, 2, , 29 We know that a body may not have, momentum but may have potential, energy by virtue of its position as in case, of a stretched or a compressed spring., But when the body does not contain, energy then its kinetic energy is zero, hence, its momentum is also zero., Dimensions of momentum, (mv ) = [MLT −1 ], Dimensions of energy, 1 mv 2 = [ML2 T −2 ], , , 2, , n×K, nK, =, 1, 1, , For upward motion, u = 2 g × e and v = 0, , 30 ∴Power = Work =, , ∴ Height upto which the ball will, rebound is, 2, u2 (2 g × e ), H =, =, 2g, 2g, , 31 Momentum p = mv or p ∝ v ,, , we get, , or, , 1, F0 × T, 2, mu, F0 = 2, T, , mu =, , 4, , =, , 4g × e 2, = 2 × 0.6 × 0.6 = 072, . m, 2g, v − 0, , Q a =, , , T , , 25 F = ma = mv, T, , Instantaneous power = Fv = mav, mv, at, =, T, mv v, mv 2, =, t=, t, T T, T2, , 26 Given, potential energy burnt by lifting, weight = mgh = 10 × 9. 8 × 1 × 1000, = 9. 8 × 104, , If mass lost by a person be m, then, 2, energy dissipated = m ×, × 3.8 × 107 J, 10, 1, ⇒, 9.8 × 104 = m × × 3.8 × 107, 5, 5, ⇒, m=, × 10−3 × 9.8, 3.8, = 12. 89 × 10−3 kg, , 20 The collision will cause an exchange of, , 27 From Newton’s second law,, , velocities. The change in momentum of, any particle = mu, which is equal to the, impulse = area under the force-time, graph, , ∆p, = F ⇒ ∆p = F∆t, ∆t, ∴, , p = ∫ dp =, , 1, , ∫0, , F dt, , Time, , i.e. momentum is directly proportional to, the velocity, so the momentum is greater, in a quicker collision between two bodies, than in a slower one. Hence, due to, greater momentum, quicker collision, between two bodies will be more violent, even if the initial and the final velocities, are identical., , 32 Energy E =, , p2, , where p is momentum,, 2m, , m is the mass moving of the particle., Maximum energy loss occurs when the, particles get stuck together as a result of, the collision., Maximum energy loss (∆E), p2, p2, =, −, 2 m 2 (m + M ), where, (m + M ) is the resultant mass, when the particles get stuck., p2 , p2 M , m , ∆E =, 1 −, =, , , 2m , m + M 2m m + M , Also, p = mv, m2 v 2, ∴ ∆E =, 2m, , M , , , m + M , mv 2 M , =, , , 2 m + M , Comparing the expression with, 1, M, ∆E = f mv 2 , f =, 2, , m+ M
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WORK, ENERGY AND POWER, , DAY SIX, W2 =, , SESSION 2, 1 Here, m = 0.5 kg, u = 10 ms −1 , θ = 30°,, , =, , µ = 0.2 , s = 10 m, , a, , a, , ∫ 0 F ⋅ d y = ∫ 0 − k ( yi +, a, , ∫ 0 − k(ai +, $, , $, , So, coefficient of restitution, e = 1, Velocity of separation, ∴ e =1=, Velocity of approach, v2 − v1, ⇒ 1=, v0 − 0, , x$j ) $j dy, , a$j )$j dy, , a, , = − ka∫ dy = − ka2, m, , 0, , 6 Mass per unit length = M, =, , 4, = 2 kg m –1, 2, , L, , 1.4 m, , 0.6 m, , θ=30°, , = 102 + 2 ( 3.268 ) × 10, = 165.36, v = 165.36, ≈ 13 ms −1, , 2 As, both surfaces I and II are frictionless, and two stones slide from rest from the, same height, therefore, both the stones, reach the bottom with same speed, 1 mv 2 = mgh ., , , 2, , As acceleration down plane II is larger, (a2 = g sinθ2 is greater than, a1 = g sin θ1 ), therefore, stone II reaches, the bottom earlier than stone I., , 3 Given, average mass of rain drop, (m ) = 3. 0 × 10−5 kg, , Average terminal velocity = (v ), = 9 ms −1, Height (h ) = 100 cm = 1m, Density of water ( ρ) = 103 kgm −3, Area of the surface ( A ) = 1 m2 = A × h, = 1×1, = 1 m3, Mass of the water due to rain (M ), = Volume×density, = V × ρ = 1 × 103, = 10 kg, 3, , ∴ Energy transferred to the surface, 1, 1, = mv 2 = × 103 × (9)2, 2, 2, = 40. 5 × 103 J, = 4.05 × 104 J, , 4 Here, mg (h + x ) = 1 kx2, , ⇒, , 5 W1 =, =, , a, , a, , ∫ 0 F dx = ∫0 − k( yi +, a, , ∫ 0 − k(0i +, $, , $, , The mass of 0.6 m of chain = 0.6 × 2, = 1.2 kg, The height of the centre of mass of the, hanging part, 0.6 + 0, h=, = 0.3 m, 2, Hence, work done in pulling the chain, on the table = work done against the, force of gravity, i.e. W = mgh = 1.2 × 10 × 0.3 = 3.6 J, P = F ⋅v, dP, dv, So,, =F⋅, dt, dt, dP, To deliver the maximum power, = 0,, dt, which gives, Pmax = 6.26 × 103 W, , x$j ) $i dx, , x$j ) $i dx = zero, , 1, 1, v2, − mv 12 + mv 20 − 0 + v 20, 9, 2, = 2, =, 1, v 20, mv 20, 2, 8, 1, = 0.88, = − + 1 =, 9, , 9, = 0.89, m, , v0, , 12m, , m, +, , v1, , 12m, , v2, , 7 As,, , 8 As no relation between k1 and k2 is, given in the question., Then, for same force, F, F2, 1, W = Fx = F, =, ⇒W ∝, k, k, k, i.e. W1 > W2 ⇒ k1 < k2, 9 Given, dk = constant ⇒ k ∝ t, dt, ⇒, , v∝ t, , Also,, , p = Fv =, , ⇒, , 1, F∝, v, , dk, = constant, dt, 1, ⇒ F∝, t, , Similarly, for neutron-carbon atom, collision;, Momentum conservation gives;, v 0 = v 1 + 12v 2 and e = 1, ⇒, v 0 = v2 − v1, 11, So,, v1 =, v0, 13, 121, + 1 = 0.28, ∴ Loss of energy = −, 169, , So,, , Pc = 0.28, , Pd = 0.89 and, , 11 Conservation of linear momentum can be, applied but energy is not conserved., Consider the movement of two particles, as shown below :, y, m, , 2v, , v, 2m, , 10 Neutron-Deuterium collision;, , 2, kx2 = 2mg (h + x ), 2mg(h + x), k =, x2, , ⇒, , …(ii), ⇒ v 0 = v2 − v1, On adding Eqs. (i) and (ii), we get, 2v 0, = v2, 2v 0 = 3v 2 ⇒, 3, So, from Eq. (i), we get, 4v 0, v 1 = v 0 − 2v 2 = v 0 −, 3, v0, v1 = −, ⇒, 3, Fractional loss of energy of neutron, −K f + K i , =, , Ki, , for neutron, , Total work done,, W = W1 + W2 = 0 − ka2 = − ka2, , Acceleration down the plane,, a = g (sin θ − µ cos θ), = 10 (sin 30° − 0.2 cos 30° ), = 3.268 ms −2, From second equation of motion,, v 2 = u2 + 2as, , (Just before collision), , m, , 2m, , m, , 2m, , v0, , v=0, , v1, , v2, , Momentum conservation gives;, mv 0 = mv 1 + 2mv 2, ⇒, v 0 = v 1 + 2v 2, Collision given is elastic ., , 75, , vy, vx, 3m, , …(i), (Just after collision), , x
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76, , DAY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , According to conservation of linear, momentum in x-direction we have, ( p1 )x = ( p1 )x or 2mv = (2m + m )v x, 2, or v 2 = v, 3, As, conserving linear momentum in, y-direction, we get, ( p i )y = ( p t )y, 2 mv = (2m + m )v x, 2, vy = v, 3, , or, or, , Initial kinetic energy of the two, particles system is, 1, 1, E = m(2v )2 + (2m )(v )2, 2, 2, 1, 1, 2, = × 4 mv + × 2mv 2, 2, 2, = 2 mv 2 + mv 2 = 3 mv 2, , ⇒, , dv, , =−, , v2, , 15 P = constant, ⇒, , k, dt, m, , ⇒, ⇒, , dv, , =−, , k t, dt, m ∫0, , [parabolic], , t, , ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , − 1 = − k t, , , v 10, m, , ⇒, , v =, , A, , s, , ds, , 1 /3, , s1 /3, 1 /3, , 3P , k = , , M , , , , t, , ∫ 0 s1 /3 = ∫ 0 k d t, s2 /3, = kt, 2/ 3, 2, s2 /3 = k t, 3, 3 /2, 2, s = k , × t 3 /2, 3 , s ∝ t 3 /2, , or, , 16 Momentum is conserved in all type of, collisions,, Final kinetic energy is 50% more than, , gx, , B, , ds, = ks1 /3, dt, , ⇒, , Change in momentum when an element, w, gx dx, dx falls is, lg, dp w, dx w, F =, =, gx, =, x, dt, lg, dt, l, Q dx = gx , , , dt, , , –v1, , P, , ⇒, , or, , or, , 2, , 3P , v = , , M, , or, , m w, =, l, lg, , x, v 2 = 2g , 2, , s, , ∫ 0 v dv = ∫ 0 M ds, , ⇒, , ⇒, , 14 Mass per unit length,, , Velocity,, , v, , v dv , , Q a = ds , , [Assuming at t = 0 it starts from rest, i.e., from s = 0], v3, P, or, =, s, 3, M, , 1, kt, = 01, . +, v, m, 1, 1, v =, =, kt, 01, ., +, 1000, k, 01, . +, m, 1, 1, × m × v 2 = mv 20, 2, 8, v, v = 0 =5, 2, 1, =5, 01, . + 1000 k, 1 = 0.5 + 5000 k, 0.5, k =, 5000, k = 10−4 kg/m, , λ=, , Fv = P [Q P = force × velocity], Ma × v = P, P, va =, M, , P, v dv , ⇒v × , =, ds M, , v, , +v1, 3t1, , dv, −k 2, =, v, dt, m, , ⇒, , v, , t1 2t1, , −k 2, v, m, , or, , v, , 12 As we know that for vertical motion,, 1 2, gt, 2, , ∴ Acceleration, a =, , ∫10 v 2, , Loss in the energy,, ∆E = E i − E f, 4, 5, = mv 2 3 − = mv 2, 3 3, , Percentage loss in the energy during the, collision, 5, mv 2, ∆E, × 100 = 3, × 100, Ei, 3 mv 2, 5, = × 100 = 56%, 9, h=, , 13 Given force, F = − kv 2, , Now, with limits, we have, , Final energy of the combined two, particles system is, 1, E t = (3 m )(v 2x + v 2y ), 2, 4 v2 4 v2 , 1, = (3 m ) , +, , 2, 9 , 9, 3 m 8 v 2 4 mv 2, =, =, , 2 9 , 3, , = Weight of half the chain, x, + Weight of length + F, 2, w, wx wx, =, +, +, 2, 2l, l, 3x , w , 1+, =, 2 , l , , Collision is perfectly elastic, then ball, reaches to same height again and again, with same velocity., Hence, option (c) is true., , initial kinetic energy, ⇒, , 1, 1, mv 22 + mv 12, 2, 2, =, , 150 1, × mv 20, 100 2, , m, , …(i), , m, v0, , A, , Before collision, m, , y, , x, , l/2, , m, v1, , h, , B, , v2, , After collision, , Conservation of momentum gives,, t, , v = − gt and after the collision,, v = gt (straight line)., [Q v = u + gt ], , x/2, , mv 0 = mv 1 + mv 2, v 0 = v2 + v1, , Tension at A, , …(ii)
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dv, = 2t [on difference above relation], dt, as dv = a = acceleration , or a = 2t, , dt, ∴ F = ma = 6t, Hence, work done, put the value of F, and dx in Eq. (ii), we get, , From Eqs. (i) and (ii), we have, v 12, , +, , v 22, , + 2v 1 v 2 =, , or, , v 02, , − v 02, 2v 1 v 2 =, 2, , ⇒, , ∴ (v 1 − v 2 )2 = (v 1 + v 2 )2 − 4 v 1 v 2 = 2v 02, or, , v rel = 2v 0, , Apply conservation of momentum, we, get, mu1 + 0 = mv x, ⇒, mv = mv x, ⇒, vx = v, 1, , Given,, F = ax + bx2, We know that work done in stretching, the rubber band by L is, |dW |=|Fdx|, , vx, , L, , |W|= ∫ (ax + bx2 )dx, , vy, After collision, , 2, , v , 2, + v =, , 3, 2, v′=, v, 3, , L, , |W|=, , 2, , have, , ∴ D = [U ( x = ∞ ) − U at equilibrium ] =, , v, , h, , 2, , b, 4a, , Time taken to fall, t =, , 19 Here, m = 3 kg, t = 2 s,, , ∫ F ⋅ dx, , u 0 cos α $ u 0 cos α $, i +, j, 2, 2, Since, both components of v are equal., Therefore, it is making 45° with, horizontal., v=, , 4π , speed 2v will rotate 240° or, while, , 3 , other particle having speed v will rotate, 2π , 120° or, . At first collision they will, , 3 , exchange their velocities. Now as shown, in figure, after two collisions they will, again reach at point A., v, , k, x, m, , …(ii), , Differentiate Eq. (i) w.r.t. time, then we, get, dx 3t 2, or dx = t 2dt, =, dt, 3, or v = t 2, as dx = v = velocity , , dt, , d = vt =, , 1st collision, 2v, , 24 According to the principle of, conservation of momentum, v, m, m1 v 1 = m2 v 2 ⇒ 1 = 2, v2, m1, 1, 2, m1 v 1, KE1, Again,, = 2, 1, KE2, m2 v 22, 2, 2, m , m, m, = 1 × 2 = 2, m1, m2 m1 , 1, ∴, KE ∝, m, r2, , ∫r, , F ⋅ dr, , (1m , 2m , 3m), , $, , $, , 4 k$ ) ⋅, , (dx i$ + dy $j + dz k$ ), (1m , 2m , 3m), 3m , 4m), , 2kh, x, mg, , = [2 x + 3 y + 4z] (2m ,, , = −9 J, , Alternate, Since, F = constant, we can also use, W = F ⋅s, Here, s = r f − ri, = ( $i + 2$j + 3k$ ) − (2$i + 3$j + 4k$ ), , j, u02 – 2gH, , m, pi = p f, , v, 2nd collision, , ∫ (2m , 3m , 4m) (2 i + 3 j +, , equation, we have,, , u0 cos a, , 2p/3, , 1, , 22 From momentum conservation, , m, , 2v, , 2p/3, , 25 W =, , 2h, g, , 2h, x=, g, , k, x, m, , A, , 2p/3, , 2v, , So, horizontal distance travelled is, …(i), , …(ii), , 2g, , v, , aL, bL, +, 2, 3, , d, 1, 1 2, 2, mv = kx ⇒ v =, 2, 2, , …(i), , u20 sin2 α, , From Eqs. (i) and (ii), we get, , 3, , 21 From law of conservation of energy, we, , x, x, U(x = ∞) = 0, dU, 12a 6b , As,, F =−, = − 13 + 7 , dx, x , x, At equilibrium, F = 0, 2a, ∴, x6 =, b, − b2, a, b, −, =, ∴ U at equilibrium =, 2, 2, a, 4a, , 2a , , , , , b, b , , t3, 3, Work done,, W = ∫ dW =, , H =, , L, , bx3 , ax2 , =, , +, 2, 0 3 0, , aL2 a × (0)2 , =, −, , 2 , 2, b × L3 b × (0)3 , +, −, , 3 , 3, , 4 2, v, 3, , 18 U ( x ) = a12 − b6, , x=, , = (2m )v, , 0, , In y-direction, apply conservation of, momentum, we get, v, v , 0+ 0= m, − mv y ⇒ v y =, 3, 3, Velocity of second mass after collision,, v′=, , ∴ m(u 0 cos α )i$ + m( u20 − 2gH )$j, , 23 At first collision one particle having, , energy of a system corresponding to a, conservative internal force as,, U f − U i = −W = − ∫ F ⋅ dr, , 2, , 2, , 2, , 20 We know that change in potential, , Before collision, v/ 3 m, 1, , 2, , , t4 , 6, dt = 6 × = [24 − 04 ], 4 0 4, , 3, = × 16 = 24 J, 2, 2, , ∫0 6 t × t, , W =, , 17 In x-direction,, , or, , 77, , WORK, ENERGY AND POWER, , DAY SIX, , i, , ∴, , = (− i$ − $j − k$ ), W = (2$i + 3$j + 4k$ )(− $i − $j − k$ ), = − 2 − 3 − 4 = − 9J
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DAY SEVEN, , System of Particles, and Rigid Body, Learning & Revision for the Day, u, , Centre of Mass, , u, , Moment of Inertia, , u, , Rigid Bodies, , u, , Theorems on Moment of Inertia, , Moment of Inertia of Some, Geometrical Oojects, , u, , Centre of Mass, Centre of mass of a system (body) is a point that moves when external forces are applied, on the body as though all the mass were concentrated at that point and all external forces, were applied there., , Centre of Mass of Two Particle System, Centre of mass of a two particles system consisting of two particles of masses m1, m2 and, respective position vectors r1 , r2 is given by, m r + m2 r2, rCM = 1 1, m1 + m2, If m1 = m2 = m (say), then rCM =, , r1 + r2, 2, , Centre of Mass of n-Particle System, Centre of mass of rCM particles system which consists n-particles of masses m1, m2 ,..., mn, with r1, r2 ,..., rn as their position vectors at a given instant of time is given by, n, , rCM =, , m1r1 + m2 r2 + K + mn rn, =, m1 + m2 + K + mn, , Σ mi ri, , i= 1, , M, , Cartesian Components of the Centre of Mass, The position vectors rCM and ri are related to their cartesian components by, rCM = xCM i$ + yCM $j + zCM k$ and ri = x i i$ + yi $j + zi k$, The cartesian components of rCM are given by, n, , xCM =, , Σ mi x i, , i= 1, , M, , n, , n, , , yCM =, , Σ mi yi, , i= 1, , M, , and zCM =, , Σ mi zi, , i= 1, , M, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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SYSTEM OF PARTICLE AND RIGID BODY, , DAY SEVEN, , 79, , Motion of Centre of Mass, , Rigid Bodies, , The position vector rCM of the centre of mass of n particle, system is defined by, m r + m2 r2 + m3 r3 + ... + mn rn, rCM = 1 1, m1 + m2 + m3 ... mn, , A rigid body is defined as that body which does not undergo, any change in shape and volume when external forces are, applied on it. When a force is applied on a rigid body, the, distance between any two particles of the body will remain, unchanged, however, larger the forces may be., Coordinates of centre of mass of a rigid body are, 1, X CM =, x dm,, M, 1, YCM =, y dm, M, 1, and, ZCM =, z dm, M, , 1, (m1r1 + m2 r2 + m3 r3 + ... + mn rn ), M, 1 , drCM, dr, dr, dr , =, m1 1 + m2 2 + ... + mn , dt, M, dt, dt, dt , 1, Velocity of centre of mass vCM =, (m1v 1 + m2 v2 + ... mn v n ), M, =, , n, , vCM =, l, , Σ mi vi, , i= 1, , M, , Similarly, acceleration of centre of mass is given by, n, , a CM =, , l, , Centre of Mass of Some Rigid Bodies, l, , Σ mi a i, , i= 1, , The centre of mass of a uniform rod is located at its, mid-point., , M, From Newton’s second law of motion,, M a CM = F1 + F2 + ... Fn ⇒ M a CM = FExt, For an isolated system, if external force on the body is, zero., d, F = Ma CM = M, (vCM ) = 0 ⇒ vCM = constant, dt, i.e. Centre of mass of an isolated system moves with, uniform velocity along a straight line path and momentum, remain conserved., , NOTE, , CM, l, , l, , position of centre of mass of the remaining portion is, obtained from the following formula, m r − m2r2, A r − A2r2, or rCM = 1 1, rCM = 1 1, m1 − m2, A1 − A2, , Centre of mass of a uniform semi-circular ring lies at a, 2R, distance of h =, from its centre, on the axis of symmetry,, π, where R is the radius of the ring., CM, , R, , Momentum Conservation, , CM, , CM, , CM, , • If some mass or area is removed from a rigid body, then the, , Let us consider a system of particles of masses m1, m2 ... mn, are respective velocities v 1, v2 ... v n . The total linear, momentum of the system would be the vector sum of the, momentum of the individual particles., i.e. p = p1 + p2 + p3 ... pn = m1v 1 + m2 v2 + .....+ mn v n, Velocity of centre of mass of a system, 1, vCM =, (m1v 1 + m2 v2 + ...... + mn v n ), M, where, M is the total mass of the system, therefore, p = MvCM, Thus, total linear momentum of a system of particles is equal, to the product of the total mass of the system and the velocity, of its centre of mass., dp, dv, Again, =M, = Ma = Fext, dt, dt, dp, If, Fext = 0, then, = 0, i.e. p = constant, dt, If external force of a system is zero, then momentum of, system of particle remain constant., , Centre of mass of a uniform rectangular, square or circular, plate lies at its centre., , 2R, π, , O, l, , l, , l, , Centre of mass of a uniform, semi-circular disc of radius R, 4R, from, lies at a distance of h =, 3π, the centre on the axis of, symmetry as shown in figure., , CM, , R, , 4R, 3p, , O, , Centre of mass of a, hemispherical shell of radius R, R, lies at a distance of h = from, 2, its centre on the axis of, symmetry as shown in figure., Centre of mass of a solid, hemisphere of radius R lies at, 3R, a distance of h =, from its, 8, centre on the axis of, symmetry., , 3R, 8
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80, , DAY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Moment of Inertia, , 2. Theorem of Perpendicular Axes, , Moment of inertia of a rotating body is its property to oppose, any change in its state of uniform rotation., If in a given rotational system particles of masses, m1, m2 , m3 ,K be situated at normal distances r1, r2 , r3 ,... from, the axis of rotation, then moment of inertia of the system, about the axis of rotation is given by, I = m1r12 + m2 r22 + m3 r32 + ... = Σ mr 2, , The sum of moment of inertia of a plane laminar body about, two mutually perpendicular axes lying in its plane is equal to, its moment of inertia about an axis passing through the point, of intersection of these two axes and perpendicular to the, plane of laminar body., Z, Planar Body, , For a rigid body having continuous mass distribution, I = ∫ dm r 2, , O, 2, , SI unit of moment of inertia is kg m . It is neither a scalar nor, a vector i.e. it is a tensor., , Radius of gyration of a given body about a given axis of, rotation is the normal distance of a point from the axis, where, if whole mass of the body is placed, then its moment of inertia, will be exactly same as it has with its actual distribution of, mass. Thus, radius of gyration, r 2 + r22 + r32 + K + rn2 , I, or K = 1, , M, n, , , , m1, m2, , r1, , r3, r4, , r2, , N, , Y, P(x, y), , X, , Radius of Gyration, , K =, , M, , 1/ 2, , Moment of Inertia of Some, Geometrical Objects, , m3, m4, , K, , If I x and I y be moment of inertia of the body about two, perpendicular axes in its own plane and I z be the moment of, inertia about an axis passing through point O and, perpendicular to the plane of lamina, then, Iz = Ix + I y, In theorem of perpendicular axes, the point of intersection of, the three axes ( x, y and z) may be any point on the plane., , l, , M, , P, , Axis of rotation, , l, , SI unit of radius of gyration is metre., Radius of gyration depends upon shape and size of the body,, position and configuration of the axis of rotation and also on, distribution of mass of body w.r.t. axis of rotation., , l, , Theorems on Moment of Inertia, There are two theorems based on moment of inertia are given, below:, , l, , 1. Theorem of Parallel Axes, Moment of inertia of a body about a, given axis I is equal to the sum of, moment of inertia of the body about a, parallel axis passing through its centre of, mass ICM and the product of mass of, body (M ) and square of normal distance, between the two axes. Mathematically,, I = ICM + Md2, , ICM, , I, , A, , CM, l, , d, , Uniform Ring of Mass M and Radius R About an axis, passing through the centre and perpendicular to plane of, 1, ring I = MR2 . About a diameter I = MR2, 2, Uniform Circular Disc of Mass M and Radius R About an, axis passing through the centre and perpendicular to plane, 1, 1, of disc I = MR2 . About a diameter I = MR2, 2, 4, Thin Uniform Rod of Mass M and Length l About an axis, passing through its centre and perpendicular to the rod,, 1, I =, Ml 2, 12, Uniform Solid Cylinder of Mass M, Length l and Radius R, About its own axis,, 1, I = MR2 . About an axis passing through its centre and, 2, l 2 R2 , perpendicular to its length I = M +, , 4, 12, 2, Uniform Solid Sphere About its diameter I = MR2 ., 5, 7, About its tangent I = MR2, 5
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81, , SYSTEM OF PARTICLE AND RIGID BODY, , DAY SEVEN, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A body A of mass M while falling vertically downwards, under gravity breaks into two parts; a body B of mass, 1, 2, M and, a body C of mass M. The centre of mass of, 3, 3, bodies B and C taken together shifts compared to that of, body A towards, (a) depends on height of breaking, (b) does not shift, (c) body C, (d) body B, , (a), , from its vertex is z 0. If the radius of its base is R and its, ª JEE Main 2015, height is h, then z 0 is equal to, (a), , h, 4R, , (b), , 3h, 4, , (c), , 5h, 8, , 2, , (d), , 3h, 8R, , 3 A circular disc of radius R is removed from a bigger, circular disc of radius 2 R, such that the circumference of, the discs coincide. The centre of mass of the new disc is, α R from the centre of the bigger disc. The value of α is, (a), , 1, 3, , 1, 2, , (b), , (c), , 1, 6, , l, 2, , (b) l, , (d), , 1, 4, , 4 Which of the following points is the likely position of the, centre of mass of the system as shown in figure?, Hollow sphere, Air, , (a) 0.273 kg- m2, (c) 0.173 kg- m2, , (b) 1.73 kg- m2, (d) 2.73 kg- m2, , 8 The surface density of a circular disc of radius a depends, on the distance as ρ(r ) = A + Br . The moment of inertia, about the line perpendicular to the plane of the disc is, A 2B , (b) πa 4 +, , 2, 5 , , A 2a , (a) πa 4 +, B, 2, 5 , A Ba , (c) 2 πa 3 +, , 2, 5 , , (d) None of these, , 9 Four point masses, each of value m, are placed at the, corners of a square ABCD of side l. The moment of inertia, of this system about an axis passing through A and, parallel to BD is, (b) 3 ml 2, , D, , (a) 1 : 2, , (b) 5 : 6, , (c) 2 : 3, , (d), , Sand, , (c) C, , (d) D, , 5 Consider a two particles system with particles having, masses m1 and m2. If the first particle is pushed towards, the centre of mass through a distance d, by what, distance should the second particle be moved, so as to, keep the centre of mass at the same position?, (a), , m2, d, m1, , (b), , (c) 3 ml 2, , (d) ml 2, , circular ring of the same radii about a tangential axis, perpendicular to plane of disc or ring is, , R/1, , (b) B, , 3l, 2, , 10 The ratio of the radii of gyration of a circular disc and a, , A, B, C, , (a) A, , (d), , of mass 5 g. The radius of the wheel is 40 cm. The, moment of inertia, , (a) 2 ml 2, , R/2, , (c) 2l, , 7 A wheel has mass of the rim 1 kg, having 50 spokes each, , 2 Distance of the centre of mass of a solid uniform cone, 2, , string passed through the hand of the man while the, cylinder reached his hands is of, , m1, m, d (c) − 1 d, m1 + m2, m2, , 3, 2, , 11 A uniform square plate has a small piece Q of an irregular, shape removed and glued to the centre of the plate, leaving a hole behind. The moment of inertia about the, z-axis, then, Y, , Y, Q, , Hole, , (d) d, X, , 6 A string of negligible thickness is wrapped several times, around a cylinder kept on a rough horizontal surface. A, man standing at a distance lm from the cylinder holds, one end of the string and pulls the cylinder towards him., There is no slipping anywhere. The length in (m ) of the, , (a) increased, (b) decreased, (c) same, (d) changed in unpredicted manner, , X
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82, , DAY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 12 Consider a uniform square plate of side a and mass m., The moment of inertia of this plate about an axis, perpendicular to its plane and passing through one of its, corners is, (a), , 5, ma 2, 6, , (b), , 1, ma 2, 12, , (c), , 7, ma 2, 12, , (d), , 2, ma 2, 3, , 13 For the given uniform square lamina ABCD, whose centre, is O., , The moment of inertia of the system of four spheres about, diagonal AB is, m, (8r 2 + 5 a 2 ), 5, m, (c) (5 r 2 + 8 a 2 ), 5, , m, (7r 2 + 4 a 2 ), 5, m, (d) (3 r 2 + 5 a 2 ), 5, , (a), , (b), , 17 From a uniform circular disc of radius R and mass 9 M, a, R, is removed as shown in the figure., 3, The moment of inertia of the remaining disc about an axis, perpendicular to the plane of the disc and passing, through centre of disc is, ª JEE Main 2018, small disc of radius, , F, , D, , C, , O, , A, , 2R, 3, , B, , E, , (a) 2 I AC = IEF, , (b) I AD = 3IEF, , (c) I AC = IEF, , (d) I AC =, , R, , 2 IEF, , 14 A solid sphere of radius R has moment of inertia I about, its geometrical axis. It is melted into a disc of radius r, and thickness t. If it’s moment of inertia about the, tangential axis (which is perpendicular to plane of the, disc), is also equal to I, then the value of r is equal to, I, , (a) 4MR 2, , (b), , 40, MR 2, 9, , (c) 10MR 2, , (d), , 37, MR 2, 9, , 18 A child with mass m is standing at the edge of a, merry-go-round having moment of inertia I, radius R and, initial angular velocity ω 0 as shown in the figure., v, ω0, , R, , (a), , 2, R, 15, , (b), , 2, R, 5, , (c), , 3, R, 15, , (d), , 3, R, 15, , 15 The moment of inertia of a system of four rods each of, length l and mass m about the axis shown is, , The child jumps off the edge of the merry-go-round with, tangential velocity v w.r.t. ground. The new angular, velocity of the merry-go-round is, 1, , (a), , 2 2, ml, 3, , I1, , I4, , I2, , I3, , (b) 2 ml 2, , (c) 3 ml 2, , I ω2 − mv 2 2, (a) 0, , I, , , (c) , , (d), , 8 2, ml, 3, , 16 Four solid spheres each of mass m and radius r are, located with their centres on four corners of a squares, ACBD of side a as shown in the figure., A, , C, , a, D, , B, , I ω0 − mvR , , , I, , 1, , (I + mR 2 )ω20 − mv 2 2, (b) , , I, , , 2, (I + mR ) ω0 − mvR , (d) , , I, , , , Direction (Q. Nos. 19-20) are the Assertion and Reason, type. Each of the these contains two Statements; Assertion, and Reason. Each of these question also has four alternative, choice,only one of which is correct. You have to select the, correct choices from the codes (a),(b), (c) and (d) given below:, (a) If both Assertion and Reason are true and the Reason is, correct explanation of the Assertion, (b) If both Assertion and Reason are true but Reason is not, correct explanation of the Assertion, (c) If Asserion is true but Reason is false, (d) If both Assertion and Reason are false
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SYSTEM OF PARTICLE AND RIGID BODY, , DAY SEVEN, 19 Assertion (A) A constant force F is applied on the two, blocks and one spring system as shown in the figure., Velocity of centre of mass increases linearly with time., m, , F, , 2m, , 83, , the axis is moved downwards, moment of inertia of the, slab will first decreases, then increases., Reason (R) Axis is first moving towards its centre of, mass, then it is receding from it., A, , Smooth, , Reason (R) Acceleration of centre of mass is constant., , 20 Assertion (A) There is a triangular plate as shown in the, , B, , C, , figure A dotted axis is lying in the plane of the slab. As, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 Mass of bigger disc having radius 2R is M. A disc of, radius R is cut from bigger disc as shown in figure., Moment of inertia of disc about an axis passing through, periphery and perpendicular to plane is, , 4 Seven identical circular planar discs, each of mass M, and radius R are welded symmetrically as shown in the, figure. The moment of inertia of the arrangement about, the axis normal to the plane and passing through the, point P is, , 2R, P, , M, , (a), , 27 MR 2, 8, , m, R, , (b), , O, , 29 MR 2, 8, , (d) 2MR 2, , (c) 3.5 MR, , 2 Four spheres of diameter 2a and mass M are placed with, their centres on the four corners of a square of side b., Then the moment of inertia of the system about an axis, along one of the sides the square is, , 55, (b), MR 2, 2, 181, (d), MR 2, 2, , 5 The moment of inertia of a uniform cylinder of length l and, , ª JEE Main 2017 (Offline), , b, , 3, 2, 3, (c), 2, (a), , C, , D, , 4, Ma 2 + 2Mb 2, 5, 8, (c) Ma 2, 5, (a), , 8, Ma 2 + 2Mb 2, 5, 4, (d) Ma 2 + 4 Mb 2, 5, , (b), , 3 From a solid sphere of mass M and radius R, a cube of, maximum possible volume is cut. Moment of inertia of, cube about an axis passing through its centre and, ª JEE Main 2015, perpendicular to one of its faces is, (a), , MR 2, 32 2 π, , (b), , 2, , (c), , 4 MR, 9 3π, , ª JEE Main 2018, , radius R about its perpendicular bisector is I. What is the, ratio l/R such that the moment of inertia is minimum?, , B, , A, , 19, (a) MR 2, 2, 73, (c), MR 2, 2, , (d), , 4 MR, 3 3π, , 3, 2, , 6 A particle moves parallel to X-axis with constant velocity v, as shown in the figure. The angular velocity of the particle, about the origin O is, Y, , v, , MR 2, 16 2 π, 2, , (d), , (b) 1, , q, O, , X, , (a) remains constant, (b) continuously increasing, (c) continuously decreasing (d) oscillates
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84, , DAY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 7 Let I1 and I 2 be the moment of inertia of a uniform square, , (a), , 1, 2, , k m2, , x0, km2, m1 + m2, km1, (d) x 0, m1 + m2, (b), , m1 + m2, m + m2 , (c) x 0k 1, , m2 , , plate about an axis as shown in the figure. Then, the ratio, I1 : I 2 is, , 10 A racing bike is travelling along a straight track at a, constant velocity of 40 m/s. A fixed CCTV camera is, recording the event as shown in the figure. In order to, keep the bike in view, in the position shown, the angular, velocity of the CCTV camera should be, , (a) 1 :, , 1, 7, , (b) 1 :, , 12, 7, , (c) 1 :, , 7, 12, , (d) 1 : 7, , 8 A thin rod of length 4l, mass 4m is bent at the points as, , 40 m/s, , shown in the figure. What is the moment of inertia of the, rod about the axis passing through O and perpendicular, to the plane of the paper?, O, 90º, , 30m, 30º, , l, 90º, , l, , ml 2, (a), 3, , CCTV, , 10 ml 2, (b), 3, , ml 2, (c), 12, , (a) 3 rad/s, (b) 2 rad/s, (c) 1 red/s, (d) 4 rad/s, , ml 2, (d), 24, , 9 Two bars of masses m1 and m2 connected by a, weightless spring of stiffness k, rest on a smooth, horizontal plane. Bar 2 is shifted by a small distance, x 0 to the left and released. The velocity of the centre of, mass of the system when bar 1 breaks off the wall is, , 1, m1, , 11 A rod of length L is placed along the X-axis between, x = 0 and x = L. The linear mass density (mass/length) ρ, of the rod varies with the distance x from the origin as, ρ = a + bx. Here, a and b are constants. The position of, centre of mass of the rod is, , 2, m2, , 3aL + 2bL2, , (a) , , 0, 0, a, +, bL, 6, 3, , , , 6aL + 3bL2, , (b) , , 0, 0, a, +, bL, 3, 2, , , , aL + bL2, , (c) , , 0, 0, 2, 3, a, +, bL, , , , (d) None of these, , ANSWERS, SESSION 1, , 1 (b), 11 (b), , 2 (b), 12 (d), , 3 (a), 13 (c), , 4 (c), 14 (a), , 5 (c), 15 (d), , 6 (c), 16 (a), , 7 (c), 17 (a), , 8 (a), 18 (d), , 9 (c), 19 (a), , 10 (d), 20 (a), , SESSION 2, , 1 (b), 11 (a), , 2 (b), , 3 (c), , 4 (d), , 5 (d), , 6 (c), , 7 (d), , 8 (b), , 9 (a), , 10 (c)
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SYSTEM OF PARTICLE AND RIGID BODY, , DAY SEVEN, , 85, , Hints and Explanations, SESSION 1, 1 The position of centre of mass remains, unaffected because breaking of mass, into two parts is due to internal forces., , 2 We know that centre of mass of a, uniform solid cone of height (h ) is at, h, height from base, therefore, 4, h, h 3h, As, h − z 0 =, or z 0 = h − =, 4, 4, 4, , 3 In this question, distance of centre of, mass of new disc from the centre of, mass of remaining disc is αR., , O2, , O1, , αR, , 6 If velocity of, , Mass of remaining disc, , 3M, M, α R+, R=0, 4, 4, 1, α =, ⇒, 3, Note In this question, the given distance, must be αR for real approach to the, solution., , 12 Moment of inertia of square plate about, ma2, . Moment of inertia about zz′, 6, can be computed using parallel axes, theorem, xy is, , y, , z¢, , that part of the system, having bigger, mass. In the given diagram, lower part, is heavier, hence CM of the system lies, below the horizontal diameter., Hence, (c) is correct option., , 5 To keep the centre of mass at the same, position, velocity of centre of mass is, zero, so, m1 v1 + m 2 v2, =0, m1 + m 2, where, v1 and v2 are velocities of, particles 1 and 2 respectively., dr, dr, ⇒ m1 1 + m2 2 = 0, dt, dt, Q v = dr1 and v = dr2 , 2, 1, dt, dt , , z, , a, , x, , 8 dm = 2 πr dr (ρ) = ( A + Br ) (2 πr dr ), π Aa4, 2 πBa5, +, 2, 5, 0, A 2a , = πa4 +, B, 2, 5 , , a, I zz ′ = I xy + m , , 2, , a, , I = ∫ dmr 2 =, , 2, , ma2 ma2, +, 6, 2, 2ma2, =, 3, =, , 9 The situation is as shown in figure., x, , 13 Let the each side of square lamina, , Q, A, , 4 Centre of mass of a system lies towards, , m1 dr1 + m2 dr2 = 0, , v, , perpendicular axes, I z = I x + I y with the, hole, I x and I y , both decreases (gluing the, removed piece at the centre of square, plate does not, affect I z )., Hence, I z decreases overall., , 3, 50 (5 × 10−3 ) (0.4)2, = 1 × (0. 4) +, 3, = 016, . (1.083), kg- m2, = 0173, ., , M 3M, =, 4, 4, , ∴ −, , ⇒, , 2v, , centre of mass, is v, then, velocity of, contact is 0, and that of the, top is 2 v,, hence when, centre of mass covers a distance l,, thread covers a distance 2l., , 11 According to the theorem of, , 2, , Mass of original disc = M, Mass of disc removed, M, M, =, × π R2 =, 2, 4, π (2R), =M −, , Let second particle has been displaced, by distance x., md, ⇒ m1 (d ) + m2 ( x ) = 0 or x = − 1, m2, , 3, 2 = 3, 2, 2, , K disc, =, K ring, , 2, 7 I = mr 2 + 50 ml, , R, , O, , dr1 and dr2 represent the change in, displacement of particles., , is d., So,, and, , B, , I EF = IGH, I AC = I BD, , D, , P, , (due to symmetry), (due to symmetry), , F, , C, , x¢, , I xx ′ = m × DP + m × BQ + m × CA, 2, 2l, = m×2× , + m × ( 2 l )2, 2 , 2, , = 3 ml 2, , ⇒, , O, , H, , 2, , A, , E, d, , B, , Now, according to theorem of, perpendicular axes,, , 10 Radius of gyration,, K =, , G, , C, , D, 2, , l, m, , I AC + I BD = I 0, ⇒, , K disc =, , 1, mR2 + nR2, 2, =, R, , K ring =, , mR + mR, = 2R, m, 2, , 2, , 3, R, 2, , and, ⇒, , 2 I AC = I 0, , …(i), , I EF + IGH = I 0, 2 I EF = I 0, , From Eqs. (i) and (ii), we get, I AC = I EF, , …(ii)
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86, , DAY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 2 MR2 = 1 Mr 2 + Mr 2, 5, , F, × t or v CM ∝ t, 3m, Hence, both Assertion and Reason are, true and the Reason is the correct, explanation of Assertion., , or v CM =, , 2, , 2, 3, MR2 = Mr 2, 5, 2, 2, ∴ r =, R, 15, or, , distance from the axis of rotation to the, centre of mass. When dotted axis moved, downward (towards centre of mass), r, decreases result moment of inertia, decrease and when dotted axis cross the, centre of mass and moved further, downwards then r increases result, moment of inertia increases. Hence,, both Assertion and Reason are true and, the Reason is the correct explanation of, Assertion., , l2, l2 , 4m , +, about, 12 12 , 2ml, 3, , Q, , 20 Moment of inertia (I ) = mr 2 , where r is, , 15 Consider a square lamina, then, , Centre of mass =, , S, , 2, , Ö2 l, l/Ö2, , R, M, , 2R R, O, , v, , Cross-sectional view of the, cube and sphere, Using geometry of the cube, 2R, PQ = 2R = ( 3 ) a or a =, 3, Volume density of the solid sphere,, Y, I, , P, R, , S, , a, O, , SESSION 2, , W, , V, Q, , Apply perpendicular axis theorem,, 2, 2ml 2, l , 8, 2, =, + 4m , = ml, 2, 3, 3, , 1 Surface density of motional disc is, σ =, , four spheres about diagonal AB is, I AB = MI of A about AB + MI of B, about AB + MI of C, about AB + MI of D about AB, 2, 2, 2, 1, = mr 2 + mr 2 + mr 2 + ma2 , 5, , 5, 5, 2, 2, 1, , 2, 2, + mr + ma , 5, , 2, 8r, , 8, mr 2 + ma2 = m , + a2 , 5, 5, , m, 2, 2, =, ( 8 r + 5a ), 5, =, , I2 = Moment of inertia due to, cutting portion, I =, , 2, , M (2R )2, m R2, , + M (2R )2 − 1, + m1 (3R )2 , 2, 2, , 19MR 2 29MR 2, 2, = 6MR −, =, 8, 8, , 2, , a, , 17 Moment of inertia of remaining solid, , = Moment of inertia of complete, solid − Moment of inertia of, removed portion, 2, 9MR2 M (R / 3)2, 2R , ∴ I =, −, + M , , 3 , 2, 2, , , 2, ⇒ I = 4MR, , 18 Since, in this condition,, Initial angular momentum, = Final angular momentum, ∴(I + mR2 ) ω 0 = (mvR ) + I ω ′, (I + mR2 )ω 0 − mvR, or, ω′ =, I, Hence, (d) is the correct option., , 19 Total mass of the system, , = m + 2m = 3m, Force applied on the system is F ., F, ∴ a CM =, 3m, = constant as F is constant, ∴ v CM = a CM × t, , b, , A, b, , D, , a, , a, , B, , b, , 3M, 8R3, 2M, ×, =, 3, 4 πR, 3 3, 3π, Moment of inertia of the cube about, given axis is, m 2, ma2, ly =, (a + a2 ) =, 12, 6, 2M, 1 4R2 4MR2, ma2, =, × ×, =, ⇒ Iy =, 6, 3, 3π 6, 9 3π, =, , 4 Moment of inertia of an outer disc about, the axis through centre is, MR2, 9, 1, =, + M (2R )2 = MR2 4 + = MR2, , 2, 2 2, , a, b, , M, , C, , Moment of inertia of each of the spheres, A and D about, 2, AD = Ma2, 5, Moment of inertia of each of the spheres, B and C about AD, 2, = Ma2 + Mb 2 , 5, , Using theorem of parallel axis, we get, Total moment of inertia, 2, 2, = I Ma2 × 2 + Ma2 + Mb 2 × 2, 5, , 5, , I =, , M, 3 M, =, , 4, 3, π R3 , 3, πR, 3, Mass of cube (m ) = (ρ)(a)3, 3, M 2R, 3, = , × 3 , 4 π R 3 , ρ=, , M, M, =, π(2R )2, 4 πR 2, , Mass of cutting portion is, M, m1 = σ × πR2 =, 4, I = I1 − I2, where,, I1 = Moment of inertia of disc about, given axis without cutting portion, , 16 The moment of inertia of the system of, , V, , 8, Ma2 + 2Mb 2, 5, , 3 Consider the cross-sectional view of a, diametric plane as shown in the, adjacent diagram., , R, , R, , For 6 such discs,, 9, MR2, 2, = 27MR2, So, moment of inertia of system, MR2, 55, =, + 27MR2 =, MR2, 2, 2, 55, Hence, I P =, MR2 + (7M × 9R2 ), 2, 181, IP =, MR2, ⇒, 2, 181, I system =, MR2, 2, Moment of inertia = 6 ×
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DAY EIGHT, , Rotational, Motion, Learning & Revision for the Day, u, , u, , u, , Concept of Rotational Motion, Equation of Rotational Motion, Moment of Force or Torque, , u, , u, , Angular Momentum, Law of Conservation, of Angular Momentum, , u, , u, , Equilibrium of a Rigid, Bodies, Rigid Body Rotation, , Concept of Rotational Motion, In rotation of a rigid body about a fixed axis, every particle of the body moves in a circle,, which lies in a plane perpendicular to the axis and has its centre on the axis., l, , l, , Rotational motion is characterised by angular displacement dθ and angular velocity, dθ, ω= ., dt, If angular velocity is not uniform, then rate of change of angular velocity is called the, angular acceleration., dω, Angular acceleration, α =, ., dt, SI unit of angular acceleration is rad/s2., , l, , Angular acceleration α and linear tangential acceleration a t are correlated, as a t = α × r., , Equation of Rotational Motion, If angular acceleration α is uniform, then equations of rotational motion may be written, as, 1, (ii) θ = ω 0 t + α t 2, (i) ω = ω 0 + α t, 2, α, 2, 2, (iii) ω − ω 0 = 2 αθ, (iv) θ nth = ω 0 + (2 n − 1), 2, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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ROTATIONAL MOTION, , DAY EIGHT, , Moment of Force or Torque, Torque (or moment of a force) is the turning effect of a force, applied at a point on a rigid body about the axis of rotation., Mathematically, torque, τ = r × F = | r × F|^, n, n = r F sin θ ^, where, ^, n is a unit vector along the axis of rotation. Torque is, an axial vector and its SI unit is newton-metre (N-m)., The torque about axis of rotation is, F, independent of choice of origin O, so long, as it is chosen on the axis of rotation AB., O r, Only normal component of force, θ A, contributes towards the torque. Radial, B, component of force does not contribute, towards the torque., A torque produces angular acceleration, in a rotating body. Thus, torque, τ = Iα, Moment of a couple (or torque) is given by product of, position vector r between the two forces and either force F., Thus, τ = r × F, If under the influence of an external torque, τ the given, body rotates by dθ, then work done, dW = τ ⋅ dθ., In rotational motion, power may be defined as the scalar, product of torque and angular velocity, i.e. Power P = τ ⋅ ω., l, , l, , 89, , Law of Conservation, of Angular Momentum, According to the law of conservation of angular momentum, if, no external torque is acting on a system, then total vector sum, of angular momentum of different particles of the system, remains constant., dL, We know that,, = τext, dt, dL, Hence, if τext = 0, then, = 0 ⇒ L = constant, dt, Therefore, in the absence of any external torque, total angular, momentum of a system must remain conserved., Comparison of Linear and Rotational Motion, , l, , l, , l, , l, , The moment of linear momentum of a given body about an axis, of rotation is called its angular momentum. If p=mv be the, linear momentum of a particle and r is its position vector from, the point of rotation, then, Angular momentum, L = r × p = r psin θ ^, n = mvr sin θ n$, where, ^, n is a unit vector in the direction of rotation. Angular, momentum is an axial vector and its SI unit is kg-m2s −1 or J-s., , l, , l, , For rotational motion of a rigid body,, angular momentum is equal to the product, of angular velocity and moment of inertia of, the body about the axis of rotation., Mathematically, L = Iω., , l, , P, , r, , According to the second law of rotational, motion, the rate of change of angular, momentum of a body is equal to the external torque, applied on it and takes place in the direction of torque., Thus,, dL d, dω, Q α = dω , τ=, =, (Iω) = I, = Iα, , dt , dt dt, dt, Total effect of a torque applied on a rotating body in a, given time is called angular impulse. Angular impulse is, equal to total change in angular momentum of the system, in given time. Thus, angular impulse,, J =∫, , ∆t, 0, , τ dt = ∆L = L f − Li, , The angular momentum of a system of particles about the, n, , origin is, , L = ∑ ri × pi, , Rotational Motion, , 2. Force, F = ma, , Torque, T = Iα, , 3. Electric energy, E = 1 mv 2, 2, , Rotational energy, E =, , 1 2, Iω, 2, , Equilibrium of a Rigid Bodies, For mechanical equilibrium of a rigid body, two condition, need to be satisfied., , Angular Momentum, , l, , Linear Motion, , 1. Linear momentum, p = mv Angular momentum, L = Iω, L = 2 IE, , 1. Translational Equilibrium, A rigid body is said to be in translational equilibrium, if it, remains at rest or moving with a constant velocity in a, particular direction. For this, the net external force or the, vector sum of all the external forces acting on the body must, be zero,, i.e., , F = 0 or F = ΣFi = 0, , 2. Rotational Equilibrium, A rigid body is said to be in rotational equilibrium, if the body, does not rotate or rotates with constant angular velocity. For, this, the net external torque or the vector sum of all the, torques acting on the body is zero., For the body to be in rotational equilibrium,, dL, = 0 or ∑ τ i = 0, τext = 0,, dt, , Rigid Body Rotation, Spinning, When the body rotates in such a manner that its axis of rotation, does not move, then its motion is called spinning motion., 1, In spinning rotational kinetic energy is given by, K R = Iω2 ., 2, Rotational kinetic energy is a scalar having SI unit joule (J)., Rotational kinetic energy is related to angular momentum as, per relation,, L2, or L = 2 IK R, KR =, 2I
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90, , DAY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , Pure Rolling Motion, , l, , Let a rigid body, having symmetric surface, about its centre of mass, is being spined at a, certain angular speed and placed on a surface,, so that plane of rotation is perpendicular to, the surface. If the body is simultaneously, given a translational motion too, then the net, motion is called rolling motion., , Velocity at the lowest point v =, , ω, v, , CM, , l, , Acceleration in motion From second equation of motion,, v2 = u2 + 2 as, By substituting u = 0, s =, , The total kinetic energy in rolling motion,, a=, , K N = K R + KT, =, KN, , K2 , 1, 1, 1, 1, mv2 + Iω2 = mv2 + mv2 2 , 2, 2, 2, 2, R , , l, , , 1, K2 , = mv2 1 + 2 , 2, R , , , g sin θ, K2, 1+ 2, R, , 2 gh, , we get, K2, 1+ 2, R, , Time of descent From first equation of motion, v = u + at, , expressions t =, , 1, , v∝, P, , Rω, , K2, R2, , 1+, , v, , ;a ∝, , 1, K2, ;t ∝ 1 + 2, 2, K, R, 1+ 2, R, , Important Terms Related to Inclined Plane, , Impure Rolling Motion, , l, , In impure rolling motion, the point of contact of the body with, the platform is not relatively at rest w.r.t. platform on which,, it is performing rolling motion, as a result sliding occurs at, point of contact., If platform is stationary, i.e. v0 = 0, then v ≠ Rω, , ⇒, A, , v = Rω, , v0, , Rolling on an Inclined Plane, R, S, , Velocity, acceleration and time of descent (for a given, K2, inclined plane) all depends on 2 . Lesser the moment of, R, K2, inertia of the rolling body lesser will be the value of 2 . So,, R, greater will be its velocity and acceleration and lesser will, be the time of descent., , Rotation about Axis through, Centre of Mass (Centroidal Rotation), , v0, , B, , K2 , Here, factor 2 is a measure of moment of inertia of a, R , body and its value is constant for given shape of the body, and it does not depend on the mass and radius of a body., , l, , For impure rolling motion, v AB ≠ 0 i.e. v − Rω ≠ v0, , When a body of mass m and radius R, rolls down on inclined plane of, height h and angle of inclination θ, it, loses potential energy. However, it, acquires both linear and angular, speeds and hence gain kinetic energy, of translation and that of rotation., , K2 , 1, 2h , 1 + 2 , sin θ g , R , , From the above expressions, it is clear that,, , If the given body rolls over a surface such, that there is no relative motion between the, body and the surface at the point of contact,, then the motion is called rolling without, slipping., , v, , h, and v =, sin θ, , By substituting u = 0 and value of v and a from above, , Rolling Without Slipping, , ω, , 2 gh, K2, 1+ 2, R, , h, , When an object of given mass is tied to a, light string wound over a pulley whose, moment of inertia is I and radius R as shown, in the figure. The wheel bearing is, frictionless and the string does not slip on, the rim, then, , R, O, T, m, , Tension in the string is, θ, , By conservation of mechanical energy, mgh =, , T =, , K2 , 1, mv2 1 + 2 , 2, R , , , and acceleration, a =, , I, × mg, I + mR2, mR2, ⋅g, I + mR2, , mg, , a
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ROTATIONAL MOTION, , DAY EIGHT, , 91, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 When a body is projected at an angle with the horizontal, in the uniform gravitational field of the earth, the angular, momentum of the body about the point of projection, as it, proceeds along the path, (a) remains constant, (b) increases continuously, (c) decreases continuously, (d) initially decreases and after reaching highest point, increases, , 2 A force of −F k$ acts on O, the origin of the coordinate, system. The torque about the point (1, − 1) is, z, , y, , O, x, , (a) F ($i − $j), , (b) −F ($i + $j) (c) F ($i + $j), , (d) −F ($i − $j), , 3 Angular momentum of the particle rotating with a central, force is constant due to, (a) constant force, (c) zero torque, , (b) constant linear momentum, (d) constant torque, , 4 A particle of mass m = 3 kg moves along a straight line, 4y − 3x = 2, where x and y are in metre, with constant, velocity v = 5 ms −1. The magnitude of angular, momentum about the origin is, (a) 12kg - m2 s−1, (c) 6.0 kg - m2 s−1, , (b) 8.0 kg - m2 s−1, (d) 4.5 kg - m2 s−1, , 5 A particle of mass m moves along line PC with velocity v, as shown in the figure. What is the angular momentum of, the particle about O?, C, , 7 A bob of mass m attached to an inextensible string of, length l is suspended from a vertical support.The bob, rotates in a horizontal circle with an angular speed, ω rad/s about the vertical support. About the point of, suspension, ª JEE Main 2014, (a) angular momentum is conserved, (b) angular momentum changes in magnitude but not in, direction, (c) angular momentum changes in direction but not in, magnitude, (d) angular momentum changes both in direction and, magnitude, , 8 A particle of mass 2 kg is moving such that at time t, its, position, in metre, is given by r (t ) = 5 $i − 2 t 2 $j. The, angular momentum of the particle at t = 2 s about the, ª JEE Main (Online) 2013, origin in kg m −2 s −1 is, (a) − 80 k$, (b) (10 $i − 16 $j ), (c) − 40 k$, (d) 40 k$, , 9 A uniform disc of radius a and mass m, is rotating freely, with angular speed ω in a horizontal plane, about a, smooth fixed vertical axis through its centre. A particle,, also of mass m, is suddenly attached to the rim of the disc, and rotates with it. The new angular speed is, , 10 If the radius of the earth contracts, , O, , (b) mv l, , (a) zero, (b) 30, (c) 30 2, (d) 30 10, , 1, of its present day, n, value, length of the day will be approximately., , r, , l, , (a) mvL, , 3 2 cm s –1 is in the xy-plane along the line y = 2 5 cm., The magnitude of its angular momentum about the origin, in g- cm 2 s – 1 is, , (a) ω / 6, (b) ω / 3, (c) ω / 2, (d) ω /5, , L, P, , 6 A particle of mass 5 g is moving with a uniform speed of, , 24, h, n, (c) 24 nh, , (a), (c) mvr, , (d) zero, , 24, h, n2, (d) 24 n 2 h, , (b)
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92, , DAY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , 11 A particle of mass m is moving along the side of a square, of side a , with a uniform speed v in the X-Y plane as, shown in the figure., Y, , a, v, , D, a v, , C, v a, , v, a, , A, , B, , R, O, , 45º, , X, , Which of the following statements is false for the angular, momentum L about the origin?, ª JEE Main 2016 (Offline), mv $, (a) L = −, R k, when the particle is moving from A to B., 2, R, (b) L = mv , + a k$ , when the particle is moving from B, 2, , to C., R, (c) L = mv , − a k$ , when the particle is moving from C, 2, , to D., mv $, (d) L =, R k, when the particle is moving from D to A., 2, , 12 A uniform rod AB of mass m and length l at rest on a, smooth horizontal surface. An impulse P is applied to the, end B. The time taken by the rod to turn through at right, angle is, (a) 2 π, , ml, P, , (b) 2, , πP, ml, , (c), , π ml, 12 P, , (d), , πP, ml, , 13 A solid cylinder of mass 20 kg rotates about its axis with, angular speed 100 rad/s. The radius of the cylinder is, 0.25 m. Then the kinetic energy associated with the, rotation of the cylinder and the magnitude of angular, momentum of the cylinder about its axis is respectively, (a) 3200 J, 62.5 J-s, (c) 3500 J, 68 J-s, , (b) 3125 J, 62.5 J-s, (d) 3400 J, 63.5 J-s, , slipping with the same linear velocity. If the kinetic energy, of the ring is 8 J, that of the disc must be, (b) 4 J, , (c) 6 J, , (d) 16 J, , 15 A round uniform body of radius R, mass M and moment, of inertia I, rolls down (without slipping) an inclined plane, making an angle θ with the horizontal. Then, its, acceleration is, (a), , g sin θ, 1 + I /MR, , 2, , (b), , g sin θ, 1 + MR /I, 2, , (c), , g sin θ, 1 − I /MR, , 2, , (d), , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 16 Statement I A disc A moves on a smooth horizontal plane, and rebounds elastically from a smooth vertical wall (Top, view is as shown in the figure) in this case about any, point on line xy. The angular momentum of the disc, remains conserved., A, y, , x, , Statement II About any point in the plane, the torque, experienced by disc is zero as gravity force and normal, contact force balance each other., , 17 Statement I A block is kept on a rough horizontal, surface, under the action of a force F as shown in the, figure. The torque of normal contact force about centre of, mass is having zero value., , F, C, , Statement II The point of application of normal contact, force may pass through centre of mass., , 14 A ring and a disc having the same mass, roll without, , (a) 2 J, , Direction (Q. Nos. 16-18) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given here., , g sin θ, 1 − MR 2 /I, , 18 Statement I Two cylinders, one hollow (metal) and the, other solid (wood) with the same mass and identical, dimensions are simultaneously allowed to roll without, slipping down an inclined plane from the same height., The solid cylinder will reach the bottom of the inclined, plane first., Statement II By the principle of conservation of energy, the, total kinetic energies of both the cylinders are equal, when, they reach the bottom of the incline.
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ROTATIONAL MOTION, , DAY EIGHT, , 93, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A thin circular ring of mass m and radius R is rotating, about its axis with a constant angular velocity ω. Two, objects each of mass M are attached gently to the, opposite ends of a diameter of the ring. The ring now, rotates with an angular velocity ω′, ω(m + 2M ), m, ωm, (c), (m + M ), , ω (m − 2M ), (m + 2M ), ωm, (d), (m + 2M ), , (a), , (b), , 2 A T shaped object with dimensions as shown in the, figure, is lying on a smooth floor. A force is applied at the, point P parallel to AB, such that the object has only the, translational motion without rotation. Find the location of, P with respect to C., , P, , 3, (b) l, 2, , (c), , 4, l, 3, , (d) l, , freely about horizontal axis passing through its end. Its, maximum angular speed is ω. Its centre of mass rises to, a maximum height of, 1 lω, (b), 6 g, , (b), , Mm, ω2R 2, (M + 2m), , (c), , Mm, ω2R 2, (M − 2m), , (d), , (M + m)M 2 2, ωR, (M + 2m), , 6 A hoop of radius r and mass m rotaing with an angular, velocity ω 0 is placed on a rough horizontal surface. The, initial velocity of the centre of the hoop is zero. What will, be the velocity of the centre of the hoop, when it ceases, to slip?, ª JEE Main 2013, (b), , rω0, 3, , (d) rω0, , rolling without slipping with a uniform angular speed. The, ratio of the forces experienced by the two particles, F, situated on the inner and outer parts of the ring, 1 is, F2, , 3 A thin uniform rod of length l and mass m is swinging, , 1 l2ω2, (a), 3 g, , m (M + 2m) 2 2, ωR, M, , 7 An annular ring with inner and outer radii R1 and R 2 is, , 2l, , C, , 2, (a) l, 3, , (a), , (a), , B, , F, , with angular velocity ω. Two identical bodies each of, mass m are now gently attached at the two ends of a, diameter of the ring. Because of this, the kinetic energy, loss will be, ª JEE Main (Online) 2013, , rω0, 4, rω0, (c), 2, , l, A, , 5 A ring of mass M and radius R is rotating about its axis, , 1 l2ω2, (c), 2 g, , 1 l2ω2, (d), 6 g, , 4 A small object of uniform density rolls up a curved, surface with an initial velocity v. It reached up to a, 3v 2, with respect to the initial, maximum height of, 4g, , (a), , R , (b) 1 , R2 , , R2, R1, , (c) 1, , (d), , 2, , R1, R2, , 8 A slender uniform rod of mass M and length l is pivoted, at one end so that it can rotate in a vertical plane (see the, figure). There is negligible friction at the pivot. The free, end is held vertically above the pivot and then released., The angular acceleration of the rod when it makes an, angle θ with the vertical, is, z, , position. The object is, , q, , v, x, , (a) ring, (c) hollow sphere, , (b) solid sphere, (d) disc, , 2g, (a), sin θ, 3l, , ª JEE Main 2017 (Offline), 3g, 2g, 3g, (b), (d), cos θ (c), cosθ, sinθ, 2l, 3l, 2l
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94, , DAY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , 9 A horizontal force F acts on the sphere at its centre as, shown. Coefficient of friction between ground and sphere, is µ. What is the maximum value of F , for which there is no, slipping?, 37º, , (a) 5.4 s, (c) 1.4 s, , F, O, , (b) 2.4 s, (d) 1.2 s, , 12 Angular acceleration of the Cylinder (B) shown in the, 5, µmg, 2, 9, (c) F ≤ µmg, 2, , figure is (all strings and pulley are ideal), , 7, µmg, 2, 3, (d) F ≤ µmg, 2, , (a) F ≤, , (b) F ≤, , A, , m, , m, , 10 A solid uniform disc of mass m rolls without slipping down, , B, , a fixed inclined plank with an acceleration a. The frictional, force on the disc due to surface of the plank is, (a), , 1, ma, 4, , (b), , 3, ma, 2, , (c) ma, , (d), , 1, ma, 2, , R, , m, , 11 A cylinder having radius 0.4 m, initially rotating (at t = 0), with ω 0 = 54 rad/s is placed on a rough inclined plane with, θ = 37° having frictional coefficient µ = 0.5. The time taken, by the cylinder to start pure rolling is [g = 10m / s 2 ]., , (a), , 2g, 3R, , (b), , 2g, 5R, , (c), , 2g, R, , (d), , g, 2R, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (c), , 3 (c), , 4 (c), , 5 (b), , 6 (d), , 7 (c), , 8 (a), , 11 (c,d), , 12 (c), , 13 (b), , 14 (c), , 15 (a), , 16 (a), , 17 (a), , 18 (b), , 1 (d), 11 (d), , 2 (c), 12 (b), , 3 (d), , 4 (d), , 5 (b), , 6 (c), , 7 (d), , 8 (d), , 9 (b), , 10 (b), , 9 (b), , 10 (d)
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ROTATIONAL MOTION, , DAY EIGHT, , 95, , Hints and Explanations, SESSION 1, , 6 L = mvr1, , 1 τ 0 = mg × r⊥, , 9 Conserving angular momentum, I1 ω1 = I2 ω2, , y, , As, r ⊥ is continuously increasing or, torque is continuously increasing on the, particle. Hence, angular momentum is, continuously increasing., , m, , ∴, , v, , r1 = y = 2Ö5 cm, x, , 2 τ =r×F, , = 30 10 g- cm2s – 1, , or, , 7 Angular momentum of the pendulum, about the suspension point O is, , O, , O, , x, , ∴, , O, , As,, , r, (1,–1), , z, , ∴, , I, , τ = ($i − $j ) × (− F k$ ), = F [(− $i × k$ ) + ($j × k$ )], , vrad, , m, , = F [ $j + $i], , O, , 3 Torque due to central force is zero,, , or, , d, τ = (L ) = 0, dt, L = constant, , ∴, , L, , 4 Equation of straight line AB is, B, 53º, 37º, A, , 0.5, , r⊥, O, , 4y − 3x = 2, 3, or, y = x + 0.5, 4, 3, ∴ Slope = tanθ =, 4, or, θ = 37°, r ⊥ = 0.5 sin 53°, = 0.4 m, ∴Angular momentum (L ) = mv r⊥, or, L = 3 × 5 × 0.4, = 6kg - m2 s−1, , 5 Angular momentum of particle about O,, L = m( r × v), |L| = mr v sin θ, = mv (r sin θ), = mvl, , L, L = m (r ´ v), , Then, v can be resolved into two, components, radial component r rad and, tangential component r tan . Due to v rad ,, L will be tangential and due to v tan , L, will be radially outwards as shown. So,, net angular momentum will be as, shown in figure, whose magnitude will, be constant (|L |= mvl ). But its direction, will change as shown in the figure., L = m (r × v), where, r = radius of circle., , 8 Given, m = 2 kg, r (t ) = 5$i − 2t 2 $j, Angular momentum (L ) = r × p, dr, d, = (5$i − 2t 2 $j ), dt dt, (at t = 2s ), = − 8 $j, , ∴Velocity, v =, , ∴, , T ∝ R2, 1, R, n, T, 24, T ′= 2 = 2 h, n, n, R′=, , the side of a square of side a. Such that, Angular momentum L about the origin, =L=r ×p, = rpsin θ n$, , Ltan, , = F [ $i + $j], , R2, = constant, T, As, I ∝ R 2 and ω ∝ 1 , , , , T, , 11 For a particle of mass m is moving along, , Lrad, , vtan v, , I1, ω, ω1 =, ω=, I2, 3, ma2 , 2, , + ma, 2 , , I ω = constant, , 10, , = (5)(3 2 )(2 5), , y, F, , ω2 =, , ma2 , , , 2 , , L = r ( p ) n$, , When a particle is moving from D, to A,, R, L=, mv (− k$ ), 2, A particle is moving from A to B,, R, L=, m v (− k$ ), 2, and it moves from C to D,, R, L = , + a mv ( k$ ), 2, , For B to C, we have, R, L = , + a mv ( k$ ), 2, , Hence, options (c) and (d) are incorrect., 12 Angular impulse = P × l, 2, = change in angular, momentum, , P, , p = mv, = 2 × (−8 $j ) = − 16 $j, , Therefore,, L =r ×p, = (5i$ − 2t 2 $j ) × (−16) $j, = − 80 k$, , v, w, , (at t = 2s )
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96, , DAY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , ml2 , Pl, = Iω = , ω, 2, 12 , 6P, ω=, ∴, ml, π/2, θ, Now, t = =, ω 6P /ml, π ml, =, 12 P, , If s be the distance covered along the, plane, then, , ∴, , h = s sinθ, , 13 Moment of inertia of the solid cylinder, about its axis of symmetry,, 1, I = MR2, 2, 1, = × 20 × (0.25)2, 2, = 10 × 0.0625, = 0.625kg-m2 = 62.5 J-s, Kinetic energy associated with the, rotation of the cylinder is given by, 1, K = Iω2, 2, 1, = × 0.625 × (100)2, 2, = 0.3125 × 10000 = 3125J, KR, 14 For ring,, =1, KT, KR 1, For disc,, =, KT 2, For ring, total kinetic energy, = 2(K T ) = 8 J, Hence,, KT = 4 J, For disc, total kinetic energy =, =, , v =, , Now,, , v 2 = 2 as, , ∴ 2 as =, , 1 + I /MR2, , 2gs sin θ, 1 + I /MR2, , M, , ⇒a=, , g sin θ, 1 + I /MR2, , gravity and normal contact force. In, addition to these, impact force (during, collision) will act on the disc along line, xy. Gravity and normal contact force, balance each other (in terms of force, and torque both), but impact force, causes non-zero torque acting on disc, about all points except the points lying, on its line of action i.e. xy. So, angular, momentum remains conserved about, any point on xy., , (given), 3, KT, 2, , equilibrium, the net torque experienced, by the body about any point has to be, zero. Here, due to F and Mg about C, the, torque is zero but friction is providing, non-zero torque in clockwise direction,, now other force is N only which can, produce non-zero force in, anti-clockwise direction to make net, torque zero and it is possible only when, N does not pass through C., , 3, × 4= 6 J, 2, , F, , f, , Mg, N, , 18 In case of pure rolling on inclined, plane,, a=, , s, h, , ∴, , C, q, , v, , 1 v2 1, 2, I, + Mv, 2 R2 2, , 1 2, I, v M + 2 = Mgh, 2 , R , M + I/R, , 2, , =, , I1 ω = I2 ω ′, mR 2ω = (mR 2 + 2MR 2 )ω ′, , ⇒, , , , m, ω′ = , ω, m + 2M , , 2 For pure translatory motion, net torque, about centre of mass should be zero., Thus, F is applied at centre of mass of, system., l, , P (0, 2l/3), Y, , (0, 2l), C, , OP =, , m1 × 0 + m2 × l, m1 + m2, , where, m1 and m2 are masses of, horizontal and vertical section of the, object. Assuming object is uniform,, m2 = 2 m1, 2l, OP =, ⇒, 3, 4l, 2l, 2l, + l = 2l − =, ∴ PC = l −, , , , 3, 3, 3, , 3 If centre of mass rises to a maximum, , g sin θ, 1 + I /mR, , X, , (0, 0), , 2, , height, then loss in KE = Gain in PE, we get, , ISolid < I Hollow, , w, , 2Mgh, , i.e., ⇒, , 17 Here, as the block is kept in, , C, , or v 2 =, , m, R, M, , m, R, , 16 The forces experienced by disc are, , against friction, the loss in potential, energy is equal to the total gain in the, kinetic energy., , or, , w¢, , w, , 15 Assuming that no energy is used up, , i.e. Mgh =, , 1 As, no external torque is acting on the, system, angular momentum of system, remains conserved., , 2gs sin θ, , ∴, , 2, , SESSION 2, , ∴ (KE) Hollow = (KE) Solid, , 2gh, 1 + I/MR, , a Solid > a Hollow, , ∴ Solid cylinder will reach the bottom, first. Further, in case pure rolling on, stationary ground, work done by, friction is zero. Therefore, mechanical, energy of both the cylinders will remain, constant., , 2, , = decrease in PE = mgh, , h, , 1 2, Iω = mgh, 2, , 1 1, 1 l2ω2, × ml2ω2 = mgh ⇒ h =, 2 3, 6 g
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DAY EIGHT, 2, , 3 v2 , 4 1 mv 2 + 1 I v = mg , , , 2, , 2, , I =, , ∴, , torque is acting on it, which is due to, the vertical component of weight of, rod., , 1, mR2, 2, , or, , and, , I2 = 2mR + MR, , …(iv), , ω, , Substituting the values from Eqs. (i), (ii), and (iii) into Eq. (iv), we get,, Mm 2 2, Change in KE = , ω R, M + 2m , R, , v, , From conservation of angular, momentum, , s, g, , q, , 9 Here, F − f = ma, , … (i), , F, O, , F2, R1, w, , F1, v, , The force exprienced by inner part,, F1 = mω2 R1 and the force experienced, by outer part,, F2 = mω2 R2, F1, R, = 1, F2, R2, , f, , a, τ = I α = I , R, 2, a, ⇒, f . R = mR2 ×, 5, R, 2, or, f = ma, 5, From Eqs. (i) and (ii), we get, 2, f= F, 7, 2, or, F ≤ µmg, 7, 7, or, F ≤ µmg, 2, Hence, (b) is the correct option., g sin θ, 10 Since, a =, 1 + I / mR2, g sin θ, or, a=, ;, 1, 1+, 2, , mR2, I, , m (3a / 2) 1, = ma, 1+ 2, 2, , Hence, (d) is the correct option., , 11, , 2, , constant. So, no net force or torque is, acting on ring. The force experienced by, any particle is only along radial, direction, or we can say the centripetal, force., , or f =, , mg sin q, , Now, torque τ = force × perpendicular, distance of line of action of force from, axis of rotation, l, = mg sin θ ×, 2, Again, torque τ = Iα, where, I = moment of inertia, ml2, =, 3, [Force and torque frequency, along axis of rotation passing through in, end], α = angular acceleration, l ml2, α, ∴ mg sin θ × =, 2, 3, 3g sin θ, ∴, α=, 2l, , 7 Since ω is constant, v would also be, , R2, , q, mg, , At any time t, , Initial condition, , v, mr ω 0 = mvr + mr ×, r, ω 0r, v =, 2, 2, , ⇒, , m, , …(iii), , Now change in,, 1, 1, KE = L1 I12 − L2 I22, 2, 2, , 6, , l/2, , …(ii), , 2, , mg sinθ, 1+, , co, , I1 = mR2, , 2, g sinθ, 3, 3, g sinθ = a, 2, , q, , …(i), 2, , 1, mR2 ], 2, , Now, force of friction ( f ) =, , l/2, , 5 By conservation of angular momentum,, where,, , [For uniform solid disc, I =, =, , Therefore, the body is a disc., I1ω1 = I2ω2, , 97, , 8 As the rod rotates in vertical plane so a, , 4g , , R, , ROTATIONAL MOTION, , and, , a = (µg cos θ) + (g sin θ), = 0.5 × 10 × 0.8 + 10 × 0.6, or a= 10 m / s 2, a, a, º, º, 37, 37 37º, mg cos 37º, sin cos, mg, mg mg, µ, 37º, , And angular acceleration (α ), (µ mg cos θ)R, =, 1, mR2, 2, 2µ g cos θ, or, α=, R, 2 × 0.5× 10 × 0.8, or, α=, = 20 rad/s2, 0.4, Pure rolling will start, when, v = Rω, where,, v = a × t and ω = (ω 0 − α t ), ∴, at = R (ω 0 − α t ), or, 10t = 0.4(54 − 20t ), or, t = 12, . s, Hence, (d) is the correct option., , 12 Linear acceleration of cylinder (B) is, , …(ii), , given by the relation, mR2, a=, ⋅g, I + m ′ R2, 1, where, I = mR2 (for solid cylinder) and, 2, m ′ = 2m, m R2, a=, ⋅g, ∴, 1, m R2 + 2m R2, 2, 2 m R2, 2g, =, ⋅g =, 5, 5m R 2, ∴ Angular acceleration (α ) of the cylinder, B is, 2g, a, or α =, α=, R, 5R, Hence, (b) is the correct option.
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DAY NINE, , Gravitation, Learning & Revision for the Day, u, , u, , u, , Universal Law of Gravitation, Acceleration due to Gravity, Gravitational Field, , u, , u, , u, , Gravitational Potential, Gravitational Potential Energy, Escape Velocity, , u, , u, , u, , Kepler’s Laws of Planetary Motion, Satellite, Geostationary Satellite, , Universal Law of Gravitation, In this universe, each body attracts other body with a force that is directly proportional, to the product of their masses and inversely proportional to the square of the distance, between them., Let m1 and m2 be the masses of two bodies and r be the separation between them., m m, F = G 12 2 ., r, The proportionality constant G is called universal gravitational constant. In SI system,, value of gravitational constant G is 6.67 × 10 −11 Nm2 kg−2 . Dimensional formula of G is, [M −1L3 T −2 ]., , Acceleration due to Gravity, The acceleration of an object during its free fall towards the earth is, called acceleration due to gravity., If M is the mass of earth and R is the radius, the earth attracts a mass, m on its surface with a force F given by, GMm, F =, R2, This force imparts an acceleration to the mass m which is known as, acceleration due to gravity (g)., By Newton’s law, we have, GMm, 2, F, GM, Acceleration (g) =, = R, = 2, m, m, R, GM, On the surface of earth, g = 2, R, Substituting the values of G, M, R, we get g = 9.81 ms −2 ., Mass of the earth m = 6 × 10, , 24, , kg and radius of the earth R = 6.4 × 10 m., 6, , m, , R, Centre, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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GRAVITATION, , DAY NINE, , Variation in g with Altitude and Depth, with respect to Earth, , l, , The value of g is variable and can vary in same cases as, mentioned below, 1. Value of acceleration due to gravity ( g ) at a height ( h ), from the surface of the earth is given by g′ =, , gR2, (R + h)2, , Gravitational field intensity due to a solid sphere, (e.g. earth) of mass M and radius R at a point distant r from, GM, its centre (r > R) is I = 2, r, GM, and at the surface of solid sphere, I = 2 ., R, GMr, However, for a point r < R, we find that I = 3, R, , 2h , , If h << R, then g′ = g 1 − , R, , , I∝r, , O, , Variation in the Value of g Due to, Rotation of the Earth, Due to rotation of the earth, the value of g decreases as the, speed of rotation of the earth increases. The value of, acceleration due to gravity at a latitude is, g′λ = g − Rω2 cos2 λ, Following conclusions can be drawn from the above, discussion, (a) The effect of centrifugal force due to rotation of the earth, is to reduce the effective value of g., (b) The effective value of g is not truely in vertical direction., (c) At the equators, λ = 0 °, (d) At the poles,, Therefore,, , g′ = g − Rω2, λ = 90 °, g′ = g, , (minimum value), (maximum value), , r<R, , I ∝ 12, r, r>R, , r, R, Variation of gravitational field intensity in solid sphere, , At the centre of the earth d = R and hence, g′ = 0., , Therefore,, , I =GM, R2, , I, , 2. Value of acceleration due to gravity (g) at a depth (d), from the surface of the earth is given by, d, , g′ = g 1 − , , R, , 99, , l, , Due to a body in the form of uniform shell gravitational, field intensity at a point outside the shell (r > R) is given, GM, by, I= 2, r, But at any point inside the shell, gravitational intensity is, zero., , l, , Gravitational intensity at a point due to the combined, effect of different point masses is given by the vector sum, of individual intensities., Thus, I = I 1 + I2 + I3 + ..., , Gravitational Potential, Gravitational potential at any point in a gravitational field is, defined as the work done in bringing a unit mass from, infinity to that point., W, Gravitational potential, V = lim, ., m 0→ 0 m0, GM, ., r, , Gravitational Field, , Gravitational potential due to a point mass is V = −, , The space surrounding a material body in which its, gravitational force of attraction can be experienced is called, its gravitational field., , Gravitational potential is always negative. It is a scalar term, and its SI unit is J kg −1., , Gravitational Field Intensity (I), , For Solid Sphere, l, , Gravitational field intensity at any point is defined as the, force experienced by any test mass devided by the magnitude, of test mass when placed at the desired point., Mathematically,, F, Gravitational field intensity, I =, m0, , l, , l, , Gravitational intensity at a point situated at a distance r, GM, from a point mass M is given by I = 2, r, , At a point inside the sphere, (r < R)., V =−, , where, m0 is a small test mass. The SI unit of gravitational, intensity is N kg −1., l, , At a point outside the solid sphere, (e.g. earth), i.e., GM, ., r > R, V = −, r, GM, At a point on the surface, V = −, ., R, , l, , GM, GM , r2 , (3 R2 − r 2 ) = −, 3 − 2 , 3, 2R , 2R, R , , At the centre of solid sphere,, 3GM, 3, V =−, = Vsurface, 2R, 2
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100, , DAY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , For Spherical Shell, l, , l, , l, , B, , At a point outside the shell,, GM, where, r > R., V =−, r, At a point on the surface of spherical shell,, GM, V =−, R, At any point inside the surface of spherical shell, GM, V =−, = Vsurface, R, , Relation between Gravitational Field, and Gravitational Potential, If r1 and r2 are position of two points in the gravitation field, with intensity (I), then change in gravitational potential, V (r2 ) − V (r1) = − ∫, ⇒, , r2, r1, , I ⋅ dr, , Thus,, , II, D, , 3. Law of Periods The square of the planet’s time period of, revolution is directly proportional to the cube of, semi-major axis of its orbit., T 2 ∝ a3, where a is the semi-major axis., , Gravitational Potential Energy, Gravitational potential energy of a body or system is negative, of work done by the conservative gravitational forces F in, bringing it from infinity to the present position., Mathematically, gravitational potential energy, U = − W = − ∫ F ⋅ dr, r, , dV = − I ⋅ dr, dr = dx $i + dy $j + dz k$, , and, , ∞, , l, , $ then, I = I x $i + I y $j + I z k,, dV = − I.dr = −I x dx − I y dy − I z dz, ∂V $ ∂V $ ∂V $, I=−, i−, j−, k, ∂x, ∂y, ∂z, , l, , Remember that partial differentiation indicates that variation, of gravitational potential in counter along the variation of, x-coordinate, then other coordinates (i.e. y and z) are, assumed to be constant., , l, , Kepler’s Laws of Planetary, Motion, l, , Kepler discovered three empirical laws which accurately, describe the motion of planets., These laws are, 1. Law of Orbits All the planets move around the sun in an, elliptical orbit with sun at one of the focus of ellipse., Minor axis, , Planet, , Swept area, Major axis, , Perihelion, or Perigee, , S, , A, , C, , Apehelion, or Apogee, a, , 2. Law of Areas The line joining the sun to the planet, sweeps out equal areas in equal intervals of time, i.e. areal, velocity of the planet w.r.t. sun is constant. This is called, the law of area, which indicates that a planet moves faster, near the sun and slowly when away from the sun., , l, , The gravitational potential energy of two particles of, masses m1 and m2 separated by a distance r is given by, Gm1m2, U=−, r, The gravitational potential energy of mass m at the surface, of the earth is, GMm, U=−, R, Difference in potential energy of mass m at a height h from, the earth’s surface and at the earth’s surface is, mgh ~, U( R + h ) − UR =, − mgh, if h << R, h, 1+, R, For three particles system,, Gm1m2 Gm1m3 Gm2 m3 , U = −, +, +, , r 13, r23 , r 12, n (n − 1), pairs form and total, For n-particles system,, 2, potential energy of the system is sum of potential energies, of all such pairs., , Escape Velocity, It is the minimum velocity with which a body must be, projected from the surface of the earth so that it escapes the, gravitational field of the earth. We can also say that a body,, projected with escape velocity, will be able to go to a point, which is at infinite distance from the earth., The value of escape velocity from the surface of a planet of, mass M, radius R and acceleration due to gravity g is, vescape =, , 2GM, = 2gR, R
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GRAVITATION, , DAY NINE, Escape velocity does not depend upon the mass or shape or, size of the body as well as the direction of projection of the, body. For earth value of escape velocity is 11.2 kms−1., , Height of Satellite in Terms of Period, , Satellites, , As the height of the satellite in terms of time period,, , Anybody that revolves around earth or any planet is called, satellite. These can be natural (e.g. Moon) or artificial. The, artificial satellites are man made satellites launched from the, earth. The path of these satellites are elliptical with the centre of, earth at a foci of the ellipse. However, as a first approximation, we may consider the orbit of satellite as circular., , Orbital Velocity of Satellite, Orbital velocity of a satellite is the velocity required to put, the satellite into its orbit around the earth. The orbital, velocity of satellite is given by, vo =, , GM, =, r, , gR2, gR2, =, (R + h), r, , If h << R or r ~, − R, then, vo =, , GM, = gR = 7.9 kms – 1, R, , 101, , The height of the satellite (from the earth planet) can be, determined by its time period and vice-versa., gR2T 2 , h=r −R=, 2 , 4π , , 1/ 3, , − R., , Energy of Satellite, 1, GMm, ., mv20 =, 2, 2r, GMm, Potential energy of satellite, U = −, r, GMm, and total energy of satellite E = K + U = −, = − K., 2r, , Kinetic energy of satellite, K =, , Binding Energy of Satellite, It is the energy required to remove the satellite from its orbit, and take it to infinity., GMm, Binding energy = − E = +, 2r, , Angular Momentum of Satellite, , vescape = 2 vorbital, , Angular momentum of a satellite, L = mv0r = m2GMr, , Period of Revolution, It is the time taken by a satellite to complete one revolution, around the earth., 2 πr, r3, 3π, r3, Revolution period, T =, = 2π, = 2π, =, vo, GM, G. e, gR2, If r ~, − R, then T = 2π, , R, = 84.6 min., g, , Geostationary Satellite, If an artificial satellite revolves around the earth in an, equatorial plane with a time period of 24 h in the same sense, as that of the earth, then it will appear stationary to the, observer on the earth. Such a satellite is known as a, geostationary satellite or parking satellite., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A mass M is split into two parts m and (M − m ), which are, separated by a certain distance. The ratio m/M which, maximizes the gravitational force between the parts is, (b) 1 : 3, (d) 1 : 1, , (a) 1 : 4, (c) 1 : 2, , (a) M will remain at rest, (b) M will move towards M, (c) M will move towards 2M, (d) M will have oscillatory motion, , 3 If one moves from the surface of the earth to the moon,, , 2 Particles of masses 2M, m and M are respectively at, 1, (BC ), m is much-much, 2, smaller than M and at time t = 0, they are all at rest as, given in figure. At subsequent times before any collision, takes place., , points A, B and C with AB =, , A, , B, , C, , 2M, , m, , M, , what will be the effect on its weight?, (a) Weight of a person decreases continuously with height, from the surface of the earth, (b) Weight of a person increases with height from the, surface of the earth, (c) Weight of a person first decreases with height and then, increases with height from the surface of the earth, (d) Weight of a person first increases with height and then, decreases with height from the surface of the earth
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102, , DAY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , g, , 4 At the surface of a certain planet acceleration due to, gravity is one-quarter of that on the earth. If a brass ball, is transported on this planet, then which one of the, following statements is not correct?, (a) The brass ball has same mass on the other planet as on, the earth, (b) The mass of the brass ball on this planet is a quarter of, its mass as measured on the earth, (c) The weight of the brass ball on this planet is a quarter of, the weight as measured on the earth, (d) The brass ball has the same volume on the other planet, as on the earth, , g, , (c), , (d), d, , d, O, , R, , O, , 11 The gravitational field, due to the ‘left over part’ of a, uniform sphere (from which a part as shown has been, ‘removed out’) at a very far off point, P located as shown,, would be (nearly), ª JEE Main 2015, , R, , by 50%, the acceleration due to gravity would, , 3R, ,where R is the radius of the, 2, earth. Assuming the gravitational pull on a mass of 1kg, on the earth’s surface to be 10 N, the pull on the satellite, will be, orbit of average radius, , (a) 880 N, (c) 885 N, , (b) 889 N, (d) 892 N, , g, becomes (where, g = the acceleration due to gravity, 9, on the surface of the earth) in terms of R, the radius of, the earth is, h, (b), 3, , h, (c), 2, , (d) 2 h, , 8 The change in the value of g at a height h above the, surface of the earth is the same as at a depth d below, the surface of the earth. When both d and h are much, smaller than the radius of the earth, then which one of the, following is correct?, 3h, (b) d =, 2, , (c) d = 2h, , (d) d = h, , 9 Average density of the earth, , distance d from centre of the Earth is best represented, ª JEE Main 2017 (Offline), by (R = Earth’s radius), g, , (b), d, O, , R, , 5 GM, 6 R2, , (b), , 8 GM, 9 R2, , (c), , 7 GM, 8 R2, , (d), , 6 GM, 7 R2, , 12 A solid sphere is of density ρ and radius R. The, gravitational field at a distance r from the centre of the, sphere, when r < R , is, (a), , ρπGR 3, r, , (b), , 4 π Gρ r 2, 3, , (c), , 4 π Gρ R 3, 3r 2, , (d), , 4 π Gρ r, 3, , dressed astronaut can jump on the earth is 0.5 m., Estimate the maximum vertical distance through which he, can jump on the moon, which has a mean density 2/3rd, that of the earth and radius one quarter that of the earth., (a) 1.5 m, , (b) 3 m, , (c) 6 m, , (d) 7.5 m, , 14 The mass of a spaceship is 1000 kg. It is to be launched, from the earth’s surface out into free space. The value of, g and R (radius of earth) are 10 m/s 2 and 6400 km, respectively. The required energy for this work will be, (a) 6.4 × 1011 J, (c) 6.4 × 109 J, , (b) 6.4 × 108 J, (d) 6.4 × 1010 J, , 15 If g is the acceleration due to gravity on the earth’s, surface, the gain in the potential energy of an object of, mass m raised from the surface of the earth to a height, equal to the radius R of the earth, is, (b), , 1, mgR, 2, , (c), , 1, mgR, 4, , (d) mgR, , 16 Two bodies of masses m and 4 m are placed at a, , 10 The variation of acceleration due to gravity g with, , (a), , (a), , (a) 2mgR, , (a) does not depend on g, (b) is a complex function of g, (c) is directly proportional to g, (d) is inversely proportional to g, , g, , x, , 13 The maximum vertical distance through which a full, , 7 The height at which the acceleration due to gravity, , h, (a) d =, 2, , R, , (b) decrease by 50%, (d) increase by 100%, , 6 A research satellite of mass 200 kg circles the earth in an, , (a) 2h, , Mass of complete, sphere = M, P, , Removed, part, , 5 If both the mass and radius of the earth, each decreases, (a) remain same, (c) decrease by 100%, , R, , d, O, , R, , distance r. The gravitational potential at a point on the line, joining them where the gravitational field is zero is, (a) −, , 4Gm, r, , (b) −, , 6Gm, r, , (c) −, , 9Gm, r, , (d) zero, , 17 A planet in a distant solar system is 10 times more, massive than the earth and its radius is 10 times smaller., Given that the escape velocity from the earth is 11 kms −1,, the escape velocity from the surface of the planet would, be, (a) 1.1 kms −1, (c) 110 kms −1, , (b) 11 kms −1, (d) 0.11 kms −1
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GRAVITATION, , DAY NINE, , 103, , 18 A projectile is fired vertically upwards from the surface, , 26 The gravitational force exerted by the sun on the moon is, , of the earth with a velocity kve , where ve is the escape, velocity and k < 1. If R is the radius of the earth, the, maximum height to which it will rise measured from the, centre of the earth will be, , about twice as great as the gravitational force exerted by, the earth on the moon, but still the moon is not escaping, from gravitational influence of the earth. Mark the option, which correctly explains the above system., , 1− k2, R, , (b), , (c) R (1 − k 2 ), , (d), , (a), , R, 1− k2, R, 1+ k2, , 19 Two cars of masses m1 and m2 are moving in circles of, radii r1 and r2, respectively. Their speeds are such that, they make complete circles in the same time t. The, ratio of their centripetal acceleration is, (a) m1r1 : m2 r2 (b) m1 : m2, , (c) r1 : r2, , (d) 1 : 1, , (a) At some point of time the moon will escape from the earth, (b) Separation between the moon and sun is larger than the, separation between the earth and moon, (c) The moon-earth system is bounded one and a minimum, amount of energy is required to escape the moon from the, earth, (d) None of the above, , 27 What is the minimum energy required to launch a satellite of, mass m from the surface of a planet of mass M and radius R, in a circular orbit at an altitude of 2R ?, , 20 What is the direction of areal velocity of the earth, around the sun?, (a) Perpendicular to positon of the earth w.r.t. the sun at, the focus, (b) Perpendicular to velocity of the earth revolving in the, elliptical path, (c) Parallel to angular displacement, (d) Both (a) and (b), , 21 The time period of a satellite of the earth is 5 h. If, the separation between the earth and the satellite is, increased to 4 times the previous value, the new time, period will become, (a) 10 h, , (b) 80 h, , (c) 40 h, , (d) 20 h, , 22 A satellite goes along an elliptical path around earth., The rate of change of area swept by the line joining, earth and the satellite is proportional to, (a) r, , 1/ 2, , (b) r, , (c) r, , 3/ 2, , (d) r, , 2, , 23 A satellite of mass m revolves around the earth of, radius R at a height x from its surface. If g is the, acceleration due to gravity on the surface of the earth,, the orbital speed of the satellite is, (a) g x, , gR, (b), R−x, , gR 2, (c), R+ x, , gR 2 , (d) , , R + x , , 1/ 2, , 24 Two particles of equal mass m go around a circle of, radius R under the action of their mutual gravitational, attraction. The speed of each particle with respect to, ª AIEEE 2011, their centre of mass is, (a), , Gm, R, , (b), , Gm, 4R, , (c), , Gm, 3R, , (d), , nth power of distance. Then the time period of a planet, in circular orbit of radius R around the sun will be, proportional to, (a) R, , (b) R, , n − 1, , , 2 , , (c) R, , n, , (d) R, , n − 2, , , 2 , , 2GmM, (b), 3R, , ª JEE Main 2013, GmM, (d), 3R, , GmM, (c), 2R, , 28 The curves for potential energy, (E P ) and kinetic energy (E K ) of, two particles system are as, shown in the figure. At what, points the system will be bound, , Energy, EK, A, , O, , an elliptical orbit as shown in the, OA, figure. The ratio, = x . The ratio, OB, of the speeds of the earth at B and, at A is, (b) x, , r, D, , EP, , 29 The earth moves around the sun in, , (a) x, , B, C, , (a) Only at point A, (b) Only at point D, (c) At points A and D, (d) At points A, B and C, , Earth, , B, , O, , (c) x 2, , Sun, , A, , (d) x x, , 30 A satellite of mass ms revolving in a circular orbit of radius rs, around the earth of mass M, has a total energy E . Then its, angular momentum will be, (a) (2E msrs )1/ 2, , (b) (2E msrs ), , (2E msrs2 )1/ 2, , (d) (2E msrs2 ), , (c), , 31 Match the term related to gravitation given in Column I with, their formula given in Column II and select the correct, option from the choices given below :, , Gm, 2R, , 25 Suppose the gravitational force varies inversely as the, , n + 1, , , 2 , , 5 GmM, (a), 6R, , Column I, , Column II, , A., , Gravitational potential at a point, outside the solid sphere, , 1., , 3 GM, 2R, , B., , Gravitational potential at a point on, the surface of sphere, , 2., , GM, r, , C., , Gravitational potential at the centre, of solid sphere, , 3., , GM, R
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104, , DAY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , (a), (b), (c), (d), , A, 1, 2, 1, 3, , B, 3, 3, 2, 2, , C, 2, 1, 3, 1, , Statement II An object moving around the earth under, the influence of the earth’s gravitational force is in a state, of ‘free-fall’., , 33 Statement I Kepler’s laws for planetary motion are, consequence of Newton’s laws., , Direction (Q. Nos. 32-33) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d)Statement I is false; Statement II is true, , 32 Statement I An astronaut in an orbiting space station, above the earth experiences weightlessness., , Statement II Kepler’s laws can be derived by using, Newton’s laws., , 34 Assertion (A) If the earth were a hollow sphere,, gravitational field intensity at any point inside the earth, would be zero., Reason (R), zero., , Net force on a body inside the sphere is, , (a) If both Assertion and Reason are correct and Reason is, the correct explanation of Assertion, (b) If both Assertion and Reason are true but Reason is not, the correct explanation of Assertion, (c) If Assertion is true but Reason is false, (d) If Assertion is false but Reason is true, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 Satellites orbitting the earth have finite life and sometimes, debris of satellites fall to the earth. This is because, (a) the solar cells and batteries in satellites run out, (b) the laws of gravitation predict a trajectory spiralling, inwards, (c) of viscous forces causing the speed of satellite and, hence height to gradually decrease, (d) of collisions with other satellites, , 2 A body is released from a point distance r from the, centre of earth. If R is the radius of the earth and r > R ,, then the velocity of the body at the time of striking the, earth will be, (a) gR, (c), , , 1, , 1, 1, 1 , (a) Gm A , − + BL (b) Gm A −, − BL , , a + L a, , a a + L, 1, , 1, , 1, 1 , (c) Gm A , − − BL (d) Gm A −, + BL , (, a, +, L, ), a, a, a, +, L, , , , , , , , , , 5 A ring having non-uniform distribution of mass having, mass M and radius R is being considered. A point mass, m0 is taken slowly from A to B along the axis of the ring. In, doing so, work done by the external force against the, gravitational force exerted by ring is, M, , (b) 2gR, , 2gRr, r −R, , (d), , R, , 2gR (r − R), r, , O, , B, , A, R, , R, , 3 Two small satellites move in circular orbits around the, earth, at distances r and r + ∆r from the centre of the, earth. Their time periods of rotation areT and T + ∆T ,, ( ∆r << r , ∆T << T ). Then, ∆T is equal to, 3 ∆r, (a) T, 2 r, , 2 ∆r, (b) T, 3 r, , − 3 ∆r, (c), T, r, 2, , ∆r, (d)T, r, , 4 A straight rod of length L extends from x = a to x = L + a ., The gravitational force, it exerts on a point mass m at, x = 0, if the mass per unit length is A + Bx 2, is, , (a), , GMm0, 2R, , (b), , GMm0 1, 1 , −, R 2, 5 , , GMm0 1, 1 , −, R 5, 2 , (d) It is not possible to find the required work as the nature, of distribution of mass is not known, (c)
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GRAVITATION, , DAY NINE, 6 With what minimum speed should, , vmin, , m be projected from point C in the, presence of two fixed masses M, each at A and B as shown in the, figure, such that mass m should, escape the gravitational attraction, M, of A and B ?, 2GM, R, , (a), (c) 2, , GM, R, , A, , 2 2 GM, R, , (b), , (d) 2 2, , m, , O, , R, , B, , GM, R, , (b) 3.33 × 10−10 J, (d) 6.67 × 10−10 J, , two strings each of length L. The distance between the, upper ends of strings is L. The angle which the strings, will make with the vertical due to mutual attraction of the, spheres is, GM , (a) tan−1 , , gL , , GM , (b) tan−1 , , 2 gL , , GM , (c) tan−1 2 , gL , , 2GM , (d) tan−1 , 2 , gL , , central force whose potential energy is given by, U = 2r 3 joule. If the body is moving in a circular orbit of, 5 m, its energy will be, (b) 125 J, (d) 650 J, , h along the normal through the centre, O of a thin circular ring of mass M and, radius r as shown in figure. If the mass, is moved further away such that OP, becomes 2h, by what factor, the force, of gravitation will decrease, if h = r ?, 5 2, 3, , (c), , 4 3, 5, , r, , P, O, , h, , m, , M, , (d), , 4 2, 5 5, , 11 An artificial satellite of the earth is launched in circular orbit, in equatorial plane of the earth and satellite is moving from, West to East. With respect to a person on the equator, the, satellite is completing one round trip in 24 h. Mass of the, earth is, M = 6 × 10 24kg. For this situation, orbital radius of the, satellite is, (a) 2.66 × 104 km, (c) 36,000 km, , (c), , −2GM, 3R, , (d), , −2GM, R, , (b) 6400 km, (d) 29,600 km, , the Earth’s surface (radius of Earth R , h < < R ). The, minimum increase in its orbital velocity required, so that, the satellite could escape from the Earth’s gravitational, field, is close to (Neglect the effect of atmosphere), ª JEE Main 2017 (Offline), , (a) 2gR, , (b) gR, , (c) gR / 2, , (d) gR ( 2 − 1), , other, move along a circle of radius R under the action of, their mutual gravitational attraction, the speed of each, particle is, ª JEE Main 2015, GM, R, GM, (c), (1 + 2 2 ), R, (a), , GM, R, 1 GM, (d), (1 + 2 2 ), 2 R, (b) 2 2, , two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 15 Statement I Three orbits are marked, , 10 A mass m is placed at P at a distance, , (b), , −GM, R, , Direction (Q. Nos. 15-16) Each of these questions contains, , 9 A body of mass 2 kg is moving under the influence of a, , 3 2, 4 3, , (b), , 14 Four particles, each of mass M and equidistant from each, , 8 Two metallic spheres each of mass M are suspended by, , (a), , −GM, 2R, , ª JEE Main 2015, , 13 A satellite is revolving in a circular orbit at a height h from, , uniform sphere of mass 100 kg and radius 10 cm. Find, the work to be done against the gravitational force, between them, to take the particle far away from the, sphere, (you may take G = 6.67 × 10−11 Nm 2 / kg2), , (a) 250 J, (c) 625 J, , Taking gravitational potentialV = 0 at, I = ∞, the potential at the centre of the, cavity thus formed is (I = gravitational, constant)., (a), , 7 A particle of mass 10 g is kept on the surface of a, , (a) 13.34 × 10−10 J, (c) 6.67 × 10−9 J, , 12 From a solid sphere of mass M and, radius R, a spherical portion of radius, R , is removed as shown in the figure., 2, , 30°, , M, R, , 105, , 2, , 3, , 1, as 1, 2 and 3. These three orbits, have same semi-major axis although, their shapes (eccentricities) are, different. The three identical, satellites are orbiting in these three orbits, respectively., These three satellites have the same binding energy., Statement II Total energy of a satellite depends on the, semi-major axis of orbit according to the expression,, − GMm, E=, 2r, 16 Statement I Two satellites are following one another in the, same circular orbit. If one satellite tries to catch another, (leading one) satellite, then it can be done by increasing, its speed without changing the orbit., Statement II The energy of the earth satellites system in, − GMm, circular orbits is given by E =, , where, r is the, 2r, radius of the circular orbit.
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106, , DAY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , ANSWERS, SESSION 1, , 1 (c), , SESSION 2, , 2 (c), , 3 (c), , 4 (b), , 5 (d), , 6 (b), , 7 (c), , 8 (c), , 9 (c), , 10 (c), , 11 (c), , 12 (d), , 13 (b), , 14 (d), , 15 (b), , 16 (c), , 17 (c), , 18 (b), , 19 (c), , 20 (d), , 21 (c), , 22 (a), , 23 (d), , 24 (b), , 25 (a), , 26 (c), , 27 (a), , 28 (d), , 29 (b), , 30 (c), , 31 (b), , 32 (a), , 33 (d), , 34 (a), , 1 (c), 11 (a), , 2 (d), 12 (b), , 3 (a), 13 (d), , 4 (d), 14 (d), , 5 (b), 15 (a), , 6 (b), 16 (d), , 7 (d), , 8 (c), , 9 (c), , 10 (d), , Hints and Explanations, SESSION 1, Gm(M − m ), 1 As, F =, x2, For maximum gravitational force, dF, G, m 1, = 2 (M − 2m ) = 0 or, =, dm, x, M 2, , 6 Here, mg = 10 or 1 × g = 10, ⇒, , g = 10 ms, R2, R2, 40, Now, g ′ = g 2 = 10 ×, =, r, (3R/2) 2, 9, Pull on satellite = m ′ g ′, , Hence, m will move towards BA, (i.e. 2M)., , 3 The gravitational attraction on a body, due to the earth decreases with height, and increases due to the moon at a, certain height., At one point it becomes zero and with, further increase in height the, gravitational attraction of the moon, becomes more than that of the earth., , 4 The mass of a body is always constant, and does not change with position., , 5 Here, g = GM /R, and g ′ =, , 2, , G (M /2) 2GM, =, = 2g, (R/2)2, R2, , g′ − g , ∴ % increase in g = , × 100, g , 2g − g , =, × 100, g , = 100%, , ⇒, , GM, (R + h )2, , R , g GM, R2, = 2 ⋅, =g , , 9, R (R + h )2, R + h, , 1 R , =, , 9 R + h, R, 1, ⇒, =, R+ h 3, , 2, , 11 We have,, M′ =, , 2, , 3R = R + h, h, R=, 2, , ⇒, , 8 g h = g 1 − 2h , , , R, , …(i), , d, and g d = g 1 − , …(ii), , R, As per statement of the problem,, g h = gd, 2h , d, , i.e. g 1 −, = g 1 − , , , R, R, ⇒, , 2h = d, GM, 9 As, g = 2 ;, R, 4, M = π R 3 ρ, 3, , ∴, , g =, , 4G πR 3, ρ, 3 R2, , M, , 3, , ×, , 4 R M, π =, 3 2, 8, , 4 3, πR, 3, ∴ Gravitational field at, GM GM, P = 2 −, R, 8R2, M , 1, 7 GM, = G × 2 1 − =, R , 8, 8 R2, , ⇒, , ⇒, , GM, d i.e. g ∝ d, R3, Outside the earth’s surface,, Gm, 1, g = 2 i.e. g ∝ 2, d, d, So, till earth’s surface g increases linearly, with distance r, shown only in graph (c)., g =, , 7 Acceleration due to gravity at height h, is, g ′ =, , ∴, ρ∝ g, (where, ρ = average density of the earth), , 10 Inside the earth’s surface,, , 40, = 200 ×, 9, ≈ 889 N, , 2 The particle B will move towards the, greater force between forces by A, and B., G (2Mm ), Force on B due to A = F BA =, ( AB )2, towards BA, GMm, Force on B due to C = F BC =, (BC )2, towards BC, As, (BC ) = 2 AB, GMm, GMm, ⇒, F BC =, =, < F BA, (2 AB )2, 4( AB )2, , 3 , ρ = , g;, 4 πGR , , ⇒, , –2, , 12 Intensity,, I =, , 4 π Gρr, GM, Gr 4, r = 3 π R3ρ =, , R3, R 3, 3, , 13 On the moon, g m = 4 π G (R/4)(2ρ/3), , 3, 14, 1, = π GRρ = g, 6, 63, , Work done in jumping, = m × g × 0.5 = m × (g /6) h 1, h 1 = 0.5 × 6 = 3.0 m, , 14 Potential energy on the earth surface is, , − mgR while in free space it is zero. So,, to free the spaceship, minimum required, energy is, K = mgR = 103 × 10 × 6400 × 103 J, = 6.4 × 1010 J
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15 Gravitational potential energy of body, on the earth’s surface is, GMe m, U =−, R, At a height h from earth’s surface, its, value is, GMe m, GMe m, Uh = −, =−, [as h = R], (R + h ), 2R, where, Me = mass of earth,, m = mass of body, and, R = radius of earth., ∴ Gain in potential energy, = Uh − U, GMe m GMe m , =−, − −, , , 2R, R , =−, , 2, , GMe m gR m, =, 2R, 2R, 1, = mgR, 2, , shown in figure., x, , r–x, , 4m, B, , r, , ∴, , Gm, G (4m ), =, x2, (r − x )2, , 4 x2 = (r − x )2 ⇒ 2x = r − x, r, x=, ⇒, 3, 9Gm, Gm G (4m ), −, =−, ∴ VP = −, x, r − x, r, ⇒, , (∴ x = r / 3), , 17 Mass of planet, M p = 10Me ,, where, Me is mass of earth., R, Radius of planet, R p = e ,, 10, where Re is radius of earth., Escape velocity is given by,, 2GM, ve =, R, 2G × M p, For planet, v p =, Rp, =, , 100 × 2GMe, Re, , = 10 × v e, = 10 × 11, = 110 kms −1, , perpendicular to the plane containing r, and v., , 21 According to Kepler’s law, T 2 ∝ r 3, 2, , or, , and h = r − R, , 3, , ⇒, , 25, r3, =, 2, 64r 3, (T ′ ), , T ′ = 1600 or T ′ = 40 h, dt, 2m, 2m, r GM, 1, =, =, GMr, 2, r, 2, , So,, , 2, , dA, ∝ r, dt, , 23 The gravitational force exerted on, satellite at a height x is, GMe m, FG =, (R + x )2, where, Me = mass of the earth., Since, gravitational force provides the, necessary centripetal force, so, GMe m, mv o2, =, 2, (R + x ), (R + x ), where, v o is orbital speed of satellite, GMe m, = mv o2, ⇒, (R + x ), , 2gR, , m, , or, , gR2 m, = mv o2, (R + x ), , i.e. F =, ⇒ v =, , R, , R, F, , m, , mv2, R, , Gm2, mv 2, =, 2, R, (2R ), Gm, 4R, , 25 The necessary centripetal force required, for a planet to move around the sun, = gravitational force exerted on it, 1 /2, GM, mv 2 GMe m, i.e., or v = n −1e , =, n, R , R, R, Now, T =, , R n −1 , 2πR, = 2 πR × , , v, GMe , , R2 × R n −1 , = 2π , , Gme, , , 1 /2, , 1 /2, , R ( n + 1 )/2 , = 2π , , (Gme )1 /2 , or, , T ∝ R( n + 1 )/2, , 26 Option (c) is correct, a minimum amount, of energy equal to|TE |of the moon-earth, system has to be given to break, (unbound) the system, the sun is exerting, force on the moon but not providing any, energy., , 27 From conservation of energy, , 22 Areal velocity = dA = L = mvr = vr, , 18 From law of conservation of energy,, mgh, 1, mv 2 =, h, 2, 1+, R, Here, v = kv e = k, , centripetal force., , sun is given by, dA, L, =, dt, 2m, where, L is the angular momentum and, m is the mass of the earth. But angular, momentum, L = r × p = r × mv, ∴ Areal velocity, dA = 1 (r × m v), , , dt 2m, 1, = (r × v), 2, Therefore, the direction of areal velocity, d A is the direction of (r × v), i.e., , , dt , , T, r, ∴ 1 = 1 , T , r , , gR2 gR2 , ( R + x ) = ( R + x ), , , , , vo =, , 24 As gravitational force provides necessary, , 20 Areal velocity of the earth around the, , as g = GMe , , R2 , , P, , 1 /2, , or, , their angular speed is also same., Centripetal acceleration is circular path,, a = ω2 r ., a, r, ω2 r, Thus, 1 = 2 1 = 1, a2, r2, ω r2, , 16 Let gravitational field is zero at P as, m, , mg (r − R ), 1, ,, m k 2 ⋅ 2gR =, r −R, 2, 1+, R, r − R, , 2, k R 1+, =r −R, , R , k2r = r − R, R, r =, ⇒, 1 − k2, , 19 As their period of revolution is same, so, , GMe m GMe m, +, 2R, R, , =, , A, , 107, , GRAVITATION, , DAY NINE, , Q g = GMe , , , , R2 , , Total energy at the planet = Total energy, at the altitude, − GMm, + (KE) surface, R, − GMm 1, …(i), =, + mv 2A, 3R, 2, In its orbit, the necessary centripetal force, is provided by gravitational force., mv 2A, GMm, =, ∴, (R + 2R ) (R + 2R )2, GM, …(ii), v 2A =, ⇒, 3R, From Eqs. (i) and (ii), we get, 5 GMm, (KE) surface =, 6 R, , 28 The system will be bound at all these, points where the total energy = (E P + E K ), is negative., In the given curve, at points A, B and C,, the E P > E K .
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108, , DAY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , 29 According to law of conservation of, angular momentum;, mv A × OA = mv B × OB ,, v B OA, =, = x, v A OB, , ∴, , 30 Total energy of satellite,, …(i), , v = 2 gR (r − R ) /r, , T 2 = kr 3, Differentiating it, we have, , … (i), , 7 The initial potential energy of the, , particle = Work done., GMm, Ui = −, r, 6.67 × 10−11 × 100 × 10−2, Ui = −, 01, ., 6.67 × 10−11, Ui = −, 01, ., = − 6.67 × 10−10 J, , 2T ∆T = 3kr 2 ∆r, , m = 10 × 10–3 kg, , Dividing it by Eq. (i), we get, , Angular momentum of satellite is given, by, , 2T ∆T, 3kr 2 ∆r, 3 ∆r, =, ⇒ ∆T = T, 2, T, kr 3, 2 r, , R = 0.1 m, M = 100 kg, , 1 /2, , = (2 Em s r s2 )1 /2, , r s = (GM m2s r s )1 /2, [from Eq. (i)], , 4 Mass of the element of length dx at a, distance x from the origin, = ( A + Bx2 ) dx, , 31 Gravitational potential,, GM, r, GM, At the surface (solid sphere)=, R, 3GM, At the centre (solid sphere)=, 2R, Hence, (b) is correct., At outside point (solid sphere) =, , dF =, , providing necessary centripetal force,, thus he feels weightlessness, as he is in, the state of free fall., , 33 Kepler’s laws are based on observations,, hit and trial method and already, recorded data but later on Newton, proved their correctness using his laws., , 34 It is clear that the net force on the body, inside the hollow sphere is zero hence,, then net gravitational field intensity, E = F at any point inside the earth, , , , m, must also be zero., , SESSION 2, 1 As the total energy of the earth satellite, − GM , bounded system is negative , ., 2a , , Where, a is radius of the satellite and M, is mass of the earth. Due to the viscous, force acting on satellite, energy, decreases continuously and radius of, the orbit or height decreases gradually., , 2 Using law of conservation of energy,, GMm 1, GMm, = mv 2 −, r, 2, R, , Gm ( A + Bx2 )dx, x2, , On integrating,, ( A + Bx2 )dx, a, x2, a+L A, , = Gm ∫, 2 + B dx, a, x, , , F = Gm ∫, , a+L, , We know that, work done = Difference in, potential energy, ∴, W = ∆U = U f − U i, ⇒, W = −Ui, [Q U f = 0], = 6.67 × 10−10 J, , 8 The metallic spheres will be at positions, as shown in the figure., , 1, , 1 , = Gm A −, + BL , a a + L, , , 32 Force acting on astronaut is utilised in, , −, , r − R, v 2 GM GM, =, −, = GM , , rR , 2, R, r, r − R, Q GM = g , = gR , , , , R2, , r , , 3 According to Kepler’s law, T 2 ∝ r 3, , GMm s, 2r s, Orbital velocity of satellite,, GM, vs =, rs, E =−, , GM , L = m sv s rs = m s , , rs , , ⇒, , T, T cos q q, , 5 Even though the distribution of mass is, unknown we can find the potential due, to ring on any axial point because from, any axial point the entire mass is at the, same distance (whatever would be the, nature of distribution)., Potential at A due to ring is,, GM, VA = −, 2R, Potential at B due to ring is,, GM, VB = −, 5R, dU = U f − U i, = U B − U A = m 0 (V B − V A ), GMm 0 1, 1 , =, −, +, R , 5, 2 , Wgr = − Wext, ⇒, Wgr = − dU = − Wext, GMm 0 1, 1 , −, ∴ Wext = dU =, R 2, 5 , , 6 Here, K i + U i = K f + U f, ∴, , T sin q, mg, , From the figure,, T sinθ = F =, , GM × M GM 2, = 2, L2, L, , and T cos θ = Mg, GM, ⇒, tanθ = 2, gL, ⇒, , GM , θ = tan −1 2 , gL , , 9 Given, U = 2r 3, ∴, , F =, , − dU, = −6r 2, dr, , mv 2, =|F |= 6r 2, r, 1, 1, or, mv 2 = (6r 3 ) = 3r 3, 2, 2, As,, TE = K + U, Now,, , = 3r 3 + 2r 3, , 2GMm, 1, mv 2 −, = 0+ 0, 2, 2R, , 2GMm, 2 2 GM, 1, or v =, mv 2 =, 2, R, 2R, Hence, (b) is the correct option., or, , F, , q, , = 5r 3, ∴ At, , r = 5m,, TE = 625J, , Hence, (c) is the correct option.
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GRAVITATION, , DAY NINE, 10 Gravitational force acting on an object, of mass m, placed at point P at a, distance h along the normal through the, centre of a circular ring of mass M and, radius r is given by, , 12 Consider cavity as negative mass and, apply superposition of gravitational, potential. Consider the cavity formed in, a solid sphere as shown in figure., V (∞ ) = 0, , r, h, , O, , P, m, , R, , ºº, P, R/2, , P R, , GMmh, (r 2 + h2 )3 /2, , When mass is displaced upto distance, 2h, then, GMm × 2h, 2GMmh, F ′= 2, =, (r + (2h )2 )3 /2 (r 2 + 4h2 )3 /2, When h = r , then, GMm × r, GMm, F = 2, =, (r × r 2 )3 /2 2 2r 2, 2GMmr, 2GMm, and F ′ = 2, =, (r + 4r 2 )3 /2, 5 5r2, ∴, , F′ 4 2, =, F, 5 5, , or, , F ′=, , 4 2, F, 5 5, , 11 Here, time period of satellite w.r.t., observer on equator is 24 h and the, satellite is moving from West to East, so, angular velocity of satellite w.r.t. the, earth’s axis of rotation (considered as, fixed) is,, 2π 2π, ω=, +, ,, Ts, Te, where, T s and Te are time periods of, satellite and the earth, respectively, π, ω = rad/h, 6, = 1.45 × 10−4 rad s –1, GM, From v =, r, GM, ⇒, rω =, r, GM, 3 /2, ⇒ r =, ω, 6. 67 × 10−11 × 6 × 1024, =, 1. 45 × 10−4, ⇒, , r = 2.66 × 107 m, = 2.66 × 104 km, , Thus, potential due to remaining part at, point P., − 11GM 3GM , V P = V3 − V2 =, − −, , 8R , 8R, (−11 + 3) GM, − GM, =, =, 8R, R, , 13 Given, a satellite is revolving in a, circular orbit at a height h from the, Earth’s surface having radius of Earth R,, i.e. h < < R., Orbital velocity of a satellite,, GM, GM, v =, =, (as h < < R ), R+ h, R, Velocity required to escape,, 1, GMm, mv ′ 2 =, 2, R+ h, 2GM, =, R+ h, , Q g = GM , , , , R2 , , gR ( 2 − 1), , 14 Net force acting on any one particle M., , According to the equation, potential at, an internal point P due to complete, sphere., 2, GM , R , V3 = − 3 3R2 − , 2, 2R , , 2, − GM 2 R − GM 11R2 , =, 3R −, 2R3 , 4 , 2R3 4 , −11 GM, =, 8R, Mass of removed part, 3, M, 4, R, M, =, × π =, 4, 2, 8, × πR 3 3, 3, Potential at point P due to removed part, −3 GM 8 −3GM, V2 =, ×, =, 2, R2, 8R, , v′ =, , 2GM, GM, −, R, R, = 2gR − gR, , v ′− v =, , R/2 -P, , M, , F =, , ∴ Minimum increase in its orbital, velocity required to escape from the, Earth’s gravitational field., , =, , +, , 109, , v, , M, , v, , R, m, , R, , v, , =, , R 45°, 45°, , O, R, M, , 2, , m, , v, , 2, , GM, GM, +, cos 45°, (2R ) 2 (R 2 ) 2, GM 2, +, cos 45°, (R 2 ) 2, , 1 , GM 2 1, +, , R2 4, 2, This force will equal to centripetal force., Mv 2 GM 2 1, 1 , So,, =, +, , R, R2 4, 2, =, , GM, (1 + 2 2 ), 4R, 1 GM, =, (2 2 + 1), 2 R, Hence, speed of each particle in a, circular motion is, 1 GM, (2 2 + 1), 2, R, v =, , 15 Total energy of earth (planet)-satellite, system is independent of eccentricity of, orbit and it depends on semi-major axis, and masses of the planet and satellite., , 16 Here, Statement I is wrong because as, , 2GM, R, (h < < R ), , speed of one satellite increases, its, kinetic energy and hence total energy, increases, i.e. total energy becomes less, negative and hence r increases, i.e. orbit, changes.
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110, , DAY TEN, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY TEN, , Unit Test 1, (Mechanics), 1 Taking into account the significant figures, what should, be the value of 9.99 + 0.0099 ?, (a) 9.9999, , (b) 10.00, , (c) 10.0, , (d) 10, , 2 The maximum error in the measurement of mass and, length of the cube are 3% and 2%, respectively. The, maximum error in the measurement of density will be, (b) 6%, , (a) 5%, , (c) 7%, , (d) 9%, , 3 If C and L denote the capacitance and inductance, then, the dimensional formula for C-L is same as that for, (a) frequency, (c) (frequency) 2, , (b) time period, (d) (time period) 2, , 4 The dimensions of (velocity) ÷ radius are the same as, 2, , that of, (a) Planck’s constant, (c) Dielectric constant, , (b) Gravitational constant, (d) None of these, , 7 A particle had a speed of 18 ms −1. After 2.4 s, its speed, was 30 ms −1 in the opposite direction. What were the, magnitude and direction of the average acceleration of, the particle during this 2.4 s interval?, (a) 10 ms −2, , (a), , ρσ, , 2, , p3, p 3σ, (c), ρ, , (b), (d), , ρp, σ, ρ, , 3, , p 3σ, , 6 An automobile travels on a straight road for 40 km at 30 km, , h−1. It then continues in the same direction for another 40, km at 60 km h−1. What is the average velocity of the car, during its 80 km trip?, (a) 30 km h−1, (c) 40 km h−1, , (b) 50 km h−1, (d) 60 km h−1, , (c) 20 ms −2, , (d) 25 ms −2, , 8 A rock is dropped from a 100 m high cliff. How long does it, take to fall first 50 m and the second 50 m?, (b) 1.5 s, 2 .5 s, (d) 3.2 s , 1.3 s, , (a) 2 s, 3 s, (c) 1.2 s, 3.2 s, , 9 Two bodies of masses M1 and M 2 are dropped from, heights H1 and H2 , respectively. They reach the ground, after time T1 and T2 , respectively. Which of the following, relation is correct?, 1/ 2, , (a), , T1 H1 , =, T2 H 2 , , (c), , T1 M1 , =, T2 M 2 , , 5 A soap bubble oscillates with time periodT , which in turn, depends on the pressure ( p ), density ( ρ ) and surface, tension ( σ ). Which of the following correctly represents, the expression for T 2 ?, , (b) 15 ms −2, , (b), , T1 H1, =, T2 H 2, , (d), , T1 M1, =, T2 M 2, , 1/ 2, , 10 As a rocket is accelerating vertically upwards, , at 9.8 ms −2 near the earth’s surface, it releases a, projectile with zero speed relative to rocket., Immediately after release, the acceleration (in ms – 2 ) of, the projectile is [take, g = 9.8 ms −2 ], (a) zero, (c) 9.8 ms−2 , down, , (b) 9.8 ms−2 , up, (d) 19.6 ms−2 , down, , 11 A bullet is fired in a horizontal direction with a muzzle, , velocity of 300 ms −1. In the absence of air resistance,, how far will it drop in travelling a horizontal distance of, 150 m? [take, g = 10 ms −2 ], (a) 1.25 cm, , (b) 12.5 cm, , (c) 1.25 m, , (d) 1.25 mm
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UNIT TEST 1 (MECHANICS), , DAY TEN, 12 A fixed mortar fires a bomb at an angle of 53° above the, , horizontal with a muzzle velocity of 80 ms −1. A tank is, advancing directly towards the mortar on level ground at, a constant speed of 5 ms −1. The initial separation (at the, instant mortar is fired) between the mortar and tank, so, that the tank would be hit is [take,g = 10 ms −2 ], (a) 678.4 m, (c) 64 m, , 18 A 20 kg block is placed on top of a 50 kg block as, shown. A horizontal force F acting on A causes an, acceleration of 3 ms −2 to A and 2 ms −2 to B as shown., For this situation mark out the correct statement., 2 ms–2, , (b) 614.4 m, (d) None of these, , Rough, , B, 20 kg, , Smooth, , A, 50 kg, , 13 A vector a is turned without a change in its length, through a small angle dθ. The value of | ∆a | and ∆a are, respectively, (b) adθ,0, (d) None of these, , (a) 0,adθ, (c) 0, 0, , 14 The vectors from origin to the points A and B are, $ respectively. The, A = 3$i − 6$j + 2k$ and B = 2$i + $j − 2k,, area of the ∆OAB be, 5, 17 sq units, 2, 3, (c), 17 sq units, 5, , 2, 17 sq units, 5, 5, (d), 17 sq units, 3, , (a), , 111, , 3 ms–2, F, , (a) The friction force between A and B is 40 N, (b) The net force acting on A is 150 N, (c) The value of F is 190 N, (d) All of the above, , 19 The figure below shows a man standing stationary w.r.t. a, horizontal conveyor belt which is accelerating at, 1 ms −2 . What is the net force on the man in this situation?, , (b), , 15 The sum of the magnitudes of the two forces acting at a, point is 18 and the magnitude of their resultant is 12. If, the resultant is at 90° with the force of smaller, magnitude, what are the magnitude of the forces?, (a) 12, 5, , (b) 14, 4, , (c) 5, 13, , (d) 10, 8, , 16 A metal sphere is hung by a string fixed to a wall. The, sphere is pushed away from the wall by a stick. The, forces acting on the sphere are shown in the second, diagram. Which of the following statements is wrong?, , Take mass of the person to be as 70 kg. If the maximum, acceleration of the belt, for which the man remains, stationary w.r.t. the belt, is 3 ms −2 , then the coefficient of, static friction between the man’s shoes and the belt, would be, (a) 70 N, 0.2, (c) 700 N, 0.1, , (b) 70 N, 0.3, (d) 700 N, 0.3, , 20 A parachutist is in free fall before opening her parachute., T, θ, , The net force on her has a magnitude F and is directed, downwards. This net force is somewhat less than, her, weight w because of air resistance. Then, she opens her, parachute. At the instant after her parachute fully inflates,, the net force on her would be, , θ, P, w, , (a) P = w tanθ, (c)T 2 = P 2 + w 2, , (b) T + P + w = 0, (d)T = P + w, , 17 Three blocks A , B and C of masses 5 kg, 3 kg and 2, kg respectively are placed on a horizontal surface., The coefficient of friction between C and surface is 0.2, while between A and surface is zero and between B, and surface is zero. If a force F = 10 N is first applied on, A as shown and then in second case on C (shown, dotted), then the ratio of normal contact force between, B and C in first with respect to the second case is, [Take, g = 10 ms −2 ], A, , F, , (a), , 13, 12, , (b), , 12, 13, , B, , (c), , C, , 1, 2, , (a) greater than F and still directed downwards, (b) less than F and still directed downwards, (c) zero, (d) directed upwards but could be more or less than F, , 21 The drive shaft of an automobile rotates at 3600 rpm and, transmits 80 HP up from the engine to the rear wheels. The, torque developed by the engine is, (a) 16.58 N-m, (c) 158.31 N-m, , 22 A disk starts rotating from rest about its axis with an, angular acceleration equal to α = 10 rads – 2 , where t is, time in seconds. At t = 0, the disk is at rest. The time, taken by the disk to make its, first complete revolution is, 1/ 3, , (d), , 2, 1, , (b) 0.022 N-m, (d) 141.6 N-m, , 6π, (a) , 5 , , 1/ 3, , 3π, (b) , 10 , , 1/ 2, , 2π, (c) , 5 , , 1/ 3, , 6π, (d) , 13
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112, , DAY TEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 29 If an object weighs 270 N at the earth’s surface, what will, , 23 A rectangular block of mass, M and height a is resting on a, 3F, smooth level surface. A force, F is applied to one corner as, a, shown in the figure. At what, a/2, point should a parallel force, F, 3 F be applied in order that, the block undergoes pure, translational motion? Assume, the normal contact force, between the block and surface, passes through the, centre of gravity of the block., a, (a) , vertically above centre of gravity, 3, a, (b) , vertically above centre of gravity, 6, (c) No such point exists, (d) It is not possible, , (c) 25 ms −1, , (b) 90 N, (d) 60 N, , 30 At its aphelion, the planet mercury is 6.99 × 1010 m from, the sun, and at its perihelion it is 4.6 × 1010 m from the sun., If its orbital speed at aphelion is 3.88 × 104 ms −1, then its, perihelion orbital speed would be, (a) 3.88 × 104 ms −1, , (b) 5.90 × 104 ms −1, , (c) 5.00 × 104 ms −1, , (d) 5.5 × 104 ms −1, , and another satellite is orbiting around the earth in a, R, circular orbit of radius , then its time period would be, 2, , zero initial velocity. In a certain time t, the pilot switches, off the engine. The sound dies away at the point of take, off in 30 s. When engine is switched off, velocity of the, helicopter is, (b) 30 ms −1, , (a) 270 N, (c) 30 N, , 31 If R is the radius of the orbit of a geosynchronous satellite, , 24 A helicopter takes off along the vertical with 3 ms −2 with, , (a) 80 ms −1, , be the weight of the object at an altitude equal to twice, the radius of the earth?, , (d)100 ms −1, , (b) 6 h, (d) Cannot be determined, , (a) 6 2 h, (c) 12 h, , 32 A cylinder of mass M and radius r is mounted on a, frictionless axle over a well. A rope of negligible mass is, wrapped around the cylinder and a bucket of mass m is, suspended from the rope. The linear acceleration of the, bucket will be, r, , 25 To maintain a rotor at an uniform angular speed of, −1, , 200 rads , an engine needs to transmit a torque of, 180 N-m. What is the power required by the engine?, (Assume efficiency of the engine to be 80%), (a) 36 kW, , (b) 18 kW, , (c) 45 kW, , w, v, , (d) 54 kW, , h, , 26 When a ball is whirled in a circle and the string, , Bucket, , supporting the ball is released, the ball flies off, tangentially. This is due to, , mg, M + 2m, mg, (c), 2M + m, , (a) its radial and tangential acceleration both are constant, (b) its radial and tangential acceleration both are varying, (c) its radial acceleration is constant but tangential, acceleration is varying, (d) its radial acceleration is varying but tangential, acceleration is constant, , quarter part of a smooth sphere of, radius R as shown in the figure. It is, released from rest at A , the normal, contact force exerted by surface on, the particle, when it reaches P is, , (c) mg ×, , 3, + mg, 2, , R, , O, , 3 mg, (b), 2, mg 3, (d), 2, , 30°, , (b), , 33 A merry-go-round, made of a ring-like platform of radius, , 27 When a particle is moving in a vertical circle,, , mg, (a), 2, , 2mg, m + 2M, 2mg, (d), M + 2m, , (a), , (a) the action of centrifugal force, (b) inertia for linear motion, (c) centripetal force, (d) some unknown cause, , 28 A particle of mass m slides on a, , M, , m, A, , P, , R and mass M, is revolving with angular speed ω. A, person of mass M is standing on it. At one instant, the, person jumps off the round, radially away from the centre, of the round (as seen from the round). The speed of the, round of afterwards is, (a) 2ω, , (b) ω, , (c) ω / 2, , d) 0, , 34 A canon ball is fired with a velocity of 200 ms − 1 at an, angle of 60° with the horizontal. At the highest point, it, explodes into three equal fragments. One goes vertically, upwards with a velocity of 100 ms − 1 and other goes, vertically downwards with 100 ms − 1. The third one moves, with a velocity of, (a) 100 ms − 1, horizontally, (b) 300 ms − 1, horizontally, (c) 200 ms − 1, at 60° with horizontal, (d) 300 ms − 1, at 60° with horizontal
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UNIT TEST 1 (MECHANICS), , DAY TEN, 35 A rocket of initial mass (including fuel) 15000 kg ejects, , 113, , P. If the velocity of the car at an instant is v, then after, travelling how much distance, it becomes double?, , mass at a constant rate of 25 kgs −1 with a constant, relative speed of 15 kms −1. The acceleration of the, rocket, 5 min after the blast is [Neglect gravity effect], , m, , F, , (b) 50 ms −2, (d) 45 ms −2, , (a) 40 ms −2, (c) 60 ms −2, , 36 An elevator of total mass (elevator + passenger), , 1800 kg is moving up with a constant speed of 2 ms ., Frictional force of 2000 N is opposing its motion. The, minimum power delivered by the motor to the elevator is, [take, g = 10 ms −2 ], (a) 36 kW, (c) 40 kW, , (a), , −1, , (b) 4 kW, (d) − 40 kW, , (b), , 4 mv 3, 3P, , (c), , mv 3, P, , (d), , 18 mv 3, 7P, , Direction, , (Q. Nos. 39-40) Each of these questions contains, two statements Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which is, the correct answer. You have to select one of the codes (a), (b),, (c) and (d) given below, , 37 A bead of mass 1 / 2 kg starts from rest from a point A to, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , B move in a vertical plane along a smooth fixed quarter, ring of radius 5 m, under the action of a constant, horizontal force F = 5 N as shown. The speed of the bead, as it reaches the point B is, [Take g = 10 ms −2 ], F, , 7mv 3, 3P, , 39 By considering the earth to be non-spherical, , A, , Statement I As, one moves from equator to the pole of, the earth, the value of accelaration due to gravity, increases., Statement II If the earth stops rotating about its own axis,, the value of accelaration due to gravity will be same at, pole and at equator., , R=5m, , B, −1, , (a) 14.14 ms (b) 7.07 ms −1 (c) 5 ms −1, , (d) 25 ms −1, , 40 Statement I Total torque on a system is independent of, the origin if the total external force is zero., , 38 A car (treat it as particle) of mass m is accelerating on a, level smooth road under the action of a single force F ., The power delivered to the car is constant and equal to, , Statement II Torque due to a couple is independent of, the origin., , ANSWERS, 1., 11., 21., 31., , (d), (c), (c), (a), , 2., 12., 22., 32., , (d), (a), (c), (d), , 3., 13., 23., 33., , (d), (b), (b), (a), , 4., 14., 24., 34., , (d), (a), (a), (b), , 5., 15., 25., 35., , (a), (c), (c), (b), , 6., 16., 26., 36., , (c), (d), (b), (c), , 7., 17., 27., 37., , (c), (a), (b), (a), , 8., 18., 28., 38., , (d), (d), (b), (a), , 9., 19., 29., 39., , (a), (b), (c), (c), , 10., 20., 30., 40., , (c), (d), (b), (a)
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114, , DAY TEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, 1 9.99 + 0.0099 = 9.9999 = 10, , The average velocity is, 80, v av =, = 40 kmh −1, 2, , 2 Here, ∆M = 3% = ± 0.03, , M, ∆l, = 2% = ± 0.02, l, ∆V, 3∆l, Hence,, =, = ± 0.06, V, l, ∆V, ∆ρ ∆ M, Now,, =, +, ρ, M, V, , 7 The average acceleration in the given, , and, , interval is, aav =, , Take, v 1 = 18 ms −1 , v 2 = − 30 ms −1 ,, , ∴ Density = Mass , Volume , , = ± 0.09 = 9%, , 3 Time period of C - L oscillations is given, by 2π LC ., Hence, [LC ] = [time period]2, , 4 Dimensional formula of (velocity) ÷, [M 0 LT −1 ]2, 0, , 0, , = [M 0 LT −2 ], , [M LT ], = [acceleration], Note In circular motion, centripetal, V2, ., acceleration is, R, , 5 Here, T = p ρ σ, 2, , a b, , c, , +c, , 1, y = ut + at 2 , at v 0 = 0, 2, 1, and solve y = v 0 t + gt 2 for t, 2, So,, t = 2 y /g, For, , …(i), , L− a − 3 b T −2 a − 2 c ], , …(ii), a+ b + c = 0, …(iii), − a − 3b = 0, and, …(iv), − 2a − 2c = 2, On solving, Eqs. (ii) (iii) and (iv), we get, a = − 3 b, b = 1 and c = 2, So, after putting the values of a, b and c, in Eq. (i), we get, ρσ 2, T2 =, p3, , Hence,, , 6 Average velocity (v av ) = ∆x, ∆t, , where, ∆x is the displacement in a given, time interval. For first part of the, journey, time taken by the car,, 40, ∆t 1 =, = 1.33 h, 30, For second part of the journey time, taken by the car, 40, ∆t 2 =, = 0.67 h, 60, Hence, the total displacement, ∆x = ∆x1 + ∆x2, = 40 + 40 = 80 km, and the total time interval, ∆t = ∆t 1 + ∆t 2 = 1.33 + 0.67 = 2 h, , 12 The situation is shown clearly in figure., Time of flight of bomb is T =, , For, , Mortar, , Tank, , =, , y = 50 m, 2(50), −, = 3.2 s, 9.8, y = 100 m, 2 (100), t2 =, = 4.5 s, 9.8, , Hence, the difference is the time taken, to fall the second 50 m, = 4.5 − 3.2 = 1.3 s, , 9 Distance of fall is independent of the, mass of the bodies, 1, 1, H 1 = gT12 and H 2 = gT22, 2, 2, , s2, , 2 × 80 × 4, = 12.8s, 10 × 5, , Distance travelled by the tank in T, seconds is, s1 = 5T = 5 × 12.8 = 64 m, The horizontal distance travelled by, bomb in T seconds is, s2 =, , =, , u2 sin 2θ, g, 802 × 2 ×, , 3 4, ×, 5 5, , 10, = 614.4 m, [Q sin 2θ = 2sin θ ⋅ cos θ], So, required separation, s = s1 + s2 = 678.4 m, , 13 From the figure|OA | = a and|OB| = a., , 1 /2, , Hence,, , 5 ms–1, , 53°, , s1, , t1 =, , 2u sin θ, g, , 80 ms–1, , The − ve sign indicates that the, acceleration is opposite to the initial, direction of motion., , So,, , Putting the dimensions of the quantities, in RHS, we get, = [ML−1 T −2 ] a [ML−3 ]b [MT −2 ] c, = [M a + b, , t 1 = 0 and t 2 = 2.4s, − 30 − 18, aav =, = −20 ms −2, 2.4, , 8 Using second equation of motion,, 2, , radius =, , (v 2 − v 1 ), (t 2 − t 1 ), , Let t be the time taken by the bullet to, cover a horizontal distance of 150 m,, then, 1, 150 = 300t ⇒ t = s, 2, 1 2 1, 1, y = gt = × 10 × = 1.25 m, 2, 2, 4, , T1 H 1 , =, T2 H 2 , , Also, from the triangle rule, B, , 10 As nothing has been mentioned that, w.r.t. which frame of reference is to be, found, it means we have to compute, w.r.t frame of reference of earth. As the, object is released, its acceleration w.r.t, ground is only due to the influence of, gravity of the earth and hence is equal, to 9.8 ms – 2 in the downward direction., , 11 Let the bullet, dropped by y metre, while covering a horizontal distance of, 150 m., , 300 ms –1, , ∆a, , a, , A, dq, , a, , O, , OB − OA = AB = ∆a, ⇒, |∆a | = AB, Arc, Using, Angle =, Radius, ⇒, AB = a⋅ dθ, So,, |∆ a | = a ⋅ dθ, ∆a means a change in magnitude of the, vector, i.e.,, , 150 m, y, , |OB| − |OA|⇒ a − a = 0, So,, ∆a = 0
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14 Given OA = a = 3$i − 6$j + 2k$ and, , ⇒, , 10 − 4, 6, =, 10, 10, , a=, , OB = b = 2 $i + $j − 2k$, , N2, , = (12 − 2)$i + (4 + 6)$j + (3 + 12) k$, , 2, , 2, , = 425 = 5 17, 1, Area of ∆OAB = |a × b|, 2, 5 17, sq units., =, 2, , ω 0 = 10 t, dθ, or, = 10 t, dt, On integrating, (for one complete, revolution), 2π, , 18 As the acceleration of A and B are, different, it means there is a relative, motion between A and B. The free body, diagram of A and B can be drawn as, B, , greater force, then according to the, problem, … (i), P + Q = 18, , ∫0, , ∫0 10 t dt, , ⇒, , 2π =, , 10T 2, 2, , ⇒, , 2π 2, T = , , 5, , 1, , 23 The free body diagram of the block can, be drawn as shown. As body has to move, in a pure translational motion, the torque, about the centre of gravity must be zero., , f, , N, , f, , a, Mg, , A, , … (iii), , F, , F, , By solving Eqs. (i), (ii) and (iii), we get, N2, , P = 5 and Q = 13, , 16, , q, T sin q, , 19 As person remains stationary w.r.t. belt,, , P, w, , As the metal sphere is in equilibrium, under the effect of the three forces, therefore, T + P + w = 0, From the figure, T cos θ = w, …(i), … (ii), T sin θ = P, From Eqs. (i) and (ii), we get, P = W tan θ, and, T 2 = P2 + w 2, [as, sin θ + cos θ = 1], 2, , 2, , 17 Friction force between C and surface is, f = µ × 2g = 0.2 × 2 × 10 = 4 N, Case I (5 + 3 + 2) a = F − f, 10 − 4, 6, ⇒, a=, =, 10, 10, a, F, , N1, f, , For C,, ⇒, Case II, , N 1 − f = 2a, , 6, 26, + 4=, N, 10, 5, F − f = (5 + 3 + 2) a, N1 = 2 ×, , so acceleration of the person w.r.t., ground is the same as that of belt w.r.t., ground. So, net force acting on the, person is, f = ma = 70 × 1 = 70 N. Let, coefficient of static friction between the, man’s shoes and belt is µ S , then, amax = µ s g, 3, µS =, = 03, ., 10, , 20 As the parachute inflates fully, the force, of air friction increases by a large, amount and the parachutist starts, decelerating, i.e. net force acting on her, is in upward direction but the, magnitude of the net force cannot be, determined from given information., , 21 From, , P = τω, P, τ =, ω, , It is given,, P = 80 HP = 80 × 746 W, = 59680 N- ms – 1, 3600, ω = 3600 rpm =, × 2 π rads – 1, 60, = 120 π rad s − 1, So, τ = 15831, . N-m, , a, ⇒, 2, , 3F × x = F ×, , Mg, , For A, F − f = Ma A = 50 × 3 = 150 N, For B, f = maB = 20 × 2 = 40 N, So, F = 150 + 40 = 190 N, , T cos q, , T, , 3F, x, , N1, , Q sin θ, tan φ =, = tan 90° = ∞, P + Q cos θ, , T, , dθ =, , N1 mg, , P 2 + Q 2 + 2PQ cos θ = 12 … (ii), , P + Q cos θ = 0, , F, , N 2 = 10 − 4 − 2 ×, , 15 Let P be the smaller force and Q be the, , ∴, , f, , 6, 24, =, N, 10, 5, N 1 26 13, Required ratio is,, =, =, N 2 24 12, , ⇒ |a × b| = 10 + 10 + 15, 2, , ⇒, , For C, F − N 2 − f = 2a, ⇒, , = 10$i + 10$j + 15k$, , ω = ω0 ± α t, , 22 As,, , a, , $i $j, k$, ∴ (a × b) = 3 − 6 2, 2 1 −2, , R=, , 115, , UNIT TEST 1 (MECHANICS), , DAY TEN, , x=, , a, 6, , 24 The altitude of the helicopter when, engine is switched off h =, not heard after t 2 = t 1 +, c = speed of sound., , at 12, . Sound is, 2, , at 12, , where, 2c, , at 12 + 2ct 1 − 2ct 2 = 0, t1 =, , ⇒, , −2 c ±, , 4c 2 + 8cat 2, 2a, , c, t1 = − +, a, , c2, a2, , +, , 2c, t2, a, , ∴ v = at 1 = − c + c 2 + 2cat 2, = − 320 +, , (320)2 + 2 × 320 × 3 × 30, , = 1600 × (10)2 − 320, = 400 − 320 = 80 ms −1, , 25 Power required for rotor, P = τ ⋅ ω, = 180 × 200 = 36 kW, P, = 45 kW, 0.8, [as efficiency is 80%], , Power of engine, P0 =, , 26 This is due to Intertia for linear motion of, ball., , 27 In a vertical circle, both radial and, tangential components of the, acceleration change direction at every, instant.
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116, , DAY TEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 28 Apply Work-Energy theorem at A and P,, , or, , 2, , mv, R, − 0 = mg ×, 2, 2, A, , R, , O, , ∴, , 30°, R/2, P, , 60°, , mg cos 60°, mg, mg sin 60°, , ⇒, , Ma, 2, mg − T, a=, m, Ma, mg −, 2 = 2mg − Ma, a=, m, 2m, , T =, , ⇒ 2ma + Ma = 2mg, 2mg, ⇒, a=, 2m + M, , 33 As no external torque acts on the, Use dynamical equations at P,, mv 2, N − mg cos 60° =, R, mg 3 mg, ⇒, N = mg +, =, 2, 2, , 29 Let m be the mass of the object and g is, the acceleration due to gravity at the, earth’s surface, then mg = 270 N., The acceleration due to gravity at an, altitude of 2Re is,, GM, GM, 1, g′ =, =, = ×g, 9, (R e + 2R e )2, 9Re2, So, required weight, = mg ′ =, , mg 270, =, = 30 N, 9, 9, , 30 From the conservation of angular, momentum,, m vA rA = mv p r p, 3.88 × 104 × 6.99 × 1010, ⇒ vp =, 4.6 × 1010, = 5.90 × 104 ms −1, , 31 From the Kepler’s law, T 2 ∝ r 3, 3 /2, , ⇒, , T1 r1 , = , T2 r2 , , ⇒, , 24 R , =, , T2 R/2 , , ⇒, , T2 =, , 24, 23 /2, , system, angular momentum should be, conserved., Hence, Iω = constant, …(i), where, I is moment of inertia of the, system and ω is angular velocity of the, system., From Eq. (i), we get, I1ω1 = I2ω2, [where ω1 and ω2 are angular velocities, before and after jumping], I, Iω = × ω2, ⇒, 2, [as mass reduced to half, hence moment, of inertia also reduced to half], ⇒, ω2 = 2ω, , 34 At the highest point, velocity before, explosion is v cos 60° along X-axis., By law of conservation of momentum,, m, (mv cos 60° ) $i =, (100$j ), 3, m, mv ′, +, (− 100 $j ) +, 3, 3, 3v 3 × 200 $, i, ⇒, v′ =, =, 2, 2, = 300 $i ms − 1, or 300 ms − 1 along X-axis or horizontally., , 35 Thrust force acting on the rocket is,, , 3 /2, , = 23 /2, , 24, =, = 6 2h, 2 2, , F = v rel, , dm, dt, , F = 15 × 1000 × 25N, , 32 Weight of bucket acts downwards while, tension T in opposite direction, mg − T = ma, Also,, τ = I α = rT, 1, 1, ⇒, Mr 2α = rT ⇒ M (rα ) = T, 2, 2, , a, , F, , Mass of rocket at t = 5min after the, blasting starts, is, m = 15000 − 25 × 5 × 60 = 7500, So,, F = ma, ⇒, , a=, =, , F, m, 15 × 25000, 7500, , = 50 ms −2, Note If gravity is not neglected, then, F − mg = ma check whether acceleration, is constant here., , 36 The net downward force on the elevator, is,, F1 = mg + f = 18000 + 2000 = 20000 N, So, the motor has to work against this, force., To move the elevator with a constant, speed, the minimum power delivered by, the motor to the elevator must be,, P = F ⋅ v = 20000 × 2 = 40 kW, , 37 Applying the work-energy theorem,, 1, × mv 2 − 0 = F × R + mg × R, 2, 1 1, 1, ⇒ × × v 2 = 5 × 5 + × 10 × 5 = 50, 2 2, 2, v = 200, = 14.14 ms −1, , 38 As, P = Fv = mv dv × v, ds, , ⇒, ⇒, , 2v, , ∫v, , mv 2dv =, s=, , s, , ∫0 P ds, 7mv 3, 3P, , 39 As one go from equator to pole of the, earth, the value of g increase due to, decrease in latitude (λ ). Also, the earth is, non-spherical, this implies the value of g,, at the poles and equitorial point on the, earth’s surface are unequal due to its, different distances from earth’s centre., , 40 If net force on the system is zero, it can, be resolved into two equal and opposite, forces which can be considered to form a, couple.
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DAY ELEVEN, , Oscillations, Learning & Revision for the Day, u, , u, , u, , Periodic Motion, Simple Harmonic Motion, Oscillations of a Spring, , u, , u, , u, , Force and Energy in SHM, Composition of Two SHMs, Simple Pendulum, , u, , Free, Damped, Forced and, Resonant Vibrations, , Periodic Motion, A motion which repeats itself over a regular interval of time is called a periodic motion., A periodic motion in which a body moves back and forth repeatedly about a fixed point, (called mean position) is called oscillatory or vibratory motion., l, , l, , l, , Period The regular interval of time after which periodic motion repeats itself is called, period of the motion., Frequency The number of times of motion repeated in one second is called frequency, of the periodic motion. Every oscillatory motion is periodic but every periodic motion, is not an oscillatory motion., Displacement as a Function of Time In a periodic motion each displacement value is, repeated after a regular interval of time, displacement can be represented as a, function of time., y = f (t ), , l, , Periodic Function A function which repeats its value after a fix interval of time is, called a periodic function., y(t ) = y(t + T ), where, T is the period of the function., Trigonometric functions sin θ and cos θ are simplest periodic functions having period, of 2π., , Simple Harmonic Motion, Simple Harmonic Motion (SHM) is that type of oscillatory motion in which the particle, moves to and fro or back and forth about a fixed point under a restoring force, whose, magnitude is directly proportional to its displacement, i.e., F ∝ x or F = − kx, where, k is a positive constant called the force constant or spring factor and x is, displacement., d2 y, Differential equations of SHM, for linear SHM, 2 + ω2 y = 0,, dt, d2 θ, for angular SHM, 2 + ω2 θ = 0, dt, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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118, , DAY ELEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Terms Related to SHM, , l, , The few important terms related to simple harmonic motion, are given as, l, , Displacement The displacement of a particle executing, SHM is, in general, expressed as y = A sin (ωt − φ)., where, A is the amplitude of SHM, ω is the angular, 2π, , , frequency where ω =, = 2 πν and φ is the initial phase, , , T, of SHM. However, displacement may also be expressed as, x = A cos (ω t − φ)., , y1 = a sin(ωt + φ1 ) and y2 = a sin(ωt + φ2 ), phase difference ∆φ = (ωt + φ2 ) − (ωt + φ1 ) = φ2 − φ1, l, , T/2, , T =, l, , T, , O, , l, , Amplitude The maximum displacement on either side of, mean position is called amplitude of SHM., Velocity The velocity of a particle executing SHM at an, instant is defined as the time rate of change of its, displacement at that instant., dy, Velocity,, = v = ω A2 − y 2, dt, At the mean position ( y = 0), during its motion, v = Aω = vmax and at the extreme positions ( y = ± A), v = 0., , 2π, | y|, Displacement, = 2π, = 2π, ω, | a|, Acceleration, , Frequency and Angular Frequency It is defined as the, number of oscillations executed by body per second. SI, unit of frequency is hertz., Angular frequency of a body executing periodic motion is, equal to product of frequency of the body with factor 2π., Angular frequency, ω = 2 πn., , Time, , –a, , l, , Time Period The time taken by a particle to complete one, oscillation is called time period. It is denoted by T., ∴ Time period of SHM,, , Displacement, , a, , Phase Difference If two particles perform S.H.M and their, equations are, , Oscillations of a Spring, If the mass is once pulled, so as to stretch the spring and is, then released, then a restoring force acts on it which, continuously tries to restore its mean position., Elongation, x1, Natural position, , ∴ Velocity amplitude, v max = Aω, , Compression, x2, , Velocity, T, T/2, O, , l, , Time, , Acceleration The acceleration of a particle executing SHM, at an instant is defined as the time rate of change of, velocity at that instant., , Restoring force, F = − k l,, where k is force constant and l is the change in length of the, spring., Here,, x1 = x2 = l, l, , d2 y, = a = − ω2 y, dt 2, The acceleration is also a variable., , Acceleration,, , At the mean position ( y = 0), acceleration a = 0 and at the, extreme position ( y = ± A), the acceleration is amax = − Aω2 ., , The spring pendulum oscillates simple harmonically, having time period and frequency given by, m, T = 2π, k, , l, , ∴ Acceleration amplitude, amax = Aω2, , T = 2π, l, , O, , T, Time, , 1, 2π, , k, m, , If the spring is not light but has a mass ms , then, , Acceleration, , T/2, , ν=, , and, , m + 1 / 3 ms, k, , If two masses m1 and m2 ,, connected by a spring, are made, to oscillate on a horizontal, surface, then its period will be T = 2π, , l, , Phase Phase is that physical quantity which tells about the, position and direction of motion of any particle at any, moment. It is denoted by φ., , where, µ =, , m1, , k, , µ, k, , m1 m2, = reduced mass of the system., m1 + m2, , m2
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OSCILLATIONS, , DAY ELEVEN, , and r1 + r2 = r = resultant position of the particle, where, m = mass of the particle., r1 , r2 = positions of the particle under two forces., , Series Combination of Springs, If two springs of spring constants k1 and k2 are joined, in series (horizontally or vertically), then their, equivalent spring constant ks is given by, k k, 1, 1, 1, =, +, ⇒ ks = 1 2, k1 + k2, ks k1 k2, ∴, , m, = 2π, ks, , T = 2π, , k1, , There are two cases, l, , k2, , m(k1 + k2 ), k1 k2, , T = 2π, , When two SHM are in same, direction the resultant is given by, , A1, , x2 = A2 sin(ωt + φ), A=, and tan β =, k2, , k1, , m, m, = 2π, kp, (k1 + k2 ), , A21 + 2 A 1 A 2 cos φ + A 22, A2 sin φ, A 1 + A 2 cos φ, , For any value of φ other than 0 and π resultant amplitude is, between| A 1 − A 2| and A 1 + A 2 ., l, , m, , When two SHM are mutually, perpendicular to each other., , Force For an object executing SHM, a force always acts on, it, which tries to bring it in mean position, i.e. it is always, directed towards mean position., The equation of motion, F = ma ,, ∴, , F = − mω x, = − kx, , [Q a = − ω x ], 2, , , Q ω =, , , k , , m, , Here, negative sign shows that direction of force is always, opposite to the direction of displacement., l, , Energy If a particle of mass m is executing SHM, then at a, displacement x from the mean position, the particle, possesses potential and kinetic energy., At any displacement x,, 1, 1, Potential energy, U = m ω2 x2 = k x2, 2, 2, 1, 1, 2, 2, Kinetic energy, K = m ω ( A − x2 ) = k ( A2 − x2 ), 2, 2, 1, Total energy, E = U + K = m ω2 A2 = 2 π 2 mν2 A2, 2, If there is no friction, the total mechanical energy,, E = K + U, of the system always remains constant even, though K and U change., , Composition of Two SHMs, If a particle is acted upon two separate forces each of which, can produce a simple harmonic motion. The resultant motion, of the particle would be a combination of two SHMs., d2 r, For which, F1 + F2 = m, dt, , y, , D, , C, , 2A2, , x, , x, y, 2 xy cos φ, + 2−, 2, A1 A2, A1 A2, 2, , Force and Energy in SHM, , 2, , A2, φ, , β, , where, x1 = A1 sin ωt ,, , m, , The resultant SHM is given by, , l, , A, , x = x1 + x2 = A sin(ωt + β), , Parallel Combination of, Springs, If two springs of spring constants k1 and k2 are, joined in parallel as shown in figure, then their, equivalent spring constant k p = k1 + k2 hence,, , 119, , 2, , A, , B, 2A1, , = sin2 φ (ellipse), where, x = A1 sin ωt and y = A2 sin(ωt + φ), Here, x is always between − A1 to + A1 and y is always, between − A2 to + A2 ., NOTE Special Cases in Composition of, , • When φ = 0, y =, , y, , D, , Two SHMs, , C, , A2, , A2, x, A1, , x, , A1, , O, A, , • When φ = π, y = −, , B, y, , D, , A2, x, A1, , then, , A12, , +, , A 22, , =1, , x 2 + y 2 = A2 (circle), , B, , y, , • When φ = π / 2. If A1 = A2 = A,, y2, , x, , A1, , O, A, , x2, , C, , A2, , D, G, , F, A2, , C, A1, , A, , H, , E, , x, , B, , Simple Pendulum, A simple pendulum, in practice, consists of a heavy but small, sized metallic bob suspended by a light, inextensible and, flexible string. The motion of a simple pendulum is simple, harmonic for very small angular displacement (θ) whose time, period and frequency are given by, l, 1 g, and ν =, T = 2π, g, 2π l, where, l is the effective length of the string and g is, acceleration due to gravity.
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120, , l, , DAY ELEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , If a pendulum of length l at temperature θ° C has a time, period T, then on increasing the temperature by ∆θ° C its, time period changes to ∆T,, ∆T 1, where,, = α ∆θ, T, 2, , where, R = radius of the earth., l, , The graphs l -T and l -T 2 intersect at T = 1 s., y, T = 1s, l–T, , where, α is the temperature coefficient of expansion of the, string., l, , l, , A second’s pendulum is a pendulum whose time period is, 2s. At a place where g = 9.8 ms −2 , the length of a second’s, pendulum is 0.9929 m (or 1 m approx)., If the bob of a pendulum (having density ρ) is made to, oscillate in a non-viscous fluid of density σ, then it can be, shown that the new period is, T = 2π, , l, , Here, positive sign is taken for an upward accelerated, motion and negative sign for a downward accelerated, motion., , l, , If a pendulum is made to oscillate in a freely falling lift or, an orbiting satellite then the effective value of g is zero and, hence, the time period of the pendulum will be infinity and, therefore pendulum will not oscillate at all., If the pendulum bob of mass m has a charge q and is, oscillating in an electrical field E, then, l, T = 2π, qE , , g ±, , , m, The positive sign is to be used if the electrical force is, acting vertically downwards and negative sign if the, electrical force is acting vertically upwards., , l, , If pendulum of charge q is oscillating in an electric field E, acting horizontally, then, T = 2π, , l, , l, , l, q 2E 2, g2 +, m2, , If the length of a simple pendulum is increased to such an, extent that l → ∞, then its time period is, R, T = 2π, = 84.6 min, g, , 2, , The graph between T and 1/g is a straight line., y, , l, , σ, g 1 − , , ρ, , l, (g ± a), , x, , O, , T2, , If a pendulum is in a lift or in some other carriage moving, vertically with an acceleration a, then the effective value of, the acceleration due to gravity becomes (g ± a) and hence,, T = 2π, , l, , l–T2, , O, l, , x, , 1/g, , The graph between T 2 and g is a rectangular hyperbola., y, T2, , O, , g, , x, , Free, Damped, Forced and, Resonant Vibrations, Some of the vibrations are described below., , Free Vibrations, If a body, capable of oscillating, is slightly displaced from its, position of equilibrium and then released, it starts oscillating, with a frequency of its own., Such oscillations are called free vibrations.The frequency, with which a body oscillates is called the natural frequency, and is given by, 1 k, ν0 =, 2π m, Here, a body continues to oscillate with a constant amplitude, and a fixed frequency.
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OSCILLATIONS, , DAY ELEVEN, , 121, , Damped Vibrations, , Forced Vibrations, , The oscillations in which the amplitude decreases gradually, with the passage of time are called damped vibrations., Damping force,, Fd = − bv, , The vibrations in which a body oscillates under the effect of, an external periodic force, whose frequency is different from, the natural frequency of the oscillating body, are called forced, vibrations., , where, v is the velocity of the oscillator and b is a damping, constant. The displacement of the oscillator is given by, x(t ) = Ae −bt /2m sin(ω′t + φ), where, ω′ = the angular frequency =, , k, b2, −, m 4 m2, , Resonant Vibrations, , The mechanical energy E of the oscillator is given, 1, by E (t ) = kA2e −bt / m, 2, , It is a special case of forced vibrations in which the, frequency of external force is exactly same as the natural, frequency of the oscillator., , amp, , ing, , A, , t, , Sm, all d, , x, , Larger, damping, w0, (b), , (a), , In forced vibrations the oscillating body vibrates with the, frequency of the external force and amplitude of oscillations, is generally small., , w, , As a result, the oscillating body begins to vibrate with a large, amplitude leading to the phenomenon of resonance to occur., Resonant vibrations play a very important role in music and, in tuning of station/channel in a radio/TV etc., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The displacement of a particle is represented by the, π, , equation y = 3 cos − 2ω t . The motion of the, 4, , particle is, (a) simple harmonic with period 2 π / ω, (b) simple harmonic with period π / ω, (c) periodic but not simple harmonic, (d) non-periodic, , 2 The displacement of a particle is represented by the, equation y = sin 3ωt . The motion is, (a) non-periodic, (b) periodic but not simple harmonic, (c) simple harmonic with period 2 π / ω, (d) simple harmonic with period π / ω, , 3 Motion of an oscillating liquid column in a U-tube is, (a) periodic but not simple harmonic, (b) non-periodic, (c) simple harmonic and time period is independent of the, density of the liquid, (d) simple harmonic and time period is directly proportional, to the density of the liquid, , 4 The relation between acceleration and displacement of, four particles are given below. Which one of the particle, is exempting simple harmonic motion?, (a) ax = +2 x, , (b) ax = +2 x 2, , (c) ax = −2 x 2, , (d) ax = −2 x, , 5 A wave travelling along the x-axis is described by the, equation y ( x , t ) = 0.005 cos (α x − β t ). If the wavelength, and the time period of the wave are 0.08 m and 2.0 s,, respectively, then α and β in appropriate units are, (a) α = 25.00 π, β = π, (c) α =, , 0.04, 10, ., ,β =, π, π, , 0.08, 2.0, ,β =, π, π, π, (d) α = 12.50 π, β =, 2.0, , (b) α =, , 6 The maximum velocity of a particle executing simple, , harmonic motion with an amplitude 7 mm, is 4.4 ms −1., The period of oscillation is, (a) 0.01 s, , (b) 10 s, , (c) 0.1 s, , (d) 100 s, , 7 A point mass oscillates along the x-axis according to the, law x = x 0 cos (ω t − π /4). If the acceleration of the, particle is written as a = A cos (ω t + δ ), then, (a) A = x 0 ,δ = −, , π, 4, , (c) A = x 0ω2 ,δ = −, , π, 4, 3π, (d) A = x 0ω2 ,δ =, 4, (b) A = x 0ω2 ,δ =, , π, 4
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122, , DAY ELEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 8 A body is executing SHM when its displacement from the, mean position are 4 cm and 5 cm and it has velocity, 10 cms − 1 and 8 cms − 1, respectively. Its periodic time t, (a), , 2π, s, 2, , (b) π s, , (c), , 3π, s, 2, , (d) 2 π s, , 9 A block rests on a horizontal table, which is executing, SHM in the horizontal direction with an amplitude a. If the, coefficient of friction is µ, then the block just starts to slip, when the frequency of oscillation is, (a), , 1 µg, 2π a, , (b) 2 π, , a, µg, , (c), , 1, a, 2 π µg, , (d), , 15 If a spring of stiffness k is cut into two parts A and B of, length lA : lB = 2 : 3, then the stiffness of spring A is given, by, ª AIEEE 2011, (a), , 5, k, 2, , (b), , undergoes horizontal SHM about a mean position O. The, coin placed on the platform does not slip, when angular, frequency of the SHM is ω. The coefficient of friction, between the coin and platform is µ. The amplitude of, oscillation is gradually increased. The coin will be begin, to slip on the platform for the first time, , (c), , k1, , k2, m, , (a), , ν, 2, , (b), , ν, 4, , 17 Two springs of force constant k and 2k are connected to, a mass as shown below. The frequency of oscillation of, the mass is, , 11 Two particles A and B are oscillating about a point O, , (a) a simple harmonic motion with amplitude (a − b), (b) a simple harmonic motion with amplitude (a + b), (c) a simple harmonic motion with amplitude a 2 + b 2, , (d) 2 ν, , (c) 4ν, , 2k, , Then, the motion of A w.r. t. B is, , (d) k, , to a mass m as shown. The frequency of oscillation of the, mass is ν. If both k1 and k 2 are made four times their, original values, the frequency of oscillation becomes, , (a) at the mean position, (b) at the extreme position of the oscillation, (c) for an amplitude of µ g /ω2, (d) for an amplitude of g /µω2, , along a common line such that equation of A is given as, x1 = a cos ω t and equation of B is given as, π, , x 2 = b sin ω t + ., , 2, , 2k, 5, , 16 Two springs of force constants k1 and k 2, are connected, , a, µg, , 10 A coin is placed on a horizontal platform, which, , 3k, 5, , (a), , 1 k, 2π m, , (b), , k, , m, , 1 2k, 2π m, , (c), , 1 3k, 2π m, , (d), , 1, 2π, , m, k, , 18 A block P of mass m is placed on a horizontal frictionless, plane. A second block Q of the same mass m is placed, on it and is connected to a spring of spring constant k,, the two blocks are pulled by a distance A. Block Q, oscillates without slipping. What is the maximum value of, frictional force between the two blocks?, , (d) not a simple harmonic motion but oscillatory motion, , 12 Two particles execute simple harmonic motion on same, straight line with same mean position, same time period, 6 s and same amplitude 5 cm. Both the particles start, SHM from their mean position (in same direction) with a, time gap of 1 s. Find the maximum separation between, the two particles during their motion., (a) 2 cm, , (b) 3 cm, , (c) 4 cm, , (d) 5 cm, , 13 A particle is acted simultaneously by mutually, perpendicular simple harmonic motion x = a cos ωt and, y = a sin ωt . The trajectory of motion of the particle will be, (a) an ellipse, (c) a circle, , A, , k, , P, , (a) k A/ 2, , (c) µ − mg, , (b) k A, , and C having equal force constant k. If the particle is, pushed a little towards any one of the springs and then, left on its own, find the time period of its oscillation., A, , 120° B, , k, , (b) 7.1 N/m, , (c) 2.2 N/m, , (d) 5.5 N/m, , C, , (a) 2 π (M /k), (c) 2 π (M / 2k), , k, , M, , 14 A silver atom in a solid oscillates in simple harmonic, , (a) 6.4 N/m, , (d) Zero, , 19 A particle of mass M is attached to three springs A, B, , (b) a parabola, (d) a straight line, , motion in some direction with a frequency of1012 per, second. What is the force constant of the bonds, connecting one atom with the other? (Take, molecular, weight of silver = 108 and Avogadro number = 6.02 × 1023, ª JEE Main 2018, g mol −1), , µs, , Q, , k, , (b) 2 π (2M /k), (d) 2 π (M / 3 k)
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OSCILLATIONS, , DAY ELEVEN, 20 A body performs SHM. Its kinetic energy K varies with, , 24 For a particle executing SHM, the displacement x is, given by x = A cos ω t . Identify the graph which, represents the variation of potential energy (PE) as a, function of time t and displacement x., , time T as indicated in the graph, K, , K, , PE, , (a), , (b), , t, , T, , 123, , PE, , t, , T, , I, , III, , II, , IV, K, , t, , K, , (a) I and III, (c) II and III, (c), , (d), , t, , T, , t, , T, , 21 A particle is executing simple harmonic motion with a, time period T . At time t = 0, it is at its position of, equilibrium. The kinetic energy-time graph of the particle, will look, like, ª JEE Main 2017 (Offline), KE, , KE, , 25 A simple pendulum performs simple harmonic motion, about x = 0 with an amplitude a, and time period T . The, speed of the pendulum at x = a / 2 will be, (a), , πa 3, T, , (b), , πa 3, 2T, , T, , O, , (b), , t, , KE, , (c), , O, , O, , T/2, , 3T/4, , T, , (d), , t, , O, , T/2, , T, , t, , (d), , πa, T, , compared to sea level. If a simple pendulum is used to, record the time, then the length must be, (b) decreased by 0.1%, (d) decreased by 0.2%, , 27 Two pendulums have time periodsT and, , KE, , T/4 T/2, , 3 π2 a, T, , 5T, . They start, 4, SHM at the same time from the mean position. What will, be the phase difference between them after the bigger, pendulum completes one oscillation?, , t, , T, , (c), , 26 The value of g decrease by 0.1% on a mountain as, , (a) increased by 0.1%, (c) increased by 0.2%, (a), , x, , (b) II and IV, (d) I and IV, , 2T, , (a) 45°, , (b) 90°, , (c) 60°, , (d) 30°, , 28 A simple pendulum of length l is suspended from the roof, 22 For a simple pendulum, a graph is plotted between its, Kinetic Energy (KE) and Potential Energy (PE) against its, displacement d. Which one of the following represents, these correctly? (graphs are schematic and not drawn to, ª JEE Main 2015, scale), E, , E, , PE, , (b), KE, , d, E, , E, , (c), , KE, , (a) 2 π l/g, , (b) 2 π l/ (a 2 + g 2 )1/ 2, , (c) 2 π l/ (a + g), , (d) 2 π l/(g − a), , 29 Two simple pendulums of length1 m and 4 m respectively, , PE, , KE, , (a), , of a train which is moving in a horizontal direction with an, acceleration a. Then, the time period T is given by, , d, , PE, , are both given small displacement in the same direction., The shorter pendulum has completed number of, ª JEE Main (Online) 2013, oscillations equal to, (a) 2, (c) 5, , (b) 7, (d) 3, , (d), KE, d, , PE, , 30 A pendulum of length 2m lift at P. When it reaches Q, it, losses 10% of its total energy due to air resistance. The, velocity of Q is, P, , 23 The total energy of a particle, executing simple harmonic, motion is, (a) ∝ x, (b) ∝ x 2, (c) independent of x, (d) ∝ x1/ 2, where, x is the displacement from the mean position., , 2m, , Q, , (a) 2 m/s, (c) 6 m/s, , (b) 1 m/s, (d) 8 m/s
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124, , 40 DAYS ~ JEE MAIN PHYSICS, , 31 Four pendulums A,B,C and D are, hung from the same elastic support, as shown alongside. A and C are of, B, C, A, the same length while B is smaller, D, than A and D is larger than A. A is, given a displacement then in steady state, (a) D will vibrate with maximum amplitude, (b) C will vibrate with maximum amplitude, (c) B will vibrate with maximum amplitude, (d) All the four will oscillate with equal amplitude, , 32 Bob of a simple pendulum of length l is made of iron. The, pendulum is oscillating over a horizontal coil carrying, direct current. If the time period of the pendulum is T ,, then, ª JEE Main (Online) 2013, (a) T < 2 π, , l, and damping is smaller than in air alone, g, , l, (b) T = 2 π, and damping is larger than in air alone, g, (c)T > 2, (d)T < 2 π, , l, and damping is smaller than in air alone, g, l, and damping is larger than in air alone, g, , 33 The amplitude of a damped oscillator decreases to 0.9, times its original magnitude is 5s. In another 10 s it will, decreases to α times its original magnitude, where α, equals, ª JEE Main 2013, (a) 0.7, , (b) 0.81, , (c) 0.729, , (d) 0.6, , Direction, , (Q. Nos. 34-38), Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d ) given below., (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, , DAY ELEVEN, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 34 If two springs S1 and S 2 of force constants k1 and k 2,, respectively are stretched by the same force, it is found, that more work is done on spring S1 than on spring S 2., Statement I If stretched by the same amount, work done, on S1, will be more than that on S 2., Statement II k1 < k2, , 35 Statement I A particle performing SHM at certain instant, is having velocity v. It again acquires a velocity v for the, first time after a time interval of T second, then the time, period of oscillation isT second., Statement II A particle performing SHM can have the, same velocity at two instants in one cycle., , 36 Statement I A particle performing SHM while crossing, the mean position is having a minimum potential energy,, this minimum potential energy could be non-zero., Statement II In the equilibrium position, the net force, experienced by the particle is zero, hence potential, energy would be zero at the mean position., , 37 Statement I A circular metal hoop is suspended on the, edge by a hook. The hoop can oscillate from one side to, the other in the plane of the hoop, or it can oscillate back, and forth in a direction perpendicular to the plane of the, hoop., The time period of oscillation would be more when, oscillations are carried out in the plane of the hoop., Statement II Time period of physical pendulum is more if, the moment of inertia of the rigid body about the, corresponding axis, passing through the pivoted point is, more., , 38 Statement I The time period of a pendulum, in a satellite, orbiting around the earth, is infinity., Statement II Time period of a pendulum is inversely, proportional to the square root of acceleration due to, gravity.
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OSCILLATIONS, , DAY ELEVEN, , 125, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A 15 g ball is shot from a spring gun whose spring has a, force constant of 600 Nm −1. The spring is compressed, by 5 cm. The greatest possible horizontal range of the, ball for this compression is (g = 10 ms −2)., (a) 10.0 m, , (b) 6.0 m, , (c) 12.0 m, , (d) 8.0 m, , 2 Two simple harmonic motions are represented by the, , π, , equations y1 = 01, . cos π t ., . sin 100π t + and y 2 = 01, , 3, The phase difference of the velocity of particle 1, with, respect to the velocity of particle 2 is (at t = 0), (a), , −π, 6, , (b), , π, 3, , (c), , −π, 3, , (d), , π, 6, , 3 A piece of wood has dimension a × b × c. It is floating in, a liquid of density ρ such that side a is vertical. It is now, pushed down gently and released. The time period is, (a) 2 π ρa /g, (c) 2 π g / ρa, , (b) 2 π abc /g, (d) 2 π bc /ρg, , 4 The length of a spring is α when a force of 4N is applied, on it. The length of a spring is β when a force of 5N is, applied on it. Then find the length of the spring when a, force of 9 N is applied on the spring., (a) 5 β − 4 α, (c) 5 α − 4 β, , (b) β − α, (d) 9(β − α), , 5 A simple pendulum of length l has a bob of mass m with, a charge q on it. A vertical sheet of charge having, surface charge density σ passes through the point of, suspension. At equilibrium, the string makes an angle θ, with the vertical. If the tension in the string is T then,, σq, 2 ε0 mg, l, (c)T > 2 π, g, , (a) tan θ =, , σq, ε0 mg, l, (d)T = 2 π, g, , (b) tan θ =, , 6 A mass m is suspended from a, massless pulley which itself is, suspended with the help of a, massless extensible spring as, shown alongside., What will be the time period of, oscillation of the mass? The force, constant of the spring is k., , 7 If x, v and a denote the displacement, the velocity and, the acceleration of a particle executing simple harmonic, motion of time period T, then which of the following does, not change with time?, aT, x, aT, (d), v, , (a) a 2T 2 + 4 π 2v 2, , (b), , (c) aT + 2 πv, , 8 A simple pendulum has time period T1. The point of, suspension is now moved upward according to the, relation y = k t 2,(k = 1ms − 2 ), where y is the vertical, displacement. The time period now becomesT2. The ratio, T2, of 12 is ( take, g = 10 ms − 2 ), T2, (a), , 6, 5, , (b), , 5, 6, , (c) 1, , A has time period T. When an additional mass M is added, to its bob, the time period changesTM . If the Young’s, modulus of the material of the wire isY , then 1/Y is equal to, ( g = gravitational acceleration), ª JEE Main 2015, T 2 A, (a) M − 1, T , Mg, 2, , A, T, (c) 1 − M , , , T, , Mg, , T 2 Mg, (b) M − 1, T , A, 2, T A, (d) 1 − , TM Mg, , , , 10 The bob of a simple pendulum is a spherical hollow ball, filled with water. A plugged hole near the bottom of the, oscillating bob gets suddenly unplugged. During, observation, till the water is coming out, the time period of, oscillation would, (a) first increase and then decrease to the original value, (b) first decrease and then increase to the original value, (c) remain unchanged, (d) increase towards a saturation value, , 11 A pendulum of length l = 1 m is released from θ 0 = 60°., The rate of change of speed of the bob at θ = 30° is, ( take, g = 10 m /s 2 )., 60°, , (b) 2 π m / k, (d) 2 π m / 2 k, , 4, 5, , 9 A pendulum made of a uniform wire of cross-sectional area, , (a) π m / k, (c) 4 π m / k, , (d), , (a) 5 3 m/s 2, (c) 10 m/s 2, , (b) 5 m/s 2, (d) 2.5 m/s 2
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126, , DAY ELEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 12 A particle at the end of a spring executes simple, , 16 A particle moves with simple harmonic motion in a, , harmonic motion with a period t1, while the corresponding, period for another spring is t 2. If the period of oscillation, with the two springs in series is T , Then,, (a)T = t1 + t 2, (c)T −1 = t1−1 + t 2−1, , (b)T 2 = t12 + t 22, (d)T −2 = t1−2 + t 2−2, , straight line. In first τ sec, after starting from rest it travels, a distance a and in next τ sec,it travels 2a, in same, direction, then, ª JEE Main 2014, (a) amplitude of motion is 3a, (b) time period of oscillations is 8τ, (c) amplitude of motion is 4a, (d) time period of oscillations is 6τ, , 13 A particle performs simple harmonic motion with, amplitude A . Its speed is tripled at the instant that it is at, 2, a distance A from equilibrium position. The new, 3, amplitude of the motion is, ª JEE Main 2016 (Offline), A, 41, 3, (c) A 3, (a), , (b) 3A, (d), , 7, A, 3, , 14 A wooden cube (density of wood d ) of side l floats in a, , 17 An ideal gas enclosed in a vertical cylindrical container, supports a freely moving piston of mass M. The piston, and the cylinder have equal cross sectional area A. When, the piston is in equilibrium, the volume of the gas isV0, and its pressure is P. The piston is slightly displaced from, the equilibrium position and released. Assuming that the, system is completely, isolated from its surrounding, the, piston executes a simple harmonic motion with frequency, , liquid of density ρ with its upper and lower surfaces, horizontal. If the cube is pushed slightly down and, released, it performs simple harmonic motion of periodT ., Then, T is equal to, (a) 2 π, , lρ, ( ρ − d) g, , (b) 2 π, , ld, ρg, , (c) 2 π, , lρ, dg, , (d) 2 π, , ld, (ρ − d) g, , (c), , the same amplitude A and frequency ω along the x-axis., Their mean position is separated by distance X 0( X 0 > A )., If the maximum separation between them is ( X 0 + A ), the, phase difference between their motion is, , 1 A 2 γ P0, 2 π MV0, , (d), , 1 MV0, 2 π A γ P0, , π, 4, π, (d), 2, , factor of 1/e of original) only in the period between t = 0 s, to t = τ s, then τ may be called the average life of the, pendulum. When the spherical bob of the pendulum, suffers a retardation (due to viscous drag) proportional to, its velocity with b as the constant of proportionality, the, average life time of the pendulum is (assuming damping, is small) in seconds, 0.693, b, 1, (c), b, , (b), , (a), , 1 V0MP0, (b), 2 π A2 γ, , 18 If a simple pendulum has significant amplitude (up to a, , 15 Two particles are executing simple harmonic motion of, , π, 3, π, (c), 6, , ª JEE Main 2013, , 1 A γ P0, (a), 2 π V0M, , (a), , (b) b, (d), , 2, b, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (b), , 3 (c), , 4 (d), , 5 (a), , 6 (a), , 7 (d), , 8 (b), , 9 (a), , 10 (c), , 11 (a), , 12 (d), , 13 (c), , 14 (b), , 15 (a), , 16 (d), , 17 (c), , 18 (a), , 19 (c), , 20 (a), , 21 (c), , 22 (b), , 23 (c), , 24 (a), , 25 (a), , 26 (b), , 27 (b), , 28 (b), , 29 (a), , 30 (c), , 31 (b), , 32 (d), , 33 (c), , 34 (d), , 35 (d), , 36 (c), , 37 (a), , 38 (a), , 1 (a), 11 (b), , 2 (a), 12 (b), , 3 (a), 13 (d), , 4 (a), 14 (b), , 5 (a), 15 (a), , 6 (a), 16 (d), , 7 (b), 17 (c), , 8 (a), 18 (d), , 9 (a), , 10 (a)
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128, , DAY ELEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 16 We know that, ν = 1, , 2π, , k1 + k2, m, , When k1 and k2 are made four times, their original value., Then, and, k + k2, 1, ν′ =, ⋅2 1, = 2ν, 2π, m, , 17 The effective spring constant is, , K = k + 2k = 3 k ., The time period of oscillation is given by, m, 1, and ν =, T = 2π, 3k, T, , 24 Potential energy is minimum (in this, , case zero) at the mean position ( x = 0), and maximum at the extreme positions, ( x = ± A )., At time t = 0, x = A, the potential energy, should be maximum. Therefore, graph I, is correct. Further in graph III, potential, energy is minimum at x = 0. Hence, this, is also correct., , 25 Since, v = ω a2 − y 2 ,, , =, , 1, ν=, 2π, , 3k, m, , 18 Angular frequency of the system,, ω=, , k, =, m+ m, , k, 2m, , Maximum acceleration of the system, kA, will be, ω2 A or, . This acceleration of, 2m, the lower block, is provided by friction., Hence, f max = ma max = m ω2 A, k A k A, = m , =, 2m , 2, , 19 When the mass m is pushed in a, downward direction through a distance, x, the effective restoring force, in, magnitude is, F = k x + k x cos 60° + k x cos 60°, = 2k x, ∴ Spring factor, k ′ = 2k, and Inertia factor = M, M, So time period, T = 2 π, 2k, , 21 KE is maximum at mean position and, minimum at extreme position, at t = T ., , , , 4, , 22 During oscillation, motion of a simple, pendulum KE is maximum at the mean, position where PE is minimum. At, extreme position, KE is minimum and, PE is maximum. Thus, correct graph is, depicted in option (b)., , 23 In a simple harmonic motion, when a, particle is displaced to a position from, its mean position, its kinetic energy is, converted into potential energy. Hence,, total energy of a particle remains, constant or the total energy in simple, harmonic motion does not depend on, the displacement x., , 2π, ×, T, , 26 As, T = 2π l / g, Taking log and differentiating the, expression, keeping T constant we have, dl dg, 01, ., =−, =, 100, l, g, ∴, , (d l / l ) × 100 = − 01, . / 100 × 100, = − 01, .%, , 27 When bigger pendulum of time period, (5T /4) completes one oscillation, the, smaller pendulum will complete (5/ 4), oscillation. It means, the smaller, pendulum will be leading the bigger, pendulum by a phase of T /4 = π /2 rad, = 90°., , 28 Effective acceleration = a + g, 2, , ∴Time period, T = 2 π, , 2, , l, (a2 + g 2 )1 /2, , 29 Let T1 and T2 be the time period of, shorter length and larger length, pendulums respectively. According to, question,, nT1 = (n − 1) T2, , 20 The frequency of kinetic energy is twice, that of a particle executive SHM., , 3 a2, a2, =ω, 4, 4, 3a π 3a, =, T, 2, , v = ω a2 −, , so, we get, , 1, = (n − 1) 2 π, 8, , So, n 2 π, , 4, 8, , or n = (n − 1) 2 = 2n − 2 ⇒ n = 2, , 30 By applying conservation of energy, between P and Q, 1, mv 2 = 0.9(mgh ), 2, ⇒ v 2 = 2 × 0.9 × 10 × 2 = 36 ⇒ v = 6 m/s, , 31 As A and C are of same length, so they, will be in resonance, hence C will, vibrate with the maximum amplitude., , 32 T < 2 π, , A = A 0e, , −, , l, g, , As, current passed through in the coil, which attracts the molecules of air, closer to it, thus density of air increases, which produces larger damping than, that in air alone., , bt, 2m, , After 5 s, 0.9 A 0 = A 0 e, 0.9 = e, , ⇒, , −, , b( 5), 2m, , −, , b ( 5), 2m, , …(i), , After 10 more second,, A = A 0e, , At, x or y = a / 2, ⇒, , 33 Amplitude of damped oscillator,, , −b, , (15), 2m, , − 5b , = A0e 2m , , , , , , 3, , …(ii), , From Eqs. (i) and (ii), we get, A = 0729, ., A0, Hence,, , ., α = 0729, , 34 As no relation between k1 and k2 is given, in the question, that is why, nothing can, be predicted about Statement I. But as in, Statement II, k1 < k2, Then, for same force, W = F⋅x= F⋅, , F, F2, =, K, K, , 1, k, i.e., W1 > W2, But for same displacement,, 1, 1, W = F ⋅ x = k x ⋅ x = k x2, 2, 2, ⇒ W ∝ k , i.e. W1 < W2, ⇒, , W ∝, , Thus, in the light of Statement II,, Statement I is false., , 35 Consider the situation as shown in the, adjoint figure. Let us say at any instant t 1 ,, the particle crosses A as shown, the, particle again acquires the same velocity,, when it crosses B let us say at instant t 2 ., According to statement I, (t 2 − t 1 ) is the, time period of SHM which is wrong., x1, B, , x1, , Equilibrium position, , A, , 36 At the mean position,, dU, =0, dx, ⇒ U = constant which can be zero or, non-zero., F = 0= −, , 37 When the hoop oscillates in its plane,, moment of inertia is, I1 = mR2 + mR2 i.e. I1 = 2mR2, While when the hoop oscillates in a, direction perpendicular to the plane of, the hoop, moment of inertia is, 3 mR2, mR2, I2 =, + mR2 =, 2, 2, Time period of physical pendulum is,, I, ; d is same in both the, T = 2π, mgd, cases.
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38 From the relation of the time period,, l, T = 2π, g, , 4 = k (α − l ), , 4, , 5 = k (β − l ), , 1, ⇒ T ∝, g, , When the satellite is orbiting around the, earth, the value of g inside it is zero., Hence, the time period of pendulum in, a satellite will be infinity and it is also, clear that time period of pendulum is, inversely proportional to square root of, acceleration due to gravity g., , 9 = k (γ − l ) ⇒, , 1 For getting horizontal range, there must, be some inclination of spring with, ground to project ball., , = 5α − 5l, = 5α − 4β, = 4γ − 4l, = 9α − 5(5α − 4β ), = 9α − 25α + 20β, = 20β − 16α, γ = 5β − 4α, , y 2 = 0.1 cos π t, dy 2, ⇒, = v 2 = − 01, . sin π t, dt, or, v 2 = 01, . sin ( π t + π ), Hence, the phase difference, ∆φ = φ 1 − φ 2, 5π , = 100 π t +, − ( πt + π ), , 6, 5π, π, (at t = 0), =, − π=−, 6, 6, 3 Force of buoyancy = b × c × ρw × g, and, , B, , (Q ρw = 1), = bc g, and mass of piece of wood = ab c ρ, So, acceleration, = − bc g / ab c ρ = − (g /aρ ), ρa, Hence, time period, T = 2π, g, , F, , T M = 2π, , then the spring is pulled by 2y and the, force with which the spring is pulled, will be F = R = mg / 2., Hence, mg / 2 = k ( 2 y ), ⇒, y / g = m / 4k, T = 2π y /g, ⇒, = 2 π m /4 k = π m / k, 2, 7 As, aT = ω xT, , ×T =, , 4 π2, T, , 8 Given, y = k t 2 ⇒ a =, , d2 y, dt 2, , MgL, AY, g, , L+, , T M = 1 + Mg, , , T , AY, , or, , Mg T M , =, −1, T , AY, 2, , 1, A T M , =, , − 1, Y Mg T , , , 10, , l, , l + Dl, G, G, , G¢, , Spherical hollow ball, filled with water, , T, = constant., , L + ∆L, ,, g, , 2, , or, , 6 If mass m moves down a distance y,, , 2, , L, g, , 2, , ⇒, , As the bob is in equilibrium,, mg, F, T, so, =, =, OC CB, BO, σq, CB, F, or, =, =, = tanθ, OC, mg 2ε0 m g, , =, , g, 10 + 2 6, =, =, 10, 5, , We know that Young modulus of the, material, Mg / A MgL, Y =, =, ∆L / L, A∆L, MgL, ∆L =, ⇒, AY, , Force on bob due to the sheet of charge,, 1 σq, F = qE =, 2 ε0, , x, 4 π2, , g + ay, , T M = 2π, , mg, Sheet of charge, 1σ, E =, 2 ε0, , x, , =, , where, ∆L is increase in length., , C, , u, =, g, , T22, , When additional mass M is added to its, bob, , l, , Rmax, , 5π , or v 1 = 10 π sin 100 π t +, , , 6, , T12, , l, g + ay, , T = 2π, , T, , 2, , 3, dy 1, . × 100 π cos, ⇒, = v 1 = 01, dt, 100 π t + π , , , , 3, π, π, or v 1 = 10 π sin 100 π t +, + , , 3, 2, , ∴, , O, , 45°, , π, 2 Given, y 1 = 01, . sin 100 π t + , , , , T2 = 2 π, , intensity at B due to the sheet of charge,, , u, , = 10 m, , and, , 9 We know that time period,, , 5 In the figure, we represent the electric, , q, , But KE acquired by ball, = PE of spring gun, 1, 1, kx2, 2, mu = kx2 ⇒ u2 =, ⇒, 2, 2, m, kx2, 600 × (5 × 10−2 )2, ⇒ R max =, =, mg, 15 × 10−3 × 10, , 4 α −l, =, 5 β−l, , 4β − 4 l, l, Now, 9α − 9l, 4 γ = 9α − 5l, , or, , SESSION 2, , R max, , 129, , OSCILLATIONS, , DAY ELEVEN, , T = 2π, , l, g, , Spherical hollow ball, half filled with water, , T1 = 2 π, , l + ∆l, g, , = 2k, , a, , l, G, , axt, , g, , or, , a y = 2 m/s2, , ∴, , T1 = 2 π, , l, g, , (as, k = 1 m/s 2 ), , Spherical hollow ball, , T2 = 2 π, , l, and T1 > T2, g, , Hence, time period first increases and, then decreases to the original value.
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DAY TWELVE, , Waves, Learning & Revision for the Day, u, , u, , u, , u, , Wave Motion, Speed of Waves, Sound Waves, , u, , u, , Displacement Relation for a, Progressive or Harmonic Wave, , Principle of, Superposition of Waves, Reflection and Transmission, of Waves, , u, , u, , u, , Standing or Stationary, Waves, Beats, Doppler’s Effect, , Wave Motion, Wave motion involves transfer of disturbance (energy) from one point to the other with, particles of medium oscillating about their mean positions i.e. the particles of the, medium do not travel themselves along with the wave. Instead, they oscillate back and, forth about the same equilibrium position as the wave passes by. Only the disturbance is, propagated., 1. Longitudinal Waves When particles of the medium vibrate parallel to the direction, of propagation of wave, then wave is called longitudinal wave. These waves, propagate in the form of compressions and rarefactions. They involve changes in, pressure and volume. The medium of propagation must possess elasticity of volume., They are set up in solids, liquids and gases., 2. Transverse Waves When the particles of the medium vibrate in a direction, perpendicular to the direction of propagation of wave, then wave is called transverse, waves. These wave propagtes in the form of crests and troughs. These waves can be, set up in solids, on surface of liquids but never in gases., , Terms Used in Wave Motion, l, , l, , l, , Angular Wave Number Number of wavelength in the distance 2π is called the wave, number or propagation constant., 2π, rad/m, K =, λ, Particle velocity It is the velocity of the particle executing simple harmonic motion., dy, i.e., v=, dt, where, y denotes displacement at any instant., Wave Velocity The velocity of transverse wave motion is given by, Distance travelled by wave, v=, Time taken, i.e., , λ 1, ω, or v = νλ, v= = λ=, , , T, T, x, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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132, , l, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Differential Equation of Wave Motion, d2 y, 1 d2 y, = 2, 2, dx, v dt 2, , Factors Affecting Speed of Sound, , Speed of Waves, , Effect of Temperature on Velocity With rise in, temperature, the velocity of sound increases as, γ RT, v, T, ; i.e. v ∝ T ; 2 = 2, v=, M, v1, T1, , Speed of waves is divided in two types as per the nature of, wave, these are given below, , Speed of sound in air increases by 0.61 m/s for every 1°C, rise in temperature., , l, , l, , 1. Speed of Transverse Wave, The expression for speed of transverse waves in a solid and in, case of a stretched string can be obtained theoretically, l, , In solids, v =, , l, , η, d, , where, η is the modulus of rigidity and d is the density of, the medium., l, , In a stretched string, v =, , T, =, m, , Mg, πr 2 d, , where, T = the tension in the string,, m = the mass per unit length of the string,, M = mass suspended from the string,, r = radius of the string and, d = density of the material of the string., , 2. Speed of Longitudinal Wave, (or Sound Wave), Following are the expressions for the speed of longitudinal, waves in the different types of media, l, , v=, , Sound Waves, The longitudinal waves which can be heard are called sound, waves., They are classified into following categories, l, , l, , l, , B+, , where B, η and ρ are values of bulk modulus, modulus of, rigidity and density of the solid respectively., If the solid is in the form of a long rod, then, v=, , Y, ρ, , Effect of Humidity When humidity in air increases, its, density decreases and so velocity of sound increases., Y, B, For solids, v =, . For liquids, v =, D, D, where, Y = Young’s modulus of elasticity, B = bulk modulus of elasticity., , If the medium is solid,, 4, η, 3, ρ, , Effect of Pressure for Gase Medium Pressure has no effect, on the velocity of sound, provided temperature remains, constant., , Infrasonics The longitudinal waves having frequencies, below 20 Hz are called infrasonics. These waves cannot be, heard. These waves can be heard by snakes., Audible waves The longitudinal waves having the, frequency between 20 Hz and 20000 Hz are called audible, waves. Human can hear these waves., Ultrasonics The longitudinal waves having the, frequencies above 20000 Hz are called ultrasonics. These, waves are also called supersonic waves or supersonics., , Displacement Relation for a, Progressive or Harmonic Wave, The equation of a plane progressive or simple harmonic wave, travelling along positive direction of x-axis is, y = a sin (ωt − kx) ⇒ y = a sin, , where, Y is the Young’s modulus of the solid material., l, , In a liquid,, v=, , B, ρ, , According to Newton’s formula, speed of sound in a gas, is obtained by replacing B with initial pressure p of the gas, i.e. B = p., p, v=, ρ, , y = a sin, , ⇒, , x, t, y = a sin 2π − ., T λ, , l, , T, , t − x ., , λ, , 2π, x, , (vt − x) ⇒ y = a sin ω t − , , λ, v, , ⇒, , where B is the bulk modulus of the liquid., l, , 2π, T, , If maximum value of y = a, i.e. a is amplitude, then, dy / dt = velocity of particle, dy, 2π, v=, = aω cos ⋅, (vt − x), dt, λ, 2πva, dy , [where, n = frequency], =, = 2πna = ωa, , dt max, λ
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WAVES, , DAY TWELVE, , l, , l, , l, , Acceleration of particle, d2 y, 2π, = − ω2 a sin, (vt − x), λ, dt 2, , Constructive and Destructive Interference, l, , For a wave, velocity of sound wave, v = frequency (n) × wavelength (λ ), ⇒, v = nλ, 2π, 2 πv, Angular speed, ω = 2 πn =, ⇒ω =, T, λ, , l, , Relation between Phase Difference,, Path Difference and Time Difference, l, , Phase difference (φ) =, , 2π, × path difference ( x), λ, , Intensity (I 1) ∝ (Amplitude A)2, , I = I 1 + I2 + 2 I 1I2 cos φ, I max = I 1 + I2 + 2 I 1I2 = ( I 1 +, , I min = ( I 1 − I2 )2, for φ = (2 n + 1) π, , and, , tan θ =, , A21 + A22 + 2 A1 A2 cos φ, A2 sin φ, A1 + A2 cos φ, , Power, If P is power of a sound source, then intensity (I ) follows, inverse square law of distance (d)., P, I =, 4πd2, , Reflection and Transmission, of Waves, , Interference of Waves, , A=, , I2 )2 , for φ = 2 πn, , and I min = I 1 + I2 − 2 I 1I2, , y = y1 + y2 + y3 + ....., , where,, , 2, , If I 1 and I2 are intensities of the interfering waves and φ is, the phase difference, then resultant intensity is given by, , Two or more waves can traverse the same space, independently of one another. The resultant displacement of, each particle of the medium at any instant is equal to the, vector sum of displacements produced by the two waves, separately. This principle is called principle of superposition, of waves., , y = y1 + y2 = A sin (ωt + φ), , I 1 A1 , = , I2 A2 , , ∴, , Principle of Superposition, of Waves, , When two waves of same frequency (or, A2, same wavelength) travelling along same, path superimpose each other, there, φ, occurs redistribution of energy in the, medium. At a given position (x being, θ, constant) displacement due to two, waves be, y1 = A1 sin ωt and y2 = A2 sin (ωt + φ), Then, resultant displacement, , When the waves meet a point with opposite phase,, destructive interference is obtained at that point., (i) Phase difference between the waves at the point of, observation φ = 180 ° or (2 n − 1)π., (ii) Resultant amplitude at the point of observation will, be minimum, Amin = A1 − A2 ., , The intensity of waves is the average amount of energy, transported by the wave per unit area per second normally, across a surface at the given point., , 2πx, φλ, ⇒x=, λ, 2π, 2π, Phase difference (φ) =, × time difference (t ), T, 2πt, Tφ, ⇒, φ=, ⇒t =, T, 2π, T, Time difference (t ) = × path difference ( x), λ, Tx, λt, ⇒, t =, ⇒x=, λ, T, φ=, , l, , When the wave meet a point with some phase,, constructuve interference is obtained at that point., (i) Phase difference between the waves at the point of, observation φ = 0 ° or 2πn., (ii) Resultant amplitude at the point of observation will, be maximum, Amax = A1 + A2 ., , Intensity, , ⇒, , l, , 133, , When sound waves are incident on a boundary separating two, media, a part of it is reflected back into the initial medium while, the remaining is partly absorbed and partly transmitted into the, second medium., , A, , Standing or Stationary Waves, A1, , Standing or stationary wave is formed due to superposition of, two progressive waves of same nature, same frequency (or, same wavelength), same amplitude travelling with same, speed in a bounded medium in mutually opposite directions., If the incident wave be represented as y1 = A sin(ω t − kx), and the reflected wave as y2 = A sin (ω t + kx),, then y = y1 + y2 = A sin(ω t − kx) + A sin(ω t + kx), ⇒ y = 2 A cos kx sin ωt, The resultant wave does not represent a progressive wave.
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134, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Standing Waves in String, , l, , First harmonic l =, , Consider a string of length L stretched under tension T, between two fixed points (i.e. clamped at its ends). Transverse, wave is set up on the string whose speed is given by v = T /µ,, , λ1, ⇒, 2, , f1 =, , A1, , A2, N, λ, l= 1, 2, (a), , where µ is the mass per unit length of the string., Different modes of vibration of stretched string are discussed, below, l, , A, Let only one anti-node A is, formed at the centre and, A, string vibrates in one, Fundamental or first harmonic, segment only, it is called, fundamental mode, then, λ, L = 1 or λ 1 = 2 L, 2, Frequency of vibration in fundamental mode, v, 1 T, ν1 =, =, λ 1 2L µ, , l, , Second harmonic or first overtone l = λ 2 ; f =, A1, , If string vibrates in two, segments, then, L = λ2, and ν2 =, , v, 1, =, λ2 L, , A, , l, , l, , T, = 2ν 1, µ, , If the string vibrates in three segments,, A, A, A, 3λ, then L = 3, 2, N, N, A, A, A, v, Third, harmonic, and ν3 =, = 3ν 1, λ3, It is called second overtone or third harmonic., , l, , In general, if a string vibrates in p segments [i.e. have, ( p + 1) nodes and p antinodes],, then ν pth =, , p, 2L, , N1, , A2, , A, , N, A, A, Second harmonic, , l, , l, , (Air Columns), l, , 1. Open Organ Pipe, In an open organ pipe, always anti-node is formed at both, open ends. Various modes of vibration of air column in an, open organ pipe are shown below, , A3, , N3, , A4, , All harmonics are present in open pipe with their, frequencies in the ratio 1 : 2 : 3 : 4 . . .. and, ratio of overtones = 2 : 3 : 4 : 5 K, λ 3λ 5λ, Position of nodes from one end x = ,, ,, K, 4 4 4, Position of anti-nodes from one end, λ, 3λ, x = 0, , λ ,, K, 2, 2, , In a chosed organ pipe, always node is formed at the closed, end. Various mode of vibration of air column in a closed, organ pipe are shown below, , Standing Waves in Organ Pipes, Organ pipes are those cylindrical pipes which are used for, producing musical (longitudinal) sounds. The standing waves, in both organ pipes (i.e. open organ pipe and closed organ, pipe) are described below., , N2, 3λ3, l=, 2, (c), , 3λ3, 3v, ;f =, 2, 2l, , 2. Closed Organ Pipe, , T, = pν 1, µ, , and it is known as pth harmonic or ( p − 1)th overtone., , A3, , N2, , Third harmonic or second overtone l =, A1, , It is known as first overtone or second harmonic., l, , A2, , N1, , 2v, 2l, , l= λ2, (b), , It is known as the fundamental frequency or first, harmonic., l, , v, 2l, , l, , First harmonic, λ, l= 1, 4, v, f =, 4l, Third harmonic, (first overtone), 3λ, l= 3, 4, 3v, f3 =, 4l, Fifth harmonic, (second overtone), 5λ 5, ;, l=, 4, 5v, f5 =, 4l, , N, , A, l=, , A1, , λ1, 4, (a), , l=, , A1, , N2, , A2, , N1, 3λ2, 4, (b), , A3, , A2, N1, , N3, , N2, l=, , 5λ3, 4, (c), , In closed organ pipe only odd harmonics are present. Ratio, of harmonic is n1 : n3 : n5 = 1 : 3 : 5.
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l, , Ratio of overtones = 3 : 5 : 7, , l, , Position of nodes from closed end x = 0,, , l, , λ, 3λ, , λ,, , ..., 2, 2, λ 3λ 5λ, Position of antinodes from closed end x = ,, ,, , ..., 4 4 4, , Beats, When two sound waves of nearly equal (but never equal) or, slightly different frequencies and equal or nearly equal, amplitudes travelling along the same direction superimpose, at a given point, the resultant sound intensity alternately rises, and falls. This alternate rise and fall of sound at a given, position is called beats., l, , l, , l, , Doppler’s Effect, The phenomena of apparent change in frequency of source, due to a relative motion between the source and observer is, called Doppler’s effect., When Source is Moving and Observer is at Rest When, source is moving with velocity vs , towards an observer at, rest, then apparent frequency, vs, v , n′ = n , , v, O, S, v − vs , l, , If source is moving away from observer, then, v , n′ = n , , v + vs , l, , Number of beats formed per second is called the frequency, of beats. If two sound waves of frequencies ν 1 and ν2, superimpose, then frequency of beats = (ν 1 ~ ν2 ), i.e. either, (ν 1 − ν2 ) or (ν2 − ν 1)., , Our perception of loudness is better co-related with the, second level measured in decibel (dB) and defined as, follows, I, β = 10 log 10 , where I 0 = 10 −12 Wm2 at 1kHz., I0, , When Source is at Rest and Observer is Moving When, observer is moving with velocity vo , towards a source at, rest, then apparent frequency., vo, v + vo , n′ = n , , S, v , O, v, , For formation of distinct beats, the difference between the, frequencies of two superimposing notes should be less than, 10 Hz., , When observer is moving away from source, then, v − vo , n′ = n , , v S, O, v, l, , v − vo , n′ = n , , v − vs , v, , S, , The tuning fork is a metallic device that produces sound of a, single frequency., Suppose, a tuning fork of known frequency nA is sounded, together with another tuning fork of unknown frequency (nB ), and x beats heard per second., There are two possibilities to know frequency of unknown, tuning fork, …(i), nA − nB = x, …(ii), nB − nA = x, We can find true frequency of tuning fork B from a pair of, tuning forks A and B, in which frequency of A is known and, where x is the beats per second., When B is filled, (its frequency increases), , (i), , If x increases, then, nB = nA − x, , (i), , If x increases, then, nB = nA + x, , (ii), , If x decreases,, then nB = nA + x, , (ii), , If x decreases, then, nB = nA − x, , (iii), , If x remains same,, then nB = nA + x, , (iii), , If x remains same, then, nB = nA − x, , (iv), , If x becomes zero,, then nB = nA + x, , (iv), , If x becomes zero, then, nB = nA − x, , vo, , When Source and Observer Both are Moving, (a) When both are moving in same direction along the, direction of propagation of sound, then, , Tuning Fork, , When B is loaded, (its frequency decreases), , 135, , WAVES, , DAY TWELVE, , O, , vs, , vo, , (b) When both are moving in same direction opposite to, the direction of propagation of sound, then, v + vo , n′ = n , , v + vs , vo, , vs, S, , O, , v, , (c) When both are moving towards each other, then, v + vo , n′ = n , , v − vs , v, , S, , O, , vs, , vo, , (d) When both are moving in opposite direction, away, from each other, then, v − vo , n′ = n , , v + vs , v, , vo, O, , S, , vs
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136, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Similarly, if the source is moving away from the observer as, shown above, with velocity component vs cos θ, then, v, ν′ =, ×ν, v + vs cos θ, , Transverse Doppler’s Effect, As shown in figure, the position of a source is S and of, observer is O. The component of velocity of source towards, the observer is v cos θ. For this situation, the approach, frequency is, θ, , l, , The Doppler’s effect in sound does not take place in the, transverse direction., , S, , v, , vc, os, , l, , P, , θ, , θ, , v, , S, , os, vc, , T, , l, , If θ = 90 °, the vs cos θ = 0 and there is no shift in the, frequency. Thus, at point P, Doppler’s effect does not occur., , Effect of Wind, If wind is also blowing with a velocity w in the direction of, sound, then its velocity is added to the velocity of sound., Hence, in this condition the apparent frequency is given by, v + w − vo , ν′ = ν , , v + w − vs , , θ, , Applications of Doppler’s Effect, The measurement of Doppler shift has been used, by police to check over speeding of vehicles., at airports to guide the aircraft., to study heart beats and blood flow in different parts of the, body., by astrophysicist to measure the velocities of plants and, stars., l, , O, , l, , ν′ =, , v, ×ν, v − vs cos θ, , l, , ν′ will now be a function of θ. So, it will no more be, constant., , l, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 Which of the following statements are true for wave, motion?, (a) Mechanical transverse waves can propagate through, all mediums, (b) Longitudinal waves can propagate through solids only, (c) Mechanical transverse waves can propagate through solids, only, (d) Longitudinal waves can propagate through vacuum, , 2 A sound wave is passing through air column in the form, of compression and rarefaction. In consecutive, compression and rarefactions,, (a) density remains constant, (b) Boyle’s law is obeyed, (c) bulk modulus of air oscillates, (d) there is no transfer of heat, , 3 A sound wave of wavelength λ is travelling in a medium, with a speed of v m/s enters into another medium where, its speed is 2v m/s. Wavelength of sound waves in the, second medium is, (a) λ, (c) 2λ, , λ, 2, (d) 4λ, , (b), , 4 When tension of a string is increased by 2.5 N, the initial, frequency is altered in the ratio of 3 : 2. The initial tension, in the string is, (a) 6 N, , (b) 5 N, , (c) 4 N, , (d) 2 N, , 5 The speed of sound in oxygen (O2 ) at a certain, , temperature is 460 ms −1. The speed of sound in helium, (He) at the same temperature will be (assume both, gases to be ideal), (a) 460 ms−1 (b) 500 ms−1, , (c) 650 ms−1, , (d) 1420 ms−1, , 6 It takes 2.0 seconds for a sound wave to travel between, two fixed points, when the day temperature is 10°C. If, the temperature rise to 30°C, the sound wave travel, between the same fixed points in, (a) 1.9 s, (c) 2.1 s, , (b) 2.0 s, (d) 2.2 s, , t, x 1 , − + , , 5 9 6 , , , 7 A wave equation is given by y = 4 sin π , , where, x is in cm and t is in sec. Which of the following is, true?, (a) λ = 18 cm, (c) a = 0.4 m, , (b) v = 4 ms −1, (d) f = 50 Hz
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WAVES, , DAY TWELVE, 8 A wave equation which gives the displacement along, y-direction is given by y = 0.001 sin [100 t + x ], where,, x and y are in metre and t is time in second. This, represents a wave, 100, Hz, π, (b) of wavelength 1 m, (c) travelling with a velocity of, , 50, ms −1 in the positive, π, , (d) travelling with a velocity of 100 ms −1 in the negative, x-direction, , 9 Which of the following is not true for progressive wave, x , t, y = 4 sin 2π , −, 0.02 100 , , (c) 0.5 N, , (d) 6.25 N, , y1 = 0.05 cos( 0.50 π x − 100πt ), y 2 = 0.05 cos( 0.46 π x − 92 πt ), where y1, y 2 and x are in metres and t in seconds. The, speed of sound in the medium is, (d) 332 m/s, , 12 In order to double the frequency of the fundamental note, emitted by a stretched string, the length is reduced to, 3, th of the original length and the tension is changed., 4, The factor, by which the tension is to be changed, is, 2, 3, , (c), , 8, 9, , (d), , 9, 4, , 13 A granite rod of 60 cm length is clamped at its middle, point and is set into longitudinal vibrations. The density of, granite is 2.7 × 103 kg/m 3 and its Young’s modulus is, 9.27 × 1010 Pa. What will be the fundamental frequency of, ª JEE Main 2018, the longitudinal vibrations?, (a) 5 kHz, , (b) 2.5 kHz, , (d) f, , (c) 10 kHz, , 7, 8, , (b), , 3, 4, , (c), , 5, 7, , (d), , 8, 7, , 16 Motion of two particles is given by, y1 = 0.25 sin ( 310 t ), y 2 = 0.25 sin ( 316 t ), Find beat frequency., (b) 3/π, , c) 6/π, , (d) 6, , (b) 3, , (c) 2, , (d) 1, , 4 beat/s. If a prong of the fork P is filled, the beats are, reduced to 2 s −1. What is the frequency of P, if Q is, 250 Hz?, (b) 250 Hz, , (c) 254 Hz, , (d) 252 Hz, , 19 16 tuning forks are arranged in the order of increasing, , medium the displacement of a particle located at X at, ª JEE Main (Online) 2013, time t is given by, , (c) 100 m/s, , (a), , (a) 246 Hz, , 11 When two sound waves travel in the same direction in a, , (b), , (c) 2f, , 18 Two tuning forks P and Q when set vibrating, give, , , t, x, , , y = 0.02 (m ) sin 2π , −, ., 0.04 (s ) 0.50 (m ) , , , 3, 8, , 3f, 4, , fourth harmonic of an open organ pipe. The ratio of the, lengths of the pipes are, , (a) 4, , 0.04 kg m −1 is given by, , (a), , (b), , ( ν − 1), ν,( ν + 1). They superimpose to give beat. The, number of beats produced per second will be, , 10 The equation of a wave on a string of linear mass density, , (b) 200 m/s, , f, 2, , 17 Three sound waves of equal amplitudes have frequencies, , (a) Its amplitude is 4 cm, (b) Its wavelength is 100 cm, (c) Its frequency is 50 Hz, (d) Its propagation speed is 50 × 10−2 cms −1, , (a) 92 m/s, , (a), , (a) 3, , where, y and x are in cm and t in second., , The tension in the string is, (b) 12.5 N, (a) 4.0 N, , of it is in water. The fundamental frequency of the air, column is now, ª JEE Main 2016 (Offline), , 15 Third overtone of a closed organ pipe is in unison with, , (a) of frequency, , x-direction, , 137, , (d) 7.5 kHz, , 14 A pipe open at both ends has a fundamental frequency, f in air. The pipe is dipped vertically in water, so that half, , frequencies. Any two successive forks give 8 beat/s,, when sounded together. If the frequency of the last fork is, twice the first, then the frequency of the first fork is, (a) 120 Hz, (c) 180 Hz, , (b) 160 Hz, (d) 220 Hz, , 20 A vehicle with a horn of frequency n is moving with a, , velocity of 30 ms −1 in a direction perpendicular to the, straight line joining the observer and the vehicle. The, observer perceives the sound to have a frequency, (n + n1). If velocity of sound in air be 330 ms −1, then, (a) n1 = 10n, , (b) n1 = 0, , (c) n1 =, , n, 11, , (d) n1 = −, , n, 11, , 21 An observer is moving with half the speed of light, towards a stationary microwave source emitting waves at, frequency 10 GHz. What is the frequency of the, microwave measured by the observer? (speed of light, ª JEE Main (Offline) 2017, = 3 × 108 ms −1), (a) 12.1 GHz, (c) 15.3 GHz, , (b) 17.3 GHz, (d) 10.1 GHz, , 22 Two trains are moving towards each other at speeds of, , 20 ms −1 and 15 ms −1 relative to the ground. The first train, sounds a whistle of frequency 600 Hz, the frequency of, the whistle heard by a passenger in the second train, before the train meet is, (Speed of sound in air = 340 ms −1), (a) 600 Hz, , (b) 585 Hz, , (c) 645 Hz, , (d) 666 Hz
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138, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , 23 A fixed source of sound emitting a certain frequency, appears as νa when the observer is approaching the, source with speed vo and vr when the observer recedes, from the source with the same speed. The frequency of, the source is, (a), , νr + νa, 2, , νr − νa, 2, , (b), , (c) νa ⋅ νb, , (d), , 2νr ⋅ νa, νr + νa, , Direction (Q. Nos. 24-31) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below:, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 24 Statement I A tuning fork is in resonance with a closed, pipe. But the same tuning fork cannot be in resonance, with an open pipe of the same length., Statement II The same tuning fork will not be in, resonance with open pipe of same length due to end, correction of pipe., , 25 Statement I In a sound wave, a displacement node is a, pressure antinode and vice-versa., Statement II Displacement node is a point of minimum, displacement., , 26 Statement I Velocity of particles while crossing mean, position (in stationary waves) varies from maximum at, antinodes to zero at nodes., , Statement II Amplitude of vibration at antinodes is, maximum and at nodes, the amplitude is zero and all, particles between two successive nodes cross the mean, position together., , 27 Statement I We can recognise our friends by listening, their voices., Statement II The quality of sound produced by different, persons are different., , 28 Statement I The basic of Laplace correction was that,, exchange of heat between the region of compression and, rarefaction in air is not possible., Statement II Air is a bad conduction of heat and velocity, of sound in air is large., , 29 Statement I If two waves of same amplitude, produce a, resultant wave of same amplitude, then the phase, difference between them will be 120°., Statement II The resultant amplitude of two waves is, equal to sum of amplitude of two waves., , 30 Statement I Two longitudinal waves given by equation, y1( x , t ) = 2a sin (ω t − kx ) and, y 2 ( x , t ) = a sin ( 2ω t − 2kx ) will have equal intensity., Statement II Intensity of waves of given frequency in, same medium is proportional to square of amplitude, only., , 31. Statement I If we see the oscillations of a stretched wire, at higher overtone mode, frequency of oscillations, increases, but wavelength decreases., 1, Statement II From v = ν ⋅ λ , λ ∝ as v = constant., ν, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 Sound of frequency f passes through a Quinck’s tube,, adjusted for intensity Im . What should be the length to, which the tube should be moved to reduce intensity to, 50% (speed of sound is v )?, (a), , v, 2f, , (b), , v, 4f, , (c), , v, 8f, , (d), , v, 16f, , 2 A racing car moving towards a cliff sounds its horn. The, driver observes that the sound reflected from the cliff has, a pitch one octave higher than the actual sound of the, horn. If v is the velocity of sound, the velocity of the car is, v, (a), 2, v, (c), 3, , v, (b), 2, v, (d), 4, , 3 A train of sound waves is propagated along an organ, pipe and gets reflected from an open end. If the, displacement amplitude of the waves (incident and, reflected) are 0.002 cm, the frequency is 1000 Hz and, wavelength is 40 cm. Then, the displacement amplitude, of vibration at a point at distance 10 cm from the open, end, inside the pipe is, (a) 0.002 cm (b) 0.003 cm (c) 0.001 cm, , (d) 0.000 cm, , 4 An engine approaches a hill with a constant speed., When it is at a distance of 0.9 km, it blows a whistle, whose echo is heard by the driver after 5 s. If the speed, of sound in air is 330 m/s, then the speed of the engine is, (a) 32 m/s, , (b) 27.5 m/s, , ª JEE Main (Online) 2013, (c) 60 m/s, (d) 30 m/s
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WAVES, , DAY TWELVE, 5 A travelling wave represented by y = A sin (ω t − kx ) is, , 9 A sonometer wire of length 114 cm is fixed at both the, ends. Where should the two bridges be placed so as to, divide the wire into three segments whose fundamental, frequencies are in the ratio 1:3:4? ª JEE Main (Online) 2013, , superimposed on another wave represented by, y = A sin (ωt + kx ). The resultant is, (a)a standing wave having nodes at, 1 λ, x = n + , n = 0, 1, 2, , 2 2, (b)a wave travelling along + x direction, (c)a wave travelling along −x direction, (d)a standing wave having nodes at x =, , 139, , nλ, , n = 0, 1, 2., 2, , (a) At 36 cm and 84 cm from one end, (b) At 24 cm and 72 cm from one end, (c) At 48 cm and 96 cm from one end, (d) At 72 cm and 96 cm from one end, , 10 The transverse displacement y ( x , t ) of a wave on a string, is given by y ( x , t ) = e − (ax, , 6 A piston fitted in cylindrical pipe is pulled as shown in the, , 2, , + bt 2 + 2 ab xt ), , . This represents, , a, , figure. A tuning fork is sounded at open end and loudest, sound is heard at open length 13 cm, 41 cm and 69 cm., The frequency of tuning fork if velocity of sound is, 350 ms −1, is, , b, a, , (a) wave moving in − x direction with speed, , (b) standing wave of frequency b, 1, b, (d) wave moving in + x direction with speed a / b, , (c) standing wave of frequency, , 11 A train is moving on a straight track with speed 20 ms −1. It, (a) 1250 Hz, (c) 417 Hz, , is blowing its whistle at the frequency of 1000 Hz. The, percentage change in the frequency heard by a person, standing near the track as the train passes him is close to, (speed of sound = 320 ms −1), ª JEE Main 2015, , (b) 625 Hz, (d) 715 Hz, , 7 A and B are two sources generating sound waves. A, listener is situated at C. The frequency of the source at A, is 500 Hz. A now, moves towards C with a speed 4 m/s., The number of beats heard at C is 6. When A moves, away from C with speed 4 m/s, the number of beats, heard at C is 18. The speed of sound is 340 m/s. The, ª JEE Main (Online) 2013, frequency of the source at B is, A, , (a) 500 Hz, (c) 512 Hz, , (a) 6%, , (c) 6, , ª JEE Main (Offline) 2016, (c) 2 2 s, (d) 2 s, , straight path at 2 ms −2. At the starting point of the motor, cycle, there is a stationary electric siren. How far has the, motor cycle gone when the driver hears the frequency of, the siren at 94% of its value when the motor cycle was at, rest? (Speed of sound = 330 ms −1), , number of possible natural oscillations of air column in, the pipe whose frequencies lie below 1250 Hz. The, ª JEE Main 2014, velocity of sound in air is 340 m/s., (b) 8, , (b) 2 s, , 13 A motor cycle starts from rest and accelerates along a, , 8 A pipe of length 85 cm is closed from one end.Find the, , (a) 12, , (d) 24%, , support. A short wave pulse is introduced at its lowest, end. It starts moving up the string. The time taken to, reach the support is (Take, g = 10 ms − 2), (a) 2 π 2 s, , (b) 506 Hz, (d) 494 Hz, , (c) 18%, , 12 A uniform string of length 20 m is suspended from a rigid, , B, , C, , (b) 12%, , (d) 4, , (a) 49 m, , (b) 98 m, , (c) 147 m, , (d) 196 m, , ANSWERS, SESSION 1, , 1 (c), , 2 (d), , 3 (c), , 4 (d), , 5 (d), , 6 (a), , 7 (a), , 8 (d), , 9 (d), , 10 (d), , 11 (b), , 12 (d), , 13 (a), , 14 (d), , 15 (a), , 16 (b), , 17 (c), , 18 (a), , 19 (a), , 20 (b), , 21 (b), , 22 (d), , 23 (a), , 24 (c), , 25 (b), , 26 (a), , 27 (a), , 28 (a), , 29 (c), , 30 (c), , 2 (c), 12 (c), , 3 (d), 13 (b), , 4 (d), , 5 (a), , 6 (b), , 7 (c), , 8 (c), , 9 (b), , 10 (a), , 31 (a), SESSION 2, , 1 (c), 11 (b)
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140, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, 1 Mechanical transverse wave can be set, up in solids but never in liquids and, gases., , 2 There is no transfer of heat from, compression to rarefaction as air is a bad, conductor of heat and time of, compression and rarefaction is too small., , 3 In the first medium, frequency, c, v, =, λ λ, It remains the same in second medium,, i.e., f′= f, v ′ 2v, v, =, =, ⇒ λ′ = 2λ, λ′, λ′, λ, f =, , 4 Since, frequency, ν ∝ T, ∴, or, , ν1, 3, T + 2.5, T1, or, =, =, ν2, 2, T, T2, 9, 2.5, 2.5 5, or, or T = 2 N, = 1+, =, 4, T, T, 4, , 5 Speed of sound is given by, γ RT, M, 7, RT, and v He =, = 5, 32, 7×3× 4, =, 5 × 32 × 5, , ∴, , vO, , 2, , v He, , or v He = 460 ×, , 5, RT, 3, 4, , 5 × 32 × 5, ≈ 1420 ms −1, 7×3× 4, , 6 Let distance between two fixed points, be d, then, , ⇒, ⇒, , 2, =, t2, , and k = 1 m −1, 100 50, Hz, ν=, =, ∴, 2π, π, 2π, λ=, = 2π m, k, ω 100 s −1, and v = =, = 100 ms −1, k, 1 m −1, Moreover, as the wave equations of the, form y = A sin (ωt + kx ) , the wave is, travelling along negative x-direction., , 9 From the given wave equation, we find, that, A = 4 cm, ω =, , 2 π −1, s = 100 πs −1, 0.02, , 2π, cm −1, 100, 2π, ∴, λ=, = 100 cm, k, ω, 100 π, ν=, =, = 50 Hz, 2π, 2π, ω 100 π, and v = =, = 50 × 10 2 cms–1, 2π, k, 100, , =, , ω2, , k2, 0.04 (2 π / 0.04)2, , …(i), , The standard wave equation can be, written as, y = a sin (ωt − kx + φ), 2π, 2π, y = a sin , t −, x + φ …(ii), T, , λ, Equating Eqs. (i) and (ii), we get, Amplitude, a = 4 cm, 1, 1, Frequency, f =, Hz = 0.1 Hz, =, T, 10, , ν1, l, = 2, ν2, l1, , T1, T2, 3, Now, ν2 = 2 ν1 , l 2 = l 1 ,, 4, Hence, we have, , (2 π / 0.50)2, , = 6.25 N, , and y 2 = 0.05 cos (0.46 π x − 92 πt ), Comparing these two equations with, y = A sin (k x − ω t ), We have, ω1 = 100 π and ω2 = 92 π, A, 0.05, Now, speeds, v 1 =, =, 100 π 100 π, A, 0.05, and, v2 =, =, 92 π 92 π, Now, the resultant speed,, 2, , v =, , 2, , 0.05 , 0.05 , , + , [ π = 3.14], 92 π , 100 π , , = 200 m/s, , 1 3, = ×, 2 4, ⇒, , T1, T2, , ⇒, , 2, =, 3, , T1, T2, , 9, T2, 9, or T2 = T1, =, 4, T1, 4, l, 4, , 13, , L, From vibration mode,, λ, =L, 2, ⇒, λ = 2L, Y, ∴ Wave speed, v =, ρ, , So, frequency, v, f =, λ, 1 Y, ⇒ f =, 2L ρ, =, , 927, . × 1010, , 1, 2 × 60 × 10, , −2, , 2.7 × 103, , ≈ 5000 Hz, f = 5 kHz, , 14 For open ends, fundamental frequency f, , 11 y 1 = 0.05 cos (0.50 π x − 100 π t ), , 7 The given equation be written as, t, x 1 , , y = 4 sin π − + , 6 , 5 9, , Hence, answer of wave speed v is, wrong, i.e. option (d) is correct., , T = µv 2 = µ, , 303, ⇒ t 2 = 1.9 s, 283, , remaining unchanged,, , standard form of wave equation,, we get, A = 0.001 m, ω = 100 s −1, , 10 Tension in the string is given by, , d, also v ∝ T, v, t1, v, T2, = 1 =, t2, v2, T1, t =, , 12 From law of string, other factors, , 8 Comparing given equation with, , and k =, , v =, , v O2, , Wavelength, λ = 2 × 9 = 18 cm, Velocity,, v = fλ, = 01, . × 18 = 1.8 cms −1, , in air, we have, λ, =l, 2, ⇒, λ = 2l, l, v= f λ, v, v, …(i), ⇒, f = =, λ 2l, When a pipe is dipped, vertically in water, so that, half of it is in water, we, l/2, have, l, λ, l, =, 4 2, ⇒ λ = 2l ⇒ v = f ′ λ, v, v, …(ii), = f, ⇒ f′= =, λ 2l, Thus, the fundamental frequency of the, air column is now,, f = f′
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WAVES, , DAY TWELVE, 15 We know that, third overtone of closed, organ pipe means seventh harmonic, ∴ (ν7 )closed = (ν 4 )open, or, or, or, , v , v , 7, , = 4, 2l o , 4l c , lo, 8, =, lc, 7, lc, 7, =, lo, 8, , 16 y 1 = 0.25 sin (310t ), and y 2 = 0.25 sin (316t ), We have, ω1 = 310, 310, unit, ⇒, f1 =, 2π, and, ⇒, , by the source will be given by the, formula,, 1 + v /c , fobserved = factual ⋅ , , 1 − v /c , v 1, Here, frequency,, =, c 2, 3 / 2, So, fobserved = factual , , 1 / 2, ∴, , …(i), …(ii), , ω2 = 316, 316, unit, f2 =, 2π, , Hence, beat frequency, 316 310, −, = f2 − f1 =, 2π, 2π, 3, = unit, π, , 17 Maximum number of beats, = ν + 1 − (ν − 1) = 2, , 18 There are four beats between P and Q ,, therefore the possible frequencies of P, are 246 or 254, (i.e. 250 ± 4) Hz., When the prong of P is filled, its, frequency becomes greater than the, original frequency., If we assume that the original frequency, of P is 254, then on filling its frequency, will be greater than 254. The beats, between P and Q will be more than 4., But it is given that the beats are reduced, to 2, therefore, 254 is not possible., Therefore, the required frequency must, be 246 Hz., , 19 As forks have been arranged in, ascending order of frequencies, hence if, frequency of Ist fork be n, then, n2 = n + 8, n3 = n + 2 × 8 = n + 16, and, n16 = n + 15 × 8, = n + 120 = 2n, ⇒, n = 120 Hz, , 20 Since, vehicle having siren is moving in, a perpendicular direction, hence there, will be no Doppler shift in frequency, and n1 = 0., , 21 As the observer is moving towards the, source, so frequency of waves emitted, , difference is maximum i.e. at the same, time, it is a pressure antinode., , 1 /2, , On the other hand, at the mid-point of a, compression or a rarefaction, the, displacement variation is maximum i.e., such a point is displacement antinode., However such a point is pressure node,, as pressure variation is minimum at such, a point., , 1 /2, , fobserved = 10 × 3 = 173, . GHz, , 26 Stationary wave is represented as shown, in figure., , −1, , 22 Here, v = 340 ms ,, , A, , −1, , v s = velocity of 1st train = 20 ms ,, v o = velocity of 2nd train = 15ms −1 and, , A, N, , N, , ν 0 = 600 Hz., As S and O are approaching each other,, hence, v + vo , ν=, ν0, v − vs , 340 + 15, × 600 = 666 Hz, =, 340 − 20 , , 23 Here, ν a = ν v + v o , , , , v, νa, vo, or, = 1+, ν, v, vo, νa, or, =, −1, v, ν, v − vo , Again, ν r = ν , , v , vo, ν, = 1− r, ∴, v, ν, From Eqs. (i) and (ii), we get, νa, ν, −1 = 1− r, ν, ν, νa νr, or, +, =2, ν, ν, νa + νr, or, =2, ν, ν + νr, or, ν= a, 2, , 141, , N, , It is quite clear from figure that at nodes, the amplitude is zero and velocity of, particle is also zero and at antinodes the, amplitude is maximum. So that the, velocity of particle is also maximum and, all particles cross mean position between, two successive nodes., , 27 Sounds coming from the different, , …(i), , sources can be recognised by virtue of, their quality which is characteristics of, sound. That is why we recognise the, voices of our friends., , 28 According to Laplace, the changes in, …(ii), , 24 If a closed pipe of length L is in, resonance with a tuning fork of, v, frequency ν, then ν =, 4L, An open pipe of same length L produces, v, vibrations of frequency, . Obviously,, 2L, it cannot be in resonance with the given, , v , tuning fork of frequency ν =, ., 4L , , 25 At the point where a compression and a, rarefaction meet, the displacement is, minimum and it is called displacement, node. At this point, the pressure, , pressure and volume of a gas, when, sound waves propagated through it, are, not isothermal but adiabatic. A gas is a, bad conductor of heat. It does not allow, the free exchange of heat between, compressed layer, rarified layer and its, surrounding., , 29 The resultant amplitude of two waves is, given by, A=, , a12 + a22 + 2a1 a2 cos θ, , Here, a1 = a2 = A = a, 1 / 2 = 1 + cos θ, or cos θ = − 1 / 2 or θ = 120°, 30 I = 1 ρω2 A2 v, 2, Here, ρ = density of medium,, A = amplitude,, ω = angular frequency and, v = velocity of wave, ∴Intensity depend upon amplitude,, frequency as well as velocity of wave, Also,, I1 = I2, , 31 ν = η v , where n = 1, 2, 3, …, , 2l, , As, η increases, frequency increases., Hence, wavelength decreases.
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142, , DAY TWELVE, , 40 DAYS ~ JEE MAIN PHYSICS, , 1 I m = 4 ka2 , I = 2ka, I ′ = k (a2 + a2 + 2a2 cos 2 θ), π, ⇒, φ=, 2, λ, Path difference = = 2x, 4, v, ⇒, = 2x, 4f, v, ⇒, x=, 8f, , 6 In a closed organ pipe in which length, of air-column can be increased or, decreased, the first resonance occurs at, λ / 4 and second resonance occurs at, 3λ / 4., , l1 =, , 2 Let n be the actual frequency of sound, of horn. If v s be the velocity of car, then, frequency of sound striking the cliff, (source is moving towards listener), n′ =, , v ×n, v − vs, , N, , …(i), , (v + v s ) n ′ (v + v s ), v ×n, =, ×, v, v, (v − v s ), n ′′ v + v s, =, =2, n, v − vs, , or, , 3 The equation of stationary wave for, open organ pipe can be written as,, 2 πx , 2 πft , y = 2 A cos , sin , ,, λ , v , where x is the open end from where, wave gets reflected., Amplitude of stationary wave is,, 2 πx , A s = 2 A cos , , λ , For x = 01, . m,, . , 2 π × 01, A s = 2 × 0.002 cos , =0, 0.4 , , 4 We have, 900 + 900 − x, = 330 × 5 = 1650, (900 – x) m, , x, , 900 m, , ∴, , x = 150 m, 150, ∴ Speed =, = 30 m/s, 5, , 5 y = y1 + y2, = A sin (ωt − kx ) + A sin(ωt + kx ), y = 2 A sin ω t cos kx, , (2n + 1) v, < 1250, 4l, , A, , (2n + 1) < 1250 ×, , l2 = 3l, 4, N, , A, , 4 × 0. 85, 340, , (2n + 1) < 12. 52, n < 5.25, So, n = 0,1,2,3,...,5, , A, , So, we have 6 possibilities., , y, Thus, at first resonance,, λ, = 13, 4, , …(i), , f, µ, , 9 ν1 = 1, , 2L1, , and at second resonance,, , v + v s = 2v − 2v s, v, vs =, 3, , or, , l, 4, , y, , The frequency of sound heard on, reflection, n ′′ =, , According to question,, Let freqency of source B is Z Hz, ∴ Z = 506 ± 6 ⇒ Z = 500 or 512, and Z = 494 ± 18 ⇒ Z = 512 or 476, Thus, required frequency = 512 Hz, 8 For closed organ pipe = (2n + 1)v, 4l, [where, n=0,1,2.....], 85 cm, , Clearly, it is equation of standing wave, for position of nodes y = 0., λ, i.e., x = (2n + 1), 4, 1 λ, , = n + , , 2 2, , SESSION 2, , 3λ, = 41, 4, Subtracting Eq. (i) from Eq. (ii),, we have, 3λ λ, − = 41 − 13, 4, 4, ∴, , L1, , …(ii), , λ = 56 cm, , Hence, frequency of tuning fork,, v, 350, n= =, = 625Hz, λ 56 × 10−2, , 7 Here, frequency of source = 500 Hz, Speed of source A = 4 m/s = u, Then, source is moving towards, stationary observer,, v, ν′ =, ν0, v −u, (where, v = speed of sound), 340, =, × 500, 340 − 4, 340, ν′ =, × 500 Hz, ⇒, 336, = 506 Hz, Now, when source is reciding from the, observer, v, ν′ =, ν0, v +u, 340, =, × 500 Hz, 344, ∴, ν′ = 494 Hz, , L2, , L3, , Let length of three segments be, L1 , L2 and L3 ,, 1, ν2 =, f /µ, 2L2, 1, ν3 =, f /µ, 2L3, ν1 L1 = ν2 L2 = ν3 L3, , So, that, , As, ν1 : ν2 : ν3 = 1 : 3 : 4, ν L, ν1 = 2 2, L1, and, , ν2 = 3 ν1 , ν3 = 4ν1, L1, ν, 1, L2 = 1 L 1 = L 1 =, ν2, 3, 3, , and L3 =, , ν1, ν, L, L1 = 1 L1 = 1, ν3, 4ν1, 4, , L 1 + L2 + L3 = 114, 1 1, Now, L 1 1 + + = 114, , 3 4, 12 + 4 + 3 , ⇒ L1 , = 114, , , 12, 19, L 1 = 114, 12 , (114 × 12), = 72 cm, ⇒ L1 =, 19, L 1 72, =, = 24 cm, L2 =, 3, 3
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WAVES, , DAY TWELVE, 10 y ( x, t ) = e − (ax, = e−(, , 2, , + bt2 + 2 abxt ), 2, , a x + b t), , It is a function of type, y = f (ω t + kx ), ∴ y ( x, t ) represents wave travelling, along − x direction., ω, b, b, Speed of wave = =, =, k, a, a, , 11 Apparent frequency heard by the person, before crossing the train., c , 320 1000, f1 = , fo = , , 320 − 20 , c − vs , , or, , 2cv , ∆f, × 100 = 2 s 2 × 100, f0, c − vs , 2 × 320 × 20, × 100, 300 × 340, 2 × 32 × 20, =, 3 × 34, = 12. 55% = 12%, =, , 12 A uniform string of length 20 m is, suspended from a rigid support. Such, that the time taken to reach the support,, T =, , mgx, l, , (l , m ) P, , [c= speed of sound in air], , x, So, velocity at point P =, , i.e., , v =, , mgx, l, m/l, gx, , dx, = gx, dt, , dx, =, x, , t, , ∫0, , g dt, , 2 20 = 10 t, t =2 2s, , 13 For motor cycle, u = 0, a = 2 ms −2, Observer is in motion and source is at, rest, then apparent frequency,, v − vo, ν′ = ν, v + vs, , ⇒, , dx, , ∆f = f1 − f2, , Percentage change in frequency heard, by the person standing near the track as, the train passes him is, , ⇒, , ⇒, , c , 320 1000, f2 = , fo = , , 320 + 20 , c + vs , , 20, , ∫0, , [2 x ]20, 0 = 10 t, , ⇒, , Similarly, apparent frequency, heard,after crossing the trains, , 2cv , = 2 s 2 f0, c − vs , , ⇒, , 143, , 330 − v o, 94, ν=ν, 100, 330, 330 × 94, 330 − v o =, 100, 94 × 33, v o = 330 −, 10, 33 × 6, =, ms −1, 10, v 2 − u2, 2a, 9 × 33 × 33, =, 100, 9 × 1089, =, ≈ 98 m, 100, , s=
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144, , DAY THIRTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY THIRTEEN, , Unit Test 2, (Waves and Oscillations), 1 A mass m attached to a spring of spring constant k is, stretched a distance x 0 from its equilibrium position and, released with no initial velocity. The maximum speed, attained by the mass in its subsequent motion and the, time at which this speed would be attained are, respectively,, (a), , k, m, x 0, π, m, k, , (b), , k x0 π m, ,, m 2 2 k, , (c), , π m, k, x 0,, 2 k, m, , (d), , m, k x0, ,π, k, m 2, , 2 The acceleration-displacement graph of a particle, executing SHM is shown in the figure. The time period of, SHM is, a(ms–2), , 37°, , harmonic motion with frequency 5Hz in horizontal plane., The maximum amplitude of the table at which block does, not slip on the surface of table is, (if coefficient of friction, between the block and surface of table is 0.6.) (Given,, g = 10 m/s 2 )., (a) 0.06 m, (c) 0.02 m, , (b) 0.006 m, (d) 0.002 m, , 5 For a particle executing SHM, determine the ratio of, average acceleration of the particle between extreme, position and the equilibrium position w.r.t. the maximum, acceleration, 4, π, 1, (c), π, (a), , 2, π, 1, (d), 2π, (b), , 6 Two springs are made to oscillate simple harmonically, x (m), , 4π, s, 3, 2π, (b), s, 3, (c) The given graph doesn’t represent SHM, (d) Information is insufficient, (a), , 3 A spring balance has a scale that can read from 0 to, 50 kg. The length of the scale is 20 cm. A body, suspended from this balance, when displaced and, released, oscillates harmonically with a time period of, 0.6 s. The mass of the body is (Take, g = 10 ms −2 ), (a) 10 kg, (c) 18 kg, , 4 A block is kept on a table which performs simple, , (b) 25 kg, (d) 22.8 kg, , when the same mass is suspended, individually. The time, periods obtained are T1 and T2 . If both the springs are, connected in series and then made to oscillate, when suspended by the same mass, the resulting time, will be, (a)T1 + T2, (c) T12 + T22, , T1T2, T1 + T2, T + T2, (d) 1, 2, (b), , 7 Find the time period of oscillations of a torsional, pendulum, if the torsional constant of wire is10 π 2 in, SI units. The moment of inertia of the rigid body is 10, kg - m 2 about the axis of rotation, (a) 1 s, (c) 4 s, , (b) 2 s, 1, (d) s, 2
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UNIT TEST 2 (WAVES AND OSCILLATIONS), , DAY THIRTEEN, 8 Acceleration-displacement graph for four particles are, shown, identify the one which represents SHM for all the, values of displacements, acceleration (a), , acceleration (a), , °, 37, , 37°, (b), displacement (x), , (a), , displacement (x), , 37, °, , acceleration (a), , 13 A spring of negligible mass having a force constant k, extends by an amount y when a mass m is hung from it., The mass is pulled down a little and then released. The, system begins to execute SHM of amplitude A and, angular frequency ω. The total energy of the mass-spring, system will be, mω 2 A 2, 2, ky 2, (c), 2, , mω2 A 2 ky 2, −, 2, 2, 2 2, mω A, ky 2, (d), +, 2, 2, , (b), , (a), , 14 Total number of independent harmonic waves in the, resultant displacement equation given by, , 20°, , (d) None of these, , (c), , 145, , y = 3 sin2, , displacement (x), , (a) 2, , (b) 1, , t, sin 800t, 2, (c) 4, , (d) 3, , 15 A load of mass m falls from a height h on the, 9 In case of simple harmonic motion, if the velocity is, plotted along the X-axis and the displacement from the, equilibrium position is plotted along the Y-axis, the, resultant curve happens to be an ellipse with the ratio, major axis (along X ), = 20π., minor axis (along Y ), What is the frequency of the simple harmonic motion?, (a) 100 Hz, (c) 10 Hz, , (b) 20 Hz, 1, (d), Hz, 10, , 10 A simple pendulum is suspended from the ceiling of a, stationary tram car. Now, the car starts accelerating, the, time period of a simple pendulum is the least when [Take, magnitude of acceleration to be same in all the cases], (a) car is accelerating up, (b) car is accelerating down, (c) car is accelerating horizontally, (d) car is stationary, , and time period T . The magnitude of its acceleration, at, T, s after the particle reaches the extreme position, would, 8, be, (a), , 4π A, , (c), , 2 π 2A, , 2T 2, 2T 2, , 4π A, 2, , (b), , T2, , (d) None of these, , 12 A string of length 1.5 m with its two ends clamped, is, vibrating in the fundamental mode. The amplitude at the, centre of the string is 4 mm. The minimum distance, between two points having amplitude 2 mm, is, (a) 1 m, (c) 60 cm, , mg, k, mg mg, (c), +, k, k, , (a), , (b) 75 cm, (d) 50 cm, , (b), 1 + 2hk, mg, , k, m, , h, , 2hk, mg, 1+, k, mg, , (d) None of these, , 16 A sound absorber attenuates the sound level by 30dB., The intensity of sound is decreased by a factor of, (a) 100, (c) 10000, , (b) 1000, (d) 10, , 17 The velocity of a particle executing a simple harmonic, , motion is 13 ms − 1 , when its distance from the equilibrium, position is 3 m and its velocity is12 ms − 1, when it is 5 m, away from equilibrium position. The frequency of the SHM, is, 5π, 8, 8π, (c), 5, , (a), , 11 A particle executes SHM about O with an amplitude A, , 2, , scale pan hung from a spring as shown. If the, spring constant is k, mass of the scale pan is zero, and the mass m does not bounce relative to the, pan, then the amplitude of vibration is, , 5, 8π, 8, (d), 5π, (b), , 18 Two identical strings A and B, have nearly the same, tension. When they both vibrate in their fundamental, resonant modes, there is a beat frequency of 3 Hz. When, string B is tightened slightly, to increase the tension, the, beat frequency becomes 6 Hz. This means, (a) that before tightening A had a higher frequency than B,, but after tightening, B has a higher frequency than A, (b) that before tightening B has higher frequency than A, but, after tightening A has higher frequency than B, (c) that before and after tightening A has higher frequency, than B, (d) that before and after tightening B has higher frequency, than A
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146, , DAY THIRTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 19 The ratio of densities of oxygen and nitrogen is 16 : 14., At what temperature, the speed of sound in oxygen will, be equal to its speed in nitrogen at 14°C?, (a) 50°C, (c) 48°C, , 26 A transverse wave on a string travelling along positive, x-axis has been shown in the figure below, , (b) 52°C, (d) 55°C, , 1, , 20 A train is passing by a platform at a constant speed of, , 40 ms −1. The horn of the train has a frequency of 320 Hz., Find the overall change in frequency detected by a, person standing on the platform, i.e. when the train, approaching and then precedes from him. (Take,, velocity of sound in air as 320 ms −1), (a) 216.4 Hz, (c) 365.7 Hz, , (b) 81.3 Hz, (d) 284.4 Hz, , 21 A string of length 0.4 m and mass 10−2 kg is clamped at, one end. The tension in the string is1.6 N. Identical wave, pulses are generated at the free end, after a time interval, ∆t. The minimum value of ∆t, so that a constructive, interference takes place between successive pulses is, (a) 01, . s, (b) 0.05 s, (c) 0.2 s, (d) Constructive interference cannot take place, , 22 A string vibrates according to the equation, , 2 πx , Y = 5 sin , × cos 20πt , where x and y are in cm, 3 , and t in second. The distance between two adjacent, nodes is, (a) 3 cm, , (b) 4.5 cm, , (c) 6 cm, , (d) 1.5 cm, , 23 A point source of sound is placed in a non-absorbing, medium. Two points A and B are at the distance of 1 m, and 2 m, respectively from the source. The ratio of, amplitudes of waves at A to B is, (a) 1 : 1, , (b) 1 : 4, , (c) 1 : 2, , (d) 2 : 1, , 24 The mathematical form of three travelling waves are, given by, y1 = ( 2 cm ) sin ( 3x − 6 t ),, y 2 = ( 3 cm ) sin ( 4x − 12 t ) ,, and y 3 = ( 4 cm ) sin ( 5x − 11 t ), of these waves,, , 25 If the maximum speed of a particle carrying a travelling, wave is v 0, then find the speed of a particle when the, displacement is half that of the maximum value, (b), , 3 v0, 4, , (c), , 3 v0, 2, , where t is in seconds and x is in cm. Find total distance, travelled by the particle at the (1), in 10 min 15 s,, measured from the instant shown in the figure and, direction of the motion of the particle at the end of this, time., (a) 6 cm, in upward direction, (b) 6 cm, in downward direction, (c) 738 cm, in upward direction, (d) 732 cm, in upward direction, , 27 A wire having a linear mass density of 5 × 10− 3 kg m − 1, is stretched between two rigid supports with a tension of, 450 N. The wire resonates at a frequency of 420 Hz. The, next higher frequency at which the wire resonates is, 490Hz. The length of the wire is, (a) 2.5 m, (c) 2.25 m, , (b) 2.14 m, (d) 2.0 m, , 28 A man stands on a weighing machine placed on a, horizontal plateform. The machine reads 50 kg. By means, of a suitable mechanism, the plateform is made to, execute harmonic vibration up and down with a, frequency of 2 vibrations per second. What will be the, effect on the reading of the weighing machine? The, amplitude of vibration of plateform is 5 cm., (Take, g = 10 ms − 2 ), (a) 11 kgf to 93 kgf, (b) 10.5 kgf to 89.5 kgf, (c) 10 kgf to 15.5 kgf, (d) 25.6 kgf to 100.5 kgf, , 29 A small trolley of mass 2.0 kg resting on a horizontal turn, , (a) wave 1 has greatest wave speed and wave has, maximum transverse string speed, (b) wave 2 has greatest wave speed and wave 1 has, greatest maximum transverse string speed, (c) wave 3 has greatest wave speed and wave 1 has, maximum transverse string speed, (d) wave 2 has greatest wave speed and wave 3 has, maximum transverse string speed, , v, (a) 0, 2, , The mathematical form of the wave is shown, 2π , , x, y = (3.0 cm) sin 2 π × 0.1 t −, 100 , , , (d) v0, , table is connected by a light spring to the centre of the, table. When the turn table is set into rotation at speed of, 360 rpm, the length of the stretched spring is 43 cm. If, the original length of the spring is 36 cm, the force, constant is, (a) 17025 Nm− 1, (c) 17475 Nm− 1, , (b) 16225 Nm− 1, (d) 17555 Nm− 1, , 30 At 16°C, two open end organ pipes, when sounded, together give 34 beats in 2 s. How many beats per, second will be produced, if the temperature is raised to, 51°C?, (Neglect increase in length of the pipes), (a) 18 s− 1, (c) 20 s− 1, , (b) 15 s− 1, (d) 10 s− 1
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UNIT TEST 2 (WAVES AND OSCILLATIONS), , DAY THIRTEEN, 31 During earthquake, both longitudinal and transverse, waves are produced having speeds 4.0 km/s and, 8.0 km/s, respectively. If the first transverse wave, reaches the seismograph 8 minutes after the arrival of, first longitudinal wave, then the distance of the position,, where the earthquake occurred is, (a) 3440 km, (c) 3840 km, , (b) 3880 km, (d) 3500 km, , oscillate in water, it acquired a time periodT ′ = 2T . The, density of the pendulum bob is equal to, (Take, density of water = 1), (b) 2, (d) None of these, , 33 On a planet a freely falling body takes 2 s when it is, dropped from a height of 8 m. The time period of simple, pendulum of length 1 m on that planet is, (a) 3.14 s, (c) 1.57 s, , (b) 16.28 s, (d) None of these, , other, average power transmitted by the wave may, change., Statement II Due to a change in the medium, amplitude,, speed, wavelength and frequency of the wave may, change., motion with amplitude A and angular frequency ω. To, change the angular frequency of simple harmonic motion, to 3 ω, and amplitude to A /2, we have to supply an extra, 5, energy of mω 2 A 2 , where m is the mass of the particle, 4, executing simple harmonic motion., Statement II Angular frequency of the simple harmonic, motion is independent of the amplitude of oscillation., , 38 Statement I Time period of spring pendulum is the same, whether in an accelerated or in an inertial frame of, reference., , Direction (Q. Nos. 34-40) Each of these questions contains, two statements Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, (c) Statement I is true, Statement II is false, (d) Statement I is false, Statement II is true, , 34 Statement I Waves on a string can be longitudinal in, nature., Statement II The string cannot be compressed or, rarified., , 35 Statement I A wave of frequency 500 Hz is propagating, with a velocity of 350 m/s. Distance between two particles, with 60° phase difference is 12 cm., Statement II x =, , 36 Statement I When a wave goes from one medium to, , 37 Statement I A particle performs a simple harmonic, , 32 A pendulum has time period T in air. When it is made to, , (a) 2, (c) 2 2, , 147, , λ, φ, 2π, , Statement II Mass of the bob of the spring pendulum and, the spring constant of spring are independent of the, acceleration of the frame of reference., , 39 Statement I The total energy of a particle executing, simple harmonic motion, can be negative., Statement II Potential energy of a system can be, negative., , 40 Statement I A circular metal hoop is suspended on the, edge, by a hook. The hoop can oscillate from one side to, the other in the plane of the hoop, or it can oscillate back, and forth in a direction perpendicular to the plane of the, hoop. The time period of oscillation would be more when, oscillations are carried out in the plane of hoop., Statement II Time period of physical pendulum is more, if, the moment of inertia of the rigid body about the, corresponding axis passing through the pivoted point, is, more., , ANSWERS, 1. (c), 11. (a), , 2. (a), 12. (a), , 3. (d), 13. (d), , 4. (b), 14. (d), , 5. (b), 15. (b), , 6. (c), 16. (b), , 7. (b), 17. (b), , 8. (b), 18. (d), , 9. (c), 19. (d), , 10. (c), 20. (b), , 21. (c), 31. (c), , 22. (a), 32. (b), , 23. (d), 33. (a), , 24. (d), 34. (d), , 25. (c), 35. (a), , 26. (c), 36. (c), , 27. (b), 37. (d), , 28. (b), 38. (a), , 29. (c), 39. (a), , 30. (a), 40. (a)
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148, , DAY THIRTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, 1 At the mean position, the speed will be, maximum., , Maximum acceleration (amax ) = ω2 A., , kx20, mv 2, =, 2, 2, ⇒, , T /4, , =, , v = v max =, , and this is attained at t =, , k, x0, m, , T, ., 4, , T = 2π, , m, k, , So, the required time is,, T, π, t =, =, 4, 2, , dx, 4, 4, On comparing with a = − ω2 x, we get, , 3, 4π, 2π, 3, ω =, ⇒, s, =, ⇒ T =, 4, T, 2, 3, 2, , 3 The scale can read a maximum of 50 kg,, for a length of 20 cm. Let spring, constant be k then,, kx 0 = mg, [for m = 50 kg, x 0 = 20 cm], k = 2500 Nm, , Let mass of the body be m 0 , then from, m0, m0, T = 2π, ⇒ 0.6 = 2 π, k, 2500, ⇒ m 0 = 22.8 kg, , 4 Maximum force on the block on the, surface of table due to simple harmonic, motion, F = mω2 A, where A →, amplitude., Friction force on the block, Fs = µmg, It will not slip on the surface of the, table, if, F = Fs, mω2 A = µmg, µg, A=, ω2, 0 ⋅ 6 × 10, =, (2 × 3 ⋅ 14 × 5)2, = 0.006 m, x = A sin ωt, Average acceleration between extreme, position and the equilibrium position,, T, i.e. from time t = 0 to t =, 4, T /4, , T / 4 × ω2 A, , =, , 2, π, , T = 2π, where,, ⇒, , keq =, , m, keq, , k1 k2, k1 + k2, , T = T12 + T22, , 7 For torisional pendulum, τ = − k θ, k, α =− θ, I, k, ω2 =, ⇒, I, I, 10, T = 2π, = 2π, = 2s, k, 10 π2, , ω2 A sin ω t dt, T /4, , Magnitude of acceleration =, , a = − ω2 x, da, = − ω2 , where ω2 is positive, dx, quantity. This means for a particle to, execute SHM, the accelerationdisplacement curve should be a straight, line having a negative slope, which is, shown in option (b)., , So,, , 9 In representation of SHM in ellipse as, velocity along x-axis and displacement, along y-axis., Major axis (along X - axis), = 20 π, Q, Minor axis (along Y - axis), ωA, = 20 π ⇒ ω = 20 π, ⇒, A, ⇒ 2 πf = 20 π, The frequency of SHM, f = 10 Hz, g eff =, , g 2 + a2, , 11 Let at t = 0,the particle be at the extreme, position, then the equation of SHM can, be written as, 2π, x = A cos (ωt ) = A cos , × t , T, , T, π, A, At t = , x = A cos, =, 8, 4, 2, , 4 π2 A, 2 T2, , 12 As = 2 A sin kx, 2 mm = 4 mm sin kx, π 5π, kx = ,, 6 6, 5π π 1 λ, −, ⇒, x2 − x1 = , × =, 6, 6 k, 3, As the string is vibrating in fundamental, mode, λ, L =, ⇒ λ = 2L = 3 m, 2, , ⇒, , So, required separation between two, points,, x2 − x1 = 1 m, , 13 From initial equilibrium position,, ky = mg, When block is at distance x below mean, position, Kinetic energy of the block,, K =, , mω2 A2, cos 2 (ωt − φ), 2, , 8 For a particle to execute SHM,, , 10 When car is accelerating horizontally, , 5 Let the equation of SHM be,, , ∫0, , ω2 A sin ωt dt, , 6 Let the spring constants of the two, , m, k, , 2 da = − tan 37° = − 3 ⇒ a = − 3 x, , ⇒ k × 0. 2 = 50 × 10 ⇒, , ∫0, , springs be k1 and k2 respectively, then,, m, 4 π2 m, and k =, T = 2π, K, T2, When the two springs are connected in, series, then, , Time period of motion is,, , Im =, , 2, , 2π , A, Acceleration = − ω2 x = − , ×, T , 2, , Then, the required ratio is,, , [From SHM theory], Elastic potential energy of, spring-block-earth system,, k ( y + x )2, Ue =, 2, where, x = A sin(ωt + φ), Gravitational potential energy of, spring-block-earth system is, U g = − mgx, Taking, mean position as reference, position for gravitation potential energy., Total energy,, E = K + Ue + U g =, , 14 y = 3sin2 t sin 800 t, , ky 2, mω2 A2, +, 2, 2, , 2, 1 − cos t , , = 3, sin 800 t, , , 2, 3, 3, = sin 800 t − sin 800 t cos t, 2, 2, 3, 3, = sin 800 t − [2sin 800 t cos t ], 2, 4, 3, 3, = sin 800 t − [sin 801 t + sin 799 t ], 2, 4, 3, 3, 3, = sin 800 t − sin 801 t − sin 799 t, 2, 4, 4, ∴Number of independent harmonic wave, = 3.
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15 From the conservation principle,, 1, mgh = kx20 − mgx 0, 2, where, x 0 is maximum elongation in, spring., 1, ⇒ kx20 − mgx 0 − mgh = 0, 2, 2mg, 2mg, ⇒ x20 −, x0 −, h=0, k, k, 2, , 2mg + 4 × 2mg h, , , k , k, ⇒ x0 =, 2, Amplitude = Elongation for lowest, extreme position − elongation for, equilibrium position., mg, mg , 2hk , ., = x 0 − x1 =, Q x1 =, 1+, k, k , mg , 2, , mg, ±, 2, , 16 If I1 be initial intensity of sound and I2, be the final intensity of sound,, I, I, then S 1 = 10 log 1 and S 2 = 10 log 2, I0, I0, S 1 − S 2 = 30 dB, I, I, 10 log 1 − 10 log 2 = 30, I0, I0, I, I , 10 log 1 / 2 , I0 I0 , I, 10 log 1, I2, I1, log, I2, I1, I2, , = 30, = 30, , If f1 − f2 = 3, then according to given, condition, when f2 increases f1 − f2 the, decreases so the frequencies of strings, are related by f2 − f1 = 3., i.e. before tightening, f2 > f1, After tightening,, f2′ − f1 = 6,, f2′ > f1, , i.e., , vN =, , ⇒, , v oxy = v N, MN, 14 + 273, =, M oxy, t + 273, , ⇒, , ρN, 14 + 273, =, ρoxy, t + 273, , As, , ⇒, ⇒, , 14 14 + 273, =, 16, t + 273, 2296 − 1911, t =, = 55° C., 7, , 20 For situation 1,, v −0, ×f, v − vs, , fap1 =, =, , 320, × 320, 320 − 40, , = 3657, . Hz, , Situation -1, , Situation -2, Train, , v 22 = ω2 (a2 − x22 ), , =, , v 12 − v 22, x22 − x12, , 169 − 144, =, 25 − 9, , =, , (13)2 − (12)2, (5)2 − (3)2, , 25 5, =, 16 4, , 5, 5, Also, ω = 2 πf =, ., ⇒ f =, 4, 8π, , v, × f, v − (− v s ), , = 284.4 Hz, ∆f = fap 1 − fap2, = 81.3 Hz, , 21 Velocity of wave on the string, , 18 Let the fundamental frequencies of A, , =, , and B, before tightening of B are f1 and, f2 , respectively., Then either f1 − f2 = 3 or f2 − f1 = 3., As tension in B increases (due to, tightening), its frequency increases and, the beat frequency also increases., , 0.4 m, , …(i), , where, a is amplitude, λ is wavelength, v, is velocity and t is time., …(ii), , Comparing Eqs. (i) and (ii), we get, 2 πx 2 πx, =, λ, 3, ⇒, λ = 3 cm, , 23 Let the power of source be P and let it be, placed at Q., Then, intensity at A and at B would be, given by, P, IA =, 4 π × 12, P, and, IB =, 4 π × 22, ⇒, , (Amp. ) A, =, (Amp. )B, , IA, =, IB, , 22, 12, , = 2:1, , 24 For the wave, y = A sin (kx − ωt ),, ω, and the maximum, k, transverse string speed is Aω., Hence option (d) is correct., the wave speed is, , 25 For the wave y = A sin (ωt − kx ),, , For situation 2,, fap2 =, , 2 πx, 2 πvt, cos, λ, λ, , Given equation, 2 πx, y = 5 sin, cos 20 πt, 3, , According to question,, v 12 = ω2 (a2 − x12 ), , ⇒ω =, , So, standard equation of standing wave is, y = 2asin, , γR(14 + 273), MN, , Train, , ⇒ v 12 − v 22 = ω2 ( x22 − x12 ), , standing wave pattern as a result of the, interference of two waves., Distance between two nodes is half of, wavelength (λ )., , Speed of nitrogen at 14°C,, , 1, I1, 1000, , ⇒ v 2 = ω2 (a2 − x2 ), , The pulse gets inverted after reflection, from the fixed end, so for constructive, interference to take place between, successive pulses, the first pulse has to, undergo two reflections from the fixed, end., 2 × 0.4 + 2 × 0.4, So, ∆t =, = 0.2 s, 8, , 22 The node and antinodes are formed in a, γRT, M, , Speed of sound in oxygen at t°C,, γR(t + 273), v oxy =, M oxy, , = 103, , 17 Speed in SHM, v = ω a2 − x2, , γP, =, ρ, , 19 Speed of sound, v =, , =3, , I2 =, , 149, , UNIT TEST 2 (WAVES AND OSCILLATIONS), , DAY THIRTEEN, , T, = 8 ms −1, µ, , v 0 = Aω, where A is the maximum displacement., For the given condition,, A, = A sin (ωt − kx ), 2, 1, sin (ωt − kx ) =, ⇒, 2, ∂y, and, = Aω cos (ωt − kx ), ∂t, 3 v0, 3, = Aω, =, 2, 2, 1, 26 We have, T =, s = 10 s, 0.1, In one complete cycle, particle, travels a distance, 4 times the, amplitude.
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150, , So, in a time interval of 10 min 15 s i.e.,, 615 s i.e., 61 full + 1 half- cycles, the, distance travelled, = (4 × 3) × 61 + (2 × 3) × 1, = 732 + 6 = 738 cm, At this instant, the particle is moving in, an upward direction., ρ T, 2l m, ρ T, 420 =, 2l m, ρ+ 1 T, 490 =, 2l, m, 420, ρ, 6, =, =, 490 ρ + 1 7, ν=, , As, and, , ⇒, ∴, ⇒, , … (i), , 1, 1 , and ν 51 , −, =m, 2l 1 2l 2 , , … (ii), , ⇒, , m=, , 2, , 22, = 4 × × (2)2 + 0 ⋅ 05, 7, Maximum force on the man, = m(g + A max ), = 50(10 + 7.9) = 895N, = 89.5 kgf, Minimum force on the man, = m(g − A max ), = 50(10 − 7.9) = 105N = 10.5 kgf, Hence, the reading of weighing machine, varies between 10.5 kgf and 89.5 kgf., 29 Here m = 2 kg; v = 360 = 6 rps, 60, Extension produced in spring,, y = 43 − 36 = 7 cm, = 7 × 10− 2 m, On rotation the required centripetal, force is provided by tension in spring, i.e. ky = mr (2 πν)2 = 4 π2 ν2 mr, 4 π2 ν2 mr, k =, y, , where, r is length of stretched spring., 4 × (22 / 7)2 × 62 × 2 × (43 × 10− 2 ), ⇒k =, 7 × 10− 2, = 17475 Nm − 1 ., , 30 Let l 1 and l 2 be the lengths of the two, pipes, then, , 18, × 17 = 18 s− 1, 17, , 37 Total initial energy of particle in SHM,, 1, mω2 A2, 2, A, Energy when amplitude is, and angular, 2, frequency is 3ω., E1 =, , 2, , E2 =, , 38 The time period of a spring pendulum is, , where, σ = density of water and, ρ = density of bob, , given by, T = 2π, , The time period of bob in air, T = 2π, , g′, =, g, , l, g, , l, g′, , σ, , g 1 − , , ρ, g, , σ, 1, = 1−, ρ, ρ, T, 1, 1, 1, Given that,, =, ⇒ =1−, T′, 2, ρ, 2, = 1−, , or, , ρ=2, , 33 The body is dropped from a height h, and takes time t, then, 1, h = ut + g pt 2, 2, 1, 2, h = g pt, (Q u = 0), 2, 2h 2 × 8, gp =, =, = 4 ms− 2, 4, t2, l, 1 2π, Time period, T = 2 π, = 2π, =, g, 4, 2, = π = 314, . s, , 1, 9, A, m(3ω )2 = mω2 A2, 2, 2, 8, , ∴ Extra energy = E2 – E1, 9, 1, = mω2 A2 – mω2 A2, 8, 2, 5, 5, = mω2 A2 = E1 (gain ), 8, 4, , σ, , g ′ = g 1 − , , ρ, , T, =, T′, , π, π, = radian, 180° 3, , 2, , water, , ∴, , φ = 60° = 60 ×, , 2 2, 36 Pav = ρvω A, , 32 The effective acceleration of bob in, , The time period in water, T ′ = 2π, , 500, , λ, φ, 2π, 07, ., π, × = 012, . m = 12 cm, ⇒ x=, 2 π 3 , , As, t 1 − t 2 = 8 min = 8 × 60 s, d d, d, − = 480 ⇒ = 480, ∴, 4 8, 8, ⇒, d = 480 × 8 = 3840 km., , = 7.9 ms − 2, , 35 λ = v = 350 = 07, . m, , As, x =, , Time taken by transverse wave,, d, d, t2 =, = s, v2, 8, , 2 2, , to compress and rarify while the string, cannot be compressed or rarified., n, , by longitudinal wave,, d, d, t1 =, = s, v1, 4, , A max = ω a = (2 πν) a = 4 π ν a, , ⇒, , 34 For longitudinal wave, the medium has, , 31 Let d be the distance, then time taken, , 28 Maximum acceleration,, 2, , ν16 ν16, −, 2l 1 2l 2, , 1, 1 , −, ν16 , = 17, 2l 1 2l 2 , , ⇒, , 7ρ = 6ρ + 6 ⇒ ρ = 6, 6, 450, 420 =, 2l 5 × 10− 3, 900, l =, = 2.14 m, 420, 2, , m = n1 − n2 =, , Divide Eq. (ii) by Eq. (i), we get, v 51, 273 + 51, m, 324, =, =, =, v 16 17, 273 + 16, 289, , 27 The frequency of vibration is, , ⇒, , DAY THIRTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , m, and hence is not, k, , affected by the acceleration of the frame, of reference., , 39 Total energy of the particle performing, simple harmonic motion is,, E = K + U = K max + U min . K is always, positive, while U could be positive,, negative or zero. If U min is negative and, its value is greater than K max , then E, would be negative., , 40 When the hoop oscillates in its plane,, moment of inertia is, I1 = mR 2 + mR 2 i .e ., I1 = 2mR 2 ., While when the hoop oscillates in a, direction perpendicular to the plane of, the hoop, moment of inertia is, I2 =, , mR2, 3mR2, + mR2 =, 2, 2, , The time period of physical pendulum,, I, ., T = 2π, mgd, Here, d is a same in both the cases.
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DAY FOURTEEN, , Properties, of Matter, Learning & Revision for the Day, u, , u, , u, , u, , u, , Elastic Behaviour, Stress, Strain, Hooke’s Law, Work Done (or Potential, Energy Stored) in a Stretched, Line, , u, , u, , u, , u, , u, , u, , Thermal Stresses and Strains, Fluid Statics, Pascal’s Law, Laws of Floatation, Viscosity, Streamline and Turbulent, Flow, , u, , u, , u, , u, , u, , u, , u, , Equation of Continuity, Bernoulli’s Principle, Surface Tension, Surface Energy, Angle of Contact, Excess Pressure Over a Liquid Film, Capillary Rise or Capillarity, , Elastic Behaviour, Elasticity is the property of body by virtue of which a body regains or tends to regain its, original configuration (shape as well as size), when the external deforming forces acting, on it, is removed., , Stress, The internal restoring force per unit area of cross-section of the deformed body is called, stress., Restoring force F, Thus,, Stress, σ =, =, Area, A, SI unit of stress is Nm−2 or pascal (Pa)., Different types of stress are given below, , PREP, MIRROR, , Your Personal Preparation Indicator, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , 1. Normal or Longitudinal Stress, If area of cross-section of a rod is A and a deforming force F is applied along the length of, the rod and perpendicular to its cross-section, then stress produced in the rod is called, normal or longitudinal stress., F, Longitudinal stress = n, A, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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152, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Longitudinal stress is of two types, (i) Tensile stress When length of the rod is increased on, application of deforming force over it, then stress, produced in rod is called tensile stress., (ii) Compressive stress When length of the rod is decreased, on application of deforming force, then the stress, produced is called compressive stress., , Stress-Strain Relationship, For a solid, the graph between stress (either tensile or, compressive) and normal strain is shown in figure., , 2. Volumetric Stress, , Elastic limit, Proportional limit, , When a force is applied on a body such that it produces a, change in volume and density, shape remaining same, (i) at any point, the force is perpendicular to its surface., , l, , l, , 3. Shearing or Tangential Stress, When the force is applied tangentially to a surface, then it is, called tangential or shearing stress., F, Tangential stress = t, A, It produces a change in shape, volume remaining same., F, , Fixed face, , An object under tangential deforming forces, , Strain, Strain is the ratio of change in configuration to the original, configuration of the body., Being the ratio of two similar quantities, strain is a unitless, and dimensionless quantity., (i) When the deforming force causes a change in length, it is, called longitudinal strain. For a wire or rod, longitudinal, strain is defined as the ratio of change in length to the, original length., Change in length (∆L), ∴ Longitudinal strain =, Original length (L), (ii) When the deforming force causes a change in volume, the, strain is called volumetric strain., Change in volume (∆V ), Volumetric strain =, Original volume (V ), (iii) When the deforming force, applied tangentially to a, surface, produces a change in shape of the body, the strain, developed is called shearing strain or shear., x, Shearing strain, φ =, L, , A, , Stress, Strain, O', A typical stress-strain curve, , (ii) at any small area, the magnitude of force is directly, proportional to its area., Then, force per unit area is called volumetric stress., F, ∴ Volumetric stress = v, A, , Plastic region, D, C, E, B, (Fracture point), , Breaking, strength, , l, , l, , In the above graph, point A is called proportional limit., Till this point, stress and strain are proportional to each, other., From point A to B, stress and strain are not proportional, B, is called elastic limit and OB is elastic region., Beyond point B, strain increases without increase in stress,, it is called plastic behaviour. Region between point C and, D is called plastic region., Finally, at point D, wire may break, maximum stress, corresponding to point D is called breaking stress., , The materials of the wire, which break as soon as stress is, increased beyond the elastic limit are called brittle., Graphically, for such materials the portion of graph between B, and E is almost zero. While the materials of the wire, which, have a good plastic range (portion between B and E) are called, ductile., , Hooke’s Law, According to the Hooke’s law, for any body, within the elastic, limit, stress developed is directly proportional to the strain, produced., Stress ∝ Strain, Stress = E × Strain, The ratio of stress to strain, within the elastic limit, is called, the coefficient (or modulus) of elasticity for the given, material., Depending on the type of stress applied and resulting strain,, we have the following three of elasticity given as,, Stress, E =, Strain, There are three modulus of elasticity., , 1. Young’s Modulus, Young’s modulus of elasticity (Y ) is defined as the ratio of, normal stress (either tensile or compressive stress) to the, longitudinal strain within a elastic limit., Young’s modules,, Normal stress, F/A, FL, Y =, =, =, Longitudinal strain ∆ L / L A∆ L
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PROPERTIES OF MATTER, , DAY FOURTEEN, , 2. Bulk Modulus, It is defined as the ratio of the normal stress to the volumetric, strain. Coefficient of volume elasticity,, F/A, pV, B=, =−, ∆V / V, ∆V, F, = the pressure or stress negative sign signifies, A, that for an increase in pressure, the volume will decrease., Reciprocal of bulk modulus is called compressibility., where, p =, , 3. Modulus of Rigidity (Shear modulus), , l, , Energy stored per unit volume (or energy density), U 1 F∆ L 1, Y, = =, = stress × strain = (strain)2, V 2 AL, 2, 2, , Thermal Stresses and Strains, When a body is allowed to expand or contract with increase or, decrease in temperature, no stresses are induced in the body., But if the deformation of the body is prevented, some stresses, are induced in the body. Such stresses are called thermal, stresses or temperature stresses. The corresponding strains are, called thermal strains or temperature strains., l, a , Y, A, , It is defined as the ratio of tangential stress to shearing stress., Tangential stress F / A, F, FL, η=, =, =, =, Shearing strain, φ, Aφ Ax, l, , Breaking force depends upon the area of cross-section of, the wire., ∴ Breaking force ∝ A, Breaking force = P × A, Here, P is a constant of proportionality and known as, breaking stress., , Poisson’s Ratio, For a long bar, the Poisson’s ratio is defined as the ratio of, lateral strain to longitudinal strain., Lateral strain, ∆ D/ D ∆ r / r, ∴ Poisson’s ratio, σ =, =, =, Longitudinal strain ∆ L / L ∆ L / L, Poisson’s ratio is a unitless and dimensionless term. Its value, depends on the nature of the material. Theoretically, value of, σ must lie between −1 and + 0.5 but for most metallic solids, 0 < σ < 0.5., , Inter-Relations Between Elastic Constants, Y = Young’s modulus, η = Rigidity modulus,, B = Bulk modulus, σ = Poisson’s ratio, The inter relation between elastic constants are, Y = 2 η (1 + σ), Y = 3 B (1 − 2 σ), 9 3 1, 9 Bη, 3B − 2 η, or Y =, or σ =, = +, Y η B, η + 3B, 6B + 2 η, , Work Done (or Potential Energy, Stored) in a Stretched Wire, Work is done against the internal restoring forces, while, stretching a wire. This work is stored as elastic potential, energy. The work done is given by, 1, Work done W = × stretching force × elongation, 2, 1, 1 YA, = F∆ L =, (∆ L)2, 2, 2 L, l, , = Energy stored in the wire (U), , 153, , B, , A, , ∆l, , A rod under thermal stress and strain, By definition, coefficient of linear expansion α =, , ∆l, lθ, , ∆l, = α ∆θ, l, So thermal stress = Yα ∆θ, Tensile or compressive force produced in the body, F = YAα ∆θ, thermal strain, , Fluid Statics, The substances which flow are called fluids, that includes, both liquid and gas. The science of fluids at rest is called fluid, statics where fluid mass is stationary w.r.t. container,, containing the fluid., , Thrust and Pressure, Normal force exerted by fluid at surface in contact is called, thrust of fluid., The thrust exerted by a fluid at rest per unit surface area of, contact surface is called the fluid pressure., Normal force (thrust) F, ∴ Pressure p =, Surface area A, Pressure is a scalar and its SI unit is Nm−2 or pascal (Pa),, where, 1 Pa = 1 N m−2 ., , Pressure due to Fluid Column, Pressure p exerted by the fluids at the bottom of container, having height h, p = hρ g, where, ρ = density of fluid., , Gauge Pressure, The pressure difference between the real hydrostatic pressure, and the atmospheric pressure is known as the gauge pressure., ∴ Gauge pressure = Real pressure (p), − Atmospheric pressure ( p0 )
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154, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Pascal’s Law, According to Pascal’s law of transmission of pressure, the, increase in pressure at any one point of the enclosed liquid in, equilibrium or at rest, is transmitted equally to all other points, of the liquid and also to the walls of the container. Hydraulic, lift, hydraulic press, hydraulic brakes etc. are based on the, Pascal’s law., , Case II The density of body is equal to the density of, liquid (i.e. ρ B = σ). In this case, W = F . so, the body, will float fully submerged in neutral equilibrium, anywhere in the liquid., Case III, , Archimedes’ Principle and Buoyancy, When a body is immersed in a fluid, it experiences an upthrust, due to the fluid and as a result the apparent weight of the body, is reduced., ∴Apparent weight of the body, = weight of the body – upthrust due to fluid, = weight of the body – weight of the fluid displaced, e.g. For a floating body, the volume of a body (V − Vs ) remaining, outside the liquid will be given by, ρ, ρ, , = V 1 − , V0 = V − Vs = V − V, , σ, σ, where, ρ = density of liquid, and, σ = density of body immersed in liquid, , Buoyant Force or Buoyancy, Buoyant force, F = hρgA = mg, where, h = height of body immersed in liquid,, m = mass of body and A = area, l, , l, , l, , l, , l, , It is an upward force acting on the body immersed in a, liquid., It is equal to the weight of liquid displaced by the immersed, part of the body., , The density of body is less than that of liquid (i.e., ρ B < σ). In this case, W < F , so the body will move, upwards and in equilibrium will float partially, immersed in the liquid such that, W = Vinσg, [Vin is the volume of body in the liquid], or, , Vρ B g = Vinρg, , or, , Vρ B = Vinσ, , Viscosity is the property of a fluid due to which it opposes, the relative motion between its different layers., Force between the layers opposing the relative motion is, called viscous force., If there are two fluid layers having surface area A and, velocity gradient dv/dr , then the viscous force is given by, dv, . where, constant η is called the coefficient of, F = −ηA, dr, viscosity of the given fluid., SI unit of coefficient of viscosity is N m–2s or Pa-s or, poiseuille (Pl)., , Terminal Velocity, , The buoyant force acts at the centre of buoyancy which is, the centre of gravity of the liquid displaced by the body, when immersed in the liquid., , If a small spherical body is dropped in a fluid, then initially it, is accelerated under the action of gravity. However, with an, increase in speed, the viscous force increases and soon it, balances the weight of the body., , The line joining the centre of gravity and centre of buoyancy, is called central line., , Now, the body moves with a constant velocity, called the, terminal velocity., , Metacentre, is a point where the vertical line passing through, the centre of buoyancy intersects the central line., , Terminal velocity vt is given by, , Laws of Floatation, When a body of density ρ B and volume V is immersed in a, liquid of density σ, the forces acting on the body are, , l, , …(i), , Viscosity, , vt =, , l, , [as, W = mg = ρ BVg], , The weight of body W = mg = Vρ B g acting vertically, downwards through the centre of gravity of the body., The upthrust F = Vσg acting vertically upwards through the, centre of gravity of the displaced liquid i.e., centre of, buoyancy., , So, the following three cases are possible., Case I, The density of body is greater than that of liquid (i.e., ρ B > σ). In this case, as weight will be more than, upthrust, the body will sink. (W > F ), , 2 r 2 (ρ − σ)g, 9, η, , where, r = radius of the falling body,, ρ = density of the falling body, and, , σ = density of the fluid., , Stokes’ Law, Stokes proved that for a small spherical body of radius r, moving with a constant speed v called terminal velocity, through a fluid having coefficient of viscosity η, the viscous, force F is given by, F = 6πηrv, It is known as Stokes’ law.
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PROPERTIES OF MATTER, , DAY FOURTEEN, , Streamline and Turbulent Flow, Flow of a fluid is said to be streamlined if each element of the, fluid passing through a particular point travels along the same, path, with exactly the same velocity as that of the preceding, element. A special case of streamline flow is laminar flow., A turbulent flow is the one in which the motion of the fluid, particles is disordered or irregular., , Energy of a Flowing Liquid, There are three types of energies in a flowing liquid., l, , = force × displacement = pAl, The volume of the liquid is Al. Hence, pressure energy, pAl, per unit volume of liquid =, =p, Al, l, , Kinetic Energy If a liquid of mass m and volume V is, 1, flowing with velocity v, then the kinetic energy = mv2, 2, ∴ Kinetic energy per unit volume of liquid, 1 m, 1, = v2 = ρv2, , , 2 V, 2, , where, r = radius of tube., , Here, ρ is the density of liquid and V = volume., , Reynold’s Number (N R ), , l, , Reynold’s number as the ratio of the inertial force per unit area, to the viscous force per unit area for a fluid., NR =, , Pressure Energy If p is the pressure on the area A of a, fluid and the liquid moves through a distance l due to, this pressure, then, Pressure energy of liquid = work done, , Critical Velocity, For a fluid, the critical velocity is that limiting velocity of the, fluid flow upto which the flow is streamlined and beyond which, the flow becomes turbulent. Value of critical velocity for the, flow of liquid of density ρ and coefficient of viscosity η, flowing, through a horizontal tube of radius r is given by, η, vc ∝, ρr, , 155, , v2ρ, ρvr, =, ηv / r, η, , A smaller value of Reynold’s number (generally N R ≤ 1000), indicates a streamline flow but a higher value (N R ≥ 1500), indicates that the flow is turbulent and between 1000 to 1500,, the flow is unstable., , Equation of Continuity, Let us consider the streamline flow of an ideal, non-viscous fluid, through a tube of variable cross-section., Let at the two sections, the cross-sectional areas be A1 and A2 ,, respectively and the fluid flow velocities are v1 and v2 , then, according to the equation of continuity, A1 v1 ρ1 = A 2 v2 ρ2, where, ρ1 and ρ2 are the respective densities of the fluid., Equation of continuity is based on the conservation of mass., , Potential Energy If a liquid of mass m is at a height h, from the reference line (h = 0), then its potential energy is, mgh., ∴ Potential energy per unit volume of the liquid, m, = gh = ρgh, V , , Bernoulli’s Principle, According to the Bernoulli’s principle for steady flow of an, incompressible, non-viscous fluid through a tube/pipe, the, total energy (i.e. the sum of kinetic energy, potential energy, and pressure energy) per unit volume (or per unit mass too), remains constant at all points of flow provided that there is, no source or sink of the fluid along the flow., Mathematically, we have, 1, p, v2, p + ρgh + ρv2 = constant or, + h+, = constant, 2, ρg, 2g, In this expression,, , p, v2, is pressure, is velocity head and, ρg, 2g, , head., , Q, , P, , v2, v1, , Velocity of Efflux, l, , A2, A1, , A liquid is flowing through a tube of non-uniform, cross-section, If the fluid which is flowing, is incompressible, then, ρ1 = ρ2, So, equation of continuity is simplified as, A1 v1 = A2 v2 or Av = constant, , If a liquid is filled in a vessel up to a height H and a small, orifice O is made at a height h, then from Bernoulli’s, theorem it can be shown that velocity of efflux v of the, liquid from the vessel is, v = 2 g (H − h), , l, , The flowing fluid describes a parabolic path and hits the, base level at a horizontal distance (called the range), R = 2 h (H − h)., H, and in that case, The range is maximum, when h =, 2, Rmax = H .
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156, , Applications Based on the Bernoulli’s, Principle, l, , l, , l, , l, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , l, , The action of carburetor, paintgun, scent sprayer, atomiser, and insect sprayer is based on the Bernoulli’s principle., , ∆U = S × 4π (nr 2 − R2 ) = S × 4πr 2 n (1 − n−1 /3 ), , The action of the Bunsen’s burner, gas burner, oil stove and, exhaust pump is also based on the Bernoulli’s principle., , Angle of Contact, , Motion of a spinning ball (Magnus effect) is based on, Bernoulli’s theorem., , Angle of contact for a given liquid-solid combination is, defined as the angle subtended between the tangents to the, liquid surface and the solid surface, inside the liquid, the, tangents are drawn at the point of contact., , Blowing of roofs by wind storms, attraction between two, parallel moving boats moving close to each other, fluttering, of a flag etc., are also based on Bernoulli’s theorem., , Angle of, contact, θ, , Surface Tension, Surface tension is the property of a liquid due to which its free, surface behaves like a stretched elastic membrane and tends, to have the least possible surface area., Force, F, Surface tension S =, =, Length l, Here, F is force acting on the unit length of an imaginary line, drawn of the surface of the liquid., SI unit of surface tension is Nm−1 or Jm−2 . It is a scalar and its, dimensional formula is [MT−2 ]., , Work done in Blowing a Liquid Drop If a liquid drop is, blown up from a radius r1 to r2 , then work done in the, process,, W = S ( A2 − A1 ) = S × 4π (r22 − r12 ), , l, , Work done in Blowing a Soap Bubble As a soap bubble, has two free surfaces, hence, work done in blowing a soap, bubble, so as to increase its radius from r1 to r2 , is given by, W = S × 8π, , l, , (r22, , −, , r12 ), , Work done in Splitting a Bigger Drop into n Smaller, Droplets If a liquid drop of radius R is split up into n, smaller droplets, all of the same size, then radius of each, droplet, r = R(n)−1 /3, and work done, W = S × 4π (nr 2 − R2 ) = S × 4πR2 (n1 /3 − 1), , θ, , Beaker, , Shape of meniscus for different media, l, , l, , Surface Energy, Surface energy of a liquid is the potential energy of the, molecules of a surface film of the liquid by virtue of its, position., When the surface area of a liquid is increased, work is done, against the cohesive force of molecules and this work is stored, in the form of additional surface energy., Increase in surface potential energy, ∆U = Work done (∆W ) = S∆ A, where, ∆ A is the increase in surface area of the liquid., , Angle, of, contact, , Beaker, , l, , l, , Coalescence of Drops If n small liquid drops of radius r, each, combine together so as to form a single bigger drop of, radius R = n1 /3 r then in the process, energy is released., Release of energy is given by, , Value of the angle of contact depends on the nature of, liquid and solid both., For a liquid having concave meniscus when adhesive, force > cohesive force angle of contact θ is acute (θ < 90 ° ), but for a convex meniscus (when cohesive force >, adhesive force) the angle of contact is obtuse (θ > 90 ° )., Value of angle of contact θ decreases with an increase in, temperature., , Excess Pressure Over a Liquid Film, If a free liquid surface film is plane, then pressure on the, liquid and the vapour sides of the film are the same, otherwise, there is always some pressure difference. Following cases, arise., l, , l, , l, , For a spherical liquid drop of radius r, the excess pressure, 2S, inside the drop p =, r, where, S = surface tension of the liquid., 2S, For an air bubble in a liquid, excess pressure p =, r, 4S, For a soap bubble in air, excess pressure p =, r, , Capillary Rise or Capillarity, Capillarity is the phenomenon of rise or fall of a liquid in a, capillary tube as compared to that in a surrounding liquid., The height h up to which a liquid will rise in a capillary tube, is given by, 2S cos θ, 2S, h=, =, rρg, Rρg, where, r = radius of the capillary tube, r, and R =, = radius of liquid meniscus., cosθ, 1, The rise in capillary tube h ∝ (Jurin’s law)., r
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PROPERTIES OF MATTER, , DAY FOURTEEN, l, , If a capillary tube, dipped in a liquid is tilted at an angle α, from the vertical, the vertical height h of the liquid column, remains the same. However, the length of the liquid, column (l ) in the capillary tube increases to, , h, , α, , l=, l, , l, l, , Beaker, , Effect of tilting capillary tube in a liquid, , 157, , h, ., cos α, , If the capillary tube is of insufficient length, the liquid rises, up to the upper end of the tube and then the radius of its, meniscus changes from R to R′ such that hR = h ′ R ′, where, h′ = insufficient length of the tube., After connection due to the weight of liquid contained in, the meniscus, the formula for the height is given by, 2s, r, h=, −, ρrg 3, This is known as ascent formula., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A man grows into a giant such that his linear dimensions, increase by a factor of 9. Assuming that his density, remains same, the stress in the leg will change by a, factor of, ª JEE Main 2017 (Offline), 1, (a), 9, 1, (c), 81, , (b) 81, (d) 9, , and is l2 when the tension isT2. The original length of, the wire is, l1 + l 2, 2, l1T2 − l 2T1, (c), T2 − T1, (a), , (b), , 1.0 × 1011 Nm −2 and 2.0 × 1011 Nm −2 . A brass wire and a, steel wire of the same length extend by 1 mm, each, under the same force. If radii of brass and steel wires are, RB and RS respectively, then, 2 RB, , (c) RS = 4RB, , RB, 2, RB, (d) RS =, 2, (b) RS =, , 3 One end of a horizontal thick copper wire of length 2L, , (d) T1T2l1l 2, , same volume. However, wire 1 has cross-sectional area A, and wire 2 has cross-sectional area 3A. If the length of, wire 1 increases by ∆x on applying force F, how much, force is needed to stretch wire 2 by the same amount?, ª AIEEE 2009, , (a) F, , (b) 4F, , (c) 6F, , (d) 9 F, , 7 If the ratio of lengths, radii and Young’s modulii of steel, and brass wires in the figure are a, b and c respectively,, then the corresponding ratio of increase in their lengths is, ª JEE Main (Online) 2013, , and radius 2R is welded to an end of another horizontal, thin copper wire of length L and radius R. When the, arrangement is stretched by applying forces at two ends,, the ratio of the elongation in the thin wire to that in the thick, wire is, (a) 0.25, (c) 2.00, , l1T2 + l 2T1, T1 + T2, , 6 Two wires are made of the same material and have the, , 2 The Young’s modulus of brass and steel are respectively, , (a) RS =, , 5 The length of a metal wire is l1 when the tension in it isT1, , Steel, M, , (b) 0.50, (d) 4.00, , Brass, , 4 The following four wires are made of the same material., Which of these will have the largest extension when the, same tension is applied ?, (a) Length = 50 cm, diameter = 0.5 mm, (b) Length = 100 cm, diameter = 1mm, (c) Length = 200 cm, diameter = 2 mm, (d) Length = 300 cm, diameter = 3 mm, , 2M, , (a), , 3c, 2ab, , 2, , (b), , 2a 2c, b, , (c), , 3a, 2, , 2b c, , (d), , 2ac, b2
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158, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 8 The pressure of a medium is changed from1.01 × 105 Pa, , (a) 204.8 × 105 Pa, (c) 51.2 × 105 Pa, , (b) 102.4 × 105 Pa, (d) 1.55 × 105 Pa, , (a), , modulus K is surrounded by a liquid in a cylindrical, container. A massless piston of area a floats on the, surface of the liquid, covering entire cross-section of, cylindrical container. When a mass m is placed on the, surface of the piston to compress the liquid, the, dr , fractional decrement in the radius of the sphere, is, r , , Y, αt, , σ − ρ, (a) , mg, σ , ρ mg, (c), σ, , ª JEE Main 2018, , (c)Yα t, , (d), , Q, , σ − ρ, (b) , mg, ρ , σ mg, (d), ρ, , 16 A ball is made of a material of density ρ where, , Water, , = applied force, Q = extension, = extension, Q = applied force, = extension, Q = stored elastic energy, = stored elastic energy, Q = extension, , 1, Yα t, , ρoil < ρ < ρ water with ρoil and ρ water representing the densities, of oil and water, respectively. The oil and water are, immiscible. If the above ball is in equilibrium in a mixture of, this oil and water, which of the following pictures, represents its equilibrium position?, ª AIEEE 2010, , length of wire in the region for which, the substance obeys Hooke’s law., P and Q represent, (a) P, (b) P, (c) P, (d) P, , (b), , string, the other end of the string is fixed to the bottom of, a tank. The tank is filled with a liquid of density σ with, σ > ρ. The tension in the string will be, , Ka, (b), 3mg, mg, (d), Ka, , 10 The graph shows the behaviour of a, , αt, Y, , 15 A wooden block of mass m and density ρ is tied to a, , 9 A solid sphere of radius r made of a soft material of bulk, , Ka, (a), mg, mg, (c), 3Ka, , 14 A metal rod of Young’s modulus Y and coefficient of, thermal expansion α is held at its two ends such that its, length remains invariant. If its temperature is raised by, ª AIEEE 2011, t °C, the linear stress developed in it is, , to 1.165 × 105 Pa and change in volume is 10% keeping, temperature constant. The bulk modulus of the medium, is, , P, , (a), , Oil, , (b), Oil, , Water, , 11 If work done in stretching a wire by 1mm is 2 J. The work, necessary for stretching another wire of same material, but with double the radius and half the length, by 1 mm, distance, is, (a) 16 J, (c) 1/4 J, , Oil, , thermal expansion α 1, α 2 and Young’s moduliiY1, Y2, respectively are fixed between two rigid massive walls., The rods are heated such that they undergo the same, increase in temperature. There is no bending of the rods., If α 1 : α 2 = 2 : 3, the thermal stresses developed in the two, rods are equal providedY1 : Y2 is equal to, (b) 1 : 1, (d) 4 : 9, , 13 The pressure that has to be applied to the ends of a steel, wire of length 10 cm to keep its length constant when its, temperature is raised by 100°C is, (For steel, Young’s modulus is 2 × 1011Nm −2 and, coefficient of thermal expansion is 1.1 × 10 −5 K −1), ª JEE Main 2014, , (a) 2 . 2 × 108 Pa, (c) 2 . 2 × 107 Pa, , Oil, , (d), , (b) 4 J, (d) 8 J, , 12 Two rods of different materials having coefficients of, , (a) 2 : 3, (c) 3 : 2, , Water, , (c), , (b) 2 . 2 × 109 Pa, (d) 2 . 2 × 106 Pa, , Water, , 17 Two mercury drops (each of radius r ) merge to form a, bigger drop. The surface energy of the bigger drop, if T is, ª AIEEE 2011, the surface tension, is, (a) 2 5/ 3 π 2T, , (b) 4 π 2T, , (c) 2 πr T, , (d) 2 8 / 3 πr 2T, , 2, , 18 A raindrop of radius 0.2 cm is falling through air with a, terminal velocity of 8.7 m/s. The viscosity of air in SI units, is (take, ρ water = 1000 kg/m 3 and ρ air = 1 kg/m 3)., (a) 10−4 poise, (c) 8.6 × 10−3 poise, , (b) 1 × 10−3 poise, (d) 1.02 × 10−3 poise, , 19 If a ball of steel (density ρ = 7.8 g cm −3 ) attains a terminal, , velocity of 10 cms −1 when falling in a tank of water, (coefficient of viscosity ηwater = 8.5 × 10−4 Pa-s) then its, terminal velocity in glycerine (ρ = 1.2 gcm −3,, ª AIEEE 2011, η = 13.2 Pa-s) would be nearly, (a) 16, . × 10−5 cms −1, (c) 6.45 × 10−4 cms −1, , (b) 6.25 × 10−4 cms −1, (d) 15, . × 10−5 cms −1
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PROPERTIES OF MATTER, , DAY FOURTEEN, 20 Water is flowing continuously from a tap having an, , internal diameter 8 × 10−3 m. The water velocity as it, leaves the tap is 0.4 ms −1. The diameter of the water, stream at a distance 2 × 10−1 m below the tap is close to, ª AIEEE 2011, , (a) 7.5 × 10−3 m, (c) 3.6 × 10−3 m, , (b) 9.6 × 10−3 m, (d) 5.0 × 10−3 m, , 21 At what speed, the velocity head of water is equal to, , tube is found to be 0.5 cm. Then, compute tension of, water using this data (take, contact angle between glass, and water as 0 and g = 9.81 m/s 2)., (a) 0.72 N/m, (b) 0.77 N/m, (c) 1.67 N/m, (d) None of the above, , 26 Match the following columns., , pressure head of 40 cm of Hg?, −1, , −1, , (a) 10.3 ms, (c) 5.6 ms −1, , (b) 2.8 ms, (d) 8.4 ms −1, , 22 A liquid X of density 3.36 g/cm 3 is poured in a U-tube,, which contains Hg. Another liquid Y is poured in the left, arm with height 8 cm and upper levels of X and Y are, same. What is the density of Y ?, Y, , 10 cm, P¢, , (b) 1.2 g /cc, , (c) 1.4 g /cc, , D., , Column I, Magnus energy, Loss of energy, Pressure is same at the, same level in a liquid, Gas burner, , Column II, 1. Pascal’s law, 2. Bernoulli’s principle, 3. Viscous force, 4. Spinning ball, , Codes, B, 4, 2, , C, 2, 4, , D, 3, 3, , A, (b) 1, (d) 4, , B, 2, 3, , C, 3, 1, , D, 4, 2, , Direction (Q. Nos. 27-30) Each of these questions, contains two statements : Statement I and Statement II. Each, of these questions also has four alternative choices, only one of, which is the correct answer. You have to select one of the codes, (a), (b), (c), (d) given below, , Hg, , (a) 0.8 g /cc, , A., B., C., , A, (a) 1, (c) 2, , X, , 8 cm, 2 cm, P, , 159, , (d) 1.6 g /cc, , 23 A thin liquid film formed between a U-shaped wire and a, light slider supports a weight of 1.5 × 10−2N (see figure)., The length of the slider is 30 cm and its weight, negligible. The surface tension of the liquid film is, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 27 Statement I Aeroplanes are made to run on the runway, before take off, so that they acquire the necessary lift., , Film, , Statement II This is as per the Bernoulli’s theorem., , 28 Statement I Finer the capillary, greater is the height to, which the liquid rises in the tube., w, , (a) 0.0125 Nm −1, (c) 0.05 Nm −1, , (b) 0.1 Nm −1, (d) 0.025 Nm −1, , ª AIEEE 2012, , 24 Work done in increasing the size of a soap bubble from, radius of 3 cm to 5 cm is nearly (surface tension of soap, ª AIEEE 2011, solution = 0.03 Nm −1), (a) 0. 2 π mJ, , (b) 2 π mJ, , (c) 0.4 π mJ, , (d) 4 π mJ, , 25 While measuring surface tension of water using capillary, rise method, height of the lower meniscus from free, surface of water is 3 cm while inner radius of capillary, , Statement II This is in accordance with the ascent, formula., , 29 Statement I The bridges are declared unsafe after a long, use., Statement II Elastic strength of bridges decreases with, time., , 30 Statement I A small drop of mercury is spherical but, bigger drops are oval in shape., Statement II Surface tension of liquid decreases with an, increase in temperature.
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160, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A light rod of length L is, suspended from a support, horizontally by means of two, vertical wires A and B of equal, lengths as shown in the figure., Cross-section area of A is half, that of B and Young’s modulus, of A is double than that of B. A, weight w is hung on the rod as, shown. The value of x, so that the, stress in A is same as that in B, is, (a), , L, 3, , (b), , L, 2, , (c), , A, , 2L, 3, , B, , A, Mg, , w, , (d), , 3L, 4, , ª JEE Main 2015, T 2, Mg, (b) M − 1, T , A, 2, , T A, (d) 1 − , TM Mg, , , , , longitudinal tension is 4 N and the length is b metres, when, the tension is 5 N. The length of the string (in metre), when the longitudinal tension is 9 N is, a, 2, , (b) 5 b − 4a, , (c) 4 a − 3 b, , (d) a − b, , two strings of equal length on two supports (figure). It, can be done in one of the following three ways:, , (b), , (c), , The tension in the strings will be, (a) the same in all cases, (c) least in (b), , (a), , 1 + sinα, 1 − sinα, , so that its volume is decreased by 0.1%. (The bulk, modulus of rubber is 9.8 × 108 N/m 2; and the density of, sea water is 103 kg/m 3.), (b) 60 m, , (c) 75 m, , (b), , 1 + cosα, 1 + tanα, (c), 1 − cosα, 1 − tanα, , a, , d2, , d1, , (d), , 1 + sinα, 1 − cosα, , 8 On heating water, bubbles being formed at, the bottom of the vessel detach and rise., Take the bubbles to be spheres of radius R, and making a circular contact of radius r, R, with the bottom of the vessel. If r <<R and, the surface tension of water is T, value of r, 2r, just before bubbles detach is (density of, ª JEE Main 2014, water is ρ), (a) R 2, , 2ρw g, ρ g, (b) R 2 w, 3T, 6T, , (c) R 2, , ρw g, T, , (d) R 2, , 3ρw g, T, , that a length of 8 cm extends above the mercury level. The, open end of the tube is then closed and sealed and the, tube is raised vertically up by additional 46 cm. What will be, length of the air column above mercury in the tube now?, ª JEE Main 2014, (Atmospheric pressure = 76 cm of Hg], (b) 22 cm, , (c) 38 cm, , (d) 6 cm, , 10 Two narrow bores of diameter 3.0 mm and 6.0 mm are, joined together to form a U-tube open at both ends. If the, U-tube contains water, what is the difference in its levels, in the two limbs of the tube? Surface tension of water at, the temperature of the experiment is 7.3 × 10−2 N/m. Take, the angle of contact to be zero and density of water to be, 1.0 × 103 kg/m 3.( take, g = 9.8 m/s 2)., (a) 6 mm, , (b) 2 mm, , (c) 5 mm, , (d) 3 mm, , 11 A long metal rod of length l and relative density σ, is, , (b) least in (a), (d) least in (c), , 5 To what depth must a rubber ball be taken in deep sea,, , (a) 100 m, , plane. Two liquids which do not mix, and of densities d1 and d 2 are filled in, the tube. Each liquid subtends 90°, angle at centre. Radius joining their, interface makes an angle α with, vertical. Ratio d1 / d 2 is ª JEE Main 2014, , (a) 16 cm, , 4 A rectangular frame is to be suspended symmetrically by, , (a), , (d) a 2 /g, , (c) ag, , 9 An open glass tube is immersed in mercury in such a way, , 3 The length of an elastic string is 1m, when the, , (a) 2 b −, , (b) a/g, , 7 There is a circular tube in a vertical, x, , area A has time period T. When an additional mass M is, added to its bob, the time period changes to TM . If the, 1, Young’s modulus of the material of the wire is Y, then, Y, is equal to (take, g = gravitational acceleration), A, Mg, , the tanker starts accelerating the free surface will be, tilted by an angle θ. If the acceleration is a ms −2, then the, slope of the free surface is, (a) g/a, , L, , 2 A pendulum made of a uniform wire of cross-sectional, , T 2, , (a) M − 1, T , , 2, , , T, (c) 1 − M , , , T, , , , 6 The free surface of oil in a tanker, at rest, is horizontal. If, , (d) 65 m, , held vertically with its lower end just touching the surface, of water. The speed of the rod when it just sinks in water,, is given by, (a) 2glσ, , (b) 2gl (2 σ − 1), , 1 , (c) 2gl 1 −, , , 2σ , , (d) 2gl
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12 A wooden wheel of radius R is made of, , fractional decrease in the radius of the wire is close to, , two semi-circular parts (see figure). The, two parts are held together by a ring, R, made of a metal strip of cross-sectional, area S and length L. L is slightly less, than 2πR. To fit the ring on the wheel, it, is heated so that its temperature rises by, ∆T and it just steps over the wheel. As it cools down to, surrounding temperature, it presses the semi-circular, parts together. If the coefficient of linear expansion of the, metal is α and its Young’s modulus is Y, the force that, one part of the wheel applies on the other part is, , ª JEE Main (Online) 2013, , (a) 1.0 × 10−4, , (c) πSYα∆T, , (d) 2SYα∆T, , 0.5 m having equal cross-sectional areas are joined end, to end. The composite wire is stretched by a certain load, which stretches the copper wire by 1 mm., If the Young’s modulii of copper and steel are respectively, 1.0 × 1011 Nm −2 and 2.0 × 1011 Nm −2, the total extension of, the composite wire is, ª JEE Main (Online) 2013, (a) 1.75 mm, , tension in it produces an elastic strain of 1%. What is the, fundamental frequency of steel, if density and elasticity, of steel are 7.7 × 103 kg/m 3 and 2 . 2 × 1011 N /m 2,, respectively?, ª JEE Main 2013, , 14 A wire of length 2L and radius, , A, , r is stretched and clamped, L, between A and B. If the, Young’s modulus of the, L2 + l 2, material of the wire be Y ,, tension in the wire, when, stretched in the position ADB will be, (a) πr 2YLt, (c) 2 πr 2YL 2 /l 2, , (d) 770 Hz, 2L, C, , B, L, , (a), , l, , (c), , Mg, k, , 1 − LAσ , , , 2M , , (d), , Mg, k, , (b), , mgl, YA, , (c), , mgl, 2YA, , (d), , 3mg, YAl, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , (b) πr 2Yl 2 / 2L2, (d) None of these, , Mg 1 − LAσ , , , k M , , 2mg, 3YA, , (Q. Nos. 19-20) Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below, , D, , (b), , (d) 1.25 mm, , Direction, , cross-sectional area A is suspended, with its length, veritcal from a fixed point by a massless spring such that, it is half submerged in a liquid of density σ at equilibrium, position. The extenstion x 0 of the spring when it is in, ª JEE Main 2013, equilibrium is, Mg, k, , (c) 1.50 mm, , ceiling. Due to its own weight it elongates, consider, cross-section area of wire as A and Young’s modulus of, material of wire asY . The elongation in the wire is, , 15 A uniform cylinder of length L and mass M having, , (a), , (b) 2.0 mm, , 18 A wire of mass m, and length l is suspended from a, , 13 A sonometer wire of length 1.5 m is made of steel. The, , (a) 188.5 Hz (b) 178.2 Hz (c) 200.5 Hz, , (b) 1.5 × 10−4 (c) 0.17 × 10−4 (d) 5 × 10−4, , 17 A copper wire of length 1.0 m and a steel wire of length, , ª AIEEE 2012, , (a) 2 πSYα∆T (b) SYα∆T, , 161, , PROPERTIES OF MATTER, , DAY FOURTEEN, , 19 Statement I In taking into account the fact that any, object which floats must have an average density less, than that of water, during world war-I, a number of cargo, vessels were made of concrete., Statement II Concrete cargo vessels were filled with air., , 20 Statement I The stream of water flowing at high speed, , 1 + LAσ , , , M, , , 16 A uniform wire (Young's modulus 2 × 1011 Nm −2 ) is, , subjected to longitudinal tensile stress of 5 × 107 Nm −2. If, the overall volume change in the wire is 0.02%, the, , from a garden hosepipe tends to spread like a fountain, when held vertically up, but tends to narrow down when, held vertically down., Statement II In any steady flow of an incompressible, fluid, the volume flow rate of the fluid remains constant., , ANSWERS, SESSION 1, , 1 (d), 11 (a), 21 (b), , 2 (b), 12 (c), 22 (a), , 3 (c), 13 (a), 23 (d), , 4 (a), 14 (c), 24 (c), , 5 (c), 15 (b), 25 (b), , 6 (d), 16 (b), 26 (d), , 7 (c), 17 (d), 27 (a), , 8 (d), 18 (b), 28 (a), , 9 (c), 19 (b), 29 (a), , 10 (c), 20 (c), 30 (b), , SESSION 2, , 1 (c), 11 (c), , 2 (a), 12 (d), , 3 (b), 13 (b), , 4 (c), 14 (b), , 5 (a), 15 (c), , 6 (b), 16 (c), , 7 (c), 17 (d), , 8 (a), 18 (c), , 9 (a), 19 (b), , 10 (c), 20 (a)
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162, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , As,, , So weight = volume × density × g will, also become (9)3 times., Area of cross-section will become (9) 2, times, =, , 93 × W 0, 92 × A 0, , W , = 9 0, A 0, , Hence, the stress increases by a factor, of 9., FL, FL, 2 ∆L =, =, Y A Yπ R2, As F , L and ∆L are the same, hence, YR 2 = constant, ∴ 2.0 × 1011 R S2 = 1.0 × 1011 R B2, R, RS = B, ⇒, 2, , 3 ∆l = FL =, AY, , ∴, ∴, , ∆l ∝, , AY, , Now,, , FL, ( πr 2 )Y, L, , ⇒, , F2 = 3F1 ×, , l, d2, , Fl, πd 2 , , Y, 4 , , or (∆l ) ∝, , ∆ ls =, , d, , 2, , is maximum in option (a)., , T1 l 2 − T1 l = T2 l 1 − T2 l, l =, , YA(∆L )2, 2L, Now, work done,, =, , T2 l 1 − T1 l 2, T2 − T1, , A1 l1 = A2 l2, A × l1, l, A l, l2 = 1 1 =, = 1, A2, 3A, 3, , W′ =, , 2, , 2b 2c, , − dp, (dV /V ), , Substituting the values, we have, (1.165 − 1.01) × 105, B =, = 1.55 × 105 Pa, (10 / 100), , 9 Q Bulk modulus,, , ∆p, Volumetric stress, =, ∆V, Volumetric strain, V, mg, K =, 3∆r , a, , r , K =, , ⇒, , Q V = 4 πr 3 , so ∆V = 3∆r , , 3, V, r , mg, ∆r, ⇒, =, r, 3aK, , 10 If Hooke’s law is being obeyed, then, force extension graph is a straight line., Stored elastic energy extension, U = 1 Fx = 1 YA x2 should be a, , , , , 2, 2 L, parabolic curve symmetric about the U, axis. Hence, in the graph P represents, extension and Q the stored elastic, energy., , Y (4 A )(∆L )2, 2(L / 2), , YA (∆L )2 , = 8, , 2L , , 3 MgL s, F × Ls, =, Ys × A s, Ys × πr s2, , 8 From the definition of bulk modulus,, B =, , 1, ∆L , × Y × , × AL, L , 2, , ⇒ 2=, , l1, = 3F1 × 3 = 9F, l2, , 3 L s Yb rb , ×, ×, × , 2 Lb Ys r s , 3a, , 1, Y × (strain)2 × volume, 2, 2, , …(ii), , [Q Mass (steel) = 2M + M = 3M ], For brass wire,, 2MgLb, F × lb, =, ∆ lb =, Yb × Ab, Yb × πrb2, , =, l, , =, , …(i), , 7 For steel wire,, , 2, , 6 As volume is same., ∴, , ⇒, , =, , ∆l 1 = l 1 − l, Similarly, the change in length of the, second wire is, ∆l 2 = l 2 − l, T, l, T, l, Now,, Y = 1 ×, = 2 ×, A, ∆l 1, A, ∆l 2, T1, T, ⇒, = 2, ∆l 1, ∆l 2, T1, T2, =, ⇒, l1 − l, l2 − l, ⇒, , ∴, , ∆x1 = ∆x2, F2, F, l2 = 1 l1, 3 Aγ, Aγ, , Here,, , 11 As work done, , [Q Mass (brass) = 2M ], ∆ ls, 3Mg L s, Y × πrb2, =, × b, ⇒, 2, ∆ lb, 2MgLb, Ys × π r s, , 5 Let l = original length of the wire,, , ⇒, , and, , r, L /R 2, ∆l 1, =2, =, ∆l 2 2L /(2R )2, , 4 ∆l = Fl =, , l1, =3, l2, F, ∆x1 = 1 × l1, Aγ, F, ∆x2 = 2 l2, 3 Aγ, , ⇒, , Weight, 1 Q Stress =, Area, Volume will become (93 ) times., , = 8 × 2 = 16 J, , 12 Thermal stress σ = Y α ∆θ, σ1 = σ2, , Given,, ∴, , Y1α1 ∆θ = Y2α2 ∆θ, Y1 α2 3, =, =, Y2 α1 2, , or, , 13 If the deformation is small, then the, stress in a body is directly proportional, to the corresponding strain., According to Hooke’s law, i.e., Young’s modulus, Tensile stress, (Y ) =, Tensile strain, F/A, So,, Y =, ∆ L/L, FL, =, A∆L, If the rod is compressed, then, compressive stress and strain appear., Their ratio Y is same as that for tensile, case., Given, length of a steel wire (L), = 10 cm, Temperature (θ ) = 100° C, As length is constant., ∆L, Strain =, = α ∆θ, ∴, L, Now, pressure = stress = Y × strain, [Given, Y = 2 × 1011 N / m2 and, α = 1.1 × 10−5 K −1 ], = 2 × 10 × 1.1 × 10, = 2.2 × 108 Pa, 11, , −5, , × 100, , 14 Change in length ∆L = αL ∆T = FL, ⇒, , AY, F, Stress =, = Yα∆T = Y α t, A, (as, ∆T = t )
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PROPERTIES OF MATTER, , DAY FOURTEEN, 15 From free body diagram of the wooden, block,, Vσg = mg + T, [ V is the volume of block], , ⇒, , D1 , , , D2 , , 4, , =1+, , =, , T, vsg, , s, , mg, , T = Vσg − mg, T =, , m, σ − ρ, σg − mg = mg , , ρ , ρ, , 16 Because ρoil < ρ < ρwater, So, it will sink through oil but will not, sink in water. So, it will stay at the, oil-water interface., , 17 Let R be the radius of the bigger drop,, then volume of bigger drop, = 2 × volume of small drop, 4, 4, πR3 = 2 × πr 3 ⇒ R = 21 /3 r, 3, 3, Surface energy of bigger drop,, E = 4 πR2T = 4 × 22 /3 πr 2T, = 28/3 πr 2T, , 18 We have,, 4 3, 4, πr gρ − πr 3 σg, 3, 3, where, ρ → ρwater and σ → ρair, 2gr 2 (ρ − σ ), ⇒ η=, 9v, 2 × 9.81 × (02, . × 10−2 )2 × 999, =, 9 × 87, ., 6 πηrv =, , r, r h + ρg, , , 3, 25 As, T =, 2 cos θ, , 2gh, v 12, D1, , D2 =, , ⇒, , , 2gh , 1 + 2 , v1 , , , 0.5, 0.5 × 10−2 3 +, × 10−2, , 3 , , 1/4, , 8 × 10−3, 1/4, , = 3.6 × 10−3 m, 2, 21 Velocity head = v, 2g, , p, ρg, , As velocity of water is equal to the, pressure head of 40 cm of Hg column,, hence, hρ g, v2, =, 2g, ρg, ⇒, ⇒, , v 2 = 2 hg, v = 2 gh, =, , 22 As shown in the figure, in the two arms, of the tube the pressure remains the, same on the surface PP ′., Hence,, 8 × ρ Y × g + 2 × ρ Hg × g = 10 × ρx × g, 8ρ Y + 2 × 13.6 = 10 × 3.36, , Y, 8 cm, 2 cm, P, , X, 10 cm, P¢, , = 1 × 10−3 poise, ρ − ρ0, η, η, v 2 ρ − ρ02, ∴, =, × 1, v 1 ρ − ρ01, η2, , Hg, , 19 v ∝, , 7.8 − 1.2 8.5 × 10−4 × 10, ×, 7.8 − 1, 13.2, = 6.25 × 10−4 cms −1, , v2 =, , 20 From Bernoulli’s theorem,, , 1, ρgh = ρ (v 22 − v 12 ), 2, 2, , 1 2 v2 , ⇒ gh = v 1 − 1, 2, , v 1 , 2, , A , 1, = v 12 1 − 1, 2, , A2 , (Q A1 v 1 = A2 v 2 ), 2, , ⇒, , A1 , 2hg, , =1+ 2, A2 , v1, , or, , ρY =, , 33.6 − 27.2, = 0. 8 g / cc, 8, , 23 At equilibrium, weight of the given, block is balanced by force due to, surface tension, i.e. 2L S = w, w, or S =, 2L, 1.5 × 10−2 N, =, 2 × 0.3 m, = 0.025 Nm −1, , 24 Work done = Change in surface, ⇒, , energy, 2, W = 2T × 4 π (R22 − R1 ), = 2 × 0.03, × 4 π [(5)2 − (3)2 ] × 10−4, W = 0.4 π mJ, , = 077, . N/m, , 26 (A) When a spinning ball is thrown it, deviates from its ususal path in flight., This is due to magnus effect., (B) Viscous forces tends to reduce the, speed of flowing fluid by virtue of, internal frictional force. Hence, the, energy of liquid flow reduced., (C) Pascal’s law states that if gravity, effect is neglected, the pressure of every, point of liquid in equilibrium of rest is, the same., (D) Gas burners are based on Bernoulli’s, principle., , 27 The shape of the aeroplane wings is, , 2 × 9.8 × 0.4, , = 2.8 ms −1, , ∴, , × 103 × 9.81, 2, , =, , , 2 × 10 × 02, . , 1 +, , , , (0.4)2, , and pressure head =, , 163, , peculier i.e. its upper surface is more, curved than its lower surface. Due to, this, when the aeroplane runs on runway, the speed of air above the wings is larger, than the speed below the wings. Thus,, according to Bernoulli’s theorem, the, pressure above wings becomes lesser, than the pressure below the wings. Due, to this difference of pressure a vertical, lift acts on aeroplane., , 28 The height of column is given by ascent, formula,, h=, , 2S cos θ r, −, r ρg, 3, , If the tube is very narrow, r/3 can be, neglected in comparison with h., 2S cos θ, ., Hence, h =, r ρg, Thus, as the value of r (radius of tube), decreases, the height increases., Q h ∝ 1 , , , , r, , 29 A bridge during its use undergoes, alternating stress and strain for a large, number of times each day, depending on, movement on it. With time it loses its, elastic strength and the amount of strain, in the bridge for a given stress becomes, large and ultimately it may collapse., , 30 In case of small drop of mercury the, force of gravity is small and the surface, tension plays a vital role. Therefore,, surface tends to have minimum surface, area and sphere has minimum area.
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164, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , In case of large mercury drop, the, gravitational pull becomes more, effective than the surface tension and, exerts downwards pull on the drop., Hence, the large drop of mercury, becomes elliptical or oval., Also, surface tension of a liquid, decreases with rise of temperature,, S t = S o (1 − αt ) ., , ∴, , T M = 2π, , T A and T B , respectively,, T A + TB = W, and, T A × x = T B (L − x ), Solving the above equations,, , TB, , TA, , MgL, AY, g, , 2, , ⇒, , T M = 1 + Mg, , , T , AY, 2, , or, , Mg T M , =, −1, T , AY, 2, , 1, A T M , =, , − 1, Y, Mg T , , , or, , SESSION 2, 1 Let tension in wire A and B be, , L +, , 3 If L is the initial length, then the, increase in length by a tension, FL, F is given by l =, πr 2 Y, 4L, Hence, a = L + l = L +, = L + 4C, πr 2 Y, 5L, and b = L +, = L + 5C, πr 2 Y, L, where, C =, πr 2 Y, Thus, on solving for L and C, we get, L = 5a − 4b and C = b − a, , x, , Hence, for F = 9 N, we get, W (L − x ), ,, L, Wx, TB =, L, TA, Stress in A =, , where A A is, AA, cross-section area of wire A., T, Stress in B = B , where A B is, AB, TA =, , cross-section area of wire B., It is given,, A, AA = B ,, 2, TA, TB, =, AA, AB, 2L, which gives x =, 3, , 2 We know that time period,, T = 2π, , L, g, , When additional mass M is added to its, bob, L + ∆L, ,, T M = 2π, g, where, ∆L is increase in length., We know that Young modulus of the, material,, Mg / A MgL, Y =, =, ∆L / L, A∆L, ⇒, , ∆L =, , MgL, AY, , x=L +, , 9L, πr 2 Y, , = L + 9C, = (5a − 4b ) + 9 (b − a), = 5b − 4 a, , 4 Consider the FBD diagram of the, rectangular frame, T sin q Tsin q, q, T cos q, m, , q, T cos q, , sin θ max = 1 ⇒ θ = 90°, Matches with option (b)., Hence, tension is least for the case (b)., Note We should be careful when, measuring the angle, it must be in the, direction as given in the diagram., , 5 Bulk modulus of rubber (K ) = 9.8 × 108, N/m2, Density of sea water ( ρ) = 103 kg/m3, Percentage decrease in volume,, ∆V × 100 = 01, ., , , V, , ∆V, 01, ., =, V, 100, ∆V, 1, or, =, V, 1000, Let the rubber ball be taken up to depth h., ∴ Change in pressure ( p ) = hρ g, ∴ Bulk modulus, p, (K ) =, ( ∆V / V ), hρ g, =, ( ∆V / V ), K × ( ∆V / V ), or, h=, ρg, 1, 9.8 × 108 ×, 1000, =, = 100 m, 103 × 9.8, or, , 6 As the tanker starts accelerating free, surface of the tanker will not be, horizontal because pseudo force acts., Consider the diagram where a tanker is, accelerating with acceleration a., dma cos q, dm, , q, , dmg sin q, , dma 90-q, q, mg, , dmg, , Balancing vertical forces, 2T sinθ − mg = 0, [T is tension in the string], …(i), ⇒, 2T sinθ = mg, Total horizontal force, = T cos θ − T cos θ = 0, Now from Eq. (i), we get, mg, T =, 2sinθ, As, mg is constant, mg, 1, ⇒ T max =, ⇒T ∝, sinθ, 2sin θ min, sin θ min = 0 ⇒ θ min = 0, No option matches with θ = 0°, mg, T min =, 2sin θ max, (since, sin θ max = 1), , a, , Consider an elementary particle of the, fluid of mass dm, The acting forces on the particle with, respect to the tanker are shown above., Now, balancing forces (as the particles is, in equilibrium) along the inclined, direction, component of weight =, component of pseudo force, i.e., dmg sinθ = dma cos θ (we have assumed, that the surface is inclined at an angle θ), where, dma is pseudo force, ⇒, g sin θ = acos θ ⇒ a = g tanθ, a, tanθ = = slope, ⇒, g
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PROPERTIES OF MATTER, , DAY FOURTEEN, 7 Equating pressure at A, we get, R sin α d 2 + R cos α d 2 + R(1 − cos α ) d 1, = R(1 − sin α ) d 1, , p 2 = patm − ρg (54 − x ) = ρg (22 + x ), V2 = A ⋅ x, ρg 76 × 8 A = ρg (22 + x )A x, x2 + 22 x − 76 × 8 = 0 ⇒ x = 16 cm, , 12 Elongation due to change in, temperature,, , 10 Given, surface tension of water, R, , R, , R sin a, , a, a R, , 90 –a, , d2, , d1, A, (sin α + cos α ) d 2 = d 1 (cos α − sin α ), , d 1 1 + tan α, =, d2, 1 − tan α, , ⇒, , 8 The bubble will detach if, Buoyant force, ≥ Surface tension force, , r, (q), ò sin q T × dl = T(2pr) sin q, , 4 3, πR ρ w g ≥ ∫ T × dl sin θ, 3, 4, ( ρw ) πR3 g ≥ (T ) (2 πr )sin θ, 3, , r, As,, sinθ =, R, Solving, r =, , Angle of contact θ = 0°, Diameter of one side, 2r1 = 3.0 mm, ∴ r1 = 1.5mm = 1.5 × 10−3 m, Diameter of other side, 2r2 = 6.0 mm, r2 = 3.0 mm, = 3.0 × 10−3 m, , ∴ Difference in levels of water rises in, both tubes,, ∆h = h1 − h2, 2S cos θ 1, 1, =, − , ρg, r2 , r1, =, , accelerating horizontally i.e. no, component of acceleration in vertical, direction. Hence, the pressure in the, vertical direction will remain, unaffected., p1 = p 0 + ρgh, , Again, we have to use the concept that, the pressure in the same level will be, same., x, , 54, 54–x, , 1.0 × 103 × 9.8, , T, , 13 Frequency, f = v = 1, 2l, , and ρ0 respectively and let x be the, length of the rod immersed in water at, an instant of time t. Then, acceleration, at that instant =apparent weight, divided by the mass of the rod, i.e., πr 2 lρg − πr 2 xρ0g, dv, =, dt, πr 2 lρ, gxρ0, x, = g 1 −, , , lρ, σl , , or, , dv dx, x, = g 1 −, , , dx dt, σl , , or, , dv, x, v, = g 1 −, , , dx, σl , v2, x2, =gx−, 2, 2σl, , p1 = patm = ρg 76, V1 = A ⋅ 8, [where, A = area of cross-section], , ⇒, , 0, , T, Ad, , Putting these values we, have, f =, , 1 2.2 × 1011 × 0.01, 2l, 7.7 × 103, , 2 103, ×, 7, 3, f ≈ 178.2 Hz, Hence, option (b) is true., F, 14 As, Y = Stress =, Strain, πr 2 × strain, f =, , ⇒ F = T = Yπr 2 × strain, Now, strain, =, , l, , T, 1, =, µ, 2l, , where, T is tension in the wire and µ is, the mass per unit length of wire., Also, Young’s modulus,, Tl, Y =, A∆l, T, Y∆l, =, ⇒, A, l, Putting this value in expression of, frequency, we have, 1 y∆ l, f =, 2l, ld, ∆l, Given, l = 1.5m,, = 0.01, l, d = 7.7 × 103 kg / m3 ,, , L2 + l 2, L, , On integrating, we get, For air trapped in tube,, p1 V1 = p 2V2, , 2l, , Y = 2.2 × 1011 N / m2, , 11 Let the densities of metal and water be ρ, , =g −, , T, , ∆l = Lα∆T, which is compensated by elastic strain,, when temperature becomes normal, i.e.,, TL, ∆l =, YS, TL, Thus,, = Lα∆T, YS, ⇒, T = YSα∆T, At equilibrium, force exerted by one half, on other,, F = 2T = 2YSα∆T, , = 4.9 mm ≈ 5mm, , p2, , 8 cm, , 2 × 73, . × 10−2 × cos 0°, , , , 1, 1, −, , −3, −3 , 1, ., 5, ×, 10, 3, ., 0, ×, 10, , , 2 − 1, 14.6, =, × 10−2 , 9.8, 3 , , 2ρw R 4 g, 2ρw g, = R2, 3T, 3T, , 9 In this question, the system is, , i.e., , Density of water ( ρ) = 1.0 × 103 kg/m3, , Height of water column rises in first and, second tubes, 2S cos θ, h1 =, r1 ρg, 2S cos θ, h2 =, r2 ρg, , qR, , F, , (S ) = 73, . × 10−2 N/m, , Acceleration due to gravity,, g = 9.8 m/s2, , 165, , l , = g l −, , , 2σ , , 1 , v = 2gl 1 −, , , 2σ , , ≈ 1+, , , l2 , − L = 1 + 2 , L , , , 1 /2, , −1, , 1 l2, l2, −1= 2, 2 L2, 2L, (by binomial expansion), , ∴T =, , Yπr 2 ⋅ l 2, 2L2
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166, , DAY FOURTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 15 In equilibrium,, upward force = downward force, kx 0 + F B = mg, kx0, FB, , 17 Here, Yc = 1 × 1011 N/m2, Ys = 2 × 10, , 11, , Mg, , 16 Given, Y = 2 × 1011 N/m2, Stress = 5 × 107 N / m2, stress, As,, =Y, strain, 5 × 107, = 2.5 × 10−4, ⇒ Strain =, 2 × 1011, It is symmetric strain., Now, strain of 2.5 × 10−4 is equivalent., As,, ∴, , ∆V, ∆r, = 3 , r , V, , mg × xdx, YAl, mgl, ., =, 2YA, , lc = 1.0 m, l2 = 0.5m, and ∆lc = 1 × 10−3 m, stress, As, (strain)c =, Yc, stress, −3, ⇒ 1 × 10 =, 1 × 1011, ⇒, , Here, kx 0 is restoring force of spring and, F B is buoyancy force., L, kx 0 + σ Ag = Mg, 2, σLAg, Mg −, 2 = Mg 1 − σLA , x0 =, , , k 2M , k, , N/m, , Total elongation,, ∆l = ∫ ∆(dx ), , 2, , =, , 19 The density of concrete of course, is, , stress = 10 N/m, stress, Now,, Ys =, strain, 108, strain =, ⇒, = 0.5 × 10−3, 2 × 1011, ∆ls, or, = 0.5 × 10−3, 1/2, ⇒, ∴, , 8, , 2, , ∆ls = 0.25 × 10−3, ∆l = ∆lc + ∆ls = 1 + 0.25, = 1.25 mm, , 18 Consider an element of wire of width dx, at a distance x from bottom end of wire., The force experienced by this element, is due to the gravitational force of, portion of wire lower to it., T /A, , where ∆(dx ) is the, So, Y =, ∆(dx ), dx, elongation in this element., mg, T, Now, ∆ (dx ) =, dx =, × xdx, YA, YAl, , 2.5 × 10−4, 3, ∆r, =, = 0.83 × 10−4, r, , ∴ Fraction decrease in radius, , dx, , = (1.00 − 0.83) 10−4, = 0.17 × 10−4, , x, , l, , ∫0, , dx, , more than that of water and a block of, concrete will sink like a stone, if dropped, into water. Concrete cargo were filled, with air and as such, average density of, cargo vessels, =, , [Mass of concrete, + Mass of air], [Volume of concrete, + Volume of air], , It follows that the average density of, cargo vessels must be less than that of, water. As a result, the concrete cargo, vessels did not sink., , 20 From the continuity equation,, Av = constant, where, Av is the volume of liquid, flowing per second., 1, or, A∝, v, As the stream falls, its speed v increases, and hence its area of cross-section a will, decrease. That is why the stream will, become narrow., When the stream will go up, its speed, decreases, hence its area of cross-section, will increase and it will become broader, and spreads out like a fountain. Hence,, option (a) is true.
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DAY FIFTEEN, , Heat and, Thermodynamics, Learning & Revision for the Day, u, , u, , u, , u, , u, , u, , u, , Heat, Thermometry, Thermal Expansion, Specific Heat Capacity, Calorimetry, Change of State, Heating Curve, , u, , u, , u, , u, , u, , Zeroth Law of, Thermodynamics, First Law of Thermodynamics, Thermodynamic Processes, Second Law of, Thermodynamics, Reversible and Irreversible, Processes, , u, , u, , u, , u, , u, , u, , u, , u, , Carnot Engine and its Efficiency, Refrigerator, Equation of State of Perfect Gas, Kinetic Theory of Gases, Degree of Freedom (f ), Specific Heat Capacities of Gases, Mean Free Path, Avogadro’s Number, , Heat, Heat is a form of energy which characterises the thermal state of matter. It is transferred, from one body to the other due to temperature difference between them., Heat is a scalar quantity with dimensions [ML2 T−2 ] and its SI unit is joule (J) while, practical unit is calorie (cal); 1 cal = 4.18 J, If mechanical energy (work) is converted into heat then, the ratio of work done (W ) to, heat produced (Q) always remains the same and constant., W, i.e., = constant = J or W = JQ, Q, The constant J is called mechanical equivalent of heat., , Temperature, The factor that determines the flow of heat from one body to another when they are in, contact with each other, is called temperature. Its SI unit is kelvin., , Thermometer, An instrument used to measure the temperature of a body is called a thermometer. For, construction of thermometer, two fixed reference point ice point and steam point are, taken. Some common types of thermometers are as follows:, 1. Liquid (mercury) thermometer Range of temperature: −50°C to 350°C, 2. Gas thermometer (Nitrogen gas) Range of temperature: −200°C to 1600°C, 3. Pyrometers Range of temperature: −800°C to 6000°C, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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168, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Scales of Temperature, Three most common scales are Celsius scale or Centigrade, scale, Fahrenheit scale and Kelvin scale (Absolute scale)., Scale, , Coefficient of superficial expansion,, ∆A, β=, A 0 × ∆T, , Ice point / lower, reference point, , Steam point / Upper, reference point, , Unit, , 0, , 100, , °C, , Final area,, , where, A0 is the area of the body at temperature T., , Celsius, Fahrenheit, Kelvin, , 2. Superficial Expansion or Areal Expansion Expansion of, solids along two dimensions of solid objects is defined as, superficial expansion., , 32, , 212, , °F, , 273.15, , 373.15, , K, , Relation between C, F and K scales is, C F − 32 K − 273.15, =, =, 5, 9, 5, Temperature of X − Ice point of X, In general,, Steam point of X − Ice point of X, =, , Temperature of Y − Ice point of Y, Steam point of Y − Ice point of Y, , A f = A0 (1 + β∆T ), , 3. Volume or Cubical Expansion Expansion of solids along, three dimensions of solids objects is defined as cubical, expansion., Coefficient of volume or cubical expansion,, ∆V, γ=, V0 × ∆T, Final volume, V = V0 (1 + γ∆ T )., where, V0 is the volume of the body at temperature T., NOTE, , • The coefficients α , β and γ for a solid are related to each, other., , Thermometry, The branch dealing with measurement of temperature is, called thermometry., Let thermometric properties at temperatures 0°C (ice point),, 100°C (steam point) and t °C (unknown temperature) are X 0 ,, X 100 , and X t , respectively. Then,, X t − X 0 X 100 − X 0, Xt − X0, t, or, =, =, t, 100, X 100 − X 0 100, Thus,, , X − X0 , t = t, × 100 ° C, X 100 − X 0 , , Thermal Expansion, Almost all substances (solid, liquid and gas) expand on, heating and contract on cooling. The expansion of a substance, on heating is called thermal expansion of substance., , Thermal Expansion of Solids, Thermal expansion in solids is of three types:, 1. Linear Expansion Thermal expansion along a single, dimension of a solid body is defined as the linear, expansion., If a rod is having length l0 at temperature T, then, elongation in length of rod due to rise in temperature by, ∆T is,, ∆l, ∆l = l0 α ∆T or α =, l0 × ∆ T, , β γ, =, 2 3, • As temperature increases, density decreases according to, relation,, ρ0, ρ=, 1 + γ∆T, α=, , or, , ρ = ρ0 ( 1 − γ ∆T ), , [valid for small ∆T], , Thermal Expansion of Liquid, Liquids do not have linear and superficial expansion but these, only have volume expansion., Liquids have two coefficients of volume expansion, 1. Coefficient of apparent expansion,, Apparent expansion in volume (∆V )a, γa =, =, Initial volume × ∆T, V × ∆T, 2. Coefficient of real expansion,, Real expansion in volume, (∆V )r, γr =, =, Initial volume × ∆T, V × ∆T, , Anomalous/Exceptional Behaviour of Water, Generally, density of liquids decreases with increase in, temperature but for water as the temperature increases from 0, to 4°C, its density increases and as temperature increases, beyond 4°C, the density decreases., The variation in the density of water with temperature is, shown in the figure given below., Density, , where, α is the coefficient of linear expansion whose, value depends on the nature of the material., Final length, l f = l0 + l0α ∆T, = l0 (1 + α ∆T ), , 4°C, , Temperature, , Anomalous behaviour of water
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, , 169, , Thermal Expansion of Gases, , Calorimetry, , Gases have no definite shape, therefore, gases have only, volume expansion., , Calorimetry means measurement of heat. When a body at, higher temperature is brought in contact with another body at, lower temperature, the heat lost by the hot body is equal to the, heat gained by the colder body and provided no heat is, allowed to escape to the surrounding., A device in which heat measurement can be made is called a, calorimeter., If temperature changes, heat exchanged is given by, Q = ms∆T, As temperature of the body increases, it means heat is taken, by the body, otherwise given out by the body., , 1. The coefficient of volume expansion at constant pressure,, ∆V, 1, α=, ×, V0 ∆T, Final volume, V ′ = V (1 + α ∆T ), 2. The coefficient of pressure expansion at constant volume,, ∆p 1, β=, ×, p ∆T, Final pressure, p′ = p (1 + β ∆T )., , Specific Heat Capacity, , Change of State, , The quantity of heat required to raise the temperature of unit, mass of a substance by 1°C is called specific heat., Q, Specific heat, s =, m × ∆T, The SI unit of specific heat is J kg −1 k −1 ., , When we supply heat (energy) to a body and its temperature, does not change, then the energy consumed by the body is, used up in changing its phase and the process is called change, of state., , l, , l, , Specific heat capacity can have any value from 0 to ∞. For, some substances under particular situations, it can have, negative values also., The product of mass of the body and specific heat is, termed as heat capacity, C = m × s., , Molar Heat Capacity, The amount of heat required to change the temperature of a, unit mole of substance by 1°C is termed as its molar heat, capacity,, Q, Cm =, µ∆T, Generally, for gases, two molar heat capacities are very, common—molar heat capacity at constant pressure (C p) and, molar heat capacity at constant volume (CV )., , Latent Heat, In case of phase change, heat is consumed during melting and, boiling while released during freezing and condensation., The heat required to change the phase of a system, is proportional to the mass of the system i.e., Q∝m, Q = mL, where, L is the latent heat, which is defined as the amount of, heat required to change the phase of the unit mass of a, substance at given temperature., l, , l, , l, , l, , where, mw = water equivalent of substance whose mass is m,, S = specific heat capacity of substance, and, , S w = specific heat capacity of water, , In case of water, the latent heat of vapourisation is, 536 cal/gm., , Sublimation, , Water Equivalent of a Substance, Water equivalent of certain amount of substance is defined as, the amount of water, which when replaced by the substance, requires the same amount of heat for the same rise in, temperature., mS, ,, mw =, Sw, , In case of ice, the latent heat of fusion of ice is 80 cal/gm., , l, , A substance can sometimes change directly from solid to, gaseous phase, this process is termed as sublimation., Corresponding latent heat is termed as latent heat of, sublimation Ls . The reverse process can also occur., Very pure water can be cooled several degrees below the, freezing temperature without freezing, the resulting, unstable state is described as supercooled. When, this supercooled water is disturbed (either by, dropping dust particles etc.), it crystallises within a second, or less., A liquid can sometimes be superheated above its normal, boiling temperature. Any small disturbance such as, agitation causes local boiling with bubble formation.
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170, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Heating Curve, If we supply energy to a body in solid state, (temperature<melting point) at a constant rate, then the curve, drawn between temperature and time is termed as the heating, curve., Temperature, E, C, , BP, MP, O, , A, , B, , t1, , t2 t3, , D, , t4, , The internal energy of an ideal gas is totally kinetic and it is, given by, 3, U = µRT, 2, and change in internal energy, 3, ∆U = µR∆T ., 2, For non-ideal gases, internal energy depends not only on the, temperature but also on the pressure., , Work, , Time, , Heating curve, OA represents heating of the solid,, 1, Ssolid ∝, Slope of OA, AB represents melting of the solid,, , Consider a system in a cylinder with movable piston, whose, volume can be changed (a gas, liquid or solid). Suppose, the, cylinder has a cross-sectional area A and pressure exerted by, system on the piston face is p. The work done by the system, on the surroundings for small displacement dx is dW = pAdx., W = ∫ dW = ∫, , l, , l, , l, , CD represents boiling (vaporisation) of the liquid, length of, CD ∝ LV , DE represents heating of the gaseous phase,, 1, S gas ∝, Slope of DE, , pdV, , i.e. work done in a finite change of volume from Vi to V f ., , length of AB ∝ L f, BC represents heating of the liquid,, 1, S liquid ∝, Slope of BC, , Vf, Vi, , Work done by the system depends on the initial and final, states., If volume of the system increases, then work is done by the, system and it is taken as positive work done., If volume of the system decreases, then work is done on, the system and it is taken as negative work done., , First Law of Thermodynamics, , Zeroth Law of Thermodynamics, , According to this law, the heat given to a system (∆Q) is equal, to the sum of increase in its internal energy (∆U) and the work, done (∆W ) by the system against the surroundings., Mathematically, ∆Q = ∆U + ∆W, , When there is no exchange of heat between two objects placed, in contact, then both are called in thermal equilibrium., , Sign Convention, , According to this law, if two systems A and B , separated by, an adiabatic wall, are separately and independently in thermal, equilibrium with a third system C, then the systems A and B, are also in a state of thermal equilibrium with each other., System, C, C, System, A, A, , ∆Q = + ve when heat supplied, = − ve when heat is ejected, ∆U = + ve when temperature increases, = − ve when temperature decreases, ∆W = + ve when work is done by the system (expansion), , System, B, B, Adiabatic wall, , Three system of thermal equilibrium, , Basic Terms Used Thermodynamics, Internal Energy, Internal energy of a system is defined as the sum of the total, kinetic energy of all its constituent particles and sum of all, the potential energies of interaction among these particles., , = − ve when work is done on system (compression), First law of thermodynamics is based on the energy, conservation., , Thermodynamic Processes, A thermodynamic process is the process of change of state of a, system involving change of thermodynamic variables, e.g., p, V , T etc. When a system undergo a thermodynamic change,, then work done either by system on surrounding or by, surroundings on system is called external work., Wext = ∫, , V2, V1, , p dV = area under p-V curve.
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, 1. Isothermal Process, , l, , Molar specific heat of a gas under adiabatic condition, , It is that process in which temperature remains constant., Here, exchange of heat with the surroundings is allowed., , C=, l, , T1>T2, , l, , p, T1, T2, V, , dp, p, =− ⋅, dV, V, Work done in an isothermal process, Vf, Vf , ∆W = ∫ pdV = nRT ln , Vi, Vi , , Slope of p-V curve at any point is, , where, n = number of moles, R = gas constant, and T = temperature., V f and Vi are final and initial volume of the gas respectively., As per first law of thermodynamics, since, ∆T = 0 , hence,, ∆U = 0 for an ideal gas and we have ∆Q = ∆W ., Thus, heat supplied to the system in an isothermal process is, entirely used to do work against external surroundings., , l, , T V γ − 1 = constant, , or, , T γ p1 − γ = constant, , Indicator diagram for an isochoric process is a straight line, parallel to p-axis., , p, , p, V, (a), , V, (b), , Graph (a) shows isometric heating graph in which pressure, increases, temperature increases, ∆Q is positive and ∆U is, positive., Similarly, Graph (b) shows isometric cooling graph in, which pressure decreases, temperature decreases, ∆Q is, negative and ∆U is negative., l, , or, , Free expansion is an adiabatic process in which ∆W = 0., Hence, in accordance with first law of thermodynamics, ∆U = 0 i.e. the final and initial values of the internal energy, are equal in free expansion., , It is that thermodynamic process in which volume remains, constant., In an isochoric process for a given mass of gas, p, p ∝ T or, = constant, T, , Q1 > Q2, , The equation of state for an adiabatic process is, pV γ = constant, , Work done in an adiabatic process, Vf, µR, ∆W = ∫ p dV =, (Ti − T f ), Vi, (γ − 1), , 3. Isochoric Process, , Q2 Q1, , p-V graph for adiabatic process, , dp, p, =−γ ., dV, V, , ∆U = − ∆W, l, , It is that process in which there is no exchange of heat of the, system with its surroundings. Thus, in an adiabatic process, p, V and T change but ∆Q = 0 or entropy remains constant, ∆Q, , , = 0 ., ∆S =, , , T, , V, , Slope of an adiabatic curve at a point is, , As per first law of thermodynamics, since, ∆Q = 0 in an, adiabatic process hence,, , 2. Adiabatic Process, , p, , ∆Q, 0, =, =0, m ⋅ ∆T, m ⋅ ∆T, , During an adiabatic expansion ∆W = + ve, hence,, temperature of gas falls, i.e. an adiabatic expansion is, always accompanied with cooling., , p-V graph of an isothermal expansion process, As temperature T remains constant in an isothermal process,, hence as per Boyle’s law, 1, p ∝ or pV = constant, V, Molar specific heat of a gas under isothermal condition, ∆Q, ∆Q, C=, =, =∞, m∆T m (0), , 171, , l, , Molar specific heat of a gas under isochoric condition, f, CV = R, where f is the number of degrees of freedom per, 2, molecule., Work done in an isochoric process, ∆W = ∫ p dV = 0, As ∆W = 0 hence, according to first law of thermodynamics,, we have, µR, ∆T, (∆Q)V = ∆U = µCV ∆T =, (γ − 1)
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172, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 4. Isobaric Process, , Cyclic Process, , It is that process in which pressure remains constant., As in an isobaric process for a given mass of gas, V ∝T, V, or, = constant, T, , In cyclic process, if the process takes the path AxB, it returns via, ByA, the initial and final points are same., , l, , p, B, x, y, , Indicator diagram for an isobaric process is a straight, line parallel to X -axis., , A, V, , p-V graph of cyclic process, p, , Carnot Engine and its Efficiency, , p, V, (a), , V, (b), , Graph (a) represent isobaric expansion, graph (b), represent isobaric compression., l, , Work done in an isobaric process, ∆W = ∫ p dV = p ∫, , Vf, Vi, , Molar specific heat of a gas under isobaric condition, f, , C p = + 1 R = CV + R, 2, , , Second Law of, Thermodynamics, Two most common statements of second law of, thermodynamics are given below, , Clausius Statement, It is impossible for a self-acting machine, working in a, cyclic process to transfer heat from a colder body to a, hotter body without the aid of an external agency., , Kelvin-Planck’s Statement, It is impossible to design an engine which extracts heat, from a reservoir and fully converts it into work without, producing any other effect., , Isothermal, expansion, , A, T1, , p, , dV, , = p(V f − Vi ) = p∆V, l, , Carnot engine is a theoretical, ideal heat engine working in a, reversible cyclic process operating between two temperatures T1, (heat source) and T2 (heat sink). The Carnot’s cycle consists of two, isothermal processes connected by two adiabatic processes as, shown in the figure., , Q1, B, , Adiabatic, compression, , D, , Adiabatic, expansion, , T2, , C, , Isothermal Q2, compression, , V, , Various process in Carnot cycle, The efficiency of a Carnot’s cycle is given by, W, Q, T, η=, =1− 2 =1− 2, Q1, Q1, T1, The efficiency does not depend on the nature or quantity of the, working substance., , Refrigerator, A refrigerator or heat pump is basically a heat engine running in, reverse direction. It takes heat from colder body (sink) and after, doing some work gives the rest heat to the hotter body (source)., An ideal refrigerator can be regarded as Carnot’s ideal heat engine, working in the reverse direction., Source, T1, , Q1, , Working, substance, , Q2, , Sink, T2, , W, , Reversible and Irreversible, Processes, A reversible process is one which can be reversed in, such a way that all changes taking place in the direct, process are exactly repeated in inverse order and in, opposite sense, and no changes are left in any of the, bodies taking part in the process or in the surroundings., Any process which is not reversible exactly is an, irreversible process., , Working of refrigerator, , Coefficient of Performance of a, Refrigerator, It is defined as the ratio of quantity of heat removed per cycle (Q2 ) to, the work done (W ) on the working substance per cycle to remove, this heat., Q, Q2, T2, 1−η, or β =, β= 2 =, =, W Q1 − Q2, T1 − T2, η
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, , 173, , Equation of State of a Perfect Gas, , Work Done on Compressing a Gas, , The equation which relates the pressure (P), volume (V ) and, temperature (T ) of the given state of an ideal gas is known as, ideal or perfect gas equation., pV, For 1 mole of gas,, = R (constant), T, pV = RT, where, R is universal gas constant. The SI unit of gas constant, is J/mol-K. Its value is 8.314 J/mol-K or 8.314 × 107 erg/mol-K, or 2 cal/mol-K. The dimensions of R are [ML2 T−1θ −1 ]., Work, Moreover, gas constant R =, Moles × Temperature, , Work done W = p ⋅ ∆V , where p = pressure of the gas and, ∆V = change in volume of the gas., When two ideal gases having molar masses M1 and M2 are, mixed, then thermodynamic variables/parameters for mixture, would be given by, n1 m1 + n2 m2, M (molar mass) =, n1 + n2, , l, , l, , l, , l, , CV (of the mixture) =, , l, , C p (of the mixture) =, , l, , γ (of the mixture) =, , The perfect gas equation for 1 molecule of gas is, pV = kT, Boltzmann’s constant is represented by per mole gas, constant, R, 8.31, i.e., k =, =, = 1.38 × 10 −23 J K −1, N 6.023 × 1023, Its dimensions are [ ML2 T −2θ −1 ]., , Kinetic Theory of Gases, Kinetic theory of gases relates the macroscopic properties of, gases (such as pressure, temperature etc.) to the microscopic, properties of gas molecules (such as speed, momentum,, kinetic energy of molecules etc)., , n1CV1 + n2CV2, n1 + n2, n1C p1 + n2C p2, n1 + n2, , n1C p + n2C p, 1, , 2, , n1CV1 + n2CV2, , or is given by, , n1 + n2, n, n2, = 1 +, γ −1, γ1 − 1 γ2 − 1, where, symbols have their usual meanings., , Kinetic Energy and Temperature, In ideal gases, the point particles can have only translational, motion and thus only translational energy., 1, 3, Translational KE of a molecule = mc2 = kT, 2, 2, 3, Mean KE per molecule = kT, 2, l, , l, , Assumptions of Kinetic, Theory of Gases, l, , l, , l, , l, , l, , l, , l, , Every gas is composed of tiny particles known as, molecules. The size of molecules is much smaller than the, intermolecular spacing., The molecules of a gas are identical, spherical, rigid and, perfectly elastic point masses., Molecules are in a state of random rapid motion. They, collide with each other. There is no loss of energy during, collision. Only the direction of motion is changed., The time spent in collision between two molecules is, negligible in comparison to time between two successive, collisions., The number of collisions per unit volume in a gas remains, constant. No attractive or repulsive force acts between gas, molecules., Gravitational attraction among the molecules is ineffective, due to extremely small masses and very high speed of, molecules., Molecules constantly collide with the walls of container, due to which their momentum changes. The change in, momentum is transferred to the walls of the container., Consequently, on the walls of container pressure is exerted, by gas molecules. The density of gas is constant at all, points of the container., , l, , Mean kinetic energy per gram mole is given by, 3, 3, 1, , KEmole = mc2 N = kTN = RT, 2, , 2, 2, , l, , Average kinetic energy of gas =, , l, , KE per molecule =, , l, , KE per mole =, , l, , 3, pV, 2, , 3 pV 3 RT, =, 2N, 2N, , 3, kT, 2, 3, KE per volume = p, 2, , Concept of Pressure, Pressure p exerted by a perfect gas on the walls of container is, given by, 1 mN c2 1 M 2, p=, =, c, 3 V, 3 V, 1, 1 2, 2 1, 2, pV = (ρc2 ) = ρc2 = ρc2 = E, 3 2, 3, 3, 3 2, Here, m = mass of each molecule, c = root mean square, velocity of molecules, ρ = density of gas, M = mass of gas, enclosed in volume V of container, and E = Total KE of the, ideal gas.
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174, , Various Speeds of Gas Molecules, l, , Root mean square speed It is defined as the square root, of mean of squares of the speed of different molecules, i.e., , vrms =, , v12 + v22 + v32 + K, = v2, N, , 1 2, From the expression of pressure, p = ρvrms, 3, vrms =, l, , 3p, =, ρ, , 3 pV, 3 RT, 3 kT, =, =, Mass of gas, M, m, , Most probable speed It is defined as the speed which is, possessed by maximum fraction of total number of, molecules of the gas., vmp =, , l, , 2p, 2 RT, 2 kT, =, =, ρ, M, m, , Average speed It is the arithmetic mean of the speeds of, molecules in a gas at given temperature., v + v2 + v3 + v4 + K, vav = 1, N, Average speed, vav =, , 8p, =, πρ, , 8 RT, 8 kT, =, π M, πm, , Degree of Freedom (f ), The term degree of freedom of a system refers to the possible, independent motions a system can have, l, , for monoatomic gas, ( f ) = 3, , l, , for diatomic gas, ( f ) = 5, , l, , for triatomic gas, ( f ) = 6(non-linear), , l, , for triatomic (linear) gas, ( f ) = 7, , l, , for N-atomic molecule ( f ) = 6 N – 3, , l, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , for N-atomic linear molecule ( f ) = 6 N – 5, , Law of Equipartition of Energy, According to law of equipartition of energy for any system in, thermal equilibrium, the total energy is equally distributed, among its various degree of freedom and each degree of, 1, freedom is associated with energy kT, 2, (where, k = 1.38 × 10 −23 J/K and T = absolute temperature of, the system)., , Specific Heat Capacities of Gases, The specific heat of gas can have many values, but out of them, following two values are important, , Specific Heat at Constant Volume, The specific heat of a gas at constant volume is defined as the, quantity of heat required to raise the temperature of unit mass, of gas through 1° C or 1 K when its volume is kept constant i.e., (∆Q)V, cV =, m∆T, , For one mole of gas,, CV = Mc V =, , M (∆Q)V 1 (∆Q)V, =, m∆T, n ∆T, , m, , Q n = M , , Specific Heat at Constant Pressure, The specific heat of a gas at constant pressure is defined as the, quantity of heat required to raise the temperature of unit mass, of gas through 1K, when its pressure is kept constant i.e., (∆Q) p, ., cp =, m∆T, For one mole of gas,, M (∆Q) p, …(i), ∴, C p = Mc p =, m∆T, (, ∆, Q, ), 1, p, [Q n = m / M ], =, n ∆T, Specific heat of a gas at constant pressure is greater than the, specific heat at constant volume i.e. C p > CV ., C p and CV are related to each other according to relation,, R, …(ii), C p − CV =, J, Eq. (ii) is called Mayer’s relation. If C p and CV are measured in, the units of work and R is also in the units of work (or energy),, then Eq. (ii) becomes C p − CV = R., , Specific Heat in Terms of Degree of, Freedom, For a gas at temperature T, the internal energy U =, , f, nRT ., 2, , f, 1, n R∆T ⇒ CV = f R, 2, 2, f, , Specific heat at constant pressure, C p = + 1 R., 2, , Change in energy, ∆U =, , f, , + 1 R, 2, , 2, Ratio of C p and CV , γ =, =, =1 + ., CV, f, f, R, 2, Cp, , Mean Free Path, The distance travelled by a gas molecule between two, successive collision is known as free path., Total distance covered, Mean free path =, Number of collisions, The mean free path of a gas molecule is the average distance, between two successive collisions. It is represented by λ., 1, λ=, 2 π σ2 n, Here, σ = diameter of the molecule and n = number of, molecules per unit volume., , Avogadro’s Number, According to Avogadro’s hypothesis, gram atomic masses of, all elements contain the same number of atoms and this, number is called Avogadro’s number (N A ) and its value is, 6.02 × 1023 .
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, , 175, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 0°C to 100°C. Its volume changes by (given that, coefficient of linear expansion for aluminium, ª AIEEE 2011, α Al = 23 × 10− 6 /°C), (a) 28.9 cc, (c) 9.28 cc, , T, , T, , 1 An aluminium sphere of 20 cm diameter is heated from, (c), , 2, , (d), , (b) 2.89 cc, (d) 49.8 cc, , constant volume, respectively. It is observed that, Cp − CV = a for hydrogen gas Cp − CV = b for nitrogen, gas. The correct relation between a and b is, ª JEE Main 2017, , 6 An ideal gas undergoes four different processes from, the same initial state. Four processes are adiabatic,, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4, which one is adiabatic., p, , (b) a = 14 b, 1, (d) a =, b, 14, , (c) a = 28 b, , p, , p, , 2 Cp and CV are specific heats at constant pressure and, , (a) a = b, , 4, 3, , 3 A copper ball of mass 100 g is at a temperature T . It is, , (b) 1250°C, , (c) 825°C, , (d) 800°C, , 4 100 g of water is heated from 30°C to 50°C. Ignoring the, slight expansion of the water, the change in its internal, energy is (Take, specific heat of water is 4184 J/kg/K), ª AIEEE 2011, , (a) 8.4 kJ, , (b) 84 kJ, , (c) 2.1 kJ, , (d) 4.2 kJ, , 2, 1, V, , (a) 4, , (b) 3, , (c) 2, , (d) 1, , 7 Equal masses of two liquids A and B contained in vessels, of negligible heat capacity are supplied heat at the same, rate. The temperature-time graphs for the two liquids are, shown in the figure. If S represents specific heat and L, represents latent heat of liquid, then, Y, Temperature, , dropped in a copper calorimeter of mass 100 g, filled, with 170 g of water at room temperature. Subsequently,, the temperature of the system is found to be 75°C. T is, (Take, room temperature = 30°C, specific heat of copper, = 01, . cal/g°C), (a) 885°C, , 2, , 1, , 1, , B, , A, , 5 Consider p-V diagram for an ideal gas shown in figure., O, , p, 1, , (c) SA < SB ; LA < LB, , 2, , (d) SA < SB ; LA > LB, , V, , Out of the following diagrams which represents the T-p, diagram?, T, , 8 p-V plots for two gases during adiabatic processes are, shown in the figure. Plots 1 and 2 should correspond, respectively to, p, , 2, 2, , (a), , X, , (a) SA > SB ; LA < LB, (b) SA > SB ; LA > LB, , p = Constant, V, , T, , Time, , 1, , (b), 2, , 1, , 1, p, , p, , V, , (a) He and O 2 (b) O 2 and He (c) He and Ar (d) O 2 and N2
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176, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 9 Figure shows the variation in temperature ( ∆T ) with the, amount of heat supplied (Q) in an isobaric process, corresponding to a monoatomic (M), diatomic (D) and a, polyatomic (P) gas. The initial state of all the gases are, the same and the scale for the axes coincide, ignoring, vibrational degrees of freedom, the lines a, b and c, respectively, correspond to, a, b, , ∆T, , ª JEE Main (Online) 2013, (b) M, D and P, (d) D, M and P, , 10 A certain amount of gas is taken through a cyclic process, ( A B C D A ) that has two isobars, one isochoric and one, isothermal. The cycle can be represented on a p -V, indicator diagram as, ª JEE Main (Online) 2013, B, , C, , (a), , p, , B, , (b), A, , p, , D, , B, , A, , D, V, , C, , (d) p B, A, , D, , (b) 30 R, (d) 20 R, , 15 A Carnot engine takes 3 × 106cal of heat from a reservoir, at 627°C and gives it to a sink at 27°C. The work done by, the engine is, , 1, has efficiency . WhenT2 is lowered by 62 K, its efficiency, 6, 1, increases to . Then,T1 and T2 are respectively, 3, ª AIEEE 2011, (a) 372 K and 330 K, (c) 310 K and 248 K, , (b) 330 K and 268 K, (d) 372 K and 310 K, , 17 A Carnot engine, whose efficiency is 40%, takes in heat, from a source maintained at a temperature of 500 K. It is, desired to have an engine of efficiency 60%. Then, the, intake temperature for the same exhaust (sink), temperature must be, ª AIEEE 2012, (a) efficiency of Carnot engine cannot be made larger, than 50%, (b) 1200 K, (c) 750 K, (d) 600 K, , C, , V, , (c), , (a) 10 R, (c) 40 R, , 16 A Carnot engine operating between temperaturesT1 and T2, , c, , p, , V = kT 2/ 3. Calculate the work done when the temperature, changes by 60 K ?, , (a) 4.2 × 106 J (b) 8.4 × 106 J (c) 16.8 × 106 J (d) 3 × 106 J, , Q, , (a) P, M and D, (c) P, D and M, , 14 A gas expands with temperature according to the relation, , 18 The temperature-entropy diagram of a reversible engine, , C, , is given in the figure. Its efficiency is, T, , D, , A, , V, , 2T0, , V, , 11 An ideal Carnot engine whose efficiency is 40%, receives, T0, , heat at 500 K. If the efficiency is to be 50%, the intake, temperature for the same exhaust temperature is, (a) 600 K, (c) 700 K, , S, , (b) 900 K, (d) 800 K, , S0, , 12 The pressure inside a tyre is 4 atm at 27°C. If the tyre, bursts suddenly, its final temperature will be, 7, , Given, r = , , 5, 7/ 2, , (b) 1/2, (d) 2/3, , 19 The expansion on of unit mass of a perfect gas at, constant pressure is shown below., a, , 2/ 7, , (a) 300 (4), (c) 300 (2)7 / 2, , (a) 1/4, (c) 1/3, , 2S0, , (b) 300 (4), (d) 300 (4)−2 / 7, , O, , 13 A refrigerator works between the temperature of melting, ice and room temperature (17°C). The amount of energy, (in kWh) that must be supplied to freeze 1kg of water at, 0°C is, (a) 1.4, , (b) 1.8, , (c) 0.058, , (d) 2.5, , (a) a = volume, b = ° C temperature, (b) a = volume, b =K temperature, (c) a = ° C temperature, b = volume, (d) a = K temperature, b = volume, , b
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, 20 The temperature of an open room of volume 30 m 3, increases from 17°C to 27°C due to the sunshine. The, atmospheric pressure in the room remains1 × 105 Pa. If ni, and nf are the number of molecules in the room before, and after heating, then nf − ni will be, ª JEE Main 2017, (a) 138, . × 1023, (c) −2.5 × 1025, , (b) 2.5 × 1025, (d) −161, . × 1023, , 21 Consider a spherical shell of radius R at temperature T., The black body radiation inside it can be considered as, an ideal gas of photons with internal energy per unit, U, 1 U , volume u = ∝ T 4 and pressure p = . If the shell, V, 3 V , now undergoes an adiabatic expansion, the relation, between T and R is, ª JEE Main 2015, (a)T ∝ e − R, , γMv 2, K, 2R, (γ − 1), (d), Mv 2K, 2 (γ + 1) R, (b), , 23 Three perfect gases at absolute temperaturesT1, T2 and, T3 are mixed. The masses of molecules are m1, m2 and m3, and the number of molecules are n1, n 2 and n 3,, respectively. Assuming no loss of energy , the final, ª AIEEE 2011, temperature of the mixture is, , (c), , n12T12 + n22T22 + n32T32, nT, 1 1 + n2T 2 + n3T 3, , nT 2 + n2T22 + n3T32, (b) 1 1, nT, 1 1 + n2T 2 + n3T 3, (d), , (T1 + T2 + T3 ), 3, , 24 The value of molar specific heat at constant volume for, 1 mole of polyatomic gas having n number of degrees of, freedom at temperature T K is, (here, R = universal gas constant), nR, 2T, nRT, (c), 2, (a), , (a) 4 R, , (b) 3 R, , (c) 4 R/3, , (d) 2.5 R, , Direction (Q. Nos. 26-30) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which is, the correct answer. You have to select one of the codes (a), (b),, (c), (d) given below:, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 27 Statement I When 1 g of water at 100°C is converted to, , molecular mass M and ratio of specific heats γ. It is, moving with speed v and suddenly brought to rest., Assuming no heat is lost to the surroundings, its, temperature increases by, ª AIEEE 2011, , nT + n2T2 + n3T3, (a) 1 1, n1 + n2 + n3, , through origin. The molar heat capacity of the gas in the, process will be, , is more than the work done by the gas in the same, expansion adiabatically., Statement II Temperature remains constant in isothermal, expansion, but not in adiabatic expansion., , 22 A thermally insulated vessel contains an ideal gas of, , (γ − 1), Mv 2K, 2γ R, (γ − 1), (c), Mv 2K, 2R, , 25 p-V diagram of a diatomic gas is a straight line passing, , 26 Statement I Work done by a gas in isothermal expansion, , (b)T ∝ e − 3R, 1, (c)T ∝, R, 1, (d) T ∝ 3, R, , (a), , 177, , (b), , nR, 2, , (d) 2nRT, , steam at 100°C, the internal energy of the system does, not change., Statement II From dU = nCV dT , if temperature of the, system remains constant, then dU = 0, i.e. internal energy, remains constant., , 28 Statement I In an isothermal process (quasistatic), the, heat exchange between the system and surroundings, takes place even though the gas has the same, temperature as that of the surrounding., Statement II There is an infinitesimal difference in, temperature between the system and the surroundings., , 29 Statement I A special type of thermometer (used to, measure very high temperatures and calibrated for an, ideal black body) measures a value lower than the, actual value of the temperature of a red hot iron piece, kept in open., Statement II As the iron piece is kept in open, it loses its, heat., , 30 Statement I The internal energy of a perfect gas is, entirely kinetic and depends only on absolute, temperature of the gas and not on its pressure or volume., Statement II A perfect gas is heated keeping pressure, constant and later at constant volume. For the same, amount of heat the temperature of the gas at constant, pressure is lower than that at constant volume., ª JEE Main (Online) 2013
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178, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A solid body of constant heat capacity 1 J/°C is being, heated by keeping it in contact with reservoirs in two, ways., (i) Sequentially keeping in contact with 2 reservoirs such, that, each reservoir supplies same amount of heat., (ii) Sequentially keeping in contact with 8 reservoirs such, that each reservoir supplies same amount of heat., In both the cases, body is brought from initial, temperature 100°C to final temperature 200°C. Entropy, change of the body in the two cases respectively, is, ª JEE Main 2015, , (a) In2, 4In2, (b) In2, In2, (c) In2, 2In2, (d) 2In2, 8In2, , 2 Consider a collection of a large number of particles each, with speed v. The direction of velocity is randomly, distributed in the collection. What is the magnitude of the, relative velocity between a pairs in the collection?, 2v, (a), π, 8v, (c), π, , v, (b), π, 4v, (d), π, , 3 An ideal gas (molar specific heat CV = 5R /2) is taken, , along paths acb, adb and ab, p2 = 2p1,V2 = 2V1. Along, ab, p = kV , where k is a constant. The various parameters, are shown in the figure. Match the Column I with the, corresponding options of Column II and mark the correct, option from the codes given below., p, p2, p1, , c, , b, d, , a, , V1, , Column I, , V2, , Codes, A B, (a) 3, 4, (c) 1, 2, , C, 1, 3, , D, 2, 4, , A, (b) 2, (d) 4, , B, 3, 3, , C, 4, 1, , D, 1, 2, , 4 Diatomic molecules like hydrogen have energies due to, both translational as well as rotational motion. From the, 2, equation in kinetic theory pV = E , E is, 3, (a) the total energy per unit volume, (b) only the translational part of energy, because rotational, energy is very small compared to the translational, energy, (c) only the translational part of the energy, because during, collisions with the wall, pressure related to change in, linear momentum, (d) the translational part of the energy, because rotational, energies of molecules can be of either sign and its, average over all the molecules is zero, , 5 An ideal monoatomic gas is confined in a cylinder by a, , spring-loaded piston of cross-section 8 × 10−3m 2. Initially,, the gas is at 300 K and occupies a volume of 2.4 × 10−3m 3, and the spring is in a relaxed state. The gas is heated by, a small heater coil H. The force constant of the spring is, 8000 Nm −1 and the atmospheric pressure is1 × 105 Pa., The cylinder and piston are thermally insulated. The, piston and the spring are massless and there is no, friction between the piston and cylinder. There is no heat, loss through heater coil wire and thermal capacity of the, heater coil is negligible. With all the above assumptions, if, the gas is heated by the heater until the piston moves out, slowly by 0.1 m, then the final temperature is, , T2, , Gas, , T1, , Spring, , H, , V, , (a) 400 K, (c) 1200 K, Column II, , A., , W acb, , 1., , 15RT1/2, , B., , W adb, , 2., , −15RT1/ 2, , C., , ∆U ab, , 3., , RT1, , D., , ∆U bca, , 4., , 2 RT1, , (b) 800 K, (d) 300 K, , 6 A diatomic ideal gas is used in a car engine as the, working substance. If during the adiabatic expansion part, of the cycle, volume of the gas increases fromV to 32V,, ª AIEEE 2010, the efficiency of the engine is, (a) 0.5, (c) 0.99, , (b) 0.75, (d) 0.25
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, 7 The specific heat capacity of a metal at low temperature, 3, , T , (T ) is given as Cp (kJK −1 kg −1) = 32 , . A 100 g, 400, vessel of this metal is to be cooled from 20 K to 4 K by a, special refrigerator operating at room temperature, (27°C). The amount of work required to cool the vessel is, ª AIEEE 2011, , (a) equal to 0.002 kJ, (b) greater than 0.148 kJ, (c) between 0.148 kJ and 0.028 kJ, (d) less than 0.028 kJ, , 8 A horizontal cylinder with adiabatic walls is closed at, both ends and is divided into two parts by a frictionless, piston that is also insulating. Initially, the value of, pressure and temperature of the ideal gas on each side, of the cylinder are V0, p0 and T0, respectively. A heating, coil in the right-hand part is used to slowly heat the gas, on that side until the pressure in both parts reaches, 64p0/27. The heat capacity CV of the gas is independent, of temperature and Cp /CV = γ = 1.5., Take, V0 = 16 m 3, T0 = 324 K, p0 = 3 × 105 Pa, Column I represents the physical parameters of the gas,, Column II gives their corresponding values, match the, Column I with Column II and mark the correct option from, the codes given below., Column I, , Column II, , A., , Final left-hand volume (in m3), , 1., , 432, , B., , Final left-hand temperature (in K), , 2., , 9, , C., , Final right-hand temperature (in K), , 3., , 1104, , D., , Work done (in kJ) on the left-hand gas, , 4., , 3200, , Codes, A B, (a) 2, 1, (c) 4, 1, , C, 3, 2, , D, 4, 3, , A, (b) 1, (d) 3, , B, 2, 4, , C, 3, 1, , coefficient of linear expansion. Suppose we want to bring, the cube to its original size by heating. The temperature, should be raised by, ª JEE Main 2017, (a), , p, αK, , (b), , (a) 2.35 × 10 N/m, (c) 2.35 × 102 N/m 2, , 2, , 10 Two moles of an ideal monoatomic gas occupies a, volume V at 27°C. The gas expands adiabatically to a, volume 2 V . Calculate (i) the final temperature of the gas, ª JEE Main 2018, and (ii) change in its internal energy., (a) (i) 189 K (ii) 2.7 kJ, (c) (i) 189 K (ii) −2.7 kJ, , (d), , p, 3α K, , chamber. As the gas undergoes an adiabatic expansion,, the average time of collision between molecules, increases asV q , where V is the volume of the gas. The, Cp , , value of q is γ =, , CV , , ª JEE Main 2015, (a), , 3γ + 5, 6, , (b), , 3γ − 5, 6, , (c), , γ+1, 2, , (d), , γ −1, 2, , 13 A pendulum clock loses 12 s a day, if the temperature is, 40°C and gains 4 s in a day, if the temperature is 20°C., The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of, the pendulum shaft are respectively, ª JEE Main 2016, (a) 25°C, α = 1.85 × 10−5 / ° C (b) 60°C, α = 1.85 × 10−4 / ° C, (c) 30°C, α = 1.85 × 10−3 / ° C (d) 55°C, α = 1.85 × 10−2 / ° C, , 14 An ideal gas undergoes a quasistatic, reversible process, in which its molar heat capacity C remains constant. If, during this process, the relation of pressure p and volume, V is given by pV n = constant, then n is given by (Here, Cp, and CV are molar specific heat at constant pressure and, constant volume, respectively), (a) n =, , Cp, CV, , (b) n =, , C − Cp, C − CV, , ª JEE Main 2016, C − CV, (c) n =, (d) n =, C − CV, C − Cp, , Cp − C, , two isochoric and isobaric lines) as shown in figure., Efficiency of this cycle is nearly (assume the gas to be, ª AIEEE 2012, close to ideal gas), , (b) (i) 195 K (ii) −2.7 kJ, (d) (i) 195 K (ii) 2.7 kJ, , 11 An external pressure p is applied on a cube at 0°C, so, that it is equally compressed from all sides. K is the bulk, modulus of the material of the cube and α is its, , B, , 2p0, , D, 4, 2, , (b) 4.70 × 10 N/m, (d) 4.70 × 102 N/m 2, 3, , (c) 3 pKα, , 15 Helium gas goes through a cycle ABCDA (consisting of, , 1023 hydrogen molecules strike per second, a fixed wall, of area 2 cm 2 at an angle of 45° to the normal and, rebound classically with a speed of103 m/s, then the, ª JEE Main 2018, pressure on the wall is nearly, 2, , 3α, pK, , 12 Consider an ideal gas confined in an isolated closed, , p0, , 9 The mass of a hydrogen molecule is 3.32 × 10−27 kg. If, , 3, , 179, , D, , A, , V0, , (a) 15.4%, , (b) 9.1%, , C, , 2V0, , (c) 10.5%2, , (d) 12.5%, , 16 An ideal gas is taken from the state A (pressure p,, volume V ) to the state B (pressure p /2, volume 2V ), a, long straight line path in the p-V diagram. Select the, correct statement from the following., (a) The work done by the gas in the process A to B,, exceeds the work that would be done by it, if system, were taken along the isothermal, (b) In the T -V diagram, the path AB becomes a part of a, hyperbola, (c) In the p-T diagram, the path AB becomes a part of a, hyperbola, (d) In going from A to B, the temperature T of the gas, first decreases to a minimum value and then increases
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180, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 17 One mole of diatomic ideal gas undergoes a cyclic, , 18 Two moles of helium are mixed with n moles of hydrogen., , process ABC as shown in the figure. The process BC is, adiabatic. The temperatures at A, B and C are 400 K,, 800 K and 600 K, respectively. Choose the correct, statement., ª JEE Main 2014, , The root mean square speed of the gas molecules in the, mixture is 2 times the speed of sound in the mixture., Then, value of n is, (a) 1, , p, B, , A, , (b) 3/2, , (c) 2, , (d) 3, , 19 n moles of an ideal gas undergoes a process A and B as, shown in the figure. The maximum temperature of the gas, during the process will be, ª JEE Main 2016, , 800 K, , 400 K, , p, , 600 K, C, V, , A, , 2p0, , (a) The change in internal energy in whole cyclic process is, 250 R, (b) The change in internal energy in the process CA is, 700 R, (c) The change in internal energy in the process AB is, − 350 R, (d) The change in internal energy in the process BC is, −500 R, , p0, , B, , V0, , (a), , 9 p0V0, 4 nR, , (b), , V, , 2V0, , 3 p0V0, 2 nR, , (c), , 9 p0V0, 2 nR, , (d), , 9p0V0, nR, , ANSWERS, SESSION 1, , SESSION 2, , 1 (a), , 2 (b), , 3 (a), , 4 (a), , 5 (c), , 6 (c), , 7 (d), , 8 (b), , 9 (c), , 10 (c), , 11 (a), , 12 (d), , 13 (c), , 14 (c), , 15 (b), , 16 (d), , 17 (c), , 18 (c), , 19 (c), , 20 (c), , 21 (c), , 22 (c), , 23 (a), , 24 (b), , 25 (b), , 26 (a), , 27 (a), , 28 (a), , 29 (c), , 30 (b), , 1 (b), 11 (d), , 2 (b), 12 (c), , 3 (d), 13 (a), , 4 (c), 14 (b), , 5 (b), 15 (a), , 6 (b), 16 (a), , 7 (c), 17 (d), , 8 (a), 18 (c), , 9 (a), 19 (a), , 10 (c), , Hints and Explanations, SESSION 1, 1 Cubical expansion, we get, ∆V = γ V∆T = 3αV∆T, 4, = 3 × 23 × 10−6 × π (10)3 , 3, , d, , × 100 Q r = = 10cm , , , 2, = 28.9 cc, , 2 By Mayer’s relation, for 1 g mole of a, , R, ; for N2, 28, From Eqs. (i) and (ii), we get, a = 14b, b = C p − CV =, , 3 Heat gained (water + calorimeter), = Heat lost by copper ball, , 5 In the diagram, T is constant and p 1 > p2 ., This situation is represented by curve (c)., In the solution figure, in which p 1 > p 2 and, straight line parallel to pressure axis, represents constant temperature., , 6, , p, , ⇒ m w sw ∆T + mc sc ∆T = m B sB ∆T, ⇒ 170 × 1 × (75 − 30) + 100, , Isothermal, , × 01, . × (75 − 30), , gas,, , C p − CV = R, So, when n gram moles are given,, R, C p − CV =, n, As per given question,, R, a = C p − C V = ; for H2, K (i), 2, , K (ii), , Adiabatic, , = 100 × 01, . × (T − 75), ∴, , T = 885° C, , 4 As, work done = 0, ∴∆U = mC∆T, = 100 × 10−3 × 4184 × (50 − 30), = 8.4 kJ, , V, , Slope of p-V curve, dp, p, =−, dV, V, dp, p, Adiabatic process,, = −γ, dV, V, Thus, (c) is correct., Isothermal process,
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80 × 1000 × 4.2, 273, 273, =, =, W, 290 − 273, 17, , 7 As temperature of A rises faster than the, temperature of B, therefore specific heat, of A is less than that of B, i.e. S A < S B ., Horizontal portions of graphs represent, conversion of liquid into vapours. The, horizontal portion is larger for liquid A,, therefore L A > L B ., , 8 As it is clear from the figure,, Slope of curve 2 > Slope of curve 1, (γp ) 2 > (γp ) 1, ⇒, , γ2 > γ1, , ⇒, , γ He > γ O2, , Adiabatic curve 2 corresponds to, helium and adiabatic curve 1, corresponds to oxygen., , 9 We know that,, Q = C p ∆T, Q Degree of freedom ∝ C p, So, slope is higher for higher degree of, freedom., , 10 From given figure, in processes BC and, DA, pressure of gas is constant, hence, these represent isobaric process., In process CD, volume is constant,, therefore it represents isochoric process., In process AB, temperature is constant,, so it represent isothermal process., T, 11 As,, η=1− 2, T1, T2, T1, T2, ⇒, 500, ⇒, T2, T2, Now,, T1 ′, ⇒, , = 627 + 273 = 900 K, T2 = 27° C = 27 + 273 = 300 K,, Q1 = 3 × 106 cal, Q, ∴, , T2 =, , ⇒, ⇒, ∴, ⇒, , (1 − 7 / 5), 7/5, , = 300 (4)−2 /7, , 13 T2 = 0° C = 273 K, T1 = 17° C = 17 + 273, = 290 K, Q2, W, T2, =, T1 − T2, , Coefficient of performance =, , T0, Q2, S0, , 2S0, , S, , Q 2 = T 0 S 0, Q 3 = 0, ⇒, , Q − Q2, W, = 1, Q1, Q1, Q2, 2 1, =1−, =1− =, Q1, 3 3, , η=, , 19 In the given graph, line has positive, slope with the X-axis and negative, intercept on the Y- axis., So, we can write the equation of lines as, …(i), y = mx − C, According to Charles’ law,, V, Vt = 0 t + V0, 273, By rewriting this equation, we get, 273 , …(ii), t =, V t − 273, V0 , , We have,, n f − ni =, ⇒ n f − ni =, …(i), , pVN A, pVN A, −, RT f, RT i, 105 × 30, × 6.02 × 1023, 83, ., 1, 1 , . , −, , 300 290 , , = − 2. 5 × 1025, …(ii), , On solving Eqs. (i) and (ii), we get, T1 = 372 K and T2 = 310 K, 17 Efficiency, η = 1 − Tsink, Tsource, T, Now,, 0.4 = 1 − sink, 500 K, ⇒, , Q3, , By comparing Eqs. (i) and (ii), we can, say that temperature is represented on, the Y-axis and volume on the X-axis., 20 From pV = nRT = N RT, NA, , 2, 2, Q 1 = × 3 × 106 cal, 3, 3, , 1, T, =1− 2, 6, T1, T2, 5, =, T1, 6, T − 62, η2 = 1 − 2, T1, T2 − 62, 1, =1−, 3, T1, , 1, 3, T0S 0 = T0S 0, 2, 2, , Q1, , T1, , p1(1 − γ ) T1γ, (1 − γ )/ γ, p , T1 1 , p2 , , 4, = 300 , 1, , Q1, 3, , 16 Q η1 = 1 − T2, , 12 In an adiabatic process,, =, , W = Q1 − Q2 = Q1 −, , = 8.4 × 106 J, , = 600 K, T2γ, , 300 1, Q 2 T2, Q, =, =, = ⇒ Q2 = 1, 900 3, Q 1 T1, 3, , W = 2 × 106 cal, , 50, 1, =, 100 2, T1′ = 2T2 = 2 × 300, , p2(1 − γ ), , 2T0, , 15 Here, T1 = 627°C, , =1−, , = 1 − η′, , Q1 = T 0 S 0 +, , T, , = 0.058 kWh, 14 dW = pdV = RT dV, …(i), V, As, V = kT 2 /3 ,, 2, dV = k T −1 /3 dT, 3, 2, k T −1 /3 dT, dV, 2 dT, =, Q, = 3 2 /3, V, 3 T, kT, From Eq. (i), we get, T2, T2, dV, 2 dT, W = ∫ RT, = ∫ RT, T1, T1, V, 3 T, 2, 2, W = R (T2 − T1 ) = R × 60 = 40R, 3, 3, , =, , 40, 3, =, 100 5, = 300 K, , 18 We have,, , 80 × 1000 × 4.2 × 17, J, ∴ W =, 273, 4, 33.6 × 17 × 10, or W =, kWh, 273 × 3.6 × 105, , =1− η, , =1−, , ⇒, , 181, , HEAT AND THERMODYNAMICS, , DAY FIFTEEN, , Tsink = 0.6 × 500 K = 300 K, 300 K, Thus,, 0.6 = 1 −, T ′source, 300 K, T ′source =, = 750 K, ⇒, 0.4, , ∴, , ∆n = − 2. 5 × 1025, , 21 According to question,, 1 U , , 3 V , nRT 1 U , [Q pV = nRT ], = , V, 3 V , nRT, 1, ∝ T4, V, 3, VT 3 = constant, 4, π R3T 3 = constant, 3, 1, TR = constant ⇒ T ∝, R, p=, , ⇒, or, or, or, or
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182, , DAY FIFTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 22 As no heat is lost., , 27 Since, 1 g of water is converted into, , Loss of kinetic energy, = Gain of internal energy of gas, 1, mv 2 = n C V ∆T, 2, 1, m, R, mv 2 =, ⋅, ∆T, ⇒, 2, M γ −1, Mv 2 (γ − 1), ⇒, ∆T =, K, 2R, 23 F n 1 kT1 + F n 2 kT2 + F n 3 kT3, 2, 2, 2, F, = (n1 + n2 + n3 ) kT, 2, n 1T1 + n 2T2 + n 3T3, T =, ⇒, n1 + n2 + n3, , 24 According to law of equipartition of, energy, average KE per molecule per, degree of freedom at temperature T is, 1, kT . The average KE per molecule of, 2, polyatomic gas, n, molecule = kT, 2, The average KE per molecule of, polyatomic gas, n, n, (E ) = kT × N = RT, 2, 2, d n, n, , CV =, RT = R, 2, dt 2, , steam at constant temperature of 100°C,, i.e. dT = 0., ∴ Change in internal energy,, dU = nC V dT = 0 i.e. U = constant, , 28 In isothermal process, the heat, exchange between system and, surrounding at constant temperature,, i.e. there is an infinitesimal difference, in temperature between the system and, the surrounding., , 29 Since, the thermometer is caliberated, with an ideal black body, the body that, emits or absorbs all the radiations, falling on it, shows a lower value of, temperature. This is because iron is not, a black body, i.e. does not absorb/emit, all radiation falling on it., , 30 The external energy depends upon, absolute temperature of gas. Also,, Statement II is correct, but both the, statements are independently true., , SESSION 2, 1 Since, entropy is a state function,, therefore change in entropy in both the, processes must be same., , 2 Consider any two particles having, angle θ between directions of their, velocities., , 25 As p-V diagram is a straight line passing, , v, , through origin, therefore p ∝ V or, pV −1 = constant., In the process, pV, , x, , B, , = constant, molar, , heat capacity is given by, R, R, C =, +, γ −1 1− x, , θ, , v, , A, →, , →, , p1V1, pV, = 2 2 ⇒ T2 = 4T1, T1, T2, 5R, 15RT1, (4T1 − T1 ) =, 2, 2, For the process,, −15RT1, ∆U bca = − ∆U ab =, 2, 2, 4 In the relation pV = E , E is only the, 3, translational part of energy of molecules., This is because during collision of, molecules with the walls, pressure, exerted relates to change in linear, momentum of gas molecules., ∆U ab =, , 5 V1 = 2.4 × 10−3 m3 , p1 = p 0 = 105 N m −2, and T1 = 300 K (given), If area of cross-section of piston is A and, it moves through distance x, then, increment in volume of the gas = Ax., If force constant of a spring is k, then, force F = kx and, pressure = F / A = kx/ A., V2 = V1 + Ax, ., = 2.4 × 10−3 + 8 × 10−3 × 01, = 3.2 × 10−3, kx, and p2 = p 0 +, A, 8000 × 01, ., = 105 +, = 2 × 105, 8 × 10−3, From ideal gas equation,, p1V1, pV, = 2 2, T1, T2, ⇒, , →, , Then, v rel = v B − v A, i.e. v rel =, , where, x = − 1 and γ = 1.4 for diatomic, gas, R, R, C =, +, 1.4 − 1 1 − (−1), 5, R, = R+, 2, 2, ∴ C = 3R, , v 2 + v 2 − 2v 2 cos θ, , = 2v 2 (1 − cos θ) = 2v sin, , θ, 2, , So, average vrel over all pairs, 2π, θ, 2π, v rel dθ ∫0 2v sin dθ, ∫, 0, 2, v =, =, rel, , 2π, , ∫0, , =, , 26, , dθ, , 2v × 2[− cos (θ, 2π, , 2π, , ∫0, , dθ, , / 2)]20π, , =, , 4v, >v, π, , 3 A → 4.; B → 3.; C → 1.; D → 2., p, , For an ideal gas,, , Isothermal, Adiabatic, V, , The slope of adiabatic curve is several, times the slope of an isothermal curve, and slope of both is negative. Thus, area, under adiabatic curve is smaller than, that under isothermal curve., , ⇒, , T2 = 800 K, , 6 The efficiency of cycle,, T2, T1, , η=1−, , For adiabatic process,, T V γ − 1 = constant, 7, For diatomic gas, γ =, 5, T1V1γ − 1 = T2V2γ − 1, , W acb = W ac + Wcb = 0 + p2 (V2 − V1 ), , V , T1 = T2 2 , V1 , , = p2V1 = 2 p1V1 = 2RT1, W adb = W ad + Wdb, , 7, , T1 = T2 (32)5, , = p1 (V2 − V1 ) + 0 = p1V1 = RT1, , γ −1, , −1, , = T2 (25 )2 / 5 = T2 × 4, , ∆U ab = U ac + U cb, = (Q ac − W ac ) + (Q cb − Wcb ), 5R, C V (Tc − T1 ) + C V (T2 − Tc ) =, (T2 − T1 ), 2, 5R, (given), CV =, 2, , 105 × 2.4 × 10−3, 300, 2 × 105 × 3.2 × 10−3, =, T2, , ∴, , T1 = 4 T2, 1, η = 1 − , , 4, 3, = = 075, ., 4
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HEAT AND THERMODYNAMICS, , DAY FIFTEEN, , Work done on the left-hand side gas is, p V − p 0V 0, W = 1 1, γ −1, 64 × 9 − 1 p V, , 0 0, 27 16, , 2, =, = p 0V 0, 3, 3, −1, 2, = 3200 kJ, , 7 Heat required to change the temperature, of vessel by a small amount dT, − dQ = mC pdT, Total heat required,, 3, , T , − Q = m ∫ 32 , dT, 20, 400 , 4, , 4, , 100 × 10−3 × 32 T 4 , , , (400)3, 4 20, ⇒ Q = 0.001996 kJ, Work done required to maintain the, temperature of sink to T2 ,, Q − Q2, Q2, W = Q1 − Q2 = 1, Q2, T, , = 1 − 1 Q 2, T2, , =, , T − T2 , ⇒ W = 1, Q2, T2 , For T2 = 20 K,, 300 − 20, × 0.001996, W1 =, 20, = 0.028 kJ, For T2 = 4 K,, 300 − 4, W2 =, × 0.001996, 4, kJ, = 0148, ., , p2, V2, T2, 2/3, , 2/3, , p , 27, V1 = V 0 0 = V 0 , 64 , p1 , 9V 0, =, = 9 m3, 16, pV, pV, Also, 0 0 = 1 1, T0, T1, pV, 4T 0, ⇒, T1 = 1 1 =, = 432 K, p 0V 0, 3, 64 p 0, ∴, p2 =, 27, p 0V 0, pV, and, = 2 2, T0, T2, ⇒, ⇒, , V2 = 2V 0 − V1, = 23 m3, 92T 0, = 1104 K, T2 =, 27, , Momentum imparted due to first, collision = 2mv sin 45° = 2mv, Q sin 45° = 1 , , 2 , n 2mv, ∴ Pressure on surface =, Area, 1023 × 2 × 332, . × 10−27 × 103, =, (2 × 10−2 )2, , 12 For an adiabatic process,, TV, , γ −1, , = constant., , We know that, average time of collision, between molecules,, 1, τ =, nπ 2 v rms d 2, where, n = number of molecules per, unit volume, and v rms = rms velocity of molecules., 1, As,, n∝, V, and v rms ∝ T, V, ⇒, τ ∝, T, Thus, we can write, n = K 1V −1, and v rms = K 2 T 1 /2, where, K 1 and K 2 are constants., For adiabatic process,, TV γ − 1 = constant., Thus, we can write, τ ∝ VT −1 /2 ∝ V (V 1 − γ )−1 /2, γ +1, , The compression in the left-hand side is, adiabatic, p 0V 0γ = p1V1γ, , T1, , 45°, , 10 For adiabatic process, relation of, , 8 A → 2; B → 1; C → 3; D → 4 ., , V1, , mv, , p = 235, . × 103 N/m2, , As temperature is changing from 20 K to, 4 K, work done required will be more, than W1 , but less than W2 ., , p1, , 9, , 183, , temperature and volume is,, T2V2γ − 1 = T1V1γ − 1, , or τ ∝ V, , 13 Time period of a pendulum,, , ⇒, , T2 (2V )2 /3 = 300(V )2 /3, 5, [γ = for monoatomic gases], 3, 300, ⇒, T2 = 2 /3 ≈ 189 K, 2, Also, in adiabatic process,, ∆Q = 0, ∆U = − ∆W, − nR(∆T ), or, ∆U =, γ −1, 3 25, = − 2 × × (300 − 189), 2, 3, ≈ −2.7 kJ, p, 11 K =, ( − ∆V / V ), p, ∆V, ⇒ −, =, V, K, pV, ⇒ − ∆V =, K, Change in volume, ∆V = γ V ∆T, where, γ = coefficient of volume, expansion., Again, γ = 3α, where, α is coefficient of linear, expansion., ∴ ∆V = V (3α ) ∆T, pV, ∴, = V (3α ) ∆T, K, p, ∴ ∆T =, 3αK, , 2, , T = 2π, , l, g, , where, l is length of pendulum and g is, acceleration due to gravity., Such as change in time period of a, pendulum,, ∆T, 1 ∆l, =, T, 2 l, When clock losses 12 s, we get, 12 1, = α (40 − θ), T, 2, , …(i), , When clock gains 4 s, we get, 4 1, = α (θ − 20), T, 2, , …(ii), , Comparing Eqs. (i) and (ii), we get, 3=, ⇒, , 40 − θ, θ − 20, , θ = 25°C, , Substituting the value of θ in Eq. (i), we, have, 12 1, = α (40 − 25), T, 2, 1, 12, = α (15), ⇒, 24 × 3600 2, α =, , 24, 24 × 3600 × 15, , α = 1.85 × 10−5/°C
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DAY SIXTEEN, , Transfer of, Heat, Learning & Revision for the Day, u, , u, , Modes of Heat Transfer, Some Common Terms and, Points, , u, , u, , u, , Perfectly Black Body, Kirchhoff’s Law of Radiation, Stefan’s Law, , u, , Newtons Law of Cooling, , u, , Wien’s Displacement Law, , Heat is a form of energy which characterises the thermal state of matter. It is transferred, from one body to the other due to temperature difference between them., Heat is a scalar quantity with dimensions [ML2 T−2 ] and its SI unit is joule (J) while, practical unit is calorie (cal); 1 cal = 4.18 J., The heat can be transferred from one body to the another body, through the following, modes, (i) Conduction, (ii) Convection, (iii) Radiation, , Conduction, The process of heat-transmission in which the particles of the body do not leave their, position is called conduction., , Thermal Conductivity, The amount of heat transmitted through a conductor is given by Q =, where, A = area of cross-section,, ∆T = temperature difference = T2 − T1 ,, t = time elapsed,, K = thermal conductivity, and, l = length of conductor, The rate of transmission of heat by conduction is given by, ∆Q, KA∆T, =, H=, ∆t, l, The unit of thermal conductivity is Wm−1 K −1 ., , KA∆T t, l, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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186, , DAY SIXTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Widemann-Franz Law, , Thermal Resistance, ∆T, ∆Q KA, |H|=, =, ⋅ ∆T =, l / KA, ∆t, l, The term, l, , l, is generally called the thermal resistance (R)., KA, , Equation for rate of heat conduction can be written as, Q, ∆T, H = =, t, Rthermal, It is equivalent/analysis to ohm’s law which states that, V, I =, R(electrical), Q, is equivalent of electric current and called as, t, heat, ∆T is equivalent of voltage (PD) and Rthermal is, equivalent of Relectrical ., where, H =, , Combination of Metallic Rods, 1. Series Combination In a series combination of two, metal rods, equivalent thermal conductivity is given, by, l +l, K1, K2, Ks = 1 2, l1, l2, +, K 1 K2, l, l, 1, , or, , Ks =, , 2K 1 K 2, K 1 + K2, , [if l1 = l2 ], , 2, , Series combination, of rods, , If temperature of the interface, combination be T, then, K T + K2T2, T = 1 1, K 1 + K2, , of, , the, , According to the Widemann-Franz law, the ratio of thermal, and electrical conductivities is same for the metals at a, particular temperature and is proportional to the absolute, temperature of the metal., K, i.e., ∝T, σ, K, or, = constant, σT, , Convection, The process of heat-transmission in which the particles of the, fluid move is called convenction., , Natural Convection, In natural convection gravity plays an important role. When a, fluid is heated, the hot part expands and becomes less dense., Consequently it rises and the upper colder part is replaced., This again gets hot, rises up and is replaced by the colder part, of the fluid., , Forced Convection, In a forced convection the material is forced to move up by a, pump or by some other physical means. Common examples of, forced convection are human circulatory system, cooling, system of an automobile engine and forced air heating system, in offices, etc., , series, , Radiation, A1, , A2, , 2. Parallel Combinations In a, parallel combination of two, metal rods, thermal conductivity, K1, K2, is given by, K A + K2 A2, Kp = 1 1, A1 + A2, Parallel combination, of rods, K 1 + K2, or K p =, [if A1 = A2 ], 2, , The process of the transfer of heat from one place to another, place without heating the intervening medium is called, radiation., , Interaction of Radiation with Matter, When radiant energy Q is incident on a body, a part of it Qa is, absorbed, another part Qr is reflected back and yet another, part Qt is transmitted such that, Q = Qa + Qr + Qt, or, , Formation and Growth of Ice on a Lake, Time required for the thickness of the layer of ice to increase, from d1 to d2 will be, ρL f 2, t =, (d2 − d12 ), 2 KT, where, ρ = density of ice,, L f = latent heat of fusion of ice, and K = thermal conductivity of ice, , Qa Qr Qt, +, +, =1, Q, Q, Q, , a + r + t =1, Qa, where, a =, = absorbing power or absorptance,, Q, Q, r = r = reflecting power or reflectance, Q, Qt, and, t =, = transmitting power or transmittance, Q, or
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TRANSFER OF HEAT, , DAY SIXTEEN, , Some Common Terms and Points, l, , Absorptive power (α) It is defined as the ratio of the, radiant energy absorbed by it in a given time to the total, radiant energy incident on it in the same interval of time., Energy absorbed, α=, Energy incident, As a perfectly black body absorbs all radiations incident on, it, the absorptive power of a perfectly black body is, maximum and unity., , l, , Spectral absorptive power (aλ ) It is the ratio of radiant, energy absorbed by a surface to the radiant energy incident, on it for a particular wavelength λ. The spectral absorptive, power aλ is related to absorptive power a through the, relation, ∞, a = ∫ aλ dλ, 0, , l, , l, , Emissive power (e) It is the total amount of energy radiated, by a body per second per unit area of surface, 1 ∆Q, e=, A ∆t, Spectral emissive power (e λ ) It is emissive power for a, particular wavelength λ. Thus,, , l, , l, , Kirchhoff’s law implies that ‘a good absorber is a good, emitter (or radiator) too’., Fraunhoffer’s lines (dark lines observed in solar, spectrum) can be easily explained on the basis of, Kirchhoff’s laws., , Stefan’s Law, According to the Stefan’s law, the emissive power of a, perfectly black body (energy emitted by black body per unit, surface area per unit time) is directly proportional to the, fourth power of its absolute temperature., Mathematically, E ∝ T 4, or, , E = σT 4, , or, , E = σ (T 4 − T04), , where, σ is a constant known as the Stefan’s constant and its, value is 5.67 × 10 −8 Wm−2 K −4 and T0 is the temperature of, surrounding of black body., l, , l, , ∞, , e = ∫ e λ dλ, , For a body, whose emissivity is ε, Stefan’s law is modified, as,, e = εσT 4, The total radiant energy Q emitted by a body of surface, area A in time t, is given by, Q = Ate = AtεσT 4, , 0, , l, , Emissivity (ε) Emissivity of a body at a given temperature, is defined as the ratio of the total emissive power of the, body (e) to the total emissive power of a perfect black body, (E) at that temperature,, e, ε=, i.e., E, , l, , l, , The radiant power (P), i.e. energy radiated by a body per, unit time is given by, Q, P = = AεσT 4, t, If a body at temperature T is surrounded by another body at, temperature T0 (where, T0 <T ), then according to Stefan’s, law of power, P = εσ A(T 4 − To4 ), , Perfectly Black Body, A perfectly black body is the one which completely absorbs, the radiations of all the wavelengths that are incident on it., Thus, absorbing power of a perfectly black body is 1 (i.e a = 1)., When perfectly black body is heated to a suitable high, temperature, it emits radiation of all possible wavelengths., e.g. temperature of the sun is very high (6000 k approx.) it, emits all possible radiations. So, it is an example of black, body., l, , For perfectly black body, a = 1, r = t = 0, , l, , For a perfect reflector, a = t = 0, r = 1, , l, , For a perfect transmitter, a = r = 0, t = 1., , Kirchhoff’s Law of Radiation, Kirchhoff’s law of radiation states that the ratio of emissive, power to absorptive power of a body, is same for all surfaces at, the same temperature and is equal to the emissive power of a, perfectly black body at that temperature., e, e, (Black body), Mathematically, 1 = 2 = K = E, a1 a2, , 187, , l, , If a body at temperature T is surrounded by another body at, temperature T0 (where, T0 < T ), then Stefan’s law is, modified as,, E = σ (T 4 − T04 ), and, , e = εσ (T, , 4, , [black body], , − T04), , [any body], , Newton’s Law of Cooling, According to the Newton’s law of cooling, rate of cooling of a, body is directly proportional to the temperature difference, between the body and the surroundings, provided the, temperature difference is small., dT, dT, Mathematically,, ∝ (T − T0 ) or −, = k (T − T0 ), dt, dt, where, k is a constant., If a body cools by radiation through a small temperature, difference from T1 to T2 in a short time t when the surrounding, temperature is T0 , then, dT ~ T1 − T2, −, dt, t, , and T =, , T1 + T2, 2
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188, , DAY SIXTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , The Newton‘s law of cooling becomes, , T + T, T1 − T2 , = k 1 2 − T0 ., , 2, t , , Variation of intensity of thermal radiation with wavelength is, shown in fig. The total area under E λ -λ curve gives the total, intensity of radiation at that temperature. The area, in, accordance with the Stefan’s law of radiation, is directly, proportional to the fourth power of the temperature., , Black Body Spectrum, Eλ, The black body, spectrum is a, T4 > T 3, continuous spectrum as, shown in the figure. At a, T3 > T 2, given temperature,, initially the intensity of, T2 > T 1, thermal radiation, increases with an, increase in wavelength, T1, and reaches a maximum, λ, λm4 λ λ, m3 m 2 λ m 1, value at a particular, wavelength λ m. On, Graph between intensity E λ and λ, increasing the, wavelength beyond λ m, the intensity of radiation E λ starts, decreasing., , Wien’s Displacement Law, According to Wien‘s law, the product of wavelength, corresponding to maximum intensity of radiation and, temperature of body is constant i.e. λ mT = constant = b, where b, is known as the Wien’s constant and its value is 2.89 × 10 −3 mK., , Solar Constant, The amount of heat received from the sun by one square, centimeter area of a surface placed normally to the sun rays at, mean distance of the earth from the sun is known as solar, constant. It is denoted by S., 2, , r, S = σT4, R, where, r is the radius of sun and R is the mean earth’s distance, from sun value of solar constant S = 1.937 cal/cm2 /min., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A cylindrical rod is having temperaturesT1 and T2 at its, ends. The rate of flow of heat is Q1. If all the linear, dimensions are doubled keeping the temperature, constant, then rate of flow of heat Q 2 will be, (a) 4 Q1, , (b) 2 Q1, , (c), , Q1, 4, , (d), , 4 The coefficient of thermal conductivity of copper is, 9 times that of steel. In the composite cylindrical bar, shown in the figure, what will be the temperature at the, junction of copper and steel?, , Q1, 2, , 100°C, , 2 A uniform metallic rod rotates about its perpendicular, bisector with constant angular speed. If it is heated, uniformly to raise its temperature slightly, (a) Its speed of rotation increases, (b) Its speed of rotation decreases, (c) Its speed of rotation remains same, (d) Its speed increases because its moment of inertia, increases, , 3 Two slabs A and B of different materials but with the, same thickness are joined as shown in the figure. The, thermal conductivities of A and B are K1 and K 2,, respectively. The thermal conductivity of the composite, slab will be, , (a), , 1, (K1 + K 2 ), 2, , l/2, , Steel, , 18 cm, , 6 cm, , (b) 67°C, , (c) 25°C, , (d) 33°C, , 5 Three objects coloured black, grey and white can, withstand hostile conditions at 2800°C. These objects are, thrown into furnace where each of them attains a, temperature of 2000°C. Which object will have the, brightest glow?, (a) The white object, (b) The black object, (c) All glow with equal brightness, (d) Grey object, , radiates heat energy at the rate Q Watt. If its surface is, smoothened, so as to lower its emissivity by 10%, what, will be the increase in its rate of radiation at double the, initial temperature?, , K2, , K1 B, , (b) K1K 2, , Copper, , 6 A black body maintained at a certain temperature, , l, A, , (a) 75°C, , 0°C, , l/2, , (c) (K1 + K 2 ), , (d), , 2K1K 2, (K1 + K 2 ), , (a) (0.9 × 2 4 − 1) Q W, (c) (0.9 × 2)4 Q W, , (b) 0.9 × 2 4 Q W, (d) (0.9)4 × 2Q W
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TRANSFER OF HEAT, , DAY SIXTEEN, 7 We consider the radiation emitted by the human body., Which of the following statement is true?, (a) The radiation is emitted during the summers and, absorbed during the winters., (b) The radiation emitted lies in the ultraviolet region and, hence is not visible., (c) The radiation emitted is in the infrared region., (d) The radiation is emitted only during the day., , 8 Parallel rays of light of intensity I = 912 Wm, , −2, , are incident, on a spherical black body kept in surroundings of, temperature 300 K. Take, Stefan constant, σ = 5.7 × 10−8 Wm −2 K −4 and assume that the energy, exchange with the surroundings is only through radiation., The final steady state temperature of the black body is, close to, ª 2014 Main, (a) 330 K, , (b) 660 K, , (c) 990 K, , (d) 1550, , 9 The spectral energy distribution of a star is maximum at, twice temperature as that of the sun. The total energy, radiated by the star is, (a) twice as that of the sun, (b) same as that of the sun, (c) sixteen times as that of the sun, (d) one-sixteenth of the sun, , temperature difference between the body and the, surroundings is, (b) more than 10°C, (d) more than 100°C, , 11 A liquid in a beaker has temperature θ(t ) at time t and θ 0, , (b), , O, , t, , (d) q, 0, , O, , O, , t, , t, , 13 A sphere, a cube and a thin circular plate, all of same, material and same mass are initially heated to same high, temperature. Then, (a) plate will cool fastest and cube the slowest, (b) sphere will cool fastest and cube the slowest, (c) plate will cool fastest and sphere the slowest, (d) cube will cool fastest and plate the slowest, , 14 Temperatures of two stars are in the ratio 3 : 2. If, wavelength for the maximum intensity of the first body is, 4000 Å, what is the corresponding wavelength of the, second body?, (a) 9000 Å, (c) 2000 Å, , (b) 6000 Å, (d) 8000 Å, , maximum around a wavelength λ 0. The temperature of, the black body is now changed such that the energy is, 3λ 0, maximum around a wavelength, . The power radiated, 4, by the two black bodies will now increase by a factor of, (a) 64/27, , (b) 256/81, , (c) 4/3, , (d) 16/9, , 16 Three discs, A, B and C having radii 2 m, 4 m and 6 m, respectively, are coated with carbon black on their outer, surfaces. The wavelengths corresponding to maximum, intensity are 300 nm, 400 nm and 500 nm, respectively., The power radiated by them are Q A , QB and QC ,, respectively, , 17 Variation of radiant energy emitted by sun, filament of, tungsten lamp and welding arc as a function of its, wavelength is shown in figure. Which of the following, options is the correct match?, , loge (θ – θ0), , (d), , (c) q, 0, , (a) QA is maximum, (b) QB is maximum, (c) Q C is maximum, (d) QA = QB = QC, , O, , t, , loge (θ – θ0), , (c), , loge (θ – θ0), , loge (θ – θ0), , is temperature of surroundings, then according to Newton’s, law of cooling, the correct graph between loge (θ − θ 0 ) and, t is, , (a), , T, , 15 The energy spectrum of a black body exhibits a, , 10 Newton’s law of cooling holds good only, if the, , (a) less than 10°C, (c) less than 100°C, , T, , E1, , O, , t, , O, , t, , 12 If a piece of metal is heated to temperature θ and then, , allowed to cool in a room which is at temperature θ 0 , the, graph between the temperature T of the metal and time, ª JEE Main 2013, will be closed to, , T3, T2, T1, , T, , (a) T, , l, , (a) Sun-T1, tungsten filament-T2 , welding arc-T3, , (b) q, 0, , (b) Sun-T2 , tungsten filament-T1, welding arc-T3, (c) Sun-T3 , tungsten filament-T2 , welding arc-T1, , O, , t, , O, , 189, , t, , (d) Sun-T1, tungsten filament-T3 , welding arc-T2
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190, , DAY SIXTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Direction (Q. Nos. 18-20) Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below., (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, , allowed to cool down in the same environment. The, hollow sphere cools faster., Statement II Rate of cooling follows the Stefan’s law, which is E ∝ T 4., , 19 Statement I A body that is a good radiator is also a good, absorber of radiation at a given wavelength., Statement II According to Kirchhoff’s law, the, absorptivity of a body is equal to its emissivity at a given, wavelength., , 20 Statement I For higher temperatures, the peak emission, , (d) Statement I is false; Statement II is true, , 18 Statement I A solid sphere of copper of radius R and a, hollow sphere of the same material of inner radius r and, outer radius R are heated to the same temperature and, , wavelength of a black body shifts towards the lower, wavelength side., Statement II Peak emission wavelength of a black body, is proportional to the fourth-power of the temperature., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A metallic sphere cools from 50° C to 40°C in 300 s. If the, room temperature is 20°C, then its temperature in the, next 5 min will be, (a) 38°C, , (b) 33.3°C, , (c) 30°C, , (d) 36°C, , 2 A pan filled with hot food cools from 94° C to 86 °C in, 2 min, when the room temperature is at 20 °C, how long, will it take to cool from 71 °C to 69 °C?, (a) 14 s, , (b) 3 s, , (c) 42 s, , over the other such that their surfaces are completely in, contact. The thickness of slab A is twice that of B. The, coefficient of thermal conductivity of slab A is twice that, of B. The first surface of slab A is maintained at 100°C,, while the second surface of slab B is maintained at 25°C., The temperature at the contact of their surfaces is, (b) 45°C, , (c) 55°C, , (d) 85°C, , 4 Assuming the sun to be a spherical body of radius R at a, temperature of T K, evaluate the total radiant power,, incident on the earth, at a distance r from the sun., (a) 4 πr02R 2 σT 4 / r 2, (c) r02R 2 σT 4 / 4 πr 2, , (b), , πr02R 2 σT 4, 2, 4, 2, , 1, 2, , (b), , /r, , 2, , (d) R σT / r, , where, r0 is the radius of the earth and σ is the Stefan’s, constant., , 5 Two identical conducting rods are first connected, independently to two vessels, one containing water at, 100°C and the other containing ice at 0°C. In the second, case, the rods are joined end to end and connected to, the same vessels. Let q1 and q 2 g/s be the rate of, melting of ice in two cases respectively. The ratio of, q1 / q 2 is, , 2, 1, , (c), , 4, 1, , (d), , 1, 4, , 6 Two circular discs A and B with equal radii are, blackened. They are heated to same temperature and are, cooled under identical conditions. What inference do you, draw from their cooling curves?, A, , (d) 13 s, , 3 Two slabs A and B of equal surface area are placed one, , (a) 62.5°C, , (a), , R, , B, , (q – q0), , (a) A and B have same specific heats, (b) Specific heat of A is less, (c) Specific heat of B is less, (d) None of the above, , 7 Three rods of copper, brass and steel are welded, together to form a Y-shaped structure. Area of, cross-section of each rod is 4 cm 2. End of copper rod is, maintained at 100°C whereas ends of brass and steel are, kept at 0°C. Lengths of the copper, brass and steel rods, are 46, 13 and 12 cm respectively., The rods are thermally insulated from surroundings, except at ends. Thermal conductivities of copper, brass, and steel are 0.92, 0.26 and 0.12 in CGS units,, respectively. Rate of heat flow through copper rod is, ª JEE Main 2014, , (a) 1.2 cals −1, (c) 4.8 cals −1, , (b) 2.4 cals −1, (d) 6.0 cals −1
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TRANSFER OF HEAT, , DAY SIXTEEN, 8 A mass of 50 g of water in a closed vessel, with, , 14, , surroundings at a constant temperature takes 2 min to, cool from 30°C to 25°C. A mass of 100g of another liquid, in an identical vessel with identical surroundings takes, the same time to cool from 30°C to 25°C. The specific, heat of the liquid is (The water equivalent of the vessel is, 30 g.), ª JEE Main (Online) 2013, (a) 2.0 kcal/kg, (c) 3 kcal/kg, , (b) 7 kcal/kg, (d) 0.5 kcal/kg, , 9 The figure shows a system of two, concentric spheres of radii r1 and r2, and kept at temperatures T1 and T2, respectively. The radial rate of flow, of heat in a substance between the, two concentric spheres, is, proportional to, (a), , (r2 − r1), (r1r2 ), , r , (b) ln 2 , r1 , , (c), , r1r2, (r2 − r1), , r1, T1, r2, , T2, , (d) (r2 − r1), , 10 A slab of stone of area 3600 cm 2 and thickness 10 cm is, exposed on the lower surface to steam at 100°C. A block, of ice at 0°C rests on the upper surface of the slab. If in, 1 h 4.8 kg of ice melted the thermal conductivity of the, stone is, (a) 1.24 W/m/k, (c) 0.24 W/m/k, , (b) 2.24 W/m/k, (d) 1.54 W/m/k, , 11 Two spherical stars A and B emit black body radiation., The radius of A is 400 times that of B and A emits, λ , 104 times the power emitted from B. The ratio A of, λB , their wavelengths λ A and λ B at which the peaks occur in, their respective radiation curves is, (a) 1, (c) 3, , (b) 2, (d) 5, , 12 A metal is heated in a furnace where a sensor is kept, above the metal surface to read the power radiated (P ), by the metal. The sensor has a scale that displays, log2(P /P0 ), where P0 is a constant. When the metal, surface is at a temperature of 487°C, the sensor shows a, value 1. Assume that the emissivity of the metallic, surface remains constant. What is the value displayed by, the sensor when the temperature of the metal surface is, raised to 2767°C?, (a) 1, , (b) 4, , (c) 7, , 18 cm) are at temperatures T1 and T2, respectively. The, maximum intensity in the emission spectrum of A is at, 500 nm and in that of B is at 1500 nm. Considering them, to be black bodies, what will be the ratio of the rate of, total energy radiated by A to that of B?, (b) 7, , (c) 5, , (d) 1, , Chamber II, real, gas, , ideal, gas, , 1, , 2, , 3, , 4, , There are two identical chambers, completely thermally, insulated from surrounding. Both chambers have a, partition wall dividing the chambers in two, compartments. Compartment 1 is filled with an ideal gas, and compartment 3 is filled with a real gas., Compartments 2 and 4 are vacuum. A small hole (orifice), is made in the partition walls and the gases are allowed, to expand in vacuum., Statement I No change in the temperature of the gas, takes place when ideal gas expands in vacuum., However, the temperature of real gas goes down, (cooling) when it expands in vacuum., Statement II The internal energy of an ideal gas is only, kinetic. The internal energy of a real gas is kinetic as well, as potential., ª JEE Main (Online) 2013, (a) Statement I is false and Statement II is true, (b) Statement I and Statement II both are true. Statement II, is the correct explanation of Statement I, (c) Statement I is true and Statement II is false, (d) Statement I and Statement II both are true, but, Statement II is not the correct explanation of Statement I, , 15 A rod AB of uniform cross-section consists of four section, AC, CD, DE and EB of different metals with thermal, conductivities K, ( 0.8) K, (1.2) K and (1.50) K,, respectively. Their lengths are respectively L,, (1.2) L, (1.5) L and (0.6) L. They are joined rigidly in, succession at C, D and E to form the rod AB. The end A, is maintained at 100 °C and the end B is maintained at, 0°C. The steady state temperatures of the joints C, D and, E are respectivelyTC ,TD and TE . Column I lists the, temperature differences (TA − TC ),(TC − TD ), (TD − TE ) and, (TE − TB ) in the four sections and column II their values, jumbled up. Match each item in column I with its correct, value in column II., A, , C, , D, , E, , Column I, , (d) 9, , 13 Two spherical bodies A (radius 6 cm ) and B (radius, , (a) 9, , Chamber I, , 191, , B, , Column II, , A., , (TA − TC ), , 1., , 9.6, , B., , (TC − TD ), , 2., , 30.1, , C., , (TD − TE ), , 3., , 24.1, , D., , (TE − TB ), , 4., , 36.2, , A, (a) 3, (c) 3, , B, 4, 4, , C, 2, 1, , D, 1, 2, , A, (b) 1, (d) 3, , B, 2, 2, , C, 4, 1, , D, 3, 4
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192, , DAY SIXTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (b), , 3 (d), , 4 (a), , 5 (b), , 6 (a), , 7 (c), , 8 (a), , 9 (c), , 10 (a), , 11 (a), , 12 (c), , 13 (c), , 14 (b), , 15 (b), , 16 (b), , 17 (c), , 18 (c), , 19 (a), , 20 (c), , 1 (b), 11 (b), , 2 (c), 12 (d), , 3 (a), 13 (a), , 4 (b), 14 (c), , 5 (c), 15 (a), , 6 (b), , 7 (c), , 8 (d), , 9 (c), , 10 (a), , Hints and Explanations, SESSION 1, , 5 An ideal black body absorbs all the, radiations incident upon it and has an, emissivity equal to 1. If a black body, and an identical body are kept at the, same temperature, then the black body, will radiate the maximum power., , KA1 (T1 - T2 ), but on, l1, doubling all dimensions l 2 = 2l 1 and, A 2 = 4 A 1., K A 2 (T1 - T2 ), Hence, Q 2 =, l2, K 4 A 1 (T1 - T2 ), K A 1 (T1 - T2 ), =, =2, 2l 1, l1, , 1 Initially, Q 1 =, , Hence, the black object at a temperature, of 2000°C will have the brightest glow., , 6 For black body,, Rate of radiation Q = sT, , = 2Q 1, , After smoothing and doubling the, temperature = Rate Q, = 0.9 s (2T )4, , 2 When a metallic rod is heated it, expands. Its moment of inertia (I ) about, a perpendicular bisector increases., According to law of conservation of, angular momentum, its angular speed, (w ) decreases, since w µ1 / I. (According, to law of conservation of angular, momentum)., , = 0.9 ´ 24 Q, Change = (0.9 ´ 24 - 1) Q W, human body have wavelength of the, order of 7.9 ´ 10-7 m to 10-3 m, which is, , length l 1 , area of cross-section A and, thermal conductivity K is given by, l, R=, KA, , 4 From temperature of interface,, q=, , K 1 q1 l 2 + K 2 q2 l 1, K 1 l2 + K 2 l1, , It is given that K Cu = 9K s ., So, if K s = K 1 = K , then, K Cu = K 2 = 9K, 9K ´ 100 ´ 6 + K ´ 0 ´ 18, Þ q=, 9K ´ 6 + K ´ 18, 5400 K, =, = 75° C, 72 K, , where, s is Stefan’s constant., Given, T = 2TS, \, E ¢ = s(2TS )4 = 16 sTS4 = 16ES, Hence, total energy radiated by star is, sixteen times as that of the sun., , 10 Newton’s law of cooling states, that,‘‘ the, rate of cooling of a body is directly, proportional to temperature difference, between the body and the surroundings,, provided the temperature difference is, small, (less than 10°C)’’ and Newton’s, law of cooling is given by, dT, µ (q - q 0 ), dt, , 11 According to Newton’s law of cooling,, , 7 The heat radiation emitted by the, , 3 The thermal resistance of a slab of, , Since, the slabs are joined in series, the, thermal resistance of the composite slab, is, \, RC = R1 + R2, l, l /2, l /2, or, =, +, KC A K1 A, K2 A, 2K 1 K 2, or, KC =, (K 1 + K 2 ), , 4, , 9 From Stefan’s law of radiation, E = sT 4, , ofcourse the range of infrared region., Hence, human body emits radiation in, infrared region., , 8 In steady state, , rate of fall in temperature is proportional, to the difference in temperature of the, body with surrounding, i.e., dq, = k (q - q 0 ), dt, dq, Þ, ò q - q0 = ò - k dt, Þ, ln (q - q 0 ) = kt + C, which is a straight line with negative, slope., , 12 According to Newton’s cooling law,, S =4πR2, , I, , πR2, Incident, Radiation, Energy incident per second = Energy, radiated per second, \ IpR2 = s (T 4 - T 04 ) 4pR2, , ⇒, ⇒ T, , I = s (T 4 - T 04 ) 4, 4, , -, , T 04, , = 40 ´ 10, , 8, , Þ T 4 - 81 ´ 108 = 40 ´ 108, ⇒, ⇒, , T 4 = 121 ´ 108, T » 330 K, , T2 = T1 + Ce - Kt, where,, C = T i - T1, (difference in temperature of body and, surrounding), Þ, T2 µ e - Kt, Thus, the graph decays exponentially., This is shown in fig. (c)., , 13 We know that, the rate of loss of heat, from a body is directly proportional to, the surface area of the body. For a given, mass of a material, the surface area of a, circular plate is maximum and of sphere, is least. Hence, plate will cool fastest and, sphere the slowest.
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TRANSFER OF HEAT, , DAY SIXTEEN, 14 According to Wien’s displacement law,, , \, , (l m )1 = 4000 Å, 4000 ´ 3, (l m ) 2 =, 2, = 6000 Å, , 15 We know that, l mT = constant and the, power radiated by a black body is, proportional to T 4 i.e. P µ T 4 , Hence,, P µ (l m )-4, Þ, , The second effect is that the total, amount of energy, the black body emits, per unit area per unit time increases, with fourth power of the absolute, temperature T., , 3 According to question, the temperature, at the contact of the surface is given by, K d q + K 2d 1 q2, = 1 2 1, K 1d 2 + K 2d 1, 2K 2d 2 ´ 100 + 2d 2 ´ K 2 ´ 25, =, 2K 2d 2 + K 2 2d 2, 200 + 50, =, = 62.5° C, 4, , 4 From Stefan’s law, the rate at which, energy is radiated by sun from its surface, is, TK, , SESSION 2, , P2 æ l m1, =ç, P1 çè l m2, 256, =, 81, , 4, , 4, , 4, ö, æ l ö, 4, ÷ = ç 0 ÷ = æç ö÷, ÷, è, ø, 3, è 3l 0 4 ø, ø, , 16 Q µ AT 4 and l mT = constant., Hence, Q µ, , 20 As the temperature of the black body, increases, two distinct behaviours are, observed. The first effect is that the, peak of the distribution shifts towards, the shorter wavelength side. This shift, is found to obey the following, relationship called the Wien’s, displacement law, which is given by, l mT = constant., , l m T = constant, (l m )1 T2, \, =, (l m )2 T1, T1 3, Here,, =, T2 2, , A, 4, , (l m ), , or Q µ, 2, , r2, (l m )4, , 2, , (2) (4) (6)2, Q A : QB : Q C = 4 : 4 : 4, (3) (4) (5), 4 1 36, =, :, :, 81 16 625, = 0.05 : 0.0625 : 0.0576, i.e. Q B is maximum., , 17 l mT = constant, From the graph T3 > T2 > T1, Temperature of sun will be maximum., Therefore, (c) is the correct option., , 18 As external radii of both the spheres are, equal, the surface areas of the two are, also equal. Therefore, when the two, spheres are heated to the same, temperature, both radiate heat at the, same rate., Now, rate of loss of heat from a sphere, dq, = Mc, dt, Therefore, rate of cooling, dq rate of loss of heat, =, dt, Mc, dq, 1, or, µ, dt, M, Since, mass of a hollow sphere is less,, its rate of cooling will be fast., , 19 According to Kirchhoff’s law of, radiation, the ratio of emissive power to, absorptive power of a body, is same for, all surfaces at the same temperature and, at a particular wavelength., Thus, Kirchhoff‘s law implies that a, good absorber is a good emitter (or, radiator) too or vice-versa., , 193, , r0, r, , Earth, , R, , 1 According to the Newton’s law of, cooling,, 50 - 40, é 50 + 40, ù, =K ê, - 20ú, 300, 2, ë, û, 10, 90, é, ù, =K, - 20 = K ´ 25, Þ, êë 2, úû, 300, 10, Þ, K =, 300 ´ 25, 1, =, 30 ´ 25, Similarly,, 40 - q, é 40 + q, ù, =K ê, - 20ú, 300, ë 2, û, q, Kq, = K é20 + - 20ù =, êë, úû, 2, 2, q, q, =, =, 2 ´ 30 ´ 25 1500, Þ, Þ, Þ, , 300 q = 1500(40 - q), = 60000 - 1500 q, 1800 q = 60000, 60000, q=, = 33.3 ° C, 1800, , 2 The average temperature of 94 °C and, 86 °C is 90 °C, which is 70 °C above the, room temperature, under these, conditions the pan cools 8°C in 2 min,, we have, Change in temperature, = kDT, Time, 8°C, …(i), = k (70° C ), 2 min, The average of 69 °C and 71°C is 70 °C,, which is 50 °C above room temperature., K is same for this situation as for the, original, 2° C, …(ii), = k(50°C ), Time, On dividing Eq. (i) by Eq. (ii), we get, 8° C/2 min k (70° C ), =, ,, 2° C/ time, k (50° C ), T = 07, . min = 42s, , Sun, P = s ´ 4pR2 ´ T 4, , [Sun is a perfect black body as it emits, radiations of all wavelengths and so for, it, e = 1], The intensity of this power at the surface, of the earth, [under the assumption r>> r 0] is, s ´ 4pR2T 4, P, sR2T 4, I =, =, =, 2, 2, 4pr, 4pr, r2, The area of the earth which receives this, energy is only one-half of the total, surface area of earth, whose projection, would be pr 02 ., \ Total radiant power as received by the, pr 2 R2 sT 4, earth = pr 02 ´ I = 0 2, r, , 5 When the rods are placed in vessels, q (T1 - T2 ), =, t, R, q, mL, (100 - 0), = æç ö÷ =, = q1 L =, è t ø1, t, R /2, , …(i), , When the rods are joined end to end, æ q ö = mL = q L = (100 - 0), …(ii), ç ÷, 2, è t ø2, t, 2R, From Eqs. (i) and (ii), we get, q1, 4, = ., q2, 1, , 6 According to Newton’s law of cooling,, rate of cooling is given by,, æ -dT ö = eAs (T 4 - T 4 ), ç, ÷, 0, è dt ø mc, where, c is specific heat of material., æ -dT ö µ 1, or, ç, ÷, è dt ø c, i.e. rate of cooling varies inversely as, specific heat. From the graph, for A,rate, of cooling is larger. Therefore, specific, heat of A is smaller.
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DAY SEVENTEEN, , UNIT TEST 3 (GENERAL PROPERTIES OF MATTER), , 195, , DAY SEVENTEEN, , Unit Test 3, (General Properties, of Matter), 1 A jar is filled with two non-mixing liquids 1 and 2 having, densities d1 and d 2 respectively. A solid ball, made of a, material of density d 3 , is dropped in the jar. It comes to, equilibrium in the position as shown in the figure. Which, of the following is true for d1, d 2 and d 3 ?, , Liquid 1, , (a) d1 > d3 > d2, (c) d1 < d3 < d2, , cooling is, (a) independent of r, (b) proportional to r, (c) proportional to r 2, (d) proportional to 1/r, , 4 A certain ideal gas undergoes a polytropic process, , d1, , pV n = constant such that the molar specific heat during, the process is negative. If the ratio of the specific heat of, the gas be γ, then the range of values of n will be, , d3, Liquid 2, , 3 A hot metallic sphere of radius r radiates heat. Its rate of, , d2, , (b) d1 < d2 < d3, (d) d3 < d1 < d2, , 2 A spherical body of volume V and density σ is, suspended from a string, the other end of the string is, connected to the roof of a sealed container filled with an, ideal fluid of density ρ., , a, , If the container accelerates towards right with a constant, acceleration a, then the force exerted by the liquid on the, body when it is in equilibrium w.r.t. fluid, is, (a)V ρ a 2 + g 2 + Vσ a, , (b)V σ a, , (c) [Vρ (g + a)]2 + [Vσ a] 2, , (d)Vρ g 2 + a 2, , (a) 0 < n < γ, (b) 1< n < γ, (c) n = γ, (d) n > γ, , 5 Pressure p, volume V and temperature T for a certain, material are related by, p=, , AT − BT 2, V, , where, A and B are constants. Find an expression for the, work done by the material if the temperature changes, from T1 to T2 reduce while the pressure remains constant., (a)W = A (T2 − T1) − B (T23 − T13 ), (b)W = A (T22 − T12 ) − B (T2 − T1), 1, (c)W = A (T2 − T1) − B T2 − T1 , , 2 , (d)W = A (T2 − T1) − B (T22 − T12 )
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196, , DAY SEVENTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 6 A soap bubble is very slowly blown on the end of a glass, tube by a mechanical pump which supplies a fixed, volume of air every minute whatever be the pressure, against which it is pumping. The excess pressure ∆p, inside the bubble varies with time is shown by which of, the graph?, Dp, , Dp, , (a), t, , t, Dp, , (c), t, , t, , 7 A small electric immersion heater is used to heat 100 g of, water for a cup of instant coffee. The heater is labelled, “200 W,” which means that it converts electrical energy, to thermal energy at this rate. Calculate the time required, to bring all this water from 23°C to 100°C, ignoring any, heat losses. [c = 4190 J kg −1 K −1 ], (b) 200 s, , (c) 190 s, , (d) 161 s, , 8 The average depth of Indian ocean is about 3000 m., Calculate the fractional compression, ∆V / V , of water at, the bottom of the ocean. Given that the bulk modulus of, water is 2.2 × 109 Nm −2 ., (Take, g = 10 ms −2 ), (a) 1. 36 × 10− 2, (c) 15, . × 10− 2, , (b) 3 × 10− 3, (d) 1. 36 × 10− 6, , 9 The temperature of the two outer surfaces of a composite, slab, consisting of two materials having coefficients of, thermal conductivity K, 2K and thickness x, 4x,, respectively are T2 and T1 (T2 > T1 ). The rate of heat, transfer through the slab in a steady state is, A(T2 − T1 )K , , f with f which is equal to, x, (a) 1, (c) 2/3, , (a) 1000°C, (c) equal to triple of water, , (b) 0°C, (d) − 273.16°C, , has an interior diameter of 2.992 cm at 25°C. At what, common temperature will the ring just slide onto the rod?, (take, α s = 11 × 10−6 ° C−1, α b = 19 × 10−6 ° C−1 ), (a) 460°C, , (b) 260°C, , (c) 500°C, , (d) 360°C, , 14 A diver is hunting for a fish with a water gun. He, accidentally fires the gun, so that bullet punctures, the side of the ship. The hole is located at a depth of, 10 m below the water surface. The speed with which, water enter in the ship is, (a) 18 ms −1, (c) 25 ms −1, , (b) 14 ms −1, (d) Cannot be determined, , 15 A material has a Poisson’s ratio 0.3. If a uniform rod of it, , suffers longitudinal strain 4 ⋅ 5 × 10− 3 , then calculate the, percentage change in its volume., (a) 0.15%, , (b) 0.25%, , (c) 0.18%, , (d) 0.5%, , 16 Compute the number of moles and in 1.00 cm 3 of an ideal, gas at a pressure of 100 Pa and at a temperature of 220 K., (a) 3.35 × 10−8 mol, (c) 5.47 × 10−8 mol, , (b) 4.57 × 10−7mol, (d) 2.75 × 10−8 mol, , 17 A slab consists of two parallel layers of copper and brass, of the same thickness same area of cross-section and, having thermal conductivities in the ratio 1 : 4. If the free, face of brass is at 100°C and that of copper is at 0°C, the, temperature of the interface is, (b) 20°C, , (c) 60°C, , (d) 40°C, , −2, , 18 On applying a stress of x Nm , the length of wire of, , (b) 1/2, (d) 1/3, , 10 An aeroplane has a mass of 1.60 × 10 kg and each wing, has an area of 40 m 2 . During level flight, the pressure on, the wings’s lower surface is 7 × 104 Pa. The pressure on, the upper surface of the wing is, (Take, p0 = 10 5 Pa and assume the pressure difference, is only on wings and not on body), (a) 10 Pa, (b) 6.8 × 104 Pa, (c) 7 × 104 Pa, (d) 6.6 × 104 Pa, , (b) Xc > Xm > X g, (d) Xm < Xc < X g, , is 1000°C. Its efficiency could be 100% only if the, temperature of the sink is, , (a) 80°C, , 4, , 5, , (a) Xc = Xm = X g, (c) Xc < Xm < X g, , 13 A steel rod is 3.00 cm in diameter at 25°C. A brass ring, , (d), , (a) 100 s, , mercury and glass are Kc , Km and K g , respectively, such, that Kc > km > K g . If the same quantity of heat is to flow, per second per unit area of each and corresponding, temperature gradients are, X c , X m and X g , respectively,, then, , 12 The temperature of the source of a Carnot’s heat engine, , (b), , Dp, , 11 The coefficients of thermal conductivity of copper,, , some material becomes double. Value of the Young’s, modulus for the material of the wire in Nm −2 , is [Assume, Hooke’s law to be valid] “Go for approx results”, (a) x, (c) x / 2, , (b) 2x, (d) Insufficient information, , 19 743 J of heat energy is added to raise the temperature of, 5 mole of an ideal gas by 2 K at constant pressure. How, much heat energy is required to raise the temperature of, the same mass of the gas by 2K at constant volume?, (Take, R = 8.3 J/K-mol), (a) 826 J, , (b) 743 J, , (c) 660 J, , (d) 620 J
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DAY SEVENTEEN, , UNIT TEST 3 (GENERAL PROPERTIES OF MATTER), , 20 Six identical conducting rods are joined as shown in, figure given below. Points A and D are maintained at, temperatures 200 °C and 20°C, respectively. The, temperature of junction B will be, , 197, , The water which comes out from the hole at the instant, when the hole is at height H above the ground, strikes, the ground at a distance x from P. Which of the following, is correct for the situation described?, 2 hH, 4 hH, (b) The value of x is, 3, 3, (c) The value of x cannot be computed from the information, provided, (d) The question is irrelevant as no water comes out from, the hole, (a) The value of x is 2, , 20°C, , 200°C, A, , (a) 120°C, , B, , C, , (b) 100°C, , D, , (c) 140 °C, , (d) 80°C, , 21 One mole of an ideal gas is taken along the process in, which pV = constant. The graph shown represent the, x, , variation of molar heat capacity of such a gas with respect, to x. The values of c′ and x′, respectively, are given by, c, , 26 Water flows through a horizontal pipe of varying, cross-section at the rate of 20 litres per minute. Then the, velocity of water at a point where diameter is 4 cm, is, (a) 0 . 25 ms− 1 (b) 0 . 26 ms− 1 (c) 0 . 22 ms− 1 (d) 0 . 4 ms− 1, , 27 Oxygen gas having a volume of 1000 cm 3 at 40.0°C and, 1.01 × 105 Pa expands until its volume is 1500 cm 3 and its, pressure is 1.06 × 105 Pa. Find the final temperature of, the sample., , c¢, (3/2)R, x, , x¢, , (a) 197°C, , (b) 220 K, , (c) 300°C, , (d) 300 K, , 28 A mercury drop of radius 1.0 cm is sprayed into 10 6, (a), , 5 5, R,, 2 2, , (b), , 5 5, R,, 2 3, , (c), , 7 7, R,, 2 2, , (d), , 5 7, R,, 2 5, , 22 A wire of length L and radius r is fixed at one end. When, a stretching force F is applied at the free end, the, elongation in the wire is l. When another wire of the same, material but of length 2L and radius 2r, also fixed at one, end is stretched by a force 2F applied at the free end,, then elongation in the second wire will be, (a) l/2, , (b) l, , (c) 2l, , (d) l/4, , 23 A Carnot engine has an efficiency of 22.0%. It operates, between constant-temperature reservoirs differing in, temperature by 75.0°C. What are the temperatures of the, two reservoirs?, , 24 A material has a Poisson’s ratio 0.50. If a uniform rod of it, suffers a longitudinal strain of 2 × 10−3 , then the, percentage change in volume is, (a) 0.6, , (b) 0.4, , (c) 0.2, , (d) zero, , 25 An open vessel full of water is falling freely under gravity., There is a small hole in one face of the vessel, as shown, in the figure., 2h/3, h, Hole, g, , H, , (a) 3.98 × 10−4 J, (c) 3.98 × 10−2 J, , 29 A chef, on finding his stove out of order, decides to boil, the water for his wife’s coffee by shaking it in a thermos, flask. Suppose that he uses tap water at 15°C and that, the water falls 30 cm in each shake, the chef making 30, shakes each minute. Neglecting any loss of thermal, energy by the flask, how long must he shake the flask, until the water reaches 100°C ?, (a) 2.25 × 103 min, (c) 4.03 × 103 min, , P, , (b) 3.97 × 103 min, (d) 5.25 × 103 min, , second by radiation by a sphere 14 cm in diameter at a, temperature of 227°C, when placed in an enclosure at, 27°C. Given, Stefan's constant = 5.7 × 10− 8 Wm − 2K − 4, (a) 45.48 cal/s, (c) 42.5 cal/s, , (b) 40 cal/s, (d) 40.5 cal/s, , 31 Four moles of an ideal gas undergo a reversible isothermal, expansion from volume V1 to volume V 2 = 2V1 at temperature, T = 400 K. Find the entropy change of the gas., (a) 8.22 × 103 J K −1, (b) 8.22 × 102 J K −1, (c) 23.1 J K −1, (d) 10.00 × 103 J K −1, , 32 A swimmer of mass m rests on top of a styrofoam slab,, which has a thickness h and density ρs . The area of the, slab if it floats in water with its upper surface just awash, is (take, density of water to be ρw ), (a), , Ground, , (b) 8.46 × 10−4 J, (d) 8.46 × 10−2 J, , 30 The maximum amount of heat which may be lost per, , (b) 78°C, −5 ° C, (d) 50°C, 0°C, , (a) 58°C, 10°C, (c) 68°C, −7 ° C, , droplets of equal sizes. The energy spent in this process is, [Surface tension of mercury is equal to 32 × 10 −2 Nm −1 ], , m, m, (b), h (ρs + ρw ), hρw, , (c), , m, m, (d), h (ρs − ρw ), h (ρw − ρs )
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198, , DAY SEVENTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 33 An ice-berg of density 900 kg -m − 3 is floating in water of, density 1000 kg -m − 3 . The percentage of volume of, ice-berg outside the water is, (a) 20%, , (b) 35%, , (c) 10%, , (d) 11%, , Direction (Q. Nos. 37-40), , Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices,, only one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below, , 34 A uniform capillary tube of length l and inner radius r with, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , its upper end sealed is submerged vertically into water., The outside pressure is p0 and surface tension of water, is γ. When a length x of the capillary is submerged into, water, it is found that the water level inside and outside, the capillary coincide. The value of x is, (a), , p r, p r, l, l, , , (b) l 1 − 0 (c) l 1 − 0 (d), , , p0r , p0r , 4γ , 2γ , , , 1 +, , 1 +, , , , 4γ , 2γ , , 37 Statement I A ship floats higher in water on a high, pressure day than on a low pressure day., Statement II Floating of ship in the water is possible, because of the buoyant force which is present due to the, pressure difference., , 35 2 moles of an ideal monoatomic gas is carried from a, state ( p0, V0 ) to state ( 2p0, 2V0 ) along a straight line path, in a p-V diagram. The amount of heat absorbed by the, gas in the process is given by, 9, p0V0, 2, 3, (d) p0V0, 2, , (a) 3 p0V0, , 38 Statement I More is the cohesive force, more is the, surface tension., , (b), , (c) 6p0V0, , Statement II More cohesive force leads to more shrinking, of the liquid surface., , 39 Statement I Water expands both when heated or cooled, from 4°C., , 36 The stress along the length of a rod (with rectangular, cross-section) is 1% of the Young’s modulus of its materials., What is the approximate percentage of change of its, volume? (Poisson’s ratio of the material of the rod is 0.3), (a) 3%, (c) 0.7%, , Statement II Density of water is minimum at 4°C., , 40 Statement I If the temperature of a star is doubled, then, the rate of loss of heat from it becomes 16 times., , (b) 1%, (d) 0.4%, , Statement II Specific heat varies with temperature., , ANSWERS, 1., 11., 21., 31., , (c), (c), (b), (c), , 2., 12., 22., 32., , (d), (d), (b), (d), , 3., 13., 23., 33., , (d), (d), (c), (c), , 4., 14., 24., 34., , (b), (b), (b), (d), , 5., 15., 25., 35., , (d), (c), (d), (c), , 6., 16., 26., 36., , (b), (c), (b), (d), , 7., 17., 27., 37., , (d), (a), (a), (d), , 8., 18., 28., 38., , (a), (a), (c), (b), , 9., 19., 29., 39., , (d), (c), (c), (c), , 10., 20., 30., 40., , (b), (c), (a), (b)
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DAY SEVENTEEN, , UNIT TEST 3 (GENERAL PROPERTIES OF MATTER), , 199, , Hints and Explanations, d1 ⇒ d1 < d3 < d2 ., , B = 22, . × 109 Nm − 2, , 2 The forces exerted by liquid on body is, , ∴Compressional strain =, =, , = 136, . × 10, , then Ds 0 + αs Ds 0 ∆T = Db 0 + αb Db 0 ∆T, Ds 0 − Db 0, So, ∆T =, αb Db 0 − αs Ds 0, , −2, , l + l2, given combination K eq = 1, l1, l, + 2, K1 K2, x + 4x, 5, =, = K, x 4x 3, +, K 2K, , So, required force, F = ρV a2 + g 2, , 3 Rate of cooling, Aεσ (T 4 − T 04 ), dθ, =, dt, mc, dθ, A, r2, dθ 1, ∝, ∝, ⇒, ∝, ⇒, dt, V, dt, r, r3, , Rc =, , =, , pV = RT , taking 1 mol of the gas for, simplicity dU = C V dt, where, C V → molar specific heat at, constant volume, Now, the molar specific heat in a, polytropic process, pV n = constant is given by, , 10 Let p1 be the, , …(i), From this equation, we see that C V will, be negative when n < γ and n > 1,, simultaneously, i.e. 1 < n < γ. Since, γ, for all ideal gases is greater than 1, if, n > γ or n < 1, then C V will be positive., , 2 p1 A + mg = 2 pA, ⇒, , 5 Work W = p( V2 − V1 ) at constant p, , R, , As, the number of moles of air, increases. Radius increases and ∆p, decreases., , 7 Heat required to raise the temperature, of water must be equal to the power, output of the heater P, multiplied by, time t., Q cm(T f − T i ), =, t =, P, P, (4190) (0.100) (100° − 23° ), =, = 161 s, 200, , 14 Applying the Bernoulli’s theorem for, any two convenient points, let us say, we are applying just at the water surface, and just inside the hole., ρv 2, p0 + 0 + 0 = p0 +, + ρg ( − h ), 2, where, v is required speed and water, surface is taken as the reference level., ⇒, v = 2 gh = 2 × 9.8 × 10, = 14 ms −1, ∆l, 15 Here, σ = 03, .,, = 4.5 × 10− 3, , l, ∆R, ∆l, ∆R / R, =−σ, σ =−, ⇒, R, l, ∆l / l, ∆R, −3, = − 03, . × (4.5 × 10, ⇒, R, = − 135, . × 10− 3, , mg, , Volume, V = πR2 l, ∆V, 2∆R, ∆l, =, +, V, R, l, , mg, 2A, , Percentage change in volume =, , 1.6 × 104 × 10, 2 × 40, , 2∆R, ∆l , = , +, × 100, R, l , , = 6.8 × 104 Pa, , BT22 ) / p, , 6 ∆p = 2T, , p1 = p −, , = 7 × 104 −, , Initial volume, V1 = ( AT1 − BT12 ) / p, , ⇒ W = A(T2 − T1 ) − B (T22 − T12 ), , The temperature is, T = 25° C + 335° C = 360° C, , 2p1A, , pressure on the 2pA, upper wing, surface, then for, the vertical, equilibrium of the, plane, , (n − γ )R, R R , =, =, −, γ − 1 n − 1 (n − 1)(γ − 1), , 11, , Rate of flow of heat per unit area, = Thermal conductivity × Temperature, gradient, Temperature gradient (X), 1, ∝, Thermal conductivity (K), As , dQ /dt = constant, , , , , A, As, K c > K m > K g , therefore, Xc < X m < X g, , 12 Efficiency of Carnot’s heat engine, , ∆V, × 100, V, , = [2 × (− 135, . ) + 4.5] × 10− 3 × 100, , dQ /dt, ∆θ, = K , ∆x , A, , = 100%, , 3.000 − 2.992, (19 × 10−6 )(2.992) − (11 × 10−6 )(3.00), , = 335° C, , Hence, rate of flow of heat through the, given combination is, θ K eq A(T2 − T1 ) 5/ 3KA(T2 − T1 ), =, =, t, ( x + 4 x), 5x, 1 / 3KA(T2 − T1 ), =, x, On comparing it with given equations,, 1, we get f = ., 3, , 4 Since, pV n = constant and also, , Final volume is V2 = ( AT2 −, , 22, . × 109, , T2 = 0 K = − 27316, . °C, , 13 If Ds = Db ,, , p, ∆V, =, V, B, , 9 Equation of thermal conductivity of the, , Forces exerted by liquid, , CV, , 3 × 107, , T1 − T2 = T1, , ⇒, , = 3 × 107 Nm − 2, , V, , Vρg, , ⇒, , p = hρg = 3000 × 103 × 10, , shown in figure, when body is in, equilibrium w.r.t. fluid., Vρa, , T1 − T2, × 100 = 100, T1, , 8 Here, h = 3000 m,, , 1 d 3 floats in d 2 and sinks in, , = 1.8 × 10− 3 × 100 = 018, . %, , 16 From the ideal gas law, pV = nRT, pV (100)(1.0 × 10−6 ), =, RT, (8.31)(220), = 5.47 × 10−8 mol, , n=, , 17, , Temperature of the interface,, K θ + K 2 θ2, θ= 1 1, K1 + K2, , , K1 1, =, ⇒ If K 1 = K , then K 2 = 4 K , Q, K2 4, , , ⇒, , θ=, , K × 0 + 4 K × 100, = 80°C, 5K
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200, , DAY SEVENTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 18 Y = Stress =, , x, x, = = x, 2l − l, 1, l, In actual, the above expression is not, exact for this much elongation., Strain, , 19 For constant pressure process,, , Q 1 = nC p ∆T = 743 J, For constant volume process,, Q 2 = nC V dT = n (C p − R ) dT, = nC pdT − nRdT, Q 2 = 743 − 5 × 8.3 × 2 = 660 J, , 20 Let the thermal resistance of each rod, be R., The two resistances connected along, two paths from B to C are equivalent to, 2 R each and their parallel combination, is R., Effective thermal resistance between B, and D = 2R, R, , R, R, A, , R, B, R, , 200°C, , R, A, , D, , C, R, q, B, , 2R, D, , 20°C, , Temperature of interface,, R1θ2 + R2θ1, R1 + R2, R × 20 + 2R × 200 420, θ=, =, = 140° C, R + 2R, 3, 21 At x = ∞,C = 3 R, from pV x = constant, 2, ⇒ p1 / x V = another constant, so at, θ=, , x = ∞, V = constant, 3, hence C = C v = R and then, 2, 5, C p = CV + R = R, 2, at x = 0, P = constant and C = C ′, 5, hence, C ′ = C p = R, 2, at x = x ′ ,C = 0, so the process is, adiabatic,hence, Cp, 5, x′ =, =, CV 3, F/A, 22 Y =, ∆ l /l, For the first wire, Y =, , F ×L, , πr 2 × l, 2F × 2L, For the second wire, Y =, π(2r )2 × l ′, From the above two equations, l ′ = l, , 23 For an ideal engine, the efficiency is, related to the reservoir temperatures by, , ε = (T H − TC )/ T H . Thus,, T H = (T H − TC )/ ε, , 28 Let r be the radius of one droplet., Now,, , = (75 K) / (0.22) = 341 K, = 68° C ., The temperature of the cold, reservoir is, TC = T H − 75 = 341 K − 75K, , A, , = 266 K = − 7° C., dV, dl, 24, = (1 + 2σ ), V, l, = 2 × 2 × 10−3 = 4 × 10−3, , 25 As vessel is falling freely under, , 2, ρv outside, + ρgH, 2, v inside = 0, p inside = poutside = p, [atmospheric pressure], Therefore,, v outside = 0, , = poutside +, , i.e. no water comes out from the hole., 20 × 1000, , m 3 s− 1, , 1, × 10− 3 m 3 s− 1, 3, 4, Radius, r = = 2 cm = 0.02 m, 2, Area of cross-section,, 22, a = πr 2 =, × (0.02)2 m 2, 7, Let v be the velocity of the flow of, water at the given point, then, V = av, 1, 22, × (0.02)2 × v, ⇒ × 10− 3 =, 3, 7, 7 × 10− 3, ⇒, v =, 3 × 22 × (0.02)2, =, , ~, = 02639, ., − 026, . ms− 1, 1.06 × 105 × 1000 × 10–6, 27 n = PV =, 831, . × 313, , RT, , = 4.07 × 10–2, Using pV = nRT, pV (1.06 × 105 ) (1500 × 10−6 ), T =, =, nR, (4.07 × 10−2 )(8.31), = 470 K, = 197° C, , = 106 × 4 πr 2, , Increase in surface energy, = S∆ A = 32 × 10−2 × 4 π × 99 × 10− 4 J, = 3.98 × 10−2 J, The increase in surface energy is at the, expense of internal energy, so energy, spent = 3.98 × 10−2 J, , 29 Time taken,, , gravity, the pressure at all points, within the liquid remains the same as, the atmospheric pressure. If we apply, Bernoulli’s theorem just inside and, outside the hole, then, 2, ρv inside, p inside +, + ρg H, 2, , 60 × (100)3, , f, , Change in area,, ∆A = A f − A i = 4 π × 99 × 10−4 m2, , Q σ = 0.5 = 1 , 2 , , ∴Percentage change in volume, = 4 × 10−1 = 0.4%, , 26 V = 20 litres/min =, , 4 3, 4, R, πR = 106 × πr 3 , r =, 3, 3, 100, 1, =, cm = 10−4 m, 100, A i = 4 πR 2, , cm(T f − T i ) c (T f − T i ), Q, =, =, Rmgh, Rmgh, Rgh, (4190) (100 − 15), =, (30)(9.8)(0.30), , t =, , = 4.03 × 103 min, , 30 Temperature of sphere, T = 227° C = 500 K, Temperature of surroundings, T 0 = 27° C = 300 K, Radius, r = 7 cm = 0.07 m, Area of sphere,, 22, A = 4 πr 2 = 4 ×, × (0.07)2, 7, = 616, . × 10− 2 m2, The energy lost from A,, AE = Aσ(T 4 − T 04 ), 616, . × 10−2 × 5.7 × 10− 8 × [(500)4 − ( 300)4 ], = 616, . × 5.7 × 10− 10 × 1004 (54 − 34 ), = 616, . × 5.7 × 10− 2 × 544 Js− 1, The heat lost per sec., 616, . × 5.7 × 10− 2 × 544, AE, H =, =, J, 42, ., = 45⋅ 48 cal / s, , 31 We have p = nR T / V . The work done by, the gas during the isothermal expansion, is, V2, V2 dV, V, W = ∫ pdV = nRT ∫, = nRT ln 2, V1, V1 V, V1, Substituting V2 = 2V1 to obtain W = nRT, = (4.00)(8.314)(400) ln 2, = 9.22 × 103 J, Since, the expansion is isothermal,, ∆E int = 0 and Q = W, Thus,, , ∆S =, =, , W, T, 9.22 × 103 J, = 23.1 J K −1, 400 K
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DAY SEVENTEEN, , 32 From equilibrium,, , mg + Ahρs × g = Ahρw × g, where, A is the required cross-sectional, area, m, ⇒, A=, h ( ρw − ρs ), , 33 Let volume of ice-berg is V and its, , density is ρ. If V in is volume inside the, water, then, V in σg = Vρg, where, σ = density of water, ρ, ⇒ V in = V, σ, σ − ρ, ⇒ V out = V − V in = , V, σ , 1000 − 900 , = , V, , , 1000, V, =, 10, V out, = 0.1 = 10%., ⇒, V, , 34 The pressure inside tube changes when, it is submerged in water. Thus, p1V1 = p2V2, p 0 (lA ) = p ′ (l − x )A, p0l, p′ =, ∴, l − x, As level of water is same inside and, outside of capillary tube, 2γ, p′ − p0 =, ∴, r, p0 l, 2γ, or, − p0 =, l − x, r, , UNIT TEST 3 (GENERAL PROPERTIES OF MATTER), , ⇒, , l, , x=, 1+, , p0r, 2γ, , 35 The internal energy, ∆U = nC V ∆T, C V = specific heat of gas at constant, volume, p V, 3R 4 p 0V 0, − 0 0 , ∆U = n ⋅, ⇒, , nR , 2 nR, 3R 3 p 0V 0, 9, = n⋅, ⋅, = p 0V 0, 2, nR, 2, Work done by the gas,, V, 3 p 0V 0, W = (2 p 0 + p 0 ) 0 =, 2, 2, From first law of thermodynamics,, ∆Q = dW + dU, 3 p 0V 0, 9, =, + p 0V 0, 2, 2, 12 p 0V 0, =, = 6 p 0V 0, 2, 36 Stress = F = 1% of Y = Y, ∆A, 100, Y, Stress F / ∆A, Also,, Y =, =, = 100, Strain, ∆l / l, ∆l / l, 1, ∆l, =, ⇒, 100, l, − ∆r / r, Poisson’s ratio, σ =, ∆l / l, ∆r, ∆l, 0.3, =−σ, =−, ⇒, r, l, 100, ∆V, ∆l , 2∆r, , × 100 = , +, ∴, × 100, r, V, l , , 201, , − 0.3, 1 , = 2 ×, +, × 100 = 0.4%, , 100, 100 , , 37 A body of weight w = mg = Vρg float in, a liquid as a upthrust F = Vσg acts, vertically upwards through the centre of, gravity of displaced liquid also called, the centre of buoyancy. It is, independence of atmospheric pressure., 38 Surface tension, S = Force = F, Length, l, ∴S ∝ F, Thus, more the force, the surface, tension is more. Also, this force tends to, have the least possible surface area., , 39 At 4°C, the volume of water is, minimum. When it is cooled below 4°C, or heated above 4°C, then it expands or, its volume increases. As volume at 4°C, is minimum, thus its density, = mass will be maximum., , , volume , , 40 From Stefan’s law, E = σT 4 or E ∝ T 4, 4, , 4, , ∴, , E 1 T1 , = , E 2 T2 , , or, , E1, 1, =, E2 16, , or, , E2 = 16E1 = 16 times, , T , = , , 2T , , Specific heat too varies with, temperature. As a matter of fact, specific, heat is zero at 0K for all the materials.
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DAY EIGHTEEN, , Electrostatics, Learning & Revision for the Day, u, , u, , u, , u, , u, , Electric Charge, Coulomb’s Law of Forces, between Two Point Charges, Superposition Principle, Electric Field, Motion of A Charged Particle, in An Electric Field, , u, , u, , u, , u, , u, , Electric Field due to, a Point Charge, , u, , u, , Continuous Charge Distribution, Electric Dipole, Electric Flux (φE ), Gauss Law, , u, , u, , u, , u, , Electric Potential, Electric Potential Energy, Equipotential Surface, Conductors and Insulators, Electrical Capacitance, Capacitor, , If the charge in a body does not move, then the fricitional electricity is known as static, electricity. The branch of physics which deals with static electricity is called, electrostatics., , Electric Charge, Electric charge is the property associated with matter due to which it produces and, experiences electric and magnetic effects., , Conservation of Charge, We can neither create nor destroy electric charge. The charge can simply be transferred, from one body to another. There are three modes of charge transfer:, (a) By friction, , (b) By conduction, , (c) By induction, , Quantisation of Charge, Electric charge is quantised. The minimum amount of charge, which may reside, independently is the electronic charge e having a value of 1.6 × 10 –19 C, i.e. Q = ± ne,, where, n is any integer., Important properties of charges are listed below, l, , l, , l, , l, , l, , Like charges repel while opposite charges attract each other., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, , Charge is invariant i.e. charge does not change with change in velocity., According to theory of relativity, the mass, time and length change with a change in, velocity but charge does not change., A charged body attracts a lighter neutral body., Electronic charge is additive, i.e. the total charge on a body is the algebraic sum of all, the charges present in different parts of the body. For example, if a body has different, charges as + 2q, + 4q, − 3 q, − q, then the total charge on the body is + 2 q., , (Without referring Explanations), u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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ELECTROSTATICS, , DAY EIGHTEEN, , Coulomb’s Law of Forces between, Two Point Charges, l, , If q 1 and q2 be two stationary point charges in free space, separated by a distance r, then the force of attraction /, repulsion between them is, F =, =, , l, , , 1 , K = 4πε , , 0, , K |q 1||q2|, 1 |q 1||q2|, =, ⋅, 4πε 0, r2, r2, , 9 × 10 9 × | q 1|| q2|, r2, , [K = 9 × 10 9 N-m2 /c2 ], , q 1q2, 1, 1 q 1q2, =, 4πε 0ε r r 2, 4πε r 2, q q, 1, =, ⋅ 1 2, 4πkε 0 r 2, , Fm =, , Electric field intensity (E) is a vector quantity., The direction of electric field is same as that of force acting on, the positive test charge. Unit of E is NC−1 or Vm −1., , Electric Field Lines, An electric field line in an electric field is a smooth curve,, tangent to which, at any point, gives the direction of the, electric field at that point., Properties of electric field lines are given below, l, , If some dielectric medium is completely filled between the, given charges, then the Coulomb’s force between them, becomes, ε, , Q ε = ε r or k , , , 0, , Forces between Multiple Charges, When a number of point charges are present in a region then, force acting between any two point charges remains, unaffected by the presence of other charges and remains same, as according to Coulomb’s law. If four identical charges of, magnitude q each are placed at the four corners of square of, side a, then the force on any one charge due to the rest of the, three charges is, 1 q2, F =, (2 2 + 1), 4πε 0 a2, , l, , l, , l, , Electric field lines come out of a positive charge and go into, the negative charge., No two electric field lines intersect each other., Electric field lines are continuous but they never form a, closed loop., Electric field lines cannot exist inside a conductor. Electric, shielding is based on this property., , Motion of a Charged Particle in, an Electric Field, Let a charged particle of mass m and charge q, enters the, electric field along X -axis with speed u. The electric field E, is along Y-axis is given by, F y = qE, and force along X -axis remains zero, i.e., Fx = 0, E, , Y, , P (x, y), , Superposition Principle, It states that, the net force on any one charge is equal to the, vector sum of the forces exerted on it by all other charges. If, there are four charges q 1, q2 , q3 and q 4, then the force on q 1, (say) due to q2 , q3 and q 4 is given by F1 = F12 + F13 + F14,, where F12 is the force on q 1 due to q2 , F13 that due to q3 and, F14 that due to q 4., , Electric Field, The space surrounding an electric charge q in which another, charge q 0 experiences a force of attraction or repulsion, is, called the electric field of charge q. The charge q is called the, source charge and the charge q 0 is called the test charge. The, test charge must be negligibly small so that it does not modify, the electric field of the source charge., , Intensity (or Strength) of Electric Field (E), The intensity of electric field at a point in an electric field is, the ratio of the forces acting on the test charge placed at that, point to the magnitude of the test charge., F, where, F is the force acting on q 0., E= ,, q0, , 203, , O, , u, , X, , ∴Acceleration of the particle along Y-axis is given by, Fy, qE, ay =, =, m, m, The initial velocity is zero along Y-axis (u y = 0)., ∴The deflection of charged particle along Y-axis after time t, 1, is given by y = u y t + a y t 2, 2, qE 2, =, t, 2m, Along X -axis there is no acceleration, so the distance, covered by particle in time t along X -axis is given by x = ut, Eliminating t , we have, qE, , y=, x2 , 2 mu2, , y ∝ x2, This shows that the path of, perpendicular field is a parabola., , charged, , particle, , in
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204, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 4. Electric Field due to a Uniformly Charged, Infinite Plane Sheet, , Electric Field due to, a Point Charge, 1. Electric Field due to a Point Charge at a, Distance, Electric field at a distance r from a point charge q is, q, 1, E =, ⋅, 4πε 0 r 2, l, , l, , If q 1 and q2 are two like point charges, separated by a distance, r, a neutral point between them is obtained at a point distant, r1 from q 1, such that, r, r1 =, , q2 , 1 +, , q1 , , If q 1 and q2 are two charges of opposite nature separated by, a distance r, a neutral point is obtained in the extended line, joining them, at a distance r1 from q 1, such that,, r, r1 =, q2, , − 1, , q, 1, , , 2. Electric Field due to Infinitely Long Uniformly, Charged Straight Wire, Electric field at a point situated at a normal distance r,, from an infinitely long uniformly charged straight wire, having a linear charge density λ, is, λ, E =, 2 πε 0 r, , 3. Electric Field due to a Charged Cylinder, For a conducting charged cylinder of linear charge density, λ and radius R, the electric field is given by, λ, , for r > R,, E =, 2 πε 0 r, E =, and, , E = 0, for r < R, , For a non-conducting charged cylinder, for r ≤ R,, λr, λ, and E =, , for r > R, E =, 2 πε 0 r, 2 πε 0R2, Emax, , r>R, , r, ∝, , r>R, , O, , r, r=R, (a) Variation of electric field, with distance for conducting, cylinder, , O, , 5. Electric Field due to a Uniformly Charged, Thin Spherical Shell, For a charged conducting sphere/ E, shell of radius R and total charge, Q, the electric field is given by, Case I E = 0, for r < R, Q, Case II E =, , for r = R, 4πε 0R 2, Case III, , E =, , r<R, , O, , R, Variation of electric field, with distance for uniformly, charge spherical shell, , Q, , for r > R, 4πε 0r 2, , Continuous Charge Distribution, The continuous charge distribution may be one dimensional,, two dimensional and three-dimensional., 1. Linear charge density (λ ) If charge is distributed along a, line, i. e., straight or curve is called linear charge, distribution. The uniform charge distribution q over a, length L of the straight rod., q, Then, the linear charge density, λ =, L, Its unit is coulomb metre −1 (Cm–1 )., 2. Surface charge density (σ) If charge is distributed over a, surface is called surface charge density, i. e.,, σ =q/A, Its unit is coulomb m, , –2, , –2, , (Cm ), , 3. Volume charge density (ρ) If charge is distributed over, the volume of an object, is called volume charge density,, q, i. e., p = . Its unit is coulomb metre −3 (Cm–3 )., V, , Electric Dipole, An electric dipole consists of two equal and opposite charges, separated by a small distance., , Emax, , E, , R, , E, , E, , l, , λ, , for r = R, 2 πε 0 R, , r<, , l, , Electric field near a uniformly charged infinite plane sheet, having surface charge density σ is given by, σ, E=, 2 ε0, , E ∝ 12, r, , r, r=R, (b) Variation of electric field, with distance for, non-conducting cylinder, , p, , A, –q, , B, +q, , 2a, , The dipole moment of a dipole is defined as the product, of the magnitude of either charges and the distance, between them. Therefore, dipole moment, p = q(2a)
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ELECTROSTATICS, , DAY EIGHTEEN, , Electric Field due to a Dipole, l, , Work Done in Rotating a Dipole, , At a point distant r from the centre of a dipole, along its, 1, 2 pr, axial line E =, ⋅, 4πε 0 (r 2 − a2 )2, , If an electric dipole initially kept in an uniform electric field E,, making an angle θ 1, is rotated so as to finally subtend an angle, θ2 , then the work done for rotating the dipole is,, W = pE (cos θ 1 − cos θ2 ), , [direction of E is the same as that of p], For a short dipole,, E=, l, , 1 2p, ⋅, 4πε 0 r 3, , [r > > a], , At a point distant r from the centre of a dipole, along its, equatorial line, 1, p, E=, ⋅, 4πε 0 (r 2 + a2 )3 / 2, [direction of E is opposite to that of p], 1, p, For a short dipole E =, ⋅, [r > > a], 4πε 0 r 3, , l, , At a point distant r from the centre of a short dipole, along, a line inclined at an angle θ with the dipole axis, 1, p, E =, ⋅, 3 cos2 θ + 1, 4πε 0 r 3, E, , Er, β, , l, , Hence, U = − pE cos θ, , +q, , E subtends an angle β from r such that, 1, tan β = tan θ, 2, , Electric Flux (φE ), It is a measure of the flow of, electric field through a surface. It, can be defined as the total, number of lines of electric field, passes through a surface placed, perpendicular to direction of, field., , When a dipole is placed in an external electric field, making, an angle θ with the direction of the uniform electric field E, it, experiences a torque given by, τ = qE × AC, τ=p×E, , ∆s, , n, , φ E = ∫ EdS cos θ = ∫ E ⋅ dS = ∫ E ⋅ n$ dS, , ds, E, θ, , where, Q enclosed Σ q is the algebraic sum of all the charges, i, i= 1, , inside the closed surface., , Electric Potential, The amount of work done in bringing a unit positive charge,, without any acceleration, from infinity to that point, along, any arbitrary path., V =, , τ = pE sin θ, or qE × 2 d sin θ = (q × 2 d) E sin θ, , W, q0, , Electric potential is a state function and does not depend on, the path followed., A, +q, , qE –, q B, , E, , i= n, , Torque on a Dipole in a Uniform, Electric Field, , 2d, , or U = − p ⋅ E, , The total electric flux linked with a, 1, closed surface is equal to, times, the, ε0, net charge enclosed by that surface., Thus,, 1, φ E = ∫ E ⋅ dS =, [Qenclosed ], S, ε0, , θ, O, 2d, , It is the amount of work done in rotating an electric dipole, from a direction perpendicular to electric field to a particular, direction., , Gauss’s Law, , z, , –q, , Potential Energy of a Dipole, , i.e., , P (r, θ), , EQ, , 205, , qE, E, , θ, , C, , 1. Electric Potential Due to a Point Charge, Potential due to a point charge Q, at a distance r is given, 1 Q, by, V =, ⋅, 4πε 0 r
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206, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 2. Electric Potential Due to a System of Charges, If a number of charges q 1, q2 , q3 ,… are present in space,, then the electric potential at any point will be, V = V1 + V2 + V3 +…, =, , , 1 q 1 q2 q3, 1, +, +, +… =, 4πε 0 r1, r2, r3, π, ε0, 4, , , n, , ∑, , i= 1, , qi , , ri , , 3. Electric Potential Due to an Electric Dipole, 1 p cos θ, 4πε 0, r2, 1, p, On the dipole axis, θ = 0 ° and V =, ⋅, 4πε 0 r 2, , At any general point, V =, , 4. Electric Potential due to Some Common, Charge Distributions, , r, r=R, For a charged conducting, sphere/shell having total charge Q and radius R, the, potential at a point distant r from the centre of the, sphere/shell is, Q, 1 Q, (i) V =, , for r > R, (ii) V =, , for r = R, 4πε 0r, 4πε 0 R, Q, , for r ≤ R, (iii) V =, 4πε 0R, , For a charged non-conducting (dielectric) sphere of radius R,, the charge Q is uniformly distributed over the entire volume., Q, Hence, (i) V =, , for r > R, 4πε 0r, Q, , for r = R, (ii) V =, 4πε 0R, Q 3 R2 − r 2 , and (iii) V =, , for r < R, , 4πε 0 2 R3 , At the centre of the sphere (r = 0), , Q , Vs = 4πε R , , 0 , , 3Q, 3, = Vs, 8 πε 0 R 2, V, , r<, , R, , r=R, , Relation between E and V, Because E is force per unit charge and V is work per unit, charge. E and V are related in the same way as work and force., , P, , infinity (reference point) to the given point VP = − ∫ E ⋅ dr volt, ∞, , where, the negative sign indicates that the work is done, against the field., , Equipotential Surface, Equipotential surface is an imaginary, surface joining the points of same, potential in an electric field. So, we, can say that the potential difference, between any two points on an, equipotential surface is zero., , +q, , The electric lines of force at each, point of an equipotential surface are, normal to the surface. Figure shows, the electric lines of force due to point charge +q. The spherical, surface will be the equipotential surface and the electrical, lines of force emanating from the point charge will be radial, and normal to the spherical surface., Regarding equipotential surface, following points are worth, noting, (i) Equipotential surface may be planar, solid etc. But, equipotential surface can never be point size., (ii) Equipotential surface is single valued. So, equipotential, surfaces never cross each other., (iii) Electric field is always perpendicular to equipotential, surface., , r>R, , O, , In general, for a system of n charges, the electric potential, energy is given by, q iq j, 1, ,i ≠ j, U= Σ, 2 4πε 0rij, , Work done against the field to take a unit positive charge from, r>R, , O, , V =, , The electric energy of a system of charges is the work that has, been done in bringing those charges from infinity to near each, other to form the system. For two point charges q 1 and q2, separated by distance r12 , the potential energy is given by, 1 q 1q2, ., U=, 4πε 0 r12, , , 1, 2 is used as each term in summation will appear twice , , On the equitorial axis, θ = 90 ° and V = 0, , Potential at a point distant r V r < R, from an infinitely long wire, having linear charge density λ,, is, λ, V =, ⋅ ln r, 2 πε 0, , Electric Potential Energy, , (iv) Work done to move a point charge q between two points, on equipotential surface is zero., r, , (v) The surface of a conductor in equilibrium is an, equipotential.
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ELECTROSTATICS, , DAY EIGHTEEN, , Conductors and Insulators, , C1, C2, C3, +Q + – –Q +Q + – –Q +Q + – –Q, , Conductors are those materials through which electricity, can pass through easily. e.g. metals like copper, silver, iron, etc. Insulators are those materials through which electricity, cannot pass through, e.g. rubber, ebonite, mica etc., , + –, + –, + –, , + –, + –, + –, , V1, , V2, , Dielectrics and Polarisation, , The equivalent capacitance Cs is given by, i= n, 1, 1, 1, 1, 1, =, +, +, +… = Σ, i = 1 Ci, Cs C1 C2 C3, , e.g. mica, glass, water etc., When a dielectric is placed in an external electric field, so, the molecules of dielectric gain a permanent electric dipole, moment. This process is called polarisation., , 2. Parallel Grouping, In a parallel arrangement, Q = Q1 + Q2 + Q3 +…, Q1 : Q2 : Q3… = C1 : C2 : C3…, , and, , Electrical Capacitance, , C1, +Q1 + – –Q1, , Capacitance of a conductor is the amount of charge needed in, order to raise the potential of the conductor by unity., Q, Mathematically, Capacitance C =, V, , + –, + –, + –, , C2, +Q2 + – –Q2, + –, + –, + –, , Sharing of Charges, , l, , V3, , V, , Dielectrics are insulating materials which transmit electric, effect without actually conducting electricity., , l, , + –, + –, + –, , –, , +, , l, , 207, , +Q3 C3 –Q3, , Let us have two charged conductors having charges, Q1 and Q2 (or potentials V1, V2 and capacitances C1, C2, respectively). If these are joined together. In such a cases, Q + Q2 C1V1 + C2V2, Common potential, V = 1, =, C1 + C2, C1 + C2, During sharing of charges, there is some loss of, electrostatic energy, which in turn reappears as heat or, light. The loss of electrostatic energy, C1C2, ∆U = Ui − U f =, (V1 − V2 )2, 2(C1 + C2 ), When charges are shared between any two bodies, their, potential become equal. The charges acquired are in the, ratio of their capacitances., , +, +, +, +, , –, –, –, –, , +, , –, , V, , The equivalent capacitance is given by, i= n, , C p = C1 + C2 + C3 + K = Σ Ci, i= 1, , Capacitance of a Parallel Plate, Capacitor, 1. Capacitor without Dielectric Medium, between the Plates, , Capacitor, A capacitor is a device which stores +Q, electrostatic energy. It consists of, conductors of any shape and size, carrying, charges, of, equal, magnitudes and opposite signs and, separated, by, an, insulating, medium., , A, , If the magnitude of charge on each plate of a parallel plate, capacitor be Q and the overlapping area of plates be A,, then, , B, –Q, l, , d, , Electric field between the plates,, σ, Q, E =, =, ε0 ε0 A, , There are two types of combination, of capacitors:, , Potential difference between the plates, σd Qd, , where d = separation between the two, V = E ⋅d =, =, ε0 ε0 A, , 1. Series Grouping, , plates., , In a series arrangement, V = V1 + V2 + V3 +…, and, , V1 : V2 : V3… =, , 1 1 1, :, :, :…, C1 C2 C3, , l, , l, , Capacitance, C =, , Q ε0 A, =, V, d
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208, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 2. Capacitor with Dielectric Medium between, the Plates, l, , l, , l, , If a dielectric medium of dielectric constant K is, completely filled between the plates of a capacitor, then its, capacitance becomes,, Kε A ′, ε 0 A′ , , C′ = 0 = KC0, where, C0 = d , d, If a dielectric slab/sheet of thickness t (where, t < d) is, introduced between the plates of the capacitor, then, ε0 A, C′ =, t , , d − t + , , K, Magnitude of the attractive force between the plates of a, parallel plate capacitor is given by, F =, , l, , σ2 A, Q2, CV 2, =, =, 2 ε 0 2 Aε 0, 2d, , The energy density between the plates of a capacitor, U, 1, u=, = ε0 E2, Volume 2, , Energy Stored in a Capacitor, If a capacitor of capacity C is charged to a potential V , the, electrostatic energy stored in it is,, 1, U = CV 2, 2, 1, = QV, 2, 1 Q2, =, 2 C, , Energy Loss During Parallel, Combination, When two capacitor of C1 capacitance charge to potential V1 ,, whereas another of C2 charge to potential of V2 , then after, parallel combination., 1 C1C2, Loss in energy =, (V1 − V2 )2, 2 C1 + C2, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 Two balls of same mass and carrying equal charge are, hung from a fixed support of length l. At electrostatic, equilibrium, assuming that angles made by each thread, is small, the separation, X between the balls is, ª JEE Main (Online) 2013, proportional, (a) l, , (b) l 2, , (c) l 2 / 3, , look like figure given in (figures are schematic and not, ª JEE Main 2015, drawn to scale), , (a), , (d) l1/ 3, , ++++, +, +, –– ––, –– ––, , +++, ++ +, ––, –, –– –––, , (b), , 2 In the basic CsCl crystal structure, + Cs and − Cl ions are, arranged in a bcc configuration as shown below. The, net electrostatic force exerted by the 8 Cs + on the the, Cl − ion is, Cs+, , Cs+, , Cs+, Cs+, Cs+, , (a), , Cs+, , 1 4e2, 1 16e 2, 1 32a 2, (b), (c), 2, 2, 4π ε0 3 a, 4π ε0 3 a, 4π ε0 3 a 2, , ++++, +, –– ––+, –– ––, , ++++, +, –– –+, –, –– ––, , (d), , 4 A positively charged particle P enters the region between, , Cs+, Cl –, , Cs+, , (c), , (d) zero, , 3 A long cylindrical shell carries positive surface charge σ, , in the upper half and negative surface charge − σ in the, lower half. The electric field lines around the cylinder will, , two parallel plates with a, velocity u, in a direction, parallel to the plates. There is, a uniform electric field in this, region. P emerges from this, region with a velocity v., Taking C as a constant, v will, depend on u as, (a) v = Cu, (c) v = u 2 +, , –, , P, , u, , E, , +, , (b) v = u 2 + Cu, C, u, , (d) v = u 2 +, , C, u2
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ELECTROSTATICS, , DAY EIGHTEEN, 5 An infinite line charge produces a field of 9 × 104 N / C at, a distance of 2 cm. Calculate the linear charge density., −3, , −4, , −5, , (a) 10 C / m (b) 10 C / m (c) 10 C / m, , −7, , (d) 10 C / m, , 12 Consider a finite insulated, uncharged conductor placed, near a finite positively charged conductor. The, uncharged body must have a potential, ª JEE Main (Online) 2013, , 6 In a uniformly charged sphere of total charge Q and, radius R, the electric field E is plotted as function of, distance from the centre. The graph which would, correspond to the above will be, ª AIEEE 2012, E, , E, , (b), , (a), R, , R, , r, , is given by φ = ar 2 + b, where r is the distance from the, centre where a, b are constants. Then the charge density, inside the ball is, ª AIEEE 2011, (a) − 6a ε0r, (c) − 6a ε0, , (d), , (c), R, , R, , r, , (a) zero, , (b), , 4 πε0r12, , (c), , Qr12, 4 πε0R 4, , (d), , Qr12, , 10 V and − 4 V, respectively. The work done in moving, 100 electrons from P to Q is, ª AIEEE 2009, , 15 The variation of electric potential with distance from a, fixed point is shown in the figure. What is the value of, electric field at x = 2 m?, , 3 πε0R 4, , Y, , 1 4P3 P2, 4 πε0 x 3, 1 6P1 P2, (c), 4 πε0 x 3, , 6, V (volt), , distance x from each other an P1 | | P2. The force between, ª JEE Main (Online) 2013, the dipoles is, 1 3 P3 P2, 4 πε0 x 3, 1 3 P1 P2, (d), 4 πε0 x 3, , 4, , (b), , 2, O, , 9 If the electric flux entering and leaving an enclosed, surface respectively are φ1 and φ 2, the electric charge, inside the surface will be, (a), , φ 2 − φ1, ε0, , (b), , φ1 + φ 2, ε0, , (c), , φ1 − φ 2, ε0, , (d) ε0 (φ1 + φ 2 ), , 10 A cylinder of radius R and length L is placed in a uniform, electric field E parallel to the axis of the cylinder, the total, electric flux for the surface of the cylinder is, (a) 2 πR 2E, , (b), , πR 2, E, , (c), , πR 2 + πR 2, (d) zero, E, , 11 A large insulated sphere of radius r, charged with Q units, of electricity, is placed in contact with a small insulated, uncharged sphere of radius r′ and is then separated. The, charge on the smaller sphere will now be, (a) Q (r + r ′), Q, (c), r′ + r, , (b) 9.60 × 10−17 J, (d) 2.24 × 10−16 J, , (a) −19 × 10−17 J, (c) −2.24 × 10−16 J, , 8 Two points dipoles of dipole moment P1 and P2 are at a, , (a), , (b) − 24 πaε0, (d) − 24 πaε0r, , 14 Two points P and Q are maintained at the potentials of, , r, , Q, 7 Let ρ(r ) =, r be the charge density distribution for a, πR 4, solid sphere of radius R and total charge Q. For a point P, inside the sphere at distance r1 from the centre of the, sphere, the magnitude of electric field is, ª AIEEE 2009, Q, , (a) less than the charged conductor and more than at, infinity, (b) more than the charged conductor and less than at, infinity, (c) more than the charged conductor and more than at, infinity, (d) less than the charged conductor and less than at infinity, , 13 The electrostatic potential inside a charged spherical ball, , r, , E, , E, , 209, , (b) Q (r − r ′), Qr ′, (d), r′ + r, , (a) Zero, (c) 6/1, , 1, , 2, , 3, x (m), , 4, , X, , (b) 6/2, (d) 6/3, , 16 The electric potentialV at any point ( x , y , z ) in space is, given by V = 4x 2 volt. The electric field at (1, 0, 2) m in, Vm −1 is, (a) 8, along the positive x-axis, (b) 8, along the negative x-axis, (c) 16, along the x-axis, (d) 16, along the z-axis, , 17 Assume that an electric field E = 30x 2 i exists in space., Then, the potential differenceVA − VO , where VO is the, potential at the origin andVA the potential at x = 2 m/s, ª JEE Main 2014, , (a) 120 J/C, (c) −80 J/C, , (b) −120 J/C, (d) 80 J/C
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210, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 18 Figure shows some equipotential lines distributed in space., A charged object is move from point A to point B., , A, , A, , B, , A, , B, , 23 A parallel plate capacitor is made of two circular plates, separated by a distance of 5 mm and with a dielectric of, dielectric constant 2.2 between them. When the electric, field in the dielectric is 3 × 104 V/m, the charge density of, the positive plate will be close to, ª JEE Main 2014, , B, , (a) 6 × 10−7 C/m 2, (c) 3 × 104 C/m 2, , (b) 3 × 10−7 C/m 2, (d) 6 × 104 C/m 2, , 24 A parallel plate capacitor of capacitance 90 pF is, 10V 20V 30V 40V 50V, Fig. (i), , 10V 30V 50V, Fig. (ii), , 10V, , 20V 40V 50V, Fig. (iii), , (a) The work done in Fig. (i) is the greatest, (b) The work done in Fig. (ii) is least, (c) The work done is the same in Fig.(i), Fig. (ii) and Fig. (iii), (d) The work done in Fig. (iii) is greater that Fig. (ii) but, equal to that in, , 19 Two condensers C1 and C2 in a circuit are joined as, shown in the figure. The potential of point A is V1 and that, of B is V2. The potential of point D will be, C1, , A, , C2, , D, , V1, , CV, 1 2 + C2V1, C1 + C2, C2V1 − CV, 1 2, (d), C1 + C2, , 1, (V1 + V2 ), 2, CV + C2V2, (c) 1 1, C1 + C2, (a), , B, V2, , (b), , 20 Two capacitors C1 and C2 are charged to 120 V and, 200 V respectively. It is found that by connecting them, together the potential on each one can be made zero., ª JEE Main 2013, Then, (a) 5 C1 = 3 C2, (c) 3 C1 + 5 C2 = 0, , (b) 3 C1 = 5 C2, (d) 9C1 = 4 C2, , 21 A combination of capacitors is set-up as shown in the, figure. The magnitude of the electric field, due to a point, charge Q (having a charge equal to the sum of the, charges on the 4 µF and 9 µF capacitors), at a point, distance 30 m from it, would equal to, 3 µF, , 4 µF, , connected to a battery of emf 20 V. If a dielectric material, 5, of dielectric constant K = is inserted between the, 3, plates, the magnitude of the induced charge will be, ª JEE Main 2018, , (a) 1.2 nC, , (b) 0.3 nC, , (c) 2.4 nC, , (d) 0.9 nC, , 25 A parallel plate capacitor having a separation between, the plates d, plate area A and material with dielectric, constant K has capacitance C0. Now one-third of the, material is replaced by another material with dielectric, constant 2K, so that effectively there are two capacitors, 1, one with area A , dielectric constant 2K and another, 3, 2, with area A and dielectric constant K. If the, 3, capacitance of this new capacitor is C then C/C0 is, (a) 1, , 4, (b), 3, , 2, (c), 3, , ª JEE Main (Online) 2013, 1, (d), 3, , 26 A parallel plate capacitor of area 60 cm 2 and separation, 3 mm is charged initially to 90 µC. If the medium between, the plate gets slightly conducting and the plate loses the, charge initially at the rate of 2.5 × 10−8 C/s, then what is, the magnetic field between the plates?, ª JEE Main (Online) 2013, , (a) 2.5 × 10−8 T, (c) 1.63 × 10−11 T, , (b) 2.0 × 10−7 T, (d) Zero, , 27 Case I Identical point charges of magnitude Q are kept, at the corners of a regular pentagon of side a., Case II One charge is now removed. Match the following, for above two cases., , 9 µF, 2 µF, , Column I, , Column II, , + –, 8V, , (a) 240 N/C, , (b) 360 N/C, , ª JEE Main 2016 (Offline), (c) 420 N/C, (d) 480 N/C, , 22 An uncharged parallel plate capacitor having a dielectric, of constant K is connected to a similar air cored parallel, capacitor charged to a potentialV . The two share the, charge and the common potential isV ′. The dielectric, constant K is, V′ −V, (a), V′ + V, , V′ −V, (b), V′, , V ′ −V, (c), V, , V −V′, (d), V′, , A. Electric field as the centre of pentagon, in case I, , 1., , 1 Q×5, 4 πε0 a, , B. Electric potential at the centre of, pentagon in case I, , 2., , Zero, , C. Electric field at the centre of pentagon, in case II, , 3., , 1 Q, 4 πε0 a 2, , D. Electric potential at the centre of, pentagon in case II, , 4., , 1 Q, ×4, 4 πε0 a
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ELECTROSTATICS, , DAY EIGHTEEN, , (d) Statement I is false, Statement II is true, , Codes, A, (a) 2, (c) 2, , 211, , B, 1, 3, , C, 3, 4, , D, 4, 2, , A, (b) 1, (d) 4, , B, 2, 1, , C, 3, 2, , D, 4, 4, , 28 Statement I For a charged particle moving from point P, , Direction, , (Q. Nos. 28-29) Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below:, (a) Statement I is true, Statement II is false, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation of Statement I, (c) Statement I is true, Statement II is true; Statement II is the, correct explanation of Statement I, , to point Q, the net work done by an electrostatic field on, the particle is independent of the path connecting point, P to point Q., Statement II The net work done by a conservative force, on an object moving along a closed loop is zero., , 29 Statement I No work is required to be done to move a, test charge between any two points on an equipotential, surface., Statement II Electric lines of force at the equipotential, surfaces are mutually perpendicular to each other., ª JEE Main (Online) 2013, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A capacitance of 2 µF is required in an electrical circuit, across a potential difference of 1kV. A large number of, 1µF capacitors are available which can withstand a, potential difference of not more than 300 V. The minimum, number of capacitors required to achieve this is, ª JEE Main 2017 (Offline), , (a) 16, , (b) 24, , (c) 32, , (a), (c), , (d) 2, , 2 Three concentric metal shells A, B and C of respective, radii a, b and c (a < b < c ) have surface charge densities, + σ, − σ and + σ, respectively. The potential of shell B is, ª JEE Main 2018, , , σ a 2 − b 2, (a) , + c, ε0 a, , , σ b 2 − c 2, (c) , + a, ε0 b, , , charge Q. The value of A such that the electric field in the, region between the spheres will be constant is, , , σ a 2 − b 2, (b) , + c, ε0 b, , , σ b 2 − c 2, (d) , + a, ε0 c, , , (d), , π(a 2 − b 2 ), , (b) 60°, , (c) 90°, , πa 2, , square. A charge q is placed at each of the other two, Q, corners. If the net electrical force on Q is zero, then, q, (b) –1, , ª AIEEE 2009, 1, (d) −, 2, , (c) 1, , 6 In the given circuit, charge Q 2 on the 2 µF capacitor, , changes as C is varied from 1 µF to 3 µF.Q 2 as a function, of C is given properly by (figures are drawn schematically, ª JEE Main 2015, and are not to scale), 1m F, , 1, , C, 2m F, , ª JEE Main 2017 (Offline), , (a) 45°, , 2 π(b 2 − a 2 ), 2Q, , 5 A charge Q is placed at each of the opposite corners of a, , (a) −2 2, , makes angle θ with respect to X-axis. When subjected to, an electric field E1 = E $i , it experiences a torque E1 = τ k$ ., When subjected to another electric field E = 3E $j, it, 2, , (b), , 2 πa 2, 2Q, , ª JEE Main 2016 (Offline), Q, , equals, , 3 An electric dipole has a fixed dipole moment p, which, , experiences a torque T2 = − T1. The angle θ is, , Q, , (d) 30°, , 4 The region between two concentric spheres of radii a, and b, respectively (see the figure),, A, has volume charge density ρ = ,, r, where A is a constant and r is the, distance from the centre. At the, centre of the spheres is a point, , E, Charge, , Q, , a, , (a), b, , Charge, , Q2, , (b), 1mF, , 3m F, , C, , Q2, 1m F, , 3mF, , C
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212, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Charge, , Charge, , (c) Q2, , (d) Q2, 1 mF, , 3 mF, , C, , 1 mF, , 3 mF, , C, , 7 A uniform electric field E exists between the plates of a, charged condenser. A charged particle enters the space, between the plates and perpendicular to E . The path of, the particle between the plates is a, ª JEE Main (Online) 2013, , (a) straight line, (c) parabola, , (b) hyperbola, (d) circle, , 8 The surface charge density of a thin charged disc of, radius R is σ. The value of the electric field at the centre, σ, of the disc is, ⋅ With respect to the field at the centre,, 2ε 0, the electric field along the axis at a distance R from the, centre of the disc, ª JEE Main (Online) 2013, (a) reduces by 71%, (c) reduces by 9.7%, , (b) reduces by 29.3%, (d) reduces by 14.6%, , 9 A uniformly charged solid sphere of radius R has, potential V0 (measured with respect to ∞) on its surface., For this sphere, the equipotential surfaces with, 3V 5V 3V, V, potentials 0 , 0 , 0 and 0 have radius R1, R 2, R 3,, 4, 2, 4, 4, and R 4 respectively. Then,, ª JEE Main 2015, (a) R1 = 0 and R 2 > (R 4 − R 3), (b) R1 ≠ 0 and (R 2 − R 1) > (R 4 − R 3), (c) R1 = 0 and R 2 < (R 4 − R 3), (d) 2R < R 4, , A, , 13 A point charge of magnitude +1 µC is fixed at ( 0, 0, 0). An, isolated uncharged spherical conductor, is fixed with its, centre at ( 4, 0, 0). The potential and the induced electric, field at the centre of the sphere is ª JEE Main (Online) 2013, (a) 1. 8 × 105 V and −5.625 × 106 V /m, (b) 0 V and 0 V / m, (c) 2.25 × 103 V and 5.625 × 102 V /m, (d) 2.25 × 105 V and 0 V / m, , 14 Two identical charged spheres are suspended by strings, of equal lengths. The strings make an angle of 30° with, each other. When suspended in a liquid of density, 0.8 g cm –3 , the angle remains the same. If density of the, material of the sphere is 16 gcm –3 , the dielectric constant, of the liquid is, ª AIEEE 2010, (a) 4, , (b) 3, , (c) 2, , 15 A circuit is shown in figure for which C1 = ( 3 ± 0.011) µF,, C2 = ( 5 ± 0.01) µF and C3 = (1 ± 0.01) µF. If C is the, equivalent capacitance across AB, then C is given by, C3, , C2, , B, C1, , C1, , (a) (0.9 ± 0114, ., ) µF, (c) (0.9 ± 0.023) µF, , (b) (0.9 ± 0.01) µF, (d) (0.9 ± 0.09) µF, , 16 A thin semi-circular ring of radius r has a positive charge, q distributed uniformly over it. The net field E at the centre, j, ª AIEEE 2010, O is, , 3Q, 4 π ε0L, Q In 2, (d), 4 π ε0L, (b), , (c) 1/ y, , i, , O, , q, j, 4 π 2 ε0r 2, q, (c) −, j, 2 π 2 ε0r 2, , (b) −, , (a), , x = a on the x-axis. A particle of mass m and charge, q 0 = q /2 is placed at the origin. If charge q 0 is given a, small displacement ( y < a ) along the y-axis, the net force, acting on the particle is proportional to ª JEE Main 2013, (b) − y, , (d) 1, , L, , 11 Two charges, each equal to q, are kept at x = − a and, , (a) y, , (d) 4 k mg tanθ, , B, , L, , Q, 8 π ε0L, Q, (c), 4 π ε0L In 2, , (b) k mg tanθ, , (c) 4 k mg / tanθ, , C1, , length L as shown in the figure. The electric potential at, ª JEE Main 2013, the point O lying at distance, , (a), , (a) 2 k mg tanθ, , A, , 10 A charge Q is uniformly distributed over a long rod AB of, , O, , electrostatic potential at the centre of line joining them will, , 1, be , = k ., , 4πε 0, ª JEE Main (Online) 2013, , (d) − 1/ y, , 12 Two small equal point charges of magnitude q are, suspended from a common point on the ceiling by, insulating massless strings of equal lengths. They come, to equilibrium with each string making angle θ from the, vertical. If the mass of each charge is m, then the, , q, , 4 π 2 ε0r 2, q, (d), j, 2 π 2 ε0r 2, , j, , 17 Two positive charges of magnitude q are placed at the, ends of a side 1 of a square of side 2a. Two negative, charges of the same magnitude are kept at the other, corners. Starting from rest, if a charge Q moves from the, middle of side 1 to the centre of square, its kinetic energy, ª AIEEE 2011, at the centre of square is, 1 2qQ , 1 −, 4 πε0 a , 1 2qQ , (c), 1 +, 4 πε0 a , , (a), , 1 , , 5, 1 , , 5, , (b) zero, (d), , 1 2qQ , 2 , 1 −, , 4 πε0, a , 5
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ELECTROSTATICS, , DAY EIGHTEEN, 18 Two identical charged spheres suspended from a, , (c), , common point by two massless strings of length l are, initially a distance d (d < < l ) apart because of their, mutual repulsion. The charge begins to leak from both, the spheres at a constant rate. As a result charges, approach each other with a velocity v. Then as a function, of distance x between them,, ª AIEEE 2011, (a) v ∝ x −1, , (b) v ∝ x1/ 2, , (d) v ∝ x −1/ 2, , (c) v ∝ x, , (a) 20 s, , (b) 10 s, , (c) 5 s, , (d) 2.5 s, , 20 A resistor R and 2 µF capacitor in series is connected, through a switch to 200 V direct supply. Across the, capacitor is a neon bulb that lights up at 120 V. Calculate, the value of R to make the bulb light up 5 s after the switch, has been closed (take, log10 2.5 = 0.4), ª AIEEE 2011, (a) 1.7 × 105 Ω, (c) 3.3 × 107 Ω, , (b) 2.7 × 106 Ω, (d) 1.3 × 104 Ω, , 21 Let there be a spherically symmetric charge distribution, 5 r , with charge density varying as ρ(r ) = ρ 0 − upto, 4 R, , 4 πρ0r 5 r , − , 3 ε0 3 R , , (b), , ρ0r 5 r , − , 3 ε0 4 R , , through a resistor R. Suppose t1 is the time taken for the, energy stored in the capacitor to reduce to half its initial, value and t 2 is the time taken for the charge to reduce to, one-fourth its initial value. Then, the ratio t1 / t 2 will be, ª AIEEE 2010, , (a) 1, (c) 1 / 4, , (b) 1 / 2, (d) 2, , 23 The question has statement I and statement II. Of the four, choices given after the statements, choose the one that, best describes the two statements., An insulating solid sphere of radius R has a uniform, positive charge density ρ. As a result of this uniform, charge distribution, there is a finite value of electric, potential, at the surface of the sphere, at the surface of, the sphere and also at a point outside the sphere. The, electric potential at infinite is zero., Statement I When a charge q is taken from the centre of, the surface of the sphere its potential energy changes by, qρ, ., 3 ε0, Statement II The electric field at a distance r ( r < R ) from, ρr, the centre of the sphere is, ., 3ε 0, ª AIEEE 2012, (a) Statement I is false, Statement II is true, (b) Statement I is true, Statement II is false, (c) Statement I is true, Statement II is true, Statement II is, the correct explanation for Statement II, (d) Statement I is true, Statement II is true, Statement II is, not the correct explanation of Statement I, , r = R , and ρ(r ) = 0 for r > R , where, r is the distance from, the origin. The electric field at a distance r (r < R ) from, the origin is given by, ª AIEEE 2010, (a), , (d), , 22 Let C be the capacitance of a capacitor discharging, , 19 Combination of two identical capacitors, a resistor R and, a DC voltage source of voltage 6 V is used in an, experiment on C-R circuit. It is found that for a parallel, combination of the capacitor, the time in which the, voltage of the fully charged combination reduces to half, its original voltage is 10 s. For series combination, the, time needed for reducing the voltage of the fully charged, series combination by half is, ª AIEEE 2011, , 4 ρ0r 5 r , − , 3 ε0 4 R , , 213, , ρ0r 5 r , − , 4 ε0 3 R , , ANSWERS, SESSION 1, , SESSION 2, , 1 (d), , 2 (d), , 3 (a), , 4 (d), , 5 (d), , 6 (c), , 7 (c), , 8 (c), , 9 (d), , 10 (d), , 11 (d), , 12 (a), , 13 (c), , 14 (d), , 15 (a), , 16 (b), , 17 (c), , 18 (c), , 19 (b), , 20 (b), , 21 (c), , 22 (d), , 23 (a), , 24 (a), , 25 (b), , 26 (d), , 27 (a), , 28 (b), , 29 (b), , 1 (c), 11 (a), 21 (b), , 2 (b), 12 (c), 22 (c), , 3 (b), 13 (c), 23 (a), , 4 (a), 14 (c), , 5 (a), 15 (c), , 6 (d), 16 (c), , 7 (c), 17 (a), , 8 (a), 18 (d), , 9 (c,d), 19 (d), , 10 (d), 20 (b)
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214, , DAY EIGHTEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , 5 Let l be the linear charge density., , q2, 1 F = 1, = T sin q, 4pe0 x2, , Given, distance r = 2 cm = 2 ´ 10-2 m, Electric field E = 9 ´ 104 N/C, , Support, , Using the formula of electric field due, to an infinite line charge., l, E =, 2pe0r, , q q, T, , T cos q, , q, T sin q, , F, , F, , X, , mg cosq, , E =, , mg, , The resultant of components mg cos q, and force of repulsion balances, the, tension in the string for the equilibrium, massive change., T cos q = mg, q2, 1, tan q =, × 2, 4pe0 x mg, q2, x, =, 2l, 4pe0 x2 mg, ql, x, µ 2 or x3 µ l or x µ l 1 /3, 2l, x, , Þ, , Dividing and multiplying by 2 to get, 1, 1, because, we have the value of, 4pe0, 4pe0, , Thus, we find the separation between, the balls is proportional to l1 /3 , where, l, is length of string., , 2. By symmetry resultant force applied by, eight charges on corners is zero., , 3 Field lines should originate from, , Putting the values, we get, 9 ´ 104 =, l=, , u, , al, u, v 2 = v 2x + v 2y, , v y = at =, As,, , v 2 = u2 +, , 2 ´ 10, , -2, , 2 ´ 9 ´ 109, , = 10-7 C/m, , Thus, the linear charge density is, 10-7 C/m., , 6 Electric field inside the uniformly, charged sphere varies linearly, as, kQ, E = 3 × r , (r £ R ), while outside the, R, sphere, it varies as inverse square of, kQ, distance, E = 2 ; (r ³ R) which is, r, correctly represented in option (c)., , 7 According to Gauss’ theorem,, dr, , P, r, , r1, , E 4pr12, , =, , E =, , ò0, , Q, pR 4, , Qr 12, , r 4pr 2 dr, e0, , 4pe0R 4, , 8 We know that, F = 1 2P13P2, 4pe0 r, P2, Pole P1, , a2 l 2, , u2, C, = u2 + 2 ,, u, C, 2, v = u + 2, u, , f1 is entering the surface., - q1, or q1 = - f1 e0, f1 =, e0, Let + q2 be the charge, due to which flux, f2 , is leaving the surface., q, or q2 = e0f2, f2 = 2, e0, Electric charge inside the surface, = q2 - q1 = e0f2 + e0f1 = e0(f2 + f1 ), , 10 As uniform electric field is parallel to the, cylindrical axis, ò E × dS =, , Pole P2, P1, With the help of this relation, we find, 1 6 P1 P2, ., the force between dipole is, 4pe0 x3, , ò E dS cos 90° = 0, , Further flux entering the cylinder at one, end = flux leaving the cylinder at other, end. Therefore, total electric flux is zero., , 11 Common potential,, V =, , Q + 0, 4pe0(r + r ¢ ), , Charge on smaller sphere,, Qr ¢, r + r¢, , 4pe0r ¢ ´ V =, , 12 The uncharged body must have a, potential less than the charged conductor, and more than at infinity., i.e., V ¥ < V or V > V ¥, d f, 13 Electric field, E =, = - 2ar, dr, By Gauss’ theorem,, q, E (4pr 2 ) =, e0, Þ, , R, , Þ, P, , 2 ´ 9 ´ 109 ´ l, , 9 ´ 104 ´ 2 ´ 10-2, , positive charge and terminate to, negative charge. Thus, (b) and (c) are, not possible. Electric field lines cannot, form corners as shown in (d). Thus,, correct option is (a)., Eq, 4 v x = u, t = l , a y =, =a, u, m, l, , l, 2, 2l, ´, =, 2 2pe0r, 4pe0r, , 9 Let - q1 be the charge, due to which flux, , q = - 8pe0ar 3, dq dq, dr, r=, =, ´, dV dr, dV, 1 ö, = (-24 pe0ar 2 ) æç, ÷, è 4pr 2 ø, = - 6 e0 a, , 14 W = QdV = Q (Vq - V p ), = -100 ´ (1.6 ´ 10-19 ) ´ (-4 - 10), = +100 ´ 1.6 ´ 10-19 ´ 14, = +2.24 ´ 10-16 J, , 15 As, E = dV and around x = 2 m,, dr, V = constant, \ dV = 0 and E = 0, , 16 As E and V are related as,, -dv -d, (4 x2 ) = -8 x, =, dx, dx, At (1, 0, 2), E = -8(1) = -8, E =
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ELECTROSTATICS, , DAY EIGHTEEN, 17 As we know that, potential difference, V A - VO is, VA, , dV = - Edx,, , 2, , ò dV, , 2, , = - ò 30 x dx, 0, , Vo, 3, , 2, , éx ù, V A - VO = - 30 ´ ê ú, ë 3 û0, = - 10 ´ [23 - (0)3 ], , of capacitor, the, eletric field, E =, , 18 The work done by a electrostatics force, is given by W12 = q(V2 - V1 ). Here, initial, and final potentials are same in all three, cases and same charge is moved, so, work done is same in all three cases., , 19 Let the potential of point D be V ., If q is charge on each condenser, then, , Divide, , Final charge = CV ¢ + KCV ¢, Since, initial charge = final charge, V - V¢, V - V¢, K =, K =, V¢, V¢, , 23 When free space between parallel plates, , = - 10 ´ 8 = - 80 J/C, , Þ, , 22 Initial charge = CV and, , V1 - V = qC1, V - V2 = qC2, V1 - V C1, =, V - V2 C2, VC1 - V2C1 = V1C2 - VC2, , When dielectric is introduced between, parallel plates of capacitor,, s, E¢=, Ke 0, Electric field inside dielectric, s, = 3 ´ 104, Ke 0, where, K = dielectric constant of, medium = 22, ., e0 = permittivity of free space, = 8.85 ´ 10-12, . ´ 8.85 ´ 10-12 ´ 3 ´ 104, Þ s = 22, , V (C1 + C2 ) = C1V2 + C2V1 ;, C V + C2V1, V = 1 2, C1 + C2, , 20, , = 6.6 ´ 8.85 ´ 10-8, ., ´ 10-7 = 6 ´ 10-7 C/m2, = 5841, , 24 Magnitude of induced charge is, , C1, + –, , given by, Q ¢ = (K - 1) CV 0, 5, = æç - 1ö÷ 90 ´ 10-12 ´ 20, è3, ø, , 120 V, C2, – +, , = 1.2 ´ 10-9C, , 200 V, For potential to be made zero after, connection, the charge on both the, capacitors are equal., i.e., q1 = q2, \, C1V1 = C2V2 , 120C1 = 200C2, Þ, 3 C1 = 5 C2, , 21 Resultant circuit,, 3mF, 9mF, 4 mF, , s, e0, , 4 mF, , 12mF, , 3 mF, , As, charge on 3mF = 3mF ´ 8V = 24 mC, \ Charge on 4m F = Charge on 12 mF, = 24 mC, Charge on 3m F = 3m F ´ 2V = 6mC, Charge on 9m F = 9 m F ´ 2V = 18 m C, Charge on 4 m F + Charge on 9mF, = (24 + 18)mC = 42mC, \ Electric field at a point distant 30 m, 9 ´ 103 ´ 42 ´ 10- 6, =, = 420N/C, 30 ´ 30, , Þ Q ¢ = 1.2 nC, , 25 As capacitance,, e0 A, d, Now equivalent capacitance,, C0 = k, , 2 K e0 A 2 k e0 A 4 k e0 A, +, =, 3 d, 3 d, 3 d, 4k e0 A, C, 4, \ Ratio =, = 3 d =, e0 A, e0, 3, k, d, C =, , 26 +, +, +, +, +, , 3 mm, , –, –, –, –, –, , 60 cm, Q=90 mC, , dQ, = 2.5 ´ 10-8 C/s, dt, But in case of capacitor, there is no, magnetic field inside the capacitor i.e., zero., Given,, , æ Q ö, Electric potential V = 5 ´ ç, ÷, è 4pe0a ø, \, B ®1, In case II, the electric field of three, identically placed charges is zero. Net, field at the centre O is due to a single, point charge and is given as, Q, E =, 4pe0a2, \, C ®3, In case II, the electric potential, being, scalar, becomes, Q, V = 4´, 4pe0a, \, , D ®4, , 28 Work done by conservative force does, not depend on the path. Electrostatic, force is a conservative force., , 29 As, W = q(V A - V B ). At equipotential, surface V A = V B so, W = 0, Now, we know that field lines makes, an angle of 90° with the equipotential, surfaces but these are parallel to, one-another., , SESSION 2, 1 As each capacitors cannot withstand, more than 300 V, so there should be four, capacitors in each row became in this, condition 1 kV i.e. 1000 V will be, divided by 4 (i.e. 250 not more than, 300 V)., Now, equivalent capacitance of one row, 1, = ´ 1 mF = 0.25mF, 4, [Q in series combination, Ceq = C / n], Now, we need equivalent of 2mF, so let, we need n such rows, \ n ´ 0.25 = 2mF, [Q in parallel combination Ceq = nC ], 2, n=, =8, 0.25, \ Total number of capacitors, = number of rows ´ number, of capacitors in each row, = 8 ´ 4 = 32, , 2 Potential of B = Potential due to charge, on A + Potential due to charge on B, + Potential due to charge on C., C, B, , A, , 27 Point charges are uniformly distributed, around the centre O. Hence, electric, field E is zero., \, A ®2, , 215, , +σ, –σ, +σ
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DAY NINETEEN, , Current, Electricity, Learning & Revision for the Day, u, , u, , u, , u, , u, , Electric Current, Ohm’s Law, Electrical Resistivity, Electrical Resistance, Resistance of Different Materials, , u, , u, , u, , u, , Series and Parallel Combinations of, Resistors, Electric Energy and Power, Electric Cell, Potential Difference and emf of a Cell, , u, , u, , u, , Kirchhoff’s Laws and their, Applications, Wheatstone’s Bridge, Meter Bridge (Special Case of, Wheatstone Bridge), , Electric Current, Electric current is defined as the amount of charge flowing across any section of wire per, unit time. If charge ∆q passes through the area in time interval ∆t at uniform rate, then, current i is defined as, ∆q, …(i), i=, ∆t, SI unit of electric current is ampere (A)., l, , l, , Conventional direction of flow of current is taken to be the direction of flow of positive, charge or opposite to the direction of flow of negative charge., Electric current is a scalar as it does not follow the law vector of addition., , Current Density, Current per unit area is termed as current density., I, J = (Am −2 ), A, It is a vector quantity., , Drift Velocity, l, , Drift velocity is the average uniform velocity acquired by conduction electrons inside, a metallic conductor on application of an external electric field., The drift velocity is given by the relation, eE, τ, vd = −, m, where, τ known as relaxation time., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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220, , l, , l, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Drift velocity per unit electric field is called the mobility of, the electrons. Thus, mobility,, vd , = e τ, µ =, E, m, In terms of drift speed, electric current flowing through a, conductor is expressed as I = nAevd, where, A = cross-section area of conductor,, n = number of conduction electrons per unit volume,, vd = drift velocity of electrons, and, e = charge of one electron., , Ohm’s Law, Ohm’s law states that the physical conditions such as, temperature, mechanical strain, etc., are kept constant, then, current (i) flowing through a conductor is directly, proportional to the potential difference across its two ends., V, i.e., = R = a constant,, i ∝ V or V ∝ i or V = Ri or, i, , Resistance of Different Materials, A perfect conductor would have zero resistivity and a perfect, insulator would have infinite resistivity. Though these are ideal, limits, the electrical resistivity of substances has a very wide, range. Metals have low resistivity of 10 −8 Ωm to 10 −6 Ωm,, while insulators like glass or rubber have resistivity, some 10 18, times (or even more) greater, Generally, good electrical, conductors like metals are also good conductors of heat, while, insulators like ceramic or plastic materials are also poor, thermal conductors., , V-I Characteristics of Ohmic and, Non-ohmic Conductors, Substances obeying Ohm’s law are called Ohmic resistors, e.g., metals and their alloys. Substances which do not obey Ohm’s, law are called non-ohmic resistors, e.g. electrolytes, gases,, thermionic tubes, transistors, rectifiers, etc., , where R depends on the nature of material and it given, dimension., I, , I, , Electrical Resistivity, The resistance of a resistor (an element in a circuit with some, resistance R) depends on its geometrical factors (length,, cross-sectional area) as also on the nature of the substance of, which the resistor is made. Electrical resistance of a, rectangular slab depends on its length (l ) and its, cross-sectional area ( A)., …(i), i. e.,, R∝l, 1, …(ii), and, R∝, A, Combining the two relations, we get, l, …(iii), R∝, A, ρl, or, …(iv), R=, A, where, ρ is a constant of proportionality called resistivity., m, ρ= 2, ne τ, , Resistance of Different Materials, Resistance offered by the conductors is minimum while, resistance offered by an insulator is maximum., Semiconductors have resistance which is intermediate to, conductor and insulator., , V, V, Non-ohmic resistors Ohmic resistors, , Colour Code for Resistors, The value of resistance used in electrical and electronic, circuits vary over a wide range. These resistances are usually, carbon resistances and a colour code is used to indicate the, value of resistance., Their value ranges from kilo-ohm to mega-ohm. Their, percentage accuracy is indicated by a colour code printed on, them. Carbon resistors are compact, inexpensive and are used, in electronic circuits., Colour Code for Carbon Resistors, Colour, , Letter of, remember, , Number, , Multiplie, r, , Tolerance, , Black, , B, , 0, , 10 0, , —, , Brown, , B, , 1, , 101, , —, , Red, , R, , 2, , 10 2, , —, , Orange, , O, , 3, , 10 3, , —, , Yellow, , Y, , 4, , 10 4, , —, , 5, , —, , Green, , G, , 5, , 10, , Blue, , B, , 6, , 10 6, , —, —, , Electrical Resistance, , Violet, , V, , 7, , 10 7, , Electrical resistance is defined as the ratio in the potential, difference (v) across the ends of the conductor to the current (i), V, flowing through it, i.e. R =, i, The SI unit of electrical restristance is Ω (ohm) and its, dimension is [ML2 T −3 A −2 ]., , Grey, , G, , 8, , 10 8, , —, , White, , W, , 9, , 10, , 9, , —, , Gold, , —, , —, , 10 −1, , 5%, , Silver, , —, , —, , 10%, , No, colour, , —, , —, , 10 −2, —, , 20%
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CURRENT ELECTRICITY, , DAY NINETEEN, Now to find the colour coding of carbon resistor, we must, remember the bold capital letters of the following sentences:, Black Brown ROY Great Britain Very Good Wife Wearing, Gold Silver Necklace, Or, Black Brown Rods of Your Gate Became Very Good When, Given Silver Colour, The colours of first two bands A and B correspond to figures, and the colour of the third band C represents multipliers,, respectively. The fourth band represents the tolerance., , A B C, , D, , e.g. Consider a carbon resistor of bands A and B of black and, red colour having figures 0 and 2. The third band C of green, colour having multiplier 10 5., ∴ Resistance of the value is given by, R = 02 × 10 5 Ω, But the fourth band D having gold colour, which represents a, tolerance of ± 5%., Hence, the value of carbon resistance is, R = 02 × 10 5 Ω ± 5%, , Series and Parallel Combinations, of Resistors, Series Combination, A series circuit is a circuit in, R1, R2, R3, which resistors are arranged in a V, chain, so the current has only one, path to take. The current is the, I, same through each resistor. The, total resistance of the circuit is found by simply adding up the, resistance values of the individual resistors. Equivalent, resistance of resistors in series, R = R1 + R2 + R3 + K, , Parallel Combination, A parallel circuit is a circuit in, I1 R1, I, I, I2, which the resistors are arranged, with their heads connected, R2, R1, together and their tails connected V, I3, R3, together. The current in a parallel, circuit breaks up, with some, flowing along each parallel branch and recombining, when, the branches meet again. The voltage across each resistor is, parallel is the same., The total resistance of a set of resistors in parallel is found by, adding up the reciprocals of the resistance values, and then,, taking the reciprocal of the total., , 221, , The equivalent resistance of resistors in parallel,, 1, 1, 1, 1, =, +, +, +K, R R1 R2, R3, , Temperature Dependence, of Resistance, Resistance and resistivity of metallic conductors increases, with increase in temperature. The relation is written as, Rθ = R0 (1 + αθ + βθ2 ) and ρ θ = ρ 0 (1 + αθ + βθ2 ), where, R0 and ρ 0 are values of resistance and resistivity at 0°C, and Rθ and ρ θ at θ°C. α and β are two constants whose value, vary from metal to metal., , Electric Energy and Power, Whenever the electric current is passed through a conductor,, it becomes hot after short time. This effect is known as, heating effect of current or Joule heating effect., H = W = I 2 Rt joule =, , I 2 Rt, cal, 418, ., , The rate at which work is done by the source of emf in, maintaining the effect of current in a circuit is called electric, power of the circuit,, P = VI watt, Other expressions for power,, P = I 2 R watt ⇒ P =, , V2, R, , Electric Cell, An electric cell is a device which maintains a continuous flow, of charge (or electric current) in a circuit by a chemical, reaction. In an electric cell, there are two rods of different, metals called electrodes., , Internal Resistance of a Cell, Thus, when a current is drawn through a source, the potential, difference between the terminal of the source is, V = E − ir, This can also be shown as below, E, , r, , A, , I, , B, , V A − E + Ir = VB, or, , V A − VB = E − Ir, , Following three special cases are possible, (i) If the current flows in opposite direction (as in case of, charging of battery), then V = E + Ir, (ii) V = E , if the current through the cell is zero., (iii) V = 0, if the cell is short circuited.
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222, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , This is because current in the circuit,, E, or E = Ir, I =, r, , Combination of Cells in Series and, in Parallel, , ∴, , A group of cells is called a battery. Two common grouping of, cells are, , E − Ir = 0 or V = 0, , 1. Series Grouping, , Short circuited, , In series grouping, if all the cells are joined so as to supply, current in the same direction, then resultant, emf,, , r, , E, , Thus, we can summarise, it was follows, E, , ⇒, , Eeq = E 1 + E2 + E3 + …, , r, I, , V = E − ir or V < E, E, , ⇒, , However, if one or more cells are joined so as to supply, current in reverse direction, then emf of that/those cells is, taken as negative, while calculating the equivalent emf., , r, , +ve, , I, , V = E + Ir or V > E, E, , ⇒, , r, , E2, , E3, , r1, , r2, , r3, , req = r1 + r2 + r3 + …, i, , E, r, , r, , V = 0 is short circuited, , Potential Difference and, emf of a Cell, Electromotive force (emf) of a cell is the terminal potential, difference of cell when it is in an open circuit, i.e. it is not, supplying any current to the external circuit. However, when, it is supplying a current to an external resistance, the voltage, across the terminals of cell is called the terminal voltage or, terminal potential difference., , If n cells, each of emf E and internal resistance r, are joined, in series, then, E eq = nE and req = nr, , 2. Parallel Grouping, In parallel grouping, if positive terminals of all cells have, been joined at one point and all negative terminals at, 1, 1, 1, 1, another point, then, =, +, +, +…, req r1 r2, r3, The equivalent emf of the parallel grouping is given by, E eq E 1 E2, E, =, +, + 3 +…, req, r1, r2, r3, E1, +, r1, , If E be the emf and r the internal resistance of a cell and a, resistance R is joined with it, then current in the circuit,, E, and terminal potential difference,, I =, R+r, , +ve, , V = E − Ir, , E, + 2, , –ve, , r2, , ER, V = IR =, (R + r ), or, , –ve, , The equivalent internal resistance of the cell,, , V = E , if I = 0, , E, , E1, , +, , E3, r3, , Internal resistance of cell,, E −V, E, , r =, R = R − 1, V , V, , , If n cells, each of emf E and internal resistance r, all joined, in parallel, then, r, req =, n, , Terminal voltage is more than emf of cell when cell is charged, and it is given by V = E + Ir ., , But, , E eq = E
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CURRENT ELECTRICITY, , DAY NINETEEN, , Kirchhoff’s Laws and their, Applications, Many electric circuits cannot be reduced to simple series, parallel combinations. For example, two circuits that cannot, be so broken down are shown in figure, A, R1, , E1, , R2, , E2, , D, , B, C, , F, , E, R3, (a), , Wheatstone’s Bridge, For measuring accurately any resistance Wheatstone bridge is, widely used. It is an arrangement of four resistances used to, measure one of them in terms of the other three., Consider four resistances P, Q, R and S are connected in the, four arms of a quadrilateral according to figure, the bridge is, said to be balanced when galvanometer gives zero deflection,, i.e. potential at point B and D is same (VB = VD )., In this condition,, P R, =, Q S, , P, , B, , C, , D, , R1, R2, , B, , T1, , A, , R3, , Q, P, , A, I2, , R4, , I1, C, I2, , G, S2, , E2, , E3, F, H, , E, I, , However, it is always possible to analyze such circuits by, applying two rules, derived by Kirchhoff., , Junction Rule, The algebraic sum of the currents at any junction is zero., Σ, , i.e., , i =0, , junction, , i2, , i1, i4, , i3, , This law can also be written as, ‘‘the sum of all the currents, directed towards a point in circuit is equal to the sum of all, the currents directed away from that point.’’, Thus, in figure,, i 1 + i2 = i3 + i 4, , I, , D, , E, , G, , R5, (b), , S, , R, , I, , E1, , S1, , Meter Bridge, (Special Case of Wheatstone Bridge), This is the simplest form of Wheastone bridge and is specially, useful for comparing resistances more accurately. The, construction of the meter bridge is shown in the figure. It, consists of one metre resistance wire clamped between two, metallic strips bent at right angles and it has two points for, connection., There are two gaps; in one of whose value is to be determined, is connected. The galvanometer is connected with the help of, jockey across BD and the cell is connected across AC. After, making connections, the jockey is moved along the wire and, the null point is found. Wheatstone bridge, wire used is of, uniform material and cross-section. the resistance can be, found with the help of the following relation, S, , R, B, , The junction rule is bases on conservation of electric charge., G, , Loop Rule, , A, , The algebraic sum of the potential difference in any loop, including those associated emf’s and those of resistive, elements, must be equal to zero., That is,, , Σ, , closed loop, , C, , D, I1, , 100, , I1, , Metre sale, , ∆V = 0, , This law represent conservation of energy., Applying Kirchhoff’s law for the following circuit, we have, Resulting equation is, Vr1 + Vr2 + Vr3 − 10 = 0., , 223, , V, , R, l1, or, =, S (100 − l1), , R =S, , K1, , l1, 100 − l1
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224, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Potentiometer, , Applications of Potentiometer, , Principle Potentiometer is an ideal device to measure the, potential difference between two points. It consists of a long, resistance wire AB of uniform cross-section in which a steady, direct current is set up by means of a battery., , (i) To find emf of an unknown battery, E1, , E1, , i, , i, , E1, , i, , i, , l1, , i, , A, , i, , i, , i, , A, , C1, , i, , B, C2, , i2 =0, , L, A, , l1, B, , B, , G, , C, , i2 =0, G, , Ek, , EU, , i2 =0, G, , E2 , r, , Potential gradient,, k =, , Potential difference across AB, Total length, , V, = AB, L, iR AB, =, = iλ, L, R AB, = resistance per unit length of, L, potentiometer wire., where, λ =, , The emf of source balanced between points B and C, E2 = kl, R, = i AB × l = iRCB, l, , We calibrate the device by replacing E2 by a source of known, emf E k and then by unknown emf Eu . Let the null points are, obtained at lengths l1 and l2 ., Then,, , Here, ρ = resistance of wire AB per unit length, , R0 = resistance inserted by means of rheostat Rh, k = potential gradient., , So, by measuring the lengths l1 and l2 , we can find the emf of, an unknown battery., , (ii) To find the internal resistance of a cell, Firstly, the emf E of the cell is balanced against a length, AD = l 1. For this, the switch S′ is left opened and S is closed. A, known resistance R is then connected to the cell as shown., The terminal voltage V is now balanced against a smaller, length AD′ = l2 . Here, now switch is opened and S′ is closed., l1, l2, S, E, , D′ D, , A, , B, , (E, r), G, , S′, R, , Then,, Since,, , L → balancing length, J → jockey., , l , E K l1, = or EU = 2 E K, EU l2, l1 , , ∴, , Here, AB is a long uniform resistance wire (length AB may be, ranging from 1 m to 10 m). E 0 is a battery whose emf is known, supplying a constant current I for flow through the, potentiometer wire. If R be the total resistance of, potentiometer wire and L its total length, then potential, gradient, i.e. fall in potential per unit length along the, potentiometer will be, V IR, k = =, L, L, E 0R, =, (R0 + R) L, where, E 0 = emf of battery,, , E K = i( ρl1) and EU = i ( ρl2 ), , or, , E l1, =, V l2, E R+r, {Q E = i(R + r ) and V = iR}, =, V, R, l, , R + r l1, =, ⇒ r = 1 − 1 R, R, l2, l2,
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CURRENT ELECTRICITY, , DAY NINETEEN, , 225, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A wire of length L and 3 identical cells of negligible, internal resistance are connected in series.Due to the, current, the temperature of the wire is raised by ∆T in a, time t. A number N of similar cells is now connected in, series with a wire of the same material and cross-section, but of length 2L. The temperature of the wire is raised by, the same amount ∆T in the same time t. The value of N is, (a) 4, , (b) 6, , (c) 8, , (d) 9, , (a) P and R, (c) Q and R, , 7 An infinite ladder network of resistances is constructed, with 1 Ω and 2 Ω resistance as shown in figure. The, 6 V battery between A and B has negligible, internal resistance, then effective resistance between, A and B is, , 2 In a large building, there are 15 bulbs of 40W, 5 bulbs of, A, , 100W, 5 fans of 80 W and 1 heater of 1kW. The voltage, of the electric mains is 220 V. The minimum capacity of, the main fuse is, ª JEE Main 2014, (a) 8A, , (b)10A, , (c) 12A, , are connected in series to a 440 V supply. Which of the, bulbs will fuse?, ª AIEEE 2012, (a) Both, , (b) 100 W, , (c) 25 W, , (d) Neither, , 4 A given resistor has the following colour scheme of the, , 1W, , 1W, , (a) 1.0 × 10 ± 10%, , (b) 1.0 × 10 ± 10%, , (c) 1.0 × 106 ± 10%, , (d) 1.0 × 107 ± 10%, , 2W, , 2W, , B, , (a) 3 Ω, , (b) 2 Ω, , (c) 6/5 Ω, , (d) 5/6 Ω, , 8 In the circuit in the figure below, the potential difference, across P and Q will be nearest to, , various strips on it brown, black, green and silver. Its, value in ohm is, 4, , 1W, , 2W, , 6V, , (d) 14A, , 3 Two electric bulbs marked 25 W-220 V and 100 W-220 V, , (b) P and Q, (d) Any two points, , 100 W, 48 V, , 5, , 80 W, 100 W, , 5 Correct set up to verify Ohm’s law is, , Q, P, , 20 W, , ª JEE Main (Online) 2013, A, V, , (a), , (b), , V, , A, , (a) 9.6 V, (c) 4 V, , (b) 6.6 V, (d) 3.2 V, , 9 The charge on the 4 µF capacitor in the steady state, is, A, (c), , 2 mF, , 5W, , V, (d), 10 V, , 5W, , A, , V, , 4 mF, , 6 Six equal resistances are connected between points P , Q, and R as shown in figure. Then net resistance will be, maximum between, ª JEE Main (Online) 2013, P, , 20, µC, 3, 5, (c) µC, 6, (a), , (b), , 10, µC, 3, , (d) 10µC, , 10 The temperature dependence of resistances of Cu and, undoped Si in the temperature range 300-400 K, is best, ª JEE Main 2016 (Offline), described by, , r, r, , r, , r, r, , Q, , r, , R, , (a) linear increase for Cu, linear increase for Si, (b) linear increase for Cu, exponential increase for Si, (c) linear increase for Cu, exponential decrease for Si, (d) linear decrease for Cu, linear decrease for Si
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226, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 11 When an electric heater is switched ON, the current, flowing through it is plotted against time (t). Taking, into account the variation of resistance with, temperature, which of the following best represents, the variations, I, , (a) 5.5 Ω, (c) 4.5 Ω, , (b) 3.5 Ω, (d) 2.5 Ω, , 15 Two batteries of emf E 1 and E 2(E 2 > E 1) and internal, resistances r1 and r2, respectively are connected in parallel, as shown in figure., E1, , I, , r1, , A, , (a), , B, , (b), , r2, , E2, , t, , t, I, , I, , (a) Two equivalent emf Eeq of the two cells is between E1 and, E 2 , i.e. E 1 < E eq < E 2, (b) The equivalent emf Eeq is smaller than E1, , (c), , (c) The Eeq is given by Eeq = E1 + E 2 always, , (d), , (d) Eeq is independent of internal resistance r1 and r2, , 16 Match the following columns., t, , t, , Column I, , 12 Two bars of radii r and 2r are kept in contact as, shown. An electric current i is passed through the, bars. Which one of the following is correct?, l/2, l/2, 2r, , i, , Smaller the resistance, greater the current, , 1. If the same voltage is, applied, , B., , Greater the resistance, smaller the current, , 2. If the same current is, passed, , C., , Greater the resistance, smaller the power, , 3. When resistances are, connected in series, , D., , Greater the resistance, greater the power, , 4. When resistances are, connected in parallel, , r, C, , A, , Column II, , A., , B, , Codes, (a) Heat produced in bar BC is 4 times the heat, produced in bar AB, (b) Electric field in both halves is equal, (c) Current density across AB is double that of across, BC, (d) Potential difference across AB is 4 times that of, across BC, , 13 Which of the four resistances P , Q , R and S generate, , A, (a) 1, (c) 4, , B, 2, 3, , C, 3, 1, , D, 4, 2, , A, (b) 4, (d) 3, , C, 2, 2, , D, 3, 1, , 17 In the circuit shown, battery, ammeter and voltmeter are, ideal and the switch S is initially closed as shown in figure., When switch S is opened, match the parameter of column I, with the effects in column II., R, , the greatest amount of heat when a current flows from, ª JEE Main (Online) 2013, A to B ?, P=2W, , B, 1, 4, , R, , V, , S, , Q=4W, A, , A, R=1W, , (a) Q, , E, , B, , (b) S, , Column I, , S=2W, , (c) P, , 14 A DC source of emf E 1 = 100 V, , (d) R, E1, , and internal resistance r = 0.5 Ω,, a storage battery of emf, E 2 = 90 V and an external, E2, resistance R are connected as, R, shown in figure. For what value, of R no current will pass through the battery?, , r=0.5 Ω, , ª JEE Main (Online) 2013, , Column II, , A., , Equivalent resistance across the battery 1. Remains same, , B., , Power dissipated by left resistance R, , 2. Increases, , C. Voltmeter reading, , 3. Decreases, , D. Ammeter reading, , 4. Becomes zero, , Codes, A, (a) 2, (c) 4, , B, 3, 1, , C, 3, 1, , D, 3, 1, , A, (b) 3, (d) 1, , B, 2, 4, , C, 2, 4, , D, 2, 4
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CURRENT ELECTRICITY, , DAY NINETEEN, 18 In the circuit shown in figure. The point F is grounded., , 22 In the below circuit, the current in each resistance is, , Which of the following is wrong statement?, , ª JEE Main 2017 (Offline), , 5W, B, , 2V, , C, , 2W, , 2V, , 2V, , 3W, 1Ω, , A, 10 V, , 2V, , (a) 0.25 A, (a) D is at 5 V, (b) E is at zero potential, (c) The current in the circuit will be 0.5 A, (d) None of the above, , 3W, A, , 3W, , (a) In a balanced Wheatstone bridge, if the cell and the, galvanometer are exchanged, the null point is disturbed, (b) A rheostat can be used as a potential divider, (c) Kirchhoff’s second law represents energy conservation, (d) Wheatstone bridge is the most sensitive when all the four, resistances are of the same order of magnitude, , 40 cm from one end of the wire when resistance X is, balanced against another resistanceY . If X < Y , then the, new position of the null points from the same end, if one, decides to balance a resistance of 3X against Y , will be, ª JEE Main (Online) 2013, close to, , (14 V, 0.8 W), , (b) 3.28 A, (d) 1.09 A, , (a) 80 cm, , 20 What is the potential difference between points A and D, of circuit as shown in figure?, 4V, , C, , A, 5V, , 3W, , 4W, D, , B, , (b) 9 V, (d) 11.4 V, , 21 In the circuit shown below, the current in the1 Ω resistor, is, , ª JEE Main 2014, , (c) 67 cm, , (d) 50 cm, , meter bridge shifts to the left by 10 cm. The resistance of, their series combination is1 kΩ. How much was the, resistance on the left slot before interchanging the, ª JEE Main 2018, resistances?, (b) 505 Ω, , (c) 550 Ω, , (d) 910 Ω, , 26 In an experiment to measure the internal resistance of a, cell by potentiometer, it is found that the balance point is, at a length of 2 m when the cell is shunted by a 4 Ω, resistance; and is at a length of 3 m when the cell is, shunted by a 8 Ω resistance. The internal resistance of, the cell is, then, (a) 12 Ω, , (b) 8 Ω, , (c) 16 Ω, , (d) 1 Ω, , 27 A 6 V battery is connected to the terminals of a 3 m long, wire of uniform thickness and resistance of 100 Ω. The, difference of potential between two points on the wire, separated by a distance of 50 cm will be, , P 2Ω, , 1Ω, , (b) 75 cm, , 25 On interchanging the resistances, the balance point of a, , (a) 990 Ω, , 6W, , 1W, , 6V, , (d) 1 A, , 24 In a metre bridge experiment, null point is obtained at, 6W, , (a) 6.56 A, (c) 2.18 A, , 2V, , (c) 0 A, , ª JEE Main 2017 (Offline), , 2W, 6W, , 2V, , (b) 0.5 A, , 23 Which of the following statements is false?, , 19 The reading of ammeter as shown in figure is,, , (a) 5 V, (c) 10.4 V, , 1Ω, , 3V, , 4W, , E, , 2W, , 1Ω, , D, , F, , 6V, , 227, , 9V, , (a) 2 V, , (b) 3 V, , (c) 1 V, , (d) 15 V, , 28 The current in the primary circuit of potentiometer is 0.2A., 3Ω, , (a) 1.3A, from P to Q, (b) 0A, (c) 0.13 A, from Q to P, (d) 0.13 A, from P to Q, , Q 3Ω, , The specific resistance and cross-section of the, potentiometer wire are 4 × 10−7 Ω m and 8 × 10−7 m 2, respectively. The potential gradient will be equal to, (a) 0.2 V/m, , (b) 1 V/m, , (c) 0.3 V/m, , ª AIEEE 2011, (d) 0.1 V/m
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228, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Direction (Q. Nos. 29-32) Each of these questions contains, two statements : Statement I and Statement II. Each of these, question also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 29 Statement I As temperature decreases, the relaxation, , higher temperature. The null point can be obtained at the, same point as before by decreasing the value of the, standard resistance., , Statement II Resistance of a metal increases with, increase in temperature., 32 Statement I In the potentiometer circuit shown in figure,, E 1 and E 2 are the emfs of cells C1 and C2 respectively, with E 1 > E 2. Cell C1 has negligible internal resistance. For, a given resistor R, the balance length is x. If the diameter, of the potentiometer wire AB is increased, the balance, length x will decrease., C1, , time of a conducting material increases., , Statement II Number of collisions per unit time of, electrons with lattice ions increases as the, temperature increases., , x, , A, , R, , D, , B, , 30 Statement I Potential difference across the terminals of a, battery can be greater than its emf., , Statement II When current is taken from battery,, V = E − Ir (Symbols have their usual meaning)., 31 Statement I In a meter bridge experiment, null point for, an unknown resistance is measured. Now, the unknown, resistance is put inside an enclosure maintained at a, , C2, , G, , Statement II At the balance point, the potential, difference between AD due to cell C1 = E 2, the emf of, cell C 2 ., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 If a wire is stretched to make it 0.1% longer, its, resistance will, , ª AIEEE 2012, , (a) increase by 0.2%, (c) decrease by 0.05 %, , (b) decrease by 0.2%, (d) increase by 0.05%, , 2 In the given circuit diagram, when the current reaches, steady state in the circuit, the charge on the capacitor of, capacitance C will be, E, , r, r1, , C, , r1, (a) CE, (r2 + r ), r1, (c) CE, (r1 + r ), , r2, , ª JEE Main 2017 (Offline)], r2, (b) CE, (r + r2 ), , (d) CE, , 3 When 5V potential difference is applied across a wire of, , length 01, . m, the drift speed of electrons is 2 . 5 × 10−4 ms −1., If the electron density in the wire is 8 × 1028m −3 the, ª JEE Main 2015, resistivity of the material is close to, (a) 1. 6 × 10−8 Ωm, (c)1. 6 × 10−6 Ωm, , (b) 1. 6 × 10−7 Ωm, (d) 1. 6 × 10−5 Ωm, , 4 A letter A is constructed of a uniform wire with resistance, 1.0 Ω per cm. The sides of the letter are 20 cm and the, cross piece in the middle is 10 cm long. The apex angle, is 60°. The resistance between the ends of the legs is, ª JEE Main (Online) 2013, close to, (a) 50.0 Ω, (c) 36.7 Ω, , (b) 10 Ω, (d) 26.7 Ω, , 5 Two batteries with emf 12 V and 13 V are connected in, parallel across a load resistor of10 Ω. The internal, resistances of the two batteries are1 Ω and 2 Ω,, respectively. The voltage across the load lies between, ª JEE Main 2018, , (a) 11.6 V and 11.7 V, (c) 11.4 V and 11.5 V, , (b) 11.5 V and 11.6 V, (d) 11.7 V and 11.8 V, , 6 In a potentiometer experiment, it is found that no current, passes through the galvanometer when the terminals of, the cell are connected across 52 cm of the potentiometer, wire. If the cell is shunted by a resistance of 5 Ω, a, balance is found when the cell is connected across, 40 cm of the wire. Find the internal resistance of the cell., ª JEE Main 2018, , (a) 1 Ω, , (b) 15, . Ω, , (c) 2 Ω, , (d) 2.5 Ω
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CURRENT ELECTRICITY, , DAY NINETEEN, 7 Two conductors have the same resistance at 0°C but, their temperature coefficients of resistance are α 1 and α 2., The respective temperature coefficients of their series, and parallel combinations are nearly, ª AIEEE 2010, α1 + α 2, , α1 + α 2, 2, α1α 2, (c) α1 + α 2 ,, α1 + α 2, , α1 + α 2, 2, α1 + α 2 α1 + α 2, (d), ,, 2, 2, (b) α1 + α 2 ,, , (a), , (a) 40 loge 4, (c) 20 loge 2, , (b) 30 loge 3, (d) 10 loge 2, , 11 The potential difference across 8Ω resistance is 48V as, shown in figure. The value of potential difference across, points A and B will be, 3Ω, , A, , 8 There are two concentric spheres of radius a and b, , 20 Ω, , respectively. If the space between them is filled with, medium of resistivity ρ, then the resistance of the, intergap between the two spheres will be, (a), , ρ, 4 π (b + a), , (b), , ρ 1 1, − , 4π b a, , (c), , ρ, 4π, , (d), , ρ, 4π, , (b) 137 Ω, , (c) 107 / 2 Ω, , (a) 62 V, , (d) 77 Ω, , 10 A source of emf E = 10 V and having negligible internal, resistance is connected to a variable resistance. The, resistance varies as shown in figure. The total charge, that has passed through the resistor R during the time, interval from t1 to t 2 is, , 60 Ω, , 8Ω, , 48 V, , 1Ω, , B, , i, galvanometer by the half, deflection method the following, R1, R2, circuit is used with resistances, R1 = 9970 Ω, R 2 = 30 Ω and, R 3 = 0. The deflection in the, galvanometer is d. With, G R3, R 3 = 107 Ω the deflection, d, changed to ⋅ The galvanometer, 2, resistance is approximately, ª JEE Main (Online) 2013, , (a) 107 Ω, , 24 Ω, , 1 − 1, , , a b , , 9 To find the resistance of a, , 30 Ω, , (b) 80 V, , (c) 128 V, , (d) 160 V, , 12 The V - I graph for a conductor at temperatures T1 and T2, , are as shown in the figure, The termT2 − T1 is proportional, to, T2, , Voltage (V ), , 1 − 1, 2, , a, b2 , , 229, , T1, q, q, Current (I), , (b) sin 2θ, , (a) cos 2θ, , (c) cot 2θ, , (d) tan 2θ, , 13 The charge supplied by source varies with time t as, Q = at − bt 2. The total heat produced in resistor 2R is, R, +, Source, –, , R, , R, , 2R, , 40 Ω, 20 Ω, , t1 = 10 s t2 = 30 s, , t, , (a), , a 3R, 6b, , (b), , a 3R, 27b, , (c), , a 3R, 3b, , (d) None of these, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (c), , 3 (c), , 4 (c), , 5 (a), , 6 (b), , 7 (b), , 8 (d), , 9 (a), , 10 (c), , 11 (b), , 12 (a), , 13 (b), , 14 (c), , 15 (a), , 16 (a), , 17 (a), , 18 (b), , 19 (c), , 20 (c), , 23 (a), , 24 (c), , 25 (c), , 26 (b), , 27 (c), , 28 (d), , 29 (a), , 30 (b), , 3 (d), 13 (b), , 4 (d), , 5 (b), , 6 (b), , 7 (d), , 8 (d), , 9 (d), , 10 (d), , 21 (c), , 22 (c), , 31 (d), , 32 (d), , 1 (a), 11 (d), , 2 (b), 12 (c)
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230, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , Thus, in order to study Ohm’s law, experimentally, voltmeter (V) should be, connected parallel to the resistor., However, ammeter (A) should be, connected in series with the resistor., , ρ2L, 1 Here, R = ρL ; R1 =, = 2R, A, A, Mass, As,, Density =, Volume, m, i.e. d =, AL, ∴, , 6 By solving this, we get net resistance as,, , m = ALd ; m1 = A 2Ld = 2m, …(i), , 7 Let R be the resistance of infinite ladder., …(ii), , On solving Eqs. (i) and (ii), we get, N =6, , The addition or subtraction of one step, in the ladder will not affect the total, resistance of network. Therefore,, equivalent circuit will be as shown in, figure., 1W, , 2 Total power (P ) consumed, , Minimum capacity should be 12 A., 2, 3 Resistance of bulb is given by R = V ., , P, As the rated power of bulb 25 W is less, than 100 W, it implies that 25 W bulb, has higher resistance. As in series, combination, current through both the, bulbs is same, so heating in 25 W bulb, is more than that of 100 W bulb. So, 25, W bulb will get fused., , 4 Numbers attached for brown, black,, green and silver are 1, 0, 5, ± 10 %., Therefore, the resistance of given, resistor, = 10 × 105 Ω ± 10%, = 1.0 × 10 Ω ± 10%, 6, , 5 Ohm’s law states that the current (I ), flowing through a conductor is directly, proportional to the potential difference, (V ) across its ends provided its physical, conditions such as temperature,, mechanical strain, etc. kept constant,, i.e., or, , I ∝ V or V ∝ I, V = RI (where, R is constant), , R, , 2W, , 6V, , Total resistance = 1 +, , Power, i.e. P = VI, 2500, 125, A=, I =, ⇒, 220, 11, = 11.3 A, , 11 The filament of the heater reaches its, , When the heater is switched ON, it, draws a larger current than its steady, state current as the filament heats up, its, resistance increases and current falls to, steady state value., , 12 Current flowing through both the bars is, equal. Now, the heat produced is given by, H = I 2 Rt, H ∝R, (1 / 2r )2, H AB, R, = AB =, H BC, R BC, (1 / r )2, , or, , = (15 × 40) + (5 × 100) + (5 × 80), + (1 × 1000), = 2500 W, As we know that,, , increase in temperature, increases the, resistance. Then, silicon (Si) is, semiconductor, so with increase in, temperature, resistance will decrease., steady resistance when the heater, reaches the steady temperature, which is, much higher than room temperature. The, resistance at room temperature is then, much lower than the resistance of its, steady state., , R PQ =, , Now, heat produced in first case, (3E )2, H1 =, × t = ms∆T, R, In the second case,, (NE )2, H2 =, × t = 2ms∆T, 2R, , 5, 4, r , RQR =, r, 11, 11, 3, and R PR =, r, 11, R PQ > RQR > R PR, Therefore, R PQ is maximum., , 10 As, we know copper is a conductor, so, , or, , 2× R, =R, R+2, , or, , R + 2 + 2R = R2 + 2R, , ⇒, , R2 − R − 2 = 0, , 1, 4, or, H BC = 4 H AB, 13 We know that, I ∝ 1 ,, R, I1 2W, 4W, =, , On solving, we get, R =2Ω, , 8 Total resistance of circuit, , = 100 + 100 + 80 + 20 = 300 Ω, 48, = 016, . A, 300, Potential difference across P, and Q = 20 × 016, . = 3.2 V, Current I =, , 9 In the steady state,, the capacitors are, fully charged and, acts as open circuit,, so the equivalent, circuit in steady, state would be as, shown alongside, figure., Steady state current I =, , Q R ∝ 1 ∝ 1 , , , , A, r2 , , A, , I2, , B, , Here,, , 2W, 2, 6, I2 =, I = I, 3+ 6, 3, , or, , I1 =, , 1W, , A, 5W, 10 V, , 5W, , B, 10, = 1A, 5+ 5, , So, potential drop across AB is, V = 5V, Sum of potential difference across 2µF, and 4 µF capacitors is 5V. As capacitors, are in series, charges on them would be, same, let us say it is q., q, q, From KVL,, + =5, 2, 4, 20, q =, µC, ⇒, 3, , 3, 1, I = I, 6+ 3, 3, , Power rate in 2 Ω of upper series, 2, 1, 2, = 2 × I = I 2, 3 , 9, Power rate in 4 Ω of upper series, 2, 1, 4, = 4 × I = I 2, 3 , 9, Power rate of 1 Ω in lower series, 2, 2, 4, = 1 × I = I 2, 3 , 9, Power rate of 2 Ω in lower series, 2, 2, 8, = 2 × I = I 2, 3 , 9, ∴ Greatest amount of heat is generated, by S.
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CURRENT ELECTRICITY, , DAY NINETEEN, 14 Given, E1 = 100 V,, , 23, −1, V =, 15, 5, −3, or, V =, = −0.13 V, 23, ∴ Current = −1 × 013, ., = − 013, . A, , 2W, , r = 0.5Ω,, E2 = 90 V, External resistance = R, For no current pass through the battery, 100, 90, =, R+ r, R, 9, 10, =, ⇒, 1, R, R+ Ω, 2, ⇒, 10R = 9R + 4.5 Ω, ∴, R = 4.5 Ω, , or, , 2W, , 2W, A, , 22 Each resistance is connected with two, , 14 V, 0.8 W, , cells combined in opposite direction, so, potential drop across each resistor is, zero. Hence, the current through each of, resistor is zero., , 2W, 4W, A, , 23 In a balanced Wheatstone bridge, there is, no effect on position of null point, if we, exchange the battery and galvanometer., So, option (a) is incorrect., , 15 The equivalent emf of this combination, 14 V, 0.8 W, , is given by, E eq =, , E2 r1 + E1 r2, r1 + r2, , This suggest that the equivalent emf, E eq of the two cells is given by, E 1 < E eq < E2 ., , 16 Ohm’s law, I = V, R, V2, R, When the resistors are connected in, series, the effective resistance is more, than that as when they are connected in, parallel., , and power = I 2 R =, , 17 When switch S is opened, then right, side resistance R which was short, circuited earlier contributes to, equivalent resistance. Hence, equivalent, resistance across the battery increases,, power dissipated by left resistance R, decreases, voltmeter reading decreases, and ammeter reading decreases., , 18 Effective emf of circuit = 10 − 3 = 7 V, Total resistance of circuit, = 2 + 5+ 3 + 4, = 14 Ω, Current, I = 7 / 14 = 0.5 A, Potential difference between, A and D = 0.5 × 10 = 5 V, Potential at D = 10 − 5 = 5V, Potential at E = 5 − 3 = 2 V, , Total resistance of the circuit, 2× 4, 12.8, 8, Ω, =, + 0.8 = + 0.8 =, 6, 6, 2+ 4, Main current in the circuit, 14, 84, =, =, A, (12.8 / 6) 12.8, Reading of ammeter, 84 2, =, × = 2.18 A, 12.8 6, from cells of emf 6 V and 4 V in the, circuits, respectively. Then,, 6, I1 =, =1A, 2+ 3+ 1, 4, and, I2 =, = 0.4 A, 6+ 4, V A − VB, V B − VC, and VC − V D, ∴ V A − VD, , = 1 × 3 = 3 V;, = 5V, = 0.4 × 6 = 2.4 V, = (V A − V B ) + ( V B − VC ), + (VC − V D ), = 3 + 5 + 2.4 = 10.4 V, , 6W, A, , 3W, , 3W, 14V, 0.8 W, , y, , 40, 40, =, 100 − 40 60, , 3x, 40, 120 2, = 3 =, =, 60 , y, 60, 1, Now, the total length = 100, 100, ∴Required length =, × 2 = 67 cm, 3, , 25 We have, X + Y = 1000 Ω, , G, , Initially,, , Y=1000 – X, , …(i), , X, G, , (l – 10), , (110 – l), , 1000 − X, X, =, l − 10, 100 − (l − 10), 1000 − X, X, =, l − 10, 110 − l, , …(ii), , From Eqs. (i) and (ii), we get, 100 − l, l − 10, =, l, 110 − l, , Q, 3Ω, , X 1000 − X, =, l, 100 − l, , When X and Y are interchanged, then, , or, 9V, , 100 – l, , l, , KCL. Consider the grounded circuit as, shown below., 6V, P, , 1Ω, , Y=1000 – X, , X, , 21 Connect point Q to ground and apply, , 5Ω, , (100 − l) (110 − l) = (l − 10) l, , 19 The equivalent circuit of the given, , 6W, , 24 As, x =, , 20 Let I1 and I2 be the currents drawn, , Hence, E cannot be at zero potential, as, there is potential drop at E., circuit will be reduced to as shown in, figure., 2W, , 231, , Applying KCL at point Q, we can write, Incoming current at Q = outgoing, current from Q, V +6 V, 9−V, +, =, ⇒, 3, 1, 5, 1 1, 9, V + + 1 = − 2, 3 5, 5, 5 + 3 + 15 9 − 10, or V , =, , , 15, 5, , or, , 11000 − 100 l − 110 l + l2 = l2 − 10 l, ⇒, 11000 = 200 l, ∴, l = 55cm, Substituting the value of l in Eq. (i), we, get, X, 1000 − X, =, 55 100 − 55, ⇒, ∴, , 20 X = 11000, X = 550 Ω
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232, , DAY NINETEEN, , 40 DAYS ~ JEE MAIN PHYSICS, , l − 2, l − 3 8, , 4 =, 3 , 2 , , 26 As, , , Q Potential drop across capacitor, = Potential drop across r2, Er2, = Ir2 =, r + r2, , ⇒, , l =6, l − 2, Therefore, r = , 4 = 8Ω, 2 , , 27 Potential gradient along the wire,, 6, V/cm, 300, Potential difference across 50 cm length, is, 6, V = k × 50 =, × 50 = 1 V, 300, K =, , ∴ Stored charge of capacitor,, Q = CV, r2, = CE, r + r2, , 0.1m, , 5V, −4, , v d = 2. 5 × 10 m/s, , 30 When the battery is undergoing, charging processes, then, V = E + Ir > E, So, Statement I is correct., , ⇒ n = 8 × 1028 / m3, We know that,, J = nev d, or I = ne v d A, , Statement II is also correct but not, explaining Statement I., , 31 With increase in temperature, the value, of unknown resistance will increase., In balanced Wheatstone bridge, condition,, R, l, = 1, X, l2, Here,, R = value of standard resistance,, X = value of unknown resistance., To take null point at same point or, remain unchanged,, , l1, to, l2, , R, should also, X, , remain unchanged., Therefore, if X is increasing R, should, also increase., , 32 If the diameter of wire AB is increased,, its resistance will decrease. Hence, the, potential difference between A and B, due to cell C1 will decrease. Therefore,, the null point will be obtained at a, higher value of x., , or, , V, = nev d, ρL, ρ=, =, , V, nev d L, 5, 8 × 1028 × 1. 6 × 10−19 × 2. 5, × 10−4 × 0.1, , ρ = 1.6 × 10−5 Ω m, , 2 In steady state, no current flows, through the capacitor. So, resistance r1, becomes ineffective., So, the current in circuit,, E, I =, r + r2 (Total Resistance), , 52 cm, , G, E, r, On balancing,, …(i), E = 52 × x, where, x is the potential gradient of the, wire., When the cell is shunted,, E¢, , 40 cm, , G, E, r, , A, 60°, , 10W, , 10W, C, , 10W, , 2, , A, V, ∆R, ∆l, =2, = + 0.2%, ∴, R, l, , E¢, , R1 + R2 = 10 + 10 = 20 Ω, , 10W, , 1 R = ρl = ρl (V = volume), , 6 With only the cell,, , 4 We have, equivalent resistance in series, , B, , SESSION 2, , 1W, , = 11.56 V, ∴ V = 11.56 V, , where, symbols have their usual, meaning., V, = nev d A, ⇒, R, VA, or, = nev d A, ρL, or, , + 12V, –, , For parallel combination of cells,, E1, E, + 2, r1, r2, Eeq =, 1, 1, +, r1, r2, 12 13, +, 2 = 37 V, ∴ Eeq = 1, 1 1, 3, +, 1 2, Potential drop across 10 Ω resistance,, E , 37 / 3, V =, × 10, × 10 =, R, , 10 + 2 , total , , , , 3, , 3 According to the question,, , . × 4 × 10−7, Iρ 02, =, . V/m, = 01, A, 8 × 10−7, , 10 W, , –, +, 2W, 13V, , 28 Potential gradient of a potentiometer,, K =, , 5, , D, , 10W, E, , 1, and in parallel =, =, RP, 20, ⇒, RP =, 3, 20, Therefore, Req =, +, 3, 20 +, =, =, , 1, 1, 3, +, =, 20 10 20, , 10 + 10, 30 + 30, 3, , 80, = 26.66 = 26.7 Ω, 3, , R=5 Ω, Similarly, on balancing,, Er, V =E−, = 40 × x, (R + r ), Solving Eqs. (i) and (ii), we get, E, 1, 52, =, =, V 1− r, 40, R+r, R+ r, 52, E, ⇒, =, =, 40, V, R, 5+ r, 52, ⇒, =, 5, 40, 3, ⇒, r = Ω ⇒ r = 1.5 Ω, 2, , …(ii)
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234, , DAY TWENTY, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY TWENTY, , Unit Test 4, (Electrostatics and, Current Electricity), 1 Two concentric spheres of radii r1 and r2 carry charges, , 4 In the given circuit, it is observed that the current I is, , q1 and q 2 , respectively. If the surface charge density ( σ ), is the same for both the spheres, the electric potential at, the common centre will be, r, σ, × 1, ε0 r 2, σ, (c), (r1 − r2 ), ε0, , independent of the value of resistance R 6. Then, the, resistance value must satisfy, R6, , σ r2, ×, ε0 r1, σ, (d), (r1 + r2 ), ε0, , (a), , (b), , I, , R1, , R5, , R3, , R4, , R2, , 2 Point charge q moves from, point P to point S along the, E, path PQRS (figure shown) in a, P, uniform electric field E pointing, Q, S, coparallel to the positive, direction of the X-axis. The, R, coordinates of the points, P , Q , R and S (a,b,0) ( 2a,0, 0) (a,−b,0) and ( 0, 0, 0), respectively.The work done by the field in the above, process is given by the expression, (b) −qEa, (d) qE [(2a)2 + b 2 ], , (a) qEa, (c) qEa 2, , 3 The variation of potential with distance R from a fixed, Potential in volts, , point is as shown below.The electric field at R = 5 m is, , (a) 2 .5 V/m, (c) 2 / 5 V/m, , X, , 1, 1, 1, 1, +, =, ×, R 5 R 6 R1 + R 2 R 3 + R 4, , (b) R1R 4 = R 2R 3, (c) R1R 2R 5 = R 3R 4R 6, (d) R1R 3 = R 2R 4 = R 5R 6, , 5 Two cells of internal resistance r1 and r2 ; and at same emf, are connected in series, across a resistor of resistance R., If the terminal potential difference across the cells of, internal resistance r1 is zero, then the value of R is, (a) R = 2 (r1 + r2 ), (c) R = r1 − r2, , (b) R = r2 − r1, (d) R = 2 (r1 – r2 ), , 6 The electric dipole moment of an electron and proton, 4.30 nm apart is, (a) 6.88 × 10−28 C-m, (c) 6.88 × 1028 C-m, , 5, 4, 3, 2, 1, 0, , (a), , (b) 5.88 × 10−28 C-m, (d) 5.88 × 1028 C-m, , 7 At what distance along the central axis of a uniformly, 1 2 3 4 5 6, Distance R in metre, , (b) −2 .5 V/m, (d) −2 / 5 V/m, , charged plastic disk of radius R is the magnitude of the, electric field equal to one-half the magnitude of the field, at the centre of the surface of the disk?, (a), , R, 2, , (b), , R, 3, , (c) 2 R, , (d) 3 R
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UNIT TEST 4 (ELECTROSTATICS AND CURRENT ELECTRICITY), , DAY TWENTY, , 8 Work done in placing a charge of 8 × 10−18 C on a, , 16 In figure, the net potential at point P due to the four point, , condenser of capacity 100 µF is, −32, , charges, if V = 0 at infinity is, −28, , (a) 16 × 10 J, (c) 64 × 10−32 J, , (b) 3.1 × 10, J, (d) 32 × 10−32 J, , d, , (a) 1.96 × 10 V, 18.6 ms, (c) 1.96 × 104 V, 19.6 ms –2, , (b) 1.86 × 10 V, 18.6 ms, (d) 2.96 × 104 V, 17.6 ms –2, , –2, , 4, , 10 What is the direction of the electric field at the centre of, the square of figure, if q = 1.0 × 10 −8 C and a = 5.0 cm?, Y, , X, , +q, , –2.0q, , a, a, , –q, , (a) 30° with X-axis, (c) 60° with X-axis, , d, +5q, , 3q, (a), 8 πε0, , a, , (d), , 9q, 8 πε0, , identical resistors are as shown in figure. Which graph, represent parallel combination?, B, , +2.0q, , I, , (b) 45° with X-axis, (d) 90° with X-axis, , (c) 1 Ω, , (a) Graph-A, (c) Both graph A and B, , (b) 60°C, (d) 80°C, , following diagram is, 2W, , A, , 2W, , 2W, , 2W, , 2W, , 4W, , 6W 2, , 6W 1, , 2W, , B, 2W, , (a) 2 Ω, , 2W, , 2W, , (b) 8 Ω, , 2W, , 2W, , (c) 9 Ω, , 2W, , (d) 10 Ω, , 19 Two cells connected in series have electromotive force of, , air near earth’s surface. A particle of charge −2 × 10−9 C, is acted on by a downward electrostatic force of, 3 × 10−6N, when placed in this field. The ratio of the, magnitude of the electrostatic force to the magnitude of, the gravitational force in the case of proton is, (a) 1.6 × 10−19 (b) 1.5 ×10−10 (c) 1.6 × 1019, , (b) Graph-B, (d) None of these, , 18 The equivalent resistance between points A and B in the, , (d) 0.5 Ω, , 13 A charged cloud system produces an electric field in the, , 1.5 V each. Their internal resistance are 0.5 Ω and 0.25 Ω, respectively. This combination is connected to a, resistance of 2.25 Ω . Potential difference across the, terminals of each cell, 1.5 V, , (d) 1.4 × 1010, , 14 An infinite non-conducting sheet has a surface charge, density σ = 0.10 µC m – 2 on one side. How far apart are, equipotential surfaces whose potentials differ by 50 V?, (b) 6.8 × 10−3 m, (d) 8.8 × 10−3 m, , 15 An electron is released from rest in a uniform electric, , field of magnitude 2.00 × 10 4 NC −1. Acceleration of the, electron is (ignore gravitation), (a) 2.51 × 1015 ms –2, (c) 3.51 × 1015 ms –2, , 7q, 8 πε0, , A, , is 60 Ω. At which temperature its resistance will be, 25Ω?, , (a) 5.8 × 10−3 m, (c) 7.8 × 10−3 m, , (c), , 17 V -I graphs for parallel and series combination of two, , 12 The resistance of a wire at 20°C is 20 Ω and at 500°C, , (a) 50°C, (c) 70°C, , 5q, (b), 8 π ε0, , a, , at length 240 cm. On shunting the cell with a resistance of, 2 Ω, the balancing length becomes 120 cm. The internal, resistance of the cell is, (b) 2 Ω, , d, , P, , V, , 11 In a potentiometer experiment, the balancing with a cell is, , (a) 4 Ω, , –5q, , d, , 2.4 × 10 −18 C is suspended between two charged, horizontal plates at a distance 1.0 cm apart. Find the, potential difference between the plates. If polarity of the, plates be changed, then calculate the instantaneous, acceleration of the drop., –2, , –5q, , +5q, , 9 A drop, having a mass of 4.8 × 10 −10 g and a charge of, , 6, , 235, , (b) 2.51 × 10−15 ms –2, (d) 3.51 × 10−15 ms – 2, , 1.5 V, 0.5 W, , 0.25 W, , r1, , r2, I, , I, 2.25 W, , (a) 1 V, 0.25 V, (c) 1.5 V, 2.25 V, , (b) 1 V, 1.25 V, (d) 1.5 V, 2.56 V, , 20 A charge of 0.8 C is divided into two charges Q1 and Q 2 ., These are kept at a separation of 30 cm. The force on Q, is maximum, when, (b) Q ≈ 0.8 C,Q2 is negligible, (a) Q1 = Q2 = 0.4 C, (c) Q1 is negligible, Q2 ≈ 0.8 C (d) Q1 = 0.2 C,Q2 = 0.6 C
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236, , DAY TWENTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 21 A particular 12 V car battery can send a total charge of, 84 A-h through a circuit, from one terminal to other. If this, entire charge undergoes a potential difference of 12 V,, how much energy is involved?, (a) 1.6 × 106 J, (c) 3.6 × 106 J, , (b) 2.6 × 106 J, (d) 4.6 × 106 J, , (a) both bulbs work properly, (b) both bulbs will fuse, (c) Only 60 W bulb will fuse, (d) Only 90 W bulb will fuse, , 27 Masses of the three wires of same material are in the ratio, of 1 : 2 : 3 and their lengths in the ratio of 3 : 2 : 1., Electrical resistance of these wires will be in the ratio of, , 22 AB is uniform resistance wire of length 1 m. A 2 V, accumulator, a Daniell cell of 1.08 V and a galvanometer, G are connected as shown. If the sliding contact is, adjusted for null deflection then the potential gradient in, AB and the balancing length, measured from end A are, respectively., , (a) 1 : 1 : 1, , (c) 9 : 4 : 1, , (d) 27 : 6 : 1, , 28 Six wires, each of resistance r1 are connected so as to, form a tetrahedron. The equivalent resistance of the, combination when current enters through one corner and, leaves through other corner is, (a) r, , 2V, , (b) 1 : 2 : 3, , (b) 2r, , (c), , r, 3, , (d), , r, 2, , 29 A unit negative charge with mass M resides at the, B, , A, , mid-point of the straight line of length 2a adjoining two, fixed charges of magnitude + Q each. If it is given a very, small displacement x ( x < < a ) in a direction, perpendicular to the straight line, it will, , 1.08 V, a, , (a) 0.02 V/cm, 54 cm, (c) 0.0092 V/cm, 49.6 cm, , (b) 0.0308 V/cm, 46 cm, (d) 0.02 V/cm, 50.4 cm, , (a) came back to its original position and stay there, Q, 1, (b) execute oscillations with frequency, 2 π 4 πε0Ma 3, , 23 A potentiometer wire of length 200 cm has a resistance, of 20 Ω. It is connected in series with a resistance of10 Ω, and an accumulator of emf 6 V having negligible internal, resistance. A source of 2.4 V is balanced against a, length L of the potentiometer wire. Find the value of L., 6V, , 10 W, , (d) execute oscillations with frequency, , 1, Q, 2 π 4 πε0Ma 2, 1, Q, 2 π 2 πε0Ma 3, , 30 A 28 µF capacitor is charged to 100 V and another 2 µ F, capacitor to 200 V, they are connected in parallel. Then,, the total final energy is, , L, , A, , (c) execute oscillations with frequency, , B, , (a) 01537, ., J, , (b) 0.0155 J, , (c) 01865, ., J, , (d) 0123, ., J, , 31 A hollow copper tube of 1m length has got external, 2.4 V, , (a) 100 cm, , (b) 120 cm, , diameter equal to 10cm and its walls are 5mm thick. Then, the resistance of tube, if its specific resistance is, 17, . × 10−8 Ω m, is, , G, , (c) 110 cm, , (d) 140 cm, , (a) 1139, ., × 10−5 Ω, (c) 1150, ., × 10−5 Ω, , 24 A charged ball A hangs from a silk thread,, which makes an angle φ with a large, charged conducting sheet B as shown in, the figure. The surface charge density of, the sheet is proportional to, (a) sin φ, (c) cos φ, , B, , φ, , 32 Manjeet’s room heater is marked as 1000 W-200V. If the, A, , (b) cot φ, (d) tan φ, , carry charges +2q and +6q, respectively. The magnitude, of electric field at a distance x outside and inside from, the surface of outer sphere is same, then the volume of x, is, 2r, (b), 3, , voltage drops to 160 V, the percentage change in the, power of the heater is, (a) 40%, , 25 Two concentric conducting thin shells of radius r and 2r, , r, (a), 2, , r, (c), 3, , (b) 1327, ., × 10−6 Ω, (d) 1125, ., × 10−4 Ω, , r, (d), 6, , 26 There are two electric bulbs rated 60 W, 120 V and 90 W,, 120 V. They are connected in parallel with 240 V supply,, then, , (b) 42%, , (c) 36%, , (d) 50%, , Direction (Q. Nos. 33-40) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true
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UNIT TEST 4 (ELECTROSTATICS AND CURRENT ELECTRICITY), , DAY TWENTY, , 33 Statement I A and B are two conducting spheres of, , 237, , V, E = 2V, , same radius. A being solid and B hollow. Both are, charged to the same potential. Then,, Charge on A = charge on B., , r = 1W, A, , Statement II Potentials on both are same., , R = 1W, , 34 Statement I A thin metallic wire is bent into semicircular, 37 Statement I In the following circuit, emf is 2V and internal, , shape, then its resistivity decreases., , resistance of the cell is1 Ω and R = 1 Ω, then reading of, the voltmeter is 1V., 2, Statement II V = E − Ir , where E = 2 V, I = = 1 A and, 2, R = 1 Ω., , Statement II On bending, the drift of electron in the wire, remains same., , 35 Statement I The circuits containing capacitor be handed, cautiously, even when there is no current., Statement II A dielectric differs from an insulator., , 38 Statement I The power delivered to a light bulb is more, just after it is switched ON and the glow of the filament is, increasing, as compared to when the bulb is glowing, steadily, i.e. after sometime of switching ON., Statement II As temperature increases, resistance of, conductor increases., , 36 Statement I In the following circuit, the net resistance, between points A and B is R, R, A, , 39 Statement I When a wire is stretched, so that its diameter, , B, R, , R, , is halved, its resistance becomes 16 times., Statement II Resistance of wire decreases with increase, in length., , R, , R, , 40 Statement I A potentiometer is preferred over that of a, , Statement II All the resistances are in parallel to each, other., , voltmeter for measurement of emf of a cell., Statement II Potentiometer is preferred as it does not, draw any current from the cell., , ANSWERS, 1. (d), , 2. (b), , 3. (a), , 4. (b), , 5. (c), , 6. (a), , 7. (b), , 8. (d), , 9. (c), , 10. (b), , 11. (b), , 12. (d), , 13. (d), , 14. (d), , 15. (c), , 16. (b), , 17. (a), , 18. (b), , 19. (b), , 20. (a), , 21. (c), , 22. (a), , 23. (b), , 24. (d), , 25. (b), , 26. (b), , 27. (d), , 28. (d), , 29. (d), , 30. (c), , 31. (a), , 32. (c), , 33. (a), , 34. (d), , 35. (b), , 36. (c), , 37. (a), , 38. (a), , 39. (c), , 40. (a), , Hints and Explanations, q1, q2, +, 4 πε0 r1, 4 πε0 r2, σ, σ, V =, × r1 +, × r2, ε0, ε0, σ, =, (r1 + r2 ), ε0, × σ, , ×σ, , 2 As electric field is a conservative field, Hence, the work done does not depend, on path, , between B and C because the slope of, BC is same throughout (i.e. electric, field between B and C is uniform)., , E, , V =, , Q q1 = 4 πr12, , 2, q2 = 4 πr2, , 3 Intensity at 5m is same as at any point, , Y, , centre, is, , (a,b,0), , P, , b, a, , Ö a2 + b 2, q, q, , O, q, , S, , √ a2 +, , b2, , Q, √ a2 + b 2, , b, R, , ∴, , W PQRS = W PQS = W PO + WOS, = Fb cos 90°+ Facos 180°, = 0 + q Ea(−1) = − q Ea, , X, , Potential in volt, , 1 Electric potential of the common, , 5, 4, 3, 2, 1, , O, , A, , B, , 2 3 4 5, 1, Distance R in metre, , C, 6
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238, , DAY TWENTY, , 40 DAYS ~ JEE MAIN PHYSICS, , Therefore electric field at R = 5m is, equal to the slope of line BC hence by, − dV, E =, ;, dr, E =−, , (0 − 5), = 2. 5V/m, 6− 4, , 1, qV, 2, 1, = × 8 × 10−18 × 8 × 10−14, 2, = 32 × 10−32, , Energy stored =, , 9 Let m and q be the mass and the charge, , 4 From figure, it can be seen that I is, independent of resistance R 5, so no, current flow through it. This require, that the R1 and R2 junction is at same, potential of the junction at R3 and R 4 ., So, according to Wheatstone bridge, condition,, R1, R, = 3 ⇒ R1 R 4 = R2 R3, R2, R4, , 5 Given, V1 = 0, V1 + V2 = IR, ⇒ V2 = IR, , of the drop and E the intensity of, electric field between the plates., Since, the drop is in equilibrium, the, electric force qE acting on it balances, its weight mg, i.e., qE = mg, If the potential difference between the, plates is V and the distance between, them is d, then E = V /d ., ∴, q (V /d ) = mg or V = mgd /q, Here, m = 4.8 × 10−10 g = 4.8 × 10−13 kg,, g = 9.8 N kg −1 ,, , − E − Ir2 = IR, But V1 = E − Ir1 ⇒ E = Ir1, ⇒ Ir1 − Ir2 = IR, or, R = r1 − r2, , [QV1 = 0], , d = 1.0 cm = 1.0 × 10−2 m, q = 2.4 × 10−18C, , and, ∴, , V =, , 6 Magnitude of a dipole moment is, p = qd, = (1.60 × 10−19 )(430, . × 10−9 ), = 6.88 × 10−28 C-m, , 7 At a point on the axis of a uniformaly, charged disk at a distance x above the, centre of the disk, the magnitude of the, electric field is, E =, , But, , σ , 1 −, 2ε0 , , , , , x + R , 2, , Then, 1 −, , x, x2 + R2, x, , or, , x +R, 2, , 2, , =, , 1, 2, , =, , 1, 2, , x, , x2 + R2, , =, , −18, , 1 2q, q , 1, q, −, , =, 4πε 0 a2 /2 a2 /2 4πε 0 a2 /2, , (9 × 109 )(1.0 × 10−8 ), , (0.050) 2, 2, = 7.19 × 104 NC –1, q , 1 2q, and E y =, −, 4 πε0 a2 /2 a2 /2 , q, 1, = 7.19 × 104 N / C, =, 4 πε0 a2 /2, , 1, 4, , The magnitude of the field is, E =, −4, , 8 Here, q = 8 × 10 C, C = 100µF = 10 F, V =, , a = 2 g = 19.6 ms –2, , =, , x2, R2, x =, +, 4, 4, R2, 2, x =, 3, R, x=, 3, , Thus,, , On changing the polarity of the plates,, the electric force q E will also be, directed downwards. Then, the, acceleration of the drop is, qE + mg, a=, m, But, qE = mg, , Ex =, , 2, , or, , = 1.96 × 104 V, , 2a, a, =, d =, 2, 2, Net field due to these two charges is, , Squaring both side, we get, 2, , 2.4 × 10−18, , 10 Since, each charge distance from centre, , σ, E, 1, such that, Ec =, =, 2 ε0, Ec, 2, , q, 8 × 10−18, = 8 × 10−14 V, =, C, 10−4, , where, l 1 = 240 cm, l 2 = 120 cm, 240 − 120, 120, × 2 = 2Ω, =, × 2=, 120, 120, R t = R 0 (1 + αt ), , 12 Use, , 20 = R 0 (1 + 20 α ), 60 = R 0 (1 + 500 α ), Here,, R t = 25 Ω, Solving, we find t = 80°C, , Here,, , 13 We have, F = qE, Thus, E =, , E 2x + E 2y = 2 (7.19 × 104 ) 2, , = 1.02 × 105 NC –1, Angle made with the x-axis is, Ey, θ = tan −1, Ex, = tan −1 (1) = 45°, , F 3 × 10−6, =, q, 2 × 10−9, , = 1.5 × 103 NC –1, Magnitude of the electrostatic force on a, proton is, Fe = eE = (1.60 × 10−19 )(1.5 × 103 ), = 2.4 × 10−16 N, , (4.8 × 10−13 ) × 9.8 × (1.0 × 10−2 ), , ∴, , x, , 2, , It is upward in the diagram, from the, centre of the square towards the centre of, the upper side., l −l, 11 Here, r = 1 2 × 2 Ω, l2, , Magnitude of the gravitational force on, the proton is, F g = mg = (1.67 × 10−27 )(9.8), = 1 . 63 × 10−26 N, The ratio of the force is, 2.4 × 10−16, Fe, =, Fg, 1. 63 × 10−26, = 1. 4 × 1010, , 14 Electric field, E = σ, , 2ε0, x, , and electric potential, V = Vs − ∫ E dx, 0, , = Vs − Ex, Here, two surfaces are separated by ∆x,, then their potentials difference in, magnitude by, σ , ∆V = E∆x = , ∆x, 2ε0 , Thus,, , ∆x =, =, , 2ε0 ∆V, σ, 2(8.85 × 10−12 )(50), . × 10−6, 010, , = 8.8 × 10−3 m, , 15 We know that, F = eE, By Newton’s second law,, F, eE, a=, =, m, m, (1.60 × 10−19 )(2.00 × 104 ), =, 911, . × 10−31, = 3.51 × 1015 ms – 2
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UNIT TEST 4 (ELECTROSTATICS AND CURRENT ELECTRICITY), , DAY TWENTY, , The force of Q 1 due to Q 2 ,, , 16 Net potential at point P,, V =, , q, 4 πε0, , 5, 5q, 5, 5, 5, − 2d − d + d + d = 8 πε, , , 0, , 17 R Parallel < RSeries . From graph A it is, clear that slope of the line A is lower, than the slope of the line B., Also, slope = resistance, so line A, represents the graph for parallel, combination., , 18 The resistances 2 Ω and 2 Ω at the last, terminals are outside the circuit and so, they may be ignored. Now, in loop 2,, the resistances (2 Ω + 2 Ω ), 4 Ω and, (2 Ω + 2 Ω ) are in series. Their, equivalent resistance is 12 Ω, which is in, parallel with 6 Ω. The equivalent, resistance is, 6 × 12, R′ =, = 4Ω, 6 + 12, , A, , 2W, , 2W, , 2W, , 2W, , 2W, , 6W 2, , 6W 1, , 2W, 4W, , B, 2W, , 2W, , 2W, , 2W, , 2W, , 2W, , Similarly, in loop 1, the resistances, (2 Ω + 2 Ω), R ′ (= 4 Ω ) and (2 Ω + 2 Ω ) are, in series and these are in parallel with, 6 Ω. Hence, their equivalent resistance, is, R ′′ = 4 Ω, Lastly, between the points A and B, the, resistances 2 Ω, R ′ ′ = (4 Ω ) and 2 Ω are in, series. Hence, their equivalent, resistance is, 2Ω + 4Ω + 2Ω = 8Ω, , 19 The arrangement is shown in the figure., The effective emf in the circuit is, E = 1.5 + 1.5 = 3.0 V, and the total resistance is, R = 0.5 + 025, . + 225, . = 3.0 Ω, Hence, the current in the circuit is, E 3.0, I =, =, = 1.0 A, R 3.0, Potential difference across the terminals, of the first cell is, V1 = E − Ir1 = 1.5 − (1.0) × (0.5) = 1.0 V, Potential difference across the terminals, of the second cell is, V2 = E − Ir2 = 1.5 −(1.0) ×(0.25) = 1.25 V, , 20 Given,, Q 1 ← → Q 2, 30 cm, , (a) Q 1 = Q 2 = 0.4 C, , F = kP, , Q 1Q 2, , 30 × 10−2, , Q Q × 100, =K 1 2, 30, , 0.4 × 0.4 × 100, 8, =, K, 30, 15, (b) When Q 1 = 0.8 C,Q 2 ≈ 0, 0.8 × 0, F =k ×, =0, 30 × 10−2, =K ×, , (d) When Q 1 = 02, . C, Q 2 = 0.6 C, 02, . × 0.6 100 2, F =k ×, = k, 30, 5, Hence, for Q 1 = Q 2 = 0.4C, the force will, be maximum., , 21 An ampere is coulomb per second, so, 84 A - h = 84 × 3600 = 3.0 × 105C, The change in potential energy is, ∆U = q∆V = 3.0 × 105 × 12 = 3.6 × 106 J, , 22 Potential difference per cm, 2V, = 0.02 V/cm, 100 cm, 100, Balancing length =, × 1.08 = 54 cm, 2, =, , 23 The current in the potentiometer wire, AB is, , The length of the wire is l = 200 cm. So,, the potential gradient along the wire is, V, 4, k =, =, l, 200, = 0.02 Vcm –1, The emf 2.4 V is balanced against a, length L of the potentiometer wire., 2.4, i.e. 2. 4 = kL or L =, k, 2.4, =, = 120 cm, 0.02, , 24 Electric field at point A, , φ, , T φ, , …(i), , T cos φ, , T sin φ A, , F=QE, , σ ∝ tan φ, , surface of outer shell, = Electric field at a distance x inside from, surface of outer shell, 2q + 6q, 2q, 1, 1, i.e., ⋅, =, ⋅, 4 πε0 (2r − x )2, 4 πε0 (2r + x )2, 4, 1, =, ⇒, (2r + x )2 (2r − x )2, 2, 1, ⇒, =, 2r + x 2r − x, ⇒, 4r − 2 x = 2r + x, ⇒, 2r = 3 x, 2r, x=, ⇒, 3, v 12 1202, =, = 240 Ω, P1, 60, , v 22 1202, =, = 160 Ω, 90, P2, P, 60, ∴ I1 = 1 =, = 0.5 A, v 1 120, P, 90, I2 = 2 =, = 075, . A, v 2 120, R2 =, , The potential difference across the, potentiometer wire is, V = current × resistance, = 0.2 × 20 = 4 V, , B, , ⇒, , 25 Electric field at a distance x outside from, , 26 R1 =, , 6, I =, = 0. 2 A, 20 + 10, , σ, ε0, , For equilibrium of forces at A,, T sin φ = Q E, σ, [from Eq. (i)], T sin φ = Q ⋅ ,, ε0, Qσ, …(ii), T sin φ =, ε0, and, …(iii), T cos φ = mg, On dividing Eq. (ii) by Eq. (iii), we get, Qσ, tan φ =, ε0 mg, ε mg, or, σ = 0, tan φ, Q, , (c) When Q 1 ≈ 01 , Q 2 = 0.8 C, 0 × 0.8, F =k ×, =0, 30 × 10−2, , E=, , 239, , When both bulbs are connected in, parallel, then equivalent resistance, R1 R2, RP =, R1 + R2, 240 × 160, = 96 Ω, =, 240 + 160, ∴ When they are connected with 240 V, supply, then, 240, I =, = 2.5 A, 96, Now, current in 60 W bulb,, 160, I1′ = I ⋅, 400, 160, = 2.5 ×, = 1A, 400, Current in 90 W bulb,, 240, I2′ = 2.5 ×, = 1.5 A, 400, Since, I1′ > I and I2′ > I2, Hence, both bulbs will fuse.
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240, , DAY TWENTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 27 Mass, M = Volume × Density, = Al × d, , or, , A=, , M, ld, , ρl2d, ρl, ρl, Resistance, R =, =, =, M, A , M, , ld , R∝, , So,, , Thus, R1 : R2 : R3 =, , M1, , :, , l 22, M2, , :, , 1 2KQ, 1, =, 2 π Ma3 2 π, , =, , 1, Q, 2 π 2 π ε0 Ma3, , C2 = 3 µ F = 3 × 10−6 F, V1 = 100V and V2 = 200V, ⇒ Charge of C1 ,, q1 = C1V1 = 28 × 10−6 × 100, , l 23, M3, , 32 22 12, =, :, :, = 27 : 6 : 1, 1 2 3, , = 28 × 10−4 C, Charge of C2, q2 = C2V2 = 3 × 10−6 × 200, , 28 Six wires each of resistance r from a, tetrahedron as shown in the following, figure., 3, r, , 1, , r, , 1, r, , 3, , r, 4, , 2, , 4, , r, In Wheatstone circuit, the equivalent, resistance of upper circuit, I, I, 1, 2, 1, =, +, =, =, R 2r, 2r 2r, r, ⇒, R=r, It will be in parallel with outer, resistance,, 1, 1 1 2, = + =, R eq, r, r, R, , ⇒, , R eq, , = 6 × 10−4 C, , r, = ., 2, , q + q2 28 × 10−4 + 6 × 10−4, V = 1, =, C1 + C2 28 × 10−6 + 3 × 10−6, , Total final energy, U =, , 1, (C1 + C2 )V 2, 2, , case the capacitor is connected in a, circuit containing a source of high, voltage, the capacitor charges itself to a, very high potential. So, if a person, handles it without discharging, he may, get a severe shock.Dielectrics and, insulators cannot conduct electricity but, in case of a dielectric, when an external, field is applied, induced charges appear, on the faces of the dielectric. In other, words, the dielectric have the property of, transmitting electric effects without, conducting., This is balanced Wheatstone bridge, hence, resistance in branch MN is not, taken into consideration. Hence, the, equivalent resistance between points A, and B is given by, M, , 1, (28 × 10−6 + 3 × 10−6 )(109.68)2, 2, = 01865, ., J, , R, , 10, = 5cm = 0.05m, 2, Internal radius r1 = r2 − thickness of tube, = 0.05 − 0.005, = 0.045 m, Area of cross-section = π (r22 − r12 ), , R, , = π [(0.05)2 − (0.045)2 ], = 1.492 × 10−3 m2, , 32 Resistance of heater,, –1C, , F, –Q, , F net = − F cos θ + (− F cos θ), = − 2F cos θ, kQ (− 1), x, = −2, ×, x2 + a2, x2 + a2, =, , Or, , F net, , 2kQ, , ( x2 + a2 )3 /2, 2 kQ , = + , ⋅ x, a3 , , Frequency of oscillation,, , V 2 (200)2, =, = 40Ω, P, 1000, Power of heater at V ′ = 160V, V ′2 (160)2, P′=, =, = 640 W, 40, R, Percentage fall in the power of the, heater,, P − P′, 1000 − 640, × 100 =, × 100 = 36%, P, 1000, R=, , ⋅x, , 34 Resistivity of metallic wire does not, [Q x << a], , B, , r2 =, , N, , or, ∴, , depend on shape of wire because it is a, material property. On bending, the, cross-sectional area of wire changes but, , 1, 1, 1, =, +, R AB, (R + R ) (R + R ), 1, 2, 1, =, =, R AB, 2R R, R AB = R, , 39 The resistance of a wire is, R =ρ, or, , F, q, q, , R, R, R, , A, , 31 External radius,, , = 1139, ., × 10−5 Ω, , +Q, , 35 A capacitor does not discharge itself. In, , =, , Resistance of copper tube,, . × 10−8 × 1, l 17, R =ρ =, a 1.492 × 10−3, , 29 From figure the net force,, , drift velocity of electron does not depend, on area of cross-section, so it remains, same., , 36 The equivalent circuit is represented as,, , Potential,, , = 109.68 V, 2, , r, , 2×1×Q, 4 π ε0 Ma3, , 30 Here C1 = 28 µ F = 28 × 10−6 F, , l2, M, l 21, , =, , or, , l, ,ρ being specific resistance, A, Al, R∝, A2, 1, R∝, (Q A = πr 2 ), r4, , Hence, when diameter is halved the, resistance of the wire is, 1, …(i), R∝, = 16 R, 4, r, , 2, Hence, its resistance will become, 16 times., Again from Eq. (i), we get, l, l2, or R ∝, or R ∝ l 2, R∝, A, Al, Therefore, on increasing the length, resistance increases.
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DAY TWENTY ONE, , Magnetic Effect, of Current, Learning & Revision for the Day, u, , u, , u, , Concept of Magnetic Field, Biot-Savart’s Law and its, Applications, Ampere’s Circuital Law, , u, , u, , Force on a Moving Charge in, Uniform Magnetic Field, Cyclotron, , u, , u, , Magnetic Force on a Current Carrying, Conductor, Moving Coil Galvanometer, , Concept of Magnetic Field, If a magnet is placed in a magnetic field, then it experiences a force on it, Also, when a, magnet is placed near a current carrying conductor, then it experiences the similar force,, it means that current carrying conductor produces a magnetic field around it. This effect, of current is called magnetic effect of current., , Biot-Savart’s Law and its Applications, The magnetic field d B at a point P, due to a current element I dl is given by, µ I (d l × r), dB = 0, 4π, r3, where, θ is the angle between d l and r., , PREP, MIRROR, , I, θ, , Your Personal Preparation Indicator, , dl, u, , r, P, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, , X, inside, , Direction of magnetic field produced due to a current carrying straight wire can be, obtained by the right hand thumb rule., , No. of Questions in Exercises (x)—, , u, , (Without referring Explanations), u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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242, , DAY TWENTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , Magnetic Field due to Circular, Current Loop, l, , l, , l, , l, , If there is a circular coil of radius R and N number of turns,, carrying a current I through the turns, then magnetic field, at the centre of coil is given by, µ NI, B= 0, 2R, If there is a circular arc of wire, subtending an angle θ at the centre, of arc, then the magnetic field at, the centre point, µ I θ, B= 0 , 2R 2π , At a point P situated at a distance, r from centre of a current, carrying circular coil along its, axial line., The magnetic field is, B =, , θ, , O, , I, , When the wire XY is of infinite length, but the point P, lies near the end X or Y, then φ1 = 0 ° and φ2 = 90 ° and, µ I, hence,, B= 0, 4πr, When point P lies on axial position of current carrying, conductor, then magnetic field at P,, B = 0., When wire is of infinite length, then magnetic field, near the end will be half, that of at the perpendicular, bisector., , 2. Magnetic Field due to a Thick (Cylindrical) Wire, I, , R, O, , P, , I, r, , If r >> R, then at a point along the axial line, B =, , µ 0NIR2, 2r3, , l, , i5, B, dI, , A current carrying, solenoid behaves as a bar, magnet. The face, where, current is flowing, clockwise behaves as, South pole and the face,, (b), (a), where current is seen, flowing anti-clockwise, behaves as North pole., For such a solenoid, the magnetic field inside it is uniform, and directed axially., , i3, , i4, , Hence, ∫ B ⋅ dl = µ 0 ⋅ (i 1 + i2 − i3 ), , Applications of Ampere’s law, 1. Magnetic field due to Straight Current, Carrying Wire, The magnetic field due to a current, carrying wire of finite length at a, point P situated at a normal distance, r is, µ I, B = 0 (sin φ1 + sin φ2 ), 4πr, , l, , X, I, r, , φ2, φ1, , l, , P, l, , Y, , Magnetic field inside a hollow current carrying, conductor is zero., , 3. Magnetic Field due to a Solenoid, , i2, i1, , Magnetic field at a point, outside the wire, µ I, B = 0 , where r is the, B, r P, 2 πr, distance of given point, I, from centre of wire and, r > R., R, Thick cylindrical wire, Magnetic field at a point, inside the wire at a, distance r from centre of wire (r < R) is, µ I r, B= 0 ⋅ 2, 2π R, l, , Ampere’s Circuital Law, , l, , l, , R, , 2(R2 + r 2 )3 /2, , Now, consider the diagram above., Here, ∑ I = i 1 + i2 − i3, , l, , R, , µ 0 NIR2, , The line integral of the magnetic, field B around any closed path is, equal to µ 0 times the net current I, threading through the area, enclosed by the closed path., Mathematically, ∫ B ⋅ dl = µ 0 ΣI, , l, , For a wire of infinite length φ1 = φ2 = 90 ° and hence, µ I, B= 0, 2 πr, , If point P lies symmetrically on, the perpendicular bisector of wire XY, then φ1 = φ2 = φ, (say) and hence, µ I, B = 0 ⋅ 2 sin φ = µ 0I sin φ, 4πr, 2 πr, , For a solenoid coil of infinite length at a point on its, axial line, the magnetic field, B = µ 0nI, where, n is number of turns per unit length., 1, At the end of solenoid, B = µ 0nI, 2, At the end field is half of at the centre, this is called, end effect., , 4. Toroidal Solenoids, For a toroid (i.e. a ring shaped closed solenoid) magnetic, field at any point within the core of toroid B = µ 0nI ,, N, where n =, , R = radius of toroid., 2πR
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Force on a Moving Charge in, Uniform Magnetic Field and, Electric Field, l, , l, , l, , If a charge q is moving with velocity v enters in a region in, which electric field E and magnetic field B both are, present, it experiences force due to both fields, simultaneously. The force experienced by the charged, particle is given by the expression, F = q (v × B) + qE, Here, magnetic force Fm = q (v × B) = Bqv sin θ and electric, force Fe = qE., The direction of magnetic force is same as v × B if charge is, positive and opposite to v × B, if charge q is negative., , Motion of a Charged Particle in a, Uniform Magnetic Field, l, , (i) If a charge particle enters a uniform magnetic field B, with a velocity v in a direction perpendicular to that of, B (i.e.θ = 90 ° ), then the charged particle experiences a, force Fm = qvB. Under its influence, the particle, describes a circular path, such that, mv, Radius of circular path, r =, qB, In general,, mv, p, 2mK, r=, =, =, qB qB, qB, 2 mqV, 1 2 mV, =, =, qB, B, q, where, p = mv = momentum of charged, particle, K = kinetic energy of charged particle, and V = accelerating potential difference., (ii) The period of revolution of charged particle T =, , 2πm, ,, qB, , qB, the frequency of revolution ν =, 2 πm, qB, or angular frequency ω =, ⋅, m, l, , 243, , MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, , (i) If a charged particle is moving at an angle θ, to the, magnetic field (where θ, is other than 0°, 90° or 180°), it, describes a helical path, where, radius of helical path,, mv sin θ, ., r=, qB, 2πm, (ii) Revolution period, T =, qB, qB, or, Frequency, ν =, 2 πm, (iii) Moreover, pitch (the linear distance travelled during, one complete revolution) of helical path is given by, 2 πmv cos θ, p = v cos θ ⋅ T =, qB, , If the direction of a v is parallel or anti-parallel to B, θ = 0, or θ = 180 ° and therefore F = 0. Hence, the trajectory of the, particle is a straight line., If the velocity of the charged particle is not perpendicular to, the field, we will break the velocity in parallel ( v|| ) and, perpendicular (v⊥ ) components., mv⊥, r=, qB, Pitch, p = (v||)T, , Cyclotron, It is a device used to accelerate positively charged particles,, e.g. proton, deuteron, α-particle and other heavy ions to, high energy of 100 MeV or more., Bq, Cyclotron frequency, ν =, ., 2 πm, Maximum energy gained by the charged particle, q 2 B2 2, E max = , r, 2m , where, r = maximum radius of the circular path followed by, the positive ion., Maximum energy obtained by the particle is in the form of, kinetic energy., , Magnetic Force on a Current, Carrying Conductor, If a current carrying conductor is placed in a magnetic field B,, then a small current element I dl experiences a force given by, dFm = Idl × B, and the total force experienced by whole current carrying, conductor will be, Fm = ∫ dFm =, , ∫ I(dl × B), , The direction of force can also be determined by applying, Fleming’s left hand rule or right hand thumb rule., , Force between Two Parallel Current, Carrying Conductors, l, , l, , Two parallel current, carrying conductors, exert magnetic force on, I1, one another., Magnetic force, experienced by length l, of any one conductor, due to the other, current carrying, conductor is, F =, , µ 0 2I1 I2, ⋅, l, 4π, r, , Force per unit length,, , 1, , 2, , I2, , F12 F21, , 1, , 2, , I2, , I1, , F12, , F21, , r, , r, , (a), , (b), , F, µ 2I I, = F0 = 0 ⋅ 1 2, I, r, 4π
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244, , DAY TWENTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , pivoted type moving coil galvanometer G. S is known as, shunt., , NOTE If the conductors carries current in same direction, then, , force between them will be attractive., If the conductor carries current in opposite direction, then, force will be repulsive., , G, , Ig, , I, , Torque, , I – Ig, , When a current carrying loop placed in a uniform magnetic, field, it experience torque,, , S, , τ = NIAB sin θ, , Ammeter, , where, NiA is defined as the magnitude of the dipole moment, of the coil, (pm ) ⋅ τ = pm B sin θ, ⇒, τ = pm × B, NOTE, , I, , Then, from circuit,, I g × G = (I − I g ) × S, Ig , G, ⇒, S =, I − I g , , • A current carrying loop (of any shape) behaves as a, magnetic dipole whose magnetic moment is given by, ( pm ) = IA, • If we have a current carrying coil having N turns, then, magnetic moment Pm of dipole will be, ( pm ) = NIA, • Magnetic moment of a current carrying coil is a vector and, its direction is given by right hand thumb rule., , So, S << G, only a small fraction of current goes through the, galvanometer., l, , Conversion of Galvanometer into Voltmeter, A voltmeter is made by connecting a resistor of high, resistance R in series with a pivoted type moving coil, galvanometer G., V, a, , Moving Coil, Galvanometer (MCG), , Ig, , MCG is used to measure the current upto nanoampere. The, deflecting torque of MCG,, , G, , A restoring torque is set up in the suspension fibre. If α is the, angle of trust, the restoring torque is, , From the circuit,, l, , where, K is galvanometer constant., , Some Important Concepts Related to, Moving Coil Galvanometer, Some of the important concepts related to galvanometer, i.e., current sensitivity, voltage sensitivity and some of, conversions used in galvanometer are given below., l, , Conversion of Galvanometer into Ammeter An ammeter is, made by connecting a low resistance S in parallel with a, , R, , Voltmeter, , τdef = NBIA, , τ restoring = KI, , b, , l, , Ig =, , V, V, ⇒R =, −G, G+ R, Ig, , Current Sensitivity The current sensitivity of a, galvanometer is defined as the deflection produced in the, galvanometer per unit current flowing through it., α NBA, Sl = =, I, C, Voltage Sensitivity Voltage sensitivity of a galvanometer is, defined as the deflection produced in the galvanometer per, unit voltage applied to it., α α S I NBA, SV = =, =, =, V IR R, RC
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MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, , 245, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A loosely wound helix made of stiff wire is mounted, vertically with the lower end just touching a dish of, mercury. When a current from a battery is started in the, coil through the mercury, , 5 The magnitude of the magnetic field (B) due to loop, ABCD at the origin (O) is, , ª AIEEE 2009, B, , a, , (a) the wire oscillates, (b) the wire continues making contact, (c) the wire breaks contact just as current is passed, (d) the mercury will expand by heating due to passes of, current, , I1, O, , A, I, , 30°, D, b, , 2 A current I flows through a closed loop as shown in, , C, , figure., , O, , I, , µ 0I b − a , 4 π ab , , 6 A current I flows in an infinity long wire with cross-section, , 2q, , R, , in the form of a semicircular ring of radius R. The, magnitude of the magnetic induction along its axis is, , I, S, , P, , The magnetic field induction at the centre O is, µ I, µ I, (b) 0 (θ + sin θ), (a) 0 θ, 4 πR, 4 πR, µ I, µ I, (c) 0 (π − θ + sin θ), (d) 0 (π − θ + tan θ), 4 πR, 2 πR, , 3 The magnetic induction at the centre O in the figure as, shown is, , R1, , 1, 1, −, , R1 R 2 , , µ 0I, 4, µ 0I, (d), 4, (b), , (R1 − R 2 ), , µ I, (b) 0, 2 πR, , µ I, (c) 0, 4 πR, , (d), , ª AIEEE 2012, µ0I, , π2 R, , 7 Two coaxial solenoids of different radii carry current I in, the same direction. Let F1 be the magnetic force on the, inner solenoid due to the outer one and F2 be the, magnetic force on the outer solenoid due to the inner, one. Then,, ª JEE Main 2015, , 8 An infinitely long conductor PQR is bent to form a right, , O, , I, , µ I, (a) 02, 2π R, , (a) F1 = F2 = 0, (b) F1 is radially inwards and F2 is radially outwards, (c) F1 is radially inwards and F2 = 0, (d) F1 is radially outwards and F2 = 0, , R2, , µ 0I, 4, µ 0I, (c), 4, , (b), , I, , (c), , (a), , µ 0 I (b − a), 24ab, µ 0I , π, (d), 2 (b − a) + (a + b), 4 π , 3, , , (a) zero, , Q, , 1, 1, +, , R1 R 2 , (R1 + R 2 ), , 4 An electron moves in a circular orbit with a uniform, , angle as shown. A current I flows through PQR. The, magnetic field due to this current at the point M is H1. Now, another infinitely long straight conductor QS is connected, at Q, so that the current is I/2 in OR as well as in QS, the, current in PQ remaining unchanged. The magnetic field at, M is now H2 . The ratio H1 : H2 is given by, M, , speed v. It produces a magnetic field B at the centre of, the circle. The radius of the circle is proportional to, v, , –¥, , e, O, , I, P, , 90°, Q, , 90°, , S, , +¥, , r, , R, , B, (a), v, , v, (b), R, , v, (c), B, , B, (d), v, , (a), , 1, 2, , –¥, , (b) 1, , (c), , 2, 3, , (d) 2
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246, , DAY TWENTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , 16 A thin flexible wire of length L is connected to two, , 9 For a positively charged particle moving in a xy-plane, , adjacent fixed points and carries a current I in the, clockwise direction, as shown in the figure. When the, system is put in a uniform magnetic field of strength B, going into the plane of the paper, the wire takes the, shape of a circle. The tension in the wire is ª AIEEE 2011, , initially along the x-axis,there is a sudden change in its, path due to the presence of electric and/or magnetic, fields beyond P. The curved path is shown in the, xy-plane and is found to be non-circular. Which one of, the following combinations is possible?, , ×, ×, ×, ×, ×, ×, , y, , P, , x, , (a) IBL, (a) E = 0, B = b $i + ck$, (c) E = 0; B = c$j + bk$, , IBL, π, , ( 4 $i + 3 $j ) × 10−10 N on a particle having a charge10−9 C, and going on the xy-plane. The velocity of the particle is, (b) −100 $i + 75 $j, (d) 2 $i + 25 $i, , (c), , ×, ×, ×, ×, ×, ×, , ××, ×, ××, ××, ×, ××, , ×, ×, ×, ×, ×, ×, , IBL, 2π, , ×, ×, ×, ×, ×, ×, , (d), , IBL, 4π, , I 2 ( < I1 ). When I1 and I 2 are in the same direction, the, magnetic field at a point mid-way between the wires is, 10 µT. If I 2 is reversed, the field becomes 30 µT. The ratio, I1 / I 2 is, (a) 1, , (b) 3, , (c) 2, , (d) 4, , 18 A loop carrying current I lies in the xy- plane as shown in, the figure. The unit vector k$ is coming out of the plane of, the paper. The magnetic moment of the current loop is, , 11 An electron, a proton and an alpha particle having the, , ª JEE Main (Online) 2013, , same kinetic energy are moving in circular orbits of radii, re , rp , rα respectively, in a uniform magnetic field B. The, relation between re , rp , rα is, ª JEE Main 2018, (a) re > rp = rα (b) re < rp = rα (c) re < rp < rα, , ×, ×, ×, ×, ×, ×, , 17 Two parallel long wires A and B carry currents I1 and, , (b) E = a $i ; B = ck$ + a $i, (d) E = a $i ; B = ck$ + b$j, , 10 A magnetic field 4 × 10−3 kT exerts a force, , (a) −75 $i + 100 $j, (c) 25 $i + 2 $j, , (b), , ×, ×, ×, ×, ×, ×, , y, , (d) re < rα < rp, , I, , 12 The cyclotron frequency of an electron gyrating in a, , a, , x, , magnetic field of 1 T is approximately, (a) 28 MHz, , (b) 280 MHz, , (c) 2.8 GHz, , a, , (d) 28 GHz, , 13 A proton and an α-particle enters a uniform magnetic, , (a) 50 µs, , (b) 25 µs, , (c) 10 µs, , π, (c) − + 1 a 2 I k$, 2, , , (d) 5 µs, , 19 Magnetic field at the centre of a circular loop of area A is B., , 14 Two long conductors separated by a distance d carry, , The magnetic moment of the loop will be, , current I1 and I 2 in the same direction. They exert a force, F on each other. Now the current in one of them is, increased to two times and its direction is reversed. The, distance is also increased to 3d. The new value of the, force between them is, (a) − 2 F, , (b), , F, 3, , (c), , 2F, 3, , 15 Two very thin metallic wires placed, along X and Y -axes carry equal, currents as shown in figure. AB and, CD are lines at 45° with the axes with, origin of axes at O. The magnetic field, will be zero on the line, (a) AB, (b) CD, (c) segment OB only of line AB, (d) segment OC only of line CD, , (d) −, , (a), , BA 2, µ0 π, , (b), , BA 3 / 2, µ0 π, , (c), , BA 3 / 2, µ 0 π1 / 2, , (d), , 2 BA 3 / 2, µ 0 π1 / 2, , 20 A rectangular loop of sides 10 cm and 5 cm carrying a, current I of 12 A is placed in different orientations as, ª JEE Main 2015, shown in the figures below., , F, 3, , z, z, , I, B, , Y, C, , B, , I, O, , I, , (i), , I, , x, , X, , I, A, , π, (b) + 1 a 2 I k$, 2, , (d) (2 π + 1) a 2 I k$, , (a) a 2 I k$, , field perpendicularly with the same speed. If proton, takes 25 µs to make 5 revolutions, then the periodic time, for the α-particle would be, , I, y, , z, , D, , I, , (iii), x, , B, , (ii), , I, , I, x, , y, I, , I, z, , I B, I, y, , (iv), , I, , B, I, , I, x, , I, , y, I
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MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, If there is a uniform magnetic field of 0.3T in the positive, z-direction, then in which orientations the loop would be, in 1. stable equilibrium and 2. unstable equilibrium?, (a) (i) and (ii) respectively, (c) (ii) and (iv) respectively, , (b) (i) and (iii) respectively, (d) (ii) and (iii) respectively, , 21 When a current of 5 mA is passed through a, galvanometer having a coil of resistance 15 Ω, it shows, full scale deflection. The value of the resistance to be put, in series with the galvanometer to convert it into a, voltmeter of range 0-10 V is, ª JEE Main 2017 (Offline), (a) 2.045 × 103 Ω, (c) 4.005 × 103 Ω, , (b) 2.535 × 103 Ω, (d) 1985, ., × 103 Ω, , 22 A galvanometer having a coil resistance of 100 Ω gives a, full scale deflection when a current of 1 mA is passed, through it. The value of the resistance which can convert, this galvanometer into ammeter giving a full scale, deflection for a current of 10 A, is ª JEE Main 2016 (Offline), (a) 0.01 Ω, , (b) 2 Ω, , (c) 0.1 Ω, , (d) 3 Ω, , Direction (Q. Nos. 23-28) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choice, only one of which, is the correct answer. you have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 23 Statement I If a charged particle is projected in a region, where B is perpendicular to velocity of projection, then, the net force acting on the particle is independent of its, mass., , 247, , Statement II The particle is performing uniform circular, mv 2, motion and net force acting on it is, ., r, , 24 Statement I A uniformly moving charged particle in a, magnetic field, may follow a path along magnetic field, lines., Statement II The direction of magnetic force, experienced by a charged particle is perpendicular to, its velocity and B., , 25 Statement I The magnetic force experienced by a, moving charged particle in a magnetic field is invariant in, nature just like any other force., Statement II Magnetic force experienced by a charged, particle is given by F = q( v × B ), where v is the velocity of, charge particle w.r.t. frame of reference in which we are, taking F., , 26 Statement I Cyclotron is a device which is used to, accelerate the positive ion., Statement II Cyclotron frequency depends upon the, velocity., , 27 Statement I Magnetic field due to a infinite straight, conductor varies inversely as the distance from it., Statement II The lines of electric force due to a straight, current carrying conductor are concentric circles., , 28 Statement I If a proton and α-particle enter a uniform, magnetic field perpendicularly with the same speed, the, time period of revolution of α-particle will be double than, that of proton., Statement II Time period of charged particle is given by, 2πm, ., Bq, , T =, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A cell is connected between two points of a uniformly, thick circular conductor. I1 and I 2 are the currents flowing, in two parts of the circular conductor of radius a . The, magnetic field at the centre of the loop will be, (a) zero, (c), , µ0, (I1 + I 2 ), 2a, , µ0, (I1 − I 2 ), 4π, µ, (d) 0 (I1 + I 2 ), a, (b), , 2 A coil having N turns is wound tightly in the form of a, spiral with inner and outer radii a and b respectively., When a current I passes through the coil, the magnetic, field at the centre is, , µ 0 NI, b, µ 0 NI, b, (c), loge, 2 (b − a), a, (a), , 2 µ 0 NI, a, µ 0I N, b, (d), loge, 2 (b − a), a, , (b), , 3 A particle of mass m and charge q moves with a constant, velocity v along the positive x-direction. It enters a region, containing a uniform magnetic field B directed along the, negative z-direction, extending from x = a to x = b. The, minimum value of v required, so that the particle can just, enter the region x > b is, (a) qbB /m, (c) qaB /m, , (b) q (b − a)B /m, (d) q (b + a)B /2m
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248, , DAY TWENTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , 4 The magnetic field normal to the plane of a wire of n turns, and radius r which carries a current I is measured on the, axis of the coil at a small distance is from the centre of, the coil. This is smaller than the magnetic field at the, centre by the fraction, (a) (2 / 3)r 2 /h 2, (c) (2 / 3)h 2 /r 2, , (b) (3 / 2)r 2 /h 2, (d) (3 / 2)h 2 /r 2, , spiral with inner and outer radii a and b respectively., When a current I passes through the coil, the magnetic, field at the centre is, ª AIEEE 2012, µ 0 NI, b, µ NI, b, (c) 0, In, 2 (b − a) a, , (d), , |B|, , R/2, , |B|, , r, , R, , R/2, , 2µ 0 NI, a, µ0I, b, (d), In, 2 (b − a) a, , gun are accelerated through a potential differenceV0, along the x-axis. These electrons emerge from a narrow, hole into a uniform magnetic field of strength B directed, along x-axis., , 6 Two long current carrying thin wires, both with current I,, , x-axis, , are held by insulating threads of length L and are in, equilibrium as shown in the figure, with threads making, an angle θ with the vertical. If wires have mass λ per unit, length, then the value of I is, (g = gravitational acceleration), , B, , V0, , (a), , q, , (c) 2, , πgL, tan θ, µ0, , I, , ª JEE Main 2015, πλgL, (b) 2 sinθ, µ 0 cos θ, , (d), , 8 π 2 mV0, 3 eB, , (b), , (d), , 4 π 2 mV0, eB 2, , z, , πλgL, tan θ, µ0, , 1.5, l, , same current I. Wire A is bent into a circle of radius R, and wire B is bent to form a square of side a. If BA and BB, are the values of magnetic field at the centres of the, B, circle and square respectively, then the ratio A is, BB, ª JEE Main 2016 (Offline), , π2, (c), 16, , π2, (b), 16 2, , 2 π 2 mV0, eB 2, , carries a fixed current of 10.0 A in −a z direction (see, figure). For a field B = 3.0 × 10−4 e −0. 2 x a y T, find the power, required to move the conductor at constant speed to, x = 2.0 m, y = 0 in 5 × 10−3 s. Assume parallel motion, ª JEE Main 2014, along the x-axis., , 7 Two identical wires A and B, each of length l, carry the, , π2, (a), 8, , 8 π 2 mV0, (c), eB 2, , 10 A conductor lies along the z-axis at −1. 5 ≤ z < 1.5 m and, , L, , πλgL, (a) sinθ, µ 0 cos θ, , π2, (d), 8 2, , B y, , 2.0, x, , (a) 1.57 W, , –1.5, , (b) 2.97 W, , (c) 14.85 W, , (d) 29.7 W, , 11 A long straight wire carries a current I. A particle of, , charge + q and mass m is projected with a speed v from, a distance x 0 as shown. The minimum separation, between the wire and particle is, , 8 An infinitely long hollow conducting cylinder with inner, radius R/2 and outer radius R carries a uniform current, density along its length. The magnitude of the magnetic, field, | B | as a function of the radial distance r from the, ª JEE Main (Online) 2013, axis is best represented by, , (a), , |B|, , (b), , x0, , R/2, , R, , I, , I, , X, , Y, , |B|, , − mv 2 πx 0, , − 2 πmv, , (a) x 0 e, r, , r, , Some electrons emerging at slightly divergent angles as, shown. These paraxial electrons are refocussed on the, x-axis at a distance., , (b), , I, , R, , 9 Electrons emitted with negligible speed from an electron, , 5 A coil having N turns is wound tightly in the form of a, , (a), , (c), , R/2, , R, , r, , µ 0 qI, , (b) x 0 e, , µ 0 qI, , − 2 πmv, , (c) x 0 e, , qI, , (d) zero
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12 A current carrying circular loop of radius R is placed in, , Region containing, Magnetic field, , the xy-plane with centre at the origin. Half of the loop with, x > 0 is now bent, so that it now lies in the yz-plane., (a) The magnitude of magnetic moment now diminishes, (b) The magnetic moment does not change, (c) The magnitude of B at (0, 0, z), z>>R increases, (d) The magnitude of B at(0, 0, z), z>>R is unchanged, , 13 A 100 turn rectangular coil ABCD (in xy-plane) is hung, from one arm of a balance (shown in figure). A mass, 500 g is added to the other arm to balance the weight of, the coil. A current of 4.9 A passes through the coil and a, constant magnetic field of 0.2 T acting inward (in, xz-plane) is switched on such that only arm CD of length, 1cm lies in the field. How much additional mass m must, be added to regain the balance?, , A, , (b) 1 g, , 1 2, , (c) 0.75 g, , C, , (d) 1.5 g, , 14 A charge Q is uniformly distributed over the surface of, non-conducting disc of radius R. The disc rotates about, an axis perpendicular to its plane and passing through its, centre with an angular velocity ω. As a result of this, rotation, a magnetic field of induction B is obtained at the, centre of the disc. If we keep both the amount of charge, placed on the disc and its angular velocity to be constant, and vary the radius of the disc then the variation of the, magnetic induction at the centre of the disc will be, ª JEE Main (Online) 2013, represented by the figure., , r2, , 3, r1, , r4, , r3, , 4, , The table below gives the masses and charges of the ions, ION, A, B, C, D, , MASS, 2m, 4m, 2m, m, , CHARGE, +e, −e, −e, +e, , The ions fall at different positions 1,2,3 and 4 as shown., Correctly match the ions with respective falling positions, , B, , D, , (a) 2 g, , A, (a) 4, (c) 4, , B, 3, 1, , Column I, A, B, C, D, , Column II, 1, 2, 3, 4, , C, 2, 2, , A, (b) 1, (d) 3, , D, 1, 3, , B, 2, 4, , C, 3, 1, , B, , carry steady equal current flowing out of the plane of the, paper as shown. The variation of the magnetic field along, ª AIEEE 2010, the line XX ′ is given by, B, , (a) X, , X′, d, , d, , B, , (b), , (b), , X, , X′, d, , R, , (d), , (c), , B, , d, B, , R, , B, , (c), , X, , X′, d, , d, , B, R, , R, , 15 A beam consisting of four types of ions a,b,c and d, enters a region that contains a uniform magnetic field as, shown in figure. The field is perpendicular to the plane of, the paper, but its precise direction is not given. All ions in, the beam travel with the same speed., , D, 4, 2, , 16 Two long parallel wires are at a distance 2d apart. They, , B, , (a), , 249, , MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, , (d), , X, , X′, d, , d
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250, , DAY TWENTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , ANSWERS, SESSION 1, , 1 (a), , SESSION 2, , 2 (d), , 3 (a), , 4 (c), , 5 (b), , 6 (d), , 7 (a), , 8 (c), , 9 (b), , 10 (a), , 19 (d), , 20 (c), , 9 (b), , 10 (b), , 11 (b), , 12 (d), , 13 (c), , 14 (c), , 15 (a), , 16 (c), , 17 (c), , 18 (b), , 21 (d), , 22 (a), , 23 (c), , 24 (d), , 25 (d), , 26 (c), , 27 (b), , 28 (a), , 1 (a), 11 (a), , 2 (c), 12 (a), , 3 (b), 13 (b), , 4 (d), 14 (a), , 5 (c), 15 (c), , 6 (b), 16 (a), , 7 (d), , 8 (d), , Hints and Explanations, SESSION 1, , =, , 1 Loosely wound helix get compressed, when possess current. So, connection of, mercury lost and demagnetisation takes, place. So, result a oscillatory motion., , 2 B 0 = B PQS + B SP, , 6 Consider the wire to be made up of, large number of thin wires of infinite, length. Consider such wire of thickness, dl subtending an angle d q at centre., , 1, m0 I, m, (2p - 2q) + 0, 4p R, 4p R cos q, (sin q + sin q), , =, , m 0I, (p - q + tan q), 2pR, , q, , q, , dB, q, q, , 3 Magnetic field due to straight parts of, wire at point O = 0. Field due to a, semicircular current loop of radius, m I, m I p, R1 , B1 = 0 æç ö÷ = 0 into the plane, 2R1 è 2p ø 4R1, of paper., Field due to semicircular current loop of, m I, radius R 2 , B2 = 0 outside the plane, 4R 2, of paper., \ Net field B = B1 - B2, m Iæ 1, 1 ö, ÷, = 0 ç, ç, 4 è R1 R 2 ÷ø, , 4 Equivalent current, I = ev, 2pr, Hence, magnetic field at centre of circle, m ev, m I m ev, B = 0 = 0×, = 0 2, 2r, 2r 2pr, 4pr, m 0 ev, v, r =, Þr µ, Þ, 4pB, B, , 5 Net magnetic field due to loop ABCD at, O is, B = B AB + B BC + BCD + BOA, m I, p, m I p, = 0+ 0 ´ + 0- 0 ´, 4pa 6, 4pb 6, , H 2 = Magnetic field at M due to QR, + magnetic field at M due to QS, + magnetic field at M due to PQ, 3, H, = 0 + 1 + H1 = H1, 2, 2, H1 2, Þ, =, H2 3, , 9 Electric field can deviate the path of the, dq, , =, , m 0I m 0I, m I, = 0 (b - a), 24a 24b 24ab, , dB, , dq, I, p, \ Magnetic field at centre due to this, portion,, m 2dI, m I, dB = 0 ×, = 02 d q, 4p R, 2p R, Net magnetic field at the centre, Current through this wire, dI =, , B =, =, =, , p/2, , ò- p/2dB cos q, m 0I, 2p2 R, m 0I, , p/2, , ò - p/2 cos q d q, , p2 R, , 7 Consider the two coaxial solenoids. Due, to one of the solenoids magnetic field at, the centre of the other can be assumed, to be constant. Due to symmetry, forces, on upper and lower part of a solenoid, will be equal and opposite and hence, resultant is zero., Therefore, F1 = F2 = 0, , 8 Magnetic field at any point lying on the, current carrying straight conductor is, zero., Here, H 1 = Magnetic field at M due to, current in PQ, , particle in the shown direction only, when it is along negative y-direction. In, the given option E is either zero or along, x-direction. Hence, it is the magnetic, field which is really responsible for its, curved path. Options (a) and (c) cannot, be accepted as the path will be helix in, that case (when the velocity vector, makes an angle other than 0°,180 or 90°, with the magnetic field, path is a helix)., Option (d) is wrong because in that case, component of net force on the particle, also comes in k direction which is not, acceptable as the particle is moving in, xy-plane. Only in option (b) the particle, can move in xy-plane., In option (d) : Fnet = qE + q(v ´ B), Initial velocity is along x-direction. So let, v = v$i, \ Fnet = qa $i + q[(v$i ) ´ (ck$ + b$j )], = qa$i - qvc$j + qvbk$, In option (b),, Fnet = q(a$i ) + q[(v$i ) ´ (ck$ + a$i ), = qa$i + qv$j, , 10 From Lorentz force, F = q(v ´ B ), Given, F = (4 i$ + 3 $j ) ´ 10-10 N,, $, q = 10-9C, B = 4 ´ 10-3 kT, 10, \ (4 i$ + 3 $j ) ´ 10, = 10-9(a $i + b $j ) ´ (4 ´ 10-3 ), Solving, we get, a = - 75, b = 100 Þ v = - 75 $i + 100 $j
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MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, 2, 11 From Bqv = mv , we have, , Þ re : r p : ra =, , me, , mp, , :, , :, , 4m p, , e, e, 2e, Clearly, r p = r a and re is least, [Q me < m p], So,, re < r p = r a, , é, ù, pa2, = I ´ ê a2 +, ´ 4ú k, ´, 4, 2, ë, û, p, p, = I ´ a2 é + 1ù k = æç + 1ö÷ a2 I k, ø, è2, ûú, ëê 2, , 19 As, B = m 0 2pI = m 0I, Þ, , Bq, 1 ´ 1.6 ´ 10-19, =, n=, 2pm, 2p ´ 91, . ´ 10-31, = 2.8 ´ 1010 Hz = 28 GHz, , 13 Time taken by to make one revolution, 25, = 5 ms, 5, q, 2pm, T, m, As T =, ; so 2 = 2 ´ 1, qB, T1, m1 q2, =, , m2 q1, 5 ´ 4 m1, q, =, =, m1q2, m1, 2q, , 4p r, 2Br, I =, m0, , 2r, , Magnetic moment, M = IA =, 1 /2, , =, , 2BA3 /2, , 2Br, A, m0, , a current carrying loop., As, t = M ´ B Þ |t| = |M||B| sin q, For orientation shown in figure (ii),, q = 00, t = 0 (stable equilibrium), and for figure (iv), q = p, t = 0, (unstable equilibrium), , 21 Suppose a resistance R s is connected in, series with galvanometer to convert it, into voltmeter., Ig, G, , RS, , d, , m 0 2 (2I1 )I2, ×, 4p, (3d ), 2, (repulsive) Þ F ¢ = F, 3, , V, , and final force F ¢ =, , 15 Along the line AB, magnetic field due to, two wires is equal but in mutually, opposite directions (as per right hand, thumb rule). Hence, net magnetic field, will be zero along the line AB., , 16, , I g (G + R s ) = V Þ R =, , 22 Maximum voltage that can be applied, across the galvanometer coil, = 100 W ´ 10-3 A = 01, . V, 1mA, , IdIB, , V, -G, Ig, , R = 1985 = 1.985 kW, R = 1.985 ´ 103 W, , Þ, or, , 100 W, , Bq, 1, =, T, 2pm, It is obvious that cyclotron frequency, does not depend upon velocity of, charged particle., , current flowing through an infinitely, long conductor is given by, m 2I, B = 0× ., 4p a, where, a is the distance of that point, from conductor. Now, according to right, hand thumb rule it follows that magnetic, field is in the form of concentric circles,, whose centres lie on the straight, conductor., , 28 We know that,, T µ, , 14 Initial force F = m 0 × 2I1 I2 (attractive), 4p, , true. Magnetic force may have different, value in different frames of reference,, that’s why it is not invariant in nature., , 27 The magnetic field at a point due to, , m 0p1 /2, , 20 Since, B is uniform only torque acts on, , = 10 ms, , 25 Statement I is false and Statement II is, , n=, , A, Also A = pr 2 or r = æç ö÷, èpø, , 2BA æ A ö, ´ç ÷, èpø, m0, , to F, so particle cannot follow magnetic, field lines (tangent to which gives the, direction of magnetic field)., , 26 Cyclotron frequency is given by, 1 /2, , =, , 12 Cyclotron frequency,, , or T2 = T1, , 24 Statement I is false, as B is perpendicular, , 18 As, M = I ´ Area of loop, , r, mv, 2mK, r =, =, Bq, Bq, where, K is the kinetic energy., As, kinetic energies of particles are, same;, m, r µ, q, , 251, , é, 2pm ù, êQ T = Bq ú, ë, û, , m, q, , For a-particle,, 4m, Ta µ, 2q, For proton,, T p+ µ, , m, q, , So, 2T p = T a, , SESSION 2, 1 The resistance of the portion PRQ will be, , Ammeter, , R1 = l 1 r, I2, , T, , dq, , RS, , T, , dq, 2T sin æç ö÷ = IdlB, è2ø, dq, Þ, 2T, = IRdqB, 2, BIL, Þ, T = BIR =, 2p, m 0 2I1 m 0 2I2, 17, = 10 mT, 4p r, 4p r, m 02I1 m 0 2I2, +, = 30 mT, 4pr, 4p r, , [Q dl = Rdq], , If R s is the shunt resistance, then, . V, R s ´ 10 A = 01, Þ, R s = 0.01 W, , 23 In this case, the charged particle, [Q L = 2pR], , On solving, I1 = 20 A and I2 = 10 A, So,, I1 / I2 = 2, , performs uniform circular motion and, magnetic force is providing the, necessary centripetal force,, mv 2, i.e., = qvB, r, 2, mv, is not the force acting on charged, r, particle it is simply equal to net force, acting on the particle., , S, , O, P, , Q, I1, , Resistance of the portion PSQ will be, R 2 = l2 r, Potential difference across P and Q, = I1 R1 = I2 R2, ...(i), I1 l 1 r = I2 l 2 r or I1 l 1 = I2 l 2, Magnetic field induction at the centre O, due to currents through circular, conductors PRQ and PSQ will be, = B1 - B 2, m I l sin 90° m 0 I2 l 2 sin 90°, = 0 1 1 2, =0, 4p, 4p, r, r2
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MAGNETIC EFFECT OF CURRENT, , DAY TWENTY ONE, , 9 The electrons will be refocussed after, distance equal to pitch., 2pm, Pitch = V x T = V, eB, =, , =, , 8p mV 0, , -2 p m v, , 10 When force exerted on a current, carrying conductor, Fext = B IL, Work done, Average power =, Time taken, 2, 2, 1, 1, P = ò Fext. . dx = ò B ( x )IL dx, t 0, t 0, =, , 2, , 1, 5 ´ 10-3, , = 9 [1 - e, , ò 3 ´ 10, , 0, -0. 4, , -4 -02, . x, , e, , ´ 10 ´ 3 dx, , ], , x0, , current I in xy-plane, the magnetic, moment M = I ´ pR2 . It acts, perpendicular to the loop along, z -direction. When half of the current, loop is bent in y z- plane, then magnetic, moment due to half current loop in, x y-plane, M1 = I (pR2 / 2) acting along, z-direction. Magnetic moment due to, half current loop in yz-plane,, M2 = I (pR2 / 2) along x-direction., Effective magnetic moment due to, entire bent current loop,, M ¢ = M12 + M22, = (I p R2 / 2)2 + (I p R 2 / 2)2, , 1, = 9 é1 - 0. 4 ù = 2.967 » 2.97 W, êë, e úû, 11 F = q($i v x + $j v y ) ´ éê m 0I k$ ùú, ë 2px û, m 0I, m I, $, $, = jq v x, - iq v y 0, 2px, 2px, Y, , I, , I, , =, , IpR2, 2, , 2<M, , i.e. magnetic moment diminishes, The magnitude of B at a point on the, axis of loop, distance z from the centre, of current loop in x y-plane is, B =, X, , m 0Iqv y, Fx, =m, 2p xm, dv x, dv x dx v x dv x, Also, ax =, =, ×, =, dt, dx dt, dx, 2, 2, 2, Since, v x + v y = v, 2v x dv x + 2v ydv y = 0, v x dv x = - v y dv y, , Hence,, m 0Iq v y, v ydv y, v x dv x, ==2p x m, dx, dx, dx dv y2p m, Þ, =, x, m 0 qI, Initially, x = x 0 and v y = 0, At minimum separation v x = 0, v y = v, x dx, 2pm - v, Thus, ò, =, dv y, x0 x, m 0 qI ò 0, , 2p / R2, m0, 2, 4p (R + z2 )3 /2, , The magnitude of B at a point distance z, from the centre of bent current loop,, whose half part is in, xy-plane and half part is in y x- plane, is, 2, , \ ax =, , or, Þ, , m 0 qI, , 12 For a circular loop of radius R, carrying, éQ eV = 1 mv 2 ù, 0, êë, úû, 2, , eB 2, , 2p m v, x, =x0, m 0 qI, x = x0 e, , Þ, , 2eV 0 2pm, ×, m, eB, 2, , log, , Þ, , B =, , =, , é m0, ù, p / R2, ê, 2, 2 3 /2 ú, p, 4, (R + z ) û, ë, 2, ém, ù, p / R2, + ê 0, 2, 2 3./2 ú, ë 4p (R + z ) û, m0, p / R2, 2<B, 2, 4p (R + z2 )3 /2, , Magnetic field B = 0.2 T (xz-plane), Length of arm CD = 1 cm,, Mass added to balance = m, Let F be the force due to magnetic field., The direction of magnetic field is inward, in xz-plane, the length vector is left, so, by using the Fleming’s left hand rule, the, direction of force is downwards in the, plane of paper, F = I (l ´ B ) = 4.9 (0.01 ´ 0.2 sin 90° ), ...(i), F = 4.9 ´ 0.01 ´ 0.2, For balancing,, mass of coil ´ g + force due to, magnetic field, = 500 ´ 10- 3 g + m ´ g, 0.5 ´ 9.8 + 4.9 ´ 0.01 ´ 0.2, = 500 ´ 10-3 ´ 9.8 + m ´ 9.8, 9.8 (0.5 + 0.001) = 9.8 (0.5 + m ), [From Eq. (i)], m = 0.001 kg = 1 g, Thus, 1 g of mass must be added to, regain the balance., m 0(dq ) æ w ö, ç ÷, 2r è 2p ø, R rdr, m w Q, B = ò dB = 0 ×, 2p, 4p pR2 ò0 r, m wQ, m wQ, 1, B = 0 2 × R, B = 0, Þ B µ, 2pR, R, 2pR, , 14 dB =, , 15 A ® 4, B ® 1, C ® 2, D ® 3, R = mv /qB, RB > R A, and R A = R C (in opposite sense), and R D is smallest., , 16 The magnetic field in between because of, each will be in opposite direction, m I, m 0I, B in between = 0 $j (- $j ), 2px, 2p (2d - x ), =, , 13 Let the mass of coil = M, Mass added other arm = 500 g, = 500 ´ 10-3 kg, Mass of coil = Mass in other arm, (for balancing), M = 500 ´ 10- 3 kg, M = 0.5 kg, When the current is switched on,, Current, I = 4.9 A, , 253, , at x = d , B in, , m 0I, 2p, , é1, 1 ù $, ê ú(j ), ë x 2d - x û, , between, , =0, , For x < d , B in, , between, , = ( $j ) and, , For x > d , B in, , between, , = (- $j ), , Towards x, net magnetic field will add, up and direction will be (- j ) and, towards x¢, net magnetic field will add, up and direction will be ( j ).
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DAY TWENTY TWO, , Magnetism, Learning & Revision for the Day, u, , u, , Current Loop as a Magnetic, Dipole, Bar Magnet, , u, , u, , u, , Magnetic Field Lines, The Earth’s Magnetism, Magnetic Behaviour of Materials, , u, , Hysteresis Curve, , u, , Electromagnet, , Current Loop as a Magnetic Dipole, A current loop is equivalent to a magnetic dipole. If A(= πa2 ) be the area of the loop, then, the magnitude of its dipole moment is, pm = iA = iπa2, where, a is radius of coil, i is current flowing through it, p, …(i), i = m2, πa, Magnetic field at the centre of a circular current loop is given by, µ i, …(ii), B= 0, 2a, Putting the value of i from Eq. (i) in Eq. (ii), we get, µ p, B= 0 m, 2 π a3, This is the expression for the magnetic field at the centre of the current loop in terms of, its dipole moment. Instead of circular loop, if there is a circular coil having n turns, its, dipole moment would be, pm = ni A = ni πa2 ., , Bar Magnet, A bar magnet may be viewed as a combination of two magnetic poles, North pole and, South pole, separated by some distance. The distance is known as the magnetic length, of the given bar magnet., A bar magnet exhibits two important properties, namely, (i) the attractive property and, (ii) the directive property., l, , l, , l, , l, , If m is the pole strength and 2l the magnetic length of the bar magnet,then its, magnetic moment is M = m (2 l )., Magnetic moment is a vector quantity whose direction is from S pole towards N pole., SI unit of magnetic pole strength (m), is ampere metre (Am) and of magnetic dipole, moment (M) is ampere metre2 (Am2 )., If a bar magnet is broken, the fragments are independent magnetic dipoles and not, isolated magnetic poles., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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MAGNETISM, , DAY TWENTY TWO, , 255, , Magnetic Field due to a Bar Magnet, , Bar Magnet as an Equivalent Solenoid, , The magnetic field in free space, at a point having distance r, from the given bar magnet (or magnetic dipole) is calculated, in two conditions, along axial line and along equatorial line., µ, 2 Mr, Along axial line B = 0 2, 4π (r − l 2 )2, , The magnetic field (axial) at a point at a distance r and, µ 0 nlIa2, radius a of solenoid is given by B =, r3, and magnetic moment of solenoid is M = n (2 l) Iπa2, , l, , and the direction of B is the same as the direction of M. For, a short dipole (or for a far away point on the axis) when, r > > l , the above relation is simplified as, µ 2M, B= 0 3, 4π r, , P, , l, r, , Along the equatorial line, M, µ, B= 0 2, 4π (r + l 2 )3 /2, If r > > l , the relation is modified as, B =, , µ0 M, 4π r 3, , • Even a single electron moving in its orbit behaves as a, magnetic dipole and has a definite magnetic moment., , • Bohr magnetron is the magnetic moment due to the orbital, motion of an electron revolving in the inner most orbit, eh, = 9.27 × 10 −24 A-m 2, ( n = 1). Its value is m B =, 4 πme, , N, , S, , l, , NOTE, , Torque on a Magnetic Dipole in a, Magnetic Field, A magnetic dipole when placed in an uniform magnetic field,, does not experience any net force. However, it experiences a, torque given by τ = M × B or τ = MB sin θ, where, θ is the angle from the magnetic field, along which the, dipole has been placed., l, , However, along the equatorial line,, the direction of B is opposite to that of M., , Work done in rotating a magnetic dipole in a uniform, magnetic field B from an initial orientation θ1 to the final, orientation θ2 , is given by W = MB(cos θ1 − cos θ2 )., q m1, , l, , Normal, , r, , I, θ1, , r, , r, , qm1, Normal, , l, , I, S, , N, , In general, in a direction making an angle θ from with the, magnetic axis, the magnetic field is given by, µ M, B = 0 3 (3 cos2 θ + 1), 4π r, In these relations, µ 0 is a constant having a value of, 4π × 10 −7 T mA −1 and it is known as the magnetic, permeability of free space., , Initial position, fig. (i), l, , l, , B = µ 0 ni, , For solenoid, , where, n is number of turns per unit length of solenoid and, i the current through it., Bθ, , B, , l, , Br, , α, , l, , P, r, , l, , θ2, I, , qm2, , l, , I, , qm2, Final position, fig. (ii), , Potential energy of a magnetic dipole placed in a uniform, magnetic field, is given by UB = − M ⋅ B = − MB cos θ, where, θ is the angle from the direction of magnetic field, and the axis of dipole., The magnetic compass (needle) of magnetic moment M, and moment of inertia I and allowing it to oscillate in the, magnetic field. Then, its time-period is T = 2 π I / MB, Behaviour of a magnetic dipole in a magnetic field, is, similar to the behaviour of an electric dipole in an electric, 1, µ, field. However, the constant, is replaced by 0 ., 4πε 0, 4π, If a magnetic dipole is in the form of a wire or a thin rod,, when bent, its magnetic dipole moment M changes because, the separation between its poles has changed., , Magnetic Field Lines, S, O, , N, , M, , The magnetic field lines is defined as the path along which, the compass needles are aligned. They are used to represent, magnetic field in a region.
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256, , l, , l, , l, , l, , DAY TWENTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , Magnetic field lines are closed continuous curves., Tangent drawn at any point on magnetic field lines gives, the direction of magnetic field at that point., Two magnetic field lines cannot intersect each other., Outside a magnet, they are directed from north to south pole, and inside a magnet they are directed from south to north., , The Earth’s Magnetism, The earth is a natural source of magnetic field, thus a, magnetic field is always present everywhere near the surface of, the earth. A freely suspended magnet always points in the, north-south direction even in the absence of any other, magnet. This suggests that the earth itself behaves as a magnet, which causes a freely suspended magnet (or magnetic needle), to point always in a particular direction : north and south. The, shape of earth’s magnetic field resembles that of a bar magnet, of length one-fifth of earth’s diameter buried at its centre., Magnetic field of earth is shown in the figure given below., Geographical, North, Magnetic, North, , α, , P, H, , δ, , O, , S, N, , M, Q, , BE, , V, R, , Magnetic Elements of Earth, l, , l, , l, , Neutral Points, A neutral point is a point at which the resultant magnetic field, is zero. Following two cases are of special interest., 1. When a bar magnet is placed along the magnetic meridian, with its North pole pointing towards geographical North,, two neutral points are obtained on either side of the, magnet along its equatorial line. If r be the distance of the, µ M, neutral point, then 0 3 = BH ., 4π r, 2. When a bar magnet is placed along the magnetic, meridian, with its North pole pointing towards the, geographical South, two neutral points are obtained on, either side of the magnet along its axial line., µ 2M, Hence, we have 0, = BH ., 4π r 3, , L, , Geographical, meridian, Magnetic, meridian, , Similarly, at the magnetic equator, BV = BE sin 0 ° = 0, and at the poles,, BV = BE sin 90 ° = BE ., Magnetic elements of the earth at a place change with time also., , Angle of Declination (α) At a given place, the acute angle, between the magnetic meridian and the geographical, meridian is called the angle of declination (or magnetic, declination) α at that place., Angle of Inclination or Dip (δ ) The angle of dip δ at a place, is the angle which the direction of the earth’s total, magnetic field BE subtends with the horizontal direction., Horizontal Component of the Earth’s Magnetic Field, ( B H ) As earth’s magnetic field, in general, is inclined at an, angle δ with the horizontal direction, it may be resolved, into horizontal component BH and a vertical component BV ,, where, BH = BE cos δ and BV = BE sin δ, B, 2, ⇒, BE = BH, + BV2 and tan δ = V, BH, , Variation of Magnetic Elements, of the Earth, At the magnetic equator, angle of dip is zero. Value of the, angle of dip gradually increases, on going from equator to, magnetic poles. At the magnetic poles, value of the dip angle, is 90°., At the magnetic equator, BH = BE cos 0 ° = BE, and at poles,, BH = BE cos 90 ° = 0., , Tangent Galvanometer, It is an instrument to measure electric current. The essential, parts are a vertical coil of conducting wire and a small, compass needle pivoted at centre of coil. The deflection, θ of, needle is given by,, B, µ IN, tan θ =, ⇒ BH tan θ = 0, BH, 2r, or, , i=, , 2 r BH, tan θ = K tan θ, µ 0N, , Magnetisation of Materials, There are some substances/materials which acquire magnetic, properties on placing them in magnetic field. This phenomena, is called magnetisation of materials., To describe the magnetic properties of material, we have to, understand the following terms:, (i) Magnetic Induction or Magnetic Flux Density (B), Whenever a piece of magnetic substance is placed in an, external magnetising field, the substance becomes, magnetised. If B0 is the magnetic field in free space, then, B = µ r B0 ., , ∫ B ⋅ dS is magnetic flux which is equal to µ 0 m inside , where, m inside is the net pole strength inside a close surface., (ii) Magnetic Permeability (µ ) It is the degree or extent to, which the magnetic lines of induction may pass through a, given distance., Magnetic permeability of free space µ 0 has a value of, 4π × 10 −7 TmA −1 . However, for a magnetic material,, absolute permeability ( µ) has a value, different than µ 0 ., µ, B, =, = µ r = relative, µ 0 B0, magnetic permeability of that substance. Relative, magnetic permeability µ r is a unitless and dimensionless, term., For any magnetic substance,
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MAGNETISM, , DAY TWENTY TWO, (iii) Intensity of Magnetisation (I) Intensity of magnetisation of a, substance is defined as the magnetic moment induced in the, substance per unit volume, when placed in the magnetising, M, field. Thus, I =, V, It is a vector quantity and its SI unit is Am−1 ., (iv) Intensity of Magnetising Field or Magnetic Intensity (H) It is a, measure of the capability of external magnetising field to, magnetise the given substance and is mathematically defined, as, B, B, H = 0 or H =, µ0, µ, , Oxygen, air, platinum, aluminium, etc., are examples of, paramagnetic materials., l, , l, , l, , l, , Magnetic intensity H is a vector quantity and its SI unit is Am−1 ., (v) Magnetic Susceptibility ( χ m ) Magnetic susceptibility of a, substance is the ratio of the intensity of magnetisation I induced, I, in the substance to the magnetic intensity H.Thus, χ m = .It is, H, a scalar quantity and it has no units or dimensions., , Paramagnetism is temperature dependent. According, to the Curie’s law, the magnetic susceptibility of a, paramagnetic substance is inversely proportional to its, temperature T., C, Mathematically, χ m = , where C is the Curie constant., T, , These are the materials which are strongly attracted by a, magnetic field and can themselves be magnetised even in, a weak magnetising field. Iron, steel,nickel and cobalt are, ferromagnetic., , B, µ, =, = µ r = relative permeability, B0 µ 0, , l, , µr = χm + 1, l, , According to behaviour of magnetic substances, they are classified, into three cases:, , Diamagnetic Materials, These are materials which show a very small decrease in magnetic, flux, when placed in a strong magnetising field. Hydrogen, water,, copper, zinc, antimony, bismuth, etc., are examples of diamagnetic, materials., , l, , The relative permeability µ r of a paramagnetic, material is slightly greater than one (µ r > 1). Magnetic, susceptibility χ m of paramagnetic materials is positive., , Ferromagnetic Materials, , Magnetic Materials, , l, , Paramagnetic materials are feebly attracted in an, external magnetic field and thus, have a tendency to, shift from the weaker to the stronger regions of, magnetic field., , B, I, , or B = µ 0 H + 1 or B = B0 (χ m + 1) or, = χm + 1, H, , B0, , ∴, , l, , In a paramagnetic material, the net magnetic moment, of every atom is non-zero., , Relation between µ r and χ m we have, B = µ 0 (I + H ), , But, , l, , 257, , In a diamagnetic material, the net magnetic moment (sum of, that due to orbital motion and spin motion of electrons) of an, atom is zero. The external magnetic field B distorts the electron, orbit and thus induces a small magnetic moment in the opposite, direction., Diamagnetic materials are feebly repelled in an external, magnetic field and thus have a tendency to shift from the, stronger to weaker regions of the magnetic field., The relative permeability of any diamagnetic substance is, slightly less than 1 (i.e. µ r < 1) and susceptibility has a small, negative value., Diamagnetism is an intrinsic property and does not vary with, magnetic field B or temperature., , Paramagnetic Materials, These are the materials which show a small increase in the, magnetic flux when placed in a magnetising field., , l, , l, , These materials show a large increase in the magnetic, flux, when placed in a magnetic field. Thus, for them, µ r > > 1. Accordingly, χ m is positive and large., Ferromagnetic materials exhibit all properties, exhibited by paramagnetic substances and by a much, larger measure., Magnetic susceptibility of ferromagnetic materials, decreases steadily with a rise in temperature. Above a, certain temperature Tc (known as Curie temperature),, the substance loses its ferromagnetic character and, begins to behave as a paramagnetic substance., Above the Curie temperature Tc , the magnetic, susceptibility of a ferromagnetic material varies as, 1, C, or χ m =, χm ∝, (T − Tc ), (T − Tc ), where, C is a constant. It is known as the Curie-Weiss, law., , Hysteresis Curve, I, The lag of intensity of, magnetisation behind the, B, magnetising field during, the process of, Retentivity, C, magnetisation and, O, F, demagnetisation of a, ferromagnetic material is, E, called hysteresis. The, D, whole graph ABCDEFA is a, closed loop and known as, Coercivity, hysteresis loop., , A, , H
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258, , DAY TWENTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , Electromagnet, Electromagnets are usually in the form of iron core solenoids., The ferromagnetic property of the iron core causes the internal, magnetic domains of the iron to line up with the smaller, driving magnetic field produced by the current in the, solenoid., , The effect is the multiplication of the magnetic field by, factors of ten to eleven thousands. The solenoid field, relationship is B = k µ 0 nI , where µ = k µ 0 and k is the, relative permeability of the iron, the figure shows the, magnetic effect of the iron core., , Permanent Magnet, Substances which at room temperature retain their, ferromagnetic property for a long period of time are called, permanent magnets. Permanent magnets can be made in a, variety of ways., An efficient way to make a permanent magnet is to place a, ferromagnetic rod in a solenoid and pass a current. The, magnetic field of the solenoid magnetises the rod., , N = North pole, N, Iron core, S, S = South pole, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 A short bar magnet placed with its axis at 30° with a, , (a) 0.2 T, , equal to 10−3 Wb, is kept in a magnetic field having, magnetic induction B equal to 4π × 10−3 T. It makes an, angle of 30° with the direction of magnetic induction. The, value of the torque acting on the magnet is, , 3 M and M 3 are the, , (a) θ = 30°, , (b) θ = 45 °, , (a) 8.89 s, (c) 8.76 s, , (b) 6.98 s, (d) 6.65 s, , 6 Hysteresis loops for two magnetic materials A and B are, as given below, B, , (b) 2 π × 10−5 N-m, (d) 0.5 × 10 2 N-m, , q, , 4 A coil of 50 turns and area1.25 × 10, , (A), , B, , MÖ3, , (c) θ = 60°, −3, , (d) θ = 15 °, 2, , B, , H, , H, , M, , magnetic dipole moments, of the two magnets, which, are joined to form a cross, figure. The inclination of, the system with the field, if, their combination is, suspended freely in a, uniform external magnetic field B is, , (d) 0.4 T, , ª JEE Main 2017 (Offline), , 2 A bar magnet of length 10 cm and having pole strength, , (a) 0.5 N-m, (c) π × 10−5 N-m, , (c) 0.5 T, , and moment of inertia 7.5 × 10−6 kg m 2 is performing, simple harmonic oscillations in a magnetic field of 0.01 T., Time taken for 10 complete oscillations is, , (b) 0.40 JT −1, (d) zero, , (a) 0.23 JT −1, (c) 0.80 JT −1, , (b) 0.3 T, , 5 A magnetic needle of magnetic moment 6.7 × 10−2 Am 2, , uniform external magnetic field of 0.16 T, experiences a, torque of magnitude 0.032 J. The magnetic moment of, the bar magnet will be, , m is pivoted, about a vertical diameter in an uniform horizontal, magnetic field and carries a current of 2 A. When the coil, is held with its plane in the N-S direction, it experiences a, couple of 0.04 Nm, and when its plane is along the, East-West direction, it experiences a couple of 0.03 Nm., The magnetic induction is, , (B), , These materials are used to make magnets for electric, generators, transformer core and electromagnet core., ª JEE Main 2016 (Offline), Then, it is proper to use, (a)A for electric generators and transformers, (b)A for electromagnets and B for electric generators, (c)A for transformers and B for electric generators, (d)B for electromagnets and transformers, , 7 The coercivity of a small magnet where the ferromagnet, , gets demagnetised is 3 × 103 Am −1. The current required, to be passed in a solenoid of length 10 cm and number, of turns 100, so that the magnet gets demagnetised when, ª JEE Main 2014, inside the solenoid is, (a) 30 mA, , (b) 60 mA, , (c) 3 A, , (d) 6 A
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MAGNETISM, , DAY TWENTY TWO, 8 The figure illustrates how B, the, , B, , flux density inside a sample of, unmagnetised ferromagnetic, material, varies with B0, the, magnetic flux density in which, the sample is kept. For the, sample to be suitable for, making a permanent magnet., , P, Q, R, , B0, , O, , cross-section. In (i) E is ideally treated as a constant, between plates and zero outside. In (ii) magnetic field is, constant inside the solenoid and zero outside. These, idealised assumptions, however contradict fundamental, laws as below, (a) case (i) contradicts Gauss’ law for electrostatic fields, (b) case (ii) contradicts Gauss’ law for magnetic fields, , S, , (c) case (i) agrees with ∫ E ⋅ d I = 0, (d) case (ii) contradicts ∫ H ⋅ d I = I en, , (a) OQ should be large, OR should be small, (b) OQ and OR should both be large, (c) OQ should be small and OR should be large, (d) OQ and OR should both be small, , 14 A paramagnetic sample shows a net magnetisation of, , 8Am −1 when placed in an external magnetic field of 0.6 T, at a temperature of 4 K. When the same sample is placed, in an external magnetic field of 0.2 T at a temperature of, 16 K, the magnetisation will be, , 9 Match the following columns., Column I, , Column II, , (a), , 32, Am−1, 3, , (b), , 2, Am−1, 3, , (c) 6 Am−1, , (d) 2.4 Am−1, , A., , Magnetic moment, , 1. [ML0T −2A −1 ], , B., , Permeability, , 2., , Vector, , 15 If the areas under the I-H hysteresis loop and B-H, , C., , Intensity of magnetisation, , 3. Nm3 / Wb, , hysteresis loop are denoted by A 1 and A2, then, , D., , Magnetic induction, , 4., , Scalar, , Codes, A, (a) 1, (c) 4, , B, 2, 3, , C, 3, 1, , D, 4, 2, , A, (b) 3, (d) 2, , B, 4, 1, , C, 2, 3, , D, 1, 4, , 10 A current carrying coil is placed with its axis, perpendicular to N-S direction. Let horizontal component, of earth’s magnetic field be H0 and magnetic field inside, the loop be H. If a magnet is suspended inside the loop,, it makes angle θ with H. Then, θ equal to, H, (a) tan−1 0 , H , H , (c) cosec −1 , , H0 , , H , (b) tan−1 , , H0 , H, (d) cot −1 0 , H , , 11 Two short bar magnets of length 1 cm each have, magnetic moments 1.20 Am 2 and 1.00 Am 2 ,, respectively. They are placed on a horizontal table, parallel to each other with their N poles pointing towards, the South. They have a common magnetic equator and, are separated by a distance of 20.0 cm. The value of, the resultant horizontal magnetic induction at the, mid-point O of the line joining their centres is close to, (Horizontal component of the earth’s magnetic induction, ª JEE Main 2013, is 3.6 × 10−5 Wb/m 2 ), (a) 3.6 × 10–5 Wb / m2, (c) 3.50 × 10–4 Wb / m2, , (b) 2.56 × 10–4 Wb / m2, (d) 5.80 × 10–4 Wb / m2, , 12 The magnetic susceptibility of a paramagnetic substance, at –73°C is 0.0060, then its value at –173°C will be, (a) 0.0030, , (b) 0.0120, , (c) 0.0180, , 259, , (d) 0.0045, , 13 Consider the two idealised systems (i) a parallel plate, capacitor with large plates and small separation and, (ii) a long solenoid of length L >> R , radius of, , (b) A 2 = A1, , (a) A 2 = µ 0 A1, A, (c) A 2 = 1, µ0, , (d) A, , 2, , = µ 02 A 1, , 16 A short magnet oscillates with a time period 0.1 s at a place,, where horizontal magnetic field is 24 µT. A downward, current of 18 A is established in a vertical wire 20 cm East, of the magnet. The new time period of oscillator, (a) 0.1 s, , (b) 0.089 s, , (c) 0.076 s, , (d) 0.057 s, , 17 Two bar magnets of the same mass, same length and, breadth but having magnetic moments M and 2M,, respectively are joined together pole to pole and, suspended by a string. The time period of the assembly, in a magnetic field of strength H is 3 s. If now the polarity, of one of the magnets is reversed and the combination is, again made to oscillate in the same field, the time of, oscillation is, (a) 3 s, , (b) 3 3 s, , (c) 3 / 3 s, , (d) 6 s, , 18 Two magnets A and B are identical and these are, arranged as shown in the figure. Their lengths are, negligible in comparision, to the separation, A, between them. A, P, magnetic needle is, q, placed between the, magnets at point P and it, d1, d2, gets deflected by an, angle θ. The ratio of, distances d1 and d 2, will be, (a) (2 cot θ)1/ 3, (c) (2 cot θ)−1/ 3, , B, , (b) (2 tan θ)1/ 3, (d) (2 tan θ)−1/ 3, , 19 The plane of a dip circle is set in the geographic meridian, and the apparent dip is δ1. It is then set in a vertical plane, perpendicular to the geographic meridian.
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260, , DAY TWENTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , The apparent dip angle is δ 2. The declination θ at the, place is, (a) θ = tan−1 (tan δ1 tan δ 2 ), tan δ1 , (c) θ = tan−1 , , tan δ 2 , , (b) θ = tan−1 (tan δ1 + tan δ 2 ), (d) θ = tan−1 (tan δ1 − tan δ 2 ), , 20 This question contains two statements : Statement I, (Assertion) and Statement II (Reason). The question also, has four alternative choices, only one of which is the, correct answer. You have to select one of the codes (a),, (b), (c), (d)., , Statement I A current carrying loop is free to rotate. It is, placed in a uniform magnetic field. It attains equilibrium, when its plane is perpendicular to the magnetic field., , Statement II The torque on the coil is zero when its, plane is perpendicular to the magnetic field., (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The dipole moment of a circular loop carrying a current I,, is m and the magnetic field at the centre of the loop is B1., When the dipole moment is doubled by keeping the, current constant, the magnetic field at the centre of the, B, loop is B2. The ratio 1 is, B2, ª JEE Main 2018, (a) 2, , (b) 3, 1, (d), 2, , (c) 2, , 400 turns and mean circumferential length is 40 cm. With, the help of search coil and charge measuring instrument, the magnetic field is found to be 1T. The susceptibility is, (b) 290, (d) 397, , 3 The length of a magnet is large compared to its width and, breadth. The time period of its oscillation in a vibration, magnetometer is 2 s. The magnet is cut along its length, into three equal parts and three parts are then placed on, each other with their like poles together. The time period, of this combination will be, (a) 2 s, , 2, s, 3, 2, s, (d), 3, (b), , (c) 2 3 s, , 4 A thin circular disc of radius R is uniformly charged with, density σ > 0 per unit area. The disc rotates about its axis, with a uniform angular speed ω. The magnetic moment of, ª AIEEE 2011, the disc is, (a) 2 πR σω, πR 4, (c), σω, 2, 4, , (a) 0° and 11.3°, (c) 11.3° and 6.5°, , (b) πR σω, πR 4, (d), σω, 4, 4, , 5 Consider the plane S formed by the dipole axis and the, axis of the earth. Let P be point on the magnetic equator, and in S. Let Q be the point of intersection of the, , (b) 0° and 0°, (d) 11.3° and 11.3°, , 6 A bar magnet has pole strength 3.6 A-m and length, 12 cm. Its area of cross-section is 0.9 cm 2. The magnetic, field B at the centre of the bar magnet is, (a) 6 × 10−3 T, (c) 2.5 × 10−2 T, , 2 The current on the winding of a toroid is 2A. It has, , (a) 100, (c) 398, , geographical and magnetic equators. The declination, and dip angles at P and Q are, , (b) 5 × 10−2 T, (d) 2.5 × 10−8 T, , 7 A bar magnet suspended by a suspension fibre, is, placed in the magnetic meridian with no twist in the, suspension fibre. On turning the upper end of the, suspension fibre by an angle of 120° from the meridian,, the magnet is deflected by an angle of 30° from the, meridian. Then, the angle by which the upper end of the, suspension fibre has to be twisted, so as to deflect the, magnet through 90° from the meridian is, (a) 270°, (c) 330°, , (b) 240°, (d) 180°, , 8 A bar magnet 8 cm long, is placed in the magnetic, meridian with the N pole, pointing towards the, geographical North. Two neutral points, separated by a, distance of 6 cm are obtained on the equatorial axis of, the magnet. If BH = 3. 2 × 10−5 T, then the pole strength of, the magnet is, (b) 10 A -cm2, , (a) 5 A -cm2, 2, , (c) 2.5 A -cm, , (d) 20 A -cm2, , 9 There are two current carrying planer coils made each, from wire of length L. C1 is circular coil (radius R) and C2, is square (side a). These are so constructed that they, have same frequency of oscillation when they are placed, in the same uniform magnetic field B and carry the same, current. The value of a in terms of R is, (a) 3 R, (c) 2R, , (b) 3R, (d) 2R
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MAGNETISM, , DAY TWENTY TWO, , 261, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), , 2 (a), , 3 (c), , 4 (d), , 5 (d), , 6 (d), , 7 (c), , 8 (b), , 9 (b), , 10 (a), , 11 (b), , 12 (b), , 13 (b), , 14 (b), , 15 (a), , 16 (c), , 17 (b), , 18 (a), , 19 (c), , 20 (c), , 1 (c), , 2 (d), , 3 (b), , 4 (d), , 5 (a), , 6 (b), , 7 (a), , 8 (a), , 9 (a), , Hints and Explanations, SESSION 1, τ, B sin θ, 0.032, =, = 0.40 JT −1, 016, . × sin 30°, , 1 As, M =, , T = 2π, ⇒ T = 2π, , −3, , 2 As, µ 0m = 10 Wb, m=, , 10−3, µ0, , I, MB, 7.5 × 10−6, 67, . × 10−2 × 0.01, , = 0.665 s, , perpendicular to each other., From figure,, H, tanθ = 0, H, H0, , Time taken to complete 10 oscillations, = 10 × 0.665 = 6.65 s, , Magnetic moment of the magnet,, 10−3, 10−4, (01, . )=, M = m × 2l =, µ0, µ0, Now, τ = MB sin θ, 10−4 , =, × 4 π × 10−3 sin 30°, µ0 , = 0.5 N-m, , 3 Torque (τ) acting on the magnet (1) is, τ1 = MB sin θ, τ2 = 3 MB sin θ, For equilibrium, τ1 = τ2, ∴ MB sin θ = 3 MB cos θ, , Hence, time for 10 oscillations, is t = 6.65 s., , 2, , 11 B net = B1 + B2 + B H, , Also,|B | → 0 when H = 0 and|H |, should be small when B → 0., , = 2.56 × 10−4 Wb/m2, 1, 12 As,, χm ∝, T, χ 2 T1, ⇒, =, χ 1 T2, , m ,, , I = 2A, M = NIA, = 50 × 2 × 1.25 × 10−3, = 0.125 A-m2, If the normal to the face of the coil, makes an angle θ with the magnetic, induction B, the torque, …(i), τ = MB cos θ = 0.04, Now, when the plane of the coil is, turned through 90°, the torque, becomes,, …(ii), τ = MB sin θ = 0.03, Squaring and adding Eqs. (i) and (ii), we, get, τ = 0.05 ⇒ MB = 0.05, 0.05, 0.05, ⇒, B =, =, = 0.4 T, M, 0125, ., , H = nI, , where, N = 100,, 10, , , l, 01, 10, =, cm, =, m, =, ., m, , , 100, N, 100, H =, I ⇒ 3 × 103 =, ×I, l, 01, ., I = 3A, , 8 For making permanent magnet, the, material should have high residual, magnetism and high coercivity i.e. OQ, and OR should be large., , 9 A - SI unit of magnetic moment is, Nm3 / Wb., B - Permeability (µ ) is a scalar., M, A, C-I =, = −1 , it is a vector., V, m, D - Magnetic induction, B = ML0T −2 A −1, , H, , H, θ = tan −1 0 , H , , ⇒, , to the net energy absorbed per unit, volume by the material, as it is taken, over a complete cycle of magnetisation., For electromagnets and transformers,, energy loss should be low., i.e. thin hysteresis curves., , to be magnetised the magnet., −3, , q, , 6 Area of hysteresis loop is proportional, , 7 For solenoid, the magnetic field needed, , ⇒ tan θ = 3 = tan 60° ⇒ θ = 60°, , 4 Here, N = 50, A = 1.25 × 10, , 10 In given case H and H 0 are, , 5 Time period of oscillation is, , µ 0 ( M1 + M2 ), + BH, 4π, r3, −7, 10 (1.2 + 1), =, + 3.6 × 10−5, (0.1)3, , B net =, , or, , χ2, 273 − 73, 200, =, =, =2, 0.0060 273 − 173 100, , or, , χ2 = 0.0120, , 13 As Gauss’ law states,, q, for electrostatic field. It does, ε0, not contradict for electrostatic fields as, the electric field lines do not form, continuous closed path. According to, Gauss’s law in magnetic field,, , ∫SE ⋅ dS =, , ∫ E⋅d S = 0, , S, , It contradicts for magnetic field, because, there is a magnetic field inside the, solenoid and no field outside the, solenoid carrying current but the, magnetic field lines form the closed path.
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DAY TWENTY THREE, , Electromagnetic, Induction, Learning & Revision for the Day, u, , u, , Magnetic Flux ( φB ), Faraday’s Law of, Electromagnetic Induction, , u, , u, , u, , Motional Emf, Rotational Emf, Self-Induction, , u, , u, , u, , Mutual Induction, Combination of Inductors, Eddy Currents, , Magnetic Flux ( φB ), The flux associated with a magnetic field is defined in a similar, manner to that used to define electric flux. Consider an element, of area ds on an arbitrary shaped surface as shown in figure. If, the magnetic field at this element is B, the magnetic flux, through the element is,, , ds, B, θ, , dφB = B ⋅ ds = Bds cos θ, Here, ds is a vector that is perpendicular to the surface and has a, magnitude equal to the area ds and θ is the angle between B and ds at that element., Magnetic flux is a scalar quantity. Outward magnetic flux is taken as positive, (i.e. θ < 90 °) and inward flux is taken as negative (i.e. θ > 90 °)., SI unit of magnetic flux is 1 weber (1 Wb)., where,, , 1 Wb = 1 T × 1 m2 = 1 T-m2, , Dimensional formula of magnetic flux is [ML2 T−2 A−1 ] ., , PREP, MIRROR, , Faraday’s Law of Electromagnetic Induction, , Your Personal Preparation Indicator, , This law states that, the induced emf in a closed loop equals the negative of the time rate, of change of magnetic flux through the loop., dφ B, Induced emf,, |e | =, dt, dφ B, For N turns,, |e | = N, dt, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , l, , However, if we consider the direction of induced emf, then, dφ B, Nd(BA cos θ) − NBA(cos θ2 − cos θ 1), =, e=−N, =−, ∆t, dt, dt, , No. of Questions in Exercises (x)—, , u, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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ELECTROMAGNETIC INDUCTION, , DAY TWENTY THREE, , l, , If the given electric circuit is a closed circuit having a total, resistance R, then the induced current,, e, N dφ B, I =, =−, R, R dt, N, Induced charge, dq = Idt = −, dφ B, R, and induced power, P =, , e2 N 2 dφ B , =, , , R, R dt , , 2, , 265, , Rotational Emf, Let a conducting rod of length l rotate about an axis passing, through one of its ends (that end may be fixed), with an angular, velocity ω in a plane perpendicular to the magnetic field B, then, an induced emf is set up between the ends of the rod, whose, magnitude is given by, |e| =, , 1 2, Bl ω, 2, , Lenz’s Law, , Self-Induction, , The negative sign in Faraday’s equations of electromagnetic, induction describes the direction in which the induced emf, drives current around a circuit. However, that direction is, most easily determined with the help of Lenz’s law. This law, states that the direction of any magnetic induction effect is, such as to oppose the cause of the effect., , Self-induction is the phenomenon due to which an induced, emf is set up in a coil or a circuit whenever the current, passing through it changes. The induced emf opposes the, change that causes it and is thus known as back emf., Induced current, (make), , Later, we will see that Lenz’s law is directly related to energy, conservation., , Induced current, (break), , Motional Emf, Let a conducting rod of length l be moving with a uniform, velocity v perpendicular to a uniform magnetic field B, an, induced emf is set up., X, , X, , X, , X, , +, , X, , X, , X, , X, , X, , X, , X, , X, , v, , X, , X, , X, , X, , X, , X, , X, , X, , –, , X, , X, , X, , X, , l, , +, , l, , l, , The magnitude of the induced emf will be, |e| = B l v, l, , If the rod is moving such that it makes an angle θ with the, direction of the magnetic field, then, | e| = B l v sin θ, Hence, for the motion parallel to B, the induced emf is zero., , l, , l, , When a conducting rod moves horizontally, then an, induced emf is set up between its ends due to the vertical, component of the earth’s magnetic field. However, at the, magnetic equator, induced emf will be zero, because, B V = 0., If during landing or taking off, the wings of an aeroplane, are along the East-West direction, an induced emf is set up, across the wings (due to the effect of BH )., , Motional Emf in a Loop, , Inductance is the inherent property of electrical circuits, and is known as the electrical inertia., An inductor is said to be an ideal inductor if its resistance, is zero., An inductor does not oppose current but opposes changes, (growth or decay of current) in the circuit., , Nφ B ∝ I or Nφ B = LI,, where the constant L is known as the coefficient of, self-induction or self-inductance of the given coil., It may be defined as the magnetic flux linked with the coil,, when a constant current of 1 A is passed through it., Induced emf due to self-induction,, dφ, dI, e=–N, =–L, dt, dt, SI unit of inductance is henry., , Magnetic Potential Energy of an Inductor, , l, , l, , Rh, , Flux linked with the coil is, , l, , B2 l2 v, Magnetic force, Fm = BIl =, R, , –, , Self-Inductance, , If a conducting rod moves on two parallel conducting rails,, then an emf is induced whose magnitude is|e | = B l v, and the direction is given by the Fleming’s right hand rule., |e | B l v, Induced current,| I | =, =, R, R, , +, , –, K, , l, , A, , l, , In building, a steady current in an electric circuit, some, work is done by the emf of the source, against the, self-inductance of the coil., 1, The work done, W =, LI 2, 2, The work done is stored as the magnetic potential energy of, that inductor., 1, Thus,, U = L I2, 2
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266, , DAY TWENTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , For a pair of two magnetically coupled coils of self-inductances, L1 and L2 respectively, the mutual inductance,, , Formulae for Self-Inductance, l, , l, , For a circular coil of radius R and N turns, the, self-inductance,, 1, L = µ0 π N2 R, 2, , l, , where, k is the coupling coefficient., , For a solenoid coil having length l, total number of turns N, and cross-sectional area A,, L=, , l, , M 12 = M21 = M = k L1L2, , µ0 N2 A, l, , Formulae for Mutual Inductance, l, , N, , where, n = l , , = µ 0 n Al, 2, , µ 0N 1N2 πr 2, 2R, where, r = radius of the coil (of small radius), and R = radius of the coil (of larger radius)., , For a toroid of radius R and number of turns N,, 1, L = µ0 N2 R, 2, , M=, , For a square coil of side a and number of turns N,, L=, , l, , 2 2, µ0 N2 a, π, , Mutual induction is the phenomenon due to which an emf is, induced in a coil when the current flowing through a, neighbouring coil changes., , Mutual Inductance, , where, a = side of the smaller coil and b = side of the larger, coil., l, , Mutual inductance of a pair of coils is defined as the magnetic, flux linked with one coil, when a constant current of unit, magnitude, flows through the other coil., Mathematically, NφB2 = MI 1, where, M is known as the mutual inductance for the given, pair of coils., Induced emf due to mutual inductance,, dφB2, dI, e2 = – N, =–M 1, dt, dt, , Combination of Inductors, l, , Hence, mutual inductance for a pair of coils is numerically, equal to the magnitude of induced emf in one coil when, current in the other coil changes at a rate of 1 As −1., , S, , P, , S, P, , (a), , (b), , (c), , Coupling coefficient is given by, Magnetic flux linked with secondary coil, k =, Magnetic flux developed in primary coil, It is observed that 0 ≤ k ≤ 1., , If two coils of self-inductances L 1 and L 2 are placed quite, far apart and are arranged in series, then their equivalent, inductance,, L s = L1 + L 2, If the coils are placed quite close to each other, so as to, mutually affect each other, then their equivalent inductance,, L s = L1 + L2 ± 2 M, Here, M has been written with ± sign depending on the fact, whether currents in the two coils are flowing in same sense, or opposite sense., , SI unit of mutual inductance M, is henry., , P, , For a given pair of coils, mutually coupled, then according, to theorem of reciprocity,, M 12 = M21 = M, , l, , Mutual inductance of a pair of coils is maximum, when the, two coils are wound on the same frame. However, mutual, inductance is negligible when the two coils are oriented, mutually perpendicular to each other (see figure). In this, context, we define a term coupling coefficient k., , For a pair of two solenoid coils, wound one over the, µ N N A, other,, M= 0 1 2, l, For a pair of concentric coplanar square coils,, 2 2 µ 0N 1N2 a2, M=, πb, , Mutual Induction, , S, , Assuming the coupling coefficient k = 1 and medium to be, a free space or air. Mutual inductance of a pair of concentric, circular coils is, , l, , If two coils of self-inductances L1 and L 2 are connected in, parallel, then equivalent inductance L p is given by, L1L 2, 1, 1, 1, =, +, ⇒ Lp =, L1 + L 2, L p L1 L 2, , Eddy Currents, Currents induced in the body of bulk of the conductors due to, change in magnetic flux linked to them, are called the eddy, currents. The production of eddy currents in a metallic, conductor leads to a loss of electric energy in the form of, heat energy., Eddy currents can be minimised by taking the metal, (generally soft iron) core in the form of a combination of thin, laminated sheets or by slotting process.
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ELECTROMAGNETIC INDUCTION, , DAY TWENTY THREE, , 267, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The magnetic field in a certain region is given by, B = ( 40 $i − 18 k$ ) gauss. How much flux passes through a, loop of area 5 cm 2, in this region, if the loop lies flat on, the xy-plane?, (a) – 900 × 10−9 Wb, (c) Zero, , (b) 900 × 10−9 Wb, (d) 9 Wb, , 2 The flux linked with a coil at any instant t is given by, φ = 10 t − 50 t + 250. The induced emf at t = 3 s is, 2, , (a) –190 V, , (b) –10 V, , (c) 10 V, , (d) 190 V, , 3 A coil having n turns and resistance R Ω is connected, with a galvanometer of resistance 4R Ω. This, combination is moved for time t seconds from a, magnetic field W1 weber to W2 weber. The induced, current in the circuit is, W2 − W1, 5 Rnt, (W − W1), (c) − 2, Rnt, , n (W2 − W1), 5 Rt, n (W2 − W1), (d) −, Rt, , (b) −, , (a), , 4 In a coil of resistance 100 Ω, a current is induced by, changing the magnetic flux through it as shown in the, figure. The magnitude of change in flux through the coil is, ª JEE Main 2017 (Offline), 10, Current, (A), , 7 Coil A is made to rotate about a, , A, , ω, , B, , vertical axis (figure). No current, flows in B, if A is at rest. The, current in coil A, when the current, in B (at t = 0) is counter clockwise, and the coil A is as shown at this instant, (t = 0), is, (a) constant current clockwise, (b) varying current clockwise, (c) varying current counter clockwise, (d) constant current counter clockwise, , 8 One conducting U-tube can slide, , B, , A, , inside another as shown in figure,, v, v, maintaining electrical contacts, between the tubes. The magnetic, C, D, field B is perpendicular to the, plane of the figure. If each tube moves towards the other, at a constant speed v, then the emf induced in the circuit, in terms of B, l and v , where l is the width of each tube, will, be, (b) −Blv, (d) 2Blv, , (a) Blv, (c) zero, , 9 A boat is moving due to East in a region where the earth’s, , magnetic field is 5 . 0 × 10−5 NA −1m −1 due to North and, horizontal. The boat carries a vertical aerial 2 m long.If the, speed of the boat is 1.50 ms −1, the magnitude of the, induced emf in the wire of aerial is, ª AIEEE 2011, (a) 1mV, (c) 0.50 mV, , (b) 0.75 mV, (d) 0.15 mV, , 10 A helicopter rises vertically with a speed of 10 ms −1., Time (s) 0.5, , (a) 225 Wb, , (b) 250 Wb, , (c) 275 Wb, , (d) 200 Wb, , 2, , 5 A coil has an area of 0.05 m and has 800 turns. After, placing the coil in a magnetic field of strength, 4 × 10−5 Wb m −2, perpendicular to the field, the coil is, rotated by 90° in 0.1 s. The average emf induced is, (a) zero, , (b) 0.016 V, , (c) 0.01 V, , (d) 0.032 V, , 6 A cylindrical bar magnet is rotated about its axis. A wire, is connected from the axis and is made to touch the, cylindrical suface through a contact. Then,, (a) a direct current flows in the ammeter A, (b) no current flows through the ammeter A, (c) an alternating sinusoidal current flows through the, 2π, ammeter A with a time periodT =, ω, (d) a time varying non- sinusoidal current flows through the, ammeter A, , If helicopter has a length of 10 m and the, horizontal component of the earth’s magnetic field is, 1.5 × 10 −3 Wbm −2 , the emf induced between the tip of the, nose and the tail of the helicopter, is, (a) 0.15 V, (c) 130 V, , (b) 125 V, (d) 5 V, , 11 The magnetic field as shown in the, , A, w, , figure is directed into the plane of, B, paper. A X C A is a semi-circular, O, X, conducting loop of radius a with, centre O. The loop rotates clockwise, C, with velocity ω about an axis fixed at, O and perpendicular to the plane of, the paper. The resistance of the loop is R. The induced, current is, (a), , ωr 2, 2R, , (b) −, , B ωr 2, 2R, , (c), , − 2R, B ωr, , (d), , 2R, ωr 2
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268, , DAY TWENTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 12 A metal rod of resistance 20 Ω is fixed along the, , 16 An inductor (L = 0.03 H) and a resistor (R = 015, . kΩ ) are, , diameter of a conducting ring of radius 0.1 m and lies, on the xy-plane. There is a magnetic field B = | 50 k|. The, ring rotates with an angular velocity ω = 20 rads −1 about, its axis. An external resistance of 10 Ω is connected, across the centre of the ring and the rim. The current, through the external resistance is, (a), , 1, A, 2, , (b), , 1, A, 3, , (c), , 1, A, 4, , connected in series to a battery of 15V emf in a circuit, shown below. The key K1 has been kept closed for a long, time. Then at t = 0, K1 is opened and key K 2 is closed, simultaneously. At t = 1ms, the current in the circuit will be, ª JEE Main 2015, (e 5 ~, = 150), 0.03H, , (d) zero, , K2, , 13 Two coils, x and y are kept in close vicinity of each other., When a varying current, I(t) flows through coil x, the, induced emf [ V(t)] in coil Y, varies in the manner shown, here. The variation if I(t ), with time can then be, represented by the graph labelled as graph., ª JEE Main (Online) 2013, (a), , (b), , I(t), , K1, 15V, , (a) 100 mA, , t, , (b) 67 mA, , (c) 6.7 mA, , dissipates it as heat at the rate of 320 W when a current, of 4 A is passed through it. The time constant of the, circuit is, (b) 0.1 s, , (c) 0.3 s, , which passes through the battery in one time constant is, L, , V0, , (d), I(t), , I(t), , t, , t, , (a), , 14 In the circuit shown here, the point C is kept connected to, point A till the current flowing through the circuit becomes, constant. Afterward, suddenly point C is disconnected, from point A and connected to point B at time t = 0. Ratio, of the voltage across resistance and the inductor at, ª JEE Main 2014, t = L / R will be equal to, A, , R, , C, , L, , e, 1− e, , (c) − 1, , (b) 1, , (d), , 1− e, e, , 15 An inductor (L = 100 mH), a resistor (R = 100 Ω ) and a, , battery (E = 100 V ) are initially connected in series as, shown in the figure. After a long time, the battery is, disconnected after short circuiting the points A and B. The, current in the circuit 1 millisecond after the short circuit is, L, R, B, , A, E, , (a) 1/e A, , (b) e A, , V0 e, Rτ, , (b), , V0 e, Rt, , (c), , Rτ, V0 e, , (d), , V0 τ, Re, , 19 An uniformly wound solenoid of inductance1.8 × 10−4H, and resistance 6 Ω is broken into two identical parts., These identical coils are then connected in parallel, across a 15 V battery of negligible resistance. The time, constant of the circuit is, (a) 3 × 10−5 s (b) 6 × 10−5 s, , (c) 15, . × 10−5 s (d) 1.8 × 10−5 s, , 20 A uniformly wound solenoidal coil of self-inductance, , B, , (a), , (d) 0.4 s, , 18 In series R -L circuit, switch is closed at t = 0. The charge, R, , (c), , (d) 0.67mA, , 17 An inductor coil stores 32J of magnetic energy and, , (a) 0.2 s, , I(t), , t, , 0.15 kW, , (c) 0.1 A, , (d) 1 A, , 1.8 × 10−4 H and a resistance of 6 Ω is broken up into two, identical coils. These identical coils are then connected in, parallel across a 120 V battery of negligible resistance., The time constant of the current in the circuit and the, steady state current through the battery is, (a) 3 × 10−5 s, 8 A, (c) 0.75 × 10−4 s, 4 A, , (b) 1.5 × 10−5 s, 8 A, (d) 6 × 10−5 s, 2 A, , 21 A coil is suspended in a uniform magnetic field with the, plane of the coil parallel to the magnetic lines of force. When, a current is passed through the coil, it starts oscillating; it is, very difficult to stop. But if an aluminium plate is placed near, ª AIEEE 2011, to the coil, it stops. This is due to, (a) development of air current when the plate is placed, (b) induction of electrical charge on the plate, (c) shielding of magnetic lines of force as aluminium is a, paramagnetic material, (d) electromagnetic induction in the aluminium plate giving, rise to electromagnetic damping
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ELECTROMAGNETIC INDUCTION, , DAY TWENTY THREE, Direction, , (Q. Nos. 22-26) Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 269, , 24 Statement I An artificial satellite with a metal surface, is, moving about the earth in a circular orbit. A current is, induced when the plane of the orbit is inclined to the, plane of the equator., Statement II The satellite cuts the magnetic field of earth., , 25 Statement I A coil A is connected to a voltmeterV and, the other coil B is connected to an alternating current, source. If a large copper sheet C is placed between the, two coils, the induced emf in the coil A is reduced., C, A, , B, , 22 Statement I The mutual inductance of two coils is, doubled if the self-inductance of the primary and the, secondary coil is doubled., Statement II Mutual inductance is proportional to the, square root of self-inductance of primary and secondary, coils., , 23 Statement I The energy stored in the inductor of 2 H,, when a current of 10 A flows through it, is 100 J., Statement II Energy stored in an inductor is directly, proportional to its inductance., , V, , Statement II Copper sheet between the coils, has no, effect on the induced emf in coil A., , 26 Statement I When a DC current is made to flow in a soft, wire loop of arbitrary shape, it tend to acquire a circular, shape., Statement II Flux linked with a wire loop is maximum, when loop is a circle., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The inductance per unit length of a double tape line as, shown in the figure., , about its vertical diameter with an angular speed of, 50 rad/s in a uniform horizontal magnetic field of, magnitude 3.0 ×10−2 T. Obtain the maximum and, average emf induced in the coil. If the coil forms a, closed-loop of resistance 10 Ω, calculate the maximum, value of current in the coil. Calculate the average power, loss due to Joule heating., , h, , d, , I, , b, I, , (a), , µ0h, b, , (b), , b, µ0h, , (c), , µ0b, h, , (d), , hb, µ0, , 2 An air-cored solenoid with length 30 cm, area of, cross-section 25 cm 2 and number of turns 500, carries a, current of 2.5 A. The current is suddenly switched off in a, brief time of 10−3 s. How much is the average back emf, induced across the ends of the open switch in the, circuit? Ignore the variation in magnetic field near the, ends of the solenoid., (a) 6.5 V, , (b) 7.4 V, , (c) 8.2 V, , (d) 9.3 V, , 3 A short circuited coil is placed in a time varying magnetic, field. Electrical power is dissipated due to the current, induced in the coil. If the number of turns are quadrupled, and the wire radius is halved, the electrical power, dissipated in the coil, would be, (a) halved, , (b) the same (c) doubled, , 4 A circular coil of radius 8.0 cm and 20 turns is rotated, , (d) quadrupled, , (a) 0.012 W, (c) 0.018 W, , (b) 0.1 W, (d) 0.42 W, , 5 A long straight solenoid with cross-sectional radius a and, number of turns per unit length n has a current varying, with time as I As −1. The magnitude of the eddy current, as a function of distance r from the solenoid axis is, (a), , µ In, − nµ 0a 2 I, (b) 0, 2a, 2r, , (c), , − na 2 I, 2µ 0 r, , (d), , µ 0I a, 2n, , 6 A circular loop of radius 0.3 cm lies parallel to a much, bigger circular loop of radius 20 cm. The centre of the, small loop is on the axis of the bigger loop. The distance, between their centres is 15 cm. If a current of 2.0 A flows, through the smaller loop, then the flux linked with bigger, loop is, (a) 9.2 × 10−11 Wb, (c) 3.3 × 10−11 Wb, , (b) 6 × 10−11 Wb, (d) 6.6 × 10−9 Wb
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270, , DAY TWENTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 7 A metallic rod of length l is tied to a string of length 2l and, A, , made to rotate with angular speed ω on a horizontal table, with one end of the string fixed.If there is a vertical, magnetic field B in the region, the emf induced across, the ends of the rod is, , L, 12 V, , R1, , R2, S, , 12 −3t, e V, t, (d) 12 e −5t V, , (a) 6e −5t V, , (b), , (c) 6(1 − e −t / 0. 2 )V, , 11 In a uniform magnetic field of induction B, a wire in the, , (a), , 2Bωl3, 2, , (b), , 3Bωl3, 2, , (c), , 4Bωl2, 2, , (d), , 5Bωl2, 2, , 8 A rectangular loop has a sliding connector PQ of length l, and resistance R Ω and it is moving with a speed v as, shown. The setup is placed in a uniform magnetic field, going into the plane of the paper. The three currents I1, I 2, and I are, P, , form of semi-circle of radius r rotates about the diameter, of the circle with angular frequency ω. If the total, resistance of the circuit is R, the mean power generated, per period of rotation is, B πr 2ω, 2R, (B πrω)2, (c), 2R, , (B πr 2ω)2, 8R, (B πrω2 )2, (d), 8R, (b), , (a), , 12 The inductance between A and D is, , l, , A, , D, 3H, , RW, , v, , RW, I, I1, , RW, I2, , Q, , B lv, 2B l v, ,I =, R, R, B lv, 2B l v, (b) I1 = I 2 =, ,I =, 3R, 3R, B lv, (c) I1 = I 2 = I =, R, B lv, Blv, (d) I1 = I 2 =, ,I =, 6R, 3R, , (b) 9 H, (d) 1 H, , 13 A loop, made of straight edges has six corners at, , region. The flux passing through the loop ABCDEFA (in, that order) is, (a) B 0L2 Wb, , (b) 2B 0L2 Wb, , (c) 2B 0L2 Wb, , (d) 4B 0L2 Wb, , 14 Magnetic flux in a circular coil of resistance10 Ω changes, , 9 An ideal coil of 10 H is connected in series with a, resistance of 5 Ω and a battery of 5 V. 2s after the, connection is made, the current flowing (in ampere) in the, circuit is, (b) e, , (a) 3.66 H, (c) 0.66 H, , 3H, , A ( 0, 0, 0), B (L, 0, 0), C (L, L, 0), D ( 0, L, 0), E ( 0, L, L ) and, F ( 0,0, L ). A magnetic field B = B0( $i + k$ ) T is present in the, , (a) I1 = − I 2 =, , (a) (1− e), , 3H, , (c) e −1, , with time as shown in figure. Cross indicates a direction, perpendicular to paper inwards. Match the following., f, 10, , (d) (1 − e −1), , 2, , 10 An inductor of inductance L = 400 mH and resistors of, resistances R1 = 4 Ω and R 2 = 2 Ω are connected to, battery of emf 12 V as shown in the figure. The internal, resistance of the battery is negligible. The switch S is, closed at t = 0. The potential drop across L as a function, ª AIEEE 2009, of time is, , –10, , 6, , 8, , 10, , 14 16, , t(s)
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ELECTROMAGNETIC INDUCTION, , DAY TWENTY THREE, , Column I, , Column II, , Column I, , A. at t = 1 s, , 1. emf induced is zero, , B. at t = 5 s, , 2. emf induced in anti-clockwise direction, , C. at t = 9 s, , 3. emf induced in clockwise direction, , D. at t = 15 s, , 4. 2 A, , Codes, A, (a) 2, (c) 4, , B, 1, 3, , C, 1, 2, , D, 2, 1, , A, (b) 1, (d) 2, , B, 2, 4, , C, 3, 1, , D, 4, 3, , Column II, , A. Loop is moved towards, right, , 1. Induced current in the loop is, clockwise, , B. Loop is moved towards, left, , 2. Induced current in the loop is, anti-clockwise, , C. Wire-1 is moved towards, left, , 3. Induced current in the loop is, zero, , D. Wire-2 is moved towards, right, , 4. Induced current in the loop is, non-zero, , Codes, , 2, , 1, , 15 A square loop is symmetrically placed, , 271, , A, (a) 1, (c) 1, , between two infinitely long current carrying, wires in the same direction. Magnitude of, currents in both the wires are same.Now, match the following two columns., , B, 2, 2, , C, 1, 3, , D, 2, 4, , A, (b) 1, (d) 4, , B, 3, 2, , C, 2, 2, , D, 4, 4, , ANSWERS, SESSION 1, , SESSION 2, , 1 (a), , 2 (b), , 3 (b), , 4 (b), , 5 (b), , 6 (b), , 7 (b), , 8 (d), , 9 (d), , 10 (a), , 11 (b), , 12 (b), , 13 (c), , 14 (c), , 15 (a), , 16 (d), , 17 (a), , 18 (d), , 19 (a), , 20 (a), , 21 (d), , 22 (a), , 23 (a), , 24 (a), , 25 (c), , 26 (b), , 1 (a), 11 (b), , 2 (a), 12 (d), , 3 (b), 13 (b), , 4 (c), 14 (a), , 5 (a), 15 (a), , 6 (a), , 7 (d), , 8 (b), , 9 (d), , 10 (d), , Hints and Explanations, SESSION 1, 1 As loop is in the xy -plane, only the, z -component of the magnetic field, is, effective., B = − 18 gauss = −18 × 10−4 T, , ∴ Induced current,, e, n dφ, I =, =−, R′, R ′ dt, Given, R ′ = R + 4R = 5R,, , φ = B A cos 0° = −18 × 10, , × 5 × 10, , [here, W1 and W2 are flux, associated with one turn], , −4, , = −90 × 10−8 Wb, = − 900 × 10−9 Wb, , 2 φ = 10t 2 − 50t + 250, From Faraday’s law of electromagnetic, induction,, e = − dφ/ dt, ∴, e = − [10 × 2t − 50], ∴e | t = 3 s = − [10 × 6 − 50] = − 10 V, , Putting the given values in Eq. (i), we, get, n (W2 − W1 ), I =−, 5R, t, , 4 Induced constant, I = e, , R, , Here, e = induced emf =, I =, , 3 The rate of change of flux or emf, induced in the coil is e = − n, , dφ, ., dt, , ∴ Here, R is constant, ∴, φ = R ∫ Idt, , ∫ I ⋅ dt, , ⇒, , e dφ 1, = ⋅, R dt R, , dφ = IRdt, φ = ∫ IRdt, , dφ, dt, , = Area under I - t graph, , 1, × 10 × 0.5 = 2.5, 2, φ = R × 2.5 = 100 × 2.5, = 250 Wb, =, , dφ = W2 − W1 ,dt = t, , A = 5 × 10−4 m2, −4, , …(i), , ∴, , 5 As from Faraday’s rule,, , dφ NAdB, =, dt, dt, 4 × 10−5, = 800 × 0.05 ×, 01, ., = 0.016 V, , e =N, , 6 When cylindrical bar magnet is rotated, about its axis,no change in flux linked, with the circuit takes place, consequently, no emf induces and hence,no current flows, through the ammeter A.
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272, , DAY TWENTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 1, Bωr 2, 2, 1, = × 50 × 20 × (01, . )2 = 5 V, 2, , V =, , Axis, A, , N, Bar, magnet, , The equivalent circuit of the, arrangement is shown in figures., , w, , A, , 5V, 10W, , 10 W, , 10 W, , E 100, =, =1A, R 100, [as inductor would be shorted in steady, state], After this decay of current starts in the, circuit, according to the equation,, I = I 0e − t / τ, I0 =, , where,, , τ = L /R, L, , 10 W, , 5V, , S, 10 W, , 7 When the current in B(at t = 0) is, counter clockwise and the coil A is, considered above to it. The counter, clockwise flow of the current in B is, equivalent to North pole of magnet and, magnetic field lines are emerging, upward to coil A. When coil A start, rotating at t = 0, the current in A is time, varying along clockwise direction by, Lenz’s rule., , 8 Relative velocity = v − (− v ) = 2v = dl, dt, , Bl dl, dφ, Now, e =, ⇒e =, dt, dt, , [Q φ = BA], Q dl = 2 v , dt, , , Induced emf, e = 2 B l v, , 9 E ind = B × v × l = 5.0 × 10−5 × 1. 50 × 2, = 10. 0 × 10−5 × 1. 5, = 15 × 10−5 = 015, . mV, , 10 e = Bl v = B H l v, = 1.5 × 10, , −3, , . V, × 10 × 10 = 015, , π, , 2, , 2, , B, , 5V, , 5W, , −3, , I = 1 × e − (1 × 10, , 2, , Current through the external resistance,, E, 5, 1, I =, =, = A, R + r 10 + 5 3, , 13 Firstly, the current decreases due to, electrical inertia goes to zero, but due to, back emf induced in the easily, the, induced current in the coil decreases off, a point when back emf is equal to the, applied emf induced in the another coil., The value of emf of two current is, zero.Then, current is regularly, increased, after that time became it is, continuously by supplied by the source, (variable)., , 14 After connecting C to B hanging the, , B, , L, , ωt r 2, 2, B ω r2, dφ, Emf, e = −, =−, 2, dt, − B ω r2, I =, 2R, , Flux, φ = BA = B, , ω tr 2, 2, will give same current but in opposite, direction., , Applying Kirchhoff’s loop equation,, VR + VL = 0 ⇒ VR = − VL, VR, ∴, = −1, VL, , 15 This is a combined example of growth, and decay of current in an L-R circuit., L=100 mH, , After half rotation A(t ) = πr 2 −, , 12 Here, resistance of rod = 20 Ω,, r = 01, . m, B = 50 T, acting along the, z-axis and ω = 20 rad s −1 ., Potential difference between the centre, of the ring and the rim is, , 1, = A, e , , and L = 100 × 10−3 H], , 10 W, , 1, 1, 1, 2, 1, =, +, ⇒, = ⇒ Rp = 5Ω, R p 10 10, 10 5, , R, θ, , )/(100 × 10−3 /100), , [Q t = 1 millisecond = 1 × 10−3 s, , switch, the circuit will act like L-R, discharging circuit., , π r 2 θ r 2 ωt r 2, 11 A = θ, =, =, , R, , 10 W, , R=100W, A, , B, E=100 V, , The current through circuit just before, shorting the battery,, , 16 After long time inductor behaves as, short-circuit., At t = 0, the inductor behaves as, short-circuited. The current, 15V, E, I0 = 0 =, = 100 mA, R, 015, . kΩ, As K 2 is closed, current through the, inductor starts decay, which is given at, any time t as, − t × 15000, , − tR, , I = I0 e, , L, , = (100 mA) e, , At t = 1 ms, I = (100 mA ) e, , −, , 3, , 1 ×10−3 ×15×103, 3, , I = (100 mA ) e − 5 = 0.6737 mA, or I = 0.67 mA, LI2, L(4)2, or 32 =, = L = 4H, 2, 2, 320, P = I 2 R, R = 2 = 20 Ω, (4), L, 4, τ =, =, = 0.2 s, ∴, R 20, 18 I = V 0 (1 − e − t / τ ), R, τV, Q = ∫ I dt = ∫ 0 (1 − e − t / τ ) dt, 0 R, V τ, V, V, ⇒ Q = 0 τ + τ (e −1 − 1) 0 = 0, R, Re, R, , 17 U =, , 19 Inductance of each part,, L, = 0.9 × 10−4 H, 2, Resistance of each part,, R, = 3Ω, R1 = R2 =, 2, Time constant,, L ||L, L1 L2, R + R2, τ = 1 2 =, × 1, R1||R2, L1 + L2, R1 R2, L1 = L2 =, , τ =, , 1.8 × 10−4 × 0.9 × 10−4, 1.8 × 10, , −4, , + 0.9 × 10, , −4, , ×, , 6+ 3, 6×3, , 1.62 × 10−4, =, = 03, . × 10−4 = 3 × 10−5s, 54, .
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ELECTROMAGNETIC INDUCTION, , DAY TWENTY THREE, 20 Since, the self-inductance in parallel is, given by, 1, 1, 1 2, = +, =, Lp L, L, L, and L =, ∴, , L, , p, , ⇒ L, , p, , =, , L, 2, , 1.8 × 10−14, = 0.9 × 10−4 H, 2, , = 0.45 × 10−4 H, , Resistance of each part, r = 6 / 2 = 3 Ω, 1, 1 1 2, Now,, = + =, rp 3 3 3, 3, ∴, rp = Ω, 2, So, the time constant of the circuit is, given by, L p 0.45 × 10−4, τ =, =, = 3 × 10−5 s, rp, 3/2, and the steady current is,, V, 12, I =, =, = 8A, rp 3/2, , 21 According to Lenz’s law,, electromagnetic induction takes place, in the aluminium plate for which eddy, current is developed. This causes loss in, energy which results in damping of, oscillatory motion of the coil., , 22 If two coils of inductances L 1 and L 2, are joined together, then their mutual, inductance is given by, M = k L1 L2, It is clear from the relation, if the, self-inductance of the primary and the, secondary coil is doubled, the mutual, inductance of the coils, will also be, doubled., , 23 The energy stored in the inductor is, , 24 It is concept of eddy current losses., 25 In the absence of the copper sheet,, induced emf will be produced in the, coil A due to the mutual induction, between the coils A and B. As a result,, voltmeter will show deflection, depending on the magnitude of the, induced emf., When the copper sheet is placed, between the two coils, eddy currents, will be setup in the coil. Since, the eddy, currents have an opposing effect, the, magnetic flux linked with A due to eddy, current will always be opposite to that, , or, , 4, , 1, P2, 1, = (4)2 × = 16 ×, =1, 2, P1, 16, , So,, , SESSION 2, , 4 Average induced emf, , 1 Neglecting end effects of magnetic field,, , e av =, , we have, , µ 0I, b, Flux φ per unit length of the plates is, µ 0I, µ hI, × h×1= 0, b, b, µ h, Also, φ = L I ⇒ L = 0, b, B =, , Number of turns N = 500, Current I1 = 2.5 A, I2 = 0, Brief time dt = 10−3 s, Induced emf in the solenoid, dφ, d, (BA), e =, =, (Q φ = BA ), dt dt, , , , , , , 1, T, , π, , ∫0, , NBA ω sin ω t dt, 2π, , M =, , µ 0R12 πR22, 2(R12 × x2 )3 /2, , Flux through the bigger coil,, M =, , µ 0 π2 R12 R22, ⋅, 4 π (R12 + x2 ), , Substituting the values, M =, , µ 0(2)(20 × 10−2 )2, 2[(0.2)2 + (015, . )2 ], , × π(0.3 × 10−2 )2, , On solving,, , proportional to the rate of change of, magnetic flux which, in turns depends, on the number of turns in the coil, i.e., V ∝ n., So, resistance of a wire is given by, ρl, R=, [ A = π r2], π r2, l, i.e. R ∝ 2, r, [ ρ is a resistivity of a wire], V2, n2, (nr )2, ∴ P =, ∝, ⇒ P =, l, R, l /r 2, , − n µ 0 a2 I, 2r, , 6 Mutual inductance of two coils,, , 3 The magnitude of the induced voltage is, , r , l , P2 n2 , = × 2 × 1, P1 n1 , r1 , l2 , , e dt =, , E (2 πr ) = − π a2 n µ 0 I ⇒ E =, , e = 500 × 25 × 10−4 × 4 × 314, . × 10−7, 2.5, 500, ×, ×, 30 × 10−2 10−3, = 6.5 V, , ∴, , 2π, , ∫0, , 1, cos ω t , ⋅ NAB ω , T, ω 0, NBA, =, [cos 2 π − cos 0° ], T, NBA, e av =, [1 − 1] = 0, T, For full cycle average emf, e av = 0, Average power loss due to heating, 0.603 × 0.0603, E I, = 0.018 W, = 0 0 =, 2, 2, 5 B = n µ 0 I and ∫ E ⋅ dl = − dφ, dt, For r < a, E (2 π r ) = − π r 2 nµ 0I, nµ 0 I r, or E = −, (for r < a), 2, dI, where, I = ., (for r > a), dt, , = 30 × 10−2 m, Area of cross-section A = 25cm2, = 25× 10−4 m2, , 2, , 1, T, , e av =, , 2 Given, length of solenoid l = 30 cm, , 2, , 4, , , n2, r, 1, = 4 and 2 = , Given,, n1, r1 2 , , , diametrically opposite section as, current flows in opposite direction., , Magnetic field induction B at a point, well inside the long solenoid carrying, current I is, B = µ 0nI, where, n = Number of turns, , N, per unit length =, , , l, dB, d N , e = NA, = A, µ 0 I , dt, dt , l , µ 0N dl, = A, ⋅, l, dt, , 2, , n , r , l1 r2 , P, = ⇒ 2 = 2 × 2, l2 r1 , P1 n1 , r1 , , 26 Each section of wire repels, , given by, 1, 1, U = L I 20= × 2 × (10) 2 = 100 J, 2, 2, It is obvious that energy stored in the, inductor, is directly proportional to its, inductance., , 2, , due to the alternating current through, B. Thus, induced current will be, reduced., , 273, , M = 9.216 × 10−11, = 9.216 × 10−11 ≈ 9.2 × 10−11 Wb, , 7 Consider a small element of length dx as, shown below emf induced due to whole, rod, 3l, [(3 l)2 − (2l)2 ], e = ∫ (ωx )Bdx = Bω, 2l, 2, 5Bl2ω, =, 2, I, w, , 2I, x, , dx
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274, , DAY TWENTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , Q Potential drop = E − I2 R2, I2 = I 0(1 − e − T / tc ), , 8 A moving conductor is equivalent to a, battery of emf = vBl (motion emf), Equivalent circuit, I = I1 + I2, P, , R, , ⇒, , [current as a function of time], 12, E, =, = 6A, I0 =, 2, R2, , and, , L, 400 × 10−3, tc =, =, = 0.2, R2, 2, , R, , R, , Q, , I 1 = I2 =, , 5Ω, , Hence, Pmean = < P >, B 2 π2 r 4ω2 1, =, ⋅, 4R, 2, Q < sin2 ωt > = 1 , , 2 , 2 2, (Bπr ω ), =, 8R, , 5V, L 10, Now, τ =, =, = 2s, R, 5, After 2s, i.e. at t = 2s, Rise of current I = (1 − e −1 ) A, , 10 I1 = E = 12 = 3 A, R1, 4, , A, , and magnetic induction B is, 1, φ = BA cos θ = Bπr 2 cos ωt, 2, Q A = 1 πr 2 , , , 2, dφ, ∴ e induced = −, dt, d 1, , 2, =−, Bπr cos ωt , , dt 2, 1, 2, = Bπr ω sin ωt, 2, e2, [Q P = V 2 / R], ∴ Power, P = induced, R, B 2 π2 r 4ω2 sin2 ωt, =, 4R, , Blv, 3R, , 12, A, , E, B, , 3H, , F, 3H, , C, , 3H, , I, A, , I1, R1, , I2, , F D, , L, , 3W, 3W, , E, R2, S, , 3W, A B, , C, , C, , E, , 11 The flux associated with coil of area A, , I = I 0(1 − e − t / τ ), E, 5, I0 =, = = 1A, R 5, , 10 H, , D, , = 12e −5 t V, , ...(i), ...(ii), , 9 Rise of current in L - R circuit is given by, where,, , Y, , I2 = 6(1 − e −5t ), , Potential drop across, L = E − R2 I2, = 12 − 2 × 6(1 − e −5 t ), , I2, , Applying Kirchhoff’s law,, I1 R + I R − v B l = 0, I2 R + I R − v B l = 0, Adding Eqs. (i) and (ii), we get, 2 IR + IR = 2v B l, 2v B l, I =, 3R, , 13 Also, the magnetic flux linked with, uniform surface of area A in uniform, magnetic field is given by, , ., I2 = 6(1 − e − t / 02, ), , ⇒, I1, , Here, inductors are in parallel., 1 1 1 1, ∴, = + +, L 3 3 3, or, L = 1H, , D, , B, , X, , F, Z, φ = B⋅A, A = A1 + A2, = (L2 k$ + L2 $i ), and B = B 0($i + k$ ) T, Now,, φ = B⋅ A, = B ($i + k$ ).(L2 k$ + L2 $i ), 0, , = 2 B 0 L2 Wb, , 14 A → 2, B → 1, C → 1, D → 2, A. At t = 1 s, flux is increasing in the, inward direction, hence induced emf will, be in anti-clockwise direction., B. At t = 5s, there is no change in, flux,so induced emf is zero., C. At t = 9 s, flux is constant, hence, induced emf will be zero., D. At t = 15s, flux is decreasing in, upward direction, so induced emf will be, in anti-clockwise direction., , 15 A - 1, B - 2, C - 1, D - 2, When loop is moved towards right,, upward flux increases, so current in loop, is clockwise. When loop is moved, towards left, downward flux increases, so, current in loop is anti-clockwise. When, wire 1 is moved left, upward flux, through loop increases, so current is, clockwise. When wire 2 is moved right,, downward flux increases, so current is, anti-clockwise.
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DAY TWENTY FOUR, , Alternating, Current, Learning & Revision for the Day, u, , u, , Peak and RMS Values of, Alternating Current/Voltage, Different Types of AC Circuits, , u, , u, , Series AC Circuits, Power in an AC Circuit, , u, , u, , AC Generator, Transformer, , An alternating current is the current (or voltage) whose magnitude keeps on changing, continuously with time, between zero and a maximum value and its direction also, reverses periodically., VI, , 0, , I0, V0, , t, T, 4, , T, 2, , 3T, 4, , T, , Peak and RMS Values of, Alternating Current/ Voltage, RMS value of alternating voltage is equal to, Similarly, I rms =, , 1, V, times of peak value. i.e. Vrms = 0, 2, 2, , I0, 2, , Mean Value or Average Value, The steady current, which when passes through a circuit for half the time period of, alternating current, sends the same amount of charge as done by the alternating current, in the same time through the same circuit, is called mean or average value of alternating, current. It is denoted by i m or i av, 2i, i m or i av = 0 = 0.636i 0, π, Mean or average value of alternating current during a half cycle is 0.636 times (or 63.6%, of) its peak value (i 0)., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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DAY TWENTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , Similarly, mean or average value of alternating emf, 2V, Vm or Vav = 0 = 0.636, π, , R, , Peak Value, RMS Value, The steady current, which when passes through a resistance, for a given time will produce the same amount of heat as the, alternating current does in the same resistance and in the, same time, is called rms value of alternating current. It is, i, denoted by i rms or i v = 0 = 0.707 i 0, 2, where, i 0 = peak value of alternating current, Similarly, rms value of alternating emf, V, V rms = 0 = 0.707 V0, 2, , Reactance and Impedance, , l, , l, , l, , The opposition offered by a pure inductor or capacitor or, both to the flow of AC, through it, is called reactance (X )., Its unit is ohm (Ω) and dimensional formula is, [ML 2 T−3 A −2 ]., Reactance is of two types, (i) Inductive reactance, X L = Lω and, 1, (ii) Capacitive reactance, X C =, Cω, Reciprocal of reactance is known as susceptance., 1, Thus,, S =, X, Total opposition offered by an AC circuit to the flow of, current through it, is called its impedance (Z). Its unit is, ohm and dimensional formula is [ML 2 T −3 A −2 ]., For an AC circuit, Z =, , l, , I, , 0, , V, , V = V0 sin wt, , The maximum value (amplitude) of alternating current and, voltage is called peak value., , l, , Irms, , 276, , X 2 + R2 = ( X L − X C )2 + R2, , Reciprocal of impedance is known as admittance., 1, Thus, Y = . Its unit is Siemens (S)., Z, , Different Types of AC Circuits, The circuit consists of only resistor or only capacitor or only, inductor are called pure resistive, pure inductive and pure, capacitive circuit., , (b), , (a), , V, , Current and voltage are in the same phase, i.e. current is, given by I = I 0 sin ωt ., , 2. Pure Inductive Circuit, Let an alternating voltage V = V0 sin ωt be applied across a, pure inductance L., π, Then, the average power = Vrms I rms cos = 0, 2, The inductance offers some opposition to the flow of AC,, known as inductive reactance X L = 2πνL = Lω., Thus, a pure inductance does not oppose the flow of DC, (ω = 0) but opposes the flow of AC., V, Current flowing, I =, XL, XL, , V, (b), , L, , (a), , V = V0 sin ωt, , (c), π, 2, , 0, , 0, , I, , V, , In pure inductive circuit, current decreases with an, π, increase in frequency, it lags behind the voltage by, 2, π, (or voltage leads the current by ) and is thus given by, 2, π, , I = I 0 sin ωt − , , 2, , 3. Pure Capacitive Circuit, Let an alternating voltage V = V0 sin ωt be applied across a, pure capacitance C. Then, the capacitance offers some, opposition to the flow of current, but allows AC to pass, through it. The opposition offered is known as the, capacitive reactance., 1, XC =, Ω, Cω, 1, =, Ω, C × 2 πν, Current flowing, I =, , V, XC, , 1. Pure Resistive Circuit, Let an alternating voltage V = V0 sin ωt be applied across a, pure resistance R. Then,, V, V, Current, I = or I rms = rms, R, R, , (c), , (a), , C, V = V0 sin ωt, , π, 2, , (b), , V, , XC, , I, (c), , V
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278, , DAY TWENTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , For X L > X C , current lags voltage., X L < X C , current leads voltage., X L = X C , current and voltage are in phase., 1, , i.e. the natural frequency of the, If X L = X C ⇒ ω 0 =, LC, circuit is equal to the applied frequency, then the circuit is, said to be in resonance., At resonance, the current in the circuit is maximum and, the impedance is minimum and equal to R., 1, Resonance frequency, ν =, 2 π LC, , Quality Factor, The Q-factor or quality factor of a resonant L-C-R circuit is, defined as ratio of the voltage drop across inductor, (or capacitor) to applied voltage. Thus,, voltage across L (or C), Q=, applied voltage, Q=, , It is known as resonance frequency., At resonance frequency admittance is minimum of the, impedance is maximum. Thus, the parallel circuit does not, allow this frequency from the source to pass in the circuit., Due to this reason, the circuit with such a frequency is known, as rejector circuit., 1, L, Dynamic resistance, Z max =, =, Ymax CR, l, , V0, V CR, = 0, L/CR, L, , l, , Peak current through the supply =, , l, , The peak current through capacitor =, , l, , Q-factor =, , V0, = V0ωC, 1/ωC, , ωL, V0ωC, =, V0CR/ L, R, , This is basically the measure of current magnification., , L-C Oscillations, An L-C circuit also called a resonant circuit, tank circuit or, tuned circuit. When connected together, they can act as an, electrical resonator, storing energy oscillating at the circuits, resonant frequency., , 1 L, R C, , Parallel Resonant Circuit, , I, , In this combination, a capacitor is connected in parallel with a, series combination of inductor and resistor., R, , L, , C, , L, , IL, IC, , C, , I, , V = V 0 sin ωt, , From the figure,, I = I L + IC, V, V, V, or, =, +, Z R + jωL − j/ωC, , Power in an AC Circuit, Let a voltage V = V0 sin ωt be applied across an AC and, consequently a current I = I 0 sin(ωt − φ) flows through the, circuit. Then,, , 1, 1, =, + jωC, Z R + jωL, , ∴, , l, , 1, is known as admittance (Y )., Z, R − jωL, 1, Thus, Y = = 2, + jωC, Z R + ω2 L2, ∴, , Y =, , The energy oscillates back and forth between the capacitor, and inductor until internal resistance makes the oscillations, die out. The oscillation frequency is determined by the, capacitance and inductance values,, ω, 1, f = 0 =, 2 π 2 π LC, , l, , R2 + (ωCR2 + ω3 L2C − ωL)2, R2 + ω2 L2, , The admittance will be minimum, when, l, , ωCR + ω L C − ωL = 0 or ω =, 2, , ∴, , 3 2, , f =, , ω, 1, =, 2π 2π, , 1, R2, − 2, LC L, , R2, 1, − 2, LC L, , Instantaneous power = V I = V0I 0 sin ωt sin(ωt − φ), and its value varies with time. Here, φ is known as phase, difference between V and I., Average power over a full cycle of AC is, 1, Pav = Vrms I rms cos φ = V0I 0 cos φ, 2, The term Vrms I rms is known as the apparent or virtual, power, but Vrms I rms cos φ is called the true power., The term cos φ is known as the power factor of the given, circuit. Thus,, true power, R, cos φ = = power factor =, apparent power, Z
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ALTERNATING CURRENT, , DAY TWENTY FOUR, , l, , For a pure resistive circuit, V and I are in phase, (φ = 0 °), hence cos φ = 1 and average power = Vrms I rms, For a pure inductive or a pure capacitive circuit, current, π, π, and voltage differ in phase by i.e. φ = ., , 2, 2, ∴, , l, , Pavg = 0, , Power loss = I 2 R =, , V2, R, , 279, , AC Generator, An electric generator or dynamo is a device used to produce, electrical energy at the expense of mechanical/thermal energy., It works on the principle of electromagnetic induction, when, a coil is rotated in a uniform magnetic field, an induced emf, is set up between its ends. The induced emf is given by, e = e 0 sin ωt = NBAω sin ωt ., The direction of the induced emf is alternating in nature., , Wattless Current, Average power is given by Pav = E rms I rms cos φ, , Transformer, , The phase difference between E rms and I rms is φ. We can, resolve I rms into two components, I rms cos φ and I rms sin φ, , It is a device which works in AC circuits only and is based on, the principle of mutual induction., , Here, the component I rms cos φ contributes towards power, dissipation and the component I rms sin φ does not contribute, towards power dissipation. Therefore, it is called wattless, current., , Choke Coil, A low resistance inductor coil used to suppress or limit the, flow of alternating current without affecting the flow of direct, current is called choke coil., Let us consider a choke coil (used in tube lights) of large, inductance L and low resistance R. The power factor for such, a coil is given by, R, R, [as, R << ωL], cos φ =, ≈, R2 + ω2 L2 ωL, As R << ωL, cos φ is very small. Thus, the power absorbed by, the coil V rms I rms cos φ is very small. On account of its large, impedance Z = R + ω L , the current passing through the, 2, , 2 2, , coil is very small. Such a coil is used in AC circuits for the, purpose of adjusting current to any required value without, wastage of energy., The only loss of energy is due to hysteresis in the iron core,, which is much less than the loss of energy in the resistance, that can also reduce the current, if placed instead of the, choke coil., , Transformer is used to suitably increase or decrease the voltage, in an AC circuit. Transformer which transforms strong AC at, low voltage into a weaker current at high alternating voltage is, called a step-up transformer. A step-down transformer, transforms weak current at a higher alternating voltage into a, strong current at a lower alternating voltage., Ip, e, V, N, For an ideal transformer s = s = s =, =k, ep Vp, Np, Is, where, k is known as the transformation ratio., For a step-up transformer, k > 1 but for a step-down, transformer, k < 1., In a transformer, the input emf and the output emf differ in, phase by π radians., The efficiency of a transformer is given by, output power Vs I s, =, η=, input power V p I p, For an ideal transformer, η = 100% or 1. However, for practical, transformers, η ≈ 85 - 90%., Possible causes of energy loss in transformer are, l, , Heating due to winding resistance, , l, , Eddy current losses, , l, , Magnetic flux leakage and, , l, , Hysteresis loss. To minimise these losses, the transformer, core is made up of a laminated soft iron strips.
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280, , DAY TWENTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The alternating current in a circuit is described by the, graph as shown in figure. The rms current obtained from, the graph would be, 3, 2, 1, , V (R 1 + R 2 ), V, at t = 0 and, at t = ∞, R2, R 1R 2, , (c), , VR1R 2, V, at t = 0 and, at t = ∞, R2, R12 + R 22, , (d), , V (R1 + R 2 ), V, at t = 0 and, at t = ∞, R1R 2, R2, , T, , 0, , I (A), , (b), , 5 An L-C circuit is in a state of resonance. If C = 01, . µF and, , 2T, , –1, , t, , L = 0 . 25 H, then neglecting the ohmic resistance of the, circuit, find the frequency of oscillations., , –2, –3, , (a) 1007 Hz, , (a) 1.414 A, (c) 1.9 A, , (b) 2.2 A, (d) 2.6 A, , (c) 109 Hz, , (d) 500 Hz, , 6 An L-C-R circuit as shown in the figure is connected to a, , 2 An alternating voltage (in volts) given by, V = 200 2 sin(100 t ), is connected to a 1µF capacitor through an AC ammeter., The reading of the ammeter will be, (a) 10 mA, (c) 40 mA, , (b) 100 Hz, , voltage sourceVAC whose frequency can be varied. The, frequency, at which the voltage across the resistor is, maximum, is, ª JEE Main (Online) 2013, V, , (b) 20 mA, (d) 80 mA, , 24 H, , 2 mF, , 15 W, , 3 The figure shows an experimental plot discharging of a, , Potential difference V, (in volt), , capacitor in an R-C circuit. The time constant τ of this, circuit lies between, ª AIEEE 2012, 25, , 15, , (d) 345 Hz, , L, , 10, 5, , C, , 50 100 150 200 250 300, Time (in second), , E, , (b) 0 and 50 s, (d) 100 s and 150 s, , 4 In the circuit shown below, the key K is closed at t = 0., The current through the battery is, V, , ª AIEEE 2010, K, , L, , R1, R2, , VR 1R 2, R, , (c) 23 Hz, , respectively, 300 V and 400 V. The voltage E of the AC, ª JEE Main Online 2013, source is, , (a) 150 s and 200 s, (c) 50 s and 100 s, , 2, 1, , (b) 143 Hz, , 7 In the circuit shown here, the voltage across L and C are, , 20, , 0, , (a), , VAC = V0 sin wt, , (a) 902 Hz, , +R, , 2, 2, , at t = 0 and, , V, at t = ∞, R2, , (a) 400 V, , (b) 500 V, , (c) 100 V, , (d) 700 V, , 8 A fully charged capacitor C with initial charge q 0 is, , connected to a coil of self-inductance L at t = 0. The time, at which the energy is stored equally between the electric, ª AIEEE 2011, and the magnetic fields is, (a), , π, LC, 4, , (b) 2 π LC, , (c) LC, , (d) π LC, , 9 In an L-C-R circuit, if V is the effective value of the, applied voltage, VR is the voltage across R, VL is the, effective voltage across L, VC is the effective voltage, across C, then, (a) V = VR + VL + VC, (c) V = V + (VL − VC ), 2, , 2, R, , (b) V 2 = VR2 + VL2 + VC2, 2, , (d)V 2 = VL2 + (VR − VC ) 2
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ALTERNATING CURRENT, , DAY TWENTY FOUR, , L-C-R series circuit. Which one of the graphs in the, figure, represents the variation of current I in the circuit, with frequency f ?, I, , 14 The plot given below is of the average power delivered to, an L-R-C circuit versus frequency. The quality factor of the, circuit is, ª JEE Main (Online) 2013, Average power (microwatts), , 10 An AC circuit of variable frequency f is connected to an, , I, , (b), , (a), , f, , f, I, , I, , 1.0, 1.5, , 0.0, 3, , (d), , (c), f, , f, , 11 In a series L-C-R circuit, C = 10−11 F, L = 10−5 H and, R = 100 Ω, when a constant DC voltage E is applied to, the circuit, the capacitor acquires a charge10−9 C. The, DC source is replaced by a sinusoidal voltage source in, which the peak voltage E 0 is equal to the constant DC, voltage E . At resonance, the peak value of the charge, acquired by the capacitor will be ª JEE Main (Online) 2013, (a) 10−15 C, (c) 10−10 C, , (b) 10−6 C, (d) 10−8 C, , 12 In a L-C-R circuit as shown below, both switches are, open initially. Now, switch S1 and S 2 , kept open, (q is charge on the capacitor and τ = RC is capacitance, time constant). Which of the following statement is, ª JEE Main 2013, correct ?, V, R, , S1, , 281, , 6, 4, 5, Frequency (kHz), , (a) 5.0, (c) 2.5, , 7, , (b) 2.0, (d) 0.4, , 15 For an R-L-C circuit driven with voltage of amplitude v m, , 1, , the current exhibits resonance., LC, The quality factor Q is given by, , and frequency ω 0 =, , ωL, (a) 0, R, (c), , ª JEE Main 2018, , ωR, (b) 0, L, , R, ω0 C, , (d), , CR, ω0, , 16 In a series L-C-R circuit, R = 200 Ω and the voltage; and, the frequency of the main supply is 220 V and 50 Hz,, respectively. On taking out the capacitance from the, circuit, the current lags behind the voltage by 30°. On, taking out the inductor from the circuit, the current leads, the voltage by 30°. The power dissipated in the L-C-R, circuit is, ª AIEEE 2010, (a) 305 W, , (b) 210 W, , (c) zero, , (d) 242 W, , 17 In an AC circuit, the voltage applied is E = E 0 sin ωt . The, , C, S2, L, , π, , resulting current in the circuit is I = I 0 sin ωt − . The, , 2, , power consumption in the circuit is given by, (a) Work done by the battery is half of the energy, dissipated in the resistor, CV, (b) At t = τ, q =, 2, (c) At t = 2 τ, q = CV (1 − e − 2 ), τ, (d) At t = , q = CV (1 − e − 1), 2, , 13 In a series resonant L-C-R circuit, the voltage across R is, 100 V and R = 1 kΩ with C = 2 µF. The resonant, frequency ω is 200 rad/s. At resonance, the voltage, across L is, (a) 2.5 × 10−2 V, (c) 250 V, , (b) 40 V, (d) 4 × 10−3 V, , E0 I0, 2, E I, (c) P = 0 0, 2, (a) P =, , (b) P = zero, (d) P =, , 2 E0 I0, , 18 Which of the following components of an L-C-R circuit,, with an AC supply, dissipates energy?, (a) L, , (b) R, , (c) C, , (d) All of these, , 19 An AC circuit consists of a 220 Ω resistance and a 0.7 H, choke. The power absorbed from 220 V and 50 Hz, source connected in this circuit, if the resistance and, choke are joined in series is, (a) 110 W, , (b) 50 W, , (c) 220 W, , (d) 440 W
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282, , DAY TWENTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , 20 The output of a step-down transformer is measured to be, 24 V when connected to a 12 W light bulb. The value of, the peak current is, (a) 1 / 2 A, , (b) 2A, , (c) 2A, , (d) 2 2A, , 21 A transformer has turn ratio 2 and input power 3600 W., Load current is 20 A. Efficiency η = 90%. The internal, resistance is, (a) 1 Ω, , (c) 19, . Ω, , (b) 0.9 Ω, , (d) 3 Ω, , Direction (Q. Nos. 22-24) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below., (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, , (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 22 Statement I Two identical heaters are connected to two, different sources one DC and other AC having same, potential difference across their terminals.The heat, produced in heater supplied with AC source is greater., Statement II The net impedance of an AC source is, greater than resistance., , 23 Statement I In a series L - C - R circuit, the resonance can, take place., Statement II Resonance takes place, if the inductive and, capacitive reactances are equal and opposite., , 24 Statement I In a series R-L-C circuit, the voltage across, the resistor, inductor and capacitor are 8V, 16V and 10V,, respectively. The resultant emf in the circuit is 10V., Statement II Resultant emf of the circuit is given by the, relation E = VR2 + (VL − VC ) 2 ., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 In an AC circuit, the instantaneous emf and current are, given by, π, , e = 100 sin 30 t , i = 20 sin 30 t − , , 4, In one cycle of AC, the average power consumed by the, circuit and the wattless current are respectively,, , different values L 1 and L 2 (L 1 > L 2 ) of L, then which of the, following represents this graph correctly? (Plots are, ª JEE Main 2015, schematic and not drawn to scale), 2, (a) Qmax, , Q2, , L1, , L1, , max, , (c), , L2, , L2, t, , t, , ª JEE Main 2018, , (a) 50 W, 10 A, , (b), , 50, W, 0 A, (c), 2, , 1000, W, 10 A, 2, , (d) 50 W, 0 A, , 2 An arc lamp requires a direct current of 10 A at 80 V to, function. If it is connected to a 220 V (rms), 50 Hz AC, supply, the series inductor needed for it to work is close, ª JEE Main 2016, to, (a) 80 H, , (b) 0.08 H, , (c) 0.044 H, , (d) 0.065 H, , 3 An L-C-R circuit is equivalent to a damped pendulum. In, an L-C-R circuit, the capacitor is charged to Q 0 and then, connected to the L and R as shown below, R, , L, , C, , If a student plots graphs of the square of maximum, 2, charge (Q max, ) on the capacitor with time (t) for two, , 2, (b) Qmax, , L2, , Q2, , (d), , L1, , max, , Q0 (For both L1and L2), , t, , t, , 4 A resistor R and 2 µF capacitor in series is connected, through a switch to 200 V direct supply. Across the, capacitor is a neon bulb that lights up at 120 V. Calculate, the value of R to make the bulb light up 5 s after the switch, ª AIEEE 2011, has been closed ( Take, log10 2 .5 = 0.4), (a) 17, . × 105 Ω, , (b) 2.7 × 106 Ω, , (c) 3.3 × 10 Ω, , (d) 13, . × 104 Ω, , 7, , 5 Let C be the capacitance of a capacitor discharging, through a resistor R. Suppose t1 is the time taken for the, energy stored in the capacitor to reduce to half its initial, value and t 2 is the time taken for the charge to reduce to, t, one-fourth its initial value. Then, the ratio 1 will be, t2, ª AIEEE 2010, (a) 1, , (b), , 1, 2, , (c), , 1, 4, , (d) 2
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ALTERNATING CURRENT, , DAY TWENTY FOUR, 6 The average current in terms of I 0 for the waveform as, , 12 An inductor of reactance1 Ω and a resistor of 2Ω are, connected in series to the terminal of a 6V(rms) AC, source.The power dissipated in the circuit is, , shown is, V, , (a) 8 W, + I0, , I0, 3, , (c), , I0, 2, , (d), , I0, 4, , 7 A bulb is rated 55 W/110 V. It is to be connected to a, 220 V/50 Hz with inductor in series. The value of, inductance, so that bulb gets correct voltage is, (a) 200 Ω, , (c)14.4 W, , (b) 110 Ω, , (c) 50 Ω, , Column I, , Column II, , (d) 220 Ω, , V1, , 8 A coil of 0.01 H inductance and1 Ω resistance is, connected to 200 V, 50 Hz AC supply. The impedance of, the circuit and time lag between maximum alternating, voltage and current would be, 1, s, 250, 1, (c) 4.2 Ω and, s, 100, , I ≠ 0, V1 is proportional to I, , 1., , LC, , (b), , 2 R 2 C 2 + 4 LC, R 2 C 2 + 4 LC, , 3 mF, , 6 mH, , V1, , I ≠ 0, V2 > V1, , B., , 2., , V2, , 6 mH, , 2W, , 2LC, R 2 C 2 + 4 LC, , V, , V1, , (d) zero, , LC, , V2, , V, , 1, s, 160, 1, (d) 2.8 Ω and, s, 120, , 9 The bandwidth in a series L-C-R circuit is, , (c), , A., , (b) 3.9 Ω and, , (a) 3.3 Ω and, , (a), , (d)18 W, , These are connected to a variable DC voltage source (the, first two circuits) or an AC voltage source of 50Hz, frequency (the next three circuits)in different ways as, shown in Column II. When a current I (steady state for DC, or rms for AC) flows through the circuit, the corresponding, voltageV1 andV2 (indicated in circuits) are related as, shown in Column I. Match the Column I with Column II and, mark the correct option from the codes given below., , t, , – I0, , (b), , (b)12 W, , 13 You are given many resistances, capacitors and inductors., T, , (a) I 0, , 283, , 10 An L - C - R circuit, consists of an inductor, a capacitor and, a resistor driven by a battery and connected by two, switches S1 and S 2 , as shown in the figure., , V1 = 0, V2 = V, , C., , 3., , V2, 2W, , 6 mH, , V, , V, R, , V1, , S1, , C, , D., , I ≠ 0, V2 is proportional to I, , 4., , V2, , 6 mH, , 3 mF, , S2, L, V, , At time t = 0, when the charge on the capacitor plates is, q, switch S1 is opened and S 2 is closed. The maximum, charge the capacitor can hold, is q 0 . Choose the correct, equation, π, t, (a) q = q0 cos, + , LC, 2, , π, t, (b) q = q0 cos, − , LC 2 , , d 2q, (c) q = − LC 2, dt, , 1 d 2q, (d) q = −, LC dt 2, , 11 An alternating emf of angular frequency ω is applied, across an inductance. The instantaneous power, developed in the circuit has an angular frequecy, (a) ω / 4, , (b) ω / 2, , (c) ω, , (d) 2ω, , 5., , V1, , V2, , 1kW, , 3 mF, , V, , Codes, (a), (b), (c), (d), , A, (2,3,4,5), (3,4,5), (1,2), (3,4,5), , B, (1,2), (2,3,4,5), (3,4,5), (1,2), , C, (2,3,4,5), (1,2), (2,3,4,5), (2,3,4,5), , D, (3,4,5), (2,3,4,5), (2,3,4,5), (2,3,4,5)
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284, , DAY TWENTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 Statement I A sinusoidal AC current flows through a, , Direction, , (Q. Nos. 14 and 15) Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below., , resistance R. If the peak current is I 0 , then the power, RI 2, dissipated is 0 ., 2, Statement II For a purely resistive circuit, the power, factor, cos φ = 1., , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 15 Statement I The nature of the impedance of L - C - R, circuit, at resonance is pure inductive., Statement II The phase angle between E and I in a, R - L - C circuit at resonance, is zero., , ANSWERS, SESSION 1, , SESSION 2, , 1 (a), , 2 (b), , 3 (d), , 4 (b), , 5 (a), , 6 (c), , 7 (c), , 8 (a), , 9 (c), , 10 (d), , 15 (a), , 16 (d), , 17 (b), , 18 (b), , 19 (a), , 20 (a), , 5 (c), 15 (d), , 6 (c), , 7 (d), , 8 (a), , 9 (c), , 10 (c), , 11 (c), , 12 (c), , 13 (c), , 14 (d), , 21 (b), , 22 (a), , 23 (a), , 24 (a), , 1 (b), 11 (d), , 2 (d), 12 (c), , 3 (a), 13 (a), , 4 (b), 14 (b), , Hints and Explanations, SESSION 1, 1 I rms =, , 4 At t = 0, inductor behaves like an, , I + I + I, 3, 2, 1, , 2, 2, , 2, 3, , 12 + 22 + 12, =, 3, = 2 = 1.414 A, =, , 6, 3, , 2 Given, V = 200 2 sin(100 t )., Comparing this equation with, V = V 0 sin ω t, we have, V 0 = 200 2 V and ω = 100 rad s −1, The current in the capacitor is, V, I = rms = V rms × ωC, ZC, Q Z = 1 , , , C, , ωC , V, = 0 × ωC, 2, 200 2, =, × 100 × 1 × 10−6, 2, = 20 × 10−3 A = 20 mA, , 3 Time constant τ is the duration when, the value of potential drops by 63% of, its initial maximum value (i.e. V 0 / e )., Here, 37% of 25 V = 9.25 V which lies, between 100 s to 150 s in the graph., , infinite resistance., V, and at t = ∞,, So, at t = 0, I =, R2, inductor behaves like a conducting, V (R1 + R2 ), V, wire I =, =, R eq, R1 R2, , 5 At the resonance, ν =, =, , 2 × 314, . ×, , 1, 2 π LC, 1, , 0.25 × 01, . × 10−6, , = 1007 Hz, , 6 As voltage across resistance is, maximum, therefore a power is, maximum which is at the resonance, frequency., At resonance,, 1, 1, =, frequency 2 π LC, 1, 1, =, 2 π 24 × 2 × 10−6, 1 1000, =, 2 π 48, 1 1000, =, 2 π 6.9V, , 1000, 2 × 314, . × 6.92, 1000, =, = 23 Hz, 43.45, , =, , 7 Since, reactances produced by inductor and, capacitor in opposite direction. So, voltage, in these elements are distributed at 180°, i.e., out of phase., Net voltage = 400 V − 300 V, = 100 V, , 8 As initially charge is maximum,, q = q 0 cos ω t, dq, I =, = − ω q 0 sin ω t, ⇒, dt, q2, 1, Given, L I 2 =, 2, 2C, (q cos ω t )2, 1, ⇒ L (ω q 0 sin ω t )2 = 0, 2C, 2, 1, But, ω=, LC, ⇒, tan ω t = 1, π, ωt =, 4, π, π, t =, =, LC, ⇒, 4ω 4
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ALTERNATING CURRENT, , DAY TWENTY FOUR, 9 Phaser diagram of L-C-R series circuit is, shown in figure., VL, (VL – VC ), VR, , I Current axis, , VC, , V 2 = V R2 + (V L - VC )2, , 10 The current in an L-C-R circuit is given, by, V, , I =, , 2, é 2 æ, 1 ö ù, ÷ ú, ê R + ç wL è, wC ø úû, êë, , 1/ 2, , where, w = 2p f, Thus, I increases with an increase in w, upto a value given by, w = wc , i.e. at w = w c , we have, 1, wL =, wC, 1, Þ, wc =, LC, , 13 At resonance, w L = 1 / w C, Current flowing through the circuit,, V, 100, I = R =, = 01, . A, R, 1000, So, voltage across L is given by, V L = I X L = Iw L, But w L = 1 / w C, I, \, VL =, = VC, wC, 01, ., =, = 250 V, 200 ´ 2 ´ 10-6, , 14 As quality factor, Q = w 0, , B, where, w 0 = resonant frequency, and, B = bandwidth., From the graph, B = 2.5 kHz,, Q = 0.4, (by observing the curve), , 15 Sharpness of resonance of a resonant, L-C-R circuit is determined by the ratio, of resonant frequency with the, selectivity of circuit. This ratio is also, called ‘‘quality factor’’ or Q-factor., w0, w L, = 0, Q-factor =, 2Dw, R, XL = X C, , At w > w c , I again starts decreasing, with an increase in w ., , Hence, power dissipated in the circuit is, V2, P =, = 242 W, R, , 2 C, Now, when AC is connected to the, 1, circuit energy speed = LI 2, 2, By equating the energies, we get, 1 q2 1 2, = LI, 2 C, 2, -9 2, (10 ), 1, = ´ 10-5 I 2, 2, 10-11, 1, I =, A, Þ, 10, Now,, V = IR, 1, =, ´ 100 = 10 V, 10, Therefore,, Q = CV, = 10-11 ´ 10 = 10-10 C, , 12 For charging of capacitor,, q = CV (1 - e - t/ t ) at t = 2t,, q = CV (1 - e -2 ), , Now, power absorbed in the circuit is, P = V rms I rms cos f, = (220) (0707, ., ) (0707, ., ), = 109.96 » 110 W, , 20 Secondary voltage, VS = 24 V, Power associated with secondary, PS = 12 W, P, 12 1, = A = 0.5 A, \ IS = S =, VS, 24 2, Peak value of the current in the, secondary,, I 0 = IS 2, = (0.5)(1.414), 1, = 0707, =, ., A, 2, , 21 As, V2 = 3600 = 180 V, , 20, \ V 0 = V2 ´ h = 180 ´ 0.9 = 162 V, Now, V 0 = V2 - I2 r, r, , 180 V, , V0, , 16 The given circuit is under resonance as, , where, I is maximum., , q2, 11 As energy stored in capacitor = 1, , 285, , 17 For given circuit, current is lagging the, p, voltage by , so circuit is purely, 2, inductive and there is no power, consumption in the circuit. The work, done by battery is stored as magnetic, energy in the inductor., , 18 In an AC circuit, only resistor R, dissipates energy. L and C do not, dissipate energy, because for both of, them current is wattless (f = 90° )., , 19 In series impedance of circuit is, Z = R2 + w2 L2 =, , R2 + (2p f L )2, , = (220 )2 + (2 ´ 314, . ´ 50 ´ 07, . )2, = 311 W, V rms, Z, 220, A, =, = 0707, ., A = 0707, ., 311, R 220, and cos f = =, = 0.707, Z 311, , \ I rms =, , \, , r =, , V2 - V 0 18, =, = 0.9 W, I, 20, , 22 For the case of DC, the frequency is zero, and the net impedance is equal to the, resistance.For the case of AC, the, impedance of the AC circuit is given by, Z =, , R2 + w2 L2, , where, R = resistance,, w = angular frequency, and, , L = inductance., , 23 At a particular value of angular, frequency, the inductive reactance and, capacitive reactance will becomes just, equal to each other and opposite in, value. So that, the impedance of circuit, is minimum, i.e. equal to R., , Þ, Þ, , X L = XC, 1, w 0C, 1, w=, LC, , w 0L =
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ALTERNATING CURRENT, , DAY TWENTY FOUR, 8 Given, inductance, L = 0.01 H,, resistance, R = 1W, voltage, V = 200 V, and frequency, f = 50 Hz, Impedance of the circuit,, Z =, , 2, , R2 + X L =, , R2 + ( 2 p f L )2, , = 12 + (2 ´ 3.14 ´ 50 ´ 0.01)2, or Z = 10.86 = 33, . W, w L 2p f L, \ tan f =, =, R, R, 2 ´ 3.14 ´ 50 ´ 0.01, =, 1, = 3.14, f = tan - 1 (3.14) = 72°, 72 ´ p, rad, Phase difference, f =, 180, Time lag between alternating voltage, and current,, f, 72p, 1, s, Dt =, =, =, w 180 ´ 2p ´ 50 250, , where, f is not equal to p / 2., Differentiating Eq. (ii) twice with, respect to t, we get, d 2q, = - w2q 0 cos(w t + f) = - w2q, dt 2, 1 d 2q, q =- 2 2, w dt, d 2q, [using Eq. (i)], = - LC 2, dt, , 11 The instantaneous value of emf and, current in inductive circuit are given by, p, E = E 0 sin wt and I = I 0 sin æç wt - ö÷, è, 2ø, respectively., So, Pinst = EI, p, = E 0 sin wt ´ I 0 sin æç wt - ö÷, è, 2ø, = E 0I 0 sin wt, æsin wt cos p - cos wt sin p ö, ç, ÷, è, 2, 2ø, , 9 At cut-off frequency, Z = 2 R, , = E 0I 0 sin wt cos wt, 1, = E 0I 0 sin 2wt, 2, (Qsin 2wt = 2sin wt cos wt ), , 2, , æ, 1 ö, 2, R 2 + ç Lw ÷ = 2R, C wø, è, 1, Lw =R, Þ, Cw, Þ, LCw2 - RCw - L = 0, Þ, , w=, , R2C 2 + 4LC, , RC ±, , 2LC, , Hence, angular frequency of, instantaneous power is 2w., , 12 Given, X L = 1W, R = 2W, , =, , 2 R2C 2 + 4LC, 2LC, 2, , =, , 2, , R C + 4 LC, , Average power dissipated in the circuit,, Pav = E rms cos f, I, E, I rms = 0 = rms, Z, 2, , LC, , 10 When S2 is closed and S1 is open, the, charge oscillates in the L - C circuit at an, angular frequency is given by, 1, ...(i), w=, LC, Now, q ¹ 0 at t = 0. Hence, options (a), and (b) are wrong. The charge q varies, with time t as, ...(ii), q = q 0 cos(w t + f), , 13 In circuit 1,, In steady state, I = 0, So, no option matches for circuit 1., In circuit 2,, V1 = 0 and V2 = 2I = V, \ B, C, D ® 2, In circuit 3,, V1 = X L I = 2pf LI, = 2p ´ 50 ´ 6 ´ 10-3 = 1.88 I, and V2 = 2I, \ A, B, D ® 3, In circuit 4,, V1 = X L I = 1.88 I, V2 = X C I = 1061 I, \ A, B, D ® 4, In circuit 5,, V1 = IR = 1000 I, V2 = X C I = 1016 I, \ A, B, D ® 5, , 14 Power dissipated is given by, P = E rms I rms cos f, We know that for a purely resistive, circuit, the power factor,, R R, cos f =, =, =1, Z, R, Hence, P = E rms ´ I rms, = (RI rms ) ´ I rms, = R (I rms ) 2, 2, , R I 20, I, = R æç 0 ö÷ =, è 2ø, 2, , E rms = 6 V, Pav = ?, , Dw = w 01 - w 02, , R2 + X 2L = 4 + 1 = 5, 6, I rms =, A, 5, R, 2, cos f =, =, Z, 5, 6, 2, Pav = 6 ´, ´, 5, 5, [from Eq.(i)], 72, 72, =, =, = 14.4 W, 5, 5 5, Z =, , \, , 287, , 15 Since, at resonance,, wL =, , 1, wC, , and impedance,, 2, , or, , Z =, , 1 ö, R2 + æç wL ÷, è, wC ø, , Z =, , R2 = R, , Hence, nature of impedance at, resonance is resistive., Also, in a L-C-R circuit, phase angle, between the emf and the alternating, current I, is zero at resonance.
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DAY TWENTY FIVE, , Electromagnetic, Waves, Learning & Revision for the Day, u, , Electromagnetic Waves and, their Characteristics, , u, , u, , Maxwell’s Equations, Transverse Nature of Electromagnetic Waves, , u, , Spectrum of Electromagnetic, Radiation, , Electromagnetic Waves and, their Characteristics, Electromagnetic waves are those waves, in which electric and magnetic fields vary, sinusoidally in space with time. The electric and magnetic fields are mutually, perpendicular to each other and each field is perpendicular to the direction of, propagation of the wave., l, , Maxwell’s theory predicted that electromagnetic waves of all frequencies (and hence, all wavelengths) propagate in vacuum, with a speed given by, 1, ., c=, µ 0ε 0, where, µ 0 is the magnetic permeability and ε 0 is the electric permittivity of vacuum., Now, for the vacuum, µ 0 = 4π × 10 −7 TmA −1 and ε 0 = 8.85 × 10 −12 C2 N −1m−2 ., Substituting these values in the above relation, we have, 1, ~ 3.0 × 10 8 ms −1, c=, −7, [(4π × 10 )(8.85 × 10 −12 )]1/ 2, , PREP, MIRROR, Your Personal Preparation Indicator, , All the electromagnetic waves are of the transverse nature whose speed depends, upon the medium, but their frequency does not depend on the medium., , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , l, , Transverse waves can be polarised., , u, , No. of Correct Questions (z)—, , l, , Energy is being transported with the electromagnetic waves., , l, , Conduction Current, It is a current in the electric circuit, which arises due to the flow of electrons in the, connecting wires of the circuit, in a definite closed path., , (Without referring Explanations), u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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ELECTROMAGNETIC WAVES, , DAY TWENTY FIVE, , In this, E 0 and B0 are the amplitudes of the fields., 1, E, Further, c = 0 =, = speed of light in vacuum, B0, ε0 µ 0, , Maxwell’s Displacement Current, It is that current which comes into play in the region,, whenever the electric field and hence the electric flux is, changing with time., dφ E, id = ε 0, dt, The generalised form of the Ampere’s law is, , , ∫ B ⋅ d l = µ 0(ic + id ) = µ 0 i c + ε 0, , dφ E , , dt , , l, , l, , where, ic is conduction current., , Maxwell’s Equations, Maxwell in 1862, gave the basic laws of electricity and, magnetism in the form of four fundamental equations, which, are known as Maxwell’s equations., l, , Gauss’s law for electrostatics This law states that the total, electric flux through any closed surface is always equal to, 1, times the net charged enclosed by that surface., ε0, q, ε0, Gauss’s law for magnetism This law also predicts that the, isolated magnetic monopole does not exist., Mathematically, ∫ E ⋅ dS =, S, , l, , i.e. net magnetic flux through any closed surface is always, zero., , l, , l, , l, , l, , Mathematically, ∫ B ⋅ dS = 0, S, , l, , l, , Faraday’s law of electromagnetic induction It states that, the induced e.m.f. produced in a circuit is numerically, equal to time rate of change of magnetic flux through it., dφ, Mathematically, ∫ E ⋅ dl = − B, dt, Ampere-Maxwell’s law At an instant in a circuit, the, conduction current is equal to displacement current., dφ , , Mathematically, ∫ E ⋅ dl = µ 0 I c + ε 0 E , , dt , These equations, equations., , are, , collectively, , called, , Maxwell’s, , Properties of, Electromagnetic Waves, l, , l, , l, , The rate of flow of energy in an electromagnetic wave, is, described by the vector S called the Poynting vector, which, is defined by the expression,, 1, S=, E×B, µ0, The time average of S over one cycle is known as the wave, intensity. When the average is taken, we obtain an, expression involving the time average of cos2 (kx − ωt ), 1, which equals ., 2, c B02, E B, E2, Wm −2, Thus,, I = Sav = 0 0 = 0 =, 2 µ 0 2 µ 0c 2 µ 0, The total average energy per unit volume is,, ε E2, B2, u = uE + uB = 0 0 =, 2, 2µ0, The radiation pressure p exerted on a perfectly absorbing, S, surface, p = ., c, If the surface is a perfect reflector and incidence is normal,, then the momentum transported to the surface in a time t is, 2u, 2S, and the radiation pressure will be, p =, ., given by, p =, c, c, Energy density of electromagnetic wave,, 1, 1 B2, ue = ε 0Eu2B =, 2, 2 µ0, u, c, 2u, p=, c, , Momentum delivered, p =, , Energy of wave =, , (absorbing surface), (reflecting surface), , hc, = hν, λ, , Transverse Nature of, Electromagnetic Waves, According to Maxwell, electromagnetic waves consist of time, varying electric and magnetic fields, which are perpendicular, to each other, as well as direction of wave propagation., Y, , If the electromagnetic wave is travelling along the positive, direction of the X -axis, the electric field is oscillating, parallel to the Y-axis and the magnetic field is oscillating, parallel to the Z-axis., E = E 0 sin(ωt − kx) ⇒ B = B0 sin(ωt − kx), , 289, , Wave propagation, E, B, X, , Z, , B, , E
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290, , DAY TWENTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , detecting the fracture of bones, hidden bullet, needle,, costly material etc. inside the body, and also used in the, study of crystal structure., , Spectrum of Electromagnetic, Radiation, The array obtained on arranging all the electromagnetic, waves in an order on the basis of their wavelength is called, the electromagnetic spectrum., , l, , These radiations are mainly used in excitation of, photoelectric effect and to kill the bacteria of many diseases., , The Electromagnetic Spectrum, Frequency, Range (Hz), , Wavelength, Range (m), , 104 to 108, , 0.1 to 600, , Oscillating electric, circuits, , Microwaves 109 to 1012, , 10−3 to 0.3, , Oscillating current in, special vacuum tubes, , Name, Radio, waves, , Source, , Infrared, , 1011 to, 5 × 1014, , 10−6 to, 5 × 10−3, , Outer electrons in atoms, and molecules, , Visible, light, , 4 × 1014 to, , 4 × 10−7 to, , Outer electrons in atoms, , Ultraviolet, , 7 × 10, , 8 × 10, , 14, , 15, , 17, , 10 to 10, , 3.5 × 10, X-rays, , Gamma, rays, , 18, , 10 to 10, , 20, , 1019 to 1024, , l, , l, , l, , −7, , 10−11 to 10−8 Inner electrons in atoms, and sudden, deacceleration of high, energy free electrons, 10−16 to 10−13 Nuclei of atoms and, sudden deacceleration of, high energy free electrons, , Various Electromagnetic Radiations, l, , l, , −7, , 1.5 × 10−7 to Outer electrons in atoms, , Gamma rays The main sources of gamma rays are the, natural and artificial radioactive substances. These rays, affect the photographic plate and mainly used in the, treatment of cancer disease., X-rays X-rays are produced, when highly energetic, cathode rays are stopped by a metal target of high melting, point. They affect the photographic plate and can penetrate, through the transparent materials. They are mainly used in, , Ultraviolet Rays The major part of the radiations received, from sun consists of the ultraviolet radiation. Its other, sources are the electric discharge tube, carbon arc, etc., , l, , Visible Light Visible light is obtained from the glowing, bodies, while they are white hot. The light obtained from, the electric bulbs, sodium lamp, fluorescent tube is the, visible light., Thermal or Infrared Waves A body on being heated, emits, out the infrared waves. These radiations have the, maximum heating effect. The glass absorbs these, radiations, therefore for the study of these radiations, rock, salt prism is used instead of a glass prism. These waves are, mainly used for therapeutic purpose by the doctors because, of their heating effect., Microwaves These waves are produced by the spark, discharge or magnetron valve. They are detected by the, crystal or semiconductor detector. These waves are used, mainly in radar and long distance communication., Radio waves They can be obtained by the flow of high, frequency alternating current in an electric conductor., These waves are detected by the tank circuit in a radio, receiver or transmitter., , Applications of, Electromagnetic Spectrum, l, , l, , l, , l, , Radio waves are used in radar and radio broadcasting., Microwaves are used in long distance wireless, communications via satellites., Infrared, visible and ultraviolet radiations are used to know, the structure of molecules., Diffraction of X-rays by crystals, gives the details of the, structure of crystals.
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ELECTROMAGNETIC WAVES, , DAY TWENTY FIVE, , 291, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 During the porpagation of electromagnetic waves in a, medium,, , ª JEE Main 2014, , (a) electric energy density is double of the magnetic energy, density, (b) electric energy density is half of the magnetic energy, density, (c) electric energy density is equal to the magnetic energy, density, (d) Both electric and magnetic energy densities are zero, , 2 A perfectly reflecting mirror has an area of 1 cm 2., Light energy is allowed to fall on it for 1 h at the rate of, 10 W cm −2. The force acting on the mirror is, (a) 67, . × 10−8 N, (c) 10−3 N, , (b) 2.3 × 10−4 N, (d) zero, , 3 The magnetic field between the plates of radius 12 cm,, separated by a distance of 4 mm of a parallel plate, capacitor of capacitance 100 pF along the axis of plates, having conduction current of 0.15 A, is, (a) zero, , (b) 1.5 T, , (c) 15 T, , (d) 0.15 T, , 4 Instantaneous displacement current of 1.0 A in the space, between the parallel plates of a1µF capacitor, can be, established by changing potential difference of, (a) 10−6 Vs–1, (c) 10−8 Vs–1, , (b) 106 Vs–1, (d) 108 Vs–1, , 5 A large parallel plate capacitor, whose plates have an, area of 1 m 2 and are separated from each other by, 1 mm, is being charged at a rate of 25 Vs −1. If the, dielectric between the plates has the dielectric constant, 10, then the displacement current at this instant is, (a) 25 µA, , (b) 11µA, , (c) 2.2 µA, , (d) 1.1 µA, , 6 A parallel plate capacitor with plate area A and, separation between the plates d, is charged by a, constant current I. Consider a plane surface of area A /2, parallel to the plates and drawn simultaneously between, the plates. The displacement current through this area is, (a) I, , I, (b), 2, , I, (c), 4, , I, (d), 8, , (a) 4 × 10−6 T, , (b) 6 × 10−8 T, , (c) 9 × 10−9 T, , (d) 11 × 10−11 T, , 9 The magnetic field in a travelling electromagnetic wave, has a peak value of 20 nT. The peak value of electric field, strength is, ª JEE Main 2013, (a) 3 V/m, , (b) 6 V/m, , (c) 9 V/m, , (d) 12 V/m, , 10 The rms value of the electric field of the light coming from, the sun is 720 NC −1. The average total energy density of, the electromagnetic wave is, (a) 4.58 × 10−6 Jm−3, (c) 81.35 × 10−12 Jm−3, , (b) 6.37 × 10−9 Jm−3, (d) 3 .3 × 10−3 Jm−3, , 11 A radiation of energy E falls normally on a perfectly, reflecting surface. The momentum transferred to the, surface is, (a), , E, c, , (b), , 2E, c, , (c) E c, , (d), , E, c2, , 12 An electromagnetic wave in vacuum has the electric and, magnetic fields E and B, which are always perpendicular, to each other. The direction of polarisation is given by X, $ Then,, and that of wave propagation by k., ª AIEEE 2012, (a) X || B and k$ || B×E, (c) X || B and k$ || E×B, , (b) X || E and k$ || E×B, (d) X || B and k$ || B×E, , 13 An electromagnetic wave travels in vacuum along, z-direction E = (E 1 $i + E 2 $j ) cos (kz − ωt ). Choose the, correct option from the following, (a) The associated magnetic field is given as, 1, B = (E1 $i − E $j) cos (kz − ωt ), C, (b) The associated magnetic field is given as, 1, B = (E1 $i − E $j) cos (kz − ωt ), C, (c) The given electromagnetic field is circularly polarised, (d) The given electromagnetic wave is plane polarised, , 14 Match List I (Electromagnetic wave type) with List II (Its, association/application) and select the correct option, ª JEE Main 2014, from the choices given below the lists., , 7 Select the correct statement from the following, List I, , ª JEE Main (Online) 2013, , (a) Electromagnetic waves cannot travel in vacuum, (b) Electromagnetic waves are longitudinal waves, (c) Electromagnetic waves are produced by charges, moving with uniform velocity, (d) Electromagnetic waves carry both energy and, momentum as they propagate through space, , 8 In an apparatus, the electric field was found to oscillate, with an amplitude of 18 Vm −1. The magnitude of the, oscillating magnetic field will be, , List II, , A., , Infrared waves, , 1., , To treat muscular strain, , B., , Radio waves, , 2., , For broadcasting, , C., , X-rays, , 3., , To detect fracture of bones, , D., , Ultraviolet waves, , 4., , Absorbed by the ozone layer of, the atmosphere, , Codes, A B C D, (a) 4 3 2 1, (c) 3 2 1 4, , A B C D, (b) 1 2 4 3, (d) 1 2 3 4
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292, , DAY TWENTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , 15 Arrange the following electromagnetic radiations per, quantum in the order of increasing energy., A. Blue light, B. Yellow light, C. X-ray, D. Radio wave, ª JEE Main 2016 (Offline), , (a) D, B, A, C, (c) C, A, B, D, , (b) A, B, D, C, (d) B, A, D, C, , Direction (Q. Nos. 16-20), , Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 16 Statement I Ultraviolet radiation being higher frequency, waves are dangerous to human being., , Statement II Ultraviolet radiations are absorbed by the, atmosphere., , 17 Statement I If the earth did not have atmosphere, its, average surface temperature would be lower than what is, now., Statement II Greenhouse effect of the atmosphere would, be absent, if the earth did not have atmosphere., , 18 Statement I Electromagnetic waves exert radiation, pressure., Statement II Electromagnetic waves carry energy., , 19 Statement I Light is a transverse wave, but not an, electromagnetic wave., Statement II Maxwell showed that speed of, electromagnetic waves is related to the permeability and, the permittivity of the medium through which it travels., , 20 Statement I Out of radio waves and microwaves, the, radio waves undergo more diffraction., Statement II Radio waves have greater frequency, compared to microwaves., ª JEE Main (Online) 2013, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 You are given a 2 µF parallel plate capacitor. How would, you establish an instantaneous displacement current of, 1 mA in the space between its plates?, (a) By applying a varying potential difference of 500 V/s, (b) By applying a varying potential difference of 400 V/s, (c) By applying a varying potential difference of 100 V/s, (d) By applying a varying potential difference of 300 V/s, , 2 A uniform but time varying magnetic field B(t ) exists in a, circular region of radius a and is, B(t), × × × × × × × × × ×, directed into the plane of the paper, × × × × × × × × × P×, r, as shown in the figure. The, × × × × × × × × × ×, × × × ×, magnitude of the induced electric× × × × × ×, × × × × ×× × × × ×, field at a point P, a distance r from× × × ×a × × × × × ×, the centre of the circular region × × ××× × × × × × ×, (a) increases with r, (b) decreases with r, 1, (c) decreases as 2, r, (d) zero, , × ×, × × × × × × ×, × × × × × × × × × ×, × × × × × × × × × ×, , magnetic field components to the intensity of an, electromagnetic wave is, (a) c:1, , (b) c 2 :1, , a flood light is given by, B = 12 × 10−8 sin (1.20 × 107 z − 3.60 × 1015 t ) T. What is the, average intensity of the beam?, (a) 1.7 W/ m2, , (b) 2.3 W/m2, , 2, , (d) 3.2 W/m2, , (c) 1:1, , (d) c :1, , 5 An FM radio station, antenna radiates a power of 10 kW, at a wavelength of 3 m. Assume the radiated power is, confined to and is uniform over a hemisphere with, antenna at its centre. E max at a distance of 10 km from, antenna is, (a) 0.62 NC −1, (c) 0.31 NC −1, , (b) 0.41 NC −1, (d) 0.10 NC −1, , 6 Assume that all the energy from a 1000 W lamp is, radiated uniformly, then the amplitude of electric field of, radiation at a distance of 2 m from the lamp is, (a) 245.01 V/m, (c) 0, , 3 The magnetic field of a beam emerging from a filter facing, , (c) 2.7 W/m, , 4 The ratio of contributions made by the electric field and, , (b) 17 V/m, (d) 2.96 V/m, , 7 A red LED emits light at 0.1W uniformly around it. The, amplitude of the electric field of the light at a distance of, ª JEE Main 2015, 1m from the diode is, (a) 1.73 V/m, (b) 2.45 V/m, (c) 5.48 V/m, (d) 7.75 V/m
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ELECTROMAGNETIC WAVES, , DAY TWENTY FIVE, 8 In a transverse wave, the distance between a crest and, , (a), , neighbouring through at the same instant is 4.0 cm and, the distance between a crest and trough at the same, place is 1.0 cm. The next crest appears at the same, place after a time interval of 0.4 s. The maximum speed, of the vibrating particles in the medium is, 3π, cm/s, 2, π, (c), cm/s, 2, , =4, , (b), , εr1, εr2, , =2, , (c), , εr1, εr2, , =, , 1, 4, , (d), , εr1, εr2, , =, , 1, 2, , 10 An electromagnetic wave of frequency ν = 3 .0 MHz, passes from vacuum into a dielectric medium with, permittivity ε = 4.0. Then,, (a) wavelength is doubled and the frequency remains, unchanged, (b) wavelength is doubled and frequency becomes half, (c) wavelength is halved and frequency remains unchanged, (d) wavelength and frequency both remain unchanged, , ª JEE Main (Online) 2013, , (b) 5 π cm/s, , (a), , εr1, εr2, , 293, , (d) 2π cm/s, , 11 The magnetic field at a point between the plates of a, , 9 An EM wave from air enters a medium. The electric, , z, , fields are E1 = E 01x$ cos 2πν − t in air and, c, , , $, E2 = E 02x cos[k ( 2z − ct )] in medium, where the wave, number k and frequency ν refer to their values in air., The medium is non-magnetic., , capacitor at a perpendicular distance R from the axis of, the capacitor plate radius R, having the displacement, current ID is given by, (a), (c), , If εr1 and εr2 refer to relative permittivities of air and, medium respectively, which of the following options is, correct?, ª JEE Main 2018, , µ IDr, , (b), , 2 πR 2, µ 0 ID, , µ 0 ID, 2 πR, , (d) zero, , πr 2, , ANSWERS, SESSION 1, , SESSION 2, , 1 (c), , 2 (a), , 3 (a), , 4 (b), , 5 (c), , 6 (b), , 7 (d), , 8 (b), , 9 (b), , 10 (a), , 11 (b), , 12 (b), , 13 (d), , 14 (d), , 15 (a), , 16 (b), , 17 (a), , 18 (b), , 19 (d), , 20 (c), , 1 (a), 11 (b), , 2 (b), , 3 (a), , 4 (c), , 5 (d), , 6 (a), , 7 (b), , 8 (b), , 9 (c), , 10 (c), , Hints and Explanations, SESSION 1, , 3 As B ∝ r , since the point is on the axis,, , 1 Both the energy densities are equal,, i.e. energy is equally divided between, electric and magnetic field., , 2 Let E = energy falling on the surface, per second = 10 J, Momentum of photons,, h, h, p= =, λ c/f, hf E, =, =, c, c, On reflection, change in momentum, per second = force, 2 × 10, 2E, = 2p =, =, c, 3 × 108, = 6.7 × 10−8 N, , = 2.2 × 10−6 = 2.2 µA, , where r = 0, so B = 0., , 4 As, Id = C V , , Q I = dQ , , , , dt , , t , , V, I, or, = d, t, C, =, , 1.0, 10−6, , = 106 Vs –1, , (8.85 × 10−12 ) × 10 × 1, 10, , −3, , = 8.85 × 10−8 F, ∴, , I =, , 6 Charge on the capacitor plates, at time t is,, q = It, Electric field between the plates at this, instant,, q, It, E =, =, A ε0, A ε0, Electric flux through the given area,, A, It, φ E = E =, 2, 2 ε0, , 5 As, C = ε0KA, d, =, , = 8.85 × 10−8 × 25, , d, dV, (CV ) = C, dt, dt, , Therefore, displacement current,, dφ, Id = ε 0 E, dt, d It I, = ε0 , =, dt 2 ε0 2
Page 306 : 294, , 7, , B, , E, , Electromagnetic wave, Direction of, Propagation, , B, , E, , As electromagnetic waves contains both, electric field and magnetic field. It carry, both energy and momentum according, to de-Broglie wave particle duality of, radiations., , 8 Here, E 0 = 18 Vm –1, ∴, , DAY TWENTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , B0 =, , E0, 18, =, c, 3 × 108, , = 6 × 10−8 T, , 9 E = B × c keep value of electric field, |E| = |B||, ⋅ c | = 20 × 10−9 × 3 × 108, = 6 V/m, , 10 Total average energy =, = 8.85 × 10, , −12, , = 4.58 × 10, , −6, , ε0 E 2rms, × (720), , 2, , Jm −3, , 11 Initial momentum of surface, p i = E /c, where, c = velocity of light (constant)., Since, the surface is perfectly reflecting,, so the same momentum will be, reflected completely. Final momentum,, [negative value], p f = E /c, ∴ Change in momentum,, ∆ p = p f − pi, 2E, E, E, =−, −, =−, c, c, c, Thus, momentum transferred to the, surface is, 2E, ∆ p ′ = |∆ p| =, c, , 12 In electromagnetic wave, the direction, of propagation of wave, electric field, and magnetic field are mutually, perpendicular, i.e. wave propagates, perpendicular to E and B or along E×B., While polarisation of wave takes place, parallel to electric field vector., , 13 Here, in electromagnetic wave, the, electric field vector is given as, E = (E1 $i+E2 $j )cos (kz − ωt ), In electromagnetic wave, the associated, magnetic field vector,, E $i+E2 $j, E, B=, = 1, cos(kz − ωt ), c, c, Also, E and B are perpendicular to each, other and the propagation of, electromagnetic wave is perpendicular, to E as well as B, so the given, electromagnetic wave is plane, polarised., , 14 (a) Infrared waves are used to treat, muscular strain., (b) Radio waves are used for, broadcasting purposes., (c) X-rays are used to detect fracture, of bones., (d) Ultraviolet waves are absorbed, by ozone., , 15 As, we know energy liberated,, hc, E =, λ, 1, i.e., E ∝, λ, So, lesser the wavelength, than greater, will be energy liberated by, electromagnetic radiations per, quantum., As, order of wavelength is given by, X-ray, VIBGYOR, Radio waves, (C), (A) (B), (D), ∴Order of increasing energy of, electromagnetic radiations per, quantum., D<B<A<C, ⇒, , 16 Ultraviolet radiations are, electromagnetic waves. The wavelength, °, of these waves ranges between 4000 A, °, , to 100 A, i.e. of smaller wavelength and, higher frequency. They are absorbed by, ozone layer of stratosphere in, atmosphere. They cause skin diseases, and they are harmful to eye and may, cause permanent blindness., , 17 Earth is heated by sun’s infrared, radiation.The earth also emits radiation, most in infrared region. These, radiations are reflected back by heavy, gases like CO2 in atmosphere. These, back radiation keep the earth’s surface, warm at night. This phenomenon is, called greenhouse effect. When the, atmosphere were absent, then, temperature of the earth falls., , 18 Electromagnetic waves have linear, momentum as well as energy. From this, we conclude that, we can exert, radiation pressure by making a beam of, electromagnetic radiation fall on an, object. Let us assume that object is free, to move and that the radiation is, entirely absorbed in the object during, time interval ∆t. The object gains an, energy ∆U from the radiation. Maxwell, showed that the object also gains linear, momentum, the magnitude ∆p of the, change in momentum of the object is, related to the energy change ∆U as, ∆U, (total absorption), ∆p =, c, , 19 In free space or vacuum, the speed of, electromagnetic waves is, 1, c =, µ 0ε0, , …(i), , Here, µ 0 = 4 π × 10−7 Ns 2 C – 2 is, permeability (constant) of free space., ε0 = 8.85418 × 10−12 C2 N –1 m – 2 is the, permittivity of free space. On substituting, the values in Eq. (i), we have, 1, c =, 4 π × 10−7 × 8.85418 × 10−12, = 2.99792 × 108 ms –1, This is same as the speed of light in, vacuum. From this we conclude that, light is an electromagnetic wave., , 20 The frequency of radio waves less than, the frequency of microwaves., Q Frequency of radio waves = 3 × 108 Hz, and frequency of microwaves = 1010 Hz, ∴ ν radio waves < ν microwaves, , SESSION 2, 1 Given, capacitance of capacitor, C = 2µF, Displacement current, Id = 1 mA, Charge, q = CV, [Q q = It ], Id dt = CdV, dV, or, Id = C, dt, dV, −3, 1 × 10 = 2 × 10−6 ×, dt, dV 1, +3, or, = × 10 = 500 V/s, dt, 2, Clearly, by applying a varying potential, difference of 500 V/s, we would produce, a displacement current of desired value., , 2 A time varying magnetic field produces, an electric field. The magnitude of the, electric field at a distance r from the, centre of a circular region of radius a,, where a time varying field B exists, is, given by, a2 dB, E =, 2r dt, At, r = a,, a dB, E = , 2 dt, This is the value of E at the edge of the, circular region. For r > a, E decreases, with r., , 3 Magnetic field, B = B 0 sin ω t ., Given equation,, B = 12 × 10−8, sin (1.20 × 107 z − 3.60 × 1015t ) T
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ELECTROMAGNETIC WAVES, , DAY TWENTY FIVE, On comparing this equation with, standard equation, we get, B 0 = 12 × 10−8, , 6 Poynting vector, S = E × H, Energy of lamp =, , The average intensity of the beam,, 1 B 20, I av =, ⋅c, 2 µ0, =, , −8 2, , 1 (12 × 10 ) × 3 × 10, ×, 2, 4 π × 10−7, , =, 8, , = 1.7 W/m2, , 4 Intensity in terms of electric field,, U av =, , 1, ε0E 20, 2, , Intensity in terms of magnetic field,, 1 B 20, U av =, 2µ0, Now, taking the intensity in terms of, electric field,, 1, (U av ) electric field = ε0E 20, 2, 1, = ε0(cB 0 )2 (Q E 0 = cB 0), 2, 1, But, c =, µ 0ε0, 1, 1, B 20, ∴ (U av ) electric field = ε 0 ×, 2, µ 0ε0, =, , 1 B 20, ⋅, 2 µ0, , Thus, the energy in electromagnetic, wave is divided equally between, electric field vector and magnetic field, vector. Therefore, the ratio of, contributions by the electric field and, magnetic field components to the, intensity of an electromagnetic wave, is 1:1., , 5 As, I =, , ρ, 2 πr, , 2, , =, , 104, 2 π (10 × 10 ), , 3 2, , =, , 10−4, 2π, , 1, ε0 E 20 c, 2, 2I, ⇒ E0 =, ε0c, , and I =, , =, , 10−4, π × 8.85 × 10−12 × 3 × 108, , or E 0 = 010, . NC −1, , = EH sin 90° = EH, 1000 W, πr 2, 1000, π × 22, , Jm −2 s −1, , S represents energy flow per unit area, per second, we have, 1000, EH =, = 79.61,, π × 22, E, = 377, H, E, EH ×, = 79.61 × 377 = 300159, ., H, ⇒, E = 300159, ., = 173.25V/m, Amplitude of electric field of radiation, is, E 0 = E 2 = 245.01 V/m, , 7 Consider the LED as a point source of, light. Let power of the LED is P., Intensity at r from the source,, P, …(i), I =, 4 πr 2, As we know that ,, 1, …(ii), I = ε0E 20c, 2, From Eqs.(i) and (ii), we can write, P, 1, = ε 0E 20c, 4 πr 2 2, or, , E 20 =, =, , 9 Speed of progressive wave is given by,, ω, k, As electric field in air is,, 2 πνz, E1 = E 01 x$ cos , − 2 πνt , c, , 2 πν, =c, ∴ Speed in air =, 2 πν , , , c , 1, Also,, c =, µ 0εr1 ε0, v =, , In medium,, E2 = E 02 x$ cos (2kz − kct ), kc c, =, ∴ Speed in medium =, 2k, 2, c, 1, Also,, =, 2, µ 0εr2 ε0, , . × 9 × 109, 2 × 01, 1 × 3 × 108, , =6, , or, , E 20, , ⇒, , E0 =, , 6 = 2.45 V /m, , 8 Given, λ = 4 cm, 4, ∴, λ = 16 cm and T = 0.4 s, As, f λ × T = 2 π, 2π, 5 π −1, ⇒, f =, =, s, 16 × 0.4 16, 5π, Now,, v = f λ=, × 16 = 5π cm/s, 16, , …(i), , …(ii), , As medium is non-magnetic medium,, µ medium = µ air, On dividing Eq. (i) by Eq. (ii), we have, εr2, ε, 1, 2=, ⇒ r1 =, εr1, εr2, 4, , 10 In vacuum, ε0 = 1, In medium, ε = 4, ∴ Refractive index,, µ =, , ε, =, ε0, , 4, =2, 1, , Wavelength,, λ′ =, , 2P, 4 πε 0r 2c, , 295, , λ λ, =, µ, 2, , and wave velocity,, c, c, v = =, µ, 2, , Q µ = c , v , , , Hence, it is clear that wavelength and, velocity will become half, but frequency, remains unchanged when the wave is, passing through any medium., , 11 According to the Ampere-Maxwell’s law,, for a closed surface,, , ∫ B⋅dl = µ 0I D, As,, ⇒, , B (2πR ) = µ 0 I D, µ I, B= 0 D, 2 πR
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296, , DAY TWENTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY TWENTY SIX, , Unit Test 5, (Magnetostatic, EMI & AC, EM Wave), 1 A solenoid of some length and radius 2 cm has a layer of, winding. A 2 cm long wire of mass 5 g lies inside the, solenoid along the axis of solenoid. The wire is, connected to some external circuit, so that a current of, 5 A flows through the wire. The value of current to be in, the winding, so that magnetic force supports the wire, weight is ( take , g = 10 ms − 2 ), (a) zero, (c) 32000A, , (b) 400A, (d) Not possible, , uniform magnetic field of induction B. A small cut is made, in the ring and a galvanometer is connected across the, ends such that the total resistance of the circuit is R., When the ring is suddenly squeezed to zero area, the, charge flowing through the galvanometer is, (a), , , , , , 2 A circular coil of 20 turns and radius 10 cm is placed in a, uniform magnetic field of 0.1 T normal to the plane of the, coil. The coil carries a current of 5 A. The coil is made up, of copper wire of cross-sectional area10−5 m 2 and the, number of free electrons per unit volume of copper is, 10 29. The average force experienced by an electron in, the coil due to magnetic field is, (a) 5 × 10−25 N, (c) 8 × 10−24 N, , (b) zero, (d) None of these, , 3 A wire AB of length 5 m carrying a current of 2A is, placed in a region of uniform magnetic field B = 0.5 T as, shown in figure. The magnetic force experienced by wire, is, B, , 5m, A, , (b) 4 N, , BR, A, , (b), , AB, R, , (c) ABR, , (d), , R2, , 6 Two infinite long current carrying wires A and B are placed, as shown in figure. Each wire carries same current I. The, resultant magnetic field intensity at point P is, P, a, B, , A, I, , a, , I, , a, , µ 0I, 2 πa, µ 0I, (c), 2 2 πa, , (b), , (a), , 2 µ 0I, , 2 πa, µ 0I, (d), 4 2 πa, , 7 An equilateral triangular loop is kept near to a current, , 2A, Y, , 37°, , I, , (c) 3 N, , B 2A, , carrying long wire as shown in figure. Under the action of, magnetic force alone, the loop, , B, , (a) 5 N, , 5 A thin circular ring of area A is held perpendicular to a, , (d) 8 N, , a, , I, , X, , a, , 4 A coil of metal wire is kept stationary in a non-uniform, magnetic field, then, (a) an emf and current both induced in the coil, (b) a current but no emf is induced in the coil, (c) an emf but no current is induced in the coil, (d) Neither emf nor current is induced in the coil, , (a) must move away from wire along X-axis, (b) must move towards wire along X-axis, (c) must move along Z-axis, (d) must move alongY-axis
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UNIT TEST 5 (MAGNETOSTATIC, EMI & AC, EM WAVE), , DAY TWENTY SIX, , 297, , 8 A current carrying loop is placed in the non-uniform, , 13 A conducting loop is placed in a magnetic field (uniform), , magnetic field whose variation in space is shown in, figure. Direction of magnetic field is into the plane of, paper. The magnetic force experienced by loop is, , as shown in the figure. For this situation mark the correct, statement., B×, , I, , R, , ×, , I, , B, α, , α, , (a) non-zero, (c) can’t say anything, , x, , (b) zero, (d) None of these, , 9 An electron is launched with velocity v in a uniform, magnetic field B. The angle θ between v and B lies, π, between 0 and . Its velocity vector v returns to its initial, 2, value in a time interval of, 2 πm, eB, 2 × 2 πm, (b), eB, πm, (c), eB, (d) depends upon angle between v and B, (a), , 10 A straight conductor carrying a direct current of 3I is, split into 2I and I as shown in the figure., 2I, , (a) The force of compression experienced by loop is IRB, (b) The force of compression experienced by loop is 2π IRB, (c) The force of expansion experienced by loop is IRB, (d) The force of expansion experienced by loop is 2 π IRB, , 14 Mark the correct statement., (a) Ideal inductor does not dissipate power in an AC circuit, (b) Ideal inductor dissipates maximum power in an AC circuit, (c) In an inductor, current lags behind the voltage by π, (d) In inductor, current leads voltage by π, , 15 Two very long straight parallel wires carry steady currents, I and − I, respectively. The distance between the wires is, d. At a certain instant of time, a point charge q is at a, point equidistant from the wires in the plane of the wires., Its instantaneous velocity v is perpendicular to this plane., The magnitude of the force due to the magnetic field, acting on the charge at this instant is, (a), , µ 0 Iqv, 2 πd, , (b), , µ 0 Iqv, πd, , (c), , 2 µ 0 Iqv, πd, , (d) zero, , 16 An aeroplane is moving North horizontally, with a speed of, 3I, , r, , 200 ms −1, at a place where the vertical component of the, earth’s magnetic field is 0.5 × 10 – 4 T . What is the induced, emf set up between the tips of the wings if they are 10 m, apart?, , 3I, , I, , Magnetic induction at centre of loop of radius r is, µ I, (b) 0, 2 πr, µ 0I, (d), 4r, , (a) 0, π I, (c) 0, 2r, , plane normal to a magnetic field B. If the wire is pulled to, take a square shape in the same plane in time t, the emf, induced in the loop is given by, πBr 2, t, πBr 2, (c), t, , 1 − π , , , , 10 , 1 − π , , , , 6, , πBr 2, t, πBr 2, (d), t, (b), , (b) 0.1 V, , 1 − π , , , , 8, 1 − π , , , , 4, , B (into page), , rings of radii r1 and r2 (with r1 > r2) placed in air is given by, µ 0 πr, 2 r1, µ 0 π (r1 + r2 ) 2, (c), 2 r1, (a), , (b), , (d) 10 V, , Ω is moved with a constant velocity v in a uniform, magnetic field B = 2 T as shown in the figure. The, magnetic field is perpendicular to the plane of the loop, and directed into the paper. The loop is connected to a, network of resistors, each equal to 3 Ω. What should be, the speed of the loop, so as to have a steady current of, 1 mA in the loop?, , 12 The mutual inductance between two planar concentric, 2, 2, , (c) 1 V, , 17 A square metal wire loop of side 10 cm and resistance 1, , 11 A wire in the form of a circular loop of radius r lies with its, , (a), , (a) 0.01 V, , µ 0 πr12, , 2r2, µ 0 π (r1 + r2 ) 2, (d), 2 r2, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , Q, , –v, , 3Ω, , P, , 3Ω, , 3Ω, , 3Ω, R, 3Ω, , S, Metal loop, , (a) 1 cm s, , −1, , (b) 2 cm s, , −1, , (c) 3 cm s −1, , (d) 4 cm s −1
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298, , DAY TWENTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 18 At a certain place, the horizontal component of earth’s, magnetic field is 3 times the vertical component. The, angle of dip at that place is, (a) 30°, , (b) 45°, , (c) 60°, , (d) 90°, , 19 A magnetic dipole is acted upon by two magnetic fields, which are inclined to each other at an angle of 75°. One, of the fields has a magnitude of 2 × 10−2 T. The dipole, attains stable equilibrium at an angle of 30° with this, field. What is the magnitude of the other fields?, (a) 0.01 T, , (b) 0.02 T, , (c) 0.03 T, , (d) 0.04 T, , 20 A charge of 4 µC is placed on a small conducting sphere, that is located at the end of thin insulating rod of length, 0.5 m. The rod rotates in horizontal plane with a constant, angular velocity of 100 rads −1 about a vertical axis that, passes through its other end. The magnetic moment of the, rotating charge is, (a) zero, (b) 0.5 × 10−4 Am2, (c) 1.25 × 10−4 Am2, (d) magnetic moment is not defined for this case, , 21 A parallel plate capacitor is moving with a velocity of, −1, , 25 ms through a uniform magnetic field of 1.5 T as, shown in figure. If the electric field within the capacitor, plates is 175 NC −1 and plate area is 25 × 10−7 m 2, then the, magnetic force experienced by positive charge plate is, , 25 The resonant frequency and Q-factor of a series L-C-R, circuit with L = 3.0 H, C = 27 µF and R = 7.4 Ω. How will, you improve the sharpness of resonance of the circuit by, reducing its full width at half maximum by a factor of 2?, (a) Resistance of circuit should be increased, (b) Resistance of the circuit remain same, (c) Resistance of circuit should be increased by 3.7 Ω, (d) Resistance of the circuit should be reduced to 3.7 Ω, , 26 About 5% of the power of a 100 W light bulb is converted, to visible radiation. What is the average intensity of visible, radiation, (i) at a distance of 1 m from the bulb?, (ii) at a distance of 10 m?, Assume that the radiation is emitted isotropically and, neglect reflection., (a) 0.2 Wm−2 , 0.002 Wm−2, (c) 0.3 Wm−2 , 0.003 Wm−2, , (b) 0.6 Wm−2 , 0.006 Wm−2, (d) 0.4 Wm−2 , 0.004 Wm−2, , 27 A charged particle oscillates about its mean equilibrium, position with a frequency of109Hz. Frequency of the, electromagnetic waves produced by the oscillator is, , 28 The amplitude of the magnetic field part of a, electromagnetic wave in vacuum is B0 = 510 nT. What is, the amplitude of the electric field part of the wave?, (a) 140 NC −1, (c) 163 NC −1, , (b) 153 NC −1, (d) 133 NC −1, , 29 Match the following column I with column II., , B, , Column I, v, , (a) 1.45 × 10−13 N, (c) 8.67 × 10−15 N, , (b) zero, (d) 3.87 × 10−15 N, , 22 In an AC circuit, the potential difference V and current I, are given respectively by V = 100 sin (100 t ) V and, π, , I = 100 sin 100 t + mA. The power dissipated in the, , 3, circuit will be, (a) 104 W, , (b) 10 W, , (c) 2.5 W, , (d) 5 W, , 23 An inductor L, a capacitor of 20 µF and a resistor of 10 Ω, are connected in series with an AC source of frequency, 50 Hz. If the current is in phase with the voltage, then the, inductance of the inductor is, (a) 2.00 H, , (b) 0.51 H, , (c) 1.5 H, , (d) 0.99 H, , Column I, , A., , Capacitor, , 1., , increases AC, , B., , Inductor, , 2., , reduces AC, , C., , Energy dissipation is, due to, , 3., , is conductor for DC, , D., , A transformer, , 4., , resistance only, , Code, (a), (b), (c), (d), , A, 2, 4, 1, 3, , B, 2, 3, 3,4, 2,3, 2, , C, 4, 2, 4, 4, , D, 1, 2, 2,3, 2, 1, , 30 Match the following of column I with column II., Column I, , Column II, , draws a power of 560 W from an AC source marked, 210 V-60 Hz. The power factor of the circuit is 0.8, the, impedance of the circuit and the inductance of the, inductor is, (b) 64 Ω, 1.0 H, (d) 50 Ω, 1.5 H, , q, ε0, , A., , Lorentz force, , 1., , ∫ E.dA =, , B., , Gauss’s law, , 2., , dB =, , C., , Biot-Savart law, , 3., , F = q [E + ( v × B )], , D., , Coulomb’s law, , 4., , F =, , 24 An AC circuit having an inductor and a resistor in series, , (a) 65 Ω, 0.2 H, (c) 63 Ω, 0.1 H, , (d) 1010 Hz, , (c) 109 Hz, , (b) 105 Hz, , (a) 10 Hz, , µ 0 IdI × r, 4π r3, , 1 q 1q 2, 4 πε0 r 2
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UNIT TEST 5 (MAGNETOSTATIC, EMI & AC, EM WAVE), , DAY TWENTY SIX, , (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , Code, A, (a) 3, (c) 4, , B, 1, 3, , C D, 2 4, 2 1, , A, (b) 1, (d) 2, , B, 2, 1, , C D, 3 4, 4 3, , 34 Statement I A flexible wire loop of irregular shape, , 31 Match the following of column I with column II., Column I, , carrying current when placed in a uniform external, magnetic field acquires circular shape., , Column II, , A., , Ultraviolet, , 1., , Radar system, , B., , Infrared, , 2., , Roengten, , C., , X-rays, , 3., , Heat radiation, , D., , Microwaves, , 4., , Water purification, , Statement II Any current carrying loop when placed in, external magnetic field tries to acquires minimum energy, and hence maximum magnetic flux and for a given, perimeter circular shape is having greatest area., , 35 Statement I A hanging spring is attached to a battery, , Code, A, (a) 4, (c) 2, , 299, , B, 3, 1, , C D, 2 1, 4 3, , A, (b) 1, (d) 3, , B, 2, 2, , and switch. On closing the switch a current suddenly, flows in the spring, as a result spring compresses., , C D, 3 4, 4 1, , Statement II When two current carrying coils are placed, close to each other in same plane, then they attract each, other if sense of current is same in both., , 32 Assertion (A) Two identical heaters are connected to two, different sources one DC and other AC having same, potential difference across their terminals. The heat, produced in heater supplied with DC source is greater., Reason (R) The net impedance of an AC source is, greater than resistance., (a) Both Assertion and Reason are true and Reason is the, correct explanation of Assertion, (b) Both Assertion and Reason are false, (c) Assertion is true but Reason is false, (d) Both Assertion and Reason are true but Reason is not, true explanation of Assertion, , 36 Statement I No current is induced in a metal loop if it is, rotated in an electric field., Statement II The electric flux through the loop does not, change with time., , 37 Statement I The power factor of an inductor is zero., Statement II In the inductor, the emf and current differ in, π, phase by ., 2, , 38 Statement I In a series R-L-C circuit the voltage, across-resistor, inductor and capacitor are 8 V, 16 V and, 10 V respectively. The resultant emf in the circuit is 10 V., Statement II Resultant emf of the circuit is given by the, relation E = VR2 + (VL − VC ) 2 ., , 33 Assertion (A) Induction coils are made of copper., Reason (B) Induced current is more in wire having less, resistance, (a) Both Assertion and Reason are true and Reason is the, correct explanation of Assertion, (b) Both Assertion and Reason are false, (c) Assertion is true but Reason is false, (d) Both Assertion and Reason are true but Reason is not, true explanation of Assertion, , 39 Statement I X-rays travel faster than light waves in, vacuum., Statement II The energy of X-rays photon is greater than, the light photon., , 40 Statement I An electron moving in the positive, x-direction enters a region where uniform electric and, magnetic fields exist perpendicular to each other. The, electric field is in the negative y-direction. If the electron, travels undeflected in this region, the direction of the, magnetic field is along the negative z-axis., , Direction (Q. Nos. 34-40) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, , Statement II If a charged particle moves in a direction, perpendicular to a magnetic field, the direction of the, force acting on it is given by Fleming’s left hand rule., , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, , ANSWERS, 1., 11., 21., 31., , (d), (d), (a), (a), , 2., 12., 22., 32., , (a), (a), (c), (d), , 3., 13., 23., 33., , (c), (d), (b), (a), , 4., 14., 24., 34., , (d), (a), (c), (a), , 5., 15., 25., 35., , (b), (d), (d), (a), , 6., 16., 26., 36., , (a), (b), (d), (c), , 7., 17., 27., 37., , (a), (b), (c), (a), , 8., 18., 28., 38., , (b), (a), (b), (a), , 9., 19., 29., 39., , (a), (a), (a), (d), , 10., 20., 30., 40., , (d), (b), (a), (b)
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300, , DAY TWENTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, 8 Each and every pair of loop elements, , 1 Whatever be the current through, solenoid winding, the direction of, magnetic field is along the axis of, solenoid and hence the magnetic force, experienced by wire is zero and hence, its weight cannot be supported by, magnetic force., , 2 Drift speed of electron,, 5, I, vd =, =, neA 1029 × 1.6 × 10−19 × 10−5, = 3.125 × 10−5 ms − 1, Average magnetic force experienced by, each electron is,, F = qvB, = 1.6 × 10−19 × 3.125 × 10−5 × 0.1, = 0.5 × 10−24 N = 5 × 10−25 N, , 3 Length of wire AB, l = 5 m, Current, i = 2 A, θ = 37° and B = 0.5 T, Magnetic force on wire, F = iBl sinθ, = 2 × 0.5 × 5 × sin 37° = 3 N, Change of flux, 5 ∴Induced charge, q =, Resistance, , returns to its initial value is same as, time period of the particle, to execute, the circle., Since, it does not depend upon θ. So,, 2 πm, time required =, ⋅, eB, µ 2I, 10 ∴B net = 0 − µ 0I = µ 0I, 4r, 4r, 4r, , 11 Induced emf, , Magnetic field × change in area, Time, B∆A, =, t, Since, the circumference of the circular, loop = 2πr,, 2 πr π r, The side of the square loop =, =, 4, 2, Therefore,, , (e ) =, , 2, , π, πr, 2, ∆ A = πr − , = πr 1 − , , 2 , 4, ∴, , B ( πr 2 ) , π, e =, 1 − , , 4, t, , its centre is, B =, , BA, 45°, , BB, , But, , 45°, , B 2A + B 2B =, , rule force of expansion, would act on it whose, magnitude is given by, dT = I (dl × B ), ⇒ T = IB ∫ dl, , µ 0I, 2 πa, , 7 The net force, acting on loop, F1, would be along, X -axis, (to determine, whether it is along, positive or, negative, X -axis, F, calculation has to be carried out) as, shown in figure., , φ = MI, µ πr 2, M = 0 2, 2r1, , =0, , other, angle between the two is zero or, 180°., F = q v × B = qv B sin θ, For, θ = 0, sin 0 = 0 ⇒ F = 0, , 16 ∴, , e = Blv = 0.5 × 10– 4 × 10 × 200 = 0.1 V, , 17 The network PQRS is a balanced, Wheatstone’s bridge. Hence, the, resistance of 3 Ω between P and R is, ineffective. The net resistance of the, network, therefore, is 3 Ω. Total, resistance R = 3 Ω + 1 Ω = 4 Ω., Now, induced emf is, e = Blv = 2 × 01, . × v = 02, . v., 0.2 v, e, =, ∴ Induced current I =, R, 4, I = 1 × 10−3 A, , Given,, , 1 × 10−3 =, , Hence,, , 0.2 v, 4, , 18 ∴B H = B cos θ and B V = B sinθ, , Given,, Therefore,, , BV, = tan θ, BH, BV, 1, =, BH, 3, 1, , i.e. θ = 30°, tan θ =, 3, , 19 Refer to figure. Let θ1 (= 30° ) be the angle, between the magnetic moment vector m, and the field vector B1 (= 1.5 × 10– 2 T) ., , 13 By right hand thumb, , a, , T, , ∫ 0 sin 2ωt dt, , 15 Since, B and v are anti-parallel to each, , Hence,, , µ 0I, 2r1, , where, I is the current in the larger coil., Flux through the inner coil is, µ I, φ = B × πr22 = 0 × πr22, 2r1, Therefore,, , So, B =, , 1 1, , − E 0i 0 , , T 2, , Which gives, v = 2 × 10−2 ms −1 = 2 cms –1, , 12 Magnetic field due to the larger coil at, , µ 0I, 2π × 2 a, , a, , P =, , 2, , R, , 1, E 0I 0 sin 2 ωt, 2, , Average power for one complete cycle is, , 9 Time interval in which its velocity v, , φ f − φi, , But final area = 0, therefore, φ f = 0., Numerically, φ i = BA. Therefore,, q = BA / R., , 6 B A = BB =, , = − E 0I 0 sin ωt cos ωt = −, , dT, , Then, as shown in figure, the angle, between m and the other field B2 will be, θ2 = 75° − 30° = 45°., B1, m, , dθ, , R, , 30°, N, , =, , located symmetrically w.r.t. central line, experiences zero net force. So, total, magnetic force experienced by loop is, zero., , 45°, , = 2 π IRB, , B2, , 14 The instantaneous, F2, , voltage and current in an AC circuit, containing an ideal inductance only are, π, E = E 0 sin ωt , I = I 0 sin ωt − , , 2, π, Pins = EI = E 0 I 0 sin ωt sin ωt − , , 2, , S, , The field B1 exerts a torque τ1 = m × B1, on the dipole and the field B2 exerts a, torque τ2 = m × B2 , where m is the, magnetic moment of the dipole.
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UNIT TEST 5 (MAGNETOSTATIC, EMI & AC, EM WAVE), , DAY TWENTY SIX, Since, the dipole is in stable, equilibrium, the net torque τ (= τ1 + τ2 ), must be zero, i.e. the two torques must, be equal and opposite. In terms of, magnitudes, we have, mB1 sin θ1 = mB 2 sin θ2, B sin θ1, or, B2 = 1, sin θ2, =, , 2 × 10−2 × sin 30°, = 0.01 T, sin 45°, , 20 A moving charge along a circle is, equivalent to a current carrying ring,, whose current is given by, , ω, , I =, ⇒ I =, , q, q, =, T, 2π / ω, 4 × 10, 2π, , −6, , × 100 = 0.64 × 10−4 A, , Magnetic moment of rotating charge is, M = IA, ⇒ M = 0.64 × 10−4 × π × (0.5)2, = 0.5 × 10−4 Am2, , 1, 1, or L = 2, ωC, ωC, Here, ω = 2 π f = 2 × 314, . × 50 s −1, = 314 s −1 and C = 20, , E =, , ∴L =, , ε0 A, , where, q is the charge on capacitor., q = ε0 A × E, = 8.85 × 10−12 × 25 × 10−7 × 175, = 3.87 × 10−15 C, Magnetic force experienced by positive, plate is,, F m = qvB = 1.45 × 10−13 N in a direction, out of plane of paper., , 22 Voltage amplitude V 0 = 100 V,, current amplitude I 0 = 100 mA, = 100 × 10−3 A, and phase difference between I and V is, π, φ = = 60°., 3, Now, power dissipated is given by, V I, P = 0 0 cos φ, 2, 100 × 100 × 10− 3, =, × cos 60° = 2.5 W, 2, , 23 In an L-C-R circuit, the current and the, voltage are in phase (φ = 0), when, 1, ωL −, ωC = 0, tan φ =, R, , = 0.4 Wm −2, , 1, (314 s −1 )2 × (20 × 10−6 F ), , = 0.51 H, , 24 The average power over a complete, cycle is, P = E rms × I rms × cos φ, where, cos φ is the power factor, P, 560, =, ∴ I rms =, E rms × cos φ 210 × 0.8, 10, A, =, 3, The impedance of the circuit is, E, 210 V, Z = rms =, = 63 Ω, I rms, (10 / 3) A, The power is consumed in R only., Therefore,, 560, = 50.4 Ω, P = (I rms )2 R or R =, 2, 10 , , 3, Now, the impedance of an L - R circuit is, Z =, , R2 + (ωL )2, , ∴ (ωL )2 = Z 2 − R2 = (63) 2 − (50.4) 2, = 1428.84 Ω 2, or, , q, , Power, P, =, Area, 4 πr 2, 5, (i) Intensity =, 4 × 314, . ×1×1, =, , µF = 20 × 10−6 F, , 21 Electric field in between the capacitor, plates is given by, , ωL =, , or, , ∴, , ωL = 1428.84 = 37.8 Ω, L =, , 37.8, 37.8, =, = 0.1 H, 2 πf, 2 × 314, . × 60, , 25 Given, L = 3.0 H,, C = 27 µF = 27 × 10−6 F, and R = 7.4 Ω, The resonant frequency is given by, 1, 1, ω0 =, =, LC, 3.0 × (27 × 10−6 ), = 111 rads −1, The Q-factor of the circuit is given by, ω L 111 × 3.0, Q = 0 =, = 45, R, 7.4, The “bandwidth” (the difference of, half-power frequencies is given by, ω2 − ω1 = R/L, Smaller the value of (ω2 − ω1 ), sharper, is the resonance. To reduce (ω2 − ω1 ) by, a factor of 2, the resistance R should be, halved that is, the resistance of the, circuit should be reduced to 37, . Ω., , 26 Power converted into visible radiation, 5, × 100 = 5W, 100, Energy, Intensity =, Area × Time, P =, , 301, , (ii), , 5, 4 × 314, . × 10 × 10, = 0.004 Wm −2, , Intensity =, , 27 Frequency of electromagnetic wave =, Frequency of oscillation of charge., 28 E 0 = c o r E 0 = cB 0, B0, , 32 Heat (AC) = V rms I rms =, , V max I max, 2, , Heat (DC) = V max I max, , 33 Since, copper consists of a very small, ohmic resistance so, inductance coils are, made of copper. A large induced current, is produced in such., , 34 Each section of irregular wire loop, experiences an outward force., , 35 Opposite polarities are produced in the, forces of spring loops., , 36 A current is induced in a loop only if, magnetic flux linked with the coil, changes., 37 ∴cos φ = cos π = 0, 2, , 38 The resultant emf in the L-C-R circuit is, given by, E = V R2 + (V L − VC )2, E = (8)2 + (16 − 10)2 =, , 64 + 36, , E = 10 V, , 39 Since, X-rays are electromagnetic waves,, we know that electromagnetic wave, travels with same velocity of light in, vacuum., Now, from the formula, hc, E =, λ, The wavelength of X-rays are small than, light waves and energy is inversely, proportional to the wavelength. Hence,, the energy of X-rays photon will be, greater than light waves., , 40 Because electron has a negative charge,, an electric field in the negative, y-direction will deflect it in the positive, y-direction. It will travel undeflected if, the magnetic field imparts an equal, deflection in the negative y-direction., Since, the magnetic force is, perpendicular to the magnetic field and, the charge of electron is negative, the, direction of the magnetic field (according, to Fleming’s left hand rule) should be, along the negative z-direction.
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DAY TWENTY SEVEN, , Ray Optics, Learning & Revision for the Day, u, , u, , u, , Reflection of Light, Mirror formula, Refraction of Light, , u, , u, , u, , Lens, Total Internal Reflection (TIR), Deviation by a Prism, , u, , u, , u, , Dispersion by a Prism, Refraction Through a Prism, Scattering of Light, , Reflection of Light, The phenomena of bouncing back of light on striking a smooth surface is called, reflection of light., l, , l, , l, , According to the laws of reflection, (i) the incident ray, reflected ray and the normal, drawn on the reflecting surface at the point of incidence lie in the same plane and, (ii) the angle of incidence ∠i = angle of reflection ∠r., Laws of reflection are true for reflection from a polished mirror or from an unpolished, surface or for diffused reflection., Whenever reflection takes place from a denser medium, the reflected rays undergo a, phase change of π., , Reflection from a Plane Mirror, l, , l, , l, , l, , l, , If a ray is incident on a plane mirror at an angle of incidence i, then it suffers a, deviation of (π − 2i) and for two inclined plane mirrors deviation is (360 °−2θ)., While keeping an object fixed, a plane mirror is rotated in its plane by an angle θ,, then the reflected ray rotates in the same direction by an angle 2θ., Focal length as well as the radius of curvature of a plane mirror is infinity. Power of a, plane mirror is zero., If two plane mirrors are inclined to each other at an angle θ, the total number of, 2π, 2π, , images formed of an object kept between them, is n =, or , − 1 , when it is odd., θ, θ, , The minimum size of a plane mirror fixed on a wall of a room, so that a person at the, centre of the room may see the full image of the wall behind him, should be 1/3rd the, size of the wall., , Reflection from a Spherical Mirror, l, , A spherical mirror is a part of a hollow sphere whose one surface is polished, so that, it becomes reflecting. The other surface of the mirror is made opaque., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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RAY OPTICS, , DAY TWENTY SEVEN, , l, , l, , l, , Image formed by a concave mirroris is virtual and erect,, when the object is placed between the pole and the, principal focus of concave mirror. In all other cases, the, image formed is real and inverted one., Image formed by a convex mirror is virtual, erect and, diminished in size irrespective of the position of the object., Moreover, image is formed in between the pole and the, principal focus of the mirror., The focal length of a spherical mirror is half of its radius of, R, curvature, i.e. f = ., 2, , Refractive Index, For a given pair of media, the ratio of the sine of angle of, incidence (i) to the sine of angle of refraction (r ) is a constant,, which is called the refractive index of second medium, w.r.t., first medium., sin i, n, Thus,, = constant = n21 = 2, sin r, n1, This is also called Snell’s law., l, , l, , Mirror Formula, Let an object be placed at a distance u from the pole of a, mirror and its image is formed at a distance v from the pole., 1 1 1, Then, according to mirror formula, + =, v u f, l, , l, , The power of a mirror (in dioptre), is given as, 1, P=, f (in metre), , l, , If a thin object of height h is placed perpendicular to the, principal axis of a mirror and the height of its image be h′,, then the transverse or lateral magnification produced is, given by, h′, v, f, f −v, m=, =− =, =, h, u f −u, f, , l, , Negative sign of magnification means the inverted image, and positive sign means an erect image., l, , When a small sized object is placed along the principal, axis, then its longitudinal (or axial) magnification is given, by, 2, , 2, , Axial magnification = −, , f f − v, dv v , = =, =, , du u , f − u f , , 2, , Refraction of Light, , This phenomenon is known as refraction., Denser, medium, , i, , Denser, medium, , (a), , r, , l, , Refractive index is a unitless, dimensionless and a scalar, quantity., The refractive index of a medium w.r.t. vacuum, (or free space) is known as its absolute refractive index. It is, defined as the ratio of the speed of light in vacuum (c) to, the speed of light in a given medium (v)., c, n=, ∴, v, Value of absolute refractive index of a medium can be 1 or, more than 1, but never less than 1., When light travels from one material medium to another,, the ratio of the speed of light in the first medium to that in, the second medium is known as the relative refractive, index of second medium, w.r.t. the first medium. Thus,, v, c / v2 n2, n21 = 1 =, =, v2 c / v1 n1, When light undergoing refraction through several media, finally enters the first medium itself, then, n31, n21 × n32 × n 13 = 1 or n32 =, n21, When the object is in denser medium and the observer is in, rarer medium, then real and apparent depth have the, Real depth, relationship,, = n21, Apparent depth, , 1 , i.e. real depth > apparent depth shift, y = h − h′ = 1 −, h, n21 , , where, h and h′ are real and apparent depths., , When light passes from one medium, say air, to another, medium, say glass, a part is reflected back into the first, medium and the rest passes into the second medium. When it, passes into the second medium, it either bends towards the, normal or away from the normal., , Rarer, medium, , 303, , i, , Rarer, medium, , Refraction from a Spherical Surface, Let an object be placed in a medium of refractive index n1 at a, distance u from the pole of a spherical surface of radius of, curvature R and after refraction, its image is formed in a, medium of refractive index n2 at a distance v, then, n2 n1 n2 − n1, −, =, v, u, R, The relation is true for all surfaces, whether the image formed, is real or virtual., , r, , (b), , Fig. (a) and (b) shows refraction of light, , Lens, A lens is part of a transparent refracting medium bound by, two surfaces, with atleast one of the two surfaces being a, curved one. The curved surface may be spherical or, cylindrical.
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304, , DAY TWENTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , The lens formula is given by, , 1 1 1, − =, v u f, , Lens Maker’s Formula, , For a thin object of height h placed perpendicular to the, principal axis at a distance u, if the height of image formed is h′,, then lateral or transverse magnification m is given by, h′ v, f, f −v, m=, = =, =, h u f +u, f, For a small sized object placed linearly along the principal axis,, its axial or longitudinal magnification is given by, 2, , 2, , Axial magnification = −, , f − v, dv v f , = =, =, , , , du, u, f + u, f , , 2, , Silvering of Lens, When one surface of a lens is silvered, it behaves as a mirror., 1 2, 1, The focal length of silvered lens is, = +, F, fl fm, where, f1 is focal length of plane convex lens and fm is focal, length of corresponding mirror., In case of plano-convex lens, When curved surface is silvered, then focal length of, R, silvered lens is F =, 2µ, l, , +, , l, , fm, , fl, , F, , where,, , For a lens having surfaces with radii of curvature R1 and R2, respectively, its focal length is given by, 1, 1, 1, = (n21 − 1) , − , f, R1 R2 , n, where, n21 = 2 = refractive index of the lens material w.r.t., n1, the surroundings., P=, , Cutting and Combination of a Lens, If a symmetrical convex, f, 2f, 2f, f, f, lens of focal length f is, cut into two parts along, its optic axis, then focal, length of each part, (a plano-convex lens) is, 2 f . However, if the two, (a), (b), (c), (d), parts are joined as shown, in the figure, the focal, length of the combination is again f ., If a symmetrical convex lens of focal length f is cut into two, parts along the principal axis, then the focal length of each, part remains unchanged, as f [Fig. f]. If these two parts are, joined with the curved ends on one side, the focal length of, f, the combination is [Fig. g]. But on joining the two parts in, 2, opposite sense, the net focal length becomes ∞ (or net power, = 0) (Fig. h)., , R = 2 fm or R = fl (µ − 1), , f, , f, , (e), , (f), , ∞, , f, 2, , When plane surface is silvered, then, R, R, and fl =, F =, , fm = ∞, 2(µ − 1), (µ − 1), f, +, l, , F, l, , fm, , fl, , When double convex lens is silvered, then, R, R, R, and fl =, F =, ⇒ fm = ., 2(2µ − 1), 2(µ − 1), 2, , (g), , (h), , The equivalent focal length of co-axial combination of two, 1, 1, 1, d, lenses is given by, =, +, −, F, f1, f2, f1 f2, f1, , f2, d < f2, , d < f1, O1, , O2, , +, , d, F, , fl, , fm, l, , Power of a Lens, l, , The power of a lens is mathematically given by the, 1, reciprocal of its focal length, i.e. power P =, f (m), SI unit of power is dioptre (D). Power of a converging lens is, positive and that of a diverging lens is negative., , If a number of lenses are in contact, then, , 1, 1 1, = + + ..., F, f1 f2, , If two thin lenses of focal lengths f1 and f2 are in contact,, 1, 1 1, then their equivalent focal length, = +, feq, f 1 f2, In terms of power, Peq = P1 + P2
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RAY OPTICS, , DAY TWENTY SEVEN, , l, , Effective diameter of light transmitting area is called, aperture. Intensity of image ∝ (Aperture)2, 2, , l, , l, , l, , f , Relation between object and image speed, vi = , vo, f + u, Newton’s formula states, f 2 = x 1 x2 where, x 1 and x2 =, distance of object; and image from first and second principlal, foci. This formula is also called thin lens formula., (aµ g − 1), (aµ g − 1), f, If lens immersed in a liquid, then l =, =, fa (l µ g − 1) aµ g, , − 1, , , aµ l, , Total Internal Reflection (TIR), When a ray of light goes from a denser to a rarer medium, it, bends away from the normal. For a certain angle of incidence, iC , the angle of refraction in rarer medium becomes 90°. The, angle iC is called the critical angle., r, i, , Rarer, n1, , 90°, ic, , i, , r, Denser, n2, , TIR in a medium, sin iC =, , n1, 1, =, n2 n21, , For the angle of incidence greater than the critical angle (i > iC ), in the denser medium, the light ray is totally internally, reflected back into the denser medium itself., , Conditions for Total Internal Reflection, l, , l, , The light ray should travel from the denser medium towards, the rarer medium., The angle of incidence should be the greater than the critical, angle., , Common Examples of Total Internal Reflection, l, l, l, , l, , A, , Deviation produced by a, prism is,, , Important Results, , Looming An optical illusion in cold countries., Mirage An optical illusion in deserts., Brilliance of diamond Due to repeated internal reflections, diamond sparkles., Optical fibre Each fibre consists of core and cladding. The, refractive index of core material is higher than that of, cladding. Light entering at small angle on one end undergoes, repeated total internal reflections along the fibre and finally, comes out., , Deviation by a Prism, A prism is a homogeneous, transparent medium bounded by, two plane surfaces inclined at an angle A with each other., These surfaces are called as refracting surfaces and the angle, between them is called angle of prism A., , n, , δ = i + i′ − A, ⇒ r + r′ = A, For grazing incidence, i = 90 °, and grazing emergence, i′ = 90°, For minimum deviation, , 305, , n′, δ, i, , i′, r, , r′, , B, , C, , δ + A, sin m, , 2 , (i) i = i′ and r = r ′, (ii) µ =, A, sin, 2, In case of minimum deviation, ray is passing through prism, symmetrically., For maximum deviation (δ max ), i = 90 ° or i′ = 90 °, For thin prism,, δ = (µ − 1) A, , Dispersion by a Prism, Dispersion of light is the phenomenon of splitting of white, light into its constituent colours on passing light through a, prism. This is because different colours have different, wavelength and hence different refractive indices., Angular dispersion = δ v − δ r = (nv − nr ) A, where, n v and n r represent refractive index for violet and red, lights., n − nr, n + nr, , where n = v, is the, Dispersive power, ω = v, n−1, 2, mean refractive index., By combining two prisms with angle A and A′ and refractive, index n and n′ respectively, we can create conditions of, (n − 1) A, Dispersion without deviation when, A′ = −, (n′ − 1), Deviation without dispersion when,, n − nr , A′ = − v, A, n′v − n′r , l, , l, , Refraction Through a Prism, A ray of light suffers two refractions at the two surfaces on, passing through a prism. Angle of deviation through a prism, δ = i + e − A. where, i is the angle of incidence, e is the, angle of emergence and A is the angle of prism., , Scattering of Light, Molecules of a medium after absorbing incoming light, radiations, emit them in all directions. This phenomenon is, called scattering. According to Rayleigh, intensity of, 1, scattered light ∝ 4 ., λ, There are some phenomenon based on scattering, Sky looks blue due to scattering., At the time of sunrise and sunset, sun looks reddish., Danger signals are made of red colour., l, l, l
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306, , DAY TWENTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 To get three images of a single object, one should have, two plane mirrors at an angle of, , 7 A ray of light is incident on the surface of separation of a, , 2 A source of light lies on the angle bisector of two plane, , medium at an angle 45° and is refracted in the medium at, an angle 30°. What will be the speed of light in the, medium?, , mirrors inclined at angle θ. The values of θ, so that the, light reflected from one mirror does not reach the other, mirror will be, , 8 A beaker contains water upto a height h1 and kerosene of, , (a) 60°, , (b) 90°, , (a) θ ≥ 120°, (c) θ ≤ 120°, , (c) 120°, , (d) 30°, , (b) θ ≥ 90°, (d) θ < 30°, , 3 A beam of light composed of red and green rays is, incident obliquely at a point on the face of a rectangular, glass slab. When coming out of the opposite parallel, face, the red and green rays emerge from, (a) two points propagating in two different non-parallel, directions, (b) two points propagating in two different parallel directions, (c) one point propagating in two different directions, (d) one point propagating in the same direction, , 4 The optical path of a monochromatic light is same if it, goes through 4.0 cm of glass or 4.5 cm of water. If the, refractive index of glass is 1.53, the refractive index of the, water is, (a) 1.30, (c) 1.42, , (b) 1.36, (d) 1.46, , 5 On a hot summer night, the refractive index of air is, smallest near the ground and increases with height from, the ground. When a light beam is directed horizontally as, ª JEE Main 2015, it travels, the light beam, (a) becomes narrower, (b) goes horizontally without any deflection, (c) bends downwards, (d) bends upwards, , 6 A transparent solid cylinder rod has a refractive index of, 2, . It is surrounded by air. A light ray is incident at the, 3, mid-point of one end of the rod as shown in the figure., θ, , The incident angle θ for which the light ray grazes along, the wall of the rod is, 1, (a) sin−1 , 2, 2 , (c) sin−1 , , 3, , , (b) sin−1 , , −1 , (d) sin , , , 3, , 2 , 1 , , 3, , (a) 196, . × 108 ms−1, (c) 3.18 × 108 ms−1, , (b) 2.12 × 108 ms−1, (d) 3.33 × 108 ms−1, , height h2 above water so that the total height of (water +, kerosene) is (h1 + h2 ). Refractive index of water is µ1 and, that of kerosene is µ 2. The apparent shift in the position of, the bottom of the beaker when viewed from above is, ª AIEEE 2011, , , (a) 1 −, , , (c) 1 −, , , , , 1, 1, h2 + 1 −, h1 (b) 1 +, µ1 , µ, , , 2, , , 1, 1, h1 + 1 −, h2 (d) 1 +, µ1 , µ2 , , , , , 1, h1 + 1 +, µ1 , , , 1, h2 − 1 +, µ1 , , , 1, h2, µ2 , 1, h1, µ2 , , 9 A printed page is pressed by a glass of water. The, refractive index of the glass and water is 1.5 and 1.33,, respectively. If the thickness of the bottom of glass is, 1 cm and depth of water is 5 cm, how much the page will, appear to be shifted if viewed from the top?, ª JEE Main (Online) 2013, , (a) 1.033 cm, (c) 1.3533 cm, , (b) 3.581 cm, (d) 1.90 cm, , 10 A fish looking up through the water sees the outside, world, contained in a circular horizon. If the refractive, 4, index of water is and the fish is 12 cm below the water, 3, surface, the radius of this circle in cm, is, (a) 36 7, , (b), , 36, 7, , (c) 36 5, , (d) 4 5, , 11 A green light is incident from the water to the air-water, interface at the critical angle (θ ). Select correct statement., ª JEE Main 2014, , (a) The spectrum of visible light whose frequency is more, than that of green light will come out to the medium, (b) The entire spectrum of visible light will come out of the, water at various angles to the normal, (c) The entire spectrum of visible light come out of the water, at an angle of 90° to the normal, (d) The spectrum of visible light whose frequency is less, than that of green light will come out to the medium, , 12 Light is incident from a medium into air at two possible, angles of incidence (a) 20° and (b) 40°. In the medium,, light travels 3.0 cm in 0.2 ns. The ray will, ª JEE Main (Online) 2013
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RAY OPTICS, , DAY TWENTY SEVEN, (a), (b), (c), (d), , suffer total internal reflection in both cases (a) and (b), suffer total internal reflection in case (b) only, have partial reflection and partial transmission in case (b), have 100% transmission in case (a), , 13 The graph between angle of deviation (δ) and angle of, incidence (i) for a triangular prism is represented by, δ, , δ, (a), , i, , O, , δ, , (d), , O, , radius r of sun in the focal plane. Then, which option is, correct?, (a) πr 2 ∝ f, (b) πr 2 ∝ f 2, (c) If lower half part is covered by black sheet, then area of, the image is equal to πr 2 / 2, , 20 In an optics experiments, with the position of the object, , i, , (c) δ, , 19 A biconvex lens of focal length f forms a circular image of, , (d) If f is doubled, intensity will increase, , (b), , O, , O, , i, , i, , 14 The refractive index of glass is 1.520 for red light and, 1.525 for blue light. Let D1 and D2 be the angles of, minimum deviation for the red and blue light respectively, in a prism of this glass. Then,, (a) D1 < D2, (b) D1 = D2, (c) D1 can be less than or greater than D2 depending upon, the angle of prism, (d) D1 > D2, , 15 Two beams of red and violet colours are made to pass, , fixed, a student varies the position of a convex lens and, for each position, the screen is adjusted to get a clear, image of the object. A graph between the object distance, u and the image distance v, from the lens, is plotted, using the same scale for the two axes. A straight line, passing through the origin and making an angle of 45°, with the x-axis meets the experimental curve at P. The, coordinates of P will be, ª AIEEE 2009, , (a) 30° for both the colours, (b) greater for the violet colour, (c) greater for the red colour, (d) equal but not 30° for both the colours, , (c) (f , f ), , 21 A student measures the focal length of a convex lens by, putting an object pin at a distance u from the lens and, measuring the distance v of the image pin. The graph, between u and v plotted by the student should look like, v (cm), , (a), , (d) 11, , 17 A wire mesh consisting of very small squares is viewed, at a distance of 8 cm through a magnifying lens of focal, length 10 cm, kept close to the eye. The magnification, produced by the lens is, (a) 5, , (b) 8, , (c) 10, , (d) 20, , 18 When the distance between the object and the screen is, more than 4F, we can obtain image of an object on the, screen for the two positions of a lens. It is called, displacement method. In one case, the image is, magnified and in the other case it is diminished. Then,, the ratio of the size of image to the diminished image is, (a), , (D + d )2, (D − d ), , 2, , (b), , D, d, , (c), , D2, d, , 2, , (b), O, , (d), , D+d, D −d, , O, , u (cm), v (cm), , u (cm), , v (cm), , (d), O, , convex lens of focal length 2.5 cm is (The least distance, of distinct vision is 25 cm), (c) 62.5, , v (cm), , (c), , 16 The maximum magnification that can be obtained with a, , (b) 0.1, , f f, (b) , , 2 2, (d) (4f , 4f ), , (a) (2f , 2f ), , separately through a prism (angle of the prism is 60°). In, the position of minimum deviation, the angle of refraction, will be, , (a) 10, , 307, , u (cm), , O, , u (cm), , 22 An object approaches a convergent lens from the left of, the lens with a uniform speed 5 m/s and stops at the, focus. The image, (a) moves away from the lens with a uniform speed 5 m/s, (b) moves away from the lens with a uniform acceleration, (c) moves away from the lens with a non-uniform, acceleration, (d) moves towards the lens with a non-uniform acceleration, , 23 In an experiment to determine the focal length (f ) of a, concave mirror by the u-v method, a student places the, object pin A on the principal axis at a distance x from the, pole P. The student looks at the pin and inverted image, from a distance keeping his/her eye in line with PA. When, the student shifts his/her eye towards left, the image, appears to the right of the object pin. Then,, (a) x < f, (c) x = 2f, , (b) f < x < 2f, (d) x > 2f
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308, , DAY TWENTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 24 The image of an illuminated square is obtained on a, screen with the help of a converging lens. The distance of, the square from the lens is 40 cm. The area of the image, is 9 times that of the square. The focal length of the lens, is, ª JEE Main (Online) 2013, (a) 36 cm, (c) 60 cm, , (b) 27 cm, (d) 30 cm, , 25 Which one of the following spherical lenses does not, exhibit dispersion? The radii of curvature of the surfaces of, the lenses are as given in the diagrams., (a), , R1, , R2, , (b), , (c), , R, , R, , (d), , ∞, , R, , R, , ∞, , 26 A double convex lens made of glass (refractive index, n = 1.5 ) has the radii of curvature of both the surfaces as, 20 cm. Incident light rays parallel to the axis of the lens, will converge at a distance L such that, , ª AIEEE 2011, , (a) not depend on colour of light, (b) increase, (c) decrease, (d) remain same, , Y′, , (b) f ′ = f , f ′′ = f, (d) f ′ = f , f ′′ = 2f, , Direction (Q. Nos. 32-36) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , Statement II Laws of reflection are strictly valid for plane, surfaces, but not for large spherical surfaces., , 34 Statement I The refractive index of diamond is 6 and, that of liquid is 3. If the light travels from diamond to the, liquid, it will be totally reflected when the angle of, incidence is 30°., 1, Statement II n =, , where n is the refractive index of, sin C, diamond with respect to liquid., , 35 Statement I A double convex lens (n = 1.5) has a focal, , 29 To make an achromatic combination, a convex lens of, focal length 42 cm having dispersive power of 0.14 is, placed in contact with a concave lens of dispersive, power 0.21. The focal length of the concave lens should, be, (b) 21 cm, (d) 14 cm, , length 10 cm. When the lens is immersed in water, (n = 4 / 3), its focal length becomes 40 cm., 1 n − nm 1, 1, Statement II = l, −, , ., f, nm R1 R 2 , , 36 A thin air film is formed by putting the convex surface of a, , 30 A convex lens is in contact with a concave lens. The, 3, magnitude of the ratio of their focal length is . Their, 2, equivalent focal length is 30 cm. What are their individual, focal lengths?, (c) 75 , 50, , X, , Statement II Optical fibres are based on the, phenomenon of total internal reflection., , (a) magnified and inverted, (b) diminished and erect, (c) of the same size and erect, (d) of the same size and inverted, , (b) −10, 15, , O, , study internal organs., , length of 20 cm and both are kept in contact. The, combination is used to view an object 5 cm long kept at a, distance of 20 cm from the lens combination. As, compared to the object, the image will be, , (a) −75 , 50, , X′, , 33 Statement I Endoscopy involves use of optical fibres to, , 28 A concave lens and a convex lens have the same focal, , (a) 63 cm, (c) 42 cm, , (a) f ′ = 2f , f ′′ = f, (c) f ′ = 2f , f ′′ = 2f, , Y, , spherical mirror is valid only for mirrors whose sizes are, very small as compared to their radii of curvature., , 27 When monochromatic red light is used instead of blue, light in a convex lens, its focal length will, , halves along (i) XOX ′ and, (ii) alongYOY ′ as shown in figure., Let f , f ′ and f ′′ be the focal lengths, of the complete lens, of each half in, case (i), and of each half in case (ii), respectively. Choose the correct, statement from the following., , 32 Statement I The formula connecting u, v and f for a, , (b) L = 10 cm, 20, cm, (d) L =, 3, , (a) L = 20 cm, (c) L = 40 cm, , 31 An equiconvex lens is cut into two, , (d) −15 , 10, , plane-convex lens over a plane glass plate. With, monochromatic light, this film gives an interference, pattern due to light reflected from the top (convex), surface and the bottom (glass plate) surface of the film., Statement I When light reflects from the air-glass plate, interface, the reflected wave suffers a phase change of π., Statement II The centre of the interference pattern is, ª AIEEE 2011, dark.
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RAY OPTICS, , DAY TWENTY SEVEN, , 309, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 Two plane mirrors are inclined at 90°. An object is placed, between them whose coordinates are (a, b ). The position, vectors of all the images formed is, , 5 The speed at which the image of the luminous point, object is moving, if the luminous point object is moving at, speed v 0 towards a spherical mirror, along its axis is, (Given, R = radius of curvature, u = object distance), (a) vi = − v 0, , (a, b), , 2u − R , (c) vi = − v 0 , , R , , O, , (a) − a $i − b $j, a $i + b $j, − a $i + b $j, (b) − a $i + b $j, − a $j − b $j, a $i − b $j, (c) a $i + b $j, − a $i − b $j, a $i − b $j, , the centre is 3 mm. If speed of light in material of lens is, 2 × 108 m/s, the focal length of lens, (a) 15 cm, , 2 A small coin is resting on the bottom of a beaker filled, with a liquid. A ray of light from the coin travels upto the, surface of the liquid and moves along its surface (see, figure)., 3 cm, , (b) 20 cm, , (c) 30 cm, , (d) 10 cm, , 7 A plano-convex lens of refractive index 1.5 and radius of, curvature 30 cm is silvered at the curved surface. Now,, this lens has been used to form the image of an object. At, what distance from this lens, an object be placed in order, to have a real image of the size of the object?, (a) 20 cm, , (b) 30 cm, , (c) 60 cm, , (d) 80 cm, , 4 cm, , 8 Monochromatic light is incident on, a glass prism of angle A. If the, refractive index of the material of, the prism is µ, a ray incident at an, angle θ, on the face AB would get, transmitted through the face AC, of the prism provided, , Coin, , How fast is the light travelling in the liquid?, (a) 18, . × 108 ms−1, (c) 3.0 × 108 ms−1, , (b) 2.4 × 108 ms−1, (d) 1.2 × 108 ms−1, , facing positive axis. What is the least value of coordinate, of the point where a ray parallel to positive x-axis, becomes parallel to positive y-axis after reflection?, π, (b) , 2 , 2, , , π, (c) , 2 , 3, , , π, (d) , 3 , 4, , , 4 A light ray is incident perpendicular to one face of a 90°, prism and is totally internally reflected at the glass-air, interface. If the angle of reflection is 45°, we conclude, that for the refractive index n as, , 45, °, , 2, , A, , −1 , , 1 , , (a) θ > sin µ sin A − sin−1 , , µ , , , , 1 , , (b) θ < sin−1 µ sin A + sin−1 , , , µ, , , 1, , , (c) θ > cos−1 µ sin A + sin−1 , , , µ, , , 1 , , (d) θ < cos−1 µ sin A − sin−1 , , µ , , , , , , 3, 2, , 9 A thin convex lens made from crown glass µ = has, focal length f. When it is measured in two different liquid, 4, 5, having refractive indices and , it has focal lengths f1, 3, 3, and f2, respectively. The correct relation between focal, ª JEE Main 2014, length is, , 45°, , (b) n >, , θ, , ª JEE Main 2015, , 3 The reflective surface is given by y = 2 sin x and it is, , 1, (a) n <, 2, , 2, , 6 Diameter of a plano-convex lens is 6 cm and thickness at, , (d) None of the above, , π, (a) , 3 , 3, , , R , (b) vi = − v 0 , , 2u − R , R , (d) vi = − v 0 , , 2u − R , , 1, (c) n >, 2, , (d) n <, , 2, , (a) f2 > f and f1 becomes one, (b) f1 and f2 both becomes one, (c) f1 = f2 < f, (d) f1 > f and f2 becomes one
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310, , DAY TWENTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , as showin in the diagram. The index of, refraction of the glass, if total internal, reflection is to occur at the vertical face,, is equal to, ª JEE Main 2013, (a), , ( 2 + 1), 2, , (b), , 5, 2, , (c), , 3, 2, , 45º, Incident ray, , Diagram, , (d), , 3, 2, , measure the angle of a prism., Main scale reading : 58.5 degree, Vernier scale reading : 09 divisions, Given that 1 division on main scale corresponds to, 0.5 degree. Total divisions on the vernier scale is 30 and, match with 29 divisions of the main scale. The angle of the, prism from the above data is, ª AIEEE 2012, (b) 58.77°, (d) 59°, , 12 A car is fitted with a convex side-view mirror of focal, length 20 cm. A second car 2.8 m behind the first car is, overtaking the first car at a relative speed of 15 m/s. The, speed of the image of the second car as seen in the, mirror of the first one is, ª AIEEE 2011, (a), , 1, m/s, 15, , (b) 10 m/s, , (c) 15 m/s, , (d), , a film 12 cm behind the lens. A glass plate 1 cm thick,, of refractive index 1.50 is interposed between lens and, film with its plane faces parallel to film. At what distance, (from lens) should object shifted to be in sharp focus on, film?, ª AIEEE 2012, (a) 7.2 m, , 11 A spectrometer gives the following reading when used to, , (a) 58.59°, (c) 58.65°, , 15 An object 2.4 m infront of a lens forms a sharp image on, , 1, m/s, 10, , 13 A diverging lens with magnitude of focal length 25 cm is, placed at a distance of 15 cm from a converging lens of, magnitude of focal length 20 cm. A beam of parallel light, falls on the diverging lens. The final image formed is, ª JEE Main 2017 (Offline), , (a) virtual and at a distance of 40 cm from convergent lens, (b) real and at a distance of 40 cm from the divergent lens, (c) real and at a distance of 6 cm from the convergent lens, (d) real and at a distance of 40 cm from convergent lens, , (c) 1.7, , (d) 5.6 m, , transparent media. Medium 1 in z ≥ 0 has a refractive, index of 2 and medium 2 with z < 0 has a refractive index, of 3. A ray of light in medium 1 given by the vector, A = 6 3 i + 8 3 j − 10 k is incident on the plane of, separation. The angle of refraction in medium 2 is, ª AIEEE 2011, , (a) 45°, , (b) 60°, , (c) 75°, , (d) 30°, 2√3m, , 17 Two plane mirrors A and, , B, B are aligned parallel to, each other, as shown in, 0.2m, the figure. A light ray is, incident at an angle 30°, A, at a point just inside one, end of A. The plane of incidence coincides with the plane of, the figure., The maximum number of times the ray undergoes, reflections (including the first one) before it emerges out is, , (a) 28, , (b) 30, , (c) 32, , (d) 34, , 18 A light ray is incident on a horizontal plane mirror at an, angle of 45°. At what angle should a second plane mirror, be placed in order that the reflected ray finally be, reflected horizontally from the second mirror as shown in, figure, is, Q, , G, C, , A, , glass of a prism by i - δ, plot, it was found that a ray, incident at an angle 35° suffers a deviation of 40° and, that it emerges at an angle 79°. In that case, which of the, following is closest to the maximum possible value of the, ª JEE Main 2016 (Offline), refractive index?, (b) 1.6, , (c) 3.2 m, , 16 Let the zx-plane be the boundary between two, , 14 In an experiment for determination of refractive index of, , (a) 1.5, , (b) 2.4 m, , 30°, , 10 A light ray falls on a square glass slab, , S, 45° 45°, P, , α, , D, , α, N, , Q, B, , (a) θ = 45 °, , (d) 1.8, , (b) θ = 60°, , (c) θ = 22.5 °, , (d) θ = 15.3 °, , ANSWERS, SESSION 1, , SESSION 2, , 1 (b), 11 (d), 21 (c), , 2 (a), 12 (b), 22 (c), , 3 (b), 13 (c), 23 (b), , 4 (b), 14 (a), 24 (a), , 5 (d), 15 (a), 25 (c), , 6 (d), 16 (d), 26 (a), , 31 (d), , 32 (c), , 33 (a), , 34 (d), , 35 (a), , 36 (b), , 1 (b), 11 (c), , 2 (a), 12 (a), , 3 (a), 13 (d), , 4 (b), 14 (a), , 5 (b), 15 (d), , 6 (c), 16 (a), , 7 (b), 17 (a), 27 (b), , 8 (c), 18 (a), 28 (c), , 9 (d), 19 (b), 29 (a), , 10 (b), 20 (a), 30 (d), , 7 (a), 17 (b), , 8 (a), 18 (c), , 9 (d), , 10 (d)
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RAY OPTICS, , DAY TWENTY SEVEN, , 311, , Hints and Explanations, SESSION 1, , sin i, c, =, sin r, v, c sin r, Hence, v =, sin i, , 11 m r < m g < m v, , 7 n=, , 1 Number of images, 360°, n=, -1, q, where, q is angle between mirrors., 360°, 3=, - 1 or q = 90°, \, q, 2 n = 360, q, Number of images, N = n, which is odd, = n-1, For the given condition, no successive, reflection takes place. So, the number, of images will be N £ 2, n-1£2, , =, , 3 ´ 108 ´ sin 30°, sin 45°, 3 ´ 108, , 12 Speed of light in medium, , 2, , =, , 8, , = 212, . ´ 10 ms, , –1, , 8 We know, apparent shift,, 1ö, æ, Dh = ç1 - ÷ h, è, mø, \ Apparent shift produced by water,, æ, 1ö, Dh1 = ç1 ÷ h1, m1 ø, è, , n£3, 360, £3, q, 120 £ q, , Þ, , and apparent shift produced by, kerosene,, æ, 1ö, Dh2 = ç1 ÷ h2, m2 ø, è, , q ³ 120°, , 3 In any medium other than air or, vacuum, the velocities of different, colours are different. Therefore, both, red and green colours are refracted at, different angles of refractions., Hence, after emerging from glass slab, through opposite parallel face, they, appear at two different points and move, in two different parallel directions., , 4 As optical paths are equal, hence, ng × x g = nw × x w, xg, nw = ng ×, Þ, xw, = 1.53 ´, , =, , Þ sin( q r ) > sin(q g ) > sin q v, q r > q g > q v [q r , q g , q v are critical angles, of corresponding colours]. Thus, all, colours from red to green will emerge out, of water., , \, , Dh = Dh1 + Dh2, æ, æ, 1ö, 1ö, = ç1 ÷ h1 + ç1 ÷ h2, m1 ø, m2 ø, è, è, , 9 Given, m = 1. 5, t 1 = 5 cm,, , æ, æ, 1 ö, 1 ö, = ç1 ÷ ´ 5 + ç1 ÷´1, 1. 5ø, 1.33 ø, è, è, , When the value of incidence angle is, greater than critical angle than total, internal reflection take place., In the second case we get total interval, reflection., angle of incidence. If we determine, experimentally the angles of deviation, corresponding to different angles of, incidence, then the plot between i and d, that we will get is shown below, δ, , » 1.90 cm, , 4.0, = 1.36, 4.5, , 10 The situation is shown in figure., , θC, , …(i), , 2, , C, , sin r = sin (90° - C ) = cos C =, sin q m2, =, sin r m1, 2, 1, sin q =, ´, 3 2, 1, q = sin -1 æç ö÷, è 3ø, , R, , A, , 14 D = (n - 1) A, θC, , O, , 1, sin qC =, m, , 1, 2, , tan qC =, \, or AB =, , i, , B, 12 cm, , 6 sin C = 3, , δm, O, , denser to a rarer medium, so it bends, upwards (away from normal)., , θ, , 0.2 ´ 10-9, 3, = ´ 108 ms -1 = 1. 5 ´ 108 ms -1, 2, m2, v, As,, = 1, m1, v2, 3 ´ 108, m, =, Þ m=2, 1 1. 5 ´ 108, 1, We have, sinC =, 2, 1, C = sin -1 æç ö÷ = 30°, Þ, è2 ø, , 13 Angle of deviation depends upon the, , m2 = 1.33 and t 2 = 1 cm, Change in path = Dt 1 + Dt 2, æ, æ, 1ö, 1ö, = ç1 ÷ ´ t2, ÷ ´ t 1 + ç1 m1 ø, m2 ø, è, è, , 5 A horizontal beam is travelling from a, , r, , 3 ´ 10-2, , AB, OA, , AB = OA tan qC, OA, 12, 36, =, =, 2, 2, 7, n -1, æ 4ö - 1, ç ÷, è3ø, , For blue light n is greater than that for, red light, so D2 > D1 ., , 15 At minimum deviation (d = d m ), r1 = r2 =, , A 60°, =, = 30°, 2, 2, (For both colours), , 16 Maximum magnification is obtained, when image is formed at near point of, eye. In that case,, m = 1+, , D, 25, = 1+, = 11, f, 2.5
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312, , 17 1 = 1 - 1, f, , DAY TWENTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , v, , …(i), , u, , 21 \ 1 - 1 = 1 = constant,, v, , u, , which is a rectangular hyperbola., , Given, f = 10 cm, (as lens is converging), u = - 8 cm, (as object is placed on left, side of the lens), On substituting these values in, Eq. (i), we get, 1, 1, 1, = 10 v - 8, 1, 1 1, =, Þ, v 10 8, 1 8 - 10, =, Þ, v, 80, 80, v =, \, = - 40 cm, -2, , 22 When object would approach the lens, then image would move away from the, lens with non-uniform acceleration., This can be seen from the fact that, when object is far away, image would, be formed at focus P. When object, approaches the lens then image would, move away such that when object, approaches to focus, then image would, approach to infinity., , 23 Since, object and image move in, opposite directions, the positioning, should be as shown in the figure. Object, lies between focus and centre of, curvature f < x < 2 f ., , Hence, magnification produced by the, lens, - 40, v, =5, m= =, u, -8, , 18 In displacement method, the ratio of, the diminished image to the object is, , Object, , and the ratio of image to the object is, D +d, D-d, Hence,, , (D + d )2, m1, I, O, = 1 ´, =, m2 O, I2, (D - d )2, , 19, f, θ, , \, , r, , r = f tan q or r µ f, pr 2 µ f 2, , 24 As magnification = 9 = v, , u, , 40 cm, Converging, lens, \ v = 9 u = 9 ´ 40 = 360, 1 1 1, Now,, - =, v u, f, 1, 1, 1, =, v, -40 f, 1, 1, 1, +, =, 360 40 f, , Þ, Þ, , 20 It is possible when object kept at centre, of curvature because then only position, of object and image would be same. i.e., u = v [which is the point of intersection, between curve and straight line], v, , Image, right, left, , D -d, I, m2 = 2 =, O, D +d, , Þ, , é 1 + 1ù 1 = 1, úû 40 f, ëê 9, 10, 1, 1, ´, =, 9, 40 f, f = 36 cm, , æ, ö, 25 1 = ( m - 1) ç 1 - 1 ÷, f, R2 ø, è R1, 1, For no dispersion,, =0, f, or R1 = R2 = R., So, (c) is correct option., , 26 Here, n = 1.5, as per sign convention, , (v), u, , u (u > f ), , followed, R1 = + 20 cm and R2 = - 20 cm, \, , u=v, u = 2f , v = 2f, , 2, 1, =, 20 20, f = 20 cm, = 0.5 ´, , f, , æ 1, 1, 1 ö, = (n - 1) ç, ÷, f, è R1 R2 ø, é 1, 1 ù, = (1.5 - 1) ê, ú, ë (+ 20) (- 20)û, , Þ, , Incident rays travelling parallel to the, axis of lens will converge at its second, principal focus., Hence, L = + 20 cm, æ, ö, 27 1 = (m - 1) ç 1 - 1 ÷, f, , è R1, , R2 ø, , Also, by Cauchy’s formula,, B, C, m = A + 2 + 4 + ..., l, l, 1, mµ, Þ, l, As l blue < l red Þ m blue > m red, Hence, f red > f blue, , 28 When a concave lens is joined in contact, with a convex lens of same focal length,, the combination behaves as a glass plate, of infinite focal length or zero power., Hence, the image of an object will be of, the same size and erect., , 29 For an achromatic combination, the, condition is, w1, w, + 2 =0, f1, f2, Here, w1 = 014, . , f1 = 42 cm, thus, we get f2 = - 63 cm, , 30 Let focal length of convex lens is + f ,, then focal length of concave lens would, 3, be - f ., 2, From the given condition,, 1, 1, 2, 1, = =, 30 f, 3f, 3f, \, , f = 10 cm, , Therefore, focal length of convex lens, = + 10 cm and that of concave lens, = - 15 cm., , 31 (i) Cutting along YOY ¢, Focal length get doubled., f ¢¢ = 2 f, (ii) Cutting along XOX ¢, Focal length unchanged., , 32 Laws of reflection are valid for plane, surfaces, irregular surface and curved, surface., , 33 Optical fibre consists of a very long and, thin fibre of quartz glass. When a light, ray is incident at one end of the fibre, making a small angle of incidence, it, suffers refraction from air to quartz and, strikes the fibre-layer interface at an, angle of incidence greater than the, critical angle.
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RAY OPTICS, , DAY TWENTY SEVEN, It therefore, suffers total internal, reflection and strikes its opposite, interface. At this interface also, the, angle of incidence is greater than the, critical angle. So, it again suffers total, internal reflection. Thus, optical fibre is, based on total internal reflection., Endoscopy is a process for viewing, internal organs of human body. This, process use a device endoscope which, is based on total internal reflection., , The centre of the circle is lying on, meeting point of mirrors (i.e. O). The, position of images from diagram is for, S 1 , r1 = - a $i + b $j, r2 = - a $i - b $j,, r = a i$ - b $j., 3, , Hence, option (b) is true., , 2 As shown in figure, a light ray from the, coin will not emerge out of liquid, if, i > C., , 34 Refraction index of diamond w.r.t., , A, , B, , R, , liquid, lnd, , =, , 1, sin C, , or, \, , A, , 45, , 36 Both statements I and II are correct but, statement II does not explain statement I., , SESSION 2, 1 The images formed by combination of, two plane mirrors are lying on a circle, whose radius is equal to OS., Y, , 90° – θ, S2, (– a, –b), , i.e. C < i, or C < 45°, (QÐ i = 45° ), 1, 1, But n =, Þ C = sin -1 æç ö÷, ènø, sin C, , C, n2 −1, , R, = tan C, h, or, R = h tan C, h, or, R=, n2 - 1, é, 1, sin C, êQ sin C = and tan C =, n, êë, 1 - sin2 C, Given, R = 3cm , h = 4cm, 3, 1, Hence,, =, 4, n2 - 1, or, , n2 =, , 25, 9, , or, , n=, , or, , ù, ú, úû, , 5, 3, , c, v, c 3 ´ 108, v = =, = 1.8 ´ 108 ms -1, n, 5/ 3, , But n =, or, , S (a , b ), , θ, , 1, , ...(i), , æ 1, 1, 1 ö, ...(ii), = (n l - 1) ç, ÷, R2 ø, fa, è R1, (1.5 - 1)(4 / 3), fw, ( n l - 1), =, =, (1.5 - 4 / 3), fa, æ nl - nm ö, ç, ÷, è nm ø, 0.5 ´ 4 / 3 (2 / 3) 6, =, =, 3 4, 1, 2 3, f w = 4 f a = 4 ´ 10 = 40cm, , θ, , 45°, 45°, , B, , n, , 35 In water,, , θ, , 1, sin -1 æç ö÷ < 45°, ènø, , or, , 1, < sin 45°, n, , or, , n>, , or, , S3, (a, – b), , 45°, 45° P, , n> 2, , 5 1+ 1=1, v, u, f, Differentiating both sides, 1 dv, 1 du, - 2, = 2, v dt, u dt, 2, , dv, v du, = v i = - æç ö÷, è u ø dt, dt, 2, , v, = - æç ö÷ v 0, èu ø, , n, , X, , 1, sin 45°, 1, n>, (1 / 2 ), , or, , 3 Let the incidence point is P ( x, y ), , 90° – θ, , O 90° – θ, θ, , C, , °, , In DSAB,, , 90° – θ, , m=, , Coin, , But here angle of incidence is lower, than critical angle, so total internal, reflection does not occur in light., , (– a, b) S1, , dy, = 2cos x = 1, dx, 1, cos x =, 2, p, x=, 3, The corresponding value of, p, y is 2 sin = 3, 3, , \, , S, , Therefore, minimum radius R, corresponds to i = C ., , \, , (from law of reflection), y = 2 sin x, , interface, critical angle C must be less, than angle of incidence., , For total internal reflection angle should, be greater than critical angle., , æ n - nm ö æ 1, 1, 1 ö, =ç l, ÷, ÷ç, f w è n m ø è R1, R2 ø, , m = tan 45° = 1, , 4 For total internal reflection from glass-air, , h, , 6, 1, =, 3 sin C, 1, sin C =, = sin 45°, 2, C = 45°, , \, , i>C, , C, , 313, , Again,, , 1 1 1 2 1 2u - R, = - =, - =, v, f, u R u, Ru, v =, , uR, 2u - R, 2, , 2, , æ R ö, v, v i = - æç ö÷ v 0 = - v 0 ç, ÷, èu ø, è 2u - R ø
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RAY OPTICS, , DAY TWENTY SEVEN, , 25 cm., As the rays are coming parallel, so the, image (I1 ) will be formed at the focus of, diverging lens i.e. at 25 cm towards left, of diverging lens., 15 cm, , I1, , 25 cm, , v, , 14 If m is refractive index of material of, prism, then from Snell’s law, sin i sin( A + d m )/ 2, m=, =, sin r, sin A / 2, , …(i), , where, A is angle of prism and d m is, minimum deviation through prism., Given, i = 35° , d = 40 °, e = 79 °., So, angle of deviation by a glass prism,, d = i + e -A, Þ 40 ° = 35° + 79° - A, i.e. Angle of prism Þ A = 74°., Such that, r1 + r2 = A = 74°., Let us put m = 1.5 in Eq. (i), we get, , d, Þ 0.9 = sin æç37° + min ö÷, è, 2 ø, (Q sin 37° » 0.6 ), d min ö, æ, sin 64° = sin ç37 °+, ÷, è, 2 ø, (Q sin 64° = 0.9), d, 37° + min = 64° Þ d min » 54°, 2, This angle is greater than the 40°, deviation angle already given. For, greater m, deviation will be even higher., Hence, m of the given prism should be, lesser than 1.5. Hence, the closest, option will be 1.5., , different, xy-plane should be boundary, between two media., Angle of incidence,, ½, ½, Az, 1, cos i = ½, ½=, ½ 2, 2, 2½, 2, ½ A x + A y + A z½, \, i = 60°, From Snell’s law,, sin i m2, 3, =, =, sin r m1, 2, 2, ´ sin 60°, 3, 2, 3, =, ´, 2, 3, 1, =, = 45°, 2, Þ, r = 45°, 0 .2, 17 d = 0.2 tan 30° =, 3, sin r =, , 15 Shift in image position due to glass, plate,, 1ö, 1 ö, æ, S = ç1 - ÷ t = æç1 ÷ ´ 1 cm, è, è, mø, 1.5ø, 1, = cm, 3, For focal length of the lens,, 1 1 1, 1, 1, = - =, f, v u 12 -240, 1 20 + 1, 240, or, cm, =, Þ f =, f, 240, 21, Now, to get back image on the film, lens, has to form image at, æ12 - 1 ö cm = 35 cm such that the glass, ç, ÷, è, 3ø, 3, plate will shift the image on the film., 1 1 1, As,, = f, v u, 1 1 1, 3, 21, Þ, = - =, u v, f, 35 240, 48 ´ 3 - 7 ´ 21, 1, ==, 560, 1680, Þ, u = - 56, . m, , l = 2√3 m, d, , 0.2 m, , 30°, , Now, the image (I1 ) will work as object, for converging lens., For converging lens, distance of object u, (i.e. distance of I1 ), = - (25 + 15) = - 40 cm, f = 20 cm, 1, 1 1, = \ From len's formula, f, v u, 1, 1, 1, 1, 1, 1, = Þ, =, 20 v, - 40, v 20 40, 1, 1, =, v, 90, Þ, v = 40 cm, v is positive so image will be real and, will form at right side of converging, lens at 40 cm., , Þ, , 74° + d min ö, sin æç, ÷, è, ø, 2, 1.5 =, sin 37°, , 16 As refractive index for z > 0 and z £ 0 is, , 30°, , A + d min ö, sin æç, ÷, è, ø, 2, 1.5 =, sin A / 2, , 13 Focal length of diverging lens is, , 315, , l, 2 3, =, = 30, d, 02, . / 3, Therefore, maximum number of, reflections are 30., , 18 From the figure CD = emergent ray, and, CD is parallel to PQ and BC is a line, intersecting these parallel lines., So,, < DCB + < CBQ = 180°, Ð DCN + ÐNCB + ÐCBQ = 180°, a + a + 45° = 180°, a = 67.5°, But, Ð NCS = 90°, So, second mirror is at angle of 22.5°, with horizontal.
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DAY TWENTY EIGHT, , Optical, Instruments, Learning & Revision for the Day, u, , Microscope, , u, , Telescope, , u, , Resolving Power of an Optical Instrument, , An optical instrument is used to enhance and analyses the light waves. The light waves are in, the form of photons, hence optical instruments also determine the characteristics properties of, light waves., , Microscope, It is an optical instrument which forms a magnified image of a small nearby object and thus,, increases the visual angle subtended by the image at the eye, so that the object is seen to be, bigger and distinct., , 1. Simple Microscope (Magnifying Glass), It consists of a single convex lens of small focal length and forms a magnified image of an, object placed between the optical centre and the principal focus of the lens., A', , PREP, MIRROR, , L, A, , A'', B', , Your Personal Preparation Indicator, , b, , a, F, , F, , B, , C, f, , , D, If the image is formed at the near point of eye, then m = 1 + , , f, But, if the image is formed at infinity, then m =, where, D = normal viewing distance (25 cm),, f = focal length of magnifying lens., , D, f, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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OPTICAL INSTRUMENTS, , DAY TWENTY EIGHT, , 2. Compound Microscope, It consists of two lenses of small focal length and small, apertures. Also, the focal length and aperture of objective lens, are smaller than that of eyepiece. The image formed by the, objective lens is real, inverted and magnified. This image acts, as the object for the eyepiece and the final image is highly, magnified, virtual and inverted w.r.t. the original object., uo, , fo, , h, , ue, , vo, , q, , Objective, , Image, , If m o and m e be the magnifications produced by the objective, and the eyepiece respectively, then total magnification of, microscope m = m o × m e ., If final image is formed at the near point (D) of the eye, then, , m=−, , vo , D, 1 + , uo , fe , L, fo, , In practical adjustment, the final image is formed at the near, point of the observer’s eye. In this arrangement,, m=−, , fo , f , 1 + e , fe , D, , Reflecting Telescope, h', , or, , In normal adjustment, the final image is formed at infinity, and magnifying power of the telescope is, f, m=− o, fe, , Eyepiece, , F, , m =−, , 317, , , D, 1 + (approx), , fe , , It consists of an objective which is a large paraboloid, concave mirror of maximum possible focal length fo and the, eyepiece is a convex lens of small focal length and small, aperture, then, f, Magnifying power,, m=− o, fe, Reflecting type telescope is considered superior as it is free, from spherical and chromatic aberrations, is easy to install, and maintain, and can produce image of greater intensity., NOTE, , • The large aperture of telescope objective, helps in, forming a brighter image., , • If diameter of pupil of human eye is d and that of, telescope be D, then image formed by telescope will be, 2, D times brighter than the image of the same object,, , d, seen directly by the unaided eye., , If final image is formed at infinity, then, v D, L D, m = − o ⋅ = − ⋅ (approx), uo fe, fo fe, , Resolving Power of an Optical, Instrument, , Length of tube of microscope, L = vo + ue ., , Resolving power of an optical instrument is its ability to, produce distinct images of two points of an object (or two, nearby objects) very close together. Resolving power of an, optical instrument is inverse of its limit of resolution., Smaller the limit of resolution of a device, higher is its, resolving power. Limit of resolution of a normal human eye, is 1′ ., The minimum distance (or angular distance) between two, points of an object whose images can be formed distinctly by, the lens of an optical instrument, is called its limit of, resolution., , NOTE, , • Huygens’ eyepiece is free from chromatic and spherical, aberration, but it cannot be used for measurement purposes., , • Ramsden’s eyepiece can be used for precise measurement as, cross wires can be fixed in this eyepiece. It slightly suffers, from spherical and chromatic aberrations., , Telescope, Telescope is an optical instrument which increases, the visual, angle at the eye by forming the image of a distant object at the, least distance of distinct vision, so that the object is seen, distinct and bigger., , Refracting Telescope, It consists of an objective lens of large focal length fo and large, aperture. The eyepiece consists of a convex lens of small, aperture and small focal length fe ., Distance between the two lenses is set as,, L = fo + fe, , Resolving Power of a Telescope, If the aperture (diameter) of the telescope objective be the D,, then the minimum angular separation (dθ) between two, distant objects, whose images are just resolved by the, telescope, is, 1. 22 λ, dθ =, D, and resolving power of the telescope,, 1, D, RP =, =, dθ 1. 22 λ
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318, , DAY TWENTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , Resolving Power of a Microscope, The least distance (d) between two points, whose images are, just seen distinctly by a microscope, is given by, 1. 22 λ, d=, 2 nm sin θ, where, λ = wavelength of light used to illuminate the object,, nm = refractive index of the medium between the object and the, objective lens, and, , θ = semi angle of the cone of light from the point object., The term nm sin θ is generally called the numerical aperture of, the microscope., ∴ Resolving power of the microscope,, 1, RP =, d, 2 nm sin θ nm sin θ, =, =, 1. 22 λ, 0.61 λ, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 To obtain a magnified image at the distance of distinct, vision with simple microscope, where should the object, be placed?, (a) Away from the focus, (b) At focus, (c) Between the focus and the optical centre, (d) None of the above, , (a) Close to the lens, (b) Half way between the focus and the optical centre, (c) Close to the focus, (d) None of the above, , 3 A man can see clearly upto 3 m. To see upto 12 m, he, must use a lens of, (b) 3 D, , (c) −3 D, , (d) −, , 1, D, 4, , 4 A magnifying glass is used as the object to be viewed, can be brought closer to the eye than the normal near, point. This results in, (a) the formation of virtual erect image and larger angle to, be subtended by the object at the eye and hence, viewed in greater detail, (b) increase in field of view, (c) infinite magnification at near point, (d) a diminished but clear image, , 5 In a compound microscope, the objective produces a, magnification of 10, while the eyepiece produces a, magnification of 5, then the over all magnification, achieved by a compound microscope is, (a) 2, , (b) 50, , (c) 0.5, , (d) 25.00, , 6 A telescope consists of two thin lenses of focal lengths, 0.3 m and 3 cm, respectively. It is focussed on moon, which subtends on angle of 0.5° at the objective. Then,, the angle subtended at the eye by the final image will be, (a) 5°, , (b) 0.25°, , (c) 0.5°, , (b) 1 m, , (c) 0.05 m, , (d) 0.25 m, , 8 The magnifying power of a telescope is 9. When it is, , microscope, where should the eye be placed?, , 3, D, 4, , and the distance between its objective and eyepiece is, 1.05 cm. The magnifying power of the telescope is 20., What is the focal length of the objective?, (a) 2m, , 2 To obtain the maximum magnification with a simple, , (a) −, , 7 An astronomical telescope is set for normal adjustment, , (d) 0.35°, , adjusted for parallel rays, the distance between the, objective and the eyepiece is found to be 20 cm . The, focal length of lenses are, (a) 18 cm, 2 cm, (c) 10 cm, 10 cm, , (b) 11 cm, 9 cm, (d) 15 cm, 5 cm, , 9 A simple telescope, consisting of an objective of focal, length 60 cm and a single eye lens of focal length 5 cm is, focussed on a distant object in such a way that parallel, rays emerge from the eye lens. If the object subtends an, angle of 2° at the objective, the angular width of the, image is, (a) 10°, , (b) 24°, , (c) 50°, , (d) (1/6)°, , 10 In a laboratory four convex lenses L1, L2, L3 and L4 of focal, lengths 2 , 4 , 6 and 8 cm respectively are available. Two of, these lenses form a telescope of length10 cm and, magnifying power 4. The objective and eye lenses are, respectively,, (a) L2 ,L3, , (b) L1,L4, , (c) L1,L2, , (d) L4 ,L1, , 11 The magnifying power of an astronomical telescope is 8, and the distance between the two lenses is 54 cm . The, focal length of eye lens and objective lens will be, respectively, (a) 6 cm and 48 cm, (c) 8 cm and 64 cm, , (b) 48 cm and 6 cm, (d) 6 cm and 60 cm, , 12 In a refracting astronomical telescope, the final image is, (a) real, inverted and magnified, (b) real, erect and magnified, (c) virtual, erect and magnified, (d) virtual, inverted and magnified
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OPTICAL INSTRUMENTS, , DAY TWENTY EIGHT, 13 The magnifying power of a telescope is high, if, (a) both the objective and eyepiece have short focal, lengths, (b) both the objective and the eyepiece have long focal, lengths, (c) the objective has a short focal length and the eyepiece, has a long focal length, (d) the objective has a long focal length and the eyepiece, has a short focal length, , 14 An astronomical telescope has a large aperture to, , Direction (Q. Nos. 17-20), , Each of these questions contains, two statements Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d ) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 17 Statement I Very large size telescopes are reflecting, , (a) reduce spherical aberration, (b) have high resolution, (c) increase span of observation, (d) have low dispersion, , telescopes instead of refracting telescopes., Statement II It is easier to provide mechanical support to, large size mirrors than large size lenses., ª JEE Main (Online) 2013, , 15 Two point white dots are 1 mm apart on a black paper., They are viewed by eye of pupil diameter 3 mm., Approximately, what is the maximum distance at which, these dots can be resolved by the eye?, (take, wavelength of light = 500 nm), (a) 5 m, (c) 6 m, , (b) 1 m, (d) 3 m, , 18 Statement I The resolving power of a telescope is more if, the diameter of the objective lens is more., Statement II Objective lens of large diameter collects, more light., , 19 Statement I The focal length of the objective of telescope, , 16 Wavelength of light used in an optical instrument are, λ 1 = 4000 Å and λ 2 = 5000 Å, then ratio of their, respective resolving powers (corresponding to, λ 1 and λ 2) is, (a), (b), (c), (d), , 319, , 16 : 25, 9:1, 4:5, 5:4, , is larger than that of eyepiece., Statement II The resolving power of telescope increases, when the aperture of objective is small., , 20 Statement I Resolving power of an optical instrument is, reciprocal to its limit of resolution., Statement II Smaller the distance between two point, objects the instrument can resolve, higher is its resolving, power., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 An observer looks at a distance tree of height 10 m with, a telescope of magnifying power of 20. To the observer, ª JEE Main 2016 (Offline), the tree appears, (a) 10 times taller, (c) 20 times taller, , (b) 10 times nearer, (d) 20 times nearer, , 2 Assuming human pupil to have a radius of 0.25 cm and a, comfortable viewing distance of 25 cm, the minimum, separation between two objects that human eye can, ª JEE Main 2015, resolve at 500 nm wavelength is, (a) 30 µm, , (b) 100 µm, , (c) 300 µm, , (d) 1µm, , 3 For compound microscope fo = 1 cm, fe = 2.5 cm . An, object is placed at distance 1. 2 cm from objective lens., What should be the length of microscope for normal, adjustment?, (a) 8.5 cm, , (b) 8 . 3 cm, , (c) 6.5 cm, , (d) 6. 3 cm, , 4 A telescope consists of two lenses of focal length 10 cm, and 1 cm. The length of the telescope when an object is, kept at a distance of 60 cm from the objective and the, final image is formed at least distant of distinct vision is, (a) 15.05 cm, (b) 12.96 cm, (c) 13.63 cm, (d) 14.44 cm, , 5 A light source is placed at a distance b from a screen., The power of the lens required to obtain k-fold, magnified image is, (a), , k+1, kb, , (b), , (c), , kb, k+1, , (d), , (k + 1)2, kb, kb, (k − 1)2
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320, , DAY TWENTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , 6 Two white dots are 1 mm apart on a black paper. They, , and 10 m. Angle subtended by the moon’s image on eye, of observer will be, , are viewed by naked eye of pupil size 3 mm diameter., Upto what distance, the dots are clearly and separately, visible?, (a) 3 m, (c) 1 m, , (a) 15°, (c) 30°, , (b) 6 m, (d) 5 m, , (b) 20°, (d) 35°, , 10 Match the following column I with column II., Column I, , 7 Resolution of human eye is about 1 min. From what, distance a normal human eye can just resolve two, objects which are 3m apart., (a) 10 km, , (b) 15 km, , (c) 20 km, , 2. D , B. Magnification of compound microscope, , , when image formed at least distance of vision, 1.22 λ , , (d) 30 km, , 8 A student make a compound microscope by using two, lenses of focal lengths 1.5 cm and 6.25 cm. She kept an, object at 2 cm from objective and forms its final image at, 25 cm from eye lens. Distance between two lenses is, (a) 6 cm, (c) 9 cm, , Column II, 1. − v 0 , D, 1 + , u0 , fe , , A. Magnification of simple microscope, , (b) 7 cm, (d) 11 cm, , C. Tube length of telescope, , 3. 1 + D , , , , f, , D. Resolving power of a telescope, , 4. f0 + fe, , Codes, A, (a) 1, (c) 3, , 9 Orbital radius of moon is 3.8 × 105 km and its diameter is, 3.5 × 10 km.It is seen from a telescope with lenses of 4 m, 2, , B, 2, 1, , C, 3, 4, , D, 4, 2, , A, (b) 2, (d) 3, , B, 1, 2, , C, 4, 4, , D, 3, 1, , ANSWERS, SESSION 1, , 1 (c), , 2 (a), , 3 (d), , 4 (a), , 5 (b), , 6 (a), , 7 (b), , 8 (a), , 9 (b), , 10 (d), , 11 (a), , 12 (d), , 13 (d), , 14 (b), , 15 (a), , 16 (d), , 17 (a), , 18 (a), , 19 (c), , 20 (c), , 1 (c), , 2 (a), , 3 (b), , 4 (b), , 5 (b), , 6 (d), , 7 (a), , 8 (d), , 9 (b), , 10 (c), , SESSION 2, , Hints and Explanations, SESSION 1, 1 Magnification is obtained in a simple, microscope when the object is close to, the lens between the focus and the, optical centre, as lens is converging., , 2 The magnification of simple, microscope is given by, D−a, m = 1+, f, where, D = least distance of vision,, f = focal length and, a = distance of lens from eye, So, lesser the distance between eye, and lens, greater is the magnification., , 3 Using, f =, ⇒ f =, , xy, x− y, , 3 × 12, = −4 m, 3 − 12, , Hence, power of lens required,, 1 −1, D, P = =, f, 4, , 4 This results in the formation of virtual,, erect image and the object subtends a, larger angle at the eye and the image is, viewed in greater detail., , 5 m = mo × me = 10 × 5 = 50, f, 6 For a telescope, β = o, ∴, , α, fe, β, 03, ., =, ⇒ β = 5°, 0.5° 0.03, , 7. Here, m = fo ⁄ fe, and, , L = fo + fe = 1.05, , or, fe = 1.05 − fo, and 20 = fo /(1.05 − fo ), This gives fo = 1 m and fe = 0.05 m ., , 8 fo + fe = 20 cm,, m = fo / fe = 9., This gives, fe = 2 cm and fo = 18 cm., , 9 It is a case of normal adjustment., Hence,, , m = fo ⁄ fe, , m =β ⁄α, f, β, Therefore,, = o, α, fe, Also,, , Here, fo = 60 cm, fe = 5 cm,α = 2°, Hence, β = 24° ., , 10 Length of tube = 10 cm, fo + fe = 10 cm, fo, =4, fe, , Magnification, m =, , fo = 4 fe, , …(i)
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OPTICAL INSTRUMENTS, , DAY TWENTY EIGHT, On putting in Eq. (i), we get, 5 fe = 10 cm, or, fe = 2 cm, and, fo = 8 cm, Hence, L 4 and L1 will be used., Note In telescope, objective always have, larger focal length than eyepiece., , 11 As L = fo + fe = 54 cm, and, , |m | =, , relaxed state is, f, m= o, fe, , fo, =8, fe, , fo = + 48 cm, , 12 The image formed by the objective lens, is real, inverted and larger object and, the eyepiece forms a second image but, virtual, inverted and larger than the, first., , 13 The magnifying power of a telescope (if, the object is at infinity) is given by, f D + fe, M = o ., fe, D, where, D = least distance of distinct, vision, where the final image is formed., , 14 An astronomical telescopic as a large, aperture to have high resolution., , 15 We know that,, , ⇒, ∴, ⇒, , y, λ, ≥ 1.22, D, d, yd, D≤, 1.22 λ, , D≤, , We have the formula, sin θ =, ∴ sin θ =, =, , 1.22 × 5 × 10− 7, , = 5m, , D max = 5m, , 16 Resolving power of an optical, instrument is inversely proportional to, λ, i.e., 1, RP ∝, λ, Resolving power at λ1, λ, ∴, = 2, Resolving power at λ 2 λ1, 5000, =, 4000, , So, for high magnification, the focal, length of objective length should be, larger than that of eyepiece., Resolving power of a telescope, d, =, 1.22 λ, For high resolving power, diameter (d ), of objective should be higher., , 20 Resolving power of an optical, instrument is its ability to produce, distinct image of two points of an object, very close together. Resolving power of, an optical instrument is inverse of its, limit of resolution. Smaller the limit of, resolution of a device higher is its, resolving power., , mechanical support to large size mirror, than size of lens. So, in order to fulfil, this mechanical support telescope is, reflecting instead of refracting, telescope., , 18 Resolving power of telescope is, =, , a, 1.22λ, , 1.22 × 5 × 10−7, 5 × 10−3, , Let Y be the minimum separation, between two objects that human eye can, Y, resolve i.e. sinθ =, D, ∴ y = D sin θ = 0.25 × 1.22 × 10−4, = 3 × 10−5 m = 30 µm, , 3 When final image is formed at normal, adjustment, then length of compound, microscope,, L = v o + ue =, =, , uo fo, fe D, +, (uo + fo ), fe + D, , − 1.2 × 1 2.5 × 25, +, − 1.2 + 1 2.5 + 25, , = 6 + 2.27, , 4 Two lenses used are eyepiece and, , magnifying power to see a 20 times, taller object, as the angular, magnification should be 20 and we, observe angular magnification. Option, (c) would not be very correct as the, telescope can be adjusted to form the, image anywhere between infinity and, least distance for distinct vision., Suppose that the image is formed at, infinity. Then, the observer will have to, focus the eyes at infinity to observe the, image. Hence, it is incorrect to say that, the image will be appear nearer to the, observer., , 2 We can write resolving angle of naked, eye as, Eye lens, θ, , 17 As very large size telescope needs, , 0.5 × 10−2, , = 8.27 ≈ 8.3 cm, , SESSION 2, , = 5: 4, , 1.22 × 5 × 10−7, , The distance of comfortable viewing is, D = 25 cm = 0.25 m, , 1 Height of image depends upon the, , 10− 3 × 3 × 10− 3, , 1.22 λ, d, , = 1.22 × 10−4, , 19 The magnifying power of telescope in, , On simplification, we get, fe = + 6 cm, and, , where, a is the diameter of objective, lens and λ is the wavelength of light, used. It is obvious that on increasing a,, more light is collected by objective lens, and so, the image formed is more bright., Thus, resolving power of telescope, increases., , 321, , Y, , 25cm, Retina, , Assuming human pupil to have a radius, r = 025, . cm or diameter, d = 2r = 2 × 025, . = 0.5 cm, the, wavelength of light λ = 500 nm, = 5 × 10−7 m, , objective., For eyepiece, fe = 1 cm,, D = v e = 25cm, , ⇒, , 1, 1, 1, −, =, ve, ue, fe, 1, 1, −, −, =1, 25 ue, , 25, cm, 26, For objective uo = − 60 cm,, fo = 10 cm, 1, 1, 1, −, =, v o uo, fo, 1, 1, 1, +, =, vo, 60 10, 1, 1, 1, =, −, ⇒, v o 10 60, 1, 5, ⇒, =, vo, 60, 60, vo =, ⇒, 5, = 12 cm, Length of telescope,, L = v o + ue, 25, = 12 +, 26, = 12.96 cm, ⇒, , ue = −
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322, , DAY TWENTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , 5 Distance of light source from lens is x 0, and distance of screen from lens is, (b − x 0 )., , 6 Using,, ⇒, , 122, . λ, x, =, a, d, d =, =, , x0, , b – x0, , xa, 122, . λ, 1 × 10−3 × 3 × 10−3, , ≈ 5m, 122, . × 500 × 10−9, Note In case, wavelength of light is not, given, we take mean value of visible, region ≈ 500 nm., 7 Using, θ = d, r, , b, , ∴ Image is formed on screen, hence, (b − x 0 ) is also the image distance., Image is formed on the screen, so m will, be negative., v, m= −k =, u, ∴, v = − ku, Here, u = − x 0, v = b − x 0, b, ∴ b − x0 = k x 0 , x0 =, 1+ k, From lens formula,, 1 1 1, − =, v u, f, 1, 1, −, =p, b − x0 − x0, ⇒, , p=, , b, x 0(b − x 0 ), , Putting the value of x 0, we have, (k + 1)2, p=, kb, , uo fo, f D, + e, uo − fo, fe + D, 2 × 1.5 6.25 × 25, =, +, = 11 cm, 2 − 1.5 6.25 + 25, =, , d, θ, r, d, θ, where, θ is in radians , , 1 ° 1 × π , = 1′ = =, , 60 , 60 180 , , 3, =, 1 × π, , 60 180, 3 × 60 × 180, =, π, ≈ 10.3 km ≈ 10 km, , We have, r =, , 8 As, fo < fe , for a compound microscope., So, fo = 1.5cm and fe = 6.25cm, Now, length of tube, = distance between lenses, , 9 Angle subtended by moon on the, objective of telescope is,, 3.5 × 103, α =, 3.8 × 105, 3.5, =, × 10−2 rad, 3.8, f, β, As, m = o =, fe, α, 400, ∴ β=, ×α, 10, 3.5, 180, = 40α = 40 ×, × 10−2 ×, ≈ 20°, 3.8, π, , 10 A - Magnification of simple microscope, is given by,, D, m = 1 + , , F, B - Magnification of compound, microscope, when image is formed at, least distance of vision is,, m=, , −v 0 , D, 1 + , u0 , fe , , C - Tube length of telescope,, L = f 0 + fe, D - Resolving power of telescope, D , =, , . λ, 122
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DAY TWENTY NINE, , Wave Optics, Learning & Revision for the Day, u, , u, , u, , Wavefront, , u, , Interference of Light, Young’s Double Slit Experiment, , u, , u, , Coherent Sources, Interference in Thin Films, Diffraction, , u, , u, , u, , u, , Polarisation of Light, Brewster’s Law, Law of Malus, Polaroids, , According to Huygens’, light is a form of energy, which travels in the form of waves, through a hypothetical medium ‘ether’. The medium was supposed to be all pervading,, transparent, extremely light, perfectly elastic and an ideal fluid., Light waves transmit energy as well as momentum and travel in the free space with a, constant speed of 3 × 10 8 ms −1. However, in a material medium, their speed varies from, medium to medium depending on the refractive index of the medium., , Wavefront, A wavefront is the locus of all those points (either particles) which are vibrating in the, same phase. The shape of the wavefront depends on the nature and dimension of the, source of light., In an isotropic medium, for a point source of light, the wavefront is spherical in, nature., For a line (slit) source of light, the wavefront is cylindrical in shape., For a parallel beam of light, the wavefront is a plane wavefront., l, , l, , l, , Huygens’ Principle, Every point on a given wavefront, acts as secondary source of light and emits secondary, wavelets which travel in all directions with the speed of light in the medium. A surface, touching all these secondary wavelets tangentially in the forward direction, gives the, new wavefront at that instant of time., Laws of reflection and refraction can be determined by using Huygens’ principle., , Interference of Light, Interference of light is the phenomenon of redistribution of light energy in space when, two light waves of same frequency (or same wavelength) emitted by two coherent, sources, travelling in a given direction, superimpose on each other. If a1 and a2 be the, amplitudes of two light waves of same frequency and φ be the phase difference, between them, then the amplitude of resultant wave is given by, AR = a21 + a22 + 2 a1a2 cos φ, , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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324, , DAY TWENTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , and in terms of intensity of light,, , Condition for Constructive Interference, If at some point in space, the phase difference between, two waves, φ = 0 ° or 2nπ or path difference between two, waves, ∆ = 0 or nλ, where n is an integer, then AR = a1 + a2, or I R = I 1 + I2 + 2 I 1I2 is maximum. Such an interference is, called constructive interference., , Condition for Destructive Interference, If at some point in space, the phase difference between two, λ, waves, φ = (2 n − 1)π or path difference, δ∆ = (2 n − 1) , then at, 2, such points AR = (a1 − a2 ) and I R = I 1 + I2 − 2 I 1I2 is, minimum leading to a destructive interference., , Fringe Width, The separation between any two consecutive bright or dark, fringes is called fringe width β., Dλ, Thus,, β=, d, and for a given arrangement, it is constant, i.e. all fringes are, uniformly spaced., (i) β ∝ D,, , 2, , I + I2 + 2 I 1 I2 I 1 + I2 , = 1, =, , I 1 + I2 − 2 I 1 I2 I 1 − I2 , , a + a2 , r + 1, = 1, =, , , a1 − a2 , r − 1, a, where, r = 1 = amplitude ratio., a2, , NOTE, , The arrangement is shown in figure monochromatic light of, one wavelength is used., P, x, d, , • If whole apparatus of Young’s double slit experiment is, immersed in a transparent medium of refractive index nm,, Dλ, then fringe width in the medium, β m =, ., nm d, , Young’s Double Slit Experiment, , Light S, source, , 1, (iii) β ∝ ., d, , If in a given field of view n1, fringes of light of wavelength, λ 1 are visible and n2 fringes of wavelength λ 2 are visible, then, n1λ 1 = n2 λ 2, , • For identical sources, I1 = I2 = I0, φ, • For constructive interference, Imax = 4 I0 and I = 4 I0 cos 2, 2, • For destructive interference, Imin = 0, , S1, , (ii) β ∝ λ and, , Angular fringe width of interference pattern,, β λ, α= =, D d, , 2, , 2, , NOTE, , where, n = 1, 2, 3, …… ., , Moreover, fringe width β is, , Amplitude Ratio, I max, I min, , xd, λ, = (2n − 1) , then we get nth dark fringe., D, 2, Hence, for nth dark fringe,, (2n − 1) Dλ, x=, 2d, , Case II If, , I R = I 1 + I2 + 2 I 1I2 cos φ., , Displacement of Fringe Pattern, When a thin transparent plate is introduced in the path of one, of the interfering waves trains, it is found that the entire fringe, pattern is shifted through a constant distance. This shift takes, place towards the wave train, in the path of which the plate is, introduced., , O, , t, , S2, , S1, , D, , Young’s experimental arrangement to produce, interference pattern, Bright and dark fringes are formed on the screen with central, point O behaving as the central bright fringe, because for O,, the path difference ∆ = 0., For light waves reaching a point P, situated at a distance x, from central point ∆, the path difference,, xd, ∆ = S2 P − S 1 P =, D, xd, Case I If, = nλ, then we get nth bright fringe. Hence,, D, position of bright fringes on the screen are given by the, relation, x = nDλ ⋅, d, , P, Shift, O, , S2, , Young’s double slit experiment, The fringe width of the patterns remains same. Effective, optical path that is equivalently covered in air is, S 1P + t (µ − 1)., Thus, the path difference = S2 P − S 1P − t (µ − 1), xd, =, − t (µ − 1), D, [Optical path = refractive index× width of the material], ∴, , Extra path = µt − t = (µ − 1) t
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WAVE OPTICS, , DAY TWENTY NINE, , Coherent Sources, Two light sources are said to be coherent, if they emit light of, exactly same frequency (or wavelength), such that the, originating phase difference between the waves emitted by, them is either zero or remains constant. For sustained, interference pattern, the interfering light sources must be, coherent one., There are two possible techniques for obtaining coherent, light sources., l, , l, , In division of wavefront technique, we divide the, wavefront emitted by a narrow source in two parts by, reflection, refraction or diffraction., In division of amplitude technique, a single extended light, beam of large amplitude is splitted into two or more waves, by making use of partial reflection or refraction., , NOTE, , • Two independent sources of light can never be coherent., Two light sources can be coherent only, if these have been, derived from a single parental light source., , NOTE, , 325, , • Fresnel’s biprism is a device to produce coherent sources, by division of wavefront,, , d = 2 a ( n − 1) α, The distance between the coherent sources and screen,, D=a+ b, λ ( a + b), Dλ, The fringe width is given by β =, =, d, 2a ( n − 1) α, , Diffraction, Diffraction of light is the phenomenon of bending of light, around the edges of an aperture or obstacle and entry of light, even in the region of geometrical shadow, when size of, aperture or obstacle is comparable to wavelength of light, used. Diffraction is characteristic of all types of waves., Greater the wavelength, more pronounced is the diffraction, effect. It is due to this reason that diffraction effect is very, commonly observed in sound., , Diffraction due to a Single Slit, Interference in Thin Films, In white light thin films, whose thickness is comparable to, wavelength of light, show various colours due to interference, of light waves reflected from the two surfaces of thin film., For interference in reflected light condition of constructive, interference (maximum intensity),, λ, ∆ = 2 nm t cos r = (2 n − 1), 2, Condition of destructive interference (minimum intensity),, λ, ∆ = 2 nm t cos r = (2 n), 2, For interference in refracted light condition of constructive, interference (maximum intensity),, λ, ∆ = 2 nmt cos r = (2 n), 2, Condition of destructive interference (minimum intensity),, λ, ∆ = 2 nmt cos r = (2 n − 1) , where n = 1, 2, 3,… ., 2, , Fraunhofer’s arrangement for studying diffraction at a single, narrow slit (width = a) is shown in adjoining figure. Lenses L1, and L2 are used to render incident light beam parallel and to, focus parallel light beam., W, , P, , θ, , A, θ, , a, , O, , θ, , S, f1, , B, , L1, , L2, f2, , Slit, , Screen, , Fresnel diffraction through a slit, As a result of diffraction, we obtain a broad, bright maxima at, symmetrical centre point O and on either side of it, we get, secondary diffraction maxima of successively falling intensity, and poor contrast, as shown in figure., I, , Shift in Interference Pattern, If a transparent thin sheet of thickness t and refractive index, nm is placed in the path of one of the superimposing waves, (say in front of slit S2 of Young’s double slit experiment), then, it causes an additional path difference due to which, interference pattern shifts., l, , l, , l, , Additional path difference due to sheet = (nm − 1) t, D, β, Fringe shift = (nm − 1)t = (nm − 1)t, d, λ, If due to presence of thin film, the fringe pattern shifts by n, fringes, then, (n − 1) t, nλ, or t =, n= m, λ, (nm − 1), Shift is independent of the order of fringe and wavelength., , O, , –ve, , θ (or x), , +ve, , Intensity curve, l, , Condition of diffraction minima is given by, a sin θ = nλ, where, n = 1, 2, 3, 4,... ., But the condition of secondary diffraction maxima is, λ, a sin θ = (2 n + 1), 2, where, n = 1, 2, 3, 4,... .
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326, , l, , l, , DAY TWENTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , Angular position of nth secondary minima is given by, λ, sin θ = θ = n, a, nDλ nf2 λ, and linear distance, x n = Dθ =, =, a, a, , • Only transverse waves can be polarised. Thus, it proved, that light waves are transverse waves., , Brewster’s Law, , Similarly, for nth maxima, we have, , According to this law, when unpolarised light is incident at an, angle called polarising angle, i p on an interface separating air, from a medium of refractive index µ, then the reflected light is, fully polarised (perpendicular to the plane of incidence),, provided, , The angular width of central maxima is 2 θ =, , 2λ, a, , 2 Dλ 2 f2 λ, or linear width of central maxima =, =, a, 2, NOTE, , µ = tan i p, This relation represents Brewster’s law. Note that the parallel, components of incident light do not disappear, but refract into, the medium, with the perpendicular components., , • The angular width of central maxima is double as compared, to angular width of secondary diffraction maxima., , Polarisation of Light, , l, , NOTE, , where, f2 is focal length of lens L2 and D = f2 ., (2 n + 1) λ, (2 n + 1) Dλ (2 n + 1) f2 λ, and x n =, sin θ = θ =, =, 2a, 2a, 2a, , l, , called plane polarised light. The vibrations in a plane, polarised light are perpendicular to the plane of polarisation., , Light is an electromagnetic wave in which electric and, magnetic field vectors very sinusoidally, perpendicular to, each other as well as perpendicular to the direction of, propagation of wave of light., The phenomenon of restricting the vibrations of light, (electric vector) in a particular direction, perpendicular to, the direction of wave motion is called polarisation of light., The tourmaline crystal acts as a polariser., A, , Law of Malus, When a beam of completely plane polarised light is incident, on an analyser, the resultant intensity of light (I ) transmitted, from the analyser varies directly as the square of cosine of, angle ( θ) between plane of transmission of analyser and, polariser., If intensity of plane polarised light incidenting on analyser is, I 0, then intensity of emerging light from analyser is I 0 cos2 θ., NOTE, , F, , G, B, , C, , Polarisation of Light, Thus, electromagnetic waves are said to be polarised when, their electric field vector are all in a single plane, called the, plane of oscillation/vibration. Light waves from common, sources are upolarised or randomly polarised., , Plane Polarised Light, The plane ABCD in which the vibrations of polarised light are, confined is called the plane of vibration. It is defined as The, light, in which vibrations of the light (vibrations of electric, vector) when restricted to a particular plane the light itself is, , • We can prove that when unpolarised light of intensity I0, gets polarised on passing through a polaroid, its intensity, 1, becomes half, i.e. I = I0 ., 2, , D, E, , H, , I ∝ cos2 θ, , i.e., , Polaroids, Polaroids are thin and large sheets of crystalline polarising, material (made artifically) capable of producing plane, polarised beams of large cross-section., The important uses are, l, , l, , l, , These reduce excess glare and hence sun glasses are fitted, with polaroid sheets., These are also used to reduce headlight glare of cars., They are used to improve colour contrast in old oil, paintings., , l, , In wind shields of automobiles., , l, , In window panes., , l, , In three dimensional motion pictures.
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WAVE OPTICS, , DAY TWENTY NINE, , 327, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The wavefront of a distant source of light is of shape, (a) spherical (b) cylindrical (c) elliptical, , (d) plane, , 2 Which of the following cannot be explained on the basis, of wave nature of light?, (i) Polarisation, , (ii) Optical activity, , (iii) Photoelectric effect, , (iv) Compton effect, , (a) III and IV, , (b) II and III, , (c) I and III, , (d) II and IV, , 3 A Young’s double slit experiment uses a monochromatic, source. The shape of the interference fringes formed on, a screen is, ª AIEEE 2005, (a) hyperbola, (c) straight line, , (b) circle, (d) parabola, , 4 Two light rays having the same wavelength λ in vacuum, are in phase initially. Then, the first ray travels a path L1, through a medium of refractive index µ1, while the, second ray travels a path of length L2 through a medium, of refractive index µ 2. The two waves are then combined, to observe interference., The phase difference between the two waves is, 2π, (µ 1 L 1 − µ 2 L2 ), λ, 2 π L 1 L2 , (c), − , , λ µ1 µ 2 , , 2π, (L2 − L 1), λ, 2π, (d), (µ 2 L 1 − µ 1 L 2 ), λ, , (a), , (b), , 5 The Young’s double slit experiment is performed with, blue and with green light of wavelengths 4360 Å and, 5460 Å respectively. If x is the distance of the 4th, maxima from the central one, then, (a) x (blue ) = x (green ), (b) x (blue ) > x (green ), (c) x (blue ) < x (green ), (d) x (blue ) / x (green ) = 5460/4360, , 7 In a Young’s double slit experiment, two coherent, sources are placed 0.90 mm apart and the fringes are, observed one metre away. If it produces the second dark, fringe at a distance of 1 mm from the central fringe, the, wavelength of monochromatic light used will be, (a) 60 × 10−4 cm, (c) 10 × 10−5 cm, , 8 In Young’s double slit experiment, the two slits act as, coherent sources of waves of equal amplitude A and, wavelength λ. In another experiment with the same, arrangement the two slits are made to act as incoherent, sources of waves of same amplitude and wavelength. If, the intensity at the middle point of the screen in the first, I, case is I1 and in the second case I 2, then the ratio 1 is, I2, ª AIEEE 2012, , (a) 4, , (b) 2, , (c) 1, , (d) 0.5, , 9 A mixture of light consisting of wavelength 590 nm and an, unknown wavelength, illuminates the Young’s double slit, and gives rise to two overlapping interference patterns on, the screen. The central maximum of both lights coincide., Further, it is observed that the third bright fringe of known, light coincides with the 4th bright fringe of the, unknown light. From this data, the wavelength of the, unknown light is, ª AIEEE 2003, (a) 885.0 nm, (c) 776.8 nm, , (b) 442.5 nm, (d) 393.4 nm, , 10 Two coherent point sources S1 and S 2 are separated by a, small distance d as shown. The fringes obtained on the, ª JEE Main 2013, screen will be, Screen, d, S1, , 6 In Young’s double slit experiment, the length of band is, 1mm. The fringe width is 0.021 mm. The number of, fringes is, , S2, D, , (a) points, (c) semi-circle, , A, , (b) straight lines, (d) concentric circles, , 11 The source that illuminates the double-slit in ‘double-slit, , 0.5 mm, C, , 1 mm, , B, , (a) 45, (c) 49, , (b) 10 × 10−4 cm, (d) 6 × 10−5 cm, , (b) 46, (d) 48, , interference experiment’ emits two distinct, monochromatic waves of wavelength 500 nm and, 600 nm, each of them producing its own pattern on the, screen. At the central point of the pattern when path, difference is zero, maxima of both the patterns coincide, and the resulting interference pattern is most distinct at, the region of zero path difference.
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328, , DAY TWENTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , But as one moves out of this central region, the two fringe, systems are gradually out of step such that maximum, due to one wave length coincides with the minimum due, to the other and the combined fringe system becomes, completely indistinct. This may happen when path, difference in nm is, ª JEE Main (Online) 2013, , (a) 2000, , (b) 3000, , (c) 1000, , (d) 1500, , 12 In Young’s double slit experiment, the intensity at a point, , λ, (λ being the wavelength of, 6, the light used) is I. If I 0 denotes the maximum intensity, I / I 0, is equal to, ª AIEEE 2007, , where the path difference is, , 1, (a), 2, , 3, (b), 2, , 1, (c), 2, , 3, (d), 4, , The percentage of incident light transmitted by the first, polariser will be, (a) 100%, , (b) 50%, , (a) 2 I 0, , (b) I 0 / 2, , A, , N, , (a) 30°, , (b) 45°, , (c) 60°, , (d) 15°, , 15 A parallel monochromatic beam of light is incident, normally on a narrow slit. A diffraction pattern is formed, on a screen placed perpendicular to the direction of, incident beam. At the first maximum of the diffraction, pattern, the phase difference between the rays coming, from the edges of the slit is, (b) π 2, , (a) 0, , (c) π, , (d) 2 π, , 16 In Fraunhofer diffraction experiment, L is the distance, between screen and the obstacle, b is the size of, obstacle and λ is wavelength of incident light. The, general condition for the applicability of Fraunhofer, diffraction is, (a), , b2, >>1, Lλ, , (b), , b2, =1, Lλ, , (c), , b2, <<1, Lλ, , (d), , b2, ≠1, Lλ, , O, , (a) the intensity is reduced down to zero and remains zero, (b) the intensity reduces down somewhat and rises again, (c) there is no change in intensity, (d) the intensity gradually reduces to zero and then again, increases, , 21 When an unpolarised light of intensity I 0 is incident on a, polarising sheet, the intensity of the light which does not, ª AIEEE 2005, get transmitted is, (a), , 1, I0, 2, , 2, (a) sin− 1 , 3, − 1, , 1, (c) sin , 4, , 3, (b) sin− 1 , 4, − 1, , 2, (d) tan , 3, , 18 An unpolarised beam of light is incident on a group of, four polarising sheets which are arranged in such a way, that the characteristic direction of each polarising sheet, makes an angle of 30° with that of the preceding sheet., , (b), , 1, I0, 4, , (c) zero, , (d) I 0, , 22 Two beams, A and B of plane polarised light with, mutually perpendicular planes of polarisation are seen, through a polaroid. From the position when the beam A, has maximum intensity (and beam B has zero intensity), a, rotation of polaroid through 30° makes the two beams, appear equally bright.If the initial intensities of the two, beams are I A and IB respectively, then I A / IB equals, (a) 3, , 17 In a Fraunhofer diffraction experiment at a single slit, using a light of wavelength 400 nm, the first minimum is, formed at an angle of 30°. The direction θ of the first, secondary maximum is given by, , B, , 33°, , 33°, , λ , −1 λ , −1 λ , (b) sin , (c) sin , (d) sin , , 2, d, 3, d, , , , , 4d , , is θ, for a light of wavelength 5000 Å. If the width of slit is, 1 × 10−4 cm. Then, the value of θ is, , (d) I 0 / 2, , a direction as shown in the figure. The reflected ray OB is, passed through a nicol prism. On viewing through a nicol, prism, we find on rotating the prism that, , −1 , , 14 The first diffraction minimum due to single slit diffraction, , (c) I 0 / 4, , 20 A beam of light AO is incident on a glass slab (µ = 1.54) in, , is 1/4 of the maximum intensity. Angular position of this, point is, ª AIEEE 2005, λ, (a) sin , d , , (d) 125%, , tourmaline crystal C1 and then it passes through another, tourmaline crystal C2 which is oriented such that its, principal plane is parallel to that of C2. The intensity of, emergent light is I 0. Now, C2 is rotated by 60° about the, ray. The emergent ray will have an intensity, , 13 In Young’s double slit experiment, the intensity at a point, , −1 , , (c) 25%, , 19 A beam of ordinary unpolarised light passes through a, , 3, (b), 2, , (c) 1, , ª JEE Main 2014, 1, (d), 3, , 23 A ray of light is incident on the surface of a glass plate of, refractive index 1.732 at the polarising angle. The angle, of refraction of the ray is, (a) 45°, , (b) 60°, , (c) 15°, , (d) 30°, , 24 A beam of unpolarised light of intensity I 0 is passed, through a polaroid A and then through another polaroid B, which is oriented, so that its principal plane makes an, angle of 45° relative to that of A. The intensity of the, ª JEE Main 2013, emergent light is, (a) I 0, (c) I 0 / 4, , (b) I 0 / 2, (d) I 0 / 8
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WAVE OPTICS, , DAY TWENTY NINE, Direction (Q. Nos. 25-28) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below:, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 25 Statement I The thick film shows no interference pattern., Take thickness of the order of a few cms., Statement II For interference pattern to be observed path, difference between two waves is of the order of few, wavelengths., , 329, , 26 Statement I To observe diffraction of light, the size of, obstacle/aperture should be of the order of10−7 m., Statement II 10−7m is the, visible light., , order of wavelength of the, , 27 Statement I For a given medium, the polarising angle is, 60°. The critical angle for this medium is 35°., Statement II µ = tan ip ., , 28 Statement I In Young’s double slit experiment, the, number of fringes observed in the field of view is small, with longer wavelength of light and is large with shorter, wavelength of light., Statement II In the double slit experiment the fringe, width depends directly on the wavelength of light., ª JEE Main (Online) 2013, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A parallel beam of light of intensity I 0 is incident on a, glass plate, 25% of light is reflected by upper surface, and 50% of light in reflected from lower surface. The, ratio of maximum to minimum intensity in interference, region of reflected ray is, , S1, , C, L=, , Air, , P, O, , θ, , A, d/4, , Screen, , S2, , AC = CO = D, S1C = S 2 C = d << D, (a), 1, +, (a) 2, 1−, , 2, , 2, , 1, 3, , +, 8 (b) 4, 3, 1−, , , 2, 8, , 2, , 3, , 8 (c) 5, 3, 8, , 8, , (d), , 8, 5, , 2 White light is used to illuminate the two slits in a Young’s, double slit experiment. The separation between the slits, is b and the screen is at a distance d ( >> b ) from the slits., At a point on the screen directly in front of one of the, slits, certain wavelengths are missing. Some of these, missing wavelengths are, 2b 2, (a) λ =, 3d, , b2, (b) λ =, 3d, , 2b 2, (c) λ =, d, , 3b 2, (d) λ =, d, , 3 A small transparent slab containing material of µ = 1.5 is, placed along AS 2 (figure). What will be the distance from, O of the principal maxima and of the first minima on, either side of the principal maxima obtained in the, absence of the glass slab., , (c), , 5, below point O, 235, 5, below point O, 220, , 5, below point O, 231, 5, (d), below point O, 110, (b), , 4 n identical waves each of intensity l0 interfere with each, other. The ratio of maximum intensities if the interference, is (i) coherent and (ii) incoherent is, ª JEE Main (Online) 2013, , (a) n, , 2, , 1, (b), n, , (c), , 1, n2, , (d) n, , 5 In a Young’s double slit experiment, one of the slit is, wider than other, so that amplitude of the light from one, slit is double of that from other slit. If Im be the maximum, intensity, the resultant intensity I when they interfere at, ª AIEEE 2012, phase difference φ is given by, Im, 9, I, (c) m, 5, , (a), , (4 + 5 cos φ), 1 + 4 cos2 φ , , , , 2, , Im, 3, I, (d) m, 9, (b), , 1 + 2 cos2 φ , , , , 2, 1 + 8 cos2 φ , , , , 2
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330, , DAY TWENTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , (a) 25 µm, (c) 75 µm, , 6 An object 2.4 m in front of a lens forms a sharp image on, a film 12 cm behind the lens. A glass plate 1 cm thick, of, refractive index 1.50 is interposed between lens and film, with its plane faces parallel to film. At what distance, (from lens) should object shifted to be in sharp focus on, film?, ª AIEEE 2012, (a) 7.2 m, , (b) 2.4 m, , (c) 3.2 m, , (d) 5.6 m, , 7 In a YDSE, bichromatic light of wavelength 400 nm and, 560 nm are used. The distance between the slits is, 0.1 mm and the distance between the plane of the slits, and the screen is 1 m. The minimum distance between, two successive regions of complete darkness is, , 11 The box of a pin hole camera of length L, has a hole of, radius a. It is assumed that when the hole is illuminated, by a parallel beam of light of wavelength λ the spread of, the spot (obtained on the opposite wall of the camera) is, the sum of its geometrical spread and the spread due to, diffraction. The spot would then have its minimum size, (say bmin ) when, ª JEE Main 2016 (Offline), (a) a =, (c) a =, , ª AIEEE 2004, , (a) 4 mm, , (b) 5.6 mm, , (c) 14 mm, , (d) 28 mm, , 8 The maximum number of possible interference maxima, for slit-separation equal to twice the wavelength in, Young’s double slit experiment, is, , (b) 50 µm, (d) 100 µm, , 2 λ2 , λ2, and bmin = , , L, L , λL and bmin =, , 2 λ2 , λL and bmin = , , L , , (b) a =, , λ2, and bmin =, L, , 4λL (d) a =, , 4λL, , 12 In the YDSE apparatus shown in figure ∆x is the path, difference between S1 P and S 2 P., P, , ª AIEEE 2004, , (a) infinite, , (b) five, , (c) three, , S1, , (d) zero, , polariser A. Another identical polariser B is placed, behind A. The intensity of light beyond B is found to be, I, . Now, another identical polariser C is placed between, 2, 1, A and B. The intensity beyond B is now found to be ., 8, The angle between polariser A and C is ª JEE Main 2018, (a) 0°, , (b) 30°, , (c) 45°, , Now a glass slab is introduced in front of S 2, then match, the following columns., Column I, , (d) 60°, , 10 The angular width of the central maximum in a single slit, diffraction pattern is 60°. The width of the slit is 1 µm. The, slit is illuminated by monochromatic plane waves. If, another slit of same width is made near it, Young’s fringes, can be observed on a screen placed at a distance 50 cm, from the slits. If the observed fringe width is 1 cm, what is, slit separation distance? (i.e. distance between the, ª JEE Main 2018, centres of each slit.), , O, , S2, , 9 Unpolarised light of intensity I passes through an ideal, , Column II, , A, , ∆x at P will, , 1., , increase, , B, , Fringe width will, , 2., , decrease, , C, , Fringe pattern will, , 3., , remain same, , D, , Number of fringes, between O and P will, , 4., , shift upward, , 5., , shift downward, , A, (a) 1, (c) 3, , B, 3, 4, , C, 5, 1, , D, 3, 3, , A, (b) 2, (d) 5, , B, 3, 5, , C, 5, 1, , D, 3, 2, , ANSWERS, SESSION 1, , SESSION 2, , 1 (d), , 2 (a), , 3 (a), , 4 (a), , 5 (c), , 6 (c), , 7 (d), , 8 (b), , 9 (b), , 10 (d), , 11 (d), , 12 (d), , 13 (c), , 14 (a), , 15 (d), , 16 (c), , 17 (b), , 18 (b), , 19 (c), , 20 (d), , 21 (a), , 22 (d), , 23 (d), , 24 (c), , 25 (a), , 26 (a), , 27 (a), , 28 (a), , 1 (a), 11 (c), , 2 (b), 12 (a), , 3 (b), , 4 (d), , 5 (d), , 6 (d), , 7 (d), , 8 (b), , 9 (c), , 10 (a)
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332, , 22 By law of Malus i.e. I = I 0 cos2 q, Initially, , SESSION 2, upper surface is, , IA, IB, , IB, Polaroid, , Polaroid, , Transmission axis, , Transmission axis, , I A¢ = I A cos 2 30°, , Now,, , I B¢ = I B cos 2 60°, As,, , I ¢A = I ¢B, I A cos 2 30° = I B cos 2 60°, 1, I, 3, 1, Þ A =, I A = IB, IB 3, 4, 4, , 25, I, = 0, 100 4, Intensity of transmitted light from, upper surface is, 3I, I, I = I0 - 0 = 0, 4, 4, \ Intensity of reflected light from lower, surface is, 3I, 3I, 50, = 0, I2 = 0 ´, 100, 8, 4, 2, I max ( I1 + I2 ), =, I min, ( I1 - I2 )2, I1 = I 0 ´ 25% = I 0 ´, , 23 From Brewster’s law,, , Þ, , æ1, ç +, è2, , =, , q p + r = 90°, r = 90° - q p = 90° - 60° = 30°, , 24 Relation between intensities is, , æ1, ç è2, , B, (I0/2), IR, , (Unpolarised), , A, , 1 I, I, I, I R = æç 0 ö÷ cos 2 (45° ) = 0 ´ = 0, è2ø, 2, 2, 4, difference between two waves is of the, order of few wavelength., , For n = 1,, , 26 For diffraction to occur, the size of an, obstacle/aperture is comparable to the, wavelength of light wave. The order of, wavelength of light wave is 10-7 m, so, diffraction occurs., , 27 From the relation,, ...(i), [from Eq. (i)], , æ 1 ö, \ C = sin -1 ç ÷ = sin -1 (0.577) = 35°, è 3ø, , 3ö, ÷, 8ø, , 28 The number of fringe is smaller in case, of larger wavelength is used while in, case of smaller wavelength is used the, number of fringe is larger., Also, fringe width is given by, lD, b=, Þ bµl, d, , point O., , 4 When interference is coherent, When two waves of intensities I1 and I2, having a phase difference f interfere, the, resultant intensity is given as, …(i), I = I1 + I2 + 2 I1 I2 cos f, The intensity will maximum, then f = 0, or cos f = 1, maximum intensity., \ I max = I1 + I2 + 2 I1 I2, I1 )2, , intensities I 0 interfere,, , refractive index m, the path difference, = 2d sin q + ( m - 1 ) L ., For the principal maxima, (path, difference is zero), i.e. 2 d sin q 0 + ( m - 1 ) L = 0, L( m - 1 ), - L (0.5), =, 2d, 2d, [\ L = d /4 ], , -D, 16, For the first minima, the path difference, l, is ±, 2, l, 2 d sin q1 + 0.5 L = ±, \, 2, or, , above point O., , In case, n identical waves each of, , 3 In case of transparent glass slab of, , -1, 16, , 3D, 247, , = ( I1 +, , 2, , b, 3l, b, or l =, =, 3d, 2d, 2, , or sin q 0 =, , =, 2, , The first principal minima on the, negative side is at distance, 5, 5, below, D tan q¢¢1 =, =, 231, 162 - 52, , 2, , b2, l, b2, = or l =, 2d, 2, d, , or sin q 0 = -, , 3, 2, , 16 - 3, , 2, , 3ö, ÷, 8ø, , or (S 2 P - S 1 P ) (S 2 P + S 1 P ) = b 2, b2, or, (S 2 P - S 1 P ) =, 2d, For dark fringes,, b2, l, = ( 2n + 1 ), 2d, 2, , 2, , [Q The diffraction occurs if the, wavelength of waves is nearly, equal to the slit width (d)], On the positive side, 3, 1 1, sin q¢1 = + =, 4 16 16, On the negative side, 5, 1 1, sin q¢¢1 = - =16, 4 16, The first principal maxima on the, positive side is at distance, sin q1¢, D tan q1¢ = D, 1 - sin2 q¢1, =D, , From the figure,, (S2 P )2 - (S 1 P )2 = b 2, , For n = 0,, , 25 For interference to occur, the path, , 1, 1, =, m, 3, , 2, , 3I 0 ö, ÷, 8 ø, , 2 Path difference = S2 P - S1 P ., 45°, , and sinC =, , 3I 0 ö, ÷, 8 ø, , I max, =, I min, æ I0, ç, è 4, , Since, the angle between i p and r is 90°, when the ray is incident at polarising, angle, then, , m = tan i p = tan 60° = 3, , 2, , æ I0, +, ç, è 4, , m = tan q p, 1.732 = tan q p, q p = tan -1 (1.732) = 60°, , I0, , ± l / 2 - 0.5 L, 2d, ± l /2 - d / 8, =, 2d, ± l /2 - l / 8, 1 1, =, =± 2l, 4 16, , or sin q1 =, , 1 The intensity of light reflected from, , Finally, , IA, , Þ, , DAY TWENTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , OP = D tan q 0 = D sin q 0 =, , I max = ( I 0 +, = (n, , I 0 )2, , I0 +, , I0 + K + n, times)2, …(i), , \ I max = n2 I 0, , …(ii), , When interference is incoherent, Since, the average value of cos f,, over a complete cycle is zero, The Eq. (i), becomes,, I = I1 + I2 + 2 I1 I2 ´ 0, …(iii), = I1 + I2, In case, n identical waves, each of, intensities I 0 interfere,, Minimum intensity,, I min = I 0 + I 0 + I 0 + K n times, …(iv), I min = nI 0, I, n2 I 0, Ratio max =, =n, \, I min, nI 0
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WAVE OPTICS, , DAY TWENTY NINE, Next 11th minima of 400 nm will, coincide with 8th minima of 560 nm., , 5 Given, a1 = 2a2, I1 = 4 I 2 = 4 I 0, , Þ, \, , I m = ( I1 +, , 2, , Location of this minima is,, , 2, , I2 ) = (3 I2 ), , = 9 I2 = 9 I 0 = I 0 =, , Im, 9, , Y2 =, , = 42 mm, , Now, resultant intensity,, , \ Required distance = Y2 - Y1 = 28 mm, , I = I1 + I2 + 2 I1 I2 cos f, = 4 I 0 + I 0 + 2 4 I 0 I 0 cos f, , 8 For possible interference maxima of the, screen, the condition is, d sin q = nl, , = 5I 0 + 4 I 0 cos f, I, = m (5 + 4 cos f), 9, Im, =, [1 + 4(1 + cos f)], 9, I, = m (1 + 8 cos 2 f / 2), 9, , Given,, , plate will shift the image on the film., 1 1 1, As,, = f, v u, 1 1 1, Þ, = u v, f, 3, 21, =, 35 240, 1 48 ´ 3 - 7 ´ 21, 1, =, =u, 1680, 560, Þ, , ...(i), , or, , 2 sin q = n, , Thus, Eq. (i) must be satisfied by 5, integer values, i.e. - 2, - 1, 0, 1, 2. Hence,, the maximum number of possible, interference maxima is 5., A, , 9, , C, I/2, , Using Malus’s law, intensity available, I, after C = ´ cos 2 a, 2, and intensity available after, I, B = cos 2 a ´ cos 2 b, 2, I, = (given), 8, I, I, So, ´ cos 2 a × cos 2 b =, 2, 8, 1, cos 2 a × cos 2 b =, Þ, 4, This is satisfied with a = 45°, and b = 45°, So, angle between A and C is 45°., , 10 Angular width of diffraction pattern = 60°, , = 14 mm, , d = 25 mm, , 11 In diffraction, first minima, we have, sin q =, , l, a, L λ/a, L, , a, , So, size of a spot,, 2 Ll, …(i), b = 2a +, a, Then, minimum size of a spot, we get, ¶b, Ll, = 0 Þ 1- 2 = 0, ¶a, a, …(ii), Þ, a = lL, So, b min = 2 lL + 2 lL, [by substituting the value of a from, Eq. (ii) in Eq. (i)], = 4 lL, So, the radius of the spot,, 4, b min =, lL = 4lL, 2, , 12 From YDSE,, 30°, , i.e. 4th minima of 400 nm coincides, with 3rd minima of 560 nm., Location of this minima is,, (2 ´ 4 - 1)(1000)(400 ´ 10-6 ), 2 ´ 01, ., , or, , β, , 7 Let nth minima of 400 nm coincides, , Y1 =, , d =, , B, , u = - 5.6 m, , with mth minima of 560 nm, then, 400 ö, æ 560 ö, (2n - 1) æç, ÷ = (2m - 1) ç, ÷, è 2 ø, è 2 ø, 2n - 1 7 14, or, = =, =K, 2m - 1 5 10, , 50, m], 100, , 10-6 ´ 50, lD, =, 1, b, 2´, ´ 100, 100, -6, = 25 ´ 10 m, , So,, , I, 8, , I, α, , 10-6, 1, m, b = 1 cm =, m,, 2, 100, , d = ? and D = 50 cm =, , n=2´1=2, , 6 Shift in image position due to glass, , For focal length of the lens,, 20 + 1, 1 1 1, 1, 1, = - =, =, f, v u 12 -240, 240, 240, cm, f =, Þ, 21, Now, to get back image on the film, lens, has to form image at, æ12 - 1 ö cm = 35 cm such that the glass, ç, ÷, è, 3ø, 3, , 2l sin q = nl, , l=, , [here, l =, , d = slit width = 2l, , \, , l = 10-6 ´ sin 30°, , 10-6, m, 2, Now, in case of interference caused by, bringing second slit,, \ Fringe width,, lD, b=, d, Þ, , The maximum value of sin q is 1, hence,, , é(1 + cos q) = 2cos 2 f ù, êë, 2 úû, plate,, 1ö, 1, 1 ö, æ, S = ç1 - ÷ t = æç1 ÷ ´ 1 cm = cm, è, è, mø, 3, 1.5ø, , (2 ´ 11 - 1)(1000)(400 ´ 10-6 ), 2 ´ 01, ., , Þ, , 333, , For first minima,, a, l, sin q = ,, 2, 2, [here, a = 10-6 m, q = 30°], , xd, D, So, as x increases, D x also, increases., Dx, (B) Fringe width (b ) =, d, independent of D x., (C) Fringe pattern will shift, downward., (A) Path difference D x =, , (D) b is constant, so number of, fringes unaffected.
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334, , DAY THIRTY, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY THIRTY, , Unit Test 6, (Optics), 1 Minimum light intensity that can be perceived by normal, , human eye is about10−10 W m −2. What is the minimum, number of photons of wavelength 660 nm that must enter, the pupil in one second, for one to see the object? Area, of cross-section of the pupil is10−4 m 2., (a) 3.3 × 10 2 (b) 3.3 × 10 3, , (c) 3.3 × 10 4, , (d) 3.3 × 10 5, , 2 A beam of light converges to a point P. A lens is placed, in path of light 1.2 cm from P. If focal length of lens is +20, cm, then image distance from lens is, (a) 4.8 cm, , (b) 20 cm, , (c) 7.5 cm, , (d) 5.2 cm, , length 100 cm. The value of µl should be, (a) 3/2, , (b) 4/3, , (c) 5/3, , (d) 2, , 8 A thin convex lens of crown glass having refractive index, 1.5 has power 1 D. What will be the power of similar, convex lens of refractive index1. 6 ?, (a) 0.6 D, , (b) 0.8 D, , (c) 1.2 D, , (d) 1.6 D, , 9 A short pulse of white light is incident from air to a glass, slab at normal incidence. After travelling through the slab,, the first colour to emerge is, (a) blue, , (b) green, , (c) violet, , (d) red, , 3 A ray of light incident at an angle θ on a refracting face of, , 10 An object approaches a converging lens from the left of, , a prism emerges from the other face normally. If the, angle of the prism is 5° and the prism is made of a, material of refractive index 1.5, the angle of incidence is, , the lens with a uniform speed 5 m/s and stops at the, focus. The image, , (a) sin−1 (013, . ) (b) sin−1 (052, . ) (c) sin−1 (017, . ) (d) sin−1 (0.86), , 4 Two coherent point sources S1 and S 2 vibrating in phase, emit light of wavelength λ. The separation between the, sources is 2λ. The smallest distance from S 2 on a line, passing through S 2 and perpendicular to S1S 2, where a, minimum of intensity occurs is, (a), , 7λ, 12, , (b), , 15 λ, 4, , (c), , λ, 2, , (d), , 3λ, 4, , 5 A thin glass prism µ = 1.5 is immersed in water µ = 1.3. If, the angle of deviation in air for particular ray be D, then, that in water will be, (a) 0.2 D, , (b) 0.3 D, , (c) 0.5 D, , (d) 0.6 D, , 6 The refractive index of the material of equilateral prism is, 3. The angle of minimum deviation for the prism is, (a) 30°, , (b) 41°, , (c) 49°, , (d) 60°, , 7 A thin convergent glass lens µ = 1. 5 has a power of, + 5.0 D. When this lens is immersed in a liquid of, refractive index µl , it acts as a diverging lens of focal, , (a) moves away from the lens with uniform speed 5 m/s, (b) moves away from the lens with uniform acceleration, (c) moves away from the lens with a non-uniform, acceleration, (d) moves towards the lens with a non-uniform acceleration, , 11 A passenger in an aeroplane shall, (a) never see a rainbow, (b) may see a primary and a secondary rainbow as, concentric circles, (c) may see a primary and a secondary rainbow as, concentric arcs, (d) shall never see a secondary rainbow, , 12 A narrow slit of width 1 mm is illuminated by, monochromatic light λ = 600 nm. The distance between, first minima on either side of center line of a screen, placed 2 m away is, (a) 1.2 cm, , (b) 1.2 mm, , (c) 2.4 mm, , (d) 2.4 cm, , 13 A myopic person having far point 80 cm uses spectacles, of power −1.0 D. How far can he see clearly?, (a) 1 m, (c) 4 m, , (b) 2 m, (d) More than 4 m
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UNIT TEST 6 (OPTICS), , DAY THIRTY, 14 The radius of curvature of the curved surface of a, plano-convex lens is 20 cm. If the refractive index of the, material of the lens be 1.5, it will, (a) act as a convex lens only for the objects that lie on its, curved side, (b) act as a concave lens for the objects that lie on its, curved side, (c) act as a convex lens irrespective of the side on which, the object lies, (d) act as a concave lens irrespective of the side on which, the object lies, , (a) at infinity, (b) at the mid-point of distance between two mirrors, (c) on the pole of concave mirror, (d) on the pole of convex mirror, , 20 A cube of side 3 m is placed in front of a concave mirror, of focal length 2 m with its face P at a distance 4 m and, face Q at a distance 7 m from the mirror. What is distance, between the images of face P and Q ?, 3m, , 4m, , 15 Monochromatic light of wavelength 800 nm is used in, double slit experiment. One of the slit is covered with a, transparent slab of thickness 2.4 × 10−5 m. The refractive, index of the material of slab is 1.4. What is the number of, fringes that will shift due to introduction of the sheet?, (a) 14, , (b) 12, , (c) 16, , 16 Two coherent sources are, , S2, , (d) 2.2 m, , The ratio of the magnitude of the power of the convex, lens to that of the concave lens is 4 : 3. If the focal length, of the convex lens is 12 cm, the focal length of the, combination will be, , 4.4 mm, , (a) 16 cm, , (b) equal to 1.33, (d) greater than 1.33, , 18 Angular width of central maximum in the Fraunhoffer’s, diffraction pattern is measured. Slit is illuminated by the, light of wavelength 6000 Å. If slit is illuminated by light of, another wavelength, angular width decreases by 30%., Wavelength of light used is, (c) 4700 Å, , (c) 2.1 m, , (b) 24 cm, , (c) 32 cm, , (d) 48 cm, , 23 The radius of curvature of a thin plano-convex lens is, , diverging lens in water. The refractive index of the, material of the lens is, , (b) 4200 Å, , (b) 2.4 m, , 21 A diminished image of an object is to be obtained on a, , 22 A convex lens and a concave lens are placed in contact., O, , 17 A lens behaves as a converging lens in air and a, , (a) 3500 Å, , (a) 1.2 m, , Q, , (a) a concave mirror of suitable focal length, (b) a convex mirror of suitable focal length, (c) a concave lens of suitable focal length, (d) a convex lens of focal length less than 0.25 m, , (a) constructive only, (b) destructive only, (c) cannot be predicted, (d) may be constructive or destructive, , (a) equal to unity, (c) between unity and 1.33, , P, , screen 1 m from it. This can be achieved by, approximately placing., , (d) 10, , 4.4 mm apart and 4.4 m from, the screen as shown in the, figure. If the sources emit, light of wavelength 440 nm, which produce an, S1, 4.4 m, interference pattern on the, screen. The pattern of the interference at point O is, , 335, , (d) 6000 Å, , 19 A convex mirror and a concave mirror of radius 10 m, each are placed 15 m apart facing each other. An object, is placed mid-way between them. If the reflection first, take place in the concave mirror and then in another, mirror, the position of the final image is, , 10 cm and the refractive index of its glass is 1.5. If the, plane surface is silvered, then it will behave like a, (a) concave mirror of focal length 10 cm, (b) concave mirror of focal length 20 cm, (c) convex mirror of focal length 10 cm, (d) convex mirror of focal length 20 cm, , 24 When the plane surface of a plano-convex lens of, refractive index 1.5 is silvered, it behaves like a concave, mirror of focal length 30 cm. When its convex surface is, silvered, it will behave like a concave mirror of focal, length, (a) 10 cm, , (b) 20 cm, , (c) 30 cm, , (d) 45 cm, , 25 Two stars are situated at a distance of 8 light years from, the earth. Their images are just resolved by a telescope of, diameter 0. 25 m. If the wavelength of light from stars is, 5000 Å, then the distance between the stars is around, (a) 3 × 1010 m, (c) 195, . × 1011 m, , (b) 3. 35 × 1011 m, (d) 4.32 × 1010 m, , 26 The refractive index of air is 1.0003. The thickness of air, 15 cm, , column which will have one more wavelength of yellow, light (λ = 6000 Å), then in the same thickness in vacuum is, (a) 2 mm, , (b) 2 cm, , (c) 2 m, , (d) 2 km
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336, , DAY THIRTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 27 A thin symmetric convex lens of refractive index 1.5 and, radius of curvature 0.3 m is immersed in water of, refractive index 4/3. Its focal length in water is, (a) 0.15 m, , (b) 0.30 m, , (c) 0.60 m, , (d) 1.20 m, , 28 A parallel beam of sodium light of wavelength 6000 Å is, incident on a thin glass plate of µ = 1.5 , such that the, angle of incidence in the plate is 60°. The smallest, thickness of the plate which will make it appear dark by, reflected light is, (a) 1260 Å, , (b) 2440 Å, , (c) 3260 Å, , (d) 4000 Å, , 29 Two polaroids are oriented with their principal planes, making an angle of 60°. The percentage of incident, unpolarised light which passes through the system is, (a) 50%, , (b) 100%, , (c) 12.5%, , (d) 37.5%, , 30 In the visible region of the spectrum the rotation of the, plane of polarisation is given by θ = a +, , b, ., λ2, , The optical rotation produced by a particular material is, found to be 30° per mm at λ = 5000 Å and 50° per mm at, λ = 4000 Å. The value of constant a will be, 50°, per mm, 9, 9°, (c) +, per mm, 50, (a) +, , 50°, per mm, 9, 9°, (d) −, per mm, 50, , (b) −, , 31 The refracting angle of a prism is A and the refractive, index of the prism is cot ( A / 2). The angle of minimum, deviation is, (a) 180°−3 A, , (b) 180° + 2 A (c) 90° − A, , (d) 180°−2 A, , 32 Cross-section of a glass prism is an isosceles triangle., One of refracting faces is silvered. A ray of light falls, normally on the other refracting face. After being, reflected twice, it emerges through the base of the prism, perpendicular to it. The angles of prism are, (a) 54°, 54°, 72°, (c) 45°, 45°, 90°, , (b) 72°, 72°, 36°, (d) 57°, 57°, 76°, , 33 A spherical surface of radius of curvature R, separates, air and glass (n air = 1.0, n glass = 1.5). The centre of, curvature is in glass. A point object P placed in air is, found to have a real image in the glass. The line PQ cuts, the surface at a point O such that PO = OQ ., Distance PQ is, (a) 5R, , (b) 3R, , (c) 2R, , (d) R, , 34 Polarising angle for water is 53°4'. If light is incident at, this angle on the surface of water and reflected, the angle, of refraction is, (a) 53 ° 4 ′, (c) 36° 56′, , (b) 126° 56′, (d) 30° 4 ′, , 35 The distance between the first and the sixth minima in the, diffraction pattern of a single slit is 0.5 mm. The screen is, 0.5 m away from the slit. If the wavelength of light used is, 5000 Å, then the slit width will be, (a) 5 mm, (c) 1.25 mm, , (b) 2.5 mm, (d) 1.0 mm, , Direction, , (Q. Nos. 36-40) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 36 Statement I Angle of deviation depends on the angle of, prism., Statement II For thin prism δ = (µ − 1) A., where, δ = angle of deviation, µ = refractive index,, A = angle of prism., , 37 Statement I Glass is transparent but its powder seems, opaque. When water is poured over it, it becomes, transparent., Statement II Light gets refracted through water., , 38 Statement I If convex lens is kept in water its convergent, power decreases., Statement II Focal length of convex lens in water, increases., , 39 Statement I Danger signals are made of red colours., Statement II Velocity of red light is maximum and thus,, more visibility in dark., , 40 Statement I The clouds in sky generally appear to be, whitish., Statement II Diffraction due to clouds is efficient in equal, measure for all wavelengths.
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UNIT TEST 6 (OPTICS), , DAY THIRTY, , 337, , ANSWERS, 1., 11., 21., 31., , (c), (b), (d), (d), , 2., 12., 22., 32., , (c), (c), (d), (b), , 3., 13., 23., 33., , (a), (c), (a), (a), , 4., 14., 24., 34., , (a), (c), (a), (c), , 5., 15., 25., 35., , (b), (b), (c), (b), , 6., 16., 26., 36., , (d), (a), (a), (a), , 7., 17., 27., 37., , (c), (c), (d), (a), , (c), (b), (b), (a), , 8., 18., 28., 38., , 9., 19., 29., 39., , (d), (d), (c), (c), , 10., 20., 30., 40., , (c), (a), (b), (c), , Hints and Explanations, 1 I = 10−10 Wm −2 = 10−10Js −1 m2 . Let the, number of photons required per second, be n., nhv, Then,, = 10−10, 10−4, Hence, n = 10−10 × 10−4 /hv, λ, = 10−14, hc, 10−14 × 660 × 10−9, =, 6.6 × 10−34 × 3 × 108, = 3.3 × 104, , 2 1=1+1, v, , f, , or, , r1, , S1, 2λ, S2, Solving this equation, we get, 7λ, x=, 12, 5 δ~, − (µ − 1) A, , For air,, , D = (1.5 − 1) A, , 1.5 , For water, δ = ( g µ w − 1) A = , − 1 A, 13, , ., Hence,, , N2, , 6, , 90°, , µ =, , δ=, , 02, ., D, ×, ≅ 0.3 D, 13, ., 0.5, , A + δm , , 2 , A, sin , 2, , sin , , , A + δm , sin , , , 2 , ⇒ 3=, ,, sin 30°, because, A = 60°, A + δm, 3, or sin, =, = sin 60°, 2, 2, or, A + δ m = 2 × 60 = 120, This gives δ m = 60°., , As the ray emerges from the, other face of prism normally,, i2 = 0°, ∴, r2 = 0°,, As, r1 + r2 = A, r1 = A − r2 = 5 − 0 = 5°, sin i1, From µ =, ,, sin r1, , where, δ m is minimum deviation., , 7 When the lens is in air, we have, , sin i1 = µ sin r1, sin i1 = 1.5 × sin 5°, = 1.5 × 0.087, sin i1 = 01305, ., i1 = sin −1 (01305, ., ), , Pa =, , 4 Path difference at S2 is 2λ. Therefore,, for minimum intensity at P., 3λ λ, ≠, S1 P − S2 P =, 2, 2, , Hence,, , …(i), , Hence,, , 1.6 − 1, P2 µ 2 − 1 , P, =, i.e. 2 =, 1.5 − 1, 1, P1 µ 1 − 1 , P2 = 1.2 D, , 9 In air, all the colours of light travel with, , 3 Here, A = 5°, i1 = ?, 5°, , 1, 1 , P = (µ − 1) , −, , R, R, 2, 1, , 8, , x, , Here, f = + 20 and u = +12, ∴, v = 7.5 cm, , N1, , 3λ, 2, P, , u, , i1, , 4λ2 + x2 − x =, , µg −µa 1, 1 , 1, =, −, , , fa, µ a R1 R2 , , When the lens is in liquid, we have, 1 µg −µl 1, 1 , Pl =, =, −, , , fl, µl, R, R, 2, 1, Here,, Pa = 5, P l = −1,µ a = 1,µ g = 1.5, On solving, we get, 5, µl =, 3, , the same velocity, but in glass, velocities, of different colours are different. Velocity, of red colour is largest and velocity of, violet colour is smallest. Therefore, after, travelling through the glass slab, red, colour will emerge first., , 10 When an object approaches a convergent, lens from the left of the lens with a, uniform speed of 5 m/s, the image moves, away from the lens with a non-uniform, acceleration. For example, f = 20 m and, u = − 50 m; − 45m, − 40 m and − 35m; we, get v = 333, . m; 36 m; 40 m and 46.7 m., Clearly, image moves away from the lens, with a non-uniform acceleration. Option, (c) is correct., , 11 In an aeroplane, a passenger may observe, a primary and a secondary rainbow as, concentric circles., , 12 For first dark fringe on either side,, d sinθ = λ, y, sinθ =, D, dy, λD, So,, = λ or y =, d, D, Distance between two minima = 2y, 2 × 600 × 10−6 × 2 × 103, mm, =, 1.0, = 2.4 mm, 1 1 1, 13 Use − =, v u, f, and, , v = −80 cm, f = −100 cm, 1, 1, 1, Hence,, − =−, − 80 u, 100, − 80 + 100, 1, 1, 1, or − = −, +, =, u, 100 80, 80 × 100, Here,, , This gives u = − 400 cm = − 4 m
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338, , DAY THIRTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 Here, µ = 1.5, , 17 The focal length f of the lens in air is, , If object lies on plane side;, R1 = ∞, R2 = −20 cm, 1, 1, 1 , = (µ − 1), −, , f, R1 R2 , 1, 1, 1, = (1.5 − 1) +, =, ∞ 20 40, f = + 40 cm., The lens behaves as convex., If object lies on its curved side,, R1 = 20 cm, R2 = ∞,, 1, 1, 1 , = (µ − 1), −, , f′, R1 R2 , 1, 1, 1, = (1.5 − 1), − =, 20 ∞ 40, f ′ = 40 cm, The lens behaves as convex., 15 The total fringe shift is H = β (µ − 1) t, λ, The number of fringes that will shift, total fringe shift, =, fringe width, β, (µ − 1)t, (µ − 1) t, or, n= λ, =, β, λ, or, , n=, , or, , n=, , (1.4 − 1) × 2.4 × 10−5, 800 × 10−9, 0.4 × 2.4 × 10−5, 8 × 10−7, , or n = 12, , 16 The path difference at point of, , 4.4 mm, , observation is given by, ∆ = S 2O − S 1O, S2, , given by, 1, 1, 1 , = (µ g − 1) −, , f, R1 R2, where, µ g is the refractive index of the, lens. If µ w is the refractive index of, water, the focal length in water ( f ′ ) is, given by, 1 µg − µw 1, 1 , =, −, , , f ′ µ w R1 R2 , Since, the lens placed in air is, convergent, f is positive. Therefore,, µ g > 1. For the lens to be divergent, when placed in water, f ′ must be, negative, i.e. µ g < µ w . Now, µ w = 1.33., Hence, µ g must lies between 1 and, 1.33., , 18 The condition for minima is given by, d sinθ = n λ, For n = 1, we have, d sinθ = λ, If angle is small, then sinθ = θ ⇒ dθ = λ, λ, Half angular width θ =, d, λ, Full angular width, 2θ = 2, d, 2λ ′, Also, ω ′ =, d, λ′ ω ′, =, ∴, λ, ω, ω′, or, λ ′ = λ = 6000 × 07, . = 4200 Å, ω, , 19 For reflection from concave mirror, Applying,, , or, , S1, , O, 4.4 m, From the figure,, S 2O = [(4.4) 2 + (4.4 × 10−3 ) 2 ]1 /2, −3 2 1 /2, , or S 2O = 4.4[1 + (10 ) ], , 1, or S 2O = 4.4 1 + (10−3 ) 2 , , , 2, 1, Therefore, ∆ = 4.4 1 + (10−3 ) 2 − 1, , , 2, −6, , 4.4 × 10, = 2.2 × 10−6 m, 2, Interference will be constructive, if path, difference is an integral multiple of, wavelength., 2.2 × 10−6, ∆, n= =, =5, λ, 440 × 10−9, =, , Hence, pattern of interference at point O, is constructive., , 1 1 1, = + , we get, f, v u, 1, 1, 1, = −, −5 v 7.5, 1, 1, 1, or v = −15m, =− +, v, 5 7.5, , The image is formed on the pole of the, convex mirror, which will be the, position of the object for convex mirror., Therefore,, u = 0 and f = 5m, Hence,, or, , 1 1 1 1, = − = − ∞ =∞, v, 5 0 5, v =0, , Therefore, final image is formed on the, pole of convex mirror., , 14, = 2.8 m, 5, Therefore, v 1 − v 2 = (4 − 2.8) = 1.2 m, , or, , v2 =, , 21 Convex mirror and concave lens do not, form real image. For concave mirror, v > u, so image will be enlarged, hence, only convex lens can be used for the, purpose., |P |, 22 Given, f1 = + 12 cm and 1 = 4, |P2 | 3, f2, 4, Since f2 is negative,, =−, 3, f1, 4, 4, Hence, f2 = − f1 = − × 12, 3, 3, = −16 cm, The focal length F of the combination is, given by, 1, 1, 1, 1, 1, 1, =, +, =, +, =, F, f1, f2 12 −16 48, F = 48 cm, , 23 When the plane surface of a, plano-convex lens is silvered, it behaves, like a concave mirror of focal length f, given by, 1 2( µ − 1), =, f, R, R, 10, or, f =, =, = 10 cm, 2(µ − 1) 2(1.5 − 1), , 24 When the plane surface is silvered the, focal length f1 is given by, 2( µ − 1), 1, =, f1, R, , But when the convex surface is silvered,, the focal length f2 is given by, 2µ, 1, ... (ii), =, f2, R, Dividing Eq. (i) by Eq. (ii), we have, f1, 1.5, µ, =, =3, =, f2 µ − 1 1.5 − 1, f, 30, or, f2 = 1 =, = 10 cm, 3, 3, , 25 Limit of resolution of the telescope, , 1.22λ d, =, a, x, 1.22λ x, d =, a, 1.22 × 5 × 10−7 × 8 × 1016, =, 0.25, , α =, or, , = 1. 95 × 1011 m, , 20 For surface P, we have, 1, 1 1, = −, v1, f u, 1 1 1, or v 1 = 4 m, = − =, 2 4 4, For surface Q, we have, 1 1, 5, v2 = − =, 2 7 14, , ... (i), , 26 Let d in cm be the thickness of air column, = thickness of vacuum column (given)., Number of waves of wavelength λ = 6000, Å = 6 × 10−5 cm in a thickness d cm in, vacuum is, nv =, , d, λ
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UNIT TEST 6 (OPTICS), , DAY THIRTY, Since, the refractive index of air, µ = 1.0003, the wavelength in air will be, λ, λa =, µ, Therefore, number of waves of, wavelength λ a in d cm of air is, d dµ, na =, =, λa λ, , θ = a+, , 30, , 50 = a +, , µ − µw 1, 1 , 27 1 = g, −, , , f, R2 , µw, R1, , 1 (1.5 − 4 / 3) 1, 1 , =, +, 0.3 0.3 , f, 4/3, , which gives f = 1. 20 m, 1, E, , cos ( A / 2), sin( A / 2), , 2, C, , r, , 90 – θ, , sin r =, , 3/2, sin 60°, =, µ, 1.5, , ∴, , r = 30°, AB = t sec r = 1.15t, AC = 2( AD ) = 2(t tan r ) = 115, . t, AE = AC cos 30° = 0.99 t, Now, net path difference between 1, and 2,, ∆ X = µ (2 AB ) − AE, = (1.5)(23, . t ) − 0.99t = 2.45 t, , θ, , θ, , B, , In ∆abc,, A + 90°+90°− i = 180°, ⇒, i = A, Also, angle θ = 2i = 2 A, This condition is satisfied by option (b), only., 33 For given condition,, , transparent. But when the glass is, powdered, the irregular reflections occur, from the surface of powdered glass and, finally the light returns back into the, same medium. Because of it the, powdered glass looks opaque. When we, pour water over the powdered glass,, refraction of light takes place and it, becomes transparent., , Here, µ g is the refractive index of lens, (glass)., The focal length of lens in air is given by, 1, 1, 1 , = (µ g − 1) , −, , f, R, R, 1, 2, , ...(ii), , From Eqs. (i) and (ii), we conclude the, focal length of lens in water ( f w ) is, greater than focal length of lens in air, ( f a ). Therefore, the focal length of lens in, water get increased consequently power, decreases., , 39 Red colour consists of longest, , For minimum intensity,, ∆X = λ, or (2.45t ) = 6000, or, t = 2448 Å ≈ 2440 Å, , P, , µ1, +, −u, 1, +, − (− x ), , 29 Intensity of polarised light from first, 100, = 50, 2, From law of Malus intensity from, second nicol, I = 50 cos 2 60°, 50, =, = 12.5 %, 4, , 37 We know very well that, glass is, , water of refractive index µ w is given by, µg − µw 1, 1, 1 , ...(i), =, −, , , f w µ w R1 R2 , , i, r, , for a thin prism an angle of prism and, refractive index of material of prism is, given by δ = ( µ − 1 ) A, , 38 The focal length f w of convex lens in, , i 90 – 2i, , d, , = 2.5 × 10−3 m, = 2.5 mm, , 36 The relation between angle of deviation δ, , 30° D, , A, t, , (4000)2, , A + δm , A, or sin , = cos , 2, , , 2, A, , = sin 90° − , , 2, A + δm, A, = 90° −, ∴, 2, 2, or, δ m = 180° − 2 A, a, 32, 90 – i, b, c, i, i, , For a symmetric lens, R2 = − R1 = − 0.3 m, , 60° 60°, , (5000)2, b, , A + δm , sin , , , , A, 2, 31 Given, µ =, = cot , 2, A, , sin , 2, =, , ad 1, D, For the sixth minimum,, ad 6, a sinθ 6 = 6λ = aθ 6 =, D, a, (6λ − λ ) = (d 6 − d 1 ), ∴, D, 5 Dλ, or a =, (d 6 − d 1 ), 5 × 0.5 × 5 × 10−7, =, 0.5 × 10−3, a sinθ1 = λ ≈ aθ1 =, , b, , Solving for a, we get, 50°, per mm, a= −, 9, , = 0.2 cm = 2 mm, , 28, , 35 For the first minimum,, , λ2, , 30 = a +, and, , Given that, n a − 1 = n v, dµ, d, Hence,, −1=, λ, λ, λ, 6 ×10−5cm, d =, =, µ − 1 1.0003 − 1, , Therefore,, , b, , 339, , polariser =, , ⇒, , O, , Q, , µ2 µ2 − µ1, =, v, R, 1.5 1.5 − 1, =, x, R, x = 5R, , 34 i p + r = 90°, or, , r = 90° − i p = 90°−53° 4′ = 36° 56′, , wavelength and it scatteres least., Therefore, signals of red colour are being, seen from the long distances. Hence,, the signals are made of red colour light., , 40 The clouds consist of dust particles and, water droplets. The scattering of sun, light by the big dust particles and water, drops is not in accordance with the, Rayleigh law. But they scatter the light of, all the colours by the same amount., Hence, the clouds are seen generally, white.
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DAY THIRTY ONE, , Dual Nature, of Matter, Learning & Revision for the Day, u, , u, , u, , Photon, Particle Nature of Light, Photoelectric Effect, , u, , u, , u, , Laws of Photoelectric Effect, de-Broglie Waves, Davisson-Germer Experiment, , Photon, A particle of light called a photon has energy E that is related to the frequency f and, wavelength λ of light wave., hc, By the Einstein equation, E = hf =, …(i), λ, where, c is the speed of light (in vacuum) and h is Planck’s constant., h = 6.626 × 10 −34 J-s = 4136, ., × 10 −15 eV-s, Since, energies are often given in electron volt (1eV = 1.6 × 10 −19 J) and wavelengths are, in Å, it is convenient to the combination hc in eV-Å. We have,, hc = 12375 eV-Å, Hence, Eq. (i), in simpler form can be written as,, 12375, …(ii), E ( in eV) =, λ ( in Å), The propagation of light is governed by its wave porperties whereas the exchange of, energy between, light with matter is governed by its particle properties. The, wave-particle duality is a general property in nature. For example, electrons (and other, so called particles) also propagate as waves and exchange energy as particles., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, , Particle Nature of Light, Photoelectric effect gave evidence to the strange fact that, light in interaction with, matter behaved as if it was made of quanta or packets of energy, each of energy hν., hν , Einstein stated that the light quantum can also be associated with momentum ⋅, c , This particle like behaviour of light was further confirmed, in 1924, by the Compton, experiment of scattering of X-rays from electrons., , No. of Questions in Exercises (x)—, , u, , (Without referring Explanations), u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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DAY THIRTY ONE, , DUAL NATURE OF MATTER, , 341, , Photoelectric Effect, l, , l, , l, , l, , l, , l, , Photoelectric current, , Photoelectric effect is the phenomenon of emission of, electrons (known as photoelectrons) from the surface of, metals when light radiation of suitable frequency are, incident on them., , ν3 > ν2 > ν1, ν3, , The minimum energy of incident radiation needed to eject, the electrons from metal surface is known as work function, (φ0) of that surface., The frequency or wavelength corresponding to the work, function is called threshold frequency or threshold, wavelength. Work function is related to threshold, frequency as,, hc, φ 0 = hν 0 =, λ0, , – V03 – V02 – V01, , V0, , O, , ν, , Cut-off voltage versus frequency of incident light, kinetic, , energy, , of, , Effect of Intensity on, Photoelectric Emission, For a light of given frequency ν > ν 0 (or given wavelength, λ < λ 0), if the intensity of light incident on photosensitive metal, surface is increased, the number of photoelectrons and, consequently the photoelectric current I increases. However, the, stopping potential V0 remain constant., Photoelectric, current, , ν0, , – φ0, , hν = φ0 + K max, , O, –V0, Stopping potential, , Collector plate, potential, , A photon may collide with a material particle. The total, energy and the total momentum remain conserved in such a, collision. Photoelectric emission is an instantaneous, phenomenon., , According to Einstein’s photoelectric equation,, , Stopping, potential, , 0, , I-V0, when frequency changed, , For photoemission to take place energy of incident light (E ), is related as,, E ≥ p0, , 1, mv2max = maximum, 2, ejected photoelectron., , ν1, , Stopping potential, , where, λ 0 = threshold wavelength., hc 12400, In electron volt units, φ (eV) =, =, eλ 0, λ (Å), , where, K max =, , ν2, , Saturation, current, , Variation of stopping potential V0 with frequency ν of, incident radiation is as shown in above figure., φ, h, As, eV0 = h(ν − ν 0) = hν − φ0 ⇒ V0 = ν − 0, e, e, h, Thus, V0- ν graph is a straight line whose slope is and, e, intercept is − φ0 eV. The graph meets the ν-axis at ν 0., 1, Photocurrent ∝ ∝ λ, ν, , Energy and Momentum of Photon, l, , I 1 > I2 > I3, , Kinetic mass of photon is m =, , I1, I2, I3, , c, , where λ is wavelength of the photon., λ, h c, h, ∴Kinetic mass of photon, m = 2 =, c λ cλ, hν, h, Kinetic mass of photon, m = 2 =, c, λ, c, , Collector plate, potential, , Photoelectric current versus stopping potential curve, l, , If keeping the intensity of incident light constant, the, frequency of incident light is increased, then the stopping, potential V0 (and hence, K max ) increases, but the photoelectric, current I remains unchanged., , hν, c2, , But ν =, , Saturation, current, , Effect of Frequency on, Photoelectric Emission, , From Einstein’s mass-energy relation E = hν = mc2, , Momentum of photon,, p = kinetic mass of photon × velocity of photon, hν, hν, = 2 ×c =, c, c, c, Also, ν =, λ, hc h, ∴Momentum of photon, p = =, c λ λ
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DAY THIRTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , Laws of Photoelectric Effect, , l, , Lenard and Millikan gave the following laws on the basis of, experiments on photoelectric effect., l, , l, , l, , l, , The rate of emission of photoelectrons from the surface of a, metal varies directly as the intensity of the incident light, falling on the surface., The maximum kinetic energy of the emitted photoelectrons, is independent of the intensity of the incident light., The maximum kinetic energy of the photoelectrons, increases linearly with increase in the frequency of the, incident light., As soon as, the light is incident on the surface of the metal,, the photoelectrons are emitted instantly, i.e. there is no, time lag between incidence of light and emission of, electrons (≈ 10 −9 s)., , λp, λe, , l, , l, , l, , l, , l, , de-Broglie Relation, According to quantum theory, energy of photon, …(i), E = hν, If mass of the photon is taken as m, then as per Einstein’s, equation, …(ii), E = mc2, From Eqs. (i) and (ii), we get, hν = mc2, c, h = mc2 ,, λ, where, λ = wavelength of photon, h, λ=, mc, de-Broglie asserted that the above equation is completely a, general function and applies to photon as well as all other, moving particles., h, h, =, mv, 2 mE, where, m is mass of particle and v is its velocity., λ=, , So,, , l, , l, , de-Broglie wavelength associated with charged particle, h, h, h, λ= =, =, p, 2 mE, 2 mqV, de-Broglie wavelength of a gas molecule, h, λ=, 3 mkT, where, T = absolute temperature, and, , k = Boltzmann’s constant = 1.38 × 10 −23 J / K, , =, , c, 2 mc2 K, 2 mK =, E, E2, , Davisson-Germer Experiment, , de-Broglie Waves, Light is said to have dual character, i.e. it behaves like matter, (particle) and wave both. Some properties like interference,, diffraction can be explained on the basis of wave nature of, light, while the phenomena like photoelectric effect, black, body radiation, etc. can be explained on the basis of particle, nature of light., In 1942, Louis de-Broglie explained that like light, matter also, show dual behaviour, there is a wave associated with moving, particle, known as matter waves or de-Broglie waves., , Ratio of wavelength of photon and electron The, wavelength of photon of energy E is given by, hc, while the wavelength of an electron of kinetic, λp =, E, h, energy K is given by λ c =, ⋅ Therefore for same, 2 mK, energy, the ratio, , l, , l, , The de-Broglie hypothesis was confirmed by, Davisson-Germer experiment. It is used to study the, scattering of electron from a solid or to verify the wave, nature of electron., A beam of electrons emitted by electron gun is made to fall, on nickel crystal cut along cubical axis at a particular, angle. Ni crystal behaves like a three dimensional, diffraction grating and it diffracts the electron beam, obtained from electron gun., The diffracted beam of electrons is received by the detector, which can be positioned at any angle by rotating it about, the point of incidence., The energy of the incident beam of electrons can also be, varied by changing the applied voltage to the electron gun., According to classical physics, the intensity of scattered, beam of electrons at all scattering angle will be same but, Davisson and Germer found that the intensity of scattered, beam of electrons was not same but different at different, angles of scattering., It is maximum for diffracting angle 50°, at 54 V potential difference., If the de-Broglie waves exist for, electrons, then these should be, diffracted as X-rays., Using the Bragg’s formula 2d sin θ = nλ ,, we can determine the wavelength of, these waves,, , Incident beam, , 342, , 54 V, , 50°, , Diffraction angle, versus intensity, where d = distance between, curve, diffracting planes,, 180 − φ, θ=, 2, = glancing angle for incident beam = Bragg’s angle., Clearly from figure, we have θ + φ + θ = 180 °, , D, , θ φ, θ d, , Reflection of electron (beam) by atoms
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DUAL NATURE OF MATTER, , DAY THIRTY ONE, , 343, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 If a source of power 4kW produces1020photons/second,, , Codes, , the radiation belong to a part of the spectrum called, , A, (a) 1, (c) 2, , (a) X-rays, (c) microwaves, , (b) ultraviolet rays, (d) γ-rays, , 2 What will be the number of photons emitted per second, by a 10 W sodium vapour lamp assuming that 90% of the, consumed energy is converted into light? (Wavelength of, sodium light is 590 nm and h = 6.63 × 10−34 J - s ), (a) 0. 267 × 1018, (c) 0. 267 × 10 20, , (b) 0. 267 × 1019, (d) 0. 267 × 1017, , 3 Two monochromatic beams A and B of equal intensity I,, hit a screen. The number of photons hitting the screen by, beam A is twice that by beam B. Then, what inference, can you make about their frequencies?, (a) νB = 2 νA, (c) νA = 2 νB, , (b) νB = νA, (d) νB > νA, , 4 The eye can detect 5 × 104 photons m −2s −1 of light of, wavelength 500 nm. The ear can hear intensity upto, 10−13 Wm −2. As a power detector, which is more, sensitive?, , B, 2, 1, , C, 3, 3, , A, (b) 3, (d) 2, , B, 2, 3, , C, 1, 1, , 8 When a point source of monochromatic light is at a, distance of 0.2 m form a photoelectric cell the cut-off, voltage and saturation current are 0.6 V and 18 mA, respectively. If the same source is placed 0.6 m away, from the photoelectric cell, then, (a) the stopping potential will be 0.2 V, (b) the stopping potential will be 0.6 V, (c) the saturation current will be 6 mA, (d) the saturation current will be 18 mA, , 9 The figure shows the variation of photocurrent (I) with, anode potential (Va) of a photosensitive surface for three, different radiations. Let la,lb and lc be the intensities and fa,, fb and fc the frequencies for the waves a,b and c, respectively, I, , (a) Sensitivity of eye is one-fifth of the ear, (b) Sensitivity of eye is five times that of the ear, (c) Both are equally sensitive, (d) Eye cannot be used as a power detector, , c, , a, , b, , 5 The threshold wavelength for photoelectric emission from, °, , a material is 5200A. Photoelectrons will be emitted when, this material is illuminated with monochromatic radiation, from a, (a) 50 W infrared lamp, (c) 50 W ultraviolet lamp, , (b) 1 W infrared lamp, (d) 1 W ultraviolet lamp, , 6 The wavelength of the photoelectric threshold for silver is, λ 0. The energy of the electron ejected from the surface of, silver by an incident light of wavelength λ ( λ < λ 0 ) will be, (a) hc (λ 0 − λ), (c), , h 1 1 , + −, , c λ λ0 , , hc, λ0 − λ, λ − λ, (d) hc 0, , λ0 λ , , (b), , 7 In photoelectric effect match the following column I with, column II., Column I, , Column II, , A., , If frequency of incident, light is increased, , 1., , Stopping potential may, increase, , B., , If intensity of incident light, is increased, , 2., , Stopping potential must, increase, , C., , If work function of metal is 3., increased, , Photo effect may stop, , Va, , O, , (a) fa = fb and Ia ≠ Ib, (c) fa = fb and Ia = Ic, , (b) fa = fc and Ia = Ic, (d) fb = fc and Ib = Ic, , 10 In a photoelectric effect measurement, the stopping, potential for a given metal is found to be V0 volt when, radiation of wavelength λ 0 is used. If radiation of, wavelength 2λ 0 is used with the same metal, then the, stopping potential (in volt) will be, (a), , V0, 2, , (c)V0 +, , (b) 2V0, hc, 2 eλ 0, , (d)V0 −, , hc, 2 eλ 0, , 11 In an experiment on photoelectric effect, a student plots, stopping potential V0 against reciprocal of the wavelength, λ of the incident light for two different metals A and B., These are as shown in the figure., Metal A, Vo, , 1/λ, , Metal B
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344, , DAY THIRTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , Looking at the graphs, you can most appropriately say, that, (a), (b), (c), (d), , Work function of metal B is greater than that of metal A, Work function of metal A is greater than that of metal B, Students data is not correct, None of the above, , 12 A copper ball of radius 1cm and work function 4.47 eV is, irradiated with ultraviolet radiation of wavelength 2500 Å., The effect of irradiation results in the emission of, electrons from the ball. Further the ball will acquire, charge and due to this there will be a finite value of the, potential on the ball. The charge acquired by the ball is, (a) 5.5 × 10−13 C, (c) 4.5 × 10−12 C, , (b) 7.5 × 10−13 C, (d) 2.5 × 10−11 C, , 13 Match List I (fundamental experiment) with List II (its, conclusion) and select the correct option from the, choices given below the list., List I, , List II, , A. Franck-Hertz experiment, , 1. Particle nature of light, , B. Photo-electric experiment, , 2. Discrete energy levels of atom, , C. Davisson-Germer experiment 3. Wave nature of electron, 4. Structure of atom, , A, (a) 1, (c) 2, , B, 4, 1, , C, 3, 3, , A, (b) 2, (d) 4, , B, 4, 3, , C, 3, 2, , 14 The anode voltage of a photocells kept fixed. The, wavelength λ of the light falling on the cathode is, gradually changed. The plate current I of photocell varies, as follows, I, , (a) 602 V, , (b) 50 V, , (c) 138 V, , (d) 812 V, , 18 The energy of photon is equal to the kinetic energy of a, proton.The energy of photon is E. Let λ 1 be the, de-Broglie wavelength, of the proton and λ 2 be the, wavelength of photon. The ratio λ 1 / λ 2 is proportional to, 1, , (a) E 0, (c) E −1, , (b) E 2, (d) E −2, , 19 An electron is moving with an initial velocity v = v 0i and is, in a magnetic field B = B0j . Then its de-Broglie, wavelength, (a) remains constant, (b) increase with time, (c) decrease with time, (d) increases and decreases periodically, , 20 Orbits of a particle moving in a circle are such that the, perimeter of the orbit equals an integer number of, de-Broglie wavelengths of the particle. For a charged, particle moving in a plane perpendicular to a magnetic, field, the radius of the nth orbital will therefore be, proportional to, (a) n 2, (c) n1/ 2, , (b) n, (d) n1/ 4, , Direction (Q. Nos. 21-27), , Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices,, only one of which is the correct answer. You have to select, one of the codes (a), (b), (c), (d) given below., , (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, , (b), λ, , λ, , I, , produce electrons of wavelength 0.50 Å is, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, , I, , (a), , 17 The voltage applied to an electron microscope to, , (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , I, , 21 Statement I As intensity of incident light (in photoelectric, (d), , (c), , effect) increases, the number of photoelectrons emitted, per unit time increases., , λ, , λ, , 15 de-Broglie wavelength of an electron accelerated by a, voltage of 50V is close to (e = 1.6 × 10−19 C,, me = 91, . × 10−31kg, h = 6.6 × 10−34 J-s)., °, , (a) 0.5 A, °, , (c) 2.4 A, , °, , (b) 1.7A, °, , (d) 1.2 A, , 16 Photons of an electromagnetic radiation has an energy, 11keV each. To which region of electromagnetic, spectrum does it belong?, (a) X-ray region, (c) Infrared region, , (b) Ultraviolet region, (d) Visible region, , Statement II More intensity of light means more energy, per unit area per unit time., , 22 Statement I The relative velocity of two photons travelling, in opposite directions is the velocity of light., Statement II The rest mass of photon is zero., , 23 Statement I Work function of copper is greater than the, work function of sodium but both have same value of, threshold frequency and threshold wavelength., Statement II The frequency is inversely proportional to, wavelength.
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DUAL NATURE OF MATTER, , DAY THIRTY ONE, 24 Statement I The de-Broglie wavelength of a molecules, varies inversely as the square root of temperature., Statement II The root mean square velocity of molecule, depends on the temperature., , 25 Statement I Davisson-Germer experiment established, , 345, , frequency incident on the surface is doubled, both the, Kmax and V0 are also doubled., Statement II The maximum kinetic energy and the, stopping potential of photoelectrons emitted from a surface, are linearly dependent on the frequency of incident light., , 27 Statement I When ultraviolet light is incident on a, , the wave nature of electrons., Statement II If electrons have wave nature, they can, interfere and show diffraction., , 26 Statement I A metallic surface is irradiated by a, monochromatic light of frequency ν > ν 0 (the threshold, frequency). The maximum kinetic energy and the, stopping potential are Kmax and V0, respectively. If the, , photocell, its stopping potential isV0 and the maximum, kinetic energy of the photoelectrons is Kmax . When the, ultraviolet light is replaced by X-rays, bothV0 and Kmax, increase., Statement II Photoelectrons are emitted with speeds, ranging from zero to a maximum value because of the, range of frequencies present in the incident light., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 An electron of mass m and charge e are initially at rest. It, gets accelerated by a constant electric field E. The rate, of change of de-Broglie wavelength of this electron at, time t is, (a), , −h, eE t 2, , (b), , −nh, eE t 2, , (c), , −h, eE, , (d), , −eht, E, , 6 The graph between 1 / λ and stopping potential (V) of, three metals having work functions φ1 , φ 2 and φ 3 in an, experiment of photoelectric effect is plotted as shown in, the figure. Which of the following statement (s) is/are, correct? (Here, λ is the wavelength of the incident ray), V, , 2 When a surface 1cm thick is illuminated with light of, , Metal 1 Metal 2 Metal 3, , wavelength λ, the stopping potential is V0, but when the, same surface is illuminated by light of wavelength 3λ,, the stopping potential is V0 / 6, the threshold wavelength, for metallic surface is, (a) 4λ, , (b) 5λ, , (c) 3λ, , (d) 2λ, , 3 A photocell with a constant potential difference of V volt, across it is illuminated by a point source from a distance, of 25 cm. When the source is moved to a distance of 1m,, the electrons emitted by the photocell, (a) carry 1/4th their previous energy, (b) are 1/16th as numerous as before, (c) are 1/4th as numerous as before, (d) carry 1/4th their previous momentum, , The maximum energy of the electron doubles when light, of wavelength 400 nm is used. Find the work function in eV., (b) 2 eV, , (c) 1.02 eV, , (d) 3.42 eV, , 5 A metallic surface is illuminated with monochromatic light, of wavelength λ, the stopping potential for photoelectric, current is 3V0 and when the same surface is illuminated, with light of wavelength 2λ, the stopping potential is V0., The threshold wavelength of this surface for photoelectric, effect is, (a) 4λ /3, , (b) 6λ, , (c) 8λ, , nm–1, 0.002, , 0.004 1/λ, , (a) Ratio of work function φ1: φ2 : φ3 = 1: 2 : 4, (b) Ratio of work function φ1: φ2 : φ3 = 4 : 2 :1, (c) tan θ is directly proportional to hc / e where, h is Planck’s, constant and c is speed of light, (d) The violet colour light can eject photoelectrons from, metal 2 and 3, , 7 Electrons are accelerated through a potential difference, , 4 Consider a metal exposed to light of wavelength 600 nm., , (a) 2.83 eV, , θ, 0.001, , (d) 4λ, , V0 and protons are accelerated through a potential, difference 4V. The de-Broglie wavelength are λ e and λ p, for electrons and protons respectively., The ratio of, , λe, is given by (Given, me is mass of, λp, , electrons and mp is mass of proton)., (a), , mp, λe, =, λp, me, , (b), , me, λe, =, mp, λp, , (c), , 1 me, λe, =, λ p 2 mp, , (d), , mp, λe, =2, λp, me
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346, , DAY THIRTY ONE, , 40 DAYS ~ JEE MAIN PHYSICS, , 8 An electron from various excited states of hydrogen atom, , 10 A particle A of mass m and initial velocity v collides with a, , emit radiation to come to the ground state. Let λ n , λ g be, the de-Broglie wavelength of the electron in the nth state, and the ground state, respectively. Let Λn be the, wavelength of the emitted photon in the transition from, the nth state to the ground state. For large n, (A, B are, constants), (a) Λn ≈ A +, Λ2n, , (c), , ≈A+, , B, Bλ2n, , (d), , Λ2n, , (a), , ≈λ, , 4, (a) > v , 3, , log V, , λA 2, =, 3, λB, , (c), , λA 1, =, 2, λB, , (d), , λA, 1, =, 3, λB, , 1/ 2, , 4, (b) < v , 3, , 1/ 2, , 4, (c) = v , 3, , 1/ 2, , 3, (d) = v , 4, , 1/ 2, , 12 The radiation corresponding to 3 → 2 transition of, hydrogen atom falls on a metal surface to produce, photoelectrons. These electrons are made to enter a, magnetic field of 3 × 10−4 T. If the radius of the largest, circular path followed by these electrons is 10.0 mm, the, work function of the metal is close to, , logλmin, , logλmin, , V to hit a metallic target to produce X-rays. It produces, continuous as well as characteristic X-rays. If λ min is the, smallest possible wavelength of X-rays in the spectrum,, the variation of log λ min with logV is correctly represented, in, , (b), , (b), , fastest emitted electron has speed v. If the wavelength is, changed to 3λ / 4, the speed of the fastest emitted, electron will be, , 9 An electron beam is accelerated by a potential difference, , (a), , λA, =2, λB, , 11 Radiation of wavelength λ is incident on a photocell. The, , (b) Λn ≈ A + Bλ2n, , λ2n, , m, which is at rest. The collision is, 2, head on, and elastic. The ratio of the de-Broglie, wavelengths λ A to λ B after the collision is, particle B of mass, , (a) 1.8 eV, , log V, , (b) 1.1 eV, , (c) 0.8 eV, , (d) 1.6 eV, , (d), , 400 nm. The kinetic energy of the ejected photoelectrons, was found to be 1.68 eV. The work function of the metal, is (hc = 1240 eV-nm), , logλmin, , (c), , logλmin, , 13 The surface of a metal is illuminated with the light of, , log V, , log V, , (a) 3.09 eV, , (b) 1.42 eV, , (c) 151 eV, , (d) 1.68 eV, , ANSWERS, SESSION 1, , 1 (a), , SESSION 2, , 2 (c), , 3 (a), , 4 (b), , 5 (c,d), , 6 (d), , 7 (c), , 8 (b), , 9 (a), , 10 (d), , 18 (b), , 19 (a), , 20 (c), , 8 (a), , 9 (d), , 10 (a), , 11 (c), , 12 (a), , 13 (c), , 14 (d), , 15 (b), , 16 (b), , 17 (a), , 21 (a), , 22 (b), , 23 (d), , 24 (b), , 25 (a), , 26 (d), , 27 (d), , 1 (a), 11 (a), , 2 (b), 12 (b), , 3 (b), 13 (b), , 4 (c), , 5 (d), , 6 (a,c), , 7 (d), , Hints and Explanations, SESSION 1, 1 4 × 10 = 10 × hf, 3, , f =, , 20, , 4 × 103, , 1020 × 6.023 × 10−34, f = 6.64 × 1016 Hz, , The obtained frequency lies in the band, of X-rays., , 2 Energy of photon,, E1 =, , 6. 63 × 3, × 10−18, 59, Light energy produced per second, 90, E =, × 10, 100, =9W, , 3 Intensity A = Intensity B, , =, , 6. 63 × 10−34 × 3 × 108, hc, =, λ, 590 × 10−9, , Number of photons emitted per sec =, =, , 9 × 59, 6. 63 × 3 × 10−18, , = 2.67 × 1019 = 0.267 × 10 20, , E, E1, , The number of photons of beam A = n A, The number of photons of beam B = n B, According to question, n A = 2n B, Let ν A be the frequency of beam A and, ν B be the frequency of beam B., ∴ Intensity ∝ Energy of photons, I ∝ (hν) × Number of photons, I A n Aν A, ∴, =, IB, nBνB
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DUAL NATURE OF MATTER, , DAY THIRTY ONE, According to question, I A = I B, ν, n, 1, ∴, n Aν A = n B ν B ⇒ A = B =, νB, nA 2, νB = 2 ν A, , So,, , 4 Sensitivity of eye = energy detected per, square meter, 5 × 104 × 6.6 × 10−34 × 3 × 108, =, 500 × 10−9, = 0.2 × 10−13 Wm −2, The sensitivity of ear = 1 × 10−13 Wm −2, , 14 As λ is increased, there will be a value, , of λ above which photoelectron will be, cease to come out, so photocurrent will, becomes zero., , 15 de-Broglie wavelength is,, λ=, , h, =, mv, , 16 As,, , E =, , and, , λ =, , Thus, the sensitivity of eye is five times, more than that of the ear., , 5 For photo emission to take place,, wavelength of incident light should be, less than the threshold wavelength of, ultraviolet light < 5200 Å while that of, infrared radiations > 5200Å., , , 6 E k = hc − hc = hc λ 0 − λ , λ, , λ0, , 9 Threshold voltage for a and b is same, , and it depends on frequency. So, f a = fb ., Photo current of a and b are unequal, and photo current depends on intensity., , So, I a ≠ Ib, 10 eV 0 = hc − W 0 and eV ′ = hc − W 0, λ0, 2λ 0, Subtracting them, we have, hc , 1, hc, e (V 0 − V ′ ) =, 1− =, λ 0 , 2 2λ 0, hc, or, V ′ = V0 −, 2 eλ 0, , 11 We have, eV 0 = hc − φ ⇒ V 0 = hc − φ, λ, eλ e, V 0 = mx + c, Q Data is not sufficient., hc, −φ, Q, λ, 12 As, 1, =, −2, 4 πε0 1 × 10, Q, ⇒, , Q = 5.5 × 10−13 C, , 13 (A) Franck-Hertz experiments is, associated with discrete energy levels of, atom., (B) Photo-electric experiment is, associated with particle nature of light., (C) Davisson-Germer experiment is, associated with wave nature of electron., , h, h, =, mv, 2mE, E = eV, h, h2, λ=, ⇒V =, 2 meV, 2meλ2, λ=, , But, , V =, , eV 0 = h (ν − ν 0 ) and hν = KE + W 0, , changes but frequency remains same, so, stopping potential remains same., , hc 6.6 × 10−34 × 3 × 108, =, E, 11 × 1 . 6 × 10−19, = 1.125 × 10−7 m, , 17 de-Broglie wavelength is, , 7 As we know,, , 8 By changing distance the intensity, , hc, λ, , Hence, UV region., , λ 0λ , , So, A → 2, B → 1,C → 3, , h, = 1 .7 Å, 2mqV, , (6.62 × 10−34 )2, (0.5 × 10−10 )2 × 2 × 91, ., × 10−31 × 1.6 × 10−19, , ⇒, , V = 601.98 V ≈ 602 V, h, hc, 18 As, λ1 =, , λ2 =, E, 2m pE, ∴, , h / 2m pE, λ1, =, =, λ2, hc / E, , ⇒, , λ1, λ, ∝ E ⇒ 1 ∝ E 1 /2, λ2, λ2, , intensity of light increases means, number of photons/area/time increases, and hence more photons take part in, ejecting the photoelectron, thus, increasing the number of, photoelectrons., , 22 Velocity of first photon = u = c, Velocity of second photon = v = − c, Now relative velocity of first photon, with respect to second photon, u−v, c − (− c ), =, =, uv, c × (− c ), 1− 2, 1−, c, c2, 2c, =, c2, 1+ 2, c, 2c, 2c, =, =, =c, 1+ 1, 2, Also, the rest of mass of photon is zero., , 23 When work function of copper is greater, than the work function of sodium,then, φCu > φ Na, (hνo )Cu > ( hνo ) Na, , …(i), , Hence, Eq. (i), becomes, E, , 2m pe 2, , Force on moving electron due to, magnetic field is, F = − e (v × B) = − e [v o i × Bo j], = − e v o Bo k, As this force is perpendicular to v and, B, so the magnitude of v will not change, i.e. momentum (mv) will remain, constant is magnitude. Hence,, de-Broglie wavelength λ = h / mv, remains constant., 20 As, 2 πr = nλ ⇒ r = nλ, 2π, h, Now, de-Broglie equation λ =, p, h, h, nh, ⇒ mvn = =, =, λ 2 πr n, 2 πr n, n, Also, for charged particle moving in a, magnetic field, mv n, nh, rn =, =, qB, (2πrn ) q B, nh, r n2 =, ⇒, 2π q B, r n ∝ n1 /2, , 21 From quantum theory of light, as, , c, But we know that, νo =, λc, , 19 Here, v = v o i B = Bo j, , ∴, , 347, , hc , hc , , >, (λo ) Na > ( λo )Cu, λo Cu λo Na, , 24 de-Broglie wavelength associated with, gas molecules varies, 1, λ∝, T, Also, root mean square velocity of gas, 3RT, molecules is v rms =, ., M, , 25 Davisson and Germer experimentally, established wave nature of electron by, observing diffraction pattern while, bombarding electrons on Ni crystal., , 26 Maximum kinetic energy (KE) max is, given by (KE) max = hν − hν 0., When frequency is increased (KE) max, will increase stopping potential is that, negative voltage given to the anode at, which photocurrent stops, hence, doubling frequency will not effect it, also, If, ν′ = 2 ν, ∴, K ′ max = eν′ 0 = h (2ν1 − ν 0 ), K ′ max = 2K max + hν 0, ∴, K ′ max > 2 K max ⇒ ν′ 0 > 2 ν 0, Hence, (KE) max and stopping potential, are linearly dependent on the frequency, of incident light.
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DAY THIRTY TWO, , Atoms, Learning & Revision for the Day, u, , u, , Scattering of α-particles, Rutherford‘s Model of an Atom, , u, , u, , Bohr’s Model, Energy Levels and Hydrogen Spectrum, , Atom is the smallest particle of an element which contains all properties of element., Molecule is a single atom or a group of atoms joined by chemical bonds. It is the smallest, unit of a chemical compound that can have an independent existence. Nuclei refers to a, nucleus of an atom, having a given number of nucleons. It is a general term referring to, all known isotopes–both stable and unstable of the chemical elements. Thus, O16 and, O17 are different nuclides., , Scattering of α-particles, In 1911, Rutherford successfully explained the scattering of α-particles on the basis of, nuclear model of the atom., Number of α-particles scattered through angle θ is given by, Z2, N (θ) ∝, sin 4 (θ /2) K 2, where, K is the kinetic energy of α-particle and Z is the atomic number of the metal., At distance of closest approach the entire initial kinetic energy is converted into, potential energy, so, 1, 1 Ze(2e), mv2 =, 2, 4πε 0 r0, ⇒, , r0 =, , 4KZe2, Ze2, =, mv2 πε 0, mv2, , Rutherford’s Model of an Atom, On the basis of scattering of α-particles, Rutherford postulated the following model of the, atom, l, , l, , l, , l, , Atom is a sphere of diameter about 10 −10 m. Whole of its positive charge and most of, its mass is concentrated in the central part called the nucleus., The diameter of the nucleus is of the order of 10 − 5 m., The space around the nucleus is virtually empty with electrons revolving around the, nucleus in the same way as the planets revolve around the sun., The electrostatic attraction of the nucleus provides centripetal force to the orbiting, electrons., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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l, , l, , Total positive charge in the nucleus is equal to the total, negative charge of the orbiting electrons., , l, , Rutherford’s model suffers from the following drawbacks, (a) stability of the atomic model., (b) nature of energy spectrum., , The total energy of the orbital electron is, me 4 Z 2 , E = − 2 2 2 , 8ε 0 h n , me 4 Z 2, = − 2 3 ch 2, 8ε 0ch n, , Bohr’s Model, , = − Rch, , Bohr added the following postulates to the Rutherford’s model, of the atom, , l, , l, , l, , The electrons revolve around the nucleus only in certain, permitted orbits, in which the angular momentum of the, electron is an integral multiple of h /2 π , where h is the, Planck’s constant, nh, , L = mvnrn =, , , 2π , , If energy of the electron in nth and mth orbits be E n and E m, respectively, and when the electron jumps from nth to mth, orbit the radiation frequency ν is emitted, such that, E n − E m = hν. This is called the Bohr’s frequency equation., • Radius of the orbit of electron in a hydrogen atom in its, , stable state, corresponding to n = 1, is called Bohr‘s radius., Value of Bohr‘s radius is r0 = 0 . 529 Å ≈ 0.53 Å ., • The time period of an electron in orbital motion in the, Bohr’s orbit is, 2 π r 2 π × 0.53Å, c , =, = 1.52 × 10 − 6 s Q v =, T=, , c, 137 , v, 137, 1, and the frequency of revolution is f = = 6.5757 × 10 15 cps, T, , 1 2 Zπ e, c Z, 6 Z , vn =, =, = 2.2 × 10 m/s, 137 n, n, 4πε 0 nh, 2, , l, , E =−, , and, , 1 Ze2, 4πε 0 r, 1, 2, , 1 Ze2 , 4πε r , 0, , , , Therefore, they are related to each other as follows, , l, , l, , KE = − E and PE = 2 E, 1, 1, For a hydrogen atom, rn ∝ n2, vn ∝ and | E | ∝ 2, n, n, The difference in angular momentum associated with the, electron in the two successive orbits of hydrogen atom is, h, nh, h, ∆L = (n + 1), −, =, 2π 2π 2π, , Energy Levels and Hydrogen, Spectrum, Hydrogen spectrum consists of spectral lines classified as five, spectral series of hydrogen atom. Out of these five, Lyman, series lies in the ultraviolet region of spectrum, Balmer series, lies in the visible region and the remaining three series, lie in, the infrared region of spectrum., n=∞, n=7, n=6, , 0, – 0.28, – 0.38, , Pfund, , – 0.54, , Brackett, , – 0.85, –1.51, –3.40, , Orbital frequency is given by, , 1, v, me 4, f = =, =, T 2πr, 4 ε20 n3 h3, , 1 1 Ze2 , 2 4πε 0 r , , PE = −, , n2 h2, n2, = 0.53, Å, 2, 2, Z, 4 π Zme, , The orbital velocity of an electron is, , me 4 Z 2, me 4 Z 2, , PE = −, 2 2 2, 8n h ε 0, 4 n2 h2 ε20, , The kinetic, potential and total energies of the electron, with r as the radius of the orbit are as follows, , The orbital radius of an electron is, rn = 4πε 0, , Z2, Z2, = −13.6 2 eV, 2, n, n, , KE =, , The energy is radiated only when the electron jumps from, an outer permitted orbit to some inner permitted orbit., (Absorption of energy makes the electron jump from inner, orbit to outer orbit)., , Some Characteristics of an Atom, , l, , l, , The electrons do not radiate energy while revolving in the, permitted orbits. That is, the permitted orbits are stationary,, non-radiating orbits., , NOTE, , l, , KE =, , E (eV), , l, , 351, , ATOMS, , DAY THIRTY TWO, , Paschen, Infrared, Balmer, Visible light, , n=5, n=4, n=3, n=2, , Lyman, series, –13.60, , n=1, Ultraviolet, , Emission spectrum of H-atom
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352, , DAY THIRTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , 1, 1, 1, = ν = R 2 − 2, λ, (5) n , , Total number of emission spectral lines from some excited state, n, to another energy n2 (< n1 ) is given by, (n1 − n2 )(n1 − n2 + 1), 2, e.g. total number of lines from n1 = n to n2 = 1 are, , where, n = 6, 7, 8, ..., n (n − 1), ., 2, , NOTE, , 1, 1, 1, 1, ∆E = E 2 − E 1 = RchZ 2 2 − 2 = 136, . Z2 2 − 2 , n1 n2 , n1 n2 , , The five spectral series of hydrogen atom are given below, , 1. Lyman Series, Spectral lines of Lyman series correspond to the transition of, electron from higher energy levels (orbits) ni = 2, 3, 4, … to, ground energy level (1st orbit) nf = 1., For Lyman series,, , 1, 1, 1, = v = R 2− 2, λ, (1) n , , where, n = 2, 3, 4, …, It is found that a term Rch = 13.6 eV = 2.17 × 10 −18 J. The term, Rch is known as Rydberg’s energy., , 2. Balmer Series, Electronic transitions from ni = 3, 4, 5, … to nf = 2, give rise to, spectral lines of Balmer series., For a Balmer series line,, 1, 1, 1, = ν = R 2 − 2, λ, n , (2), where, n = 3, 4, 5, …, , 3. Paschen Series, Lines of this series lie in the infrared region and correspond to, electronic transition from ni = 4, 5, 6,… to nf = 3., 1, 1, 1, = ν = R 2 − 2, λ, 3, (, ), n, , , where, n = 4, 5, 6, ..., , 4. Brackett Series, Spectral lines in the infrared region which corresponds to, transition from ni = 5, 6, 7, … to nf = 4., For Brackett series,, 1, 1, 1, = ν = R 2 − 2, λ, n , (4), where, n = 5, 6, 7, …, , • Frequency, ν =, , For a spectral line in Pfund series,, , 1, ∆E, 1, = RcZ 2 2 − 2 , h, n1 n2 , , Ionisation Energy and Potential, Ionisation energy of an atom is defined as the energy, required to ionise it i.e. to make the electron jump from its, present orbit to infinity., Thus, ionisation energy of hydrogen atom in the ground, state = E ∞ − E1, = 0 − (−13.6 eV) = + 13.6 eV, The potential through which an electron is to be accelerated, so that it acquires energy equal to the ionisation energy is, called the ionisation potential., Therefore, ionisation potential of hydrogen atom in its, ground state is 13.6V., z2, E, In general, E ion = 13.6 2 eV or Vion = ion, e, n, , Excitation Energy and Ionisation, Potential, Excitation energy is the energy required to excite an, electron from a lower energy level to a higher energy level., The potential through which an electron is accelerated so as, to gain requisite ionisation energy is called the ionisation, potential., Thus, first excitation energy of hydrogen atom, = E2 − E1, = − 3.4 − (− 13.6) eV, = + 10.2 eV, Similarly, second excitation energy of hydrogen atom, = E3 − E1, = − 1.51 − (−13.6), = 12.09 eV, , 5. Pfund Series, It lies in the far infrared region of spectrum and corresponds to, electronic transitions from higher orbits, ni = 6, 7, 8, ... to orbit having nf = 5., , • Energy of emitted radiation,, , NOTE, , • Total energy of a closed system is always negative and, its magnitude is the binding energy of the system., , • Kinetic energy of a particle can’t be negative, while the, potential energy can be zero, positive or negative.
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ATOMS, , DAY THIRTY TWO, , 353, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 An α-particle of energy 5 MeV is scattered through 180°, by a fixed uranium nucleus. The distance of the closest, approach is of the order of, (a) 1Å, , (b) 10− 10 cm, , (c) 10− 12 cm, , (d) 10−15 cm, , 2 To explain theory of hydrogen atom, Bohr considered, (a) quantisation of linear momentum, (b) quantisation of angular momentum, (c) quantisation of angular frequency, (d) quantisation of energy, , 3 For the Bohr’s first orbit of circumference 2πr, the, de-Broglie wavelength of revolving electron will be, (a) 2 πr, , (b) πr, , (c), , 1, 2 πr, , (d), , 1, 4 πr, , 4 Which of the following transitions in hydrogen atoms emit, photons of highest frequency?, (a) n = 2 to n = 6, (c) n = 2 to n = 1, , (b) n = 6 to n = 2, (d) n = 1to n = 2, , 5 In the Bohr’s model of the hydrogen atom, let r, V and E, represents the radius of the orbit, the speed of electron, and the total energy of the electron, respectively. Which, of the following quantities is proportional to the quantum, number n?, (a) E/ v, , (b) r/E, , (c) vr, , (d) rE, , 6 The ratio of the kinetic energy to the energy of an, electron in a Bohr’s orbit is, (a) − 1, (c) 1 : 2, , the ground state of a hydrogen like atom/ion, ª JEE Main 2015, , (a) its kinetic energy increases but potential energy and, total energy decrease, (b) kinetic energy, potential energy and total energy, decrease, (c) kinetic energy decreases, potential energy increases but, total energy remains same, (d) kinetic energy and total energy decrease but potential, energy increases, , 11 An electron jumps from the 4th orbit to the 2nd orbit of, hydrogen atom. Given the Rydberg’s constant, R = 107 cm −1, the frequency (in hertz) of the emitted, radiation will be, (a), , 3, × 105, 16, , radius of last orbit of 100Fm 257 is n times the Bohr’s, radius, then find the value of n ?, (c) 4, , (d) 1/4, , 8 Taking the Bohr’s radius as a 0 = 53 pm, the radius of Li 2+, ion in its ground state, on the basis of Bohr’s model, will, be about, (a) 53 pm, (c) 18 pm, , (b) 27 pm, (d) 13 pm, , 9 In a hypothetical Bohr’s hydrogen atom, the mass of the, electron is doubled. The energy E 0 and radius r0 of the, first orbit will be (a 0 is the Bohr radius), (a) E 0 = − 27. 2 eV ; r0 = a0 / 2, (b) E 0 = − 27. 2 eV ; r0 = a0, (c) E 0 = − 13.6 eV ; r0 = a0 / 2, (d) E 0 = − 13 .6 eV ; r0 = a0, , 3, × 1015, 16, , (c), , 9, × 1015, 16, , (d), , 3, × 1015, 4, , 12 In figure, the energy levels of the hydrogen atom have, been shown along with some transitions marked, A, B and C. The transitions A, B and C, respectively, represents, Continuum, n=5, n=4, , C, , n=3, , (b) 2, (d) None of these, , (b) 200, , (b), , B, , 0 eV, –0.54 eV, –0.85 eV, –1.51 eV, –3.40 eV, , n=2, , 7 If the atom 100Fm 257 follows the Bohr’s model and the, , (a) 100, , 10 As an electron makes a transition from an excited state to, , A, n=1, , –13.60 eV, , (a) the first member of the Lyman series, third member of, Balmer series and second member of Paschen series, (b) the ionisation potential of H, second member of Balmer, series and third member of Paschen series, (c) the series limit of Lyman series, second member of, Balmer series and second member of Paschen series, (d) the series limit of Lyman series, third member of Balmer, series and second member of Paschen series, , 13 The given diagram indicates the, , 2E, 4/3E, E, , energy levels of a certain atom., When the system moves from 2E, 0, level to E, a photon of wavelength, λ is emitted. The wavelength of, photon produced during its transition from 4E/3 level to E, is, (a) λ / 3, , (b) 3 λ / 4, , (c) 4 λ / 3, , (d) 3λ
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354, , DAY THIRTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , 14 Energy E of a hydrogen atom with principal quantum, 13.6, number n is given by E = − 2 eV. The energy of a, n, photon ejected when the electron jumps from n = 3 state, to n = 2 state of hydrogen, is approximately, (a) 1.5 eV, , (b) 0.85 eV, , (c) 3.4 eV, , (d) 1.9 eV, , 15 The ionisation potential of H-atom is 13.6 V. When it is, excited from ground state by monochromatic radiations, of 970.6 Å, the number of emission lines on deexcitation, will be (according to Bohr’s theory), (a) 10, , (b) 3, , (c) 6, , (d) 4, , 16 In a hydrogen like atom electron makes transition from an, energy level with quantum number n to another with, quantum number (n −1). If n >> 1, the frequency of, radiation emitted is proportional to, ª JEE Main 2013, 1, (a), n, , (b), , 1, n2, , (c), , 1, n3 / 2, , (d), , 1, n3, , 17 Hydrogen atom is excited from ground state to another, state with principal quantum number equal to 4. Then,, the number of spectral lines in the emission spectra will, be, ª AIEEE 2012, (a) 2, , (b) 3, , (c) 5, , (d) 6, , 18 The wavelength of the first spectral line in the Balmer, series of hydrogen atom is 6561 Å. The wavelength of, the second spectral line in Balmer series of single, ª AIEEE 2011, ionised helium atom is, (a) 1215 Å, , (b) 1640 Å, , (c) 2450 Å, , (d) 4687 Å, , 19 The transition from the state n = 4 to n = 3 in a, hydrogen-like atom results in ultraviolet radiation., Infrared radiation will be obtained in the transition from, ª AIEEE 2010, , (a) 2 → 1, , (b) 3 → 2, , (c) 4 → 2, , 20 Some energy levels of a, , –E, , molecule are shown in the, –4/3E, figure. The ratio of the, wavelengths r = λ 1 / λ 2 is given, –2E, ª JEE Main 2017 (Offline), by, , (d) 5 → 3, , 2, 3, 1, (c) r =, 3, , 3, 4, 4, (d) r =, 3, , (a) r =, , (b) r =, , 21 Energy required for the electron excitation in Li2+ from the, first to the third Bohr orbit is, (a) 36.3 eV, (c) 122.4 eV, , ª AIEEE 2011, , (b) 108.8 eV, (d) 12.1 eV, , 22 In hydrogen atom, if the difference in the energy of the, electron in n = 2 and n = 3 orbits is E, the ionisation, energy of hydrogen atom is, (a) 13.2 E, (c) 5.6 E, , (b) 7.2 E, (d) 3.2 E, , 23 Excitation energy of a hydrogen like ion in its first, excitation state is 40.8 eV. Energy needed to remove the, electron from the ion in ground state is, (a) 54.4 eV, (c) 40.8 eV, , (b) 13.6 eV, (d) 27.2 eV, , Direction (Q. Nos. 24-25), , Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true; Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true; Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 24 Statement I Bohr had to postulate that the electrons in, stationary orbits around the nucleus do not radiate., Statement II According to classical physics all moving, electrons radiate electromagnetic radiation., , 25 Statement I The different lines of emission spectra (like, , λ2, λ1, , Lyman, Balmer, etc) of atomic hydrogen gas are, produced by different atoms., Statement II The sample of atomic hydrogen gas, consists of millions of atoms., , –3E, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 If the series limit frequency of the Lyman series is νL,, then the series limit frequency of the Pfund series is, (a) 25 νL, , (b) 16 νL, , ν, (c) L, 16, , ª JEE Main 2018, ν, (d) L, 25, , 2 In Rutherford’s experiment, the number of alpha particles, scattered through an angle of 90° is 28 per minute. Then,, , the number of particles scattered through an angle of 60°, per minute by the same nucleus is, (a) 28 per minute, (c) 12.5 per minute, , (b) 112 per minute, (d) 7 per minute, , 3 A hydrogen like ion having wavelength difference, between first Balmer and Lyman series equal 593 Å has, Z equal to, (a) 2, , (b) 3, , (c) 4, , (d) 1
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ATOMS, , DAY THIRTY TWO, 4 The largest wavelength in the ultraviolet region of the, hydrogen spectrum is 122 nm. The smallest wavelength, in the infrared region of the hydrogen spectrum (to the, nearest integer) is, (a) 802 nm, , (b) 823 nm, , (c) 1882 nm, , (d) 1648 nm, , 5 Energy levels A, B and C of a certain atom, corresponding to increasing values of energy i.e., E A < E B < E C . If λ 1, λ 2 and λ 3 are the wavelengths of, radiations corresponding to the transitions C to B, B to A, and C to A respectively, which of the following, statements is correct?, C, , λ1, , B, λ2, , 10 A hydrogen atom moves with a velocity u and makes a, head on inelastic collision with another stationary H-atom., Both atoms are in ground state before collision.The, minimum value of u if one of them is to be given a, minimum excitation energy is, (a) 2.64 × 104 ms −1, (c) 2.02 × 106 ms −1, , 11 In the Bohr’s model an electron moves in a circular orbit, around the proton. Considering the orbiting electron to be a, circular current loop, the magnetic moment of the hydrogen, atom, when the electron is in nth excited state, is, e n 2h, e nh, (b) , (a) , , 2m π, m 2π, , (b) λ 3 =, , (c) λ1 + λ 2 + λ 3 = 0, , (d) λ23 = λ21 + λ22, , (a), , 2, , 6 Hydrogen (1H ), deuterium (1H ), singly ionised helium, 6 2+, , ( 2He ) and doubly ionised lithium ( 3Li ) all have one, electron around the nucleus. Consider an electron, transition from n = 2 to n = 1. If the wavelengths of, emitted radiations are λ 1, λ 2, λ 3 and λ 4 respectively, then, approximately which one of the following is correct?, ª JEE Main 2014, , (a) 4 λ1 = 2 λ 2 = 2 λ 3 = λ 4, (c) λ1 = λ 2 = 4 λ 3 = 9λ 4, , (b) λ1 = 2 λ 2 = 2 λ 3 = λ 4, (d) λ1 = 2 λ 2 = 3 λ 3 = 4 λ 4, , 7 The potential energy between a proton and an electron is, PE =, (a), (c), , e2, , then the radius of the Bohr’s orbit is, 4πε 0( 3R 3 ), 4 π 2e 2m, , (b), , 4 πε0n h, e 2m, , 2 2, , (d), , 4 πε0h 3, , 6 π 2e 2m, 4 πε0n 3h 3, 2 πe 2m, 4 πε0nh 3, , 8 A small particle of mass m moves such that potential, 1, mr 2ω 2. Assuming Bohr’s model of, 2, quantisation of angular momentum and circular orbit,, radius of nth orbit is proportional to, , energy PE =, , (a) n, , (b) n 3, , (c), , 1, n, , (d), , 1, n3, , 9 The electric potential between a proton and an electron, is given by V = V0 ln, , r, , where r0 is a constant. Assuming, r0, , Bohr’s model to be applicable, write variation of rn with n,, n being the principal quantum number., (a) rn ∝ n, (c) rn ∝ n 2, , 1, (b) rn ∝, n, 1, (d) rn ∝ 2, n, , e nh, (c) , , 2m 2 π, , e n 2h, (d) , m 2π, , which are separated by a distance r. If we calculate its, rotational energy by applying Bohr’s rule of angular, momentum quantisation, its energy will be given by (n is, an integer), ª AIEEE 2012, , λ1λ 2, λ1 + λ 2, , (a) λ 3 = λ1 + λ 2, , 4 +, , (b) 6.24 × 104 ms −1, (d) 6.24 × 108 ms −1, , 12 A diatomic molecule is made of two masses m1 and m2, , λ3, A, , 1, , 355, , (c), , (m1 + m2 )2 n 2h 2, 2m12m22r 2, 2 2, , 2n h, , (m1 + m2 )r, , 2, , (b), (d), , n 2h 2, 2 (m1 + m2 )r 2, (m1 + m2 )n 2h 2, 2m1m2r 2, , 13 In the Bohr’s model of hydrogen-like atom the force, between the nucleus and the electron is modified as, e2 1, β, F =, + , where β is a constant. For this atom,, 4πε 0 r 2 r 3 , the radius of the nth orbit in terms of the Bohr’s radius, , ε h2 , a 0 = 0 2 is, m πe , , ª AIEEE 2010, (a) rn = a0 n − β, (c) rn = a0 n 2 − β, , (b) rn = a0 n 2 + β, (d) rn = a0 n + β, , Direction (Q. Nos. 14-15) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true; Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true; Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 14 Statement I Balmer series lies in the visible region of, electromagnetic spectrum., 1, 1, 1, Statement II, = R 2 − 2 , where n = 3 , 4 , 5 ,……, 2, λ, n , , 15 Statement I The ionisation potential of hydrogen is found, to be 13.6 eV, the ionisation potential of doubly ionised, lithium is 122.4 eV., Statement II Energy in the nth state of hydrogen atom is, 13.6, En = − 2 ., n
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356, , DAY THIRTY TWO, , 40 DAYS ~ JEE MAIN PHYSICS, , ANSWERS, SESSION 1, , SESSION 2, , 1 (c), , 2 (b), , 3 (a), , 4 (b), , 5 (c), , 6 (a), , 7 (d), , 8 (c), , 9 (a), , 10 (a), , 16 (d), , 17 (d), , 18 (a), , 19 (d), , 20 (c), , 6 (c), , 7 (a), , 8 (a), , 9 (a), , 10 (b), , 11 (c), , 12 (d), , 13 (d), , 14 (d), , 15 (c), , 21 (b), , 22 (b), , 23 (a), , 24 (b), , 25 (b), , 1 (d), 11 (c), , 2 (b), 12 (d), , 3 (b), 13 (c), , 4 (b), 14 (a), , 5 (b), 15 (b), , Hints and Explanations, SESSION 1, , 6 The kinetic energy and total energy of, , q q, 1 Here, 1 mv 2 = 1 1 2, 2, 4p e0 r, 9 ´ 109 ´ (2e ) ´ (92e ), 5 MeV =, r, æQ 1 mv 2 = 5 MeV ö, ç, ÷, è 2, ø, , \, , Þr =, , 9 ´ 109 ´ 2 ´ 92 ´ (1.6 ´ 10-19 )2, 5 ´ 106 ´ 1.6 ´ 10– 19, , r = 53, . ´ 10- 14 m, » 10, , -12, , cm, , atom, Bohr considered quantisation of, angular momentum as the essential, condition for the stationary orbits., , 3 According to Bohr’s first postulate,, nh, 2p, h ö, 2pr = n æç, ÷ = nl, è mv ø, mvr =, , For n = 1, l = 2pr, , 4 Emission spectrum would rises when, electron makes a jump from higher, energy level to lower energy level., Frequency of emitted photon is, proportional to change in energy of two, energy levels, i.e., æ1, 1 ö, n = RcZ 2 ç 2 - 2 ÷, n2 ø, è n1, , 5 According to the Bohr’s postulate., mvr =, Þ, , nh, h ö, Þ vr = n æç, ÷, è 2pm ø, 2p, , vr µ n, , Thus, vr is directly proportional to the, principal quantum number., , ö, æ, 11 n = c = c . R ç 12 - 12 ÷, l, , è Z ø, , m2, =n, Z, m = 5 (for 100 fm257 the outermost shell), , \, , and Z = 100, n=, , (5) 2, 1, =, 100 4, , 8 On the basis of Bohr’s model,, r =, Let Li 2+, , n2 l2, 2, , 2, , 4p mKZe, ion, Z = 3,, , =, , a0n2, Z, , n = 1 for ground state., Given, a 0 = 53 pm, r =, , 53 ´ (1)2, 3, , ~, - 18 pm, 9 As, r µ 1, m, 1, \, r 0 = a0, 2, As, E µ m, \, E 0 = 2 (- 13.6), = - 27.2 eV, , 10 As we know that, kinetic energy of an, electron is KEµ (Z /n )2 . When the, electron makes transition from an, excited state to the ground state, then n, decreases and KE increases. We know, that PE is lowest for ground state., Also, TE = - KE. TE also decreases., , è n1, , n2 ø, , 1, 1, = 3 ´ 108 ´ 107 æç 2 - 2 ö÷, è 2, 4 ø, 9, =, ´ 1015 Hz, 16, , æ 2ö, 7 rm = ç m ÷ (0.53 Å ) = (n ´ 0.53) Å, , \, , 2 While proposing his theory of hydrogen, , \, , an electron are related as, KE = -E, KE, = -1, \, E, , 12 A represents series limit of Lyman series,, B represents third member of Balmer, series and C represents second member, of Paschen series., 13 l µ 1, DE, l¢, (2E - E ), 1, =, =, =3, \, l, (4E /3 - E ) 1/3, \, , l ¢ = 3l, , 14 Given, E n = - 132.6 eV, n, 13.6, , 13.6, eV, 9, 13.6, and E2 = eV = eV, 4, (2) 2, 13.6 æ 13.6 ö, So, DE = E3 - E2 = - ç÷, è, 4 ø, 9, (approximately), = 1.9 eV, hc 12400, 15 Using E =, =, eV = 1277, . eV, l, 970.6, So, electron is excited upto n = 4, \, n2 = 4, On deexcitation number of emission, n (n - 1), lines produced =, =6, 2, E3 = -, , (3) 2, 13.6, , eV = -, , 16 D E = hn, é 1, 1 ù, DE, = Kê, - 2ú, 2, h, n û, ë (n - 1), K 2n, 2K, = 2, =, n (n - 1)2, n3, , n=
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DAY THIRTY THREE, , Nuclei, Learning & Revision for the Day, u, , u, , Concept of Nucleus, Radioactivity, , u, , u, , Mass Energy Relation, Mass Defect and Binding Energy, , u, , u, , Nuclear Fission, Nuclear Fusion, , Concept of Nucleus, In every atom, the positive charge and mass is densely concentrated at the centre of the, atom forming its nucleus. In nucleus, the number of protons is equal to the atomic number, of that element and the remaining particles to fulfil the mass number are the neutrons., , Composition of Nucleus, Nucleus consists of protons and neutrons. Electrons cannot exist inside the nucleus., A proton is a positively charged particle having mass (m p) of 1.007276 u and charge, (+ e) = +1.602 × 10 –19 C., For a neutral atom, Number of proton ( Z ) = Number of electron, This number is called the atomic number. A neutron is a neutral particle having mass, mn = 1.008665 u. The number of neutrons in the nucleus of an atom is called the neutron, number N. The sum of the number of protons and neutrons is called the mass number A., Thus, A = N + Z ., , Properties of Nucleus, Nuclear size, (a) Size of the nucleus is of the order of fermi (1 fermi = 10 −15 m )., (b) The radius of the nucleus is given by R = R0 A1 /3 ,, where, R0 = 1.3 fermi and A is the mass number., , Your Personal Preparation Indicator, , Volume, The volume of nucleus is V =, , PREP, MIRROR, , 4, π (R0 A1 /3 )3 , where, R0 = radius of the nucleus., 3, , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , Density, Am p, mp, Mass of nucleus, =, =, 4, Volume of the nucleus 4 π (R A1 /3 )3, πR03, 0, 3, 3, where, m p = 1.6 × 10 −27 kg = mass of proton and R0 = 1.3 fermi., , (a) Density =, , (b) Density of nuclear matter is of the order of 1017 kg/m3 ., (c) Density of nuclear matter is independent of the mass number., , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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360, , DAY THIRTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , Isotopes, Isobars and Isotones, Isotopes, Isotopes of an element are nuclides having same atomic, number Z, but different mass number A (or different neutron, number N) is called isotopes. 11 H, 21H, 31H and 611 C, 612C, 614C, etc.,, are isotopes., , Isobars, Nuclides having same mass number A, but different atomic, number Z are called isobars. In isobars number of protons Z, as well as number of neutrons N differ but total nucleon (or, mass) number A = N + Z is the same. 31 H, 32He and 614 C, 14, 7 N are, isobars., , Isotones, Nuclides with different atomic number Z and different mass, number A, but same neutron number are called isotones., 3, 4, 198, 197, 1 H, 2He and 80 Hg, 79 Au are examples of isotones., , dN, dN, ∝ N or, = − λN, dt, dt, Here, λ is a proportionality constant, known as the decay, constant (or disintegration constant). Unit of λ is s–1 or day –1, or year –1, etc., , Mathematically,, , It can be shown that number of nuclei present after time t is, given by, N = N 0 e − λt, where, N 0 = number of nuclei present at time t = 0., Again, number of nuclei decayed in time t will be, N − N 0 = N 0 [e − λt − 1], = number of daughter nuclei produced at time t ., , Half-Life Period (T1/2 ), It is the time in which, activity of the sample falls to one-half of, its initial value., T, N, R, Thus, for t = , N = 0 and R = 0, 2, 2, 2, l, , Radioactivity, Radioactivity is the phenomenon of spontaneous emission of, radiations by heavier nucleus. Some naturally occurring, radioactive substances are uranium, thorium, polonium,, radium, neptunium, etc. In fact, all elements having atomic, number Z > 82 are radioactive in nature., Radiations emitted by radioactive substances are of three, types, namely (i) α-particles, (ii) β-particles and (iii) γ- rays., l, , l, , l, , α-particles are positively charged particles with charge, q α = + 2 e and mass mα = 4m p. Thus, α-particles may be, considered as helium nuclei (or doubly charged helium, ions). Ionising power of α-particles is maximum, but their, penetrating power is minimum., β-particles are negatively charged particles with rest mass, as well as charge same as that of electrons. But origin of, β-particles is from the nucleus. Their ionising power is, lesser than that of α-particles, but speed as well as, penetrating power is much greater than that of α-particles., Generally, β-decay means β − - decay., γ-rays are electromagnetic radiations of extremely short, wavelengths. Thus, γ-rays travel with the speed of light., Their ionising power is least, but penetrating power is, extremely high. These are not deflected either in an, electric or a magnetic field., , Law of Radioactive Decay, According to Rutherford-Soddy’s law for radioactive decay,, ‘The rate of decay of a radioactive material at any instant is, proportional to the quantity of that material actually present, at that time.’, , l, , The half-life period is related to decay constant λ as, 0.693, T1 /2 =, λ, After n half-lives, the quantity of a radioactive substance left, intact (undecayed) is given by, n, , t, , 1 T, 1, N = N 0 = N 0 1 /2, 2, 2, , Mean Life Period ( τ ), l, , l, , Mean life of a radioactive sample is the time, at which both, 1, N and R have been reduced to or e −1 or 36.8% of their, e, 1, initial values. It is found that τ = ., λ, Half-life T1 /2 and mean life τ of a radioactive sample are, correlated as, T1/2 = 0.693 τ or τ = 1.44T1/2 ., , Activity, The activity of a radioactive substance is defined as the rate of, disintegration (or the count rate) of that substance., Mathematically, activity is defined as, dN, R=−, = λN = λN 0 e − λt = R0 e − λt, dt, where, R0 = λN 0 = initial value of activity., Units of activity are, l, , 1 becquerel = Bq = 1 disintegration per second (SI unit), , l, , 1 curie = 1 Ci = 3 . 7 × 1010 Bq, , l, , 1 rutherford = 1 Rd = 10 6 Bq
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NUCLEI, , DAY THIRTY THREE, , Mass Energy Relation, , l, , In nuclear physics, mass is measured in unified atomic mass, units (u), 1 u being one-twelfth of the mass of carbon-12 atom, and equals 1.66 × 10–27 kg. It can readily be shown using, E = mc2 that, 1 u mass has energy 931.5 MeV, , Thus, 1 u ≡ 931.5 MeV or, , l, , E 0 = m0 c = 9.1 × 10, , −31, , × (3 × 10 ) J = 0.51 MeV, 82, , Therefore, an energy of atleast 1.02 MeV is needed for pair, production., , Mass Defect and Binding Energy, , l, , l, , NOTE, , The difference in mass of a nucleus and its constituent, nucleons is called the mass defect of that nucleus. Thus,, Mass defect, ∆M = Zm p + ( A − Z )mn − M, where, M is the mass of a given nucleus., , A characteristic nuclear fission reaction equation for 235, 92 U is, 1, 0 n(slow ), , l, , Building energy per nucleon (MeV), , 8, , 12, , C, , 4, , He, 6, , 14, , 6, , 32, 18, , S, , 56, , Fe, , 100, , l, , l, , Mo, , 12, , I, 184, , O, , N, , W, , 197, , An, , 238, , U, , Li, , 3, , H, , 2, , 0, , H, , 0, , →, , 144, 56Ba, , +, , 89, 36 Kr, , + 3 10 n + Q, , In the fission of one nucleus of, , 235, 92 U,, , on an average,, , Neutrons formed as a result of fission have an energy of, about 2 MeV, whereas for causing further fission, we need, slow thermal neutrons having an energy of about 0.3 eV., For this purpose, suitable material called a moderator is, used, which slow down the neutrons. Water, heavy water, and graphite are commonly used as moderators., A nuclear reactor is an arrangement in which nuclear, fission can be carried out through a sustained and a, controlled chain reaction and can be employed for, producing electrical power, for producing different isotopes, and for various other uses., nE, Power of a reactor, P =, , where n = number of atoms, t, undergone fission in time t seconds and E = energy, released in each fission., , Reproduction Factor, , 4, 2, , 236, 92 U, , 1, neutrons are released. These released neutrons may, 2, further, trigger more fissions causing more neutrons being, formed, which in turn may cause more fission. Thus, a self, sustained nuclear chain reaction is formed. To maintain, the nuclear chain reaction at a steady (sustained) level, the, extra neutrons produced, are absorbed by suitable neutron, absorbents like cadmium or boron., , 10, O, , →, , 2, , = [Zm p + ( A − Z )mn − M ] × 931.5 MeV, , 16, , 235, 92 U, , Controlled Chain Reaction, and Nuclear Reactor, , l, , Binding energy per nucleon (∆E bn ) is the average energy, needed to separate a nucleus into its individual nucleons., ∆Eb ∆M × 931 MeV, Thus,, ∆E bn =, =, A, A, Nucleon, , +, , In the fission of uranium, the percentage of mass converted, into energy is about 0.1% ., , If masses are expressed in atomic mass units, then, ∆Eb = ∆M × 931.5 MeV, l, , distance of 10 –14 m., • The density of nucleus is of the order of 10 17 ., , Nuclear fission is the process of splitting of a heavy nucleus, 239, (235, 92 U or 94 Pu) into two lighter nuclei of comparable, masses along with the release of a large amount of energy, (~, − 200 MeV ) after bombardment by slow neutrons., , Packing fraction of an atom is the difference between mass, of nucleus and its mass number per nucleon. Thus,, M−A, ., Packing fraction =, A, The energy equivalent of the mass defect of a nucleus is, called its binding energy., Thus, binding energy, ∆Eb = ∆M c2, = [Zm p + ( A − Z )mn − M ] c2, , • Nucleons attract each other when they are separated by a, , Nuclear Fission, , At the rest, mass energy of each of electron and positron, is, 2, , l, , The shown figure show binding energy per nucleon versus, mass number. The nuclides showing binding energy per, nucleon greater than 7.5 MeV/nucleon are stable., , c 2 = 931.5 MeV/u, , A unit of energy may therefore be considered to be a unit of, mass. For example, the electron has a rest mass of about, 0.5 MeV., If the principle of conservation of energy is to hold for nuclear, reactions it is clear that mass and energy must be regarded as, equivalent. The implication of E = mc2 is that any reaction, producing an appreciable mass decrease is a possible source, of energy., , 361, , 50, , 100, 150, Mass number (A), , 200, , 250, , Reproduction factor (k ) of a nuclear chain reaction is defined, as, Rate of production of neutrons, k =, Rate of loss / Absorption of neutrons
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362, , l, , l, , l, , DAY THIRTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , If k = 1, then the chain reaction will be steady and the, reactor is said to be critical., If k > 1, then the chain reaction is accelerated and it may, cause explosion in the reactor. Such a reactor is called, super-critical., If k < 1, then chain reaction gradually slows down and, comes to a halt. Such a reactor is called sub-critical., , The reactors giving fresh nuclear fuel which often exceeds the, nuclear fuel used is known as breeder reactor., , Nuclear Fusion, Nuclear fusion is the process, in which two or more light, nuclei combine to form a single large nucleus., , The mass of the single nucleus, so formed is less than the sum, of the masses of parent nuclei and this difference in mass,, results in the release of tremendously large amount of energy., The fusion reaction going on in the central core of sun is a, multistep process, but the net reaction is, 4 11H + 2 ε – → 42He + 2 ν + 6γ + 26.7 MeV, When two positively charged particles (protons or deuterons), combine to form a larger nucleus, the process is hindered by, the Coulombian repulsion between them., To overcome the Coulombian repulsion, the charged particles, are to be given an energy of atleast 400 keV., For this, proton/deuterons must be heated to a temperature of, about 3 × 10 9 K. Nuclear fusion reaction is therefore, known as, thermo nuclear fusion reaction., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 Two nucleons are at a separation of 1 fm. The net force, between them is F1 if both are neutrons, F2 if both are, protons and F3 if one is a proton and the other is a, neutron., (a) F1 > F2 > F3, (c) F1 = F3 > F2, , (b) F2 > F1 > F3, (d) F1 = F2 > F3, , number Z ) emits 3 α-particles and 2 positrons. The ratio of, number of neutrons to that of protons in the final nucleus, ª AIEEE 2010, will be, A−, (b), Z, A−, (d), Z, , Z, −, Z, −, , −4, 8, −4, 2, , β, , α, , α, , N0 → N1 → N2 → N3 → N4. If nucleon, number and atomic number of N2 are 176 and 71, respectively, then what are their values for N4 and N0?, (a) 168, 67 and 180, 71, (c) 180, 67 and 72, 180, , (b) 67, 168 and 180, 72, (d) None of these, , 4 A radioactive nucleus undergoes a series of decays, according to the scheme, α, , β, , γ, , α, , A → A1 → A2 → A3 → A4, If the mass number and atomic number of A are 180 and, 72 respectively, these numbers of A 4 are, (a) 172, 69, , (b) 177, 69, , (c) 171, 69, , (d) 172, 68, , 5 If Nt1 = N0 e −λ t1 , then the number of atoms decayed, , during the time interval from t1 and t 2 ( t1 > t 2 ), will be, (a) Nt 1 − Nt 2 = N 0 [e −λt 1 − e −λt 2 ], (b) Nt 2 − Nt 1 = N 0 [e −λt 2 − e −λt 1 ], (c) Nt 2 − Nt 1 = N 0 [ e −λt 2 − e λt 1 ], (d) None of the above, , (a) t1 + t 2, , (b) t1 − t 2, , (c) (t1 + t 2 ) / 2 (d), , t1 t 2, t1 + t 2, , probability that a nucleus will decay in two half-lives is, (a), , 1, 4, , (b), , 3, 4, , (c), , 1, 2, , (d) 1, , 8 The half-life of a radioactive substance is 20 min. The, approximate time interval ( t 2 − t1) between the time t 2, 2, 1, when of it has decayed and time t1 when of it had, 3, 3, ª AIEEE 2011, decayed is, , 3 The sequence of decay of a radioactive nucleus is, α, , Half-life for the first process is t1 and for the second, process is t 2. The effective half-life is, , 7 Half-life of a radioactive substance A is 4 days. The, , 2 A radioactive nucleus (initial mass number A and atomic, , A−Z −8, (a), Z −4, A − Z − 12, (c), Z −4, , 6 A radioactive sample decays by two different processes., , (a) 14 min, , (b) 20 min, , (c) 28 min, , (d) 7 min, , 9 A sample of a radioactive element has a mass of 10 g at, an instant t = 0. The approximate mass of this element in, the sample after two mean lives is, (a) 1.35 g, , (b) 2.50 g, , (c) 3.70 g, , (d) 6.30 g, , 10 When uranium is bombarded with neutrons, it undergoes, fission. The fission reaction can be written as, 235, + 0 n1 → 56 Ba 141 + 36 Kr 92 + 3X + Q (energy), 92 U, where three particles names X are produced and energy, Q is released. What is the name of the particle X ?, ª JEE Main (Online) 2013, , (a) electron, , (b) α-particle (c) neutron, , 11 On fission of one nucleus of U, , 235, , (d) neutrino, , , the amount of energy, obtained is 200 MeV. The power obtained in a reactor is, 1000 kW. Number of nuclei fissioned per second in the, reactor is, (a) 3.125 × 1016, (c) 3.125 × 1032, , (b) 6.25 × 1010, (d) 6.25 × 1020
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NUCLEI, , DAY THIRTY THREE, 12 If Mo is the mass of an oxygen isotope 8O17,M p and Mn, are the masses of a proton and a neutron, respectively the, nuclear binding energy of the isotope is, (a) (Mo − 8M p )c 2, (c) Mo c 2, , (b) (M o − 8 M p − 9Mn )c 2, (d) (Mo − 17Mn )c 2, , 13 The binding energies per nucleon of Li 7and He 4 are, 5.6 MeV and 7.06 MeV respectively, then the energy of, the reaction Li 7 + p = 2 [ 2 He 4 ] will be, (a) 17.28 MeV (b) 39.2 MeV (c) 28.24 MeV (d) 1.46 MeV, , 14 The below is a plot of binding energy per nucleon E b ,, against the nuclear mass M; A, B, C, D, E, F correspond, to different nuclei., , 363, , 15 Statement I A certain radioactive substance has a, half-life period of 30 days. Its disintegration constant is, 0.0231 day −1., Statement II Decay constant varies inversely as half-life., , 16 Statement I Half-life of a certain radioactive element is, 100 days. After 200 days, fraction left undecayed will be, 50%., n, , Statement II, , N 1, = , where symbols have usual, N 0 2, , meaning., , 17 Statement I In a decay, daughter nucleus shifts two, places to the left from the parent nucleus., Statement II An alpha particle carries four units of mass., , B, , Eb, , C, , D, , 18 Statement I Energy is released in nuclear fission., , E, , A, , F, M, , Consider four reactions, (i) A + B → C + ε, (iii) D + E → F + ε and, , ª [AIEEE 2010], , (ii) C → A + B + ε, (iv) F → D + E + ε, , where ε is the energy released. In which reactions is ε, positive?, (a) (i) and (iv), (c) (ii) and (iv), , (b) (i) and (iii), (d) (ii) and (iii), , Statement II Total binding energy of the fission fragments, is larger than the total binding energy of the parent, nucleus., , 19 Statement I If half-life period and the mean-life of a, radioactive element are denoted by T and Tm, respectively, then T < Tm ., 1, Statement II Mean-life =, decay constant, , 20 Statement I Energy is released when heavy nuclei, undergo fission or light nuclei undergo fusion., , Direction (Q. Nos. 15-21) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , Statement II For heavy nuclei, binding energy for per, nucleon increases with increasing Z while for light nuclei., It decreases with increasing Z., , 21 Statement I A nucleus having energy E 1 decays by β −, emission to daughter nucleus having energy E 2, but β −, rays are emitted with a continuous energy spectrum, having end point energy E 1 − E 2., , Statement II To conserve energy and momentum in, β-decay, atleast three particles must take part in the, ª AIEEE 2011, transformation., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, −α, , −α, , 1 Consider x → y → z , where half-lives of x and y are, z year and one month. The ratio of atoms of x and y, when transient equilibrium [T1/ 2( x ) > T1/ 2( y )] has been, established is, (a) 1 : 22, , (b) 1 : 26, , (c) 26 : 1, , (d) 23 : 1, , 2 In a nuclear reactor, U 235 undergoes fission liberating, 200 MeV of energy per fission. The reactor has 10%, efficiency and produces 1000 MW power. If the reactor is, , to function for 10 yr, the total mass of uranium required is, (Avogadro’s number = 6.02 × 1026/K-mol,1 eV=1.6 × 10−19 J), (a) 3.84 × 104 kg, (c) 3.84 × 108 kg, , (b) 9.28 × 106 kg, (d) 9.28 × 104 kg, , 3 The half-life of a radioactive sample is 10 h. The total, number of disintegration in 10th hour measured from a, time when the activity was one Ci is, (a) 053, . × 10−3, (c) 2.63 × 10−3, , (b) 6.91 × 1013, (d) 9.91 × 1013
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364, , DAY THIRTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , 4 A piece of wood from the ruins of an ancient building was, found to have a 14C activity of 12 disintegrations per, minute per gram of its carbon content. The 14C activity of, the living wood is 16 disintegrations per minute per gram., How long ago did the tree, from which the wooden, sample came, die? Given, half-life of 14C is 5760 yr ?, (a) 2391 yr, , (b) 2300 yr, , (c) 2250 yr, , calculate how much bigger that B must E be for such a, process to happen?, (a), , 2 mc, , 2, , (b), , B, 2 mc, , 2, , (c), , B2, 4 mc, , (d), , 2, , 3B, 4 mc 2, , 8 Assume that a neutron breaks into a proton and an, electron. The energy released during this process is, (mass of neutron = 1.6725 × 10−27 kg, mass of proton, = 1.6725 × 10−27 kg, mass of electron = 9 × 10 −31 kg), , (d) 2261 yr, , 5 A radioactive sample S1 having an activity of 5 µCi has, twice the number of nuclei as another sample S 2 which, has an activity of 10 µCi. The half-lives of S1 and S 2 can be, (a) 20 yr and 5 yr, respectively, (b) 20 yr and 10 yr, respectively, (c) 10 yr each, (d) 5 yr each, , B2, , ª AIEEE 2012, , (a) 0.9 MeV, , (b) 7.10 MeV (c) 6.30 MeV (d) 5.4 MeV, , 9 A radioactive nucleus A with a half-lifeT , decays into a, nucleus B. At t = 0, there is no nucleus B. After sometime, t, the ratio of the number of B to that of A is 0.3. Then, t is, given by, ª JEE Main 2017 (Offline), , 6 The half-life period of a radioactive element X is same as, , log 13, ., loge 2, T, (c) t =, log 13, ., , the mean life time of another radioactive element Y., Initially, they have the same number of atoms. Then,, (a) X will decay faster than Y, (b) Y will decay faster than X, (c) Y and X have same decay rate initially, (d) X and Y decay at same rate always, , (a) t = T, , (b) t = T log 13, ., (d) t =, , T loge 2, 2 log 13, ., , 10 Half-lives of two radioactive elements A and B are, 20 min and 40 min, respectively. Initially, the samples, have equal number of nuclei. After 80 min, the ratio of, decayed numbers of A and B nuclei will be, , 7 Deuteron is a bound state of a neutron and a proton with, a binding energy B = 2.2 MeV. A γ-ray of energy E is, aimed at a deuteron nucleus to try to break it into a, (neutron + proton) such that the n and p move in the, direction of the incident γ-ray. Where E ≠ B. Then,, , ª JEE Main 2016 (Offline), , (a) 1 : 16, , (b) 4 : 1, , (c) 1 : 4, , (d) 5 : 4, , ANSWERS, SESSION 1, , 1 (c), 11 (a), 21 (a), , 2 (b), 12 (b), , 3 (a), 13 (a), , 4 (a), 14 (a), , 5 (a), 15 (a), , 6 (d), 16 (c), , 7 (b), 17 (b), , 8 (b), 18 (a), , 9 (a), 19 (b), , 10 (c), 20 (c), , SESSION 2, , 1 (d), , 2 (a), , 3 (b), , 4 (a), , 5 (a), , 6 (b), , 7 (c), , 8 (a), , 9 (a), , 10 (d), , Hints and Explanations, SESSION 1, 1 Nuclear force of attraction between, any two nucleons (n - n, p - p, p - n ) is, same. The difference comes up only, due to electrostatic force of repulsion, between two protons., ∴, F1 = F3 ≠ F2, As,, F2 < F3 or F1, ∴, F1 = F3 > F2, , 2 In positive β-decay a proton is, transformed into a neutron and a, positron is emitted., , p + → n 0 + e +, Number of neutrons initially was, A−Z, Number of neutrons after decay is, (A − Z)− 3 × 2, (due to α-particles) − 2 × 1, (due to positive β-decay), As [3 × 2 (due to α-particles) + 2, (due to positive β-decay)], Hence, atomic number reduces by 8., , So, the ratio of number of neutrons to that, A−Z −4, of protons =, Z −8, , 3 As mass number of each α-particle is 4, units and its charge is 2 units, therefore for, N4, A = 176 − 8 = 168, and Z = 71 − 4 = 67, Now, the charge of β is −1 and its mass, number is zero., So, A = 176 + 0 + 4 = 180, and Z = 71 − 2 + 2 = 71
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NUCLEI, , DAY THIRTY THREE, 4 As the mass number of each a-particle, is 4 units and its charge is 2 unit., Therefore for A 4 ,, Mass number = 180 - 8 = 172, and Z = 72 - 4 + 1 (due to b - ) = 69, , 5 Since, N t1 = N 0e -lt1 and N t2 = N 0e -lt2, Then, the number of atoms decayed, during the time interval t 1 to t 2 is, = N t1 - N t2 = N 0 [e -lt1 - e -lt2 ], , t =, , t1 t2, t1 + t2, , 7 After two half-lives 1/4 th fraction of, nuclei will remain undecayed. Or, 3/4th, fraction will decay. Hence, the, probability that a nucleus decays in two, half-lives is 3 / 4 ., 8 N 1 = N 0 - 1 N 0 = 2 N 0,, 3, 3, 2, 1, N2 = N 0 - N 0 = N 0, 3, 3, n, N2 æ1 ö, We have,, =ç ÷, N1 è2 ø, Here, n = 1, \ t 2 - t 1 = one half-life = 20 min, , 9 The relation of mean-life and decay, constant is,, , Then we get from the equation,, m = m 0e -lt, m = 10 ´ e -l ´2/l = 10 ´ e -2, , Þ, , = 10 ´ 0135, ., = 135, . g, , 10 The fission of, 235, , 92 U, , = 2 (4 ´ 7.06) - 7 ´ 5.6, = 56.48 - 39.2 = 17.28 MeV, , 14 Both fusion and fission reaction results, into tremendous amount of energy, release and nucleus/nuclei which has, higher binding energy per nucleon than, parent nuclei. So, option (a) is correct., , 235, 92 U, , 1, , + 0 n ® 56 Ba, , is represented by, 141, , + 36 Kr, , 16 Number of half-lives, t, 200, =, =2, T, 100, The fraction left undecayed is given by, n=, , n, , \, , 92, , +3 0 n1 + Q, The name of the particle X is neutron, ( 0 n1 )., , 11 Power received from the reactor is, P = 1000 kW, = 1000 ´ 1000 = 106 J s -1, Also, 1 MeV = 1.6 ´ 10-13 J, Number of nuclei fissioned per second, 106, =, 200 ´ 1.6 ´ 10-13, , 17 On adecay, charge number of parent, nucleus decreases by 2 units. As, classification or grouping of elements is, based on charge number, hence, daughter nucleus shifts two places to, the left from the parent nucleus., fission can occur because the total mass, energy will decrease; that is DEbn, (binding energy) will increase. We see, that for high mass nuclide ( A = 240), the, binding energy per nucleon is about, 7.6 MeV/nucleon. For the middle weight, nuclides ( A = 120), it is about, 8.5 MeV/nucleon. Thus, binding energy, of fission fragments is larger than the, total binding energy of the parent, nucleus., , 19 We know that half-life period T and, decay constant l are related by the, equation., 0.6931, …(i), T =, l, While mean-life T m is related with l by, the equation, 1, …(ii), Tm =, l, From Eqs. (i) and (ii), we get, , = 3.125 ´ 1016 s -1, or, , 12 Binding energy, BE = (M, =, , nucleons )c, (Mo - 8 M p - 9 M n )c 2, nucleus, , -M, , 2, , 2, , N æ1 ö æ1 ö, =ç ÷ =ç ÷, N 0 è2 ø è2 ø, 1, = = 25%, 4, , 18 According to concept of binding energy,, , 2, 1, , where t =, l, l, , t = 2t =, , E p = 2E (2 He ) - E ( Li), , nuclear reaction are related by a, relation, which is, 0.693, T1 /2 =, l, 0.693 0.693, Þ l=, =, = 0.0231 day -1, T1 /2, 30, , 1 1, 1 t2 + t1, =, +, =, t t1, t2, t1 t2, , or, , \, , 4, , 15 Half-life and decay constant for a, , 6 As, l = l1 + l2, Þ, , 13 The reaction is 3 Li7 + 1 p1 ¾® 2 (2 He 4 ), , T = 0.6931 T m, T < Tm, , 20 Here, Statement I is correct and, Statement II is wrong, which can be, directly concluded from binding energy, nucleon curve., , 365, , 21 In practicle situation, atleast three, particles take place in transformation, so, Energy of b-particle + Energy of third, particle = E1 - E2, Hence, energy of b-particle £ E1 - E2, , SESSION 2, 1 N2 =, , l1 N 1, l2 - l1, , (e, , - l1 t, , - e - l2 t ), , When (T1 /2 )1 > (T1 /2 )2 at transient, equilibrium, l1 < l2, e - l2 t < < e, \, , \, , N2 =, , l1 N 1e, , - l1 t, , - l1 t, , l2 - l1, l1 N 1, =, l2 - l1, l - l1, N1, = 2, N2, l1, 0.693, 0.693, 1, 2 ´ 12 23, =, =, 0.693, 1, 2 ´ 12, , 2 Energy generated by the reactor, 1000 ´ 106 W = 109 Js -1, Total energy generated in 10 yr is, E = (109 Js -1 ) ´ 10 ´ 365, ´ 24 ´ 60 ´ 60, = 1.97 ´ 1030 MeV, In the reactor 200 MeV energy is, liberated in the fission of nucleus of U235, atom., \ Total number of U235 atoms required is, 1.97 ´ 1030, = 0.985 ´ 1028, 200, 1 kmol that is 235 kg of U235 has, 6.02 ´ 1026 atoms Therefore, total mass of, U235 having 0.985 ´ 1028 atoms is, 235, ´ (0.985 ´ 1028 ), 6.02 ´ 1026, = 3.84 ´ 103 kg, Since, efficiency of reactor is 10%, actual, mass of U235 required is, 100, (3.84 ´ 103 ) ´, = 3.84 ´ 104 kg, 10, 3 As, - dN = lN ,, dt, 37, . ´ 1010 ´ 3.6 ´ 104, N =, 0.693, DN = - (N 0 e - l ´ 10 ´ 3600, - N 0 e - l ´ 9 ´ 3600 ), (Q DN = N 1 - N 2 ), 37, . ´ 1014 ´ 3.6, =, [0.535 - 0.5], 0.693, = 6.91 ´ 1013
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366, , DAY THIRTY THREE, , 40 DAYS ~ JEE MAIN PHYSICS, , From conservation of momentum,, E, …(ii), pn + p p =, c, 2, 2, As E = B , Eq. (i), p n + p p = 0, , 4 Given, R = 12 dis/min per g,, R 0 = 16 dis/min per g, T1 /2 = 5760 yr, Let t be the time span of the tree., According to radioactive decay law,, R, R = R 0e - l t or e l t = 0, R, Taking log on both the sides, R, lt log e e = log e 0, R, 16 ö, æ, lt = ç log 10 ÷ ´ 2.303, è, 12 ø, 2.303 (log 4 - log 3), l, = 2391.20 yr » 2391 yr, , t =, , 5 We know that,, Activity ( A ) = lN 0, For S 1 , A s1 = 5m Ci = l1 2 N 0, For S 2 , A s2 = 10m Ci = l2 N 0, As we know,, 0.693, T s1 1 /2 =, l1, 0.693, and, T s2 1 /2 =, l2, , Þ, , T s1 1 /2, , 2, , æE, ö, ç - p p÷, p2p, èc, ø, X =, +, 2m, 2m, 2 Ep p E 2, or 2 p2p + 2 - 2mX = 0, c, c, Using the formula of quadratic, equation, we get, , …(i), …(ii), pp =, , 2E, +, c, , 16 m X =, X =, , =4, , c2, , æ E2, ö, - 8 ç 2 - 2m X ÷, èc, ø, , 4 E2, c2, E2, 4 mc 2, , B2, , 4 mc, , 2, , neutron and proton are equal which do, not permit the breaking up of neutron, and proton. But if we take standard, mass of neutron as 1.6750 ´ 10-27 kg,, , 6 According to question,, T1 /2 ( X ) = t (l ), 0.693 1, =, lX, lY, lX, or, lY =, 0.693, Þ, lY > l X, So, Y will decay faster than X., Þ, , then, Energy released = mass defect ´ c 2, = (m n - m p - me ) ´ c 2, , =, , 7 Binding energy B = 2.2 MeV, , (1.6750 ´ 10-27 - 1.6725 ´ 10-27, - 9 ´ 10-31 ), , From the energy conservation law,, p2p, p2, = n +, 2m 2m, , 1.66 ´ 10-27, ´ 931.5 MeV, , …(i), , » 0.9 MeV, , N atoms, of B, A, , A, B, , Let N atoms decays, into B in time t, No, No – N, at t=0, atoms of A, NB, 3, Given,, . =, = 03, NA, 10, NB, 30, =, Þ, N A 100, , So,, N 0 = 100 + 30 = 130 atoms, By using, N = N 0 e - lt, 100 = 130e - lt, 1, = e - lt, 13, ., log 13, . = lt, is half-life, then, log e 2, l=, T, log e 2, log 13, . =, ×t, T, T × log (13, . ), t =, log e 2, , We have,, Þ, Þ, If T, , Þ, \, , 10 Given, 80 min = 4 half-lives of A = 2, , 8 According to given data, mass of, , So, only option (a) can be satisfied., , E - B = Kn + K p, , 4 E2, , 4, For the real value p p¢ the discriminant, is positive, æE, ö, 4 E2, = 8 ç 2 - 2mX ÷, 2, è c2, ø, c, , Therefore, by dividing Eqs. (i), and (ii), we get, 5 T s2 1 /2 2N 0, =, 10 T s1 1 /2 N 0, T s2 1 /2, , It only happen if p n = p p = 0, So, the Eq. (ii) cannot satisfy and the, process cannot take place., Let E = B + X , where X << B for the, process to take place., Put value of p n from Eq. (ii) in, Eq. (i), we get, , 9 Decay scheme is ,, , half-lives of B., Let the initial number of nuclei in each, sample be N., For radioactive element A,, N, N A after 80 min = 4, 2, Þ Number of A nuclides decayed, N, 15, =N =, N, 16 16, For radioactive element B,, N, N B after 80 min = 2, 2, Þ Number of B nuclides decayed, N, 3, =N = N, 4, 4, \ Ratio of decayed numbers of A and B, nuclei will be, (15/ 16)N, 5, =, (3 / 4)N, 4
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DAY THIRTY FOUR, , Electronic, Devices, Learning & Revision for the Day, u, , u, , u, , Energy Bands in Solids, Semiconductors, Semiconductor Diode, , u, , u, , I-V Characteristics Semiconductor Diode in, Forward and Reverse Bias, Diode as a Rectifier, , u, , Special Purpose Diodes, , u, , Transistor, , Energy Bands in Solids, According to band theory of solids, in a crystalline solid due to mutual interaction, among valence electrons of neighbouring atoms, instead of sharp energy levels, energy, bands are formed. Energy bands are of the following three types, (i) Valence band It is the energy band formed by a series of energy levels of valence, electrons actually present. Ordinarily, valence band is completely filled and, electrons in this band are unable to gain energy from external electric field. The, highest energy level in a valence band at 0 K is called fermi energy level., (ii) Conduction band The energy band having just higher energy than the valence band, is called conduction band. Electrons in conduction band are commonly called the, free electrons., (iii) Forbidden band The energy gap between the valence band and the conduction band, of a solid is called the forbidden energy gap E g or forbidden band. Width of, forbidden energy gap depends upon the nature of substance., , Semiconductors, l, , l, , l, , In semiconducting solids, the valence band is completely filled but conduction band, is completely empty and the energy gap between them is small enough (E g < 3 eV). At, absolute zero temperature, it behaves as an insulator., A pure semiconductor, in which no impurity of any sort has been mixed, is called, intrinsic semiconductor. Germanium (E g = 0.72 eV) and silicon (E g = 1.1 eV) are, examples of intrinsic semiconductors., Electrical conductivity of pure semiconductor is very small. To increase the, conductivity of a pure semiconducting material, it is doped with a controlled quantity, (1 in 10 5 or 10 6) of suitable impurity. Such a doped semiconductor is called an, extrinsic semiconductor., , PREP, MIRROR, Your Personal Preparation Indicator, u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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368, l, , DAY THIRTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , The number of electrons reaching from valence band to, conduction band,, − E / 2 kT, , n = AT 3 / 2e g, where, k = Boltzmann’s constant,, temperature and A = atomic weight., , T =, , absolute, , Superconductors, When few metals are cooled, then below a certain critical, temperature, their electrical resistance suddenly becomes, zero. In this state, these substances are called, superconductors, and, this, phenomena, is, called, superconductivity. Mercury become superconductor at 4.2 K,, lead at 7.25 K and niobium at 9.2 K., , impurities,, , The potential difference developed across the p-n junction, due to diffusion of electrons and holes is called the potential, barrier Vb (or emf of fictitious battery). For germanium diode, barrier potential is 0.3 V, but for Si diode, its value is 0.7 V., The barrier electric field developed due to it, is of the order of, 10 5 Vm–1., , Mobility of Charge Carriers, The mobility of a charge carrier is defined as the velocity, gained by its per unit electric field, i.e. µ = Vd / E ., l, , Types of Extrinsic Semiconductor, According to type of doping, semiconductor are of two types, , This layer containing immobile ions is called depletion layer., The thickness of depletion layer is approximately of the order, of 10 −6 m., , extrinsic, , Current in semiconductor is, i = ie + i h = eA(ne Ve + nhVh ), J, i, Conductivity, σ = =, = e(ne µe + nhµ h ), E AE, (where, J = current density = nqV ), , 1. n-type Semiconductor, To prepare an n-type semiconductor, a pentavalent, impurity, e.g. P, As, Sb is used as a dopant with Si or Ge., Such an impurity is called donor impurity, because each, dopant atom provides one free electron., In n-type semiconductor ne > > nh , i.e. electrons are, majority charge carriers and the holes are minority charge, carriers, such that ne ⋅ nh = n2i . An n-type semiconductor is, electrically neutral and is not negatively charged., Conductivity,, σ ≈ ne µe e, , 2. p-type Semiconductor, To prepare a p-type semiconductor, a trivalent impurity,, e.g. B, Al, In, Ga, etc., is used as a dopant with Si or Ge., Such an impurity is called acceptor impurity as each, impurity atom wants to accept an electron from the crystal, lattice. Thus, effectively each dopant atom provides a hole., In p-type semiconductor nh > > ne , i.e. holes are majority, charge carriers and electrons minority charge carriers, such, that nh ⋅ ne = n2i . A p-type semiconductor is electrically, neutral and is not positively charged., The number of free electrons in a semiconductor varies, with temperature as T 3 / 2 ., Conductivity,, σ ≈ nhµ he, , Semiconductor Diode, A p-n junction is obtained by joining a small p-type crystal, with a small n-type crystal without employing any other, binding material in between them. Whenever a p-n junction is, formed, electrons from n-region diffuse through the junction, into p-region and the holes from p-region diffuse into, n-region., As a result of which neutrality of both n and p-regions is, disturbed, and a thin layer of immobile negative charged ions, appear near the junction in the p-crystal and a layer of, positive ions appear near the junction in n-crystal., , I-V Characteristics of Semiconductor, Diode in Forward and Reverse Bias, When we join an external potential source, such that p-side of, p-n junction is joined to positve terminal of voltage source and, n-side to negative terminal of voltage source, the junction is, said to be forward biased and applied electric field E opposes, the barrier electric field Eb ., As a result, width of depletion layer is reduced and on, applying a voltage V > Vb , a forward current begins to flow., Resistance offered by p-n junction in forward bias is small, (about 10-50 Ω)., Eb, , E, , sr, s, r, r, s, s sr r, p r sr s n, + –, (a) FB, , E, Eb, r ss rr s, s ss rr r, r, s, p r ss rr s n, –, +, (b) RB, , IF, (mA), VR, , VF, IR (µA), (c), , If connections of potential source are reversed [Fig. (b)], i.e., p -side is connected to negative terminal of battery and n-side, to positive terminal, the junction is said to be reverse biased, and in this case E and Eb , being in same direction, are added, up. So, the depletion layer broadens and potential barrier is, fortified. Consequently, an extremely small leakage current, flows across the junction due to minority charge carriers and, junction resistance is extremely high (~, − 10 5 Ω). For a, sufficiently high reverse bias voltage (25 V or even more), the, reverse current suddenly increases. This voltage is called, Zener voltage or breakdown voltage or avalanche voltage., NOTE, , • A p-n junction behaves as a voltage controlled switch. In, forward bias, it acts like ON switch and in reverse bias as, OFF switch., • The p-n junction can be presumed as a capacitor, in which, the depletion layer acts as dielectric.
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ELECTRONIC DEVICES, , DAY THIRTY FOUR, , 369, , Diode as a Rectifier, , Light Emitting Diode (LED), , Junction diode allows current to pass only when it is forward, biased. So, if an alternating voltage is applied across a diode,, the current flows only in that part of the cycle, when the diode, is forward biased., This property is used to rectify alternating voltages and the, circuit used for this purpose is called a rectifier, and the, process is known as rectification., There are two types of rectifier diode as given below, 1. Half Wave Rectifier A rectifier, which rectifies only, one-half of each AC input supply cycle, is called a half, wave rectifier., , It is a specially designed diode made of GaAsP, GaP, etc., When used in forward biased, it emits characteristic, almost, monochromatic light. In reverse biased, it works like a normal, diode., l, , I-V Characteristics LEDs are current dependent devices, , 2. Full Wave Rectifier A rectifier, which rectifies both, halves of each AC input cycle is called a full wave, rectifier., The output of a full wave rectifier is continuous, but, pulsating in nature. However, it can be made smooth by, using a filter circuit., As output is obtained corresponding to both the half, cycles of the AC input supply, its efficiency is more than, that of half wave rectifier., NOTE, , • The ripple factor is defined as the ratio of rms value of AC, component in the output of the rectifier to the DC, component in the input., , There are few diodes which are designed to serve some special, purpose and application., , Zener Diode, , l, , n, p, , Zener Diode as Voltage Regulator The following circuit is, used for stabilising voltage across a load RL . The circuit, consists of a series voltage-dropping resistance R and a, Zener diode in parallel with the load RL ., +, , p, , n, , I, , Fluctuating, DC input, voltage, , –, , R, , +, n, , Zener, diode, , VZ, p, , Blue, , Green, , Yellow, , Forward voltage, , VF, , Photodiode, It is a special diode used in reverse bias which conducts only, when light of suitable wavelengths is incident on the junction, of diode. The energy of incident light photon must be greater, than the band gap of semiconductor (i.e. hν > E g ). Materials, used are Cds, Se, Zns, etc., , Solar Cell, , Special Purpose Diodes, , It is a highly doped p-n junction diode which is not, damaged by high reverse current. It is always used in, reverse bias in breakdown voltage region and is, chiefly used as a voltage regulator., , Amber, , Forward current, , A half wave rectifier gives discontinuous and pulsating, DC output. As no output is obtained corresponding to, alternate half cycles of the AC input supply, its efficiency, is quite low., , Red, , I (mA), , Infrared, , with its forward voltage drop (VF ) depending on the, forward biased LED current. Characteristics of light, emitting diode I-V are shown below, , Constant, RL DC output, voltage, –, , The Zener diode is selected with Zener voltage Vz equal to, the voltage desired across the load., , It is a special p-n junction, in which one of the, semiconductors is made extremely thin, so that solar radiation, falling on it reaches junction of diode without any absorption., A solar cell directly converts, solar energy into electrical, energy. Popularly used solar cells, Ni-cd, PbS cell, etc., , Transistor, A transistor is a combination of two p-n junctions joined in, series. A junction transistor is known as Bipolar Junction, Transistor (BJT). It is a three terminal device., Transistors are of two types, (i) n-p-n transistor,, (ii) p-n-p transistor, A transistor has three regions, (i) An emitter (E), which is most heavily doped, and is of, moderate size. It supplies large number of charge carriers,, which are free electrons in an n-p-n transistor and holes, in a p-n-p transistor., (ii) A base (B), which is very lightly doped and is very thin, (thickness ~ 10 −5m)., (iii) A collector (C), which is moderately doped and is, thickest.
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370, , DAY THIRTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , A transistor is symbolically represented as shown in figures., n-p-n, E, , l, , p-n-p, E, , C, , C, , CE, , l, , B, , B, , l, , (b), , (a), , (ii) Common Base (CB) configuration, (iii) Common Collector (CC) configuration., Generally, we prefer common emitter configuration, because, power gain is maximum in this configuration., , Characteristics of a Transistor, In common emitter configuration, variation of current on the, input side with input voltage (I B versus VBE ) is known as the, input characteristics, and the variation in the output current, with output voltage (IC versus VCE ) is known as output, characteristics. From these characteristics, we obtain the, values of following parameters, ∆VBE, Input resistance, ri =, ∆I B V = constant, , l, , l, , ∆ IC, ∆I B, , An electronic oscillator is a device that generates electrical, oscillations of constant amplitude and of a desired frequency,, without any external input., The circuit providing such oscillation, is known as a tank, oscillator, is using positive feedback., , I B = constant, , L′, C, B, , n - p- n, N, , C, , VCE = constant, , The current gain for common-emitter configuration, β ranges from 20 to 200., ∆ IC, β, Transconductance, g m =, =, ∆VBE ri, A transistor can be used as an amplifier. The voltage gain, of an amplifier will be given by, V, R, A V = o = β⋅ C, Vi, RB, where, RC and RB are net resistances in collector and base, circuits, respectively., , β, 1+ β, , Transistor as an Oscillator, , Inducting coupled, , β=, , or α =, , n- p -n C, A transistor consisting of, two p-n junctions, one, B, forward biased and the, other reverse biased can, E, Output, be used to amplify a weak Input, signal., The, forward, biased junction has a low, + –, +, –, resistance path, whereas, the, reverse, biased, junction has a high, resistance path. The weak input signal is applied across the, forward biased junction, and the output signal is taken across, the reverse biased junction., Since, the input and output currents are almost equal, the, output signal appears with a much higher voltage. The, transistor, thus acts as an amplifier. Common-emitter, configuration of transistor amplifier is given alongside., , CE, , ∆VCE, Output resistance, ro =, ∆ IC, , α, 1−α, , Transistor as an Amplifier, , l, , AC current gain,, , • Current gains α and β are correlated as, β=, , For proper functioning of a transistor, the emitter-base, junction is forward biased, but the collector-base junction is, reverse biased. In an n-p-n transistor, electrons flow from, emitter towards the base and constitute a current I E ., Due to larger reverse bias at base-collector junction, most of, these electrons further pass into the collector, constituting a, collector current IC . But a small percentage of electrons (less, than 5%) may combine with holes present in base. These, electrons constitute a base current I B . It is self evident, that, I E = IC + I B ., Action of p-n-p transistor is also same, but with one difference, that holes are moving from emitter to base and then to, collector., A transistor can be connected in either of the following three, configurations, (i) Common Emitter (CE) configuration, , l, , Value of α is slightly less than 1. In fact, 0.95 ≤ α ≤ 1., ∆P, Power gain = o = β2AC × Resistance gain, ∆Pi, , NOTE, , Transistor Action, , l, , In common base configuration, AC current gain is defined, ∆I, as α = C, ∆I V = constant, , L, , B1, , K, , Some of the properties of the oscillator are, l, , l, , l, , Oscillator is using positive feedback., To work as an oscillator,, |Aβ| = 1; β → feedback factor, 1, 1, ., f = frequency of oscillation =, ×, 2π, LC, , B2
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ELECTRONIC DEVICES, , DAY THIRTY FOUR, , 371, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The conductivity of a semiconductor increases with, increase in temperature because, (a) number density of free current carriers increases, (b) relaxation time increases, (c) Both number density of carriers and relaxation time, increases, (d) number density of current carriers increases, relaxation, time decreases but effect of decrease in relaxation time, is much less than increase in number of density, , 2 Carbon, silicon and germanium atoms have four valence, electrons each. Their valence and conduction bands are, separated by energy band gaps represented by, (E g ) C , (E g ) Si and (E g ) Ge , respectively. Which one of the, following relationship is true in their case?, (a) (E g ) C > (E g ) Si, (c) (E g ) C < (E g ) Ge, , (b) (E g ) C = (E g ) Si, (d) (E g ) C < (E g ) Si, , 3 In n-type silicon, which of the following statement is true, (a) Electrons are majority carriers and trivalent atoms are, the dopants, (b) Electrons are minority carriers and pentavalent atoms, are the dopants, (c) Holes are minority carriers and pentavalent atoms are, the dopants, (d) Holes are majority carriers and trivalent atoms are the, dopants, , (c) Ec and Ev decreases, but E g increases, (d) All Ec , E g and Ev decreases, , 6 In an unbiased p-n junction, holes diffuse from the, p -region to n-region because, (a) free electrons in the n- region attract them, (b) they move across the junction by the potential difference, (c) hole concentration in p-region is more as compared to, n- region, (d) All of the above, , 7 Application of a forward bias to a p-n junction, (a) increases the number of donors on the n-side, (b) increases the electric field in the depletion zone, (c) increases the potential difference across the depletion, zone, (d) widens the depletion zone, , 8 When forward bias is applied to a p-n junction, what, happens to the potential barrierVB and the width of, charge depleted region x ?, (a)VB, (b)VB, (c)VB, (d)VB, , 9 The temperature (T ) dependence of resistivity (ρ) of a, semiconductor is represented by, , (a) The number of free conduction electrons is significant in, C but small in Si and Ge, (b) The number of free conduction electrons is negligibly, small in all the three, (c) The number of free electrons for conduction is, significant in all the three, (d) The number of free electrons for conduction is, significant only in Si and Ge but small in C, , 5 If the lattice constant of this semiconductor is decreased,, then which of the following is correct?, Conduction, bandwidth, , ª AIEEE 2010, , ρ, , ρ, , 4 Carbon, silicon and germanium have four valence, electrons each. At room temperature, which one of the, ª AIEEE 2012, following statements is most appropriate?, , increases, x decreases, decreases, x increases, increases, x increases, decreases, x decreases, , (a), , (b), O, , ρ, , Eg, , Valence, bandwidth, , (c), , (d), O, , T, , (a) All Ec , E g and Ev increases, (b) Ec and Ev increases, but E g decreases, , O, , T, , 10 In figure , V0 is the potential barrier across a p-n junction,, when no battery is connected across the junction., 1, 2, 3, V0, , Ev, , T, , ρ, , Ec, Band gap, , O, , T
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372, , DAY THIRTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , (a) 1 and 3 both corresponds to forward bias of junction, , 15 The circuit has two oppositely connected ideal diodes in, parallel. What is the current flowing in the circuit?, , (b) 3 corresponds to forward bias of junction and 1, correspond to reverse bias of junction, , 4Ω, , (c) 1 corresponds to forward bias and 3 corresponds to, reverse bias of junction, , D1, D2, 12 V, , (d) 3 and 1 both corresponds to reverse bias of junction, , 2Ω, , 3Ω, , 11 For the given circuit of p-n junction diode, which of the, following statements is correct?, (a) 1.71 A, , p, , R, , n, , (c) 2.31 A, , (d) 1.33 A, , 16 In the following circuits, which one of the diodes is, reverse biased?, , –, , +, , (b) 2.00 A, , – 12V, , + 10 V, , V, (a), , (a) In forward biasing the voltage across R is V, (b) In forward biasing the voltage across R is 2 V, (c) In reverse biasing the voltage across R is V, (d) In reverse biasing the voltage across R is 2 V, , (b), , R, , R, , +5V, , – 10 V, + 5V, , 12 In the case of forward biasing of p-n junction, which one, of the following figures correctly depicts the direction of, the flow of charge carriers?, pp, , V Vb, , (d), , (c), , R, R, , VV Bb, n, n, , pp, , (a), , nn, , – 10V, , (b), V, , 17. The forward biased diode connection is, , b, , (a), , V VV BB b, pp, , nn, (d) None of these, , (c), , 13 A junction diode is connected to a 10 V source and103 Ω, rheostat figure. The slope of load line on the, characteristic curve of diode will be, , 2V, , +4V, , (b) –2V, , +2V, , (c) +2V, , –2V, , (d) –3V, , –3V, , ª JEE Main 2014, , 18 The reading of the ammeter for a silicon diode in the, given circuit is, , ª JEE Main 2018, 200Ω, , 103 Ω, 10 V, , 3V, , + –, −2, , (a) 10 AV, , −1, , −3, , (b) 10 AV, , −1, , −4, , (c) 10 AV, , −1, , −5, , (d) 10 AV, , −1, , 14 In a forward biased p-n junction diode, the potential, barrier in the depletion region willp be of the form, nn, , p, p, , Potential, P o t e n t ia l, b a r r ie r, barrier, , (a), , n, , p, , (b), , n, , P Potential, o t e n t ia l, b barrier, a r r ie r, , (a) 0, , (b) 15 mA, , (c) 11.5 mA, , (d) 13.5 mA, , 19 In a full-wave rectifier circuit operating from 50 Hz, mains frequency, the fundamental frequency in the, output would be, (a) 50 Hz, , (b) 25 Hz, , (c) 100 Hz, , (d) 70.7 Hz, , 20 A p-n junction (D) shown in the figure can act as a, rectifier. An alternating current source (V ) is connected in, ª AIEEE 2013, the circuit., , p, , p, (c), , n, , p, , n, P o t Potential, e n t ia l, b a r barrier, r ie r, , (d), , p, , n, , n, , P Potential, o t e n t ia l, b barrier, a r r ie r, , D, V~, , R
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ELECTRONIC DEVICES, , DAY THIRTY FOUR, I, , I, , (a), , (b), t, , t, , I, , I, , (c), , t, , 21 Zener breakdown in a semiconductor diode occurs,, , n-p-n transistor, the phase difference between the input, and the output voltages will be, ª JEE Main 2017 (Offline), (a) 90°, , (b) 135°, , (c) 180°, , (d) 45°, , 25 When A is the internal stage gain of an amplifier and B is, , when, , the feedback ratio, then the amplifier becomes as, oscillator if, , (a) forward current exceeds certain value, (b) reverse bias exceeds certain value, (c) forward bias exceeds certain value, (d) potential barrier is reduced to zero, , 22 The I-V characteristics of an LED is, R Y GB, , ª AIEEE 2013, , V, , V, , O, , O, , V, R, (d) Y, G, B, , I, , V, , O, , I, , 23 Identify the semiconductor devices whose, characteristics are as given below, in the order, ª JEE Main 2016 (Offline), (p),(q),(r),(s)., I, , (a) B is negative and magnitude of B = A / 2, (b) B is negative and magnitude of B = 1/ A, (c) B is negative and magnitude of B = A, (d) B is positive and magnitude of B = 1/ A, , Direction (Q. Nos. 26-30), , R, Y, G, (b), B, , I, , O, , (c), , (c) Solar cell, Light dependent resistance, Zener diode,, Simple diode, (d) Zener diode, Solar cell, Simple diode, Light dependent, resistance, , 24 In a common emitter amplifier circuit using an, , (d), t, , (a), , 373, , I, , Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below., (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 26 Statement I If forward current changes by 1.5 mA when, forward voltage in semiconductor diode is changed from, 0.5 V to 2 V, the forward resistance of diode will be 1 Ω., ∆Vf, Statement II The forward resistance is given by Rf =, ∆If, , 27 Statement I A Zener diode is used to get constant, (p), , V, , (q), , V, , voltage at variable current under reverse bias., Statement II The most popular use of Zener diode is as, voltage regulator., , I, Dark, (r), , 28 Statement I Light Emitting Diode (LED) emits, , I, Resistance, V, , (s), , V, Intensity, of light, , Illuminated, , Choose the correct order, (a) Simple diode, Zener diode, Solar cell, Light dependent, resistance, (b) Zener diode, Simple diode, Light dependent resistance,, Solar cell, , spontaneous radiation., Statement II LED are forward biased p-n junctions., , 29 Statement I When base region has larger width, the, collector current increases., Statement II Electron-hole combination in base results in, increases of base current., , 30 Statement I In a common-emitter transistor amplifier the, input current is much less than output current., Statement II The common-emitter transistor amplifier has, very high input impedance.
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374, , DAY THIRTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The input resistance of a common-emitter transistor, amplifier, if the output resistance is 500 k Ω, the current, gain α = 0.98 and the power gain is 6.0625 × 106 is, (a) 198 Ω, , (b) 300 Ω, , (c) 100 Ω, , (d) 400 Ω, , 10 V, , (a), , (b), , 2 If the resistivity of copper is 1.7 × 10−6 Ω-m, then the, mobility of electrons in copper, if each atom of copper, contributes one free electron for conduction is, [the atomic weight of copper is 63.54 and density is, 8.96 g/cc], (a) 23.36 cm2 / Vs, (c) 43.25 cm2 / Vs, , 5V, (c), , (b) 503.03 cm2 / Vs, (d) 88 cm2 / Vs, , 3 A red LED emits light at 0.1 W uniformly around it. The, amplitude of the electric field of the light at a distance of, 1 m from the diode is, ª JEE Main 2015, , (d), –5 V, , 7 To plot forward characteristic of p-n junction diode, the, correct circuit diagram is, , E, , +, V – (0 – 1) V, , A, , +, –, , (b), , E, , +, V – (0 – 2) V, , A, , +, –, , (c), , E, , +, V – (0 – 2) V, , A, , –, +, , (d), , E, , +, V – (0 – 2) V, , A, , –, +, , (0 – 1000) mA, , R is tested using a multimeter., , (a), , (0 – 1000) mA, , 4 A working transistor with its three legs marked P, Q and, , (0 – 1000) mA, , (b) 2.45 V/m, (d) 7.75 V/m, , (0 – 20) A, , (a) 1.73 V/m, (c) 5.48 V/m, , –10 V, , No conduction is found between P and Q. By conneting, the common (negative) terminal of the multimeter to R, and the other (positive) terminal to P or Q. Some, resistance is seen on the multimeter. Which of the, following is the true for the transistor?, ª AIEEE 2013, (a) It is an n-p-n transistor with R as base, (b) It is a p-n-p transistor with R as collector, (c) It is a p-n-p transistor with R as emitter, (d) It is an n-p-n transistor with R as collector, , 5 A piece of copper and another of germanium are cooled, from room temperature to 77 K, the resistance of, (a), (b), (c), (d), , each of them increases, each of them decreases, copper decreases and germanium increases, copper increases and germanium decreases, , 6 If in a p-n junction diode, a square input signal of 10 V is, applied as shown., 5V, RL, , –5 V, , Then, the output signal across RL will be
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ELECTRONIC DEVICES, , DAY THIRTY FOUR, , 375, , 10 The length of germanium rod is 0.928 cm and its area of, , 8 A figure is given below, D3, , cross-section is 1 mm 2. If for germanium, , 5Ω, , D1, , ni = 2.5 × 1019m −3, µh = 019, . m 2 V −1 s −1,, µe = 0.39 m 2V −1s −1, then resistance is, , 10Ω, , 20Ω, , (a) 2.5 k Ω, , (c) 5.0 k Ω, , (d) 10.0 k Ω, , 11 In the figure, potential difference between A and B is, , 5Ω, , D2, , (b) 4.0 k Ω, , A, 10 k Ω, , 10 V, , 30 V, , (a) 0.5 A, (c) 1.5 A, , 10 k Ω, , 10 k Ω, , The current through the battery is, (b) 1 A, (d) 2 A, , 9 In the given circuit, the current through the zener diode is, , B, , (a) 10 V, , I, , (b) 20 V, , (c) 30 V, , (d) 40 V, , 12 In a common-base mode of a transistor, the collector, R1, , current is 5.488 mA for an emitter current of 5.60 mA. The, value of the base current amplification factor (β) will be, , 500Ω, , 15 V, , ª AIEEE 2011, R2, , (a) 10 mA, (c) 5 mA, , (a) 49, 1500Ω, , (b) 50, , (c) 51, , (d) 48, , 13 For a common emitter configuration, if α and β have their, , Vz=10 V, , usual meanings, the incorrect relationship between α and, ª JEE Main 2016 (Offline), β is, , (b) 6.67 mA, (d) 3.33 mA, , (a), , 1 1, β, = + 1 (b) α =, 1− β, α β, , (c) α =, , β, 1+ β, , (d) α =, , β2, 1+ β2, , ANSWERS, SESSION 1, , SESSION 2, , 1 (d), , 2 (a), , 3 (c), , 4 (d), , 5 (c), , 6 (c), , 7 (a), , 8 (d), , 9 (c), , 10 (b), , 11 (a), , 12 (c), , 13 (b), , 14 (d), , 15 (b), , 16 (d), , 17 (c), , 18 (c), , 19 (c), , 20 (c), , 21 (b), , 22 (a), , 23 (a), , 24 (c), , 25 (d), , 26 (d), , 27 (a), , 28 (b), , 29 (d), , 30 (c), , 1 (a), 11 (a), , 2 (c), 12 (a), , 3 (b), 13 (a,c), , 4 (a), , 5 (c), , 6 (d), , 7 (b), , 8 (c), , 9 (d), , 10 (b), , Hints and Explanations, SESSION 1, , (E g ) C = 5.2 eV,, (E g ) Si = 1.21 eV, , 1 Based on the theory discussed, we can, conclude that when temperature, increases, number density of current, carriers increases in the, semiconductor, relaxation time, decreases but effect of relaxation time, will be ignored., , 2 Carbon, silicon and germanium are, semiconductors., , and, , (E g ) Ge = 0.75eV, , Thus, (E g ) C > (E g ) Si, and, , (E g ) C > (E g ) Ge, , 3 n-type is obtained by doping the Ge or, Si with pentavalent atoms. In n-type, semiconductor, electrons are majority, carriers and holes are minority, carriers, hence option (c) is correct., , 4 The number of free electrons for, conduction is significant only in Si and Ge, but small in C, as C is an impurity., , 5 If lattice constant of semiconductor is, decreased, then Ec and E v decreases but E g, increases., , 6 In an unbiased p -n junction, the diffusion, of charge carriers across the junction takes, place from higher concentration to lower, concentration. Thus, option (c) is correct.
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376, , DAY THIRTY FOUR, , 40 DAYS ~ JEE MAIN PHYSICS, , 7 In forward biasing more number of, , 16 For reverse biasing of an ideal diode,, , electrons enter in n-side from battery, thereby increasing the number of, donors on the n-side., , the potential of n-side should be higher, than potential of p-side. Only option (d), is satisfying the criterion for reverse, biasing., , 8 In a p-n junction in forward bias, potential barrier V B as well as the width, of charge depleted region x decreases., , 17 For forward biased condition of a p-n, junction, p-junction should be at higher, potential and n-junction should be at, lower potential. So, option (c) is correct., , 9 The resistivity of a semiconductor, decreases with increase in temperature, exponentially. Hence, option (c) is, correct., , 10 When p-n junction is forward biased, it, opposes the potential barrier across, junction. When is reverse biased, it, supports the same., , 11 In forward biasing for an ideal diode, resistance of diode is zero and whole, resistance in the circuit is R. Hence,, voltage across R is V., , 12 Forward bias is obtained when the, negative terminal of the battery is, connected to the n-side and the positive, terminal to the p -side of the, semiconductor. Then, the negative, terminal will repel free electrons in the, n-section towards the junction and the, positive terminal on the p-side will, push the holes towards the junction., , 18 Potential drop in a silicon diode in, forward bias is around 0.7 V., In given circuit, potential drop across, 200 Ω resistor is, 3 − 07, ., ∆V net, =, I =, R, 200, ⇒ I = 0.0115 A ⇒ I = 11.5 mA, 19 Given, f = 50 Hz and T = 1, 50, T, 1, For full-wave rectifier, T1 =, =, 2 100, and f1 = 100 Hz, , and I is the circuit current, then, E V, V, E, V + IR = E or I = − = − +, R R, R R, Slope of load line, 1, 1, =− =, = 10−3 AV −1, R 1000, potential barrier decreases and becomes, less. Though in both the options (c) and, (d). The diode is forward biased but in, (d ) the barrier width is less., , diode occurs when reverse bias exceeds, certain value, which is known as, breakdown or Zener or Avalanche, voltage., , 22 For same value of current higher value, , D2, 2Ω, , 12 V, , Vo, t, , t, (Output), , In a CE n-p-n transistor amplifier output, is 180° out of phase with input., , 25 The condition for a circuit to oscillate, are, (i) feedback should be positive, (ii) output voltage feedback B =, , Apply KVL to get current flowing, through the circuit, − 12 + 4 I + 2I = 0, or, , I =, , 12, = 2A, 6, , 26 R f =, , ∆V f, ∆I f, , =, , (2 − 0.5), 1.5 × 10−3, , = 103, , 1 Power gain = Current gain × Voltage, gain, α, 0.98, Current gain = β =, =, 1 − α 1 − 0.98, 500 × 103 , = 49 , , R1, , , , , R2 , As, A v = β R , 1, , Power gain = 6.0625 × 106, , region., So, Simple diode → (p), Zener diode → (q), Solar cell → (r), Light dependent resistance → (s), , (Input), , SESSION 2, , AV, , 23 Zener diode works in breakdown, , Vi, , has input resistance equal to 1 kΩ, (approx.) and output resistance equal to, 10 kΩ (approx.). The output current in, CE amplifier is much larger than the, input current., , = 49, , of voltage is required for higher, frequency., , 15 In the given circuit diode D1 is reverse, biased while D2 is forward biased, so the, circuit can be redrawn as, I 4Ω, , 30 The common-emitter transistor amplifier, , 21 Zener breakdown in a semiconductor, , 24, , energy is released at the junction due to, recombination of electrons and holes. In, the junction diode made up of gallium, arsenide or indium phosphide, the, energy is released in visible region. Such, a junction diode is called Light Emitting, Diode or LED. The radiated energy, emitted by LED is equal or less than band, gap of semiconductor., electron-hole combination increases the, base current. The output collector, current decreases by the relation,, I E = I B + IC ., , conducts only for positive half-cycle., Hence, output waveform is obtained for, half cycle only as in figure (c)., , 14 The diode is forward biased, hence the, , 28 When a junction diode is forward biased, , 29 When base region has larger width,, , 20 Given figure is half wave rectifier as diode, , 13 If V is the voltage across the junction, , continuously without being damaged., The Zener diode is used as a voltage, regulator as constant voltage at variable, current under reverse bias is obtained, from it., , 1, A, , Ω, , = 1 kΩ, , 27 Zener diodes are specially designed, junction diodes, which can operate in, the reverse breakdown voltage region, , 500 × 103 , = 49 × , × 49, R1, , , R1 = 198 Ω, 2 Mobility of electron (µ ) = σ, …(i), ne, 1, Resistivity (ρ) =, …(ii), σ, From Eqs. (i) and (ii), we get, 1, …(iii), µ =, neρ, where, n = number of free electrons per, unit volume, N0 × d, n=, atomic weight, 6.023 × 1023 × 8.96, 63.54, = 8.5 × 1022, , =, , …(iv), , From Eqs. (iii) and (iv), we get, 1, µ =, . × 10−6, 8.5 × 1022 × 1.6 × 10−19 × 17, = 43.25cm2 / Vs
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ELECTRONIC DEVICES, , DAY THIRTY FOUR, 3 Consider the LED as, a point source of, light., Let power of the, LED is P., Intensity at r from, the source, P, I =, 4 πr 2, , r, , P, , ...(i), , 1, ...(ii), ε0E 20 c, 2, From Eqs. (i) and (ii), we can write, P, 1, = ε0E 20 c, 4 πr 2 2, 2P, 2 × 01, . × 9 × 109, or E 20 =, =, 2, 4 πε0r c, 1 × 3 × 108, As we know that, I =, , or E 20 = 6 ⇒ E 0 =, , 6 = 2.45 V/m, , 4 Since, no conduction is found when, multimeter is connected across P and Q,, it means either both P and Q are n-type, or p-type. So, it means R is base, when, R is connected to common terminal and, conduction is seen when other terminal, is connected to P or Q. So, it means, transistor is n-p-n with R as base., , 5 We know that resistance of conductor is, directly proportional to temperature, (i.e. R ∝ ∆t ), while resistance of, semiconductor is inversely proportional, 1, to temperature i.e. R ∝, ., , ∆t , Therefore, it is clear that resistance of, conductor decreases with decrease in, temperature or vice-versa, while in case, of semiconductor, resistance increases, with decrease in temperature or, vice-versa. Since, copper is pure, conductor and germanium is a, semiconductor, hence due to decrease, in temperature, resistance of conductor, decreases while that of semiconductor, increases., , 6 During − ve cycle, diode will not allow, the signal to pass., 5V, , RL Vo, , Vi, , For V i < 0, the diode is reverse biased, and hence offer infinite resistance, so, circuit would be like as shown in Fig., (ii) and Vo = 0., For V i > 0, the diode is forward biased, and circuit would be as shown in Fig., (iii) and Vo = V i ., Hence, the option (d) is correct., , 7 For forward bias mode, the p-side of, diode has to be at higher potential than, n-side. The meters used are DC, so we, have to be careful while connecting, them w.r.t. polarity., Last point is to decide the range of, meters, the range of meters has to be in, such a way that we can have the, readings which leads to plot on realistic, scale. If we take 0-20 A ammeter, then, reading we read from this is tending to 0, to 5 divisions which is not fruitful., In options (c) and (d), the polarity of, ammeter is not correct. Hence, (b) is, correct circuit., , 8 In the given circuit, diode D1 is reverse, biased, so it will not conduct. Diode D2, and D3 are forward biased, so they, conduct. The equivalent circuit is as, shown below:, D3, 5Ω, , 20Ω, , 5Ω, , D2, , 10 V, Now, the equivalent resistance of circuit, is, (5 + 5) × 20 10 × 20, Req =, =, (5 + 5) + 20 10 + 20, 200 20, =, =, Ω, 30, 3, Then, current through the battery,, V, 10, 3, I =, =, =, Req, 20 / 3 2, , = 1.5A, , 9 The voltage drop across R2 is, –5 V, , V R2 = V2 = 10 V, , Fig. (i), , I, , R1, , Vo, , Vi, , Vo, , Vi, , 15 V, , IR2, R2, , Fig. (ii), , Fig. (iii), , 500Ω, , 1500Ω, , Iz, Vz=10 V, , 377, , The current through R2 is, V, 10, I R2 = R2 =, R2, 1500, = 0.667 × 10−2 A, = 6.67 mA, The voltage drop across R1 is, V R1 = 15V − V R2, = 15 − 10 = 5V, The current through R1 is, V, 5, I R1 = R1 =, = 10−2 A, R1, 500, = 10 mA, The current through the zener diode is, I z = I R1 − I R2 = (10 − 6.67)mA, = 333, . mA, , 10 ∴ R = ρl =, A, , =, , L, Q ρ = 1 , , , n i e (µ e + µ h )A , σ, 0.928 × 10−2, , [2.5 × 1019 × 1.6 × 10−19, (0.39 + 0.19) × 10−6], , = 4000 Ω or 4 k Ω, , 11 For forward biased p-n junction diodes, its resistance is zero., So, net resistance of circuit, 10 × 10, = 10 +, = 15kΩ, 10 + 10, V, 30, Net current I =, =, R 15 × 103, = 2 × 10−3 A, So, potential difference across, AB = 2 × 10−3 × 5 × 103 = 10V, 12 ∴ β = IC and I E = IC + I B, IB, IC, ∴ β=, I E − IC, =, , 5488, ., = 49, 560, . − 5488, ., , 13 As, we know, in case of a, common-emitter configuration, DC, current gain,, I, α = c., Ie, where, Ic is collector current and Ie is, emitter current, and AC current gain,, I, β = c., Ib, where, Ib is base current., Also,, Ie = Ib + Ic, Dividing whole equation by Ic , we get, Ie, I, ⇒, = b +1, Ic, Ic, 1 1, = +1, ⇒, α β, β, ⇒, α =, 1+ β
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DAY THIRTY FIVE, , Gate Circuit, Learning & Revision for the Day, u, , Logic Gates and Truth Table, , u, , The OR Gate, , u, , u, , The AND Gate, The NOT Gate, , u, , u, , Combination of Logic Gates, Transistor as a Switch, , Logic Gates and Truth Table, Logic Gates The digital circuit that can be analysed with the help of Boolean algebra is, called logic gate or logic circuit. A logic gate has two or more inputs but only one input., There are primarily three logic gates namely the OR gate, the AND gate and the NOT, gate., Truth Table The operation of a logic gate or circuit can be represented in a table which, contains all possible inputs and their corresponding outputs is called the truth table. To, write the truth table, we use binary digits 1 and 0., , The OR Gate, The OR gate is a device has two or more inputs and one output. This device combines, two inputs to give one output. The logic symbol of OR gate is, A, , Y, , B, , The Boolean expression for OR gate is, Y =A+B, This indicates Y equals A OR B., Truth table for OR gate (Y = A + B ), , PREP, MIRROR, Your Personal Preparation Indicator, , A, , B, , Y, , 0, , 0, , 0, , u, , 0, , 1, , 1, , u, , No. of Questions Attempted (y)—, , u, , No. of Correct Questions (z)—, , 1, , 0, , 1, , 1, , 1, , 1, , The output of an OR gate assumes 1 if one of more inputs assume 1. The output is high, when either of inputs A or B is high, but not if both A and B are low., , No. of Questions in Exercises (x)—, , (Without referring Explanations), u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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GATE CIRCUIT, , DAY THIRTY FIVE, , The logic symbol of NAND gate is shown as, , The AND Gate, The AND gate a device has also two or more inputs and one, output. The output of AND gate is equal to product of its, inputs. The logic symbol of AND gate is, A, B, , A, Y, B, , The Boolean expression of NAND gate is Y = A ⋅ B, which, indicates ‘A AND B are negated’., , Y, , The logic symbol of AND gate is given as under. The Boolean, expression for AND gate is Y = A ⋅ B, this indicates Y equals to, A AND B., , Truth table for NAND gate, A, 0, 1, 0, 1, , Truth table for AND gate (Y = A ⋅ B ), A, 0, 0, 1, 1, , 379, , Y, 0, 0, 0, 1, , B, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Y′, 0, 0, 0, 1, , Y, 1, 1, 1, 0, , 2. NOR Gate, , The output of an AND gate is 1 only, when all the inputs, assume 1., , In this type of gate, the output of OR gate is fed to input of the, NOT gate and final output is obtained at output of the NOT, gate., y′, , A, , The NOT Gate, , y, , B, , The NOT gate is a device which has only one input and only, one output. Its output is complement of input. The logic, symbol of NOT gate is as shown in figure., A, , The logic symbol of NOR gate is shown as, A, , y=A+B, , B, , Y, , The Boolean expression for NOR gate is Y = A + B, which, The Boolean expression for NOT gate is, Y = A,, which indicates Y equals NOT A., , indicates that ‘A OR B are negated’, Truth table for NOR gate, A, 0, 1, 0, 1, , Truth Table for NOT gate (Y = A ), A, , Y, , 0, , 1, , 1, , 0, , The output of a NOT gate assumes 1, if input is 0 and, vice-versa. These basic gates (OR, AND and NOT) can be, combined in various ways to provide large number of, complicated digital circuits., , Combination of Logic Gates, NAND gate and NOR gate are used to make any gate., , 1. NAND Gate, In this type of gate, the output of AND gate is fed to input of a, NOT gate and final output is obtained at output of NOT gate., A, B, , Y′, , Y, , NOTE, , B, 0, 0, 1, 1, , Y′, 0, 1, 1, 1, , Y, 1, 0, 0, 0, , • NAND and NOR gates are known as universal gate., • The Boolean expressions obey the commutative law,, associative law as well as distributive law., Commutative law, (i) A + B = B + A, (ii) A ⋅ B = B ⋅ A, Associative law, (iii) A + (B + C ) = ( A + B ) + C, (iv) ( A ⋅ B ) ⋅ C = A ⋅ (B ⋅ C ), Distributive law, (v) A ⋅ (B + C ) = A ⋅ B + A ⋅ C, (vi) A + A ⋅ B = A + B, (vii) A + A ⋅ B = A, (viii) A ⋅ ( A + B ) = A, (ix) A ⋅ ( A + B ) = A ⋅ B, (x) A ⋅ B = A + B, (xi) A + B = A ⋅ B, (xii) A = A
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380, , DAY THIRTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , (Vce ) across it. In both the cut-off and saturation regions, the, power dissipated by the transistor is at its minimum., , Transistor as a Switch, The circuit resembles that of the Common-Emitter circuit., The difference this time is that to operate the transistor as a, switch the transistor needs to be turned either fully ‘‘OFF’’, (Cut-off) fully ‘‘ON’’ (Saturated)., An ideal transistor switch would have an infinite resistance, when turned ‘OFF’ resulting in zero current flow and zero, resistance, when turned ‘‘ON’’, resulting in maximum current, flow., In practice, when turned ‘‘OFF’, small leakage currents flow, through the transistor and when fully “ON” the device has, a low resistance value causing a small saturation voltage, , Vcc, Load, Output, , Relay, , Flyback, Diode, , Ic, , ia, , RB, , Vce, , +, Vin Ω, –, , β, , R, , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The output of a two input OR gate is fed to a NOT gate,, , 6 Which of the following gates will have an output of 1 ?, , the new gate obtained is, (a) OR gate, (c) NOR gate, , (a), , (b) NOT gate, (d) NAND gate, , 2 Which of the following is the truth table for NOT gate?, 1 1, (a) , , 0 0, , 1 0, (b) , , 0 0, , 0 1, (c) , , 1 0, , 0 1, (d) , , 1 1, , (c), , 1, , (b), , 0, 0, , 0, 1, , (d) None of these, , 1, , 7 The following truth table is for, , 3 The output of OR gate is high, (a) if either or both the inputs are 1, (b) only if both inputs are 1, (c) if either input is zero, (d) if both inputs are zero, , 4 To get an output 1 from the circuit as shown in the figure,, the input must be, (a) NAND, , A, , A, , B, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , (b) AND, , Y, , (c) XOR, , (d) NOT, , 8 Which of the following is the truth table for the circuit, , B, Y, , below?, A, , Y, , C, , (a) A = 0, B = 1, C = 0, (b) A = 1, B = 0, C = 0, (c) A = 1, B = 0, C = 1, (d) A = 1, B = 1, C = 0, , (a), , 5 Digital circuit can be made by the repetitive use of, (a) OR gates, (c) NOT gates, , (b) AND gates, (d) NAND gates, , (c), , A, , Y, , 1, 0, , 0, 1, , A, , Y, , 1, 0, , 1, 1, , (b), , (d), , A, , Y, , 0, 1, , 0, 1, , A, , Y, , 0, 0, , 1, 0
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GATE CIRCUIT, , DAY THIRTY FIVE, 9 The circuit as shown in figure below will act as, , (c) NOR gate, Y, , B, , (a) OR gate, (c) XOR gate, , (d) NAND gate, , 15 The diagram of a logic circuit is given below. The output, , Y′, , A, , F of the circuit is given by, , (b) AND gate, (d) None of these, , W, X, , 10 The circuit as shown below will acts as, , F, , A, , W, Y, , Y, , (a)W ⋅ (X + Y ), (c)W + (X ⋅Y ), , B, , (a) AND gate, (c) NAND gate, , (b) OR gate, (d) NOR gate, , A, , NAND, , B, , NOT, , Y, , A, , Y, , B, , (a) AND gate, (c) NOR gate, , (b) NAND gate, (d) OR gate, , C, , 12 The output of an OR gate is connected to both the inputs, of a NAND gate. The combination will serve as a, ª AIEEE 2011, , (a) OR gate, (c) NOR gate, , figure is, , ª AIEEE 2012, , Y, B, , (c), , B, , Y, , 0, 0, 1, 1, , 0, 1, 0, 1, , 0, 1, 1, 0, , A, , B, , 0, 0, 1, 1, , 0, 1, 0, 1, , (b) A ⋅ (B ⋅ C), (d) A + B + C, , 17 What will be the input of A and B for the Boolean, (a) 0, 0, (c) 1, 0, , (b) 0, 1, (d) 1, 1, , 18 Which of the following is not equal to 1 in Boolean, algebra?, , A, , A, , (a) A ⋅ (B + C), (c) (A + B) ⋅ (A + C), , expression ( A + B ) ⋅ ( A ⋅ B ) = 1 ?, , (b) NOT gate, (d) AND gate, , 13 Truth table for system of four NAND gates as shown in, , (a), , (b)W ⋅ (X ⋅Y ), (d)W + (X + Y ), , 16 The output, Y of given logic circuit is, , 11 The circuit is equivalent to, NOR, , 381, , A, , B, , Y, , 0, 0, 1, 1, , 0, 1, 0, 1, , 0, 0, 1, 1, , Y, , A, , B, , Y, , 1, 1, 0, 0, , 0, 0, 1, 1, , 0, 1, 0, 1, , 1, 0, 0, 1, , (b), , (d), , (a) A + 1, , (b) A ⋅ A, , (c) A + A, , (d) A ⋅ A, , Direction (Q.Nos. 19-21) Each of these questions contains, two statements: Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes, (a),(b), (c),(d) given below., (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 19 Statement I The logic gate NOT cannot be built using, diode., , 14 The combination of gates shown below yields, A, , Statement II The output voltage and the input voltage of, the diode have 180° phase difference., , 20 Statement I NOT gate is also called invertor., X, , Statement II NOT gate inverts the input signal., , 21 Statement I NAND or NOR gates are called digital, , B, , ª AIEEE 2010, , (a) OR gate, , (b) NOT gate, , building blocks., Statement II The repeated use of NAND or NOR gates, can produce all the basic or complicated gates.
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382, , DAY THIRTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 The output of an OR gate is connected to both the inputs, , A, , B, , Y, , A, , B, , Y, , 1, (a), , 0, 1, 0, , 0, (b), , 1, 0, , 0, 1, 1, , 0, 1, 0, 1, , 1, 0, 0, 0, , A, , B, , Y, , 0, 0, 1, 1, , 0, 1, 0, 1, , 0, 0, 0, 1, , (c), , 4 In the following combinations of logic gates, the outputs, of A, B and C are respectively, , of a NAND gate , the truth table is, , 0, 1, 1, , (A) 1, 1, , 0, , (B) 0, 1, (C), , 1, 1, , 0, , (d) None of these, , 2 For the given combination of gates, if the logic states of, inputs A , B and C are as follows. A = B = C = 0 and, A = B = 1, C = 0, then the logic states of output D are, , (a) 0,1,1, (b) 0,1, 0, (c) 1,1, 0, (d) 1, 0,1, , 5 The truth table of the following combination of gates is, , A, Y, , A, , B, , B, , D, , C, , Inputs, , (a), , (a) 0, 0, , (b) 0, 1, , (c) 1, 0, , (d) 1, 1, , A, , 3 The real time variation of input signals A and B are as, shown below.If the inputs are fed into NAND gate, then, select the output signal from the following, , A, , A, , Y, , B, , B, , Outputs, , 0, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , Inputs, , Outputs, , A, , Y, , Y, , (a), , A⋅ B, , B, , Y, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , (b), , Inputs, , (c), 0, , (c), , Y, , 0, , (b), , t(s), , A⋅B, , B, , 2, , 4, , 6, , 8 t(s), , Y, , 0, , (d), , 0, , 2, , 4, , 6, , 8 t(s), , 2, , 4, , 6, , 8 t(s), , Y, , 0, , 2, , 4, , 6, , 8 t(s), , Outputs, , A, , B, , A+ B, , Y = A ⋅ ( A + B), , 0, , 0, , 0, , 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , 1, , (d) None of the above
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GATE CIRCUIT, , DAY THIRTY FIVE, , 383, , 6 The logic circuit shown below has the input waveforms A, and B as shown. Pick out the correct output waveform., , (a), , A, Y, , (b), , B, , (c), , Input A, , C, , A, B, A, B, , C, , A, B, , C, , Input B, (d), , (a), , A, B, , C, , 9 Consider two n-p-n transistors as shown in figure. If 0 V, corresponds to false and 5 V corresponds to true, then the, output at C corresponds to, , (b), , 5V, , (c), , C, 1, A, , (d), , 2, B, , 7 In the adjacent circuit, A and B represent two inputs and, C represents the output,, A, C, , ª JEE Main (Online) 2013, (b) A OR B, (d) A NOR B, , (a) A NAND B, (c) A AND B, , B, , 10 If a, b, c and d are inputs to a gate and x is its output, every time, then as per the following time graph, the gate, ª JEE Main 2016 (Offline), is, , The circuit represents, (a) NOR gate, (c) NAND gate, , (b) AND gate, (d) OR gate, , 8 Which of the following circuits has given outputs?, A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , a., , b., , c., , d., , x., , C, 0, 0, 1, 0, , (a) NOT, (b) AND, (c) OR, (d) NAND, ª JEE Main (Online) 2013, , ANSWERS, SESSION 1, , 1 (c), , 2 (c), , 3 (a), , 4 (c), , 5 (d), , 6 (c), , 7 (a), , 8 (a), , 9 (a), , 10 (a), , 11 (c), , 12 (c), , 13 (a), , 14 (a), , 15 (d), , 16 (c), , 17 (a), , 18 (b), , 19 (c), , 20 (a), , 2 (d), , 3 (b), , 4 (c), , 5 (c), , 6 (a), , 7 (d), , 8 (c), , 9 (a), , 10 (c), , 21 (a), SESSION 2, , 1 (b)
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384, , DAY THIRTY FIVE, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, SESSION 1, , = (A + B)+ (A + B) = A + B, Output of NOT gate,, Y = Y2 = A + B, , 1 The combination of OR and NOT gates, is NOR gate., , written as, Y = ( A + B ) × ( AB ), , which is output of NOR gate., , 2 Truth table of NOT gate is, A, , 12 A, , B, , B, 1, 0, , = ( A × B )× ( A + B ), , Y′, , B, , A, 0, 1, , 17 The given Boolean expression can be, , = A×B + A×B, = A×B, , Y ¢= A + B and Y = Y ¢ = A + B, , So, the truth table is, , i.e. output of a NOR gate., , 13 Boolean expression for the given circuit, , 3 OR gate output is high, if anyone or, both input are high., , Y = (( A × ( A × B ))) × (B × ( A × B )), , 4 The Boolean expression for the given, , = ( A + A × B ) × (B + A × B ), , combination is Y = ( A + B ) × C, The truth table is, , = ( A + A × B ) + (B + ( A × B )), , A, , B, , C, , A+ B, , Y = (A + B) C, , 0, 0, 0, 0, 1, 1, 1, 1, , 0, 0, 1, 1, 0, 0, 1, 1, , 0, 1, 0, 1, 0, 1, 0, 1, , 0, 0, 1, 1, 1, 1, 1, 1, , 0, 0, 0, 1, 0, 1, 0, 1, , A, , B, , A, , B, , 0, 0, 1, 1, , 0, 1, 0, 1, , 1, 1, 0, 0, , 1, 0, 1, 0, , B× A, , Y, , gate gives digital circuits., , A, 0, 0, 1, 1, , 6 For option (c), it is a NAND gate, its, output = 01, . = 0=1, , 7 For NAND gate, Y = AB, 8 The output of the NAND gate is, , 0, 1, 0, 0, , 0, 1, 1, 0, , B, 0, 1, 0, 1, , X, 0, 1, 1, 1, , 1, , 0, , other complex gates., , 1 When two inputs of a NAND gate are, joined together, it works as a NOT gate., The OR gate connected to this NOT gate, results is a NOR gate., , =W + X + Y, , 16 The gate circuit is given as, A, , (digital) building blocks because using, these two types of gates we can produce, all the basic gates namely OR, AND or, , SESSION 2, , =W + X +W + Y, , NOR gate., , 2 The output D for the given combination, , A+B, , D = ( A + B )× C = ( A + B ) + C, , B, , two points A and B., , If A = B = C = 0, then, Y, , NAND, , NOT, Y2, , B, , Output of NOR gate, Y1 = A + B, Output of NAND gate,, Y2 = Y1 × Y1 = A + B × A + B, = A+ B+ A+ B, , 0, , 1, , 21 NAND or NOR gates are called universal, , = (W + X ) + (W + Y ), , 10 The output of two NOT gate is input for, , Y1, , 1, , input is 1 then output will be zero or, vice-versa. Therefore, it is called as, invertor. NOT gate inverts the input, order means that for low input, it gives, high output or for high input, it gives low, output., , 15 The output F = (W + X ) × (W + Y ), , Y = A + B= A+ B, , NOR, , 0, , 0, , 20 NOT gate inverts the input signal i.e. if, , This comes out to be truth table of OR, gate., , Y = A× A = A + A = A, , 11 The gate circuit can be shown by given, , 0, , But in diode, the input and output are in, same phase. Thus, NOT gate cannot be, built using diode., , 14 Truth table for given combination is, , Hence, Y = A + B = A × B = A × B, (AND gate), , 1, , 1, , 19 NOT gate inverts the signal applied to it., A× B, 0, 0, 1, 0, , So, option (a) is correct., , for NOT gate., , Y, , 0, , A = 0 or 1., , = A×B + B× A, , 9 The output of NOR gate is made input, , B, , 0, , 18 Here, A × A = 0 always when either, , = A ×(A + B) + B ×(A + B), , 5 The repetitive use of NAND and NOR, , A, , = A ×(A × B) + B ×(A × B), , Hence, A = 1, B = 0 and C = 1, , A, , = ( A × A ) × B + A( B × B ), , Y, , A+C, , Y, , C, Y =(A+B) (A+C), For this circuit, output, , Y = (A + B)(A + C), , D = (0 + 0) + 0, = 0+ 0, =1+ 1=1, If A = B = 1, C = 0, then, D = (1 + 1) + 0, =1+ 0, = 0+ 1 = 1
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GATE CIRCUIT, , DAY THIRTY FIVE, , 3 From real time variation of input, , Inputs, , signals,we can from truth table for A, and B and conclude output from NAND, gate., Inputs, A, 0, 1, 0, 1, 0, , and NOT gate, , Outputs, , A, , B, , A+B, , Y = A × (A + B), , 0, , 0, , 0, , 0, , Output, , 0, , 1, , 1, , 0, , Y, , 1, , 0, , 1, , 1, , 1, 1, 1, 0, 1, , 1, , 1, , 1, , 1, , B, 0, 0, 0, 1, 0, , V=6V, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , 0, , Clearly, the function X = NOT (A AND, B) of the logical variables A AND B is, called NAND gate., , 10 Output of OR gate is 0 when all inputs, are 0 and output is 1 when atleast one of, the inputs is 1., , 7 If we give the following inputs to A and, , Y, , B, then corresponding output is shows, in table, t(s), , A, 0, 0, 1, 1, , 4 A. NAND operation on (1, 1) = 0, NOT operation on (0) = 1, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 9 From the figure of AND gate, , 0, , 1, , 1, , 0, , 1, , A, , 0, , 1, , 1, , 1, , 1, , 1, , 0, , 0, , 0, , 1, , that gate would be same as given in, option in which., The values A = 0, B = 0 gives output 0, The values A = 0, B = 1 gives output 0, The values A = 1, B = 0 gives output 1, The value A = 1, and B = 1 gives output 0, , AND operation on (1, 0) = 0, , 5 Let us draw the given combination, , OR, B, , A+B, , Y, A·(A + B), , OR Gate, x, , 8 Observing the given gate we observe, , The inputs of the AND gate are, A and A + B and its output is, A × ( A + B ) that is A AND (A or B ). The, truth table for the output is Y, = A.( A + B ) is as follows, , Alternative Method, , d, , C. OR operation on (1, 1) = 1, , pointing out that the first gate is OR, gate second gate is AND gate. The, inputs of the OR gate, are A and B ,, and its output is A + B that is A OR B., , \ The gate is OR., , c, , The above table is similar to OR gate., , NAND operation on (1, 0) = 1, , AND, , C, 0, 1, 1, 1, , b, , B. NOT operation on (0, 1) = (1, 0), , A, , B, 0, 1, 0, 1, , Observing output x It is 0 when all, inputs are 0 and it is 1 when atleast one, of the inputs is 1., , a, , OR operation on (1, 0) = 1, , A, , X = NOT A, Rs, , 6 Truth table, , From output, we can show real time, variation of output signal as below., , 385, , X = A and B, B, V = 5V, V=0
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DAY THIRTY SIX, , Communication, Systems, Learning & Revision for the Day, u, , u, , Basic Elements of a, Communication System, Line Communication, , u, , u, , u, , Optical Communication, Modulation, Demodulation or Detection, , u, , u, , Propagation of, Electromagnetic Waves, Satellite Communication, , Basic Elements of a Communication System, A communication system is a set up used to transmit information from one point to, another. The essential parts of a communication system are transmitter, communication channel and receiver as shown in the following block diagram., Encoding, modulation, Information, source, , Transmitter, , Decoding demodulation, (distortion), Channel, , Receiver, , Destination, , Noise, , (i) Transmitter Transmitter converts the message signal produced by information, source into a form (e. g., electrical signal) that is suitable for transmission through, the channel to the receiver., (ii) Communication Channel Communication channel is medium (transmission line,, an optical fibre or free space) which connects a receiver and a transmitter. It carries, the modulated wave from the transmitter to the receiver., (iii) Receiver This set up receives the signals from the communication channel and, converts these signals into their original form., , PREP, MIRROR, Your Personal Preparation Indicator, , Important Terms Used in Communication System, Signal Signal represents the electrical analog of the information. It can be analog or, digital., (a) Analog Signal, analog signal., , u, , No. of Questions in Exercises (x)—, , u, , No. of Questions Attempted (y)—, , u, , Signal which varies continuously with respect to time is called, , y, t, , No. of Correct Questions (z)—, (Without referring Explanations), , u, , Accuracy Level (z / y × 100)—, , u, , Prep Level (z / x × 100)—, , In order to expect good rank in JEE,, your Accuracy Level should be above, 85 & Prep Level should be above 75.
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COMMUNICATION SYSTEMS, , DAY THIRTY SIX, , n2 , i = sin −1 , n1 , , (b) Digital Signal A digital signal has two levels of, current or voltage represented by 0 or 1. It is usually in, the form of pulses., 1, , Buffer coating, , 10-100 mm, 100 to, 400 µm, , y, 0, , t, , Transducer A device that converts the message signal, into electrical signal before feeding it to transmitter. In, other words, transducer converts one form of energy into, another., , n2 − n2, 2, (θ 0 ) max = sin −1 1, n0, , , The wires that are most popular for wired communication, or line communication are, (i) Co-axial (ii) Parallel wire lines (iii) Twisted pair cables, , Optical Communication, Optical communication uses light waves in the frequency, range 1012 to 1016 Hz as the guided wave medium for, propagation of audio frequency signal., Main advantage of optical communication system lies in, the fact that here very high bandwidths of MHz and even, GHz are possible. Consequently, a large number of, messages can be transmitted through a single cable, without any risk of their intermixing., Moreover due to very little line loss the quality of, reception is vastly superior., An optical communication system consists of mainly three, parts which are, (i) optical source and modulator, (ii) optical fibre cable, and (iii) optical signal detector., , Optical Fibre, Optical fibre make use of the principle of total internal, reflection of light. The refractive index n1 of central core is, higher than refractive index n2 of cladding., Total internal reflection will take place at core-cladding, interface if angle of incidence there is, , , , , , where, n0 is the refractive index of the outer medium. For air,, n0 = 1 and then, , (θ 0 ) max = sin −1 ( n21 − n22 ), NOTE, , In line communication, there is a physical connection, between source and destination. The wired connections, between two points are known as transmission lines., , Core, , Cladding, , Amplification It is the process of increasing the, amplitude and thus strength of an electrical signal., , Line Communication, , n1, , n2, , For above condition to be fulfilled the light ray must enter the, optical fibre at a maximum acceptance angle θ 0 from the axis of, fibre such that, , Noise It refers to the unwanted signals that tend to, disturb the transmission and hence a distorted version of, the transmitted signal reaches at receiver., Bandwidth Bandwidth refers to the range over, which, the frequencies in a signal vary., , 387, , • Numerical Aperture (NA) = n0 sin(θ0 ) max = n12 − n22, • All the information in optical fibre is carried out by the principal, of total internal reflection and all the information is carried in, core of the optical fibre., • If angle of incidence is greater than (θo )max , then total internal, reflection will not take place and some information will be, lossed., , Modulation, The phenomenon of superposition of information signal over a, high frequency carrier wave is called modulation. In this process,, amplitude, frequency or phase angle of a high frequency carrier, wave is modified in accordance with the instantaneous value of, the low frequency information., , Need for Modulation, Digital and analog signals to be transmitted are usually of low, frequency and hence, cannot be transmitted as such. These, signals require some carrier to be transported., (i) Frequency of Signal The audio frequency signals (20 Hz, to 20 kHz) cannot be transmitted without distortion over, long distances due to less energy carried by low frequency, audio waves. The energy of a wave is directly proportional, to square of its frequency. Even if the audio signal is, converted into electrical signal, the later cannot be sent, very far without employing large amount of power., (ii) Height of antenna For efficient radiation and reception, the, height of transmitting and receiving antennas should be, comparable to a quarter wavelength of the frequency used., Wavelength =, , velocity, 3 × 10 8, =, metre, frequency frequency (Hz), , For 1 MHz it is 75 m and for 15 kHz frequency, the height of, antennas has to be about 5 km which size is unthinkable.
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388, , DAY THIRTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , (iii) Number, of, channels Audio, frequencies, are, concentrated in the range 20 Hz to 20 kHz. This range is, so narrow that there will be overlapping of signals. In, order to separate, the various signals it is necessary to, convert all of them to different portions of the, electromagnetic spectrum., There are two types of modulation, , 1. Amplitude Modulation, l, , l, , l, , l, , Total bandwidth of modulated signal = 2n ⋅ ω m,, where, n = number of significant side band pairs., FM technique is more complex and costly. However,, efficiency is more and audio quality is vastly improved., Noise level is negligible., δ, In the FM wave, modulation index m f =, νm, where, δ = maximum frequency of deviation, , (AM), , Amplitude Modulation (AM) is the process of changing the, amplitude (A c) of a carrier wave linearly in accordance, with the amplitude of message signal ( Am)., A, A, − Amin, The ratio µ = m = max, is called the modulation, Ac, Amax + Amin, index and in practice we maintain µ ≤ 1, so as to avoid, distortion., , l, , l, , In AM modulated wave signal we have carrier wave of, frequency ωc and two side bands of frequencies (ωc − ω m), and (ωc + ω m), respectively. Thus, total bandwidth of AM, signals is 2ω m., , = (νc − ν min ) or (ν max − νc ), ν m = modulation frequency., , Demodulation or Detection, The process of recovering the original audio signal from the, modulated wave is called demodulation., Demodulation can be done by using a diode and a capacitor, filter as shown under, A, , Upper Side Band Frequency (USB) = νc + ν m, , AM, Input, , where, νc = carrier frequency, ν m = signal frequency., l, , Lower Side Band Frequency (LSB) = νc − ν m, where, νc = carrier frequency, ν m = signal frequency, Bandwidth = ν USB − ν LSB = (νc + ν m) − (νc − ν m) = 2 ν m, , l, , Power of carrier, PC =, , l, , Power of side band, Psb =, , ( A c / 2)2 A c2, =, R, 2R, 2, , l, , l, , l, , l, , l, , 2, , 1 µ A c, 1 µ A c µ2 A c2, , + , =, R 2 2, R 2 2, 8R, , AM technique is simpler and cost effective. However, it, suffers from noisy reception, low efficiency, small, operating range and poor audio quality., , µ2 , Power dissipated in AM wave, P = Pc 1 + where,, 2, , 2, A, Pc = c is power dissipated by unmodulated carrier, 2R, wave and µ = modulation index., , 2. Frequency Modulation, , R, B, , l, , (FM), , Frequency modulation is the process of changing the, frequency of a carrier wave in accordance with the, frequency of message signal., In FM modulated wave the amplitude of wave and total, transmitted power remains constant., Frequency of modulated signal consists of central band of, frequency ωc and side bands of frequencies, (ωc ± ω m),(ωc ± 2ω m),(ωc ± 3 ω m) ....., The number of side bands depends on the modulation, index., , A′, , Output, , B′, , Working of R-C Circuits The value of R-C is so selected such, 1, << RC, where fc = frequency of carrier wave., fc, , that, , For AM modulated wave, a p-n junction diode or a vacuum, tube diode is used as a demodulator. A diode basically acts as, a rectifier and thus, reduces the modulated carrier wave into, positive envelope only. This positive envelope is sent through, a R-C circuit. Carrier wave passes through the capacitor and, AF signal is regenerated across R., , Propagation of Electromagnetic, Waves, It is that category of communication in which no line or cable, is used as a communication channel and the modulated signal, is propagated through free space., Different types of propagation depending upon properties are, , 1. Ground Wave Propagation, In this type of communication, transmitting and receiving, antenna are close to surface of the earth. This type of, propagation can be sustained only at low frequency (≈ 500 kHz, to 1500 kHz). Due to such less frequency range, it is also called, medium wave propagation., , 2. Sky Wave Propagation, Sky wave is the radiowave which is directed towards the sky, and reflected back by the ionosphere towards the desired, location on the earth. Radiowaves of frequencies 2 MHz to 20, MHz can be reflected by the ionosphere.
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COMMUNICATION SYSTEMS, , DAY THIRTY SIX, , l, , Critical Frequencies It is the maximum frequency of the, radiowaves which can be reflected from ionosphere and, returns to the earth. The radiowave will penetrate the, ionosphere above this frequency. It is given by, , l, , fc = 9(N max )1 /2, , 3. Space Wave Propagation, , The sky waves being electromagnetic in nature, changes the, dielectric constant and refractive index of the ionosphere., The effective refractive index of ionosphere is, , The transmitted signal is received by the direct interception, of the signal by receiving antenna. The, frequency range is, (100 MHz to 220 MHz). The maximum range of this, transmission depends upon the height of transmitting antenna, and is given by,, , 81.45 N, v2, , d = 2 hR, R >> h., , where, N = electron density of ionosphere,, ν = frequency of electromagnetic wave in Hz and, µ 0 = refractive index of free space, , l, , Skip distance When the sky wave is reflected back from, the ionosphere having a constant frequency, but less than, that of the critical frequency, then the smallest distance, from the transmitter to the earth’s surface covered by the, sky wave is known as skip distance., , where, N max is the maximum electron density of the, ionosphere., , µ = µ0 1 −, , l, , 389, , Bandwidth of a Communication Channel The difference, between the highest and lowest frequencies that a, communication channel allows to pass through it is called, its bandwidth., Total bandwidth of channel, Number of channels =, Bandwidth per channel, Maximum Usable Frequency (MUF) In this the sky waves, with maximum frequencies are sent at some angles towards, the ionosphere. Then these rays will again be reflected by, the ionosphere to the earth., Critical Frequency (CF), MUF =, = CF sec θ, cosθ, This is the angle which is formed along the direction of the, incident wave and the normal., , where, h = height of the antenna and R = radius of the earth., , Satellite Communication, l, , l, , l, , l, , It is mainly done with the help of a geostationary satellite, orbiting the earth in the equatorial plane from West to East, at a height of about 36000 km above the surface of the, earth, so that its revolution time is 24 h., The transmitted signal from the earth station is uplinked to, satellite. The satellite receives it, demodulate it, amplify it, and remodulate it and transmit it back. Now it is, downlinked to the earth station. A radio transponder does, all these jobs in a satellite., Uplink frequency and downlink frequency are kept, widely different, so as to avoid their interference in free, space., For world wide coverage three geostationary satellites are, required at 120° apart from each other., , DAY PRACTICE SESSION 1, , FOUNDATION QUESTIONS EXERCISE, 1 The minimum length of antenna required to transmit a, radio signal of frequency 20 MHz is, (a) 5 m, , (b) 7.5 m, , (c) 2 m, , (d) 3.75 m, , 2 Repeaters are required for transmitting microwave, terrestrial communication system over a 40-50 km distance, because, (a) microwave power decreases rapidly with distance, (b) the curvature of the earth limits the distance over which, the line of sight can be established, (c) signal to noise ratio decreases rapidly with distance, (d) signal distortion creeps in rapidly with distance, , 3 The characteristic impedance of a coaxial cable is of the, order of, (a) 50 Ω, (c) 270 Ω, , (b) 200 Ω, (d) None of these, , 4 If µ1 and µ 2 are the refractive indices of the materials of, core and cladding of an optical fibre, then the loss of, light due to its leakage can be minimised by having, (a) µ 1 > µ 2, (b) µ 1 < µ 2, (c) µ 1 = µ 2, (d) None of the above, , 5 Which of the following four alternatives is not correct?, We need modulation, , ª AIEEE 2011, , (a) to increase the selectivity, (b) to reduce the time lag between transmission and, reception of the information signal, (c) to reduce the size of antenna, (d) to reduce the fractional band width, i.e. the ratio of the, signal band width to the centre frequency
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390, , DAY THIRTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 6 A message signal of frequency ωm is superposed on a, carrier wave of frequency ωc to get an Amplitude, Modulated wave (AM). The frequency of the AM wave, will be, (a) ωm, , (b) ωc, , (c), , ωc + ωm, 2, , (d), , ωc − ωm, 2, , 7 A speech signal of 3 kHz is used to modulate a carrier, signal of frequency 1 MHz, using amplitude modulation., The frequencies of the side bands will be, (a) 1⋅ 003 MHz and 0 ⋅ 997 MHz, (b) 3001 kHz and 2997 kHz, (c) 1003 kHz and 1000 kHz, (d) 1 MHz and 0.997 MHz, , (b) 100 cycle/s, (d) 50000 cycle/s, , 9 For what value of ma will the total power per cycle be, maximum in the modulated wave?, (b) 1, , (c) 1/2, , (d) greater than 1, , 10 In amplitude modulation, sinusoidal carrier frequency used, is denoted by ωc and the signal frequency is denoted by, ωm . The bandwidth ( ∆ωm ) of the signal is such that, ∆ωm << ωc . Which of the following frequencies is not, contained in the modulated wave? ª JEE Main 2017 (Offline), (a) ωc, (c) ωc − ωm, , (b) ωm + ωc, (d) ωm, , 11 Choose the correct statement., , ª JEE Main 2016 (Offline), , (a) In amplitude modulation, the amplitude of the high, frequency carrier wave is made to vary in proportion to, the amplitude of the audio signal., (b) In amplitude modulation, the frequency of the high, frequency carrier wave is made to vary in proportion to, the amplitude of the audio signal., (c) In frequency modulation, the amplitude of the high, frequency carrier wave is made to vary in proportion to, the amplitude of the audio signal., (d) In frequency modulation, the amplitude of the high, frequency carrier wave is made to vary in proportion to, the frequency of the audio signal., , 12 A signal of 5 kHz frequency is amplitude modulated on a, carrier wave of frequency 2MHz. The frequencies of the, ª JEE Main 2015, resultant signal is/are, (a) 2 MHz only, (b) 2005 kHz and 1995 kHz, (c) 2005 kHz, 2000 kHz and 1995 kHz, (d) 2000 kHz and 1995 kHz, , 13 An EM wave of maximum frequency 300 kHz and critical, frequency 100 kHz is to be transmitted to a height equal, to 150 km. Calculate the skip distance., (a) 624 km, (c) 636 km, , (a) 2 : 4 : 3, , (b) 4 : 3 : 2, , (c) 2 : 3 : 4, , (d) 3 : 2 : 4, , 15 On a particular day, the maximum frequency reflected from, the ionosphere is 9 MHz. On another day, it was found to, increase by 1MHz. What is the ratio of the maximum, electron densities of the ionosphere on the two days?, (b) 1.0, , (c) 1.43, , (d) 0.75, , 16 Maximum Usable Frequency (MUF) in F-region layers is, , 500 cycle/s, the appropriate carrier frequency will be, , (a) 0, , 2 × 1011m −3, 5 × 1011m −3 and 8 × 1011 m −3, respectively., What is the ratio of critical frequency for reflection of, radiowaves?, , (a) 1.23, , 8 In an amplitude modulated wave for audio frequency of, (a) 50 cycle/s, (c) 500 cycle/s, , 14 The electron density of E , F1 and F2 layers of ionosphere is, , (b) 849 km, (d) 942 km, , x, when the critical frequency is 60 MHz and the angle of, incidence is 70°, then x is, (a) 150 MHz, , (b) 170 MHz, , (c) 175 MHz, , (d) 190 MHz, , 17 Frequencies higher than 10 MHz were found not being, reflected by the ionosphere on a particular day at a place., The maximum electron density of the ionosphere on the, day was near to, (a) 1.5 × 1010 m−3, (c) 3 × 1012 m−3, , (b) 1.24 × 1012 m−3, (d) None of these, , 18 How the sound waves can be sent from one place to, another in space communication?, (a) Through wires, (b) Through space, (c) By superimposing it on undamped electromagnetic waves, (d) By superimposing it on damped electromagnetic waves, , 19 To cover a population of 20 lakh, a transmitter tower, should have a height of (Given radius of the, earth = 6400 km, population per square km = 1000) is, (a) 25 m, , (b) 50 m, , (c) 75 m, , (d) 100 m, , 20 The TV transmission tower in Delhi has a height of 240 m., The distance up to which the broadcast can be received., (Taking the radius of the earth to be 6.4 × 106 m) is, (a) 100 km, , (b) 60 km, , (c) 55 km, , (d) 50 km, , 21 A radar has a power of 1 kW and is operating at a, frequency of 10 GHz. It is located on a mountain top of, height 500 m. The maximum distance upto which it can, detect object located on the surface of the earth (Radius, ª AIEEE 2012, of earth = 6.4 × 106 m) is, (a) 80 km, , (b) 16 km, , (c) 40 km, , (d) 64 km, , Direction (Q. Nos. 22-25), , Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true
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COMMUNICATION SYSTEMS, , DAY THIRTY SIX, 22 Statement I Optical fibre communication has immunity to, , 391, , 24 Statement I FM broadcast is preferred over AM, , cross-talk., , broadcast., , Statement II Optical interference between fibres is zero., , Statement II Process of combining the message signals, with carrier wave is called demodulation., , 23 Statement I Transducer in communication system, converts electrical signal into a physical quantity., Statement II For information signal is to be transmitted, directly to long distances, modulation is necessary., , 25 Statement I Modem is a demodulator., Statement II It works only in a transmitting and receiving, mode., , DAY PRACTICE SESSION 2, , PROGRESSIVE QUESTIONS EXERCISE, 1 A telephonic communication service is working at carrier, frequency of 10 GHz. Only 10% of it is utilised for, transmission. How many telephonic channels can be, transmitted simultaneously, if each channel requires a, bandwidth of 5 kHz?, ª JEE Main 2018, (a) 2 × 103, , (b) 2 × 104, , (c) 2 × 105, , (d) 2 × 106, , 2 A diode AM detector with the output circuit consisting of, R = 1 kΩ and C = 1µF would be more suitable for, detecting a carrier signal of, (a) 10 kHz, , (b) 1 kHz, , (c) 0.75 kHz, , (d) 0.5 kHz, , 3 In optical communication system operating at 1200 nm,, only 2% of the source frequency is available for TV, transmission having a bandwidth of 5 MHz. The number, of TV channels that can be transmitted is, (a) 2 million (b) 10 million (c) 0.1 million (d) 1 million, , 4 If sky wave with a frequency of 50 MHz is incident on D, region at an angle of 30°, then angle of refraction is, (a) 15°, , (b) 30°, , (c) 60°, , (d) 45°, , 5 Three waves A , B and C of frequencies 1600 kHz, 5 MHz, and 60 MHz, respectively are to be transmitted from one, place to another. Which of the following is the most, appropriate mode of communication?, (a) A is transmitted via space wave while B and C are, transmitted via sky wave, (b) A is transmitted via ground wave, B via sky wave and C, via space wave, (c) B and C are transmitted via ground wave while A is, transmitted via sky wave, (d) B is transmitted via ground wave while A and C are, transmitted via space wave, , 6 Consider telecommunication through optical fibres., Which of the following statements is not true?, (a) Optical fibres can be graded refractive index, (b) Optical fibres are subjected to electromagnetic, interference from outside, (c) Optical fibres have extremely low transmission loss, (d) Optical fibres may have homogeneous core with a, suitable cladding, , 7 A diode detector is used to detect an amplitude, modulated wave of 60% modulation by using a, condenser of capacity 250 pF in parallel with a load, resistance 100 kΩ. Find the maximum modulated, frequency which could be detected by it. ª JEE Main 2013, (a) 10.61 MHz (b) 10.61 kHz (c) 5.31 MHz (d) 5.31 kHz, , 8 What is the modulation index if an audio signal of, amplitude one half of the carrier amplitude is used in AM?, (a) 1, (c) 0.5, , (b) 0, (d) greater than 1, , 9 For 100% modulation, the power carried by the side, bands (PSB ) is given by, (a) PSB = P, , (b) PSB = 3 P, , (c) PSB =, , 1, P, 3, , (d) PSB = 0, , Direction (Q. Nos. 10-12) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 10 Statement I Sky wave signals are used for long, distance radio communication. These signals are in, general, less stable than ground wave signals., Statement II The state of ionosphere varies from hour to, ª AIEEE 2011, hour, day to day and season to season., , 11 Statement I Higher the modulation index, the reception, will be strong and clear., Statement II The degree, to which the carrier wave is, modulated is called modulation index., , 12 Statement I Television signals are received through sky, wave propagation., Statement II The ionosphere reflects electromagnetic, waves frequencies less than a certain critical frequency.
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392, , DAY THIRTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , ANSWERS, SESSION 1, , SESSION 2, , 1 (d), , 2 (b), , 3 (a), , 4 (a), , 5 (b), , 6 (b), , 7 (a), , 8 (d), , 9 (b), , 10 (d), , 16 (c), , 17 (b), , 18 (c), , 19 (b), , 20 (c), , 6 (b), , 7 (b), , 8 (c), , 9 (c), , 10 (b), , 11 (a), , 12 (c), , 13 (b), , 14 (c), , 15 (a), , 21 (a), , 22 (a), , 23 (d), , 24 (c), , 25 (d), , 1 (c), 11 (b), , 2 (a), 12 (d), , 3 (d), , 4 (b), , 5 (b), , Hints and Explanations, SESSION 1, 1 ν = 20 MHz = 20 × 106 Hz, Wavelength of antenna is,, 3 × 108, c, λ= =, = 15 m, ν 20 × 106, The minimum length of antenna, λ 15, = =, = 3.75 m, 4, 4, , 2 To increases the range of transmission, of microwaves, a number of antennas, are erected in between the transmitting, and receiving antennas. Such antennas, in between the transmitting and, receiving antennas are known as, repeaters., , 3 Coaxial cable have a characteristic, , impedance from 40 Ω to 150 Ω. So,, option (a) is correct., , 4 Refractive index of core is always, greater than refractive index of, cladding, to minimise the loss of light., , 5 Modulation does not change time lag, between transmission and reception., , 6 In amplitude modulation the frequency, of modulated wave is equal to the, frequency of carrier wave. Thus, option, (b) is correct., , 7 Here, ∆ν = 3 kHz = 0.003 MHz, Using amplitude modulation, the, frequencies of the side band, = (ν + ∆ν) and (ν − ∆ν), = (1 + 0.003 ) and (1 − 0.003), = 1. 003 MHz and 0.997 MHz, Thus, option (a) is correct., , 8 Carrier frequency is always greater than, modulating frequency (i.e. audio, frequency), so option (d) is appropriate, carrier frequency., , 2, , 9 Since, P = Pc 1 + m a , , 2, , 2 , , Power will be maximum, if m a = 1, 1, Therefore, Pmax = Pc 1 + , 2, , 3, = Pc = 1.5 Pc, 2, , , 13 ∴ D skip = 2h ν − 1, νc , , 2, , 300 , = 2 × 150 , −1, 100 , = 2 × 150 × 2 2, = 300 × 2 × 1.414, = 2.828 × 300 = 848.4 ~ 849 km, , 10 Frequency spectrum of modulated wave, is, , 14 Critical frequency for reflection of, LSB, , ωm, , ωc–ωm, , USB, ωc, , ωc+ωm, , Bandwidth, , Clearly, ω m is not included in the, spectrum., , 11 In amplitude modulation, the amplitude, of the high frequency carrier wave is, made to vary in proportion to the, amplitude of the audio signal., In frequency modulation, the frequency, of the high frequency carrier signal, varies with the frequency of audio, signal., , 12 Frequency associated with AM are, fc − f m , f , fc + f m, According to the question, fc = 2 MHz = 2000 kHz, f m = 5kHz, Thus, frequency of the resultant, signal is/are carrier frequency, fc = 2000 kHz,, LSB frequency, fc − f m = 2000 kHz −5 kHz, = 1995 kHz, and USB frequency, fc + f m = 2005kHz, , radiowaves is given by, νc ∝ N 1/2, νCE : νCF1 : νCF2, = (2 × 1011 )1/ 2 : (5 × 1011 )1 / 2 : (8 × 1011 )1/ 2, = 2:3: 4, 2, , 2, , , 9 + 1, 15 ∴ N ′max = V ′c = , , , , , N max, , Vc , , 9, , 2, , 10 , = = 1.23, 9, , 16 ∴ MUF = v c / cos i = 60 × 106 / cos 70°, = 60 × 106 ×, , 1, 0342, ., , = 17543, . × 106, = 175.43 MHz ≈ 175MHz, , 17 The critical frequency of a sky wave for, reflection from a layer of atmosphere is,, νc = 9(N max )1 /2, N max = number density of, ionosphere, ν2c, (10 × 106 )2 1014 −3, ⇒ N max =, =, =, m, 81, 81, 81, = 124, . × 1012 m −3, , 18 In space communication signals are sent, directly from transmitting antenna to, receiving antenna by superimposing it on, undamped electromagnetic waves.
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COMMUNICATION SYSTEMS, , DAY THIRTY SIX, 19 Area of region covered = π(2hR ), In 1 km = 1000 people, 1, × 20 × 106 = 2 × 103 = A, 1000, 2 × 103 = π (2 × h × 6400), , 6 Some of the characteristics of an optical, , SESSION 2, , fibre are as follows, , 2, , ⇒ h=, , 2 × 103, = 0.0497 km, π × 2 × 6400, , = 497, . m ≈ 50 m, , = 2 × 6.4 × 106 × 240 m = 55km, , 21 Range of radar on earth surface (optical, distance, for space wave, i.e. line of, view)., h, , Ra, , ng, , e, , R, R, , Range = (R + h )2 − R2, = 2Rh + h2 ~, − 2Rh, = 2 × 6400 ×, , 1 Only 10% of 10 GHz is utilised for, , (i) It works on the principle of total, internal reflection., (ii) It consists of core made up of, glass/silica/plastic with refractive, index n1 , which is surrounded by, a glass or plastic cladding with, refractive index n2 (n2 > n1 ). The, refractive index of cladding can, be either changing abruptly or, gradually changing (graded, index fibre)., (iii) There is a very little transmission, loss through optical fibres., (iv) There is no interference from, stray electric and magnetic fields, to the signals through optical, fibres., , transmission., ∴ Band available for transmission, 10, =, × 10 × 109 Hz, 100, = 109 Hz, Now, if there are n channels each using 5, kHz, then, n × 5 × 103 = 109, , 20 ∴ d = 2Rh, , 1, km = 80 km, 2, , 22 Optical communication is a system by, which we transfer the informations on, any distance from one location to other, through optical range of frequency, using optical fibre. The optical, interference between fibres is zero., Hence, optical fibre communication has, immunity to cross-talk., , 23 In any communication system, information (a physical quantity) is first, converted into an electrical signal by a, device called transducer. Most of the, speech or information signal cannot be, directly transmitted to long distances., For this an intermediate step of, modulation is necessary in which the, information signal is loaded or, superimposed on a high frequency wave, which acts as a carrier wave., , ⇒, , 2 Given, R = 1 k Ω, R = 1 × 103 Ω, C = 1µF = 1 × 10−6 F, In this condition frequency of carrier, 1, signal,, < < fc, RC, 1, < < fc, 1 × 103 × 10−6, ⇒, f c > > 1 kHz, Because frequency is greater than 1, fc = 10 kHz, , 3 The frequency optical communication, , ⇒ ν=, , Demodulation is the process in which, the original modulating voltage is, recovered from the modulated wave., , 25 Modem is a modulating and, demodulating device. It acts as a, modulator in transmitting mode and as, demodulator in receiving mode., , c, λ, , 1200 × 10−9, , = 25 × 10 Hz, But only 2% of the source frequency is, available for TV transmission, ν ′ = 2.5 × 1014 × 2%, 2, ν ′ = 2.5 × 1014 ×, 100, ν ′ = 5 × 1012 Hz, ν′, Number of channels =, bandwidth, 5 × 1012, Number of channels =, = 106, 5 × 106, = 1 million, , 4 For D-region, N = 109 m − 3, µ = 1−, = 1−, , 81.45 N, ν2, 81.45 × 109, (50 × 106 )2, , ≈1, , sin i, µ =, =1, sin r, or sin r = sin i, or, , 7, , D, , Signal, , C, , R, , τ = RC = 100 × 103 × 250 × 10−12 s, , 3 × 108, 13, , 24 In AM modulation, the amplitude of the, carrier signal varies in accordance with, the information signal. AM signals are, noisy because electrical noise signals, significantly affect this. In FM, modulation, amplitude of carrier wave, is fixed while its frequency is changing., FM gives better quality transmission. It, is preferred for transmission of music., , n = 2 × 105, , ν=, , 393, , r = i = 30°, , 5 For ground wave propagation, the, frequency range is 530 kHz to 1710 kHz., For sky wave propagation, the, frequency range is 1710 kHz to 40 MHz., For space wave propagation, the, frequency range is 54 MHz to 4.2 GHz., Thus, option (b) is correct., , = 2.5 × 107 × 1012 s, = 2.5 × 10−5 s, The highest frequency which can be, detected with tolerable distortion is, 1, f =, 2πm aRC, [where, m a is modulation], 1, Hz, =, 2 π × 0.6 × 2.5 × 10−5, 4, 100 × 104, × 104 Hz, =, Hz =, 1.2 π, 25 × 1.2 π, = 10.61 kHz, This condition is obtained by applying the, condition that rate of decay of capacitor, voltage must be equal or less than the rate, of decay modulated signal voltage for, proper detection of modulated signal., , 8 Here, E m = 1 Ec, 2, , Therefore,, 1, Ec = 1.5 Ec, 2, 1, E min = Ec − E m = Ec − Ec = 0.5 Ec, 2, E, − E min, Also, m a = max, E max + E min, 1.5 Ec − 0.5 Ec, =, 1.5Ec + 0.5 Ec, Ec, ma =, = 0.5, 2.0Ec, E max = Ec + E m = Ec +
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394, , DAY THIRTY SIX, , 40 DAYS ~ JEE MAIN PHYSICS, , 2, , 9 PSB = 1 m aEc + 1 m aEc , , , , , R, , PSB =, , R, , 2 2, , m2a, , Ec2, , 4R, , =, , m2aPc, , =, , 2, , 2 2, , Ec2 , Q Pc =, , 2R , , , Pc, 2, , 2, , m2 , Also, P = Pc 1 + a , 2 , , Here,, ma = 1, P = Pc 1 +, ⇒, , PSB, P /2, Hence,, = c, =, 3, P, Pc, 2, 1, or, PSB = P, 3, , [∴m a = 1], , 1 3, = Pc, 2 2, 1, 3, , 10 In radio communication, sky wave refers, to the propagation of radio waves, , reflected or refracted back towards earth, from the ionosphere., Since, it is not limited by the curvature, of the earth, sky wave propagation can, be used to communicate beyond, horizon. Ionosphere is a region of upper, atmosphere and induces the, thermosphere and parts of mesosphere, and exosphere. It is distinguished, because it is ionised by solar radiation., It plays an important part in, atmospheric electricity., , 11 The modulation index determines the, strength and quality of the transmitted, signal., If the modulation index is small the, amount of variation in the carrier, amplitude will be small consequently, the audio signal being transmitted will, not be strong., , Hence, for high modulation index or, greater degree of modulation, the audio, signal reception will be clear and strong., , 12 In sky wave propagation, the radiowaves, which have frequency between 2 MHz to, 30 MHz, are reflected back to the ground, by the ionosphere. But radio waves, having frequency greater than 30 MHz, cannot be reflected by the ionosphere, because at this frequency they penetrates, the ionosphere. It makes the sky wave, propagation less reliable for propagation, of TV signal having frequency greater, than 30 MHz., Critical frequency is defined as the, highest frequency that is returned to the, earth by the ionosphere. Thus, about this, frequency a wave whether it is, electromagnetic will penetrate the, ionosphere and is not reflected by it., Hence, option (d) is correct.
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DAY THIRTY SEVEN, , Unit Test 7, (Modern Physics), 1 The wavelength of incident light falling on a, , V (Volts), , photosensitive surface is charged from 2000 Å to 2100 Å, the corresponding change in stopping potential is, (a) 0. 03 V, (c) 3 V, , Input B, , 16 V, , (b) 0. 3 V, (d) 3 . 3 V, , 8V, , 2 Ultraviolet light of wavelength 350 nm and intensity, 1.00 Wm – 2 is incident on a potassium surface. If 0.5% of, the photons participate in ejecting the photoelectrons,, how many photoelectrons, are emitted per second, if the, potassium surface has an area of1 cm 2 ?, (a) 1.76 × 1018 photoelectrons/s, (b) 1.76 × 1014 photoelectrons/s, (c) 8.8 × 1011photoelectrons/ s, (d) The value of work function is required to complete the, value of emitted photoelectrons/s, , 2, , 4, , rad, $, , 8 rad, 1.8 m y + 5.4 × 10 s . t i, , , Wavelength of this wave as it passes through a medium, 3, of refractive index will be, 2, (a) 3.55 m, (c) 1.44 m, , (b) 2.33 m, (d) 3.22 m, , V, , Li + + ion in its ground state, on the basis of Bohr’s model,, will be about, (a) 53 pm, , (b) 27 pm, , (c) 18 pm, , (d) 13 pm, , 5 Consider inputs A and B;, V (Volts), , Input A, , 12, , t(s), , (b), 6 8 10 12, , t, , (c), , (d), 6 8 10 12, , t, , 6 8 10 12, , 2, , 4, , 6, , 8, , 10, , 12, , t(s), , t, , 6 A neutron collides with a hydrogen atom in its ground, state and excites it to n = 3. The energy given to, hydrogen atom in this inelastic collision is (Neglect the, recoiling of hydrogen atom and assume that energy is not, absorbed as KE of H-atom), (b) 12.1 eV, , (c) 12.5 eV (d) None of these, , 7 An X-ray tube operates at 50 kV. Consider that at each, collision, an electron converts 50% of its energy into, photons and 10% energy would be dissipated as thermal, energy due to the collision then the wavelength of emitted, by photons during 2nd collision is, (Take, hc = 1242 eV-nm), (c) 4.968 nm, , (d) 4.968 Å, , 226, → 86 Rn 222 + 2 He 4 the, 88 Ra, radium nucleus is initially at rest and the alpha particle, carries the energy 5.3 MeV. The energy released in the, reaction is, , 8 For the nuclear reaction,, , 5V, , t, , 6 8 10 12, V, , V, , (a) 1.242 nm (b) 1.242 Å, , 10 V, , 10, , V, , (a), , (a) 10.2 eV, , 4 Taking the Bohr radius as a 0 = 53 pm, the radius of, , 8, , Output of a NAND gate on these inputs will be, , 3 Electric field of an electromagnetic wave in vacuum is;, N, E = 31, . . cos, , C, , 6, , (a) 5.4 MeV, , (b) 5.0 MeV, , (c) 300 MeV, , (d) 286 MeV
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396, , DAY THIRTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 9 The combinations (M) and (N) of the NAND gates are as, shown below., , Point C is kept at a constant voltage of 5 V and point D is, earthed. If input given at A is, VA, , A, Y, B, , 2, , (M), , t(s), , 5, , Then output obtained at B is, , A, , Y, , VB, , VB, , B, (a), , (N), , (b), , The output (Y ) of (M) and (N) are equivalent to the, output of, (a) OR gate and AND gate respectively, (b) AND gate and NOT gate respectively, (c) AND gate and OR gate respectively, (d) OR gate and NOT gate respectively, , 2, , This last expression is not correct because, (a) n would not be integral, (b) Bohr-quantisation applies only to electron, (c) the frame in which the electron is at rest is non-inertial, (d) the motion of the proton would not be in circular orbits,, even approximately, , 11 Light strikes a sodium surface, causing photoelectric, emission. The stopping potential for the ejected electrons, is 5.0 V, and the work function of sodium is 2.2 eV. What, is the wavelength of the incident light?, (a) 100 nm, (c) 150 nm, , (b) 170 nm, (d) 200 nm, , C, VCC =+ 5V, , B, , 2, , 5, , 2, , 5, , t(s), , VB, , (c), , (d), , 2, , 5, , t(s), , t(s), , 13 The simple Bohr’s model cannot be directly applied to, calculate the energy levels of an atom with many, electrons. This is because, (a) of the electrons not being subject to a central force, (b) of the electrons colliding with each other, (c) of screening effects, (d) the force between the nucleus and an electron will no, longer be given by Coulomb’s law, , 14 An electron is trapped in a one dimensional infinite well of, width 250 pm and is in its ground state. What is the, longest wavelengths of light that can excite the electron, from the ground state via a single photon absorption?, (a) λ =, (c) λ =, , 4 mL2c, , (b) λ =, , h (ni2 − nf2 ), 8 mL2c, , (d) λ =, , h (nf2 − ni2 ), , 2 mLc, h (nf2 − ni2 ), 8 mL, h (nf2 − ni2 ), , 15 What is the ratio of the shortest wavelength of the Balmer, series to the shortest wavelength of the Lyman series?, (b) 2 : 4, (d) 1 : 1, , (a) 1 : 4, (c) 4 : 1, , 12 Consider a given circuit, , t(s), , VB, , 10 The binding energy of a H-atom, considering an electron, me 4, moving around a fixed nuclei (proton), is B = −, 4 n 2 ε 02 h 2, (m = electron mass)., If one decides to work in a frame of reference where the, electron is at rest, the proton would be moving around it., By similar arguments, the binding energy would be, Me 4, = (M = proton mass), B=, 8 n 2 ε 20 h 2, , 5, , 16 In the ground state of the hydrogen atom, the electron, has a total energy of −13.6 eV, its kinetic energy is, (a) 12.5 eV, (c) 14.9 eV, , (b) 13.6 eV, (d) −27. 2 eV, , 17 A particle of mass m at rest decays into two particles of, m1 and m2 having non-zero velocities. The ratio of, de-Broglie wavelengths of particles λ 1 / λ 2 is, , A, , (a), D, , m1, m2, , (b), , m2, m1, , (c) 1, , (d) m2 / m1
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UNIT TEST 7 (MODERN PHYSICS), , DAY THIRTY SEVEN, 18 If there are 2 bulbs of same power, one of them gives, red colour light, while other gives blue colour light., If nr and nb are the number of photons per unit time, emitted by bulbs, then choose correct option;, (r : red; b : blue), (a) nr = nb, , (b) nr < nb, , (c) nr > nb, , (d) nr ⋅ nb = c, , 2, , 19 An α-particle makes an elastic head-on collision with a, , 23 Let potential energy of electron in Bohr’s first orbit of, hydrogen atom is zero., Then, total energy of electron in IInd orbit is, (b) 27.20 eV, , (c) 13.6 eV, , (d) 26.25 eV, , 24 Consider the process;, , Ratio of de-Broglie wavelength associated with α-particle, and proton after collision will be, (a) 2 : 1, , (b) Readings of A are correct, (c) Readings of B are correct, (d) Reading of Both A and B are incorrect, , (a) 23.80 eV, , proton initially at rest., , 397, , (b) 4 : 3, , (c) 1 : 2, , (d) 2 : 3, , 20 A beam of light is allowed to fall over cathode of a, photocell after passing through two polaroids. None of, the polaroid is rotated keeping other fixed., , 4, → 228, 90 Th + 2He, Energy released in above process is 5.40 MeV. If this, energy remains mainly with ‘α’ and daughter nucleus, then, kinetic energy of 228, 90 Th, nucleus will be, 232, 92 U, , (a) 5.4 MeV, , (b) 5.3 MeV, , (c) 0.1 MeV, , (d) 0.4 MeV, , 25 The following circuit represents, , Variation of photocurrent is best given by, A, , I, , I, , B, A, , (b), , (a), , B, B, , θ, , θ, , (a) OR gate, I, , I, , (c), , number of active nuclei at t = 0. Then, probability that a, randomly choosen nucleus is disintegrated in time t is, , θ, , θ, , 21 Two particles A1 and A2 of masses m1, m2 (m1 < m2 ) have, (a) then masses are the same, (b) thin energies are the same, (c) energy of A1 is less than the energy of A2, (d) energy of A1 is more than the energy of A2, , 22 Two students makes observations of stopping potentials, (V0 ) and frequencies (f ) and plotted their observations, graphically as shown below, 14, , B (10×10 , 5), Observations, of student, A, f (Hz), , A (5×1014, 0), , Q (15×1014, 2.5), , (c), , e −λt, N0, , (d) e −λt, , 27 Match Column I and Column II and mark correct option., A., B., , Column I, α-decay, β + decay, , p., q., , Column II, Large nucleus., More neutrons in nucleus., , C., , β − decay, , r., , More protons in nucleus., , D., E., , γ - decay, k - capture, , s., t., , More energy in nucleus., Proton number is more than 83 in, nucleus., , (a), (b), (c), (d), , A, p, p,t, p, t, , B, q, r,t, t, s, , C, r, q,t, q, r, , D, s, s, s, q, , E, t, r,t, r, p, , 28 On a particular day, the maximum frequency reflected, from the ionosphere is 10 MHz. On another day, it was, found to decrease to 8 MHz. What is the ratio of the, maximum electron densities of the ionosphere on the two, days?, (a) 20 : 10, , Observations, of student, B, , (b) N 0 e −λt, , (a) 1 − e − λt, , the same de-Broglie wavelength. Then, , V0, (volts), , (b) XOR gate (c) AND gate (d) NAND gate, , 26 Let a sample of a radioactive substance contains N0, , (d), , V0, (volts), , Y, , (b) 30 : 15, , (c) 25 : 16, , (d) 24 : 11, , 29 A transmitting antenna at the top of a tower has a height, P (10×1014, 0), , Now, choose the correct option;, (a) Both students gives accurate observation, , f (Hz), , 32 m and that of the receiving antenna is 50 m. What is, the maximum distance between them for satisfactory, communication in line of sight mode? Given radius of the, earth 6.4 × 106m., (a) 50.0 km, , (b) 45.5 km, , (c) 35.5 km, , (d) 30.2 km
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398, , DAY THIRTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , Direction (Q. Nos. 30-31) According to Einstein, when a, photon or light of frequency ν or wavelength λ is incident on, photosensitive metal surface of work function φ 0, where, φ 0 < hν (here h is planck’s constant) then the emission of, photoelectrons takes place.The maximum kinetic energy of, the emitted photoelectrons is given by K max = hν − φ 0 If the, frequency of the incident light is ν 0 called threshold, frequency. The photoelectrons are emitted from metal without, any kinetic energy.So hν 0 = φ 0, 30 Stopping potential of emitted photoelectron is given by, (a), , hν − φ0, e, , (b) hν − φ0, , (c), , hν, e, , (d), , φ0 + h ν, e, , 31 The variation of maximum kinetic energy (Kmax ) of the, , emitted photoelectrons with frequency ( ν ) of the incident, radiations can be represented by, Kmax, , Kmax, , (a), , melting point metals., Statement II Most of the energy of striking electrons is lost, into collisions which simply appears as thermal energy., , 35 Statement I The different lines of emission spectra, (like Lyman, Balmer etc) of atomic hydrogen gas are, produced by different atoms., Statement II The sample of atomic hydrogen gas consists, of millions of atoms., , Direction (Q. Nos. 36-37) A beam of light has three, wavelengths 440 nm, 495 nm and 660 nm with a total, intensity of 3.24 × 10−3 Wm−2 equally distributed amongst the, three wavelengths. The beam falls normally on an area of, 1.0 cm2 of a clean metallic surface of work function 2.2 eV., Assume that there is no loss of light by reflection and each, energetically capable photon ejects one electron and take,, h = 6.6 × 10−34 J-s., 36 Photoelectric emission is caused by, , (b), , ν, , ν, , Kmax, , 34 Statement I Targets in X-ray tubes are made from high, , Kmax, , (a) light of wavelength 440 nm alone, (b) light of wavelength 660 nm alone, (c) lights of wavelengths 440 nm and 495 nm, (d) lights of wavelengths 495 nm and 660 nm, , 37 The incident energy (in Js −1) of each wavelength is, (c), , (a) 3.24 × 10−7, (c) 108, . × 10−7, , (d), , ν, , ν, , 32 Frequencies higher than 10 MHz are found not to be, reflected by the ionosphere on a particular day at a place., What is the maximum electron density of the ionosphere?, 14, , (a), , 10, em−3, 9, , (b) 1014 em−3, , (c), , 1014, em−3, 81, , (d), , 1014, em−3, 7, , Direction, , (Q. Nos. 33-35), Each of these questions, contains two statements : Statement I and Statement II., Each of these questions also has four alternative choices, only, one of which is the correct answer. You have to select one of, the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 33 Statement I As intensity of incident light, (in photoelectric effect) increases, the number of, photoelectrons emitted per unit time increases., Statement II More intensity of light means more energy, per unit area per unit time., , (b) 162, . × 10−7, (d) 0.81 × 10−7, , Direction (Q. Nos. 38-40) Carbon-14 (symbol 14, 6 C) is, produced by the bombardment of atmospheric nitrogen with, high energy neutrons according to the equation., 14, 1, 14, 1, 7 N + 0 n → 6 C +1 H, Radiocarbon is unstable and decays to nitrogen with a half-life, of 5600 yr. The carbon-14 is incorporated into atmospheric, carbon dioxide molecules which are taken in by plants when, they breathe in carbon dioxide. Animals which eat the plants, also take in carbon-14. By measuring the ratio of the, concentration of 14C to 12C in any ancient organism, say a tree,, one can determine the date when the organism died., , 38 A capsule contains 8 g of 146 C whose half-life is, , 5600 yr. After 16800 yr, the amount of 146 C left in the, capsule will be, (a) 4 g, , (b) 2 g, , (c), , 8, g, 3, , (d) 1 g, , 39 Radiocarbon is produced in the atmosphere as a result of, (a) collisions between fast neutrons and nitrogen nuclei, (b) the action of cosmic rays on atmospheric oxygen, (c) the action of X-rays on carbon, (d) lighting discharge in atmosphere, , 40 Choose the only incorrect statement. In radioactive decay, of an element, (a) α-particles may be emitted, (b) β-particles may be emitted, (c) γ-rays may be emitted, (d) the nucleus does not undergo any change
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UNIT TEST 7 (MODERN PHYSICS), , DAY THIRTY SEVEN, , 399, , ANSWERS, 1., 11., 21., 31., , (b), (b), (d), (c), , 2., 12., 22., 32., , (c), (d), (d), (c), , 3., 13., 23., 33., , (b), (a), (a), (b), , 4., 14., 24., 34., , (c), (c), (c), (a), , 5., 15., 25., 35., , (a), (c), (c), (d), , 6., 16., 26., 36., , (b), (d), (a), (c), , 7., 17., 27., 37., , (b), (c), (b), (c), , 8., 18., 28., 38., , (a), (c), (c), (d), , 9., 19., 29., 39., , (a), (d), (b), (a), , 10., 20., 30., 40., , (c), (a), (a), (d), , Hints and Explanations, 1 Given, λ1 = 2000 Å = 2 × 10−7 m, −7, , λ2 = 2100Å = 2.1 × 10 m, hc, …(i), = W + eV1, λ1, hc, …(ii), = W + eV2, λ2, Subtracting Eq. (ii) from Eq. (i), 1, 1 , hc , −, = e (V1 − V2 ), λ1 λ2 , Change in stopping potential,, hc 1, 1 , ∆V = V1 − V2 =, −, , , e λ1 λ2 , =, , 6. 6 × 10−34 × 3 × 108, 1. 6 × 10−19, , 1, 1, , , −, , , 2 × 10−7 2.1 × 10−7 , 6 . 6 × 3 × 0 .1, V = 0 .3 V, =, 1 . 6 × 2 × 2 .1, , 2 Energy of photon, E =, , hc, 1242, =, eV = 3.55 eV, λ, 350, = 5.68 × 10−19 J, , Let n photons, per unit area per unit, time are reaching the potassium surface,, then, 1.00, n=, = 1.76 × 1018, 5.68 × 10−19, So, number of photons received by, potassium surface per unit time is,, n × Area of potassium surface, = 1.76 × 1018 × 1 × 10− 4 = 1.76 × 1014, Required number of photoelectrons, emitted per unit time, 0.5, = 1.76 × 1014 ×, = 8.8 × 1011, 100, , 3 Phase of wave is ‘ky + ωt ’,, So,, ⇒, , k = 1.8 or, λ1 =, , 2π, 1.8, , 2π, = 1.8, λ, , When this wave passes through a, 3, medium of refractive index, ; its, 2, wavelength will be, 22 10, λ, 2 π /1.8 2, = ×2×, ×, λ2 = 1 =, n, 3, 7, 18, 3/2, = 233, . m, , 1242, nm, 10000, = 1.242 Å, , So, required wavelength, λ =, , 8 As parent nucleus is at rest and emitted, particle (α ) carries some energy, daughter, nucleus (Rn) recoils to conserve the, momentum., The energy released in the reaction, appears in the form of kinetic energy of, α-particle and the daughter nucleus., Q = Kα + KD, From momentum conservation,, pα = pD, Solving above equation, we have, MD, Kα =, ×Q, MD + Mα, M + Mα, ⇒, Q = D, × Kα, MD, 222 + 4, =, × 5.3, 222, = 5.4 MeV, , 4 On the basis of Bohr’s model,, r =, , n2 h2, 2, , 4π m KZe, , 2, , = a0, , n2, Z, , ++, , For Li ion, Z = 3; n = 1 for ground, state., Given, a0 = 53 pm, 53 × 12, r =, = 18 pm, ∴, 3, , 5 For a NAND gate, output is, A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Y, 1, 1, 1, 0, , 9 It follows from the logic symbol (A) that, , So, output waveform is like option (a)., , X = AB, , 6 The energy taken by hydrogen atom, corresponds to its transition from n = 1, to n = 3 state., ∆E (given to hydrogen atom), 1, 8, = 13.6 1 − = 13.6 × = 12.1 eV, , 9, 9, , 7 During first collision, Initial energy of, electron = 50 keV, Energy appearing as photon, = 50% of 50 keV = 25keV, Energy lost in collision, = 10% of 50 keV = 5 keV, Energy left for second collision, = (50 − 25 − 5) keV = 20 keV, For second collision, Initial energy, = 20 keV, Energy of the emitted photon, = 50% of 20 keV = 10000 eV, , for which the truth table is as follows, A, , B, , A, , B, , A⋅B, , X = AB, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , 0, , 1, , This truth table satisfies the Boolean, expression X = A + B , which is the OR, gate. Hence, the logic symbol (A) is, equivalent to an OR gate. It follows from, logic symbol (B) that, X = A⋅B = A⋅B, which is the Boolean expression for AND, gate.
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400, , DAY THIRTY SEVEN, , 40 DAYS ~ JEE MAIN PHYSICS, , 10 If electron is considered at rest, then, photons are accelerating around COM, of system. Thus, with respect to COM, electrons are also accelerating and, hence frame is non-inertial., , 11 The energy of an incident photon, , E = hf = hc /λ, kinetic energy of the, most energetic electron emitted, K m = E − φ = (hc /λ ) − φ, eV 0 is related, to kinetic energy by eV 0 = K m ,, so, eV 0 = (hc /λ ) − φ and, hc, λ=, eV 0 + φ, 6.6 × 10-34 × 3 × 108, , =, , 5 × 1.6 × 10−19 + 22, . × 1.6 × 10−19, −9, = 171.8 × 10 m, = 171.8 nm, ≈ 170 nm, , 12 Given circuit is a transistorized ‘NOT’, gate. When A is made positive,, transistor is ON and it draws maximum, current to collector., So, V B = 0 for the time A remains, positive., , 14 Energy levels are E n = n2 h2 8 / mL2 ,, f = ∆E / h = (h/8mL2 )(n2f − n2i ) and the, wavelength of the light is, c, 8mL2c, λ= =, f, h(n2f − n2i ), , 15 The energy E of the photon emitted, 1, 1 , E = 13.6 2 − 2 , n2 , n1, Frequency f of the electromagnetic, wave f = E /h and the wavelength, λ = c/f ., f, 1, E, Thus,, = =, λ c, hc, 13.6 1, 1 , =, − 2, , hc n12, n2 , For which n2 = ∞. For the Balmer, series, n1 = 2 and the shortest, wavelength is λ B = 4hc /13.6. For the, Lyman series, n1 = 1 and the shortest, wavelength is λ L = hc / 13.6. The ratio is, λ B/ λ L = 4, , 17 Initial momentum = 0, ∴ Final momentum = 0, p+ p = 0, (numerically), p1 = p2, h, 1, λ=, ∝, p, p, p2, λ1, =1, ⇒, =, λ2, p1, , 18 Power, P =, , Energy radiated, time, , 23 PE of electron in Ist orbit = − 27.20 eV, , So, number of photons,, E, P, n=, =, hf × ∆t, hf, As,, ⇒, , Now, in IInd orbit,, Total energy of electron = −, , f blue > f red, n red > n blue, , 19 Let initial speed of α is v 0 and final, speeds of α and proton are v α and v p ., Then, momentum conservation gives, 4m pv i = 4m pv α + m pv p, {Q m α = 4m p}, Also, e = 1; elastic collision, ⇒ vi − 0 = v p − vα, Elimination of v i gives,, 3, v α = vp, 8, m pv p, λα, h, Now,, =, ×, λ p mα v α, h, =, , mp, , ., , vp, , mα v α, 1 8 2, = × =, 4 3 3, , 20 In figure (a) following Malus’ law,, intensity reduces upto θ =, , π, and then, 2, , increases., Also, intensity ∝ photocurrent., , 21 Masses of particles A1 and A2 are, m1 and m2 ., where, m1 < m2, m1, ⇒, <1, m2, , 22 As slope of V 0 and f graph = h, e, So, h = e × slope of graph., From readings of A,, 1.6 × 10−19 × (5 − 0), h=, (10 − 5) × 1014, = 16 × 10−34 J-s, From readings of B,, 1.6 × 10−19 × (2.5 − 0), h=, (15 − 10) × 1014, = 8 × 10−34 J-s, Hence, both of the readings are, incorrect., , n2, 13.6, , 22, = − 3.4 eV, In IInd orbit, KE = 3.4 eV and, PE = − 2 × 3.4 = − 6.8 eV, To make PE = 0 in Ist orbit, energy must, be increased by 27.20 eV., So, PE in IInd orbit = 2720, . + (−6.8), = 20.40 eV, Hence total energy = PE + KE, = 20.40 + 3.4, = 23.80 eV., , 24 Energy is distributed in inverse ratio of, masses of products,, k, 228, So, α =, k Th, 4, where (228 + 4)x = 540, . MeV, 540, ., ∴ x=, 232, 228, So, k α =, × 5.40, 232, = 5.30 MeV, So, k Th = 01, . MeV, , 25 Output of upper AND gate = AB, Output of lower AND gate = BA,, Output Y = A B + BA, …(i), , Since, both particles have same, de-Broglie wavelength., h, Hence, momentum, p = will be same, λ, for both particles., p2, ∴ Energy, E =, 2m, E2, m, = 1, E1, m2, E2, [from Eq. (i)], <1, E1, E2 < E1, , =−, , 13.6, , 26 Probability of surviving after time t, Number undecayed in time (t ), Total number, N 0 e − λt, =, = e − λt, N0, , =, , ∴ Probability of decay, = 1 − Probability of survival, = 1 − e − λt, , 27 For α-decay, electrostatic repulsion must, be greater than nuclear force; this, happens when nucleus is large., In β + decay, a proton is changed to a, neutron. This occurs when protons are, more., In β − decay, a neutron is changed to a, proton. This happens when neutrons are, more., In k-capture, if protons are large, a, proton and an electrons forms a neutron., 28 f c = 10 MHz, f ′c = 8 MHz, 2, , 2, , f , 10, N max, = c = = 25 : 16, 8, N ′ max f ′c , , 29 d m = 2 × 64 × 105 × 32 +, 2 × 64 × 105 × 50 m, = 64 × 102 × 10 + 8 × 103 × 10 m, = 144 × 102 10 m = 45.5 km
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UNIT TEST 7 (MODERN PHYSICS), , DAY THIRTY SEVEN, , 35 In one particular sample, atoms can be, , 32 Critical frequency,, fc = 10 MHz = 10 Hz, 7, , fc = 9(N max )1 /2, f, or (N max )1 /2 = c, 9, or, , N max, , f, = c , 9, , 2, , 107 , −3, N max = , em, 9 , =, , 36 The threshold wavelength is, λ0 =, =, , 2, , or, , excited only upto a particular level., , 1014, em −3, 81, , 33 Intensity ∝ number of photons., 34 When energy lost in a collision is less, it, appears in form of heat radiation., , hc, W0, (6.6 × 10−34 ) × (3 × 108 ), , 22, . × 1.6 × 10−19, = 6 × 10−7 m, = 600 nm, Out of the three given wavelength, two, wavelengths λ1 = 440 nm and λ2 = 495, nm will cause photoelectric emission as, these wavelengths are less than λ 0., , 37 Intensity of each wavelength is, I =, , 1, × 324, . × 10−3, 3, , 401, , = 1.08 × 10−3 Wm −2 ., Area of metal surface is, A = 1cm2, = 1 × 10−4 m2 ., Therefore, energy of each wavelength is, E =I×A, = 1.08 × 10−7 J - s −1, , 38 Number of half-lives n = 16800 = 3, 5600, Therefore, the amount of C - 14 left, 8g, = n, (2), 8g, = 3, (2), =1g
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402, , DAY THIRTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY THIRTY EIGHT, , Mock Test 1, (Based on Complete Syllabus), Instructions, 1. The test consists of 30 questions., 2. Candidates will be awarded marks for correct response of each question. 1/4 (one-fourth) marks will be deducted for indicating incorrect response, of each question. No deduction from the total score will be made, if no response is indicated for an item in the answer sheet., , 3. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response., , 1 A running man has half the kinetic energy of a boy of half, , his mass. The man speeds up by 1.0 ms −1 and then has, the same kinetic energy as the boy. The original speed of, the boy was, (a) 2.4 ms −1, , (b) 9.6 ms −1, , (c) 4.8 ms −1, , (d) 7.2 ms −1, , 2 The length of the string of a simple pendulum is, measured with a meter scale, is found to be 92.0 cm, the, radius of the bob plus the hook is measured with the help, of vernier calliper to be 2.17 cm. Mark out the correct, statement., (a) Least count of meter scale is 0.1 cm, (b) Least count of vernier callipers is 0.01 cm, (c) Effective length of simple pendulum is 94.2 cm, (d) All of the above, , 3 Two bodies A and B of equal mass are suspended from, two separate massless springs of spring constants k1, and k 2 respectively. If the bodies oscillate vertically such, that their maximum velocities are equal, the ratio of the, amplitudes of A to that of B is, k, (a) 1, k2, (c), , k2, k1, , k, (b) 1, k2, (d), , k2, k1, , 4 The length of a simple pendulum executing simple, harmonic motion is increased by 21%. The percentage, increase in the time period of the pendulum of increased, length is, (a) 11%, , (b) 21%, , (c) 42%, , (d) 10%, , 5 As the object moves from infinity to focus, then which is, true, about the image formed by a single concave mirror?, (a) Always real and speed of image continuously increases, (b) Always real and speed is initially smaller and finally, larger than object speed, (c) Initially real and moving with speed smaller than object, speed but later on image becomes virtual and moving, with speed of object, (d) Always virtual and speed is less than object speed, , 6 Two bodies of different masses has been released from, the top of tower. One is thrown in the horizontal direction, while other is dropped, then which will reach the ground, first?, (a) The body which has been thrown horizontally, (b) The body which has been dropped, (c) Both will reach the ground simultaneously, (d) Depends on the velocity with which the first body has, been projected horizontally
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MOCK TEST 1, , DAY THIRTY EIGHT, 7 A raft of wood of density 600 kgm −3 and mass 120 kg, , δ, , floats in water. How much weight can be put on the raft, to make it just sink?, (a) 200 kg, , (b) 120 kg, , (c) 80 kg, , 403, , δ0, δm, , (d) 40 kg, , 8 A ‘double star’ is a composite system of two stars, rotating about their centre of mass under their mutual, gravitational attraction. Let us consider such a ‘double, star’ which has two stars of masses m and 2m at, separation l. If T is the time period of rotation about their, centre of mass, then, (a)T = 2 π, , l3, mG, , (b)T = 2 π, , l3, 2mG, , (c)T = 2 π, , l3, 3mG, , (d)T = 2 π, , l3, 4mG, , 9 In a hall, a person receives direct sound waves from a, source 120 m away. He also receives waves from the same, source which reach him after being reflected from the 25 m, high ceiling at a point half-way between them. The two, waves interfere constructively for wave lengths (in metre) of, 5, (a) 10, 5, , …, 2, (c) 30, 20, 10, …, , 20 20, ,, , ..., 3 5, (d) 35, 25, 15, …, , (b) 20,, , 10 An AC source producing emf, e = e 0 [cos (100 π s −1) t + cos ( 500 π s −1) t ], is connected in series with a capacitor and resistor. The, steady state current in the circuit is found to be, I = I1 cos [(100 πs −1 ) t + φ1] + I2 cos [(500 π s −1 ) t + φ 2 ], (a) I1 > I 2, (b) I1 = I 2, (c) I1 < I 2, (d) The information is insufficient to find the relation, between I1 and I 2, , 11 A SONAR system fixed in a submarine operates at a, frequency 40.0 kHz. An enemy submarine moves, towards the SONAR with a speed of 360 km/h. What is, the frequency of sound reflected by the submarine? Take, the speed of sound in water to be 1450 m/s., (a) 40.00 kHz (b) 50.53 kHz (c) 45.93 kHz (d) 55.63 kHz, , 12 If electric potential due to some charge distribution is, given by V = 3 /r 2, where r is radial distance, then find, electric field at (1, 1, 1), (a), (c), , 2, 3, 2, $, 8 ( i + $j + k$ ), , i, , (a) The range of deviation for which two angles of incidence, are possible for same deviation is δ 0 − δm, (b) The curve is unsymmetrical about i 0, (c) For a given δ, i is unique, (d) Both (a) and (b) are correct, , 14 A photon of 10.2 eV energy collides with hydrogen atom, in ground state inelastically. After few microseconds one, more photon of energy 15 eV collides with same, hydrogen atom. Then what can be detected by a suitable, detector?, (a) 1 photon of 10.2 eV and an electron of energy 1.4 eV, (b) 2 photons of energy 10.2 eV, (c) 2 photons of energy 3.4 eV, (d) 1 photons of energy 3.4 eV and 1 electron of 1.4 eV, , 15 A non-conducting plate (infinite plane plate) is given a, charge in such a way that Q1 appears on one side and Q 2, on other side. The face area of plate is A. Find the electric, field at points 1 and 2., Q2, , 1, , Q1 + Q2 Q2 − Q1, ,, 2 ε0 A, 2 ε0 A, Q1 + Q2 Q2 − Q1, (c), ,, ε0 A, ε0 A, , Q1, , 2, , Q1 − Q2 Q1 + Q2, ,, 2 ε0 A, 2 ε0 A, Q1 − Q2 Q1 + Q2, (d), ,, ε0 A, ε0 A, (b), , (a), , 16 The emf and internal resistance of the battery as shown in, figure are 4.3 V and 1 Ω respectively. The external, resistance R is 50 Ω. The resistance of the voltmeter and, ammeter are 200 Ω and 2 Ω respectively. Find the, readings of the two meters., , 2 ($i + $j + k$ ), 3, 3, (d), 2 ($i + $j + k$ ), , 4.3 V, , (b), , 50 Ω, , 13 In the diagram, a plot between δ (deviation) versus i, (angle of incidence) for a triangular prism is given. From, the observed plot, some conclusions can be drawn., Mark out the correct conclusions., , π/2, , i0, , 1Ω, , A, 2Ω, , V, 200 Ω, , (a) 0.1 A, 2 V, (c) 0. 4 A, 1 V, , (b) 0.1 A, 4 V, (d) 0.4 A, 4 V
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404, , DAY THIRTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , 17 All the accelerations as shown in figure are with respect, to ground, find acceleration of B., 2 ms–2, , P, , thermal conductivity K 2. The two ends of the combined, system are maintained at two different temperatures., There is no loss of heat across the cylindrical surface and, the system is in steady state. Find the effective thermal, conductivity of the system., (a) K1 + K 2, (c), , A, , B, , (b) 5 ms −2 , upward, (d) None of these, , 18 Three dielectric slabs of thickness d/4 , d/ 7 and d/2, having dielectric constants 2, 8/7 and 4 respectively are, inserted between the plates of a parallel plate capacitor, having plate separation d and plate area A. Find the, capacitance of the system., 118 ε0 A, (a), 75 d, , 88 ε0 A, (b), 63 d, , 226 ε0 A, (c), 135 d, , 284 ε0 A, (d), 75 d, , 19 Light is incident at an angle α on one planer end of a, transparent cylindrical rod of refractive index n. The least, value of n for which the light entering the rod will not, emerge from the curved surface of rod, irrespective of, value of α is, (a), , 1, 2, , (b) 2, , (c), , 1, 3, , (d) 3, , 20 An electron of hydrogen atom is considered to be, revolving around the proton in the circular orbit of radius, h2, 2 πe 2, with velocity, . The equivalent current due, 2, 2, h, 4π me, to circulating charge is, (a), (c), , 4 π 2me 4, , (b), , 3, , h, 4 π 2m 2e 4, , (a) 20%, (c) 30%, , 25 A hole is bored along the diameter of the earth and a, particle is dropped into it. If R is the radius of the earth, and g is the acceleration due to gravity at the surface of, the earth, then the time period of oscillation of the particle, is, (a) 2 π, , R, g, , (b) 2 π, , R, 2g, , (c) 2 π, , 2R, g, , (d) 2 π, , R, 3g, , 26 Five rods of same dimensions are arranged as shown in, the figure. They have thermal conductivities K1, K 2, K 3, K 4, and K 5. When the points A and B are maintained at, different temperatures, no heat flows through the central, rod if, C, , 4 π 2me 5, , (c) 30°, , (d) 15°, , 22 A galvanometer has resistance100 Ω and it requires, current 100 µA for full scale deflection. A resistor 0 .1 Ω is, connected to make it ammeter. The smallest current in, circuit to produce the full scale deflection is, (a) 1000.1 mA (b) 1.1 mA, , (c) 10.1 mA, , K2, , K1, , h3, , D-region of earth’s atmosphere at 45°. The angle of, refraction is (electron density for D-regions is 400 electron/, cm 3), (b) 45°, , (K1 + 3K 2 ), 4, , (b) 25%, (d) 45%, , (d) 100.1 mA, , 23 A cylinder of radius R made of a material of thermal, conductivity K1 is surrounded by a cylindrical shell of, inner radius R and outer radius 2 R made of a material of, , K5, , A, , 21 A sky wave with a frequency 55 MHz is incident on, , (a) 60°, , (d), , then comes to rest inside a second plate of mass, m2 = 2.98 kg. It is found that the two plates, initially at rest,, now move with equal velocities. The percentage loss in, the initial velocity of bullet when it is between m1 and m2, (neglect any loss of material of the bodies, due to action, of bullet) will be, , (d) None of these, , h3, , K1K 2, (K1 + K 2 ), , 24 A 20 g bullet pierces through plate of mass m1 = 1kg and, , 3 ms–2, , (a) 3 ms −2 , upward, (c) 3 ms −2 , downward, , (3K1 + K 2 ), 4, , (b), , B, K4, , K3, D, , (a) K1 K 4 = K 2 K 3, K, K, (c) 1 = 2, K4 K3, , (b) K1 = K 4 and K 2 = K 3, (d) K1K 2 = K 3K 4, , 27 A block of wood has a mass of 25 g. When a 5 g metal, piece with a volume of 2 cm 3 is attached to the bottom of, the block, the wood barely floats in water. What is the, volume V of the wood?, (a) 28 cm3, (c) 48 cm3, , (b) 35 cm3, (d) 12 cm3
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MOCK TEST 1, , DAY THIRTY EIGHT, 28 A body dropped from a height H reaches the ground, with a speed of 1.2 gH . Calculate the work done by, air-friction., (a) 2.8 mgH, (c) 1.3 mgH, , (b) –1.3 mgH, (d) − 0.28 mgH, , 405, , (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 29 Statement I The rocket works on the principle of, conservation of linear momentum., , Direction (Q. Nos. 29-30) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c), (d) given below, (a) Statement I is true, Statement II is true; Statement II is the, correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is not, the correct explanation for Statement I, , Statement II Whenever there is the change in, momentum of one body, the same change occurs in the, momentum of the second body of the same system, (having two bodies only) but in opposite direction., , 30 Statement I If the half-life of a radioactive substance is, 40 days, then 75% substance decays in 20 days., n, , Time elapsed, 1, Statement II N = N0 , where n =, 2, Half -life period, , ANSWERS, 1. (c), 11. (c), 21. (b), , 2. (c), 12. (b), 22. (d), , 3. (d), 13. (d), 23. (d), , 4. (d), 14. (a), 24. (a), , 5. (b), 15. (a), 25. (a), , 6. (c), 16. (b), 26. (a), , 7. (c), 17. (d), 27. (a), , 8. (c), 18. (d), 28. (d), , 9. (b), 19. (b), 29. (a), , 10. (c), 20. (b), 30. (d), , Hints and Explanations, 1, , 1, (KE) boy, 2, 1 1 m 2, = ×, vb , , 2 2, 2, vb, =, 2, , (KE) man =, ⇒, , 1, mv 2m, 2, , ⇒, , vm, , …(i), , Further,, 1, 1 m, m (v m + 1)2 = v b2, 2, 22, v, ⇒, vm + 1 = b, 2, From Eqs. (i) and (ii), we get, v b = 2( 2 + 1) = 4.82 ms −1, , …(ii), , v m = 2 + 1 = 2.41 ms −1, , 2 Effective length of the simple pendulum, is (92.0 + 2.17) cm, , = 94.2 cm after rounding off to 3, significant digits., , 3 Maximum velocity = aω = a k, , m, , Given that,, ⇒, , 4 Q T = 2π, , a1, , k1, k2, = a2, m, m, a1, k2, =, a2, k1, , l 2, l, g , , T = 4π2 , l = 2 T 2, 4π , g, g, , ∴ % change = x + y +, , xy, 100, , Vaild only for two variables in terms of, percentage., x → % change in first variable, x → % change in second variable, x2, % increase in length = x + x +, 100, x2, 21 = 2 x +, 100, On solving, x = 10%, [by cross-check method], , v, , h, , So,, , 5 As the object moves from infinity to, centre of curvature, the image formed, by a concave mirror would be real and, is moving from focus to centre of, curvature, but as the object crosses, centre of curvature and moves towards, focus the image is still real but moves, from centre of curvature towards, infinity and when the object is at focus, the real image would be formed at, infinity., So, image speed is smaller in beginning, when the object is moving from infinity, to centre of curvature and increases, thereafter., , 6 Since, vertical displacement is same,, as well as initial velocity in vertical, downward direction is zero for both the, bodies., , ∴, , g t 12, 2, gt 22, h=, 2, t1 = t2, h=, , (for horizontal throwing), (for dropping), , 7 Volume of raft = 120 = 1 m3, , 600 5, Fraction of volume inside water is, ρwood, 600, 3, = Relative density =, =, ρwater, 1000 5, So, fraction of volume outside water is, 3 2, , = 1 − =, , 5 5, ⇒ Volume outside water is,, 2 1, 2 3, Vout = × =, m, 5 5 25, When the raft just sinks, the additional, upthrust is, 2, U =, × 103 × g, 25
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406, , DAY THIRTY EIGHT, , 40 DAYS ~ JEE MAIN PHYSICS, , The weight m put on the raft is, 2, mg =, × 103 × g, 25, ∴, m = 80 kg, , 8 The system will revolve/rotate about an, axis passing through the centre of mass, of the combined system. Considering, origin at the particle of mass 2m, we, have the centre of mass at a distance l/3, 2l, from 2l and from m, 3, , 2m, , CM, , m, , λ, ; where, n = 0, 1, 2, ..., 2, ⇒ λ = 20 (2n + 1) ; where, n = 0, 1, 2, ..., 20 20, or λ = 20, , , ..., 3 5, e0, e, 10 I1 =, = 0,, 2, Z1, 1 , R2 + , , ω1C , ⇒ 10 = (2n +1), , where, ω1 = 100 π, e0, e, I2 =, = 0,, 2, Z2, 1 , R2 + , , ω2C , , Q + Q2, 15 At 1; E1 = σ 2 + σ 1 = 1, 2 ε0, , At 2; E2 =, , Q − Q1, σ2, σ, − 1 = 2, 2 ε0 2 ε0, 2 A ε0, , where, σ 2 =, , Q2, A, , ⇒, ⇒, , ω=, , 3Gm, l, , 3, , T = 2π, , v + ve , ν′ = , ν, v s, 1450 + 100 , 3, =, × 40 × 10, , , 1450, , diagram, current flowing through circuit, 4.3, = 0.1 A, (50 || 200 + 2 +1), 4.3 V, 50 Ω, , l, 3Gm, , 9 ∴Path difference,, ∆x = (SA + AP ) − SP = (65 + 65) − 120, ⇒ ∆x = 10 m, A, , 1Ω, , Voltmeter reading = 4.3 − 0.1 × 3 = 4 V, , 17 Consider downward direction as positive, a AP = − a BP, a AG = a AP + a PG, 3 = a AP − 2, a AP = 5 ms −2, , ⇒, ∴, , a BG = a BP + a PG, = − 5 − 2 = − 7 ms −2, , 18 Here, three slabs are in series, , 65m, 25m, 60m 60m, S, P, 120m, , Now, the reflected waves have a, different frequency,, v , ν′′ = , ν′, v − vs , Here, v s = 100 m/s is velocity of enemy, submarine,, , 1450 , . × 103, ν′′ = , × 4276, 1450 − 100 , , 65m, , But at A, the wave suffers reflection at, the surface of rigid/fixed end or denser, medium. Hence, the wave must suffer, λ, an additional path change of or a, 2, phase change of π., λ, , ⇒ Net path difference = 10 − , , 2, For maxima (constructive interference),, λ, Net path difference = (2n ) ;, 2, where, n = 0, 1, 2, ..., λ, λ, ⇒ 10 − = 2n ;, 2, 2, where, n = 0, 1, 2, ..., , 2Ω, 200 Ω, , . × 103 Hz, = 4276, 3, , = 4593, . × 103 Hz = 4593, . kHz, , 12 As, V =, , 3, , r2, ∂ 3, 6, dV , ∴ E=−, r = 3 r, r =−, dr , ∂ r r 2 , r, ( $i + $j + k$ ), 2 $ $ $, =, (i + j + k), ⇒ E =6, 3, ( 3 )3, , 14 When photon strickes the hydrogen, atom, the photon is absorbed and H, atom reaches in (n=2 state) or first, excited state, emitting a photon of, energy 10.2 eV. Ionisation energy of, H-atom =13.6 eV, so the second photon, of energy 15 eV will ionise the H atom, and extra energy (15 -13.6)eV = 1.4 eV, will be retained by the electron. Thus, finally we have one photon of energy, 10.2 eV and one electron of energy, 1.4eV., , Q1, A, , =, , νs = 40 kHz = 40 × 103 Hz, , Speed of sound in water = 1450 m/s, Apparent frequency received by the, submarine,, , and σ 1 =, , 16 First of all draw the equivalent circuit, , 11 SONAR frequency,, , l/3, 2l/3, The gravitational force of attraction, between 2m and m provides the, necessary centripetal force to the mass, 2l, to revolve in a circle of radius for m, 3, l, or for 2m., 3, Gm (2m ), 2l , ⇒ m ω2 =, 3, l2, , 2 A ε0, , towards right, , where, ω2 = 500 π, So, Z1 > Z2 , therefore I1 < I2 ., , Speed of enemy submarine, 5, ve = 360 km / h = 360 ×, m/s, 18, 5, , , m / s, = 100 m / s, Q 1 km / h =, , , 18, , 2 ε0, , towards left, , 2 8 4, 7, , d d d, 4 7 2, , C1 =, , ε0 A, d /4 d / 7 d /2 , +, +, , , 8/ 7, 4 , 2, , ε0 A, 8ε A, = 0, d, d, d, 3d, + , +, 8, 8, 8, ε0 A, C2 =, d, d , d, , d − 4 + 7 + 2 , , , 28 ε0 A, =, 25 d, 284 ε0 A, Now, Ceq = C1 + C2 =, 75d, =, , 19 Q r + i = 90°, ⇒, , i = 90° − r, , α, , r, , i, n
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MOCK TEST 1, , DAY THIRTY EIGHT, For ray not to emerge from curved, surface,, i>C, ⇒, sin i > sin C, ⇒ sin (90° − r ) > sin C, ⇒, cos r > sin C, 1, Q sin C = 1 , 1 − sin2 r >, ⇒, , n, n , 2, 1 − sin i, 1, ⇒, > 2, n2, n, 1, 1 > 2 (1 + sin2 i ), ⇒, n, ⇒, n2 > 1 + sin2 i ⇒ n > 2, , 24 The situation is as shown in figure., , =, , 21 neff = n0 1 − 80.52 N , , , v, v, Now, take second sheet and bullet as, the system,, mu1 = (m1 + m2 ) v, Solving this equation, we get, Percentage loss in, (u − u1 ), u=, × 100% = 20%, u, , ⇒, ig, , F =−, , x2, 4, , G π x3ρ m, 3, , ⇒F = −, x2, 4, , = − πGρm x, 3, , , m, , r = i = 45°, , M′, , = 100.1 × 10−3 A = 100.1 mA, , = 2π, , 23 This can be considered as a parallel, combination of two, one the inner, cylinder and the other surrounding, cylinder., 2, 1, , ⇒, , T2, , A2 = π (4R2 − R2 ) = 3 πR2 ,, L1 = L2 = L, Heat is flowing only along the length of, tube, H eq = H 1 + H 2, K A (T − T2 ) K 2 A2 (T1 − T2 ), = 1 1 1, +, L, L, ⇒ K eq × 4 = K 1 + 3K 2, K eq =, , (K 1 + 3K 2 ), 4, , T = 2π, , 3, 4 πGρ, R, g, Q g = Gm = 4 πRρG , 3, , , R2, , 2R, , A1 = πR2 ,, , ⇒, , x, O, , Time period of oscillation,, x, T = 2π, | x ′|, , 2R, , where all the quantities are in CGS unit, and ρ is the density of water., ⇒, , (V + 2) × 1 = 30, , ∴, , V = 28 cm3, , 28 The forces acting on the body are force of, gravity and air-friction, According to work-energy theorem,, total work done on the body = Gain in, Kinetic energy, 1, W = mv 2, 2, 1, = m (12, . gH ) 2, 2, = 072, . mgH, As work done by gravity,, W1 = mgH, ∴ Work done by friction,, W2 = W − W1, = 0.72 mgH − mgH, = −0.28 mgH, , 29 Since, in the rocket fuel is undergoing, , S, 22, =, i, S+G, S+G, i =, ⋅ ig, S, 0.1 + 100, =, × 100 × 10−6 A, 0 .1, , T1, , GM ′ m, , 4, , ⇒ x ′ = − π Gρ x, 3, , , 80.5 × (400 × 106 ), , =1, (55 × 106 )2, sin i, =, ⇒ sin i = sin r, sin r, , = 1 1−, Also, n eff, , , , ν, , u1, , u, , 25, , 4π2 me 4, e, 4 π2 me 5, =, ∴ Current, I =, T, h3, , m2, , m, , 20 As, T =, , h3, , Water, Metal, , mu = mu1 + m1 v, , [Q sin i = 1 for i = 90° ], ∴ Least value is 2., 2 πr 2 π × h2, h, =, ×, v, 4 π2 me 2 2 πe 2, , Wood, , Firstly take first sheet and bullet as the, system,, m1, , 407, , 26 The arrangement of rods is analogous to, the arrangement of resistances in a, Wheatstone bridge balanced condition., Thus, no heat flows through the rod, conductivity K 5, then, K1 K2, =, K3 K 4, ⇒, , K1K 4 = K2K3, , 27 Let volume of wood is V cm3 , then total, volume of displaced water is, (V + 2) cm3 , then for translational, equilibrium, (V + 2) ρg = (25g + 5g ), , combustion, the gases produced in this, process leave the body of the rocket with, large velocity and produce upthrust to, the rocket. Let us assume that the fuel is, undergoing combustion at the constant, rate, then rate of change of momentum of, the rocket will be constant. Since, more, and more fuel will be burnt the mass of, rocket will go on decreasing, so it will, lead to increase the velocity of the rocket, more and more rapidly., , 30 Here,, or, , 1, N = N0 , 2, N, 1, = , N 0 2, , t/T, , t/T, , …(i), , N, is, N0, fraction of atoms left after time t. Here,, N, 25, T = 40 days and, =, N 0 100, 1, = = 0.25, 4, N, Putting the values of T and, in Eq. (i),, N0, we get, where, T is the half-life period and, , t / 40, , 1 1, or, = , 4 2, t, or, = 2 or t, 40, , 2, , 1, 1, = , 2, 2, = 80 days, , t / 40
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408, , DAY THIRTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY THIRTY NINE, , Mock Test 2, (Based on Complete Syllabus), Instructions, 1. The test consists of 30 questions., 2. Candidates will be awarded marks for correct response of each question. 1/4 (one-fourth) marks will be deducted for indicating incorrect, response of each question. No deduction from the total score will be made, if no response is indicated for an item in the answer sheet., , 3. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response., 1 A particle projected with velocity v 0 strikes at right angles, a plane passing through the point of projection and, having inclination β with the horizontal. Find the height, (from horizontal plane) of the point, where the particle, strikes the plane, (a) y =, (c) y =, , 2v02, , (b) y =, , g (4 + cot β), v 02, 2, , (d) y =, , (4 + cot 2 β), , v 02, g (4 + cot 2 β), v 02, 2g (4 + cot 2 β), , 2 If the charge of 10 µC and −2 µC are given to two plates, of a capacitor, which are connected across a battery of, 12 V, find the capacitance of the capacitor., (a) 0.33 µF, , (b) 05, . µF, , (c) 0.41 µF, , (d) 0.66 µF, , 3 A man can swim with a speed of 4 Kmh−1 in still water., How long does he take to cross a river 1 km wide, if the, river flows steadily 3 Kmh−1 and he makes his strokes, normal to the river current. How far down the river does, he go when he reaches the other bank?, A, , 4 A body when projected vertically up, covers a total, distance s, during its time of flight. If we neglect gravity, then how much distance the particle will travel during the, same time. Will it fall back?, (a), (b), (c), (d), , s, Yes, s, No, 2s, Yes, 2s, No, , 5 Number of neutrons in a gold nucleus with A =197 and, Z = 79 is (gold,, , Z, , AuA ), , (a) 79, (b) 197, (c) 118, (d) None of these, , 6 The density of hydrogen nucleus with Z =1 is, 2.29 × 1017 kgm −3. The density of gold nucleus Z = 79, would be, 2. 29, × 1017 kgm−3, 79, (b) 2. 29 × 79 × 1017 kgm−3, (c) 2. 29 × 1017 kgm−3, , β, O, , (b) 750 km, (d) 850 km, , (a), , D, , β, , C, , (a) 600 km, (c) 800 km, , B, , (d), , 2. 29, × 1017 kgm−3, 79
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MOCK TEST 2, , DAY THIRTY NINE, 7 If mass of a proton is 1.007825 amu and mass of a, neutron is 1.008665 amu, then mass of 3 Li 7 nucleus, approximately be, (a) 7.058075 amu, (c) 7.023475 amu, , (b) 7.000000 amu, (d) 7.034600 amu, , of rods as shown in the figure. Every rod has the same, length l and cross-sectional area A. Thermal, conductivities are mentioned in figure. Assume that there, is no heat loss due to radiation or convection., 1, , K, , 2, , 3, , (a), , l, 4KA, , (b), , 4K, T2, , 7l, 4KA, , (c), , 7l, 12 K A, , (d), , l, 12K A, , 9 Two steel balls A and B are, placed inside a right circular, A, P, cylinder, of diameter 54 cm, R, B, making contacts at points P,Q, Q, and R as shown in the figure. The, 54 cm, radius rA = 12 cm and rB = 18 cm., The masses are mA = 15 kg and, mB = 60 kg. The force exerted by the floor at the point Q, and the wall at R are respectively (taking, g = 10 ms −2), (a) 600 N,150 N, (c) 600 N, 200 N, , 2.0 m wide. It has a mass of 180 kg. If it is originally lying, on the flat ground, how much work is needed to stand it, on one end?, (b) 3.0 J, , (c) 3.0 kJ, , Direction (Q. Nos. 14-15) Each of these questions contains, two statements : Statement I (Assertion) and Statement II, (Reason). Each of these questions also has four alternative, choices, only one of which is the correct answer. You have to, select one of the codes (a), (b), (c) and (d) given below., , (b) 750 N, 150 N, (d) 750 N, 200 N, , 14 Statement I When the range of projectile is maximum,, the time of flight is the largest., Statement II Range, projection is 45°., , is, , maximum,, , (b) 642 Vm−1, (d) 644 Vm−1, , 11 A wire of length 100 cm is connected to a cell of emf 2V, and negligible internal resistance. The resistance of the, wire is 3 Ω. The additional resistance required to produce, a potential difference of 1 mV/cm is, (a) 47 Ω, (c) 60 Ω, , A, m, , m, 53°, , (a) 0.45 m, (c) + 4.8 m, , (b) 440 Hz, 436 Hz, (d) 449 Hz, 440 Hz, , 37°, , W, , (b) 0.9 m, (d) Question is irrelevant, , 17 As shown in figure, a uniform solid sphere rolls on a, , horizontal surface at 20 ms −1. It then rolls up the incline, shown. What will be the value of h, where the ball stops?, v = 20 ms–1, h, 30°, , 12 A particular piano string is supposed to vibrate at a, , (a) 444 Hz, 436 Hz, (c) 444 Hz, 440 Hz, , of, , B, , (b) 57 Ω, (d) 55 Ω, , frequency of 440 Hz. In order to check its frequency, a, tuning fork known to vibrate at a frequency of 440 Hz is, sounded at the same time the piano key is struck, and a, beat frequency of 4 beats/s is heard. Find the possible, frequencies at which the string could be vibrating., , angle, , Statement II Energy equivalent (E ) or mass (m ) is, E = mc 2, 16 The two blocks as shown in the figure, have equal, masses and µs = µk = 0.3 for both blocks. Wedge W is, fixed and block A is given initial speed of 1 ms −1, down, the plane. How far will it move before coming to rest, if, inclines and strings are quite long? (take, g = 10 ms −2), , 1m, s –1, , 320 V is accelerating the electron. The electron beam is, entering a region having uniform magnetic field, 6 × 10−5 T, acting perpendicular to it. Find the value of, electric field in this region, so that the electron does not, experience any deflection. (Take, me = 9.1 × 10−31 kg ), , when, , 15 Statement I 1 amu is equivalent to 931 MeV., , 10 In J J Thomson’s experiment, a potential difference of, , (a) 640 Vm−1, (c) 637 Vm−1, , (d) 3000 kJ, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 2K, , T1, , 13 A uniform rectangular marble slab is 3.4 m long and, , (a) 2.0 kJ, , 8 Find the equivalent thermal resistance of the combination, , 409, , (a) 28.6 m, , (b) 8.6 m, , (c) 6 m, , (d) 18.6 m, , 18 A 1.6 kg block on a horizontal surface is attached to a, , spring with a spring constant of 1. 0 × 103 Nm −1. The, spring is compressed to a distance of 2.0 cm, and the, block is released from rest. Calculate the speed of the, block as it passes through the equilibrium position, x = 0,, if the surface is frictionless., (a) 0.75 ms−1 (b) 0.50 ms−1 (c) 0.25 ms−1 (d) 2.25 ms−1
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410, , DAY THIRTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , 19 A small conducting circular loop is placed inside a long, solenoid carrying a current. The plane of the loop, contains the axis of the solenoid. If the current in the, solenoid is varied, the current induced in the loop is, (a) clockwise, (b) anti-clockwise, (c) zero, (d) clockwise or anti-clockwise depending on whether the, current is increased or decreased, , t1 t2, , t3 t4, , t5, , t6, , t5, , t6, , (a), t1, , t2, , t3, , t4, , (b), t1, , t2, , t3, , t4, , t5, , t6, , t1, , t2, , t3, , t4, , t5, , t6, , 20 An inductor coil joined to a 6 V battery draws a steady, current of 12 A. This coil is connected to a capacitor and, an AC source of rms voltage 6V in series. If the current in, the circuit is in phase with the emf, the rms current will be, (a) 16.9 A, (c) 8 A, , (b) 12 A, (d) 9.87 A, , (c), , (d), , 21 Two wires, each having a weight per unit length of, , 1.0 × 10−4 Nm −1, are strung parallel to one another above, earth’s surface, one directly above the other. The wires, are aligned in a North-South direction, so that earth’s, magnetic field will not affect them. When their distance of, separation is 0.10 m, what must be the current in each in, order for the lower wire to levitate the upper wire?, Assume that the wires carry the same currents, travelling, in opposite directions., (a) 2.7 A, (c) 3.5 A, , (b) 0.1 A, (d) 7.1 A, , 22 Calculate the minimum thickness of a soap-bubble film, (n = 1.33) that will result in constructive interference in the, reflected light, if the film is illuminated by light with a, wavelength in free space of 602 nm., (a) 98 nm, (c) 125 nm, , (b) 113 nm, (d) 25 nm, , 23 A ground receiver station is receiving a signal at, 100 MHz, transmitted from a ground transmitter at a height, of 300 m located at a distance of 100 km. Then,, (Nmax = 1012 per m3 ), (a), (b), (c), (d), , down on a frictionless hill through a vertical distance of, 10.0 m. Use conservation of mechanical energy to find, the speed of the sledge at the bottom of the hill,, assuming the rider pushes off with an initial speed of, 5.00 ms −1. Neglect air resistance,, (a) 21.5 ms−1 (b) 14.9 ms−1 (c) 4.9 ms−1, , (d) 20.3 ms−1, , 26 Satellite dishes do not have to change directions in order, to stay focussed on a signal from a satellite. This means, that the satellite always has to be found at the same, location with respect to the surface of the earth. For this, to occur, the satellite must be at a height such that its, revolution period is the same as that of earth, 24 h. At, what height must the satellite be so to achieve this?, T2, , (a) , GMe , 2, 4π, , , 1/ 3, , T2, , (b) , GMe , 2, 4π, , , T2, , (c) , GMe , 2, 4π, , , 1/ 2, , T, , (d) , GMe , 2, 4π, , , 1/ 3, , 27 Water with a mass of 2.0 kg is held at constant volume in, , signal is coming via space wave, signal is coming via sky wave, signal is coming via satellite transponder, None of the above, , 24 The output waveform (Y ) of AND gate for the following, inputs A and B given below is, t1 t2, , 25 A sledge and its rider together weight 800 N. They move, , t3, , t5, t4, , t6, , a container while 10.0 kJ of energy is slowly added by a, flame. The container is not well insulated, and as a result, 2.0 kJ of energy leaks out to the surroundings. What is, the temperature increase of water?, (a) 0.28° C, , (b) 27° C, , (c) 0.96° C, , (d) 1.27° C, , 28 A 20 g bullet is fired horizontally with a speed of 600 ms −1, into a 7 kg block on a table top. The bullet b lodges in the, block B. If the coefficient of kinetic friction between the, block and the table top is 0.4, what is the distance the, block will slide?, , A, , b, , inputs, , B, , µ = 0.4, , B, , (a) 0.5 m, , (b) 1.2 m, , (c) 0.37 m, , (d) 0.85 m
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MOCK TEST 2, , DAY THIRTY NINE, 29 A uniform rope, of mass m per unit length, hangs, , 30.0 cm, , vertically from a support, so that the lower end just, touches the table top. If it is released, then at the time a, length y of the rope has fallen, the force on the table is, equivalent to the weight of the length k y of the rope. Find, the value of k., (a) 1, , (b) 2, , (c) 3, , 411, , 20.0 cm, , Object, , (d) 3.5, , 30 Two converging lenses are placed 20.0 cm apart, as, , f1 = 10.0 cm f2 = 20.0 cm, , shown in the figure. If the first lens has a focal length of, 10.0 cm and the second has a focal length of 20.0 cm,, locate the final image formed of an object 30.0 cm in, front of the first lens., , (a) 6.67 cm left, (c) 15.0 cm left, , (b) 6.67 cm right, (d) 15.0 cm right, , ANSWERS, 1. (a), 11. (b), 21. (d), , 2. (b), 12. (a), 22. (b), , 3. (b), 13. (c), 23. (c), , 4. (d), 14. (d), 24. (b), , 5. (c), 15. (a), 25. (b), , 6. (c), 16. (a), 26. (a), , 7. (a), 17. (a), 27. (c), , 8. (c), 18. (b), 28. (c), , 9. (d), 19. (c), 29. (c), , 10. (c), 20. (b), 30. (a), , Hints and Explanations, 1 Let α be the angle between the velocity, of projection and the inclined plane, v0, , =, , 2, , v0, 2cot β, g (4 + cot2 β ), , ∴ y = y tan β =, , Y′, X′, , =, , α, , β, , 2, , 2v 0, , 4 Let particle is projected with speed u, so, , g(4 + cot2 β ), , 2 Charge of capacitor is the charge on, , X′, , v 0x ′ = v 0 cos α, v 0 y′ = v 0 sin α, ax ′ = − g sin β, a y ′ = −g cos β, ⇒ v x ′t = v 0 cos α − g sin β t, At the point of impact v x ′ = 0, v cos α, …(i), t = 0, ⇒, g sin β, Also, y at this point is zero, ⇒ v 0 sin α t − 1 / 2g cos β t 2 = 0, 2v 0 sin α, t2 =, g cos β, , 2, , v 0 2cot β, ⋅ tan β, g 4 + cot2 β, , facing surfaces of the plates of capacitor, q − q2 [10 − (− 2)], Q = 1, = 6 µC, =, , 2 , 2, Potential difference across the capacitor, = 12 V, Q (6 × 10−6 ), So, C =, =, F = 0.5 µF, V , 12, , , 3 Given, speed of man (v m ) = 4 kmh −1, …(ii), , From Eqs. (i) and (ii), we get, cot β, tan α =, 2, x = v 0 cos (α + β )t, 2, , , , 2, , , cot β, 2 , , 2 , v, = 0 4 + cot β , , g , , cot β, 2, ⋅, , −, 4 + cot2 β, 4 + cot2 β , , , Speed of river (v r ) = 3 kmh, , −1, , Width of the river (d ) = 1 km, B, , Distance travelled along the river = v r × t, 1 3, = 3 × = km, 4 4, 3000, =, = 750 m, 4, , vr, , C, , total time of flight,, 2u , T = , g , and s = 2 × maximum height, u2, u2, =2×, =, 2g, g, If there is no gravity, then s ′ = u × T, 2u2, =, = 2s, g, If gravity is not there, it will never fall, back., , 5 N = A − Z = 197 − 79 = 118, 6 Density of every nucleus is same, = 2.29 × 1017 kg m −3, , 1 km, , vm β, , v, , Time taken by the man to cross the river, 1 km, Width of the river, t=, =, Speed of the man 4 Kmh −1, 1, 1, = h = × 60 = 15min, 4, 4, , 7 In 3 Li7 , Z = 3 ; N = A − Z = 7 − 3 = 4, ∴ Mass of nucleus = Z mp + ( A − Z ) m n, = 3 × 1. 007825 + 4 × 1. 008665, = 7. 058075 amu, The actual mass of nucleus is slightly, less than this calculated value.
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412, , DAY THIRTY NINE, , 40 DAYS ~ JEE MAIN PHYSICS, , KA, , Req, , = 10. 61512 × 106 × 6 × 10−5, , l, l, , R3 =, 2K A, 4KA, , 8 R1 = l , R2 =, , R1 R2, , =, + R3 , R1 + R2, , , = 636. 9072 Vm, , , l ⋅ l, , K A 2K A, l, , Req = , +, l, 4K A , l, +, , , , K A 2K A, l2, , 2 2, , l , 2K A, =, +, 4K A , 2l + l, , 2K A, , , , 4l+ 3l, l, l, 7l, =, +, =, =, 3K A 4K A, 12K A 12K A, (as rods 1 and 2 are in parallel and, equivalent is in series with 3)., 9 Free body diagrams of balls A and B,, NP, NR, , 600, , 16 Find acceleration,, , = 63. 69072 × 101, , 150, , −1, , ≈ 637 Vm, , −1, , 11 Let a resistance R is connected in series, with the battery and wire., 100 cm, 3Ω, , 12, , 17 The rotational and translational kinetic, energy of the ball at the bottom will be, changed to gravitational potential energy,, when the sphere stops. We therefore, write, Mv 2, Iω2 , , , +, = (Mgh ) end, 2 start, 2, , R, , 2V, , Voltage drop across wire, = 1 × 10−3 × 100 = 01, . V, , For a solid sphere, I =, , Let current in circuit is I., 2, ⇒ R = 57 Ω, R+3, , ∴ 0.1 = l × 3 = 3 ×, , 12 The number of beats per second is equal, to the difference in frequency between, the two sound sources. In this case,, because one of the source frequencies is, 440 Hz, 4 beats s −1 would be heard, if, the frequency of the string (the second, source) were either 444 Hz or 436 Hz., done, as if all the mass were, concentrated at the centre of mass. The, work necessary to lift the object can be, thought of as the work done against, gravity and is just W = mgh, where h is, the height through which the centre of, mass is raised., W = 180 (9.8) (1.7) = 3.0 kJ, , cm, 18, , cm, , u2 sin 2θ, g, 2u sin θ, Time of flight, T =, g, , θ, , 14 The horizontal range, R =, , 24cm, cos θ = 24/30 = 4/5, Free body diagram of ball B,, , N, , NR, , 600, , NQ = 750, , N sinθ + 600 − 750 = 0, 3N, = 150, 5, 150 × 5, ⇒, N =, = 250, 3, N cos θ − N R = 0, N R = 200 N, , 10 E =, =, , Range is maximum, when θ = 45°, So that, sin 2θ = sin 90° = 1, u2, R max =, g, Time of flight is maximum when, θ = 90°, So that, sin θ = sin 90° = 1, 2u, T max =, g, , 15 Substituting m = 1 amu = 1. 67 × 10−27 kg, and c = 3 × 108 ms −1 in the energy-mass, equivalence relation, E = mc 2, = 1. 67 × 10, , 2q v, ×B, M, , −27, , × (3 × 10 ), , 82, , = 1. 67 × 10−27 × 9 × 1016 J, , 2 × 1. 60218 × 10−19 × 320, 9.1 × 10, , = 112. 680 × 10, , 12, , −31, , × 6 × 10, , −5, , × 6 × 10−5, , u = 1 ms −1, On solving, we get s = 0.45 m, , 13 The work done by gravity is the work, Na, NQ=750=0, , 2 ma = mg sin 53° − µ mg cos 53°, − mg sin 37° − µ mg cos 37°, Then, use v 2 = u2 + 2as, v = 0,, , =, , 1. 67 × 10−27 × 9 × 1016, 1. 6 × 10−13, (Q 1 MeV = 1. 6 × 10−13 V ), , = 931 MeV, , Also, ω =, , 2 Mr 2, 5, , v, . Then, above equation, r, , becomes, 2, , 1, 1 2, v, Mv 2 + Mr 2 = Mgh, r, 2, 2 5, 1 2 1 2, or, v + v = (9. 8 ) h, 2, 5, Using v = 20 ms–1, gives h = 28. 6 m, , 18 The initial elastic potential energy of the, compressed spring is, 1, PE s = k x 2i, 2, Because the block is always at the same, height above earth’s surface, the, gravitational potential energy of the, system remains constant. Hence, the, initial potential energy stored in the, spring is converted to kinetic energy at, x = 0. i.e., 1, 1, k x2i = m v 2f, 2, 2, k, Solving for v f gives, v f =, xi, m, =, , 1.0 × 103, (2.0 × 10−2 ) = 0.50 ms −1, 1.6, , 19 The angle between magnetic field and, area vector is 90°, so the flux associated, with coil is zero. Although magnetic field, is changing but flux is remaining, constant equal to zero, so emf induced, and hence, current in the loop is equal to, zero., 20 Resistance of coil = V DC = 6 = 0.5 Ω, I DC, 12, In an AC circuit, the current is in phase, with emf. This means that the net, reactance of the circuit is zero. The, impedance is equal to the resistance, i.e., Z = 0.5 Ω, rms voltage, Rms current =, Z, 6, =, = 12 A, 0.5
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21 If the upper wire is to float, it must be, in equilibrium under the action of two, forces : the force of gravity and, magnetic repulsion. The weight per unit, length here1. 0 × 10−4 Nm −1 must be, equal and opposite the magnetic force, per unit length. Because the currents, are the same, we have, mg µ 0I 2, F1, =, =, I, I, 2 πd, ⇒ 1.0 × 10−4 =, , −7, , (4 π × 10 ) (I ), (2 π )(0.10), 2, , We solve for the current to find, I = 71, . A, λ, 22 Because 2 n t = , we have, 2, λ, 602, t =, =, = 113 nm, 4 n (4)(133, . ), , 23 Maximum distance covered by space, wave communication, = 2 Rh, = 2 × 6. 4 × 106 × 300 = 62 km, Since, receiver-transmitter distance is, 100 km, this is ruled out for signal, frequency., Further fc for ionospheric propagation is, fc = 9 (N max )1 /2 = 9 × (1012 )1 /2 = 9 MHz, So, the signal of 100 MHz (7fc ) comes, via the satellite mode., , 24 For t ≤ t 1 ; A = 0, B = 0; Hence Y = 0, For t 1 to t 2 ; A = 1, B = 0 ; Hence Y = 0, For t 2 to t 3 ; A = 1, B = 1; Hence Y = 1, For t 3 to t 4 ; A = 0, B = 1; Hence Y = 0, For t 4 to t 5; A = 0, B = 0; Hence Y = 0, For t 5 to t 6; A = 1, B = 0; Hence Y = 0, For t > t 6; A = 0, B = 1; Hence Y = 0, Based on the above, the output, waveform for AND gate can be drawn as, given below., t2, t1, , t3, t4, , t5, , 413, , MOCK TEST 2, , DAY THIRTY NINE, , t6, , 25 The initial energy of the, sledge-rider-earth system includes, kinetic energy because of the initial, speed, 1, 1, m v 2i + m g y i = m v 2f + m g y f, 2, 2, 1 2, 1, or, v i + g y i = v 2f + g y f, 2, 2, If we set the origin of our coordinates at, the bottom of the incline, the initial and, final y-coordinates of the sledge are, y i = 10.0 m and y f = 0., , Thus, we get, 1 2, 1, v i + g y i = v 2f + 0, 2, 2, v 2f = v 2i + 2g y i, = (5.00)2 + 2 (9.80)(10.0), v f = 14. 9 ms −1, , 26 The force that produces the centripetal, acceleration of the satellite is the, gravitational force, so, M m m v2, ∴, G e2 =, r, r, , From work-energy theorem,, ∆K = W f, 0 − K = − µ k (m + M ) g s, Substituting values in Eqs.(i) and (ii), we, get, ∴, s = 0.37 m, Also the rest mass of photon is zero., , 29 The descending part of the rope is in free, fall. It has speed v = 2 gy at the instant, all its points have descended a distance, y. The length of the rope which lands on, the table during an interval dt following, this instant is v dt. The increment of, momentum imparted to the table by this, length in coming to rest is m (v dt ) v ., Thus, the rate at which momentum is, transferred to the table is, , …(i), , where, Me is earth’s mass and r is the, satellite’s distance from the centre of, the earth., Also, we find the speed of the satellite, to be, 2π r, d, …(ii), v =, =, T, T, , v, , where, T is the orbital period of the, satellite., Solving Eqs. (i) and (ii) simultaneously, for r yields,, T2, , r = , G Me , 2, 4π, , , 1 /3, , 27 Recall that an isovolumetric process is, one that takes place at constant volume., In such a process the work done is, equal to zero because there is no change, in volume. Thus, the first law of, thermodynamics gives, ∆U = Q, This indicates that the net energy Q, added to the water goes into increasing, the internal energy of the water. The net, energy added to the water is, Q = 10.0 − 2.0 = 8.0 kJ, Because Q = m c ∆T , the temperature, increase of the water is, 8.0 × 103, Q, ∆T =, =, mc, (2.0)(4.186 × 103 ), , F, , dp, = m v 2 = (2 m y ) g, dt, and this is the force arising from, stopping the downward fall of the rope., Since, a length of rope y of the weight, (my ) g , already lies on the tabletop, the, total force on the tabletop is, (2 m y ) g + (m y ) g = (3 m y ) g , or the, weight of a length 3y of rope., So,, k =3, , 30 First we make ray diagrams roughly to, scale to see where the image from the, first lens falls and how it acts as the, object for the second lens. The location, of the image formed by the first lens is, found via the thin lens equation, 1, 1, 1, , v 1 = + 15.0 cm, +, =, 30.0 v 1 10.0, , = 0.96° C, , 28 By conservation of momentum, the, momentum of the block bullet system, just after the interaction is p = mv ., where, m is the mass of bullet and v is, its velocity before striking the block., Hence, the kinetic energy of the system, just after the lodging of bullet into the, block is, p2, (mv ) 2, ...(i), K =, =, 2 (M + m ) 2 (M + m ), The friction force does work, ...(ii), W f = − f s = − µ k (m + M ) g s, in stopping the block where s is the, distance traversed by block-bullet, system on the table top., , y, , F 1 I2 I1, O1, , F1, , F2, , F2, , 10.0 cm, , 15.0 cm, , 6.67 cm, 30.0 cm, 20.0 cm, Lens 2, Lens 1, , The image formed by this lens becomes, the object for the second lens. Thus, the, object distance for the second lens is, 20.0 cm − 15.0 cm = 5.00 cm. We again, apply the thin lens equation to find the, location of the final image., 1, 1, 1, , v 2 = − 6. 67 cm, +, =, 5. 00 v 2 20.0, Thus, the final image is 6.67 cm to the, left of the second lens.
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414, , DAY FOURTY, , 40 DAYS ~ JEE MAIN PHYSICS, , DAY FOURTY, , Mock Test 3, (Based on Complete Syllabus), Instructions, 1. The test consists of 30 questions., 2. Candidates will be awarded marks for correct response of each question. 1/4 (one-fourth) marks will be deducted for indicating incorrect, response of each question. No deduction from the total score will be made, if no response is indicated for an item in the answer sheet., , 3. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response., , 1 An organ pipe of length L 0, open at both ends is bound, , 4 Choose the correct alternative., , to vibrate in its first harmonic, when sounded with a, tuning fork of 480 Hz. What should be the length of a, pipe closed at one end, so that it also vibrates in its first, harmonic with the same tuning fork?, (a) Lc = 2L 0, , (b) Lc =, , L0, 3, , (c) Lc =, , L0, 2, , (d) Lc =, , (a) Gravitational potential at curvature centre of a thin, hemispherical shell of radius R and mass M is equal to, GM /R, (b) Gravitational field strength at a point lying on the axis of, a thin uniform circular ring of radius R and mass M is, GM, equal to 2, , where x is distance of that point, (R + x 2 ) 3 / 2, , 2L 0, 3, , 2 A hill is 500 m high. Supplies are to be sent across the, hill using a canon that can hurl packets at a speed of 125, ms −1 over the hill. The canon is located at a distance of, 800 m from the foot of hill and can be moved on the, ground at a speed of 2 ms −1, so that its distance from the, hill can be adjusted. What is the shortest time in which a, packet can reach on the ground across the hill? (Take,, g = 10 ms −2), (a) 31 s, , (b) 27 s, , (c) 37 s, , 3 A motor cyclist starts from the, , (d) 45 s, A, , bottom of a slope of angle 45° and, 45°, travels along the slope to jump, O, clear of the valley AB as shown in, figure. The width of the valley is, 160 m and the length of the slope is 160 2 m. The, minimum velocity with which he should leave the bottom, O, so that he can clear the valley, is (nearest to in ms −1), (a) 50, , (b) 56, , (c) 60, , (d) 70, , from centre of the ring, (c) Newton’s law of gravitation for gravitational force, between two bodies is applicable only, when bodies, have spherically symmetric distribution of mass, (d) None of the above, , 5 A thin wire of length L and uniform linear mass density ρ is, bent into a circular loop with centre at O as shown. The, moment of inertia of the loop about the axis XX ′ is, X, , B, , X′, 90°, O, , (a), , ρL3, , (b), , 8 π2, 5 ρL, , 3, , (c), , 16 π 2, , (d), , ρL3, 16 π 2, 3 ρL3, 8 π2
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MOCK TEST 3, , DAY FOURTY, 6 If a drop of liquid breaks into smaller droplets, it results, in lowering of temperature of the droplets. Let a drop of, radius R, break into N small droplets each of radius r., Estimate the temperature in drop., S 1, ρs R , 3 S 1 1, (c), −, ρs R r , , 2S, ρs, 2S, (d), ρs, , (a), , (b), , 1 − 1 , r R , 1 − 1, R r , , 7 A man beats a drum at a certain distance from a, mountain. He slowly increase the rate of beating and, finds that the echo is not heard distinctly, when the drum, beating is at the rate of 40 per min. He moves by 80 m, towards the mountain and finds that the echo is again not, heard distinctly, when the rate of beating of the drum is, 1 per sec. What is the original distance of the man from, the mountain?, (a) 120 m, , (b) 240 m, , (c) 270 m, , 415, , 10 Two non-ideal batteries of unequal emf ’s are connected in, parallel. Consider the following statements., , (A) The equivalent emf is smaller than either of the, two emf ’s., (B) The equivalent internal resistance is smaller than, either of the two internal resistances., (a) Both A and B are correct, (b) A is correct but B is incorrect, (c) B is correct but A is incorrect, (d) Both A and B are incorrect, , 11 A solid conducting sphere of radius a has a net positive, charge 2 Q. A conducting spherical shell of inner radius b, and outer radius c is concentric with the solid sphere and, has a net charge − Q., The surface charge density on the inner and outer, surfaces of the spherical shell will be, , (d) 340 m, , 8 The smiling face in figure consists of three items (i), A, thin rod of charge –3.0 µC that forms a full circle of radius, 6 cm. (ii) second thin rod of charge 2.0 µC that forms a, circular arc of radius 4.0 cm, subtending an angle of 20°, about the centre of the full circle and (iii) electric dipole, with a dipole moment that is ⊥ to a radial line and has a, magnitude 1. 28 × 10−21cm, what is the net electric, potential of the centre., , a, , b, , c, , (a) −, (c) 0,, , 2Q, Q, ,, 4 πb 2 4 πc 2, Q, 4 πc 2, , (b) −, , Q, Q, ,, 4 πb 2 4 πc 2, , (d) None of these, , Direction (Q. Nos. 12-14) Each of these questions contains, two statements : Statement I and Statement II. Each of these, questions also has four alternative choices, only one of which, is the correct answer. You have to select one of the codes (a),, (b), (c), (d) given below., −3, , (b) 2 . 3 × 10, (d) None of these, , (a) zero, (c) 1. 28 × 1021, , 9 Pressure versus temperature graphs of an ideal gas are, as shown in figure. Choose the incorrect statement., p, , p, , (a) Statement I is true, Statement II is true; Statement II is, the correct explanation for Statement I, (b) Statement I is true, Statement II is true; Statement II is, not the correct explanation for Statement I, (c) Statement I is true; Statement II is false, (d) Statement I is false; Statement II is true, , 12 Statement I Time period of oscillation of two magnets,, when like poles are in same direction (in a vibration, magnetometer) is smaller, than the period of vibration, when like poles are in opposite direction., , T, , T, , (i), , (ii), p, , Statement II Moment of inertia increases in same, position., , 13 Statement I It is impossible for a ship to use the internal, energy of sea water to operate its engine., Statement II Refrigerator is a type of heat engine., T, (iii), , (a) Density of gas is increasing in graph (i), (b) Density of gas is increasing in graph (ii), (c) Density of gas is constant in graph (iii), (d) None of the above, , 14 Statement I A tennis ball bounces higher on hills than in, plains., Statement II Acceleration due to gravity on the hill is, greater than that on the surface of the earth.
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416, , DAY FOURTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 15 In the circuit shown, the coil has inductance and, resistance. When X is joined to Y , the time constant is τ, during growth of current. When the steady state is, reached, heat is produced in the coil at a rate P. X is, now joined to Z, , 20 The circuit shown in figure, contains a resistance of, R = 6 Ω connected with a battery of emf 6 V., R, I, , 6V, , Y, X, Z, , (a) the total heat produced in the coil is P τ, 1, (b) the total heat produced in the coil is P τ, 2, (c) the total heat produced in the coil is 2P τ, (d) the data given is not sufficient to reach a conclusion, , 16 Nickle shows ferromagnetic property at room, temperature. If the temperature is increased beyond, curie temperature, then it will show, (a) anti-ferromagnetism, (b) no magnetic property, (c) diamagnetism, (d) paramagnetism, , Given, n = number of electrons per volume = 1029 / m 3,, length of circuit = 10 cm, cross-section A = 1 mm 2. The, energy absorbed by electrons from initial state of no, current (ignore thermal motion) to the state of drift, velocity is, (a) 15, . × 10−18 J, (c) 2 × 10−17 J, , (b) 3.5 × 10−19 J, (d) 3 × 10−15 J, , 21 Two long parallel wires are at a distance 2d apart. They, carry steady equal currents flowing out of the plane of the, paper, as shown. The variation of the magnetic field B, along the line XX ′ is given by, B, (a), , B, D, , X, , X′, , C, , A, d, , (b), , D, , X, A, d, , d, , X′, , C, d, , 17 A glass prism ABC (refractive index 1.5), immersed in, water (refractive index 4/3). A ray of light is incident, normally on face AB. If it is totally reflected at face AC,, then, , (c), , X, , X′, A, , C, d, , B, , θ, , A, , B, , B, , (c) sinθ =, , (b) sinθ ≥, 3, 2, , (d), , 2, 3, , 2, 8, < sinθ <, 3, 9, , 18 A metal wire of linear mass density of 9.8 gm −1 is, stretched with a tension of 10 kg-wt between two rigid, supports which are 1m apart. The wire passes through, the middle points between the poles of a permanent, magnet and it vibrates in resonance, when carrying on, alternating current of frequency n. The frequency n of the, alternating current is, (a) 25 Hz, , (b) 50 Hz, , (c) 200 Hz, , (d) 100 Hz, , 19 In a given process of an ideal gas, dW = 0 and dQ < 0., Then, for the gas, (a) the temperature will decrease, (b) the volume will increase, (c) the pressure will remain constant, (d) the temperature will increase, , A, , D, , X′, , C, d, , d, , d, , 22 The potential energy of a particle of mass m is given by, , C, , 8, 9, , X, , D, , E, U(x) = 0, 0, , (a) sinθ ≥, , (d), , 0 ≤ x ≤1, x >1, , λ 1 and λ 2 are the de-Broglie wavelengths of the particle,, when 0 ≤ x ≤ 1 and x > 1, respectively. If the total energy of, λ, particle is 2 E 0, the ratio 1 will be, λ2, (a) 2, , (b) 1, , (c) 2, , (d), , 1, 2, , 23 Two radioactive nuclei A and B have their disintegration, constant λ A and λ B , respectively. Initially, NA and NB, number of nuclei are taken, then the time after which their, undisintegrated nuclei are same is, (a), , N , λ A λB, ln B , (λ A − λB ) N A , , (b), , N , 1, ln B , (λ A + λB ) N A , , (c), , N , 1, ln B , (λB − λ A ) N A , , (d), , N , 1, ln B , (λ A − λB ) N A
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MOCK TEST 3, , DAY FOURTY, 24 A student constructed a vernier callipers as shown in the, , 27 The fundamental frequency of a sonometer wire of length, l is f0. A bridge is now introduced at a distance of ∆l from, the centre of the wire ( ∆l << l ′ ). The number of beats, heard, if both sides of the bridges are set into vibration in, their fundamental modes, are, , figure. He used two identical inclines and tried to, measure the length of line PQ. For this instrument, determine the least count., , 8f0 ∆l, l, 2f ∆l, (c) 0, l, , f0 ∆l, l, 4f ∆l, (d) 0, l, (b), , (a), , P, Q, , l, , 417, , 28 A mason is supplied with bricks by his assistant who is, 3 m below him, the assistant is tossing the brick vertically, up. The speed of the brick, when it reaches the mason is, 2 ms −1. What percentage of energy used up by the, servant serves no useful purpose ?, , l (1 − cos θ), units, cos θ, l, units, (b), cos θ, (c) l (1 − cos θ) units, 1 − cos θ, (d), units, l, (a), , (a) 9.8%, , (b) 4.9%, , (c) 5.6%, , (d) 10%, , 29 To plot forward characteristic of p-n junction diode, the, correct circuit diagram is, , mg, 2, , V– (0-2)V A–+, , (c) E, , +, V– (0-2)V A+, , –, , +, , +, , 30 A block of mass m moves with a velocity vo on a smooth, , (d) mg(1 − µ ), , 26 A double star consists of two stars having masses, M and 2 M . The distance between their centres is equal, to r. They revolve under their mutual gravitational, interaction. Then, which of the following statement(s), is/are correct?, , horizontal surface. If that passes over a cylinder of radius, R and mass m, capable of rotating about its own fixed, axis through O the block, while passing over, slips on the, cylinder. The slipping stops before, it loses contact. The, block then moves on similar smooth horizontal surface, with a velocity v. then the velocity v is, m, , (a) Heavier star revolves in orbit of radius 2r/3, (b) Both of the stars revolve with the same period which is, 2π, equal to, r 3/ 2, 2 GM /3, (c) Kinetic energy of heavier star is twice that of the other, star, (d) None of the above, , vo, , m, , v, , O, R, m, , (a), , 2, vo, 5, , (b), , 2, vo, 3, , (c) vo, , (d), , vo, 2, , ANSWERS, 1. (c), 11. (a), 21. (b), , 2. (d), 12. (c), 22. (c), , 3. (d), 13. (b), 23. (c), , +, , (0-1000) mA, , (c), , mg, 4, , (d) E, , (0-1)V A+–, V+, –, , L, , (b), , V– (0-2)V A –, , (a) E, , µ, , (a) infinitesimal, , (b) E, , (0-1000) mA, , F, , (0- 20) A, , surface with coefficient of friction µ. A horizontal force F, is applied on the block as shown in the figure. If the, coefficient of friction is sufficiently high, so that the block, does not slide before toppling, the minimum force, required to topple the block is, , (0-1000) mA, , 25 A cubical block of side L rests on a rough horizontal, , 4. (c), 14. (c), 24. (a), , 5. (d), 15. (b), 25. (c), , 6. (c), 16. (d), 26. (b), , 7. (b), 17. (a), 27. (a), , 8. (a), 18. (b), 28. (c), , 9. (a), 19. (a), 29. (b), , 10. (a), 20. (c), 30. (b)
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418, , DAY FOURTY, , 40 DAYS ~ JEE MAIN PHYSICS, , Hints and Explanations, 1 The fundamental frequency for an open, ν, pipe, ν 0 =, 2L 0, , 3 Velocity to take-off from A to clear the, valley is given by, u2, R=, sin 2α, g, , Lc, L0, Closed organ pipe, , α = 45° , u =, , gR = 40 ms −1, , Velocity to start from lowest point (due, to retardation on inclined plane,, g sin α), v 20 = u2 + 2 g sin α × s, ∴, , v0 =, , (40) 2 + 2 × 10 ×, , 1, 2, , Energy released = S × ∆ A, (where, S = surface tension), = S × 4 π(R2 − Nr 2 ), Due to release of this energy, the, temperature is lowered., If ρ is the density and s is specific heat of, liquid and its temperature is lowered by, ∆θ then,, Energy released = ms∆θ, 4π, S × 4 π (R2 − N r 2 ) = , × R3 × ρ s∆θ, , 3, , × 160 × 2, Open organ pipe, Fundamental frequency for closed pipe,, ν, ν 0′ =, 4L c, , According to the question,, ν 0 = ν ′0, ν, ν, L, =, ⇒ Lc = 0, 2L 0, 4L 0, 2, , 2 Given, height of the hill (h) = 500 m, Velocity of canon, u = 125 ms −1, To cross the hill, the vertical, component of the velocity should be, sufficient to cross such height., ∴ u y ≥ 2gh ≥ 2 × 10 × 500, ≥ 100 ms −1, But u =, 2, , u2x, , +, , u2y, , ∴ Horizontal component of initial, velocity,, u x = u2 − u2y = (125)2 − (100)2, = 75ms −1, Time taken to reach the top of the hill,, 2 × 500, 2h, t =, =, = 10 s, g, 10, Time taken to reach the ground from, the top of the hill,, t ′ = t = 10 s, Horizontal distance travelled in 10 s,, x = ux × t, = 75 × 10 = 750 m, ∴ Distance through which canon has to, be removed = 800 − 750 = 50 m, Speed with which canon can move, = 2 ms −1, 50, ∴ Time taken by canon =, 2, t ′′ = 25 s, Hence, total time taken by a packet to, reach on the ground = t ′′ + t + t ′, = 25 + 10 + 10, = 45s, , =, , S × 4 π (R2 − Nr 2 ), 4 3, πR ρ × s, 3, 3S R 2, Nr 2 , =, 3 − 3 , ρs R, R , , ∆θ =, , 4800 ~ 70 ms −1, , 4 Because every element of hemispherical, shell is at a distance R from centre of, curvature, therefore gravitational, GM, potential at its centre = −, ,, R, , 3S 1 ( R 3 / r 3 ) × r 2 , −, , ρs R, R3, , 3S 1 1 , =, −, ρs R r , =, , i.e. option (a) is incorrect., Gravitational field strength at a point,, lying on the axis of a thin uniform, GMx, circular ring of radius R is 2, (R + x2 )3 /2, So, option (b) is incorrect., Newton’s law of gravitation is, applicable to only those bodies which, have spherically symmetric distribution, of mass. So, option (c) is correct., , 5 Mass of the ring, M = ρL, Let R be the radius of the ring., L, Then, L = 2 πR or R =, 2π, Moment of inertia about X X ′ (from, parallel axis theorem) will be given by, 1, 3, I XX ′ = MR2 + MR2 = MR2, 2, 2, Putting values of M and R,, L2 3 ρL3, 3, I XX ′ = ( ρL ) 2 =, 2, 4 π 8 π2, , 6 When a big drop of radius R, break into, N droplets each of radius r, the volume, remains constant., ∴ Volume of big drop, = N × volume of small drop, 4 3, 4, πR = N × πr 3, 3, 3, R3, or, R3 = N r 3 or N = 3, r, Now, change in surface area, = 4 πR 2 − N 4 πr 2, = 4 π (R2 − Nr 2 ), , 7 The echo is not heard distinctly, when, the echo and the next beat fall on the ear, simultaneously, i.e. time per beat = time, taken by the reflected beat to reach the, man., 2d, 60 3, Hence,, =, =, v, 40 2, 2(d − 80), and, =1, v, This gives,, d = 240 m, , 8 The dipole potential (θ = 90° ) is given by, the equation,, p cos θ, p cos 90°, V =, =, =0, 4 πεo r 2, 4 πεο r 2, [Q cos 90° = 0], Also, the potential due to the short arc is, q1 / 4 πε0 r1 and that caused by the long, arc is q2 / 4 πε 0r 2 ., Since, q1 = + 2µC, r1 = 4cm, q2 = −3µC, and r2 = 6 cm, the potential of the arcs, cancel., Thus, the result is zero., pM, RT, Density ρ remains constant, when p / T, or volume remains constant. In graph (i), volume is decreasing, hence density is, increasing; while in graphs (ii) and (iii), temperature is increasing, hence, density, is decreasing. Note that, volume would, have been constant in case the straight, line in graph (iii) had passed through, origin., , 9 As, ρ =
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14 Suppose that the tennis ball bounces, , 10 In parallel,, Eeq, , E /r + E2 /r2, = 1 1, 1/r1 + 1/r2, r + (E2 /E1 ) ⋅ r1 , = E1 2, , r1 + r2, , , r1 + (E1 /E2 ) ⋅ r2 , = E2 , , r1 + r2, , , , with a velocity u. It will go up, till its, velocity becomes zero. If h is the height, upto which it rises on the hill, then, (0) 2 − u2 = 2(− g ′ ) h, where, g ′ is acceleration due to gravity, on the hill, u2, h=, ∴, 2g ′, , Now, if E1 = E2 ., Then, Eeq = E1 if E2 > E1 ., , Since, the acceleration due to gravity on, the hill (g ′ ) is less than that on earth, (effect of height), it follows that tennis, ball will bounce higher on hills than in, plains., , Then, Eeq > E1 and Eeq < E2 ., Similarly, if E1 > E2 ., Then, Eeq > E2 but Eeq < E1 ., , 15 As, P = (I 0 ) 2 ⋅ R, , 11 Due to induction, inner surface of, spherical shell has charge − 2Q,, , ∴, , a, , 16 The curie temperature is defined as the, , b, , temperature beyond which the, ferromagnetic material shows, paramagnetic behaviour., , c, , 17 For total internal reflection at AC-face, So, surface charge density on inner side, −2Q, σ inner =, 4 πb 2, and surface charge density on outer, Q, side σ outer =, 4 πc 2, , 12 Case I When the like poles of two, magnets are placed in same direction,, then the time period of vibration is, expressed as, T′ = 2π, , I1 + I2, ( M1 + M2 ) B, , …(i), , Case II When the like poles of two, magnets are placed in opposite, direction, then period of vibration is, expressed as, T ′′ = 2 π, , I1 + I2, ( M1 − M2 ) B, , …(ii), , It is clear from Eqs. (i) and (ii) that,, T ′ < T ′ ′., , 13 For using the internal energy of sea, water to operate the engine of a ship,, the internal energy of the sea water has, to be converted into mechanical energy., Since, whole of the internal energy, cannot be converted into mechanical, energy, a part has to be rejected to a, colder body (sink). As, no such body is, available, the internal energy of the sea, water cannot be used to operate the, engine of the ship. Note that a, refrigerator is a heat engine working in, the reverse direction., , sin i ≥, , µw, µg, , 4, 3 × 1.5, 8, sinθ ≥, 9, sinθ ≥, , 1, × 10−4 ms −1, 1.6, , The energy of electrons, (KE) =, =, , 1, mv 2, 2, , 1, × me × v d2 × volume × number of, 2, electrons per volume, 2, , =, , 10−4 , 1, × 91, . × 10−31 × , × A×I ×n, 2, 1.6 , (Q Mass of electron m e = 91, . × 10−31 ), , =, , 91, . × 10−39, × 10−6 × 01, . × 1029, 2 × 1.6 × 1.6, , = 2 × 10−17 J, magnetic field at points to the right of the, wire will be upwards and to the left will, be downward. Now, magnetic field at C, is zero. The field in the region DX ′ will, be upwards (+ve), because all points, existing in this region are to the right of, both the wires. Similarly, magnetic field, in the region AX will be downwards, (–ve). The field in the region AC will be, upwards (+ve), because points are closer, to A as compared to D. Similarly,, magnetic field in region DC will be, downward (–ve)., , 22 KE = 2 E 0 − E 0 = E 0, , 18 Since, the tension,, T = 10 kg-wt = 10 × 9.8 = 98 N, and m = 9.8 × 10−3 kgm −1 , L = 1m, So, we get, 1 T, 1, n=, =, ×, 2L m 2 × 1, = 50 Hz, , =, , 21 If the current flows out of the paper, the, , P, R, 1 2 1, P, 1, U = LI 0 = (τR ) = P τ, R 2, 2, 2, , i.e. (I 0 ) 2 =, , –2Q, +2Q, , 419, , MOCK TEST 3, , DAY FOURTY, , 98, 9.8 × 10−3, , 19 From first law of thermodynamics, dQ = dU + dW, dQ = dU, (Q dW = 0), Since,, dQ < 0, Therefore,, dU < 0, or, U final < U initial, or temperature will decrease., , 20 Given, V = 6 V, R = 6 Ω ,, A = 1 × 10−6 m2, and, l = 10 cm = 01, . m, The current in the circuit,, V, 6, I =, = = 1A, R 6, Use the relation I = ne A v d, Drift velocity of electrons,, I, vd =, neA, 1, = 29, 10 × 1.6 × 10−19 × 1 × 10−6, , So,, , λ1 =, , Again, , KE = 2 E 0, , ∴, , λ2 =, , (for 0 ≤ x ≤ 1 ), , h, 2 mE 0, , …(i), (for x > 1 ), , h, 2 m2 E 0, , …(ii), , From Eqs. (i) and (ii), we get, λ1, = 2, λ2, , 23 After disintegration,, , N A e − λ A t = N B e − λ B t (for 0 ≤ x ≤ 1), N, or e ( λ B − λ A ) t = B, …(i), NA, , N , ∴ (λ B − λ A ) t = In B , NA, ⇒, , t =, , N , 1, In B , λB − λ A, NA, , 24 Let θ be the angle of incline. Here, the, incline kept horizontally is working as, main scale while the other incline kept, on horizontally placed incline is treated, as vernier scale., From the figure, it is clear that,, l, unit and 1 VSD = l unit, 1 MSD =, cos θ, So, LC of instrument is,, LC = 1MSD − 1 VSD, l, l (1 − cos θ), units, =, − l =, cos θ, cos θ,
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420, , DAY FOURTY, , 40 DAYS ~ JEE MAIN PHYSICS, , 25 At the critical condition, normal, reaction N will pass through point P., , KE of a star =, , KE of heavier star, , N, , E1 =, , F, G, , L, , L/2, f, , The block will topple when, L, 2, , mg, 2, Therefore, the minimum force required, mg, to topple the block is F =, ., 2, , ∴, , F>, , 26 The centre of mass of the double star, system remains stationary and both the, stars revolve round in circular orbits,, which are concentric with the centre of, mass., The distance of centre of mass from the, Mr + 2 M × 0 r, heavier star =, =, M + 2M, 3, Hence, the heavier star revolves in a, r, circle of radius while the lighter star, 3, 2r, in a circle of radius ., 3, M ⋅2M, Reduced mass of the system =, M + 2M, 2M, =, 3, Period of revolution of the double star, 2π, system =, r 3 /2, 2GM, 3, where, r is the distance between two, stars., , 2, , and that of lighter star, , P, , or FL > (mg ), , 1, r, × 2M × ω , 3 , 2, , 1, 2r , ω, E2 = M , 3 , 2, , mg, , τ F > τ mg, , 1, mv 2, 2, , 2, , So, kinetic energy of lighter star is two, times that of heavier star., 27 As, f 0 = v, 2l, Beat frequency = f1 − f2, v, v, =, −, l, l, , , , 2 − ∆l 2 + ∆l, 2, , 2, , 2, 2 , = ( f 0l) , −, , ∆, l, l, −, +, l, 2, 2 ∆l , , 8 f ∆l, 4∆l, = 2 f 0 l 2 ≈ 0, l, l , , 28 Once the bricks leave the assistant’s, hands the only force that acts on them, is gravitational force. Since this, produces a constant acceleration, a = − g = − 32 fts −2 , the kinematic, equation, v 2 = v 20 − 2 a( x − x 0 ), can be, used to describe the motion. The initial, velocity v 0 is found by putting known, values in the above equation,, v 20 = 36 + 2 × 32 × 10 = 676, ⇒, , v 0 = 26 fts −1, , The kinetic energy given to each brick, and supplied by the assistant is, 1, E1 = mv 20, 2, 1, = × m × 676, 2, = 338 m ft 2s −1, , If the brick assistant supplied is only just, enough energy to reach the required level, and no more, the initial velocity being u,, they would have zero velocity at the, Mason’s hand., ∴ u2 = 0 + 2 g ( x − x 0 ), = 2 × 32 × 10 = 640, ⇒ u = 8 10 fts −1, KE supplied in this case,, E2 =, , 1, mu2 = 320 m ft2s −2, 2, , ∴ Wasted energy = E1 − E2, E1 − E2, × 100, E1, 338 − 320, =, × 100, 320, = 5.6%, , % waste =, , 29 For forward bias mode the p-side of, diode has to be at higher potential than, n-side. The meters used are DC, so we, have to be careful while connecting them, w.r.t. polarity., Last point is to decide the range of, meters, the range of meters has to be in, such a way that we can have the readings, which leads to plot on realistic scale. If, we take 0-20 A ammeter, then reading we, read from this is tending to 0 to, 5 divisions which is not fruitful., , 30 Using conservation of angular, momentum about the axis of cylinder for, the (block + cylinder) system, MR2ω, mvo R = mvR +, 2, 3, mv o R = mvR, ⇒, (Q v = ωR ), 2, (when slipping stops), 2v o, ⇒, v =, 3
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ONLINE QUESTION PAPERS, , JEE Main 2019 (April & January Attempt), APRIL ATTEMPT, 8 April, Shift-I, 1 A steel wire having a radius of, , 2.0 mm, carrying a load of 4 kg, is, hanging from a ceiling. Given that, g = 3.1 π ms−2, what will be the tensile, stress that would be developed in the, wire?, (a) 6.2 × 106 Nm −2, (c) 31, . × 106 Nm −2, , (b) 5.2 × 106 Nm −2, (d) 4.8 × 106 Nm −2, , 2 If 1022 gas molecules each of mass, , 10−26 kg collide with a surface, (perpendicular to it) elastically per, second over an area 1 m2 with a speed, 104 m/s, the pressure exerted by the, gas molecules will be of the order of, (a) 104 N / m 2, (c) 103 N / m 2, , (b) 108 N / m 2, (d) 1016 N / m 2, , 3 The bob of a simple pendulum has, , mass 2g and a charge of 5.0 µC. It is, at rest in a uniform horizontal electric, field of intensity 2000 V/m. At, equilibrium, the angle that the, pendulum makes with the vertical is, (take g = 10m/s2), (a) tan −1 (2.0), (c) tan −1 (5.0), , (b) tan −1 (0.2), (d) tan −1 (0.5), , 4 A boy’s catapult is made of rubber, , cord which is 42 cm long, with 6 mm, diameter of cross-section and of, negligible mass. The boy keeps a, stone weighing 0.02 kg on it and, stretches the cord by 20 cm by, applying a constant force. When, released the stone flies off with a, velocity of 20 ms−1 . Neglect the, change in the area of cross-section of, the cord while stretched. The Young’s, modulus of rubber is closest to, (a) 106 Nm −2, (c) 108 Nm −2, , (b) 104 Nm −2, (d) 103 Nm −2, , 5 A plane electromagnetic wave travels, , in free space along the x-direction., The electric field component of the, wave at a particular point of space, and time is E = 6Vm−1 along, y-direction. Its corresponding, magnetic field component, B would be, (a), (b), (c), (d), , −8, , 2 × 10 T along z - direction, 6 × 10−8 T along x - direction, 6 × 10−8 T along z - direction, 2 × 10−8 T along y - direction, , 6 Ship A is sailing towards north-east, with velocity v = 30$i + 50$j km/h,, where $i points east and $j north. Ship, B is at a distance of 80 km east and, 150 km north of Ship A and is sailing, towards west at 10 km/h. A will be at, minimum distance from B in, (a) 4.2 h (b) 2.6 h, , (c) 3.2 h (d) 2.2 h, , 7 A thin strip 10 cm, , long is on an, 10 cm, U-shaped wire of, B, negligible, resistance and it, is connected to a spring of spring, constant 0.5 Nm−1 (see figure). The, assembly is kept in a uniform, magnetic field of 0.1 T. If the strip is, pulled from its equilibrium position, and released, the number of, oscillations it performs before its, amplitude decreases by a factor of e is, N. If the mass of the strip is 50, grams, its resistance 10 Ω and air, drag negligible, N will be close to, (a) 1000 (b) 50000 (c) 5000, , (d)10000, , 8 Four particles, , Y, , a, A , B , C and D with, a, B, C, masses mA = m,, mB = 2m, mC = 3m, X, and mD = 4m are at, D, a A, the corners of a, square. They have, a, accelerations of, equal magnitude with directions as, shown. The acceleration of the centre of, mass of the particles (in ms−2 ) is, , (a), , a $ $, ( i − j), 5, , (c) zero, , (b) a($i + $j), a, (d) ($i + $j), 5, , 9 A solid conducting sphere, having a, , charge Q, is surrounded by an, uncharged conducting hollow, spherical shell. Let the potential, difference between the surface of the, solid sphere and that of the outer, surface of the hollow shell be V . If the, shell is now given a charge of −4Q,, the new potential difference between, the same two surfaces is, (a) −2V, (c) 4 V, , (b) 2 V, (d) V, , i, , 10 A 20 H inductor, , 10 W, , coil is connected to, 20 H, E, a 10 ohm, resistance in series, as shown in figure. The time at which, rate of dissipation of energy (Joule’s, heat) across resistance is equal to the, rate at which magnetic energy is, stored in the inductor, is, (a), , 2, ln 2, , (b), , 1, ln 2 (c) 2ln 2 (d) ln 2, 2, , 11 A thin circular plate of mass M and, , radius R has its density varying as, ρ(r ) = ρ0 r with ρ0 as constant and r is, the distance from its centre. The, moment of inertia of the circular plate, about an axis perpendicular to the, plate and passing through its edge is, I = aMR 2. The value of the coefficient, a is, (a), , 1, 2, , (b), , 3, 5, , (c), , 8, 5, , (d), , 12 In SI units, the dimensions of, (a) A−1 TML3, (c) AT −3ML3/ 2, , 3, 2, , ε0, is, µ0, , (b) AT 2M−1 L−1, (d) A2T 3M−1 L−2, , 13 A thermally insulated vessel contains, , 150 g of water at 0°C. Then, the air, from vessel is pumped out adiabatically., A fraction of water turns into ice and, the rest evaporates at 0°C itself. The, mass of evaporated water will be, closest to (Latent heat of vaporisation, of water= 2.10 × 106 J kg −1 and latent, heat of fusion of water = 3.36 × 105 J kg −1 ), , (a) 150 g (b) 20 g, , 14 The reverse, , (c) 130 g (d) 35g, 200 W, , breakdown, IZ, voltage of a, 9V, Zener diode, is 5.6 V in, the given circuit. The current I z, through the Zener is, (a) 10 mA, (c) 15 mA, , 800 W, , (b) 17 mA, (d) 7 mA, , 15 In an interference experiment, the, , ratio of amplitudes of coherent waves, a, 1, is 1 = . The ratio of maximum and, a2 3, minimum intensities of fringes will be, (a) 2, , (b) 18, , (c) 4, , (d) 9
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2 40 DAYS ~ JEE MAIN PHYSICS, 16 Water from a pipe is coming at a rate, of 100 liters per minute. If the radius, of the pipe is 5 cm, the Reynolds, number for the flow is of the order of, (density of water = 1000 kg /m3 ,, coefficient of viscosity of water = 1, mPa s), (b) 104, , (a) 103, , (c) 102, , (d) 106, , 17 Two identical beakers A and B, contain equal volumes of two different, liquids at 60°C each and left to cool, down. Liquid in A has density of, 8 × 102 kg / m3 and specific heat of, 2000 J kg −1 K−1 while liquid in B has, density of 103 kg m−3 and specific, heat of 4000 J kg −1 K−1 . Which of the, following best describes their, temperature versus time graph, schematically? (Assume the, emissivity of both the beakers to be, the same), (a) 60°C, , (b) 60°C, , T, , T, , A, B, , t, , (d) 60°C, T, A, , A and B, , B, t, , t, , 18 Voltage rating of a parallel plate, , capacitor is 500 V. Its dielectric can, withstand a maximum electric field of, 106 V/m. The plate area is 10−4 m2., What is the dielectric constant, if the, capacitance is 15 pF?, (Take, ε0 = 8.86 × 10−12 C2 / N- m2), (a) 3.8, (c) 4.5, , (b) 8.5, (d) 6.2, , each other. Their de-Broglie, wavelengths are λ1 and λ 2,, respectively. The particles suffer, perfectly inelastic collision. The, de-Broglie wavelength λ of the final, particle, is given by, 1, λ2, , =, , (b) λ =, , 1, λ12, , +, , 1, , (a) 100 Ω, (c) 300 Ω, , (b) 400 Ω, (d) 500 Ω, , 27, , 21 Four identical particles of mass M are, located at the corners of a square of, side a. What should be their speed, if, each of them revolves under the, influence of other’s gravitational field, in a circular orbit circumscribing the, square ?, , GM, a, GM, (c) 121, ., a, , GM, a, GM, (d) 141, ., a, , (a) 135, ., , (b) 116, ., , 22 The wavelength of the carrier waves, (b) 1500 nm, (d) 900 nm, , 23 An upright object is placed at a, , distance of 40 cm in front of a, convergent lens of focal length 20 cm., A convergent mirror of focal length 10, cm is placed at a distance of 60 cm on, the other side of the lens. The position, and size of the final image will be, , (a) 20 cm from the convergent mirror,, same size as the object, (b) 40 cm from the convergent mirror,, same size as the object, (c) 40 cm from the convergent lens,, twice the size of the object, (d) 20 cm from the convergent, mirror,twice size of the object, , radius r carries a current I. It is held, in the XZ-plane in a magnetic field, B$i. The torque on the coil due to the, magnetic field (in N-m) is, Br 2I, πN, Bπr 2I, (c), N, , (b) Bπr 2IN, , (a), , λ1 + λ 2, 2, 2, 1, 1, (d), =, +, λ λ1, λ2, , (c) λ =, , B, , L, , L, , (b) 2.2 ms, (d) 3.3 ms, , (b) n = 1 → n = 4, (d) n = 2 → n = 4, , A wire of length 2L, is made by, joining two wires A and B of same, length but different radii r and 2r and, made of the same material. It is, vibrating at a frequency such that the, joint of the two wires forms a node. If, the number of antinodes in wire A is, p and that in B is q, then the ratio, p : q is, (a) 3 : 5 (b) 4 : 9, , (c) 1 : 2, , optical fibre, d, q2, is l = 2 m, long and has 40° q1, a diameter of, d = 20 µ m. If a ray of light is incident, on one end of the fibre at angle, θ1 = 40°, the number of reflections it, makes before emerging from the other, end is close to (refractive index of, fibre is 1.31 and sin 40° = 0.64), (a) 55000, (c) 45000, , (b) 66000, (d) 57000, , 29 A particle moves, , 3, in one dimension, from rest under, 2, the influence of Force, (in N) 1, a force that, varies with the, distance, 1, 2, Distance, travelled by the, (in m), particle as, shown in the figure. The kinetic, energy of the particle after it has, travelled 3 m is, , 2. (*), 12. (d), 22. (b), , 3. (d), 13. (b), 23. (*), , * No option is correct, , 4. (a), 14. (a), 24. (b), , 5. (a), 15. (c), 25. (d), , 6. (b), 16. (b), 26. (d), , 7. (c), 17. (b), 27. (c), , 8. (a), 18. (b), 28. (d), , 9. (d), 19. (a), 29. (c), , 10. (c), 20. (d), 30. (d), , 3, , (b) 2.5 J (c) 6.5 J (d) 5 J, R1, , a, , R1, , shown with, E3, R2, R1 = 10, . Ω,, E1, R1, R2 = 2.0Ω, E1 = 2, E2, V and E2 = E3 = 4, b, R1, V, the potential, difference between the points a and b, is approximately (in volt), (a) 2.7, (c) 3.7, , (b) 2.3, (d) 3.3, , ANSWERS, 1. (c), 11. (*), 21. (b), , (d) 1 : 4, , 28 In figure, the, , 30 For the circuit, , (d) Zero, , (a) 5 ms, (c) 7.2 ms, , A, , (a) 4 J, , V (t ) = 220 sin 100 πt volt is applied to, a purely resistive load of 50 Ω. The, time taken for the current to rise from, half of the peak value to the peak, value is, , λ1 λ 2, , n = 2 to n =1of hydrogen atoms fall on, He+ ions in n = 1and n = 2 states. The, possible transition of helium ions as, they absorb energy from the radiation, is, (a) n = 2 → n = 3, (c) n = 2 → n = 5, , 25 An alternating voltage, , λ 22, , 26 Radiation coming from transitions, , a, , 24 A circular coil having N turns and, , 19 Two particles move at right angle to, , (a), , code. If one replaces the red colour by, green in the code, the new resistance, will be, , (a) 2400 nm, (c) 600 nm, , B, A, , T, , 20 A 200 Ω resistor has a certain colour, , in a modern optical fibre, communication network is close to, , t, , (c) 60°C, , ONLINE JEE Main 2019, , For Detailed Solutions, Visit : http://tinyurl.com/y277m5o6, Or Scan :
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ONLINE JEE Main 2019, 8 April, Shift-II, 1 In a simple pendulum, experiment for, , determination of acceleration due to, gravity ( g ), time taken for 20 oscillations, is measured by using a watch of 1, second least count. The mean value of, time taken comes out to be 30 s. The, length of pendulum is measured by, using a meter scale of least count 1, mm and the value obtained 55.0 cm., The percentage error in the, determination of g is close to, (a) 0.7% (b) 6.8% (c) 3.5% (d) 0.2%, , 2 A nucleus A, with a finite de-Broglie, , wavelength λ A , undergoes, spontaneous fission into two nuclei B, and C of equal mass. B flies in the, same directions as that of A, while C, flies in the opposite direction with a, velocity equal to half of that of B. The, de-Broglie wavelengths λB and λC of, B and C respectively, λ, (b) A , λ A, 2, λ, (d) λ A , A, 2, , (a) 2λ A , λ A, (c) λ A , 2λ A, , 3 In the figure shown, what is the, , current (in ampere) drawn from the, battery? You are given :, R1 = 15 Ω, R2 = 10 Ω, R3 = 20 Ω,, R4 = 5 Ω, R5 = 25 Ω, R6 = 30 Ω, E = 15 V, R3, R1, E, , +, –, , R2, , R6, , (a) 13/24 (b) 7/18, , R4, , R5, , (c) 20/3, , 4 Two magnetic, , (d) 9/32, d, , dipoles X and, S, θ, N, Y are placed at S, P, N, X, a separation d,, (M), Y, with their axes, (2 M), perpendicular, to each other. The dipole moment ofY, is twice that of X. A particle of charge, q is passing through their mid-point, P, at angle θ = 45° with the horizontal, line, as shown in figure. What would, be the magnitude of force on the particle, at that instant? (d is much larger, than the dimensions of the dipole), µ, M, (b) 0, (a) 0 , × qv, 4π d 3, , 2, µ, , M × qv, 0, (c) 2 , , 4π d 3, , 2, µ 0 2M, , (d) , × qv, , 4π d 3, , 2, , APRIL ATTEMPT, 5 A rocket has to be launched from, , earth in such a way that it never, returns. If E is the minimum energy, delivered by the rocket launcher,, what should be the minimum energy, that the launcher should have, if the, same rocket is to be launched from, the surface of the moon? Assume that, the density of the earth and the moon, are equal and that the earth’s volume, is 64 times the volume of the moon., (a), , E, 64, , (b), , E, 16, , (c), , E, 32, , (d), , E, 4, , 6 The electric field in a region is given, , by E = (Ax + B )$i, where E is in NC−1, and x is in metres. The values of, constants are A = 20 SI unit and, B = 10 SI unit. If the potential at x = 1, is V1 and that at x = − 5 is V 2, then, V1 − V 2 is, (a) − 48 V (b) − 520 V (c) 180 V (d) 320 V, , 7 Let |A1| = 3, |A 2| = 5 and |A1 + A 2| = 5., The value of (2A1 + 3A 2 ) ⋅ (3A1 − 2A 2 ), is, (a) −106.5, (c) −99.5, , (b) −112. 5, (d) −118. 5, , 8 The temperature, at which the root, , mean square velocity of hydrogen, molecules equals their escape velocity, from the earth, is closest to :, [Boltzmann constant kB = 138, . × 10−23, J/K, Avogadro no. N A = 6.02 × 1026 /kg,, Radius of earth = 6.4 × 106 m,, Gravitational acceleration on, earth = 10 ms −2], (a) 104 K, (c) 3 × 105 K, , (b) 650 K, (d) 800 K, , 9 Two very long,, , y, , straight and, d P, I, x, insulated wires, d, are kept at 90°, I, angle from each, other in xy-plane, as shown in the, fig. These wires carry currents of, equal magnitude I, whose directions, are shown in the figure. The net, magnetic field at point P will be, , (a) zero, (c) −, , µ 0I $, (x + y$ ), 2 πd, , +µ 0I $, (b), ( z), πd, µ 0I $, (d), (x + y$ ), 2 πd, , 10 An electric dipole is formed by two, , equal and opposite charges q with, separation d. The charges have same, mass m. It is kept in a uniform, electric field E. If it is slightly rotated, from its equilibrium orientation, then, its angular frequency ω is, (a), , 2qE, qE, (b) 2, (c), md, md, , qE, md, , (d), , qE, 2md, , 3, , 11 A cell of internal resistance r drives, , current through an external, resistance R. The power delivered by, the cell to the external resistance will, be maximum when, (a) R = 2r, (c) R = 0.001 r, , (b) R = r, (d) R = 1000 r, , 12 A body of mass m1 moving with an, unknown velocity of v1 $i, undergoes a, collinear collision with a body of mass, m2 moving with a velocity v2$i. After, collision, m1 and m2 move with, velocities of v3 $i andv4 $i , respectively., If m2 = 0.5m1 and v3 = 0.5 v1 , then v1 is, v2, 4, (d) v4 − v2, (b) v4 −, , (a) v4 + v2, (c) v4 −, , v2, 2, , 13 A circuit connected to an AC source of, emf e = e0 sin(100t ) with t in seconds,, π, gives a phase difference of between, 4, the emf e and current i. Which of the, following circuits will exhibit this?, , (a) RC circuit with R = 1kΩ andC = 1 µF, (b) RL circuit with R = 1kΩ and L = 1mH, (c) RC circuit with R = 1kΩ andC = 10 µF, (d) RL circuit with R = 1kΩ and L = 10, mH, , 14 In the circuit shown, a four-wire, , potentiometer is made of a 400 cm, long wire, which extends between A, and B. The resistance per unit length, of the potentiometer wire is r = 0.01, Ω/cm. If an ideal voltmeter is, connected as shown with jockey J at, 50 cm from end A, the expected, reading of the voltmeter will be, 1.5 V, 1.5 V,, 0.5 Ω 0.5 Ω, A, , V, , J, 50 cm, , 1Ω, B, , (a) 0.20 V, (c) 0.25 V, , 100 cm, , (b) 0.75 V, (d) 0.50 V, , 15 A solid sphere and, , solid cylinder of, identical radii, approach an, incline with the, same linear velocity (see figure). Both, roll without slipping all throughout., The two climb maximum heights hsph, and hcyl on the incline., h, The ratio sph is given by, hcyl, , (a), , 2, 5, , (b), , 14, 15, , (c) 1, , (d), , 4, 5
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4 40 DAYS ~ JEE MAIN PHYSICS, 16 The given diagram, , shows four processes,, i.e. isochoric, isobaric,, isothermal and, adiabatic. The correct, assignment of the, processes, in the same, order is given by, (a) d a b c, (c) d a c b, , a, p, , ONLINE JEE Main 2019, (a = acceleration, v = velocity,, x = displacement, t = time), , b, d, , c, , 40, , V, , (b) a d b c, (d) a d c b, , O, , (b) 2, , (c) 0.1, , l, , box of length 0.3 m is, held horizontally,, with one of its sides, h, on the edge of a, platform of height 5, m. When released, it, slips off the table in a very short time, τ = 0.01 s, remaining essentially, horizontal. The angle by which it, would rotate when it hits the ground, will be (in radians) close to, (a) 0.02 (b) 0.3, , (c) 0.5, , (d) 0.28, , 19 In a line of sight radio, , communication, a distance of about 50, km is kept between the transmitting, and receiving antennas. If the height, of the receiving antenna is 70 m, then, the minimum height of the, transmitting antenna should be, (Radius of the earth = 6.4 × 106 m), , (a) 20 m (b) 32 m (c) 40 m (d) 51 m, , 20 A positive point, , charge is released, from rest at a, distance r0 from a, r0, positive line charge, with uniform density., The speed (v) of the, point charge, as a, function of instantaneous distance r, from line charge, is proportional to, +, , t, , (D) x, , O, , t, , O, , t, , (b) (A), (B), (C), (d) (A), (B), (D), , (a) (A), (c) (B), (C), , (d) 1, , 18 A rectangular solid, , O, , t, , (C) x, , Ca and 16 O is close to, , (a) 5, , (B) v, , (A) a, , 17 The ratio of mass densities of nuclei, of, , 26 A convex lens (of focal length 20 cm), , 22 Calculate the limit of resolution of a, , telescope objective having a diameter, of 200 cm, if it has to detect light of, wavelength 500 nm coming from a, star., (a) 610 × 10−9 rad (b) 305 × 10−9 rad, (c) 457.5 × 10−9 rad (d) 152.5 × 10−9 rad, , 23 If surface tension (S), moment of, inertia (I ) and Planck’s constant (h ),, were to be taken as the fundamental, units, the dimensional formula for, linear momentum would be, (a) S1/ 2I 1/ 2h −1, (c) S1/ 2I 1/ 2h 0, , (b) S 3/ 2I 1/ 2h 0, (d) S1/ 2I 3/ 2h −1, , 24 A damped harmonic oscillator has a, , frequency of 5 oscillations per second., The amplitude drops to half its value, for every 10 oscillations. The time it, 1, of the, will take to drop to, 1000, original amplitude is close to, (a) 20 s, (c) 100 s, , (b) 50 s, (d) 10 s, , ABCD of mass M has length a and, breadth b, as shown in the figure. If, the shaded portion HBGO is cut-off,, the coordinates of the centre of mass, of the remaining portion will be, (a , b ), B, , H, a, b, 2 2, , r, r0, , r, (a) v ∝ , r0 , , (b) v ∝ e, , r, (c) v ∝ ln , r0 , , r, (d) v ∝ ln , r0 , , E, , 21 A particle starts from origin O from, , rest and moves with a uniform, acceleration along the positive X-axis., Identify all figures that correctly, represent the motion qualitatively., , D, (0, 0), , electromagnetic wave is given by, B = 16, . × 10−6 cos(2 × 107 z + 6 × 1015 t ), (2$i + $j) Wbm−2, The associated electric field will be, (a) E = 4.8 × 102 cos(2 × 107 z − 6 × 1015 t ), (−2$j + $i ) Vm −1, 2, (b) E = 4.8 × 10 cos(2 × 107 z − 6 × 1015 t ), (2$j + $i ) Vm −1, (c) E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t ), ($i − 2$j) Vm −1, 2, 7, (d) E = 4.8 × 10 cos(2 × 10 z + 6 × 1015 t ), (− i$ + 2$j)Vm −1, , 28 A parallel plate capacitor has 1µF, capacitance. One of its two plates is, given + 2µC charge and the other, plate + 4µC charge. The potential, difference developed across the, capacitor is, (a) 1 V, , (b) 5 V, , (c) 2 V, , built using an n -p - n transistor, is, shown in the figure. Its DC current, gain is 250, RC = 1kΩ and VCC = 10 V., What is the minimum base current, for VCE to reach saturation?, RB, , RC, VCC, , VB, , (a) 40 µA (b) 10 µA (c) 100 µA (d) 7 µA, C, (a, 0), , F, , 5a 5b , (b) , ,, , 12 12 , 5a 5b , (d) , ,, , 3 3, , 30 Young’s moduli of two wires A and B, , are in the ratio 7 : 4. Wire A is 2 m long, and has radius R. Wire B is 1.5 m long, and has radius 2 mm. If the two wires, stretch by the same length for a given, load, then the value of R is close to, , (a) 1.3 mm, (c) 1.9 mm, , (b) 1.5 mm, (d) 1.7 mm, , ANSWERS, 1. (b), 11. (b), 21. (d), , 2. (b), 12. (d), 22. (b), , 3. (d), 13. (c), 23. (c), , 4. (b), 14. (c), 24. (a), , 5. (b), 15. (b), 25. (b), , 6. (c), 16. (a), 26. (c), , 7. (d), 17. (d), 27. (c), , 8. (a), 18. (c), 28. (a), , (d) 3 V, , G, , O, , 2a 2b , (a) , ,, , 3 3, 3a 3b , (c) , ,, , 4 4, , (a) 25 cm (b) 20 cm (c) 10 cm (d) 30 cm, , 27 The magnetic field of an, , 29 A common emitter amplifier circuit,, , 25 A uniform rectangular thin sheet, , (0, b), A, , and a concave mirror, having their, principal axes along the same lines,, are kept 80 cm apart from each other., The concave mirror is to the right of, the convex lens. When an object is, kept at a distance of 30 cm to the left, of the convex lens, its image remains, at the same position even if the concave, mirror is removed. The maximum, distance of the object for which this, concave mirror, by itself would, produce a virtual image would be, , 9. (a), 19. (b), 29. (a), , 10. (a), 20. (d), 30. (d), , For Detailed Solutions, Visit : http://tinyurl.com/yxjbp89y, Or Scan :
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ONLINE JEE Main 2019, , APRIL ATTEMPT, , 09 April, Shift-I, 1 A stationary horizontal disc is free to, , rotate about its axis. When a torque is, applied on it, its kinetic energy as a, function of θ, where θ is the angle by, which it has rotated, is given as kθ2. If, its moment of inertia is I, then the, angular acceleration of the disc is, k, (a), θ, 2I, , k, (b) θ, I, , k, (c), θ, 4I, , 2k, (d), θ, I, , rms speed of the molecules is 200 m/s, at 127° C. At 2 atm pressure and at, 227° C, the rms speed of the molecules, will be, (b) 80 m/s, (d) 80 5 m/s, , 3 A wire of resistance R is, , (a), , 7, 3, R (b) R, 64, 4, , (c) R, , (d), , B, , E, , C, , 1, R, 16, , 4 A simple pendulum oscillating in air, , has period T . The bob of the, pendulum is completely immersed in, a non-viscous liquid. The density of, 1, the liquid is th of the material of, 16, bob. If the bob is inside liquid all the, time, its period of oscillation in this, liquid is, , 1, (a) 2T, 10, 1, (c) 4T, 14, , 1, (b) 2T, 14, 1, (d) 4T, 15, , as shown in the figure, , (b) 1.60 m, (d) 0.24 m, , 8 The stream of a river is flowing with a, speed of 2 km/h. A swimmer can swim, at a speed of 4 km/h. What should be, the direction of the swimmer with, respect to the flow of the river to cross, the river straight ?, (b) 120°, , (c) 90°, , Q, , d, , If D >> d, the potential energy of the, system is best given by, 1 q2, 2qQd , (a), −, +, 4 πε0 d, D 2 , 2, qQd , 1 q, (b), +, +, 4 πε0 d, D 2 , 2, qQd , 1 q, (c), −, −, , 4 πε0 d, 2D 2 , 1 q2 qQd , (d), −, −, 4 πε0 d, D 2 , , 6 A body of mass 2 kg makes an elastic, , collision with a second body at rest, and continues to move in the original, , the speed of the centre of mass at the, bottom of the incline is same, the, ratio of the maximum height they, climb is, (a) 10 : 15 : 7, (c) 14 : 15 : 20, , (d) 150°, , 9 A string is clamped at both the ends, , and it is vibrating in its 4th, harmonic. The equation of the, stationary wave is, Y = 0.3 sin(0157, ., x) cos(200 πt ). The, length of the string is (All quantities, are in SI units), (a) 60 m (b) 40 m (c) 80 m (d) 20 m, , a capillary tube of radius r, then mass, of water which will rise in a capillary, tube of radius 2r is, (b) 4M, , (a) 40 V (b) 10 V (c) 15 V (d) 20 V, , 11 A capacitor with capacitance 5 µF is, , (a) 6.25 × 10−6 J, (c) 2.55 × 10−6 J, , (b) 216, . × 10−6 J, (d) 3.75 × 10−6 J, , the mass and edge length are, measured as (10.00 ± 010, . ) kg and, (010, . ± 0.01) m, respectively. The error, in the measurement of density is, 3, , (a) 0.01 kg/m, (c) 0.07 kg/m 3, , 3, , (b) 0.10 kg/m, (d) 0.31 kg/m 3, , 13 The total number of turns and, , cross-section area in a solenoid is, fixed. However, its length L is varied, by adjusting the separation between, windings. The inductance of solenoid, will be proportional to, (a) 1 / L, , (b) L2, , (c) L, , (d) 1 / L2, , 14 The following bodies are made to roll, , up (without slipping) the same, inclined plane from a horizontal plane:, (i) a ring of radius R, (ii) a solid, cylinder of radius R/2 and (iii) a solid, sphere of radius R/4. If in each case,, , (c), , M, 2, , (d) M, , 16 Following figure, , shows two processes, f, p, A, A and B for a gas. If, ∆QA and ∆QB are, B, the amount of heat, i, absorbed by the, V, system in two cases,, and ∆U A and ∆UB are changes in, internal energies respectively, then, (a), (b), (c), (d), , ∆Q A > ∆QB ,, ∆Q A < ∆QB ,, ∆Q A > ∆QB ,, ∆Q A = ∆QB ;, , ∆U A > ∆U B, ∆U A < ∆U B, ∆U A = ∆U B, ∆U A = ∆U B, , 17 Determine the charge on the, , capacitor in the following circuit, , 10 A moving coil galvanometer has resistance 50 Ω and it indicates full deflection at 4 mA current. A voltmeter is, made using this galvano- meter and a, 5 kΩ resistance. The maximum voltage, that can be measured using this, voltmeter, will be close to, , (b) 4 : 3 : 2, (d) 2 : 3 : 4, , 15 If M is the mass of water that rises in, , (a) 2M, , 12 In the density measurement of a cube,, D, , –q, , focal length of 0.4 m. The distance at, which you hold the mirror from your, face in order to see your image, upright with a magnification of 5 is, , charged to 5 µC. If the plates are, pulled apart to reduce the capacitance, to 2 µF, how much work is done?, , 5 A system of three charges are placed, , +q, , 7 A concave mirror for face viewing has, , (a) 60°, , A, , bent to form a square, ABCD as shown in the, figure. The effective, resistance between E and, C is [E is mid-point of, D, arm CD], , (a) 1.5 kg (b) 1.2 kg (c) 1.8 kg (d) 1.0 kg, , (a) 0.16 m, (c) 0.32 m, , 2 For a given gas at 1 atm pressure,, , (a) 100 5 m/s, (c) 100 m/s, , direction but with one-fourth of its, original speed. What is the mass of, the second body ?, , 5, , 6Ω, 72V, , (a) 2µC, (c) 10 µC, , 2Ω, 4Ω, , 10Ω, , 10µF, , (b) 200µC, (d) 60 µC, , 18 A uniform cable of mass M and length, L is placed on a horizontal surface, 1, such that its th part is hanging, n, below the edge of the surface. To lift, the hanging part of the cable upto the, surface, the work done should be, (a), (c), , 2MgL, , n2, MgL, n2, , (b) nMgL, (d), , MgL, 2n 2, , 19 The electric field of light wave is, given as, , 2 πx, E = 10−3 cos , − 2 π × 6 × 1014 t , 5 × 10−7, , $ NC−1 ., x, This light falls on a metal plate of, work function 2eV. The stopping, potential of the photoelectrons is, 12375, Given, E (in eV) =, °), λ(in A, , (a) 0.48 V, (c) 2.0 V, , (b) 0.72 V, (d) 2.48 V, , 20 A ball is thrown vertically up (taken, as + Z-axis) from the ground. The, correct momentum-height (p-h), diagram is
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6 40 DAYS ~ JEE MAIN PHYSICS, , (a), , h (b), , O, , h, , O, , p, , p, , (c), , h (d), , O, , h, , O, , 21 A signal A cosωt is transmitted using, , v0 sinω0 t as carrier wave. The correct, amplitude modulated (AM) signal is, , (a) (v0 sin ω 0t + A cosωt, (b) (v0 + A ) cosωt sin ω 0t, (c) v0 sin[ω 0 (1 + 0.01A sin ωt )t ], A, (d) v0 sin ω 0t + sin(ω 0 − ω )t, 2, A, +, sin(ω 0 + ω )t, 2, , translational and vibrational motions., If the rms velocity of HCl molecules in, its gaseous phase is v, m is its mass, and kB is Boltzmann constant, then, its temperature will be, (a), , mv, 3kB, , (b), , 2, , mv, 7kB, , (c), , 2, , mv, 5kB, , Young’s double slit, Screen, experimental, a, setup. It is, observed that when, D, a thin transparent, sheet of thickness t and refractive, index µ is put in front of one of the, slits, the central maximum gets, shifted by a distance equal to n fringe, widths. If the wavelength of light, used is λ, t will be, 2nDλ, a ( µ − 1), Dλ, (c), a ( µ − 1), , 2Dλ, a ( µ − 1), nDλ, (d), a ( µ − 1), , (a), , (b), , 25 A rectangular coil (dimension 5 cm × 2.5, , 22 An HCl molecule has rotational,, , 2, , (c) zero, , 24 The figure shows a, , p, , p, , ONLINE JEE Main 2019, , (d), , 2, , mv, 6kB, , 23 The pressure wave, , p = 0.01sin[1000t − 3x]Nm−2,, corresponds to the sound produced by, a vibrating blade on a day when, atmospheric temperature is 0° C. On, some other day when temperature is, T , the speed of sound produced by, the same blade and at the same, frequency is found to be 336 ms −1 ., Approximate value of T is, (a) 15° C (b) 11° C (c) 12° C (d) 4° C, , cm) with 100 turns, carrying a current, of 3A in the clockwise direction, is, kept centred at the origin and in the, X-Z plane. A magnetic field of 1 T is, applied along X-axis. If the coil is, tilted through 45° about Z-axis, then, the torque on the coil is, (a) 0.27 N-m, (c) 0.42 N-m, , (b) 0.38 N-m, (d) 0.55 N-m, , 26 A rigid square, , (d) repulsive and equal to, , 27 Taking the wavelength of first Balmer, line in hydrogen spectrum (n = 3 to n, = 2) as 660 nm, the wavelength of the, 2nd Balmer line (n = 4 to n = 2) will be, (a) 889.2 nm, (c) 642.7 nm, , I1, , I, , µ 0I1 I 2, 2π, µ I I, (b) attractive and equal to 0 1 2, 3π, , (a) repulsive and equal to, , (b) 388.9 nm, (d) 488.9 nm, , 28 An n-p-n transistor is used in, , common emitter configuration as an, amplifier with 1 kΩ load resistance., Signal voltage of 10 mV is applied, across the base-emitter. This, produces a 3 mA change in the, collector current and 15 µA change in, the base current of the amplifier. The, input resistance and voltage gain are, (a) 0.67 kΩ , 200, (c) 0.67 kΩ, 300, , (b) 0.33 kΩ, 1.5, (d) 0.33 kΩ, 300, , 29 A solid sphere of mass M and radius a, is surrounded by a uniform concentric, spherical shell of thickness 2a and, 2M. The gravitational field at, distance 3a from the centre will be, (a), , 2, loop of side a and, carrying current, a, I 2 is lying on a, horizontal surface, a, near a long, current I1 carrying wire in the same, plane as shown in figure. The net, force on the loop due to the wire will be, , µ 0I1 I 2, 4π, , GM, 9a 2, , (b), , 2GM, 9a 2, , (c), , GM, 3a 2, , (d), , 2GM, 3a 2, , 30 The magnetic field of a plane, , electromagnetic wave is given by, B = B0[cos(kz − ωt )]i$ + B1 cos(kz + ωt )$j, , where, B0 = 3 × 10−5 T and, B1 = 2 × 10−6 T. The rms value of the, force experienced by a stationary, charge Q = 10−4 C at z = 0 is closest to, (a) 0.1 N, (c) 0.6 N, , (b) 3 × 10−2 N, (d) 0.9 N, , ANSWERS, 1. (d), 11. (d), 21. (d), , 2. (a), 12. (*), 22. (b), , 3. (a), 13. (a), 23. (d), , 4. (d), 14. (c), 24. (*), , 5. (d), 15. (a), 25. (a), , A′, , (refractive index, A, = 1.5) is placed on a, plane mirror M., When a pin is, L, placed at A, such, M, O, that OA = 18 cm, its, real inverted image is formed at A, itself, as shown in figure. When a, liquid of refractive index µ l is put, between the lens and the mirror, the, pin has to be moved to A′, such that, OA′ = 27 cm, to get its inverted real, image atA′ itself. The value of µ l will be, (a), , 3, , (b), , 2, , (c), , 4, 3, , 7. (c), 17. (b), 27. (d), , 8. (b), 18. (d), 28. (c), , 9. (c), 19. (a), 29. (c), , 10. (d), 20. (d), 30. (c), , 2 A massless spring (k = 800 N/m),, , 9 April Shift-II, 1 A thin convex lens L, , 6. (b), 16. (c), 26. (d), , (d), , 3, 2, , attached with a mass (500 g) is, completely immersed in 1 kg of water., The spring is stretched by 2 cm and, released, so that it starts vibrating., What would be the order of, magnitude of the change in the, temperature of water when the, vibrations stop completely? (Assume, that the water container and spring, receive negligible heat and specific, heat of mass = 400 J/kg K, specific, heat of water = 4184 J/kg K), (a) 10−4 K, (c) 10−1 K, , (b) 10−3 K, (d) 10−5 K, , 3 Two materials having coefficients of, , thermal conductivity ‘3K’ and ‘K’ and, , For Detailed Solutions, Visit : http://tinyurl.com/y6x7rxu3, Or Scan :, , thickness ‘d’ and ‘3d’ respectively, are, joined to form a slab as shown in the, figure. The temperatures of the outer, surfaces are ‘θ2’ and ‘θ1 ’ respectively,, (θ2 > θ1 ). The temperature at the, interface is, θ + θ1, (a) 2, 2, θ, 5θ, (c) 1 + 2, 6, 6, , d 3d, θ2 3K K, , θ1, θ1 2θ 2, (b), +, 3, 3, θ, 9θ 2, (d) 1 +, 10, 10, , 4 A test particle is moving in a circular, orbit in the gravitational field, K, produced by mass density ρ(r ) = 2 ., r, Identify the correct relation between
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ONLINE JEE Main 2019, the radius R of the particle’s orbit and, its period T, (a), (b), , T2, R3, T, R2, , is a constant, is a constant, , 5 A convex lens of focal length 20 cm, , produces images of the same, magnification 2 when an object is, kept at two distances x1 and x2, (x1 > x2 ) from the lens. The ratio of x1, and x2 is, (c) 4 : 3, , (d) 3 : 1, , 6 A wedge of mass M = 4 m lies on a, , frictionless plane. A particle of mass, m aproaches the wedge with speed v., There is no friction between the, particle and the plane or between the, particle and the wedge. The, maximum height climbed by the, particle on the wedge is given by, (a), , 2 v2, v2, (b), g, 7g, , (c), , 2 v2, 5g, , (d), , v2, 2g, , 7 The parallel combination of two air, , filled parallel plate capacitors of, capacitance C and nC is connected to, a battery of voltage, V . When the, capacitors are fully charged, the, battery is removed and after that a, dielectric material of dielectric, constant K is placed between the two, plates of the first capacitor. The new, potential difference of the combined, system is, (a), , (n + 1)V, nV, V, (b), (c) V (d), (K + n ), K+n, K+n, , 8 The area of a square is 5.29 cm 2. The, area of 7 such squares taking into, account the significant figures is, , (a) 37.030 cm 2, (c) 37.03 cm 2, , (b) 37.0 cm 2, (d) 37 cm 2, , 9 The logic gate equivalent to the given, logic circuit is, A, Y, B, , (a) NOR, (c) OR, , (b) NAND, (d) AND, , 10 The resistance of a galvanometer is, 50 ohm and the maximum current, which can be passed through it is, 0.002 A. What resistance must be, connected to it in order to convert it, into an ammeter of range 0-0.5 A?, (a) 0.2 ohm, (c) 0.002 ohm, , 11 The position vector of particle, , changes with time according to the, relation r (t ) = 15t 2$i + (4 − 20t 2 )$j., What is the magnitude of the, acceleration (in ms−2 ) at t = 1?, (a) 50, , T, (c) TR is a constant (d), is a constant, R, , (a) 5 : 3 (b) 2 : 1, , APRIL ATTEMPT, , (b) 0.5 ohm, (d) 0.02 ohm, , (b) 100, , (c) 25, , (d) 40, , 12 A particle P is formed due to a, , completely inelastic collision of, particles x and y having de-Broglie, wavelengths λ x and λ y , respectively., If x and y were moving in opposite, directions, then the de-Broglie, wavelength of P is, (a) λ x − λ y, (c), , λ xλ y, λx + λy, , (b), , λx λy, , λx − λy, , (d) λ x + λ y, , 13 A moving coil galvanometer has a coil, with 175 turns and area 1 cm 2. It, uses a torsion band of torsion, constant 10−6 N-m/rad. The coil is, placed in a magnetic field B parallel, to its plane. The coil deflects by 1º for, a current of 1 mA. The value of B (in, Tesla) is approximately, (a) 10−3, , (b) 10−4, , (c) 10−1, , (d) 10−2, , 14 Four point charges − q, + q, + q and −q, are placed on Y -axis at y = −2d,, y = − d, y = + d and y = +2d,, respectively. The magnitude of the, electric field E at a point on the, X-axis at x = D, with D >> d, will, behave as, (a) E ∝, (c) E ∝, , 1, D, 1, D2, , (b) E ∝, (d) E ∝, , 1, D3, 1, D4, , 15 The physical sizes of the transmitter, and receiver antenna in a, communication system are, , (a) proportional to carrier frequency, (b) inversely proportional to modulation, frequency, (c) independent of both carrier and, modulation frequency, (d) inversely proportional to carrier, frequency, , 7, , natural frequency of the sound source, in car B ? (speed of sound in air = 340, ms −1 ), (a) 2060 Hz, (c) 2300 Hz, , (b) 2250 Hz, (d) 2150 Hz, , 18 The position of a particle as a, , function of time t, is given by, x(t ) = at + bt 2 − ct3, where a , b and c are constants. When, the particle attains zero acceleration,, then its velocity will be, , b2, 2c, b2, (c) a +, 3c, (a) a +, , b2, 4c, b2, (d) a +, c, (b) a +, , 19 In a conductor, if the number of, , conduction electrons per unit volume, is 8.5 × 1028 m −3 and mean free time is, 25 fs (femto second), it’s approximate, resistivity is (Take, me = 91, . × 10−31 kg), , (a) 10−7 Ω-m, (c) 10−6 Ω-m, , (b) 10−5 Ω-m, (d) 10−8 Ω-m, , 20 The specific heats, C p and CV of a gas, of diatomic molecules, A are given (in, units of J mol −1 K−1 ) by 29 and 22,, respectively. Another gas of diatomic, molecules B, has the corresponding, values 30 and 21. If they are treated, as ideal gases, then, (a) A has a vibrational mode but B has, none, (b) Both A and B have a vibrational, mode each, (c) A has one vibrational mode and B has, two, (d) A is rigid but B has a vibrational, mode, , 21 50 W/m 2 energy density of sunlight is, normally incident on the surface of a, solar panel. Some part of incident, energy (25%) is reflected from the, surface and the rest is absorbed. The, force exerted on 1 m 2 surface area, will be close to (c = 3 × 108 m/s), , (a) 20 × 10−8 N, (c) 15 × 10−8 N, , (b) 35 × 10−8 N, (d) 10 × 10−8 N, , 22 A string 2.0 m long and fixed at its, , telescope is 250 cm. For light of, wavelength 600 nm coming from a, distant object, the limit of resolution, of the telescope is close to, , ends is driven by a 240 Hz vibrator., The string vibrates in its third, harmonic mode. The speed of the, wave and its fundamental frequency, is, , (a) 3.0 × 10−7 rad, (c) 15, . × 10−7 rad, , (a) 180 m/s, 80 Hz (b) 320 m/s, 80 Hz, (c) 320 m/s, 120 Hz (d) 180 m/s, 120 Hz, , 16 Diameter of the objective lens of a, , (b) 2.0 × 10−7 rad, (d) 4.5 × 10−7 rad, , 17 Two cars A and B are moving away, , from each other in opposite directions., Both the cars are moving with a speed, of 20 ms −1 with respect to the ground., If an observer in car A detects a, frequency 2000 Hz of the sound, coming from car B, what is the, , 23 A thin smooth rod of length L and, , mass M is rotating freely with, angular speed ω0 about an axis, perpendicular to the rod and passing, through its centre. Two beads of mass, m and negligible size are at the centre, of the rod initially. The beads are free
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8 40 DAYS ~ JEE MAIN PHYSICS, , ONLINE JEE Main 2019, the equivalent resistance between, these two points will be, , to slide along the rod. The angular, speed of the system, when the beads, reach the opposite ends of the rod,, will be, M ω0, M + 3m, M ω0, (c), M + 2m, , (a), , M ω0, M + m, M ω0, (d), M+ 6m, , (a), , (b), , speed 2v and collides with a mass 2m, moving with speed v in the same, direction. After collision, the first, mass is stopped completely while the, second one splits into two particles, each of mass m, which move at angle, 45º with respect to the original, direction.The speed of each of the, moving particle will be, (a), , 2 v (b), , (c), , v, (2 2 ), , (b), , 5, Ω, 2, , (c), , 12, 5, Ω (d) Ω, 5, 3, , 26 A very long solenoid of radius R is, , 24 A particle of mass m is moving with, , v, 2, , 7, Ω, 2, , 28 A wooden block floating in a bucket, , (d) 2 2 v, , carrying current I (t ) = kte−αt (k > 0), as, a function of time (t ≥ 0). Counter, clockwise current is taken to be, positive. A circular conducting coil of, radius 2R is placed in the equatorial, plane of the solenoid and concentric, with the solenoid. The current, induced in the outer coil is correctly, depicted, as a function of time, by, I, , I, , (a)t=0, , t, , (b) t=0, , I, , (c) t=0, , t, , (d) t=0, , (c) 0.7, , (d) 0.5, , some distance. When a current of 3 A, flows through coil P, a magnetic flux, of 10−3 Wb passes through Q. No, current is passed through Q. When no, current passes through P and a, current of 2 A passes through Q, the, flux through P is, (b) 6.67 × 10−4 Wb, (d) 3.67 × 10−4 Wb, , given axis is 1.5 kg m 2. Initially, the, body is at rest. In order to produce a, rotational kinetic energy of 1200 J,, the angular acceleration of 20 rad/s 2, must be applied about the axis for a, duration of, , 27 A He+ ion is in its first excited state., Its ionisation energy is, (a) 54.40 eV, (c) 48.36 eV, , (b) 0.8, , 30 Moment of inertia of a body about a, , t, , 25 A metal wire of resistance 3 Ω is, elongated to make a uniform wire of, double its previous length. This new, wire is now bent and the ends joined, to make a circle. If two points on this, circle make an angle 60º at the centre,, , (a) 0.6, , 29 Two coils P and Q are separated by, , (a) 6.67 × 10−3 Wb, (c) 3.67 × 10−3 Wb, , I, t, , of water has 4/5 of its volume, submerged. When certain amount of, an oil is poured into the bucket, it is, found that the block is just under the, oil surface with half of its volume, under water and half in oil. The, density of oil relative to that of water is, , (b) 13.60 eV, (d) 6.04 eV, , (a) 5 s, , (b) 2 s, , (c) 3 s, , (d) 2.5 s, , ANSWERS, 1. (c), 11. (a), 21. (a), , 2. (d), 12. (b), 22. (b), , 3. (d), 13. (a), 23. (d), , 4. (d), 14. (d), 24. (d), , 5. (d), 15. (d), 25. (d), , 1 A uniformly charged ring of radius 3a, and total charge q is placed in, xy-plane centred at origin. A point, charge q is moving towards the ring, along the Z-axis and has speed v at, z = 4a. The minimum value of v such, that it crosses the origin is, 1/ 2, , 2 1 q2 , , , m 5 4 πε0a , , (b), , 2 4, q2 , , , m 15 4 πε0a , , (c), , 2 1, q2 , , , m 15 4 πε0a , , (d), , 2 2, q2 , , , m 15 4 πε0a , , 2 Figure shows, , 7. (a), 17. (b), 27. (b), , 8. (c), 18. (c), 28. (a), , 3 Two wires A and B, , 10 April Shift-I, , (a), , 6. (c), 16. (a), 26. (d), , 1/ 2, , 9. (c), 19. (d), 29. (b), , A, , 10. (a), 20. (a), 30. (b), , C, , B, , are carrying, currents I1 and I 2 as, shown in the figure. I1, I2, The separation, x, between them is d. A, third wire C carrying, d, a current I is to be, kept parallel to them at a distance x, from A such that the net force acting, on it is zero. The possible values of x, are, I2 , I2 , (a) x = , d and x = , d, I1 + I 2 , I1 − I 2 , , 1/ 2, , I1 , I2 , (b) x = , d and x = , d, I1 − I 2 , I1 + I 2 , , 1/ 2, , I1 , I2 , (c) x = , d and x = , d, I, +, I, 1, I1 − I 2 , 2, I1 d, (d) x = ±, (I1 − I 2 ), , charge (q) versus, voltage (V ) graph, for series and, parallel combination, of two given, capacitors. The, capacitances are, , q(µC), , A, , 500, , 4 A message signal of frequency 100, B, , 80, 10 V, , V(volt), , (a) 60 µF and 40 µF (b) 50 µF and 30 µF, (c) 20 µF and 30 µF (d) 40 µF and 10µF, , MHz and peak voltage 100 V is used, to execute amplitude modulation on a, carrier wave of frequency 300 GHz, and peak voltage 400 V. The, modulation index and difference, between the two side band, frequencies are, , For Detailed Solutions, Visit : http://tinyurl.com/y4n74r62, Or Scan :, , (a) 0.25 ; 1 × 108 Hz (b) 4 ; 1 × 108 Hz, (c) 0.25 ; 2 × 108 Hz (d) 4 ; 2 × 108 Hz, , 5 Two coaxial discs, having moments of, I1, are rotating with, 2, respective angular velocities ω1 and, ω1, , about their common axis. They, 2, are brought in contact with each, other and thereafter they rotate with, a common angular velocity. If Ef and, Ei are the final and initial total, energies, then (Ef − Ei ) is, inertia I1 and, , (a) −, (c), , I1ω12, 24, , 3, I1ω12, 8, , I1ω12, 12, I1ω12, (d), 6, , (b) −, , 6 An n-p-n transistor operates as a, , common emitter amplifier, with a, power gain of 60 dB. The input circuit, resistance is 100 Ω and the output, load resistance is 10 kΩ. The common, emitter current gain β is, (a) 102, , (b) 6 × 102 (c) 104 (d) 60, , 7 The displacement of a damped, , harmonic oscillator is given by, x(t ) = e− 0.1 t cos (10 πt + φ)., Here, t is in seconds.
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ONLINE JEE Main 2019, The time taken for its amplitude of, vibration to drop to half of its initial, value is close to, (a) 27 s, , (b) 13 s, , (c) 4 s, , (d) 7 s, , 8 The ratio of surface tensions of, , mercury and water is given to be 7.5, while the ratio of their densities is, 13.6. Their contact angles with glass, are close to 135° and 0°, respectively., It is observed that mercury gets, depressed by an amount h in a, capillary tube of radius r1 , while, water rises by the same amount h in, a capillary tube of radius r2. The ratio, (r1 / r2 ), is then close to, (a) 3/5, , (b) 2/3, , (c) 2/5, , (d) 4/5, , 9 A ray of light AO in vacuum is, , incident on a glass slab at angle 60°, and refracted at angle 30° along OB, as shown in the figure., The optical path length of light ray, from A to B is, A, 60°, , a, , (c) 2a + 2b, , 2b, (b) 2a +, 3, 2b, (d) 2a +, 3, , 10 n moles of an ideal gas with constant, , volume heat capacity CV undergo an, isobaric expansion by certain volume., The ratio of the work done in the, process, to the heat supplied is, 4nR, (a), CV + nR, nR, (c), CV − nR, , 2, , has mass per unit area σ( r ) = kr ,, where r is the distance from its, centre. Its moment of inertia about an, axis going through its centre of mass, and perpendicular to its plane is, MR 2, MR 2, MR 2, 2MR 2, (b), (c), (d), (a), 2, 6, 3, 3, , 12 A particle of mass m is moving along, a trajectory given by, x = x0 + a cosω1 t and, y = y0 + b sinω2t., , The torque acting on the particle, about the origin at t = 0 is, (a) zero, $, (b) m (− x0b + y0a ) ω12 k, $, (c) − m (x0bω 22 − y0aω12 ) k, 2 $, (d) + my a ω k, 1, , (b) re > rp = rHe, (d) re > rp > rHe, , (a) re < rp = rHe, (c) re < rp < rHe, , 2, , 20 In a meter bridge experiment, the, , circuit diagram and the, corresponding observation table are, shown in figure, , 1, (c), 9λ, , G, , (a) R (T ) = R0e−T, (c) R (T ) = R0e, , 2, , /T02, , −T02 /T 2, , (b) R (T ) = R0eT, R, (d) R (T ) = 02, T, , /T02, , the threshold wavelength of light is, 380 nm. If the wavelength of incident, light is 260 nm, the maximum kinetic, energy of emitted electrons will be, 1237, Given, E (in eV) =, λ( in nm), (a) 15.1 eV, (c) 1.5 eV, , (b) 3.0 eV, (d) 4.5 eV, , 18 A moving coil galvanometer allows a, , full scale current of 10−4 A. A series, resistance of 2 MΩ is required to, convert the above galvanometer into a, voltmeter of range 0-5 V. Therefore,, the value of shunt resistance required, to convert the above galvanometer, into an ammeter of range 0.10 mA is, (a) 100 Ω (b) 500 Ω (c) 200 Ω (d) 10 Ω, , 19 A current of 5 A passes through a, , copper conductor (resistivity, = 17, . × 10−8 Ω-m) of radius of, cross-section 5 mm. Find the mobility, of the charges, if their drift velocity is, 11, . × 10−3 m/s., , S. No., , R (Ω), , l (cm), 60, , 1., , 1000, , 2., , 100, , 13, , 3., , 10, , 1.5, , 4., , 1, , 1.0, , Which of the readings is inconsistent?, (a) 3, , (b) 2, , (c) 1, , (d) 4, , 21 A ball is thrown upward with an, , initial velocity v0 from the surface of, the earth. The motion of the ball is, affected by a drag force equal to mγv2, (where, m is mass of the ball, v is its, instantaneous velocity and γ is a, constant). Time taken by the ball to, rise to its zenith is, (a), , 2, , K, , E, , 15 A 25 × 10−3 m3 volume cylinder is, , material is plotted as a function of, temperature (in, some range). As, InR(T), shown in the, figure, it is a, 1/T2, straight line., One may conclude that, , Unknown, resistance, , l, , 1, (d), 10 λ, , filled with 1 mole of O2 gas at room, temperature (300 K). The molecular, diameter of O2 and its root mean, square speed are found to be 0.3 nm, and 200 m/s, respectively. What is, the average collision rate (per second), for an O2 molecule?, , X, , R, Resistance, box, , have decay constants 10 λ and λ,, respectively. If initially they have the, same number of nuclei, then the ratio, of the number of nuclei of A to that of, B will be 1/e after a time, 11, (b), 10 λ, , (d) 1.8 m 2 / V-s, , (c) 1.0 m / V-s, , 14 Two radioactive materials A and B, , 1, (a), 11 λ, , 9, , (b) 1.3 m 2 / V-s, , (a) 1.5 m 2 / V-s, , (b), (c), , 17 In a photoelectric effect experiment,, , 4nR, (b), CV − nR, nR, (d), CV + nR, , 11 A thin disc of mass M and radius R, , 0, , nucleus, have the same energy. They, are in circular orbits in a plane due to, magnetic field perpendicular to the, plane. Let rp , re and rHe be their, respective radii, then, , 16 In an experiment, the resistance of a, B, , 2 3, (a), + 2b, a, , 13 A proton, an electron and a helium, , (a) ~ 1010 (b) ~ 1012 (c) ~ 1011 (d) ~ 1013, , Vacuum, Glass, , 30°, , b, , O, , APRIL ATTEMPT, , (d), , 2γ, , 1, tan −1 , v0 , 2γg, g, , , , 1, γ, tan −1 , v0 , γg, g, , , , γ, 1, sin −1 , v0 , γg, g, , , , γ, 1, ln 1 +, v0 , g, γg, , , , 22 One plano-convex and one, , plano-concave lens of same radius of, curvature R but of different materials, are joined side by side as shown in, the figure. If the refractive index of, the material of 1 is µ 1 and that of 2 is, µ 2, then the focal length of the, combination is, 1, , µ2, , µ1, , 2, , 2R, µ1 − µ 2, R, (c), 2(µ 1 − µ 2 ), , (a), , R, 2 − (µ 1 − µ 2 ), R, (d), µ1 − µ 2, , (b), , 23 Given below in the left column are, , different modes of communication, using the kinds of waves given in the, right column.
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10 40 DAYS ~ JEE MAIN PHYSICS, A. Optical fibre, P. Ultrasound, communication, B. Radar, , Q. Infrared light, , C. Sonar, , R. Microwaves, , D. Mobile phones, , S. Radio waves, , From the options given below, find, the most appropriate match between, entries in the left and the right, column., (a), (b), (c), (d), , A-Q, B-S, C-R, D-P, A-S, B-Q, C-R, D-P, A-Q, B-S, C-P, D-R, A-R, B-P, C-S, D-Q, , E0, c, E, (c) B = 0, c, E, (d) B = 0, c, (b) B =, , (c) 3.2 m/s and 12.6 m/s, (d) 6.5 m/s and 6.3 m/s, , $j sin (kz ) cos (ωt ), $ sin (kz ) cos (ωt ), k, , 28 The value of acceleration due to, , $j cos (kz ) sin (ωt ), , 26 A transformer consisting of 300 turns, , in the primary and 150 turns in the, secondary gives output power of 2.2, kW. If the current in the secondary coil, is 10 A, then the input voltage and, current in the primary coil are, (a) 440 V and 5 A (b) 220 V and 20 A, (c) 220 V and 10 A (d) 440 V and 20 A, , 24 A stationary source emits sound, , waves of frequency 500 Hz. Two, observers moving along a line passing, through the source detect sound to be, of frequencies 480 Hz and 530 Hz., Their respective speeds are in ms−1 ,, (Take, speed of sound = 300 m/s), , (a) 12, 16, (c) 16, 14, , ONLINE JEE Main 2019, , 27 Two particles of masses M and 2M,, , moving as shown, with speeds of 10, m/s and 5 m/s, collide elastically at, the origin. After the collision, they, move along the indicated directions, with speed v1 and v2 are nearly, M, , 2M, , (b) 12, 18, (d) 8, 18, , 10 m/s, , v1, , 30°, , 30°, , 25 The electric field of a plane, , electromagnetic wave is given by, E = E0 i$ cos (kz ) cos (ωt ), The corresponding magnetic field B is, then given by, E, (a) B = 0 $j sin (kz ) sin (ωt ), c, , gravity at earth’s surface is 9.8 ms−2., The altitude above its surface at, which the acceleration due to gravity, decreases to 4.9 ms−2, is close to, (Take, radius of earth = 6.4 × 106 m ), (a) 9.0 × 106 m, (c) 6.4 × 106 m, , (b) 2.6 × 106 m, (d) 16, . × 106 m, , 29 A cylinder with fixed capacity of 67.2, L contains helium gas at STP. The, amount of heat needed to raise the, temperature of the gas by 20°C is, [Take, R = 8.31 J mol −1 K−1 ], (a) 700 J, (c) 374 J, , (b) 748 J, (d) 350 J, , 30 In the given circuit, an ideal, , voltmeter connected across the 10 Ω, resistance reads 2 V. The internal, resistance r, of each cell is, 15 Ω, 2Ω, , 45°, , 45°, , 10 Ω, , 5m/s, , v2, M, , 2M, , 1.5 V, 1.5 V,, rΩ rΩ, , (a) 6.5 m/s and 3.2 m/s, (b) 3.2 m/s and 6.3 m/s, , (a) 1.5 Ω (b) 0.5 Ω (c) 1 Ω, , (d) 0 Ω, , ANSWERS, 1. (d), 11. (d), 21. (b), , 2. (d), 12. (d), 22. (d), , 3. (d), 13. (c), 23. (c), , 4. (c), 14. (c), 24. (b), , 5. (a), 15. (*), 25. (a), , 10 April 2019 Shift-II, 1 A plane is inclined at an angle α = 30º, with respect to the horizontal. A, particle is projected with a speed, u = 2 ms− 1 , from the base of the, plane, making an angle θ = 15º with, respect to the plane as shown in the, figure. The distance from the base, at, which the particle hits the plane is, close to [Take, g = 10 ms− 2], , 6. (a), 16. (c), 26. (a), , 7. (d), 17. (c), 27. (d), , 8. (c), 18. (*), 28. (b), , (a) 12 N (b) 16 N (c) 8 N, , (a) 26 cm, (c) 18 cm, , 2 Two blocks A, , (d) 40 N, , 3 The time dependence of the position, , of a particle of mass m = 2 is given by, r (t ) = 2t$i − 3t 2$j. Its angular, momentum, with respect to the origin,, at time t = 2 is, $, (c) − 48 k, , °, 15, θ=, α=30°, , 10. (d), 20. (d), 30. (b), , of the table is also 0.2. The maximum, force F that can be applied on B, horizontally, so that the block A does, not slide over the block B is, [Take, g = 10 m / s2], , $, (a) 36 k, u, , 9. (c), 19. (c), 29. (b), , $ − $i ), (b) − 34 (k, (d) 48 ($i + $j), , 4 A spaceship orbits around a planet at, (b) 20 cm, (d) 14 cm, A, , and B of masses, F, B, mA = 1 kg and, mB = 3 kg are, kept on the, table as shown in figure. The, coefficient of friction between A and B, is 0.2 and between B and the surface, , a height of 20 km from its surface., Assuming that only gravitational field, of the planet acts on the spaceship,, what will be the number of complete, revolutions made by the spaceship in, 24 hours around the planet? [Take,, mass of planet = 8 × 1022 kg, radius of, planet = 2 × 106 m, gravitational, constant G = 6.67 × 10− 11 N- m2 /kg 2], (a) 11, , (b) 17, , (c) 13, , (d) 9, , For Detailed Solutions, Visit : http://tinyurl.com/y2kdgeua, Or Scan :, , 5 In free space, a particle A of charge, , 1µC is held fixed at a point P. Another, particle B of the same charge and, mass 4 µg is kept at a distance of 1, mm from P. If B is released, then its, velocity at a distance of 9 mm from P is, [Take,1/ 4 πε0 = 9 × 109 N- m2C− 2 ], (a) 15, . × 102 m / s, (c) 1.0 m/s, , (b) 3.0 × 104 m / s, (d) 2.0 × 103 m / s, , 6 A submarine experience a pressure of, 5.05 × 106 Pa at a depth of d1 in a sea., When it goes further to a depth of d2,, it experiences a pressure of, 8.08 × 106 Pa, then d2 − d1 is, approximately (density of water, = 103 kg / m3 and acceleration due to, gravity = 10 ms− 2), , (a) 500 m(b) 400 m (c) 600 m (d) 300 m, , 7 Light is incident normally on a, , completely absorbing surface with an, energy flux of 25 W cm− 2. If the, surface has an area of 25 cm2, the, momentum transferred to the surface, in 40 min time duration will be, (a) 3.5 × 10− 6 N - s (b) 6.3 × 10− 4 N - s, (c) 14, . × 10− 6 N - s (d) 5.0 × 10− 3 N - s
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ONLINE JEE Main 2019, 8 A square loop is carrying a steady, , current I and the magnitude of its, magnetic dipole moment is m. If this, square loop is changed to a circular, loop and it carries the same current,, the magnitude of the magnetic dipole, moment of circular loop (in A-m) will be, 4m, (a), π, , 3m, (b), π, , 2m, (c), π, , m, (d), π, , 9 A 2 mW laser operates at a, , wavelength of 500 nm. The number of, photons that will be emitted per, second is, [Given, Planck’s constant, h = 6.6 × 10− 34 Js, speed of light, c = 3.0 × 108 m / s], (b) 5 × 1015, (d) 2 × 1016, , (a) 1 × 1016, (c) 15, . × 1016, , R is divided into two unequal parts., 7M, The first part has a mass of, and, 8, is converted into a uniform disc of, radius 2R. The second part is, converted into a uniform solid sphere., Let I1 be the moment of inertia of the, disc about its axis and I 2 be the, moment of inertia of the new sphere, about its axis., The ratio I1 / I 2 is given by, (b) 185, , (c) 65, , (d) 140, , 11 The correct figure that shows, , schematically, the wave pattern, produced by superposition of two, waves of frequencies 9Hz and 11 Hz, is, y, , (a) 0, y, , t(s), 1, , y, , t(s), 1, , 2, , (c) 0, y, , t(s), 1, , (a) 807 Hz, (c) 750 Hz, , 2, , t(s), 1, , speed of 1ms− 1 , just before it starts, penetrating a mud wall of thickness, 20 cm. If the wall offers a mean, resistance of 2.5 × 10− 2 N, the speed of, the bullet after emerging from the, other side of the wall is close to, (a) 0.3 ms− 1, (c) 01, . ms− 1, , 13 A cubical block of side 0.5 m floats on, water with 30% of its volume under, water. What is the maximum weight, that can be put on the block without, fully submerging it under water?, (a) 30.1 kg, (c) 87.5 kg, , at the centre of an equilateral, triangular loop of side 1 m which is, carrying a current of 10 A is, [Take, µ 0 = 4 π × 10− 7 NA− 2], (a) 9 µT, (c) 3 µT, , (b) 46.3 kg, (d) 65.4 kg, , (b) [M− 2L0 T − 4 A− 2 ], (d) [M− 2L− 2 T 6A3 ], , 15 In a Young’s double slit experiment,, , the ratio of the slit’s width is 4 : 1. The, ratio of the intensity of maxima to, minima, close to the central fringe on, the screen, will be, (a) 4 : 1, (c) 9 : 1, , (b) 25 : 9, (d) ( 3 + 1) 4 : 16, , 16 The graph shows, , b2, (b), ac, , a, (c), c, , 2, , b, (d), c, A, , and radius 1 cm is fixed, to a thin stick AB of, 1cm, negligible mass as shown, in the figure. The system, is initially at rest. The, constant torque, that will, B, make the system rotate, about AB at 25 rotations per second, in 5s, is close to, (a) 4.0 × 10− 6 N- m (b) 2.0 × 10− 5 N- m, (c) 16, . × 10− 5 N- m (d) 7.9 × 10− 6 N- m, , 12 A source of sound S is moving with a, velocity of 50 m/s towards a, stationary observer. The observer, , conducting spheres of radii a and, b (b > a ) is filled with a medium of, resistivity ρ. The resistance between, the two spheres will be, , ρ, (a), 2π, ρ, (c), 2π, , 1 + 1, , , a, b, 1 − 1, , , a b, , ρ, (b), 4π, ρ, (d), 4π, , wires of length 1 m each with areas of, cross-section 1 mm2 are used. The, wires are connected in series and one, end of the combined wire is connected, to a rigid support and other end is, subjected to elongation. The stress, requires to produce a net elongation, of 0.2 mm is, [Take, the Young’s modulus for steel, and brass are respectively, 120 × 109 N /m2 and 60 × 109 N /m2], (a) 12, . × 106 N / m 2 (b) 0.2 × 106 N / m 2, (c) 18, . × 106 N / m 2 (d) 4.0 × 106 N / m 2, , 22 A coil of self inductance 10 mH and, , m, , how the, magnification m, c, produced by a thin, lens varies with, a, b, image distance v., What is the focal length of the lens, used?, b2c, (a), a, , (b) 1 µT, (d) 18 µT, , 21 In an experiment, brass and steel, , have dimensions of capacitance and, magnetic field, respectively. What are, the dimensions of Y in SI units?, (a) [M− 1 L− 2 T 4 A2 ], (c) [M− 3L− 2 T 8A4 ], , (b) 0.4 ms− 1, (d) 0.7 ms− 1, , 20 The magnitude of the magnetic field, , [Take, density of water = 103 kg / m3 ], , 18 Space between two concentric, , (d) 0, , 19 A bullet of mass 20 g has an initial, , (b) 1143 Hz, (d) 857 Hz, , 17 A metal coin of mass 5g, , 2, , (b) 0, , measures the frequency of the source, as 1000 Hz. What will be the, apparent frequency of the source, when it is moving away from the, observer after crossing him? (Take,, velocity of sound in air is 350 m/s), , 14 In the formula X = 5YZ 2, X and Z, , 10 A solid sphere of mass M and radius, , (a) 285, , APRIL ATTEMPT 11, , 1 − 1, , , a b, 1 + 1, , , a, b, , v, , resistance 01, . Ω is connected through, a switch to a battery of internal, resistance 0.9 Ω. After the switch is, closed, the time taken for the current, to attain 80% of the saturation value, is [Take, ln 5 = 16, . ], (a) 0.002 s, (c) 0.103 s, , 23 The figure, , (b) 0.324 s, (d) 0.016 s, Ri, , represents, a voltage, regulator, √B, RL, circuit using, a Zener, diode. The, breakdown voltage of the Zener diode, is 6 V and the load resistance is, RL = 4 kΩ. The series resistance of the, circuit is Ri = 1 kΩ. If the battery, voltage VB varies from 8V to 16V, what, are the minimum and maximum values, of the current through Zener diode?, (a) 1.5 mA, 8.5 mA (b) 1 mA, 8.5 mA, (c) 0.5 mA, 8.5 mA (d) 0.5 mA, 6 mA, , 24 A simple pendulum of length L is, , placed between the plates of a, parallel plate capacitor having, electric field E, as shown in figure. Its, bob has mass m and charge q. The, time period of the pendulum is given, by
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12 40 DAYS ~ JEE MAIN PHYSICS, +, +, +, +, +, +, +, +, +, +, +, , (a) 2 π, , qE , g 2 + , , m, (c) 2π, , 26 One mole of an ideal gas passes, , –, –, L –, –, –, m –, –, q, –, –, –, –, E, L, , L, g + qE , , , , m, , 2, , through a process, where pressure, and volume obey the relation, 2, , 1 V, p = p0 1 − 0 . Here, p0 and V 0, 2 V , , , are constants. Calculate the change, in the temperature of the gas, if its, volume changes from V 0 to 2V 0 ., , g2 −, , 1 p0V 0, 2 R, 3 p0V 0, (c), 4 R, , 1, 4, 5, (d), 4, (b), , (a), , L, , (b)2 π, , q2E 2, m2, , L, (d) 2π, g − qE , , , , m, , 25 When heat Q is supplied to a diatomic, gas of rigid molecules, at constant, volume, its temperature increases by, ∆T . The heat required to produce the, same change in temperature, at a, constant pressure is, 2, (a) Q, 3, 3, (c) Q, 2, , ONLINE JEE Main 2019, , 5, (b) Q, 3, 7, (d) Q, 5, , 2. (b), 12. (c), 22. (d), , downwards with an initial speed of, 10, . ms− 1 . The cross-sectional area of, the tap is 10− 4 m2. Assume that the, pressure is constant throughout the, stream of water and that the flow is, streamlined. The cross-sectional, area of the stream, 0.15 m below the, tap would be [Take, g = 10 ms− 2], 2 × 10− 5 m 2, 1 × 10− 5 m 2, 5 × 10− 4 m 2, 5 × 10− 5 m 2, , 4. (a), 14. (c), 24. (a), , 5. (d), 15. (c), 25. (d), , gun with an initial speed u such that, it hits the target on the ground at a, distance R from it. If t1 and t2 are the, values of the time taken by it to hit, the target in two possible ways, the, product t1 t2 is, R, g, 2R, (d), g, , (b), , 2 The trajectory of a projectile near the, surface of the earth is given as, y = 2x − 9x2., , If it were launched at an angle θ0 with, −2, , speed v0 , then (Take, g = 10 ms ), , , (a) θ 0 = sin −1 , , −1 , (b) θ 0 = cos , , −1 , (c) θ 0 = cos , , −1 , (d) θ 0 = sin , , , (a) 0.90 mm, (c) 1.16 mm, , (b) 1.00 mm, (d) 1.36 mm, , 30 Two radioactive substances A and B, , have decay constants 5 λ and λ,, respectively. Att = 0, a sample has the, same number of the two nuclei. The, time taken for the ratio of the number, 2, 1, of nuclei to become will be, e, (a) 2 / λ, , 6. (d), 16. (d), 26. (d), , 7. (d), 17. (b), 27. (d), , 8. (a), 18. (b), 28. (c), , 9. (b), 19. (d), 29. (c), , 10. (d), 20. (d), 30. (b), , (b) 1 / 2 λ (c) 1 / 4 λ (d) 1 / λ, , 1 , 5, −1, and v0 = ms, 3, 5, 2 , 3, −1, and v0 = ms, 5, 5, 1 , 5, −1, and v0 = ms, 3, 5, 2 , 3, −1, and v0 = ms, 5, 5, , a shell made of a, conductor. It has inner, radius a and outer, radius b and carries, charge Q. At its centre, is a dipole p as shown., , the figure is, , p, , In this case,, (a) surface charge density on the inner, Q , , 2, surface is uniform and equal to, 4 πa 2, (b) electric field outside the shell is the, same as that of a point charge at the, centre of the shell, (c) surface charge density on the outer, surface depends on p, (d) surface charge density on the inner, surface of the shell is zero everywhere, , 4 When M1 gram of ice at −10°C (specific, heat = 0.5 cal g −1 °C −1 ) is added to, M2 gram of water at 50°C, finally no, ice is left and the water is at 0°C. The, value of latent heat of ice, in cal g −1 is, , 50M 2, −5, M1, 50M 2, (c), M1, (a), , 50M1, − 50, M2, 5M 2, (d), −5, M1, (b), , Visit : http://tinyurl.com/y67eggxe, Or Scan :, , 5 The truth table for the circuit given in, , 3 Shown in the figure is, , 1 A shell is fired from a fixed artillery, , R, 4g, R, (c), 2g, , What should be the minimum, diameter of a brass rod, if it is to, support a 400 N load without, exceeding its elastic limit?, , For Detailed Solutions, 3. (c), 13. (c), 23. (c), , 12 April, Shift-I, , (a), , (b) 12.3 nm, (d) 11.4 nm, , 29 The elastic limit of brass is 379 MPa., , ANSWERS, 1. (b), 11. (a), 21. (*), , excited to a level by a radiation of, wavelength λ. When the ion gets, de-excited to the ground state in all, possible ways (including intermediate, emissions), a total of six spectral lines, are observed. What is the value of λ?, [Take, h = 6.63 × 10− 34 Js;, c = 3 × 108 ms− 1 ], (a) 9.4 nm, (c) 10.8 nm, , p0V 0, R, p0V 0, R, , 27 Water from a tap emerges vertically, , (a), (b), (c), (d), , 28 In Li + + , electron in first Bohr orbit is, , Y, , A, B, , A B Y, 0 0 1, (a) 0 1 1, 1 0 1, 1 1 1, A B Y, 0 0 1, (c) 0 1 1, 1 0 0, 1 1 0, , 6 A circular disc of, , A, 0, (b) 0, 1, 1, A, 0, (d) 0, 1, 1, , B Y, 0 1, 1 0, 0 0, 1 0, B Y, 0 0, 1 0, 0 1, 1 1, , b, radius b has a hole, of radius a at its, centre (see figure)., a, If the mass per, unit area of the, disc varies as, σ 0 , then the, , , r , radius of gyration, of the disc about its axis passing, through the centre is
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ONLINE JEE Main 2019, (a), , a 2 + b2 + ab, 2, , (b), , (c), , a 2 + b2 + ab, 3, , a+ b, (d), 3, , If V 0 is almost zero, then identify the, correct statement., , a+ b, 2, , 7 The stopping potential V 0 (in volt) as, a function of frequency (ν) for a, sodium emitter, is shown in the, figure. The work function of sodium,, from the data plotted in the figure,, will be (Take, Planck’s constant, (h ) = 6.63 × 10−34 J-s, electron charge,, e = 16, . × 10−19 C], 3.0, 2.0, V0, , 1.0, 2, , 4, , 6, 8, ν(1014Hz), , (a) 1.82 eV, (c) 1.95 eV, , T(x), , x, , l, , x, , T(x), , (c), , (d), x, , l, , l, , 9 To verify Ohm’s law, a student, , x, , connects the voltmeter across the, battery as shown in the figure. The, measured voltage is plotted as a, function of the current and the, following graph is obtained, V, , Internal, resistance, , Ammeter, , $ + 3$j, 3$i − 4$j, − 4k, (b) s$ =, (a) s$ =, 5, 5, $j − 3k, $, $, − 3$j + 4k, 4, (d) s$ =, (c) s$ = , , , 5, 5, , , , 30 times per minute at a place, where, the dip is 45° and 40 times per, minute, where the dip is 30°. If B1 and, B2 are respectively, the total magnetic, field due to the earth at the two, B, places, then the ratio 1 is best given, B2, by, (a) 1.8, (c) 3.6, , (b) 0.7, (d) 2.2, , 13 At 40°C, a brass wire of 1 mm radius, , is hung from the ceiling. A small mass, M is hung from the free end of the, wire. When the wire is cooled down, from 40°C to 20°C, it regains its, original length of 0.2 m. The value of, M is close to [Coefficient of linear, expansion and Young’s modulus of, brass are 10−5 /°C and 1011 N/m 2, respectively, g =10 ms −2], (b) 0.5 kg, (d) 0.9 kg, , 14 A galvanometer of resistance 100 Ω, , 1.5V, , I, , 1000 mA, , R1, , G, , R2, , has 50 divisions on its scale and has, sensitivity of 20 µA/division. It is to be, converted to a voltmeter with three, ranges of 0-2 V, 0-10 V and 0-20 V., The appropriate circuit to do so is, , R3, R1 = 2000 Ω, R2 = 8000 Ω, R3 = 10000 Ω, 20 V, , (a), 2V, G, , (b), , R1, , 10 V, , R2, 2V, , G, , R1, , R3, 10 V, , R2, , 2V, G, , (d), , R1, , R1 = 1900 Ω, R2 = 8000 Ω, R3 = 10000 Ω, 20 V, , 10 V, , R2, 20 V, , R1 = 1900 Ω, R2 = 9900 Ω, R3 = 19900 Ω, 20 V, , R3, , (c), , R3, 10 V, , R1 = 19900 Ω, R2 = 9900 Ω, R3 = 1900 Ω, 2V, , 15 A progressive wave travelling along, , the positive x-direction is represented, by y(x, t ) = A sin (kx − ωt + φ). Its, snapshot at t = 0 is given in the figure., y, , $ sin[ωt + (6 y − 8z )]. Taking, E = E0 n, unit vectors in x, y and z- directions to, $ the direction of propagation, be i$ , $j, k,, s$ , is, , (a) 9 kg, (c) 1.5 kg, , R, V, , V0, , (b) 3 × 10−5 C, (d) 7 × 10−6 C, , 12 A magnetic compass needle oscillates, , (b), l, , uniformly distributed charge. The, ring rotates at a constant angular, speed of 40π rad s −1 about its axis,, perpendicular to its plane. If the, magnetic field at its centre is, 3.8 × 10−9 T, then the charge carried, by the ring is close to (µ 0 = 4 π × 10−7, N/A 2)., , represented by the electric field, , rotated in a horizontal plane with a, constant angular speed about an axis, passing through one of its ends. If the, tension generated in the rod due to, rotation is T (x) at a distance x from, the axis, then which of the following, graphs depicts it most closely?, , T(x), , 10 A thin ring of 10 cm radius carries a, , 11 An electromagnetic wave is, , 8 A uniform rod of length l is being, , (a), , (a) The emf of the battery is 1.5 V and, its internal resistance is 1.5 Ω, (b) The value of the resistance R is 1.5 Ω, (c) The potential difference across the, battery is 1.5 V when it sends a, current of 1000 mA, (d) The emf of the battery is 1.5 V and, the value of R is 1.5 Ω, , (a) 2 × 10−6 C, (c) 4 × 10−5 C, , 10, , (b) 1.66 eV, (d) 2.12 eV, , T(x), , APRIL ATTEMPT 13, , A, x, , For this wave, the phase φ is, (a) −, , π, 2, , (b) π, , (c) 0, , (d), , π, 2, , 16 The value of numerical aperture of, , the objective lens of a microscope is, 1.25. If light of wavelength 5000Å is, used, the minimum separation, between two points, to be seen as, distinct, will be, (a) 0.24 µm, (c) 012, . µm, , (b) 0.38 µm, (d) 0.48 µm, , 17 A point dipole p = − p0 x$ is kept at, the origin. The potential and electric, field due to this dipole on the Y-axis, at a distance d are, respectively, [Take, V = 0 at infinity], p, − p, p, (b) 0,, ,, 4 πε0d 3, 4 πε0d 2 4 πε0d 3, p, −p, p, (c) 0,, (d), ,, 3, 4 πε0d, 4 πε0d 2 4 πε0d 3, , (a), , 18 The resistive network shown below is, , connected to a DC source of 16 V. The, power consumed by the network is, 4 W. The value of R is, 4R, , 6R, R, , 4R, , R, 12 R, E = 16 V, , (a) 6 Ω, (c) 1 Ω, , (b) 8 Ω, (d) 16 Ω
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14 40 DAYS ~ JEE MAIN PHYSICS, 19 The transfer, , (400, 20), , characteristic, (300, 15), Ic, curve of a, (200, 10), (mA), transistor,, having input, (100, 5), and output, Ib (µA), resistance 100, Ω and 100 kΩ respectively, is shown, in the figure. The voltage and power, gain, are respectively, , (a), (b), (c), (d), , ONLINE JEE Main 2019, 23 Two identical parallel plate capacitors, , of capacitance C each, have plates of, area A, separated by a distance d. The, space between the plates of the two, capacitors, is filled with three, dielectrics of equal thickness and, dielectric constants K1 , K 2 and K3 ., The first capacitor is filled as shown, in Fig. I, and the second one is filled, as shown in Fig. II. If these two, modified capacitors are charged by, the same potential V , the ratio of the, energy stored in the two, would be (E1, refers to capacitor (I) and E2 to, capacitor (II)) :, , 2.5 × 104 , 2.5 × 106, 5 × 104 , 5 × 106, 5 × 104 , 5 × 105, 5 × 104 , 2.5 × 106, , 20 Which of the following combinations, , has the dimension of electrical, resistance (ε0 is the permittivity of, vacuum and µ 0 is the permeability of, vacuum)?, (a), , µ0, ε0, , (b), , µ0, ε0, , (c), , ε0, µ0, , (d), , K1, K2, K3, (I), , ε0, µ0, , 21 A sample of an ideal, , c, gas is taken through p, the cyclic process, a, abca as shown in, b, the figure. The, V, change in the, internal energy of the gas along the, path ca is −180 J. The gas absorbs 250, J of heat along the path ab and 60 J, along the path bc. The work done by, the gas along the path abc is, , (a) 120 J (b) 130 J (c) 100 J (d) 140 J, , 22 The figure shows a square loop L of, , side 5 cm which is connected to a, network of resistances. The whole, setup is moving towards right with a, constant speed of 1 cm s −1 . At some, instant, a part of L is in a uniform, magnetic field of 1 T, perpendicular to, the plane of the loop., If the resistance of L is 1.7 Ω, the, current in the loop at that instant will, be close to, v=1 cm/s, L, 1Ω, , B, , A, 1Ω, , B 2Ω, 3Ω, D, , K3, , (II), , (a), , E1, K 1K 2 K 3, =, E 2 ( K 1 + K 2 + K 3 )( K 2 K 3 + K 3 K 1 + K 1K 2 ), , (b), , E1 ( K 1 + K 2 + K 3 )( K 2 K 3 + K 3 K 1 + K 1K 2 ), =, E2, K 1K 2 K 3, , (c), , 9 K 1K 2 K 3, E1, =, E 2 ( K 1 + K 2 + K 3 )( K 2 K 3 + K 3 K 1 + K 1K 2 ), , (d), , E1 ( K 1 + K 2 + K 3 )( K 2 K 3 + K 3 K 1 + K 1K 2 ), =, E2, 9 K 1K 2 K 3, , 24 A person of mass M is sitting on a, , swing to length L and swinging with, an angular amplitude θ0 . If the person, stands up when the swing passes, through its lowest point, the work, done by him, assuming that his centre, of mass moves by a distance l(l << L),, is close to, (a) Mgl (1 − θ 20 ), (c) Mgl, , (b) Mgl (1 + θ 20 ), , θ2 , (d) Mgl 1 + 0 , 2, , , 25 Two moles of helium gas is mixed, , with three moles of hydrogen, molecules (taken to be rigid). What is, the molar specific heat of mixture at, constant volume?, [Take, R = 8.3 J/mol-K], , C, , 2Ω, , (a) 19.7 J/mol-K, (c) 17.4 J/mol-K, , (b) 15.7 J/mol-K, (d) 21.6 J/mol-K, , 26 A submarine A travelling at 18 km/h, , is being chased along the line of its, velocity by another submarine B, travelling at 27 km/h. B sends a sonar, , 5 cm, , (a) 60 µA, (c) 150 µA, , K2, , K1, , (b) 170 µA, (d) 115 µA, , signal of 500 Hz to detect A and, receives a reflected sound of, frequency ν. The value of ν is close to, (Speed of sound in water = 1500 ms −1 ), (a) 504 Hz, (c) 499 Hz, , (b) 507 Hz, (d) 502 Hz, , 27 A man (mass = 50 kg) and his son, , (mass = 20 kg) are standing on a, frictionless surface facing each other., The man pushes his son, so that he, starts moving at a speed of 0.70 ms −1, with respect to the man. The speed of, the man with respect to the surface is, (a) 0.28 ms −1, (c) 0.47 ms −1, , (b) 0.20 ms −1, (d) 0.14 ms −1, , 28 A concave mirror has, , (a) 6.7 cm, (c) 8.8 cm, , (b) 13.4 cm, (d) 11.7 cm, , 29 An excited He+ ion emits two photons, , in succession, with wavelengths 108.5, nm and 30.4 nm, in making a, transition to ground state. The, quantum number n corresponding to, its initial excited state is [for photon, 1240 eV, of wavelength λ, energy E =, ], λ (in nm), (a) n = 4, (c) n = 7, , (b) n = 5, (d) n = 6, , 30 In a double slit experiment, when a, , thin film of thickness t having, refractive index µ is introduced in, front of one of the slits, the maximum, at the centre of the fringe pattern, shifts by one fringe width. The value, of t is (λ is the wavelength of the light, used), 2λ, (µ − 1), λ, (c), (µ − 1), (a), , λ, 2(µ − 1), λ, (d), (2µ − 1), (b), , ANSWERS, 1. (d), 11. (c), 21. (b), , 2. (c), 12. (b), 22. (b), , 3. (b), 13. (a), 23. (d), , 4. (a), 14. (c), 24. (b), , 5. (c), 15. (b), 25. (c), , 6. (c), 16. (a), 26. (d), , 7. (b), 17. (b), 27. (b), , 8. (b), 18. (b), 28. (c), , 9. (a), 19. (d), 29. (b), , 10. (b), 20. (a), 30. (c), , Particle, , radius of curvature of, 40 cm. It is at the, 5 cm, bottom of a glass that, has water filled up to, 5 cm (see figure). If a, small particle is, floating on the surface of water, its, image as seen, from directly above the, glass, is at a distance d from the, surface of water. The value of d is, close to [Refractive index of water, = 133, . ], , For Detailed Solutions, Visit : http://tinyurl.com/y2a2w6sp, Or Scan :
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ONLINE JEE Main 2019, 12 April, Shift-II, 1 Two particles are projected from the, , same point with the same speed u, such that they have the same range, R, but different maximum heights h1, and h2. Which of the following is, correct?, (a) R 2 = 4 h1 h 2, (c) R 2 = 2 h1 h 2, , (b) R 2 = 16 h1 h 2, (d) R 2 = h1 h 2, , 2 In an amplitude modulator circuit,, the carrier wave is given by, , C (t ) = 4 sin(20000 πt ) while, modulating signal is given by,, m(t ) = 2 sin(2000 πt ). The values of, modulation index and lower side band, frequency are, (a) 0.5 and 10 kHz (b) 0.4 and 10 kHz, (c) 0.3 and 9 kHz, (d) 0.5 and 9 kHz, , 3 Two sources of sound S1 and S2, produce sound waves of same, frequency 660 Hz., A listener is moving from source S1, towards S2 with a constant speed u, m/s and he hears 10 beats/s. The, velocity of sound is 330 m/s. Then, u, equal to, (a) 5.5 m/s, (c) 2.5 m/s, , (b) 15.0 m/s, (d) 10.0 m/s, , 4 Three particles of masses 50 g, 100 g, , and 150 g are placed at the vertices of, an equilateral triangle of side 1 m (as, shown in the figure). The (x, y), coordinates of the centre of mass will be, Y, , 0, , (a) n, , m2=100 g, X, 0.5 m 1.0 m, , 60°, , 3, , 5, (a) , m,, m, 12 , 4, 7, 3 , (c) , m,, m, 4, 12, , , 7, 3 , (b) , m,, m, 8, 12, , 3, , 7, (d) , m,, m, 12 , 8, , 5 A Carnot engine has an efficiency of, , 1/6. When the temperature of the sink, is reduced by 62°C, its efficiency is, doubled. The temperatures of the, source and the sink are respectively,, (a) 62°C, 124°C, (c) 124°C, 62°C, , (b) 99°C, 37°C, (d) 37°C, 99°C, , 6 A spring whose unstretched length is, l has a force constant k. The spring is, cut into two pieces of unstretched, lengths l1 and l2 where, l1 = nl2 and n, is an integer. The ratio k1 / k2 of the, corresponding force constants k1 and, k2 will be, , 1, , (b), , (c), , n2, , 1, n, , 7 A transparent cube of, , 13 A solid sphere of radius R acquires a, , (d) n 2, E, , B, , C, , side d, made of a, material of refractive, q, m2, index µ 2, is immersed, m1, in a liquid of, refractive index, A, D, µ 1 (µ 1 < µ 2 ). A ray is, incident on the faceAB at an angle θ, (shown in the figure). Total internal, reflection takes place at point E on, the face BC., Then, θ must satisfy, (a) θ < sin −1, , µ1, µ2, , (c) θ < sin −1, , µ 22, µ 12, , µ 22, , (b) θ > sin −1, , µ 12, , − 1 (d) θ > sin −1, , −1, , µ1, µ2, , 8 A tuning fork of frequency 480 Hz is, , used in an experiment for measuring, speed of sound (v) in air by resonance, tube method. Resonance is observed, to occur at two successive lengths of, the air column l1 = 30 cm and l2 = 70, cm. Then, v is equal to, (a) 332 ms −1, (c) 338 ms −1, , (b) 384 ms −1, (d) 379 ms −1, , 9 The electron in a hydrogen atom first, , jumps from the third excited state to, the second excited state and, subsequently to the first excited state., The ratio of the respective, wavelengths λ1 / λ 2 of the photons, emitted in this process is, (a) 20/7, , m3=150 g, , 50 g = m1, , APRIL ATTEMPT 15, , (b) 27/5, , (c) 7/5, , (d) 9/7, , 10 A diatomic gas with rigid molecules, , does 10 J of work when expanded at, constant pressure. What would be the, heat energy absorbed by the gas, in, this process?, (a) 25 J, , (b) 35 J, , (c) 30 J, , (d) 40 J, , 11 Let a total charge 2Q be distributed in, a sphere of radius R, with the charge, density given by ρ(r ) = kr, where r is, the distance from the centre. Two, charges A and B, of − Q each, are, placed on diametrically opposite, points, at equal distance a, from the, centre. If A and B do not experience, any force, then, (a) a = 8−1/ 4 R, (c) a = 2−1/ 4 R, , (b) a =, , 3R, , 21/ 4, (d) a = R / 3, , 12 Consider an electron in a hydrogen, , atom, revolving in its second excited, state (having radius 4.65 Å). The, de-Broglie wavelength of this electron, is, (a) 3.5 Å (b) 6.6 Å (c) 12.9 Å (d) 9.7 Å, , terminal velocity v1 when falling (due, to gravity) through a viscous fluid, having a coefficient of viscosityη . The, sphere is broken into 27 identical solid, spheres. If each of these spheres, acquires a terminal velocity, v2 when, falling through the same fluid, the, ratio (v1 / v2 ) equals, (a) 9, , (b) 1/27, , (c) 1/9, , (d) 27, , 14 A smooth wire of length 2πr is bent, , into a circle and kept in a vertical, plane. A bead can, w, slide smoothly on the, wire. When the circle, A, is rotating with, r, angular speed ω, about the vertical, O, diameter AB, as, shown in figure, the, r/2, P, bead is at rest with, respect to the, B, circular ring at, position P as shown. Then, the value, of ω2 is equal to, 3g, 2r, (c) ( g 3 ) / r, , (a), , (b) 2g / (r 3), (d) 2g / r, , 15 A particle is moving with speed, , v = b x along positive X-axis., Calculate the speed of the particle at, time t = τ (assume that the particle is, at origin at t = 0)., (a), , b2τ, 4, , (b), , b2τ, 2, , (c) b2τ, , (d), , b2τ, 2, , 16 The ratio of the weights of a body on, the earth’s surface, so that on the, surface of a planet is 9 : 4. The mass, 1, of the planet is th of that of the, 9, earth. If R is the radius of the earth,, what is the radius of the planet?, (Take, the planets to have the same, mass density), (a), , R, 3, , (b), , R, 4, , (c), , R, 9, , (d), , R, 2, , 17 A system of three polarisers P1 , P2 , P3, is set up such that the pass axis of P3, is crossed with respect to that of P1 ., The pass axis of P2 is inclined at 60°, to the pass axis of P3 . When a beam of, unpolarised light of intensityI 0 is, incident on P1 , the intensity of light, transmitted by the three polarisers is, I. The ratio (I 0 / I ) equals (nearly), (a) 5.33 (b) 16.00 (c) 10.67 (d) 1.80, , 18 A uniform cylindrical rod of length L, , and radius r, is made from a material, whose Young’s modulus of elasticity, equals Y . When this rod is heated by, temperature T and simultaneously, subjected to a net longitudinal
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16 40 DAYS ~ JEE MAIN PHYSICS, , (a) 9F / ( πr 2 YT ), (c) 3F / ( πr 2 YT ), , (b) 6F / ( πr 2 YT ), (d) F / (3 πr 2 YT ), , 19 The number density of molecules of a, , gas depends on their distance r from, 4, the origin as, n (r ) = n0 e− ar .Then, the, total number of molecules is, proportional to, (a) n 0α −3/ 4, (c) n 0α1/ 4, , (b) n 0 α1/ 2, (d) n 0α −3, , output. At what distance from the, speaker will one detect 120 dB, intensity sound? [Take, reference, intensity of sound as 10−12 W/m 2], , (a) 40 cm (b) 20 cm (c) 10 cm (d) 30 cm, , 21 Half lives of two radioactive nuclei A, , and B are 10 minutes and 20 minutes,, respectively. If initially a sample has, equal number of nuclei, then after, 60 minutes, the ratio of decayed, numbers of nuclei A and B will be, (b) 1 : 8, , (c) 8 : 1, , (d) 9 : 8, , heated in an electric kettle whose, heating element has a mean, (temperature averaged) resistance of, 20 Ω. The rms voltage in the mains is, 200 V. Ignoring heat loss from the, kettle, time taken for water to, evaporate fully, is close to [Specific, heat of water = 4200 J/(kg°C), Latent, heat of water = 2260 kJ/kg], , resistance G, produces full scale, deflection when a current I g flows, through it. This galvanometer can be, converted into (i) an ammeter of, range 0 to I 0 (I 0 > I g ) by connecting a, shunt resistance RA to it and (ii) into, a voltmeter of range 0 to V (V = GI 0 ), by connecting a series resistance RV, to it. Then,, , 23 In the given circuit, the charge on 4, µF capacitor will be, , , Ig, RA , =, , RV (I 0 − I g ) , , RA I g , =, , RV I 0 − I g , Ig , (c) RARV = G 2 , and, I0 − Ig , , 5 mF, , 2, , (b) RARV = G 2 and, , RA I 0 − I g , =, , RV I g , , E, , S, , i, , (a), (c), , 2.7 EL, , (b), , R2, 7.3 EL, , (d), , R2, , (d) RARV = G 2 and, , EL, 2.7 R 2, EL, 7.3 R 2, , Rs=2kΩ, IL, IZ, RL=4kΩ, , Ig, RA, =, RV, (I 0 − I g ), , (a) 2.5 mA, (c) 7.5 mA, , 26 Find the magnetic, , (b) 1.5 mA, (d) 3.5 mA, , 30 A block of mass 5 kg is (i) pushed in, , P, , 5c, , field at point P due to, a straight line, segment AB of length, 6 cm carrying a, current of 5 A (See, figure)., A, (Take, µ 0 = 4 π × 10−7, −2, N- A ), , B, 6 cm, , (b) 15, . × 10−5 T, (d) 2.5 × 10−5 T, , a frequency ν = 23.9 GHz propagates, along the positive z -direction in free, space. The peak value of the electric, field is 60 V/m. Which among the, following is the acceptable magnetic, , 10 V, , R, , Is, , 27 A plane electromagnetic wave having, , 3 mF, , L, , 2, , (a) 2.0 × 10−5 T, (c) 3.0 × 10−5 T, , 1 mF, 4 mF, , figure. If the switch S is closed at, t = 0, then the amount of charge that, passes through the battery between, L, t = 0 and t = is, R, , circuit, with a Zener diode of, breakdown voltage = 6 V. If the, unregulated input voltage varies, between 10 V to 16 V, then what is, the maximum Zener current?, , 2, , m, , (b) 12.87 cm, (d) 2.25 cm, , 28 Consider the L-R circuit shown in the, , 29 Figure shows a DC voltage regulator, , I − Ig , (a) RARV = G 2 0, , Ig , , 5c, , (Take, electron’s charge = 16, . × 10−19, C, mass of electron = 91, . × 10−31 kg), , (b) 22 min, (d) 10 min, , (a) B = 2 × 107 sin ( 05, . × 103 z + 15, . × 1011 t ) i$, (b) B = 2 × 10–7 sin ( 05, . × 103 z − 15, . × 1011 t ) $i, 3, $, (c) B = 60sin ( 05, . × 10 x + 15, . × 1011 t ) k, (d) B = 2 × 10–7 sin (15, . × 102 x + 05, . × 1011 t ) $j, , 25 A moving coil galvanometer, having a, , Q, , moving along the, X-axis with an, d, initial energy of, 100 eV, enters a, P, region of magnetic S, field, $, 2 cm, B = (15, . × 10−3 T) k, 8 cm, at S (see figure)., The field extends between x = 0 and, x = 2 cm. The electron is detected at, the point Q on a screen placed 8 cm, away from the point S. The distance d, between P and Q (on the screen) is, , field component in the, electromagnetic wave?, , 24 One kilogram of water at 20°C is, , and, , 22 An electron, , (a) 11.65 cm, (c) 1.22 cm, , (b) 9.6 µC, (d) 24 µC, , (a) 16 min, (c) 3 min, , 20 A small speaker delivers 2 W of audio, , (a) 3 : 8, , (a) 5.4 µC, (c) 13.4 µC, , m, , compressional force F, its length, remains unchanged. The coefficient of, volume expansion of the material of, the rod, is (nearly) equal to, , ONLINE JEE Main 2019, , case (A) and (ii) pulled in case (B), by, a force F = 20 N, making an angle of, 30° with the horizontal, as shown in, the figures. The coefficient of friction, between the block, the floor is µ = 0.2., The difference between the, accelerations of the block, in case (B), and case (A) will be, (Take, g = 10 ms−2), F=20 N, , 30°, 30°, , F=20 N, (A), , (a) 0.4 ms−2, (c) 0.8 ms−2, , (B), , (b) 3.2 ms−2, (d) 0 ms−2, , ANSWERS, 1. (b), 11. (a), 21. (d), , 2. (d), 12. (d), 22. (*), , 3. (c), 13. (a), 23. (d), , 4. (c), 14. (b), 24. (b), , 5. (b), 15. (b), 25. (b), , 6. (c), 16. (d), 26. (b), , 7. (c), 17. (c), 27. (b), , 8. (b), 18. (c), 28. (b), , 9. (a), 19. (a), 29. (d), , 10. (b), 20. (a), 30. (c), , For Detailed Solutions, Visit : http://tinyurl.com/y36jgozj, Or Scan :
Page 449 : JANUARY ATTEMPT, /, , 9 January, Shift-I, 1 A bar magnet is demagnetised by, , inserting it inside a solenoid of length, 0.2 m, 100 turns and carrying a, current of 5.2 A. The coercivity of the, bar magnet is, (a) 1200 A/m, (c) 2600A/m, , (b) 285 A/m, (d) 520A/m, , 2 A rod of length L at room temperature, , and uniform area of cross-section A, is, made of a metal having coefficient of, linear expansion α /°C. It is observed, that an external compressive force F,, is applied on each of its ends,, prevents any change in the length of, the rod, when its temperature rises by, ∆T K. Young’s modulus, Y for this, metal is, , F, 2Aα ∆T, 2F, (c), Aα∆T, (a), , F, Aα (∆T − 273), F, (d), Aα∆T, , (b), , 3 Three charges + Q , q, + Q are placed, d, and d, 2, from the origin on the X-axis. If the, net force experienced by +Q placed at, x = 0 is zero, then value of q is, , respectively at distance 0,, , +Q, 2, −Q, (c), 2, , (a), , +Q, 4, −Q, (d), 4, , (b), , 4 Two masses m, , m, are, 2, l, connected at the, m, two ends of a, massless rigid rod m/2, of length l. The rod is suspended by a, thin wire of torsional constant k at, the centre of mass of the rod-mass, system (see figure). Because of, torsional constant k, the restoring, torque is τ = kθ for angular, displacement θ. If the rod is rotated by, θ0 and released, the tension in it, when it passes through its mean, position will be, and, , (a), , 2kθ 20, kθ 20, (b), l, l, , 5 An infinitely long, , (c), , 3kθ 20, kθ 20, (d), l, 2l, , current-carrying wire, and a small, d, current-carrying loop, are in the plane of the, paper as shown. The radius of the, loop is a and distance of its centre, from the wire is d (d >> a ). If the loop, applies a force F on the wire, then, , a2 , (a) F ∝ 3 , d , , (b) F = 0, , a, (c) F ∝ , d, , a, (d) F ∝ , d, , minimum intensity is 16. The, intensity of the waves are in the ratio, 2, , 6 A plane electromagnetic wave of, , frequency 50 MHz travels in free, space along the positive x - direction., At a particular point in space and, time, E = 6.3 $j V/m. The corresponding, r, magnetic field B, at that point will be, , $ T, (a) 18.9 × 108 k, −8 $, (c) 18.9 × 10 k T, , $ T, (b) 6.3 × 10−8 k, −8 $, (d) 21, . × 10 k T, , 7 Drift speed of electrons, when 1.5 A of, , current flows in a copper wire of, cross-section 5 mm2 is v. If the, electron density in copper is, 9 × 1028 / m3 , the value of v in mm/s is, close to (Take, charge of electron to be, = 16, . × 10−19 C), (a) 0.02 (b) 0.2, , (c) 2, , semiconductor is defined as the ratio, of their drift velocity to the applied, electric field. If for an n - type, semiconductor, the density of, electrons is 1019 m−3 and their mobility, is 16, . m2 (V-s), then the resistivity of, the semiconductor (since, it is an, n-type semiconductor contribution of, holes is ignored) is close to, (b) 0.2 Ω-m, (d) 4 Ω-m, , 9 A copper wire is stretched to make it, 0.5% longer. The percentage change, in its electrical resistance, if its, volume remains unchanged is, , (a) 2.0% (b) 1.0% (c) 0.5% (d) 2.5%, , 10 A gas can be taken from A to B via, two different processes ACB and, ADB., p, , C, , B, , A, , D, , v = k ( y$i + x$j), where k is a constant., The general equation for its path is, (a), (b), (c), (d), , y = x 2 + constant, y 2 = x + constant, xy = constant, y 2 = x 2 + constant, , 13 Temperature difference of 120°C is, , maintained between two ends of a, uniform rod AB of length 2L. Another, bent rod PQ, of same cross-section as, 3L is connected across, AB and length, 2, AB (see figure). In steady state,, temperature difference between P and, Q will be close to, L, 4, A, , L, , P, , (a) 45°, , B, , Q, , (b) 35°C (c) 75°C (d) 60°C, , 14 Surface of certain metal is first, , illuminated with light of wavelength, λ1 = 350 n-m and then by light of, wavelength λ 2 = 540 n-m. It is found, that the maximum speed of the, photoelectrons in the two cases differ, by a factor of 2. The work function of, the metal (in eV) is close to, 1240, (energy of photon =, eV), λ (in n - m), (a) 5.6, , (b) 2.5, , (c) 1.8, , (d) 1.4, , 15 A parallel plate capacitor is made of, , two square plates of side ‘a’ separated, by a distance d (d<<a). The lower, triangular portions is filled with a, dielectric of dielectric constant k, as, shown in the figure. Capacitance of, this capacitor is, , K, , d, , V, , When path ACB is used 60 J of heat, flows into the system and 30 J of work, is done by the system. If path ADB is, used work done by the system is 10 J, the heat flow into the system in path, ADB is, (a) 80 J (b) 40 J, , (c) 25 : 9 (d) 4 : 1, , 12 A particle is moving with a velocity, , (d) 3, , 8 Mobility of electrons in a, , (a) 2 Ω-m, (c) 0.4 Ω-m, , (a) 16 : 9 (b) 5 : 3, , (c) 100 J (d) 20 J, , 11 Two coherent sources produce waves, , of different intensities which, interfere. After interference, the ratio, of the maximum intensity to the, , a, 2, , (a), , Kε0a, ln K, d, , (b), , Kε0a 2, ln K, d (K − 1), , (c), , Kε0a 2, 2d (K + 1), , (d), , 1 . Kε0a 2, d, 2, , 16 A current loop, having two circular, , arcs joined by two radial lines as, shown in the figure. It carries a, current of 10 A. The magnetic field at, point O will be close to
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18 40 DAYS ~ JEE MAIN PHYSICS, O, , m, , θ=45, , 3c, , 3c, m, , °, , friction between the plane and the, block is 0.6. What should be the, minimum value of force F, such that, the block does not move downward ?, (Take, g = 10ms−2), , R, m, 2c, , 2c, m, , Q, , P, , S, , (b) 1. 0 × 10−7 T, (d) 1. 5 × 10−5 T, i, , made of glass, (refractive index is, 1.5) with a thick, bottom. It is filled, n =1.5, with a liquid of, refractive index µ. A, student finds that, irrespective of, what the incident angle i (see figure), is for a beam of light entering the, liquid, the light reflected from the, liquid glass interface is never, completely polarised. For this to, happen, the minimum value of µ is, 3, 5, , (b), , 5, 3, , (c), , 4, 3, , 5, 3, , (d), , 18 A convex lens is put 10 cm from a, , light source and it makes a sharp, image on a screen, kept 10 cm from, the lens. Now, a glass block, (refractive index is 1.5) of, 1.5 cm thickness is placed in contact, with the light source.To get the sharp, image again, the screen is shifted by a, distance d. Then, d is, (a), (b), (c), (d), , 0, 1.1 cm away from the lens, 0.55 cm away from the lens, 0.55 cm towards the lens, , 19 A sample of radioactive material A,, , that has an activity of 10 mCi (1 Ci, = 3.7 × 1010 decays/s) has twice the, number of nuclei as another sample of, a different radioactive material B, which has an activity of 20 mCi. The, correct choices for half-lives of A and, B would, then be respectively, (a), (b), (c), (d), , 20 days and 10 days, 5 days and 10 days, 10 days and 40 days, 20 days and 5 days, , from the ceiling of a car by a light, string of mass m(m << M ). When the, car is at rest, the speed of transverse, waves in the string is 60 ms−1 . When, the car has acceleration a, the wave, speed increases to 60.5 ms−1 . The, value of a, in terms of gravitational, acceleration g is closest to, (a), , g, 20, , (b), , g, 5, , (c), , g, 30, , (d), , g, 10, , A, object made of, z, θ, thin rods of, uniform mass, 90°, B, density is, x, suspended with a, string as shown, C, in figure. If, AB = BC and the angle is made by AB, with downward vertical is θ, then, , 2, 3, 1, (c) tan θ =, 2, , 1, 2 3, 1, (d) tan θ =, 3, , (a) tan θ =, , (b) tan θ =, , 23 Three blocks A , B, , A, , B, , (b) 2, , (c) 3, , Red, , F, , C, , (d) 5, , Orange, , Violet Silver, , (a) 270 Ω, 5%, (c) 27 k Ω, 10%, , B, , 4Ω, , i, 2Ω, S, , V=0, , (a) 4A, , (b) 3A, , (c) 2A, , (b) 27 k Ω, 20%, (d) 270 k Ω, 10%, , ANSWERS, , 26 For a uniformly charged ring of, , radius R, the electric field on its axis, has the largest magnitude at a, distance h from its centre. Then,, value of h is, R, 2, , (b) R 2 (c) R, , 4. (b), 14. (c), 24. (c), , 5. (d), 15. (b), 25. (d), , 6. (d), 16. (a), 26. (a), , 7. (a), 17. (a), 27. (d), , 8. (c), 18. (c), 28. (c), , 9. (b), 19. (d), 29. (c), , (d), , R, 5, , 27 A conducting circular loop is made of, , a thin wire has area 3.5 × 10−3 m2 and, resistance 10 Ω. It is placed, perpendicular to a time dependent, magnetic field B (t ) = (0.4T) sin(0.5 π t )., The field is uniform in space. Then, the net charge flowing through the, loop during t = 0 s and t = 10 ms is, close to, (b) 21 mC, (d) 14 mC, , (atomic mass = 4u) and 1 mole of, argon gas (atomic mass = 40u) is kept, at 300 K in a container. The ratio of, vrms(helium) , their rms speeds , is close to, vrms(argon) , (a) 0.32 (b) 2.24, , (c) 3.16, , 10. (b), 20. (a), 30. (b), , (d) 0.45, , 29 If the angular momentum of a planet, of mass m, moving around sun in a, circular orbit is L about the centre of, the sun , its areal velocity is, (a), , 4L, m, , (b), , 2L, m, , (c), , L, 2m, , (d), , L, m, , 30 A block of mass, , m lying on a, m, F, smooth, horizontal surface is attached to a, spring (of negligible mass) of spring, constant k. The other end of the, spring is fixed as shown in the figure., The block is initially at rest in its, equilibrium position. If now the block, is pulled with a constant force F, the, maximum speed of the block is, (a), , πF, (b), mk, , F, (c), mk, , 2F, F, (d), mk, π mk, , For Detailed Solutions, 3. (d), 13. (a), 23. (a), , (d) 5A, , 28 A mixture of 2 moles of helium gas, , m, m, M, and C are lying, on a smooth, horizontal surface as shown in the, figure. A and B have equal masses m, while C has mass M. Block A is given, an initial speed v towards B due to, which it collides with B perfectly, inelastically. The combined mass, collides with C, also perfectly, 5, inelastically th of the initial kinetic, 6, energy is lost in whole process. What, M, is value of ?, m, , (a) 4, , i2 10 V, , C, , (a) 6 mC, (c) 7 mC, , Its value and tolerance are given, respectively by, , kg is kept on a, kg, rough inclined, 10, plane as shown in, the figure. A force, 3 N 45°, of 3 N is applied on, the block. The coefficient of static, , 2. (d), 12. (d), 22. (d), , 20 V i1, A 2Ω, , (a), , 22 An L-shaped, , 24 A resistance is shown in the figure., , 20 A block of mass 10, , 1. (c), 11. (c), 21. (b), , shown is closed, then the value of, current i will be, , 21 A heavy ball of mass M is suspended, , 17 Consider a tank, , (a), , 25 When the switch S in the circuit, , (a) 32 N (b) 25 N (c) 23 N (d) 18 N, , i=10 A, , (a) 1.0 × 10−5 T, (c) 1. 5 × 10−7 T, , ONLINE JEE Main 2019, , Visit : http://tinyurl.com/y4mssqfe, Or Scan :
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ONLINE JEE Main 2019, 9 January, Shift-II, , 6 The energy required to take a, , 1 In form of G (universal gravitational, constant), h (Planck constant) and c, (speed of light), the time period will, be proportional to, (a), , Gh, c, , 5, , (b), , hc5, G, , (c), , c3, Gh, , (d), , Gh, c3, , 2 A parallel plate, , K1, K2 L/2, capacitor with, square plates is, filled with four, K4, K4 L/2, dielectrics of, dielectric, d/2, d/2, constants, K1 , K 2 , K3 , K 4 arranged as shown in, the figure. The effective dielectric, constant K will be:, , ( K1 + K 2 ) ( K 3 + K 4 ), 2 ( K1 + K 2 + K 3 + K 4 ), ( K1 + K 2 ) ( K 3 + K 4 ), (b) K =, K1 + K 2 + K 3 + K 4, ( K1 + K 3 ) ( K 2 + K 4 ), (c) K =, K1 + K 2 + K 3 + K 4, ( K1 + K 4 ) ( K 2 + K 3 ), (d) K =, 2 ( K1 + K 2 + K 3 + K 4 ), (a) K =, , 3 A series AC circuit containing an, , inductor (20 mH), a capacitor (120 µF), and a resistor (60 Ω ) is driven by an, AC source of 24 V/50 Hz. The energy, dissipated in the circuit in 60 s is, (b) 5.65 × 102 J, (d) 517, . × 102 J, , (a) 3.39 × 103 J, (c) 2.26 × 103 J, , 4 In the given circuit, the internal, , resistance of the 18 V cell is, negligible. If R1 = 400 Ω, R3 = 100 Ω, and R4 = 500 Ω and the reading of an, ideal voltmeter across R4 is 5 V, then, the value of R2 will be, R3, R1, , JANUARY ATTEMPT 19, , R4, , satellite to a height ‘h’ above earth, surface (where, radius of earth, = 6.4 × 103 km) is E1 and kinetic, energy required for the satellite to be, in a circular orbit at this height is E2., The value of h for which E1 and E2 are, equal is, (b) 1.28 × 104 km, (d) 1.6 × 103 km, , (a) 3.2 × 103 km, (c) 6.4 × 103 km, , 7 At 0.3V and 0.7 V, the diodes Ge and, Si become conductor respectively. In, given figure, if ends of diode Ge, overturned, the change in potential, V 0 will be, Ge, V0, 12 V, , Si, , (a) 0.2 V (b) 0.6V, , (c) 0.4 V (d) 0.8V, , 8. A particle having the same charge as, , of electron moves in a circular path of, radius 0.5 cm under the influence of a, magnetic field of 0.5 T. If an electric, field of 100 V/m makes it to move in a, straight path, then the mass of the, particle is (Take, charge of electron, = 16, . × 10−19 C), (a) 1.6 × 10−19 kg, (c) 9.1 × 10−31 kg, , (b) 1.6 × 10−27 kg, (d) 2.0 × 10−24 kg, , 9. A mass of 10 kg is suspended vertically, , by a rope from the roof. When a, horizontal force is applied on the mass,, the rope deviated at an angle of 45° at, the roof point. If the suspended mass is, at equilibrium, the magnitude of the, force applied is (Take, g = 10 ms−2), (b) 200 N, (d) 140 N, , 10 A 15 g mass of nitrogen gas is enclosed, 18 V, , (a) 550 Ω, (c) 300 Ω, , (b) 230 Ω, (d) 450 Ω, , 5 In a car race on a straight path, car A, , takes a time t less than car B at the, finish and passes finishing point with, a speed ‘v’ more than that of car B., Both the cars start from rest and, travel with constant acceleration a1, and a2 respectively. Then ‘v ’ is equal, to, (a), , 2a1 a2, t, a1 + a2, , (c) a1 a2 t, , (b) 2a1 a2 t, (d), , a1 + a2, t, 2, , suspended at its middle by a wire. It, exhibits torsional oscillations. If two, masses each of ‘m’ are attached at, distance ‘L/2’ from its centre on both, sides, it reduces the oscillation, frequency by 20%. The value of ratio, m/M is close to, , (a) 0.57, , (b) 0.37, , (c) 0.77, , (d) 0.17, , 13 Two Carnot engines A and B are, , operated in series. The first one, A, receives heat at T1 (= 600 K) and, rejects to a reservoir at temperature, T2. The second engine B receives heat, rejected by the first engine and in, turn rejects to a heat reservoir at, T3 ( = 400 K). Calculate the, temperature T2 if the work outputs of, the two engines are equal., (a) 600 K, (c) 400 K, , (b) 500 K, (d) 300 K, , 14 A musician produce the sound of, , (a) 70 N, (c) 100 N, , R2, , 5 kW, , 12 A rod of mass ‘M’ and length ‘2L’ is, , in a vessel at a temperature 27°C., Amount of heat transferred to the gas,, so that rms velocity of molecules is, doubled is about, (Take, R = 8.3 J / K - mole), (a) 10 kJ (b) 0.9 kJ (c) 14 kJ (d) 6 kJ, , 11 A power transmission line feeds input, power at 2300 V to a step-down, transformer with its primary, windings having 4000 turns. The, output power is delivered at 230 V by, the transformer. If the current in the, primary of the transformer is 5A and, its efficiency is 90%, the output, current would be, (a) 45 A, , (b) 50 A, , (c) 25 A, , (d) 35 A, , second harmonics from open end flute, of 50 cm. The other person moves, toward the musician with speed 10, km/h from the second end of room. If, the speed of sound 330 m/s, the, frequency heard by running person, will be, (a) 666 Hz, (c) 753 Hz, , (b) 500 Hz, (d) 333 Hz, , 15 The plane mirrors (M1 and M2) are, inclined to each other such that a ray, of light incident on mirror M1 and, parallel to the mirror M2 is reflected, from mirror M2 parallel to the mirror, M1 . The angle between the two mirror, is, (a) 45°, , (b) 75°, , (c) 90°, , (d) 60°, , 16 In free space, the energy of, , electromagnetic wave in electric field, is UE and in magnetic field is UB ., Then, (a)UE = UB, (c)UE < UB, , (b)UE > UB, U, (d)UE = B, 2, , 17 Charge is distributed within a sphere, of radius R with a volume charge, A, density ρ(r ) = 2 e−2r / a , where A and a, r, are constants. If Q is the total charge, of this charge distribution, the radius, R is, , , , , 1, (a) a log , , Q, 1 −, , , 2 πaA , Q , (b) a log 1 −, , , 2 πaA
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20 40 DAYS ~ JEE MAIN PHYSICS, a, Q , log 1 −, , , 2, 2 πaA , , , , , a, 1, (d) log , , Q, 2, 1 −, , , 2 πaA , , 23 A particle is executing simple, , (c), , harmonic motion (SHM) of amplitude, A, along the X-axis, about x = 0. when, its potential energy (PE) equals, kinetic energy (KE), the position of, the particle will be, , 18 In a Young’s double slit experiment,, the slits are placed 0.320 mm apart., Light of wavelength λ = 500 n-m is, incident on the slits. The total number, of bright fringes that are observed in, the angular range − 30° ≤ θ ≤ 30° is, , (a) 320, , (b) 321, , (c) 640, , (d) 641, , 19 In communication system, only one, percent frequency of signal of, wavelength 800 nm can be used as, bandwidth. How many channal of, 6MHz bandwidth can be broadcast, this?, −34, , 8, , (c = 3 × 10 m / s, h = 6.6 × 10 J - s), (a) 3.75 × 106, (b) 3.86 × 106, 5, (c) 6.25 × 10, (d) 4.87 × 105, , 20 In given time t = 0, Activity of two, , radioactive substances A and B are, equal. After time t, the ratio of their, R, activities B decreases according to, RA, e−3 t . If the half life of A is In 2, the, half-life of B will be, , (a) 4 ln 2 (b), , ln 2, 4, , (c), , ln 2, 2, , (d) 2 ln 2, , 21 In three dimensional system, the, , position coordinates of a particle (in, motion) are given below, x = a cosωt, y = a sinωt, z = aωt, The velocity of particle will be, (a) 2 aω (b) 2 aω, , ONLINE JEE Main 2019, , (c) aω, , (d) 3 aω, , 22 A force acts on a 2 kg object, so that its, position is given as a function of time, as x = 3t 2 + 5. What is the work done, by this force in first 5 seconds?, , (a) 850 J (b) 900 J (c) 950 J (d) 875 J, , (a) A, , (b), , A, 2, , (c), , A, 2 2, , A, 2, , (d), , 24 The magnetic field associated with a, , light wave is given at the origin, by, B = B0, [sin (314, . × 107 ) ct + sin (6.28 × 107 )ct ]., If this light falls on a silver plate, having a work function of 4.7 eV,, what will be the maximum kinetic, energy of the photoelectrons?, (Take, c = 3 × 108 ms−1 and, h = 6.6 × 10−34 J-s), (a) 7.72 eV, (c) 8.52 eV, , (b) 6.82 eV, (d) 12.5 eV, , 25 Two point charges q1 ( 10 µC) and, q2 (− 25 µC) are placed on the x-axis at, x = 1m and x = 4 m, respectively. The, electric field (in V/m) at a point y = 3, m onY -axis is, , , 1, = 9 × 109 N- m2C−2 , Take,, 4 πε0, , , (a) (63 $i − 27$j) × 102, (b) (81 $i − 81$j) × 102, , wires of length L is bent in the form of, a circular loop and the other one into, a circular coil of N identical turns. If, the same current is passed in both,, the ratio of the magnetic field at the, centre of the loop (BL ) to that, B, at the centre of the coil (BC ), i.e. L, BC, will be, (a), , 1, N, , (b) N, , (c), , 1, , N2, , color code. What is the value of the, resistance?, , G O Y, , (a) 5.3 MΩ ± 5%, (c) 6.4 MΩ ± 5%, , Golden, , (b) 64 kΩ ± 10%, (d) 530 kΩ ± 5%, , 29 The pitch and the number of, , divisions, on the circular scale for a, given screw gauge are 0.5 mm and, 100, respectively. When the screw, gauge is fully tightened without any, object, the zero of its circular scale, lies 3 divisions below the mean line., The readings of the main scale and, the circular scale for a thin sheet are, 5.5 mm and 48 respectively, the, thickness of this sheet is, (b) 5.725 mm, (d) 5.740 mm, , 30 A rod of length, , 26 The top of a water tank is open to air, , and its water level is maintained. It is, giving out 0.74 m3 water per minute, through a circular opening of 2 cm, radius is its wall. The depth of the, centre of the opening from the level of, water in the tank is, close to, (b) 6.0 m, (d) 9.6 m, , 50 cm is pivoted, at one end. It is, raised such that, 30º, if makes an, angle of 30° from the horizontal as, shown and released from rest. Its, angular speed when it passes through, the horizontal (in rad s−1 ) will be, (Take, g = 10 ms−2 ), (a), , 30, 2, , (b) 30, , (c), , 20, 3, , ANSWERS, 1 (a), 11 (a), 21 (a), , 2 (*), 12 (b), 22 (b), , * No option is correct, , 3 (d), 13 (b), 23 (d), , 4 (c), 14 (a), 24 (a), , 5 (c), 15 (d), 25 (a), , 6 (a), 16 (a), 26 (a), , 7 (c), 17 (d), 27 (c), , 8 (d), 18 (d), 28 (d), , (d) N 2, , 28 A carbon resistance has a following, , (a) 5.950 mm, (c) 5.755 mm, , (c) (−81 $i + 81$j) × 102, (d) (−63 $i + 27$j) × 102, , (a) 4.8 m, (c) 2.9 m, , 27 One of the two identical conducting, , 9 (c), 19 (c), 29 (c), , 10 (a), 20 (b), 30 (b), , For Detailed Solutions, Visit : http://tinyurl.com/yy9v5wgb, Or Scan :, , (d), , 30, 2
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ONLINE JEE Main 2019, 10 January, Shift-I, 1 A potentiometer, , D (ε,r), , + –, wire AB having, length L and, J, resistance 12r A, B, is joined to a, cell D of EMF ε, + –, G, C, and internal, ε,, 3r, resistance r. A, 2, cell C having, ε, emf and internal resistance 3r is, 2, connected. The length AJ at which, the galvanometer as shown in figure, shows no deflection is, , (a), , 5, 11, 13, 11, L (b), L (c), L (d), L, 12, 12, 24, 24, , 2 In a Young’s double slit experiment, , with slit separation 0.1 mm, one, 1, observes a bright fringe at angle, 40, rad by using light of wavelength λ1 ., When the light of the wavelength λ 2, is used a bright fringe is seen at the, same angle in the same set up. Given, that λ1 and λ 2 are in visible range, (380 n-m to 740 n-m), their values are, (a), (b), (c), (d), , 380 n-m, 525 n-m, 400 n-m, 500 n-m, 380 n-m, 500 n-m, 625 n-m, 500 n-m, , 3 An insulating thin rod of length l has, , x, on, l, it. The rod is rotated about an axis, passing through the origin (x = 0) and, perpendicular to the rod. If the rod, makes n rotations per second, then, the time averaged magnetic moment, of the rod is, a linear charge density ρ(x) = ρ0, , (a) n ρ l 3, π, (c), n ρl3, 3, , (b) π n ρ l 3, π, (d), n ρl3, 4, , 4 A 2 W carbon resistor is color coded, , with green, black, red and brown, respectively. The maximum current, which can be passed through this, resistor is, (a) 0.4 mA, (c) 20 mA, , (b) 63 mA, (d) 100 mA, , 5 In the given circuit, the cells have, , zero internal resistance. The currents, (in Ampere) passing through, resistances R1 and R2 respectively are, R1 20 Ω R2, – +, , + –, , 10 V, , 10 V, , (a) 0.5, 0 (b) 1, 2, , 20 Ω, , (c) 2, 2, , (d) 0, 1, , JANUARY ATTEMPT 21, , 6 In an electron microscope, the, , resolution that can be achieved is of, the order of the wavelength of, electrons used. To resolve a width of, 7.5 × 10−12 m, the minimum electron, energy required is close to, (a) 500 keV, (c) 100 keV, , (b) 1 keV, (d) 25 keV, , 7 To get output ‘1’ at R, for the given, logic gate circuit, the input values, must be, X, Y, , P, , R, , Q, , (a) X = 0, Y = 0, (c) X = 1, Y = 1, , 8 Two guns A and B can fire bullets at, , speeds 1 km/s and 2 km/s, respectively., From a point on a horizontal ground,, they are fired in all possible, directions. The ratio of maximum, areas covered by the bullets on the, ground fired by the two guns is, (d) 1 : 2, , 9 A satellite is moving with a constant, , speed v in circular orbit around the, earth. An object of mass ‘m’ is ejected, from the satellite such that it just, escapes from the gravitational pull of, the earth. At the time of ejection, the, kinetic energy of the object is, , 3, 1, (a) mv 2 (b) 2 mv 2 (c) m v 2 (d) m v 2, 2, 2, , 10 A heat source at T = 103 K is, connected to another heat reservoir at, T = 102 K by a copper slab which is 1, m thick. Given that the thermal, conductivity of copper is 0.1 WK −1 m −1 ,, the energy flux through it in the, steady state is, (a) 90 Wm −2, (c) 120 Wm −2, , (b) 65 Wm −2, (d) 200 Wm −2, , 11 A magnet of total magnetic moment, , 10−2 $i A-m 2 is placed in a time varying, magnetic field, B$i (cosωt ), where B = 1, T and ω = 0125, rad/s. The work done, ., for reversing the direction of the, magnetic moment at t = 1s is, (a) 0.01 J, (c) 0.014 J, , 2, µFR, (a) µFR (b), 3, 6, , (c), , µ FR, µ FR, (d), 3, 2, , 13 A homogeneous solid cylindrical roller, of radius R and mass m is pulled on a, cricket pitch by a horizontal force., Assuming rolling without slipping,, angular acceleration of the cylinder is, 2F, 3mR, F, (d), 3mR, , F, 2mR, 3F, (c), 2mR, , (b), , (a), , 14 Using a nuclear counter, the count, , (b) X = 1 , Y = 0, (d) X = 0, Y = 1, , (a) 1 : 4 (b) 1 : 16 (c) 1 : 8, , the torque applied by the machine on, the mop is, , (b) 0.007 J, (d) 0.028 J, , 12 To mop-clean a floor, a cleaning, , machine presses a circular mop of, radius R vertically down with a total, force F and rotates it with a constant, angular speed about its axis. If the, force F is distributed uniformly over, the mop and if coefficient of friction, between the mop and the floor is µ,, , rate of emitted particles from a, radioactive source is measured. At, t = 0, it was 1600 counts per second, and t = 8 s, it was 100 counts per, second. The count rate observed as, counts per second at t = 6 s is close to, (a) 400, , (b) 200, , (c) 150, , (d) 360, , 15 A charge Q is distributed over three, concentric spherical shells of radii, a , b, c (a < b < c) such that their, surface charge densities are equal to, one another. The total potential at a, point at distance r from their common, centre, where r < a would be, (a), (b), , Q ( a 2 + b2 + c 2 ), , 4 πε0 (a 3 + b3 + c3 ), Q ( a + b + c), , 4 πε0 (a 2 + b2 + c2 ), Q, (c), 4 πε0 (a + b + c), ab + bc + ca, Q, (d), ⋅, 12 π ε0, abc, , 16 A parallel plate capacitor is of area, , 6 cm 2 and a separation 3 mm. The, gap is filled with three dielectric, materials of equal thickness, (see figure) with dielectric constants, K1 = 10, K 2 = 12 and K3 = 14. The, dielectric constant of a material which, give same capacitance when fully, inserted in above capacitor, would be, , K1, , (a) 4, , K2, , (b) 36, , K3, , (c) 12, , 3 mm, , (d) 14, , 17 A solid metal cube of edge length 2 cm, is moving in a positiveY -direction at, a constant speed of 6 m/s. There is a, uniform magnetic field of 0.1 T in the, positive Z-direction. The potential, difference between the two faces of, the cube perpendicular to the X-axis is, (a) 2 mV (b) 12 mV (c) 6 mV (d) 1 mV
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22 40 DAYS ~ JEE MAIN PHYSICS, 18 In the cube of side ‘a’ shown in the, , figure, the vector from the central, point of the face ABOD to the central, point of the face BEFO will be, z, B, , E, , A, , a, , H, , G, , F, , y, , a, , 1 $ $, a (i − k ), 2, 1 $ $, (c) a ( j − k ), 2, , 1 $ $, a ( j − i), 2, 1 $ $, (d) a (k − i ), 2, , (a), , (b), , 19 If the magnetic field of a plane, electromagnetic wave is given by, B = 100 × 10, , −6, , x , , sin 2 π × 2 × 1015 t − , , , c , , then the maximum electric field, associated with it is (Take, the speed of, light = 3 × 108 m/s), , (a) 6 × 104 N/C, (c) 3 × 104 N/C, , observer with speed 34 m/s. The train, sounds a whistle and its frequency, registered by the observer is f1 . If the, speed of the train is reduced to 17, m/s, the frequency registered is f2. If, speed of sound is 340 m/s, then the, f, ratio 1 is, f2, 19, 18, , (b), , (b) 4 × 104 N/C, (d) 4.5 × 104 N/C, , (c), , 20, 19, , (d), , 18, 17, , of 140 m and the height of the, receiving antenna is 40 m. What is, the maximum distance upto which, signals can be broadcasted from this, tower in LOS (Line of Sight) mode?, (Take, radius of earth = 6.4 × 106 m)., , (a) 65 km, (c) 40 km, , (b) 80 km, (d) 48 km, , series between a heat source at a, temperature T1 and a heat sink at, temperature T4 (see figure). There are, two other reservoirs at temperatures, T2 and T3 , as shown with, T1 > T2 > T3 > T4 . The three engines, are equally efficient if, T1, , mg 2t 2, 8, , T3, , (a) T 2 = (T13T 4 )1/ 4 ; T 3 = (T1T 43 )1/ 4, , 3mg 2t 2, 8, mg 2t 2, (d) −, 8, , (c) 0, , (b) T 2 = (T12T 4 )1/ 3 ; T 3 = (T1T 42 )1/ 3, , 21 The density of a material in SI units, −3, , is 128 kg m . In certain units in, which the unit of length is 25 cm and, the unit of mass is 50 g, the, numerical value of density of the, material is, (a) 40, , (b) 16, , (c) 640, , (d) 410, , (a) 20 m (b) 30 m (c) 10 m (d) 40 m, , 27 A uniform metallic wire has a, , resistance of 18 Ω and is bent into an, equilateral triangle. Then, the, resistance between any two vertices of, the triangle is, (c) 2 Ω, , 28 Two electric dipoles, A , B with, respective dipole moments, d A = − 4 qa $i and dB = − 2 qa $i are, placed on the X-axis with a, separation R, as shown in the figure, R, X, A, , B, , 2R, 2+1, R, 2+1, , (b), (d), , 2 (d), 12 (a), 22 (a), , 3 (d), 13 (b), 23 (a), , 4 (c), 14 (b), 24 (b), , 5 (a), 15 (b), 25 (d), , (c) T 2 = (T1 T 4 )1/ 2 ; T 3 = (T12T 4 )1/ 3, (d) T 2 = (T1 T 42 )1/ 3 ; T 3 = (T12T 4 )1/ 3, , 25 A string of length 1 m and mass 5 g is, fixed at both ends. The tension in the, string is 8.0 N. The string is set into, vibration using an external vibrator, of frequency 100 Hz. The separation, between successive nodes on the, string is close to, , 6 (d), 16 (c), 26 (d), , 7 (b), 17 (b), 27 (d), , 8 (b), 18 (b), 28 (a), , 9 (c), 19 (c), 29 (c), , 2R, 2−1, R, 2−1, , flat bottom at the rate of 10−4 m3 s −1 ., Water is also leaking out of a hole of, area 1 cm 2 at its bottom. If the height, of the water in the tank remains, steady, then this height is, (a) 4 cm, (c) 5.1 cm, , (b) 2.9 cm, (d) 1.7 cm, , 30 A plano-convex lens of refractive, , index µ 1 and focal length f1 is kept in, contact with another plano-concave, lens of refractive index µ 2 and focal, length f2. If the radius of curvature of, their spherical faces is R each and, f1 = 2f2, then µ 1 and µ 2 are related as, (a) 3 µ 2 − 2 µ 1 = 1, (c) 2 µ 1 − µ 2 = 1, , (b) 2 µ 2 − µ 1 = 1, (d) µ 1 + µ 2 = 3, , ANSWERS, 1 (c), 11 (c), 21 (a), , (d) 4 Ω, , 29 Water flows into a large tank with, , ε3, , (b), , dropped from the top of a 100 m, height building. At the same time, a, bullet of mass 0.02 kg is fired, vertically upward with a velocity 100, ms −1 from the ground. The bullet gets, embedded in the wood. Then, the, maximum height to which the, combined system reaches above the, top of the building before falling, below is (Take, g = 10 ms −2), , (c), , ε2, , T4, , (a), , 26 A piece of wood of mass 0.03 kg is, , (a), , T2, , g, a=, 2, , (b) 33.3 cm, (d) 20.0 cm, , The distance from A at which both of, them produce the same potential is, , ε1, , platform which starts from rest with, g, constant acceleration upwards as, 2, shown in figure. Work done by normal, reaction on block in time t is, , (a) 16.6 cm, (c) 10.0 cm, , (a) 12 Ω (b) 8 Ω, , 24 Three Carnot engines operate in, , 20 A block of mass m is kept on a, , m, , 21, 20, , 23 A TV transmission tower has a height, , a, D, , 22 A train moves towards a stationary, , (a), , O, , x, , ONLINE JEE Main 2019, , 10 (a), 20 (b), 30 (c), , For Detailed Solutions, Visit : http://tinyurl.com/y2jls3lb, Or Scan :
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ONLINE JEE Main 2019, , JANUARY ATTEMPT 23, 11 A hoop and a solid cylinder of same, , 10 January, Shift-II, , R, , 1 A particle which is experiencing a, , force, is given by F = 3$i − 12$j,, undergoes a displacement of d = 4$i. If, the particle had a kinetic energy of 3, J at the beginning of the, displacement, what is its kinetic, energy at the end of the displacement ?, (a) 9 J, , (b) 15J, , (c) 12 J, , (d) 10 J, , 2 An unknown metal of mass 192 g, , heated to a temperature of 100°C was, immersed into a brass calorimeter of, mass 128 g containing 240 g of water, at a temperature of 8.4°C. Calculate, the specific heat of the unknown, metal, if water temperature stabilises, at 215, . °C. (Take, specific heat of brass, is 394 J kg −1 K−1 ), , (a) 916 J kg −1 K −1 (b) 654 J kg −1 K −1, (c) 1232 J kg −1 K −1 (d) 458 J kg −1 K −1, , 3 A current of 2 mA was passed, , through an unknown resistor which, dissipated a power of 4.4 W., Dissipated power when an ideal, power supply of 11 V is connected, across it is, (a) 11 × 10−4 W, (c) 11 × 105 W, , (b) 11 × 10−5 W, (d) 11 × 10−3 W, , 4 A cylindrical plastic bottle of, , negligible mass is filled with 310 mL, of water and left floating in a pond, with still water. If pressed downward, slightly and released, it starts, performing simple harmonic motion at, angular frequency ω . If the radius of, the bottle is 2.5 cm, then ω is close to, (Take, density of water = 103 kg/m3 ), , (a) 2. 50 rad s −1, (c) 1.25 rad s −1, , (b) 5.00 rad s −1, (d) 3.75 rad s −1, , 5 Two stars of masses 3 × 1031 kg each, 11, , and at distance 2 × 10 m rotate in a, plane about their common centre of, mass O. A meteorite passes through O, moving perpendicular to the star’s, rotation plane. In order to escape, from the gravitational field of this, double star, the minimum speed that, meteorite should have at O is (Take,, gravitational constant,, G = 6.67 × 10−11 N-m 2 kg −2), 5, , 4, , (a) 2. 8 × 10 m/s, , (b) 3.8 × 10 m/s, , (c) 2. 4 × 104 m/s, , (d) 1. 4 × 105 m/s, , 6 Two identical spherical balls of mass, , M and radius R each are stuck on, two ends of a rod of length 2R and, mass M (see figure)., The moment of inertia of the system, about the axis passing, perpendicularly through the centre of, the rod is, , 2R, , 137, MR 2, 15, 17, (c), MR 2, 15, (a), , R, , 209, MR 2, 15, 152, (d), MR 2, 15, , (b), , 7 The modulation frequency of an AM, , radio station is 250 kHz, which is 10%, of the carrier wave. If another AM, station approaches you for license, what, broadcast frequency will you allot ?, (a) 2000 kHz, (c) 2900 kHz, , (b) 2250 kHz, (d) 2750 kHz, , 8 The actual value of, , V, , resistance R , shown, A, in the figure is 30 Ω., R, This is measured in, an experiment as, shown using the standard formula, V, R = , where V and I are the, I, readings of the voltmeter and, ammeter, respectively. If the, measured value of R is 5% less, then, the internal resistance of the, voltmeter is, (a) 600 Ω, (c) 350 Ω, , refracting surface. The radius of, curvature of this surface is equal to, that of cornea (7.8 mm). This surface, separates two media of refractive, indices 1 and 1.34. Calculate the, distance from the refracting surface, at which a parallel beam of light will, come to focus., (c) 3.1 cm (d) 1 cm, , 10 Charges –q and + q, , P, , located at A and, B, respectively,, Q P′, constitute an, electric dipole., O, B, Distance AB = 2a, A –q, +q, O is the mid point, of the dipole and, OP is perpendicular to AB. A charge, Q is placed at P, where OP = y and y, > > 2a. The charge Q experiences an, electrostatic force F., If Q is now moved along the, equatorial line to P ′ such that, y, OP ′ = , the force on, 3, y, Q will be close to >> 2a , 3, , F, 3, (c) 9F, , (a), , (a) T h = 0.5 T c, (c) T h = 2 T c, , (b) 3F, (d) 27F, , (b) T h = T c, (d) T h = 15, . Tc, , 12 Half-mole of an ideal monoatomic gas, , is heated at constant pressure of 1, atm from 20° C to 90° C . Work done by, gas is close to (Take, gas constant,, R = 8.31 J/mol-K), (a) 291 J (b) 581 J (c) 146 J (d) 73 J, , 13 A rigid massless rod of length 3l has, , two masses attached at each end as, shown in the figure. The rod is, pivoted at point P on the horizontal, axis (see figure). When released from, initial horizontal position, its, instantaneous angular acceleration, will be, l, , (b) 570 Ω, (d) 35 Ω, , 9 The eye can be regarded as a single, , (a) 4.0 cm (b) 2 cm, , mass and radius are made of a, permanent magnetic material with, their magnetic moment parallel to, their respective axes. But the, magnetic moment of hoop is twice of, solid cylinder. They are placed in a, uniform magnetic field in such a, manner that their magnetic moments, make a small angle with the field. If, the oscillation periods of hoop and, cylinder are Th and Tc respectively, then, , (a), , 2l, , 5 Mo, , P, , g, 13l, , (b), , 2 Mo, , g, 2l, , (c), , 7g, 3l, , (d), , g, 3l, , 14 Consider the nuclear fission, Ne20 → 2He4 + C12, Given that the binding, energy/nucleon of Ne20 , He4 and C12, are respectively,, 8.03 MeV, 7.07 MeV and 7.86 MeV,, identify the correct statement., (a), (b), (c), (d), , Energy of 3.6 MeV will be released., Energy of 12.4 MeV will be supplied., 8.3 MeV energy will be released., Energy of 11.9 MeV has to be, supplied., , 15 For the circuit shown below, the, , current through the Zener diode is, 5 kΩ, , 120 V, , (a) 14 mA, (c) 5 mA, , 50 V, , 10 kΩ, , (b) zero, (d) 9 mA, , 16 Two forces P and Q of magnitude 2F, and 3F, respectively,are at an angle θ, with each other. If the force Q is, doubled, then their resultant also gets, doubled. Then, the angle θ is, (a) 60°, (c) 30°, , (b) 120°, (d) 90°
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24 40 DAYS ~ JEE MAIN PHYSICS, 17 A closed organ pipe has a, , fundamental frequency of 1.5 kHz., The number of overtones that can be, distinctly heard by a person with this, organ pipe will be (Assume that the, highest frequency a person can hear is, 20,000 Hz), (a) 7, , (b) 4, , (c) 5, , (d) 6, , 18 At some location on earth, the, , horizontal component of earth’s, magnetic field is, 18 × 10−6 T. At this location, magnetic, needle of length 0.12 m and pole, strength 1.8 A-m is suspended from, its mid point using a thread, it makes, 45° angle with horizontal in, equilibrium. To keep this needle, horizontal, the vertical force that, should be applied at one of its ends is, (a) 6. 5 × 10−5 N, (c) 1. 3 × 10−5 N, , (b) 3. 6 × 10−5 N, (d) 1. 8 × 10−5 N, , experiment as shown in figure., P, , First minima, , d, Source S2, 2d, , λ, (b), (5 − 2 ), λ, (d), ( 5 − 2), , motion with an amplitude of 5 cm., When the particle is at 4 cm from the, mean position, the magnitude of its, velocity in SI units is equal to that of, its acceleration. Then, its periodic, time (in seconds) is, (b), , 8π, 3, , Q2, (c), 2 2πε0, , Q2, 4 πε0, , 1 , , 1 +, , , 3, , Q2 , 1 , (d), 1 +, , 4 πε0 , 5, , 22 Two vectors A and B have equal, , magnitudes. The magnitude of (A + B), is ‘n’ times the magnitude of (A − B)., The angle between A and B is, n 2 − 1, (a) sin −1 2, , n + 1, n 2 − 1, (c) cos−1 2, , n + 1, , n − 1, (b) sin −1 , , n + 1, n − 1, (d) cos−1 , , n + 1, , pressure of 4 × 104 N/m 2. The density, of the gas is 8 kg/m3 . What is the, order of energy of the gas due to its, thermal motion ?, (a) 106 J (b) 103 J, , (c), , 7, π, 3, , (d), , 3, π, 8, , 21 Four equal point charges Q each are, placed in the xy-plane at (0, 2), (4, 2),, (4, −2) and (0, − 2). The work required, to put a fifth charge Q at the origin of, the coordinate system will be, , 27 A metal plate of area 1 × 10−4 m 2 is, illuminated by a radiation of intensity, 16 m W/m 2. The work function of the, metal is 5 eV. The energy of the, incident photons is 10 eV and only, 10% of it produces photoelectrons., The number of emitted photoelectrons, per second and their maximum, energy, respectively will be, (Take, 1 eV = 1.6 × 10−19 J), (a) 1011 and 5 eV, , (b) 1012 and 5 eV, , (c) 1010 and 5 eV, , (d) 1014 and 10 eV, , 28 The self-induced emf of a coil is 25 V., , 23 2 kg of a monoatomic gas is at a, , (c) 104 J (d) 105 J, , When the current in it is changed at, uniform rate from 10 A to 25 A in 1s,, the change in the energy of the, inductance is, , (a) 437.5 J, (c) 637.5 J, , (b) 740 J, (d) 540 J, , 29 The electric field of a plane polarised, , 24 A parallel plate capacitor having, , (b) 508 pJ, (d) 600 pJ, , are measured by a meter scale to be, 12.6 ± 01, . cm and 34.2 ± 01, . cm,, respectively. What will be the value of, its volume in appropriate significant, figures ?, (a), (b), (c), (d), , 20 A particle executes simple harmonic, , 4π, 3, , (b), , 25 The diameter and height of a cylinder, , What should be the slit separation d, in terms of wavelength λ such that, the first minima occurs directly in, front of the slit (S1 ) ?, , (a), , Q2, 4πε0, , (a) 560 pJ, (c) 692 pJ, , Screen, , λ, (a), 2(5 − 2 ), λ, (c), 2( 5 − 2), , (a), , capacitance 12 pF is charged by a, battery to a potential difference of 10, V between its plates. The charging, battery is now disconnected and a, porcelain slab of dielectric constant, 6.5 is slipped between the plates. The, work done by the capacitor on the slab is, , 19 Consider a Young’s double slit, , S1, , ONLINE JEE Main 2019, , 3, , 4300 ± 80 cm, 4260 ± 80 cm 3, 4264.4 ± 81.0 cm 3, 4264 ± 81 cm 3, , 26 A particle starts from the origin at, , time t = 0 and moves along the, positive X-axis. The graph of velocity, with respect to time is shown in, figure. What is the position of the, particle at time t = 5s ?, , v, (m/s), 4, , electromagnetic wave in free space at, time t = 0 is given by an expression., E (x, y) = 10$j cos[(6x + 8z )], The magnetic field B (x, z , t ) is given, by (where, c is the velocity of light), 1 $, (6k −, c, 1 $, (b) (6k, −, c, 1 $, (c) (6k, +, c, 1 $, (d) (6k, +, c, , (a), , 8$i ) cos[(6x + 8z + 10ct )], 8$i ) cos[(6x + 8z − 10ct )], 8$i ) cos[(6x − 8z + 10ct )], 8$i ) cos[(6x + 8z − 10ct )], , 30 The Wheatstone bridge shown in, , figure here, gets balanced when the, carbon resistor is used as R1 has the, color code (orange, red, brown). The, resistors R2 and R4 are 80 Ω and 40, Ω, respectively., Assuming that the color code for the, carbon resistors gives their accurate, values, the color code for the carbon, resistor is used as R3 would be, R1, , R2, G, , R3, , R4, , 3, + –, , 2, 1, 0, , 1, , (a) 6 m, , 2, , 3, , 4, , 5 6, , (b) 3 m, , 7, , 8, , 9 10, , t (s), , (c) 10 m (d) 9 m, , (a) brown, blue, black, (b) brown, blue, brown, (c) grey, black, brown, (d) red, green, brown, , ANSWERS, 1. (b), 11. (b), 21. (d), , 2. (a), 12. (a), 22. (c), , 3. (b), 13. (a), 23. (c), , 4. (*), 14. (*), 24. (b), , 5. (a), 15. (d), 25. (b), , 6. (a), 16. (b), 26. (d), , 7. (a), 17. (d), 27. (a), , 8. (b), 18. (a), 28. (a), , 9. (c), 19. (c), 29. (b), , 10. (d), 20. (b), 30. (b), , For Detailed Solutions, Visit : http://tinyurl.com/y3zolw34, Or Scan :
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ONLINE JEE Main 2019, 11 January, Shift-I, 1 A liquid of density ρ is coming out of a, , hose pipe of radius a with horizontal, speed v and hits a mesh. 50% of the, liquid passes through the mesh, unaffected 25% losses all of its, momentum and, 25% comes back with, the same speed. The resultant, pressure on the mesh will be, (a) ρv 2, , (b), , 1 2, 1, 3, ρv (c) ρv 2 (d) ρv 2, 2, 4, 4, , JANUARY ATTEMPT 25, 6 In an experiment, electrons are, , accelerated, from rest by applying a, voltage of 500 V. Calculate the radius, of the path, if a magnetic field 100 mT, is then applied. (Take, charge of the, electron = 1.6 × 10−19 C and mass of, the electron = 91, . × 10−31 kg), (a) 7.5 × 10−2 m, (c) 7.5 × 10−3 m, , (b) 7.5 × 10−4 m, (d) 7.5 m, , 7 Three charges Q, + q, , Q, , harmonic motion has time dependent, πt, displacement given by x(t ) = A sin ., 90, The ratio of kinetic to potential, energy of this particle at t = 210 s will be, , and + q are placed at, the vertices of a right, angle isosceles triangle, as shown below. The +q, +q, net electrostatic, energy of the configuration is zero, if, the value of Q is, , (a) 2, , (a) −2q, , 2 A particle undergoing simple, , (d) 3, , 3 An electromagnetic wave of intensity, , 50 Wm −2 enters in a medium of, refractive index ‘n’ without any loss., The ratio of the magnitudes of electric, fields and the ratio of the magnitudes, of magnetic fields of the wave before, and after entering into the medium, are respectively, given by, (b) ( n , n ), 1 , (d) n ,, , , n, , 4 A slab is subjected to two forces F1, and F2 of same magnitude F as shown, in the figure. Force F2 is in xy-plane, while force F1 acts along Z-axis at the, point (2$i + 3$j). The moment of these, forces about point O will be, , a convex lens of focal length 0.3 m., The lens forms an image of the object., If the object moves away from the, lens at a speed of, 5 m/s, the speed and direction of the, image will be, (a), (b), (c), (d), , −3, , 3.22 × 10 m/s towards the lens, 0.92 × 10−3 m/s away from the lens, 2.26 × 10−3 m/s away from the lens, 116, . × 10−3 m/s towards the lens, , 9 The variation of refractive index of a, , crown glass thin prism with, wavelength of the incident light is, shown. Which of the following graphs, is the correct one, if Dm is the angle of, minimum deviation ?, , 1.530, , z, F1, , 30°, , y, , 4m, , 1.520, , x, , $ )F (b) (3$i − 2$j + 3k, $ )F, (a) (3$i + 2$j − 3k, $, $, $, $, $, $ )F, (c) (3i − 2 j − 3k )F (d) (3i + 2 j + 3k, , Dm, , (a), , (b), , n 2 r1, ⋅, n1 r2, n, (c) 2, n1, , n 2 r22, ⋅, n1 r12, n, (d) 1, n2, , (b), , λ (n-m), , Dm, , Dm, , (c), , (d), 400, 500, 600, 700, , of same length l. The inner and outer, coils have radii r1 and r2 and number, of turns per unit length n1 and n2,, respectively. The ratio of mutual, inductance to the self-inductance of, the inner coil is, , 400 500 600 700, , Dm, , 5 There are two long coaxial solenoids, , 4 µF, , (a) + 12 µC, (c) − 12 µC, , 2 µF, , (b) + 18 µC, (d) − 18 µC, , 11 If the de-Broglie wavelength of an, , electron is equal to 10−3 times, the, wavelength of a photon of frequency, 6 × 1014 Hz, then the speed of electron, is equal to (Take, speed of light, = 3 × 108 m/s, Planck’s constant, = 6.63 × 10−34 J-s and mass of electron, = 91, . × 10−31 kg), (a) 1.45 × 106 m/s, (c) 11, . × 106 m/s, , (b) 1.8 × 106 m/s, (d) 1.7 × 106 m/s, , 12 A satellite is revolving in a circular, , orbit at a height h from the earth, surface such that h << R, where R is, the radius of the earth. Assuming, that the effect of earth’s atmosphere, can be neglected the minimum, increase in the speed required so that, the satellite could escape from the, gravitational field of earth is, (a), (c), , gR, 2, 2 gR, , (b), , gR, , (d), , gR ( 2 − 1), , 13 Equation of travelling wave on a, , stretched string of linear density 5, g/m is y = 0.03 sin(450t − 9x), where, distance and time are measured in SI, units. The tension in the string is, (a) 5 N, (c) 7.5 N, , (b) 12.5 N, (d) 10 N, , 14 A hydrogen atom, initially in the, , (a) 4 a0, , 1.515, 1.510, , 6m, , (a), , n2, , 1.525, , F2, , O, , 10 µF, , ground state is excited by absorbing a, photon of wavelength 980 Å. The, radius of the atom in the excited state, in terms of Bohr radius a0 will be, (Take hc = 12500 eV-Å), , 1.535, , λ (n-m), , (b) 9 a0, , (c) 16 a0 (d) 25 a0, , 15 A body is projected at t = 0 with a, , λ (n-m), , 400, 500, 600, 700, , 1, , , n, , n, 1, 1 , ,, , n, n, , 8 An object is at a distance of 20 m from, , 400, 500, 600, 700, , , (a) , , , (c) , , , −q, − 2q, (c) + q (d), (b), 2+1, 1+ 2, , λ (n-m), , velocity 10 ms −1 at an angle of 60°, with the horizontal. The radius of, curvature of its trajectory at t = 1s is, R. Neglecting air resistance and, taking acceleration due to gravity, g = 10 ms −2, the value of R is, (a) 10.3 m, (c) 5.1 m, , (b) 2.8 m, (d) 2.5 m, , 16 The force of interaction between two, , 400, 500, 600, 700, , (b) 1, , 1, (c), 9, , 6 µF, , λ (n-m), , 10 In the figure shown below, the charge, , on the left plate of the 10 µF capacitor, is − 30 µC. The charge on the right, plate of the 6 µF capacitor is, , x2 , ;, atoms is given by F = αβ exp −, αkT , where x is the distance, k is the, Boltzmann constant and T is, temperature and α and β are two, constants. The dimension of β is, , (a) [MLT −2], , (b) [M0L2T −4 ], , −4, , (d) [M2L2T −2], , 2, , (c) [M LT ]
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26 40 DAYS ~ JEE MAIN PHYSICS, 17 In a Young’s double slit experiment,, , the path difference at a certain point, on the screen between two interfering, 1, waves is th of wavelength. The ratio, 8, of the intensity at this point to that at, the centre of a bright fringe is close to, (a) 0.80 (b) 0.74, , (c) 0.94, , ONLINE JEE Main 2019, The switch S1 is closed at time t = 0, and the switch S2 is kept open. At, some later time (t0 ), the switch S1 is, opened and S2 is closed. The, behaviour of the current I as a, function of time ‘t’ is given by, I, , (d) 0.85, (a), , 18 A rigid diatomic ideal gas undergoes, an adiabatic process at room, temperature. The relation between, temperature and volume for this, process is TV x = constant, then x is, 2, 5, , (a), , (b), , 2, 3, , (c), , 5, 3, , (d), , 3, 5, , R2, , 1500 Ω, , to, , t, , to, , t, , I, , (c), , (d), to, , t, , 23 The given graph shows variation, (with distance r from centre) of, , (a) 6.0 mA, (c) 0, , R2, , ro, , (b) 6.7 mA, (d) 4.0 mA, , 20 The resistance of the meter bridge AB, in given figure is 4Ω. With a cell of, emf ε = 0.5 V and rheostat resistance, Rh = 2Ω . The null point is obtained at, some point J. When the cell is, replaced by another one of emf ε = ε2,, the same null point J is found for, Rh = 6Ω. The emf ε2 is, ε, , ro, , r, , (a) electric field of a uniformly charged, spherical shell, (b) potential of a uniformly charged, spherical shell, (c) electric field of a uniformly charged, sphere, (d) potential of a uniformly charged, sphere, , 24 An amplitude modulates signal is, A, , given by, v (t ) = 10[1 + 0.3 cos(2.2 × 104 t )], sin(5.5 × 105 t )., , B, , J, Rh, , 6V, , (a) 0.6 V (b) 0.3 V (c) 0.5 V (d) 0.4 V, , 21 Two equal resistances when, , connected in series to a battery, consume electric power of 60 W. If, these resistances are now connected, in parallel combination to the same, battery, the electric power consumed, will be, (a) 60 W (b) 30 W (c) 240 W(d) 120 W, , 22 In the circuit shown,, R, , L, S2, , Here, t is in seconds. The sideband, frequencies (in kHz) are, Take, π = 22 , , , , 7, (a) 892.5 and 857.5 (b) 89.25 and 85.75, (c) 178.5 and 171.5 (d) 1785 and 1715, , 25 Ice at − 20°C is added to 50 g of water, at 40°C. When the temperature of the, mixture reaches 0°C, it is found that, 20 g of ice is still unmelted. The, amount of ice added to the water was, close to (Take, specific heat of water, = 4.2 J/g/°C specific heat of ice = 21, ., J/g/°C and heat of fusion of water at, 0°C = 334 J/g), (a) 40 g, (c) 60 g, , S1, , 27 A gas mixture consists of 3 moles of, oxygen and 5 moles of argon at, temperature T. Considering only, translational and rotational modes,, the total internal energy of the, system is, , (a) 12 RT (b) 15 RT (c) 20 RT (d) 4 RT, , 28 A particle is moving along a circular, path with a constant speed of 10, ms −1 . What is the magnitude of the, change in velocity of the particle,, when it moves through an angle of, 60° around the centre of the circle?, , 12 V, V2=10 V, , height of 100 m on a platform of mass, 3 kg which is mounted on a spring, having spring constant k = 125, . × 106, N/m. The body sticks to the platform, and the spring’s maximum, compression is found to be x. Given, that g = 10 ms −2, the value of x will be, close to, , (a) 8 cm (b) 4 cm (c) 40 cm (d) 80 cm, , t, , I, , through zener diode is close to, 500 Ω, , (b), to, , 19 In the given circuit, the current, R1, , I, , 26 A body of mass 1 kg falls freely from a, , (b) 50 g, (d) 100 g, , (a) 10 2 m/s, (c) 10 3 m/s, , 29 An equilateral, , A, , triangle ABC is cut, from a thin solid, D, E, sheet of wood. (see, G, figure) D , E and F, are the mid points of, B, C, F, its sides as shown, and G is the centre of the triangle., The moment of inertia of the triangle, about an axis passing through G and, perpendicular to the plane of the, triangle is I 0 . If the smaller triangle, DEF is removed from ABC, the, moment of inertia of the remaining, figure about the same axis is I. Then, 3, I0, 4, I0, (c) I ′ =, 4, , (a) I ′ =, , 15, I0, 16, 9, (d) I ′ =, I0, 16, (b) I ′ =, , 30 In a Wheatstone, , B, , bridge (see figure),, P, Q, resistances P and Q, G, C, A, are approximately, K2, equal. When, X, R, R = 400Ω, the bridge, D, is balanced. On, K1, interchanging P and, Q , the value of R for balance is 405Ω., The value of X is close to, (a) 404.5 Ω, (c) 402.5 Ω, , ε, , (b) 10 m/s, (d) Zero, , (b) 401.5 Ω, (d) 403.5 Ω, , ANSWERS, 1 (d), 11 (a), 21 (c), , 2 (*), 12 (d), 22 (b), , 3 (d), 13 (b), 23 (b), , 4 (b), 14 (c), 24 (b), , 5 (c), 15 (b), 25 (a), , 6 (b), 16 (c), 26 (*), , 7 (d), 17 (d), 27 (b), , 8 (d), 18 (a), 28 (b), , 9 (c), 19 (c), 29 (b), , 10 (b), 20 (b), 30 (c), , For Detailed Solutions, Visit : http://tinyurl.com/y579lef2, Or Scan :
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ONLINE JEE Main 2019, 11 January, Shift-II, 1 An amplitude modulated signal is, plotted below, V(t), , 10 V, 8V, , JANUARY ATTEMPT 27, 6. The circuit shown below contains two, ideal diodes, each with a forward, resistance of 50 Ω. If the battery, voltage is 6 V, the current through, the 100 Ω resistance (in ampere) is, D1, , 150Ω, , t, , 75Ω, D2, , 8 µs, , 100 µs, , 100Ω, 6V, , Which one of the following best, describes the above signal?, , (a) 0.027 (b) 0.020 (c) 0.030 (d) 0.036, , (a) [1 + 9 sin(2π × 10 4 t )]sin(2.5π × 10 5 t ) V, (b) [ 9 + sin(2π × 10 4 t )]sin(2.5π × 10 5 t ) V, (c) [ 9 + sin(4 π × 10 4 t )]sin(5π × 10 5 t ) V, (d) [ 9 + sin(2.5π × 10 5 t )]sin(2π × 10 4 t ) V, , 2 Two rods A and B of identical, , dimensions are at temperature 30ºC., If A is heated upto 180ºC and B upto, T º C, then new lengths are the same., If the ratio of the coefficients of linear, expansion of A and B is 4 : 3, then the, value of T is, (a) 230ºC (b) 270ºC (c) 200ºC (d) 250ºC, , 3 A galvanometer having a resistance of, 20 Ω and 30 divisions on both sides, has figure of merit 0.005, ampere/division. The resistance that, should be connected in series such, that it can be used as a voltmeter, upto 15 volt is, (a) 100 Ω, (c) 120 Ω, , 7 When 100 g of a liquid A at 100ºC is, , added to 50 g of a liquid B at, temperature 75ºC, the temperature of, the mixture becomes 90ºC. The, temperature of the mixture, if 100 g of, liquid A at 100ºC is added to 50 g of, liquid B at 50ºC will be, (a) 60ºC (b) 80ºC (c) 70ºC (d) 85ºC, , 8 A string is wound around a hollow, , cylinder of mass 5 kg and radius 0.5, m. If the string is now pulled with a, horizontal force of 40 N and the, cylinder is rolling without slipping on, a horizontal surface (see figure), then, the angular acceleration of the, cylinder will be (Neglect the mass and, thickness of the string), 40 N, , (a) 10 rad / s2, (c) 20 rad / s2, , (b) 125 Ω, (d) 80 Ω, , 4 A circular disc D1 of mass M and, radius R has two identical discs D2, and D3 of the same mass M and, radius R attached rigidly at its, opposite ends (see figure). The, moment of inertia of the system about, the axis OO ′ passing through the, centre of D1 , as shown in the figure, will be, , 9 A particle moves from the point, (2.0 $i + 4.0 $j ) m at t = 0 with an initial, velocity (5.0 $i + 4.0 $j) ms− 1 . It is acted, upon by a constant force which, produces a constant acceleration, (4.0 $i + 4.0 $j) ms− 2. What is the, distance of the particle from the, origin at time 2 s?, (a) 5 m, (c) 10 2 m, , O′, , (b) 16 rad / s2, (d) 12 rad / s2, , (b) 20 2 m, (d) 15 m, , 10 A monochromatic light is incident at a, D2, , certain angle on an equilateral, triangular prism and suffers, minimum deviation. If the refractive, index of the material of the prism is, 3, then the angle of incidence is, , D3, , O, D1, , 2, MR 2, 3, (c) 3MR 2, , 4, MR 2, 5, (d) MR 2, , (a), , (b), , 5 The magnitude of torque on a particle, of mass 1 kg is 2.5 N-m about the, origin. If the force acting on it is 1 N, and the distance of the particle from, the origin is 5 m, then the angle, between the force and the position, vector is (in radian), (a), , π, 8, , (b), , π, 4, , (c), , π, 3, , (d), , π, 6, , (a) 45º, (c) 60º, , (b) 90º, (d) 30º, , 12 The mass and the diameter of a, , planet are three times the respective, values for the earth. The period of, oscillation of a simple pendulum on, the earth is 2 s. The period of, oscillation of the same pendulum on, the planet would be, (a), , 2, 3, s, s (b), 2, 3, , (c) 2 3 s (d), , 13 A particle of mass m and charge q is, in an electric and magnetic field is, given by, $., E = 2$i + 3$j , B = 4$j + 6k, The charged particle is shifted from, the origin to the point P (x = 1; y = 1), along a straight path. The magnitude, of the total work done is, (a) (0.35) q, (c) (2.5) q, , (b) (015, . )q, (d) 5 q, , 14 In a double-slit experiment, green, , light (5303 Å) falls on a double slit, having a separation of 19.44 µ-m and a, width of 4.05 µ-m. The number of, bright fringes between the first and, the second diffraction minima is, (a) 5, , (b) 10, , (c) 9, , (d) 4, , 15 If speed (V ), acceleration (A ) and force, (F ) are considered as fundamental, units, the dimension of Young’s, modulus will be, (b) [V−2A2 F2 ], (d) [V−4 A2 F], , (a) [V−4 A− 2F], (c) [V−2A2 F− 2 ], , 16 In a hydrogen like atom, when an, , electron jumps from the M-shell to, the L-shell, the wavelength of emitted, radiation is λ. If an electron jumps, from N-shell to the L-shell, the, wavelength of emitted radiation will, be, (a), , 27, 25, 20, 16, λ (b), λ (c), λ (d), λ, 20, 16, 27, 25, , 17 In the experimental set up of meter, , bridge shown in the figure, the null, point is obtained at a distance of 40, cm from A. If a 10 Ω resistor is, connected in series with R1 , the null, point shifts by 10 cm., The resistance that should be, connected in parallel with (R1 + 10) Ω, such that the null point shifts back to, its initial position is, R1, , R2, , 11 A paramagnetic substance in the form, of a cube with sides 1 cm has a, magnetic dipole moment of, 20 × 10− 6 J / T when a magnetic, intensity of 60 × 103 A / m is applied., Its magnetic susceptibility is, (a) 3.3 × 10− 4, (c) 4.3 × 10− 2, , (b) 3.3 × 10− 2, (d) 2.3 × 10− 2, , 3, s, 2, , G, A, , B, (), , (a) 60 Ω, (c) 30 Ω, , (b) 20 Ω, (d) 40 Ω
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28 40 DAYS ~ JEE MAIN PHYSICS, 18 The region between y = 0 and y = d, , ONLINE JEE Main 2019, 22 A metal ball of mass 0.1 kg is heated, , qvB 3 $, 1 , i + $j, , m 2, 2 , qvB 1 $, 3 $, (b), j, i−, m 2, 2 , $ $, $ $, qvB − j + i , qvB i + j , (d), (c), , , , m , m 2 , 2 , , (a), , (a) 25% (b) 15%, , 1, kR∆T, 2, 1, (c), R∆ T, 2, , (a), , (a) 1 kV/m, (c) 2 kV/m, , (a) 35, , (c) 40, , (d) 25, , (b) 10− 5 rad / s, (d) 10− 1 rad / s, , (b) − 10 × 10− 29 J, (d) − 7 × 10− 27 J, , 26 In the circuit shown, the potential, difference between A and B is, M, 5Ω, A, , 1Ω, , 1V, , 1Ω, , 2V, , D, N, , 1Ω, , (a) 3 V, (c) 6 V, , 3V, , 10Ω, C, , p, k, k, (d) 2, p, (b) 2, , 2 µF are to be connected in a, configuration to obtain an effective, 6, capacitance of µF. Which of the, 13 , combinations shown in figures below, will achieve the desired value?, , applied to an electric dipole at angle, of 45º. The value of electric dipole, moment is 10− 29 C-m. What is the, potential energy of the electric dipole?, (a) − 9 × 10− 20 J, (c) − 20 × 10− 18 J, , 2p, k, 2k, p, , 28 Seven capacitors, each of capacitance, , 25 An electric field of 1000 V/m is, , (b) 0.7 kV/m, (d) 1.4 kV/m, , increases by a factor of 3, decreases by a factor of 9 3, increases by a factor of 27, decreases by a factor of 9, , (b) 60, , (a) 1 rad/s, (c) 10− 3 rad / s, , 21 A copper wire is wound on a wooden, , (a), (b), (c), (d), , (c), , oscillating with an angular frequency, 10 rad/s. The support of the, pendulum starts oscillating up and, down with a small angular frequency, of 1 rad/s and an amplitude of 10− 2 m., The relative change in the angular, frequency of the pendulum is best, given by, , 2, , frame, whose shape is that of an, equilateral triangle. If the linear, dimension of each side of the frame is, increased by a factor of 3, keeping the, number of turns of the coil per unit, length of the frame the same, then, the self-inductance of the coil, , (d) 20%, , 24 A simple pendulum of length 1 m is, , 2k, ∆T, 3, 3, (d), R∆ T, 2, , cross-sectional area of 10 mm . The, magnitude of the maximum electric, field in this electromagnetic wave is, given by [Take, permittivity of space,, ε0 = 9 × 10− 12 SI units and speed of, light, c = 3 × 108 m / s], , (a), , to a linear scale reads a value x0 ,, when in contact with boiling water, and x0 / 3, when in contact with ice., What is the temperature of an object, in ºC, if this thermometer in the, contact with the object reads x0 / 2 ?, , (b), , 20 A 27 mW laser beam has a, , (c) 30%, , straight line with momentum p., Starting at time t = 0, a force F = kt, acts in the same direction on the, moving particle during time interval, T , so that its momentum changes, from p to 3p., Here, k is a constant. The value of T, is, , 23 A thermometer graduated according, , 19 In a process, temperature and volume, , of one mole of an ideal monoatomic, gas are varied according to the, relation VT = k, where k is a, constant. In this process, the, temperature of the gas is increased by, ∆T . The amount of heat absorbed by, gas is (where, R is gas constant), , 27 A particle of mass m is moving in a, , upto 500ºC and dropped into a vessel, of heat capacity 800 JK− 1 and, containing 0.5 kg water. The initial, temperature of water and vessel is, 30ºC. What is the approximate, percentage increment in the, temperature of the water?, [Take, specific heat capacities of, water and metal are respectively, 4200 Jkg − 1 K− 1 and 400 Jkg − 1 K− 1 ], , $. A, contains a magnetic field B = Bk, particle of mass m and charge q enters, the region with a velocity v = v$i. If, mv, then the acceleration of the, d=, 2qB, charged particle at the point of its, emergence at the other side is, , B, , (a), , (b), , (c), , (d), , 29 In a photoelectric experiment, the, , wavelength of the light incident on a, metal is changed from 300 n-m to 400, n-m. The decrease in the stopping, hc, potential is close to , = 1240 n-mV, e, , (a) 0.5 V, (c) 1.5 V, , (b) 2.0 V, (d) 1.0 V, , 30 A pendulum is executing simple, , harmonic motion and its maximum, kinetic energy is K1 . If the length of, the pendulum is doubled and it, performs simple harmonic motion, with the same amplitude as in the, first case, its maximum kinetic, energy is K 2., Then, (a) K 2 = 2K1, , (b) 1 V, (d) 2 V, , (c) K 2 =, , K1, 4, , (b) K 2 =, , (d) K 2 = K1, , ANSWERS, 1. (b), 11. (a), 21. (c), , 2. (a), 12. (c), 22. (d), , 3. (d), 13. (d), 23. (d), , 4. (c), 14. (a), 24. (c), , 5. (d), 15. (d), 25. (d), , 6. (b), 16. (c), 26. (d), , 7. (b), 17. (a), 27. (b), , 8. (b), 18. (*), 28. (c), , 9. (b), 19. (c), 29. (d), , 10. (c), 20. (d), 30. (b), , K1, 2, , For Detailed Solutions, Visit : http://tinyurl.com/y56dxvwz, Or Scan :
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ONLINE JEE Main 2019, , 5, , conductivity of the material of the, inner cylinder is K1 and that of the, outer cylinder is K 2. Assuming no loss, of heat, the effective thermal, conductivity of the system for heat, flowing along the length of the, cylinder is, , 4, , (a), , 12 January, Shift-I, 1 For the given cyclic process CAB as, shown for a gas, the work done is, 6.0, , p (Pa), , C, , A, , K1 + K 2, 2, 2K1 + 3K 2, (c), 5, , 3, 2, 1, , B, 1, , (a) 5 J, , 2, , (b) 10 J, , 2 As shown in the, , 3, , (c) 1 J, S, , (d) 30 J, y, , 40 A, perpendicular out of the page, 20 A, perpendicular into the page, 20 A, perpendicular out of the page, 40 A, perpendicular into the page, , 3 A person standing on an open ground, hears the sound of a jet aeroplane,, coming from north at an angle 60°, with ground level. But he finds the, aeroplane right vertically above his, position. If v is the speed of sound,, then speed of the plane is, (a), , 3, v (b) v, 2, , 2v, (c), 3, , 4 The galvanometer, , v, (d), 2, K2, , deflection, when R1=220Ω, key K1 is closed, but K 2 is open, G, equals θ0 (see, figure). On closing, K 2 also and, K1, adjusting R2 to, 5Ω, the deflection, θ, in galvanometer becomes 0 . The, 5, resistance of the galvanometer is, given by (neglect the internal, resistance of battery) :, (a) 22 Ω, , (b) 5 Ω, , (c) 25 Ω, , (b), , K1 + 3K 2, 4, , (d) K1 + K 2, , 6 Two light identical, , 5 3, V (m ), , 4, , figure, two, infinitely long,, O Q, identical wires, M x, L, P, are bent by 90°, and placed in, N, such a way that, the segments LP and QM are along, the X-axis, while segments PS and, QN are parallel to theY -axis. If, OP = OQ = 4 cm and the magnitude of, the magnetic field at O is 10−4 T and, the two wires carry equal currents, (see figure), the magnitude of the, current in each wire and the direction, of the magnetic field at O will, be (Take, µ 0 = 4 π × 10−7 NA−2 ), (a), (b), (c), (d), , JANUARY ATTEMPT 29, , A, , springs of spring, y, constant k are, attached horizontally, x, O, at the two ends of an, uniform horizontal, rod AB of length l and, B, mass m. The rod is, pivoted at its centre ‘O’ and can rotate, freely in horizontal plane. The other, ends of the two springs are fixed to, rigid supports as shown in figure., The rod is gently pushed through a, small angle and released. The, frequency of resulting oscillation is, 1 2k, 2π m, 1 6k, (c), 2π m, , (a), , 1, 2π, 1, (d), 2π, (b), , 3k, m, k, m, , 7 Two electric bulbs rated at 25 W, 220 V, and 100 W, 220 V are connected in, series across a 220 V voltage source. If, the 25 W and 100 W bulbs draw powers, P1 and P2 respectively, then, , (a) P1, (b) P1, (c) P1, (d) P1, , = 16 W , P2 = 4W, = 4 W , P2 = 16W, = 9 W , P2 = 16W, = 16 W , P2 = 9W, , 8 The output of the given logic circuit is, A, Y, , R2, , (d) 12 Ω, , 5 A cylinder of radius R is surrounded, , by a cylindrical shell of inner radius R, and outer radius 2R. The thermal, , B, , (a) AB, (c) AB + AB, , (b) AB, (d) AB + AB, , 9 The least count of the main scale of a, screw gauge is 1 mm. The minimum, number of divisions on its circular, scale required to measure 5 µm, diameter of a wire is, (a) 50, (c) 500, , (b) 200, (d) 100, , 10 A point source of light, S is placed at a, distance L in front of the centre of, plane mirror of width d which is, hanging vertically on a wall. A man, walks in front of the mirror along a, line parallel to the mirror, at a, distance 2L as shown below., , The distance over, which the man, d, can see the, image of the light, source in the, mirror is, d, (a), 2, , (b) d, , S, L, 2L, , (c) 3d, , (d) 2d, , 11 A passenger train of length 60 m, , travels at a speed of 80 km/hr., Another freight train of length 120 m, travels at a speed of, 30 km/hr. The ratio of times taken by, the passenger train to completely, cross the freight train when : (i) they, are moving in the same direction and, (ii) in the opposite direction is, (a), , 3, 2, , (b), , 25, 11, , (c), , 11, 5, , (d), , 5, 2, , 12 An ideal gas occupies a volume of, , 2 m3 at a pressure of 3 × 106 Pa. The, energy of the gas is, (a) 6 × 104 J, (c) 9 × 106 J, , (b) 108 J, (d) 3 × 102 J, , 13 A simple pendulum is made of a, , string of length l and a bob of mass m,, is released from a small angle θ0 . It, strikes a block of mass M, kept on a, horizontal surface at its lowest point, of oscillations, elastically. It bounces, back and goes up to an angle θ1 . Then,, M is given by, θ + θ1 , (a) m 0, , θ 0 − θ1 , θ − θ1 , (c) m 0, , θ 0 + θ1 , , m θ 0 − θ1 , , , 2 θ 0 + θ1 , m θ 0 + θ1 , (d), , , 2 θ 0 − θ1 , (b), , 14 A particle of mass m moves in a circular, orbit in a central potential field, 1, U (r ) = kr 2. If Bohr’s quantization, 2, conditions are applied, radii of possible, orbitals and energy levels vary with, quantum number n as, (a) rn ∝ n , En ∝ n (b) rn ∝ n 2 , En ∝, , 1, , n2, 1, (c) rn ∝ n , En ∝ n (d) rn ∝ n , En ∝, n, , 15 A proton and an α-particle (with their, masses in the ratio of 1 : 4 and, charges in the ratio of 1 : 2) are, accelerated from rest through a, potential difference V. If a uniform, magnetic field B is set up, perpendicular to their velocities, the, ratio of the radii rp : rα of the circular, paths described by them will be, (a) 1 : 2 (b) 1 : 3 (c) 1 : 3, , (d) 1 : 2, , 16 In the figure shown, a circuit contains, two identical resistors with resistance, R = 5Ω and an inductance with, L = 2 mH. An ideal battery of 15 V is, connected in the circuit. What will be
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30 40 DAYS ~ JEE MAIN PHYSICS, the current, through the, battery long after, the switch is closed?, , S, , L, R, , 15 V, , R, , (a) 6 A, (b) 3 A, (c) 5.5 A (d) 7.5 A, , glass slab of refractive index 1.5. If, 4% of light gets reflected and the, amplitude of the electric field of the, incident light is 30 V/m, then the, amplitude of the electric field for the, wave propogating in the glass, medium will be, , (a) 30 V/m (b) 6 V/m (c) 10 V/m (d) 24 V/m, , 18 A straight rod of length L extends, , from x = a to x = L + a. The, gravitational force it exerts on a point, mass m at x = 0, if the mass per unit, length of the rod is A + Bx2 , is given by, , 1, 1, (a) Gm A , − , +, a, L, a, , , , 1, 1, (b) Gm A , − , a + L a, 1, 1 , (c) Gm A −, , a, a, +, L, , 1, 1 , (d) Gm A −, , a a + L, , , − BL , , , + BL , , , + BL , , , − BL , , , between 160 V and 40 V by a, modulating signal. What is the, modulation index?, (c) 0.6, , 20 In the figure, , (d) 0.3, A, , B, , shown, after the, S, switch ‘S ’ is, ε, 3C, turned from, C, position ‘A’ to, position ‘B ’, the energy dissipated in, the circuit in terms of capacitance ‘C ’, and total charge ‘Q ’ is, (a), , (d) 14.14, , 23 Let the moment of inertia of a hollow, , cylinder of length 30 cm (inner radius, 10 cm and outer radius 20 cm) about, its axis be I. The radius of a thin, cylinder of the same mass such that, its moment of inertia about its axis is, also I, is, (a) 16 cm (b) 14 cm (c) 12 cm (d) 18 cm, , 24 A travelling harmonic wave is, , represented by the equation, y (x, t ) = 10−3 sin (50t + 2x), where x, and y are in metre and t is in second., Which of the following is a correct, statement about the wave?, , (a) The wave is propagating along the, negative X -axis with speed 25 ms−1 ., (b) The wave is propagating along the, positive X -axis with speed 25 ms−1 ., (c) The wave is propagating along the, positive X -axis with speed 100 ms−1 ., (d) The wave is propagating along the, negative X -axis with speed 100 ms−1 ., , 25 There is uniform spherically, , 19 A 100 V carrier wave is made to vary, , (b) 0.5, , accelerated by a potential difference, of 2500 V. The ratio of de-Broglie, λ, wavelengths A is close to, λB, (a) 4.47, (b) 10.00, (c) 0.07, , 17 A light wave is incident normally on a, , (a) 0.4, , ONLINE JEE Main 2019, , 3 Q2, 5 Q2, 1 Q2, 3 Q2, (b) ., (c) ⋅, (d) ⋅, ⋅, 4 C, 8 C, 8 C, 8 C, , symmetric surface charge density at a, distance R0 from the origin. The, charge distribution is initially at rest, and starts expanding because of, mutual repulsion. The figure that, represents best the speed v[R (t )] of, the distribution as a function of its, instantaneous radius R (t ) is, , (a), , (b), Ro, , (a) 395 Ω (b) 495 Ω (c) 490 Ω (d) 480 Ω, , 22 A particle A of mass ‘m’ and charge ‘q’, is accelerated by a potential, difference of 50 V. Another particle B, of mass ‘4m’ and charge ‘q’ is, , Ro, , R(t), v [R(t)], , R(t), , vo, , Ro, , R(t), , (a) 3 ql, (c) −, , $j − $i, 2, , 3 ql $j, , in the same circular orbit of radius R, in an elliptical orbit, such that it escapes to infinity, in a circular orbit of a different radius, , 28 What is the position and nature of, , image formed by lens combination, shown in, 2 cm, figure?, (where,, A, B, f1 and f2 O, 20 cm, are focal, f1 = + 5 cm f2 = –5 cm, lengths), , 20, cm from point B at right, real, 3, (b) 70 cm from point B at right, real, (c) 40 cm from point B at right, real, (d) 70 cm from point B at left, virtual, (a), , 29 In a meter bridge,, , the wire of length, R′, R′, 1m has a, non-uniform, G, P, cross-section such, that the variation A, B, l, 1–l, dR, of its resistance, dl, 1, dR, ∝ . Two equal, R with length l is, dl, l, resistance are connected as shown in, the figure. The galvanometer has zero, deflection when the jockey is at point, P. What is the length AP ?, , vector of the, 2m, centre of mass rcm, of an asymmetric L, m, m, uniform bar of, 2L, 3L, L, negligible area of, cross-section as shown in figure is, , R(t), , 26 Determine the electric, , dipole moment of the, y, system of three, charges, placed on, the vertices of an, +q, equilateral triangle, as shown in the figure., , (a), (b), (c), (d), , 30 The position, , (d), Ro, , orbit of radius R about the centre of, the earth. A meteorite of the same, mass falling towards the earth, collides with the satellite completely, inelastically. The speeds of the, satellite and the meteorite are the, same just before the collision. The, subsequent motion of the combined, body will be, , (a) 0.3 m (b) 0.25 m (c) 0.2 m (d) 0.35 m, , v [R(t)], , (c), , 21 An ideal battery of 4 V and, , resistance R are connected in series, in the primary circuit of a, potentiometer of length 1 m and, resistance 5 Ω. The value of R to, give a potential difference of 5 mV, across 10 cm of potentiometer wire is, , v [R(t)], , v [R(t)], , 27 A satellite of mass M is in a circular, , –2q, l, , l, l, , 13 $, 5, L x + L y$, 8, 8, 11 $, 3, (b) r =, L x + L y$, 8, 8, 3 $, 11 $, (c) r = L x, +, Ly, 8, 8, 5 $, 13 $, (d) r = L x +, Ly, 8, 8, , (a) r =, +q, x, , (b) 2ql $j, $i + $j, (d) (ql), 2, , ANSWERS, 1. (b), 11. (c), 21. (a), , 2. (b), 12. (c), 22. (d), , 3. (d), 13. (a), 23. (a), , 4. (a), 14. (c), 24. (a), , 5. (b), 15. (a), 25. (c), , 6. (c), 16. (a), 26. (c), , 7. (a), 17. (d), 27. (b), , 8. (a), 18. (c), 28. (b), , 9. (b), 19. (c), 29. (b), , 10. (c), 20. (d), 30. (a), , For Detailed Solutions, Visit : http://tinyurl.com/y5uq4233, Or Scan :
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ONLINE JEE Main 2019, The new value of increase in length of, the steel wire is, , 12 January, Shift-II, 1 To double the covering range of a TV, , transmission tower, its height should, be multiplied by, (a), , (b) 4, , 2, , (c) 2, , (d), , JANUARY ATTEMPT 31, , 1, 2, , 2 The charge on a capacitor plate in a, , circuit as a function of time is shown, in the figure., , (a) zero, (c) 4.0 mm, , sphere, about an axis parallel to its, diameter and at a distance of x from, it, is ‘I (x)’. Which one of the graphs, represents the variation of I (x) with x, correctly?, I(x), , 5, q(µC) 4, , 2, , 4, t(s), , 6, , and 2m respectively. A is in a circular, orbit of radius R and B is in a circular, orbit of radius 2R around the earth., The ratio of their kinetic energies,, TA / TB is, 1, 2, , (c), , (d) 1, , 4 A simple harmonic motion is, , represented by, y = 5(sin 3 πt + 3 cos 3 πt ) cm. The, amplitude and time period of the, motion are, (a) 10 cm,, (c) 5 cm,, , 3, s, 2, , 2, s, 3, 2, (d) 10 cm, s, 3, (b) 5 cm,, , 3, s, 2, , 5 In the given circuit diagram, the, , currents I1 = − 0.3 A, I 4 = 0.8 A and, I5 = 0.4 A, are, flowing as shown. The currents I 2, I3, and I 6 respectively, are, I6 Q, , P, I3, , S I4, , I2, , I1, , R, , 1.1 A, 0.4 A, 0.4 A, 1.1 A, − 0.4 A, 0.4 A, 0.4 A, 1.1 A, 0.4 A, − 0.4 A, 0.4 A, 1.1 A, , 6 A load of mass M kg is suspended, , from a steel wire of length 2 m and, radius 1.0 mm in Searle’s apparatus, experiment. The increase in length, produced in the wire is 4.0 mm. Now,, the load is fully immersed in a liquid, of relative density 2. The relative, density of the material of load is 8., , x, , O, , x, , 8 An ideal gas is enclosed in a cylinder, , at pressure of 2 atm and temperature,, 300 K. The mean time between two, successive collisions is 6 × 10− 8 s. If, the pressure is doubled and, temperature is increased to 500 K,, the mean time between two, successive collisions will be close to, , (a) 4 × 10− 8 s, (c) 2 × 10− 7 s, , (b) 3 × 10− 6 s, (d) 0.5 × 10− 8 s, , 9 A vertical closed cylinder is separated, , into two parts by a frictionless piston, of mass m and of negligible thickness., The piston is free to move along the, length of the cylinder. The length of, the cylinder above the piston is l1 and, that below the piston is l2, such that, l1 > l2. Each part of the cylinder, contains n moles of an ideal gas at, equal temperature T . If the piston is, stationary, its mass m, will be given, by (where, R is universal gas constant, and g is the acceleration due to, gravity), nRT l1 − l2 , (a), g l1 l2 , , (c), , I5, , (d), O, , 3 Two satellites A and B have masses m, , (b) 2, , x, , 8, , (a) 2 µA (b) 15, . µA (c) Zero (d) 3 µA, , 1, 2, , Screen, 2f, , O, I(x), , I(x), , What is the value of current at t = 4 s?, , (a), (b), (c), (d), , x, , (c), , 0, , biconvex lens is shown below. If the, whole set up is immersed in water, without disturbing the object and the, screen positions, what will one, observe on the screen?, , (b), O, , 3, , 12 Formation of real image using a, , I(x), , (a), , 2, , (a) 250 Ω (b) 6200 Ω (c) 200 Ω (d) 6250 Ω, , 7 The moment of inertia of a solid, , 6, , (a), , (b) 5.0 mm, (d) 3.0 mm, , division. To use this galvanometer as, a voltmeter of range 2.5 V, it should, be connected to a resistance of, , RT 2l1 + l2 , g l1 l2 , , nRT 1, 1, (b), + , g l2, l1 , (d), , RT l1 − 3l2 , ng l1 l2 , , 10 A parallel plate capacitor with plates, of area 1 m2 each, are at a separation, of 0.1 m. If the electric field between, the plates is 100 N/C, the magnitude, of charge on each plate is, , C2 , Take, ε0 = 8.85 × 10− 12, , N − m2 , , (a) 9.85 × 10− 10 C, (c) 7.85 × 10− 10 C, , (b) 8.85 × 10− 10 C, (d) 6.85 × 10− 10 C, , 11 A galvanometer whose resistance is, , 50 Ω, has 25 divisions in it. When a, current of 4 × 10− 4 A passes through, it, its needle (pointer) deflects by one, , f, , 2f, , f, , (a) No change, (b) Magnified image, (c) Image disappears, (d) Erect real image, , 13 Let l, r , c, and v represent inductance,, resistance, capacitance and voltage,, respectively. The dimension of l in, rcv, SI units will be, (a) [LT 2 ], (c) [A− 1 ], , (b) [LTA], (d) [LA− 2 ], , 14 A particle of mass 20 g is released, , with an initial velocity 5 m/s along, the curve from the point A, as shown, in the figure. The point A is at height, h from point B. The particle slides, along the frictionless surface. When, the particle reaches point B, its, angular momentum about O will be, (Take, g = 10 m / s2), O, a = 10 m, A, h=10 m, B, , (a) 8 kg - m 2 / s, (c) 2 kg - m 2 / s, , (b) 3 kg - m 2 / s, (d) 6 kg - m 2 / s, , 15 In a Frank-Hertz experiment, an, , electron of energy 5.6 eV passes, through mercury vapour and emerges, with an energy 0.7 eV. The minimum, wavelength of photons emitted by, mercury atoms is close to, (a) 250 nm, (c) 1700 nm, , (b) 2020 nm, (d) 220 nm, , 16 A block kept on a rough inclined, , plane, as shown in the figure, remains, at rest upto a maximum force 2 N, down the inclined plane. The, maximum external force up the, inclined plane that does not move the, block is 10 N. The coefficient of static, friction between the block and the, plane is (Take, g = 10 m / s2)
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32 40 DAYS ~ JEE MAIN PHYSICS, 10, , ONLINE JEE Main 2019, pump at the mouth of a tube,, increases in volume, with time, at a, constant rate. The graph that correctly, depicts the time dependence of, pressure inside the bubble is given by, , 2 N30°, , (a), , 2, 3, , (b), , 3, 2, , (c), , 3, 4, , 1, 2, , (d), , p, , 17 The mean intensity of radiation on, , (b), , (c), , one-dimensional elastic collision with, a nucleus at rest of unknown mass. It, is scattered directly backwards losing, 64% of its initial kinetic energy. The, mass of the nucleus is, (a) 1.5 m (b) 4 m, , (d), , p, , t, , log (t), , 23, , (a), , I2, R1 I1, , 19 In a radioactive decay chain, the, , (a) A = 202; Z = 80 (b) A = 208; Z = 82, (c) A = 200; Z = 81 (d) A = 208; Z = 80, , 20 A 10 m long horizontal wire extends, , from North-East to South-West. It is, falling with a speed of 5.0 ms− 1 at, right angles to the horizontal, component of the earth’s magnetic, field of 0.3 × 10− 4 Wb / m2. The value, of the induced emf in wire is, −3, , −3, , (a) 15, . × 10 V, (c) 0.3 × 10− 3 V, , (b) 11, . × 10 V, (d) 2.5 × 10− 3 V, , 21 Two particles A and B are moving on, two concentric circles of radii R1 and, R2 with equal angular speed ω. At, t = 0, their positions and direction of, motion are shown in the figure, Y, , (b) 60º, , (c) 0º, , (d) 90º, , 24 A long cylindrical vessel is half-filled, with a liquid. When the vessel is, rotated about its own vertical axis,, the liquid rises up near the wall. If, the radius of vessel is, 5 cm and its rotational speed is 2, rotations per second, then the, difference in the heights between the, centre and the sides (in cm) will be, (a) 0.1, , (b) 1.2, , (c) 0.4, , (d) 2.0, 28, , 25 A paramagnetic material has 10, , atoms/m3 . Its magnetic susceptibility, at temperature 350 K is 2.8 × 10− 4 . Its, susceptibility at 300 K is, , A, , R2, , 3, In the above circuit, C =, µF,, 2, 3, H and R1 = 10 Ω., R2 = 20 Ω, L =, 10, Current in L - R1, path is I1 and in C - R2 path is I 2.The, voltage of AC source is given by, V = 200 2 sin(100t ) volts. The phase, difference between I1 and I 2 is, (a) 30º, , −4, , X, , R1, , (a) 3.726 × 10, (c) 2.672 × 10− 4, , B, , −4, , (b) 3.672 × 10, (d) 3.267 × 10− 4, , 26 In the figure, given that VBB supply, π, The relative velocity vA − vB at t =, 2ω, is given by, (a) ω(R1 + R2 )$i, (c) ω(R1 − R2 )i$, , (b) − ω(R1 + R2 )$i, (d) ω(R2 − R1 )$i, , VBB, , v0, IE, , VCC, , 6, µF (b) 4 µF, 5, , (c), , C, , A, , 2, 2, , 2. (c), 12. (c), 22. (b), , 3. (d), 13. (c), 23. (a), , 4. (d), 14. (d), 24. (d), , 5. (a), 15. (a), 25. (d), , can vary from 0 to 5.0 V, VCC = 5 V,, β DC = 200, RB = 100 k Ω, RC = 1 k Ω, and VBE = 10, . V. The minimum base, current and the input voltage at, which the transistor will go to, saturation, will be, respectively, , 6. (d), 16. (b), 26. (b), , 7. (b), 17. (c), 27. (d), , 8. (a), 18. (b), 28. (a), , 9. (a), 19. (b), 29. (c), , 1, , 2, 2, , 2, B, , 7, 7, µF (d), µF, 10, 11, , jagged end. It is still used in the, laboratory to determine velocity of, sound in air. A tuning fork of, frequency 512 Hz produces first, resonance when the tube is filled with, water to a mark 11 cm below a, reference mark. near the open end of, the tube. The experiment is repeated, with another fork of frequency 256 Hz, which produces first resonance when, water reaches a mark, 27 cm below the reference mark. The, velocity of sound in air, obtained in, the experiment is close to, (a) 328 ms− 1, (c) 322 ms− 1, , (b) 341 ms− 1, (d) 335 ms− 1, , 29 When a certain photosensitive surface, , is illuminated with monochromatic, light of frequency v, the stopping, potential for the photocurrent is, − V 0 / 2. When the surface is, illuminated by monochromatic light of, frequency ν / 2, the stopping potential, is − V 0 . The threshold frequency for, photoelectric emission is, (a), , 4, ν, 3, , (b) 2 ν, , (c), , 3ν, 2, , 10. (b), 20. (a), 30. (b), , (d), , 5ν, 3, , 30 A plano-convex lens (focal length f2,, refractive index µ 2, radius of, curvature R) fits exactly into a, plano-concave lens (focal length f1 ,, refractive index µ 1 , radius of, curvature R). Their plane surfaces are, parallel to each other. Then, the focal, length of the combination will be, (a) f1 − f2, (c) f1 + f2, , R, µ 2 − µ1, 2f1 f2, (d), f1 + f2, (b), , ANSWERS, 1. (b), 11. (c), 21. (d), , 2, , 28 A resonance tube is old and has, , R2, , C, L, , (c) 3.5 m (d) 2 m, , 232, initial nucleus is 90, Th. At the end,, there are 6 α-particles and 4, β-particles which are emitted. If the end, nucleus is AZ X, A and Z are given by, , RC, , E, , find C if the effective, capacitance of the, whole circuit is to be, 0.5 µF. All values in, the circuit are in µF., , 1, t, , p, , 18 An α-particle of mass m suffers, , B, vi, , 27 In the circuit shown,, , 1, t3, , (b) 102 T, (d) 10− 2 T, , C, , RB, , (a) 25 µA and 2.8 V (b) 25 µA and 3.5 V, (c) 20 µA and 3.5 V (d) 20 µA and 2.8 V, , p, , (a), , the surface of the sun is about, 108 W / m2. The rms value of the, corresponding magnetic field is, closest to, (a) 1 T, (c) 10− 4 T, , IC, , 22 A soap bubble, blown by a mechanical, , N, , For Detailed Solutions, Visit : http://tinyurl.com/y22wuljv, Or Scan :
